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Is $ \frac{x^2 -4}{x-2} $ a polynomial? The expression $ \frac{x^2 -4}{x-2}$ doesn’t looks like a polynomial because of $ x $ in the denominator. But it can be factorized. After factorizing and canceling the common factor $$\frac{x^2 -4}{x-2} = \frac{(x+2)(x-2)}{x-2} = x+2, $$ and $ x+2 $ is a polynomial which satisfies the definition of the polynomial.
Is the original expression $ \frac{x^2 -4}{x-2} $ is a polynomial or not?
| It is not because it is not equivalent to $x+2$. You can evaluate $x+2$ for any real $x$ if you are working in the reals. You cannot evaluate $\frac {x^2-4}{x-2}$ at $x=2$ because of the division by $0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Angle chasing: Finding the $\angle{NRQ}$ in a quarter circle
Let $MON$ be a quarter circle. $P$ is the midpoint of $OM$ and $PQ$ is the angle bisector of $\angle{OPN}$. If $QR\parallel{OM}$, find $\angle{QRN}$.
Here is an angle-chasing problem which I am trying to solve. I couldn't make a good progress solving the problem. Though here is my approach:
If we assume $\angle{OPQ}=\theta$ and $NP$ intersects $QR$ at $A$, then we have: $$\angle{OPQ}=\angle{NPQ}=\theta,$$ $$\angle{ONP}=\frac{\pi}{2}-\theta,$$ $$\angle{NAQ}=2\theta,$$ $$\angle{PQR}=\theta.$$So, $\Delta{PAQ}$ is isosceles.
This doesn't help very much. How to solve the problem?
| Here is a solution which tries to not use the algebraic knowledge of some trigonometric function (sine, cosine, tangent) of the angle of $36^\circ$ or some related angle (half or double of it). However, since there are "only a few" angles in the picture which are in measure a rational multiple of $\pi$, mainly those in $R$, we still need to involve some algebraic part.
The idea is to show that the points $N,R,M$ from the OP are among the vertices of the $20$-gon. I will also use for them alternative notations $X_0$, $X_3$, and respectively $X_4$. To start with, let $S$ be the reflection of $O$ w.r.t. $Q$, so $OQ=QS$. I will work with a figure where the radius of the circle is
$$
4 = OM=ON\ .
$$
Then $PN=\sqrt{2^2+4^2}=2\sqrt 5$, and in $\Delta NOP$ the angle bisector theorem gives $OQ:QN=2:2\sqrt 5$, so $\displaystyle\frac{OQ}4=\frac{OQ}{OQ+QN}=\frac2{2\sqrt 5+2}=\frac1{\sqrt 5+1}$, so $\displaystyle OQ=\frac 4{\sqrt 5+1}=\sqrt 5-1$. Some further computations are:
$$
\begin{aligned}
QS &= OQ=\sqrt 5-1\ ,\\
SN &= ON-2OQ=6-\sqrt 5=(\sqrt 5-1)^2\ ,\\
QR^2 &=OR^2-OQ^2=16-(5-2\sqrt 5+1)=10+2\sqrt 5=2\sqrt 5(\sqrt 5+1)\ ,\\
NR^2 &= QR^2+QN^2=(10+2\sqrt 5) + (5-\sqrt 5)^2=40-8\sqrt 5=8\sqrt 5(\sqrt 5-1)\ ,\text{ so}\\
\frac{NR^2}{QR^2}
&
= \frac{8\sqrt 5(\sqrt 5-1)}{2\sqrt 5(\sqrt 5+1)}
= \frac{4(\sqrt 5-1)}{(\sqrt 5+1)}
=(\sqrt 5-1)^2 =\left(\frac{NS}{SQ}\right)^2\ .
\end{aligned}
$$
The reciprocal of the angle bisector theorem (its usage here being the idea of this proof), insures now that the line $RS$ bisects the angle in $R$ in $\Delta QRN$. Let $x$ be the measure of $\widehat{ROM}$. Then we have the following situation with further angles computed in terms of $x$, using parallelism and an isosceles triangle:
We obtain an equation involving $x$ from
$$
3x +\frac 12(180^\circ-x)=\widehat{NRM}=\frac 12\cdot (360^\circ-\overset\frown {MN})=\frac 12\cdot 270^\circ\ .
$$
We double, and solving for $x$ in $6x+(180^\circ-x)=270^\circ$, we get
$$
\color{blue}{\boxed{
\qquad
\begin{aligned}
x &=\frac 15\cdot90^\circ=18^\circ\ ,\\
\widehat {QRN}&=2x =36^\circ \ .
\end{aligned}
\qquad}}
$$
$\square$
We are done, but it may be a good impression to also see the $20$-gon in the picture...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$ $$\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$$
I can reduce it to the general term,
$$\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)2^r}$$
I don't know how to go about this any further though. I also ran this in python and the sum is exceeding $1.5$ for $10,000$ terms, which is weird since it should converge to $1.5$, so it makes me doubt if the general term I've written is correct.
| Here is a longer and more painstaking way that uses calculus:
For $|x|\lt 1$, $$\sum_1^{\infty}(2r^2 +7r+4) x^{r-1}\\ = 2\sum_1^{\infty} r^2 x^{r-1} +7\sum_1^{\infty} rx^{r-1} +4\sum_1^{\infty}x^{r-1} \\ = 2\frac{d}{dx} \left( x \frac{dS}{dx}\right ) +7\frac{dS}{dx} +4S \\ = \frac{-2(1+x)}{(1-x)^3} +\frac{7}{(1-x)^2} +\frac{4}{1-x} $$
where $S=\sum_0^{\infty} x^r =\frac{1}{1-x}$. Integrating both sides thrice, dividing by $x^2$ and setting $x=\frac 12$ will yield the exact sum you mention.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Calculate integral $\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$ I recently saw the integral problem
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$$
and tried to solve it. Below is what I did.
Interesting to look at other easier solutions.
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=4\int_{0}^{\pi /2}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}\\\overset{t=\operatorname{tg} x}{=}\int\limits_{0}^{\infty }\frac{1+t^2}{\left ( 1+\left ( 1+n^2 \right )t^2 \right )^2}dt\\
\overset{t=\frac{y}{\sqrt{1+n^2}}}{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int\limits_{0}^{\infty }\frac{1+n^2+y^2}{\left ( 1+y^2 \right )^2}dy\\ \overset{y=\operatorname{tg} \theta }{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int_{0}^{\pi /2}\left ( 1+n^2\cos^2 \theta \right )d\theta \\
=\frac{\pi \left ( 2+n^2 \right )}{\left ( 1+n^2 \right )\sqrt{1+n^2}}$$
| Alternatively
$$I(n)= \int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=\frac4{n^4}\int_{0}^{\pi /2}\frac{dx}{\left ( a-\cos^2 x \right )^2},\>\>\>\>\>a= 1+\frac1{n^2}$$
Note that
$$J(a)= \int_{0}^{\pi/2}\frac{dx}{ a- \cos^2 x }
= \int_{0}^{\pi/2}\frac{d(\tan x)}{ a\tan^2x +(a-1)}dx= \frac\pi{2\sqrt{a(a-1)}}
$$
Then
$$I(n)= -\frac4{n^4} \frac{dJ(a)}{da}\bigg|_{a= 1+\frac1{n^4}}=\frac{\pi(2+n^2)}{ (1+n^2)^{3/2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Evaluate:- $\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
Evaluate:- $\dfrac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
What I Tried:- Let $a = 4 , b = \sqrt{15} , c = 6, d= \sqrt{35}$ . Then I get :-
$$\rightarrow \frac{[a + b]^{3/2} + [a - b]^{3/2}}{[c + d]^{3/2} - [c - d]^{3/2}}$$
Now I can put the formulas $(a^3 + b^3)$ and $(c^3 - d^3)$
$$\rightarrow \frac{[(a + b)^{1/2} + (a - b)^{1/2}][(a + b) - \sqrt{a^2 - b^2} + (a - b)]}{[(c + d)^{1/2} - (c - d)^{1/2}][(c + d) + \sqrt{c^2 - d^2} + (c - d]}$$
$$\rightarrow \frac{7[(a + b)^{1/2} + (a - b)^{1/2}]}{13[(c + d)^{1/2} - (c - d)^{1/2}]}$$
From here, I do not know how to proceed. Can anyone help me?
| $$\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$$
$$=\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}\times\frac{2^{3/2}}{2^{3/2}}$$
$$=\frac{[8+2\sqrt{15}]^{3/2} + [8-2\sqrt{15}]^{3/2}}{[12+2\sqrt{35}]^{3/2} - [12-2\sqrt{35}]^{3/2}}$$
$$=\frac{[\sqrt5+\sqrt3]^3 + [\sqrt5-\sqrt3]^3}{[\sqrt7+\sqrt5]^3 - [\sqrt7-\sqrt5]^3}$$
$$=\frac{28\sqrt{5}}{52\sqrt{5}}$$
$$=\frac{7}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Matrix Multiplication - Express a row as a linear combination $$ Let \ A \ = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 3 & 6 \\ \end{bmatrix} and
\ let \ B = \begin{bmatrix} 0 & 1 & -3 \\ -3 & 1 & 4 \end{bmatrix} $$
Express the third row of AB as a linear combination of the rows of B
$$ AB \ = \ \begin{bmatrix} -6 & 3 & 5 \\ -15 & 9 & 8 \\ -18 & 9 & 15 \end{bmatrix} $$
3rd row of AB would be
$$ \begin{bmatrix} 0 \ (-18) & 1 \ (9) & -3 \ (15) \\ -3 \ (-18) & 1 \ (9) & 4 \ (15) \end{bmatrix} $$
ANS: So the third row represented as a linear combination of the rows of B is given by:
$$ -18 \ \begin{bmatrix} \ \ \ 0 \\ -3 \end{bmatrix} \ \
+ \ \ 9 \ \begin{bmatrix} \ \ 1 \ \\ \ \ 1 \ \end{bmatrix} \ \
+ \ \ 15 \ \begin{bmatrix} -3 \\ \ \ 4 \end{bmatrix} \ \ $$
Is my answer correct?
If not any suggestion or help would be appreciated.
Thanks for your time and cooperation from now.
| From the definition of matrix multiplication it follows that:
$$\underbrace{\begin{bmatrix}-18&9&15\end{bmatrix}}_{\text{third row of }AB} = 3\cdot \underbrace{\begin{bmatrix}0&1&-3\end{bmatrix}}_{\text{first row of }B}+6\underbrace{\begin{bmatrix}-3&1&4\end{bmatrix}}_{\text{second row of }B}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving that for integers $a \ge 5, a > b, b \ge 1$, $2^a - 27 \nmid 2^b + 15$ I am trying to show that for integers $a \ge 5, a > b, b \ge 1$:
$$2^a - 27 \nmid (2^b + 15)$$
Is the following inductive argument valid?
Please let me know if I made any mistake or if there is a more standard way to prove the same point.
(1) Base Case: $a=5$:
$32 - 27 = 5 \nmid (2^b + 15)$ since $2^b \not\equiv 0 \pmod 5$
(2) Assume that the assumption is true for any $a \ge 5$
(3) Inductive Case:
*
*Assume that $2^{a+1} - 27 | (2^b + 15)$ with $a+1 > b$
*$2^{a+1} - 42$ is not a power of $2$
*
*Assume that $2^{a+1} - 42 = 2^c$
*$2^c(2^{a+1-c} - 1) = 42$
*Since $2^c | 42$, $c=1$
*$2^{a+1-c} = 22$ which is impossible since $11 \nmid 2^{a+1-c}$
*We can reject the original assumption.
*$2^b \equiv 2^{a+1}-42 \pmod {2^{a+1}-27}$
*$2^b > 2^{a+1} - 27$ Since $2^{a+1}-42$ is not a power of $2$
*Since $2^a > 27$, it follows that $2^b > 2^{a}$ Since $2^a < 2^a + (2^a - 27)$
*But then $b > a$ which contradicts the assumption that $b < a+1$
*So we reject the assumption that $2^{a+1} - 27 | (2^b + 15)$
| Your proof appears to be basically correct. Nonetheless, a minor point is your inductive case doesn't actually technically involve induction as there's no actual use of the inductive hypothesis, i.e., that $2^a - 27 \not\mid 2^b + 15$, so induction was not required.
Alternatively, using size comparisons similar to what you did, here is what I consider to be a somewhat simpler & more direct proof. First, note that
$$2^a - 27 \mid 2^b + 15 \implies 2^a - 27 \le 2^b + 15 \implies 2^a \le 2^b + 42 \tag{1}\label{eq1A}$$
Next, with $a \gt b \implies a \ge b + 1 \implies 2^a \ge 2^{b+1} = 2(2^b)$, then \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned}
2(2^{b}) & \le 2^{a} \le 2^b + 42 \\
2^{b} & \le 42 \\
2^{b} + 42 & \le 84
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Using this with the right side inequality in \eqref{eq1A} gives
$$2^a \le 84 \implies a \le 6 \tag{3}\label{eq3A}$$
You've already shown $a = 5$ doesn't work. With $a = 6$, then $2^a - 27 = 64 - 27 = 37$. However, $2^b + 15$ is not a multiple of $37$ for any $b \lt 6$. This therefore shows your conjecture is always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$A^ {1/A} =B^ {1/B} =C^ {1/C} ,A^ {BC} +B^ {AC} +C^ {AB} =729$ If $A^
{1/A}
=B^
{1/B}
=C^
{1/C}
,A^
{BC}
+B^
{AC}
+C^
{AB}
=729$
Which of the following equals $A^
{1/A}$?
I tried solving it and in the end got $A^{BC}= B^{AC}=C^{BA}= 3^5$ Thus $A^{1/A}=\ ^{ABC}\sqrt {3^5}$
Now I have two doubts. In some of the solutions I have seen, they have concluded that since $A^{BC}= 3^5, A= 3 \ and\ BC = 5$ Thus $ A^{1/A}= 3^{1/3}$. Is this even a correct method? Because using this logic, we can conclude the same that $B^{AC}= 3^5, thus, B= 3 \ and\ AC = 5$ and so does for $C$ making all three $A,B,C= 3 $ which contradicts the rest.
And my second doubt is the answer to my question however is given as $\sqrt2$. How does one get this answer?
| Let $A^{1/A} = B^{1/B} = C^{1/C} = k$. Now, raise each term to the $ABC$ power and you get:
$A^{BC} = B^{AC} = C{AB} = k^{ABC}$ We know that $A^{BC} + B^{AC} + C^{AB} = 729$.
So
it becomes,
$k^{ABC}\times k^{ABC}\times k^{ABC}=729$
$(k^{ABC})^3 = 729$
$k^{ABC} = 9$
$k = 9^{\frac{1}{ABC}}$
So, $A^{1/A} = 9^{1/ABC}$
Hope it helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the rank of $\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$ Found this exercise in Serge Lang's Introduction to Linear Algebra:
Find the rank of the matrix $$\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$$
So my process to solve it is as follows. First, I set a system
$$ x \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 2 \\ 2 \\ 1 \end{pmatrix} + z \begin{pmatrix} 3 \\ 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} $$
$$\begin{cases}
x + y + 3z = 0 \\
x + 2y + 4z = 0 \\
2y + 2z = 0 \\
x + y +3z = 0
\end{cases}$$
Immediately we see that we can ignore the last equation. Then we can subtract the first one from the second one so we get
$$\begin{cases}
x + y +3z = 0 \\
y + z = 0 \\
2y + 2z = 0
\end{cases}$$
The third one is $2$ times the second one so we can remove it. Finally we have
$$\begin{cases}
x + y +3z = 0 \\
y + z = 0 \\
\end{cases}$$
Since this is a system of two equations in three unknowns, it has a non-trivial solution and thus, both are linearly dependent. Therefore, the rank is $1$
I must have made some mistake since the answers at the back of the book state that the solution is $2$ but I don't see where I'm wrong
| Thanks to the comments I found that I just have to reduce it to row echelon form, I guess I misunderstood ranks, the answer is
$$\begin{aligned}
\begin{pmatrix}
1 &1 &0 &1 \\
1 &2 &2 &1 \\
3 &4 &2 &3 \end{pmatrix} &\overset{(2) - (1)}{\implies} \\
\begin{pmatrix}
1 &1 &0 &1 \\
0 &1 &2 &0 \\
3 &4 &2 &3
\end{pmatrix} &\overset{(3)-3\cdot (1)}{\implies} \\
\begin{pmatrix}
1 &1 &0 &1 \\
0 &1 &2 &0 \\
0 &1 &2 &0
\end{pmatrix} &\overset{(3) - (2)}{\implies} \\
\begin{pmatrix}
1 &1 &0 &1 \\
0 &1 &2 &0 \\
0 &0 &0 &0
\end{pmatrix}
\end{aligned}$$
Then, since it's in row echelon form it's clear that these are two linearly independent equations
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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The derivative of the cumulative distribution of $\min(X_1, a_1) + \min(X_2, a_2)$ Assume $X_1, X_2$ are continuous random variables in $L^{\infty}$. Let $a_1, a_2$ be real numbers and
$$Y := \min(X_1, a_1)+ \min(X_2, a_2).$$
The dependency structure between $X_1$ and $ X_2$ is not known, but we assume that the joint and marginal distributions are well-defined.
Is there a way to calculate
\begin{align}
\frac{d}{da_i} F_{Y}(t),
\end{align}
where $F_{Y}(t) := P(\min(X_1, a_1)+ \min(X_2, a_2) \leq t)$?
I have tried to find an explicit expression for $Y$, with the idea to express the cumulative distribution function as a double integral, with the hope of appealing to the fundamental theorem of calculus, but I cannot seem to find an expression that makes sense.
| We can write, for the first term
$$
\eqalign{
& \Pr \left( {\min \left( {X_{\,1} ,a_{\,1} } \right) \le t_{\,1} } \right)
= F_{\,m} \left( {t_{\,1} ;\,a_{\,1} } \right) = \cr
& = \left[ {t_{\,1} < a_{\,1} } \right]\Pr \left( {X_{\,1} \le t_{\,1} } \right)
+ \left[ {a_{\,1} \le t_{\,1} } \right]1 = \cr
& = \left( {1 - \left[ {a_{\,1} \le t_{\,1} } \right]} \right)\Pr \left( {X_{\,1} \le t_{\,1} } \right)
+ \left[ {a_{\,1} \le t_{\,1} } \right] = \cr
& = \Pr \left( {X_{\,1} \le t_{\,1} } \right)
+ \left[ {a_{\,1} \le t_{\,1} } \right]\left( {1 - \Pr \left( {X_{\,1} \le t_{\,1} } \right)} \right) = \cr
& = \left[ {t_{\,1} < a_{\,1} } \right]F_{\,1} \left( {t_{\,1} } \right)
+ \left[ {a_{\,1} \le t_{\,1} } \right] = \cr
& = F_{\,1} \left( {t_{\,1} } \right)
+ \left[ {a_{\,1} \le t_{\,1} } \right]\left( {1 - F_{\,1} \left( {t_{\,1} } \right)} \right) = \cr
& = F_{\,1} \left( {t_{\,1} } \right)
+ H\left( {t_{\,1} - a_{\,1} } \right)\left( {1 - F_{\,1} \left( {t_{\,1} } \right)} \right) \cr}
$$
where
*
*the square brackets denote the Iverson bracket ;
*H is the Heaviside step function ($H(0)=1$);
and where we consider $a_1$ as a parameter.
Analogously for the other term in $X_2$.
Now, for computing the CDF of the sum of two terms we need the PDF of at least one of the distributions,
i.e. the derivative of the expression above.
This shall be done with caution, considering that there is a finite jump at $t_1 = a_1$.
Only at that discontinuity we are using the Heaviside function, and its formal derivative as a Dirac Delta.
Thus we write
$$
\eqalign{
& P\left( {\min \left( {X_{\,1} ,a_{\,1} } \right) = t_{\,1} } \right)
= f_{\,m} \left( {t_{\,1} ;\,a_{\,1} } \right) = \cr
& = {d \over {dt_{\,1} }}F_{\,m} \left( {t_{\,1} ;\,a_{\,1} } \right) = \cr
& = \left[ {t_{\,1} < a_{\,1} } \right]f_{\,1} \left( {t_{\,1} } \right)
+ \left( {1 - F_{\,1} \left( {a_{\,1} } \right)} \right)\delta \left( {t_{\,1} - a_{\,1} } \right) = \cr
& = f_{\,1} \left( {t_{\,1} } \right)
+ \delta \left( {t_{\,1} - a_{\,1} } \right)\left( {1 - F_{\,1} \left( {t_{\,1} } \right)} \right)
- H\left( {t_{\,1} - a_{\,1} } \right)f_{\,1} \left( {t_{\,1} } \right) = \cr
& = \left( {1 - H\left( {t_{\,1} - a_{\,1} } \right)} \right)f_{\,1} \left( {t_{\,1} } \right)
+ \delta \left( {t_{\,1} - a_{\,1} } \right)\left( {1 - F_{\,1} \left( {t_{\,1} } \right)} \right) \cr}
$$
We can check that the integral of the above properly returns the $F_m$
$$
\eqalign{
& \int\limits_{x = - \infty }^t {f_{\,m} \left( {x;\,a} \right)dx} = \cr
& = \int\limits_{x = - \infty }^t {\left[ {x < a} \right]f\left( x \right)dx}
+ \int\limits_{x = - \infty }^t {\left( {1 - F\left( a \right)} \right)\delta \left( {x - a} \right)dx} = \cr
& = \left[ {t < a} \right]\int\limits_{x = - \infty }^t {f\left( x \right)dx}
+ \left[ {a \le t} \right]\int\limits_{x = - \infty }^a {f\left( x \right)dx}
+ \left( {1 - F\left( a \right)} \right)\int\limits_{y = - \infty }^{t - a} {\delta \left( y \right)dy} = \cr
& = \left[ {t < a} \right]F(t) + \left[ {a \le t} \right]F(a)
+ \left( {1 - F\left( a \right)} \right)\left[ {0 \le t - a} \right] = \cr
& = \left[ {t < a} \right]F(t) + \left[ {a \le t} \right] = F_{\,m} \left( {t;\,a} \right) = \cr
& = \int\limits_{x = - \infty }^t {\left( {1 - H\left( {x - a} \right)} \right)f\left( x \right)dx}
+ \int\limits_{x = - \infty }^t {\left( {1 - F\left( x \right)} \right)\delta \left( {x - a} \right)dx} = \cr
& = \left[ {t < a} \right]F(t) + \left[ {a \le t} \right]F(a)
+ \left( {1 - F\left( a \right)} \right)\left[ {0 \le t - a} \right] = \cr
& = \left[ {t < a} \right]F(t) + \left[ {a \le t} \right] = F_{\,m} \left( {t;\,a} \right) \cr}
$$
Then the CDF of the sum of the two terms will be
$$
\eqalign{
& P\left( {\min \left( {X_{\,1} ,a_{\,1} } \right) + \min \left( {X_{\,2} ,a_{\,2} } \right) \le s} \right)
= F_s \left( {s;\,a_{\,1} ,a_{\,2} } \right) = \cr
& = \int\limits_t {P\left( {\min \left( {X_{\,1} ,a_{\,1} } \right) \le t} \right)
P\left( {\min \left( {X_{\,2} ,a_{\,2} } \right) = s - t} \right)dt} = \cr
& = \int\limits_t {\left( \matrix{
\left( {F_{\,1} \left( t \right) + H\left( {t - a_{\,1} } \right)
\left( {1 - F_{\,1} \left( t \right)} \right)} \right) \cdot \hfill \cr
\left( \matrix{
\left( {1 - H\left( {s - t - a_{\,2} } \right)} \right)f_{\,2} \left( {s - t} \right) + \hfill \cr
+ \delta \left( {s - t - a_{\,2} } \right)\left( {1 - F_{\,2} \left( {s - t} \right)} \right) \hfill \cr} \right)
\hfill \cr} \right)dt} \cr}
$$
in the case that $X1$ and $X2$ are independent.
Otherwise we shall introduce the appropriate expressions for the CDF and PDF.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many ways are there to color a $3 × 3$ grid in the same way if five squares have to be red and four squares have to be blue? When configurations after rotations and flips are considered the same, how many ways are there to color a $3 × 3$ grid in the same way if five squares have to be red and four squares have to be blue?
We can color a $3 × 3$ grid in $$\frac{n^9 + 4n^6 + n^5 + 2n^3}{8}$$ ways using $n$ colors if configurations after rotations and flips are considered the same.
But I couldn't find how many ways are there when the grid has five squares have to be red and four squares have to be blue. How can I calculate this?
| Enumerating the symmetries of the grid for the cycle index so that we
may apply PET we obtain
The identity, $a_1^9.$
Two rotations by $90$ and $270$ degrees which fix
the central slot, $2 a_1 a_4^2$
One rotation by $180$ degrees which also fixes the central slot,
$a_1 a_2^4$
A vertical and a horizontal flip, which fixes the central column /
row: $2 a_1^3 a_2^3$
A flip about one of the two diagonals which are being fixed,
$2 a_1^3 a_2^3.$
We get for the cycle index
$$Z(Q) = \frac{1}{8}
(a_1^9 + 2 a_1 a_4^2 + a_1 a_2^4 + 4 a_1^3 a_2^3).$$
Just to check we get for colorings using at most $N$ colors
$$\frac{1}{8} (N^9 + 4 N^6 + N^5 + 2 N^3)$$
which gives the sequence
$$1, 102, 2862, 34960, 252375, 1284066, 5105212, \ldots$$
which points to OEIS A217331
where these data are confirmed.
Now we use PET to compute the number of colorings with five red
squares and four blue ones, essentially performing
$$[R^5 B^4] Z(Q; R+B),$$
which yields (substitution is $a_d = R^d B^d$)
$[R^5 B^4] (R+B)^9 = {9\choose 5} = 126$
$[R^5 B^4] 2 (R+B) (R^4 + B^4)^2
= 2 [R^4 B^4] (R^4 + B^4)^2 = 4$
$[R^5 B^4] (R+B) (R^2 + B^2)^4
= [R^4 B^4] (R^2 + B^2)^4
= [R^2 B^2] (R + B)^4 \\ = {4\choose 2} = 6$
$[R^5 B^4] 4 (R+B)^3 (R^2 + B^2)^3
\\ = [R^5 B^4] 4 (R^3 + 3 R^2 B + 3 R B^2 + B^3)
(R^2 + B^2)^3
\\ = 4 ( {3\choose 1} + 0 + 3 {3\choose 2} + 0)
= 48.$
We thus get for our answer $\frac{1}{8} (126 + 4 + 6 + 48)
= 23.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4118792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $3^{n-2} \geq n^5$ for $n\geq20$ I am new to induction proofs and wanted to know if my reasoning behind the following proof is correct. Here are my steps :
*
*$n=18 \Rightarrow 3^{18} \geq 20^5$
*Given that $3^{n-2} \geq n^5$ is true for n, show $3^{n-1} \geq (n+1)^5$
$3^{n-1} = 3\times(3)^{n-2} \geq 3n^5$
If $3n^5 \geq (n+1)^5$, then $3\times (3)^{n-2} \geq 3n^5 \geq (n+1)^5 \Leftrightarrow 3^{n-1} \geq (n+1)^5$
I show this by induction for all $n \geq 20$.
*
*$n=18 \Rightarrow 3(20)^{5} \geq 21^5$
*$3(n+1)^{5} = 3n^5+15n^4+30n^3+30n^2+15n+3 \geq (n+1)^5+15n^4+30n^3+30n^2+15n+3$
$3n^5+15n^4+30n^3+30n^2+15n+3 \geq (n+2)^5-10n^4-40n^2-60n-28$
$(3n^5+25n^4+70n^3+30n^2+75n+31) = (n+2)^5 + (2n^5+15n^4+30n^3+10n^2+10n+1)\geq (n+2)^5$
Since $(2n^5+15n^4+30n^3+10n^2+10n+1) > 0$ for $n\geq20$, it follows that $(n+2)^5+(2n^5+...+1) \geq (n+2)^5$.
Therefore, $3\times (3)^{n-2} \geq 3n^5 \geq (n+1)^5 \Leftrightarrow 3^{n-1} \geq (n+1)^5$
| It might be easier to bypass the second induction and instead show directly that, for $n\ge20$, we have
$$\begin{align}
(n+1)^5&=n^5+5n^4+10n^3+10n^2+5n+1\\
&\le n^5+40n^4\\
&=n^5+2(20n^4)\\
&\le n^5+2(n\cdot n^4)\\
&=3n^5
\end{align}$$
where the first inequality (introducing the $40n^4$) holds for $n\ge1$ while the second inequality invokes the assumption $n\ge20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right) \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
| Considering the more general case $$S_a=\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^2 - a^2}$$
$$\frac{\log (k)}{k^2 - a^2}=\sum_{n=0}^\infty \frac{\log (k)}{k^{2n+2}} a^{2n}$$
$$\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^{2n+2}}=-\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$
$$S_a=-\sum_{n=0}^\infty a^{2n}\,\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$ converges very fast.
If the summation had started at $k=1$ instead of $k=a+1$, we would have obtained
$$-\sum_{n=0}^\infty a^{2n}\,\zeta '(2 (n+1))$$
Considering the case of $a=2$ and the partial sums up to $p$, the value $0.920492$ is obtained for $p=17$.
$$\left(
\begin{array}{cc}
p & -\sum_{n=0}^p \\
1 & 0.866620 \\
2 & 0.898968 \\
3 & 0.911406 \\
4 & 0.916559 \\
5 & 0.918769 \\
6 & 0.919732 \\
7 & 0.920155 \\
8 & 0.920343 \\
9 & 0.920426 \\
10 & 0.920462 \\
11 & 0.920479 \\
12 & 0.920486 \\
13 & 0.920489 \\
14 & 0.920491 \\
15 & 0.920491 \\
16 & 0.920491 \\
17 & 0.920492
\end{array}
\right)$$ while the orginal summation requires tens of thousands.
