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Value of $y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$ I was given this problem on series by a friend.
If
$$y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$$
then how to solve such problem.
I don’t want the full answer, rather, insights, mathematical facts, theorems, and relationships that would help me solve it on my own.
My efforts: I thought that the whole thing inside the square bracket must be a perfect square so we have [$4~+$ something] should be a positive perfect square but that would be like finding a trivial solution by trial and error method so I don't know how to solve it.
I also tried by squaring and checking like this
$$y^2 - 4=\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}$$
so we get two factors $y-2$ and $y+2$, but still it was like same trial and error method of finding factors . So can any one help.
Thanks in advance.
Edit: The only reasonable interpretation is the recurrence $y_n=\sqrt[n]{4+y_{n+1}}$ (Thanks @par)
| Too long for a comment, so
$\begin{array}\\
y
&=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac14\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac14\sqrt[3]{4}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4}\sqrt[5]{1+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{4^{1/20}}\sqrt[5]{1+...............}}}}\\
\end{array}
$
It looks like
there is a pattern of
$\dfrac1{4^{1/n-1/(n+1)}}
=\dfrac1{4^{1/(n(n+1))}}
$
which might make it easier
to get a more solvable recurrence.
And, of course,
$4$ can be replaced
by any value,
probably preferably a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
To prove the given inequality Question:-
If $a,b,c$ are positive real numbers which are in H.P. show that $$\dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b} \ge 4$$
Attempt at a solution:-
I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-
$$\dfrac{ \dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b}}{2} \ge \sqrt{\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b}}$$
Evaluating the right hand side of the inequality,
$$\left(\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b} \right) = \left( \dfrac{b^2+\left(a+c \right)b+ac}{b^2-2b\left(a+c \right)+4ac}\right)$$
Now, as $a,b,c$ are in H.P., hence we get the relation $$b=\dfrac{2ac}{a+c}$$ On substituting this result in the equation, we get
$$\dfrac{10 a^2c^2 + 3ac(a^2+c^2)}{4a^2c^2}=\dfrac{5}{2}+\dfrac{3(a^2+c^2)}{4ac}$$
Now from the above calculations, what I observed is that, that for the proof to be valid, the condition $\dfrac{3(a^2+c^2)}{4ac}=\dfrac{3}{2}$ should be satisfied. The final conclusion to which I arrived was $a=c$. Now although the numbers would still be in H.P. but it doesn't prove the result for the numbers which are in H.P. and are not equal.
Any other approach to the proof is invited too.
| So, why don't we rewrite the HP condition in a nicer way, as
$$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
and let's multiply the numerator and denominator of the first fraction by $2/b$. We get
$$ \frac{a+b}{2a-b} = \frac{a(1/a+1/c)+2}{2a(1/a+1/c)-2} = \frac{3+a/c}{2a/c}=\frac{3c+a}{2a}$$
Similarly, the other one becomes $\frac{3a+c}{2c}$, so what we want to show reduces to
$$\frac{3c+a}{2a} + \frac{3a+c}{2c}\geq 4$$
which reduces to
$$\frac{c}{a}+\frac{a}{c}\geq 2$$
which follows directly from AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve the system for the given parameter a \begin{align}
ax+y+z&=1,\\
2x+2ay+2z&=3\\
x+y+az&=1
\end{align}
I tried forming the system matrix and discuss it using its rank, but I'm not sure how to row reduce:
$$\begin{pmatrix}a&1&1&1\\ 2&2a&2&3\\ 1&1&a&1\end{pmatrix}$$
| Consider the matrix $$A=\begin{pmatrix}
a&1&1\\2&2a&2\\1&1&a
\end{pmatrix}.$$ We have that $\det(A)=2(a-1)^2(a+2)$. Hence if $a\neq 1$ and $a\neq -2$, then $A$ is invertible and the solution of the system of equations is $A^{-1}b$, where $b=\begin{pmatrix}1\\3\\1
\end{pmatrix}$.
If $a=1$, then the system of equations is false, hence there are no solutions. If $a=-2$, then again the system of equations has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Constant such that $\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$ What is the greatest constant $k>0$ such that
$$\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$$
for all $0\leq b\leq 1$ and $0\leq c\leq d\leq 1$?
The right-hand side looks like a weighted sum of the two terms on the left-hand side, but not quite. If we plug in $b=1$ and $c=d$, then all three terms are equal, so $k\leq 1$.
On the other hand, we have $k\geq 3/5$. Indeed, we will show that $$\frac{5}{5-3c}\geq\frac35\cdot\frac{2+3b}{5-c-2d}.$$
Note that $d\leq 1$ and $2+3b\leq 5$, so it suffices to show $$\frac{1}{5-3c}\geq\frac35\cdot\frac{1}{3-c},$$
or $$5(3-c)\geq 3(5-3c)$$
or $$15-5c\geq 15-9c$$
which is true. But the bound is not tight here, since we must have $b=d=1$ and $c=0$, and we have $\max(5, 5/2)\geq 5/3$. (The term $\frac{5b}{5-3d}$, which we did not use at all, is large.)
Update: By dividing the cases into whether $b\leq 3/5$ (and compare with the first term in the $\max$ if so) or $b\geq 3/5$ (and compare with the second term in the $\max$ if so), we can show that $k\geq 15/19$. Moreover, as Aravind pointed out in the comments, we have $k\leq 15/16$. So the gap is now between $15/19$ and $15/16$.
Update 2: WolframAlpha confirms that $k=15/16$ is the right answer. The question is now how to prove it:
http://www.wolframalpha.com/input/?i=find+minimum+of+max(5%2F(5-3c),(5b)%2F(5-3d))*(5-c-2d)%2F(2%2B3b)+for+0%3C%3Db%3C%3D1+and+0%3C%3Dc%3C%3Dd%3C%3D1
| The original problem is equivalent to finding the minimum value of
$$
\max (\frac{5}{5-3c}, \frac{5b}{5-3d}) \cdot \frac{5-c-2d}{2+3b}
$$
in the region $\{(b, c, d) \mid 0 \leq b \leq 1 \wedge 0 \leq c \leq d \leq 1\}$. We consider two cases below, namely,
*
*Case 1: $\frac{5}{5-3c} \leq \frac{5b}{5-3d}$;
*Case 2: $\frac{5}{5-3c} \geq \frac{5b}{5-3d}$.
Case 1: The problem can be stated as
\begin{align}
\text{minimize}\quad & f(b, c, d) = \frac{5-c-2d}{5-3d}\cdot \frac{5b}{2+3b} \\
\text{subject to}\quad & b \geq \frac{5-3d}{5-3c} \quad\quad (\text{by the condition of Case 1}) \\
& 0 \leq b \leq 1 \\
& 0 \leq c \leq d \leq 1
\end{align}
Note that the term $\frac{5b}{2+3b} = \frac{5}{3} - \frac{10}{6+9b}$ in $f(b, c, d)$ is a strictly increasing function of $b$. To minimize $f(b, c, d)$, $b$ should be minimized, for fixed $c$ and $d$. Therefore,
$$b = \frac{5-3d}{5-3c} \tag{1}$$
Substituting $b$ with (1) in $f$, we obtain that
$$
f(b, c, d) = \frac{25-5c-10d}{25-6c-9d} = \frac{5}{6} + \frac{\frac{25}{6} - \frac{15}{6}d}{25 - 6c - 9d}
$$
For fixed $d$, $f(b, c, d)$ is minimized when $c$ is minimized, i.e., $c = 0$. We further obtain that
$$
f(b, c, d) = \frac{5}{6} + \frac{\frac{25}{6} - \frac{15}{6}d}{25 - 9d} = \frac{5}{6} + \frac{5}{18} - \frac{\frac{50}{18}}{25-9d}
$$
Setting $d = 1$, we obtain the minimum value $\frac{5}{6} + \frac{5}{18} - \frac{50}{18\cdot16} = \frac{15}{16}$.
Case 2: The problem can be stated as:
\begin{align}
\text{minimize}\quad & g(b, c, d) = \frac{5-c-2d}{5-3c}\cdot \frac{5}{2+3b} \\
\text{subject to}\quad & b \leq \frac{5-3d}{5-3c} \quad\quad (\text{by the condition of Case 2}) \\
& 0 \leq b \leq 1 \\
& 0 \leq c \leq d \leq 1
\end{align}
and the solution is similar to that of Case 1. First, the term $\frac{5}{2+3b}$ is a strictly decreasing function of $b$. Thus, we set $b = \frac{5-3d}{5-3c}$ and obtain
$$
g(b, c, d) = \frac{25-5c-10d}{25-6c-9d}
$$
which is the same function as that in Case 1. So the minimum value is still $\frac{15}{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Find the value of $a$ for which v is in the set H = span
Find the value of $a$ for which $ v = \begin{bmatrix}10\\-6\\6\\a\end{bmatrix}$ is in the set $H = span \{ \begin{bmatrix}-5\\2\\-5\\-3\end{bmatrix}, \begin{bmatrix}0\\-2\\-1\\1\end{bmatrix}, \begin{bmatrix}0\\0\\-3\\-1\end{bmatrix}\}$
$a = ?$
I found:
$10 = -5x \implies x = -2$
$-6 = 2x - 2y \implies -4 - 2(1) \implies y = 1$
$6 = -5x - y -3z \implies -5(-2) - 1 - 3z \implies z = 1$
$a = 3x + y - z \implies a = -6+1-1 = -6$ but I keep getting the answer wrong, how is $a \ne -6$?
| Note that the augmented system
$$
\left[
\begin{array}{rrr|r}
-5 & 0 & 0 & 10 \\
2 & -2 & 0 & -6 \\
-5 & -1 & -3 & 6 \\
-3 & 1 & -1 & a
\end{array}
\right]
$$
can be row-reduced with the following steps:
*
*scale row 1 by $-1/5$
*add $-2$ times row 1 to row 2
*add $5$ times row 1 to row 3
*add $3$ times row 1 to row 4
*scale row 2 by $-1/2$
*add $1$ times row 2 to row 3
*add $-1$ times row 2 to row 4
*scale row 3 by $-1/3$
*add $1$ times row 3 to row 4
Performing these operations gives
$$
\DeclareMathOperator{rref}{rref}\rref
\left[\begin{array}{rrr|r}
-5 & 0 & 0 & 10 \\
2 & -2 & 0 & -6 \\
-5 & -1 & -3 & 6 \\
-3 & 1 & -1 & a
\end{array}\right]
=
\left[\begin{array}{rrr|r}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & a - 6
\end{array}\right]
$$
This shows that $v\in H$ if and only if $a=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= 16 + 8\cos\theta $
[Adding on both the sides]
=> $64\sin^{2}\theta + 64\cos^{2}\theta= 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $64 = 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $48 = 8\cos\theta + 65\cos^{2}\theta$
=> $48 = \cos\theta(65\cos\theta + 8)$
I can't figure out what to do next !
| $$8\sin(\theta)- \cos(\theta) =4$$
Dividing by $\sqrt{1^2 +8^2}$
$$\frac{8\sin(\theta)}{\sqrt{65}}- \frac{\cos(\theta)}{\sqrt{65}} =\frac{4}{\sqrt{65}}$$
Let $\sin(\alpha) = \frac{1}{\sqrt{65}}$, we $\cos(\alpha) = \frac{8}{\sqrt{65}}$.
Thus
$$\sin(\theta)\cos(\alpha)-\cos(\theta)\sin(\alpha) = \frac{4}{\sqrt{65}}$$
$$\sin(\theta -\alpha) = \frac{4}{\sqrt{65}}$$
$$\theta = \sin^{-1}(\frac{4}{\sqrt{65}}) + \alpha $$
$$= \sin^{-1}(\frac{4}{\sqrt{65}}) + \sin^{-1}(\frac{1}{\sqrt{65}})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| Another approach:
Let:
$$I(c)=\int \frac{x^2-2}{x^2+c} dx$$
$$=\int \frac{x^2+c-c-2}{x^2+c} dx$$
The above step is can be avoided by long division, if it you see it as coming out of the blue. Anyways continuing:
$$=\int \left(1-\frac{c+2}{x^2+c} \right) dx$$
$$=x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1$$
Whet $c$ and $C_1$ are two different things. $C$ here represents the constant gained when integrating. Now find
$$\frac{I''(c)}{2}+C_2=\frac{1}{2} \frac{\partial^2}{\partial c^2} \left(x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1 \right)+C_2 =\int \frac{x^2-2}{(x^2+c)^3} dx$$
Where $C_2$ is another constant gained when integrating. You can more specifically find:
$$\frac{I''(2)}{2}+C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Circle Geometry Question 1) In triangle $ABC$, $AB = 10$, $AC = 8$, and $BC = 6$. Let $P$ be the point on the circumcircle of triangle $ABC$ so that $\angle PCA = 45^\circ$. Find $CP$.
Diagram(1)
2) Let $B$, $C$, and $D$ be points on a circle. Let $\overline{BC}$ and the tangent to the circle at $D$ intersect at $A$. If $AB = 4$, $AD = 8$, and $\overline{AC} \perp \overline{AD}$, then find $CD$. Diagram(2)
I think both of them have to do with power of a point
| 1) Since the angle $\angle PCA = 45^\circ$ and the $\triangle ACB$ is clearly Pythagorean (double of the minimal of sides $3,4,5$) the diameter of the circle is equal to
$\overline{AB}$ and because the angle $\angle APB$ subtends an arc of $90^\circ$ the quadrilatere $ACBP$ is a rectangle. Hence $$\overline{CP}=\overline{AB}=\color{red}{10}$$
2) By the power of the point $A$ respect to the given circle we have $$4(4+x)=8^2\iff x=12$$ It follows
$$\overline{CD}=\sqrt{8^2+16^2}=\color{red}{8\sqrt 5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
| Integrals of this sort can be done using some useful identities that are immediate corollaries of the angle-sum formulas $\cos A \cos B-\sin A \sin B=\cos (A+B)$ and $\sin A \cos B+\cos A \sin B=\sin (A+B),$ in conjunction with $\sin (-A)=-\sin A$ and $\cos (-A)=\cos A.$
We have $\cos (A+B)=\cos A \cos B -\sin A \sin B\;, $ and $\; \cos (A-B)=\cos A \cos B+\sin A \sin B.$ Adding these , we have $$\cos (A+B)+\cos (A-B)=2\cos A \cos B.$$ Subtracting them , we have $$\cos (A+B)-\cos (A-B)=-\sin A \sin B.$$ Similarly from the formulas for $\sin (A\pm B)$ we have $$\sin (A+B)+\sin (A-B)=2\sin A \cos B$$ and $$\sin (A+B)-\sin (A-B)=2\cos A \sin B.$$ From the angle-sum formulas, with $A=B$ and from $\cos^2A+\sin^2A=1,$ we have $$\sin^2 A=(1-\cos 2A)/2$$ and $$\cos^2 A=(1+\cos 2A)/2.$$ In particular $\sin^2 A \cos 4A=\frac {1}{2}(1-\cos 2A)\cos 4A=\frac {1}{2}(\cos 4A-\cos 2A\cos 4A)=\frac {1}{2}(\cos 4A-\frac {1}{2}(\cos 6A +\cos 2A)).$
For a lot more of this, see Trigonometry, by Hobson. (Dover Press re-print.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Solving a system of linear congruences
Find all positive integer solutions to \begin{align*}x &\equiv -1 \pmod{n} \\ x&\equiv 1 \pmod{n-1}. \end{align*}
I rewrote the system as $x = nk_1-1$ and $x = (n-1)k_2+1$. Thus, we have $nk_1-1 = (n-1)k_2+1$ and so $n(k_1-k_2) = 2-k_2 \implies n = \frac{2-k_2}{k_1-k_2}$. How do I solve it from here?
| As $n,n-1$ are relatively prime the Chinese Remainder Theorem guarantees a unique solution $\pmod {n(n-1)}$ . Indeed we have, $$x\equiv 2n-1 \pmod {n(n-1)}$$
To see this, solve the first congruence to get $x=-1+mn$. Substituting that into the second gives $$-1+mn\equiv 1 \pmod {n-1}\implies mn\equiv 2 \pmod {n-1}\implies m\equiv 2 \pmod {n-1}$$
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Problem calculating the sine of a matrix Given the matrix $A=\begin{pmatrix}-\frac{3\pi}{4} & \frac{\pi}{2}\\\frac{\pi}{2}&0\end{pmatrix}$, I want to calculate the sine $\sin(A)$.
I do so by diagonalizing A and plugging it in the power series of the sine:
\begin{align}
\sin (A) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} A^{2k+1}.
\end{align}
The diagonalization leads to:
\begin{align}
A = \frac{1}{5}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\end{align}
and thus:
\begin{align}
A^n = \frac{1}{5}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}^n
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}.
\end{align}
Hence:
\begin{align}
\sin (A) &= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}\sin(-\pi) & 0\\0&\sin(\frac{\pi}{4})\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\
&= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}0 & 0\\0&\frac{1}{\sqrt{2}})\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\
&= \frac{1}{5}\begin{pmatrix}\frac{1}{\sqrt{2}} & \sqrt{2}\\\sqrt{2}&2\sqrt{2}\end{pmatrix},
\end{align}
which differs from "Wolfram Alpha's result"
\begin{align}
\sin(A) &= \begin{pmatrix}-\frac{1}{\sqrt{2}} & 1\\ 1 & 0 \end{pmatrix} .
\end{align}
How can this happen?
| The problem is that Wolfram Alpha interprets "sin(A)" for a matrix A (or array of however many dimensions, or list of list of lists, or what have you) as meaning simply the result of applying sin component-wise.
This is not what you intended, and you did your intended calculation perfectly fine.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Number of ways of getting same number when four people throw a die once Four people are rolling a die once. How many ways
*
*None of them get same number
*Exactly two of them get same number
*Two of them get the same number
*Three of them get same number
*All of them get same number
Solution:
*
*None of them get same number$= 6 \cdot 5 \cdot 4 \cdot 3= 360$ ways
*Exactly two of them get same numbers$= 4C2 \cdot 6 \cdot 5 \cdot 4= 720$ ways
*Two of them same numbers$= 4C2 \cdot 6 \cdot 5 \cdot 5= 900$ ways
*All of them get same number$= 4C4 \cdot 6 \cdot 1 \cdot 1= 6$ ways
Is my approach correct? Can anyone clarify in detail?
| Your answers to the first two questions are correct, as is your answer to the last question.
If we interpret the third question to mean at least two of them get the same number, then we can compute the answer by subtracting the number of cases in which none of them get the same number from the total number of outcomes. Since there are six possible outcomes for each of the four times the die is thrown, the total number of possible outcomes is $6^4$. You found that the number of outcomes in which no two people get the same number is $6 \cdot 5 \cdot 4 \cdot 3$. Hence, the number of outcomes in which at least two people get the same number is
$$6^4 - 6 \cdot 5 \cdot 4 \cdot 3 = 1296 - 360 = 936$$
We can verify this by considering the cases in which at least two people obtain the same number. Since the partitions of $4$ are
\begin{align*}
4 & = 4\\
& = 3 + 1\\
& = 2 + 2\\
& = 2 + 1 + 1\\
& = 1 + 1 + 1 + 1
\end{align*}
the set of possible cases is
*
*no two people get the same number
*exactly two people get the same number
*two distinct numbers occur, with two people apiece getting each outcome
*exactly three people get the same number
*all four people get the same number
You correctly calculated that the number of cases in which exactly two people get the same number is
$$\binom{6}{2} \cdot 6 \cdot 5 \cdot 4$$
The number of cases in which two distinct numbers occur, with two people apiece getting each outcome is
$$\binom{6}{2}\binom{4}{2} = 90$$
since there are $\binom{6}{2}$ ways of selecting two numbers and $\binom{4}{2}$ ways for exactly two of the four people to get the larger of those numbers.
