Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$
$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$
Find $x$.
I am not sure how to proceed.
Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?
Any help would be sincerely appreciated.
| I doubt a closed form exists for this number, but I'll show convergence and give some bounds on the values.
Let $f_m^n = \sqrt{m\pm\sqrt{(m+1)\mp \sqrt{\cdots\pm \sqrt n}}} $
be the the version where we go from $m$ to $n$ (don't take the $\pm$'s too seriously). We have the recursive expression
$$
f_m^n = \begin{cases}
\sqrt{m + (-1)^{m+1}f_{m+1}^n} & m<n \\
\sqrt n & m=n
\end{cases}
$$
Realness and general bounds:
Let $3 \le m < n$ with $n$ large enough. I claim that
$$
f_m^n - \sqrt{m-1} \in \begin{cases}
[0.5, 0.7] & m \textrm{ odd} \\
[-0.4, -0.6] & m \textrm{ even}
\end{cases}
$$
These constants can surely be refined, and maybe the ranges can be made to depend on $m$ as well to sharpen it further. Furthermore, it shouldn't be hard to show that $f_m^\infty-\sqrt m \to \pm \frac12$ for $m\to \infty$.
In particular everything is real in these cases (which is easy to check). We can show it by backwards induction on $m$. You can check that the base case $m=n-1$ holds for $n\ge8$.
For $m<n-1$ we must check for odd $m$ that
$$
f_m^n \in \left[ \sqrt{m + \sqrt{m} - 0.4}, \sqrt{m + \sqrt{m} - 0.6} \right]
\subset \left[ \sqrt{m-1} + 0.5, \sqrt{m-1} + 0.7 \right]
$$
and for even $m$ that
$$
f_m^n\in \left[ \sqrt{m - \sqrt{m} - 0.7}, \sqrt{m - \sqrt{m} - 0.5} \right]
\subset \left[ \sqrt{m-1} - 0.6, \sqrt{m-1} - 0.4 \right]
$$
These are quadratic inequalities, so we can run the numbers and see that they all hold for $m\ge3$.
Bounds on $f_1$:
Assume for now that $f_m^n$ converges as $n\to\infty$, which I'll prove below. Then the same bounds as above hold for $f_m := \lim_{n\to\infty}f_m^n$. Let's get some better bounds on $f_3$. We have
$$
f_3 = \sqrt{3+\sqrt{4-f_5}}
\in \left[\sqrt{3+\sqrt{4-\sqrt{4}-0.7}}, \sqrt{3+\sqrt{4-\sqrt{4}-0.5}}\right]
\approx [2.035, 2.055]
$$
We could get increasingly better bounds by using a higher term than $f_5$. Assuming the principal squareroot for complex numbers, we get approximately
$$
f_2 = \sqrt{2-f_3}
\in [0.186i, 0.235i]
$$
This finally gives us bounds on $f_1$. If $f_1=x+iy$, then $f_1$ lies on the curve $x^2-y^2=1$ and between the values approximately $1.00430 + 0.0928i$ and $1.00681 + 0.117i$.
Convergence:
We want to use that
$$
\lvert \sqrt{a+x} - \sqrt a \rvert \le \frac{|x|}{2}
$$
or
$$
\sqrt{a+x} = \sqrt a + \varepsilon\left(\frac{|x|}{2}\right)
$$
where $\varepsilon(x)$ means some number that is smaller than $x$ in absolute value.
This holds for all $x$ as long as $a\ge4$. We only need to show convergence for one value of $m$, so choose $m_0$ large enough that $(f_m^n)^2 \ge 4$ for all $n \ge m \ge m_0$ (we can do this using the bounds above). We unravel the radical from the inside out and get inductively (with the base case $k=0$ evident):
$$ \begin{split}
f_{n-k}^n
&= \sqrt{(n-k) \pm f_{n-k+1}^{n}} \\
&= \sqrt{(n-k) \pm f_{n-k+1}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= \sqrt{\left(f_{n-k}^{n-1}\right)^2 + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= f_{n-k}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k}}\right) \\
\end{split}
$$
In particular
$$
f_{m_0}^n = f_{m_0}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which means that
$$
f_{m_0} = \sum_{n=m_0}^\infty \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which is absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
How to solve $\int \csc^{4} \frac{3}{2} x \,dx$ using integration by parts How to solve $\int \csc^{4} \frac{3}{2} x \,dx$
This is what I have so far:
$$\int \csc^{4} \frac{3}{2} x \,dx=\frac{3}{2} \int \csc^{2} x \csc^{2}x\,dx.$$
Let $u=\csc^{2}$ and $dv=\csc^{2} \,dx$. Then we have $du=-2\csc^{2}x\cot x\,dx$ and $v=-\cot x$.
\begin{align}
\frac{3}{2} \int \csc^{2} x \csc^{2} x \,dx&=\frac{3}{2}\left(-\csc^{2}x\cot x-\int-\cot x(-2\csc^{2}x\cot x\,dx)\right)\\
&= \frac{3}{2}\left(-\csc^{2}x\cot x-2\int \csc^{2}x\cot^{2}x\,dx\right)
\end{align}
At this point, I am uncertain about how to continue. Do I find the integral in the right hand side again? Or perhaps I missed something? Thank you.
| Let $t=\frac32x$ and integrate by parts as follows
\begin{align}
\int \csc^{4} \frac{3}{2} x \,dx
&= \frac23\int \csc^{4} t \,dt=- \frac29\int \sec^{2} t \>d(\cot^3 t)\\
&= -\frac29\csc^2 t\cot t+\frac49 \int\csc^2 t dt\\
&= -\frac29\csc^2 t\cot t-\frac49 \cot t +C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
locus of points with sum of distance
Find the quadratic equation for the locus of points whose sum of its distance from $(1, 0)$ and $(−1, 0)$ is $6$.
I know to start with $\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$, then square both sides. I'm not sure about the algebra after squaring both sides as I got that part wrong.
| $$\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$$
$$\sqrt{(x+1)^2+y^2}=6-\sqrt{(x-1)^2+y^2}$$
square both sides ($LHS>0;\;RHS>0$)
$$(x+1)^2+y^2=\left(6-\sqrt{(x-1)^2+y^2}\right)^2$$
$$x^2+2 x+y^2+1=36+x^2-2 x+1+y^2-12 \sqrt{(x-1)^2+y^2}$$
bring every term but the square root in the RHS
$$12 \sqrt{(x-1)^2+y^2}=36-4 x$$
simplify dividing both sides by $4$
$$3 \sqrt{(x-1)^2+y^2}=9-x$$
Notice that both sides are positive, so square again
$$9 \left((x-1)^2+y^2\right)=(9-x)^2$$
expand and bring variables into LHS and constants in the RHS
$$8 x^2+9 y^2=72$$
divide both sides by $72$
$$\frac{x^2}{9}+\frac{y^2}{8}=1$$
Which is the equation of an ellipse center $(0,0)$ and semiaxes $a=3;\;b=2\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$, I keep getting the same wrong answer For the conditions $x\left(t\right)=t^{\frac{1}{2}},\:y\left(t\right)=t^{\frac{1}{3}},\:z\left(t\right)=0,\:t>0$
I get this:
$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
$$(I)\:\:\: \frac{dx}{3y^2}=\frac{dz}{3y^2} $$
$$x=z+C_1$$
$$C_1 = x-z$$
$$(II)\:\:\: \frac{dy}{-z+x}=\frac{dz}{3y^2} $$
$$3y^2dy\:=\:\left(-z+x\right)dz$$
$$y^3=-\frac{z^2}{2}+zx+C_2$$
$$C_2=y^3+\frac{z^2}{2}-zx$$
Putting in conditions
$$C_1=t^{\frac{1}{2}}$$
$$C_2=t$$
$$C_2=C_1^2$$
This means that
$$y^3+\frac{z^2}{2}-zx = \left(x-z\right)^2$$
$$2y^3+z^2-2zx = 2 \left(x-z\right)^2$$
$$2y^3+\left(z^2-2zx+x^2-x^2\right)\:=\:2\:\left(x-z\right)^2$$
$$2y^3-x^2+\left(x-z\right)^2\:=\:2\:\left(x-z\right)^2$$
$$2y^3-x^2=\:\left(x-z\right)^2$$
$$\sqrt{2y^3-x^2}=x-z$$
$$x-\sqrt{2y^3-x^2}=z$$
But it's still wrong
So where's the mistake?
| $$\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
You should use the constant $C_1$ instead. Since $C_1=x-z$. Your integration of this differential equation is not correct.
$$3y^2{dy}=C_1{dz}$$
Integrate.
$$y^3=C_1z+C_2$$
$$y^3=(x-z)z+C_2$$
$$C_2=y^3-xz+z^2$$
This line seems not correct to me:
$$y^3=-\frac{z^2}{2}+zx+C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Different answers after different methods in solving a limit
Evaluate $$L=\lim_{x \to 0} \frac{e^{\sin(x)}-(1+\sin(x))}{(\arctan(\sin(x)))^2}$$
Method $1$: $$\frac{h^2\left(\frac{e^{h}-1}{h^2}-\frac1{h}\right)}{(\arctan(h))^2}=1^2\left(\frac{1*1}{h}-\frac1{h}\right)=\frac1{h}-\frac1{h}=0$$ Therefore $L=0$.
The identities I have used here to simplify the expression are $$\lim_{x \to 0} \frac{\arctan\left(x\right)}{x}=1$$ $$\lim_{x \to 0} \frac{a^{x}-1}{a}=\ln\left(a\right) \implies \lim_{x \to 0} \frac{e^{x}-1}{x}=\ln\left(e\right)=1$$
Method $2$: $$L=\frac{e^{h}-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(1+h+\frac{h^{2}}{2!}\right)-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(\frac{h^{2}}{2!}\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{1}{2}$$ Therefore $L=\frac12$.
I am not at all familiar with $O(n)$ notation BTW.
| My (personal) favored approach is composition of series one piece at the time.
$$y=\frac{e^{\sin(x)}-(1+\sin(x))}{\big[\arctan(\sin(x))\big]^2}$$ Since the denominator is $\sim x^2$, let us use expansions to $O(x^4)$. Working one piece at the time
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$
$$e^{\sin(x)}=1+x+\frac{x^2}{2}+O\left(x^4\right)$$
$$\arctan(\sin(x))=x-\frac{x^3}{2}+O\left(x^4\right)$$
$$\big[\arctan(\sin(x))\big]^2=x^2-x^4+O\left(x^5\right)$$ Putting all together
$$y=\frac { \frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)} {x^2-x^4+O\left(x^5\right) }=\frac{1}{2}+\frac{x}{6}+O\left(x^2\right)$$ which shows not only the limit but also how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$? I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I have to integrate it. I would think that the $-x$ would equal to $\frac{-x^2}{2}$ but the answer on the back of my book is: $\displaystyle \frac{y^2}{2} - \frac{2}{5} y^{5/2} =\frac{x^3}{3}-x-\frac{54}{5}$
What is the particular solution given $y=4$ and $x=3$?
My Solution
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}\\
y(1-\sqrt y)\frac{dy}{dx}=\frac{1}{x}\cdot (x^3-x)\\
\int (y-y^\frac{3}{2}) \frac{dy}{dx}=\int{x^2-x}\\=\frac{y^2}{2}-\frac{2y^{5/2}}{5}=\frac{x^3}{3}-\frac{x^2}{2}+c\\
\frac{y^2}{2}-\frac{2y^{5/2}}{5}-\frac{x^3}{3}+\frac{x^2}{2}=c$$
| The ODE is $$\displaystyle \small xy\frac{dy}{dx} = \frac{x^3-x}{1- \sqrt y},$$ which can be rewritten as
$$\displaystyle \small (y-y^{3/2}) \ dy = (x^2-1) \ dx. \ $$ This is the step where you have a mistake! By integrating both sides we get:
$$\displaystyle \small \frac{y^2}{2} - \frac{2}{5} y^{5/2} = \frac{x^3}{3} - x + C,$$
Now, plug in $\small y = 4, x = 3$ and so you'll get $\small C = \displaystyle - \frac{54}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find length BC of triangle with incircle and circumcircle
Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be able to figure the angle at A, and once I have that probably I can use yet another theorem on chords to determine the length BC from the length PQ...idk
| Here are 2 solutions 1) an analytical solution 2) a solution inspired by the solution of @Quanto.
*
*First solution; Let us take coordinate axes as displayed in Fig. 1 below.
Line $PQ$ is taken as the $x$-axis, with origin and unit such that $P,D,E,Q$ have resp. abscissas $-3,-1,1,2$.
Triangle $ADE$ is isosceles, due to the fact that $A$ is the intersection point of tangents in $D$ and $E$ to the incircle ; therefore $A$ belongs to the ordinate axis with coordinates $(0,\tan a)$, where $a$:=angle($ODA$).
Fig. 1.
As a consequence:
$$AD=AE=\dfrac{1}{\cos a} \ \implies \ \vec{AD}=\binom{-1}{-\tan a} \ \text{and} \ \vec{AE}=\binom{1}{-\tan a}\tag{1}$$
A classical theorem on intersecting chords in a circle gives:
$$AD.DB=PD.PQ=6 \ \implies \ DB=\dfrac{6}{AD}=6 \cos a\tag{2}$$
For a similar reason:
$$EC=4 \cos a\tag{3}$$
Therefore, using (1) and (2) and the trigonometric formula $1+\tan^2 a=1/\cos^2 a$:
$$\vec{AB}=\dfrac{\dfrac{1}{\cos a}+6 \cos a}{\dfrac{1}{\cos a}} \vec{AD}=(1+6 \cos^2 a)\binom{-1}{-\tan a}$$
implying
$$B=\binom{-1-6 \cos^2 a}{- 6 \sin a\cos a}\tag{4}$$
For completely similar reasons, using (3):
$$C=\binom{1+4 \cos^2 a}{- 4\sin a \cos a}\tag{5}$$
As a consequence of (4) and (5):
$$\vec{BC}=\binom{2+10 \cos^2 a}{2 \cos a \sin a}\tag{6}$$
It remains to express that line segment $BC$ must be tangent to the incircle. This will be done by the following condition:
$$BC=BD+CE\tag{7}$$
Squaring (7) and using (2), (3) and (6):
$$\underbrace{(2+10 \cos^2 a)^2+(2 \cos a \sin a)^2}_{BC^2}
=\underbrace{(10 \cos a)^2}_{(BD+CE)^2}\tag{8}$$
Setting $x=\cos^2 a$ in (8), we get the following quadratic equation:
$$(1+5x)^2+x(1-x)=25x \ \ \iff \ \ 24 x^2-14x+1=0\tag{9}$$
whose roots are
$$x_1=(\cos a_1)^2=1/2 \ \ \text{and} \ \ x_2=(\cos a_2)^2=1/12\tag{10}$$
Therefore, as we are looking for positive values of $\cos a$:
$$\cos a_1=\dfrac{\sqrt{2}}{2} \ \ \text{and} \ \ \cos a_2=\dfrac{1}{2\sqrt{3}}\tag{11}$$
Consequently (see (8)), the two solutions are:
$$BC=10 \cos a_1=5 \sqrt{2} \ \ \text{and} \ \ BC=10 \cos a_2=\dfrac{5}{\sqrt{3}}\tag{12}$$
Remarks :
a) Figure 1 corresponds to the second case in (12).
b) This solution doesn't use any specific triangle formula.
*Second solution:
Let $a:=AD$; due to chord theorem in circles, we have :
$$DB=6/a \ \ \ \ \& \ \ \ \ EC=4/a\tag{1}$$
Consider Fig. 2.
Fig. 2.
In rectangular triangles $AHD$ and $ADI$, we have :
$$\sin(\tfrac12 A) = \dfrac 1a \ \ \ \ \& \ \ \ \ \tan(\tfrac12 A) = \dfrac{r}{a}\tag{2}$$
Using relationship $1+\dfrac{1}{\tan^2 x}=\dfrac{1}{\sin^2x}$, we deduce from (2) that:
$$r^2=\dfrac{a^2}{a^2-1}\tag{3}$$
Besides, we have this triangle formula:
$$\tan(\tfrac12 A)\tan(\tfrac12 B)\tan(\tfrac12 C)=\dfrac{r}{p}\tag{4}$$
where $p$ is the semiperimeter (half of perimeter).
Relationship (4) can be written:
$$\dfrac{r}{a}\dfrac{r}{(6/a)}\dfrac{r}{(4/a)}=\dfrac{r}{a+\tfrac6a+\tfrac4a}\tag{5}$$
that can be transformed into
$$r^2=\dfrac{24}{a^2+10}\tag{6}$$
Equating (3) and (6) gives a second degree polynomial equation in $x=a^2$:
$$x(x+10)=24(x-1)$$
whose two real positive roots are $x_1=2=a_1^2$ and $x_2=12=a_2^2$.
And we find back the two solutions for $BC=\dfrac{10}{a}$:
$$BC=\dfrac{10}{a_1}=\dfrac{10}{\sqrt{2}} \ \ \ \text{and} \ \ \ BC=\dfrac{10}{a_2}=\dfrac{5}{\sqrt{3}} $$
Remark: relationship (4) could have been obtained using Heron's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.
Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 - 4p
\\
-1 - p = - 16 - 4p
\\
3p = -15
\\
p = -5
\\
q = -1 - (-5)
\\
q = 4
$$
According to the book the answer should be $-3$ & $-4$
| The Vieta's formulas tells you that
$$p=-(r+s),\\q=rs$$ (where $r,s$ are the roots).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
| HINT:
$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$
Separate odd and even terms.
$\frac{1}{3} +\frac{1}{3^{3}} + \frac{1}{3^{5}} + ...$
$ \frac{2}{3^{2} } + \frac{2}{3^{4}} +\frac{2}{3^{6}} ...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Find eigenvalues of $T$ and find a basis $\beta$ for $P_2(\mathbb{R})$ such that$[T]_\beta$ is a diagonal matrix. $T(f(x)=3f(x)+xf'(x)+xf''(x)$ what I did was, $T(1)=3+0+0=3,T(x)=3x+x+0=4x,T(x^2)=3x^2+2x^2+2=5x^2+2$, then $[T]_\beta=\begin{pmatrix}3&0&0\\0&4&0\\0&2&5\end{pmatrix}$ how should I continue after this?
| Were you given a problem like this without knowing what eigenvalues and eigenvectors are? That's very strange!
The eigenvalues satisfy the equation $\left|\begin{array}{ccc}3-\lambda & 0 & 0 \\ 0 & 4-\lambda & 0 \\ 0 & 2 & 5- \lambda \end{array}\right|= (3- \lambda)(4- \lambda)(5- \lambda)= 0$ so are 3, 4, and 5.
An eigenvector corresponding to eigenvalue 3 satisfies the equation $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 2 & 5\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3x \\ 4y \\ 2y+ 5z\end{bmatrix}= \begin{bmatrix} 3x \\ 3y \\ 3z \end{bmatrix}$ so 3x= 3x, 4y= 3y, and 2y+ 5z= 3z. The first equation, 3x= 3x, is t ue for all x. 4y= 3y is only true for y= 0, and 2y+ 5z= 5z- 4z is only true for z= 0. So the eigenvectors corresponding to eigevalue 3 are of the form $\begin{bmatrix}x \\ 0 \\ 0 \end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$.
Do the same to find eigenvectors corresponding to eigenvalues 4 and 5.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
| Noticing that
by which $$I= \left.\frac{\partial}{\partial a} I(a)\right|_{a=0} $$
where$$I(a)= \int_0^{\infty} \frac{x^a}{\left(1+x^2\right)^2} d x =\frac{1}{2} \int_0^1 y^{\frac{1-a}{2}}(1-y)^{\frac{a-1}{2}} d y\\ \quad \qquad =\frac{1}{2} B\left(\frac{3-a}{2},\frac{a+1}{2}\right) =\frac{1}{2} \Gamma\left(\frac{3-a}{2}\right) \Gamma\left(\frac{a+1}{2}\right) $$
Differentiating both sides w.r.t. to $a$ at $a=0$ yields $$
\begin{aligned}
I=\left.\frac{\partial}{\partial a} I(a)\right|_{a=0} &=\frac{1}{2} I\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)\left(-\frac{1}{2} 4\left(\frac{3}{2}\right)+\frac{1}{2} \psi\left(\frac{1}{2}\right)\right) \\
&=-\frac{1}{8} \Gamma^2\left(\frac{1}{2}\right)\left(4\left(\frac{3}{2}\right)-4\left(\frac{1}{2}\right)\right) \\
&=-\frac{\pi}{8} \cdot 2 \\
&=-\frac{\pi}{4}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?
This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!
| Note that $\sin x\le 1$ and $\cos \ge -1.$
$$\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge{e^{x^2}-5}.$$
So
$$\lim_{x\to\infty}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge \lim_{x\to\infty}{e^{x^2}-5}.\to\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
| We want to show that $$\displaystyle \lim_{n \to \infty} \dfrac{H_n}{n}=0$$.
We have $\log n = \displaystyle \int_{1}^{n} \dfrac{1}{t} dt$. Also, $H_n = \displaystyle \sum_{r=1}^{n}\dfrac{1}{r} = \dfrac{1}{n}\sum_{r=1}^{n}\dfrac{1}{\frac{r}{n}}$
Thus, by comparison with the Riemann sums of $\log n$, we see that $\log n < H_n < 1 + \log n$.
So we get that, $$\dfrac{\log n}{n} < \dfrac{H_n}{n} < \dfrac{1+\log n}{n}$$
But both the left and right sides of the above inequality tend to $0$ as $n \to \infty$, and hence, by the Sandwich theorem (or Squeeze principle), $$\lim_{n \to \infty} \dfrac{H_n}{n} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Finding the orthogonal trajectories of the family of cissoids Find the orthogonal trajectories of the cissoids, $y^2=\frac{x^3}{a-x}$.
My solution:
$\left( a-x\right)y^2=x^3$
$\left( a-x\right)2y dy-(2x^2+y^2)dx=0$
$\left( a-x\right)2ydx + (2x^2+y^2)dy=0$
$a=\frac{x^3}{y^2}+x$
$2y \left( \frac{x^3}{y^2}\right)dx+(2x^2+y^2)dy=0$
$2x^3dx+(2x^2+y^2)dy=0$
I do not know how to go further. What method could be used to solve this differential equation?
| Taking a derivative of your equation
\begin{align}
y&=\sqrt{\frac{x^3}{a-x}},\\
\frac{\mathrm dy}{\mathrm dx}&=\frac{3a-2x}{2}\sqrt{\frac{x}{(a-x)^3}},
\end{align}
and substituting $a=x^3/y^2+x$ gives
\begin{align}
\frac{\mathrm dy}{\mathrm dx}&=\frac{3x^3/y^2+x}{2}\sqrt{\frac{x}{(x^3/y^2)^3}}.
\end{align}
The orthogonal curves have slope that's the negative of the inverse, i.e.,
\begin{align}
\frac{\mathrm dy_{\bot}}{\mathrm dx}=\frac{-2}{3x^3/y_{\bot}^2+x}\sqrt{\frac{(x^3/y_{\bot}^2)^3}{x}}.
\end{align}
Substituting $y_{\bot}=xu$ gives that
\begin{align}
x\frac{\mathrm du}{\mathrm dx}+\frac{2+u(3+u^2)}{3+u^2}=0.
\end{align}
This is separable,
\begin{align}
\int\left( \frac{3u+u^3}{2+u(3u+u^3)}\frac{\mathrm du}{\mathrm dx}+\frac{1}{x}\right)\mathrm dx&=\int 0\mathrm dx\\
\ln(u^2+1)-\frac{1}{2}\ln(u^2+2)+\ln x&=k\\
\frac{u^2+1}{\sqrt{u^2+2}}&=\frac{k}{x}.
\end{align}
Solving for $y_{\bot}=u/x$,
\begin{align}
y_{\bot}^2=k\pm\sqrt{k(2x^2+k)}-x^2.