This is explained by the fact that
$$\zeta '(2 (n+1)) \sim -\exp\Big[-\frac{1106 }{797}n-\frac{713}{430} \Big]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 0
} |
Rewrite $\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$ in a common base then solve for $x$ I am to rewrite $\frac{125}{(\frac{1}{625})^{-x-3}}=5^3$ and then solve for x. My textbooks solutions section says the solution is -3. I gave it a shot and got 3.25. Here is my working:
$$\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$$
$$\frac{5^3}{\left(\frac{1}{5^4}\right)^{-x-3}}=5^3$$
(Here's where I get pretty confused. I think I can pull the denominator component $\frac{1}{5^4}$ up above and transform from a fraction to a full number, but I'm unsure how to handle the exponent $-x-3$)
$$5^3 \times 5^{4(x-3)}=5^3$$
$$5^{12x-36}=5^3$$
$$12x-36=3$$
$$12x=39$$
$$x=\frac{39}{12}=3.25$$
Where did I go wrong and how can I arrive at 3? I suspect I went stray between the second and 3rd lines. If this is so, rather than just provide a solution please please do also explain the logic explicitly... how to I 'pull up' $\left(\frac{1}{5^4}\right)^{-x-3}$, what are the rules for each component?
| $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large -}x-3}}=5^3$
Here you have $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large\text{negative sign}}x-3}}=5^3$
Here you correctly surmised that we can "pull" the $\frac 1{5^4}$ "up" to get that $\frac 1{(\frac 1{5^4})^M} = (5^4)^M$. That is good. So you wrote
$5^3 \times 5^{4(\color{green}{\large \text{NO NEGATIVE SIGN}}x-3)}=5^3$
What happened to the negative sign before the $x$?
You ought to have $5^3 \times 5^{4(-x-3)} = 5^3$
Your second mistake is you figured:
$5^3 \times 5^{4(\color{green}{\large \text{NO NEGATIVE SIGN}}x-3)}= 5^{3\color{red}{\large\times} 4(x-3)}$
It should be $5^3 \times 5^{4(-x-3)} = 5^{3 \color{green}{\large +} 4(-x-3)}$.
From there you'd get everything right.
$5^{3 \color{green}{\large +} 4(-x-3)}= 5^3$
$5^{-4x - 12 + 3} = 5^3$
$5^{-4x -9} = 5^3$
$-4x -9 = 3$
$-4x = 12$
$x = \frac {12}{-4} = -3$.
But note: We didn't have to expand $3+4(-x-3)$ out to $-4x -9$. We could have just left it as is and get $3+4(-x-3) = 3$ and subtract three from both sides to get $4(-x-3) = 0$ to get $x+3 = 0$ and $x =-3$.
Also from $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large -}x-3}}=5^3$ we could have divided both sides byt $5^3$ to get $\frac 1{(\frac 1{5^4})^{-x-3}} = 1$
Also if we are being more systematic we could have done either:
$\frac {5^3}{(\frac 1{5^4})^{-x-3}} = \frac {5^3}{(5^{-4})^{-x-3}}=\frac {5^3}{5^{-4(-x-3)}}=\frac {5^3}{5^{4(x+3)}} = 5^{3- 4(x+3)}$
or
$\frac {5^3}{(\frac 1{5^4})^{-x-3}} = 5^3 \times (\frac 1{5^4})^{x+3} =5^3 \times 5^{-4(x+3)}$
There's a plethora of options.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
I need a sanity check to compute the limit of a function in two variables So I have $g(x,y) = \tfrac{f(x,y) - y}{\sqrt{x^2 + y^2}}$, where $f(x,y) = \tfrac{y^3 -x^8y}{x^6 + y^2}$ for $(x,y) \neq (0,0)$.
I want to find $\lim_{(x,y) \to (0,0)} g(x,y)$ if it exists.
My question is if after doing some algebraic manipulation, I can get g as a product of $\tfrac{-y}{\sqrt{x^2 + y^2}}$ and $\tfrac{x^6(1+x^2)}{x^6 + y^2}$, can I bound the former by $1$ in absolute value? What can I do with the second part of the product? I feel that the numerator would dominate but want to be rigorous with it. Any help would be appreciated.
| call the ratio $G,$ I got
$$ |G| = \frac{|y| x^6 (1+x^2)}{r (x^6 + y^2)} $$
Demand $c = | \cos \theta | $ and $s = | \sin \theta | $
$$ |G| = \frac{r^7 c^6 s (1+r^2 c^2 )}{r^3 (r^4c^6 +s^2)} $$
$$ |G| = \frac{r^4 c^6 s (1+r^2 c^2 )}{ r^4c^6 +s^2} $$
We are allowed to demand $r < 1,$ so that $1+r^2 c^2 < 2,$
$$ |G| < 2 \frac{r^4 c^6 s }{ r^4c^6 +s^2} $$
From $$ 0 \leq (r^2 c^3 \pm s)^2 $$
we get $$ r^4 c^6 + s^2 \geq 2 r^2 c^3 s $$ and
$$ \frac{1}{r^4 c^6 + s^2} \leq \frac{1}{2 r^2 c^3 s} $$
$$ |G| < 2 \frac{r^4 c^6 s }{ r^4c^6 +s^2} \leq 2 \frac{r^4 c^6 s }{2 r^2 c^3 s } \leq r^2 c^3 \leq r^2$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ | $$\lim_{(x,y)\to(0,0)}\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}} \\
= \lim_{(x,y)\to(0,0)}\frac{(xy)^3 }{(x^4+y^4)\sqrt{x^2+y^2}}\frac{\sin(xy^2)}{xy^2}$$
Let $y=x\tan\theta$
$$\lim_{(x,y,x\tan\theta)\to(0,0,0)}\frac{(x^{{6} }\tan^3\theta)}{x^4(1+\tan^4\theta)x\sqrt{1+\tan^2\theta}}\frac{\sin(xy^2)}{xy^2}\\
=\lim_{(x,y,x\tan\theta)\to(0,0,0)}\frac{(x\tan^3\theta)}{(1+\tan^4\theta)\sqrt{1+\tan^2\theta}}\frac{\sin(xy^2)}{xy^2}$$
The denominator involving the $\tan\theta$ would never be less than 1. This implies that the above limit shrinks to "0" due to $x$ in the numerator. Hence,
$$\lim_{(x,y)\to(0,0)}\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Simplifying the sequence defined by $A_0=6$ and $A_n=\frac{8}{9}A_{n-1}+\frac{6}{5}(\frac{20}{9})^n$, using sigma notation
I have $A_0 = 6$ and
$$A_n = \left(\frac{8}{9}\right)A_{n-1} + \left(\frac{6}{5}\right) \left(\frac{20}{9}\right)^n$$
and I want to simplify it with sigma notation.
I have it up to
$$\begin{align}A_4 = &6 \left(\frac{8}{9}\right)^4 + {\left(\frac{8}{9}\right)^3} \left(\frac{6}{5}\right) {\left(\frac{20}{9}\right)^{1}} + {\left(\frac{8}{9}\right)^2} \left(\frac{6}{5}\right){\left(\frac{20}{9}\right)^{2}} \\
&+ {\left(\frac{8}{9}\right)^1}\left(\frac{6}{5}\right){{\left(\frac{20}{9}\right)^{3}}} + \left(\frac{6}{5}\right) {\left(\frac{20}{9}\right)^4}
\end{align}$$ because I wanted to find the pattern
Apparently the answer is
$$4\left(\frac{8}{9}\right)^n + 2\left(\frac{20}{9}\right)^n$$ and the simplification into sigma notation is $$6\left(\frac{8}{9}\right)^n + \sum_{k=0}^{n}\left(\frac{8}{9}\right)^{n-k}\left(\frac{6}{5}\right)\left(\frac{20}{9}\right)^k$$
but I can seem to can seem to justify that. It's has to be some sort of algebraic manipulation that I did wrong but I don't see how the $k=0$ term works? To me, it seems that the sigma notation adds an extra term when $k=0$
(btw, this is based on the second-last and last pages of the PowerPoint presentation "Mathematics Geometry: Menger Sponge" (PPT link via ubc.ca) by the University of British Columbia Science and Mathematics Education Research Group.)
--
Personally, I think it's a typo since on Wolfram Alpha, if I use my $k=1$ then I get the right result but with $k=0$ it doesn't work
| For $ n\ge 1$,
$$A_n=aA_{n-1}+b_n$$
$$aA_{n-1}=a^2A_{n-2}+ab_{n-1}$$
$$a^2A_{n-2}=a^3A_{n-3}+a^2b_{n-2}$$
$$a^kA_{n-k}=a^{k+1}A_{n-k-1}+a^kb_{n-k}$$
...
$$a^{n-1}A_1=a^nA_0+a^{n-1}b_1$$
the sum gives
$$A_n=6a^n+\sum_{k=0}^{n-1}a^kb_{n-k}$$
$$=6(\frac 89)^n+\frac 65\sum_{k=0}^{n-1}(\frac 89)^k(\frac{20}{9})^{n-k}$$
$$=6(\frac 89)^n+\frac 65(\frac {20}{9})^n\sum_{k=0}^{n-1}(\frac 25)^k$$
$$=6(\frac{20}{9})^n\Bigl((\frac 25)^n+\frac 15\sum_{k=0}^{n-1}(\frac 25)^k\Bigr)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$ $$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
I tried $$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\leq\frac{n}{\sqrt{n^2+n+1}}$$
But$$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+n}}\right)^n=\lim_{n\to\infty}\left(1+\frac2n\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac2n)}=\frac1e$$
And in the same way,i got $$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+1}}\right)^n=\lim_{n\to\infty}\left(1+\frac1n+\frac{1}{n^2}\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac1n+\frac{1}{n^2})}=\frac{1}{\sqrt e}$$
So i only got $$\frac1e\leq I\leq\frac{1}{\sqrt e}$$
Could someone help me get the value of $I$. Thanks!
| The natural log of this expression is $$ K= n\ln( 1+ (x-1) )$$ where $$x=\sum_{r=1}^n \frac{1}{\sqrt{n^2+n+r}} $$ Due to the bounds you have shown, $K$ must tend to a finite non-zero number, which can only happen if $x\to 1$. Now, $$K = n(x-1) +n O((x-1)^2)$$ The first term is $$n \sum_1^n \left( \frac{1}{\sqrt{n^2+n+r}} -\frac 1n \right) \\ = \sum_1^n \left( \frac{1}{\sqrt{ 1+\frac 1n +\frac{r}{n^2}}} -1 \right) \\ = \sum_1^n \left( -\frac 12 \left( \frac 1n +\frac{r}{n^2} \right) +O(\frac{1}{n^2}) \right) \\ = -\frac 12 -\frac{n(n+1)}{4n^2} +O(\frac 1n ) \\ \to -\frac 12 -\frac 14 \\ =-\frac 34$$ The rest of the terms in $K$ can be shown to go to zero in a similar way. Hence, your limit is $$\Large{ \color{blue}{e^{-\frac 34 }} }$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Perfect Square With Two Integer Variables I am trying to solve a number theory problem in general form. However, I got stuck in the following step:
$a,b,n \in \mathbb Z^{+}$ for which values of $n$, this equation is solvable $\frac{(n+1)(n+2a)}{2} = b^2$ ?
Can we make a general statement about $n$ ? By the way I have tried to apply some inequalities with cases $n>a$, $n<a$ and $n=a$ . Also I think modular arithmetic doesn't work so good for this problem.
| $\frac{1}{2}(n+1)(n+2a) = b^2$
$(2n+1+2a)^2-8b^2 = (2a-1)^2$
Let $x = \frac{\large{2n+1+2a}}{\large{2a-1}}, y = \frac{\large{2b}}{\large{2a-1}}$ then we get
$x^2-2y^2 = 1\tag{1}$
This equation is Pell equation.
The fundamental solution of equation $(1)$ is $(x,y)=(3,2).$
Thus, all of the solutions are given by $x+\sqrt{2}y=\pm(3+2\sqrt{2})^k.$
Hence n and b are given as follows.
$n = (a-\frac{1}{2})x-\frac{1}{2}-a$
$b = \frac{1}{2}y(2a-1)$
For example, some $(x,y)$ is given below.
$(k,x,y)$
$(2,17,12)$
$(3,99,70)$
$(4,577,408)$
$(5,3363,2378)$
$k=2$: take $x=17$ and $y=12$ then we get $(n,b)=(-9+16a, 12a-6).$
| {
"language": "en",
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Finding the Taylor series for $\frac{1}{1-x^2}$ Where I'm stuck
$f(x)=\frac{1}{1-x^2}=\frac{1}{1-x}\cdot\!\frac{1}{1+x}=(1-x)^{-1} \cdot(1+x)^{-1}$
$(1+x)^{-1}= 1-x+x^2-x^3+...+(-1)^{2n}x^{2n}+o(x^{2n})$
$(1-x)^{-1}= 1+x+x^2+x^3+...+x^{2n}+o(x^{2n})$
At this point I have no idea how to multiply all this together.Could you please help?
The answer is $1+x^2+...+x^{2n}$.
| For every $x\in(-1,1), \frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$. So we replace $x$ with $x^2$, and since $x\in(-1,1)\rightarrow x^2\in(-1,1)$ we get $\frac{1}{1-x^2}=\sum_{i=0}^{\infty}x^{2i}$.
| {
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"source": "stackexchange",
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Limit of $\left(y+\frac23\right)\mathrm{ln}\left(\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$ We consider the limit for large $y$ of the following expression :
$$\left(y+\frac23\right)\mathrm{ln}\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}.$$
Many references state that the large $y$ behaviour of this last expression is $y^{-3/2}$. I have been trying to show this in any way possible but I'm only hitting dead ends. Wolframalpha gives a Puiseux series of that expression which makes this behaviour explicit, but I do not understand how to get to this series either. Could you please help me on this?
| Set $x=\sqrt{1+y}-1$. Then $y=(x+1)^2-1=x^2+2x$, and your expression simplifies to
$$ \bigl(x^2+2x+\tfrac23 \bigr)\log\bigl(1+\tfrac2x \bigr) - 2x - 2 $$
Expanding the logarithm in powers of $2/x$ gives us
$$ \bigl(x^2+2x+\tfrac23 \bigr)\Bigl(\frac2x-\frac2{x^2}+\frac{8}{3x^3}-\frac{4}{x^4}+\frac{32}{5x^5}+o(x^{-5}) \Bigr) - 2x - 2 $$
We can then multiply out and collect terms to get
$$ \begin{align} (2-2)&x^1 + {}
\\ (-2 + 4 - 2)&x^0 +{}
\\ (\tfrac83 -4 + \tfrac43)&x^{-1} +{}
\\ (-4+\tfrac{16}3 - \tfrac43)&x^{-2} +{}
\\ (\tfrac{32}5 -8 + \tfrac{16}{9})&x^{-3} + o(x^{-3})
= \tfrac{8}{45}x^{-3} + o(x^{-3})
\end{align}
$$
Since asymptotically $x\sim y^{1/2}$ this is also
$$\frac{8}{45}y^{-3/2} + o(y^{-3/2}).$$
| {
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Proving that $\lim\limits_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist I want to prove that $\displaystyle \lim_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist
I denoted $f(x, y):=\frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$. I tried looking at $f(\frac{1}{n}, \frac{1}{n})$ and other similar sequences, but no result. No matter what sequence I chose I got that the limit is $0$. However, WA says that this doesn't exist. What sequence should I pick?
| As Maxim pointed out the limit actually exists and is equal to 0. I give an alternative proof for that now. It suffices to show, that for |x|,|y| sufficiently small there holds
$$\left| \frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\right| \le \sqrt{2}\max(\sqrt{|y|},|x|^{\frac{1}{3}}).$$
This we see at follows. First by Cauchy-Schwart and the triangle inequality there holds
$$\frac{1}{\sqrt{2}}(|x|+|y|)\le \sqrt{x^2+y^2}\le |x|+|y|.$$
We assume in the following w.o.l.g. that $1>>x,y>0$. Then,
$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^3}{(x^4+y^6)(x+y)}.$$
First assume, that $x\ge y$. Then
$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^3}{(x^4+y^6)(x+y)}\le \sqrt{2}\frac{x^2y^3}{x^4} =\le \sqrt{2} \frac{y^3}{x^2}\le \sqrt{2} x$$
and the above statement is fulfilled. We now assume in the following, that $x\le y$. Then
$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^2}{x^4+y^6}.$$
Now we consider the two cases $x^4\le y^6$ and $x^4\ge y^6$. In the first case,
$$\frac{x^3y^2}{x^4+y^6} \le \frac{x^3y^2}{y^6}\le \frac{x^3}{y^4} \le \frac{y^\frac{3*6}{4}}{y^4} = \sqrt{y}.$$
In the remaining case,
$$\frac{x^3y^2}{x^4+y^6} \le \frac{x^3y^2}{x^4} = \frac{y^2}{x}\le x^\frac{1}{3}.$$
Therefore, the statement is proved.
The following reasoning does not yield the statement, as Maxim pointed out:
Set $y=y(x)=x^q$ with $q>0$. You can try to tune $q$, such that the numerator dominates the denominator. By the above, $q$ has to be smaller than 1. For the denominator there holds
$$(x^4+x^{6q}) \sqrt{x^2+x^{2q}} = x^{\min(4,6q)}(1+O(1)) x^{min(1,q)} (1+O(1))$$
for $x\to0$. E.g., for $q=4/6$ we have
$$x^{3+3q} = x^5 > x^{28/6}(1+o(1)) = x^{7q}(1+o(1)).$$
Thus for that choice, the sequence will diverge (This last deduction is wrong).
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ Using the generating function of Legendre polynomials, show
\begin{equation}
\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(x)=\frac{1}{2}\ln
\left(\frac{1+x}{1-x}\right)
\end{equation}
My attempt
I know the generating function of Legendre polynomials is given by
$G (x, t) = \sum_{n = 0}^{\infty} P_n (x) t^n \text{ with } | t | < 1,
| x | \leqslant 1. $
Also we can see to $G$ as
$G (x, t) = \frac{1}{\sqrt{1 - 2 x t + t^2}} . $
Then
\begin{equation}
G (x, x) = \frac{1}{\sqrt{1 - x^2}} = \sum_{n = 0}^{\infty} P_n (x) x^n \tag{2}.
\end{equation}
Now I want to know to where $\underset{n = 0}{\overset{\infty}{\sum}}
\dfrac{P_n (x) x^n}{n + 1}$ converges?
Since if we find $f (x)$ from (2) such as $\underset{n =
0}{\overset{\infty}{\sum}} \dfrac{P_n (x) x^n}{n + 1} \rightarrow f (x)$, then
$x f (x)$ should be equal to $\frac{1}{2} \ln \left( \frac{1 + x}{1 - x}
\right)$.
Any suggestion will be welcome.
| Write \begin{equation}
f(x,y)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_n(y)
\end{equation}
Differentiate this to obtain $f_x(x,y) = \sum_{n=0}^{\infty} x^n P_n(y)$. This is known to be $\frac{1}{\sqrt{1 - 2 x y + x^2}}$. Integrating we arrive at $f(x,y) = \log(\sqrt{-2yx+x^2+1} -y+x)$. Subtitute $x=y$ to obtain the answer.
| {
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Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
&= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
| You can use Young's inequality to prove it
$$(\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}}(\frac{2}{1})^{\frac{1}{2N+2}}\leq \frac{1}{\frac{2N+2}{2N+1}}((\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}})^{\frac{2N+2}{2N+1}}+\frac{1}{2N+2}((\frac{2}{1})^{\frac{1}{2N+2}})^{2N+2}=1$$
the two sides are equal iff $((\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}})^{\frac{2N+2}{2N+1}}=((\frac{2}{1})^{\frac{1}{2N+2}})^{2N+2}$, which is impossible
| {
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A closed expression for this sequence of integers Say we have $a_k=n-p$ for $n,p \in \mathbb{Z}_{+}$ where $n-p >0$ and $a_1=n-p-1$, and in general:
$a_k=n-p$
$a_1=n-p-1$
$a_{k-1}=n-p-2$
$a_2=n-p-3$
$a_{k-2}=n-p-4$
$a_{3}=n-p-5$
$\vdots$
and so on. The idea is that if the sequence $(a_1,a_2,...,a_k)$ is ordered from greatest to smallest, we have $(a_k,a_1,a_{k-1},a_{2},a_{k-2},a_3,...)$ where all the numbers are consecutive.
I want to find a general closed expression for $a_i$ for $i\in\{1,...,k\}$, but I cannot seem to find a way to come up with it.
| It is convenient to split the problem into two cases with even and odd $k$. Here we consider the case $k=2K$ even:
\begin{align*}
\left(a_j\right)_{1\leq j\leq 2K}=\left(a_1,a_2,\ldots,\color{blue}{a_K,a_{K+1}},\ldots,a_{2K}\right)
\end{align*}
We list the elements $a_j, 1\leq j\leq k=2K$ in a detailed way to better see what's going on. We obtain
\begin{align*}
\begin{array}{ll}
\qquad a_{2K}\ \ \ =a_k\quad=n-p&\qquad\qquad\; a_1=n-p-1\\
\qquad a_{2K-1}=a_{k-1}=n-p-2&\qquad\qquad\; a_2=n-p-3\\
\qquad a_{2K-2}=a_{k-2}=n-p-4&\qquad\qquad\; a_3=n-p-5\\
\qquad\quad\quad\ \ \ \vdots\qquad&\qquad\qquad \quad\ \ \;\vdots\\
a_{2K-(K+1)}=a_{k-(K-1)}=n-p-2(K-1)&\qquad\qquad a_K=n-p-(2K-1)\\
\qquad\qquad\ =a_{K+1}\quad\ \ =n-p-2K+2&\qquad\qquad \quad\ = n-p-2K+1
\end{array}
\end{align*}
From the list above we obtain for $\left(a_j\right)_{1\leq j\leq k}=\left(a_j\right)_{1\leq j\leq 2K}$:
\begin{align*}
\color{blue}{a_j}&=
\begin{cases}
n-p-2j+1&\qquad\quad\ \ \ \;1\leq j\leq K\\
n-p-2K+2(j-K)&\qquad K+1\leq j\leq 2K
\end{cases}\\
&\,\,\color{blue}{=
\begin{cases}
n-p-2j+1&\qquad\qquad\qquad\quad 1\leq j\leq K\\
n-p-4K+2j&\qquad\qquad\quad K+1\leq j\leq 2K
\end{cases}}
\end{align*}
The odd case can be obtained similarly.
| {
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Prove by induction if $H_n=\sum\limits_{i=1}^{n}\frac{1}{i}$ then $H_{2^n}\ge 1+\frac{n}{2}$? Suppose $H_n=\sum\limits_{i=1}^{n}\frac{1}{i}$ and we want to prove that $H_{2^n}\ge 1+\frac{n}{2}$?
*
*Since $\sum\limits_{i=1}^{2^1}\frac{1}{i}\ge1+\frac{1}{2}$ note that $H_{2^n}\ge 1+\frac{n}{2}$ for $n= 1$
*Assume $H_{2^n}\ge1+\frac{n}{2}$ for all $n\le k$.
Note $$\sum\limits_{i=1}^{2^{n}}\frac{1}{i}\ge1+\frac{n}{2}$$
However, I'm having trouble showing $\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \frac{1}{2}$ inorder to show that
$$H_{2^{n+1}}=\left(\sum\limits_{i=1}^{2^n}\frac{1}{i}\right)+\left(\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\right)\ge \left(1+\frac{n}{2}\right)+\left(\frac{1}{2}\right)=1+\frac{(n+1)}{2}$$
How do we show by induction that:
$\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \frac{1}{2}$
Is there a better method to prove by induction that:
$\sum\limits_{i=1}^{2^{n+1}}\frac{1}{i}\ge \frac{(n+1)}{2}+1$
| You see
$$\sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{i}\ge \sum\limits_{i=2^{n}+1}^{2^{n+1}}\frac{1}{2^{n+1} } = \frac{1}{2}$$
| {
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Prove that $~~a=b=c~~$ if $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are integers.
Prove that $~~a=b=c~~$ if $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are integers.
My first attempt was to add $~\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $~\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ together and I got that $\frac{(a+b)(b+c)(c+a)}{abc}$ is an integer. I was not able make any further progress from here.
Then I tried by multiplying them both.
I got $~\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right) \left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)=3+\frac{a^2}{bc}+\frac{bc}{a^2}+\frac{b^2}{ac}+\frac{ac}{b^2}+\frac{c^2}{ab}+\frac{ab}{c^2}$.
I was able to prove $~\frac{a^2}{bc}+\frac{bc}{a^2}+\frac{b^2}{ac}+\frac{ac}{b^2}+\frac{c^2}{ab}+\frac{ab}{c^2}$ is a integer but still could not make any progress.
Then I tried squaring $~\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$.
I got $~~\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)$ .
So I was able to prove $~\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} $ and $~\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}$ is an integer.
Using the same method I proved $~\frac{a^{2^k}}{b^{2^k}}+\frac{b^{2^k}}{c^{2^k}}+\frac{c^{2^k}}{a^{2^k}} $ and $~\frac{a^{2^k}}{c^{2^k}}+\frac{b^{2^k}}{a^{2^k}}+\frac{c^{2^k}}{b^{2^k}}$ is an integer although it was of no use.
I don't know how to proceed further. Any help would be appreciated.
| Note that this is false when $a,b,c$ aren’t integers, so we need to use that fact.
Let $p$ be prime. Let $x,y,z$ be the largest so that $p^x|a,p^y|b,p^z|c$.
If $a,b,c$ are all different then there’s some $p$ such that the corresponding $x,y,z$ aren’t the same. Either one is larger than the other two or one is smaller than the other two. In the first case, WLOG $x>y \geq z$. Then, $p$ doesn’t divide the numerator of $p^{x-y} b/a$ but does for $p^{x-y} c/b$ and $p^{x-y} a/c$, so it doesn’t divide the numerator of $p^{x-y}(b/a+c/b+a/c)$, so p is in the numerator of a reduced $b/a+c/b+a/c$, so it’s not an integer.
The case when one is smaller than the others is similar.
Thus, $a=b=c$.
| {
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Expressing $\cos(\frac{\pi}{2}+i)$ and $\sin(\frac{\pi}{2} +i)$ in the exponential function
Is $\cos(\frac{\pi}{2}+i)$ equal to $i\frac{-1+e^2}{2e}$ and $\sin(\frac{\pi}{2} +i)$ to $\frac{1+e^2}{2e}$?
Trying to answer a test online. Apparently the answer in the question is incorrect or I am not expressing it correctly. Is there a different way to express it with the exponential function?
| You can use the formulas
$$
\cos x = \frac 1 2 (e^{ix}+e^{-ix}) \qquad \sin x = \frac{1}{2i}(e^{ix}-e^{-ix})
$$
After inserting the values $\frac \pi 2 + i$ into both equations you get the results
$$
\cos \left(\frac \pi 2 + i\right) = \frac 1 2(e^{i\pi/2 + i^2}+e^{-i\pi/2-i^2} ) \\
=\frac 1 2\left(\frac i e + \frac{e}{i}\right) = \frac{i}{2e} - \frac{ie}{2}= i\frac{1-e^2}{2e}
$$
and
$$
\sin\left(\frac \pi 2 + i\right) = \frac{1}{2i}(e^{i\pi/2 + i^2}-e^{-i\pi/2-i^2} ) \\
=\frac{1}{2i}\left(\frac i e - \frac{e}{i}\right) = \frac{1}{2e} + \frac{e}{2}= \frac{e^2+1}{2e}
$$
using the fact that $e^{i\pi/2}=i$ and $i^2=-1$, thus $1/i=-i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
| $$1+x+x^2+x^3+\dots$$
$$=(1+x)(1+x^2+x^4+\dots)$$
$$=(1+x)(1+x^2)(1+x^4+\dots)$$
$$=(1+x)(1+x^2)(1+x^4)\dots$$
$$=\frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^8}{1-x^4}\dots$$
$$=\frac{1}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Looking for other approaches to find the height $AH$ in triangle $ ABC $ where $A(1,5)$ , $B(7,3)$, $C(2,-2)$
We have a triangle with vertices $A(1,5)$ , $B(7,3)$, $C(2,-2)$. What
is the length of the height $AH$ in the triangle $ABC$ ?
$1)4\qquad\qquad2)3\sqrt2\qquad\qquad3)5\qquad\qquad4)4\sqrt2$
This is a problem from a timed exam.
Here is my approach:
The equation of the line passes through $B$ and $C$ is $y_=x-4$ so slope of $AH$ is $-1$ and its a line passes through $A(1,5)$ so the equation for this line becomes $y=-x+6$. in order to find coordinate of $H$ we should equate the formulas of those lines:
$$x-4=-x+6\quad\to\quad x=5\quad\text{and $\quad y=1\quad$ So$\quad H (5,1)$}$$
$$\text{Distance from $A$ to $H$:$\qquad$}AH=\sqrt{(5-1)^2+(1-5)^2}=4\sqrt2$$
Although I believe my answer is quick but I'm looking for other ideas to solve this problem.
| Since the vertices have integer coordinates we can use Pick's theorem.
In the border of the triangle there are the three vertices, and also $(4,4)$ on $AB$, the points $(6,2)$, $(5,1)$, $(4,0)$ and $(3,-1)$ on $BC$ and no other points in $AC$. These make $8$ points in the border.
On the other hand the interior of the triangle is given by
$$\begin{cases}
x+3y<16\\
x-y<4\\
7x+y>12
\end{cases}$$
The left-most point of the triangle is $A(1,5)$. For $x=2$ we have six interior points $(2,-1),\ldots,(2,4)$. For $x=3$ we have from $(3,0)$ to $(3,4)$. For $x=4$ we have from $(4,1)$ to $(4,3)$. For $x=5$, only $(5,2)$ and $(5,3)$, and for $x=6$, $(6,3)$. That makes $17$ interior points.
If you relax the mathematical preciseness a bit, you can easily count them in a good graph:
By Pick's theorem, the area of the triangle is $20$. The length of $BC$ is $5\sqrt 2$, so
$$5\sqrt 2\cdot AH=40$$
$$10\cdot AH=40\sqrt 2$$
$$AH=4\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4160028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Does $\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)$ have a closed form? I am wondering if anyone has a nice way of approaching the following definite integral $\newcommand{\dilog}{\operatorname{Li}_2}$
$$\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)\,.$$
Here $C$ is a positive, real constant that satisfies the constraint $C>1$. So far I have tried a simple $u$ substitution, $u = \cos\left(\phi\right)/C$. However, this doesn't get me anywhere. I have also tried performing a series expansion in small $1/C$ in the second log, performing the integration, and then summing in powers of $1/C$. However, the sum does not simplify nicely.