The number of cases in which exactly three people get the same number is
$$\binom{4}{3} \cdot 6 \cdot 5 = 120$$
since there are $\binom{4}{3}$ ways for exactly three of the people to get the same number, six choices for the repeated number, and five choices for the remaining number.
Since there are six possible outcomes in which all four people get the same number, the number of outcomes in which at least two people get the same number is
$$\binom{4}{2} \cdot 6 \cdot 5 \cdot 4 + \binom{6}{2}\binom{4}{2} + \binom{4}{3} \cdot 6 \cdot 5 + \binom{4}{4} \cdot 6 = 720 + 90 + 120 + 6 = 936$$
as we found above.
If we interpret the fourth question to mean at least three people get the same number, we can find the answer by adding the cases in which exactly three people get the same number and exactly four people get the same number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I find the derivative of $(1 +1/x)^x $ I tried one approach but the correction in the book shows me a total different answer.
Here's what I did:
$(1+ 1/x)^x=xln(1+1/x)$
Thus, now we try to find the derivative of a multiplication:
$ u(x)=x$
$(u(x))'=1$
$v(x)=ln(1+1/x)$
$(v(x))'= -1/(x^2) +1/x$
And so:
$(uv)'=u'v+uv'$
which gives us:
$(uv)'=xln(1 +1/x) +(-1/x^2 -1/x)$
Yet, the correction gives me this as an answer:
$ln(1+1/x)-1/(1+x)$
| Let $f(x) = (1+1/x)^x$. Then $$\ln(f(x)) = x \ln(1 + 1/x) = x \ln( (x+1)/x) = x \ln(x+1) - x \ln(x).$$
Taking derivatives on both sides gives:
$$\frac{f'(x)}{f(x)} = \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1.$$
We wish to find $f'(x)$, so we solve for it and subsitute the value of $f(x)$.
$$f'(x) = \left(1 + \frac{1}{x} \right)^x \left[ \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1 \right]=\left(1 + \frac{1}{x} \right)^x \left[ \ln\left( 1 + \frac{1}{x} \right) - \frac{1}{x+1} \right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $S$, $D$, and $S^{-1}$ such that $A = SDS^{-1}$
Let $A = \begin{bmatrix}18&12\\-40&-26\end{bmatrix}$Find $S$, $D$, and $S^{-1}$ such that $A = SDS^{-1}$
So I did $\det(A-\lambda I)$ to get the char. poly. eqn. and got eigenvalues $\lambda_1 = -6, \lambda_2 = -2$ then:
I did $(A - \lambda_1 I)$ to get $\begin{bmatrix}1&1/2\\0&0\end{bmatrix}$ and similarly $(A - \lambda_2 I)$ to get $\begin{bmatrix}1&3/5\\0&0\end{bmatrix}$
so I got $S = \begin{bmatrix}1/2&3/5\\1&1\end{bmatrix}$, $D = \begin{bmatrix}-6&0\\0&-2\end{bmatrix}$, and $S^{-1} = \begin{bmatrix}-10&6\\10&-5\end{bmatrix}$ but my answer is incorrect for some reason
| The characteristic polynomial of $A=\left[\begin{array}{rr}
18 & 12 \\
-40 & -26
\end{array}\right]$ is
$$
\chi_A(t)=\det(tI-A)=\det\left[\begin{array}{rr}
t - 18 & -12 \\
40 & t + 26
\end{array}\right]=t^{2} + 8t + 12=(t + 2) \cdot (t + 6)
$$
The eigenvalues of $A$ are thus $\lambda=-2$ and $\mu=-6$.
To compute a basis of $E_\lambda$, note that
$$
E_\lambda=\DeclareMathOperator{Null}{Null}\Null(\lambda I-A)
=\Null\left[\begin{array}{rr}
-20 & -12 \\
40 & 24
\end{array}\right]\overset{\circledast}{=}\Null\left[\begin{array}{rr}
1 & \frac{3}{5} \\
0 & 0
\end{array}\right]=\DeclareMathOperator{Span}{Span}\Span\left\{
\left[\begin{array}{rr}
3 \\ -5
\end{array}\right]
\right\}
$$
where the equality marked $\circledast$ comes from row-reducing.
So, one of the columns of $S$ should be a nonzero scalar multiple of $\left[\begin{array}{rr}
3 \\ -5
\end{array}\right]
$.
To compute a basis of $E_\mu$, note that
$$
E_\mu=\Null(\mu I-A)=\Null
\left[\begin{array}{rr}
-24 & -12 \\
40 & 20
\end{array}\right]\overset{\circledast}{=}
\Null
\left[\begin{array}{rr}
1 & \frac{1}{2} \\
0 & 0
\end{array}\right]=\Span\left\{
\left[\begin{array}{rr}
1 \\ -2
\end{array}\right]\right\}
$$
So the other column of $S$ should be a nonzero scalar multiple of $\left[\begin{array}{rr}
1 \\ -2
\end{array}\right]$.
Putting this together, we have $A=SDS^{-1}$ where
\begin{align*}
D &=
\left[\begin{array}{rr}
\color{red}{-2} & 0 \\
0 & \color{blue}{-6}
\end{array}\right] &
S &=
\left[\begin{array}{rr}
\color{red}{3} & \color{blue}{1}\\ \color{red}{-5} & \color{blue}{-2}
\end{array}\right]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-sinxdx$
$$-\int \frac{(1-u^2 )u^2}{2-u^2}du=\int \frac{u^4-u^2 }{2-u^2}du$$
Polynomial division give us
$$\int-u^2-1+\frac{2}{u^2-2}du=-\frac{u^3}{3}-u=2\int \frac{1}{u^2-2}du$$
Using partial fraction we get to:
$$\frac{1}{u^2-2}=\frac{1}{(u+\sqrt{2})(u^2-\sqrt{2})}=\frac{A}{u^2+\sqrt{2}}+\frac{B}{u^2-\sqrt{2}}$$
So we get to:
$$1=(A+B)u^2+\sqrt{2}(A-B)$$
So $A=B$ but $0*\sqrt{2}\neq 1$, where did it go wrong?
| Observe that, $$\displaystyle\int\dfrac{\sin^3 x\cos^2 x}{1+\sin^2 x}dx=\displaystyle\int\dfrac{\sin^3 x}{1+\sin^2 x}dx-\displaystyle\int\dfrac{\sin^5 x}{1+\sin^2 x}dx$$
Transformation of $\displaystyle\int\dfrac{\sin^3 x}{1+\sin^2 x}dx$
Observe that $\cos^2 x=z\implies \sin x\ dx=-\dfrac{dz}{2}$. Then the integral becomes, $$\displaystyle\int\dfrac{\sin^3 x}{1+\sin^2 x}dx=-\dfrac{1}{2}\displaystyle\int\dfrac{1-z}{2-z}dz$$
Transformation of $\displaystyle\int\dfrac{\sin^5 x}{1+\sin^2 x}dx$
Observe that $\cos^2 x=z\implies \sin x\ dx=-\dfrac{dz}{2}$. Then the integral becomes, $$\displaystyle\int\dfrac{\sin^5 x}{1+\sin^2 x}dx=-\dfrac{1}{2}\displaystyle\int\dfrac{(1-z)^2}{2-z}dz$$
I hope you can evaluate both the integrals easily.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How would I calculate points on the Peano curve? The Peano curve is often given as an example of a space filling curve which maps the unit line to the unit square. So, it is a function of the form $[0,1] \rightarrow [0,1]^2$? In which case can I evaluate it for given numbers like 0.2, 0.5 and 0.7? How would I calculate those points?
| The approach I would take is to define the curve construction by a set of affine similarity transformations, each corresponding to a mapping from the unit square to part of the curve. Write the length parameter fraction in base(number of transformations) to index a sequential compositions of transformations, and take the limit to find the corresponding point in the unit square.
Using the Hilbert curve as an example because it has fewer and simpler transformations (the same approach applies to the Peano curve with $9$ transformations):
$$
\begin{aligned}
f_0 \begin{pmatrix}x \\ y\end{pmatrix} &= \begin{pmatrix}0 & \frac{1}{2} \\ \frac{1}{2} & 0\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\
f_1 \begin{pmatrix}x \\ y\end{pmatrix} &= \begin{pmatrix}\frac{1}{2} & 0\\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix}0 \\ \frac{1}{2}\end{pmatrix}\\
f_2 \begin{pmatrix}x \\ y\end{pmatrix} &= \begin{pmatrix}\frac{1}{2} & 0\\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix}\frac{1}{2} \\ \frac{1}{2}\end{pmatrix}\\
f_3 \begin{pmatrix}x \\ y\end{pmatrix} &= \begin{pmatrix}0 & -\frac{1}{2}\\ -\frac{1}{2} & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix}1 \\ \frac{1}{2}\end{pmatrix}\\
\end{aligned}
$$
Weight the transformations equally across the length parameter, because each maps to the same area. Now there are $4$ transformations, so write your parameter fraction in base $4$:
$$l = 0 . l_1 l_2 l_3 l_4 \ldots \text{ where } l_i \in \{0,1,2,3\}$$
which corresponds to a composition of transfomations:
$$f = f_{l_1} \circ f_{l_2} \circ f_{l_3} \circ f_{l_4} \circ \ldots$$
This infinite limit can be evaluated for rational length parameters using fixed point arguments, for example:
$$l = \frac{7}{12} = 0.2\overline{1}_4 \\ p_1 = f_1 (p_1) = \begin{pmatrix}0 \\ 1\end{pmatrix} \\ p = f_2 (p_1) = \begin{pmatrix}\frac{1}{2} \\ 1 \end{pmatrix}$$
For irrational length parameters, I don't know how to find the limit.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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On real part of the complex number $(1+i)z^2$
Find the set of points belonging to the coordinate plane $xy$, for which the real part of the complex number $(1+i)z^2$ is positive.
My solution:-
Lets start with letting $z=r\cdot e^{i\theta}$. Then the expression $(1+i)z^2$ becomes $$\large\sqrt2\cdot|z|^2\cdot e^{{i}\left(2\theta+\dfrac{\pi}{4}\right)}$$
Now, as $\sqrt2\cdot|z|^2\gt0$, so $\Re{((1+i)z^2)}\gt 0 \implies\cos{\left(2\theta+\dfrac{\pi}{4}\right)}\gt 0$. So, we get
$$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}
\implies-\dfrac{3\pi}{4}\lt2\theta\lt\dfrac{\pi}{4}
\implies-\dfrac{3\pi}{8}\lt\theta\lt\dfrac{\pi}{8}$$
Now, lets find the equation of the lines which would help us show these inequalities in the coordinate plane.
The inequality can be represented by
$$\begin{equation}
y\lt \tan{\dfrac{\pi}{8}}x\implies y\lt(\sqrt2-1)x \tag{1}
\end{equation}$$
$$\begin{equation}
y\gt \tan{(-\dfrac{3\pi}{8})}x \implies y\gt-(\sqrt2+1)x \tag{2}
\end{equation}$$
So, the inequality can be represented in the coordinate plane as in the following portion of the graph with the cross-hatched part.
My deal with the question:-
The book I am solving gives the answer as the (cross-hatched part + un-hatched part), so what am I missing in my solution. And, as always more elegant solutions are welcome.
| I would find the general solutions of the inequation first:
\begin{align*}
\cos\Bigl(2\theta+\frac\pi4\Bigr)>0&\iff-\frac\pi2<2\theta+\frac\pi4<\frac\pi2\iff-\frac{3\pi}4<2\theta<\frac\pi4\color{red}{\mod2\pi}\\
&\iff-\frac{3\pi}8<\theta<\frac\pi8\color{red}{\mod\pi}
\end{align*}
Now that if conventionally, we choose $\;-\pi<\theta\le\pi$,
we obtain
\begin{alignat*}{2}&\bullet\quad&-\dfrac{3\pi}8&<\theta&&<\dfrac\pi8,\\
&\bullet\quad&-\pi&<\theta&&<-\dfrac{7\pi}8,\\
&\bullet\quad&\dfrac{5\pi}8&<\theta&&<\pi.
\end{alignat*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find Area Enclosed by Curve I want to find the area enclosed by the plane curve $x^{2/3}+y^{2/3}=1$. My attempt was to set $x=\cos^3t, \ y=\sin^3t$ so:$$x^{2/3}+y^{2/3}=\cos^2t+\sin^2t=1$$
Then the area is $$2A=\oint_Cxdy-ydx=3\oint_C\cos^3ty'dy+\sin^3tx'dx=3\int_0^{2\pi}\cos^2t\cdot \sin^2tdt=\frac{3\pi}{4}\implies A=\frac{3\pi}{8}$$
However, when I did a level curve plot I got the following figure:
so does "area enclosed by figure" even make sense? For the graph above, my calculator gives me $A=\frac{3\pi}{32}$.
| In simple cartesian coordinates, through the substitution $x=z^{3/2}$ and Euler's Beta function,
$$ A = \int_{0}^{1}(1-x^{2/3})^{3/2}\,dx = \frac{3}{2}\int_{0}^{1}z^{1/2}(1-z)^{3/2}\,dz=\frac{3}{2}\,B\left(\frac{3}{2},\frac{5}{2}\right)\tag{1}$$
hence:
$$ A = \frac{3\,\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{5}{2}\right)}{2\,\Gamma(4)}=\color{red}{\frac{3\pi}{32}}\tag{2}$$
as wanted.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Series question involving a cubic polynomial The question asks: Consider the polynomial
$\displaystyle{\,\mathrm{f}\left(X\right) = X^{3} -6X^{2} + mX - 6}$, where $m$ is a real parameter.
a. Show that:
$\displaystyle{{1 \over x_{1}x_{2}} + {1 \over x_{1}x_{3}} + {1 \over x_{2}x_{3}} = 1}$ where $\displaystyle{x1,x2,x3}$ are the roots of the polynomial $\,\mathrm{f}$ .
b. Determine the parameter $m$ such that the roots of polynomial $f$ are three consecutive integer numbers
For part a., I have attempted to solve it as follows:
$\displaystyle{\,\mathrm{f} = X\left(X^{2} - 6X + m\right) - 6 = 0}$
$(X-6)(X^2-6X+m)=0$
$x1=6$
$X^2-6X+m=0$
Solving quadratically, $x2=3+\sqrt{36-4m}$,
$x3=3-\sqrt{36-4m}$
Then I substituted these roots into the equation to be proved. However, I get into really messy workings, ending up with the wrong answer.
Could someone please guide me on solving both part (a) and (b)? It would be highly appreciated!
| Let a,b,c the roots. Note that $\frac{1}{ab}+ \frac{1}{ac}+ \frac{1}{bc}= \frac{a+b+c}{abc}=*$.
Now, $f=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$.
And then $*=6/6=1$.
For b, note that $m=ab+bc+ca$
If (suposse) $n, n+1, n+2$ are the roots, their sum is 6 by a), and then $3n+3=6\to n=1$.
$m=1*2+2*3+3*1=11$
| {
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"source": "stackexchange",
"question_score": "1",
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Prove that $a_n$ is a perfect square
*
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
$$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
*Prove that $a_{n}$ is a perfect square for every $n$.
We have $a_{3} = 4, a_{4} = 25, a_{5} = 169,\ldots$
Is there a way we can simplify the recursion or get its closed form in order to get that it is a perfect square ?.
| A brute-force approach
$$b_{n} = 3b_{n-1} - b_{n-2}$$
Squaring on both sides,
$$b_n^2 = 9b_{n-1}^2+b_{n-2}^2 -6b_{n-1}b_{n-2} $$
Similarly,
$$b_{n-1}^2 = 9b_{n-2}^2+b_{n-3}^2 -6b_{n-2}b_{n-3} $$
Subtracting,
$$b_n^2 - b_{n-1}^2 = 9b_{n-1}^2 - 8b_{n-2}^2 - b_{n-3}^2 -6a_{n-2}(b_{n-1}-b_{n-1}) $$
Since,
$$3b_{n-2} = b_{n-1} + b_{n-3}$$
$$b_n^2 - b_{n-1}^2 = 9b_{n-1}^2 - 8b_{n-2}^2 - b_{n-3}^2 -2(b_{n-1} + b_{n-3})(b_{n-1}-b_{n-3}) $$
$$b_n^2 - b_{n-1}^2 = 9b_{n-1}^2 - 8b_{n-2}^2 - b_{n-3}^2 -2(b_{n-1}^2 - b_{n-3}^2) $$
$$b_n^2 = 8(b_{n-1}^2 - b_{n-2}^2) + b_{n-3}^2 $$
Summing up both sides from 3 to n,
$$\sum_{i=3}^{n}(b_i^2-b_{i-3}^2)=8\sum_{i=3}^{n}(b_{i-1}^2 - b_{i-2}^2)$$
$$\sum_{i=3}^{n}b_i^2-\sum_{j=0}^{n-3}b_{j}^2=8(\sum_{k=2}^{n-1}b_{k}^2 - \sum_{l=1}^{n-2}b_{l}^2)$$
$$b_n^2 + b_{n-1}^2 +b_{n-2}^2 - b_{2}^2 -b_{1}^2-b_{0}^2= 8(b_{n-1}^2 - b_1^2) $$
$$b_n^2 = 7b_{n-1}^2 - b_{n-2}^2 -2 $$
$\therefore a_n=b_n^2$ satisfies the recurrence.
Hence,proved
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an equation involving an integral: $\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$ Determine a pair of number $a$ and $b$ for which
$$\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$$
I tried putting $x$ as $1-x$ as the integral wouldn't change but could not move forward from there so can you please suggest me what should I do next.
| Hint. One may observe that
$$
(x^2+3x+2)=(x+1)(x+2)
$$ leading to the following partial fraction decomposition
$$
\begin{align}
\frac{ax+b}{(x^2+3x+2)^2}&=\frac{ax+b}{(x+1)^2(x+2)^2}
\\\\&=\frac{-a+b}{(x+1)^2}+\frac{3 a-2 b}{x+1}+\frac{-2 a+b}{(x+2)^2}+\frac{-3 a+2 b}{x+2}.
\end{align}
$$ Integrating each term gives
$$
\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx= \frac{-5a+4b}6+(3a-2b)(2\ln 2-\ln 3)
$$
Can you take it from here?
Edit. Partial fraction decomposition.
$$
\begin{align}
\frac{ax+b}{(x+1)^2(x+2)^2}&=\frac{A_1}{(x+1)^2}+\frac{B_1}{x+1}+\frac{A_2}{(x+2)^2}+\frac{B_2}{x+2}.
\end{align}
$$ By multiplying throughout by $(x+1)^2$ One has
$$
\begin{align}
\frac{ax+b}{(x+2)^2}&=A_1+(x+1)B_1+(x+1)^2\left(\frac{A_2}{(x+2)^2}+\frac{B_2}{x+2}\right).
\end{align}
$$ Putting $x=-1$ gives $A_1=-a+b$. Similarly by multiplying throughout by $(x+2)^2$ putting $x=-2$ gives $A_2=-2a+b$. Then by multiplying throughout by $(x+1)$ letting $x \to \infty$ gives $B_1+B_2=0$, finally $x=0$ gives $B_1=3a-2b=-B_2$. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Legendre symbol simplification I saw a simplification using the Legendre symbol which said $$\left(\dfrac{19}{29}\right) = \left(\dfrac{10}{19}\right) = \left(\dfrac{2}{19}\right) \cdot \left(\dfrac{5}{19}\right) = -1.$$ My question is how did they get $\left(\dfrac{19}{29}\right) = \left(\dfrac{10}{19}\right)$ and finally that $\left(\dfrac{2}{19}\right) \cdot \left(\dfrac{5}{19}\right) = -1$?
| For the first question
$$
\left(\dfrac{19}{29}\right) = \left(\dfrac{29}{19}\right)
= \left(\dfrac{10}{19}\right)
$$
since $29$ is $1$ mod $4$.