\end{align}
Fun problem!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solving $\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$ I am trying to solve this equation
$$\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$$
By using Mathematica, I know that, the equation has unique solution $x=1$.
I tried to write the equation in the form
$$\sqrt{(x-1) (x+1) \left(2x^2-1\right)} + \sqrt{x^2\left(2x^2-1\right)} =4 x-3.$$
From here, I can not solve it. How can I solve it?
| let $a=2x^4-3x^2+1,b=2x^4-x^2$ where $ a,b\ge 0$ now $\sqrt{a+b}=2x^2-1$ so $$\sqrt{a}+\sqrt{b}-4x+3=0$$ $$\to \sqrt{a}+\sqrt{b}-\sqrt{a+b}+\sqrt{a+b}-4x+3=0$$ $$\to \sqrt{a}+\sqrt{b}-\sqrt{a+b}+2{(x-1)}^2=0$$ because $\sqrt{a}+\sqrt{b}\ge \sqrt{a+b}$ $$\to x=1 \space \text{and} \space \sqrt{a}+\sqrt{b}=\sqrt{a+b}$$ $$\to x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$ The expression I have here is part of the upper bound given to the minimum distance of any 2 points out of $n$ points embedded in the unit sphere $\mathbb{S}^2$ in $\mathbb{R}^3$. If you have $n$ points on this unit sphere then we must have 2 points with a distance $d$ such that
$$d \le \sqrt{4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)}$$
This result was proven by Fejes Tóth. What I wish to show is that $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$.
We can see in the limit as $n \to \infty$, $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \to 0^+$. Also using basic numerical simulation for many and large $n$ I can visually verify that this statement seems to be true. The point now would be to show that for any $n$, then there exists some constant $k$ such that
$$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{k}{n}$$
This would be good enough. I'll be working on it here just thought the community may have some ideas :-)
| We wish to show that $$f(x)=\left(4-\frac kx\right)\sin^2\frac{\pi x}{6(x-2)}\le1$$ for all positive real $x$. This is evidently the case when $4-k/x\le0$ so we consider $x>k/4$ with $$f'(x)=\frac k{x^2}\sin^2\frac{\pi x}{6(x-2)}-\left(4-\frac kx\right)\frac{24\pi}{(6x-12)^2}\sin\frac{\pi x}{6(x-2)}\cos\frac{\pi x}{6(x-2)}.$$ We claim that $f'(x)>0$ for all $x>k/4$. This is equivalent to showing that $$\tan\frac{\pi x}{6(x-2)}-\frac{2\pi x(4x-k)}{3k(x-2)^2}>0.$$ The second term is at most $8\pi/3k$ so the inequality holds whenever $$\frac x{x-2}>\frac6\pi\arctan\frac{8\pi}{3k}\implies x>2+\frac2{\frac6\pi\arctan\frac{8\pi}{3k}-1}.$$ This is true whenever $k>8\pi/\sqrt3$ so $$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{\frac{8\pi}{\sqrt3}+\epsilon}{n}$$ for any $\epsilon>0$. It follows that $d<3.81n^{-1/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to resolve this integral $J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$? $$J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$$
I'm already try to do this step in below:
$$\overset{split}{=} \int{\frac{2}{2+\cos{x}}dx}+\int{\frac{d(\cos{x}+2)}{2+\cos{x}}} = \int{\frac{2}{1+2\cos^2{\frac{x}{2}}}dx} + \ln({2+\cos{x}})$$
| Note
$$\int{\frac{2}{1+2\cos^2{\frac{x}{2}}}dx}=\int{\frac{2\sec^2\frac x2}{\sec^2\frac x2+2}dx}
=4\int{\frac{d(\tan\frac x2)}{\tan^2\frac x2+3}dx}
= \frac4{\sqrt3}\tan^{-1}\frac{\tan\frac x2}{\sqrt3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
what result I should consider for GCD with polynomials? I need to calculate the GCD of
$$x^4+3x^3+2x^2+x+4 \ \text{and } x^3+3x+3 \ \text{in} \ \mathbb{Z}_5$$
Using Euclid algorithm:
$$x^4+3x^3+2x^2+x+4 = (x^3+3x+3)(x+3)-3x\\ x^3+3x+3 = (-3x)(\frac{1}{3}x^2 - \frac{2}{3})+3 \\-3x = (3)(-x)+0 $$
Now I should consider the last not-null remainder (so 3) that (I think) it should be the GCD, actually it seems that the GCD is 1 according to wolfram
So my question is: what result should I consider as GCD when GCD is calculated with polynomials?
| Over the field $\Bbb F_5$ we have
$$
x^4+3x^3+2x^2+x+4=(x^3 + 4x^2 + x + 1)(x + 4)
$$
and $x^3+3x+3$ has no root in $\Bbb F_5$ and hence is irreducible.
So is the other polynomial of degree $3$. Here we have used that a cubic polynomial over a field is reducible if and only if it has a root.
From this it is clear that the gcd is equal to one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\sin\angle HIO$ given $\triangle ABC$ with $(\overline{AB},\overline{BC},\overline{CA})=(8,13,15)$? In triangle $ABC, AB = 8, BC = 13$ and $CA = 15$.
Let $H, I, O$ be the orthocenter, incenter and circumcenter of triangle $ABC$ respectively. Find $\sin$ of angle $HIO$ .
My attempt: By using law of cosine I can see angle $A = 60^{\circ}$. I also observed that angle $BHC$, $BIC$ and $BOC$ equals $120^{\circ} $. How do I proceed after this?
| $\displaystyle \small \angle IAO = \angle IAH = \frac{\angle B - \angle C}{2}$
Now use the identity, $\small AH = 2R \cos \angle A \implies AH = R = AO \ $ (as $\angle A = 60^0$)
That means $\small \triangle HAO$ is an isosceles triangle. As $\small AI$ is angle bisector of $\small \angle HAO, IH = OI$.
Now remember that, $\small OI^2 = R(R-2r), OH^2 = 9R^2 - (a^2+b^2+c^2)$
$\small R = \displaystyle \frac{a}{2 \sin A} = \frac{13}{\sqrt3}$
$\small \displaystyle r = \frac{\Delta}{s} = \frac{(bc \sin A) / 2}{(a+b+c)/2} = \frac{5}{\sqrt3}$
So $\small OI = \sqrt {13} = IH, \ OH = 7$
$\small \displaystyle \cos \angle HIO = - \frac{23}{26} \implies \fbox {$\sin \angle HIO = \frac{7 \sqrt3}{26}$}$
This is just an addendum showing how $\displaystyle \small \angle IAO = \angle IAH = \frac{B - C}{2}$
$\small \displaystyle \angle BAH = 90^0 - B, \angle IAH = \angle IAB - \angle BAH = \frac{A}{2} - 90^0 + B$
As $\small \displaystyle \frac{A+B+C}{2} = 90^0, \angle IAH = \frac{B-C}{2}$
Now $\small \angle AOB = 2C, \angle OAB = \angle OBA = 90^0 - C$
$\small \displaystyle \angle IAO = \angle OAB - \angle IAB = \frac{A+B+C}{2} - C - \frac{A}{2} = \frac{B-C}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the area of three quadrilaterals inside an equilateral triangle $\triangle ABC$ is an equilateral triangle with side length of $1$. $D,E,F$ are $\frac{1}{3}$ away from $C,A,B$. What is the total area of the three quadrilaterals 1,2 and 3 enclosed by the orange sides? I can think about finding the coordinates of the vertices of the quadrilateral and use formula to calculate the area (see note below), but could not find an easy geometric way to solve this.
Note:
Solution by coordinates calculation
AF intercepts EC at G, BD at H, and BD intercept EF at I.
Given $A(0, \frac{\sqrt 3}{2}, B(-\frac{1}{2},0), C(\frac{1}{2},0)$,
$E(x_1,y_1)=E(-\frac 16,\frac{\sqrt3}{3}),G(x_2,y_2)=G(-\frac 1{14},\frac{2\sqrt3}{7}),H(x_3,y_3)=H(-\frac 17,\frac{\sqrt3}{14}),I(x_4,y_4)=I(-\frac 16,\frac{\sqrt3}{15})$
$S_1=S_2=S_3=\frac {1}{2} ⋅ {(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)
- (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)}\approx 0.02612$
$S=S_1+S_2+S3 \approx 0.07835$
$\frac{S}{\triangle ABC} \approx 0.18$
| Denote area of $\triangle ABC$ by $\Delta$. Let areas of the three quadrilaterals each be $X$ and the blue triangles in following diagram, each be $Y$.
We'll take ratios approach, which is fun, form two linear equations in $X,Y$ and solve for it.
First notice $\triangle$s $AED$, $BFE$, $CDF$ are all $30^\circ-60^\circ-90^\circ$. Each of these e.g., $\triangle BFE$ has base one-third and height two-third the original, hence area of equilateral $\triangle DFE$ is
$$[DFE] = \Delta \left(1-3\cdot \frac{1}{3} \cdot \frac{2}{3}\right)=\frac{\Delta}{3}$$
If you don't know the hatched triangle is the one-seventh area triangle, we can also activate Routh's theorem to obtain its area as $\Delta/7$. We write
$$[DFE] = \frac{\Delta}{3} = 3X + 3Y + \frac{\Delta}{7} \tag{1}$$
Next $\triangle AFE$ has same height but one-third base of $\triangle ABF$ which in turn has one-third base of $\triangle ABC$, so $[AFE] = \Delta/9$.
Drop $FP \perp AB$. $\triangle BPF$ is also $30^\circ-60^\circ-90^\circ$. Let $BF=a$. $\triangle AQE \sim \triangle AFP$ with ratio $AE/AP=AE/(AB-BP)=2/5$. Thus $AQE$ has same base but two-fifths the height of $\triangle AFE$. We can write
$$[AFE]=\frac{\Delta}{9}=X+2Y+\frac{2}{5}\cdot \frac{\Delta}{9} \tag{2}$$
On simplifying $(1),(2)$,
$$3X+3Y=\frac{4}{21}\Delta \tag{3}$$
$$X+2Y=\frac{1}{15}\Delta \tag{4}$$
Eliminating $Y$, we get
$$3X=\left(2\cdot \frac{4}{21} - 3\cdot \frac{1}{15} \right) \Delta$$
That is
$$\boxed{\text{Area of quadrilaterals} = \frac{19}{105}\Delta}$$
which is about $18 \%$ of area of $\triangle ABC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?
| Using the quadratic formula is a good approach, and everything you have done so far is correct. The modulus/absolute value can be replaced by normal brackets as there is a $\pm$ preceding it, and the numbers involved here are real. Notice that:
$$\frac {a^2 +b^2 + (a^2 - b^2)}{2ab} = \frac ab, \quad \frac {a^2 +b^2 - (a^2 - b^2)}{2ab} = \frac ba$$
hence the roots are $\dfrac ab$ and $\dfrac ba$, and we are looking at the factors $(bx- a)$ and $(ax - b)$. Now:
$$(bx-a)(ax- b) = abx^2 - a^2 x - b^2 x + ab$$
which is exactly what we are trying to factorize.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Minimum of $E(z)=|z+1|+|z^{2n}+1|$
Let $n \in \mathbb{N}$. Find the minimum of the expression $E(z)=|z+1|+|z^{2n}+1|$ over $\mathbb{C}$.
I found this problem in a Romanian magazine with all sorts of math problems.
(Edit: ${\color{blue}{\textrm{This was suggested for a contest for 10th graders}}}$, According to @alexanderv's comment.)
What I've done so far(WRONG):
$E \geq 0$ so if we can find $z$ with $E(z)=0$ it's enough. If $n=0$, the minimum is achieved in $-1$, $E(z) =2$. If $n \geq 1$, let's take $z= \cos \theta + i\sin \theta$. Using $|1+\cos \alpha + i\sin \alpha|=2| \cos (\frac{\alpha}{2})|$, we get $E(z)=2(|cos(\frac{\theta}{2})|+|cos(n\theta)|)$. Then, taking the modulus and using $\cos(\pi-\alpha)=-\cos(\alpha)$ and the $\cos(x) + \cos(y)$ formula, we take $z$ so one of the product terms equals to $0$, achieving the minimum. But I'm not sure if it's right, can someone approve?
What's wrong above is $E(z)=0$ implies $z=-1$, that means $E(z)=2$, contradiction. I don't have any idea how to find the minimum.
| shymilan@AoPS's elegant proof (I rewrote it):
Let $w_k = \mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi}, ~ k = 1, 2, \cdots, 2n$. Then $Q(z) := z^{2n} + 1
= (z - w_1)(z - w_2)\cdots (z - w_{2n})$.
We have the partial fraction decomposition
$$\frac{1}{z^{2n} + 1} = \frac{1}{Q(z)}
= \sum_{k=1}^{2n} \frac{1}{Q'(w_k)}\frac{1}{z - w_k}
= \sum_{k=1}^{2n} \frac{-w_k}{2n}\frac{1}{z - w_k}. \tag{1}$$
(easy to prove, e.g. see Partial_fraction_decomposition).
For $k = 1, 2, \cdots, 2n$, we have
\begin{align*}
|z + 1| + |z^{2n} + 1| &=
|z - w_k + w_k + 1| + |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\
&\ge |w_k + 1| - |z - w_k|
+ |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\
&= |w_k + 1| + |z - w_k|
\left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right). \tag{2}
\end{align*}
Using (1), we have
\begin{align*}
\sum_{k=1}^{2n} \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right)
&= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k}\right| - 2n \\
&= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\
&\ge \left|\sum_{k=1}^{2n} \frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\
&= 0. \tag{3}
\end{align*}
From (2) and (3), there exists $k \in \{1, 2, \cdots, 2n\}$ such that $\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1 \ge 0$ and
$$|z + 1| + |z^{2n} + 1| \ge |w_k + 1|.$$
For $k = 1, 2, \cdots, 2n$, we have
$$|w_k + 1| = |\mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi} + 1| = 2\left|\sin \frac{(2n - 2k + 1)\pi}{4n}\right|
\ge 2 \sin \frac{\pi}{4n}.$$
Thus, we have
$$|z + 1| + |z^{2n} + 1| \ge 2 \sin \frac{\pi}{4n}.$$
Also, if $z = \mathrm{e}^{\mathrm{i}\frac{2n - 1}{2n}\pi}$,
we have $|z+1| + |z^{2n} + 1| = 2 \sin \frac{\pi}{4n}$.
Thus, the minimum of $|z+1| + |z^{2n} + 1|$ is $2 \sin \frac{\pi}{4n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Order of a binomial coefficient
I am trying to find the Big-O order of $\binom{n-1}{\frac{n}{2}-1}$ where $n$ is even.
I know that $\binom{n}{\frac{n}{2}}=O\left(\frac{2^n}{\sqrt{n}}\right)$ using Stirling's approximation. I know that $\binom{n-1}{\frac{n}{2}-1}$ is smaller than $\binom{n}{\frac{n}{2}}$, so I'm guessing that the order of $\binom{n-1}{\frac{n}{2}-1}$ is close to $O\left(\frac{2^n}{\sqrt{n}}\right)$ like maybe $O\left(\frac{2^n}{n}\right)$?
My attempt:
\begin{align}
\binom{n-1}{n/2-1} &= \frac{(n-1)!}{(n/2)! (n/2-1)!} \\
&= \frac{(n-1)^{n-1}e^{-(n-1)}\sqrt{n-1}}{(n/2)^{n/2}e^{-n/2}\sqrt{n}\times \left(n/2-1\right)^{n/2-1}e^{-(n/2-1)}\sqrt{n/2-1}} \text{ By Stirling's } \\
&\approx\frac{(n-1)^{n-1}\sqrt{n-1}}{(n/2)^{n/2}\left(n/2-1\right)^{n/2-1}\sqrt{n(n-2)}} \\
&\approx\frac{(n-1)^{n-1}\sqrt{n-1}\times2^n}{n^{n/2}\left(n-2\right)^{n/2-1}\sqrt{n(n-2)}} \\
&\approx\frac{(n-1)^{n-1}\sqrt{n-1}\times (n-2)}{\left(n(n-2)\right)^{n/2}\sqrt{n(n-2)}}\times2^n \\
\end{align}
So I am not able to find the order the first term above. Any ideas?
| Write this as
$$
2^n\times \frac{\sqrt{n-1}}{\sqrt{n}\cdot \sqrt{n-2}}\times \frac{(n-1)^{n-1}(n-2)}{(n(n-2))^{n/2}}\\
\\\approx 2^n\times \color{#2B2}{\frac1{\sqrt{n}}}\times { (1-\frac1n)^{n-1}(1-\frac2n)\over (1-\frac1{n/2})^{n/2}}
$$
and use the limit definition of $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding integer solutions to the system $xy=6(x+y+z)$, $x^2+y^2=z^2$ How do you go about solving a system of equations like below for integer solutions?
$$xy=6(x+y+z)$$
$$x^2+y^2=z^2$$
Would you first try and list out a number of Pythagorean triples, then try and see which ones when multiplied are divisible by 6 (by 2 and 3)? Since the other side is just the sum of the 3 of them left.
I mean is there an elegant way to find these triplets, apart from a brute-force or any exhaustion method? Or even using Graphs perhaps? And why is there not infinite solutions to this? (there are actually 6 triplets).
P.S. As a bonus, replace the first equation above with:
$$x^2 + (z-510)x + 1020p = 0$$
while in the 2nd equation:
$$y=1020p/x$$
where $p$ is Prime.
Mathematica gives me only 2 solutions for $p$ for this. But how??
| Let $t = x+y+z \implies z^2 = x^2+y^2 = (x+y)^2 - 2xy = (x+y+z - z)^2 - 2\cdot 6(x+y+z)= t^2-2tz+z^2-12t\implies t^2-2tz-12t=0\implies t(t-2z-12)=0\implies t = 0$ or $t = 2z+12$. Thus $x+y=-z$ or $x+y+z=2z+12$ or $x+y=z+12$. Thus first case yields $xy = 0 \implies x = 0 $ or $y = 0$ or $x = y = 0=z$. Thus $x = 0, y = -z$ is a solution, and $x = -z, y = 0$ is another. For if $x+y = z+12 \implies x^2+y^2=(x+y-12)^2= x^2+y^2+144+2xy-24x-24y\implies 24x+24y=144+2xy\implies 24x-2xy=144-24y\implies 12x-xy=72-12y\implies (12-y)x=72-12y\implies x = \dfrac{72-12y}{12-y}= 12- \dfrac{72}{12-y}\implies (12-y)\mid 72\implies 12-y = \pm \{1, 2,3,4,6,8,9,12,18,24,36,72\}$. You can now check case by case to pick the one that is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.
I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-3)^2}=3PA+PB$$
Where $P$ is a generic point on $x^2+y^2=4$ and $A(1,0),B(0,3)$.
So the problem essentially means which point on the circle $x^2+y^2=4$ minimizes $3PA+PB$.
I am really struggling to find the geometrical notion of this and hence unable to solve.
| We can parametrize the constraint condition as $$(x,y) = (2 \cos t, 2 \sin t), \quad t \in [0,2\pi),$$ hence we are interested in the extrema of $$f(t) = 3\sqrt{5-4\cos t} + \sqrt{13-12\sin t}.$$ Differentiating to locate critical points, we find $$\sin t \sqrt{13 - 12 \sin t} = \cos t \sqrt{5 - 4 \cos t},$$ or $$\tan^2 t = \frac{5 \sec t - 4}{13 \sec t - 12 \tan t}.$$ Collecting like terms yields $$(12 \tan^3 t - 4)^2 = (13 \tan^2 t - 5)^2 \sec^2 t,$$ hence with $z = \tan t$,
$$\begin{align}
0 &= (13 \tan^2 t - 5)^2 (1 + \tan^2 t) - (12 \tan^3 t - 4)^2 \\
&= 25 z^6 + 39 z^4 + 96 z^3 - 105 z^2 + 9 \\
&= (5z^2 + 6z - 3)(5z^4 - 6z^3 + 18z^2 - 6z - 3).
\end{align}$$
Consequently, the critical values correspond to$$\tan t = z = \frac{-3 \pm 2 \sqrt{6}}{5}.$$ The roots of the quartic factor are all complex-valued. It follows that $$(\cos t, \sin t) = \left\{ \left( \frac{3 \sqrt{6} - 2}{10}, \frac{3 \sqrt{6} + 2}{10} \right), \left( - \frac{6 + \sqrt{6}}{10}, \frac{6 - \sqrt{6}}{10} \right) \right\},$$ and $$f(t) = \left\{ \frac{\sqrt{101 + 6 \sqrt{6}} + 9 \sqrt{3} - 3 \sqrt{2}}{\sqrt{5}}, 2 \sqrt{10} \right\}.$$ The first one is the maximum and the second is the minimum. Attached is a plot of $f$:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4098030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$ Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$
I solved this integral by euler substitution by replacing
$\sqrt{x^2+x-1}=x+t$
but it's not allowed by the problem.
p.s Is there any other method to solve with?
Thank you in advance :)
| Note
\begin{align}
\int \sqrt{x^2+x-1} \>dx
&\overset{IBP}=\frac12 x \sqrt{x^2+x-1} +\frac14\int \frac {x-2}{\sqrt{x^2+x-1}}dx\\
\end{align}
Then
\begin{align}
&\int\frac{3x^2-1}{\sqrt{x^2+x-1}}\>dx\\
=& 3 \int \sqrt{x^2+x-1} dx - \int \frac {3x-2}{\sqrt{x^2+x-1}}dx\\
=& \frac32 x \sqrt{x^2+x-1} -\frac98 \int \frac {2x+1}{\sqrt{x^2+x-1}}dx + \frac{13}8 \int \frac {1}{\sqrt{x^2+x-1}}dx \\
=& \frac32 x \sqrt{x^2+x-1} -\frac94 \sqrt{x^2+x-1}
+ \frac{13}8 \cosh^{-1}\frac{2x+1}{\sqrt5}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expressing a determinant in the form $k(a^3 + b^3 + c^3 - 3abc)^n$ I have been given a determinant $\begin{vmatrix} bc - a^2 & ac - b^2 & ab - c^2 \\ ac - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ac - b^2\end{vmatrix}$ and I want to compute it in the form $k(a^3 + b^3 + c^3 - 3abc)^n$
Now I know that for determinants like these this property exists that $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}$ can be written as $-(a^3 + b^3 + c^3 - 3abc)$ but I am not able to figure out a way to do some operations so as to express the matrix above in this form. Ofcourse I could compute the entire determinant and then possibly factories it or so, but that process looks very tedious and definitely doesn't seem like the right approach to it. Any hints/explanations are appreciated on how to approach this.
| The determinant is fully symmetric and homogeneous of degree $6$, and vanishes if $c=-a-b$ since then all entries are $-a^2-ab-b^2$, and also if $a=b=c$ since then all entries are $0$, so is divisible by both $a+b+c$ and$$\tfrac12((a-b)^2+(b-c)^2+(c-a)^2)=a^2+b^2+c^2-ab-bc-ca,$$and hence by their product, $a^3+b^3+c^3-3abc$. The determinant is also invariant under changing the signs of all variables, so it's of the desired form with $n=2$. Then $k$ is just the determinant for $a=1,\,b=c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating a limit. Is WolframAlpha wrong or am I wrong? What I'm trying to solve:
$$\lim _{x\to -\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}$$
What I put into WolframAlpha:
(sqrt(x^2+14)+x)/(sqrt(x^2-2)+x)
My result: $1$, which I get by simply dividing bot the numerator and the denominator by $x$, then letting it go towards $-\infty$
$$\frac{\frac{\left(\sqrt{x^2+14}+x\right)}{x}}{\frac{\left(\sqrt{x^2-2}+x\right)}{x}}=\frac{\left(\sqrt{\frac{x^2}{x^2}+\frac{14}{x^2}}+\frac{x}{x}\right)}{\left(\sqrt{\frac{x^2}{x^2}-\frac{2}{x^2}}+\frac{x}{x}\right)}=\frac{\left(\sqrt{1+\frac{14}{x^2}}+1\right)}{\left(\sqrt{1-\frac{2}{x^2}}+1\right)}\:=\:\frac{\left(\sqrt{1}+1\right)}{\left(\sqrt{1}+1\right)}\:=\:\frac22 \ = \ 1$$
WolframAlpha's result: $-7$. It has a long, complicated 25 step solution.