I have also tried relating the expression to $\dilog\left(1-\frac{\cos^2\phi}{C^2}\right)$ and $\dilog\left(\frac{\cos^2\phi}{C^2}\right)$ and performing the integration of these polylog functions using their series representation. However I have trouble performing the summation for the $\dilog\left(1-\frac{\cos^2\phi}{C^2}\right)$ term.
| It seems that a series solution is possible
$$I=\int_0^{2\pi} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx=4\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx$$ Expanding the second logarithms
$$I= -4\sum_{n=1}^\infty \frac 1 {n\,c^{2n}}\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ The integrals
$$J_n=\int\log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ are relatively simple. One integration by parts gives
$$J_n=-\frac{\cos ^{2 n+1}(x) \log \left(\frac{\cos ^2(x)}{c^2}\right) \,
_2F_1\left(\frac{1}{2},n+\frac{1}{2};n+\frac{3}{2};\cos ^2(x)\right)}{2 n+1}-$$
$$\frac{1}{2} \Gamma \left(n+\frac{1}{2}\right)^2 \cos ^{2 n+1}(x) \,
_3\tilde{F}_2\left(\frac{1}{2},n+\frac{1}{2},n+\frac{1}{2};n+\frac{3}{2},n+\frac {3}{2};\cos ^2(x)\right)$$
Using the bounds
$$K_n=\int_0^{\frac \pi 2}\log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$
$$K_n=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right) \left(H_{n-\frac{1}{2}}-H_n-2 \log
(c)\right)}{2 \Gamma (n+1)}$$ Then
$$I=2 \sqrt{\pi }\sum_{n=1}^\infty \frac{ \Gamma \left(n+\frac{1}{2}\right) }{n \,c^{2 n}\,\Gamma (n+1)}\left(2 \log
(c)+H_n-H_{n-\frac{1}{2}}\right)$$
$$\color{blue}{I=8 \pi \log (c) \left(\log (2 c)-\cosh ^{-1}(c)\right)+2 \sqrt{\pi }\sum_{n=1}^\infty \frac{ \Gamma \left(n+\frac{1}{2}\right) }{n \,c^{2 n}\,\Gamma (n+1)}\left(H_n-H_{n-\frac{1}{2}}\right)}$$
If we denote the summand by $a_n$ we have, for large values of $n$
$$\frac{a_{n+1}}{a_n}=\frac 1{c^2}\left(1-\frac{5}{2 n} \right)+O\left(\frac{1}{n^2}\right)$$ which leads to a quite fast convergence.
For $c=2$, the partial sums (from $n=0$ to $n=p$)
$$\left(
\begin{array}{cc}
p & S_p \\
0 & 1.20789 \\
1 & 1.39219 \\
2 & 1.51043 \\
3 & 1.52617 \\
4 & 1.52870 \\
5 & 1.52914 \\
6 & 1.52923 \\
7 & 1.52925
\end{array}
\right)$$
We ca have save calculation time using for large enough $n$ the expansion
$$a_n=\frac{\sqrt \pi} {c^{2n}\, n^{\frac 52}}\left(1-\frac{3}{8 n}+\frac{5}{128 n^2}+\frac{35}{1024
n^3}+O\left(\frac{1}{n^4}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$?
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$
Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac x2)}{1+\tan^2(\frac{x}2)}=-\frac35\quad$. by testing the options we can find out $\tan(\frac x2)=-3$ works (although by solving the quadratic I get $\tan(\frac x2)=-\frac13$ too. $-3$ isn't the only possible value.)
I wonder is it possible to solve the question with other (quick) approaches?
| You have\begin{align}\frac{1-\sin(x)}{1+\sin(x)}=4&\iff\frac{\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}=4\\&\iff\left(\frac{\sin\left(\frac x2\right)-\cos\left(\frac x2\right)}{\sin\left(\frac x2\right)+\cos\left(\frac x2\right)}\right)^2=4\\&\iff\frac{\sin\left(\frac x2\right)-\cos\left(\frac x2\right)}{\sin\left(\frac x2\right)+\cos\left(\frac x2\right)}=\pm2\\&\iff\frac{\tan\left(\frac x2\right)-1}{\tan\left(\frac x2\right)+1}=\pm2\end{align}The only solution of the equation $\frac{\tan\left(\frac x2\right)-1}{\tan\left(\frac x2\right)+1}=2$ is $\tan\left(\frac x2\right)=-3$, which is on that list, whereas the only solution of the equation $\frac{\tan\left(\frac x2\right)-1}{\tan\left(\frac x2\right)+1}=-2$ is $\tan\left(\frac x2\right)=-\frac13$, which is not on that list.
So, the problem has two solutions, but only one of them is on the list of options.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Probability that the sum of two integers in $\{1,\dots,n\}$ equals a perfect square I found the following question in MIT OCW's Mathematical Problem Solving and I'd like to know if my solution is ok:
"Let $p_n$ be the probability that $c+d$ is a perfect square when the integers $c$ and $d$ are selected independently at random from the set $\{1,\dots,n\}$. Show that $\lim_{n\to \infty} p_n\sqrt{n}$ exists, and express this limit in the form $r(\sqrt{s}-t)$ where $s$ and $t$ are integers and $r$ is a rational number."
To me, it looks like that
$$p_n= \sum_{k=2}^{\lfloor\sqrt{n}\rfloor} \sum_{i=1}^{k^2-1} P\left(c=i,d=k^2-i\right)=\sum_{k=2}^{\lfloor\sqrt{n}\rfloor}\sum_{i=1}^{k^2-1} \frac{1}{n^2}=\frac{\lfloor\sqrt{n}\rfloor(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)}{6n^2}$$
for $n\ge 4$ and hence $\lim p_n\sqrt{n}=\frac{1}{3}$ using $\lfloor\sqrt{n}\rfloor\le \sqrt{n}<\lfloor\sqrt{n}\rfloor+1$ and sandwich theorem.
However, the form the problem tells to write the answer makes me think I may have made some mistake. Can someone say whether my solution is correct or not and in negative case, where I missed? Thanks in advance!
EDIT: I realized that the question does not say that the perfect square must be in $\{1,\dots,n\}$. So I must add to the above a sum with $\lfloor \sqrt{n}\rfloor<k\le \lfloor\sqrt{2n}\rfloor$, being careful to the restrictions that $1\le c,d\le n$. Hence:
$$p_n=\sum_{k=2}^{\lfloor\sqrt{n}\rfloor}\sum_{i=1}^{k^2-1} \frac{1}{n^2}+\sum_{k=\lfloor\sqrt{n}\rfloor+1}^{\lfloor\sqrt{2n-1}\rfloor} \sum_{i=k^2-n}^{n-1} \frac{1}{n^2}=$$
$$=\frac{m_n(2m_n^2+3m_n-2)}{3n^2}-\frac{q_n(q_n+1)(2q_n+1)}{6n^2}+\frac{2q_n}{n}-\frac{2m_n}{n}$$
where $m_n=\lfloor \sqrt{n}\rfloor$ and $q_n=\lfloor \sqrt{2n-1}\rfloor$. We can compute
$$\lim_n \frac{m_n(2m_n^2+3m_n-2)}{3n^{3/2}}=\frac{2}{3}$$
$$\lim_n \frac{q_n(q_n+1)(2q_n+1)}{6n^{3/2}}=\frac{2\sqrt{2}}{3}$$
$$\lim_n \frac{2q_n}{\sqrt{n}}=2\sqrt{2}$$
$$\lim_n \frac{2m_n}{\sqrt{n}}=2$$
Therefore
$$\lim_{n\to \infty}\ p_n\sqrt{n}=\frac{4}{3}(\sqrt{2}-1)\approx 0.552284749831$$
| When you choose two integers uniformly at random in $\{1, 2, ..., n\}$, the probability that the sum is $i$ is
$$
P(i) = \frac{n - |n+1 - i|}{n^2} = \frac{1}{n^2}\begin{cases}i -1 & 2 \le i \le n+1 \\ 2n+1-i & n+1 \le i \le 2n \\ 0 & \mathrm{else}\end{cases}
$$
So you get
$$P(sq) = \sum_{k = 2}^{\lfloor\sqrt{n}\rfloor}\frac{k^2-1}{n^2} + \sum_{k = \lfloor\sqrt{n}\rfloor + 1}^{\lfloor \sqrt{2n}\rfloor}\frac{2n+1 - k^2}{n^2} $$
And doing the sums gives
$$
P(sq)= \frac{\lfloor\sqrt{n}\rfloor(2\lfloor\sqrt{n}\rfloor^2 + 3\lfloor\sqrt{n}\rfloor - 6n - 5)}{3n^2}-\frac{\lfloor \sqrt{2n}\rfloor(2\lfloor \sqrt{2n}\rfloor^2 + 3\lfloor \sqrt{2n}\rfloor - 12n - 5)}{6n^2}.
$$
Letting $\epsilon = \sqrt{n} - \lfloor\sqrt{n}\rfloor$ and $\delta = \sqrt{2n}-\lfloor \sqrt{2n}\rfloor$, where $0 \le \epsilon,\delta < 1$, we have
$$
P(sq) = \frac{4(\sqrt{2}-1)}{3\sqrt{n}} + O\left(\frac{1}{n^{3/2}}\right).
$$
So the limit exists and is equal to $(4/3)(\sqrt{2}-1)$, as you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{k}{n^2}\right )^{\frac{k}{n^2} +1} $ Compute limit of sum :
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{k}{n^2} \right)^{\dfrac{k}{n^2} +1} $$
My Attempt :
$$\Big( \dfrac{k}{n^2} \Big)^{\frac{k}{n^2} +1} = e^{\Big( \frac{k}{n^2} +1 \Big) \log\Big(\frac{k}{n^2}\Big)} = \Big( e^{\Big( \frac{k}{n^2} +1 \Big) }\Big)^{\log\Big(\frac{k}{n^2}\Big)} = \dfrac{1}{n} \cdot \dfrac{k}{n} \Big( e^{\frac{k}{n^2}} \Big)^{\log (\frac{k}{n^2})} $$
I am unable to move ahead, please help.
| Trying to solve my problem moving along with Winther's idea (thanks a lot) :
$\Big( \dfrac{k}{n^2} \Big)^{\frac{k}{n^2} +1} = e^{\Big( \frac{k}{n^2} +1 \Big) \log\Big(\frac{k}{n^2}\Big)} = \Big( e^{\Big( \frac{k}{n^2} +1 \Big) }\Big)^{\log\Big(\frac{k}{n^2}\Big)} = \dfrac{1}{n} \cdot \dfrac{k}{n} \Big( e^{\frac{k}{n^2}} \Big)^{\log (\frac{k}{n^2})} $
$= \dfrac{1}{n} \cdot \dfrac{k}{n} \Bigg(1 + \mathcal{O} \Big( \frac{k}{n^2}\log (\frac{k}{n^2}) \Big) \Bigg) $
we want to use the given series expansion for small $x$ :
$x^x \sim \sum_{k = 0}^{\infty} \frac{1}{k!} (x \log x)^k = 1 + x \log x + \frac{1}{2} x^2 \log^2 x + O(x^3 \log^3 x) .$
first term obviously gives :
$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \dfrac{1}{n} \cdot \dfrac{k}{n} = \int_0^1 xdx = \dfrac{1}{2}$
for second term we use :
$ \dfrac{k}{n^2} \mathcal{O} \Big( \frac{k}{n^2}\log (\dfrac{k}{n^2}) \Big) = \Big[ \Big(\dfrac{k}{n^2}\Big)^2 \log \Big( \dfrac{k}{n^2} \Big) + ... \Big]$
where, $ n \cdot \Big(\dfrac{1}{n^2}\Big)^2 \log \Big( \dfrac{1}{n^2} \Big) < \displaystyle\sum_{k=1}^n \Big(\dfrac{k}{n^2}\Big)^2 \log \Big( \dfrac{k}{n^2} \Big) < n \cdot \Big(\dfrac{n}{n^2}\Big)^2 \log \Big( \dfrac{n}{n^2} \Big)$
using squeez theorem, we see, both sides sum will go to $0$ as $n \rightarrow \infty$ and similarly the higher order terms as well.
hence, answer is = $\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Complex integration $\int_{0}^{1}\frac{\sqrt[3]{4x^{2}\left(1-x\right)}}{\left(1+x\right)^{3}}dx$ I have integral
$$I_1=\int_{0}^{1}\frac{\sqrt[3]{4x^{2}\left(1-x\right)}}{\left(1+x\right)^{3}}dx.$$
I tried something like this:
$$f\left(z\right)=\frac{\sqrt[3]{4z^{2}\left(1-z\right)}}{\left(1+z\right)^{3}}=\frac{\left(1-z\right)}{\left(1+z\right)^{3}}\left(\frac{2z}{1-z}\right)^{\frac{2}{3}}$$
Using the figure I have
$$I=\int_{C_{R}}f\left(z\right)dz+\int_{C_{r1}}f\left(z\right)dz+\int_{z_{1}}f\left(z\right)dz+\int_{C_{r2}}f\left(z\right)dz+\int_{z_{2}}f\left(z\right)dz.\tag1$$
Since I have only one singularity in my contour:
$$\operatorname {Res}\left(f,z=-1\right) =\frac{1}{\left(3-1\right)!}\cdot\lim_{z\rightarrow\left(-1\right)}\left[\left(z+1\right)^{3}\cdot f\left(z\right)\right]^{''}\\
=\frac{1}{2}\cdot\lim_{z\rightarrow\left(-1\right)}\frac{-2\cdot4^{\frac{1}{3}}z^{2}}{9\left(-z^{2}\left(-1+z\right)\right)^{\frac{5}{3}}}\\
=\frac{1}{2}\cdot\frac{-2\cdot4^{\frac{1}{3}}\left(-1\right)^{2}}{9\left(-\left(-1\right)^{2}\cdot\left(-1-1\right)\right)^{\frac{5}{3}}}\\
=-\frac{2^{\frac{2}{3}}\cdot1}{9\left(2\right)^{\frac{5}{3}}}=-\frac{1}{9}2^{\frac{2}{3}-\frac{5}{3}}=-\frac{1}{9}2^{-1}=-\frac{1}{18}.$$
Thus I got
$$I=2\pi i\cdot\left(-\frac{1}{18}\right)=-\frac{\pi i}{9}.$$
Now, I use Jordan's lemma and get:
$$\lim_{z\rightarrow0}z\cdot f\left(z\right)=0$$
$$\lim_{z\rightarrow1}\left(z-1\right)f\left(z\right)=0$$
$$\lim_{z\rightarrow\infty}z\cdot f\left(z\right)=0$$
So I conclude that
$$\lim_{r\rightarrow0}\int_{C_{r1}}f\left(z\right)dz=0$$
$$\lim_{r\rightarrow0}\int_{C_{r2}}f\left(z\right)dz=0.$$
$$\lim_{R\rightarrow\infty}\int_{C_{R}}f\left(z\right)dz=0.$$
When I put $r\rightarrow0$ and $R\rightarrow\infty$ into (1) and put that on $z_1$ the argument of the function is 0, and on $z_2$ the argument is $2\pi \cdot \frac{2}{3}$ then I have:
$$I_{1}\left(1-e^{i\frac{4\pi}{3}}\right)=-\frac{\pi}{9}i.$$
Now I have the problem because on the left side I have real and imaginary part and on the right hand side I only have imaginary part and I don't know what to do.
| You already showed that the integral over the three circles vanishes in the limit, and we are left with (since the argument of $1 - z$ increases by $-2\pi$ when moving from the upper to the lower edge)
$$I_{1}\left(1 - e^{-2\pi i / 3}\right) = 2\pi i\operatorname{Res}(f,-1).$$
Since we are dealing with the third root, the difficult part will be to get the argument right.
Write
$$g(z) = 4^{1/3}z^{2/3}(1 - z)^{1/3}$$
so
$$f(z) = \frac{g(z)}{(1 + z)^3}$$
and
$$\operatorname{Res}(f,-1) = \frac12 g''(-1).$$
Now our definition of $f$ determines how $g$ is defined on this branch, namely in such a way that for $x \in (0,1)$ we have that $g(x)$ is positive and real. In order to be able to determine the branch of its derivatives, we have to express those in $g$ itself and we see:
$$g'(z) = \left(\frac2{3z} - \frac1{3(1 - z)}\right)g(z) = q(z)g(z)$$
if we set
$$q(z) = \frac2{3z} - \frac1{3(1 - z)},$$
hence also
$$g''(z) = q'(z)g(z) + q(z)g'(z) = \left(q'(z) + q(z)^2\right)g(z).$$
Let's write out $q'(z)$:
$$q'(z) = -\frac2{3z^2} - \frac1{3(1 - z)^2}.$$
Then
$$g''(-1) = \left(q'(-1) - q(-1)^2\right)g(-1) = \left(-\frac56 - \left(-\frac34\right)^2\right)g(-1) = -\frac{g(-1)}{18}.$$
We clearly have that $|g(-1)| = 2$, so we only need its argument. The argument of $g(x)$ for $x \in (0,1)$ is 0, and when going from $x$ to $-1$ the argument of $1 - z$ doesn't change, while that of $z$ goes from 0 to $\pi$. It follows that the argument of $g(z)$ goes up by $\frac23\pi$, so
$$g''(-1) = -\frac{e^{2\pi i/3}}9$$
and
$$\operatorname{Res}(f,-1) = -\frac{e^{2\pi i/3}}{18}.$$
Finally
$$I_{1} = 2\pi i\frac{\operatorname{Res}(f,-1)}{1 - e^{-2\pi i / 3}} = -\frac{2\pi i}{18}\frac{e^{2\pi i/3}}{1 - e^{-2\pi i / 3}} = \frac{\pi\sqrt3}{27}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
| Observe,
$$a^2+b^2=2\implies (a+b)^2=2(ab+1)$$
It remains to show that
$$3(a+b)+(ab+1)+4\ge0\;\Longleftrightarrow\;3(a+b)+\frac{(a+b)^2}{2}+4\ge0$$
Let $a+b=x$, we have,
$$3x+\frac{x^2}{2}+4\ge0\;\Longleftrightarrow\;x^2+6x+8\ge0\;\Longleftrightarrow\;(x+2)(x+4)\ge0$$
$$\therefore\;x\le-4 \;\;\text{or}\;\;x\ge -2$$
Since you have shown that $a+b\ge-2$, the inequality must be true.
Alternate Solution:
Using AM-GM inequality,
$$\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}\implies 1\ge ab$$
Adding the inequalities,
$$3a+3b+ab\ge-5$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\ge ab$$
It remains to show that $a+b\ge-2$ which we know is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Diagonalizing the matrix $A$ then finding $A^{10}$ Diagonalize the matrix
$$A= \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}$$
Then find $A^{10}.$
We have the characteristic polynomial of $A:$
$$\left | A- \lambda I \right |= \left | \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}- \lambda\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \right |= \begin{vmatrix} -3- \lambda & -14 & -10\\ 2 & 13- \lambda & 10\\ -2 & -7 & -4- \lambda \end{vmatrix}=$$
$$= \left ( 6- \lambda \right )\left ( 1+ \lambda \right )\left ( 1- \lambda \right )$$
which has the roots $\lambda_{1}= 6, \lambda_{2}= -1, \lambda_{3}= 1.$ These roots are the diagonal elements as well as the eigenvalues of $A.$ These three eigenvalues correspond to the eigenvectors $v_{1}= \left ( 2, -2, 1 \right ),$
$v_{2}= \left ( 2, -1, 1 \right ), v_{3}= \left ( 1, -1, 1 \right ).$ Hence, $A= SJS^{-1}$ with $S= \begin{pmatrix} 2 & 1 & 2\\ -1 & -1 & -2\\ 1 & 1 & 1 \end{pmatrix},$ $J= \begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 6 \end{pmatrix}, S^{-1}= \begin{pmatrix} 1 & 1 & 0\\ -1 & 0 & -2\\ 0 & -1 & -1 \end{pmatrix}.$ I have no plans to find $A^{10}.$ I need to the helps
| $A^{10}=(SJS^{-1})^{10}=SJ^{10}S^{-1}$ and it is easy to find $J^{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove that $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $ $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $
the only idea I have is using the series expansion of $x^{-x} \approx 1 - x\log x + \dfrac{(x\log x)^2}{2!} - \ldots$. But it ends up little complicated, any idea?
| Yes go ahead
$$S=\int_{0}^{1} x^{-x} dx= \int_{0}^{1} e^{-x\ln x} dx =\int_{0}^{1} \sum_{k=0}^{\infty} \frac{(-x \ln x)^k}{k!}dx.$$
Let $x=e^{-t}$,and use $\int_{0}^{\infty} t^n e^{-at}dt=\frac{n!}{a^{n+1}}.$
$$\implies S=\sum_{k=0}^{\infty} \int_{0}^{\infty}\frac{t^ke^{-(k+1)t}}{k!} dt=\sum_{k=0}^{\infty}\frac{1}{(k+1)^{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For $a>0$, $\displaystyle\int_{\arctan a}^{\frac{\pi}{2}}\int_{0}^{a\csc\theta}\frac{r}{(1+r^2)^2}\,dr\,d\theta=?$ I found an answer, $\displaystyle\int_{\arctan a}^{\frac{\pi}{2}}\int_{0}^{a\csc\theta}\frac{r}{(1+r^2)^2}\,dr\,d\theta=\frac{a}{2\sqrt{a^2+1}}\arctan\left(\frac{1}{\sqrt{a^2+1}}\right)$ for $a>0$.
But I am not sure how to get the result.
First of all, $\displaystyle\int_{0}^{a\csc\theta}\frac{r}{(1+r^2)^2}\,dr=-\frac{1}{2}\left(\frac{1}{1+a^2\csc^2\theta}-1\right)$ and using an integral calculator, $\displaystyle\int-\frac{1}{2}\left(\frac{1}{1+a^2\csc^2\theta}-1\right)\,d\theta=\frac{a}{2\sqrt{a^2+1}}\arctan\left(\frac{\sqrt{a^2+1}\tan \theta}{a}\right)+C$. But I can't evaluate $\tan \theta$ at $\theta=\frac{\pi}{2}$. How can I get to the answer above?
| So, your second integral is
$$\frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2 \csc^2 \theta}{1 + a^2\csc^2\theta}d\theta$$
Multiplying the numerator and denominator by $\sin^2\theta$, you get
$$\frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2}{\sin^2\theta + a^2}d\theta$$
The next step is very generic for solving such integrals.
Hint: Divide both numerator and denominator by $\cos^2 \theta$. Should be doable after this. I will write the complete solution anyway, but you can skip from here and try solving on your own.
$$\frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2\sec^2
\theta}{\tan^2\theta + a^2\sec^2\theta}d\theta = \frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2\sec^2
\theta}{(a^2+1)\tan^2\theta + a^2}d\theta$$
Now, let $\tan \theta = x$, you get
$$\frac{1}{2}\int_a^\infty \frac{a^2}{(a^2+1)x^2+a^2}dx = \frac{a}{2\sqrt{a^2+1}} \left[\tan^{-1}\frac{\sqrt{a^2+1}}{a}x\right]_a^\infty$$
This gives us
$$\frac{1}{2}\int_{\tan^{-1}a}^{\frac{\pi}{2}}\frac{a^2 \csc^2 \theta}{1 + a^2\csc^2\theta}d\theta = \frac{a}{2\sqrt{a^2+1}}\left[\frac{\pi}{2} - \tan^{-1}\sqrt{a^2+1}\right] = \frac{a}{2\sqrt{a^2+1}} \tan^{-1}\frac{1}{\sqrt{a^2+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the area of the part of the sphere $x^2+y^2+z^2=2$ that lies inside the cylinder $y^2+z^2=1$ This is what I have tried so far. I am just not sure whether I got it right or not.
We have $y^2+z^2=1$, then $r=1,\quad x=\sqrt{2-y^2-z^2}\quad\text{and}\quad x=\sqrt{2-r^2}$
$$\frac{\partial{x}}{\partial{y}}=-\frac{y}{\sqrt{2-y^2-z^2}}\quad\text{and}\quad
\frac{\partial{x}}{\partial{z}}=-\frac{z}{\sqrt{2-y^2-z^2}}
$$
$$\begin{align}
A(S)
&=2\iint\limits_{R}\sqrt{1+\left(-\frac{y}{\sqrt{2-y^2-z^2}}\right)^2+\left(-\frac{z}{\sqrt{2-y^2-z^2}}\right)^2}\,\mathrm dA\\
&=2\iint\limits_{R}\sqrt{1+\frac{y^2+z^2}{2-y^2-z^2}}\,\mathrm dA\\&=2\sqrt{2}\iint\limits_{R}\frac{1}{\sqrt{2-y^2-z^2}}\,\mathrm dA
\end{align}
$$
Converting to polar coordinate $R=\{(r, \theta)\mid0\leq r \leq 1, 0\leq \theta \leq 2\pi\}$
$$A(S)= 2\sqrt{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\frac{r}{\sqrt{2-r^2}}\,\mathrm dr\,\mathrm d\theta=4\pi(2-\sqrt{2})$$
| My calculations are easy, but I use the
arc length of parametric curves. The question is to
find the area of the sphere $x^2+y^2+z^2=2$ inside the cylinder $x^2+y^2=1.$ (The changed variables give the same area.)
The total area of two caps equals $8A,$ where $A$ is one-quarter of the area
of the top cap. I use cylindrical coordinates and
consider an arbitrary point $P=(r\cos\theta, r\sin\theta, \sqrt{2-r^2})$ on the top cap. Observe
that the length of
$${\partial\over\partial r}P=(\cos\theta, \sin\theta, {{-r}\over{\sqrt{2-r^2}}})$$
equals $\sqrt{2\over2-r^2}$ and is independent of $\theta.$
(Geometrically, this length times $dr$ is the length of a piece of arc on the great circle
for fixed $\theta.$ The small piece sweeps out an area when rotated about the $Z$-axis.)
$$A=\int_0^{\pi\over2}\int_0^1 \sqrt2 \,(2-r^2)^{-{1\over2}} \,r\,dr\,d\theta$$
$$={\sqrt2\pi\over2}\int_0^1 (2-r^2)^{-{1\over2}} \,r\,dr\,.\ \ \ \ \text{Thus,}$$
$$8A={4\sqrt2\pi}\sqrt{2-r^2}\big|_1^0={4\sqrt2\pi}(\sqrt2-1)={4\pi}(2-\sqrt2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equation}
\begin{equation}
(y+1)^{8}=y^{8}+a_{7} y^{7}+a_{6} y^{6}+\cdots+a_{1} y+1
\end{equation}
\begin{equation}
(y-1)^{8}=y^{8}-a_{7} y^{7}+a_{6} y^{6}+\cdots-a_{1} y+1
\end{equation}
\begin{equation}
\begin{array}{l}
2\left(y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}+1\right)=2 \Leftrightarrow \\
y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}=0 \Leftrightarrow y^{2}=0 \Leftrightarrow y=0
\end{array}
\end{equation}
y=0, x= -1
Maybe there is another way of solving this...
| I encourage you to keep working on your solution using derivatives. However, solving this equation does not need derivatives. Here I write a hint and if you need more help I'll give you another one!
Hint1:
Rearrange the equation: $(x+2)^8 - 1 = 1 - x^8$
Edit: Ah! Now that you posted your work I see you don't need help. That is a good solution. Well done!
Now, here is an alternative solution. Not as nice and concise as yours, but still might be a good exercise. Like you, I too felt more comfortable with the variable change $y = x + 1$ . In fact this would be the hint number two! We have:
$$(y+1)^8 - 1^8 = 1^8 - (y-1)^8$$
$$[(y+1)^4+1][(y+1)^2+1](y+2)y = [1+(y-1)^4][1+(y-1)^2]y(2-y)$$
Clearly, one answer is $y=0$, which means $x=-1$ .
Below, we show that the equation does not have any other real answers. Assuming that $y \ne 0$ we can divide both sides by $y$ :
$$[(y+1)^4+1][(y+1)^2+1](y+2) = [1+(y-1)^4][1+(y-1)^2](2-y)$$
Without any more factorisation, we can find several useful things in the above equation:
Firstly, we can see that if $y>2$ or $y<-2$ then the two sides of the equation have different signs. Therefore the equation does not have any solutions in those value ranges.
Secondly, at $y=2$ and at $y=-2$ one side of the equation becomes zero while the other side is nonzero. So these two are not solutions of the equation either.
Finally, for $-2 < y < 2$ where both sides are positive, we can see that if $y > 0$ then each term of the left hand side is greater than its corresponding term in the right hand side (therefore left hand side $ > $ right hand side); and if $y < 0$ then each term of the left hand side is less than its corresponding term in the right hand side (therefore left hand side $ < $ right hand side). Either way, the two sides cannot be equal.
So we are left with the only case that $y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
Here is a procedure how to do it analiticaly.
*
*If $T(x_0,y_0)$ is a touching point, then $x_0+y_0=5$
*The equation of ellipse is $${(x_0-{3\over 2})^2\over a^2} +{(y_0+1)^2 \over b^2}=1$$
*Since $2e=1$ we have $a^2-b^2 = {1\over 4}$
*Since the slope of tangent is $-1$ we have $${2(x_0-{3\over 2})\over a^2} -{2(y_0+1)\over b^2}=0$$
And now we have to solve this tedious system. How to do it more geometrical?
| We have the coordinates of any point on the ellipse centered at the origin $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in parametric form as $P(a\cos\theta,b\sin\theta)$, where $\theta$ is a parameter. Now, by using basic calculus, we get the equation of tangent to the ellipse at this point as $\frac{x}{a^2}a\cos\theta+\frac{y}{b^2}b\sin\theta=1$.
Now, by the concept of shifting of origin, we can obtain similar results for the ellipse given in the question.
$\Rightarrow$ Equation of ellipse: $$\frac{(x-\frac32)^2}{a^2}+\frac{(y+1)^2}{b^2}=1$$
$\Rightarrow$ Coordinates of any point on the ellipse in parametric form: $Q(a\cos\theta+\frac32,b\sin\theta-1)$
$\Rightarrow$ Equation of tangent to the ellipse at this point: $$\frac{(x-\frac32)}{a^2}a\cos\theta+\frac{(y+1)}{b^2}b\sin\theta=1$$
$$\Rightarrow \frac xa \cos\theta+\frac yb \sin\theta=\frac{3\cos\theta}{2a}-\frac{\sin\theta}{b}+1$$
Comparing this with the given equation of tangent $x+y=5$, we get $$\frac{\cos\theta}{a}=\frac{\sin\theta}{b}=\frac{\frac{3\cos\theta}{2a}-\frac{\sin\theta}{b}+1}{5}\ldots(1)$$
Also, we have $$a^2-b^2=\frac14\ldots(2)$$
Now, we can assume $\frac{\cos\theta}{a}$ and $\frac{\sin\theta}{b}$ as two single variables and solve for them to get $$\Rightarrow\frac{\cos\theta}{a}=\frac{\sin\theta}{b}=\frac29\ldots(3)$$
$(2)$ and $(3)$ can easily be solved and we get $\cos\theta=\sqrt\frac{41}{81}$ and $ \sin\theta=\sqrt\frac{40}{81}$.