For the second,
$$
\left(\dfrac{5}{19}\right) =
\left(\dfrac{19}{5}\right) =
\left(\dfrac{-1}{5}\right) = 1
$$
since $5$ is $1$ mod $4$
and
$$
\left(\dfrac{2}{19}\right) = -1
$$
because $19$ is $3$ mod $8$. (http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-quadratic-character-of-2)
| {
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Show that $a_n$ is decreasing
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.
First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_n) - 1}{3 - a_n} = \frac{8 - 3a_{n}}{3 - a_n} > 0$, which means showing $8 > 3a_{n}$, but I'm having trouble showing it.
| Approach $\boldsymbol{1}$:
Note that
$$
\begin{align}
a_{n+1}-a_n
&=\frac1{3-a_n}-a_n\\
&=\frac{a_n^2-3a_n+1}{3-a_n}\\
&=\frac{\left(a_n-\frac32\right)^2-\frac54}{3-a_n}\tag{1}
\end{align}
$$
If $\frac{3-\sqrt5}2\le a_n\le\frac{3+\sqrt5}2$, then $(1)$ is negative and
$$
\frac{3-\sqrt5}2\le\frac1{3-a_n}\le\frac{3+\sqrt5}2\tag{2}
$$
Furthermore, $\frac{3-\sqrt5}2\le2\le\frac{3+\sqrt5}2$.
Approach $\boldsymbol{2}$:
Note that $\frac1{3-x}$ is an increasing function of $x$ on $(0,3)$. Furthermore, if $a_n\lt3$, then $\frac1{3-a_n}\gt0$.
Since
$$
a_1=2\gt a_2=1\tag{3}
$$
by repeatedly applying $\frac1{3-x}$ to $(3)$, we get that
$$
a_n\gt a_{n+1}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
| Let $n\ge5$ and suppose $$ \sum_{k=3}^{n+2}k^n<(n+3)^n.$$Then $$ \sum_{k=3}^{n+3} k^n <2(n+3)^n,$$ and we deduce \begin{align} \sum_{k=3}^{n+3} k^{n+1}<2(n+3)^{n+1}&<(n+4)^{n+1} \\ \left(1+\frac1{n+3}\right)^{n+1}\ge\left(\frac98\right)^6&>2, \end{align} where he have used the increasingness of $f(x)=\left(1+\frac1{x+3}\right)^{x+1}$ for $x>-3$: $f'(x)>0$ follows from $$ (x^2+6x+12)\log\left(1+\frac1{x+3}\right)>(x+3)^2((x+3)^{-1}-(x+3)^{-2})=x+2>x+1.$$
By induction, LHS < RHS for all $n\ge5$, and as you noted, it also holds for $n=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to factorize the polynomial $a^6+8a^3+27$?
I would like to factorize $a^6+8a^3+27$.
I got different answers but one of the answers is
$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$
Can someone tell me how to get this answer? Thanks.
| $$a^6+8a^3+27= (a^2)^3+3^3+(-a)^3-3\cdot a^2\cdot3(-a)$$
Now $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
See : If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Integration by means of partial fraction decomposition I'm trying to solve this indefinite integral by means of partial fraction decomposition:
$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.
The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the integrand function in this form:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax+B}{\left(x^2+4x+5\right)}+\dfrac{Cx+D}{\left(x^2+4x+5\right)^2}$.
I get:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{\left(Ax+B\right)\left(x^2+4x+5\right)+Cx+D}{\left(x^2+4x+5\right)^2}$.
Multiplying the right term:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax^3+4Ax^2+5Ax+Bx^2+4Bx+5B+Cx+D}{\left(x^2+4x+5\right)^2}$.
Now i collect the terms with the same pwer of $x$:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D+ 5B}{\left(x^2+4x+5\right)^2}$.
Now i equate the two numerators:
$x+1=Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D$
and equate term by term: i get:
$A=0$, $B=0$, $C=1$, $D=1$.
With these values i get a correct identity:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{x+1}{\left(x^2+4x+5\right)^2}$
but this is unuseful in order to solve the integral.
Where is my mistake ?
| As suggested in the comment directly use Trigonometric substitution
As $x^2+4x+5=(x+2)^2+1^2,$ let $\arctan(x+2)=y\implies x+2=\tan y$
$$2I\int\dfrac{x+1}{(x^2+4x+5)^2}dx=\int(\tan y-1)\cos^2y\ dy$$
$$4I=2\int(\sin2y-1-\cos2y)dy=-\sin2y-2y-\cos2y+K$$
Now $\sin2y=\dfrac{2\tan y}{1+\tan^2y}=\cdots$
and $\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac2{1+\tan^2y}-1=\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac{\pi}{12} $
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$
$$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$
Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$
How to solve this ?
| We have for all $x >0$, $\tan^{-1} x + \tan^{-1} \frac1{x} = \pi/2$ (hint: take the derivative of LHS). Hence, the obtained expression is just:
$$\frac{\pi}2 - \tan^{-1} \sqrt 3 = \frac{\pi}2 - \frac{\pi}3 = \frac{\pi}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
how can I prove this $\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$ How can I prove the following equation?:
$$s=\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
Simplifying both terms of the equation:
$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k} = \sum_{k=1}^{n}\frac{1}{(2k-1)2k} =(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})$$
$$\sum_{k=n+1}^{2n}\frac{1}{k}=(\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k})=(\sum_{k=1}^{n}\frac{1}{2k-1}+\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k})$$
Now we have:
$$s=(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})=(\sum_{k=1}^{n}\frac{1}{2k-1}\color{blue}{+}\sum_{k=1}^{n}\frac{1}{2k}\color{blue}{-\sum_{k=1}^{n}\frac{1}{k}})$$
How can I continue?
| The core part of the induction argument spelled out more explicitly:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{2i(2i-1)}&= \sum_{i=1}^k\frac{1}{2i(2i-1)}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+1}^{2k}\frac{1}{i}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+2}^{2k+2}\frac{1}{i}+\frac{1}{k+1}-\frac{1}{2k+1}-\frac{1}{2k+2}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+2}^{2k+2}\frac{1}{i}+\frac{2(2k+1)-2(k+1)-(2k+1)}{2(k+1)(2k+1)}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+2}^{2k+2}\frac{1}{i}-\frac{1}{2(k+1)(2k+1)}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+2}^{2k+2}\frac{1}{i}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\tan^{-1}x$, $\tan^{-1}y$, $\tan^{-1}z$ are in arithmetic progression, as are $x$, $y$, $z$. Show ...
$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:
*
*$x$, $y$, $z$ are in geometric progression
*$x$, $y$, $z$ are in harmonic progression.
*$x=y=z$
*$(x-y)^2 +(y-z)^2+(z-x)^2 =0$
My attempt:
$$\text{A}=\tan^{-1}x \qquad \text{B}=\tan^{-1}y \qquad \text{C}=\tan^{-1}z$$
$$x=\tan A \qquad y=\tan B \qquad z=\tan C $$
$$x+z=2y$$
$$A+C=2B$$
$$\tan(A + B + C)=\frac{\tan A +\tan B +\tan C - \tan A\tan B\tan C }{1-\tan A\tan B -\tan B\tan C -\tan C\tan A}$$
$$\tan(3B)=\frac{x +y +z - xyz }{1-xy -yz -zx}$$
How do I continue from here?
| HINT:
$$\tan^{-1}z-\tan^{-1}y=\tan^{-1}y-\tan^{-1}x$$
$$\implies\tan(\tan^{-1}z-\tan^{-1}y)=\tan(\tan^{-1}y-\tan^{-1}x)$$
$$\iff\dfrac{z-y}{1+yz}=\dfrac{y-x}{1+xy}$$
As $x,y,z$ are in A.P.,$y-x=z-y$
Case$\#1:$If If $y-x=z-y=0$
Case$\#2:$ If $y-x=z-y\ne0, 1+yz=1+xy\iff y(z-x)=0$
$\implies y=0$ or $z-x=0\iff z=x$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inequality with triangle sides Let $a,b,c$ be the sides of a triangle. Show that:
$$(\sqrt a + \sqrt b - \sqrt c)(\sqrt a - \sqrt b + \sqrt c)(-\sqrt a + \sqrt b + \sqrt c) \ge \sqrt {(a+b-c)(a-b+c)(-a+b+c)}.$$
| Let $\sqrt{a}=x$, $\sqrt{b}=y$ and $\sqrt{c}=y$.
Hence, we need to prove that $\prod\limits_{cyc}(x+y-z)^2\geq\prod\limits_{cyc}(x^2+y^2-z^2)$.
We'll prove that the last inequality is true for all reals $x$, $y$ and $z$.
Indeed, we can assume that $\prod\limits_{cyc}(x^2+y^2-z^2)>0$ and from here
we obtain $x^2+y^2-z^2>0$, $x^2+z^2-y^2>0$ and $y^2+z^2-x^2>0$ because
if for example, $x^2+y^2-z^2<0$ and $x^2+z^2-y^2<0$ so $x^2<0$, which is a contradiction.
Now easy to see that $$(x+y-z)^2(x+z-y)^2-(x^2+y^2-z^2)(x^2+z^2-y^2)=2(y^2+z^2-x^2)(y-z)^2\geq0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}[\ln(\tan \ x)^{\frac{2}{3}}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}-\ln(1)^\frac{2}{3}]=\frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}] = \bf \color{red}{\ln(\sqrt{3})}$$
However, the answer given is not this value, but instead $= \frac{3\sqrt[3]{3}-3}{2} \approx 0.663... $ while $\ln(\sqrt{3}) \approx 0.549...$
| Your solution was fine. Just needed to have put $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}f'(x)dx$$ instead of $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$which makes your solution $$\frac{3}{2}(f(\frac{\pi}{3})-f(\frac{\pi}{4}))$$where f(x) is the function you defined, namely $$f(x) = (\tan \ x)^{\frac{2}{3}}$$This indeed yields the correct answer,
$$\frac{3}{2}(3^{\frac{1}{3}}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| $$ \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)} = \frac{x \cos x}{ x \sin x + \cos x } + \frac{x \sin x}{x \ cos x - \sin x} $$
Why I get this:
$$\frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)} = \frac{A(x)}{ x \sin x + \cos x } + \frac{B(x)}{x \ cos x - \sin x} \\\ \implies A(x)( x \cos(x)- \sin(x))+ B(x) (x \sin x + \cos x) = x^2$$
So $ A(x) = ax+b, B(x) = cx + d$.
$(ax+b)(x \cos(x) - \sin(x)) + (cx+d) ( x \sin x + \cos x) = x^2\\\ \implies ax^2 \cos(x) - ax \sin(x) -b\sin(x) + bx \cos(x) + cx ^2 \sin(x) +\\ cx \cos(x) + dx \sin(x) +d \cos(x) = x^2 \implies$
$$ a\cos(x) +c\sin(x) = 1 \tag{1}$$
$$d \cos(x) - b\sin(x) = 0 \tag{2}$$
$$ -a\sin(x)+b\cos(x)+c \cos(x)+ d \sin(x) = 0 \tag{3}$$
$\forall x \in R$
By $(2)$, $d = b = 0$,By $(1)$ and (3), $\cos(x) = a, c = \sin(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Row and Column rotation matrices Today I started studying Rotation matrices and derived a rotation matrix like this:
\begin{equation*}
\begin{pmatrix}
x & y
\end{pmatrix}
\begin{pmatrix}
\cos(\beta) & \sin(\beta) \\
-\sin(\beta) & \cos(\beta)
\end{pmatrix}
\end{equation*}
When I was googling for the solution to check if this was ok, I found almost everybody expressed it like this:
\begin{equation*}
\begin{pmatrix}
\cos(\beta) & -\sin(\beta) \\
\sin(\beta) & \cos(\beta)
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
\end{equation*}
If I'm not mistaken, this leads to the same equations, but is there anything else special about expressing it in these 2 different forms? Has this something to do with Handedness?
| The difference is that you used a row vector and the other expression uses a column vector. And they are the same if you transpose one of them.
\begin{align}
\left(
\begin{pmatrix}
x & y
\end{pmatrix}
\begin{pmatrix}
\cos(\beta) & \sin(\beta) \\
-\sin(\beta) & \cos(\beta)
\end{pmatrix}
\right)^T
&=
\begin{pmatrix}
\cos(\beta) & \sin(\beta) \\
-\sin(\beta) & \cos(\beta)
\end{pmatrix}
^T
\begin{pmatrix}
x & y
\end{pmatrix}
^T
\\ &=
\begin{pmatrix}
\cos(\beta) & -\sin(\beta) \\
\sin(\beta) & \cos(\beta)
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
show that $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ is a unit in $\mathbb{Z}[\sqrt[3]{2}]$ I found by numerical experiment the norm of $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ (result of multiplying conjugates) is: $$0.9999999999989706-4.4408920985006262 \times 10^{-16}i$$ but I am betting this is just $1$. How can I show this is a unit in $\mathbb{Z}[\sqrt[3]{2}]$? Here is another possible unit:
$$ 521-62\sqrt[3]{2}-279\sqrt[3]{4} $$
| Or,
Given integers $a,b,c,$ and cubic form
$$ f(a,b,c) = a^3+2 b^3-6 a b c+4 c^3 = \left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|, $$
That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where
$$ X = \begin{bmatrix} 0 & 0 & 2\\1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}. $$ Then $X^3 = 2 I$ and $X^4 = 2 X.$
The elements with norm $1$ and coefficients below $600$ in absolute value are
-161 -99 180
-35 24 3
-7 -2 6
-1 1 0
1 -2 1
1 0 0
1 1 1
1 3 -3
1 100 -80
5 4 3
19 -5 -8
19 15 12
41 -59 21
73 58 46
281 223 177
521 -62 -279
You want $a = 19, b = -5, c = -8.$ The resulting determinant is $1,$ both times.
? f = 19 * id - 5 * x - 8 * x2
%5 =
[19 -16 -10]
[-5 19 -16]
[-8 -5 19]
? matdet(f)
%6 = 1
? x
%7 =
[0 0 2]
[1 0 0]
[0 1 0]
? x^2
%9 =
[0 2 0]
[0 0 2]
[1 0 0]
? f = 521 * id - 62 * x - 279 * x2
%10 =
[521 -558 -124]
[-62 521 -558]
[-279 -62 521]
? matdet(f)
%11 = 1
?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
An inequality involving hyperbolic sine functions The following problem arised in my research work and has been challenging me for several days: Prove that for all $r\in (0,1)$ and for all $x>0$, we have $$\frac{\sinh(r(2-r)x)}{r(2-r)x} \bigg[ \frac{\sinh((1-r)x)}{(1-r)x} \bigg]^2 > \frac{\sinh(x)}{x} \frac{\sinh((1-r)^2x)}{(1-r)^2x}.$$
Below are my thoughts. The inequality is obviously true when $x$ is large enough, thanks to the equivalence $\sinh(y)\sim \frac{1}{2}\exp(y)$, $y\rightarrow +\infty$, and to the fact that $$r(2-r)+2(1-r) > 1 + (1-r)^2$$ for $r\in(0,1)$. The inequality also holds true when $x$ is small enough, since it can be readily shown that the Taylor expansion of the difference between the LHS and the RHS has $$\frac{1}{45}r^2(1-r)^2(2-r)^2 x^4$$ as the leading term. However, the difficulty lies in establishing the claim for all $x>0$, which is "testified" by various graphic display softwares (Maple, Gnuplot, WolframAlpha).
A first natural transformation is to take the logarithm of both sides and to invoke the convexity of the function $x\mapsto\ln(\sinh(x)/x)$. But this leads nowhere: Jensen's inequality alone cannot help us deriving $$f(r(2-r)x)+2f((1-r)x) > f(x)+f((1-r)^2x)$$ for any convex function $f$ (consider a linear function and its opposite).
As a second attack, we can multiply both sides by $4r(2-r)(1-r)^2x^3$ and use the linearization formulae $$2\sinh(a)\sinh(b) = \cosh(a+b)-\cosh(a-b)$$ $$2\sinh(a)\cosh(b) = \sinh(a+b) + \sinh(a-b)$$ in order to obtain the equivalent inequality $$\sinh((2-r^2)x) -\sinh((r^2-4r+2)x) - 2\sinh(r(2-r)x) > 2r(2-r)x \big( \cosh((r^2-2r+2)x)- \cosh(r(2-r)x)\big).$$ The direct study of the difference between the new LHS and the new RHS appears to be unwieldy. Nevertheless, it is now straightforward to compute the Taylor series of this new difference. This series (whose radius of convergence is infinite) is equal to $$\sum_{n=0}^{\infty} \frac{a_n(r)}{(2n+1)!} x^{2n+1},$$ with $$a_n(r)= (2-r^2)^{2n+1} - (r^2-4r+2)^{2n+1} + 4n (r(2-r))^{2n+1} - (4n+2) r(2-r)(r^2-2r+2)^{2n}.$$ The first three coefficients are easily shown to vanish, i.e., $a_0(r)=a_1(r) = a_2(r) = 0$. Pushing further the calculations yields $$a_3(r)=448 r^3(1-r)^4(2-r)^3$$ $$a_4(r)=768 r^3(1-r)^4(2-r)^3 [3(1-r)^4 + 2(1-r)^2 + 3]$$ $$a_5(r)= 1408 r^3(1-r)^4(2-r)^3 [5(1-r)^8 + 12(1-r)^6 + 6(1-r)^4 + 12(1-r)^2 + 5].$$ These suggest that $a_n(r)$ is the product of $r^3(1-r)^4(2-r)^3$ with some even and symmetric polynomials in $1-r$ whose coefficients are all non-negative. The conjecture $a_n(r)> 0$ for $r\in(0,1)$ implies the desired result, but again I cannot prove it for $n\geq 6$.
Is there a general technique to prove that the power series coefficients of a given function are all non-negative? According to what I could find in the litterature, this seems to be a delicate issue.