Is the solution $1$, $-7$ or neither?
Edit: of course, I set it $x$ to go towards $-\infty$ in WolframAlpha too
| Your statement is correct for the limit as $ \ x \ $ approaches "positive infinity",
$$\lim _{x \ \to \ +\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)} \ \ = \ \ 1 \ \ , $$
and the means by which you calculated it is valid in that direction.
A basic issue which arises in the expression for which we are taking limits is that the function $ \ \sqrt{x^2+a} \ $ (with any real $ \ a \ $) has even symmetry on its domain, while $ \ x \ $ has odd symmetry, so the numerator, denominator, and the rational function produced are not symmetric about the $ \ y-$ axis. Rather than having a single horizontal asymptote, this function has two distinct asymptotes.
We can examine the general function $ \ x \ + \sqrt{x^2+a} \ . $ The asymptotes for $ \ \sqrt{x^2+a} \ $ are given by $ \ y \ = \ |x| \ , $ so as $ \ x \rightarrow +\infty \ , $ we expect that $ \ x \ + \ \sqrt{x^2+a} \ \rightarrow \ 2x \ . $ So a ratio of two such functions has the limit
$$\lim _{x \ \to \ +\infty \:}\frac{\left( \ x \ + \ \sqrt{ x^2+a } \ \right)}{\left( \ x \ + \ \sqrt{ x^2+b} \ \right)} \ \ \rightarrow \ \ \frac{2x}{2x} \ \ = \ \ 1 \ \ , $$
independently of the values of $ \ a \ $ and $ \ b \ , $ as you demonstrated.
For the limit as $ \ x \rightarrow -\infty \ , $ however, we can see that $ \ x \ + \ \sqrt{x^2+a} \ \rightarrow \ x \ + \ |x| \ \rightarrow \ x \ - \ x \ \rightarrow \ 0 \ . $ But it does so in a way that does depend on the sign and value of $ \ a \ , $ since $ \sqrt{x^2+a} \ $ approaches $ \ -x \ $ "from above" for $ \ a > 0 \ $ and "from below" for $ \ a < 0 \ \ . $ Applying the "conjugate-factor" method, we produce
$$ \lim _{x \ \to \ -\infty \:} \ \frac{( \ x \ + \ \sqrt{x^2+a} \ ) · ( \ x \ - \ \sqrt{x^2+a} \ ) }{ x \ - \ \sqrt{x^2+a} } \ \ = \ \ \lim _{x \ \to \ -\infty \:} \ \ \frac{ x^2 \ - \ (x^2+a)}{ x \ - \ \sqrt{x^2+a} } $$ $$ = \ \ \lim _{x \ \to \ -\infty \:} \ \ \frac{- \ a}{ x \ - \ \sqrt{x^2+a} } \ \ . $$
We can "extract" a factor of $ \ x \ $ in the denominator, but as mentioned more than once among the comments, because we have $ \ x < 0 \ , $ we must work "under the square-root" as $ \ x \ = \ - |x| \ = \ - \sqrt{x^2} \ : $
$$ \lim _{x \ \to \ -\infty \:} \ \ \frac{- \ a}{ x \ - \ \sqrt{x^2+a} } \ \ = \ \ \lim _{x \ \to \ -\infty \:} \ \ \frac{- \ a}{ x · ( \ 1 \ - \ \left[ \frac{1}{-\sqrt{x^2}} \right] · \sqrt{x^2+a} \ ) } $$ $$ = \ \ \lim _{x \ \to \ -\infty \:} \ \ \frac{- \ a}{ x · ( \ 1 \ + \ \sqrt{\frac{x^2}{x^2} + \frac{a}{x^2}} \ ) } \ \ \rightarrow \ \ \frac{- \ a}{ x · ( \ 1 \ + \ \sqrt{1 \ + \ 0} \ ) } \ = \ -\frac{a}{2x} \ \ \rightarrow \ \ 0 \ \ . $$
The "negative-infinity" limit for $ \ x \ + \sqrt{x^2+a} \ $ leaves the value of $ \ a \ $ "exposed", so the corresponding limit for our ratio becomes
$$\lim _{x \ \to \ -\infty \:}\frac{\left( \ x \ + \ \sqrt{ x^2+a } \ \right)}{\left( \ x \ + \ \sqrt{ x^2+b} \ \right)} \ \ \rightarrow \ \ \frac{-a}{-b} \ · \ \frac{2x}{2x} \ \ = \ \ \frac{a}{b} \ \ . $$
For the particular function in question then,
$$\lim _{x \ \to \ -\infty \:}\frac{x \ + \ \sqrt{ x^2+14 } }{x \ + \ \sqrt{ x^2-2 } } \ \ = \ \ \frac{14}{-2} \ \ = \ -7 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
When we express $\sin x - \cos x$ as $A \sin (x+c)$, how many solutions are there for $c \in [0, 2\pi)$? This is a problem from problem set 1 of MIT OCW 18.01SC:
express $\sin x - \cos x$ in the form $A \sin (x+c)$.
Their solution is $\sqrt{2} \sin (x - \frac{\pi}{4})$
I found two solutions (for $c \in [0, 2\pi]$):
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
Are there two solutions or is there something amiss with the second solution above?
Here is how I got the solutions:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be written in the form $A \sin(x+c)$.
We have two equations in two unknowns (A and c) and so we can solve for these variables. Square both sides of each equation:
$$A^2 \cos^2 c = 1$$
$$A^2 \sin^2 c = 1$$
Sum the two equations
$$A^2(\cos^2 x + \sin^2 x) = 2$$
$$A^2 = 2 \Rightarrow A = \pm \sqrt{2}$$
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
$$\cos c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
$$\sin c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{5\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{7\pi}{4}$, which is the same as $-\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
$$\cos c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{3\pi}{4} \text{ or } c = \frac{5 \pi}{4}$$
$$\sin c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{3 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{3\pi}{4}$.
| Both of your answers are correct. Both are valid solutions. Remember this: for any trig-problem, there exists exactly one solution in a interval of length $\pi$. So, for an interval of length $2\pi$, there will be two solutions.
Usually, such problems arise in physics while studying waves and oscillations. There, $A$ must be positive. That's why the first solution is preferred.
Hope this helps. Ask anything if not clear :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$? For numbers $a,b,c$ we know that $\frac{a-c}{b+c}+\frac{b-a}{c+a}+\frac{c-b}{a+b}=1$ What is the value of $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$ ?
$1)3\qquad\qquad2)4\qquad\qquad3)6\qquad\qquad4)2$
Because it is a multiple choice question at first I tried to find $a,b,c$ so that each fraction be equal to $\frac13$ and their sum be $1$. but unfortunately I couldn't find such numbers.
I think adding the fractions by finding common denominator is not a good idea because the numerator will be so ugly. but I can rewrite it as:
$$\frac{a}{b+c}-\frac c{b+c}+\frac b{c+a}-\frac{a}{c+a}+\frac{c}{a+b}-\frac{b}{a+b}=1$$
But I can't continue from here.
| \begin{align}&\frac{a-c}{b+c}+\frac{b-a}{c+a}+\frac{c-b}{a+b} = 1\\
\implies&\frac{a+b-b-c}{b+c}+\frac{b+c-c-a}{c+a}+\frac{c+a-a-b}{a+b} = 1\\
\implies&\frac{a+b}{b+c}-1+\frac{b+c}{c+a}-1+\frac{c+a}{a+b}-1 = 1\\
\implies&\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}=4\\\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
k-th smallest number of standard normal distribution There are $n$ random variables obey the standard normal distribution $N(0 ,1)$.
What is the expectation of the k-th smallest number?
My attempt
I tried to use conditional probability to solve the problem, but the situation of $n=4$ is too complex to handle.
for n = 2
$$p_1(x) = \frac{e^{-\frac{x^2}{2}} }{\sqrt{2 \pi }}\text{erfc}\left(\frac{x}{\sqrt{2}}\right)$$
$$p_2(x) = \frac{e^{-\frac{x^2}{2}} }{\sqrt{2 \pi }}\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)$$
$E_1 = -\dfrac{1}{\sqrt{\pi }}, E_2 = \dfrac{1}{\sqrt{\pi }}$
for n = 3
$$p_1(x) = \frac{3 e^{-\frac{x^2}{2}} }{4 \sqrt{2 \pi }}\text{erfc}\left(\frac{x}{\sqrt{2}}\right)^2$$
$$p_2(x) = -\frac{3 e^{-\frac{x^2}{2}} }{2 \sqrt{2 \pi }}\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)^2-1\right)$$
$$p_3(x) = \frac{3 e^{-\frac{x^2}{2}} }{4 \sqrt{2 \pi }}\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)^2$$
$E_1 = -\dfrac{3}{2 \sqrt{\pi }}, E_2 = 0, E_3 = \dfrac{3}{2 \sqrt{\pi }}$
Is there any good way to deal with this problem?
| The density of the $k$-th smallest number among $n$ i.i.d standard normal random variables is given by
$$
f_{X_{(k)}}(x) = \frac{n!}{(r-1)!(n-r)!} \phi(x) \left[\Phi(x)\right]^{k-1}
\left[1-\Phi(x)\right]^{n-k},$$
where $\phi(x)$ and $\Phi(x)$ are PDF and CDF of standard normal distribution respectively. Then the task is to compute$\int x f_{X_{(k)}}(x) \,dx$. Thr approximation of this integral is done here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Function with many values in a sequence Cool problem I was reading but couldn't solve!
Problem:
Let $f(n)$ be a function satisfying the following conditions:
a) $f(1) = 1$.
b) $f(a) \leq f(b)$ where $a$ and $b$ are positive integers with $a \leq b$.
c) $f(2a) = f(a) + 1$ for all positive integers a.
Let $M$ denote the number of possible values of the $2020$-tuple $(f(1), f(2), f(3), ..., f(2020))$ can take. Find $M ($mod $1000)$.
$\\$
My solution (incomplete):
I started listing a few values of $f(n)$, such as $f(1) = 1, f(2) = 2, f(3) = 2, 3, f(4) = 3, $etc. When $n$ is not a power of $2$, $f(n)$ has many values. For example, $f(3) = 2, 3$ and $f(5), f(6), f(7) = 3, 4.$ Using condition b, there are $2$ ways to select a value for $f(3); 2, 3$. $f(5), f(6), f(7)$ has $4$ ways to select values for those $3$; $(3, 3, 3), (3, 3, 4), (3, 4, 4), (4, 4, 4)$ respectively for $f(5), f(6), f(7).$ For the next "group"; $f(9), f(10), ..., f(15)$, there are $8$ ways. This process keeps happening for when a group starts with $f(n)$ such that $n = 2^x + 1$ for some positive integer $x$ and the group ends with $f(n) = 2^{x+1} - 1$, there are $2^x$ ways to assign values to that group. However, from here, I don't know how to calculate how many total possible values the $2020$-tuple can take. Do I add or multiply the values in these groups because they are independent events? Please help. Thanks in advance to those who help.
By the way, the correct answer is $\boxed{502}$ but I don't know how to get this.
| (My final answer differs from yours. I believe Ross and I are right.)
Define $ k_n$ to be the maximal $k$ such that $ f(k) = n$.
Show that
*
*$f (k_n ) = f(k_n - 1 ) = \ldots = f(k_{n-1} + 1) = n$
*$k_1 = 1$
*$k_{n+1} = 2k_n $ or $ 2k_{n}+1$.
*$k_n = \lfloor \frac{k_{n+1} }{2} \rfloor $
*A sequence $f(1), f(2), \ldots $ is uniquely defined by $k_1, k_2, k_3, \ldots $
*For this problem, we only care about $k_n \leq 2020$.
*A sequence is uniquely defined by the largest $k_n$, as all previous values are uniquely determined. Let's denote this $K$.
*We have $1011 \leq K \leq 2020$, so there are $2020-1011 + 1 = 1010$ such sequences.
As an example, if we care about the first 10 numbers, then we have $ 6 \leq K \leq 10$ giving the corresponding sequences:
$\begin{array} {l l l}
K = 6 \rightarrow & 1, 3, 6 \rightarrow & 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 \\
K = 7 \rightarrow & 1, 3, 7 \rightarrow & 1, 2, 2, 3, 3, 3, 3, 4, 4, 4 \\
K = 8 \rightarrow & 1, 2, 4, 8 \rightarrow & 1, 2, 3, 3, 4, 4, 4, 4, 5, 5 \\
K = 9 \rightarrow & 1, 2, 4, 9 \rightarrow & 1, 2, 3, 3, 4, 4, 4, 4, 4, 5 \\
K = 10 \rightarrow & 1, 2, 5, 10 \rightarrow & 1, 2, 3, 3, 3, 4, 4, 4, 4, 4\\
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$ Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$.
Where do I start with this integral? I can easily see that it is possible to fator $x^{2}$ out on the denominator and use partial fractions. The numerator is also factorable but it does not have any integer roots. Can someone show me how to integrate this (either using what I said about partial fractions), or another method? Thanks
| Using partial fractions:
$$\frac{x^3+4x^2 +x-1 }{x^3 +x^2} = \frac ax +\frac{b}{x^2} +\frac{c}{x+1}+d \\ \implies x^3 +4x^2 +x-1 = ax(x+1) +b(x+1) +cx^2 +dx^2(x+1)$$ Put $x=0$ to get $b=-1$ and put $x=-1$ to get $c=1$. Then, compare the coefficient of $x^3$ to get $d=1$ and then that of $x^2$ to get $ a=2$.
All in all, this integrates to $$2\ln|x| +\frac 1x +\ln|x+1|+x +C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\frac{1}{a^2-ac-ab+bc} + \frac{2}{b^2-ab-bc+ac} + \frac{1}{c^2-ac-bc+ab}$ for $a=1.69$, $b=1.73$, and $c=0.48$ An interesting question on algebra:
If $a=1.69$, $b=1.73$, and $c=0.48$, find the value of$\\$
$$\frac{1}{a^2-ac-ab+bc} + \frac{2}{b^2-ab-bc+ac} + \frac{1}{c^2-ac-bc+ab}$$
By substituting the values, I got the answer, but I want to know if there is any short algebraic way because putting the values was a lot time consuming.
Thanks.
| The expression $L$ is equal to
\begin{align}
L &= \frac{1}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)} \\
&= \frac{-(b-c)-2(c-a)-(a-b)}{(a-b)(b-c)(c-a)} \\
&= \frac{a-c}{(a-b)(b-c)(c-a)} \\
&=- \frac{1}{(a-b)(b-c)} \\
\end{align}
and it's easy now to calculate $L = 20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sums of Squares algorithm I want to express a large number as the sum of two squares, given that it is possible and given its prime factors. Let's say the number is $273097$. It's prime factors are $11^2, 37$ and $61$. Here is is easy to see $11^2=11^2+0^2$ and $37=6^2+1^2$. Through trial and error I found that $61=5^2+6^2$, but having read Efficiently finding two squares which sum to a prime, it looks like I just needed to notice that $11^2\equiv-1\pmod{61}$ and I could then apply the Euclidean algorithm to $61$ and $11$ to find the first two remainders below $\sqrt{61}$, which are of course $5$ and $6$. Then I have $273097=11^2(6\times5-1\times6)^2+11^2(6\times6+1\times5)^2$. Is this the best way going about a question like this?
| Prime number $p$ can be expressed as sum of two (non-zero) squares if $p\equiv 1 \pmod{4}$.
We know $37 = 1^2 + 6^2, 61 = 5^2 + 6^2.$
Hence $37$ and $61$ are expressed using Gaussian integers below.
Norm$(1+6i) = (1+6i)(1-6i) = 1^2 + 6^2 = 37$
Norm$(5+6i) = (5+6i)(5-6i) = 5^2 + 6^2 = 61$
Since $273097 = 11^2\cdot37\cdot61$ = Norm$(11(1\pm6i)(5\pm6i))$, then we get
$11(1+6i)(5+6i) = -341+396i \implies 273097 = 341^2 + 396^2.$
$11(1-6i)(5+6i) = 451-264i \implies 273097 = 451^2 + 264^2.$
$11(1+6i)(5-6i) = 451+264i \implies 273097 = 451^2 + 264^2.$
$11(1-6i)(5-6i) = -341-396i \implies 273097 = 341^2 + 396^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\dfrac{x-13}{x-14}-\dfrac{x-15}{x-16}=-\dfrac{1}{12}.$$
For $x\ne14$ and $x\ne 16$ by multiplying the whole equation by $$12(x-14)(x-16)$$ we get: $$12(x-16)(x-13)-12(x-14)(x-15)=-(x-14)(x-16).$$ This doesn't look very nice. Can we do something else at the beginning? $$x-14=(x-13)-1\\x-15=(x-14)-1\\..?$$
| Hint: Notice that if we visually erase the symbols "$x - 1$" in both numerators and denominators we are left with $\frac34$ and $\frac56$ respectively, which is nice because
$$
\frac34 = \frac{9}{12}\quad\text{and}\quad\frac56 = \frac{10}{12}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$? We know that hyperbolic sine is:
$$\sinh \theta={\frac {e^{\theta}-e^{-\theta}}{2}}$$
and that hyperbolic cosine is
$$\cosh \theta={\frac {e^{\theta}+e^{-\theta}}{2}}$$
Let $n\in\mathbb N$.
If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$?
| $\begin{align}\sinh(\theta)=\frac{e^\theta-e^{-\theta}}{2}=x &\Rightarrow e^{2\theta}-2xe^{\theta}-1 =0\\
&\Rightarrow e^{\theta}= \frac{2x\pm\sqrt{4x^2+4}}{2} = x\pm\sqrt{x^2+1} \\
&\Rightarrow \boxed{e^{\theta} = x+\sqrt{x^2+1}} ~~~\text{ as } e^\theta \ge0,\theta \in \Bbb R\end{align}$
So, $$\sinh(n\theta) = \frac{e^{n\theta}-e^{-n\theta}}{2} = \frac{(x+\sqrt{x^2+1})^n-(x+\sqrt{x^2+1})^{-n}}{2}$$
and
$$\cosh(n\theta) = \frac{e^{n\theta}+e^{-n\theta}}{2} = \frac{(x+\sqrt{x^2+1})^n+(x+\sqrt{x^2+1})^{-n}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Range of $\frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$? I have the following problem:
Let $f(x) = \frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$ with $c>0$ and $x \in \mathbb{R}$. How to prove that $0 \leq f(x) \leq 1$?
I'm not sure how to check the range, I tried plotting the function but that's not as formal as the problem asks. Though, at least I'm sure that the range is correct. Thanks for the help.
| First to check if $0 \leq f(x)$ everywhere: $c>0,$ anything squared is $\geq 0$, so yes. The minimum is reached at $x=0$ only, because the numerator of a fraction must be $0$ for the fraction to be $0$.
If $c=1$ then the function is $0$ everywhere. If $c$ is large we have roughly $c^3/c^4 = 1/c$ so that will be $\leq 1$. We could try small values of $c$, but I don't see any values of $x$ that will bring the function over $1$.
The calculus to check for extrema expanded out is straightforward but messy: $21$ terms. So we simplify by pulling out constants and expand the denominator to get $$\frac{x^2}{x^4 + x^2(c^3+c) + c^4}$$ and then we set $y=x^2$ and relabel to get
$$\frac{y}{y^2+ay+b}$$ which is far more tractable, and has extrema at $y^2 = b$. Substituting $x^2 = y$ and $c^4 = b$, we get $x = \pm c$ as locations of the extrema.
For the values there, we get $$\frac{(1-c)^2}{(1+c)^2} = \frac{(c-1)^2}{(1+c)^2}$$ which is less than $1$ for all $c>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt79+\sqrt81}$ Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}$
I've thought about multiplying every fraction by 1, but like this $\frac{1} {\sqrt1+\sqrt3} *\frac{\sqrt1+\sqrt3}{\sqrt1+\sqrt3}$, $\frac{1} {\sqrt3+\sqrt5} *\frac{\sqrt3+\sqrt5}{\sqrt3+\sqrt5}...$
After multiplying I end up with $\frac{\sqrt1+\sqrt3} {(\sqrt1+\sqrt3)^2} + \frac{\sqrt3+\sqrt5} {(\sqrt3+\sqrt5)^2} ...$ but I don't have any more ideas than this, am I on the right path?
I apologize for the question's simplicity, but I'm a younger math enthusiast so questions like this are harder for me.
| $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}=\frac{1}{\sqrt3+\sqrt1} *\frac{\sqrt3-\sqrt1}{\sqrt3-\sqrt1}+ \frac{1}{\sqrt5+\sqrt3}*\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3} + \frac{1}{\sqrt7+\sqrt5}*\frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5} + ... \frac{1}{\sqrt{81}+\sqrt{79}} *\frac{\sqrt81-\sqrt79}{\sqrt81-\sqrt79}=\frac{\sqrt3-\sqrt1}{2}+\frac{\sqrt5-\sqrt3}{2}+\frac{\sqrt7-\sqrt5}{2}+...+\frac{\sqrt81-\sqrt79}{2} =(\frac{1}{2})((\sqrt3-\sqrt1)+(\sqrt5-\sqrt3)+(\sqrt7-\sqrt5)+...(\sqrt81-\sqrt79))=(\frac{1}{2})(\sqrt81-\sqrt1)=4$
So the problem is done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the equation of the locus of the mid point of AB as m varies I am working through a pure maths book as a hobby. This question puzzles me.
The line y=mx intersects the curve $y=x^2-1$ at the points A and B. Find the equation of the locus of the mid point of AB as m varies.
I have said at intersection:
$mx = x^2-1 \implies x^2 - mx - 1=0$
Completing the square:
$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$
$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$
$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$
x-coordinates for points of intersection are
$x= \frac{-\sqrt(4+m^2) +m}{2}$ and $x= \frac{\sqrt(4+m^2)+m}{2}$
$\implies$ x-co-ordinate of P is mid-way between the two above points, namely $\frac{2m}{2} = m$
So $x=m, y = mx = m^2\implies y = x^2$
But my book says $y=2x^2$
| You made a small mistake near the end: the midpoint's $x$-coordinate is not the sum of those of the intersection points, but half that sum, which is still $m/2$. Then the corresponding $y$-coordinate is $m^2/2$ and $y=2x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Area of a Triangle in $\mathbb{R}^4$ I find this question quite tricky, and I don't know if this kind of treatment is right.
I was asked about the area of a triangle formed by the points: $A:(1,2,-3,3)$; $B:(3,-6,-4,2)$; and $C:(-3,-16,-4,0)$. The only way I could make a reason out of this would be getting all the $2\times2$ matrices within the matrix M whose columns are AB and AC, getting their determinants, and finally dividing it by two. Is this procedure right? I will write the processes below:
Firstly, get vectors AB and AC:
$AB = (2,-8,-1,-1)$
$AC = (-4,-18,-1,-3)$
Secondly, construct matrix M:
\begin{equation}
M_{4,2}=
\begin{pmatrix}
2 & -4 \\
-8 & -18 \\
-1 & -1 \\
-1 & -3
\end{pmatrix}
\end{equation}
Thirdly, get all the $2\times2$ determinants within M:
\begin{equation}
S_{2}=
\begin{vmatrix}
2 & -4 \\
-8 & -18
\end{vmatrix}
+
\begin{vmatrix}
2 & -4 \\
-1 & -1
\end{vmatrix}
+
\begin{vmatrix}
2 & -4 \\
-1 & -3
\end{vmatrix}
+
\begin{vmatrix}
-8 & -18 \\
-1 & -1
\end{vmatrix}
+
\begin{vmatrix}
-8 & -18 \\
-1 & -3
\end{vmatrix}
+
\begin{vmatrix}
-1 & -1 \\
-1 & -3
\end{vmatrix}
\end{equation}
Forthly, dividing by two, and getting the absolute value (Area=$|\frac{S_{2}}{2}|$):
Area$=43$
| Given $n$-vectors $x_1,\dots,x_n\in\mathbb R^m$, the $n$-dimensional volume of the parallelepiped spanned by $x_1,\dots,x_n$ is given by $\sqrt{\det(G)}$, where $G$ is the Gramian matrix given by $G=(\langle x_i, x_j\rangle)_{ij}$.