Now you can easily obtain the coordinates of $Q$ by substituting $a$ and $b$ in terms of $\cos\theta$ and $\sin\theta$.
Therefore, $$\Rightarrow Q=(\frac{34}{9},\frac{11}{9})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that $3 \mid n^4-n^2 \forall n \in \mathbb{Z^+}, n \ge 2$. Proposition: $3 \mid n^4-n^2$ for all $n \in \mathbb{Z^+}, n \ge 2$
My attempt
Lemma: $3 \mid (m-1)m(m+1)$
Proof.
Suppose $m \in \mathbb{Z}$. By the QRT, we have $m=3q+r$ $\,\ni\, r \in \{0,1,2\}$, and $q=\lfloor{\frac{m}{3}}\rfloor$.
We want to show that $(m-1)m(m+1)$ is divisible by 3.
Clearly, if any one of the three factors in $(m-1)m(m+1)$ is divisible by 3, then the product must also be divisible by 3.
$\cdot$ Case 1 ($r=0$): If $r=0$, then $m=3q$ is divisible by 3.
$\cdot$ Case 2 ($r=1$): If $r=1$, then $m-1=(3q+1)-1=3q$ is divisible by 3.
$\cdot$ Case 3 ($r=2$): If $r=2$, then $m+1=(3q+2)+1=(3q+3)=3(q+1)$ is divisible by 3.
In either case, one of the three factors in $(m-1)m(m+1)$ is divisible by 3. Thus $3 \mid (m-1)m(m+1)$. Therefore, the product of any 3 consecutive integers is divisible by 3.
Proof.
Base case: Let $n=2$. Then $\exists k \in \mathbb{Z}$ such that $12=3k$, namely, $k=4$. Thus $3 \mid 12$ and hence the claim holds for the base case.
Assume for positive integer $m \ge 2$ that $m^4-m^2=3k$ for some $k \in \mathbb{Z}$. We need only to show that this implies that $3 \mid (m+1)^4-(m+1)^2$.
$(m+1)^4-(m+1)^2= \sum_{i=0}^{4} {4\choose i} m^i \cdot 1^{4-i}-(m+1)^2=(m^4-m^2)+4m^3+6m^2+2m=(m^4-m^2)+3(m^3+2m^2+m)+(m^3-m)$.
By the inductive hypothesis, $3 \mid(m^4-m^2)$. And since $(m^3+2m^2+m) \in \mathbb{Z}$, it follows that $3 \mid 3(m^3+2m^2+m)$. Note that $(m^3-m)=m(m^2-1)=(m-1)m(m+1)$. By the Lemma, $3 \mid (m^3-m)$.
Thus we have $3 \mid (m^4-m^2)$, $3 \mid 3(m^3+2m^2+m)$, and $3 \mid (m^3-m)$.
To show that their sum is divisible by 3, let $k_1,k_2,k_3 \in \mathbb{Z} \,\ni\, m^4-m^2=3k_1, 3(m^3+2m^2+m)=3k_2$, and $(m^3-m)=3k_3$. Thus $3k_1+3k_2+3k_2=3(k_1+k_2+k_3)=m^4-m^2+3(m^3+2m^2+3m)+(m^3-m)$, where $(k_1+k_2+k_3) \in \mathbb{Z}$.
Hence $3 \mid (m+1)^4-(m+1)^2$.
Therefore, by induction, $3 \mid n^4-n^2, \forall n \in \mathbb{Z^+}, n \ge 2$.
| Note that $n^4 - n^2 = (n^2 - n) (n^2 + n) = (n - 1) n^2 (n + 1)$, one of the three factors is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$5^x - y^2 = 4$ Diophantine Equation I wrote a Diophantine equation and tried solving it. Then I got stuck at a stage of the solution.
Problem: Find all $(x,y)$ positive integer pairs that satisfy the equation $5^x - y^2=4$.
My Partial Solution: If $x$ is an even number then $x=2n\quad$ ($n \geq 1$ an integer). Therefore
$$ (5^n - y)(5^n + y) = 4$$
and we can write
\begin{equation*}
\left\{
\begin{split}
5^x-y & = 1 \\
5^x + y & = 4
\end{split}
\right.
\end{equation*}
Thus, $2\cdot 5^x = 5$ and there is no positive integer solution in this case.
If $x$ is an odd number,
$\bullet$ For $x=1$; $\quad 5^1 -y^2=4 \implies y=1$.
$\bullet$ For $x=3$; $\quad 5^3 -y^2=4 \implies y=11$.
$\bullet$ For $x\geq 5$; I thought of finding a contradiction using modular arithmetic. For example; $x=2k + 3 , \quad$ ($k\geq 1 $ an integer) $125\cdot 25^k - y^2 = 4$. In $\mod 24$,
$$5 - y^2 \equiv 4 \pmod{24}$$
But this is not a contradiction. So, I failed. How can I tell if the equation has a solution for $x>3$ or not? Thanks for your interest.
| $$5^{x} - y^2 = 4\tag{1}$$
We take the three cases $x=3a, x=3a+1$, and $x=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet\ x=3a$
Let $X=5^{a}$, then we get
$y^2 =X^3 - 4.$
According to LMFDB, this elliptic curve has integral solutions $(X,y)=(2,\pm 2), (5,\pm 11).$
From $(5,\pm 11),$ we get $(x,y)=(3,11).$
$\bullet\ x=3a+1$
Let $X=5\cdot5^{a}, Y=5y$, then we get
$Y^2 =X^3 - 100.$
This elliptic curve has integral solutions $(X,Y)=(5,\pm 5),(10,\pm 30),(34,\pm 198).$
From $(5,\pm 5)$, we get $(x,y)=(1,1).$
$\bullet\ x=3a+2$
Let $X=25\cdot5^{a}, Y=25y$, then we get
$Y^2 =X^3 - 2500.$
This elliptic curve has integral solution $(X,Y)=(50,\pm 350).$
We get no solution $(x,y).$
Hence there are only integral solutions $(x,y)=(1,1),(3,11).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Minimum percentage of a random variable within two bounds
Random variable $Z$ has a mean of $15$ and a standard deviation of $2$. What is the minimum percentage of $Z$ values that lie between $8$ and $17$?
I have tried the following:
Here on the right side the value is $17$ that is $1$ sd, on the left side it is $(8-15)/2 =3.5$ sd. Since it has asked for minimum percentage I tried Chebyshev rule, which states that the probability is no less than $1-1/k^2$, where $k$ is number of standard deviations from mean.
For $1$ sd the minimum percentage of $Z$ values is $0$. For $3.5$ sd the minimum percentage of $Z$ values is $91.8$. But number of sd is unequal on both side of mean.
Am I missing something? Please help in solving this.
|
$\def\F{\mathscr{F}}\def\R{\mathbb{R}}\def\d{\mathrm{d}}\def\brace#1{\left\{#1\right\}}\def\paren#1{\left(#1\right)}$Proposition: For any $σ > 0$, define$$
\F_σ = \brace{ F: \R → [0, 1] \,\middle|\, F\ \text{is c.d.f.}, \int_\R x \,\d F(x) = 0, \int_\R x^2 \,\d F(x) = σ^2 }.
$$
If $a, b > 0$, then$$
\min_{F \in \F_σ} \int_{(-a, b)} \d F(x) = \begin{cases}
0; & σ^2 > ab\\
\dfrac{4(ab - σ^2)}{(a + b)^2}; & \dfrac{1}{2} (ab - \min(a^2, b^2)) < σ^2 \leqslant ab\\
\dfrac{\min(a^2, b^2)}{\min(a^2, b^2) + σ^2}; & σ^2 \leqslant \dfrac{1}{2} (ab - \min(a^2, b^2))
\end{cases}.
$$
Proof: For $F \in \F_σ$, suppose $F$ is the c.d.f. of random variable $X_F$.
Case 1: $σ^2 > ab$. Obviously $P(X_F \in (a, b)) \geqslant 0$ for any $F \in \F_σ$, and the equality is attained if$$
P(X_F = -a) = \frac{σ^2}{a^2 + σ^2}, \quad P\paren{ X_F = \frac{σ^2}{a}} = \frac{a^2}{a^2 + σ^2}.
$$
Case 2: $\dfrac{1}{2} (ab - \min(a^2, b^2)) < σ^2 \leqslant ab$. Note that for $F \in \F_σ$,$$
E((a + X_F)(b - X_F)) = ab + (b - a) E(X_F) - E(X_F^2) = ab - σ^2,
$$
and $(a + x)(b - x) \leqslant 0$ for $x \in (-a, b)^c$, thus\begin{gather*}
E((a + X_F)(b - X_F) I_{(-a, b)}(X_F)) \geqslant E((a + X_F)(b - X_F)) = ab - σ^2. \tag{1}
\end{gather*}
By AM-GM inequality,$$
(a + x)(b - x) \leqslant \paren{ \frac{1}{2} ((a + x) + (b - x)) }^2 = \frac{1}{4} (a + b)^2, \quad \forall x \in (a, b)
$$
therefore$$
\frac{1}{4} (a + b)^2 P(X_F \in (-a, b)) \geqslant E((a + X_F)(b - X_F) I_{(-a, b)}(X_F)) \geqslant ab - σ^2,
$$
which implies that$$
P(X_F \in (-a, b)) \geqslant \frac{4(ab - σ^2)}{(a + b)^2}.
$$
The equality is attained if\begin{align*}
\small P(X_F = -a) = \frac{b^2 - ab + 2σ^2}{(a + b)^2}, \quad P\paren{ X_F = \frac{1}{2}(b - a) } = \frac{4(ab - σ^2)}{(a + b)^2}, \quad P(X_F = b) = \frac{a^2 - ab + 2σ^2}{(a + b)^2}.
\end{align*}
Case 3: $σ^2 \leqslant \dfrac{1}{2} (ab - \min(a^2, b^2))$. Without loss of generality, suppose $a \leqslant b$. Note that for $F \in \F_σ$,$$
E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } } = (a^2 + 2σ^2) + \frac{2σ^2}{a} E(X_F) - E(X_F^2) = a^2 + σ^2,
$$
and $a + \dfrac{2σ^2}{a} \leqslant b$ implies that $(a + x) \paren{ a + \dfrac{2σ^2}{a} - x } \leqslant 0$ for $x \in (-a, b)^c$, thus\begin{gather*}
{\small E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } I_{(-a, b)}(X_F) } \geqslant E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } } = a^2 + σ^2.} \tag{2}
\end{gather*}
By AM-GM inequality,$$
\small (a + x) \paren{ a + \frac{2σ^2}{a} - x } \leqslant \paren{ \frac{1}{2} \paren{ (a + x) + \paren{ a + \frac{2σ^2}{a} - x } } }^2 = \paren{ a + \frac{σ^2}{a} }^2, \quad \forall x \in (-a, b)
$$
therefore$$
\small \paren{ a + \frac{σ^2}{a} }^2 P(X_F \in (-a, b)) \geqslant E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } I_{(-a, b)}(X_F) } \geqslant a^2 + σ^2,
$$
which implies that$$
P(X_F \in (-a, b)) \geqslant \frac{a^2}{a^2 + σ^2}.
$$
The equality is attained if\begin{gather*}
P(X_F = -a) = \frac{σ^2}{a^2 + σ^2}, \quad P\paren{ X_F = \frac{σ^2}{a} } = \frac{a^2}{a^2 + σ^2}. \tag*{$\Box$}
\end{gather*}
Now return to the question. Since $E(Z) = 15$ and $D(Z) = 2^2 = 4$, define $X = Z - 15$, then $E(X) = 0$ and $E(X^2) = 4$. Because $8 - 15 = -7$, $17 - 15 = 2$, and $\dfrac{1}{2} (7 · 2 - 2^2) > 4$, so the proposition shows that$$
P(Z \in (8, 17)) = P(X \in (-7, 2)) \geqslant \frac{2^2}{2^2 + 4} = \frac{1}{2},
$$
and the equality is attained if $P(X = -2) = P(X = 2) = \dfrac{1}{2}$, i.e. $P(Z = 13) = P(Z = 17) = \dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Triple integral between two spheres Evaluate
$$
\iiint_{E}(2.1 z) d V
$$
where $E$ is bounded between two spheres:
$$
x^{2}+y^{2}+z^{2}=8.5^{2} \text { and } x^{2}+y^{2}+(z-8.5)^{2}=8.5^{2} .
$$
Region between two spheres
I am supposed to convert the integral to spherical coordinate $(\rho, \phi, \theta)$ where $\phi$ is the azimuthal angle and $\theta$ is the polar angle. I don't know how to determine the boundaries of the three variables. Please help me.
Here is my attempt
We have
$\rho = \sqrt{x^2 + y^2}$
The region is in the upper half of the first sphere so
$$x^2 + y^2 + z^2 = 8.5^2 \implies z = \sqrt{8.5^2 - \rho^2} \quad (1)$$
For the second sphere, it is in the lower half so I took the minus sign instead
$$x^2 + y^2 + (z-8.5)^2 = 8.5^2 \implies z = -\sqrt{8.5^2 - \rho^2} + 8.5 \quad (2)
$$
From (1) and (2), I found the intersection $\displaystyle z = 4.25, \rho = \frac{17\sqrt3}{4}$
Therefore the integral is
$$\int_0^{2\pi}\int_{0}^{\frac{17\sqrt3}{4}}\int_{-\sqrt{8.5^2 - \rho^2} + 8.5}^{\sqrt{8.5^2 - \rho^2}} (2.1z\rho) dzd\rho d\phi$$
| Your working is correct for cylindrical coordinates. If you were to do it in spherical coordinates,
$x = \rho \cos\phi \sin \theta, y = \rho \sin\phi \sin\theta, z = \rho \cos\theta$
$x^2 + y^2 + (z-8.5)^2 = 8.5^2 \implies x^2 + y^2 + z^2 = 17 z$
So $ \ \rho = 17 \cos\theta, 0 \leq \theta \leq \frac{\pi}{2}$
The sphere centered at the origin is $\rho = \dfrac{17}{2}$
At intersection of both sphere, $\rho = 17 \cos\theta = \dfrac{17}{2} \implies \theta = \dfrac{\pi}{3}$
Now if we integrate wrt $\rho$ first and then $\theta$, we need to split it into two integrals. For $0 \leq \theta \leq \dfrac{\pi}{3}$, $\rho$ is bound above by the sphere centered at the origin whereas for $\dfrac{\pi}{3} \leq \theta \leq \dfrac{\pi}{2}$, $\rho$ is bound above by the sphere $\rho = 17 \cos\theta$.
But if we integrate wrt $\theta$ first, we can set it up in one integral.
$\displaystyle \int_0^{2\pi} \int_0^{17/2} \int_0^{\arccos(\rho/17)} 1.05 \rho^3 \sin (2\theta) \ d\theta \ d\rho \ d\phi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Confused about order of operations to simplify $\frac15\div\frac15\div\frac15\div\frac15\div\frac15\div5\div5\div5$ (and others) Here is the question that confused me:
$$\text{What is the value of}\;\frac15\div\frac15\div\frac15\div\frac15\div\frac15\div5\div5\div5\;? \tag1$$
If the signs stay the same, is the operation done LTR or RTL? I always assumed that it would be LTR. However, divide reverses numerator and denominator so how do we handle a complex chain of operations like the above example?
What about the following operations?
$$5\div5\div5\div5\div5\div5\times5\div5\div5\div5\div5\div5 \tag2$$
and
$$5\div5\div5\div5-5\div5\div5\div5\div5\times5\div5\div5+5\div5\div5 \tag3$$
Here is the solution in the book:
| As far as I know, when dealing with consecutive multiplication and division, one can simply go from left to right. This means that your example problem is solved as follows:
$$
\begin{array}
\div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5}\div 5 \div 5 \div 5 &= 1 \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5}\div 5 \div 5 \div 5 \\
&= 5 \div \frac{1}{5} \div \frac{1}{5}\div 5 \div 5 \div 5 \\
&= 25 \div \frac{1}{5}\div 5 \div 5 \div 5 \\
&= 125\div 5 \div 5 \div 5 \\
&= 25 \div 5 \div 5 \\
&= 5 \div 5 \\
&= 1 \\
\end{array}
$$
If you wanted to, you could turn the division into multiplication as you suggested:
$$
\begin{array}
\div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5} \div \frac{1}{5}\div 5 \div 5 \div 5 &= \frac{1}{5} \times 5 \times 5 \times 5 \times 5 \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \\
&= \frac{5 \times 5 \times 5 \times 5}{5 \times 5 \times 5 \times 5} \\
&= 1
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$ As seen in the title I'm interested in a way to evaluate
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$$
But I'm not sure what to do, I did attempt something but ended up with a wrong result
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k-\sqrt{2}n\right)\left(k+\sqrt{2}n\right)}$$
then applying partial fraction decomposition
\begin{align*}
&=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k-\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k+\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^2n^2}\\
&=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k-\sqrt{2}n-1}\:\mathrm{d}x+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k+\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{-\sqrt{2}n-1}\:\mathrm{d}x-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x-\int _0^1\frac{x^{\sqrt{2}}\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=-\frac{7}{8}\zeta (4)+\frac{5}{4}\zeta (4)\\
&=\frac{3}{8}\zeta (4)
\end{align*}
Which seems to be a wrong answer due to numerical methods and approximations, what did I do wrong? how else could I approach this problem?
| $$\frac 1{n^2\left(k^2-2n^2\right)}=\frac 1{n^2\left(k-n \sqrt 2\right)\left(k+n \sqrt 2\right)}$$
Using partial fracion decomposition
$$\frac 1{n^2\left(k^2-2n^2\right)}=\frac{1}{2 \sqrt{2} n^3}\left(\frac{1}{k-\sqrt{2} n}-\frac{1}{k+\sqrt{2} n}\right)$$
$$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k-\sqrt{2} n}=\frac{1}{2}
\left(H_{-\frac{n}{\sqrt{2}}}-H_{-\frac{n}{\sqrt{2}}-\frac{1}{2}}\right)$$
$$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+\sqrt{2} n}=\frac{1}{2}
\left(H_{\frac{n}{\sqrt{2}}}-H_{\frac{n}{\sqrt{2}}-\frac{1}{2}}\right)$$
$$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k-\sqrt{2} n}-\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+\sqrt{2} n}=\pi \csc \left(\sqrt{2} \pi n\right)-\frac{1}{\sqrt{2} n}$$
$$\frac{1}{2 \sqrt{2}}\sum_{k=1}^\infty \frac 1 {n^3}\left(\pi \csc \left(\sqrt{2} \pi n\right)-\frac{1}{\sqrt{2} n}\right)=\frac{1}{2 \sqrt{2}}\left(-\frac{13 \pi ^3}{360 \sqrt{2}}-\frac{\pi ^4}{90 \sqrt{2}}\right)=-\frac{\pi ^4}{360}-\frac{13 \pi ^3}{1440}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
What does the tilde ("$\sim$") mean in $\tan^{-1}\frac{M_1\sim M_2}{M_1+M_2}$?
$$ \theta_{} \leq \pi/2 $$
$~ H_{x} ~$ can be positive or negative but $~ H_{y} ~$ can be assumed which takes only upward vertical value.
$$ \tan \left( \theta_{} \right) = \frac{ H_{x} }{ H_{y} } = \frac{ \left( M_{1}-M_{2} \right) }{ \left( M_{1}+M_{2} \right) } $$
The problem has been provoked from the below equation.
$$ \therefore ~~ \left| \theta_{} \right| = \tan^{-1} \left( \frac{ M_{1} \sim M_{2} }{ \left( M_{1}+M_{2} \right) } \right) $$
What it this "$~ \sim ~$"? misprint? Or a new conception for me? How should I interpret it ?
Moreover, I've been confused of bars of $~ \theta_{} ~$ . How do I interpret it?
p.s
I have to go to bed in 10 minutes.
| $$ \tan \left( -\theta_{} \right) =-\tan \left( \theta_{} \right) $$
$$ \tan \left( \theta_{} \right) =\frac{ M_{1}-M_{2} }{ M_{1}+ M_{ 2 } } $$
$$ \therefore ~~ \theta_{} = \tan^{-1} \left( \frac{ M_{ 1 } -M_{ 2 } }{ M_{ 1 } +M_{ 2 } } \right) $$
$$
\theta_{} =
\begin{cases}
\tan^{-1} \left( \frac{ M_{ 1 } -M_{ 2 } }{ M_{ 1 } +M_{ 2 } } \right) &~ \left( M_{ 1 } \geq M_{ 2 } \right) \\\\
\tan^{-1} \left( \frac{ -\left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 }+ M_{ 2 } } \right) &~ \left( M_{ 1 } \leq M_{ 2 } \right)
\end{cases}
$$
$$
=
\begin{cases}
\tan^{-1} \left( \frac{ \left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 } +M_{ 2 } } \right) &~ \left( M_{ 1 } \geq M_{ 2 } \right) \\\\
\tan^{-1} \left( \frac{ -\left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 }+ M_{ 2 } } \right) &~ \left( M_{ 1 } \leq M_{ 2 } \right)
\end{cases}
$$
$$
=
\begin{cases}
\tan^{-1} \left( \frac{ \left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 } +M_{ 2 } } \right) &~ \left( M_{ 1 } \geq M_{ 2 } \right) \\\\
-\tan^{-1} \left( \frac{ \left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 }+ M_{ 2 } } \right) &~ \left( M_{ 1 } \leq M_{ 2 } \right)
\end{cases}
$$
$$ = \pm \tan^{-1} \left( \frac{ \left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 } + M_{ 2 } } \right) $$
$$ \therefore ~~ \left| \theta_{} \right| =\tan^{-1} \left( \frac{ \left| M_{ 1 } -M_{ 2 } \right| }{ M_{ 1 } + M_{ 2 } } \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fibonacci generating function - quick way to see formula I know that
$$
\frac{1}{1-z-z^2} = \sum_{k = 1}^\infty a_nz^n
$$
in $B_{\varepsilon}(0) \subseteq \mathbb{C}$ where $(a_n)_{n \in \mathbb{N}}$ denotes the Fibonacci series and $\varepsilon > 0$ is small. This can be obtained by simply comparing coefficients.
Then I wanted to find the power series in a direct way. So I carried out partial fraction decomposition and used the geometric series. I got
$$
\frac{1}{1-z-z^2} = \sum_{k = 0}^\infty\frac{1}{\sqrt{5}}\left(\varphi_{+}^{-k-1} - \varphi^{-k-1}_-\right)z^k, ~ \quad \varphi_+ := \frac{-1+\sqrt{5}}{2},~\varphi_- := \frac{-1-\sqrt{5}}{2}.
$$
So how can I see quickly confirm the well known explicit Fibonacci-formula, i.e.
$$
\varphi_+^{-k-1} = \left(\frac{1+\sqrt{5}}{2} \right)^{k+1}
$$
for $k \in \mathbb{N}_0$ and the analogue for $\varphi_-$?
| Let $\sum_{n=0}^\infty a_n z^n$ be the Macluarin series for $\frac{1}{1 - z - z^2}$. We can easily prove that $a_n$ is the Fibonacci sequence. Note that, for sufficiently small $z$,
\begin{align*}
1 &= (1 - z - z^2)\sum_{n=0}^\infty a_n z^n \\
&= \sum_{n=0}^\infty a_n z^n - \sum_{n=0}^\infty a_n z^{n+1} - \sum_{n=0}^\infty a_n z^{n+2} \\
&= a_0 + a_1 z + \sum_{n=2}^\infty a_n z^n - a_0 z - \sum_{n=1}^\infty a_n z^{n+1} - \sum_{n=0}^\infty a_n z^{n+2} \\
&= a_0 + (a_1 - a_0)z + \sum_{n=0}^\infty a_{n + 2} z^{n + 2} - \sum_{n=0}^\infty a_{n+1} z^{n+2} - \sum_{n=0}^\infty a_n z^{n+2} \\
&= a_0 + (a_1 - a_0)z + \sum_{n=0}^\infty (a_{n + 2} - a_{n+1} - a_n)z^{n + 2}.
\end{align*}
From this, we see $a_0 = 1$, $a_1 - a_0 = 0$, hence $a_1 = 1$, and $a_{n+2} - a_{n+1} - a_n = 0$ for all $n \ge 0$, i.e. $a_n = a_{n-1} + a_{n-2}$, for all $n \ge 2$. This is precisely the (shifted) Fibonacci sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $\frac{k}{8}$ when the rest of the lengths are given 6 points $A,B,C,D,E,F$ are on a circle in this order and line segments $AD,BE,CF$ intersect at one point. If $AB=1,BC=2,CD=3,DE=4,EF=5,FA=\frac{k}{8}$ find the value of $k$.
I tried to sum the total lengths of the arc:
$1+2+3+4+5+\frac{k}{8}=2\pi r$
$15+\frac{k}{8}=2\pi r$
$\frac{120+k}{8}=2\pi r$
$\frac{120\cdot 7+k}{8\cdot 2\cdot 22}=r$
$\frac{840+k}{352}=r$
So I tried substituting the values in $2\pi r$
$2\pi \frac{840+k}{352}=\frac{120+k}{8}$
$\frac{36,960+44k}{2464}=\frac{120+k}{8}$
$295,680+352k=295,680+44k$
$352k=44k$
And if I cut $k$ from both the side, $352=44$ would remain which is absurd.
What went wrong here? And how can I get the value of $k$?
If I missed something obvious, please be gentle
| I think they are chord lengths and not arc lengths.
By Intersecting Chords theorem or by Inscribed Angle theorem, $\triangle GAB \sim \triangle GED$
So, $GD = 4y, GE = 4x$
Similarly, $\triangle GBC \sim \triangle GFE$
So, $GE = \frac{5z}{2}, GF = \frac{5y}{2}$
$\triangle GCD \sim \triangle GAF$
So, $GF = \frac{5y}{2} = \frac{k}{24} \cdot GD = \frac{ky}{6} \implies k = 15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$? I was trying to solving this question:
If roots of the equation $a x^{2}+b x+c=0$ are $\alpha$ and $\beta$, find the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}$
I was not able to solve it so I looked to the solution given in the book, it was as follows:
Let $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{1+y}$
.
Now $\alpha$ is root of the equation $a x^{2}+b x+c=0$ $\Rightarrow a \alpha^{2}+b \alpha+c=0$
$\Rightarrow \quad a\left(\frac{1-y}{1+y}\right)^{2}+b\left(\frac{1-y}{1+y}\right)+c=0$
. Hence required equation is $a(1-x)^{2}+b\left(1-x^{2}\right)+c(1+x)^{2}=0$
In the first step the auther used componendo dividend rule as follows:
$\begin{aligned} & \frac{1-\alpha}{1+\alpha}=y \\ \Rightarrow & \frac{1-\alpha+1+\alpha}{1-\alpha-(1+\alpha)}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{2}{-2 \alpha}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{-1}{\alpha}=\frac{y+1}{y-1} \Rightarrow \alpha=\frac{1-y}{y+1} \end{aligned}$
But I don't understand how he could use it, as we are not sure whether $y$ could be - 1 and hence $1 +y$ can be zero. So is this an error? If so then how can we solve this question and if not then why not?
| Notice that $y$ cannot be equal to $-1$ in the first place, if $y = \dfrac{1-\alpha}{1+\alpha}$
Range 0f $f(\alpha) = \dfrac{1-\alpha}{1+\alpha}$ is $\mathbb{R} \setminus \{-1\} ~\forall ~\alpha \in \mathbb{R}$
This is because if $\dfrac{1-\alpha}{1+\alpha} = -1$ then $$ 1-\alpha = -1-\alpha$$ $$ 1=-1$$
Which is clearly meaningless.
In general, $\dfrac{ax+b}{cx+d}$ cannot be equal to $\dfrac{a}{c} $ if $\dfrac{a}{c} \ne \dfrac{b}{d}$
| {
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"url": "https://math.stackexchange.com/questions/4194946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Find x in the figure (Answer: 40 degrees)
I found the following mathematical relationships but some are still missing.
$\triangle ABC: 80^o+4\theta+2\alpha = 180^o \rightarrow \boxed{2\theta+\alpha=50^o}(i)\\
\triangle CAH:40^o+4\theta+m = 180^o\rightarrow \boxed{m=140^o - 4\theta}(II)\\
\triangle CBG: n+4\theta+\alpha=180^o\rightarrow \boxed{n = 140+ \alpha}(III)\\
\triangle DJE:180^o-40^o-\theta + x + n = 180^o\rightarrow \boxed{x = \theta-n+40^o(IV)}$
| $\angle BFE = 90^0 - 2 n = \angle BAF + \angle ABF = 40^0 + \alpha $
So, $\alpha + 2n = 50^0$ ...$(i)$
Now $\angle AFC = 140^0 - 2\theta$, $\angle BFC = 180^0 - \alpha - 2\theta \ $ and
$\angle AFB = 140^0 - \alpha$
But they should add to $360^0$,
So we get $100^0 = 4 \theta + 2\alpha \ $ or $\alpha + 2 \theta = 50^0$ ...$(ii)$
From $(i)$ and $(ii)$,
$\theta = n$
So $\angle ACD = \angle AED$ but as they are both on segment $AD$, $ADEC$ must be cyclic.
Then, $\angle CDE = x = \angle CAE = 40^0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Curvature and Torsion for a Space Curve Calculate the curvature and torsion of
x= $\theta - \sin \theta, y = 1 - \cos \theta, z = 4 \sin (\theta/2)$
$\vec r = (θ - \sin θ)i+(1-\cos θ)j + 4 \sin (θ/2) \ k$
$\vec dr/dt = (1- \cos θ)i + \sin θ \ j+2 \cos (θ/2) \ k$
$d^2\vec r/dt^2 = \sin θ \ i + \cos θ \ j - sin (θ/2) \ k$
Since $k = |\vec dr/dt \times d^2\vec r/dt^2| \ / \ |\vec dr/dt|^3$
I calculated $\vec dr/dt \times d^2\vec r/dt^2$, which upon simplification gave,
$\vec dr/dt \times d^2\vec r/dt^2 = (\sin θ \sin (θ/2) - 2 \cosθ \cos(θ/2)) \ i + (3 \cos θ + 1) \sin (θ/2) \ j + (\cos θ-1) \ k$
In calculating |$\vec dr/dt$ x $d^2$$\vec r/d$$t^2$| is where I'm starting to face complications. I'm not sure if I am doing this correctly or not. But I understand the application of the formulas of torsion and curvature. Kindly guide me for the same.
| $\vec r (\theta) = (\theta - \sin \theta, 1 - \cos \theta, 4 \sin \frac{\theta}{2})$
There is another formula for curvature that is easier here.