Is there any fresh approach to the initial question? Many thanks for any help.
| First, let's replace your $r$ with $y=1-r$ $\qquad - \quad$ this makes the arguments a little nicer:
$$\frac{\sinh((1-y^2)x)}{(1-y^2)x} \left[ \frac{\sinh(yx)}{yx} \right]^2 > \frac{\sinh(x)}{x} \frac{\sinh(y^2x)}{y^2x}$$
Now we have the imaginary $\operatorname{sinc}$ functions here, and it makes sense to use the infinite product formula instead of the series:
$$\frac{\sinh(\pi z)}{\pi z}=\prod_{n=1}^\infty \left(1+\frac{z^2}{n^2} \right)$$
Let's set $x=\pi z$ then our inequality will become:
$$\prod_{n=1}^\infty \left(1+\frac{(1-y^2)^2z^2}{n^2} \right) \prod_{n=1}^\infty\left(1+\frac{y^2z^2}{n^2} \right)^2>\prod_{n=1}^\infty \left(1+\frac{z^2}{n^2} \right) \prod_{n=1}^\infty\left(1+\frac{y^4z^2}{n^2} \right)$$
All of the products converge, thus we can multiply them by term:
$$\prod_{n=1}^\infty \left(1+\frac{(1+y^4)z^2}{n^2}+\frac{y^2(2-3y^2+2y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} \right)> \\ > \prod_{n=1}^\infty \left(1+\frac{(1+y^4)z^2}{n^2}+\frac{y^4z^4}{n^4}\right)$$
Now it is enough to prove that each term on the left is larger than corresponding term on the right. Then the product on the left will be larger as well:
$$1+\frac{(1+y^4)z^2}{n^2}+\frac{y^2(2-3y^2+2y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} > 1+\frac{(1+y^4)z^2}{n^2}+\frac{y^4z^4}{n^4}$$
$$\frac{y^2(2-3y^2+2y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} > \frac{y^4z^4}{n^4}$$
$$\frac{2y^2(1-2y^2+y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} > 0$$
$$\frac{2y^2(1-y^2)^2z^4}{n^4}+\frac{y^4(1-y^2)^2z^6}{n^6} > 0$$
The last inequality is obviously true, thus we can trace all of the steps back and prove the original inequality.
| {
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"url": "https://math.stackexchange.com/questions/1882399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$ $$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$
$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$
$$1=Ax+A+Bx-2B$$
$$1=(A+B)x+A-2B$$
$A+B=0\iff A=-B$
$-3B=1$
$B=-\frac{1}{3}$, $A=\frac{1}{3}$
$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\frac{1}{3}\int_{3}^{\infty}\frac{dx}{x-2}-\frac{1}{3}\int_{3}^{\infty}\frac{dx}{(x+1)}=|\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1)|_{3}^{\infty}$$
$$=lim_{t\to \infty}(\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1))-\frac{1}{3}ln(1)+\frac{1}{3}ln(4)=\infty-\infty-0+\frac{ln(4)}{3}$$
But the answer is $\frac{2ln(2)}{3}$, What have I done wrong?
| $$\frac{1}{3}\int_{3}^{+\infty}\left(\frac{1}{x-2}-\frac{1}{x+1}\right)\,dx =\frac{1}{3}\int_{1}^{4}\frac{dz}{z}=\frac{\log 4}{3}=\color{red}{\frac{2}{3}\log 2}\tag{1}$$
since:
$$ \int_{3}^{M}\frac{dx}{x-2}=\int_{1}^{M-2}\frac{dz}{z},\qquad \int_{3}^{M}\frac{dx}{x+1}=\int_{4}^{M+1}\frac{dz}{z}\tag{2} $$
and:
$$ \left|\int_{M-2}^{M+1}\frac{dz}{z}\right|\leq \frac{1}{M-2}\int_{M-2}^{M+1}dz = \frac{3}{M-2}\to 0.\tag{3}$$
| {
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"url": "https://math.stackexchange.com/questions/1882556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does this trick gives precisely the formulas for the sum of the $n$ first naturals and the $n$ first squares? I've learned a cool trick several days ago. Suppose I want to find a polynomial that gives me:
$$f(1)=1, \quad f(2)=5,\quad f(3)=14,\quad f(4)=30\tag{1}$$
I could do the following: Take a polynomial of degree $3$ and make the following system with it:
\begin{eqnarray*}
{ax^3+bx^2+cx+d}&=&{1} \\
{ax^3+bx^2+cx+d}&=&{5} \\
{ax^3+bx^2+cx+d}&=&{14} \\
{ax^3+bx^2+cx+d}&=&{30}
\end{eqnarray*}
I just need to substitute the $x^n$'s according to $(1)$. This gives me:
\begin{eqnarray*}
{a1^3+b1^2+c1+d}&=&{1} \\
{a2^3+b2^2+c2+d}&=&{5} \\
{a3^3+b3^2+c3+d}&=&{14} \\
{a4^3+b4^2+c4+d}&=&{30}
\end{eqnarray*}
Now this is easy, we need only to find the coefficients $a,b,c,d$, that is, find the inverse of a matrix $A$ in the matricial equations $Ax=b$:
$$\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
2 & 4 & 8 & 16 \\
3 & 9 & 27 & 81 \\
4 & 16 & 64 & 256 \\
\end{array}
\right)$$
And then: $x=A^{-1}b$. This gives me:
$$a=\frac{1}{3}\quad\quad b= \frac{1}{2}\quad\quad c= \frac{1}{6}\quad\quad d= 0$$
That is, our polynomial is: $\cfrac{x^3}{3}+\cfrac{x^2}{2}+\cfrac{x}{6}$ and this is - at least to me - surprising because it is the formula for the sum of the first $n$ square numbers. What baffles me more is that if I do the same with a polynomial of degree $4$ and a $5\times 5$ matrix and $f(5)=55$ it will give me the same formula. So I have three questions:
*
*When I did this trick, I used only $4$ and $5$ values of the sum of the first $n$ squares. So why does it give me exactly the polynomial for the sum of the first $n$ squares instead of any other polynomial? There is an infinite abyss of possible polynomials, why exactly this one? For example: Why didn't it give a polynomial that gives me:
$$f(1)=1, \quad f(2)=5,\quad f(3)=14,\quad f(4)=30 \quad f(5)=55\quad f(6)=2\quad ?$$
*Why does the result holds for a higher $n$, that is: Why doing the same thing with $n+1$ gives me the same polynomial it gave me for $n$ instead of any other polynomial in the behemothic chasm of possible polynomials?
*I've tried for other formulas in the past, just like the sum of the first $n$ positive integers. Why does it hit exactly the polynomial for the sum of the first $n$ positive integers instead of any other polynomial in the humongous crevasse of possible polynomials?
Sorry if the question is too stupid, but I've figured this trick some time ago and couldn't find a clue to why this is hapenning. The main drama of the thing for me is that it seems as if I were doing incomplete induction and then BAM! it just hits exactly where I wanted it to hit.
| This is an example of working out Ahmed's answer, which is good and contains a lot of information.
This trick works best if you start from $0$. To find the polynomial $f(x)$ with $$f(0) = 0, \; f(1) = 1, \; f(2) = 5, \; f(3) = 14, \; f(4) = 30,$$ form a triangle of differences
\begin{align*} &0\quad \quad 1 \quad\quad 5 \quad\quad 14 \quad\quad 30 \\ &\quad 1 \quad\quad 4 \quad\quad 9 \quad\quad 16 \\ &\quad\quad 3 \quad\quad 5 \quad\quad 7 \\ &\quad\quad\quad 2 \quad\quad 2 \\ &\quad\quad \quad\quad 0\end{align*}
and take the left side of the triangle as coefficients in a Newton series:
\begin{align*} f(x) &= 0 \binom{x}{0} + 1 \binom{x}{1} + 3 \binom{x}{2} + 2 \binom{x}{3} + 0 \binom{x}{4} \\ &= 0 + x + \frac{3x (x-1)}{2} + \frac{2 x(x-1)(x-2)}{6} \\ &= \frac{x}{6} + \frac{x^2}{2} + \frac{x^3}{3}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040} \right)$.
How do we proceed further?
| Your approach is correct, but without determining the order of the pole $z=0$ you are stuck because each coefficient of $f \cdot \sin$ will, in principle, be a sum with infinitely many terms. Finding the order of the pole of $f$ will allow us to kill infinitely many of these, leaving only finitely many of them, as you are going to see.
Remember that the Taylor series of $\sin$ around $z=0$ begins with $z^1$, so $0$ is a zero of order $1$ for $\sin$, therefore it is a pole of order $1$ for $\frac 1 \sin$.
For even greater clarity, remember that $z_0$ is a pole of order $n>0$ for $f$ if and only if $n$ is smallest with the property $\lim \limits _{z \to z_0} (z-z_0)^n f(z) \ne 0$. Let's check this for your $f$ and $n=1$:
$$\lim _{z \to 0} z \ \frac {z^2 + 1} {\sin z} = \lim _{z \to 0} \frac z {\sin z} (z^2 + 1) = \lim _{z \to 0} \frac z {\sin z} \lim _{z \to 0} (z^2 + 1) = 1 \cdot 1 = 1 .$$
Is $n=1$ the smallest such strictly positive number? Yes, because below it there is only $0$, and $0$ is not allowed to be the order of a pole, by definition.
Having found the order of the pole of $f$ is essential, since this allows us to restrict the lower bound for the summation index:
$$f(z) = \sum _{n = -1} ^\infty c_n z^n$$
(notice that $n$ starts now at $-1$, not at $-\infty$).
Plugging this into your own approach (and keeping only terms up to and including $z^3$, because you want only the first 4 terms) gives
$$z^2 + 1 = \left( \sum _{n =-1} ^\infty c_n z^n \right) \left( z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \dots \right) = \\
\left( c_{-1} - \frac {c_{-1}} {3!} z^2 + \dots \right) + \left( c_0 z - \frac {c_0} {3!} z^3 + \dots \right) + \left( c_1 z^2 + \dots \right) + \left( c_2 z^3 + \dots \right) + \dots ,$$
so equating the coefficients of the corresponding powers of $z$ in both hands of the equation gives
$$c_{-1} = 1, \quad c_0 = 0, \quad -\frac {c_{-1}} 6 + c_1 = 1, \quad -\frac {c_0} 6 + c_2 = 0 ,$$
whence it follows that
$$c_{-1} = 1, \quad c_0 = 0, \quad c_1 = \frac 7 6, \quad c_2 = 0 ,$$
so
$$f(z) = \frac 1 z + \frac 7 6 z + \dots .$$
Notice that there is no surprise in the fact that $c_0 = c_2 = 0$, because
$$f(-z) = \frac {(-z)^2 + 1} {\sin (-z)} = \frac {z^2 + 1} {-\sin z} = -f(z) ,$$
so $f$ is odd, and the coefficients of even order of an odd function are always $0$.
| {
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"url": "https://math.stackexchange.com/questions/1883785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$? In the process of my numerical computations I have found a very special identity:
*
*$\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$
*$
-1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2}
= 9.313225746154785 \times 10^{-10}$
Therefore if we subtract these two equations one should find the difference is zero. On my computer I found:
$$
2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{
\begin{array}{cl} 0 & \text{by hand} \\
-1.862645149230957 \times 10^{-9} & \text{by computer} \end{array}\right.
$$
Subtracting these two numbers - which might be the same - we have gotten twice the number!
| To expand on the comments of Fabio and Jyrki:
You're claiming that $\sqrt[3]7$ is a root of the quadratic equation
$$2764976 + 3734045x -2707603x^2 = 0$$
but that leads to a contradiction.
The solutions of a quadratic with integer coefficients can be written as
$$x = \frac{p \pm \sqrt q}{r} $$
for some integers $p, q, r$.
If $x^3 = 7$ then
$$\begin{align}
\left(\frac{p \pm \sqrt q}{r}\right)^3 = 7\\
\frac{p^3 \pm 3p^2\sqrt q + 3pq +q\sqrt q}{r^3} = 7\\
\frac{(p^3 + 3pq) \pm \sqrt q(3p^2 +q)}{r^3} = 7\\
\end{align}$$
Now the only way the expression on the LHS can be rational (let alone integer) is if $\sqrt q$ is rational. But $q$ is an integer, so its square root can only be rational if $q$ is a perfect square. But that would make $x = \frac{p \pm \sqrt q}{r}$ rational, and we know that $x = \sqrt[3]7$ is not rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of this series $\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{4{{n}^{2}}-1}}=??$ I start first to use $u=n+1$ then
$$\sum_{u=2}^{\infty}
{\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$
| Note $\sum_{u=2}^{\infty}
{\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}=\sum_{n=1}^{\infty}
{\frac{{{\left( -1 \right)}^{n+1}}}{4{n^{2}}-1}}$. Let
$$ f(x)=\sum_{n=1}^{\infty}
{\frac{{{\left( -1 \right)}^{n+1}}}{4n^2-1}}x^{2n+1}. $$
Then $f(1)=\sum_{n=1}^{\infty}
{\frac{{{\left( -1 \right)}^{n+1}}}{4n^2-1}}, f(0)=0$ and
\begin{eqnarray}
f'(x)&=&\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}x^{2n}
\end{eqnarray}
and $f'(0)=0$.
So
$$ (\frac{f'(x)}{x})'=\sum_{n=1}^{\infty}(-1)^{n+1}x^{2n-2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}=\frac{1}{1+x^2} $$
Thus
$$ \frac{f'(x)}{x}=\int\frac{1}{1+x^2}dx=\arctan x+C. $$
and
$$ f'(x)=x\arctan x+Cx.$$
But $f'(0)=0$ from which we have $C=0$.
So
$$f(1)=\int_0^1x\arctan xdx=\frac{\pi}{4}-\frac{1}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove a series that equals to $\frac{e}{e-1}$ Prove that
$$
\lim_{N\to\infty}\sum_{k=0}^\infty \left(1+\frac{k}{N}\right)^{-N}=\frac{e}{e-1}
$$
I think $\sum_{k=0}^\infty \left(1+\frac{k}{N}\right)^{-N}$ should be a Riemann sum of a function but could find it. What is the trick in this question?
In addition, the equation holds true when it could interchange the limits, but how to prove it?
| Note that
$$\left(1+\frac{k}{N}\right)^N = \sum\limits_{j=0}^{N}{{N\choose j}\left(\frac{k}{N}\right)^j} \ge 1 + {N\choose 1}\frac{k}{N} + {N\choose 2}\frac{k^2}{N^2} = 1 + k + \frac{N-1}{2N}k^2 \ge 1 + k + \frac{k^2}{4}$$
for $N\ge 2$. It follows that for any $M\ge 1$ we have
$$\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}} = \sum\limits_{k=M+1}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}\le\sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}.$$
Note that the RHS goes to zero as $M\rightarrow\infty$. Since $\lim\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}} = \sum\limits_{k=0}^{M}{e^{-k}}$, it follows that
\begin{align} &\limsup\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{e^{-k}}\right)} \\
&= \limsup\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}\right)}\\
&\le\sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}
\end{align}
i.e. $\limsup\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\le \sum\limits_{k=0}^{M}{e^{-k}} + \sum\limits_{k=M+1}^{\infty}{\frac{1}{1+k+k^2/4}}$. Now clearly
\begin{align} &\liminf\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{e^{-k}}\right)} \\
&= \liminf\limits_{N\rightarrow\infty}{\left(\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}} - \sum\limits_{k=0}^{M}{\left(1+\frac{k}{N}\right)^{-N}}\right)}\\
&\ge 0
\end{align}
i.e. $\liminf\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\ge\sum\limits_{k=0}^{M}{e^{-k}}$. Letting $M\rightarrow\infty$ yields
$$\limsup\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\le\sum\limits_{k=0}^{\infty}{e^{-k}}\le\liminf\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}}\\\implies \lim\limits_{N\rightarrow\infty}{\sum\limits_{k=0}^{\infty}{\left(1+\frac{k}{N}\right)^{-N}}} = \sum\limits_{k=0}^{\infty}{e^{-k}} = \frac{e}{e-1} $$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
| $$
\begin{align}
I_n
&=\int_0^1x^n\arctan(x)\,\mathrm{d}x\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^{n+2k+1}}{2k+1}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty(-1)^k\frac1{(n+2k+2)(2k+1)}\\
&=\frac1{n+1}\sum_{k=0}^\infty(-1)^k\left(\frac1{2k+1}-\frac1{n+2k+2}\right)\\
&=\frac\pi{4(n+1)}-\frac1{n+1}\sum_{k=0}^\infty\frac{(-1)^k}{n+2k+2}
\end{align}
$$
Thus,
$$
\begin{align}
(n+1)I_n-\frac\pi4
&=-\sum_{k=0}^\infty\frac{(-1)^k}{n+2k+2}\\
&=\color{#00A000}{-\sum_{k=0}^\infty\left(\frac1{n+4k+2}-\frac1{n+4k+4}\right)}
\end{align}
$$
and because
$$
\color{#C00000}{-\frac2{(n+4k)(n+4k+4)}}
\le\color{#00A000}{-\frac2{(n+4k+2)(n+4k+4)}}
\le\color{#0000F0}{-\frac2{(n+4k+2)(n+4k+6)}}
$$
we get
$$
\color{#C00000}{-\frac12\underbrace{\sum_{k=0}^\infty\left(\frac1{n+4k}-\frac1{n+4k+4}\right)}_{\large\frac1n}}
\le(n+1)I_n-\frac\pi4
\le\color{#0000F0}{-\frac12\underbrace{\sum_{k=0}^\infty\left(\frac1{n+4k+2}-\frac1{n+4k+6}\right)}_{\large\frac1{n+2}}}
$$
Therefore, by the telescoping series above,
$$
-\frac{n}{2n}\le n\left((n+1)I_n-\frac\pi4\right)\le-\frac{n}{2n+4}
$$
and thus, by the Squeeze Theorem, we get
$$
\lim_{n\to\infty}n\left((n+1)I_n-\frac\pi4\right)=-\frac12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Gauss circle problem : a simple asymptotic estimation.
Find the number of integer points lying in or inside a circle of radius $n\in \mathbb N$ centered at the origin.
The problem asks for all $(a,b)\in \mathbb Z^2$ such that $a^2+b^2\leq n^2$. Looking at such points lying strictly in the North-East quadrant, pick an $a\in \{1,\ldots,n\}$ and let $b$ run through $\{1,\ldots,\lfloor \sqrt{n^2-a^2}\rfloor\}$ : there are exactly $\displaystyle \sum_{k=1}^n \lfloor \sqrt{n^2-k^2} \rfloor$ integer points strictly in the quadrant.
Multiply that by $4$ to account for all $4$ quadrants, add the origin and points on the $x$ and $y$ axis and the total number is $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor$$
I'm interested in asymptotics of this sum. A trivial estimate is the following:$$\begin{align}\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor &= \sum_{k=0}^n \sqrt{n^2-k^2} + O(n)\\ &= n^2\cdot \left(\frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}}\right) + O(n) \\ &=n^2 \left( \int_0^1 \sqrt{1-t^2}dt + o(1)\right) + O(n) \\ &= \frac \pi 4n^2 + o(n^2)\end{align} $$
Hence $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + o(n^2)$$
Is there a way to refine the estimate above and get $\pi n^2 + O(n)$ instead (or something better) ?
| The estimate $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + o(1)$ can easily be refined .
Indeed, $$\begin{align}0\leq \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} - \int_0^1 \sqrt{1-t^2}dt &= \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-t^2}\right)dt\\ &\leq\sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-\frac{(k+1)^2}{n^2}}\right)dt \\ &= \frac 1n \end{align}$$
Hence $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + O(\frac 1n)$ and $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + O(n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this determinant
Question Statment:-
Show that
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\
\end{vmatrix}
=2abc(a+b+c)^3
\end{align*}
My Attempt:-
$$\begin{aligned}
&\begin{vmatrix}
\\(a+b)^2 & ca & bc \\
\\ca & (b+c)^2 & ab \\
\\bc & ab & (c+a)^2 \\\
\end{vmatrix}\\
=&\begin{vmatrix}
\\a^2+b^2+2ab & ca & bc \\
\\ca & b^2+c^2+2bc & ab \\
\\bc & ab & c^2+a^2+2ac \\\
\end{vmatrix}\\
=&\dfrac{1}{abc}\begin{vmatrix}
\\ca^2+cb^2+2abc & ca^2 & b^2c \\
\\ac^2 & ab^2+ac^2+2abc & ab^2 \\
\\bc^2 & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\\\\
&\qquad (C_1\rightarrow cC_1, C_2\rightarrow aC_2, C_3\rightarrow bC_3)\\\\
=&\dfrac{2}{abc}\times\begin{vmatrix}
\\ca^2+cb^2+abc & ca^2 & b^2c \\
\\ab^2+ac^2+abc & ab^2+ac^2+2abc & ab^2 \\
\\bc^2+a^2b+abc & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\\\\
&\qquad (C_1\rightarrow C_1+C_2+C_3)\\\\
=&\dfrac{2abc}{abc}\left(\begin{vmatrix}
\\a^2+b^2 & a^2 & b^2 \\
\\b^2+c^2 & b^2+c^2+2bc & b^2 \\
\\c^2+a^2 & a^2 & c^2+a^2+2ac \\\
\end{vmatrix}+
\begin{vmatrix}
\\1 & ca^2 & b^2c \\
\\1 & ab^2+ac^2+2abc & ab^2 \\
\\1 & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\right)
\end{aligned}$$
The second determinant in the last step can be simplified to
\begin{vmatrix}
\\1 & ca^2 & b^2c \\
\\0 & ab^2+ac^2+2abc-ca^2 & ab^2-b^2c \\
\\0 & a^2b-ca^2 & bc^2+a^2b+2abc-b^2c \\\
\end{vmatrix}
I couldn't proceed further with this, so your help will be appreciated and if any other simpler way is possible please do post it too.
| Let me try. You have $$LHS = [(a+b)(b+c)(c+a)]^2 + 2(abc)^2 - \sum b^2c^2(b+c)^2.$$
Note that $(a+b)(b+c)(c+a) = \sum bc(b+c) + 2abc$.