Hence, we get
$$
A = \frac{\sqrt{\det(X^T X)}}{2}=35,\quad \text{where}\quad X=\begin{pmatrix}2&-4\\-8&-18\\-1&-1\\-1&-3\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Circle passing through two points and tangent to a line
"Find the equation of the circle passing through the origin $(0,0)$, the point $(1,0)$ and tangent to the line $x-2y+1=0$."
What I have done:
The equation of a circle with radius $R$ and center $(x_0,y_0)$ is $(x-x_0)^2+(y-y_0)^2=R^2$. Since the circle passes through $(0,0)$ and $(1,0)$ it must be $$\tag{1} x_0^2+y_0^2=R^2$$ $$ \tag{2} (1-x_0)^2+y_0^2=R^2$$ and since it is tangent to the line $x-2y+1=0$ the distance from the center to this line must equal the radius $R$ hence (using the fact that the distance $d$ from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is $d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$) we have $$R=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}=\frac{|1x_0-2y_0+1|}{\sqrt{1^2+(-2)^2}}=\frac{|x_0-2y_0+1|}{\sqrt{5}} \tag{3}$$
From (1) and (2) we have
$$x_0^2 + y_0^2 = (1 - x_0)^2 + y_0^2 \Rightarrow x_0 = \dfrac{1}{2} \tag{4}$$
Equating (1) , (3) and (4) we have
$$\dfrac{1}{4}+y_0^2 = \dfrac{(\frac{3}{2}-2y_0)^2}{5} \Rightarrow y_0 = -3 \pm \sqrt{10}$$
So there are two solutions,
$$R^2 = \dfrac{1}{4}+(-3+\sqrt{10})^2 \Rightarrow (x - \dfrac{1}{4})^2+(y + 3 - \sqrt{10})^2 = \dfrac{1}{4} + (-3 + \sqrt{10})^2$$
and
$$R^2 = \frac{1}{4} + (-3 - \sqrt{10})^2 \Rightarrow (x - \frac{1}{4})^2 + (y + 3 + \sqrt{10})^2 = \dfrac{1}{4} + (-3 - \sqrt{10})^2$$.
Now, when I plot these solutions they appear to be wrong, so I would like to know which mistake(s) I have made.
| I plotted it on desmos, your answer does seem right. Here is the link
The idea is that the point on which the circle touches the line can be varied. This leads to two circles satisfying the other two constraints of passing through the given points.
The mistake in OP's plot: They put $x=1/4$ instead of $x=1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$? (Note: This question is an offshoot of this closely related one.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers likely needs no introduction.
The initial question is as is in the title:
If $p^k m^2$ is an odd perfect number with special prime $p$, then is there a constant $D$ such that $$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$
(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)
MY ATTEMPT
Since $\gcd(p^k,\sigma(p^k))=1$, we know that
$$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}=\frac{D(m^2)}{s(p^k)},$$
where $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
Now, let $D > 0$ be a constant such that
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}.$$
This is equivalent to
$$D > \frac{p^k(m^2 - p^k)}{\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{\sigma(p^k)\sigma(m^2)} = \frac{p^k \sigma(p^k) (m^2 - p^k)}{2p^k m^2}$$
$$= \frac{\sigma(p^k)}{2} - \frac{p^k \sigma(p^k)}{2 m^2} = \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)}.$$
Using mathlove's answer to a closely related question, we have the lower bound
$$\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375.$$
Hence, we have
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p - 1)} - \frac{p^k}{3375} = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$
Let
$$f(k) = \frac{3373p^{k+1} + 2p^k - 3375}{6750(p - 1)}.$$
Then the first derivative
$$\frac{\partial f}{\partial k} = \frac{(3373p + 2)p^k \log(p)}{6750(p - 1)}$$
is positive for $p \geq 5$. This implies that $f$ is an increasing function of $k$.
Therefore, since $k \geq 1$, we obtain
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2}.$$
But we know that $p$ is at least $5$, since $p$ is the special prime satisfying $p \equiv 1 \pmod 4$. Therefore, we have
$$D > \frac{\sigma(p^k)}{2} - \frac{p^k \cdot p^k}{\sigma(m^2)} \geq f(k) \geq f(1) = \frac{3373p}{6750} + \frac{1}{2} \geq \frac{2024}{675} = 2.9985\overline{185}.$$
Here are my final questions:
Does this mean that we can take $D=3$ in the inequality
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}?$$
If $D=3$ does not work, then what value of $D$ works?
| Using your idea, one has$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}\implies p^k\lt \frac{6750D}{3373}$$So, I think that for any given $D\gt 0$, $\dfrac{\sigma(m^2)}{p^k} > \dfrac{m^2 - p^k}{D}$ does not hold in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Derive that a sum equals $\log(1+x)-\log(1-x)$ Show that $$\sum_{n=1}^{\infty}\frac{x^{2n-1}}{n-1/2}=\log(1+x)-\log(1-x)$$
I got a series for $\log(1+x)$ so by replacing the $x$ by $-x$ I got that
$$\left (\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{x^n}{n} \right )-\left ( \sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(-x)^n}{n} \right )\\=\log(1+x)-\log(1-x)$$
all of this is for $|x|<1$. I do not know how to proceed with this because I should be able to rewrite the two sums into the one in the first equation or is there an easier approach I am overseeing?
| First observe that:
$$\begin{aligned}
\ln(1-x)
&=
\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x)^n}{n}
\\ \\ &=
\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-1)^n x^n}{n}
\\ \\ &=
\sum_{n=1}^{\infty}(-1)^{2n+1}\frac{x^n}{n}
\\ \\ &=
-\sum_{n=1}^{\infty}\frac{x^n}{n}
\end{aligned}$$
Thus we can write the sum as:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{x^{2n-1}}{n-1/2}
&=
2\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n-1}
\\ \\ &=
2\left(\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{x^{2n}}{2n}\right)
\\ \\ &=
2\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{(x^2)^n}{n}
\\ \\&=
-2\ln(1-x)+\ln(1-x^2)
\\ \\ &=
\ln\left(\frac{(1-x)(1+x)}{(1-x)^2}\right)
\\ \\ &=
\ln\left(\frac{1+x}{1-x}\right)
\\ \\ &=
\ln(1+x)-\ln(1-x)
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5}$, then $\dfrac{a}{b} \leq \dfrac{91}{150}$ and if $\dfrac{a}{b}<\dfrac{3}{5}$, then $\dfrac{a}{b} \geq \dfrac{89}{150}$.
Hence the maximum value of $\dfrac{a}{b}$ is $\dfrac{91}{150}$. And $\dfrac{a}{b}$ is maximum when $b$ is the smallest.
So, I thought $150$ would be the smallest value of $b$.
But my friend said that the smallest value of $b$ would be $32$ as $\left\lvert \dfrac{19}{32}- \dfrac{3}{5}\right\rvert= \dfrac{1}{160}$ works actually. Though he said that he found this intuitively.
So, how to solve the problem with proper procedure?
| Clearing denominators, we are effectively asked to find minimal $b$ s.t. $0 < |150a-90b| \leqslant b$, for $a, b \in \mathbb N$.
Note as the gcd $(150,90)=30$, the middle term will at least be $30$ (or a larger multiple), so $b\geqslant 30$.
It remains to to check if $150 \mid (30\pm 90b)$ or equivalently if $b \equiv \pm2 \pmod 5$ for $b\geqslant 30$ to conclude $b_{min}=32$.
The following constructive approach avoids brute force.
From Bézout's identity $150\cdot(-1)+90\cdot2=30\tag{1}$ and the obvious $150\cdot(-3)+90\cdot5=0\tag{2}$ we can use the latter equation to scale up $b$ above $30$ without adding to the RHS error term. Further, as we noted $b \equiv \pm2 \pmod 5$, all solutions will be of form equation $(2)\times n \,\pm$ equation $(1)$. If there is any solution which is not of this form, subtract $(2)$ multiple times from it to get a contradiction with $b \equiv \pm 2\pmod 5$.
Now as $5n\pm2 \geqslant 30$ has minimal solution $n=6$, equation $(2)\times 6\,+$ equation $(1)$ gives $150\cdot(-19)+90\cdot(32)=30 < 32 \implies b_{min}=32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
limit of a quotient of sequences Consider the sequence $u_n=\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})$. I want to prove that $\lim_{n\to+\infty}\frac{u_{n+1}}{u^2_n}=+\infty$. I did not manage to prove it. For this, I write
$$\ln(u_{n+1})-2\ln(u_n)=2^{n+1}\ln(2)+\sum_{k=0}^n\ln\left(1-\frac1{2^{2^{n+1}-2^k}}\right)-2\sum_{k=0}^n\ln\left(1-\frac1{2^{2^n-2^k}}\right)$$
I do not know how to conclude. Thanks in advance for any help.
| Using that $\;2^{2^{n+1}}-2^{2^{k+1}}=\left(2^{2^n}-2^{2^{k}}\right)\left(2^{2^n}+2^{2^{k}}\right)\,$:
$$
\require{cancel}
\begin{align}
\frac{u_{n+1}}{u_n^2} &= \frac{\prod_{k=0}^{n}(2^{2^{n+1}}-2^{2^k})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= \left(2^{2^{n+1}}-2^{2^0}\right) \cdot \frac{\prod_{k=1}^{n}(2^{2^{n+1}}-2^{2^k})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= \left(2^{2^{n+1}}-2\right) \cdot \frac{\prod_{k=0}^{n-1}(2^{2^{n+1}}-2^{2^{k+1}})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= 2 \left(2^{2^{n+1}-1}-1\right) \cdot \frac{\prod_{k=0}^{n-1}\cancel{(2^{2^{n}}-2^{2^{k}})}(2^{2^{n}}+2^{2^{k}})}{\prod_{k=0}^{n-1}\cancel{(2^{2^n}-2^{2^k})}(2^{2^n}-2^{2^k})}
\\ &= 2 \left(2^{2^{n+1}-1}-1\right) \prod_{k=0}^{n-1}\left(1 + \frac{2}{2^{2^n-2^k}-1}\right)
\\ &\ge 2 \left(2^{2^{n+1}-1}-1\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to use polar coordinate in ODE? I don't understand how to use polar coordinate.
\begin{cases} \frac{dx(t)}{dt}=2x-y \\ \frac{dy(t)}{dt}=5x-2y \\ \end{cases}
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \begin{pmatrix} 2 & -1\\ 5 & -2 \end{pmatrix} \left( \begin{array}{c} x \\ y \end{array} \right)$$
let $A = \begin{bmatrix} 2 & -1\\ 5 & -2 \end{bmatrix}$, we have $\det(A) = 1, \operatorname{tr}(A) = 0, P(\lambda) = \lambda^2+1 = 0 \Rightarrow \lambda_{1,2}=\pm i $
\begin{cases} 2x-y-ix = 0 \\ 5x-2y-iy = 0 \\ \end{cases}
x = 1, y = 0
$\vec v_1 = (x,y) = (x, x(2-i)) = x(1,(2-i)) = \binom{1}{2-i} $
$E_{\lambda1} = \binom{1}{0}+i \binom{0}{-1} = \begin{pmatrix} 1 & 0\\ 2 & -1 \end{pmatrix} = P $ which give us $P^{-1}= \begin{pmatrix} -1 & 0\\ -2 & 1 \end{pmatrix}$ and $J = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$
$X(t)= e^{0t} \begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
$Y(t) = \begin{pmatrix} 1 & 0\\ 2 & -1 \end{pmatrix}\begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
It's a center stable
So the solution is pretty simple but i want to try to use the polar coordinate and i can't do it
$r^2= x^2 + y^2$ and $x = r\cos(\theta)$, $y = r \sin(\theta)$
$2rr'= 2xx'+ 2yy' = 2x(2x-y) + 2y(5x-2y) = 4x^2 +8xy -4y^2 $
$\theta ' = \frac{y(2x-y)+x(5x-2y)}{x^2+y^2} = \frac{-y^2+5x^2}{x^2+y^2}$
maybe if the exercise doesnt ask polar coordinate, we don't do it ?
i've check this : Polar coordinates differential equation but i still can't do it
| If you define, following the rows of $P^{-1}$, $u=x$ and $v=y-2x$, then the system becomes
$$
u'=x'=-v
\\
v'=y'-2x'=x=u
$$
which is the standard circle system, or rotation with constant angular speed of $-1$. That you get this structure you also found in the matrix $J$.
With this you get the constant radius
$$
R^2=u^2+v^2=x^2+(y-2x)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the inequality $\frac{3-x}{x^2-2x-3}\le\frac{3-x}{x^2+2x-3}$ Solve the inequality $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
We have $D: \begin{cases}x^2-2x-3\ne0\Rightarrow x\ne -1;3 \\ x^2+2x-3\ne0\Rightarrow x\ne-3;1\end{cases}$
Is the given equality equivalent (in $D$) to $$x^2-2x-3\ge x^2+2x-3\\\iff x\le0$$ So the solutions are $x\le0\cap D$? It seems like the inequalities I wrote aren't equivalent, because I don't get the answer.
| $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
$$\implies(3-x)\Bigl(\dfrac1{x^2-2x-3}-\dfrac1{x^2+2x-3}\Bigr)\le0$$
$$\implies(3-x)\Bigl(\dfrac{x^2+2x-3-x^2+2x+3}{(x-3)(x+1)(x+3)(x-1)}\Bigr)\le0$$
$$\implies (3-x)\Bigl(\dfrac{4x}{(x-3)(x+1)(x+3)(x-1)}\Bigr)\le0$$
$$\implies \dfrac{4x}{(x+1)(x+3)(x-1)}\ge0$$
$$\implies x\in(\infty, -3)\cup(-1,0]\cup(1,3)\cup(3,\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Power series of a function around a point ≠ 0 Using the power series $ \sum_{k \geq 0}z^k = \frac{1}{1-z} $ for $ |z|<1 $, where it's centered around $0$. How would the series look like if someone wanted to let, for example $ z_0 = \frac{1}{2}$?
Also, how would one compute the power series of $ \frac{1}{1+z^2} $ around $z_0 = 1$?
I know that, centered around 0 we can manipulate the geometric series to obtain $ \frac{1}{1+z^2} = \sum_{k \geq 0} (-1)^k z^{2k} $. However, I want to write it on the form $ \sum_{k \geq 0} c_k (z-1)^k $.
| Usually, when I want a series about a point $z_0 \ne 0$, I change variables and write $w = z-z_0, z = w+z_0$ and do the series in powers of $w$.
Example
$\frac{1}{1+z^2}$ centered at $z=1$. Write $w=z-1, z=w+1$, so
$$
\frac{1}{1+z^2} = \frac{1}{1+(w+1)^2} = \frac{1}{2+2w+w^2}
$$
For this I could use partial fractions
$$
\frac{1}{2+2w+w^2} = \frac{i/2}{w+(1+i)} - \frac{i/2}{w+(1-i)}
$$
and each of these is a geometric series. It turns out (as expected) that
the imaginary parts all cancel. We can see a pattern:
$$
\frac{1}{2} -\frac{1}{2}(z-1)
+\frac{1}{4}(z-1)^2
-\frac{1}{8}(z-1)^4 +\frac{1}{8}(z-1)^5
-\frac{1}{16}(z-1)^6
+\frac{1}{32}(z-1)^8 -\frac{1}{32}(z-1)^9
+\frac{1}{64}(z-1)^{10} \dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Family of generators for congruence subgroup $\Gamma_0(11)$ Consider the congruence subgroup
$$\Gamma_0(11)=\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in M(2,\mathbb{Z}): ad-11bc=1\right\}$$
I want to prove that the family
$$\Omega=\{-\text{Id}\}\cup\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in \Gamma_0(11): c>0,\ gcd(c,6)=1\right\}$$
generates the whole group.
Here is my attempt: let's start by a generic
$$\gamma=\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in\Gamma_0(11)$$
My idea is to act with matrices in $\Omega$ until I find an element of this set. If $c\leq 0$
I can act with $-\text{Id}$ to get a matrix with $c\geq 0.$ If I have $c=0$ I also have $ad=1,$ hence $a=d=\pm1.$ Using $-\text{Id}$ I can suppose $a=d=1,$ and acting on the right, for instance, with the matrix
$$\begin{pmatrix}2&-1\\11&-5\end{pmatrix}\in\Omega$$
I get
$$\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}2&-1\\11&-5\end{pmatrix}=\begin{pmatrix}2+11b&-1-5b\\11&-5\end{pmatrix}$$
The issue is with the other condition. Suppose $gcd(c,6)\neq1:$ then $2$ or $3$ divide $c:$ how can I manage these cases?
EDIT
Writing $c$ in the form $c=2^m\cdot 3^n\cdot C,$ if $m,n>0$ (i.e. $6\mid c$) I can write $\gamma$ using only elements of $\Omega$ in the following way: the determinant condition implies
$$(11\cdot2^m\cdot3^n c,a)=1$$
and in particular $a$ is odd and is not a multiple of $3.$ The matrix
$$\begin{pmatrix}
d&2^m3^n b\\
11c& a
\end{pmatrix}\in\Omega,$$
and
$$\begin{pmatrix}
d&2^m3^n b\\
11c& a
\end{pmatrix}\begin{pmatrix}
a&b\\
11\cdot 2^m\cdot 3^n c& d
\end{pmatrix}=\begin{pmatrix}
ad+2^{m}3^n(11c+b)&bd(1+2^m3^n)\\
11ac(2^m3^n+1)&11bc+ad
\end{pmatrix}\in\Omega $$
Maybe I can generalize this approach even for the other cases
| A calculation taking a tiny fraction of a second in Sage reveals that
$$\{ -\mathrm{Id}\} \cup \left\{\left(\begin{array}{rr}
18 & -5 \\
11 & -3
\end{array}\right), \left(\begin{array}{rr}
7 & -2 \\
11 & -3
\end{array}\right), \left(\begin{array}{rr}
8 & -3 \\
11 & -4
\end{array}\right)\right\}$$
is a generating set of $\Gamma_0(11)$. Since these all have $c = 1$ you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expected number of steps until game stops A game starts with some $N$ and a player chooses randomly an integer in range $\left[0,N-1\right]$:
*
*If the chosen number, let's denote it by $K$, is $0$ the game stops, otherwise we move to the next round, but shrink the range to $\left[0,K-1\right]$.
*What is the expected number of rounds the game will be played before it's $\mbox{stopped}~?$.
I have tried writing it as a Markov chain with the transition matrix of the form:
\begin{bmatrix}
0 & \frac{1}{N} & \frac{1}{N} & ... & \frac{1}{N} \\
0 & 0 & \frac{1}{N-1} & ... & \frac{1}{N-1} \\
& & ... \\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 &... & 1 \\
\end{bmatrix}
that gives for, say, $N=3$
\begin{align}
Q & =
\begin{bmatrix}
0 & \frac13 & \frac13 \\
0 & 0 & \frac12 \\
0 & 0 & 0
\end{bmatrix}, \quad N=(I-Q) = \begin{bmatrix}
1 & \frac23 & \frac23 \\
1 & 1 & \frac12 \\
1 & 1 & 1
\end{bmatrix},
\\[3mm] N^{-1} & = \begin{bmatrix}
3 & 0 & -2 \\
-3 & 2 & 1 \\
0 & -2 & 2
\end{bmatrix}
\end{align}
that would result in 1 expected step to reach 0, however it does not look right and computing same expectation manually gives
\begin{align*}
E_3 &= \frac13 \cdot 1 + \frac13 (E_2 + 1) + \frac13 (E_1 + 1) \\
&= 1 + \frac13\frac32 + \frac13 = \frac{11}{6}
\end{align*}
Could you please tell me what I am missing or if there is a better approach to solving this?
| Your $N$ matrix should be
$$N = (I-Q) =
\begin{bmatrix}
1 & -1/3 & -1/3 \\
0 & 1 & -1/2 \\
0 & 0 & 1
\end{bmatrix}.$$
Remember, $I$ is the identity matrix, not the all $1$s matrix. The inverse of this is
$$(I-Q)^{-1} =
\begin{bmatrix}
1 & 1/3 & 1/2 \\
0 & 1 & 1/2 \\
0 & 0 & 1
\end{bmatrix},$$
and the sum of the first row is $11/6$ as you found.
Another way to do this is to solve the recurrence relation for $E_N$.
$$E_N = \frac{1}{N}+\frac{1}{N}\sum_{i=1}^{N-1}(1+E_i) = 1+ \frac{1}{N}\sum_{i=1}^{N-1}E_i.$$
This can be solved by substituting $S_N = \sum_{i=1}^N E_i$, solving the recurrence relation for $S_N$, and then finally finding $E_N = S_N - S_{N-1}$. This leads to an answer of $E_N = 1 + \sum_{i=1}^N \frac{1}{i}$ if you go through all the work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ where $x$ is a real number.
Background: Doing Olympiad question and got one from the book.
Attempt:
Let $3^x$ be $u$.
\begin{align*}
3^{2x+1} + 4 \cdot 3x - 15 &= 0 \\
3^{2x} \cdot 3^1 + 4 \cdot 3^x - 15 &= 0 \\
3^{2x} \cdot 3 + 4 \cdot 3^x - 15 &= 0 \\
3 \cdot 3^{2x} + 4 \cdot 3^x - 15 &= 0 \\
3u^2 + 4u - 15 &= 0
\end{align*}
Factoring the equation we get
$$ u=\frac{5}{3} $$
(We eliminate $u=-3$ as $x$ is real.)
$$ 3^x=\frac{5}{3} $$
Taking $\log$ both sides
$$ x \log 3 = \log\left(\frac{5}{3}\right) $$
Now I want to know how to further solve this. Also is there any easier way to solve this?
| $$3^{2x+1}+4\cdot 3^x=15
\\3\cdot(3^x)^2+4\cdot 3^x=15$$
let $u=3^x$:
$$3u^2+4u-15=0$$
solve the quadratic then you have:
$$x_{1,2}=\log_3(u_{1,2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Given that $\int \frac{1}{x\sqrt{x^2-1}}dx=\arccos(\frac{1}{x})+C$, what is $\int\frac{1}{x\sqrt{x^2-a^2}}dx$? The following is clear:
$x\sqrt{x^2-a^2}=x\sqrt{a^2}\sqrt{\frac{x^2}{a^2}-1}=ax\sqrt{\frac{x^2}{a^2}-1}= a^2\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}$.
So I get that $$\int\frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a^2}\int\frac{1}{\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}}dx=\frac{1}{a^2}\arccos(\frac{a}{x})+C$$
Is this correct?
| You are almost there. Consider $x/a =z.$ This gives $dx = a dz$. That means one of a's will be cancelled. Note that I am assuming that the given number $a$ is nonnegative. Anyway, your solution is correct. You just need to adjust the coefficient with $cos^{-1}(x/a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
does this nested radicle converge? $1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}$ Let $a_0=1$ and $a_n=1+\sqrt{a_{n-1}}+\sqrt{1+\sqrt{a_{n-1}}}$
I want to know if the limit of $a_n$ as n goes to infinity.
$$1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}}$$
When I tried to find a possible answer I tried to solve $1+\sqrt{x}+\sqrt{1+\sqrt{x}}=x$. I tried for an hour so I gave up and asked WolframAlpha and it got $x≈5.04891733952231$
I'm not sure how to find out if $a_n$ converges. I think it does
| Starting from @Will Jagy first hint $$w^3 - 2 w^2 - w + 1 = 0\qquad \text{with} \qquad x=w^2$$ the cubic shows three real roots $(\Delta=49)$. Using the trigonometric method, they are
$$w_k=\frac{2}{3} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \left(2 \pi k-\sec ^{-1}\left(2
\sqrt{7}\right)\right)\right)\right)\qquad \text{with} \qquad k=0,1,2$$ and the solution of concern is $w_0$ (not prooved).