$r'(\theta) = (1 - \cos \theta, \sin\theta, 2 \cos \frac{\theta}{2})$
$||r'(\theta)|| = \sqrt{(1-\cos\theta)^2 + \sin^2\theta + 4 \cos^2 \frac{\theta}{2}}$
$ = \sqrt{(1 + \cos^2\theta - 2 \cos\theta + \sin^2\theta + 4 \cos^2 \frac{\theta}{2}}$
$ = \sqrt{2 - 2 \cos\theta + 2 + 2 \cos\theta} = 2$
So unit tangent vector, $ \ T(\theta) = \frac{1}{2} (1 - \cos \theta, \sin\theta, 2 \cos \frac{\theta}{2})$
Derivative of unit tangent is $\ T'(\theta) = \frac{1}{2} (\sin \theta, \cos\theta, - \sin \frac{\theta}{2})$
$||T'(\theta)|| = \dfrac{\sqrt{3 - \cos\theta}}{2 \sqrt2}$
Curvature $ \ k = \dfrac{||T'(\theta)||}{||r'(\theta)||} = \dfrac{\sqrt{3 - \cos\theta}}{4 \sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197860",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$? How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$ ?
Here is my work:
By expanding the expression we get, $$(x^2-x-2)(x^2+15x+54)+108=x^4+14x^3+37x^2-84x$$
So I got $x(x^3+14x^2+37x-84)=0$. One root is zero which doesn't lie in the interval $(-10,-1)$. But I don't know how many roots the cubic equation has in that interval.
| The midpoints of $-2,9$ and $1,6$ are the same: $3.5$. Thus let $x = t - 3.5$:
$$(t-5.5)(t+5.5)(t - 2.5)(t + 2.5) + 108 = 0$$
$$(t^2 - 30.25)(t^2 - 6.25) + 108 = 0$$
$$t^4 - 36.5t^2 + \text{constant term} \ (297.0625) = 0$$
Since $t \to -t$ results in the same polynomial, it is symmetric across the line $x = -3.5$. Also, since $x = 0$ is one root, $x = -7$ is another.
The equation of symmetry in between the pairs of roots is:
$$t^2 = -\frac{-36.5}{2} \implies x_{\text{positive line of symmetry}} < -3.5 + \sqrt{50/2} = 1.5$$
hence the other negative root is strictly greater than $-7-2(1.5) = -10$. This implies that there are only $2$ roots that lie in $x \in (-10,-1)$: all without a calculator (if you work out the multiplications by hand).
Here is a graph of the polynomial and the axes of symmetry:
| {
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"source": "stackexchange",
"question_score": "3",
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} |
With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$, so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y^5&=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)\\
&=x^3+y^3-x^2y^2(x+y)\\
&=(x+y)(x^2-xy+y^2)-x^2y^2(x+y)\\
&=(x+y)(1-xy-x^2y^2)\\
&=\left( {x + y} \right)\left( {1+\frac{1}{2} - {{\left( {x + y} \right)}^2} - \frac{{{{\left( {x + y} \right)}^4}}}{4} + \frac{{{{\left( {x + y} \right)}^2}}}{2} - \frac{1}{4}} \right)\\
&=t\left( {\frac{5}{4} - \frac{{{t^4}}}{4}} \right)\\
\end{aligned}
This problem is for those who have not studied derivatives so I dont have any idea for next step. Anyone can help me for the hint or other solutions? Very Thanks
| If $x^2+y^2=1$, then $|x|$ and $|y|$ are each no larger than $1$. Consequently, $|x|^5\le x^2$ and $|y|^5\le y^2$. It follows that
$$|x^5+y^5|\le|x|^5+|y|^5\le x^2+y^2=1$$
| {
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If $\dfrac1{200}\sum_{n=1}^{399}\dfrac{5^{200}}{5^n+5^{200}}=\dfrac ab$, then find $\vert a-b\vert$ (where $a$ & $b$ are relatively prime) The following question is taken from the practice set of JEE exam.
If $\dfrac1{200}\sum_{n=1}^{399}\dfrac{5^{200}}{5^n+5^{200}}=\dfrac ab$, then find $|a-b|$ (where $a$ & $b$ are relatively prime).
I tried something but I don't think it's the right way to go:
$$\dfrac1{200}\sum_{n=1}^{399}\dfrac{1}{5^{n-200}+1}\\=\dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{1}{5^{n-200}+1}+\dfrac12+\sum_{n=201}^{399}\dfrac{1}{5^{n-200}+1}\right)\\=\dfrac1{200}\left(\sum_{n=1}^{199}\dfrac{1}{5^{n-200}+1}+\dfrac12+\sum_{n=1}^{199}\dfrac{1}{5^{n}+1}\right)$$
| a very common trick among JEE summation questions is: pair up the first and last terms
$$\sum_{n=1}^{399} \frac{5^{200}}{5^n+5^{200}}=\frac{1}{2}+\sum_{n=1, n\not=200}^{399} \frac{5^{200}}{5^n+5^{200}}=\frac{1}{2}+ \sum_{n=1}^{199} \frac{5^{200}}{5^{200}+5^n}+\frac{5^{200}}{5^{200}+5^{400-n}}$$ and now an amazing thing happens $\frac{5^{200}}{5^{200}+5^n}+\frac{5^{200}}{5^{200}+5^{400-n}}=1$. So our sum is $\frac{1}{2}+199 \cdot 1=\frac{399}{2}$ $$\implies \frac{a}{b}=\frac{1}{200} \cdot \frac{399}{2}=\frac{399}{400} \implies |a-b|=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find greatest and least values of $|z-2-3i|$ if $|z-5-7i|=9$ Context: conceptual question in jee
I got the answer by using properties of modulus but i don't know if the procedure is correct
If $|z-5-7i|=9$ then find the greatest and least values of $|z-2-3i|$
Answer: $14$, $4$
here's what I did:
given,
$$||z|- |-5-7i||\leq |z-5-7i| \leq |z| + |-5-7i|$$
$$||z|- \sqrt{74}|\leq 9 \leq |z| + \sqrt{74}$$
idk if what i'm about to do next is correct I did this
$$9\leq|z|+\sqrt{74}=> |z| \geq 9-\sqrt{74}$$
I assumed $|z| = 9-\sqrt{74}$
$$||z|-|-2-3i|| \leq|z-(2+3i)|\leq|z|+|-2-3i|$$
$$|9-\sqrt{74}-5| \leq |z-(2+3i) \leq 9+5$$
$$|-4.60| \leq |z-(2+3i) \leq 14$$
Again I assumed $|-4.60|$ approx $4$
$$4\leq|z-(2+3i) \leq 14$$
Lookig at the question $|z-(5+7i)| = 9$ resembles equation of circle whose center at $5+7i$with radius 9, is there any way to solve this question using circle equation? and how does equation of cirlce co-relate to finding greatest and least value of $|z-2-3i|$
Is what i did correct above cause I got that answer by randomly by plugging value of $|z|$ as I can't find solid solution to this question in any module/internet/book.
| You could also treat this as an extremization problem in which you want the minimum and maximum values of $ \ (x - 2)^2 \ + \ (y - 3)^2 \ $ subject to the constraint $ \ (x - 5)^2 \ + \ (y - 7)^2 \ = \ 9^2 \ \ . $ The method of Lagrange multipliers (for example) gives us the equations
$$ 2·( x \ - \ 2) \ - \ \lambda \ · \ 2·(x \ - \ 5) \ \ = \ \ 0 \ \ , \ \ 2·( y \ - \ 3) \ - \ \lambda \ · \ 2·(y \ - \ 7) \ \ = \ \ 0 $$
$$ \lambda \ \ = \ \ \frac{x \ - \ 2}{x \ - \ 5} \ \ = \ \ \frac{y \ - \ 3}{y \ - 7} \ \ \Rightarrow \ \ xy \ - \ 7x \ - \ 2y \ + \ 14 \ \ = \ \ xy \ - \ 3x \ - \ 5y \ + \ 15 $$
$$ \Rightarrow \ \ 4x \ - \ 3y \ \ = \ \ -1 \ \ , \ \ x \neq 5 \ , \ y \neq \ 7$$
(this is the line depicted in José Carlos Santos's graph). Inserting this into the equation for the constraint circle yields
$$ 16·(x - 5)^2 \ = \ 16·81 \ - \ 16·(y - 7)^2 \ \ \rightarrow \ \ (4x - 20)^2 \ = \ 16·81 \ - \ 16·(y - 7)^2 $$
$$ \Rightarrow \ \ (3y - 21)^2 \ = \ 16·81 \ - \ 16·(y - 7)^2 \ \ \Rightarrow \ \ 9·( y - 7)^2 \ = \ 16·81 \ - \ 16·(y - 7)^2 $$
$$ \Rightarrow \ \ 25·( y - 7)^2 \ = \ 16·81 \ \ \Rightarrow \ \ y \ \ = \ \ 7 \ \pm \ \sqrt{\frac{16·81}{25}} \ \ = \ \ 7 \ \pm \ \frac{36}{5} $$
$$ \Rightarrow \ \ x \ \ = \ \ \frac{3y \ - \ 1}{4} \ \ = \ \ \frac{3·\left(7 \ \pm \ \frac{36}{5} \right) \ - \ 1}{4} \ \ = \ \ \frac{20 \ \pm \ \frac{108}{5} }{4} \ \ = \ \ 5 \ \pm \ \frac{27}{5} \ \ . $$
We now have the coordinates of the two points of tangency for the circle centered on $ \ 2 + 3i \ $ with the circle centered on $ \ 5 + 7i \ $ when the former circle has its maximum and minimum radii, respectively. Using these coordinates in the "radius-squared" function gives us
$$ \left(5 \ \pm \ \frac{27}{5} \ - \ 2 \right)^2 \ + \ \left(7 \ \pm \ \frac{36}{5} \ - \ 3 \right)^2 \ \ = \ \ \left(3 \ \pm \ \frac{27}{5} \right)^2 \ + \ \left(4 \ \pm \ \frac{36}{5} \right)^2 $$ $$ = \ \ (3^2 \ + \ 4^2) · \left(1 \ \pm \ \frac{9}{5} \right)^2 \ \ = \ \ 5^2 · \left(1 \ \pm \ \frac{9}{5} \right)^2 \ \ . $$
Hence, the maximum and minimum values we've sought are $ \ 5 · \frac{14}{5} = \ 14 \ \ $ and $ \ 5 · \frac{4}{5} = \ 4 \ \ . $
| {
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"url": "https://math.stackexchange.com/questions/4206738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
I don't know how to exactly compute this determinant I've tried to compute this determinant by row transformations and column transformations, but it gives me a formula that doesn't work. The determinant is:
\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & x
\end{vmatrix}
I thought I could start doing row 5 - row 4, row 4 - row 3, row 3 - row 2 and row 2 - row 1 and then you get this determinant:
\begin{vmatrix}
x & a & b & c & d\\
a-x & x-a & 0 & 0 & 0\\
0 & b-x & x-b & 0 & 0\\
0 & 0 & c-x & x-c & 0\\
0 & 0 & 0 & d-x & x-d
\end{vmatrix}
Then I made column 1 - column 2, column 2 - column 3, column 3 - column 4, column 4 - column 5 and you get:
\begin{vmatrix}
x-a & a-b & b-c & c-d & d\\
0 & x-a & 0 & 0 & 0\\
0 & 0 & x-b & 0 & 0\\
0 & 0 & 0 & x-c & 0\\
0 & 0 & 0 & 0 & x-d
\end{vmatrix}
And, as it is triangular, you can multiply the diagonal elements, so you get that the determinant is:
$(x-a)^2(x-b)(x-c)(x-d)$
But this isn't correct and I don't know what to do, could someone please help me? I'd really appreciate.
| Notice that the sum of every row equals to $(a+b+c+d+x).$ You can take this approach:
$$\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & x
\end{vmatrix} \xrightarrow{\text{$C_1=C_1+C_2+C_3+C_4+C_5$}} $$
$$
\begin{vmatrix}
(a+b+c+d+x) & a & b & c & d\\
(a+b+c+d+x) & x & b & c & d\\
(a+b+c+d+x) & b & x & c & d\\
(a+b+c+d+x) & b & c & x & d\\
(a+b+c+d+x) & b & c & d & x
\end{vmatrix} $$ $$= (a+b+c+d+x) \begin{vmatrix}
1 & a & b & c & d\\
1 & x & b & c & d\\
1 & b & x & c & d\\
1 & b & c & x & d\\
1 & b & c & d & x
\end{vmatrix}$$
Now, we can substract the first row from all of the other rows:
$$= (a+b+c+d+x) \begin{vmatrix}
1 & a & b & c & d\\
0 & x-a & 0 & 0 & 0\\
0 & b-a & x-b & 0 & 0\\
0 & b-a & c-b & x-c & 0\\
0 & b-a & c-b & d-c & x-d
\end{vmatrix} \xrightarrow{\text{Expand $C_1$}} $$
$$(a+b+c+d+x) \begin{vmatrix}
x-a & 0 & 0 & 0\\
b-a & x-b & 0 & 0\\
b-a & c-b & x-c & 0\\
b-a & c-b & d-c & x-d
\end{vmatrix} = $$ $$=(a+b+c+d+x)(x-a)(x-b)(x-c)(x-d)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
| Try using Cauchy-Schwarz on Attempt 2. You wanted only AM-GM, so you can use this post: Prove Cauchy-Schwarz with AM-GM for three variables
| {
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"source": "stackexchange",
"question_score": "8",
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Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}}dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^4}{\sqrt{a^2-x^2}}\,dx$$
Focusing on the denominator and using a triangle, I found that $\sin\theta=\frac{x}{a}$, therefore $x=a\sin\theta$ and $dx=a\cos\theta \, d\theta$
Plugging this into the given problem, I got $\int\frac{x^4}{\sqrt{a^2-x^2}}\, dx$ ${}= \int\frac{a^4 \sin^4\theta}{a\cos\theta}\cdot a\cos\theta \, d \theta $= $a^4\int (\sin^4 \theta ) \, d \theta$
Since $\sin^2\theta=\frac{1-\cos2\theta}{2}$, I can say
$$a^4\int (\sin^4 \theta \,d \theta =a^4\int\frac{1-2\cos2\theta+\cos^22\theta}{4}\,)d\theta$$ Pulling out the 4 from the denominator and substituting $\cos^22\theta=1-\sin^22\theta$, I get
$$\frac{a^4}{4} \int({1-2\cos2\theta+1-\sin^22\theta})\, d\theta=\frac{a^4}{4}\int({2-2\cos2\theta-\sin^22\theta}) \, d\theta$$
Substituting $\sin^22\theta for \frac{1-\cos4\theta}{2}$ gets me
$$\frac{a^4}{4}\int(2-2\cos2\theta-\frac{1-\cos4\theta}{2})d\theta=\frac{a^4}{8}\int(4-4\cos2\theta-1+\cos4\theta) \, d\theta=\frac{a^4}{8}\int(3-4 \cos2\theta + \cos4\theta)\, d\theta$$
$$=\frac{a^4}{8}[3\theta-\frac{4\sin2\theta}{2}+\frac{\sin4\theta}{4}]+C$$
From here, there are some substitutions that we must do.
We know that $sin\theta=\frac{x}{a}$ So, $\theta=sin^{-1}(\frac{x}{a})$
Also, $sin2\theta=2sin\theta cos\theta$=$2*\frac{x}{a}*\frac{\sqrt{x^2-a^2}}{a}$ and $sin4\theta=4cos^3\theta sin\theta-4sin^3\theta cos\theta=4*({\frac{\sqrt{a^2-x^2}}{a}})^3*\frac{x}{a}-4*(\frac{x}{a})^3*\frac{\sqrt{a^2-x^2}}{a}$
Substituting this gives,
$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{8x\sqrt{a^2-x^2}}{a^2}+\frac{4x(\sqrt{a^2-x^2}^3}{a^2}-\frac{4x^3\sqrt{a^2-x^2}}{a^2}]+C$
And then finally, $\frac{3a^4}{8}sin^{-1}\frac{x}{a}-\frac{a^2x\sqrt{a^2-x^2}}{2}+\frac{a^2x(\sqrt{a^2-x^2}^3}{2}-\frac{ax^3\sqrt{a^2-x^2}}{2}]+C$
Is this correct? Please critique, comment, and help clarify. I really need to understand this problem.
| There is a typo, though it was not critical:
$$sin2\theta=2sin\theta cos\theta=2*\frac{x}{a}*\frac{\sqrt{\color{red}x^2-\color{red}a^2}}{a} \text{ must be } \\
\sin2\theta=2\sin\theta \cos\theta=2*\frac{x}{a}*\frac{\sqrt{\color{red}a^2-\color{red}x^2}}{a}$$
You made some minor errors while substituting:
$$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{\color{red}8x\sqrt{a^2-x^2}}{a^2}+\frac{\color{red}4x(\sqrt{a^2-x^2}^3}{a^\color{red}2}-\frac{\color{red}4x^3\sqrt{a^2-x^2}}{a^\color{red}2}]+C \text{ must be}\\
\frac{a^4}{8}[3\sin^{-1}\frac{x}{a}-\frac{\color{red}4x\sqrt{a^2-x^2}}{a^2}+\frac{x(\sqrt{a^2-x^2}\color{red})^3}{a^\color{red}4}-\frac{x^3\sqrt{a^2-x^2}}{a^\color{red}4}]+C=\\
\frac{3a^4}{8}\sin^{-1}\frac{x}{a}-\frac{xa^2\sqrt{a^2-x^2}}{2}+\frac{x(a^2-x^2)\sqrt{a^2-x^2}}{8}-\frac{x^3\sqrt{a^2-x^2}}{8}+C=\\
\frac{3a^4}{8}\sin^{-1}\frac{x}{a}-\frac{3xa^2\sqrt{a^2-x^2}}{8}-\frac{x^3\sqrt{a^2-x^2}}{4}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^4}{\sqrt{a^2-x^2}}\,dx$$
Focusing on the denominator and using a triangle, I found that $\sin\theta=\frac{x}{a}$, therefore $x=a\sin\theta$ and $dx=a\cos\theta \, d\theta$
Plugging this into the given problem, I got $\int\frac{x^4}{\sqrt{a^2-x^2}}\, dx$ ${}= \int\frac{a^4 \sin^4\theta}{a\cos\theta}\cdot a\cos\theta \, d \theta $= $a^4\int (\sin^4 \theta ) \, d \theta$
Since $\sin^2\theta=\frac{1-\cos2\theta}{2}$, I can say
$$a^4\int (\sin^4 \theta \,d \theta =a^4\int\frac{1-2\cos2\theta+\cos^22\theta}{4}\,)d\theta$$ Pulling out the 4 from the denominator and substituting $\cos^22\theta=1-\sin^22\theta$, I get
$$\frac{a^4}{4} \int({1-2\cos2\theta+1-\sin^22\theta})\, d\theta=\frac{a^4}{4}\int({2-2\cos2\theta-\sin^22\theta}) \, d\theta$$
Substituting $\sin^22\theta for \frac{1-\cos4\theta}{2}$ gets me
$$\frac{a^4}{4}\int(2-2\cos2\theta-\frac{1-\cos4\theta}{2})d\theta=\frac{a^4}{8}\int(4-4\cos2\theta-1+\cos4\theta) \, d\theta=\frac{a^4}{8}\int(3-4 \cos2\theta + \cos4\theta)\, d\theta$$
$$=\frac{a^4}{8}[3\theta-\frac{4\sin2\theta}{2}+\frac{\sin4\theta}{4}]+C$$
From here, there are some substitutions that we must do.
We know that $sin\theta=\frac{x}{a}$ So, $\theta=sin^{-1}(\frac{x}{a})$
Also, $sin2\theta=2sin\theta cos\theta$=$2*\frac{x}{a}*\frac{\sqrt{x^2-a^2}}{a}$ and $sin4\theta=4cos^3\theta sin\theta-4sin^3\theta cos\theta=4*({\frac{\sqrt{a^2-x^2}}{a}})^3*\frac{x}{a}-4*(\frac{x}{a})^3*\frac{\sqrt{a^2-x^2}}{a}$
Substituting this gives,
$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{8x\sqrt{a^2-x^2}}{a^2}+\frac{4x(\sqrt{a^2-x^2}^3}{a^2}-\frac{4x^3\sqrt{a^2-x^2}}{a^2}]+C$
And then finally, $\frac{3a^4}{8}sin^{-1}\frac{x}{a}-\frac{a^2x\sqrt{a^2-x^2}}{2}+\frac{a^2x(\sqrt{a^2-x^2}^3}{2}-\frac{ax^3\sqrt{a^2-x^2}}{2}]+C$
Is this correct? Please critique, comment, and help clarify. I really need to understand this problem.
| Let $x=a\cos\theta$ then ${dx/d\theta} = -a\sin\theta$ and $\sqrt{a^2-x^2}=a(1-(x/a)^2)^{1/2}=a\sin\theta$, so the integral
$$
\int {x^4\over\sqrt{a^2-x^2}} dx = -a^4\int\cos^4\theta \,d\theta.
$$
Repeated substitution of the formula $\cos\phi = 1/2 (\cos 2\phi+1)$ gives
$$
\int\cos^4\theta\,d\theta = {1\over 8}\int(\cos4\theta+4\cos2\theta+3)\,d\theta.
$$
This integral is simply
$$
{1\over4}(\sin4\theta+8\sin 2\theta+ 12\theta) + C
$$
Use $\sin 2 \phi=2\sin\phi\cos\phi$ and $\cos 2\phi=2\cos^2\phi-1$ to get
$$
2\sin\theta\cos^3\theta+3\sin\theta\cos\theta+3\theta.
$$
The answer for the whole integral is then
$$
I={-a^4\over 8}(2qp^3+3qp+3\cos^{-1}(x/a))+C
$$
where I've defined $p=x/a$ and $q=(1-p^2)^{1/2}$ to simplify.
To check this, differentiate with respect to x. Note that
$$
{dq\over dx}={-x\over a^2 q}
$$
and
$$
{d\over dx}\cos^{-1}(x/a) = -{1\over aq}
$$
Differentiating $I$ by $x$ gives
$$
{dI\over dx}={-a^4\over 8}{1\over q}\left(-{2xp^3\over a^2}+{6x^2\over a^3}q^2-3{x^2\over a^3}+{3\over a}q^2-{3\over a}\right)
$$
Then expand out $p=x/a$ and $q^2=1-(x/a)^2$. You'll find it all cancels out leaving the original integrand. I haven't included all the steps here since the post would get too long, but if you get stuck then leave a comment.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is multiplying and dividing inequalities valid? I am facing certain problems in solving inequalities. I do not know whether multiplying and dividing inequalities works or not but it has helped me solve many problems. Here are some problems which cannot be solved using multiplying and dividing. Why does the rule not apply here?
$1$. If $a$,$b$, $c$ are positive real numbers then prove that $$(a+1)^7 (b+1)^7 (c+1)^7 > 7^7 a^4b^4c^4$$
My approach:
Applying AM-GM to $a$ and $1$ we get $(a+1) > 2\sqrt{a}$ or $(a+1)^7 > 2^7 a^\frac{7}{2}$. Proceeding similarly, we get $$(a+1)^7 (b+1)^7 (c+1)^7 > 2^{21} (abc)^\frac{7}{2}$$
Actual answer:
Apply AM-GM to $a$,$b$,$c$,$ab$,$bc$,$ca$,$abc$
$2$. If $x$ and $y$ are positive real numbers such that $x+y = 8$, then find the minimum value of $$\left(1+ \frac{1}{x}\right)\left(1+\frac{1}{y}\right)$$.
My approach:
Applying AM-GM to $1$ , $\frac{1}{x}$ and $1$ , $\frac{1}{y}$ and multiplying the inequalities we get $$\left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right) ≥ 4 \sqrt{\frac{1}{xy}}$$ Substituting $y = 8 - x$ and solving the quadratic we get $$\sqrt{\frac{1}{x(8-x)}} ≥ \frac{1}{4}$$ or the minimum value of the required expression is $1$
Actual Answer: $25/16$
| By AM-GM $$(a+1)^7(b+1)^7(c+1)^7=\prod_{cyc}\left(4\cdot\frac{a}{4}+3\cdot\frac{1}{3}\right)^7\geq$$
$$\geq\left(7^7\right)^3\prod_{cyc}\left(\left(\frac{a}{4}\right)^4\left(\frac{1}{3}\right)^3\right)=\frac{7^{21}}{4^{12}3^9}a^4b^4c^4>7^7a^4b^4c^4.$$
The last inequality we can prove by the following way.
$$\frac{7^{21}}{4^{12}3^9}=7^7\cdot\frac{7^{14}}{2^{24}3^9}=7^7\cdot\frac{7^6\cdot7^8}{2^{24}3^9}>7^7\cdot\frac{(2^4\cdot3^8)\cdot\left(2^{11}\right)^2}{2^{24}3^9}=7^7\cdot\frac{4}{3}>7^7.$$
Now, we see why your work does not help:
we need to get $\sqrt[7]{a^4b^4c^4}$, which I got, but you got $\sqrt{a^7b^7c^7},$ which is bad.
The second inequality.
In your way you did not save the case of the equality occurring: $x=y=4$.
By your way we need $1=\frac{1}{x}$ and $1=\frac{1}{y},$ which is impossible for $x+y=8.$
Now, we can get a solution.
For $x=y=4$ we obtain a value $\frac{25}{16}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\geq\frac{25}{16}$$ or $$1+8+xy\geq\frac{25}{16}xy$$ or
$$xy\leq16,$$ which is true by AM-GM:
$$xy\leq\left(\frac{x+y}{2}\right)^2=16.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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if $x, y, z$ are $3 \leq x \leq y \leq z$ and they are odd numbers, show that $1 < (xyz - 1) / (x-1)(y-1)(z-1) < 3$ There are three question. Maybe there are related.
(1) if x, y, z are integers greater than 2, showing that
$$ \frac{xyz - 1}{(x-1)yz} < \frac{xyz}{(x-1)(y-1)(z-1)}$$
(2) if $x, y, z$ are $3 \le x \le y \le z$ and they are odd numbers, showing that $$1 < \frac{xyz - 1}{(x-1)(y-1)(z-1)} < 3$$
(3) if function $g(x) = x^3 + Ax^2 + Bx + C$ satisfy the following two conditions, get $A, B, C$
*
*Equation $g(x) = 0$ have three different integers in their roots, and all of roots are odd numbers greater than 3
*$\frac{A+B+2C+2}{A+B+C+1}$ is an integer.
(1) was easy.. $xyz - 1 < xyz$ and $(x-1)yz > (x-1)(y-1)(z-1)$ so it is correct.
But I'm stuck with (2), especially $\frac{xyz - 1}{(x-1)(y-1)(z-1)} < 3$
| Presumably part 3 wants to use part 2 to conclude that $(C+2)/(A+B+C+1)=2$.
If so, I believe that part 2's condition should be $ 3 \leq x < y < z$, reflecting the "three different integers in their roots". As such, I will make this change to the question, and proceed accordingly.
Hints/guide. If you're stuck, show what you've tried.
(2) $1 < \frac{xyz-1}{(x-1)(y-1)(z-1)}$ is obvious by expanding it out to get $ xy + yz + zx > x + y + z$.
For the other inequality, show that $ f(x,y,z) =3(x-1)(y-1)(z-1) - (xyz-1)$ is increasing in each variable.
The coefficient of $z$ is $2xy-3x-3y+3 = x(y-3)+y(x-3)+3 > 0$.
Similarly for the coefficient of $x$, $y$.
Hence we just need to verify that $ f(3,5,7)>0$ .
(3) Let $g(X) = X^3+AX^2+BX + C = (X-x)(X-y)(X-z)$.
We are given that $\frac{A+B+2C+2}{A+B+C+1} = 1 + \frac{-xyz+1}{(1-x)(1-y)(1-z)}$ is an integer.
The previous part tells us that the second term is bounded strictly between 1 and 3, so has to be 2. Thus: $xyz-1 = 2 (x-1)(y-1)(z-1)$.
Show that has a unique solution $(3, 5, 15)$, hence $A, B, C$ can be determined.
You could modify the $f(x,y,z)$ approach above to consider $h(x,y,z) = 2 (x-1)(y-1)(z-1) - (xyz-1)$ and look at the finitely many cases where it is non-positive.
As before, the coefficient of $z$ is $xy-2x-2y+2 =(x-2)(y-2)-2 > 0$.
$h(3,7,9) >0$, so any solution is restricted to $ x = 3, y = 5$.
We verify that this gives $z=15$.
Note: We can actually remove the "odd numbers" requirement.
(2) holds because $f(3,4,5) =13 > 0$.
(3) holds after checking the additional cases to verify that there are no extra solutions.
The only use of odd numbers was to force the coefficient $xy-2x-2y+2$ to be positive.
In the case of $ x = 3, y = 4$, the coefficient is actually $1\times 2 - 2 = 0 $, so $h(3,4, z) = -11 \, \forall z$. This clearly yields no solutions.
The rest follows in a similar manner as before: $h(3,6,10) > 0, h(3,7,8) > 0, h(4,5,6) > 0 $
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this proof regarding the nonexistence of odd perfect numbers correct? Preamble: This post is an offshoot of this earlier question, which was not so well-received in MathOverflow.