Then, $$LHS = \left(\sum bc(b+c)\right)^2 + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right) - \sum b^2c^2(b+c)^2$$
$$=2\sum a^2bc(a+b)(a+c) + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right)$$
$$ =2abc\left(\sum a(a+b)(a+c) + 3abc + 2\sum bc(b+c)\right) $$
$$ = 2abc \left(\sum a^3 + 6abc + 3\sum bc(b+c)\right) $$
$$ = 2abc \left(a+b+c\right)^3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
finding square root I want to know how to find the square root/cube root of a number ending with zero.
For instance, $\sqrt{1024}$ : though it's a perfect square, there is a shortcut. Like wise how can I find the $\sqrt{2240}$ or any number ending with zero?
Thanks in advance!
|
By $$(10a+b)^2 = a\times 100a+(10\times 2a+b)\times b$$
We can use long division,
\begin{array}{rrrr}
& & 3 & 2 \\
& & -- & -- \\
3 & \sqrt{} & 10 & 24 \\
& & 9 & \\
& & -- & -- \\
& & 1 & 24 \\
62 & & 1 & 24 \\
& & -- & -- \\
& & & 0 \\
\end{array}
which means $32^2=(10\times 3+2)^2=100\times 9+62 \times 2=1024$,
thus $\sqrt{1024}=32$.
By $$(10a+b)^3 = a^2\times 1000a+(100\times 3a^2+10\times 3ab+b^2)\times b$$
We can use long division,
\begin{array}{rrr}
& & 2 & 3 \\
& & -- & -- \\
4 & \sqrt[3]{} & 12 & 167 \\
& & 8 & \\
& & -- & -- \\
& & 4 & 167 \\
1389 & & 4 & 167 \\
& & -- & -- \\
& & & 0 \\
\end{array}
where $4=\color{red}{2}^2$ and
$1389=100\times 3\times \color{red}{2}^2+10\times 3\times \color{red}{2} \times \color{blue}{3}+\color{blue}{3}^2$
therefore $\sqrt[3]{12167}=23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Expectation value of trials needed to get $k$ consecutive outcomes Suppose that independent trials, each of which is equally likely to have any of $m$ possible outcome, are performed until the same outcome occurs $k$ consecutive times. If $N$ denotes the number of trials, show that
$$E[N] = \frac{m^k-1}{m-1}$$
This is a homework question. I was trying to reverse engineer this into a GP but without success.
I also tried an induction approach to $k$. For getting one outcome, the number of trials needed is always $1$, so $k=1$ is a trivial case. Now assume that the expectation value of trials needed for $k$ consecutive outcomes is $E[N]$, then how do I find the update to $E[N]$ for $k+1$.
I am very confused on the approach. Looks like a one-liner would do. Please help.
| We use the following generating function for $k\ge 2:$
$$G(z, u) = z^k\times \sum_{q\ge 0} u^q (z+z^2+\cdots+z^{k-1})^q
\\ = z^k \sum_{q\ge 0} u^q z^q (1+z+\cdots z^{k-2})^q
\\ = z^k
\sum_{q\ge 0} u^q z^q \frac{(1-z^{k-1})^q}{(1-z)^q}
\\ = z^k \frac{1}{1-uz(1-z^{k-1})/(1-z)}
= z^k \frac{1-z}{1-z-uz(1-z^{k-1})}.$$
As a sanity check we should get one when we sum the probabilities
of first getting $k$ consecutive outcomes after $n$ trials. This is
$$m \sum_{n\ge 0} m^{-n} [z^n]
\left. z^k \frac{1-z}{1-z-uz(1-z^{k-1})} \right|_{u=m-1}
\\ = m \sum_{n\ge 0} m^{-n} [z^n]
z^k \frac{1-z}{1-z-(m-1)z(1-z^{k-1})}
\\ = m \sum_{n\ge 0} m^{-n} [z^n]
z^k \frac{1-z}{1-mz+(m-1)z^k}
\\ = m \frac{1}{m^k} \frac{1-1/m}{(m-1)/m^k}
= \frac{1}{m^k} \frac{m-1}{(m-1)/m^k} = 1$$
and the sanity check goes through. Now for the expectation we obtain
$$m \sum_{n\ge 1} n \frac{1}{m^n} [z^n]
\left. z^k \frac{1-z}{1-z-uz(1-z^{k-1})} \right|_{u=m-1}
\\ = \sum_{n\ge 1} n \frac{1}{m^{n-1}} [z^n]
z^k \frac{1-z}{1-z-(m-1)z(1-z^{k-1})}
\\ = \left.
\left(z^k \frac{1-z}{1-mz+(m-1)z^{k}}\right)'\right|_{z=1/m}
\\ = \left.\left(k z^{k-1} \frac{1-z}{1-mz+(m-1)z^{k}}
- z^k \frac{1}{1-mz+(m-1)z^{k}}
\\ - z^k \frac{1-z}{(1-mz+(m-1)z^{k})^2} (k(m-1)z^{k-1}-m)
\right)\right|_{z=1/m}.$$
Using $\left.1-mz+(m-1)z^{k}\right|_{z=1/m} = (m-1)/m^{k}$
we get
$$\frac{k}{m^{k-1}} (1-1/m) \frac{m^{k}}{m-1}
- \frac{1}{m^k} \frac{m^{k}}{m-1}
- \frac{1}{m^k} (1-1/m) \frac{m^{2k}}{(m-1)^2}
\left(\frac{k(m-1)}{m^{k-1}}-m\right)
\\ = k - \frac{1}{m-1}
- \frac{1}{m^{k+1}} \frac{m^{2k}}{m-1}
\frac{k(m-1)}{m^{k-1}}
+ \frac{1}{m^{k+1}} \frac{m^{2k}}{m-1} m
\\ = k - \frac{1}{m-1} - k + \frac{m^k}{m-1}.$$
We thus obtain the end result
$$\bbox[5px,border:2px solid #00A000]{
\frac{m^k-1}{m-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of the sequence involving integral
Let $f:[0,1] \rightarrow [0,1]$ increasing function and
$a_n=\int_{0}^{1} \frac {1+(f(x)^n}{1+(f(x))^{n+1}} dx \tag 1$
Prove $a_n$ is convergent and find the limit.
It's easy to prove $a_n \ge 1$ and $a_n$ is decreasing, therefore is convergent. By taking $f$ identical zero, we get $a_n=1, \forall n$ hence the limit is $1$. If the limit is not dependent on $f$ then I have to prove the limit is $1$ for all functions, but I can't figure out how to do it.
| A different approach, not using dominated convergence:
First, I claim that $r^n - r^{n+1} \le \frac{(n-1)^{n-1}}{n^n} $ for all $r\in [0,1]$. To see this, we note that $r^n - r^{n+1} = (n-1)^{n-1}\left(\frac{r}{n-1}\right)^{n-1}(r-1)$, and by AM-GM
$$ \left(\frac{r}{n-1}\right)^{n-1}(1-r) = \left(\sqrt[n]{\underbrace{\left(\frac{r}{n-1}\right)\dots\left(\frac{r}{n-1}\right)}_{n-1}(1-r)}\right)^n\le\left(\frac{\overbrace{\frac{r}{n-1}+\dots+\frac{r}{n-1}}^{n-1} + 1-r}{n}\right)^n = \frac{1}{n^n}.$$
Now
$$\frac{1+f(x)^n}{1+f(x)^{n+1}} = \frac{1+f(x)^{n+1} + f(x)^n - f(x)^{n+1}}{1+f(x)^{n+1}} = 1 + \frac{f(x)^n - f(x)^{n+1}}{1+f(x)^{n+1}}$$
and hence
\begin{align} \int\limits_{0}^{1}{\left(\frac{1+f(x)^n}{1+f(x)^{n+1}}-1\right)\text{ d}x} &= \int\limits_{0}^{1}{\frac{f(x)^n-f(x)^{n+1}}{1+f(x)^{n+1}}\text{ d}x} \\
&\le\int\limits_{0}^{1}{\frac{\frac{(n-1)^{n-1}}{n^n}}{1}\text{ d}x} \\
&=\frac{(n-1)^{n-1}}{n^n}.
\end{align}
Since $\frac{(n-1)^{n-1}}{n^n}\le\frac{n^{n-1}}{n^n} = \frac{1}{n}$ and $\frac{1}{n}\rightarrow 0$ as $n\rightarrow\infty$, it follows that
$$\limsup\limits_{n\rightarrow\infty}{(a_n - 1)}\le 0.$$
Since, as noted in the question, $a_n\ge 1$ for all $n$, it follows that $\lim\limits_{n\rightarrow\infty}{a_n} = 1$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many possible ways of organizing the first round ($5$ matches) are there?
$10$ players play table tennis. How many possible ways of organizing the first round ($5$ matches) are there?
My attempt:
we can choose 2 players form 10 in the first match and 2 players from 8 (expect the first 2 players), and 2 players from 6 (expect the first 4 players)etc...
So the answer should be $\binom{10}{2}\times \binom{8}{2}\times \binom{6}{2}\times \binom{4}{2}\times \binom{2}{2}=113400 $
But the right answer is: $9\times7\times5\times3\times1=945$ which I don't understand
and I don't understand where is the mistake in my solution
I need an explanation to my mistake and the right solution please
| Double Factorial
Number the players from $1$ to $2n$. Player $1$ can play $2n-1$ people. For each of those choices, the next lowest unpaired player can play $2n-3$ people. For each of the choices so far, the next lowest unpaired player can play $2n-5$ people. Etc. Thus, there are
$$
\begin{align}
(2n-1)(2n-3)(2n-5)\cdots1
&=(2n-1)!!\\
&=\frac{(2n)!}{2^nn!}
\end{align}
$$
ways to arrange the first round. See Double Factorial.
Overcounting Using Binomial Coefficients
If we count the pairs that can be formed from $10$ people, $\binom{10}{2}$ and then the number of pairs that can be formed from $8$ people, $\binom{8}{2}$, etc., we have counted the arrangement
$$
\{(1,2),(3,4),(5,6),(7,8),(9,10)\}
$$
once for each permutation of the pairs. That is $5!$ times. Note that
$$
\begin{align}
\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}
&=\color{#C00000}{\frac{10}2}\cdot9\cdot\color{#C00000}{\frac{8}2}\cdot7\cdot\color{#C00000}{\frac{6}2}\cdot5\cdot\color{#C00000}{\frac{4}2}\cdot3\cdot\color{#C00000}{\frac{2}2}\cdot1\\
&=\color{#C00000}{5}\cdot\color{#C00000}{4}\cdot\color{#C00000}{3}\cdot\color{#C00000}{2}\cdot\color{#C00000}{1}\cdot9\cdot7\cdot5\cdot3\cdot1\\
&=\color{#C00000}{5!}\cdot9!!
\end{align}
$$
That is, $5!$ times the correct number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve this $6$-th degree polynomial equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$ The question is as follows:
Solve the equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$, one root being $\sqrt{2}+i$
It's trivial that another root is $\sqrt{2}-i$. But, I can go no further. Can anyone please help me how to solve this?
| In addition to $\sqrt{2} -i$, we have that $-\sqrt{2}+i$ and $-\sqrt{2}-i$ are also guaranteed to be roots. (These are the remaining Galois conjugates, which always take the specified forms in the case of a sum of square roots like $\sqrt{2}+i$.)
To determine the other roots, divide your given polynomial $f(x)$ by $$\begin{align}(x-\sqrt{2}-i)(x-\sqrt{2}+i)(x+\sqrt{2}-i)(x+\sqrt{2}+i)
&= (x^2-2\sqrt{2}x+3)(x^2+2\sqrt{2}x+3) \\
&= x^4-2x^2+9 \end{align}$$
This yields $f(x) = (x^2-x-6)(x^4-2x^2+9)$, so the remaining roots are the roots of $x^2-x-6 = (x-3)(x+2)$, namely -2 and 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The rank of a matrix, dependent on the value of $t$ I'm trying to analyse the rank of the following matrix, for $t\in \mathbb{R}$.
$$\begin{bmatrix}
t+3 & 5 & 6 \\
-1 & t-3 & -6 \\
1 & 1 & t+4
\end{bmatrix}$$
With $R_1\leftrightarrow R_3$, $-(t+3)R_1+R_3 \rightarrow R_3$, $R_1+R_2 \rightarrow R_2$, and finally $-R_2+R_3\rightarrow R_3$. I get
$$\begin{bmatrix}
1 & 1 & t-4 \\
0 & t-2 & t-2 \\
0 & 0 & -t^2-8t-4
\end{bmatrix}$$
And this makes that when $t=2$ or $t=2(-2 \pm \sqrt{3})$. However, these values are not what they are supposed to give(-2,2,4). Where did I make a mistake? Any help would be appreciated.
| \begin{align}
\begin{bmatrix}
t+3 & 5 & 6 \\
-1 & t-3 & -6 \\
1 & 1 & t+4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & t+4 \\
-1 & t-3 & -6 \\
t+3 & 5 & 6 \\
\end{bmatrix}
&&R_1\leftrightarrow R_3
\\
&\to
\begin{bmatrix}
1 & 1 & t+4 \\
0 & t-2 & t-2 \\
t+3 & 5 & 6 \\
\end{bmatrix}
&&R_2\gets R_2+R_1
\\
&\to
\begin{bmatrix}
1 & 1 & t+4 \\
0 & t-2 & t-2 \\
0 & 2-t & -t^2-7t-6 \\
\end{bmatrix}
&&R_3\gets R_3-(t+3)R_1
\\
&\to
\begin{bmatrix}
1 & 1 & t+4 \\
0 & t-2 & t-2 \\
0 & 0 & -t^2-6t-8 \\
\end{bmatrix}
&&R_3\gets R_3+R_1
\end{align}
The roots of $-t^2-6t-8$ are $-2$ and $-4$.
*
*If $t\notin\{2,-2,-4\}$ the rank is $3$.
*If $t=2$, the rank is $2$.
*If $t=-2$ or $t=-4$, the rank is $2$.
The result you're given is wrong. The numbers $-2$, $2$ and $4$ are the eigenvalues of
$$
A=\begin{bmatrix}
3 & 5 & 6 \\
-1 & -3 & -6 \\
1 & 1 & 4
\end{bmatrix}
$$
and your case is finding the eigenvalues of $-A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit $\lim_{n\rightarrow\infty}(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2...(1-\frac{1}{n(n+1)/2})^2$ Any hints on how to find the following limit ?.
I haven't been able to figure it out still.
$$
\lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2}
\left(\, 1 - \frac{1}{6}\, \right)^{2}\left(\, 1 - \frac{1}{10}\, \right)^{2}\ldots\left[\, 1 - \frac{1}{n\left(\, n + 1\, \right)/2}\, \right]^{2}
$$
Pardon me if it's a silly question.
Edit: Solved upto this much
$$
\prod_{n = 2}^{\infty}\left[\, 1-\frac{2}{n\left(\, n + 1\, \right)}\, \right] =
\prod_{n = 2}^{\infty}\left(\, \frac{n + 2}{n + 1}\,\frac{n - 1}{n}\, \right)
$$
I did'nt notice most of the terms cancel out.
| It may be interesting to see that, using the the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ we can prove that $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d
$$ so $$\prod_{n\geq2}\frac{\left(n+2\right)\left(n-1\right)}{\left(n+1\right)n}=\prod_{n\geq0}\frac{\left(n+4\right)\left(n+1\right)}{\left(n+3\right)\left(n+2\right)}$$ $$=\frac{\Gamma\left(3\right)\Gamma\left(2\right)}{\Gamma\left(4\right)\Gamma\left(1\right)}=\color{red}{\frac{1}{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.
| If $a$ is the smallest leg then you can find others in very simple way-
If $a$ is odd then -
$$b = \dfrac{n^2-1}{2};\; c = \dfrac{n^2+1}{2}\\
[\dfrac{n^2-1}{2}]^2 + n^2 = [\dfrac{n^2+1}{2}]^2$$
If $a$ is even then -
$$b = \dfrac{n^2}{4}-1;\; c = \dfrac{n^2}{4}+1\\
[\dfrac{n^2}{4}-1]^2 + n^2 = [\dfrac{n^2}{4}+1]^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2}}$ which evaluates to $1<0$Do you know what to do here ?
| An easier proof without induction:
Let $r$ be a positive integer, whence we have by the AM-GM inequality,
$\dfrac{(2r-1)+(2r+1)}{2}>\sqrt {(2r-1)(2r+1)}$
$\Rightarrow \dfrac{2r}{2r-1}>\sqrt {\dfrac{2r+1}{2r-1}}$
$\Rightarrow \displaystyle\prod_{r=1}^n\left( \frac{2r}{2r-1}\right)>\prod_{r=1}^n\left(\sqrt {\frac{2r+1}{2r-1}}\right)$
$\Rightarrow \dfrac{2.4.6...(2n)}{1.3.5...(2n-1)}>\sqrt {2n+1}$
$\Rightarrow \dfrac{1.3.5...(2n-1)}{2.4.6...(2n)}<\dfrac{1}{\sqrt{2n+1}}<\dfrac{1}{\sqrt {2n}}. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $\tan B$ but that doesn't help in getting the value of $\tan A + \tan B$. The value comes in terms of $\tan A$ or $\tan B$ but the expected answer is independent of any variable .
Also
$$\frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \tan(A) + \tan(B)$$
We could get a value only if instead of $\cos A$ there was $\sin B$ in the relation(which we get on adding the ratios)
| Something is wrong. I calculated
\begin{align}
\sin A &= \dfrac{\sqrt 3}{2}\sin B \\
\cos A &= \dfrac{\sqrt 5}{3}\cos B \\
\sin^2 A + \cos^2 A &= \dfrac 34 \sin^2 B + \dfrac59 \cos^2 B\\
1 &= \dfrac 34 - \dfrac{7}{36} \cos^2 B\\
\cos^2 B = -\dfrac{9}{7}
\end{align}
Which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Square root of a matrix. Determine all $A,B \in \mathbf{M}_{2}(\mathbf{R})$ such that $A^2+B^2=\begin{pmatrix} 22 & 44\\
14 & 28 \end{pmatrix}$ and $AB+BA=\begin{pmatrix} 10 & 20\\
2 & 4 \end{pmatrix}$.
I have tried to assume that $A=\begin{pmatrix} a & b\\
c & d \end{pmatrix}$ and $B=\begin{pmatrix} e & f\\
g & h \end{pmatrix}$. And we know that $(A+B)^{2}=A^{2}+B^{2}+AB+BA$ and $(A-B)^{2}=A^{2}+B^{2}-(AB+BA)$. Then, I will end up with $8$ equations and $8$ which can be solve, but it is really complicated. I am just wondering if someone could show me how to take the square root of an matrix. Please if you show that please make sure that you will prove it.
Would any one have an idea what is the input code for solving that in MATLAB
| Here is a general approach to find all square root of a $2\times2$ matrix $A$ which is not a multiple of the identity matrix.