This gives
$$x=\frac{4}{9} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \sec ^{-1}\left(2
\sqrt{7}\right)\right)\right)^2$$ which is another expression of the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$$
This is my attempt.
We have
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}=\frac{a}{a+\sqrt{(a+b+c)a+bc}}+\frac{b}{b+\sqrt{(a+b+c)b+ca}}+\frac{c}{c+\sqrt{(a+b+c)c+ab}}=\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}=1-\frac{\sqrt{(a+b)(a+c)}}{a+\sqrt{(a+b)(a+c)}}+1-\frac{\sqrt{(b+c)(b+a)}}{b+\sqrt{(b+c)(b+a)}}+1-\frac{\sqrt{(c+a)(c+b)}}{c+\sqrt{(c+a)(c+b)}}$$
| Using the Cauchy-Schwarz inequality, we have
$$\sqrt{(b+a)(a+c)} \geqslant \sqrt{ab}+\sqrt{ac}.$$
Therefore
$$\frac{a}{a+\sqrt{2013a+bc}} = \frac{a}{a+\sqrt{a(a+b+c)+bc}} = \frac{a}{a+\sqrt{(b+a)(a+c)}} $$
$$\leqslant \frac{a}{a+\sqrt{ab}+\sqrt{ac}} = \frac{\sqrt {a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$$
So
$$\sum \frac{a}{a+\sqrt{2013a+bc}} \leqslant \sum \frac{\sqrt {a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} =1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$
Let $a,b,c\in\Bbb N$ such that $a>b>c$. Then $K:=(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$.
My attempt : Since each $a,b,c$ are either even or odd, WLOG we may assume $a,b$ are both even or odd. For both cases, $a+b$ and $a-b$ are divisible by $2$ so $K$ is divisible by $4$. Note that any $n\in\Bbb N$ is one of $\overline{0},\overline{1},\overline{2}$ in $\operatorname{mod}3$. Well from this, I can argue anyway but I want to show $K$ is divisible by $3$ more easier or nicer way. Could you help?
| It is possible to answer the question by looking directly at modulo $12$.
First, notice that $\forall n,\ n^2 \equiv \{0,1,4,9\} \bmod 12$. So each of $a,b,c$ must have a square with one of those residues. If any two such squares have the same residue, their difference will be $\equiv 0 \bmod 12$, and the product of the differences will be $\equiv 0 \bmod 12$, and we would be done. That leaves to be addressed those cases in which $a^2,b^2,c^2$ all have different residues $\bmod 12$.
There are four ways to select three different residues from four possibilities:
Case 1: Residues of the squares are $9,4,1$. Then the differences are $(9-4)(9-1)(4-1)=5\cdot 8\cdot 3=120 \equiv 0 \bmod 12$
Case 2: Residues of the squares are $9,4,0$. Then the differences are $(9-4)(9-0)(4-0)=5\cdot 9\cdot 4=180 \equiv 0 \bmod 12$
Case 3: Residues of the squares are $9,1,0$. Then the differences are $(9-1)(9-0)(1-0)=8\cdot 9\cdot 1=72 \equiv 0 \bmod 12$
Case 4: Residues of the squares are $4,1,0$. Then the differences are $(4-1)(4-0)(1-0)=3\cdot 4\cdot 1=12 \equiv 0 \bmod 12$
Note that changing the order in which the residues occur in the terms might change the sign of their product, but not its absolute value, and hence not its divisibility by $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Trouble solving exercise Exercise. Evaluate
\begin{equation*}
\int_{-\infty}^{+\infty}\frac{x^{2k}}{1+x^{2n}}dx
\end{equation*}
, where $0\leq k < n$ and $n,k \in \mathbb{N_0}$
My attempt. I have tried to solve the integral through the residues method,i.e., completing the segment $[-R,R]$ with a curve $C_R$ such that $|z| = R, Im(z)\geq 0$ and then declaring that
\begin{equation*}
\int_D \dots = \int_{-R}^{R}\dots + \int_{C_R} \dots
\end{equation*}
, where
\begin{equation*}
\int_D f(x)dx= 2\pi i \sum_{a_j \in int(D)}Res(f,a_j)
\end{equation*}
By calculating $f$ singularities I've that $x=\sqrt[n]{i}$ is a pole of order $n$.
But how I am stuck calculating the associated residues. Is there some alternative method to solve this? Thanks for ali the helpin advance.
| I got nerdsniped and ended up solving the problem. Hope it helps you.
Let $f(z) = \dfrac{z^{2k}}{z^{2n}+1}$. If $\omega$ is any $2n$th root of $-1$, then $f$ has simple pole at $\omega$ and
\begin{align*}
\operatorname{Res}(f,\omega)
&= \lim_{z\to\omega} \frac{z^{2k}(z-\omega)}{z^{2n} - \omega^{2n}}
\\&= \frac{\omega^{2k}}{2n\omega^{2n-1}} = - \frac{\omega^{2k+1}}{2n}
\end{align*}
Set $\omega = \exp\left(\frac{\pi i}{2n}\right)$. The $2n$th roots of $-1$ that are in the upper half-plane are $\omega, \omega^3, \omega^5, \dots, \omega^{2n-1}$. Using your contour notation and assuming $R > 1$,
\begin{align*}
\int_D f(z)\,dz
&= 2\pi i \sum_{j=0}^{n-1} \operatorname{Res}(f,\omega^{2j+1})
\\&= -\frac{\pi i}{n} \sum_{j=0}^{n-1} (\omega^{2j+1})^{2k+1}
\\&= -\frac{\pi i}{n} \sum_{j=0}^{n-1} (\omega^{2k+1})^{2j+1}
\end{align*}
Let $\zeta = \omega^{2k+1}$. Then
\begin{align*}
\int_D f(z)\,dz
&= -\frac{\pi i}{n} \sum_{j=0}^{n-1} \zeta^{2j+1}
= -\frac{\pi i \zeta}{n} \sum_{j=0}^{n-1} (\zeta^2)^j
\\&= - \frac{\pi i\zeta}{n} \left(\frac{1 - \zeta^{2n}}{1-\zeta^2}\right)
= -\frac{2\pi i}{n} \cdot \frac{\zeta}{1-\zeta^2}
\end{align*}
(since $\zeta^{2n} = (\omega^{2k+1})^{2n} = (\omega^{2n})^{2k+1} = 1$).
Since $|\zeta| = 1$, $\zeta^{-1} = \bar\zeta$. So
\begin{align*}
\frac{\zeta}{1-\zeta^2}
&= \frac{\zeta}{1-\zeta^2} \cdot \frac{\bar\zeta}{\bar\zeta}
\\&= \frac{1}{\bar\zeta - \zeta} = \frac{i}{2}\cdot \frac{1}{\Im(\zeta)}
\end{align*}
Therefore,
$$
\int_D f(z)\,dz = \frac{\pi}{n \Im(\zeta)}
$$
The integral over $C_R$ is bounded by $\pi R^{2(k-n)+1}$, which (since $0 \leq k < n$) tends to zero as $R\to\infty$. Also, by Euler's formula,
$$
\zeta = \exp\left(\frac{(2k+1)\pi i}{2n}\right)
= \cos \left(\frac{(2k+1)\pi}{2n}\right)
+ i \sin \left(\frac{(2k+1)\pi}{2n}\right)
$$
Therefore,
$$
\int_{-\infty}^\infty \frac{x^{2k}\,dx}{1+x^{2n}} = \frac{\pi}{n} \csc \left(\frac{(2k+1)\pi}{2n}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Consider the solid region defined by $S = \{{ (x, y, z) \in \mathbb{R} ^3: x^2 + y^2+z^2 \leq 1 \wedge z ^ 2 \geq 3 (x ^ 2 + y ^ 2)\wedge z \geq 0}\}$ Consider the solid region defined by $S = \{{ (x, y, z) \in \mathbb{R} ^ 3: x ^ 2 + y ^ 2 + z ^ 2 \leq 1 \wedge z ^ 2 \geq 3 (x ^ 2 + y ^ 2)\wedge z \geq 0}\}$
I have to graph the region and calculate the volume. But I am having trouble visualizing the region. Is a kind of 1/2 cone formed?
What I have so far considering its a cone:
\begin{align*}
z^2=\sqrt{3} y\\
\phi = arctg(\frac{1}{\sqrt3})\\
\theta=\frac{\pi}{6}
\end{align*}
Using polar coordinates:
\begin{align}
0 \leq r\leq1\\0\leq\phi\leq\frac{\pi}{6}\\0\leq\phi\leq 2\pi\\
Vol(S)=\int_{0}^{2 \pi}\int_{0}^{\frac{\pi}{6}}\int_{0}^{1}sin\phi dr d\phi d\theta\\
Vol(S)=\theta\mid_{0}^{2\pi}\frac{r^3}{3}_{0}^{1}\int_{0}{\frac{\pi}{6}} sin\phi d\phi = \frac{2\pi}{3}(-cos \phi)\mid_{0}^{\frac{\pi}{6}}=-\frac{2\pi}{3}((\frac{\sqrt3}{2})-1)
\end{align}
$VS= \frac{2\pi}{3}(1-\frac{\sqrt3}{2})=0.2806 u^3$
| Projection of intersection of sphere $x ^ 2 + y ^ 2 + z ^ 2 \leqslant 1$ with cone $z ^ 2 = 3 (x ^ 2 + y ^ 2)$ gives circle $x ^ 2 + y ^ 2 = \frac{1}{4}$ on $OXY$ plane, so volume can be found by integral
$$4\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{\sqrt{\frac{1}{4}-x^2}}\int\limits_{\sqrt{3 (x ^ 2 + y ^ 2)}}^{\sqrt{1-x ^ 2- y ^ 2}}dzdydx$$.
Using cylindrical coordinates $x=\rho \cos \phi, y = \cos \phi, z=z$ with Jacobian $J=\rho$ gives "sphere" $\rho^ 2+z^2=1$ and "cone" $z \geqslant 3 \rho$, so, volume can be found as
$$4\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{1}{2}}\int\limits_{3\rho}^{\sqrt{1-\rho^ 2}}\rho dzd\rho d\phi$$
and for spherical coordinates $x=r\sin\phi \cos\theta, y=r\sin\phi \sin\theta, z=r\cos\phi$ with $r \geqslant 0, \theta \in [0, 2\pi], \phi \in [0, \pi]$ and with Jakobian $J=r^2 \sin{\phi}$ we should have "sphere" $r \leqslant 1$ and "cone" $\tan \phi = \frac{1}{\sqrt{3}}$
$$4\int\limits_0^{\frac{\pi}{2}} \int\limits_0^{\frac{\pi}{6}} \int\limits_0^1 r^2 \ \sin{\phi} \ dr \ d\phi \ d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing$ \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$ Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$
We can show this by re-writing $I$ as
$$
\implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx,
$$
which is Frullani Integral.
$$J=\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(\infty)-f(0)]\ln(a/b).$$ Here, $f(x)=\frac{1-\cos(x)}{x},$ hence $I=0.$
So the question is how to show (1), otherwise?
| The result can be found almost immediately from the work in the question statement.
Note that
\begin{align*}
\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx
&= 6 \underbrace{\int_0^\infty\frac{1-\cos 2x}{2x^2}dx}_{\textrm{let }x=t/2} -
6 \underbrace{\int_0^\infty\frac{1-\cos 3x}{3x^2} dx}_{\textrm{let }x=t/3} \\
&= 6\int_0^\infty\frac{1-\cos t}{t^2} - 6\int_0^\infty\frac{1-\cos t}{t^2} \\
&= 0.
\end{align*}
It remains to show that
$(1-\cos 2x)/(2x^2)$ and
$(1-\cos 3x)/(3x^2)$ are integrable on $[0,\infty)$, which is a straightforward exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$M_t = \frac{1}{\sqrt{T_1}} \mathbb{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t}, t\geq 0$ is a martingale I try to solve an old exam question, but I find it difficult. Maybe someone can suggest a hint.
Let $\{ T_i | i\in \mathbb{N} \}\subseteq \mathbb{R} _{\geq 0}$ be a homogeneous Poission point process with intensity 1, where we assume $T_1 < T_2 < \dots$. Let $M_t = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t}$. Let $N_t = \# \{ i\in \mathbb{N} | T_i \leq t \}$. Let $\mathcal{F} _t = \sigma ( (N_s) _{s\leq t} )$. Prove that $(M_t)$ is an $(\mathcal{F} _t)$-martingale.
I already showed that $M_t$ is $\mathcal{F} _t$-adapted by taking a careful look at events of the form $\{ \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) \leq x \}$ and $\{ \sqrt{T_1 \wedge t} \leq x \}$. I proved integrability by using upper bounds and the integration by parts technique. Now I have to show that $M_t$ satisfies the martingale property. So let $s<t$. Then:
\begin{align*}
E [ M_t | \mathcal{F} _s ] & = E \left [\frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ]\\
& = E \left [\frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) \ \mid \ \mathcal{F}_s \right ] - 2 E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = E \left [\frac{1}{\sqrt{T_1}} (\mathbf{1} (T_1 \leq s) + \mathbf{1} (s < T_1 \leq t)) \ \mid \ \mathcal{F}_s \right ] - 2 E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = E \left [\frac{1}{\sqrt{T_1}} (\mathbf{1} (N_s \geq 1) + \mathbf{1} (N_s = 0) \cdot \mathbf{1} (N_t - N_s \geq 1))\ \mid \ \mathcal{F}_s \right ] - 2 E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s \geq 1) + \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s = 0) \cdot P (N_t - N_s \geq 1)\ - 2 E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s \geq 1) + \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s = 0) \cdot (1-\exp(s-t))\ - 2 E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\ - 2 E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\ - 2 E \left [ \sqrt{T_1} \cdot \mathbf{1} (T_1 \leq s) + \sqrt{T_1} \cdot \mathbf{1} (s < T_1 \leq t) + \sqrt{t} \cdot \mathbf{1} (T_1 > t) \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\ - 2 E \left [ \sqrt{T_1} \cdot \mathbf{1} (N_s \geq 1) + \sqrt{T_1} \cdot \mathbf{1} (N_s = 0) \cdot \mathbf{1} (N_t - N_s \geq 1) + \sqrt{t} \cdot \mathbf{1} (N_t = 0) \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\ - 2 \sqrt{T_1} \cdot \mathbf{1} (N_s \geq 1) - 2\sqrt{T_1} \cdot \mathbf{1} (N_s = 0) \cdot P (N_t - N_s \geq 1) - 2 \sqrt{t} \cdot P (N_t = 0) \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\ - 2 \sqrt{T_1} \cdot \mathbf{1} (N_s \geq 1) - 2\sqrt{T_1} \cdot \mathbf{1} (N_s = 0) \cdot (1-\exp(s-t)) - 2 \sqrt{t} \cdot \exp(-t) \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\ - 2 \sqrt{T_1} \cdot \mathbf{1} (T_1 \leq s) - 2\sqrt{T_1} \cdot \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t)) - 2 \sqrt{t} \cdot \exp(-t) \\
\end{align*}
I do not know how to proceed further? Can someone give a hint? Thank you in advance.
EDIT: I think I need to use the memoryless property somewhere. Also, I possibly made one or more mistakes with $\mathcal{F} _s$-measurability and conditioning on $\mathcal{F}_s$.
| Note that we can write
$$ 2\sqrt{T_1 \wedge t} = \int_{0}^{t} \frac{1}{\sqrt{u}} \mathbf{1}(T_1 > u) \, \mathrm{d}u. $$
From this, for $0 \leq s \leq t$,
\begin{align*}
\mathbb{E}[ M_t - M_s \mid \mathcal{F}_s]
&= \mathbb{E}\biggl[ \frac{1}{\sqrt{T_1}} \mathbf{1}(s < T_1 \leq t) \,\biggm|\, \mathcal{F}_s \biggr] - \int_{s}^{t} \frac{1}{\sqrt{u}} \mathbb{E}[ \mathbf{1}(u < T_1) \mid \mathcal{F}_s] \, \mathrm{d}u \tag{*}
\end{align*}
The memoryless property tells that, given $\{T_1 > s\}$, the extra waiting time $T_1 - s$ is also $\operatorname{Exp}(1)$ and independent of $\mathcal{F}_s$. From this, the right-hand side of $\text{(*)}$ simplifies to
\begin{align*}
&\biggl( \mathbb{E}\biggl[ \frac{1}{\sqrt{s + T_1}} \mathbf{1}(T_1 \leq t-s) \biggr] - \int_{s}^{t} \frac{1}{\sqrt{u}} \mathbb{P}(u-s < T_1) \, \mathrm{d}u \biggr) \mathbf{1}(s < T_1) \\
&= \biggl( \int_{0}^{t-s} \frac{1}{\sqrt{s+u}} e^{-u} \, \mathrm{d}u - \int_{s}^{t} \frac{1}{\sqrt{u}} e^{-(u-s)} \, \mathrm{d}u \biggr) \mathbf{1}(s < T_1).
\end{align*}
It is easy to verify that the two integrals in the last line are equal, hence the desired conclusion follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$ I found a proof that $\lim \sqrt[n]{1^2+2^2+...+n^2}=1$ by $\varepsilon$-language, but I think it's quite complicated and not sure that it's correct.
My question is:
1- Is my proof correct?
2- Is there another simpler proof in the sense of $\varepsilon$-language?
Please help me. Thanks.
Solution:
Let $a_n=\sqrt[n]{1^2+2^2+...+n^2}-1$. Then $a_n>0$ and:
$$ (a_n+1)^n= 1^2+2^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}$$
By the binomial theorem we have:
$ (a_n+1)^n = 1 + na_n+\frac{n(n-1)}{2}a_n^2+\frac{n(n-1)(n-2)}{6}a_n^3+\frac{n(n-1)(n-2)(n-3)}{24}a_n^4+\cdots+a_n^n $
Since $a_n>0$, then $(a_n+1)^n>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$ and therefore:
$$\dfrac{n(n+1)(2n+1)}{6}>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$$
This is equivalent to
$$(n+1)(2n+1)>\dfrac{(n-1)(n-2)(n-3)}{4}a_n^4 \Leftrightarrow a_n^4<\dfrac{4(n+1)(2n+1)}{(n-1)(n-2)(n-3)}$$
$$\Rightarrow a_n^4< \dfrac{8(n+1)^2}{(n-2)^2(n-3)}=\dfrac{8}{n-3}\Big(1+\dfrac{3}{n-2}\Big)^2<\dfrac{128}{n-3}.$$
Thus: $a_n<\sqrt[4]{\dfrac{128}{n-3}}$. For every $\varepsilon >0$, take $N>3+\dfrac{128}{\varepsilon^4}$, then for all $n\ge N$,
$$ n-3 \ge N -3 > \dfrac{128}{\varepsilon^4} $$
Thus: $ \varepsilon^4 > \dfrac{128}{n-3} \Rightarrow \varepsilon > \sqrt[4]{\dfrac{128}{n-3}} >a_n $.
Hence, $\lim a_n=0$ or equivalently, $\lim \sqrt[n]{1^2+2^2+...+n^2} =1$.
| A computation without $\epsilon$ but assuming a closed formula for the sum of the first $n$ squares.
For $n\ge 1$,
$a_n=\sqrt[n]{\frac{n(n+1)(2n+1)}{6}}=\exp\left(\frac{1}{n}\ln\left(\frac{n(n+1)(2n+1)}{6}\right)\right) \to \exp(0)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the maximum "alive density" of cells in Conway's Game of Life when played on a torus? I've read that Conway's Game of Life (CGOL) can have unbounded growth from a finite initial number of alive cells (e.g. a glider gun). However, if CGOL is played on a torus, space (the number of cells) becomes finite, and glider guns are guaranteed to eventually destructively interfere with themselves.
Because of this, I wondered about the maximum proportion of alive cells on a torus. To be specific,
What is the maximum stable density of alive cells in CGOL on a torus, where density means proportion of alive cells out of all cells, and stable means this density occurs infinitely many times (as opposed to the recurrence of position: a glider might have nonperiodic position, but stable density)
| In the specific case of a pattern that is fixed under the rules of the Game of Life, the maximum density is 1/2, as shown by Noam Elkies in The still-Life density problem and its generalizations.
Elkies also presents (on page 22) a simple example of a period 6 oscillator with maximum density 3/4. Here are all its phases:
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
1&1&0&0&0&0&0&0\\
\hline
1&1&0&0&0&0&0&0\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
0&0&1&0&0&0&0&1\\
\hline
0&0&1&0&0&0&0&1\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
1&1&1&1&0&0&1&1\\
\hline
1&1&1&1&0&0&1&1\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
0&0&0&0&1&1&0&0\\
\hline
0&0&0&0&1&1&0&0\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
0&0&0&1&0&0&1&0\\
\hline
0&0&0&1&0&0&1&0\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
0&0&1&1&1&1&1&1\\
\hline
0&0&1&1&1&1&1&1\\
\hline
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_{x=1}^nb_x$$
And I want an explicit formula for $c_n$. After some research, I found the explicit formula for $b_n=\frac{n(n+1)(n+2)}{6}$. Seeing the patterns from $a_n$ and $b_n$, I figured the explicit formula for $c_n$ would be $\frac{n(n+1)(n+2)(n+3)}{24}$ or $\frac{n(n+1)(n+2)(n+3)}{12}$.
Then I tried to plug in those two potential equations,
If $n=1$, $c_n=1$, $\frac{n(n+1)(n+2)(n+3)}{24}=1$, $\frac{n(n+1)(n+2)(n+3)}{12}=2$. Thus we can know for sure that the second equation is wrong.
If $n=2$, $c_n=1+4=5$, $\frac{n(n+1)(n+2)(n+3)}{24}=5$. Seems correct so far.
If $n=3$, $c_n=1+4+10=15$, $\frac{n(n+1)(n+2)(n+3)}{24}=\frac{360}{24}=15$.
Overall, from the terms that I tried, the formula above seems to have worked. However, I cannot prove, or explain, why that is. Can someone prove (or disprove) my result above?
| The easiest way to prove your conjecture is by induction. You already checked the case $n=1$, so I won’t do it again. Let’s assume your result is true for some $n$. Then:
$$c_{n+1}=c_n+b_{n+1}$$
$$=\frac{n(n+1)(n+2)(n+3)}{24} + \frac{(n+1)(n+2)(n+3)}{6}$$
$$=\frac{n^4+10n^3+35n^2+50n+24}{24}$$
$$=\frac{(n+1)(n+2)(n+3)(n+4)}{24}$$
and your result holds for $n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:
If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
S \space$ then if $a = b\cdot c\space$ prove that $c \in S$
My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\space$ and $b = 1 + 3 \cdot y\space$ and $x = y + j\space$ for some $j >= 1$
If $a = bc \implies c \mid a \space \And \space b\mid a$
Now $b \mid a \implies (1 + 3y)\mid (1 + 3x) \implies (1 + 3y) \mid (1 + 3(y + j)) \implies (1 + 3y) | (1 + 3y + 3j) \implies (1 + 3y) | (1 + 3y) + 3j$
Since $1 + 3y \mid (1 + 3y) \implies (1 + 3y) \mid 3j$ and we keep that.
Now we know that
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = \frac{1+3(y + j)}{1 + 3y} = \frac{1 + 3y + 3j}{1 + 3y} = \frac{1 + 3y}{1 + 3y} + \frac{3j}{1+3y} = 1 + \frac{3j}{1 + 3y}$$
But we have shown that $1 + 3y \mid 3j \implies 3j = k(1 + 3y)\space$ where $1 + 3y$ is a positive integer.
Hence $$c = 1 + \frac{k(1 + 3y)}{1 + 3y} = 1 + k$$
But now I am stuck because I need to show that $k$ is a multiple of $3$ to finish the proof and I am not sure how to do that.