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
If $m$ is odd and $\sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form
$$m = q^k n^2$$
where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
As proved in MO, we have the following equation:
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{1}{q^{k+1}}\cdot{I(n^2)},$$
whereupon, starting from the lower bound
$$I(n^2) > \frac{2(q - 1)}{q}$$
we get the recursive estimates
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\cdot\frac{2(q - 1)}{q}$$
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\bigg(\frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\cdot\frac{2(q - 1)}{q}\bigg)$$
$$= \frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2\Bigg)$$
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\Bigg(\frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2\Bigg)\Bigg)$$
$$= \frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2 + \bigg(\frac{1}{q^{k+1}}\bigg)^3\Bigg)$$
$$\ldots$$
$$\ldots$$
$$\ldots$$
Repeating the process ad infinitum, we get:
$$I(n^2) > \frac{2(q - 1)}{q}\cdot\Bigg(\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i}\Bigg).$$
But
$$\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i}$$
is an infinite geometric series, with sum
$$\frac{a_0}{1 - r}$$
where the first term $a_0 = 1$ and the common ratio
$$r = \frac{1}{q^{k+1}}.$$
Hence, we obtain
$$\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i} = \frac{1}{1 - \frac{1}{q^{k+1}}},$$
from which we finally get
$$I(n^2) > \frac{2(q - 1)}{q}\cdot\frac{q^{k+1}}{q^{k+1} - 1}.$$
But we can simplify the RHS of the last inequality as follows:
$$\frac{2(q - 1)}{q}\cdot\frac{q^{k+1}}{q^{k+1} - 1} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = \frac{2q^k}{\sigma(q^k)} = \frac{2}{I(q^k)} = I(n^2).$$
We have therefore finally arrived at the contradiction
$$I(n^2) > I(n^2).$$
We therefore conclude that there cannot be any odd perfect numbers.
Here is my:
QUESTION: Does this proof hold water? If it does not, where is the error in the argument, and can it be mended so as to produce a valid proof?
| In your mathoverflow post, you write
$$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \gcd(n^2, \sigma(n^2))$$
because $\gcd(\sigma(q^k), q^k) = 1.$ Why is this true? I do not see any reason it should hold.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with $\int_{0}^{\infty} \frac{\cos^k (x)}{a^2 + x^2} \, dx$ I'm having trouble evaluating the following integral:
$$\int_{0}^{\infty} \frac{\cos^k(x)}{a^2+x^2} \, dx \,\text{ where } k \in \mathbb{N} \text{ and } a >0$$
I've tried to follow this question which approaches tan instead.
$$I = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos^k(x)}{a^2 + x^2} \, dx$$
$$I = -i \int_{0}^{i \infty} \frac{\cosh^k(x)}{(a-x)(a+x)} \, dx$$
From here, I'm not sure how to progress.
I am aware of the following expansions of $\cos^n$:
$$\cos^{2n} (z) = 2^{-2n} \binom{2n}{n} + 2^{1-2n} \sum_{k=0}^{n-1} \binom{2n}{k} \cos(2(n-k)z)$$
$$\cos^{2n} (z) = 1 + 2^{1-2n} \sum_{k=0}^{n-1} \binom{2n}{k}( \cos(2(n-k)z)-1)$$
$$\cos^{2n+1} (z) = 2^{-2n} \sum_{k=0}^{n} \binom{2n+1}{k} \cos((2n-2k+1)z)$$
$$\cos^{n} (z) = 2^{-n} \binom{n}{\frac{n}{2}}(1- \, n \text{ mod } 2) + 2^{1-n} \sum_{k=0}^{ \lfloor{\frac{n-1}{2}\rfloor}} \binom{n}{k} \cos((n-2k)z)$$
Based on @Claude Leibovici response, it is possible to evaluate the following two equations:
\begin{equation}
\int_{0}^{\infty} \frac{\cos^{2n} (x)}{a^2+x^2} \, dx = \frac{\sqrt{\pi } \Gamma \left(n+\frac{1}{2}\right) \left(1-2 \,
_2F_1\left(1,-n;n+1;-e^{2 a}\right)\right)}{2 a \Gamma (n+1)}+\frac{\pi 4^{-n}
e^{-2 a n} \left(e^{2 a}+1\right)^{2 n}}{a}
\end{equation}
\begin{equation}
\int_{0}^{\infty} \frac{\cos^{2n+1} (x)}{a^2+x^2} \, dx = \frac \pi a e^{-a (2 n+1)} (1-\tanh (a))^{-(2 n+1)}-
\frac{\sqrt{\pi } e^a \Gamma \left(n+\frac{3}{2}\right)}{a \Gamma (n+2)}{}_2F_1\left(1,-n;n+2;-e^{2 a}\right)
\end{equation}
| At least, for odd values of the exponent, the expansion given by @aaaaaaaaabbbbbbbbbcccccc in comments is very useful since
$$\int_0^\infty \frac {\cos \big[ (2 n+1-2k)x\big] }{x^2+a^2}\,dx=\frac \pi{2a}\,\exp\big[-(2n+1-2k)a\big]$$ Summing from $k=0$ to $k=n$ then gives for
$$I_n=\int_{0}^{\infty} \frac{\cos^{2n+1} (x)}{x^2+a^2 } \, dx$$
$$I_n=\frac \pi a e^{-a (2 n+1)} (1-\tanh (a))^{-(2 n+1)}-
\frac{\sqrt{\pi } \,e^a\, \Gamma \left(n+\frac{3}{2}\right) \,
}{a \,\Gamma (n+2)}\,_2F_1\left(1,-n;n+2;-e^{2 a}\right)$$ where appears the gaussian hypergeometric function.
Now,
$$\,_2F_1\left(1,-n;n+2;-e^{2 a}\right)=\frac{P_n(t)}{\binom{2 n+1}{n+1}} \qquad \text{where} \qquad t=e^{2a} $$ The first terms are
$$\left(
\begin{array}{cc}
n & P_n(t) \\
0 & 1 \\
1 & t+3 \\
2 & t^2+5 t+10 \\
3 & t^3+7 t^2+21 t+35 \\
4 & t^4+9 t^3+36 t^2+84 t+126 \\
5 & t^5+11 t^4+55 t^3+165 t^2+330 t+462 \\
6 & t^6+13 t^5+78 t^4+286 t^3+715 t^2+1287 t+1716 \\
7 & t^7+15 t^6+105 t^5+455 t^4+1365 t^3+3003 t^2+5005 t+6435 \\
8 & t^8+17 t^7+136 t^6+680 t^5+2380 t^4+6188 t^3+12376 t^2+19448 t+24310
\end{array}
\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find minimum value of $\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$
Let $x,y,z>$ and $x+y+z=xy+yz+zx$
. Find the minimum value of $$P=\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$$
My solution: I know the minimum value is $\frac{3}{2}$ when $x=y=z=1$
So $$P=\frac{x^4}{\sqrt{x^2.2(y^4+1)}}+\frac{y^4}{\sqrt{y^2.2(z^4+1)}}+\frac{z^4}{\sqrt{z^2.2(x^4+1)}}$$
$$\ge\frac{(x^2+y^2+z^2)^2}{\sqrt{2x^2(y^4+1)}+\sqrt{2y^2(z^4+1)}+\sqrt{2z^2(x^4+1)}}$$
$$\ge\frac{(x^2+y^2+z^2)^2}{\sqrt{2(x^2+y^2+z^2)(y^4+1+z^4+1+x^4+1)}}$$
$$\ge\sqrt{\frac{(x+y+z)^3}{2(x^4+y^4+z^4+3)}}$$ because $(x^2+y^2+z^2)\ge(x+y+z)$
So now i need prove $x^4+y^4+z^4+3 \le \frac{2}{9}(x+y+z)^3$ but i stuck here for a hour. So please help me, thnank
| We have
\begin{align*}
P &= \frac{x^3}{\sqrt{2(y^4+1)}} + \frac{y^3}{\sqrt{2(z^4+1)}} + \frac{z^3}{\sqrt{2(x^4+1)}}\\
&\ge \frac{x^3}{2(y^2 - y + 1)} + \frac{y^3}{2(z^2 - z + 1)} + \frac{z^3}{2(x^2 - x + 1)} \tag{1}\\
&\ge \frac{(x^2 + y^2 + z^2)^2}{2x(y^2 - y + 1) + 2y(z^2 - z + 1) + 2z(x^2 - x + 1)} \tag{2}\\
&= \frac{(x^2 + y^2 + z^2)^2}{2(xy^2 + yz^2 + zx^2) + 2(x + y + z - xy - yz - zx)}\\
&= \frac{(x^2 + y^2 + z^2)^2}{2(xy^2 + yz^2 + zx^2)}\\
&\ge \frac{(x^2 + y^2 + z^2)^2}{2\sqrt{(x^2y^2 + y^2z^2 + z^2x^2)(y^2 + z^2 + x^2)}} \tag{3}\\
&\ge \frac{(x^2 + y^2 + z^2)^2}{2\sqrt{\frac{(x^2 + y^2 + z^2)^2}{3}(y^2 + z^2 + x^2)}} \tag{4}\\
&= \frac{1}{2}\sqrt{3(x^2 + y^2 + z^2)}\\
&\ge \frac{1}{2}\sqrt{3(xy + yz + zx)}\\
&\ge \frac{3}{2}. \tag{5}
\end{align*}
Explanations:
(1) the identity $4(u^2 - u + 1)^2 = 2(u^4 + 1) + 2(u - 1)^4$;
(2) the Cauchy-Bunyakovsky-Schwarz inequality;
(3) the Cauchy-Bunyakovsky-Schwarz inequality;
(4) $(a + b + c)^2 \ge 3(ab + bc + ca)$ for all reals $a, b, c$;
(5) $(xy + yz + zx)^2 = (x + y + z)^2 \ge 3(xy + yz + zx)$, so $xy + yz + zx \ge 3$.
Also, if $x = y = z = 1$, then $P = 3/2$. Thus, the minimum of $P$ is $3/2$.
| {
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"url": "https://math.stackexchange.com/questions/4216372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the value of $f(0)$? Let $f$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in R$, then $f(0)$ is $?$
So this is what I've tried so far,
$f(f(0))=1$ so $f(f(0))=4f(0)+1$ (*)
Also, $f(f(1))=5$,this implies $f(5)=16f(0)+5$ Now how should I find the value of $f(5)$ in terms of $f(1)$ so that I can solve (*)? Any help is appreciated .
The answer given is $1/2$
| Let $f(x) = a_kx^k + a_{k-1}x^{k-1} + ..... + a_0$.
Then $f(f(x) = a_k(a_kx^k + a_{k-1}x^{k-1} + ..... + a_0)^k + a_{k-1}(a_kx^k + a_{k-1}x^{k-1} + ..... + a_0)^{k-1} + ..... + a_0$
If we expand that out we will get a very large polynomial of degree $k^2$.
But $f(f(x)) = 4x + 1$ which is of degree $1$.
So $k^2 = 1$. And as $k \ge 0$ we have $k = 1$.
So let $f(x) =ax+b$. And $f(f(x)) = a(ax+b) + b = a^2 +(ab + b) = 4x + 1$.
So we have $a^2 =4$ and $ab+b = b(a+1) = 1$.
So $a = \pm 2$. $a+1=-1,3$ and $b = \frac 1{a+1}= -1, \frac 13$.
As $f(0) = a\cdot 0 + b = b > 0$ and $-1 < 0; \frac 13 > 0$ we must have $b > 0$ so $b = \frac 13$ and $f(x) = 2x + \frac 13$ and $f(0) = 2\cdot 0 + \frac 13 = \frac 13$.
[Note $f(f(x)) = 2(2x + \frac 13) + \frac 13 = 4x + \frac 23 + \frac 13 = 4x + 1$.
| {
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A difficult Vietnamese university entrance exam problem minimize/maximize $P = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}}+\frac{z}{{z + x}}$? A difficult problem from the Vietnamese Entrance exam of 2011 for Natural Science and Technology group.
Note that the original problem only involve minimization.
How to minimize and maximize $P = \frac{x}{{2x + 3y}} + \frac{y}{{y + z}} + \frac{z}{{z + x}}$ when $x,y,z \in \left[ {1,4} \right]$, $x \geqslant y$ and $x \geqslant z$?
The answer to the maximization problem is $P = \frac{6}{5}$ when $x = y$.
A nice solution without Lagrange multipliers/calculus is preferred, just like the solution for the minimization problem in @Toby Mak's link.
The answer to the minimization problem is $P = \frac{{34}}{{33}}$ when $x = 4,y = 1,z = 2$ but the approach from the official government answer sheet was quite unfair.
To be particular, solving by knowing the answer before hand. It requires student to first proof something from thin air into existence
$\frac{1}{{1 + a}} + \frac{1}{{1 + b}} \geqslant \frac{2}{{1 + \sqrt {ab} }}$.
| The maximization problem:
Since $P$ is homogeneous, WLOG, assume $x = 4$.
If $y = 4$, then $P = \frac{6}{5}$ for any $z$.
Indeed, $\frac65$ is the maximum of $P$, since
\begin{align*}
P &= \frac{4}{8 + 3y} + \frac{y}{y + z} + \frac{z}{z + 4}\\
&= \frac{6}{5} - \frac{(4 - y)[3(z - 1)(4y - z) + 25z]}{5(8 + 3y)(y + z)(z + 4)}\\
&\le \frac{6}{5}
\end{align*}
with equality if and only if $y = 4$.
| {
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"url": "https://math.stackexchange.com/questions/4218242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
calculate cubic equation discriminant Let $p(x)=x^3+px+q$ a real polynomial.
Let $a,b,c$ the complex root of $p(x)$.
What is the easiest way to calculate $\Delta=(a-b)^2(b-c)^2(a-c)^2$ in function of $p,q$?
The result is $\Delta=-4p^3-27q^2$
Ps. $f=(x-a)(x-b)(x-c)$ and so $a+b+c=0$; $ab+bc+ac=p$ and $abc=-q$; how to proceed?
| Note that
$$f(x)=(x-a)(x-b)(x-c)$$
$$f'(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$$
$$\boxed{f'(a)=(a-b)(a-c)}$$
and its cyclic variants.
We have that
$$\prod_\text{cyc}f'(a)=\prod_\text{cyc} (a-b)(a-c)$$
$$\prod_\text{cyc} f'(a)=-(a-b)^2(b-c)^2(a-c)^2$$
Hence, our expression is equivalent to $-f'(a)f'(b)f'(x)$
Since $f'(x)=3x^2+p$, this is equivalent to
$$-(3a^2+p)(3b^2+p)(3c^2+p)$$
$$\boxed{-(27a^2b^2c^2+9p(a^2b^2+b^2c^2+a^2c^2)+3p^2(a^2+b^2+c^2)+p^3)}$$
Using viete's, we get the following identites
1.
$$a^2b^2c^2$$
$$=(abc)^2$$
$$=(-q)^2$$
$$=q^2$$
2.
$$a^2b^2+b^2c^2+a^2c^2$$
$$=(ab+bc+ac)^2-abc(a+b+c)$$
$$=p^2-0q$$
$$=p^2$$
3.
$$a^2+b^2+c^2$$
$$=(a+b+c)^2-2(ab+bc+ac)$$
$$=0^2-2p$$
$$=-2p$$
Substituting these 3 identities into our expression gives
$$-(27q^2+9p^3-6p^3+p^3)$$
$$\boxed{-4p^3-27q^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Maximum value of a expression Problem: If $\alpha+\beta+\gamma=20$, then what is $\max(\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5})$?
My attempt: Assume $\alpha \geq \beta \geq \gamma$. Then $\alpha+\beta+\gamma \leq 3\alpha$ so $25 \leq 3\alpha+5$.
Also $\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\leq 3\sqrt{3\alpha+5}$
At this point, I don't have idea what to do next. What should I do next?
Am I doing it incorrectly?
| $\alpha+\beta+\gamma=20$
$\implies (3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5) = 75$
By AM-QM inequality,
$\cfrac {\sqrt{3 \alpha + 5} + \sqrt{3 \beta + 5} + \sqrt{3 \gamma + 5}}{3} \leq \sqrt{\cfrac{(3 \alpha + 5) + (3 \beta + 5) + (3 \gamma + 5)}{3}}$
leads to $\sqrt{3 \alpha + 5} + \sqrt{3 \beta + 5} + \sqrt{3 \gamma + 5} \leq 15$
Equality occurs at $\alpha = \beta = \gamma = \cfrac{20}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate $ \intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx $ (Using line integral) (complex analysis) I want to calculate the integral $$ \intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx $$ using a line integral $\varGamma $ which is the boundary of an arc of a circle of radius $ R $ and $ 0\leq Arg(z) \leq 2\pi/3 $
Such a path $\gamma $ is sort of a slice of pizza.
Now I'm gonna denote this path $ \gamma $ and decompose it into 3 parts.
One part - the part that lies on the real axis, given by $ x, R\leq x \leq 0 $.
Second part - the arc of the circle which we'll parametrize $ Re^{i\theta},\thinspace\thinspace\thinspace0\leq\theta\leq\frac{2\pi}{3} $
And the last part, a ray from the origin with phase $2\pi/3 $ and length R (the radius of this pizza slice), which we'll parametrize $ xe^{i\frac{2\pi}{3}}\thinspace\thinspace\thinspace0\leq x\leq R $
So we get $$ \intop_{\gamma}\frac{1}{z^{6}+z^{3}+1}dz=\intop_{0}^{R}\frac{1}{x^{6}+x^{3}+1}dx-\intop_{0}^{R}\frac{e^{i\frac{2\pi}{3}}}{x^{6}+x^{3}+1}dx+\intop_{0}^{\frac{2\pi}{3}}\frac{Rie^{i\theta}}{R^{6}e^{6i\theta}+R^{3}e^{3i\theta}+1}d\theta $$
And notice that $$ \intop_{0}^{\frac{2\pi}{3}}\frac{Rie^{i\theta}}{R^{6}e^{6i\theta}+R^{3}e^{3i\theta}+1}d\theta \to 0 $$
When $ R \to \infty $.
Also, by the Residue theorem of Cauchy, we get $$ \intop_{\gamma}\frac{1}{z^{6}+z^{3}+1}dz=2\pi i\text{Res}\left(\frac{1}{z^{6}+z^{3}+1};z_{k}\right) $$
So when $ R \to \infty $, the line integral would be the summation of all the poles of the integrand inside $\gamma$, that is, all the poles with the argument between $ 0 $ and $2\pi/3$.
Now I have checked, and found that the poles of $f $ which satisfies this conditions are $ e^{i\frac{2\pi}{9}},e^{i\frac{4\pi}{9}} $
So that $$ 2\pi i\left(\text{Res}\left(\frac{1}{z^{6}+z^{3}+1};e^{i\frac{2\pi}{9}}\right)+\text{Res}\left(\frac{1}{z^{6}+z^{3}+1};e^{i\frac{4\pi}{9}}\right)\right)=\left(1-i\frac{2\pi}{3}\right)\intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx $$
Now, for a function which has a simple pole at $z_k$ we can write the function as $ \frac{A\left(z\right)}{B\left(z\right)} $ where $B$ has a zero at the pole and $ A $ does not have a zero. And the residue is given by $$\lim_{z\to z_{k}}\left(z-z_{k}\right)\frac{A\left(z\right)}{B\left(z\right)}=\frac{A\left(z\right)}{B'\left(z\right)} $$ evaluated at $z_{k}$
So in order to find the residue of my poles, I found the derivative of the denominator and let z= the pole. So finally I got after all the calculations that the integral should be $$ \left(\frac{1}{6e^{i\frac{10}{9}\pi}+3e^{i\frac{4}{9}\pi}+1}+\frac{1}{6e^{i\frac{20}{9}\pi}+3e^{i\frac{4}{9}\pi}+1}\right)2\pi i\frac{1}{1-e^{i\frac{2}{3}\pi}} $$
Which is not real (I checked, unfortunately)
What am I missing? what went wrong?
Thanks in advance.
| I had a few calculations mistakes: the correct calculation would be the following:
The residue at the poles I mentioned given by
$$ \text{Res}\left(\frac{1}{z^{6}+z^{3}+1};e^{i\frac{2\pi}{9}}\right)=\frac{1}{\left(z^{6}+z^{3}+1\right)'}|_{z=e^{i\frac{2\pi}{9}}}=\frac{1}{6e^{i\frac{10}{9}\pi}+3e^{i\frac{4\pi}{9}}} $$
And $$ \text{Res}\left(\frac{1}{z^{6}+z^{3}+1};e^{i\frac{4\pi}{9}}\right)=\frac{1}{\left(z^{6}+z^{3}+1\right)'}|_{z=e^{i\frac{4\pi}{9}}}=\frac{1}{6e^{i\frac{20}{9}\pi}+3e^{i\frac{8\pi}{9}}} $$
So we get on the LHS
$$ 2\pi i\left(\frac{1}{6e^{i\frac{10}{9}\pi}+3e^{i\frac{4\pi}{9}}}+\frac{1}{6e^{i\frac{10}{9}\pi}+3e^{i\frac{2\pi}{9}}}\right) $$
and on the RHS $$ \intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx+\intop_{0}^{\infty}\frac{-e^{i\frac{2\pi}{3}}}{x^{6}+x^{3}+1}dx $$
So eventually $$ \intop_{0}^{\infty}\frac{1}{x^{6}+x^{3}+1}dx=\left(\frac{1}{6e^{i\frac{10}{9}\pi}+3e^{i\frac{4}{9}\pi}+1}+\frac{1}{6e^{i\frac{20}{9}\pi}+3e^{i\frac{8}{9}\pi}+1}\right)2\pi i\frac{1}{1-e^{i\frac{2}{3}\pi}} $$
Which apparently is a real number which is approximated form is $\approx$ 0.8975
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Solution Verification for Problem Finding constant $a$ and $b$ such that $\lim\limits_{x\to\frac{\pi}{2}}\dfrac{a(2x-\pi)\cos x+b}{\sin x-1}=1.$
Find $a$ and $b$ such that
$$\lim\limits_{x\to\frac{\pi}{2}}\dfrac{a(2x-\pi)\cos x+b}{\sin
x-1}=1.$$
I try answer as follows.
The denumerator is zero when $x=\dfrac{\pi}{2}$. Now I assumed numerator is equal to $0$ (because if the numerator not equal $0$, the RHS is infinity). Then $b=0$ must hold. Now I try to using L'Hospital rule as below.
\begin{align}
\lim\limits_{x\to\frac{\pi}{2}}\dfrac{2a\cos x-a(2x-\pi)\sin x}{\cos x}&=1\\
\iff \lim\limits_{x\to\frac{\pi}{2}}\dfrac{-2a\sin x-2a\sin x-a(2x-\pi)\cos x}{-\sin x}&=1\\
\iff \lim\limits_{x\to\frac{\pi}{2}}\left(4a+a(2x-\pi)\cot x\right)&=1\\
\iff 4a&=1\\
\iff a&=\dfrac{1}{4}.
\end{align}
So, we have $a=\dfrac{1}{4}$ and $b=0$. Is it correct answer? I'm not sure with my answer.
| Without l'Hospital, to verify we can also proceed by standard limits, by $y=\frac \pi 2-x \to 0$
$$\dfrac{a(2x-\pi)\cos x+b}{\sin x-1}=\dfrac{-2ay\sin y+b}{\cos y-1}=\frac{y^2}{1-\cos y}\dfrac{2ay\sin y-b}{y^2}$$
from which we obtain the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a technique in how to write certain expression in certain times of integration as in case of $\int \frac {(3x + 2)} {(5x + 1)^2}dx$
I can solve this integration $$\frac{3} {5}\int\frac{1}{5x + 1}dx + \frac{7} {5}\int\frac{1}{(5x + 1)^2}dx$$ but thing is I dont under how that online calculator wrote $$3x + 2$$ as $$\frac{3}{5}(5x + 1) + \frac{7}{5}$$
I have difficulty understanding that step. I know we have to make numerator and denominator same but is there a method to not guess that I have to multiply by 3/5 and add 7/5 but a standard way of doing some mathematical magic and getting a value that makes it easy to change $3x + 2$ to $\frac{3}{5}(5x + 1) + \frac{7}{5}$
| One way to get at such things is to use a substitution. Let $u= 5x +1$ in your example, then $x=(u-1)/5$ and your integrand goes like this:
$$\frac{3x+2}{(5x+1)^2} = \frac{ 3\left(\frac{u-1}{5}\right)+2}{u^2} = \frac{ \frac{3}{5}u - \frac{3}{5}+2}{u^2} = \frac{ \frac{3}{5}u+\frac{7}{5}}{u^2}. $$
Now put $5x+1$ back in for $u$.
You want to express the numerator in terms of the denominator, so make the denominator just one variable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is integral of $(x+y)^2 \,dx$ different than integral of $(x^2+ 2xy + y^2) \,dx$? I calculated the integral of $(x + y)^2 \,dx$ using the substitution method and got $\frac{1}{3}(x+y)^3$ as result.
Then I applied the distributive property to $(x + y)^2$, calculated the integral of $(x^2+2xy+y^2) \,dx$ and I got $\frac{x^3}{3}+x^2y+xy^2$ as result, which is not equal to $\frac{1}{3}(x+y)^3$.
But, knowing that $(x+y)^2$ = $x^2+2xy+y^2$, why aren't the results of these integrals the same?
| The meaning of the word "constant" gets too little attention in our pedagogy. "Constant" means not changing as something else changes, but what is the thing in the role of "something else"?
\begin{align}
& \int (x^2 + 2xy+y^2)\, dx \\[10pt]
= {} & \frac{x^3}3 + x^2 y + xy^2 + \text{constant} \\[10pt]
= {} & \frac 1 3 \left( x^3 + 3x^2y +3xy^3 \right) + \text{constant} \tag 1 \\[10pt]
= {} & \frac 1 3 \left( x^2 + 3x^2y + 3xy^2 \right) + \Big( \tfrac 1 3 y^3 + \text{constant} \Big) \tag 2 \\
& \text{In this context, “constant” means not depending} \\
& \text{on $x,$ i.e. not changing as $x$ changes, because $x$ is} \\
& \text{the variable with respect to which we are integrating.} \\
& \text{The function whose antiderivative we seek is a} \\
& \text{function of $x$ with $y$ held constant. The “constant”} \\
& \text{In line $(2)$ is the “constant” in line $(1)$ minus $\tfrac13y^3.$} \\
& \text{Get used to that. And $\tfrac 13y^3$ is “constant” since it} \\
& \text{doesn't change as $x$ changes.} \\[10pt]
= {} & \frac 1 3 (x^2 + 3x^2y+3xy^2 + y^3) + \text{constant} \\[10pt]
= {} & \frac 1 3 (x+y)^3 + \text{constant.}
\end{align}
Appendix: Whoever hesitates to think the meaning of “constant” deserves somewhat more attention in the classroom than it usually gets should consider this:
\begin{align}
& \frac d {dx} 2^x = \lim_{h\,\to\,0} \frac{2^{x+h} - 2^x} h \\[10pt]
= {} & \lim_{h\,\to\,0} \left( 2^x \cdot\frac{2^h-1} h \right) & & (\text{just algebra}) \\[10pt]
= {} & 2^x \lim_{h\,\to\,0} \frac{2^h-1} h \\
& \text{because $2^x$ is constant} \\
& \text{and “constant” means} \\
& \text{not depending on $h$} \\[10pt]
= {} & \big( 2^x\cdot\text{constant} \big) \\
& \text{and “constant” means} \\
& \text{not depending on $x.$}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/4227617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying $\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$ I was trying to simplify
$$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$$
I change all $\tan, \cot, \sec, \operatorname{cosec}$ into $\cos$ and $\sin$ using conversion formulas (I find it easier). Here, I get this:
$$\frac{\sin A + 1 - \cos A}{\sin A - 1 + \cos A}$$
from which I can not simplify into anything. I have tried "rationalization technique" by multiplying both numerator and denominator by $\sin A \pm (1 - \cos A)$ but to no yield.
I was then surprised to know the model solution, which is very clever, we replace the $1$ in the numerator with $$\sec^2 A - \tan^2 A$$ to get $$\frac{(\tan A + \sec A) - (\sec^2 A -\tan^2 A)}{\tan A - \sec A + 1}$$ after which it is just one step factoring and cancelling and answer is $$\frac{1 + \sin A}{\cos A}$$
For a person who has only solved these problems by first converting into $\sin, \cos$, I find it difficult to comprehend. Moreover, I also really find it difficult how to think this solution in an exam. I know that that I can not think of such clever solution.
So, I was wondering if there is any easier solution. Any help would be really appreciated.
| Another way.
$$LS=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}+2\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}-2\sin^2\frac{A}{2}}=\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}-\sin\frac{A}{2}}=$$
$$=\frac{\left(\cos\frac{A}{2}+\sin\frac{A}{2}\right)^2}{\cos^2\frac{A}{2}-\sin^2\frac{A}{2}}=\frac{1+\sin A}{\cos A}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a$ is a root of $x^2(x-5)+2$, then what is $[a^4]$? If $a$ is the largest root of $f(x)=x^2(x-5)+2$, then what is $[a^4]$ where $[\ ]$ is a Gauss bracket ? (i.e. $[x]$ is a largest integer not strictly greater than $x$)
Here $f(5)=2,\ f(4)=-14$. By considering $f'(x)$ we can know that $a$ is in an open interval $(4,5)$.
If $b_0=4$, then we define $b_n$ : $$ \frac{f(5)- f(b_n) }{5-b_n } (b_{n+1} - 5) +f(5) =0 \ (i.e. b_{n+1} = 5- \frac{2}{b_n^2} ) $$
Note that an increasing sequence $b_n$ s.t. $|\dfrac{b_{n}-b_{n+1} }{b_{n-1} - b_{n} } |\leq \frac{1}{12}$ goes to $a$ since $f''(x)>0$ on $[4,5]$. And $$ a\in [b_3,b_3+C],\ C = \frac{7}{8\cdot 11 \cdot (12)^2} $$
Here $[(b_3)^4] =[(b_3+C)^4]=584$ But this is a numerical way. How can we calculate $[a^4]$ in the highschool level ?
| Combining two ideas from the comments.
Let $a,b,c$ be the zeros with $a$ the largest. By expanding we see that
$$f(x)f(-x)=4-20x^2+25x^4-x^6=-g(x^2),$$
where
$$g(x)=x^3-25x^2+20x-4$$
has $a^2,b^2$ and $c^2$ as its zeros. By the Vieta relations it follows that
$$
s_1:=a^2+b^2+c^2=25
$$
and the second symmetric polynomial
$$
s_2:=a^2b^2+a^2c^2+b^2c^2=20.
$$
Therefore
$$
a^4+b^4+c^4=s_1^2-2s_2=585.
$$
But $f(-3/4)<0<f(0)$ and $f(3/4)<0$, so $|b|$ and $|c|$ are both $<3/4$.
Therefore $0<b^4+c^4<1$ and hence
$$
584<a^4<585.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that uncorrelatedness implies independence for bivariate joint normal random variables Question
Let $X$ and $Y$ be independent standard normal random variables and consider the following linear transformations, $$U = aX + bY$$ and $$V = cX + dY,$$ where $a, b, c, d \in \mathbb{R}$. Find the joint density of $U$ and $V$ and prove that, if $U$ and $V$ are uncorrelated, then $U$ and $V$ are independent.