Assume that $X^2 = A$. If we write $t = \operatorname{tr}(A)$ and $d = \det(A)$, then by the Cayley-Hamilton theorem we have
$$ X^2 - tX + dI = 0. $$
Plugging this to $X^2 = A$, we have $A = tX - dI$. Since $A$ is not a multiple of $I$, $t$ is automatically non-zero and we can write
$$X = \frac{1}{t}(A + dI). \tag{1}$$
So it suffices to determine $d$ and $t$. For $d$, we know that
$$d^2 = \det(X)^2 = \det(X^2) = \det(A) \qquad \Rightarrow \qquad d = \pm\sqrt{\det(A)}, \tag{2}$$
which can be computed. Similarly,
$$ \operatorname{tr}(A) = \operatorname{tr}(tX - dI) = t^2 - 2d \qquad \Rightarrow \qquad t = \pm \sqrt{2d + \operatorname{tr}(A)}. \tag{3}$$
This shows that if $X^2 = A$, then $X$ must be of the form $\text{(1)}$ with $d$ and $t$ as above. Conversely, if $d$ and $t$ are as above, then $X$ given by $\text{(1)}$ satisfies
\begin{align*}
X^2
&= \frac{1}{t^2}(A^2 + 2dA + d^2I) \\
&= \frac{1}{t^2}(\operatorname{tr}(A)A - \det(A)I + 2dA + d^2I) \\
&= \frac{1}{t^2}((t^2 - 2d)A - d^2I + 2dA + d^2I) \\
&= A.
\end{align*}
Therefore $X^2 = A$ if and only if $X$ is given by $\text{(1)}$ with $d$ and $t$ given by $\text{(2)}$ and $\text{(3)}$.
Now in your case,
$$ (A+B)^2 = \begin{pmatrix} 32 & 64 \\ 16 & 32 \end{pmatrix}
\quad \Rightarrow \quad d = 0
\quad \Rightarrow \quad t = \pm 8
\quad \Rightarrow \quad A+B = \pm \begin{pmatrix} 4 & 8 \\ 2 & 4 \end{pmatrix} $$
and likewise
$$ (A-B)^2 = \begin{pmatrix} 12 & 24 \\ 12 & 24 \end{pmatrix}
\quad \Rightarrow \quad d = 0
\quad \Rightarrow \quad t = \pm 6
\quad \Rightarrow \quad A-B = \pm \begin{pmatrix} 2 & 4 \\ 2 & 4 \end{pmatrix} $$
This will give you all the possible 4 choices of the pair $(A, B)$.
Remark. If you want to solve an equation of the form $X^2 = kI$, then the general form of $X$ is
$$ X = \begin{pmatrix}a & b \\ c & -a \end{pmatrix} \quad \text{or} \quad \pm\sqrt{k} I, $$
with $a^2 + bc = k$. So there may be an infinitely many solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the geometric series for q Is there a general way to find the $q > 0$ solving the equation from the geometric series $$1+q+q^2+q^3+\ldots + q^n = a$$ or $$\frac{1-q^{n+1}}{1-q} = a\quad\text{with } q \neq 1$$ for $a > 1$ and $n\in\mathbb N$?
My thoughts: Since polynomials aren't solvable in general for degree 5 or higher, I guess the above equation has no explicit solution for $n\ge 5$. In this case numerical approximations can be used. For $n=5$ also this method can be used.
| Re-arrange
\begin{align*}
q &= \frac{a-1}{a-q^n} \\
&= \frac{a-1}{a-\left(\frac{a-1}{a-q^n}\right)^n} \\
&= \frac{a-1}
{a-\left(
\frac{a-1}
{a-\left(
\frac{a-1}{a-\ddots}
\right)^n}
\right)^n} \\
&= \frac{a-1}{a}
\left \{
1+\frac{(a-1)^n}{a^{n+1}}+\ldots+
\frac{(nk+k)!}{k! (nk+1)!}
\left[ \frac{(a-1)^n}{a^{n+1}} \right]^{k}+\ldots
\right \}
\end{align*}
Lagrange inversion formula
\begin{align*}
f(q) &= aq-q^{n+1} \\
q &= f^{-1}(a-1) \\
&= \sum_{i=1}^{\infty} \frac{1}{i!} \frac{d^{i-1}}{dx^{i-1}} \left. \left(\frac{x-0}{f(x)-f(0)} \right)^{i} \right|_{x=0} [(a-1)-f(0)]^{i} \\
&= \sum_{i=1}^{\infty} \frac{1}{i!} \frac{d^{i-1}}{dx^{i-1}} \left. \left(\frac{1}{a-x^{n}} \right)^{i} \right|_{x=0} (a-1)^{i}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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What is the expansion $(x+a)^b$? I don't have strong math background. Whats is the expansion of the following equation: $(x+a)^b$.
| This seems a difficult problem for someone without a strong mathematical background. One way to approach this is to arrange $b$ copies of $(x+a)$, as follows:
$$
\underbrace{(x+a)(x+a) \cdots (x+a)}_{\text{$b$ copies}}
$$
The product of all of these sums $(x+a)$ will be a bunch of terms of the form $x^ka^{b-k}$, where $k$ is some integer between $0$ and $b$, inclusive. Each of the copies of $(x+a)$ can contribute either a factor of $x$, or a factor of $a$. We then count up the number of ways to choose $k$ factors of $x$ and $b-k$ factors of $a$, and that will give us the coefficient of $x^ka^{b-k}$. This works, ultimately, because multiplication distributes over addition: $u(v+w) = uv+uw$.
So, for instance, there is only one way to choose the $x$ from each of $b$ copies of $x+a$; those $b$ factors of $x$ multiply to $x^b$, so the coefficient of $x^b$ is just $1$. On the other hand, since there are $b$ copies, there are $b$ ways to choose the $a$ from one of those $b$ copies (and consequently choose the $x$ from the remaining $b-1$ copies), so the coefficient of $x^{b-1}a$ is $b$. Therefore, the expansion of $(x+a)^b$ must begin
$$
x^b+bx^{b-1}a+\cdots
$$
We can continue along the rest of the terms in the same way. It turns out that there is a fairly simple formula for the number of ways to choose $k$ factors of $x$ and $b-k$ factors of $a$ from a total of $b$ copies of $(x+a)$; that formula is
$$
\binom{b}{k} = \frac{b!}{k!(b-k)!}
$$
which is in fact read "$b$ choose $k$", and where $b!$ is read "$b$ factorial" and represents the product $b \times (b-1) \times (b-2) \times \cdots \times 2 \times 1$. Thus, the overall formula for $(x+a)^b$ is
$$
\binom{b}{b}x^ba^0 + \binom{b}{b-1}x^{b-1}a^1 + \binom{b}{b-2}x^{b-2}a^2 +
\cdots + \binom{b}{1}x^1a^{b-1} + \binom{b}{0}x^0a^b
$$
which we write in shorthand as simply
$$
(x+a)^b = \sum_{k=0}^b \binom{b}{k} x^ka^{b-k}
$$
| {
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"url": "https://math.stackexchange.com/questions/1904633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the limit of $1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)$
Compute the limit: $$\lim_{n \to \infty} \left(1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)\right) $$
Can someone help me to solve this limit? I forgot how to manupulate fractions in limit calculus.
| If you want to go beyond the limit itself, write $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=1-n \log\left(1+ \frac{2}{2n - 1}\right)$$ Now, remembering that, for small values of $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ relace $x$ by $\frac{2}{2n - 1}$ which makes $$\log \left(\dfrac{2n + 1}{2n - 1}\right)=\log\left(1+ \frac{2}{2n - 1}\right)=\frac{2}{2n - 1}-\frac 12 \left(\frac{2}{2n - 1} \right)^2+\cdots$$ Now, use long division to get $$\log \left(\dfrac{2n + 1}{2n - 1}\right)=\frac{1}{n}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ So, for large values of $n$ $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=-\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$ Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.
I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers will, when plotted in 3D space, generate a hypotenuse whose length is also an integer. For example; $\sqrt {6^2 + 42^2 + 85^2} = 95$
What I've found is quite interesting. It seems that there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x=2^y$ or $x=5 \times 2^y$ and $y$ is an integer.
$$\begin{array}{c|c|c}
y & 2^y & 5 \times 2^y \\ \hline
0 & 1 & 5 \\ \hline
1 & 2 & 10 \\ \hline
2 & 4 & 20 \\ \hline
3 & 8 & 40 \\ \hline
4 & 16 & 80 \\ \hline
5 & 32 & 160 \\ \hline
6 & 64 & \text{see below} \\ \hline
7 & 128 & \text{see below} \\
\end{array}$$
So, there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x$ equals any value from the following series: $1,2,4,5,8,10,16,20,32,40,64,80,128,160...$
I wrote a program which brute-force checked this for values up to $250$ for $a, b, \text {and } c$ (which is why I show no solutions to $5 \times 2^x$ using $6$ and $7$ above, since they should be $320$ and $640$ respectively). The program found at least 1 solution for all other integers from $1 \text{ to }250$.
Here's the code I used (excel VBA): http://pastebin.com/jVA27jYp
| WLOG, $a,b,c,y$ are positive. If there exists a solution to $a^2+b^2+c^2\in \{4^y, 25\cdot 4^y\}$ then for some $y$ then there is a solution with a least value of $\min (a,b,c).$ Such a solution cannot have $a,b,c$ all even, else $(a/2)^2+(b/2)^2+(c/2)^2\in \{4^{y-1},4^{y-1}\cdot 25\},$ which contradicts the minimality of $\max (a,b,c)$, unless $y=1$.But there is no solution with $y=1.$
But if $a,b,c$ are not all even then $a^2+b^2+c^2$ is congruent to $1,2,$ or $3$ (modulo $4$) so it cannot be divisible by $4,$ so it can't belong to $\{4^y,25\cdot 4^y\}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $0Prove that if $0<a<b$, then
$a<\sqrt{ab}<\dfrac{a+b}{2}<b.$
I have seen that the inequality $\sqrt{ab}\leq (a+b)/2$ is verified for all $a,b\geq 0$.
| This are $3$ inequalities, and $1$ is already solved. $2$ to go, lets solve them by equivalence transformations:
The first one ($a < \sqrt{ab}$):
\begin{align}
a &< \sqrt{ab} \\
a^2 &< ab \\
a &< b
\end{align}
The last one ($\dfrac{a+b}{2} < b$):
\begin{align}
\dfrac{a+b}{2} &< b \\
a + b &< 2 \cdot b \\
a &< b
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$
1.$\frac{3}{2}$
2.$\frac{\sqrt[3]{65}}{4}$
3.$\sqrt[3]{2}$
4.$1$
I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$
| Note your number $\alpha$
It turns out that (after some calculation) that $\alpha^3=10-9\alpha$
1 seems to do the trick...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-4^n=9m,\text{where $m$ is an integer.}$$
What I am thinking in the $n+1$ step is,
\begin{align}
& 5^{n+2}-4^{n+2} \\
= {} & 5^2(5^n-4^n)+5^24^n-4^{n+2} \\
= {} & 5^2(5^n-4^n)+4^n(5^2-4^2) \\
= {} & 5^29m+4^n9 \\
= {} & 9(5^2m+4^n)
\end{align}
Does this approach make sense?
*Show that $a+b$ is a factor of $a^n+b^n$ where $n$ is a positive odd number.
I am thinking this in the $n+1$ step.
$$a^{2n+1}+b^{2n+1}$$
But then I cannot get it further.
| I'll give you a hint for the first one. Since you have asked three different questions, I wait to see if they tell you to split your question in there questions...
1
You already have a formula:
$$1^3 + 2^3 + 3^3 + \ldots + n^3 = \frac{n^2((n+1)^2)}{4}$$
Hence in your case $n = 30$ but your sum starts from $2^3$ and not from $1^3$, hence you have to subtract $1^3$ (namely one):
$$2^3 + 3^3 + \ldots + 30^3 = \frac{30^2(30+1)^2}{4} - 1 = \frac{30^2\cdot 31^2}{4} - 1 = 216224$$
Edit because I understood nothing
Since you want only the sum of even numbers, then you have:
$$2^3 + 4^3 + 8^3 + \ldots + 30^3 = 2^3(1^3 + 2^3 + 3^3 + \ldots + 15^3) = 115200$$
Thanks to Crostule for notifying me.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can we show that $\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)$ $$\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)\tag1$$
Any hints?
| This is a result of theory of theta functions. Let $0 < q < 1$ and $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})}$$ Next we can see that \begin{align}
b(q) &= \sum_{n = 1}^{\infty}(-1)^{n - 1}\cdot\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= \sum_{n \text{ odd}}\frac{q^{n}}{n(1 - q^{n})} - \sum_{n \text{ even}}\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} - 2\sum_{n \text{ even}}\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= a(q) - a(q^{2})\notag
\end{align}
Similarly we can prove that $$c(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})} = a(q) - 2a(q^{2})$$ and $$d(q) = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{q^{n}}{n(1 + q^{n})} = c(q) - c(q^{2})$$ i.e $$d(q) = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{q^{n}}{n(1 + q^{n})} = a(q) - 3a(q^{2}) + 2a(q^{4})$$ The sum in question is $$S = 12b(q) + 4d(q) = 16a(q) - 24a(q^{2}) + 8a(q^{4})\tag{1}$$ where $q = e^{-\pi}$. We can see from this post that
\begin{align}
a(q) &= -\frac{\log kk'^{4}}{12} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{24K}\tag{2}\\
a(q^{2}) &= -\frac{\log kk'}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{3}\\
a(q^{4}) &= -\frac{\log k^{4}k'}{12} + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{4}
\end{align}
Since $q = e^{-\pi}$ it follows from theory of theta functions and elliptic integrals that $K' = K$ and $k = k' = 1/\sqrt{2}$. Thus $$a(q) = -\frac{\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{24}$$ and $$a(q^{2}) = -\frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{12}$$ and $$a(q^{4}) = \frac{3\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{6}$$ and then from $(1)$ we get $S = \log 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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when does the following system has infinite solutions I was trying to solve the following exercise but I couldn't get anywhere:
Find the values of a and b for which the following system has infinite solutions:
$2x+3y-z=a$
$3x-by+z=1$
$ax-y-z=2$
I tried to solve the problem using the determinant, I mean we know that $Ax=b$ has infinite solutions if $|A|=0$ but at the end I get an expression that I'm not able to reduce.
I would really appreciate any help or advice you could give me.
Many thanks
| Idea
Write $$A=\begin{pmatrix} 2 & 3 & -1 \\ 3 & -b & 1 \\ a & -1 & -1\end{pmatrix}, B=\begin{pmatrix} 2 & 3 & -1 & a \\ 3 & -b & 1 & 1 \\ a & -1 & -1 & 2\end{pmatrix}.$$
The system has infinitely many solutions if and only if $ran(A)=ran(B)<3.$ Since $A$ has a 2-order minor that non vanishes, $$A=\begin{pmatrix} \color{red}{2} & 3 & \color{red} {-1} \\ \color{red}{3} & -b & \color{red}{1} \\ a & -1 & -1\end{pmatrix},\begin{vmatrix}2 & -1 \\ 3 & 1\end{vmatrix}\ne 0,$$ the only posibility is $ran(A)=ran(B)=2.$ So, that happens if and only if
$$\begin{vmatrix} 2 & 3 & -1 \\ 3 & -b & 1 \\ a & -1 & -1\end{vmatrix}=\begin{vmatrix} 2 & -1 & a \\ 3 & 1 & 1 \\ a & -1 & 2\end{vmatrix}=0.$$
I hope you can finish.
| {
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Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything simpler. Also tried expressing $x^3$ as $\sqrt{x^6}$, but that doesn't really help either.
One would think that, as $x^6$ grows more quickly than $x^3$ by a factor of $x^3$, the contribution of the $x^3$ term to the term in the square root would be dwarfed by the contribution of the the $x^6$ term, so the overall behavior of the first term in the limit would "behave" like $x^3$, as x gets bigger and bigger, so I would think intuitively that the limit would evaluate to 0.
| $$\sqrt{x^6-9x^3}-x^3=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3}$$
As $x$ goes to $\infty$, $\sqrt{x^6-9x^3} \approx x^3.$
More, precisely, divide the numerator and denominator by $x^3$:
\begin{align}
\sqrt{x^6-9x^3}-x^3&=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}\\&=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3} \\ &
=\frac{-9}{\sqrt{1-9x^{-3}}+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Computing $\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \frac{(x+2)^{2017}}{(x-2)^{2015}}$ I'm studying for an exam, but I have trouble with computing the following limit:
$$\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \frac{(x+2)^{2017}}{(x-2)^{2015}}$$
I tried directly plugging in a $0$, but that just results in $\frac{0}{0}$. Using L'Hospital's rule doesn't seem like it would help simplify this. Any help would be appreciated.
| As pointed out, you can pull out the terms on the right to get a factor of $-4$.
Next, $$\sqrt{2-e^{x^2}} = \sqrt{2 - 1 - x^2 - \frac{x^4}{2} + O(x^6)} = 1 - \frac{x^2}{2} - \frac{3x^4}{8} + O(x^6)\\
\cos x - \sqrt{2-e^{x^2}} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - 1 + \frac{x^2}{2} + \frac{3x^4}{8} + O(x^6) = \frac{5x^4}{12} + O(x^6)
$$
For the denominator:
$$
\log(\cos(x)) = \log \left(1-\frac{x^2}{2} +\frac{x^4}{24} + O(x^6) \right) = \left(-\frac{x^2}{2} +\frac{x^4}{24} \right)- \frac12 \left(-\frac{x^2}{2} +\frac{x^4}{24} \right)^2+O(x^6)\\=-\frac{x^2}2 - \frac{x^4}{12}+ O(x^6) \\
\log(\cos(x)) +\frac12x\sin x =-\frac{x^2}2 - \frac{x^4}{12} + \frac{x^2}2 - \frac{x^4}{12}+ O(x^6) = -\frac{x^4}{6} + + O(x^6)
$$
Putting everything together, the $x^4$ cancel each other and we get
$$
\frac{(-4)(\frac5{12})}{ (-\frac16) }= 10
$$
Please give Dr. Graubner the credit, he got to his answer before I got to the correct value for my answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920673",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
| There's also another more algebraic way. You can easily show (by expansion) that if $$a+b+c=0$$ then $$a^3+b^3+c^3=3abc$$
Since, in your problem, for every $A$ $$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$ Then you can use the above identity
$$\cos^3{A}+\cos^3{(A+2\pi/3)}+\cos^3{(A-2\pi/3)}=3\cos{A}\cos{(A+2\pi/3)}\cos{(A-2\pi/3)}$$
By using the previously mentioned identity
$$\cos{x}\cos{y}=\frac{1}{2}\left(\cos{(x+y)}+\cos{(x-y)}\right)$$
you can simplify the RHS to this
$$\frac{3}{2}\cos{A}\left(\cos{(2A)}-\frac{1}{2}\right)$$
and then (using it again)
$$\frac{3}{4}\left(\cos{3A}+\cos{(A)}\right)-\frac{3}{4}\cos{A}=\frac{3}{4}\cos{(3A)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
Symmetric polynomials with Vieta's and Newton's theorems
Let $ x_{1}, x_{2}, x_{3}$ be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$
and also
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$
| For the first case, using Vieta's Formula $$x_1x_2x_3=1\implies\dfrac1{x_1x_2}=x_3$$
For the second,
Method$\#1$: observe that $$x_r^3-3x_r^2+x_r-1=0\implies x_r^3=3x_r^2-x_r+1\ \ \ \ (r)$$ for $r=1,2,3$
$(1)+(2)+(3)=?$
Now $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_3x_1)$
Method$\#2$:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$a^2+b^2+c^2-ab-bc-ca=(a+b+c)^2-3(ab+bc+ca)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Help on a very interesting integral: $\int_0^1\frac{x^5(1-x)^4}{1+x^3}\,dx$ I've been starting to work on problems with difficult integrals, and I came to
one of my problems where I end up having to integrate
$$\int_0^1 \frac{x^5(1-x)^4}{1+x^3} dx.$$
Wolfram seems to time out on integrals like these, so I decided to tackle this
integral by myself.