I know that $$k = 3\frac{j}{1 + 3y}$$ but the $\frac{j}{1 + 3y}$ is not an integer I think so I a messing up somewhere.
Update:
Following the comment of @Infinity_hunter
Let $c = r + 3k\space$ for some $r >= 1$. We have:
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = 1 + \frac{3j}{1 + 3y}$$
This was shown earlier.
But
$$c = r + 3k = 1 + \frac{3j}{1 + 3y} $$
We know that $$\frac{3j}{1 + 3y}$$ since we have shown that $1 + 3y \mid 3j$ so let $\frac{3j}{1 + 3y} = p$
So we have:
$c = r + 3k = 1 + p \implies c = r - 1 - p + 3k \implies c = (r - 1 - p) + 3k$
And I am not sure how to progress from here.
| Straightforward multiplication affords the easiest visualization. Any one member of the set which is formed by the given sequence can be represented by $3a+1$, and any second member of the set can be represented by $3b+1$. Multiplying:
$$(3a+1)(3b+1)=9ab+3b+3a+1=3(3ab+b+a)+1$$
We can readily let $c=3ab+b+a$, from which the product is $3c+1$, and hence a member of the set. This demonstrates closure under multiplication.
Added in response to comment. I originally answered the question posed in the title. The slightly distinct internal question (involving numbers $a,b,c$) can be addressed by the same basic method. Let $a=3A+1, b=3B+1, c=3C+r$ where $r$ can be $0,1,2$, which covers all possibilities.
$$3A+1=(3B+1)(3C+r)=9BC+3Br+3C+r=3(3BC+Br+c)+r$$
Plainly $3$ times something ($A$) plus $1$ equals $3$ times something else ($3BC+Br+c$) plus $r$ can only hold if $r=1$, and specifically not the other possibilities $r=0,2$. Thus $c=(3C+1) \in S$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Probability of Y, where $Y=X^2$ and $X \sim U[-1, 1]$ I'm trying to prove that if $X \sim U[-1, 1]$ and $Y = X^2$ then $p_Y(y)=\frac{1}{2} y^{-\frac{1}{2}}$, using the change of variables theorem.
The theorem states one can show that $p_Y(y) = \frac{dx}{dy} p_X(x)$.
Given the problem above, I find that $p_X(x) = \frac{1}{b-a} = \frac{1}{2}$. And that $\frac{dx}{dy} = \frac{d}{dy}\sqrt{y} = \frac{1}{2}y^{-\frac{1}{2}}$.
Therefore,
$$ p_Y(y) = \frac{dx}{dy}p_X(x) = \frac{1}{4}y^{-\frac{1}{2}} $$
But I know that it's not true, since $p_Y(y) = \frac{1}{2}y^{-\frac{1}{2}}$. So
where is the problem?
| \begin{align}
F_Y(y) = P( Y \le y) = P(X^2 \le y) = P( - \sqrt{y} \le X \le \sqrt{y}) = \frac{\sqrt y + 1}{2} - \frac{-\sqrt y + 1}{2} = \sqrt y,
\end{align}
thus
$$
f_Y(y) = \frac{\partial }{\partial y}F_Y(y) = \frac{1}{2}y^ {-1/2}
$$
Namely, your formula works only for one-to-one functions. In this case, using the transformation formula is equivalent to considering only the positive side, i.e.,
$
P(X\le \sqrt y)
$, that gives you the $1/4$ instead of $1/2$ when you deriving the density function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $yy' = \sqrt{y^2+y'^2}y''-y'y''$ Solve $$yy' = \sqrt{y^2+y'^2}y''-y'y''$$
First I set $p = y'$ and $p' = \frac{dp}{dy}p$ to form:
$$yp=\sqrt{y^2+p^2}\frac{dp}{dy}p-p\frac{dp}{dy}p \rightarrow y=\sqrt{y^2+p^2}\frac{dp}{dy}-\frac{dp}{dy}p$$
I am trying to come up with a clever substitution to deal with the square root and $p$ of :
$$y=\frac{dp}{dy}(\sqrt{y^2+p^2}-p)$$
| After OP's work
$$y=\frac{dp}{dy}(\sqrt{y^2+p^2}-p)$$
This is a homogeneous equation for $p(y)$
$$\frac{dp}{dy}=\frac{y}{\sqrt{y^2+p^2}-p}=\frac{\sqrt{y^2+p^2}+p}{y}$$
Let $p=vy$, then get
$$v+y\frac{dv}{dy}=\sqrt{1+v^2}+v \implies \int \frac{dv}{\sqrt{1+v^2}}= \int \frac{dy}{y}$$
Let $v=\tan t$, thenwe get
$$\ln(\sec t+ \tan t)=\ln cy \implies cy=\sqrt{1+v^2}+v \implies c^2y^2-1=2cp$$
$$\implies \frac{dy}{dx}=\frac{c^2y^2-1}{2c}\implies \int\frac{2cdy}{c^2y^2-1}=\int dx+b$$
$$\implies \ln \left(\frac{cy-1}{cy+1} \right)=x+b \implies y=\frac{1}{c}\left(\frac{1+ae^x}{1-ae^x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $a$ & $b$ , s.t. $a+b=0$ and $ab=-3$? I've a problem in this question :
In the polynomial identity $x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)$ , find $ab$ $? $
MY APPROACH
We have : $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$
Now according to the Problem : $$x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)=(x^2+1)(x^4-x^2+1)$$ Or $$x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1=x^4-x^2+1$$ $$(a+b)x^3+(ab+2)x^2+(a+b)x=-x^2$$ In order to satisfy this , $a+b=0$ and $ab+2=-1$
Hence , $$ab=-3$$
but what'll be the value of $a$ & $ b$ , which'll satisfy these conditions $?$
| So we have $a+b=0$ and $ab=-3$
$$ a+b=0 => a=-b $$
Put this value of $a$ in the second equation
$$ -b \cdot b=-3 $$
$$ b^2=3 => b=\pm\sqrt{3} $$
Can you now find $a$ with help of first equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving $X^2-6Y^2=Z^3$ in positive integers I’m trying to solve the Diophantine equation
$$X^2-6Y^2=Z^3 \tag{$\star$}$$
in positive integers $x,y,z$.
Brute force calculations confirm the naïve intuition that there are many [read: surely infinite!] primitive solutions; numerical observation suggests the solutions have determinable [perhaps even “descent”- or “recurrence”-based?] patterns. For example, one “stream” of solutions is
$$(x,y,z) \in \{(5,2,1),\ (49,20,1),\ (485,198,1),\ (4801,1960,1),\dots\},$$
where $z=1$ and the $(x_n,y_n)$ satisfy the recurrence $t_n = 10t_{n-1}-t_{n-2}$. Another such stream exists for $z=19$, another for $z=25$, etc.
Evidently, ($\star$) is related to [but different from] the Pellian equation
$$X^2-6Y^2=Z^2,$$
so I’m wondering:
Question #1: Is a complete solution already known, either for the equation or for one of the variables (e.g. characterization of $z$)?
Question #2: If not, what are the most promising ways to attack the problem?
| We can always construct a linear relation between x and y such as $x=ay+b; a, b \in \mathbb z$ to convert pell equations $x^2-Dy^2=1$ to a single unknown equation. We can find numerous families of solutions by this method. For Pell like equations we use rational solutions. I try to show this bellow by an example:
Let $x=y+a$ we have:
$x^2-6y^2=1\Rightarrow 5y^2-2ay-a^2+1=0$
$\Delta'=6a^2-5$
which have number of solutions:
$(a, \Delta')=(1, 1), (3, 7^2), (7, 17^2), \cdot \cdot\cdot$
which gives:
$(x, y)=(\frac 25, 1), (\frac 4 5, 1), (8, 5), (\frac {11}5, -\frac 45)\cdot\cdot\cdot $
For example take $(x, y)=(\frac {11}5, -\frac 45)$ we have:
$\big(\frac {11}5\big)^2-6\big(\frac 45\big)^2=1\Rightarrow\big(\frac {11^2}{25}\big)-6\big(\frac{4^2} {25}\big)=1$
We can rewrit this relation as:
$(11\times 25)^2-6(4\times 25)^2=25^3$
So: $(x, y, z)=(675, 100, 25) $
Similarly we can find more solutions where $z=25$ as a family of solutions.
Generally there can alway be a solution for equation $x^2-Dy^2=z^{2k+1}$ . for example for $x^2-6x^2=z^5$ we can write:
$(11\times625)^2-6(4\times 625)^2=25^5$
For D=19 we have:
$x=\frac{35} 3$ , $y=\frac 83$ and we have:
$(35\times 9)^2-19(6\times 9)^2=9^3$
$(35\times 81)^2-19(8\times 81)^2=9^5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
What is my mistake in finding this pythagorean triplet? Since Project Euler copyright license requires that you attribute the problem to them, I'd like to add that this is about question 9 there.
I am trying to solve this problem on only two brain cells and can't figure out what am I doing wrong. Here is the system for $(a, b, c) \in \mathbb{N}^3$,
\begin{align*}
a^2 +b^2 &= c^2 \\
a+ b + c &= 1000 \\
a &< b < c
\end{align*}
Here is my approach,
\begin{align*}
a + b + c &= 1000 \\
a + b &= 1000 - c &&\text{Subtract } c\\
a^2 + b^2 + 2ab &= 1000^2 + c^2 - 2000c &&\text{Square both sides}\\
c^2 + 2ab &= 1000^2 + c^2 - 2000c &&\text{Since }a^2 +b^2 = c^2\\
2ab &= 1000^2 - 2000c &&\text{Subtract } c^2\\
\frac{ab}{500} &= 1000 - 2c &&\text{Divide } 1000\\
2c &= 1000 - \frac{ab}{500} &&\text{Rearrange}\\
\end{align*}
Now let $a=5,b=200$,
\begin{align*}
2c &= 1000 - 2\\
2c &= 998 \\
c &= (998 \div 2) = 499 \\
\end{align*}
But certainly these values do not work. I can't see why.
| One problem is the assumption that $\quad 5^2+200^2=795^2.\quad $ There are no "small" triples with side differences of two orders of magnitude. The $b$-value is good but the following solution provides the $a$-value to "fit" your logic and the $c$-value will be shown to match your "solution" in the last equations below.
If we assume that $a$ is odd and $b$ is even, then the rules are that $a$ is any odd number greater that $1,\space$ $b$ is a multiple of $4,\space$ and $c$ must be of the form $4n+1$. Experimentally, a spreadsheet can show that one solution is $(375,200,425).\quad$ Alternatively, the following works if only paper and a calculator are available.
We begin with Euclid's formula where
$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$
Since $a$ is every $2nd$ integer but $b$ is every $4th$, it is faster to try $b$-values using the following formula, where any integer solution yields a valid triple.
\begin{equation}
B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$
$$B=200\implies\qquad \bigg\lfloor \frac{1+\sqrt{400+1}}{2}\bigg\rfloor =10 \le m \le \frac{200}{2}=100\\
\land \quad m\in\{20,25\}\implies k\in\{5,4\}$$
$$f(20,5)=(375,200,425)\qquad f(25,4)=(609,200,641)$$
We can see that $f(20,5)$ is the only solution where $a+b+c=1000.\quad$ This required testing only $11$ $m$-values for $b=200.$
If we try these "assumed" values in your equations we get
$$2c=1000-\frac{ab}{500}
=\frac{500,000-75000}{500}
=850$$
$$c=425$$
If you require that $a<b<c$, use $(200,375,425).\quad$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If $a,b,c$ are rational numbers and if $\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$
If $a,b,c$ are rational numbers and if $\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$
My attempt
Binomial expansion is not a good idea because there will be $27$ terms so I tried to prove using factorization.
$$\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3-\left(c\sqrt[3]{4}\right)^3
\\=\left(a+
b\sqrt[3]{2}\right)\left[\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^2+\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)\left(c\sqrt[3]{4}\right)+\left(c\sqrt[3]{4}\right)^2\right]$$
This again leads to complicated calculations. Then I tried to equate it to a rational number $r$.
\begin{align*}
\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3&=r\\
\implies a+b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}\\
\implies b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}-a\\
\implies(b\sqrt[3]{2}+c\sqrt[3]{4})^3&=(r^{1/3}-a)^3\\
\implies2b^3+6\sqrt[3]{2}~b^2c+6\sqrt[3]{4}~bc^2+4c^3&=r-3r^{2/3}a+3r^{1/3}a^2-a^3
\end{align*}
When I got stuck here I wrote the equation $a+b\sqrt[3]{2}+c\sqrt[3]{4}=r^{1/3}$ in $3$ different ways, each time multiplying with $\sqrt[3]{2}$
\begin{align*} \displaystyle a+b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}\\
a\sqrt[3]{2}+b\sqrt[3]{4}+2c&=\sqrt[3]{2}r^{1/3}\\
a\sqrt[3]{4}+2b+2\sqrt[3]{2}c&=\sqrt[3]{4}r^{1/3}\end{align*}
I tried adding the above three equations but it wasn't helpful.
Can someone help me in solving the question. Thanks in advance.
Is it possible to generalize the question as $\left(a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\right)^3$ where $n$ is a non-square integer?
| By using $$(x+y+z)^3=x^3+y^3+z^3+3xy(x+y)+3xz(x+z)+3yz(y+z)+6xyz$$ we get $a^2b+2b^2c+2ac^2=0$ and $ab^2+a^2c+2bc^2=0$. Now multiply the first equation by $b$ and the second by $a$ and then subtract them. We obtain $c=0$ (and then $a=0$ or $b=0$) or $a=b=0$.
Edit. I forgot to mention that I used the following: if $p+q\sqrt[3]{2}+r\sqrt[3]{4}=0$, with $p,q,r$ rational numbers, then $p=q=r=0$.
The generalization is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Is the polynomial $x^5 - 5x^4 + 7x^3 + x^2 + x - 1$ irreducible in $\mathbb{Z} [x]$ or in $\mathbb{Q} [x]$? Is the following polynomial irreducible in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$?
$x^5 - 5x^4 + 7x^3 + x^2 + x - 1$
If it's reducible, there should be a linear factor with degree $1, 2$ or $3.$
I try $\mathbb{Z}_2[x]$ and we get $x^5 +x^4 + x^3 + x^2 + x + 1$ looking for roots $f(0) = 1$ and $f(1) = 6 = 6 \mod 2 = 0$. So it's reducible in $\Bbb Z_2[x]$?
Does this help me?
I see the polynomial is also primitive, so if I prove that it's irreducible in $\mathbb{Q}[x]$ or $\mathbb{Z}[x]$, then it's also irreducible in $\mathbb{Z}[x]$ or $\mathbb{Q}[x]$.
If I use rational root system, there is no root for $x^5 - 5x^4 + 7x^3 + x^2 + x - 1$, so there can only be a decomposition in degree $2+3$.
I don't know how to show now, that it's irreducible.
| By the Rational Root Theorem, the only possible linear factors are $(x\pm1)$, but evaluation at $\mp1$ shows that this is not the case.
Modulo $2$, we have the factorization
$$x^5+x^4+x^3+x^2+x+1\equiv (x^2+x+1)(x^3+1)\equiv (x^2+x+1)^2(x+1)$$
Hence any quadratic factor in $\Bbb Z[x]$ must $\equiv x^2+x+1\pmod 2$, i.e, all coefficients odd, moreover leadeing $1$, and of course constant term $\pm1$, in other words: $ x^2+ax\pm1$ with odd $a$. Similarly, the corresponding cubic factor must be $x^3+bx^2+cx\mp1$ with $b,c$ even.
We compute
$$(x^2+ax\pm1)(x^3+bx^2+cx\mp1)
=x^5+(a+b)x^4+(ab+c\pm1)x^3+(ac\pm b\mp1)x^2+(\pm c\mp a)x-1 $$
so from the linear term
$$ c=a\pm1$$
and from the fourth power term
$$ b= -a-5.$$
Plug this into the cubic term
$$ ab+c=7\mp1$$
to arrive at
$$ a(a+4)=\pm2-7=-5\text{ or }-9,$$
which has no integer solution (two factors of $-5$ or $-9$ must differ by at least $6$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Converting $\frac{\sqrt{2-\sqrt{3}}}{2}$ into $\frac{\sqrt{3}-1}{2\sqrt{2}}$ A few days ago I had a quiz question on calculating the sine of $15^\circ$. Using the half-angle identity for sine, I did the following:
$$\sin15^\circ=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$$
But the lecturer give this:
$$\frac{\sqrt{3}-1}{2\sqrt{2}}$$
This is also the first answer given by WolframAlpha.
So my question is: how does one convert between the two forms?
$$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$
Edit: and why would one prefer the second over the first, if the the first is so straightforward to reach?
| $$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given $x+y+z=3$, how do I maximize $xy+yz+zx-xyz$? Given non-negative numbers $x,y,z$ such that $x+y+z=3$. How do I maximize $xy+yz+zx-xyz$. I found out that the maximum is 2 and equality holds when one of the numbers is 1 and the other two sum up to 2 but had no idea how to prove it. Can someone help please?
| Since the expression is symmetric, we can assume $z \leq 1$. Note that $xy+yz+zx-xyz=xy(1-z) + z(x+y) = xy(1-z)+z(3-z)$. It means that for given fixed $z \leq 1$, the maximum is reached when $xy$ is maximized (as $1-z\geq0$), and since $x+y=3-z$ is fixed, it happens when $x=y=\frac{3-z}{2}$. Therefore the expression now becomes a polynomial in $z$: $$P(z)=(\frac{3-z}{2})^2(1-z)+z(3-z)=(3-z)(\frac{(3-z)(1-z)}{4}+z)=(3-z)(\frac{z^2}{4}+\frac{3}{4})=\frac{-z^3+3z^2-3z+9}{4}$$ And we want to find it's maximum where $z \in [0,1]$. However it is easy to see that $P$ is positive and decreasing on $[0,1]$, therefore the maximum is when $z=0$ and $P(0)=\frac{9}{4}$ is the maximum. From the proof it follows that the maximum is reached when two of $x,y,z$ are equal to $\frac{3}{2}$ and the other is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Expected value of area of a triangle
Two (possibly equal) integers $x$ and $y$ are chosen such that $1 \le x \le 5$ and $1 \le y \le 5$. What is the expected value of the area of the triangle with vertices $(0, 0)$, $(x, 0)$, and $(0, y)$?
I ended up getting$${1\over2}\left({{1+5}\over2}\right)\left({{1+5}\over2}\right)= {9\over2}$$Is this correct?
| I assume, you have two independent random variables $x$ and $y$ with uniform distribution on the integers between 1 and 5. Each integer has a probability of $1/5$, so the probability of the pair $(x,y)$ is
$$p(x,y) = \frac{1}{25}$$
The Triangle size $ A(x,y) = \frac{xy}{2}$ is also a random variable. Its expectation value is
\begin{align}
\mathbb{E}(A) &= \sum_{x,y = 1}^5 A(x,y) p(x,y) = \sum_{x,y = 1}^5 \frac{xy}{2} \frac{1}{25} = \frac{1}{50} \left( \sum_{x = 1}^5 x \right) \left( \sum_{y = 1}^5 y \right) = \frac{15^2}{50} = \frac{9}{2}
\end{align}
So your result is apparently correct!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x\to 1^+}{\frac{1}{x^2-1}}$(with epsilon-delta) I want to prove that $\lim_{x\to 1^+}{\frac{1}{x^2-1}}=+\infty$
So I tried to define the $\delta$:
$\frac{1}{x^2-1}=\frac{1}{(x+1)(x-1)}>\frac{1}{\delta (x+1)}(\because 0<x-1<\delta)$
My problem is How to evaluate $\frac{1}{x+1}$ and How to proceed with the discussion after this.
Thanks,for any help.
| By definition of $\lim_{x\to 1^{+}}\dfrac{1}{x^{2}-1}=+\infty$ we have proof
$$
(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})
\left(( 0<\color{red}{x-1}<\delta)\implies \left(\dfrac{1}{x^{2}-1}>\color{red}{M}\right) \right)
$$
The secret here is to go working on the expression $\left(\dfrac{1}{x^{2}-1}>\color{red}{M}\right)$ until You get the expression $\left(0<x-1<\delta\right)$.
\begin{align}
\dfrac{1}{x^{2}-1}>\color{red}{M}
\implies &
{x^{2}-1}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{(x+1)\color{red}{(x-1)}}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{(\color{red}{(x-1)}+2)\color{red}{(x-1)}}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{\color{red}{(x-1)}^{2}+2\cdot\color{red}{(x-1)}}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{\color{red}{(x-1)}^{2}+2\cdot\color{red}{(x-1)}}+1<\dfrac{1}{\color{red}{M}}+1
\\
\implies &
(\color{red}{(x-1)} +1)^{2}<\dfrac{1}{\color{red}{M}}+1
\\
\implies &
|\color{red}{(x-1)} +1|<+\sqrt{\dfrac{1}{\color{red}{M}}+1}
\\
\implies &
0<\color{red}{(x-1)} +1<+\sqrt{\dfrac{1}{\color{red}{M}}+1}
\\
\implies &
0<\color{red}{(x-1)} <+\sqrt{\dfrac{1}{\color{red}{M}}+1} - 1
\end{align}
Now the job is with you. Show that
$$
0<x-1<\delta = \sqrt{\dfrac{1}{\color{red}{M}}+1} - 1 \mbox{ implies } \dfrac{1}{x^{2}-1}>\color{red}{M}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that, for every $a\neq0$, the equations $ax^3 - x^2 -x -(a +1)= 0$ and $ax^2 - x -(a +1)=0$ have a common root
Show that, for every real number $a\neq0$, the equations
$ax^3 - x^2 -x -(a +1)= 0$ and $ax^2 - x -(a +1)=0$
have a common root.
I want to know whether there are any mistakes in it.
Proof: If the system of equations
$$ax^3-x^2-x-(a+1)=0\cdots(1)$$ and $$ax^3-x^2-x=ax^2-x\cdots(2)$$ is considered,let us see what roots are obtained. Now,Subtracting (1) and (2) we get,
$$-(a+1)=x-ax^2\Rightarrow ax^2-x-(a+1)=0$$ This equation has roots $-1$ and $(1+2a+1)\times \frac{1}{2a}=\frac {a+1}{a}$.We see that $\frac {a+1}{a}$ is the root of (1) and (2) but (-1) is not(we can verify that). Thus, we found a common root for $a$ for all $a\in \mathbb R$ with $a\neq0$.
| You have almost finished the solution. But you're missing one step.
Because, we must check that the solution satisfies at least one equation.
You can also reach the result as follows:
$$\begin{align}&\begin{cases}ax^3-x^2-x-(a+1)=0\\ ax^2-x-(a+1)=0\end{cases} \\\\ \implies &\begin{cases}ax^3-x^2-x-(a+1)=0\\ ax^3-x^2-x(a+1)=0\end{cases}\\
\implies &x(a+1-1)=a+1\\
\implies &x=\frac{a+1}{a} \end{align}$$
Finally, to conclude the proof, it is sufficient to check whether the root we found satisfies the quadratic equation:
$$\frac{(a+1)^2}{a}-\frac{a+1}{a}-(a+1)=0.$$
This means, you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $a^5 - b^5 - c^5 = 30abc$ and $a^2 = 2(b + c)$ over positive integers. I have found this system of equations in a Romanian Mathematical magazine:
Solve over positive integers:
$a^5 - b^5 - c^5 = 30abc$ and $a^2 = 2(b+c)$. I tried multiplying the first equation with an $a$ and then substituting $a^6$ with $8(b+c)^3$, but then it becomes very messy. I tried some factoring, but couldn't get rid of the others $a$. Do you have any other ideas?
| By Fermat's little theorem $a^5 \equiv a \pmod {5}$. And since $a^5 - a = a(a^2 + 1)(a + 1)(a - 1)$, $a^5 - a \equiv 0 \pmod 3$, $\pmod 2$. Since $2,3,5$ are all coprime, this yields $a^5 \equiv a \pmod {30}$.