Hint
The bivariate normal PDF for jointly normal random variables $W$ and $Z$ is given by $$f_{W, Z}(w, z) = \frac 1 {2\pi\sqrt{\lvert \Sigma \rvert}} \exp\left\{-\frac 1 2 \begin{pmatrix} z & w \end{pmatrix} \Sigma^{-1} \begin{pmatrix} z \\ w \end{pmatrix}\right\}.$$
My working
$\begin{aligned}
U \sim \mathcal{N}(0, a^2 + b^2)\\[1 mm]
V \sim \mathcal{N}(0, c^2 + d^2)
\end{aligned}$
$\begin{aligned}
Cov(U, V) & = \mathbb{E}(UV) - \mathbb{E}(U)\mathbb{E}(V)\\[1 mm]
& = \mathbb{E}[(aX + bY)(cX + dY)]\\[1 mm]
& = ac + bd
\end{aligned}$
$\implies \Sigma =
\begin{pmatrix}
a^2 + b^2 & ac + bd\\[1 mm]
ac + bd & c^2 + d^2
\end{pmatrix}$
$\implies \lvert \Sigma \rvert = (ad - bc)^2$
$\implies \Sigma^{-1} = \dfrac 1 {(ad - bc)^2}
\begin{pmatrix}
c^2 + d^2 & -ac - bd\\[1 mm]
-ac - bd & a^2 + b^2
\end{pmatrix}$
$\begin{aligned}
\therefore f_{U, V}(u, v) & = \frac 1 {2\pi(ad - bc)} \exp\left\{-\frac 1 {2(ad - bc)^2}
\begin{pmatrix}
u & v
\end{pmatrix}
\begin{pmatrix}
c^2 + d^2 & -ac - bd\\[1 mm]
-ac - bd & a^2 + b^2
\end{pmatrix}
\begin{pmatrix}
u\\[1 mm]
v
\end{pmatrix}
\right\}\\[1 mm]
& = \frac 1 {2\pi(ad - bc)} \exp\left\{-\frac 1 {2(ad - bc)^2} [(c^2 + d^2)u^2 - 2(ac + bd)uv + (a^2 + b^2)v^2]\right\}
\end{aligned}$
For uncorrelated $U$ and $V$, we have
$\begin{aligned}
f_{U, V}(u, v) & = \frac 1 {2\pi(ad - bc)} \exp\left\{-\frac 1 {2(ad - bc)^2} [(c^2 + d^2)u^2 + (a^2 + b^2)v^2]\right\}
\end{aligned}$
However, I am having issues showing that $$f_{U, V}(u, v) = f_U(u)f_V(v)$$ and any intuitive explanations will be greatly appreciated :)
| I did not check your calculations but, assuming that
*
*$U,V$ are jointly Gaussian,
*EDIT: $a,b,c,d \ne 0$
Your covariance calculation is right, thus $\rho^2=\frac{(ac+bd)^2}{(a^2+b^2)(c^2+d^2)}$ and the joint density is the following
$$f_{UV}(u,v)=\frac{1}{2\pi\sigma_u\sigma_v\sqrt{1-\rho^2}}\text{exp}\left\{-\frac{1}{2(1-\rho^2)}\left[\frac{(u-\mu_u)^2}{\sigma_u^2}-2\rho\frac{(u-\mu_u)(v-\mu_v)}{\sigma_u^2\sigma_v^2} +\frac{(v-\mu_v)^2}{\sigma_v^2} \right] \right\}$$
Your case is even simplier, having $\mu_u,\mu_v=0$ thus
$$f_{UV}(u,v)=\frac{1}{2\pi\sigma_u\sigma_v\sqrt{1-\rho^2}}\text{exp}\left\{-\frac{1}{2(1-\rho^2)}\left[\frac{u^2}{\sigma_u^2}-2\rho\frac{uv}{\sigma_u^2\sigma_v^2} +\frac{v^2}{\sigma_v^2} \right] \right\}$$
So just substitute. This expression, which is equivalent to yours, is faster to manage with $n=2$
P.S.: in the expression I wrote $\mu_u,\sigma_u$ instead of the more correct notation $\mu_U,\sigma_U$ for a better formula visualization.
prove that, if U and V are uncorrelated, then U and V are independent.
To answer the question, simply set $\rho=0$ in the above formula getting
$$f_{UV}(u,v)=\frac{1}{\sqrt{(a^2+b^2)2\pi}}e^{-u^2/(2(a^2+b^2))}\times \frac{1}{\sqrt{(c^2+d^2)2\pi}}e^{-v^2/(2(c^2+d^2))}=f_U(u)\times f_V(v) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric elimination between two variables Eliminate $\theta$ and $\phi$ between the following equations: $$\begin{cases}\sin \theta + \sin \phi = x \\ \cos \theta + \cos \phi = y \\ \tan \frac {\theta}{2} \tan \frac {\phi}{2} = z\end{cases}$$
What I've done so far
I've established that $$\tan \left(\frac {\theta+\phi}{2}\right) = \frac {\sin \theta + \sin \phi}{\cos \theta + \cos \phi}$$ so that $$\tan \frac {\theta+\phi}{2} = \frac {x}{y}.$$
I then used the trigonometric identity $$\tan \left(\frac {\theta+\phi}{2}\right) = \frac {\tan \frac {\theta}{2} + \tan \frac {\phi} {2}}{1-\tan \frac {\theta}{2}\tan \frac {\phi} {2}}$$ and with a little manipulation got to $$\tan \frac {\theta}{2} + \tan \frac {\phi} {2} = \frac {x(1-z)}{y}$$
I'm stumped on the next steps...would I have to find the difference of the roots also (i.e. $\tan \frac {\theta}{2} - \tan \frac {\phi} {2}$)? Or is there a simpler way? (Side note: I also tried $\tan \frac {\theta}{2}$ substitution but that went nowhere.)
| By sum to product and product to sum formulas we have
$$\begin{cases}
\sin \theta + \sin \phi = 2\sin\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)=x\\
\cos \theta + \cos \phi = 2\cos\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)=y \\
\tan \frac {\theta}{2} \tan \frac {\phi}{2} =\frac{\cos\left(\frac{\theta-\phi}2\right)-\cos\left(\frac{\theta+\phi}2\right)}{\cos\left(\frac{\theta-\phi}2\right)+\cos\left(\frac{\theta+\phi}2\right)} =z
\end{cases} \implies\begin{cases}
2ab=x\\
2cb=y\\
\frac{b-c}{b+c}=z
\end{cases}$$
and since $a^2+c^2=1$ we obtain
*
*$b=\pm\frac12\sqrt{x^2+y^2}$
*$c=\pm \frac{y}{\sqrt{x^2+y^2}}$
and then
$$z=\frac{\pm\frac12\sqrt{x^2+y^2}\mp\frac{y}{\sqrt{x^2+y^2}}}{\pm\frac12\sqrt{x^2+y^2}\pm\frac{y}{\sqrt{x^2+y^2}}}=\frac{x^2+y^2-2y}{x^2+y^2+2y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ & $Z = {z_1} + {z_2} + {z_3} = 2\left( {{e^{i{\theta _1}}} + {e^{i{\theta _2}}} + {e^{i{\theta _3}}}} \right)$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {{e^{i{\theta _1}}} - {e^{i{\theta _2}}}} \right| = \left| {{e^{i{\theta _1}}} - {e^{i{\theta _3}}}} \right|$
Let ${\theta _1} = 0$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {1 - \left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {1 - \left( {\cos {\theta _3} + i\sin {\theta _3}} \right)} \right|$
$ \Rightarrow \left| {1 - \cos {\theta _2} - i\sin {\theta _2}} \right| = \left| {1 - \cos {\theta _3} - i\sin {\theta _3}} \right| \Rightarrow \left| {2{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| {2{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right|$
$\Rightarrow \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right| \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {\cos \frac{{{\theta _2}}}{2} + i\sin \frac{{{\theta _2}}}{2}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {\cos \frac{{{\theta _3}}}{2} + i\sin \frac{{{\theta _3}}}{2}} \right)} \right|$
$ \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right|$
$ \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right|$
$\Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|$
${\theta _2} \ne {\theta _3}$
How do I proceed further?
| WLOG, it can be assumed that $z_1$ is on Imaginary axis(or y-axis).
Since $\left|\frac{z_1+z_2+z_3}{3}\right|=\frac{2}{3}$
It means that distance of the Centroid of triangle formed by joining $z_1,z_2,z_3$ from the centre of circle is $\frac{2}{3}$ and consequently distance of centroid from $z_1$ is $\frac{4}{3}$.
So perpendicular distance of $z_1$ from line segment joining $z_2$ and $z_3$ is
$\frac{1}{2}(\frac{4}{3})+\frac{4}{3}=2$
(since centroid divides median in ratio 2:1)
which is radius of the circle on which $z_1,z_2,z_3$ lie.
Thus $z_2$ and $z_3$ lie on the x-axis and are thus end-points of the diameter of the circle.
So clearly $z_2=-z_3$ and $z_3=-z_2$.
So, $|z_1+z_2||z_1+z_3|=|z_1-z_3||z_1-z_2|=(2\sqrt{2})(2\sqrt{2})=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Completeness of continuous functions space under $L^{1}$ norm Prove that the space of continuous functions $C([0,1])$ is not complete under the norm
$\|\cdot\|_{L^{1}([0,1])}$.
I have tried the following sequence of functions
$f_{n}(x)=\left\{\begin{array}{cl}1 & \text { if } & 0 \leq x \leqslant \frac{1}{2} \\ 1-n\left(x-\frac{1}{2}\right) & \text { if } \frac{1}{2} \leqslant x \leqslant \frac{1}{2}+\frac{1}{n} \\ 0, & \text { if } \frac{1}{2}+\frac{1}{n} \leqslant x \leqslant 1\end{array}\right.$
I got
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left|f_{n}(x)-f_{m}(x)\right| d x .$
$=\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left|\left(1-n\left(x-\frac{1}{2}\right)\right)-\left(1-m\left(x-\frac{1}{2}\right)\right)\right| d x=\left|\frac{m-n}{2 n^{2}}\right|$
Is the last term bounded by $\frac{1}{n}$? why? and is it ok that my integration limits deals with $n$ and not also $m$?
As last step I thought of taking limit of n to $\infty$ and since 2 limits are obtained - $0,1$ this space is not complete.
| Assuming $m<n$, we first show $(f_n)$ is cauchy's sequence.
$\|f_n-f_m\|= \int_0^1 |f_n-f_m|dx$
RHS is equal to
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}|m(x-\frac{1}{2})-n(x-\frac{1}{2})|dx+$
$\int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}|1-m(x-\frac{1}{2})|$
Try to simplify, if needed you can use $\frac{n-m}{2n^2} \leq \frac{n+m}{2n^2} <\frac{n+n}{2n^2}=\frac{1}{n}$.
Once you prove it is cauchy's, you can easily show it does not coverge in given space.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Nicolaus Mercator's extension of logarithmic power series from $(-1,1) \to \mathbb{R}$ I was reading this Quora answer in which the expansion of $\log(1+x)$ was extended in such a way that one could approximate any real number's natural logarithm quickly. The following was written in the answer as an identity:
$$ \ln(Z) - \ln(z)= 2 \left[ Q+ \frac{1}{3} Q^3... O(Q^5) \right] \tag{1}$$
With $Q = \frac{Z-z}{Z+z}$
How do we derive $(1)$ starting from : $\ln(1+x) = x- \frac{x^2}{2!} +O(x^3)$? In the post it is mentioned that is done by $x= \frac{X-1}{X+1}$.. I suppose you put $X= \frac{Z}{z}$ but I can't see how the even powers on left side cancel and how one will get $\ln(Z) - \ln(z)$ on LHS.
It maybe noted that for finding $\ln(2), \ln(3)$ just by clever application of the formula.
$\ln(2)$
$$ \ln(2) - \ln(1) \approx 2 \left[ \frac{1}{3} \right] = .66$$
$\ln (3)$
$$ \ln(3) = \ln\left(2(1.5) \right)
= \ln(2) + \ln(1.5) = .66 + 2 \left[ \frac{1.5-1}{1.5+1}\right]=1.06$$
$\ln(4)$
$\ln(4) = 2 \ln(2) \approx 1.32$
$\ln(5)$
$$ \ln(5) - \ln(4) \approx 2 \frac{1}{9} \approx .\overline{22}$$
Hence, $\ln(5) \approx 1.54$
And so on with more clever manipulation.
| Hint:
For $|x|\lt 1$,
$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+O(x^5)$
and $\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4+O(x^5)$
Subtract these to get: $\ln(1+x)-\ln(1-x)=2(x+\frac{x^3}3+O(x^5))$
Put $x=Q$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Calculate the closed form of the following series $$\sum_{m=r}^{\infty}\binom{m-1}{r-1}\frac{1}{4^m}$$
The answer given is $$\frac{1}{3^r}$$ I tried expanding the expression so it becomes $$\sum_{m=r}^{\infty}\frac{(m-1)!}{(r-1)!(m-r)!}\frac{1}{4^m}$$but I do not know how to follow.
Any help will be appreciated, thanks.
| Consider the series
$$S := \displaystyle \sum_{m=r}^{\infty}\binom{m-1}{r-1}x^m $$
We have the following factorial relation: $$\displaystyle \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$$
So that $\displaystyle \binom{m-1}{r-1} = \frac{r}{m}\binom{m}{r}. $ Hence we have
$$S := \sum_{m \ge r}\binom{m-1}{r-1}x^m = \sum_{m \ge r}\frac{r}{m}\binom{m}{r}x^m = \sum_{k \ge 0}\frac{r}{k+r}\binom{k+r}{r}x^{k+r} $$
For $\beta \in \mathbb C$ we have the binomial series $\displaystyle \frac{1}{(1-z)^{\beta+1}} = \sum_{k \ge 0} {k+\beta \choose k}z^k$.
Writing $\displaystyle \frac{1}{k+r} = \int_0^1 y^{k+r-1} \, \mathrm dy $ we have:
$$\begin{aligned} S & = rx^r\sum_{k \ge 0}\binom{k+r}{r}x^{k} \int_0^1 y^{r+k-1} \, \mathrm dy \\& = rx^r \int_0^1 \sum_{k \ge 0}\binom{k+r}{r}x^{k}y^{r+k-1} \, \mathrm dy \\& = rx^r \int_0^1 y^{r-1} \sum_{k \ge 0}\binom{k+r}{r}(xy)^k \, \mathrm dy \\& = rx^r \int_0^1 y^{r-1} \frac{1}{(1-xy)^{r+1}} \, \mathrm dy \\& = \frac{rx^r}{r(1-x)^r} \\& = \frac{x^r}{(1-x)^r}.
\end{aligned} $$
The case where $\displaystyle x = \frac{1}{4}$ gives $\displaystyle S = \frac{1}{3^r}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Simple way to compute the finite sum $\sum\limits_{k=1}^{n-1}k\cdot x^k$ I'm looking for an elementary method for computing a finite geometric-like sum,
$$\sum_{k=1}^{n-1} k\cdot3^k$$
I have a calculus-based solution: As a more general result, I replace $3$ with $x$ and denote the sum by $f(x)$. Then
$$f(x) = \sum_{k=1}^{n-1} k\cdot x^k = x\sum_{k=1}^{n-1}k\cdot x^{k-1} = x\frac{\mathrm d}{\mathrm dx}\left[\sum_{k=1}^{n-1}x^k + C\right]$$
for some constant $C$. I know that
$$\sum_{k=1}^{n-1}x^k = \frac{x-x^n}{1-x}$$
so it follows that
$$\begin{align}
f(x) &= x\frac{\mathrm d}{\mathrm dx}\left[\frac{x-x^n}{1-x}+C\right] \\[1ex]
&= x\cdot\frac{(1-x)\left(1-nx^{n-1}\right) + x-x^n}{(1-x)^2} \\[1ex]
&= \frac{x-nx^n(1-x)-x^{n+1}}{(1-x)^2}
\end{align}$$
which means the sum I started with is
$$\sum_{k=1}^{n-1}k\cdot3^k = \frac{3+2n\cdot3^n-3^{n+1}}4$$
I am aware of but not particularly adept with summation by parts, and I was wondering if there was another perhaps simpler method I can employ to get the same result?
| Let $$S=\sum_{k=1}^{n-1} k\cdot3^k$$
Then $$3S=\sum_{k=1}^{n-1} k\cdot3^{k+1}$$
Subtracting , we get $$2S=\sum_{k=1}^{n-1} k\cdot3^{k+1}-\sum_{k=1}^{n-1} k\cdot3^k$$
$$2S=(1.3^2+2.3^3+3.3^4+\cdots+(n-1)3^n)-(3+2\cdot3^2 + 3\cdot3^3+\cdots+(n-1)3^{n-1})$$
$$-2S=(3+2\cdot3^2+3\cdot3^3+\cdots+(n-1)3^{n-1})-(1\cdot3^2 + 2\cdot3^3 + 3\cdot3^4+\cdots+(n-1)3^n))$$
$$-2S=3+(2-1)3^2+(3-2)3^3+\cdots+((n-1)-(n-2))3^{n-1})-(n-1)3^n$$
$$-2S=3+3^2+3^3+3^4+\cdots+3^{n-1}-(n-1)3^n$$
$$2S=(n-1)3^n-3\frac{3^{n-1}-1}{3-1}$$
$$2S=(n-1)3^n-\frac{3}{2}({3^{n-1}-1})$$
$$S=\frac{2(n-1)3^n-3({3^{n-1}-1})}{4}$$
This type of series is called $\text{ Arithmetico-Geometric Series }$ https://en.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int\frac{3\cot3x-\cot x}{\tan x-3\tan3x}\,dx$ To evaluate:
$$I= \int_{ }^{ }\frac{3\cot3x-\cot x}{\tan x-3\tan3x}dx$$
My approach:
Convert $\cot x$ and $\cot 3x$ terms into $\tan x$ and $\tan3x$ respectively and use $$\displaystyle \tan3x=\frac{\left(3\tan x-\tan^{3}x\right)}{1-3\tan^{2}x}$$
Further simplification gives me $$\displaystyle I=\int_{ }^{ }\frac{\tan x}{\tan3x}dx$$
How do I proceed further? Any hints are welcome!
Edit:
For anyone wondering how $\int{\tan x/\tan(3x)}dx=\int{(1-3\tan^2x)/(3-\tan^2x)}dx$ don't forget $$\displaystyle \tan3x=\frac{\left(3\tan x-\tan^{3}x\right)}{1-3\tan^{2}x}$$
| Lets take $\int\tan(x)\cot(3x)\, dx$.
We write $\tan(x) \cot(3 x)$ as $\frac{\sin(4 x)-\sin(2 x)}{\sin(2 x)+\sin(4 x)}$ which leads to $\int\frac{\sin(4 x)-\sin(2 x)}{\sin(2 x)+\sin(4 x)}\, dx$
Expanding the integrand leads to:
$$
\int\frac{\sin(4 x)}{\sin(2 x)+\sin(4 x)}-\frac{\sin(2 x)}{\sin(2x)+\sin(4 x)}\, dx
=
\int\frac{2\cos(2x)}{2\cos(2 x)+1}-\frac{\sin(2x)}{\sin(2x)+\sin(4x)}\, dx
$$
For the first integrand, we can substitute $u=2x$ and $du=2dx$ which leads to:
$$
\int\frac{\cos(u)}{2\cos(u)+1}\, du-\int\frac{\sin(2x)}{\sin(2x)+\sin(4x)}\, dx
$$
Then substituting for the first integrand $s=\tan\frac{u}{2}$ and $ds=\frac{1}{2}du\sec^2\frac{u}{2}$. Then $\sin u=\frac{2s}{s^2+1}$ and $\cos u=\frac{1-s^2}{s^2+1}$ and $du=\frac{2ds}{s^2+1}$. Then we have:
$
2\int\frac{s^2-1}{s^4-2s^2-3}\, ds-\int\frac{\sin(2x)}{\sin(2x)+\sin(4x)}\, dx=
\int\frac{1}{s^2+1}\, ds-\frac{1}{2\sqrt{3}}\int\frac{1}{s+\sqrt{3}}\, ds-\frac{1}{2\sqrt{3}}\int\frac{1}{\sqrt{3}-s}\, ds-\int\frac{\sin(2x)}{\sin(2x)+\sin(4x)}\, dx
$
The integral of $\frac{1}{s^2+1}$ is $\tan^{-1}(s)$ and for $\frac{1}{s+\sqrt{3}}$ we substitute $p=s+\sqrt{3}$ and $dp=ds$. Since the integral of $\frac{1}{p}$ is $\log p$ we have:
$$
\tan^{-1}(s)-\frac{\log(p)}{2\sqrt{3}}-\frac{1}{2\sqrt{3}}\int\frac{1}{\sqrt{3}-s}\, ds-\int\frac{\sin(2x)}{\sin(2x)+\sin(4x)}\, dx
$$
For the integrand $\frac{1}{\sqrt{3}-s}$ we substitute $w=\sqrt{3}-s$ and $dw=-ds$. Thus our resulting integral is so far:
$
\tan^{-1}(s)-\frac{\log(p)}{2\sqrt{3}}+\frac{\log(w)}{2\sqrt{3}}-\int\frac{\sin(2x)}{\sin(2x)+\sin(4x)}\, dx
=
\tan^{-1}(s)-\frac{\log(p)}{2\sqrt{3}}+\frac{\log(w)}{2\sqrt{3}}-\int\frac{1}{2\cos(2x)+1}\, dx
$
Now lets substitute $v=2x$ and $dv=2dx$:
$$
\tan^{-1}(s)-\frac{\log(p)}{2\sqrt{3}}+\frac{\log(w)}{2\sqrt{3}}-\frac{1}{2}\int\frac{1}{2\cos(v)+1}\, dv
$$
After simplifying the second integrand and perform the resubstitutions we obtain:
$$
x-\frac{2 \tanh ^{-1}\left(\frac{\tan (x)}{\sqrt{3}}\right)}{\sqrt{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int_0^\infty \frac{e^{-x}}{x}\ln\big(\frac{1}{x}\big) \sin(x)dx$ I'm having trouble with this integral
Evaluate $$\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx$$
$$I=\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx=\Im\left[\int_0^\infty \frac{e^{-x+ix}}{x}\ln\left(\frac{1}{x}\right) dx\right]$$
How would one proceed from here ? $u$ substitute $u=\ln(x)$? How do you solve this integral? Thank you for your time.
| We'll use the following two crucial facts to calculate the integral. The first is that:
$$\boxed{\tan^{-1}x\ln(1+x^2)=-2\sum_{ n \ge 1}\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}}$$
which is shown in here. The second is that
$$\displaystyle \boxed{\Gamma'(2k-1) = (2k-2)! \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n} \bigg)} $$
Which follows from logarithmic differentiation of $$\Gamma(2z-1)=e^{-\gamma (2z-2)}\cdot\prod_{n\geq 1}\left(1+\frac{2z-2}{n}\right)^{-1}e^{\frac{{2z-2}}n}$$
i.e. taking log of both sides and differenting w.r.t. $z.$
Let $\displaystyle I = \int_0^{\infty} \frac{1}{x} \log\left(\frac{1}{x}\right)\sin x \, \mathrm dx. $ Using the taylor series for sine, we have:
$$\begin{aligned} I & = -\int_0^{\infty} \frac{1}{x} e^{-x} \log x \sin{x}\, \mathrm dx \\& = -\int_0^{\infty} e^{-x} \log x \sum_{k \ge 1} \frac{(-1)^{k-1} x^{2k-2}}{(2k-1)!}\, \mathrm
dx \\& = -\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!}\int_0^{\infty} x^{2k-2}e^{-x} \log x \, \mathrm
dx \\& = -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!} \frac{\partial}{\partial k } \int_0^{\infty} x^{2k-2}e^{-x} \, \mathrm
dx \\&= -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!} \frac{\partial \Gamma (2k-1)}{\partial k } \\&= -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!} \cdot (2k-2)! \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n} \bigg) \\&= -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)} \cdot \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n}\bigg) \\& = \gamma \sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)} -\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)}\sum_{n=1}^{2k-2}\frac{1}{n} \\&= \frac{\gamma \pi }{4}- \sum_{k \ge 1} \frac{(-1)^{k-1} H_{2k-2}}{2k-1} \\& = \gamma \frac{\pi}{4}- \sum_{k \ge 1} \frac{(-1)^k H_{2k}}{2k+1} \\& = \gamma \frac{\pi}{4}-\left( -\frac{1}{2} \cdot \tan^{-1}x \cdot \ln(1+x^2)\right)\bigg|_{x=1} \\& = \gamma \frac{\pi}{4}+ \frac{1}{2} \cdot \frac{\pi}{4} \cdot \log{2} \\& = \frac{\pi}{8} \left(2\gamma + \log 2\right). \end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Compressibility factor (Z) of the Redlich-Kwong Equation I've been having a problem trying to calculate the compressibility factor of the Redlich-Kwong equation:
\begin{equation}
P = \frac{RT}{v-b}-\frac{a}{\sqrt{T} \cdot (v^2+vb)}
\end{equation}
The compressibility factor is calculated as:
\begin{equation}
z = \frac{P \cdot v}{RT}
\end{equation}
This factor is calculated if we substitute the molar volume ($v$), but I can't express the Redlich-Kwong equation in terms of $P$ and $T$, that's why I would like to know if some of you guys could help me to isolate the $v$ of this Redlich-Kwong equation. Thanks! :)
| \begin{align*}
P &= \frac{RT}{v-b} - \frac{a}{\sqrt{T}(v^2+vb)}\\
&=\frac{RT\sqrt{T}(v^2+vb)-a(v-b)}{\sqrt{T}(v^2+vb)(v-b)}\\
\implies P\sqrt{T}(v^3 - b^2 v)
&-(RT\sqrt{T}(v^2+bv)-av+ab)=0\\
\implies P \sqrt{T} v^3 - b^2 P \sqrt{T} v
&-a b + a v - b R \sqrt{T^3} v - R \sqrt{T^3} v^2=0 \\
\implies P \sqrt{T} v^3- R \sqrt{T^3} v^2
&- \big(b^2 P \sqrt{T}+b R \sqrt{T^3} -a\big)v - a b =0 \\
\end{align*}
We now have a cubic of the form
$\quad ax^3+x^2+cx+d=0\quad$ where
$$a=P \sqrt{T}\quad
b=- R \sqrt{T^3}\quad
c=- \big(b^2 P \sqrt{T}+b R \sqrt{T^3} -a\big)\quad
d=-a$$
and these may be plugged into the
The Cubic Formula
to obtain one real root. The cubic will be a product of this "factor" and a quadratic equation. It's not pretty. The cubic formula looks like this.
\begin{align*}
n&=\sqrt[\huge{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
&+\sqrt[\huge{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
&-\frac{b}{3a} \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
| $x, y$ and $z$ are roots of the polynomial $x^3-x+1=0$
In the usual notation ,we have $$\Sigma x^2=(\Sigma x)^2-2\Sigma xy=0-2(-1)=2$$
Then since the relation $x^3=x-1$ holds for all three roots, summing these gives:
$$\Sigma x^3=\Sigma x - \Sigma 1=0-3=-3$$
Likewise,
$$\Sigma x^4=\Sigma x^2-\Sigma x=2$$
$$\Sigma x^5=\Sigma x^3-\Sigma x^2=-3-2=-5$$
$$\Sigma x^6=\Sigma x^4-\Sigma x^3=2--3=5$$
And finally,
$$\Sigma x^8=\Sigma x^6-\Sigma x^5=5--5=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 3
} |
Integrate $\int^{\pi}_{0}\left\{\frac{\tan^2\left(\frac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$ $\displaystyle\int^{\pi}_{0}\left\{\dfrac{\tan^2\left(\dfrac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$
$\sf{\color{blue}{My\,\,approach\,}:}$
$=\displaystyle\int^{\pi}_{0}\left\{\dfrac{\tan^2\left(\dfrac{x}{2}\right)}{\dfrac{4\,\tan^2(\frac{x}{2})}{\sec^4(\frac{x}{2})}\cdot\,\dfrac{(1-\tan^2(\frac{x}{2}))^2}{\sec^4(\frac{x}{2})}}\right\}^{\frac{1}{9}}\,dx$
$=\displaystyle\int^{\pi}_{0}\left\{\dfrac{\sec^8\left(\dfrac{x}{2}\right)}{4\cdot\,\left(1-\tan^2\left(\dfrac{x}{2}\right)\right)^2}\right\}^{\frac{1}{9}}\,dx$
$=\displaystyle\int^{\pi}_{0}\dfrac{\sec^{\frac{8}{9}}\left(\dfrac{x}{2}\right)}{2^{\frac{2}{9}}\cdot\,\left(1-\tan^2\left(\dfrac{x}{2}\right)\right)^{\frac{2}{9}}}\,dx$
$=\displaystyle\int^{\pi}_{0}\dfrac{\sec^{\frac{-10}{9}}\left(\dfrac{x}{2}\right)\cdot\dfrac{1}{2}\sec^2\left(\dfrac{x}{2}\right)dx}{2^{\frac{-7}{9}}\cdot\,\left(1-\tan^2\left(\dfrac{x}{2}\right)\right)^{\frac{2}{9}}}$
If I substitute $\color{orange}{\tan\left(\dfrac{x}{2}\right)=t,}$ a weird expression occurs..
Help me to figure out this
| As @person commented, using $x=2\tan^{-1}(t)$,ws have to compute
$$I=2^{7/9}\int_0^\infty\frac{dt}{\left(1-t^2\right)^{2/9} \left(1+t^2\right)^{5/9}}$$
The antiderivative is not so bad
$$\int_0^ u\frac{dt}{\left(1-t^2\right)^{2/9} \left(1+t^2\right)^{5/9}}=u\, F_1\left(\frac{1}{2};\frac{2}{9},\frac{5}{9};\frac{3}{2};u^2,-u^2\right)$$ where appears the Appell hypergeometric function of two variables.