It seems like to me that there should be a meaningful integration by parts
that occurs here, in particular a relationship that could be developed between
$x^5$ and $\frac{(1-x)^4}{1+x^3}.$ The problem is, I'm having some difficulties
with the difficulty of building a meaningful integral of $dv = \frac{(1-x)^4}{
1+x^3} dx.$ any recommendations on how to go about starting with direction on
this problem?
| In general, any integral of the form $\int_{0}^{1}\frac{p(x)}{1+x^3}\,dx $ with $p(x)\in\mathbb{Q}[x]$ can be written as a linear combination with rational coefficients of $1$ and
$$ I_1=\int_{0}^{1}\frac{x}{1+x^3}\,dx=\frac{1}{9} \left(\pi \sqrt{3} -3\log 2\right),\qquad I_2= \int_{0}^{1}\frac{x^2}{1+x^3}\,dx=\frac{1}{3}\log(2) $$
$$ I_0=\int_{0}^{1}\frac{1}{1+x^3}\,dx = \frac{1}{9} \left(\pi\sqrt{3} +3\log 2\right)$$
and we just need to perform a polynomial division betwen $p(x)$ and $(1+x^3)$ to compute the original integral. Notice that $I_2$ is trivial and
$$ I_1+I_0 = \int_{0}^{1}\frac{dx}{1-x+x^2}=\frac{2\pi}{3\sqrt{3}}, $$
$$ I_1-I_0 = \int_{0}^{1}\frac{x-1}{x^3+1}=-\sum_{n\geq 0}\left(\frac{1}{6n+1}-\frac{1}{6n+2}-\frac{1}{6n+4}+\frac{1}{6n+5}\right)=-\frac{2\log 2}{3}$$
are quite simple to prove.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that
| Diophantus has already shown that
$$(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}.$$ This proves the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to integrate $ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $? I am having a little problem with my maths homework. The problem is as follows:
\begin{equation}
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
\end{equation}
I tried to do the following but got stuck halfway:
Let $\ \ x \ = asin\theta, \ hence, \ dx = acos\theta \ d\theta $
$
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
$
$
= \int^\frac{\pi}{2}_0{\cfrac{acos\theta}{asin\theta \ + \ \sqrt{a^2 \ - \ a^2sin^2\theta}}}\ d\theta
$
$
= a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ \sqrt{a^2cos^2\theta}}}\ d\theta
$
$
= a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ acos\theta }}\ d\theta
$
$
= \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{sin\theta \ + \ cos\theta }}\ d\theta \\ \\
$
$
= \int^\frac{\pi}{2}_0{\left(1 \ + \ \cfrac{(cos\theta \ - \ sin\theta)}{sin\theta \ + \ cos\theta } - \cfrac{cos\theta}{sin\theta \ + \ cos\theta}\right)}\ d\theta
$
Could someone please advise me how to solve this problem?
| Try Euler substitutions. The second one might do the job.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
} |
Basic Complex inequalities It is given that $|z+2+3i| \leq 1$. Show that $3 \leq |2z+1+2i| \leq 7$. I really do not have a clue how to start on this but have proceeded by noting that
$$(2z+1+2i) = z+2+3i + z-1+i$$
and so
$$|2z+1+2i| \leq |z+2+3i| + |z-1+i|$$
But this doesn't get me too far. Can anyone help me complete this proof please ?
| We have that
$$ |2z+1+2i|=2\left|z+\frac{1}{2}+i\right|=2\left|z+\frac{1}{2}+i\pm(\frac{3}{2}+2i)\right|\\=2\left|(z+2+3i)+\left(-\frac{3}{2}-2i\right)\right|.$$
Note that $|-\frac{3}{2}-2i|=\frac{5}{2}$. Now recall that $||u|-|w||\leq |u+w|\leq |u|+|w|$.
Hence if $|z+2+3i|\leq 1$, then
$$3=2\left(\frac{5}{2}-1\right)\leq |2z+1+2i|\leq 2\left(1+\frac{5}{2}\right)=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n$ for all $n \in \mathbb{N}$ By calculating for $n\in \{1,2,3,4,5,6,7\}$, I've formulated the rule
\begin{equation}
F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n,
\end{equation}
where $F_n$ is the $n$th fibonacci number. I want to show that this is true for all $n \in \mathbb{N}$.
I tried using induction, with $n=1$ as the basis step, but didn't get very far:
For the induction step, we assume the formula holds for a $n = k$, and checks for $n=k+1$:
\begin{align*}
F_{k} \cdot F_{k+2} &= F_{k} \cdot (F_{k+1} + F_{k}) \\
&= F_k \cdot F_{k+1} + F_k^2 \\
\end{align*}
If somehow $F_k \cdot F_{k+1} = (-1)^k$, then I would be done. But I don't see how that's possible.
Is there a better way of proving this, maybe without using induction? Or am I just going about it the wrong way?
| We can prove the rule without induction using Binet's formula
$$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$
where
$$\varphi = \frac{1 + \sqrt{5}}{2} \qquad \psi = \frac{1 - \sqrt{5}}{2} = -\frac{1}{\varphi}$$
so that
$$\varphi\psi = -1 \qquad \frac{\varphi}{\psi} + \frac{\psi}{\varphi} = -3 = 2 - 5$$
Then
\begin{align}
F_{n+1} F_{n-1} & = \frac{\varphi^{n+1} - \psi^{n+1}}{\sqrt{5}} \cdot \frac{\varphi^{n-1} - \psi^{n-1}}{\sqrt{5}}\\
& = \frac{1}{5}\left(\varphi^{2n} - \varphi^{n+1}\psi^{n-1} - \varphi^{n-1}\psi^{n+1} + \psi^{2n}\right)\\
& = \frac{1}{5}\left(\varphi^{2n} - \left(\frac{\varphi}{\psi} + \frac{\psi}{\varphi}\right)\varphi^n\psi^n + \psi^{2n}\right)\\
& = \frac{1}{5}\left(\varphi^{2n} - (2-5)\varphi^n\psi^n + \psi^{2n}\right)\\
& = \frac{1}{5}\left(\varphi^{2n} - 2\varphi^n\psi^n + \psi^{2n} + 5\varphi^n\psi^n\right)\\
& = \left(\frac{\varphi^n - \psi^n}{\sqrt{5}}\right)^2 + \frac{1}{5}\cdot5(\varphi\psi)^n\\
& = F_n^2 + (-1)^n
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}\le3\sqrt{\frac{2(a^2+b^2+c^2)}3}$$
| Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sum\limits_{cyc}\sqrt{2x^2+y^2+z^2+2xy+2xz}\leq(2+\sqrt2)(x+y+z)$$
By C-S $$\left(\sum\limits_{cyc}\sqrt{2x^2+y^2+z^2+2xy+2xz}\right)^2\leq\sum\limits_{cyc}\frac{2x^2+y^2+z^2+2xy+2xz}{\sqrt2x+y+z}\sum\limits_{cyc}(\sqrt2x+y+z)$$
Thus, it remains to prove that
$$\sum\limits_{cyc}\frac{2x^2+y^2+z^2+2xy+2xz}{\sqrt2x+y+z}\leq(2+\sqrt2)(x+y+z)$$ or
$$\sum\limits_{sym}((1+\sqrt2)x^3y+2.5x^2y^2+(3\sqrt2+1.5)x^2yz)\geq0$$
which is obvious. Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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How many ways are there to distribute $12$ red jelly beans to four children, a pair of identical female twins and a pair identical male twins?
Question: How many ways are there to distribute $12$ red jelly beans to four children, a pair of identical female twins and a pair identical male twins?
I attempted to solve this problem. Here is my solution:
Distribute $n$ jelly beans to female twins and $12-n$ to male twins(Maybe twins are supposed to be identical if their sex is same). If $n$ is even, then the number of ways to distribute $n$ jelly beans to female twins is $\frac{n}{2}+1$ and if $n$ is odd, the number of ways to distribute $n$ jelly beans to female twins is $\frac{n+1}{2}$. Thus the number of ways to distribute 12 jelly beans to a pair of identical female twins and a pair of identical male twins is
\begin{align}
&\sum_{n\text{ even}\le 12}\left(\frac{n}{2}+1\right)\left(\frac{12-n}{2}+1\right)+\sum_{n\text{ odd}\le 12}\left(\frac{n+1}{2}\right)\left(\frac{13-n}{2}\right)\\
&=\sum_{k=0}^6 (k+1)(7-k) + \sum_{k=1}^6 k(7-k)\\
&=7+\sum_{k=1}^6 (7+6k-k^2)+\sum_{k=1}^6 (7k-k^2)\\
&=7+\sum_{k=1}^6 (7+13k-2k^2)\\
&=7+ 7\cdot 6 + 13\cdot \frac{6\cdot 7}{2} - 2 \cdot \frac{6\cdot 7\cdot 13}{6}\\
&=7+42+273-182\\
&=140.
\end{align}
However, I want to solve it by applying Polya enumeration theorem, but I have no idea.
| Instead of Polya enumeration theorem, I found an answer using Burnside's lemma. Define sets
$$
X=\{(f_1,f_2,m_1,m_2):f_1,f_2,m_1,m_2\text{ are integers and }f_1+f_2+m_1+m_2=12\}
$$
and
$$
G=\{e,g_1,g_2,g_3\},
$$
where $e,g_1,g_2,g_3$ are functions from $X$ to $X$ such that
\begin{align}
e(f_1,f_2,m_1,m_2)&=(f_1,f_2,m_1,m_2),\\
g_1(f_1,f_2,m_1,m_2)&=(f_2,f_1,m_1,m_2),\\
g_2(f_1,f_2,m_1,m_2)&=(f_1,f_2,m_2,m_1),\\
g_3(f_1,f_2,m_1,m_2)&=(f_2,f_1,m_2,m_1).
\end{align}
$G$ is a group under function composition, because the Cayley table of $G$ is
$$
\begin{array}{c|cccc}
\circ & e & g_1 & g_2 & g_3\\
\hline
e & e & g_1 & g_2 & g_3\\
g_1 & g_1 & e & g_3 & g_2\\
g_2 & g_2 & g_3 & e & g_1\\
g_3 & g_3 & g_2 & g_1 & e
\end{array}$$
Thus we can define a group action of $G$ on $X$ by $g\cdot x=g(x)$. By Burnside's lemma, the number of orbits is
\begin{align}
&\frac{1}{|G|}(|F_{e}|+|F_{g_1}|+|F_{g_2}|+|F_{g_3}|)\\
&= \frac{1}{4}\left(\left(\binom{4}{12}\right)+2\left(\left(\binom{2}{12}\right)+\left(\binom{2}{10}\right)+\left(\binom{2}{8}\right)+\left(\binom{2}{6}\right)+\left(\binom{2}{4}\right)+\left(\binom{2}{2}\right)+\left(\binom{2}{0}\right)\right)+\left(\binom{2}{6}\right)\right)\\
&=\frac{1}{4}\left(\binom{15}{12}+2\left(\binom{13}{12}+\binom{11}{10}+\binom{9}{8}+\binom{7}{6}+\binom{5}{4}+\binom{3}{2}+\binom{1}{0}\right)+\binom{7}{6}\right)\\
&=\frac{1}{4}(455+2(13+11+9+7+5+3+1)+7)\\
&=140,
\end{align}
where $F_g=\{x\in X:g\cdot x=x\}$ and $\left(\binom{n}{r}\right)$ the number of combinations with repetition.
| {
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"timestamp": "2023-03-29T00:00:00",
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Not getting the correct answer for an infinite limit problem The problem:
$\lim _{ x\to -\infty } \frac { \sqrt { 9x^{ 6 }-x } }{ x^{ 3 }+8 }$
My answer: $3$
What I did:
$\\ \lim _{ x\to -\infty } \frac { \sqrt { x^{ 6 }(9-\frac { 1 }{ { x }^{ 5 } } ) } }{ x^{ 3 }(1+\frac { 8 }{ { x }^{ 3 } } ) } \\ \\ \lim _{ x\to -\infty } \frac { { x }^{ 3 }\sqrt { 9-\frac { 1 }{ { x }^{ 5 } } } }{ x^{ 3 }(1+\frac { 8 }{ { x }^{ 3 } } ) } \\ \lim _{ x\to -\infty } \frac { \sqrt { 9-\frac { 1 }{ { x }^{ 5 } } } }{ 1+\frac { 8 }{ { x }^{ 3 } } } \\ $
And since
$\\ \lim _{x\to-\infty} \frac{1}{{x}^{5}}\quad =\quad 0\\ \\ \lim _{x\to -\infty} \frac{8}{{x}^{3}} \quad =\quad 0\\ \\ $
plugging in $0$ for above values...
$\\ \frac {\sqrt{9}}{1} \quad =\quad 3\\$
The correct answer is apparently $-3$, I don't understand why.
| Hint: use that $\sqrt{x^6}=|x^3|$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$ The main question is:
Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$
My approach:
Initially, nothing clicked, so I went forward with simplifying the series.
So, we get, after simplifying,
$S$(The sum of series) = $1+3+6+10+15+21+\dots$
So, I try to write $S$ as,
$1+(4-1)+(9-3)+(16-6)+\dots$
So what I finally get is,
$S$=$(1+4+9+16+25+\dots) - (1+3+6+10+\dots)$
Therefore,
$$2S=\sum_{n=1}^n{n^2}$$
Thus,
$S$ = $\frac{n(n+1)(2n+1)}{12}$
But, the answer given in my textbook is
$\frac{n(n+1)(n+2)}{6}$
My answer is not matching. Please help me by pointing out my mistake or providing a new approach and solution altogether.
| Hint
$$1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4} \\
1+2+3+..+n=\frac{n(n+1)}{2}$$
Edit:
You can then use the identity (thanks to Winther for fixing the typo)
$$n(n+1)= \frac{1}{3} \left[(n+1)^3-n^3 \right]-\frac{1}{3}$$
to evaluate the new sum.
| {
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"url": "https://math.stackexchange.com/questions/1931771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Initial-value problem $y' = Ax^py$ with initial condition $y(2)=1$
Assuming $p\neq-1$, find the solution $y=y(x)$ to the initial-value problem:
$$\frac{dy}{dx}= Ax^py$$
$$y(2)=1$$
in terms of $A$ and $p$.
So far I have, after rearranging the problem and integrating, came up with this equation:
$$\ln(y) = \frac{Ax^{p+1}}{p+1} + C$$
Simplifying it becomes:
$$ y = e^{\frac{Ax^{p+1}}{p+1} + C}$$
In which case if you plugged in $y(2) = 1$, $\frac{Ax^{p+1}}{p+1} + C$ must $ = 0$. In which case C must be $=-\frac{Ax^{p+1}}{p+1}$ and the entire solution would be $$y = \frac{Ax^{p+1}}{p+1} -\frac{Ax^{p+1}}{p+1}$$
However this does not appear to be the correct answer. What am I doing wrong?
| You have almost correctly solved the problem except the last step.
$$\ln(y) = \frac{Ax^{p+1}}{p+1} + C$$
Plugging in the value $y(2) = 1$, we get $$\ln(1) = \frac{A\cdot2^{p+1}}{p+1} + C \implies C=-\frac{A\cdot(2)^{p+1}}{p+1}$$
So the required solution is $$\ln(y) = \frac{Ax^{p+1}}{p+1}-\frac{A\cdot2^{p+1}}{p+1}= \frac{A(x^{p+1}-2^{p+1})}{p+1}$$
$$\implies y=\exp\left[\frac{A(x^{p+1}-2^{p+1})}{p+1}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-3a^2+9$ a prime number?
Here is what I did: $$a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a)$$ To find $a$, I looked at the following situations:
$$a^4-3a^2+9=1$$
$$a^4-3a^2+9=3$$
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$
| HINTS
Your factorisation is the key: $a^4 - 3a^2 + 9 \equiv (a^2-3a+3)(a^2+3a+3)$.
Since $a$ is an integer, so are both $a^2-3a+3$ and $a^2+3a+3$.
If both $a^2-3a+3$ and $a^2+3a+3$ are bigger than $1$ then $a^4 - 3a^2 + 9$ will have two positive integer factors larger than one, i.e. $a^4 - 3a^2 + 9$ won't be prime. (Consider only positive $a$.)
You could prove this by exhaustion. Look at when $a^2-3a+3 > 1$ and when $a^2+3a+3>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do this?
My method:
$$
{f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x
$$
and
$$
f''(x)=(-x)'=-1\
$$
This took me much more than 1.5 hours just to type into LaTex :'(
| Take advantage of logarithm properties.
$$\ln\left(\frac{x}{\sqrt{x^2 + 1}}\right) = \ln(x) - \frac12\ln(x^2 + 1)$$
Then the derivative is easy:
$$f'(x) = \frac1x - \frac{x}{x^2 + 1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
| Also we can us C-S and $uvw$:
We need to prove that $$\sum\limits_{cyc}\left(\frac{a}{a+b^2}-1\right)\leq\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}$$ or
$$\sum\limits_{cyc}\frac{b^2}{a+b^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$
Now by C-S:
$$\sum\limits_{cyc}\frac{b^2}{a+b^2}\geq\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}=\frac{a+b+c}{(a+b+c)^2+a^2+b^2+c^2}$$
Id est, it remains to prove that
$$\frac{a+b+c}{(a+b+c)^2+a^2+b^2+c^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a decreasing function,
which says that it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$.
Since $a$, $b$ and $c$ are positive roots of the equation
$$(x-a)(x-b)(x-c)=0$$ or
$$w^3=x^3-3ux^2+3v^2x$$
we see that $w^3$ gets a maximal value for equality case of two variables
and since $f(w^3)\geq0$ is an homogeneous inequality, it's enough to prove it for $b=c=1$, which gives
$$(a-1)^2(2a^2+3a+6)\geq0$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.
I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=xy \Rightarrow 12 \sqrt{x^2+y^2} = xy$
Substitute $y=5+x$ and square both sides:
$144 (x^2 + (5+x)^2)=x^2 (5+x)^2 \Rightarrow x^4+10x^3-263x^2-1440x-3600=0$
Which only positive solution is $x=15$ and therefore $y=20$ and the area is $\frac{15 \cdot 20 }{2}=150$
Thanks in advance.
| you have to ways to arrive at the area.
$A = \frac 12 xy = 6\sqrt{(x^2 + y^2)}$
and you know: $(x-y) = 5$
$(x-y)^2 = 25\\
x^2 + y^2 - 2xy = 25\\
x^2 + y^2 = 25 + 4A$
Plug this into the equation above for areas.
$A = 6\sqrt{25 + 4A}$
square both sides and solve the quadratic
$A^2 = 36(25 + 4A)\\
A^2 - 144A - 900 = 0\\
(A - 150)(x+6) = 0\\
A = 150$
| {
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"timestamp": "2023-03-29T00:00:00",
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why the sequence $\{b_{n}\}$ can't periodic sequence Let $$a_{1}=1,a_{n+1}-a_{n}=a_{\lfloor\frac{n+1}{2}\rfloor}$$
if $a_{n}=\left(\overline{b_{1}b_{2}\cdots b_{n}}\right)_{10}$
show that the sequence $\{b_{n}\}$ can't periodic sequence
A sketch of my thoughts:
$a_{1},$so $ b_{1}=1$
$a_{2}-a_{1}=a_{1},a_{2}=2,$ so $b_{2}=2$
$a_{3}=a_{2}+a_{1}=3$ so $b_{3}=3$
$a_{4}=a_{3}+a_{2}=3+2=5$,so $b_{4}=5$
$a_{5}=a_{4}+a_{2}=7,b_{5}=7$
$a_{6}=a_{5}+a_{3}=7+3=10$ so $b_{6}=0$
$a_{7}=a_{6}+a_{3}=10+3=13$ so $b_{7}=3$
$a_{8}=a_{7}+a_{4}=13+5=18$,so $b_{8}=8$
$a_{9}=a_{8}+a_{4}=18+5=23$,so $b_{9}=3$
$a_{10}=a_{9}+a_{5}=23+7=30$,so $b_{10}=0$
$a_{11}=a_{10}+a_{5}=30+7=37$,so $b_{11}=7$
$$\cdots\cdots$$
| Assume $a_n\bmod 10$ is (eventually) periodic with period $p$.