If $a-b-c = 0$, then $a^2 = 2(b+c) = 2a \implies a = 2$. $(a,b,c) = (2,1,1)$ is the only case to check given the condition $a^2=2(b+c)$, which is false.
Now let $a>b≥c$, or otherwise the left-hand side becomes negative or zero while the RHS is always positive. From the above, $a-b-c ≥ 30 \implies a-30≥b+c$. So $a-30≥a^2/2$, which is never true as $\Delta < 0$.
Hence there are no positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$O$ is the circumcenter of non-right $\triangle ABC$. $\frac{|AB \cdot CO|}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$. Find $\tan A$ Problem: $O$ is the circumcenter of $\triangle ABC$, which is not a right triangle. $$\frac{| AB \cdot CO |}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$$. Find $\tan A$. Here $AB \cdot CO$ represents the dot product of vector $\overrightarrow{AB}$ and $\overrightarrow{CO}$
My thoughts: WLOG let the radius of the excircle be $1$. Let $AB = c$, $BC = a$, $AC
= b$. Then $AB \cdot CO = c \cdot 1 \cdot \cos( 90^{\circ} + \angle B- \angle A ) = -c \cdot \sin( \angle B- \angle A)$
$AC \cdot BO = -b \cdot \sin( \angle C- \angle A)$
$AB \cdot BO = c \sin C$
$AC \cdot CO = b \sin B$
But none of these look very trivial to solve..
| Let us record as in the OP the values for the scalar products, then start the computation using the formulas $a=2R\sin A$ (and the similar ones for $b,c$) to express the appearing sines, and $b^2+c^2-a^2=2bc\cos A$ (and the similar ones) to express the appearing cosines.
$$
\begin{aligned}
|AB\cdot CO|
&= AB\cdot CO\cdot\cos(\widehat{AB,CO})=cR\; \cos\left(\frac \pi2-A+B\right)
=cR\sin(A-B)\ ,
\\
|AC\cdot BO|
&= AC\cdot BO\cdot\cos(\widehat{AC,BO})=bR\; \cos\left(\frac \pi2-A+C\right)
=bR\sin(A-C)\ ,
\\[4mm]
|AB\cdot BO|
&= AB\cdot BO\cdot\cos(\widehat{AB,BO})=cR\; \cos\left(\frac \pi2-C\right)
=cR\sin C=2R^2\sin^2 C\ ,
\\
|AC\cdot CO|
&= AC\cdot CO\cdot\cos(\widehat{AC,CO})=bR\; \cos\left(\frac \pi2-B\right)
=bR\sin B=2R^2\sin^2 B\ .
\end{aligned}
$$
From the last two equalities, and the given second proportion, we get
$$
3=\frac{|AB\cdot BO|}{|AC\cdot CO|}
=\frac{\sin^2 C}{\sin^2 B}=\frac {c^2}{b^2}\ .
$$
So $c=b\sqrt 3$.
We may want to norm $b=2$, so then $c=2\sqrt 3$.
Let us use the first condition...
$$
\begin{aligned}
\pm 3
&=
\frac{|AB\cdot CO|}{|AC\cdot BO|}
\\
&=\frac{cR\sin(A-B)}{bR\sin(A-C)}
=\frac cb\cdot\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos C-\cos A\sin C}
\\[2mm]
&=
\frac cb\cdot
\frac
{\displaystyle a\cdot\frac {a^2+c^2-b^2}{2ac} - \frac {b^2+c^2-a^2}{2bc} \cdot b}
{\displaystyle a\cdot\frac {a^2+b^2-c^2}{2ab} - \frac {b^2+c^2-a^2}{2bc} \cdot c}
\\[2mm]
&=
\frac
{(a^2+c^2-b^2) - (b^2+c^2-a^2)}
{(a^2+b^2-c^2) - (b^2+c^2-a^2)}
\\[2mm]
&=
\frac
{a^2-b^2}
{a^2-c^2}\ .
\end{aligned}
$$
Recall the norming making $b^2=4$, $c^2=3b^2=12$.
The two chances for $a^2$ are now $10$ and $16=4+12$, where the above becomes
$-3=\frac{10-4}{10-12}$
and
$+3=\frac{16-4}{16-12}$. The second case gives rise to a triangle with $A=90^\circ$, which was excluded in the title. So we have to work with the triangle with sides:
$$
\color{blue}{
\boxed{\qquad a=\sqrt{10}\ ,\ b=2\ ,\ c=2\sqrt 3=\sqrt{12}\ .\qquad}
}
$$
Let us compute $\tan A$ by explicitly computing $\cos A$, $S$, $R$, $\sin A$ in this order:
$$
\begin{aligned}
\cos A&=\frac{b^2+c^2-a^2}{2bc}=\frac{4+12-10}{2\cdot 2\cdot 2\sqrt 3}=\frac 6{8\sqrt 3}=\color{blue}{\frac{\sqrt 3}4}\ ,
\\[2mm]
S^2&=s(s-a)(s-b)(s-c)\qquad\text{(Heron)}
\\
&=\frac 1{16}(a+b+c)(b+c-a)(a+c-b)(a+b-c)
=\frac 1{16}\Big((b+c)^2-a^2\Big)\Big(a^2-(b-c)^2\Big)
\\
&=\frac 1{16}\Big(2bc + (b^2+c^2-a^2)\Big)\Big(2bc - (b^2+c^2-a^2)\Big)
=\frac 1{16}\Big(4b^2c^2 -(b^2+c^2-a^2)^2\Big)
\\
&=\frac 1{16}\Big(4\cdot 4\cdot 12 -6^2\Big)
=\frac{39}4\ ,
\\
S&=\frac {\sqrt{39}}2\ ,
\\[2mm]
R&=\frac{abc}{4S}=2\sqrt \frac{10}{13}\ ,
\\[2mm]
\sin A&=\frac a{2R}=\frac {\sqrt {10}}{4\sqrt {10/13}} =\color{blue}{\frac{\sqrt{13}}4}\ ,
\\
\tan A&=\frac{\sin A}{\cos A}=\frac{\sqrt{13}/4}{\sqrt 3/4}
=\color{blue}{\boxed{\ \sqrt{\frac{13}3}}\ }\ .
\end{aligned}
$$
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4205469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving $2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)=3(a^2+b^2+c^2+(a+b+c)^2)^2$ in another way? How do I prove the following identity without expanding both sides directly.
$$2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)\\=3(a^2+b^2+c^2+(a+b+c)^2)^2$$
I expanded both sides directly and it is true. However, I was hoping there could be another way to prove it, like that of Candido's identity which could be proved using diagrams.
| Why do this by hand?
Mathematica:
Simplify[2 ((a + b)^4 + (a + c)^4 + (b + c)^4) + 4 (a^4 + b^4 + c^4 + (a + b + c)^4) == 3 (a^2 + b^2 + c^2 + (a + b + c)^2)^2]
(* True *)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
| Write $a_n=2^n + 5^n + 56\cdot1^n$ and write $(x-2)(x-5)(x-1)=0$ as $x^3=8 x^2 - 17 x + 10$. Then
$$
a_{n+3} = 8 a_{n+2} -17 a_{n+1}+ 10 a_n
$$
Then $a_n \bmod 9$ is
$$
4,0,4,0,\color{red}{4,0,4},\dots
$$
Because of the linear recurrence, the sequence repeats as soon as it repeats three consecutive terms, $4,0,4$ in this case.
Bottom line $a_n \equiv 0 \bmod 9$ iff $n \equiv 1 \bmod 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
} |
Find the size of the segment joining the foot of the perpendiculars of a scalene triangle The sides of a scalene triangle measure 13, 14, and 15 units. Two outer bisectors of different angles are drawn and the third vertex is drawn perpendicular to these bisectors. Calculate the size of the segment joining the foot of these perpendiculars.(Answer:21)
My progress ..I thought of using sine theorem, cosine theorem and Pythagoras but it will be very complicated. There is probably a simpler solution
$D ~é ~excentro \therefore AD ~é~ bissetriz \triangle ABC\\
Teorema Bissetriz: \frac{BK}{AB}=\frac{KC}{AC}\rightarrow\frac{14-KC}{13}=\frac{KC}{15}\\
\therefore KC = 7,5 ~e~BK = 6,5\\
\triangle ABK\sim \triangle AML: Razão~Semelhança=\frac{13}{6,5} = 2\\
\therefore LM = \frac{6,5}{2}=3,25\\
De~forma~análoga: LN = 3,75\\
\triangle AHB: M(ponto~médio)\rightarrow HM = MB = 6,5\\
\triangle ACI:N(ponto ~médio) \rightarrow NI = NC = 7,5\\
\therefore \boxed{\color{red}x = 6,5+3,25+3,75+7,5 = 21 }$
|
As $D$ is intersection of external bisectors of $\angle B$ and $\angle C$, $AD$ must be internal bisector of $\angle A$.
So, $\angle ADB = 90^0 - \cfrac{\angle B}{2} - \cfrac{\angle A}{2} = \cfrac{\angle C}{2}$
Similarly, $\angle ADC = \cfrac{\angle B}{2}$
Now notice that quadrilateral $AIDH$ is cyclic.
So, $\angle AHI = \angle ADC = \cfrac{\angle B}{2}$
Also, $\angle BAH = \cfrac{\angle B}{2} = \angle AHI$. So $M$ is the midpoint of $AB$.
And similarly, it follows that $N$ is the midpoint of $AC$.
That leads to $HI = HM + MN + NI$
$= AM + \cfrac{BC}{2} + AN = 6.5 + 7 + 7.5 = 21$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion.
It is given that $(x^2 +y +2t +3k)^{10}$. What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion. For example the terms are $x^2y^4t^3k^2$ etc.
I said that let $(x^2 +2t)$ be $"a"$ and $(y +3k)$ be $"b"$. Then, $(x^2 +y +2t +3k)^{10}=(a+b)^{10}$.
If we are looking for $x^2t^3$, then the exponential of $a$ must be $4$. Then, the exponential of $b$ is $6$.
Then, the coefficient of $x^2t^3$ is $4 \times 2^3=32$. Now, we have $(y +3k)^6$ as $b$. If we find the sum of the all coefficients in $(y +3k)^6$, we can handle the question.
The sum of the all coefficients in $(y +3k)^6$ is $4^6$.
Then, answer is $32 \times 4^6$
Is my solution correct ?
| Here's a hint to simplify the calculation. Suppose you were to fully expand $(x^2+y+2t+3k)^{10}$ and then drop all but those terms containing $x^2 t$. What results is an expression of the form
$$ax^2 t+bk x^2 t+cy x^2 t+dk^2 x^2 t+\cdots=(a+bk+cy+dk^2+\cdots)x^2 t$$
Summing up these coefficients yields $a+b+c +d+\cdots$.
What's the relation between this sum and the polynomial in $y,k$ written above? (This reduces the problem to finding the coefficient of $x^2 t$ in some polynomial in $x,t$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Baby Rudin Theorem 7.18 Here is Theorem 7.18 from Baby Rudin:
There exists a real continuous function on the real line which is nowhere differentiable.
Here is a proof of the theorem:
Define
$$\tag{34} \varphi(x) = \lvert x \rvert \qquad \qquad (-1 \leq x \leq 1) $$
and extend the definition of $\varphi(x)$ to all real $x$ by requiring that
$$ \tag{35} \varphi(x+2) = \varphi(x). $$
Then, for all $s$ and $t$,
$$\tag{36} \lvert \varphi(s) - \varphi(t) \rvert \leq \lvert s-t \rvert. $$
In particular, $\varphi$ is continuous on $\mathbb{R}^1$. Define
$$ \tag{37} f(x) = \sum_{n=0}^\infty \left( \frac{3}{4} \right)^n \varphi \left( 4^n x \right). $$
Since $0 \leq \varphi \leq 1$, Theorem 7.10 shows that the series (37) converges uniformly on $\mathbb{R}^1$. By Theorem 7.12, $f$ is continuous on $\mathbb{R}^1$.
Now fix a real number $x$ and a positive integer $m$. Put
$$ \tag{38} \delta_m = \pm \frac{1}{2} \cdot 4^{-m} $$
where the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. This can be done, since $4^m \left\lvert \delta_m \right\rvert = \frac{1}{2}$. Define
$$ \tag{39} \gamma_n = \frac{ \varphi \left( 4^n \left( x + \delta_m \right) \right) - \varphi \left( 4^n x \right) }{ \delta_m }. $$
When $n > m$, then $4^n \delta_m$ is an even integer, so that $\gamma_n = 0$. When $0 \leq n \leq m$, (36) implies that $\left\lvert \gamma_n \right\rvert \leq 4^n$.
Since $\left\lvert \gamma_m \right\rvert = 4^m$, we conclude that
$$
\begin{align}
\left\lvert \frac{ f \left( x + \delta_m \right) - f(x) }{ \delta_m } \right\rvert &= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&\geq 3^m - \sum_{n=0}^{m-1} 3^n \\
&= \frac{1}{2} \left( 3^m + 1 \right).
\end{align}
$$
As $m \to \infty$, $\gamma_m \to 0$. It follows that $f$ is not differentiable at $x$.
I have two questions regarding the proof.
*
*At $(38)$ it's stated that the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. Why is it necessary that no integer should be within that range and what would happen if they were?
*At the later part of the proof we have:
$$
\begin{align}
\left\lvert \frac{ f \left( x + \delta_m \right) - f(x) }{ \delta_m } \right\rvert &= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&\geq 3^m - \sum_{n=0}^{m-1} 3^n \\
&= \frac{1}{2} \left( 3^m + 1 \right).
\end{align}
$$
I want to know why is the following part true:
$$
\begin{align}
\left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert
&\geq 3^m - \sum_{n=0}^{m-1} 3^n \\
\end{align}
$$
Any help is appreciated!
| *
*If you look at a picture of $\varphi$, it is a saw blade with a slope of $1$ in intervals of the form $(2k,2k+1)$, and a slope of $-1$ in intervals of the form $(2k+1,2(k+1))$. The approach to the proof is essentially to take an arbitrary $x$ and show that there exists a sequence $\{\delta_m\}$ such that $\delta_m \to 0$, but
$$
\left|\frac{f(x+\delta_m)-f(x)}{\delta_m}\right| \to +\infty.
$$
This method sufficiently demonstrates that the difference quotient cannot converge. The reason we guarantee that no integer lies between $4^mx$ and $4^m(x+\delta_m)$ is to guarantee that we are comparing points on the "saw blade" that correspond to the same branch (either one with slope of $1$ or $-1$, but no mixing). There's no reason to not do this because $\delta_m \to 0$ anyways, and it makes the proof easier.
*Here's the full work for that step. The first step of reducing the sum from an infinite sum to one up to $m$ is because $\gamma_n=0$ for $n>m$. Later, we use that $\gamma_m=4^m$.
\begin{align}
\left\lvert \frac{ f \left( x + \delta_m \right) - f(x) }{ \delta_m } \right\rvert &= \left\lvert \sum_{n=0}^\infty \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&= \left\lvert \left( \frac{3}{4} \right)^m \gamma_m+ \sum_{n=0}^{m-1}\left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&\geq \left\lvert \left(\frac{3}{4}\right) ^m \gamma_m \right\rvert-\left\lvert \sum_{n=0}^{m-1} \left( \frac{3}{4} \right)^n \gamma_n\right\rvert \\
&= 3^m -\left\lvert \sum_{n=0}^{m-1} \left( \frac{3}{4} \right)^n \gamma_n\right\rvert \\
&\geq 3^m - \sum_{n=0}^{m-1} \left( \frac{3}{4} \right)^n |\gamma_n| \\
&\stackrel{|\gamma_n|\leq 4^n}{\geq} 3^m - \sum_{n=0}^{m-1} 3^n \\
&= \frac{1}{2} \left( 3^m + 1 \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Show that $ \frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z) $ Show that for positive reals $x,y,z$ the following inequality holds and that the constant cannot be improved
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z)
$$
Background: I was digging through some old correspondence and found a letter from a very young me to Professor Love at the University of Melbourne. I had apparently ask via a letter (yes it was back when we wrote letters) how one could prove the above inequality (my version had $1/\sqrt{3}$ in it). He kindly wrote back but without a full proof. I just found the correspondence today and thought that this was a good question for this site.
Based on his letter and my old writings you can transform the above inequality as follows. First note that
$$
\text{g.l.b.}f(x,y,z) = \text{g.l.b.}f(x,z,y) = k \quad (say)
$$
where g.l.b is the greatest lower bound and $f(x,y,z)$ is the function
$$
\left(\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}}\right)\bigg/(x+y+z).
$$
and so
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq 2k (x+y+z)
$$
Thus we need to prove that for positive reals $x,y$ and $x$ the following is true and tight:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \sqrt{2} (x+y+z)
$$
One approach to prove this (used by Prof. Love) was to apply Hölder's inequality but this unfortunately only gives:
$$
\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \frac{2}{\sqrt{3}} (x+y+z)
$$
Geometric View: From a geometric view point this inequality can be viewed as stating that the perimeter of $\Delta PQR$ is not less than $\sqrt{2}$ times the sum of the the three edge-lengths of the box of sides $x,y,$ and $z$ and the points $P,Q$ and $R$ are three corners of the box that are not adjacent to each other.
I suspect that this is a "well known" inequality in the right circles but it is still not known to me. Thought that is was a nice problem for lovers of inequalities.
| Another way.
By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{2(xy+xz+yz)^2}{\sum\limits_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}}=\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}}$$ and it's enough to prove that:
$$\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}\geq\frac{(x+y+z)^2}{2}$$ or
$$\sum_{cyc}\left(\frac{x^4}{x^2+y^2}-\frac{3x^2-y^2}{4}\right)+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}\geq\frac{(x+y+z)^2}{2}-\frac{x^2+y^2+z^2}{2}$$ or
$$\sum_{cyc}\frac{(x^2-y^2)^2}{4(x^2+y^2)}\geq xy+xz+yz-\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}$$ or
$$\sum_{cyc}(x-y)^2\left(\frac{(x+y)^2}{x^2+y^2}-\frac{2(xy+xz+yz)}{x^2+y^2+z^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-y)^2(x^2+y^2-xz-yz)^2}{x^2+y^2}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots? A number $a$ is chosen at random within the interval $(-1, 1)$. What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots?
For it to have its real roots, we must guarantee that $1-4a \geq 0$, or $a\leq \frac{1}{4}$.
It is no longer clear to me what I have to do.
| $$a \in(-1,1) \land a \leq \frac{1}{4} \iff a \in(-1,\frac{1}{4}]$$
because $(-1,1) \cup (-\infty,\frac{1}{4}] = (-1,\frac{1}{4}]$
We can therefore, divide the length of both intervals to get the probability:
$$P(a \in(-1,\frac{1}{4}]) = \frac{|\frac{1}{4} - \left(-1\right)|}{|1 - \left(-1\right)|} = \frac{\frac{5}{4}}{2} = \frac{5}{8}$$
The fact that $(-1,\frac{1}{4}]$ doesn't include $-1$ and $\left(-1,1\right)$ doesn't include either of its endpoints doesn't make a difference because they are single points that can be considered insignificant compared to the infinite number of reals in the intervals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can one integral have two answers? I am currently taking a calculus class. My teacher while teaching a specific type of indefinite integral told one mysteriously beautiful of the solving the integral. The general form of the integral was -
$\int \sin^m x \cos^n x dx $
He said when both m and n are odd, for example -
$\int \sin^3x \cos^5xdx$
There are 2 methods to solve this which gave 2 completely different answers and both were correct. Let me elaborate.
Method 1 -
$\int \sin^3x \cos^5xdx = \int \sin^2x \cos^5x \sin xdx$
Now let, $\cos x = u$
$\implies -\sin x dx = du$
$\implies -\int u^5(1-u^2)du$
$=\int \big(u^7 - u^5 \big)du$
$= {u^8\over 8} - {u^6\over 6}+c$
$= {\cos ^8x \over 8} - {\cos ^6x \over 6} + c$
Method 2 -
$\int \sin^3x \cos^5xdx = \int \sin^3x\cos^4x\cos xdx$
Now this time let, $\sin x=v$
$\implies \cos xdx = dv$
$\implies \int v^3\big(1-v^2)^2dv$
$= \int \big( v^7 - 2v^5 +v^3\big)dv$
$= {\sin^8x \over 8}+{\sin^4x \over 4} - {1\over 3}\sin^6 x+c_1$
As said before, my teacher said both these answers are correct. My questions are -
*
*How are both the answers correct?
*How do we interpret this result?
| if you substitute $cos^2\theta=1-sin^2\theta$ into your first expression you get (for tidiness I'm letting $cos\theta=C$ and $sin\theta=S$)
$$ \frac{C^8}{8}-\frac{C^6}{6}+c_0$$
$$ =\frac{C^6}{2}(\frac{C^2}{4}-\frac{1}{3})+c_0$$
$$ =\frac{(1-S^2)^3}{2}(\frac{1-S^2}{4}-\frac{1}{3})+c_0$$
$$ =\frac{(1-S^2)^3}{2}(\frac{-1-3S^2}{12})+c_0$$
$$ =-\frac{1}{24}((S^2)^3-3(S^2)^2+3(S^2)-1)(1+3S^2)+c_0$$
$$ =-\frac{1}{24}((S^6)-3(S^4)+3(S^2)-1)(1+3S^2)+c_0$$
$$ =-\frac{1}{24}((S^6)-3(S^4)+3(S^2)-1+3S^8-9S^6+9S^4-3S^2)+c_0$$
which simplifies to your second expression
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{a^2}{a^2+bc}+\frac{(a+b+c)^3+9abc}{\prod\limits_{cyc}(a+b)}\geq6.$ There is the following anhduy98's problem.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}+\frac{(a+b+c)^3+9abc}{(a+b)(a+c)(b+c)}\geq6.$$
I solved this problem by BW(about BW see here: https://artofproblemsolving.com/community/c6h522084)
and I am looking for an alternative solution.
The equality occurs also for $c=0$ and $a=b$.
I tried $uvw$(about $uvw$ see here https://artofproblemsolving.com/community/c6h278791).
It gives $$4w^9+(22u^3-39uv^2)w^6+(27u^6-99u^4v^2+81u^2v^4+13v^6)w^3+9uv^6(3u^2-4v^2)\geq0$$ and I did not get solution for variations of $u$, of $v^2$ and of $w^3$.
Also, I tried $SOS$ and I did not get a solution.
Thank you!
PS. The River Li's solution we can write in the following form.
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}+\frac{(a+b+c)^3+9abc}{(a+b)(a+c)(b+c)}-6=$$
$$=\frac{\sum\limits_{cyc}\left(4a^2b+4a^2c-\frac{1}{3}abc\right)\prod\limits_{cyc}(a-b)^2+\left(\sum\limits_{cyc}(2a^3-a^2b-a^2c)\right)^2abc}{4\prod\limits_{cyc}(a+b)\prod\limits_{cyc}(a^2+bc)}\geq0$$
| Remarks: The pqr method works well, furthermore, it leads to a SOS solution directly.
Let $p = a + b + c, ~ q = ab + bc + ca, ~ r = abc$.
We need to prove that $F(p, q, r) \ge 0$ where
\begin{align*}
F(p, q, r) &= -4\,p{q}^{4}+ \left( {p}^{3}+13\,r \right) {q}^{3}+27\,{p}^{2}r{q}^{2} \\
&\quad
- \left( 11\,{p}^{4}r + 117\,p{r}^{2} \right) q+{p}^{6}r+22\,{p}^{3}{r}
^{2}+108\,{r}^{3}.