The problem is that I cannot evaluate the limit when $u\to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
| Maybe this identity for ${}_{2}F_{1}$ helps:
$\displaystyle I=\int_{0}^\frac{\pi}{2} \frac{\sqrt{\cos\theta}}{1+\cos^2 \theta}d\theta = \sum_{n=0}^{\infty} (-1)^n\int_{0}^{\frac{\pi}{2}} \cos^{2n+\frac{1}{2}}\theta d\theta = \sum_{n=0}^{\infty} (-1)^n\frac{\sqrt{\pi}\Gamma\left(n+\frac{3}{4}\right)}{2\Gamma\left(n+\frac{5}{4}\right)}$
Now using
$\displaystyle (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$ and $(1)_{n}=n!$
where $(x)_{n}$ is the rising factorial
$\displaystyle I = \sum_{n=0}^{\infty} (-1)^n\frac{\sqrt{\pi}\Gamma\left(n+\frac{3}{4}\right)}{2\Gamma\left(n+\frac{5}{4}\right)}= \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}\sum_{n=0}^{\infty} (-1)^n\frac{\left(\frac{3}{4}\right)_{n}}{\left(\frac{5}{4}\right)_{n}} =\frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}\sum_{n=0}^{\infty} \frac{(1)_{n}\left(\frac{3}{4}\right)_{n}}{\left(\frac{5}{4}\right)_{n}}\frac{(-1)^n}{n!} = \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}{}_{2}F_{1}\left(1,\frac{3}{4};\frac{5}{4};-1\right)$
With this identity:
$ \displaystyle {}_{2}F_{1}(a,b;a-b+1;-1) = \frac{\Gamma(a-b+1)\Gamma\left(\frac{1}{2}a+1\right)}{\Gamma(a+1)\Gamma\left(\frac{1}{2}a-b+1\right)}$
we have
$\displaystyle I= \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{2}\right)}{2\Gamma\left(\frac{5}{4}\right)\Gamma\left(2\right)\Gamma\left(\frac{3}{4}\right)} = \frac{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}{2} =\frac{\pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding a closed formula for a simple integer sequence sum I'm trying to compute the average path lengths of a path graph.
I made the following observations:
There are $n$ nodes and $n-1$ edges and:
$n-1$ paths of length $1$
$n-2$ paths of length $3$
$n-3$ paths of length $4$
...
$1$ path of length $n-1$
I'd like to compute this sum, so that I could then find the average by dividing it by $n(n-1)$. It seems very easy yet I'm having trouble finding a closed formula for :
$$
S = \sum_{i = 1}^{n-1}{i (n-i)}
$$
I feel like it's related to the binomial formula.
| The answer that's already been given is no doubt the "right" one, but maybe you still find the generating function approach interesting. It's a quite simple idea that you can apply to these kinds of problems very often and that may yield results even if you can't directly find a solution via known formulas as you're able to do in this case. The basic idea is to view as your sequence $$S_n = \sum_{i=1}^{n-1} i(n-i), n \geq 2$$ as the coefficient in a formal power-series defining some function $F(x) = \sum_{n=2}^\infty S_n x^n$. You then try to find a closed form expression for the function and then taylor-expand (or expand in other ways) this closed-form to get back the coefficients which you've defined to be exactly the elements of your sequence. This is basically a discrete integral transformation. To start applying this technique it's often useful to find a recurrence, so we start
$$
S_{n+1} = \sum_{i=1}^n i (n+1 - i) = \sum_{i=1}^n i (n-i) + \sum_{i=1}^n i = \sum_{i=1}^{n-1} i (n-i) + n(n-n) + \frac{n(n+1)}{2} = S_n + \frac{n(n+1)}{2}.
$$
So we have found a recurrence in the sequence. Now you multiply both sides by a power (I like to go with $x^{n+1}$, some authors prefer just always doing $x^n$) of the formal variable $x$ and sum over all possible $n$. So we get
$$
\sum_{n=2}^\infty S_{n+1} x^{n+1} = \sum_{n=2}^\infty S_n x^{n+1} + \frac{n(n+1)}{2} x^{n+1}
\iff \sum_{n=2}^\infty S_{n} x^{n} - S_2 x^2 = x \sum_{n=2}^\infty S_n x^n + \sum_{n=2}^\infty \frac{n(n+1)}{2}x^{n+1}
$$
At this point we bring in our function $F$ which we defined exactly as this power series that we now have. So we have
$$
F(x) - S_2 x^2 = x F(x) + \sum_{n=2}^\infty \frac{n(n+1)}{2}x^{n+1}
\iff F(x) = \frac{1}{1-x} (S_2 x^2 + \sum_{n=2}^\infty \frac{n(n+1)}{2}x^{n+1}).
$$
We readily find $S_2=1$ by just substituting $2$ into the definition of $S_n$ and thus have
$$
F(x) = \frac{1}{1-x} (x^2 + \sum_{n=2}^\infty \frac{n(n+1)}{2}x^{n+1}).
$$
Lets try to find a closed form expression for $\sum_{n=2}^\infty \frac{n(n+1)}{2}x^{n+1}$ by focusing on $\sum_{n=0}^\infty n(n+1)x^{n+1}$. To do this note that the geometric series is $\sum_{i=0}^\infty x^i = \frac{1}{1-x}$. Take the derivative on both sides to find $\frac{d}{dx} \frac{1}{1-x} = \sum_{i=0}^\infty ix^{i-1} = \sum_{i=1}^\infty i x^{i-1}$ and again to get
$$
\frac{d^2}{dx^2} \frac{1}{1-x} = \sum_{i=1}^\infty i(i-1)x^{i-2} = \sum_{i=2}^\infty i(i-1) x^{i-2} = \sum_{i=1}^\infty (i+1)i x^{i-1} \\
\iff \sum_{i=1}^\infty (i+1)ix^i = x \frac{d^2}{dx^2} \frac{1}{1-x} = \frac{2x}{(1-x)^3}.
$$
Now we just have to do some "bookkeeping" to get this into the form we need. We calculate
$$
\sum_{n=2}^\infty \frac{n(n+1)}{2}x^{n+1} = \frac{x}{2} (\sum_{i=1}^\infty n(n+1) x^n - 1(1+1)x^1) = \frac{x}{2}(\frac{2x}{(1-x)^3} - 2x) = \frac{x^2}{(1-x)^3}-x^2
$$
and thus
$$
F(x) = \frac{1}{1-x} (x^2 + \frac{x^2}{(1-x)^3} - x^2) = x^2 \frac{1}{(1-x)^4} = x^2 \sum_{n=0}^\infty \frac{(n+1)(n+2)(n+3)}{6}x^n = \sum_{n=0}^\infty \frac{(n+1)(n+2)(n+3)}{6} x^{n+2} = \sum_{n=2}^\infty (\frac{(n-1)n(n+1)}{6} - 1)x^n
$$
and since we defined $F$ via such an exact sequence where the coefficients were $S_n$ we have
$$
S_n = \frac{(n-1)n(n+1)}{6}.
$$
Note that this may seem terribly complicated/roundabout but that's because I've chosen to do everything explicitly here. You can also use a "calculus of generating functions" that allows you to skip a lot of steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$
and I wanted to try and prove it.
Inspired by this answer by Jack D'Aurizio, I took the Weierstrass product for $\cosh(x)$ to obtain
$$
\cosh\left(\frac{\pi x}{2} \right) = \prod_{n \ge 1}\left(1 + \frac{x^2}{(2n-1)^2} \right)
$$
And by logarithmically differentiating twice we get
$$
\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi x}{2} \right) = \sum_{n \ge 1} \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}
$$
Which means we get
\begin{align*}
\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx & =\frac{4}{\pi^2}\sum_{n\ge 1} \int_{0}^{\infty} \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}\right)\, dx
\end{align*}
However, after this, I couldn't figure out how to evaluate the resulting integral.
Does anyone know how I could continue this method? Or alternatively, does anyone know another way in which the result can be proven? Thank you very much!!
Edit:
Per jimjim's request, I'll add that I found this integral on the Wikipedia article for $\zeta(3)$. I believe the reference is to this text where the following formula is given
$$
(s-1) \zeta(s) = 2\pi \int_{\mathbb{R}}\frac{\left(\frac{1}{2} + xi \right)^{1-s}}{\left(e^{\pi x} +e^{-\pi x} \right)^2}\, dx
$$
which for the case of $s=3$ reduces to the surprisingly concise
$$
\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx = \frac{\zeta(3)}{\pi}
$$
And I presume that one can modify the previous equation to get to the original integral from the question, but it is not apparent to me how this may be done.
Edit 2:
Random Variable has kindly posted in the comments how to go from $\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx$ to $ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx$. Thank you very much!
| 1)Integral evaluation applying Differentiation Under the Integral Sign:
First, let's consider the following function:
$$f( \alpha ) =\int _{0}^{\infty }\frac{1-x^{2}}{\left( 1+x^{2}\right)^{2}}\frac{\tanh( \alpha x)}{x} dx$$
Then similarly to OP's series expansion for $\operatorname{sech}^2\left(\frac{\pi x}{2}\right)$, let's employ the series expansion of $\tanh(\alpha x)$ presented in Section 1.421 of Gradshteyn and Ryzhik
$$f( \alpha )=\frac{2}{a}\sum _{n\geqslant 1}\int _{0}^{\infty }\frac{1-x^{2}}{\left[ x^{2} +\left(\frac{\pi ( 2n-1)}{2a}\right)^{2}\right]\left( 1+x^{2}\right)^{2}} dx$$
This integral is pretty straight foward and just requires partial fractions:
$$\begin{align} &\int _{0}^{\infty }\frac{1-x^{2}}{\left( x^{2} +\beta ^{2}\right)\left( 1+x^{2}\right)^{2}} dx \\\ &= \int _{0}^{\infty }\left[\frac{\beta ^{2} +1}{\left( \beta ^{2} -1\right)^{2}\left( x^{2} +\beta ^{2}\right)} -\frac{\beta ^{2} +1}{\left( \beta ^{2} -1\right)^{2}\left( 1+x^{2}\right)} +\frac{2}{\left( \beta ^{2} -1\right)\left( 1+x^{2}\right)^{2}}\right] dx\\\ &=\left[\frac{\left( \beta ^{2} +1\right)\arctan\left(\frac{x}{\beta }\right)}{\beta \left( \beta ^{2} -1\right)^{2}} -\frac{\left( \beta ^{2} +1\right)\arctan( x)}{\left( \beta ^{2} -1\right)^{2}} +\frac{\frac{x}{1+x^{2}} +\arctan( x)}{\left( \beta ^{2} -1\right)}\right]_{0}^{\infty }\\\ &=\frac{\pi }{2\beta ( \beta +1)^{2}}\end{align}$$
Then
$$\begin{align}f(\alpha)=\frac{\pi }{a}\sum _{n\geqslant 1}\frac{1}{\frac{\pi ( 2n-1)}{2a}\left( 1+\frac{\pi ( 2n-1)}{2a}\right)^{2}} &=\sum _{n\geqslant 1}\left[\frac{1}{n-\frac{1}{2}} -\frac{1}{n-\frac{1}{2} +\frac{a}{\pi }} -\frac{a/\pi }{\left( n-\frac{1}{2} +\frac{a}{\pi }\right)^{2}}\right]\\&=\psi ^{( 0)}\left(\frac{a}{\pi } +\frac{1}{2}\right) -\psi ^{( 0)}\left(\frac{1}{2}\right) -\frac{a}{\pi } \psi ^{( 1)}\left(\frac{a}{\pi } +\frac{1}{2}\right)\end{align}$$
$\require{cancel}$
Differentiating and setting $\alpha=\frac{\pi}{2}$
$$\begin{align}\frac{d}{d\alpha}f\left(\frac{\pi}{2}\right) &=\int _{0}^{\infty }\frac{1-x^{2}}{\left( 1+x^{2}\right)^{2}}\operatorname{sech}^2\left(\frac{\pi x}{2}\right) dx\\&=\cancel{{\frac{1}{\pi } \psi ^{( 1)}( 1)} }-\cancel{\frac{1}{\pi } \psi ^{( 1)}( 1)} -\frac{1}{2\pi } \psi ^{( 2)}( 1)\\&=\frac{\zeta(3)}{\pi}\end{align}$$
2) Completing OP's solution:
To prove that the arc contribution vanishes as the raidius of the contour approaches infinity, let's apply the Estimation Lemma.
$$\begin{align}\left|\int _{\Gamma } f( z) dz\right| &\leqslant \int _{0}^{\pi }\left|\frac{1-\left( Re^{it}\right)^{2}}{\left( 1+\left( Re^{it}\right)^{2}\right)^{2}}\operatorname{sech}^{2}\left(\frac{\pi Re^{it}}{2}\right)\right|\left| Rie^{it} dt\right|\\&\leqslant \frac{R^3}{R^4}\int _{0}^{\pi }\frac{dt}{\left|\operatorname{sech}^{2}\left(\frac{\pi Re^{it}}{2}\right)\right|}\leqslant \frac{R^3}{R^4}\int _{0}^{\pi }dt\leqslant \frac{\pi}{R}\end{align}$$
Which means that:
$$\lim_{R\rightarrow \infty}\left|\int _{\Gamma } f( z) dz\right|\leqslant \lim_{R\rightarrow \infty}\frac{\pi}{R}\leqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 2
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Explicit estimate of sum of primes Let $p$ be prime. In this post, Charles gave the following answer
$$\sum_{p\leq x}p=\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{3x^2}{8\log^4 x}+O\left(\frac{x^2}{\log^5 x}\right). $$
I have two problems which I don't understand well. The first is how can we get an such estimate as above? The second is that can we get an explicit form of the estimate above, that is can we compute a concrete number $c$ such that for $x>100$,
$$\left|\sum_{p\leq x}p-\left(\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{3x^2}{8\log^4 x}\right)\right|\leq c\left(\frac{x^2}{\log^5 x}\right)? $$
Also if we take fewer terms to compute $k$ such that for $x>100$,
$$\left|\sum_{p\leq x}p-\left(\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}\right)\right|\leq k\left(\frac{x^2}{\log^3 x}\right)? $$
Are there references on these problems?
| I quote from Christian Axler, New bounds for the sum of the first $n$ prime numbers, https://arxiv.org/abs/1606.06874
Let $\pi(x)$ denote the number of primes not exceeding $x$. de la
Vallée-Poussin estimated the error term in the Prime Number Theorem by showing that $$\pi(x)={\rm li}(x)+O(xe^{-a\sqrt{\log x}})\tag1$$ where $a$ is a positive absolute constant and the logarithmic integral ${\rm li}(x)$ is defined for every real $x\ge0$ as $${\rm li}(x)=\int_0^x{dt\over\log t}=\lim_{\epsilon\to0+}\left\{\int_0^{1-\epsilon}{dt\over\log t}+\int_{1+\epsilon}^x{dt\over\log t}\right\}\tag2$$ Denoting the sum of the first prime numbers not exceeding $x$ by $S(x)$, Szalay [24, Lemma 1] used (1) to find $$S(x)={\rm li}(x^2)+O(x^2e^{-a\sqrt{\log x}})\tag3$$ Using (3) and integration by parts in (2), we get the asymptotic expansion [given in the first line of the question].
[The Szalay reference is M. Szalay, On the maximal order in $S_n$ and $S_n^*$, Acta Arith. 37 (1980) 321–331.]
Then Axler proves various results like
For every $x \ge 110 118 925$, we have $$S(x)<\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{5.3x^2}{8\log^4 x} $$ and
For every $x \ge 905 238 547$, we have$$S(x)>\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{1.2x^2}{8\log^4 x} $$
| {
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"url": "https://math.stackexchange.com/questions/4260101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluating $(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$ I'm trying to evaluate the following expression$$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$$
I'm not really used to these types of problems, so I first tried using logarithms but I'm not sure what to do from there. See:
Let $P = (2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$.
Then we have:
$$\log_2{P} = \log_2((2^1 + 3^1)...(2^{64} + 3^{64}))$$
$$\log_2{P} = \log_2(2^1 + 3^1) + \log_2(2^2 + 3^2) + ... + \log_2(2^{64} + 3^{64})$$
$$P = 2^{2^1 + 3^1} + 2^{2^2 + 3^2} +... + 2^{2^{64} + 3^{64}}$$
$$P = 2^{2^1}2^{3^1} + 2^{2^2}2^{3^2} +... + 2^{2^{64}}2^{3^{64}}$$
Would factoring $2^2$ be of any use here?
| Hint: Multiply by $(3^1-2^1)=1$
Explanation-
$$(3^1-2^1)(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})\\ =(3^2-2^2)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})\\=(3^4-2^4)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})$$
Complete the rest.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3$$
$\bullet$ We need to prove $$3+2(a-1)(b-1)(c-1)\ge3$$ or $$abc-(ab+bc+ca)+a+b+c-1\ge0$$
$\bullet$ Note that $ab+bc+ca\le \dfrac{(a+b+c)^2}{3}=3 $ so we need to prove: $$abc-3+3-1\ge0$$ or $$abc\ge1$$
But I have no idea from here, please help me
| WLOG assume $a\leq 1.$ Then, let $f(a,b,c) = a^8+b^8+c^8 +2(a-1)(b-1)(c-1)$ and $\dfrac{b+c}{2} = t$ consider:
$$f(a,b,c) - f\left(a,t,t\right) = \left(b^8+c^8 - 2\left(\dfrac{b+c}{2}\right)^8\right)+\dfrac{1}{2}(1-a)(b-c)^2\geq 0.$$
This reduces your inequality into single variable case:
$$g(t) = (3-2t)^8+2t^8-4(t-1)^3-3\geq 0$$
which should be bashable.
EDIT: In fact, it turns out that:
$$g(t)\geq (3-2t)^4+2t^4 - 4(t-1)^3-3 = 2(t-1)^2(9t^2-32t+41)\geq 0,$$
so your inequality even holds with $a^4,b^4,c^4$ etc instead of the exponent $8.$ To see why it is stronger, one can simply observe that:
$$\dfrac{a^8+b^8+c^8}{3}\geq\left(\dfrac{a^4+b^4+c^4}{3}\right)^2\geq\dfrac{a^4+b^4+c^4}{3}\cdot\left(\dfrac{a+b+c}{3}\right)^4=\dfrac{a^4+b^4+c^4}{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Range of $\frac{x-y}{1+x^2+y^2}=f(x,y)$ I have a function $\frac{x-y}{1+x^2+y^2}=f(x,y)$. And, I want to find the range of it. I analyzed this function by plotting it on a graph and found interesting things. Like if the level curve is $0=f(x,y)$, then I get $y=x$ which is a linear function. But if the level curve is something not 0, then the level curve becomes a circle. And for big values of level curves, the circle disappears. Is there something I can use to find the range of this function?
| Let $k\in\mathbb{R}$. It is enough to find the $(x,y)$ in the domain of $f$ (all $\mathbb{R}^2$) for which the equality is true
$$\frac{x-y}{1+x^2+y^2}=k$$
It is clear that the equality is true when $k=0$ (in this case $\frac{x-y}{1+x^2+y^2}=0\Longleftrightarrow x-y=0\Longleftrightarrow x=y$). Suppose that $ k\neq 0$. Then
\begin{align}
\frac{x-y}{1+x^2+y^2}=k &\Longleftrightarrow x-y= k(1+x^2+y^2)\\ &\Longleftrightarrow x-y=k+kx^2+ky^2\\&\Longleftrightarrow kx^2+ky^2-x+y=-k \\&\Longleftrightarrow (kx^2-x)+(ky^2+y)=-k\\&\Longleftrightarrow k\left(x^2-\frac{x}{k}\right)+k\left(y^2+\frac{y}{k}\right)=-k\\&\Longleftrightarrow k\left(x^2-\frac{x}{k}+\frac{1}{4k^2}-\frac{1}{4k^2}\right)+k\left(y^2+\frac{y}{k}+\frac{1}{4k^2}-\frac{1}{4k^2}\right)=-k\\&\Longleftrightarrow k\left(x^2-\frac{x}{k}+\frac{1}{4k^2}\right)-\frac{1}{4k}+k\left(y^2+\frac{y}{k}+\frac{1}{4k^2}\right)-\frac{1}{4k}=-k\\& \Longleftrightarrow k\left(x-\frac{1}{2k}\right)^2+k\left(y+\frac{1}{2k}\right)^2=-k+\frac{1}{4k}+\frac{1}{4k}\\& \Longleftrightarrow k\left(x-\frac{1}{2k}\right)^2+k\left(y+\frac{1}{2k}\right)^2=\frac{1-2k^2}{2k}\\& \Longleftrightarrow \left(x-\frac{1}{2k}\right)^2+\left(y+\frac{1}{2k}\right)^2=\frac{1-2k^2}{2k^2}
\end{align}
since the left side of this equality is the sum of two squares, it makes sense if and only if \begin{align}
\frac{1-2k^2}{2k^2}\geq 0 & \Longleftrightarrow 1-2k^2\geq 0 \\& \Longleftrightarrow 1\geq 2k^2 \\& \Longleftrightarrow \frac{1}{2}\geq k^2 \\& \Longleftrightarrow |k|\leq \frac{1}{\sqrt{2}}.
\end{align}
It is concluded that the range of $ f $ is $\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right].$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"solved" Confusion calculating $\mathop {\lim }\limits_{x\to 0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}$ I can get the correct answer through one solution, but when I try the second method, it shows an obvious error, and I can't find where and why. Can someone know the reason, or can provide some useful suggestions? thanks, : )
Solution2
$$\begin{align*}
\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}
& = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{({\rm{1 + }}\cos 2x{\rm{ - 1}})}^{\frac{1}{2}}}{{({\rm{1 + }}\cos 3x{\rm{ - 1}})}^{\frac{1}{3}}}}}{{{x^2}}}\\
& =\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x({\rm{1 + }}\frac{{\cos 2x{\rm{ - 1}}}}{{\rm{2}}})({\rm{1 + }}\frac{{\cos 3x{\rm{ - 1}}}}{{\rm{3}}})}}{{{x^2}}}\\
&=\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x}}{{{x^2}}}\\
&=\frac{{\rm{1}}}{{\rm{2}}}
\end{align*}$$
Maybe Solution1 is a bit informal, but all I want about it is just to talk about ideas
Solution1
$$\begin{align*}
\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}
& = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x + \cos x(1 - {{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\
&=\frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(\cos 2x)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\
&= \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(1 + \cos 2x - 1)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(1 + \cos 3x - 1)}^{\frac{1}{3}}})}}{{{x^2}}}\\
& = \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 2x}}{{2{x^2}}}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 3x}}{{3{x^2}}}\\
&= \frac{1}{2} + 1 + \frac{3}{2} = 3
\end{align*}$$
I find my mistake is "forget considering the infinitesimal term when replacing"
thanks help for clear answer for @user
a wonderful and general answer for @CHAMSI
also, Parthib Ghosh's opinion is also useful
| It should be noted that
$\color{blue}{\left(1+x\right)^{\frac{1}{2}}=1+\dfrac{x}{2}+\cdots}$
is true for $\color{orange}{|x|\le1}$
In your assumption, $\mathtt{\big(1+cos(2x)-1\big)^{\frac{1}{2}}=1+\dfrac{cos(2x)-1}{2}},$
$\mathtt{\color{violet}{|cos(2x)-1|\le2}}$
So, I think it should not be used.
Instead, use binomial expansion for $\cos(x)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many sequences are there? I saw this problems in an combinatorics exam:
\begin{array}{l}For\;all\;n\geq0,\;find\;the\;number\;of\;integer\:sequences\;a_0,...,a_{2n}\;,\;a_0=a_{2n}=1\;\\and\;for\;all\;1\leq i\leq2n\;we\;have\;\frac{a_i}{a_{i-1}}\in\{\frac16,\frac23,\frac32,6\}\end{array}
It reminds me similar problems regarding Catalan numbers, but i didn't find a connection in this case.
Any help will be appriciated.
| The hint of @TeresaLisbon is nice. We generalise the set
\begin{align*}
\left\{\frac{1}{2\cdot3},\frac{2}{3},\frac{3}{2},2\cdot 3\right\}
\end{align*}
and denoting with $[z^n]$ the coefficient of $z^n$ of a series we consider
\begin{align*}
\color{blue}{[x^0y^0]}&\color{blue}{\left(xy+\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}\right)^{2n}}\\
&=[x^0y^0]\left(x\left(y+\frac{1}{y}\right)+\frac{1}{x}\left(y+\frac{1}{y}\right)\right)^{2n}\\
&=[x^0y^0]\left(x+\frac{1}{x}\right)^{2n}\left(y+\frac{1}{y}\right)^{2n}\\
&=\left([x^0]\frac{1}{x^{2n}}\left(1+x^{2}\right)^{2n}\right)^2\\
&=\left([x^{2n}]\sum_{j=0}^{2n}\binom{2n}{j}x^{2j}\right)^2\\
&\,\,\color{blue}{=\binom{2n}{n}^2}\tag{1}
\end{align*}
We conclude from (1) the number of valid sequences of length $2n+1$ is the square of the central binomial coefficient $\color{blue}{\binom{2n}{n}^2}$.
The Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n-1}$ can be written as the difference of the central binomial coefficient with a shifted binomial coefficient. The relationship with (1) don't be to seem very close.
Lattice paths: Let's look at a graphic with lattice paths for small values $n=1$ and $n=2$ to better see what's going on.
We consider logarithmically scaled axis with unit $2$ in $x$-direction and unit $3$ in $y$-direction.
*
*The top left part in grey shows an element $a_i=\left(2^2,3^6\right)$ and the four directed edges weighted with $\left\{6,\frac{3}{2},\frac{1}{6},\frac{2}{3}\right\}$ which point to a candidate $a_{i+1}$.
*The bottom blue part with element $a_j=\left(2^3,3^2\right)$ in the center shows all valid subsequences of length $2$ starting and ending in $a_j$. We start in $a_j$ and there are $4$ ways to make a step of length $1$ according to the grey part and then we go back to $a_j$. We obtain this way
\begin{align*}
\color{blue}{\binom{2n}{n}^2=\binom{2}{1}^2=4}
\end{align*}
valid paths and therefore $4$ valid sequences.
*The right-hand green part with element $a_k=\left(2^7,3^4\right)$ in the center shows all valid subsequences of length $4$ starting and ending in $a_k$. We see for instance there are $9$ valid paths starting in $a_k=\left(2^7,3^4\right)$ and ending in $\left(2^8,3^5\right)$ followed by one final step back to $a_k$.
*
*These $9$ paths are weighted with
\begin{align*}
\begin{array}{cccccc}
6&\frac{1}{6}&6&\qquad xy&\frac{1}{xy}&xy\\
\frac{2}{3}&\frac{3}{2}&6&\qquad\frac{x}{y}&\frac{y}{x}&xy\\
\frac{3}{2}&\frac{2}{3}&6&\qquad\frac{y}{x}&\frac{x}{y}&xy\\
\frac{1}{6}&6&6&\qquad xy&\frac{1}{xy}&xy\\
\\
\frac{3}{2}&6&\frac{2}{3}&\qquad \frac{y}{x}&xy&\frac{x}{y}\\
6&\frac{3}{2}&\frac{2}{3}&\qquad xy&\frac{x}{y}&\frac{y}{x}\\
\\
\frac{3}{2}&6&\frac{2}{3}&\qquad \frac{y}{x}&xy&\frac{x}{y}\\
\frac{2}{3}&6&\frac{3}{2}&\qquad \frac{x}{y}&xy&\frac{y}{x}\\
\\
6&6&\frac{1}{6}&\qquad xy&xy&\frac{1}{xy}
\end{array}
\end{align*}
and $4$ times this set of paths results in
\begin{align*}
\color{green}{\binom{2n}{n}^2=\binom{4}{2}^2=36}
\end{align*}
valid paths and therefore $36$ valid sequences.
| {
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"url": "https://math.stackexchange.com/questions/4272173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\lim_{n→∞} \left( \frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1} \right)$ where $k∈\mathbb{N}$ I had to find the limit of the sequence
$$ a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}$$, where $ k $ is a natural number.
After applying the Stolz theorem, I was able to get here.
$$\lim \:_{n\to \:\infty \:}\left(\frac{\left(k+1\right)\left(n+1\right)^k-\left(\left(n+1\right)^{k+1}-n^{k+1}\right)}{\left(k+1\right)\left(\left(n+1\right)^k-n^k\right)}\right)$$
I tried different ways on continuing from here but unfortunately I am unable to get anywhere. I appreciate any kind of help.
Edit:
For users who may want a solution without the Big O notation, I was able to solve this limit by using the binomial theorem to calculate largest coefficient, since the numerator and denominator are polynomials in $n$, as a user below suggested.
| We have
$$a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=a_n =\frac{(k+1)(1^k+2^k+\cdot \cdot \cdot +n^k)-n^{k+1}}{(k+1)n^k}$$
and then by Stolz-Cesaro
$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k}$$
and since by binomial expansion
*
*$(k+1)(n+1)^{k}=(k+1)n^k+k(k+1)n^{k-1}+O(n^{k-2})$
*$(n+1)^{k+1}= n^{k+1}+(k+1)n^k+\frac{k(k+1)}2n^{k-1}+O(n^{k-2})$
we have
$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k} = \frac{k(k+1)n^{k-1}-\frac{k(k+1)}2n^{k-1}+O(n^{k-2})}{k(k+1)n^{k-1}+O(n^{k-2})}=$$
$$\frac{\frac12+O(1/n)}{1+O(1/n)}\to \frac12$$
Edit
As an alternative, directly by Faulhaber's formula since
$$1^k+2^k+\cdot \cdot \cdot +n^k = \frac{n^{k+1}}{k+1}+\frac12 n^k+O(n^{k-1})$$
we have
$$\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=\frac{n}{k+1}+\frac12+O(1/n) -\frac{n}{k+1}=\frac12+O(1/n) \to \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Inequality without BW and Uvw, just AM GM Problem: Let $a,b,c\ge0: ab+bc+ca=1.$ Prove that: $$(a+b+c)(3-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge4$$
I guess equality holds: two of them equal $1$ and one equal to $0$.
My approach: The equality equivalent to
$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-a-b-c)+3(a+b+c)\ge5$
I tried to use Pqr method but it is quite complicated. Any one help me to get nice proof? Thanks.
| $uvw$ helps!
Let $\sqrt{ab}=z$, $\sqrt{ac}=y$ and $\sqrt{bc}=x$.
Thus, $x^2+y^2+z^2=1$ and we need to prove that:
$$\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\right)(3-x-y-z)+2(xy+xz+yz)\geq4.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $w^3$, but
$$\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}=\frac{9v^4-6uw^3}{w^3}=\frac{9v^4}{w^3}-6u$$ decreases respect to $w^3,$ which by $uvw$ says
(about $uvw$ see here:
https://artofproblemsolving.com/community/c6h278791 ) that it's enough to prove the last inequality for equality case of two variables.
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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