\end{align*}
Note that
$$\Delta = -4\,{q}^{3}+{p}^{2}{q}^{2}+18\,prq-4\,{p}^{3}r-27\,{r}^{2} = (a - b)^2(b - c)^2(c - a)^2 \ge 0.$$
We have
$$F - (pq - 13r/4)\Delta = \frac{1}{4}r(2p^3 - 7pq + 9r)^2.$$
Since $pq \ge 9r \ge 13r/4$, we have $F \ge 0$. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
Find the derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
I'm learning differentiation and this is an exercise problem from my book. I used chain rule and got the following:
$\begin{align}
\dfrac d{dx}\left[\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac x2\right)\right] &= \dfrac{1}{1+\frac{a-b}{a+b}\tan^2\frac x 2}\cdot\dfrac{d}{dx}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)\\ &= \dfrac{1}{1+\frac{a-b}{a+b}\tan^2\frac x 2}\cdot\frac 1 2\sqrt{\frac{a-b}{a+b}}\sec^2\frac x 2
\end{align}$
But this doesn't match the answer in the book. The given answer is $\frac{\sqrt{a^2-b^2}}{2(a+b\cos x)}$. So, where did I go wrong and what is the correct solution?
| Just keep going from where you stopped: Taking things one or two steps at a time, we have
$$\begin{align}
{1\over1+{a-b\over a+b}\tan^2{x\over2}}\cdot{1\over2}\sqrt{a-b\over a+b}\sec^2{x\over2}
&={a+b\over a+b+(a-b)\tan^2{x\over2}}\cdot{1\over2}\sqrt{a-b\over a+b}\sec^2{x\over2}\\
&={1\over2}\cdot{\sqrt{(a+b)(a-b)}\over a+b+(a-b)\tan^2{x\over2}}\cdot{1\over\cos^2{x\over2}}\\
&={1\over2}\cdot{\sqrt{a^2-b^2}\over(a+b)\cos^2{x\over2}+(a-b)\sin^2{x\over2}}\\
&={1\over2}\cdot{\sqrt{a^2-b^2}\over a\left(\cos^2{x\over2}+\sin^2{x\over2}\right)+b\left(\cos^2{x\over2}-\sin^2{x\over2}\right)}\\
&={1\over2}\cdot{\sqrt{a^2-b^2}\over a+b\cos x}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find k in a curve equation when equation of a line tangent to curve is given
The equation of the curve is
$$y=x\left(\frac{k}{\sqrt{x}} - 1\right)$$
Does the problem mean the curve has a slope of zero at $y = 25/4$? The problem asks to find the value of $k$ and equation of line "l" which can be seen in the graph.
| Note that the equation of the curve
$$y = x\left(\frac{k}{\sqrt x} - 1\right) \tag 1$$
is undefined at $x=0$. We can change it to the more convenient
$$y = k\sqrt x - x \tag 2$$
which is equivalent, except that it is defined at zero.
As you mention in the comments,
$$y' = \frac{dy}{dx} = \frac{k}{2\sqrt x} - 1 \tag 3$$
We're given that the curve is tangent to the horizontal line
$$y = 25/4 \tag 4$$
We can find where the curve has a horizontal tangent by setting $y'=0$.
$$\frac{k}{2\sqrt x} = 1$$
$$\sqrt x = \frac{k}{2}$$
$$x = \frac{k^2}{4}$$
Now we can plug that $\sqrt x$, $x$, and $y=25/4$ into eqn (2).
$$25/4 = k \frac{k}{2} - \frac{k^2}{4}$$
$$25/4 = \frac{k^2}{2} - \frac{k^2}{4}$$
$$25/4 = \frac{k^2}{4}$$
Thus $k=5$
For part (ii) of the question, we can see from eqn (3) that for large $x$, the slope of the tangent line to the curve approaches -1 because $\frac{k}{2\sqrt x}$ is small. Eg, at $x=10000$, $y'=\frac{1}{200} - 1$.
However, the tangent line is displaced from the line $y = -x$. In fact, at $x = x_0$, the equation of the tangent line is
$$y = \left(\frac{5}{2\sqrt {x_0}}-1\right)x + \frac52\sqrt{x_0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$
$=>S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)}$
$\leq \frac{a^2}{4}.(\frac{1}{2a+b}+\frac{1}{2a+c})+\frac{b^2}{4}.(\frac{1}{2b+c}+\frac{1}{2b+a})+\frac{c^2}{4}.(\frac{1}{2c+a}+\frac{1}{2c+b})$
$=\frac{1}{4}.[a^2.(\frac{1}{2a+b}+\frac{1}{2a+c})+b^2.(\frac{1}{2b+a}+\frac{1}{2b+c})+c^2.(\frac{1}{2c+a}+\frac{1}{2c+b})]$
$\leq \frac{1}{4}.[\frac{a^2}{9}.(\frac{2}{a}+\frac{1}{b}+\frac{2}{a}+\frac{1}{c}) +\frac{b^2}{9}.(\frac{2}{b}+\frac{1}{c}+\frac{2}{b}+\frac{1}{a}) +\frac{c^2}{9}.(\frac{2}{c}+\frac{1}{a}+\frac{2}{c}+\frac{1}{b})]$
$=\frac{1}{36}.[a^2.(\frac{4}{a}+\frac{1}{b}+\frac{1}{c})+b^2.(\frac{1}{a}+\frac{4}{b}+\frac{1}{c})+c^2.(\frac{1}{a}+\frac{1}{b}+\frac{4}{c})]$
$= \frac{1}{36}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) $
We need prove that: $S \le \frac{1}{3}$
$=> S \le \frac{1}{12}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) \le 1 $ <- in there, I don't know how to do that :<
I'm trying to find a solution to continue for me or another solution, Can you help me?
| The inequality looks familiar. In fact, I solved the inequality a few weeks ago from an inequality handout. Here is the solution:
We have $$\begin{align}\frac{9a^2}{(2a+b)(2a+c)} &= \frac{(2a+a)^2}{2a(a+b+c)+2a^2+bc}\\ & \leq \frac{(2a)^2}{2a(a+b+c)}+\frac{a^2}{2a^2+bc} \\ &= \frac{2a}{a+b+c}+\frac{a^2}{2a^2+bc}
\end{align}$$
from Cauchy-Schwarz inequality.
Now, $$\begin{align}9\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)} &\leq 2\sum_{cyc}\frac{a}{a+b+c}+\sum_{cyc}\frac{a^2}{2a^2+bc}\\ &= 2+\sum_{cyc}\frac{a^2}{2a^2+bc}\leq 3
\end{align}$$
which is the desired result.
Here the last inequality follows because $$\sum_{cyc}\frac{a^2}{2a^2+bc}\leq1\iff \sum_{cyc}\frac{bc}{2a^2+bc}\geq1$$ is true by Cauchy-Schwarz.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How to solve for when this trigonometric function intersects the line $y=1$? How can I solve for $\alpha$ in
$$4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
on the domain $0\leq\alpha\leq\pi$? Clearly, one solution is when $\alpha=\pi$, but through plotting, it seems to only hold true when $t-r$ is less than a value around $0.3$. When $t-r$ is greater than this value, it seems to have different solutions.
| Solving for $\alpha$
$$4\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)^{3}\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
$$
4\sin{\left(\frac{\alpha}{2}\right)}\cos{\left(\frac{\alpha}{2}\right)}\left(1-\sin{\left(\frac{\alpha}{2}\right)^2}\right)\left(t-r\right)+\sin{\left(\frac{\alpha}{2}\right)}=1
$$
Let $x = \sin(\frac{\alpha}{2})$
$$
4x\cos(\frac{\alpha}{2})(1-x^2)(t-r)+x=1
$$
Because $\cos(\frac{\alpha}{2})=\sqrt{1-\sin^2(\frac{\alpha}{2}})=\sqrt{1-x^2}$
$$
\Longrightarrow4x\sqrt{1-x^2}(1-x^2)(t-r)+x=1
$$
$$
\Longrightarrow4x(1-x^2)^{\frac{3}{2}}(t-r)+x=1
$$
$$
\Longrightarrow(1-x^2)^3=\frac{\left(1-x\right)^2}{16(t-r)^2x^2}
$$
$$
\Longrightarrow x^2(1+x)^3(1-x)=\frac{1}{16(t-r)^2}
$$
$$
\Longrightarrow -x^6-2x^5+2x^3+x^2-\frac{1}{16(t-r)^2} = 0
$$
Solving for $\alpha$ amounts to solving for the root of the equation above with dependencies on the value of $t,r$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $T_1=7^7,T_2=7^{7^{7}},T_3=7^{7^{7^{7}}}$ and so on, what will be the tens digit of $T_{1000}$? $7^4$ ends with $2301$, so $7^{4k+r}$ ends with $7^r$ digits
$7^2\equiv1 \mod(4) $, and $7^7\equiv 3 \mod(4) $
Can we use the modulo function in exponent form? I think we will use these two properties, how can I proceed further?
| As said in comments observe that $T_n = 7^{T_n}$.
We use induction to find tens digits of $T_n$.
Observe that $T_1 = 43 \mod 100$.
Assume that $T_n \equiv 43\mod 100$. So write $T_n = 43 + 100k$ for some $k$
So $T_{n+1} \mod 100 = 7^{T_n} \mod 100 = 7^{43} 7^{100k} \mod 100$.
Now see that $7^4 = 1 \mod 100$ so we have $7^{100k} = 1 \mod 100$ and $7^{43} \equiv 7^3 \equiv 43 \mod 100$. So we get $T_{n+1} \equiv 43 \mod 100$
So we have $T_n \equiv 43 \mod 100$ for all $n$. This means that tens digit of $T_n$ will be $43$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Remainder Theorem Technique
Determine the remainder when $(x^4-1)(x^2-1)$ is divided by $1 + x + x^2$ (HMMT 2000, Guts Round)
A. Write the division in the form:
$$(x^4-1)(x^2-1)= (1 + x + x^2)Q(x) + R(x)$$
B. Multiply both sides by $x-1$:
$$(x-1)(x^4-1)(x^2-1)= (x^3-1)Q(x) + R(x)(x-1)$$
C. Substitute $x^3=1,x\neq1$, and reduce the resulting equation:
$$(x-1)(x-1)(x^2-1)= R(x)(x-1)$$
D. Divide both sides by $x-1$:
$$R(x)=(x-1)(x^2-1)=x^3 -x - x^2 + 1=-(x^2+x+1)+3=3$$
For someone who knows the method, is it valid to skip Steps B and D, directly substitute $x^3=1,x\neq1$ and use the fact that $x$ is a cube root of unity to get $x^2+x+1=0$.
| Alternate method:
In the spirit of my answer to your linked question, we recognize that $ P (x) = x^2 + x + 1$ can be "simplified" with $ Q(x) = x-1$ to give us $P(x) Q(x) = x^3 -1 $, and hence
$$ \begin{align} & ( x^4 - 1 ) (x^2 -1 ) \\
= & (x^3 -1 ) A(x) + (x-1)(x^2 - 1) & (1)\\
= & (x^3 -1 ) B(x) + (-x^2 - x + 2 ) & (2) \\
= & (x^2 + x + 1 ) C(x) + 3. \\
\end{align}$$
Notes:
*
*In step (1), this uses that $ x^4 - x = (x^3 -1)D(x)$
*In step (2), this uses $(x-1)(x^2 -1) = x^3 - x^2 - x + 1 = (x^3-1) E(x) - x^2 -x + 2 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}
I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?
| One way that is bound to succeed is to notice that this expression depends only on $n\bmod \varphi(9)$, which is $6$. This is because if $n\equiv m \bmod \varphi(9)$ then $10^n \equiv 10^m$ and $4^{n+2} \equiv 4^{m+2}$. This is a consequence of euler's theorem, so this would allow us to only need to check $6$ values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \right)} \right)^4} + {\left( {f\left( x \right)} \right)^8} + ..$
Let $T = f\left( x \right);U = {\left( {f\left( x \right)} \right)^2};V = {\left( {f\left( x \right)} \right)^4};W = {\left( {f\left( x \right)} \right)^8}$
Let the coefficient of $x^{10}$ in $T $ is zero
Taking U
${\left( {f\left( x \right)} \right)^2} = {\left( {x + {x^2} + {x^4} + {x^8} + ..} \right)^2}$
Hence the coefficient of $x^{10}$ in $U$ is $2x^{10}$ hence $2$
For $V$ and $W$ it is getting complicated hence any short cut or easy method to solve it
| We have
$$f(f(x))=f(x)+f(x)^2+f(x)^4+f(x)^8+f(x)^{16}+\ldots$$
To find the coefficient of $x^{10}$ in this expansion, note that the smallest power of $f(x)^k$ for some $k$ is $x^k$.
Hence, to find the coefficient of $x^{10}$, we only need to consider
$$f(x)+f(x)^2+f(x)^4+f(x)^8$$
The coefficient of $x^{10}$ in $f(x)^k$ represents the number of solutions to
$$\sum_{i=1}^k a_i=10$$
Where $a_i$ are integral powers of $2$.
We can systematically find these solutions using the binary representation of $10_{10}=1010_2$
If we are a little hand wavy with how we use binary representation, we can say that $10_{10}$
$$=1010_2$$
$$=210_2=1002_2$$
$$=130_2=202_2$$
Note that the expressions in the $i^\text{th}$ line represent the different ways to represent $10$ as a sum of $i+1$ integral powers of $2$ e.g. in the second line we have $210_2\implies 10=4+4+2$. We created the expressions in a line by recursively taking the expressions from the previous line and splitting a single power of $2$ in half.
We then have to account for permutations. Clearly $f(x)$ has no coefficient of $x^{10}$
The coefficient of $x^{10}$ in $f(x)^2$ is the number of permutations of $\{8,2\}$, which is $2!=\boxed{2}$.
The coefficient of $x^{10}$ in $f(x)^4$ is the number of permutations of $\{4,2,2,2\}$ and $\{4,4,1,1\}$, which is $\frac{4!}{3!}+\frac{4!}{2!2!}=\boxed{10}$.
The coefficient of $x^{10}$ in $f(x)^8$ is not as easily found using the recursive method we used before. However, it is easily found that the only way to express $10$ as the sum of $8$ powers of $2$ is $1+1+1+1+1+1+2+2$. There are $\frac{8!}{6!2!}=\boxed{28}$ permutations of this.
Hence, the total coefficient of $x^{10}$ is $2+10+28=\boxed{40}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Sum of binomial coefficient with 'skips'
Find
$$\sum_{k=0}^{n}
{(-4)}^k{n+k\choose2k}$$
I have dealt with such summations with variable $n$ in ${n \choose k}$ before, but never have seen one in which some $k$'s were skipped (like $1,3,5$...).
Can someone please give me a hint as to how to proceed?
| Snake oil:
\begin{align}
\sum_{n=0}^\infty\left(\sum_{k=0}^n (-4)^k \binom{n+k}{2k}\right)z^n
&= \sum_{k=0}^\infty (-4)^k \sum_{n=k}^\infty \binom{n+k}{2k} z^n \\
&= \sum_{k=0}^\infty (-4z)^k \sum_{n=0}^\infty \binom{n+2k}{2k} z^n \\
&= \sum_{k=0}^\infty (-4z)^k \frac{1}{(1-z)^{2k+1}} \\
&= \frac{1}{1-z}\sum_{k=0}^\infty \left(\frac{-4z}{(1-z)^2}\right)^k \\
&= \frac{1}{1-z}\cdot\frac{1}{1-\frac{-4z}{(1-z)^2}} \\
&= \frac{1-z}{(1+z)^2} \\
&= (1-z)\sum_{n=0}^\infty (n+1)(-z)^n \\
&= \sum_{n=0}^\infty (n+1)(-1)^n z^n - \sum_{n=0}^\infty (n+1)(-1)^n z^{n+1} \\
&= \sum_{n=0}^\infty (n+1)(-1)^n z^n -\sum_{n=1}^\infty n(-1)^{n-1} z^n \\
&= \sum_{n=0}^\infty (2n+1)(-1)^n z^n \\
\end{align}
So
$$\sum_{k=0}^n (-4)^k \binom{n+k}{2k} = (2n+1)(-1)^n$$
for $n \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The way to prove that $σ(a) = 3^k$ has no solution? $\sigma(n)$ = sum of divisors of n is a divisors function.
How to prove there are no such $a$ and $k \ge 2$ satisfy $\sigma(a) = 3^k$.
This proplem can be simplify to the case when $a$ is a power of prime ($a=p^\alpha$) because
if $a = p_0^{\alpha_{0}}p_1^{\alpha_{1}}...p_n^{\alpha_{n}}$, then
$$\sigma(a) = \sigma(p_0^{\alpha_{0}})\sigma(p_1^{\alpha_{1}})...\sigma(p_n^{\alpha_{n}})=\prod_{i=0}^n \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}$$
| As you stated, since $\sigma()$ is a multiplicative function, we only need to check the prime powers, i.e., for prime $p$, $e \ge 1$ and $j \ge 1$, that
$$\sigma(p^e) = \sum_{i=0}^{e}p^{i} = \frac{p^{e + 1} - 1}{p - 1} = 3^{j} \tag{1}\label{eq1A}$$
First, $p = 2$ and $e = 1$ gives $\sigma(2) = 2 + 1 = 3$, so $j = 1$ works. For $e \gt 1$, then
$$\sigma(2^e) = 2^{e + 1} - 1 = 3^{j} \; \; \to \; \; 2^{e + 1} - 3^{j} = 1 \tag{2}\label{eq2A}$$
Here, $j \gt 1$, but Mihăilescu's theorem shows there's no such solution (also, more simply, with $2^{e+1} = 3^j + 1$, then $e + 1 \ge 3$ means $3^j \equiv 7 \pmod{8}$, but $3^j \equiv 1 \pmod{8}$ or $3^j \equiv 3 \pmod{8}$ only). This means $a$ has at most $1$ factor of $2$ and there must be odd prime factors. Since $p = 3$ gives $\sigma(3^e) \equiv 1 \pmod{3}$, this means $p \ge 5$.
With the summation in \eqref{eq1A}, as there are $e + 1$ terms, getting an odd sum requires $e + 1$ to be odd, i.e., $e + 1 = 2r + 1$ for some integer $r \ge 1$. Using the fraction part in \eqref{eq1A}, if $p \equiv 2 \pmod{3}$, then $p^{e + 1} - 1 \equiv (2^2)^r(2) - 1 \equiv (1)^r(2) - 1 \equiv 1 \pmod{3}$ and $p - 1 \equiv 1 \pmod{3}$, so the result would be $\equiv 1 \pmod{3}$. Thus, this means $p \equiv 1 \pmod{3}$.
Once again from the summation in \eqref{eq1A}, as each term is $\equiv 1 \pmod{3}$, there must be a multiple of $3$ terms, i.e., $e + 1 = 3m$ for some integer $m \ge 1$. The fractional part of \eqref{eq1A} becomes
$$\frac{p^{e + 1} - 1}{p - 1} = \frac{p^{3m} - 1}{p - 1} = \left(\frac{p^{m} - 1}{p - 1}\right)(p^{2m} + p^{m} + 1) \tag{3}\label{eq3A}$$
Note that $\frac{p^{m} - 1}{p - 1} = \sum_{i=0}^{m-1}p$ is an integer. Also, since $p \equiv 1 \pmod{3}$, then $p^m = 3q + 1$ for some integer $q \ge 1$. Thus,
$$\begin{equation}\begin{aligned}
p^{2m} + p^{m} + 1 & = (3q + 1)^2 + (3q + 1) + 1 \\
& = (9q^2 + 6q + 1) + (3q + 1) + 1 \\
& = 9q^2 + 9q + 3 \\
& = 3(3q^3 + 3q + 1)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, $3q^3 + 3q + 1 \gt 1$ and is not a power of $3$, so \eqref{eq1A} cannot hold. This therefore proves what is requested, i.e., there's no integer $a$ such that $\sigma(a) = 3^k$ for $k \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$ $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
If we can prove that $\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$ is always greater than $0$ we can find the minima of the function. How can we prove that the term is always positive?
And how can we prove that $\displaystyle \int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$
| Hint $:$ For maxima observe that $x \mapsto \frac {x} {x+1}$ is increasing on $[0,1].$ For minima observe that $x \mapsto \frac {1} {x+1}$ is decreasing on $[0,1].$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that the product of the lengths of the perpendiculars drawn from two points to a straight line is $b^2$ Question:
Show that the product of the lengths of the perpendiculars drawn from the points $(\pm c,0)$ to the straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
My attempt:
Let, the length of the perpendicular on drawn upon $bx\cos\theta+ay\sin\theta-ab=0$ from $(c,0)$ is $d_1$ and the length of that drawn upon the same from $(-c,0)$ is $d_2$. Now,
$$d_1=\frac{|bc\cos\theta+0-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1=\frac{|bc\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
Again,
$$d_2=\frac{|-bc\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
Now,
$$d_1d_2=\frac{|-(bc\cos\theta-ab)(bc\cos\theta+ab)|}{{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2(1-\cos^2\theta)}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2-a^2\cos^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{a^2-(a^2-b^2)\cos^2\theta}$$
$$\implies d_1d_2=|\frac{b^2(a^2-c^2\cos^2\theta)}{a^2-(a^2-b^2)\cos^2\theta}|$$
[$a^2-(a^2-b^2)\cos^2\theta$ is positive, so bringing it inside the modulus sign doesn't make a difference]
$$\implies d_1d_2=|\frac{b^2(a^2-(a^2-b^2)\cos^2\theta)}{(a^2-(a^2-b^2)\cos^2\theta)}|$$
[Given, $a^2=b^2+c^2\implies c^2=a^2-b^2$]
$$d_1d_2=|b^2|$$(shown?)
The question asked me to show that $d_1d_2=b^2$, but I showed them that $d_1d_2=|b^2|$. So, is there something wrong with my process, or is there something wrong with the question?
| Comment
Btw, the product of the length of the perpendiculars drawn from the focal points of an ellipse $(\pm c,0)$ to the tangent straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
Can be also derived from Newton's ellipse canonical form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $ax+by = a^n + b^n$ then $\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$
Let a,b,n be positive integers such that $(a,b) = 1$.
Prove that if $(x,y)$ is a solution of the equation $ax+by = a^n + b^n$ then
$$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$$
So I was trying this question which is a Romania TST problem.
$$ax+by=a^n+b^n\implies a^n-ax=b^n-by\implies \frac{a^n-ax}{ab}=\frac{b^n-by}{ba}$$
$$\frac{a^{n-1}-x}{b}=\frac{b^{n-1}-y}{a}\implies \frac{a^{n-1}}{b}-\frac{x}{b}=\frac{b^{n-1}}{a}-\frac{y}{a}. $$
I don't think so we can proceed more.
We also have $a(a^n-ax)=b(b^n-y)\implies a^{n-1}\equiv x\mod b, b^{n-1}\equiv y\mod a.$
Any hints on how to proceed?
| We will prove John's final observation:
If $a, b$ are co-prime positive integers, and $ c, d, x, y$ are integers such that $ax+by = ac + bd$, then
$$\left\lfloor \frac{x}{b} \right\rfloor + \left\lfloor \frac{y}{a} \right\rfloor = \left\lfloor \frac{c}{b} \right\rfloor + \left\lfloor \frac{d}{a} \right\rfloor. $$
As observed by Teresa, the statement is true without the floor brackets, so with the floor brackets, we know that the values are very close.
Subtracting from $ \frac{x}{b} + \frac{y}{z} = \frac{c}{b} + \frac{d}{a}$, the observation is equivalent to demonstrating:
$$\left\{ \frac{x}{b} \right\} + \left\{ \frac{y}{a} \right\} = \left\{ \frac{c}{b} \right\} + \left\{ \frac{d}{a} \right\}. $$
Since $ a (x-c) = b (y-d)$ and $ \gcd(a, b) = 1$, so $ b \mid x - c$, and thus $\left\{ \frac{x}{b} \right\} = \left\{ \frac{c}{b} \right\}. $ In fact, it is equal to $\frac{ x \pmod{b} } { b}$.
Likewise, $ a \mid y-d$, and thus $\left\{ \frac{y}{a} \right\} = \left\{ \frac{d}{a} \right\}$.
Adding up these equalities, we get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.