Then so is $a_{n+1}-a_n\bmod 10$, i.e., $a_{\lfloor(n+1)/2\rfloor}\bmod 10$.
If $p$ is even, this implies that $a_n\bmod 10$ is eventually periodic with period $\frac p2$.
Therefore, if the sequence is eventually periodic then its minimal periof $p$ is odd.
This odd $p$ is also a period of $a_{n+2}-2a_{n+1}+a_{n}\bmod 10 = a_{\lfloor (n+2)/2\rfloor}-a_{\lfloor (n+1)/2\rfloor}\bmod 10$. As this latter sequence is $0$ for all odd $n$, we conclude that it is zero for almost all $n$.
This means that $a_{n+1}-a_n\bmod 10$ is eventually constant, and hence $a_{\lfloor(n+1)/2\rfloor}\bmod 10$ is eventually zero.
But in order to have $a_n\equiv 0\pmod{10}$ for all $n\ge N$, we need $a_n\equiv 0\pmod{10}$ for all $n\ge \lfloor \frac{N+1}2\rfloor$. Hence the minimal such $N$ has the property $N\le \lfloor \frac{N+1}2\rfloor$, or $N\le 1$. We verify that $a_1\not\equiv 0\pmod{10}$, hence the original assumption of eventual periodicity is false.
| {
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"timestamp": "2023-03-29T00:00:00",
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How i can integrate this expression? I've read about integration, and i believe i understood concept correctly. But, unfortunately, the simplest exercise already got my stumbled. I need to find an integral of $x{\sqrt {x+x^2}}$. So i proceed as follows,
By the fundamental theorem of calculus:
$f(x)=\int[f'(x)]=\int[x\sqrt{x+x^2}]$,
First I've tried to apply chain rule and i end up with:
$f'(x)=xu^\frac{1}{2}\frac{du}{2x+1}$ , not sure how i can proceed in this case.
Next I've tried to apply product rule:
If $f(x)=i(x)j(x)$, then $x\sqrt{x+x^2}=i'(x)j(x)+j'(x)i(x)$,
Using sum rule i could assume that, $f(x)=i(x)j(x)-\int[j'(x)i(x)]$,
Now, finding that $i'(x)=x, i(x)=\frac{x^2}{2}, j(x)=\sqrt{x+x^2}$ and $j'(x)=\frac{2x+1}{2\sqrt{x+x^2}}$,
$f(x)$ should be of form $f(x)=\frac{x^2\sqrt{x+x^2}}{2}-\int\frac{x^2(2x+1)}{ 4\sqrt{x+x^2}}$, so now i should find integral of this fraction,
If i can assume, that $p(x)=\frac{a(x)}{b(x)}$, then $\frac{x^2(2x+1)}{ 4\sqrt{x+x^2}}=\frac{a'(x)b(x)-a(x)b'(x)}{(b(x))^2}$, hence:
$b(x)=2(x+x^2)^\frac{1}{4}, b'(x)=\frac{1}{2(x+x^2)^\frac{3}{4}}$ and as a result $a(x)$, should be $a(x)=\frac{(x+x^2)^\frac{3}{4}(4a'(x)(x+x^2)^\frac{1}{4}-4x^3-2x^2)}{2x+1}$, but now i don't now how substitute $a'(x)$, if i differentiate this expression i will get $a''(x)$.
So my question is, what substitution i shall perform to obtain a(x), a'(x)?
Thank you! And forgive me my ignorance.
| I hope you do not mind if I prefer to start from scratch. We have
$$ \int (2x+1)\sqrt{x^2+x}\,dx = C+\frac{2}{3}(x^2+x)^{3/2} \tag{1}$$
and the problem boils down to computing $\int\sqrt{x^2+x}\,dx$. Integration by parts gives
$$ \int \sqrt{x^2+x}\,dx = x\sqrt{x^2+x}-\int\frac{x+2x^2}{2\sqrt{x+x^2}}\,dx \tag{2}$$
hence
$$ 2\int\sqrt{x^2+x}\,dx = x\sqrt{x^2+x}+\frac{1}{2}\int \frac{x}{\sqrt{x^2+x}}\,dx \tag{3}$$
and the problem boils down to computing $\int\frac{x}{\sqrt{x^2+x}}$. With an argument similar to $(1)$, we have:
$$ \int\frac{2x+1}{\sqrt{x^2+x}}\,dx = C+2\sqrt{x^2+x}\tag{4} $$
so it is enough to compute $\int\frac{dx}{\sqrt{x^2+x}}$ and here it comes an interesting trick. Since
$$ \frac{1}{\sqrt{x^2+x}}=2\cdot\frac{\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}}{\sqrt{x}+\sqrt{x+1}}\tag{5}$$
we have
$$ \int\frac{dx}{\sqrt{x^2+x}}=2\log\left(\sqrt{x}+\sqrt{x+1}\right)\tag{6}$$
and by putting everything together
$$ \int x\sqrt{x^2+x}\,dx = \color{red}{C+\frac{x\left(8x^3+10x^2-x-3\right)}{24\sqrt{x+x^2}}+\frac{\log(\sqrt{x}+\sqrt{x+1})}{8}}.\tag{7}$$
It is not that simple, indeed!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let n=0. Then $\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$. Thus for $n=0$ our equation is satisfied.
Inductive Step. Let $n\ge0$. Assume $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$. Now observe that
$$\begin{align*}\sum_{k=0}^{n+1}k{m+k \choose m}&=\\
\sum_{k=0}^{n}k{m+k \choose m}+(n+1){m+n+1\choose m}&=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}+(n+1){m+n+1\choose m} \end{align*}$$
This is where I'm getting stuck... using Pascal's Identity to combine some of these terms seems ideal. However, the factors of $n$ and $n+1$ are making a clever manipulation difficult for me.
| $$\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$$
Inductive Step. 1: $n\longrightarrow0$
$\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$.
the equation is satisfied.
Inductive Step. 2: $n\longrightarrow n$
$$\sum_{k=0}^nk{m+k \choose m}=$$
By Pascal's identity: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} \Rightarrow \binom{n}{k} - \binom{n-1}{k}= \binom{n-1}{k-1} \Rightarrow \binom{n+1}{k+1} - \binom{n-1+1}{k+1}= \binom{n-1+1}{k-1+1}= $
$\binom{n+1}{k+1} - \binom{n}{k+1}= \binom{n}{k}$
$$\sum_{k=0}^nk\left[\binom{m+k+1}{m+1}-\binom{m+k}{m+1}\right]=$$
$$\sum_{k=0}^nk\binom{m+k+1}{m+1}-\sum_{k=0}^nk\binom{m+k}{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n-1}(k+1)\binom{m+(k+1)}{m+1} + 0\binom{m+ 0 }{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n-1}(k+1)\binom{m+k+1}{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}(k-(k+1))\binom{m+k+1}{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}-1\binom{m+k+1}{m+1}=$$
By the Hockey-Stick identity:
$\sum_{k=r}^n\binom{k}{r}=\binom{n+1}{r+1}$
$$n\binom{m+n+1}{m+1}-\binom{m+(n-1)+1 +1}{m+2}=$$
$$n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}$$
The equation is satisfied.
Inductive Step. 3: $n\longrightarrow (n+1)$
$$\sum_{k=0}^{(n+1)}k{m+k \choose m}=$$
$$\sum_{k=0}^{(n+1)}k\left[\binom{m+k+1}{m+1}-\binom{m+k}{m+1}\right]=$$
$$\sum_{k=0}^{(n+1)}k\binom{m+k+1}{m+1}-\sum_{k=0}^{(n+1)}k\binom{m+k}{m+1}=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n}(k+1)\binom{m+(k+1)}{m+1} + 0=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n}(k+1)\binom{m+k+1}{m+1}=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}(k-(k+1))\binom{m+k+1}{m+1}=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}-1\binom{m+k+1}{m+1}=$$
By the Hockey-Stick identity:
$\sum_{k=r}^n\binom{k}{r}=\binom{n+1}{r+1}$
$$(n+1)\binom{m+(n+1)+1}{m+1}-\binom{m+(n+1) +1}{m+2}$$
The equation is satisfied.
q.e.d
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$ Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$.
How do i do this by using these identities.
$C^r_r + C^{r+1}_r + C^{r+2}_r + ....+C^{n}_r = C^{n+1}_{r+1}$
Or
$C^r_0 + C^{r+1}_1 + C^{r+2}_2 + ....+C^{r+k}_k = C^{r+k+1}_{k}$
| Note that $2\binom{k}{2}=(k-1)k$. Hence
$$(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))=\sum_{k=2}^{n+1}2\binom{k}{2}=2\binom{n+2}{3}.$$
| {
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"url": "https://math.stackexchange.com/questions/1939593",
"timestamp": "2023-03-29T00:00:00",
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If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold? PROBLEM STATEMENT
If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold?
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<\frac{B}{D}<\frac{B}{C}<\frac{C}{D}<1$$
$$1<\frac{D}{C}<\frac{C}{B}<\frac{D}{B}<\frac{B}{A}<\frac{C}{A}<\frac{D}{A}$$
MY ATTEMPT
Since $A, B, C, D \in \mathbb{N}$, we obtain
$$\frac{A}{D} < \frac{B}{D} < \frac{C}{D} < 1$$
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C}$$
$$\frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B}$$
$$1 < \frac{B}{A} < \frac{C}{A} < \frac{D}{A}$$
Summarizing the first and the second:
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C} < \frac{D}{B} < \frac{D}{A}$$
Summarizing the third and the fourth:
$$\frac{A}{D} < \frac{A}{C} < \frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B}$$
Summarizing the first and the third:
$$\frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B} < \frac{D}{A}$$
Summarizing the second and the fourth:
$$\frac{A}{D} < \frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C}$$
Summarizing the first and the fourth:
$$\frac{A}{D} < \frac{B}{D} < \frac{C}{D} < 1 < \frac{B}{A} < \frac{C}{A} < \frac{D}{A}$$
Summarizing the second and the third:
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{C}{B} < \frac{D}{B}.$$
Following a different approach, we also have
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<1$$
$$\frac{B}{D}<\frac{B}{C}<1<\frac{B}{A}$$
$$\frac{C}{D}<1<\frac{C}{B}<\frac{C}{A}$$
$$1<\frac{D}{C}<\frac{D}{B}<\frac{D}{A}.$$
CONCLUSION
We are therefore sure about the validity of the series of inequalities
$$\frac{A}{D} < \frac{A}{C} < \frac{B}{C} < 1 < \frac{C}{B} < \frac{D}{B} < \frac{D}{A}.$$
(Note that $\min(A,B,C,D)=A$ and $\max(A,B,C,D)=D$. Therefore, $A/D$ is the minimum possible fraction with numerator and denominator (distinct from the numerator) coming from the set $\{A,B,C,D\}$. Reciprocally, $D/A$ is the maximum possible fraction.)
However, we cannot conclusively "decide" which member of the (unordered) pairs
$$\left\{\frac{D}{C},\frac{C}{B}\right\}$$
and
$$\left\{\frac{D}{B},\frac{B}{A}\right\}$$
is larger (or equivalently, smaller).
QUESTION
Is my analysis of the problem correct? Or is the problem "decidable" using some other approach/in another context?
| You are correct; here's a simpler way to analyze the problem.
From $C<D$ we get $1/D<1/C$, so $A/D<A/C$. Similarly, $A/C<A/B$ and $B/D<B/C$. Also $C/D<1$, since $C<D$. Moreover $B/D<C/D$, because $B<C$.
Note also that taking reciprocals of positive numbers reverses inequalities, so the second set is essentially the same as the first one.
The ones that could go wrong are
$$
\frac{A}{B}<\frac{B}{D}
$$
and
$$
\frac{B}{C}<\frac{C}{D}
$$
(together with the similar ones between the reciprocals).
The first one is equivalent to $AD<B^2$, the second one to $BD<C^2$.
So we'd like to find $AD\ge B^2$, which is obtained, for instance, with $A=2$, $D=8$ and $B=4$. Now $BD=32$, so we can choose $C=5$ and $BD>C^2$.
You may try and find examples where those inequalities both hold, or just one of them.
| {
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"url": "https://math.stackexchange.com/questions/1940125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate the number of positive integral solutions for the equations $\frac{1}{x} + \frac{1}{y} = \frac{1}{ab}$? Given $a$, $b$.Calculate the number of positive integral solutions for the eqations $\frac{1}{x} + \frac{1}{y} = \frac{1}{ab}$, where $a$, $b$ can be up to $1000000.$
| $\frac{1}{x}+\frac{1}{y}=\frac{1}{ab}\implies ab(x+y)=xy\implies(x-ab)(y-ab)=a^2b^2$
Thus, if $a^2b^2=pq$, we can take $x=ab+p,y=ab+q$, and the number of solutions is thus equal to the number of factorisations of $a^2b^2$, also known as $d(a^2b^2)$.
One way of seeing that the only solutions with positive $x,y$ are the ones listed is by sketching $(x-ab)(y-ab)=a^2b^2$, and noting that none of the bottom-left branch of the hyperbola are in the first quadrant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$ $$k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
It is easy to show $k=1$ and $k=2$
$k=1$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}=1$$
$k=2$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}$$
If we collect $x^n$ terms;
$$\frac{d^n}{dx^n} (1+2x+....+(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})\frac{x^n}{n!}+.....)|_{x=0}$$
$$\frac{d^n}{dx^n} (1+2x+....+(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})\frac{x^n}{n!}+.....)|_{x=0}=(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})=2^n$$
I thought to use Leibniz's method to prove general case but it seems we need to know $$\frac{d^{n-r}}{dx^{n-r}} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
Therefore I am stuck .
Is there an easy way to prove the general case?
Thanks a lot for helps
| Notice that
$$ \frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} \Bigg)^k \Bigg|_{x=0} =
\frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} + x^{n+1}f(x) \Bigg)^k \Bigg|_{x=0}$$
for any $n$-times differentiable function $f$. So we may choose $f$ at our convenience. Now if we choose
$$f(x) = \sum_{j=n+1}^{\infty} \frac{x^{j-n-1}}{j!}$$
then we get
$$ \frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} \Bigg)^k \Bigg|_{x=0}
= \frac{d^n}{dx^n} (e^x)^k |_{x=0}
= k^n e^{kx} |_{x=0}
= k^n.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int\frac{x^5}{x^3-2x^2-5x+6} dx$? None of substitution, partial fraction, and integration by parts seems to work here.
| First note that by long division,
$$
\frac{x^5}{x^3 - 2x^2 - 5x + 6} = x^2 + 2x + \frac{243}{10(x-3)} - \frac{1}{6(x-1)} - \frac{32}{15(x+2)} + 9.
$$
Now, you can integrate the sum term by term, factoring out the constants to get as your answer
$$
\int \frac{x^5}{x^3 - 2x^2 - 5x + 6} dx = \frac{x^3}{3} + x^2 + 9x - \frac{1}{6} \log|1-x| + \frac{243}{10}\log|x-3| - \frac{32}{15} \log|x + 2| + c,
$$
where $c$ is a constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the numerator of $1+\frac12 +\frac13 +\cdots +\frac1{96}$ is divisible by $97$
Let $\frac{x}{y}=1+\frac12 +\frac13 +\cdots +\frac1{96}$ where $\text{gcd}(x,y)=1$. Show that $97\;|\;x$.
I try adding these together, but seems very long boring and don't think it is the right way to solving. Sorry for bad english.
| What we need to do is as follows:
$$
\begin{equation}\begin{split}
1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{96} & = \Big(1+\frac{1}{96}\Big) + \Big(\frac{1}{2}+\frac{1}{95}\Big) + \ldots + \Big(\frac{1}{48} + \frac{1}{49}\Big) \\
& = \frac{97}{96} + \frac{97}{2*95} + \frac{97}{3*94} + \ldots + \frac{97}{48*49} \\
& = 97 \Big( \frac{1}{96} + \frac{1}{2*95} + \frac{1}{3*94} + \ldots + \frac{1}{48*49}\Big)
\end{split}\end{equation}
$$
Hence, the numerator is a multiple of $97$. To see that the denominator is not a multiple of $97$, note that $97$ is a prime, and the denominator contains natural numbers less than $97$ (in fact, the reduced denominator is a factor of $96$!), hence cannot possibly be a multiple of $97$. Thus, in it's reduced form, the numerator of the expression is a multiple of $97$.
In fact, Wolstenhomme's theorem says that the numerator is a multiple of $97^2 = 9409$!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I see that every left ideal of a square matrix ring over a field is principal? As the question title suggests, how do I see that every left ideal of a square matrix ring over a field is principal?
| Let $I$ be a left ideal. The first step is that if a matrix is in the ideal then all the rows can be obtained separately:
Say $$\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} \in I$$
then
$$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}1&2&3\\0&0&0\\0&0&0\end{pmatrix} \in I$$
and
$$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}4&5&6\\0&0&0\\0&0&0\end{pmatrix} \in I$$
$$\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}7&8&9\\0&0&0\\0&0&0\end{pmatrix} \in I$$
Further we can reconstitute the original matrix from the above three
by permuting the rows and adding. For example
$$\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix} \begin{pmatrix}7&8&9\\0&0&0\\0&0&0\\\end{pmatrix} =\begin{pmatrix}0&0&0\\0&0&0\\7&8&9\end{pmatrix} \in I$$
Thus it suffices to consider matrices of the form,
$$\begin{pmatrix}*&*&*\\0&0&0\\0&0&0\\\end{pmatrix}$$
There may be many such matrices in $I$ but since this is basically an $n$ vector there are at most $n$ linearly independent such such vectors, thus having chosen a basis, we can amalgamate this basis into one matrix. For example say we have
$$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\\\end{pmatrix},\begin{pmatrix}0&2&1\\0&0&0\\0&0&0\\\end{pmatrix}\in I.$$
The we have
$$\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}0&2&1\\0&0&0\\0&0&0\\\end{pmatrix} =\begin{pmatrix}0&0&0\\0&2&1\\0&0&0\\\end{pmatrix} \in I.$$
So in this example,
$$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\\\end{pmatrix}+\begin{pmatrix}0&0&0\\0&2&1\\0&0&0\\\end{pmatrix}=\begin{pmatrix}1&0&0\\0&2&1\\0&0&0\\\end{pmatrix}\in I$$ will generate the ideal.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show that $\inf({\frac{mn}{1+m+2n}}) = 1/4$? I am trying to prove the result:
$\inf({\frac{mn}{1+m+2n}}) = \frac{1}{4}$, where $m, n \in\mathbb{N}$.
I proved existence of infimum, but then I do not know how to prove that $1/4$ is a lower bound and then can not prove that $1/4$ is the greatest lower bound.
| To show it achieves this lower bound, let $m = n = 1$. Then
$$\frac{mn}{1+m+2n} = {1 \over 1+1+2} = \frac{1}{4}.$$
Now we need to show for all other $m,n \in \Bbb N$,
$$\frac{mn}{1+m+2n} \ge \frac14.$$
If $n \ge 2$, then
$$\frac{mn}{1+m+2n} \ge \frac{2m}{5+m} \ge \frac{2}{5+1} = \frac{1}{3} > \frac14.$$
If $m \ge 2$, then
$$\frac{mn}{1+m+2n} \ge \frac{2n}{3+2n} \ge \frac{2}{3+2} = \frac25 > \frac14.$$
Thus we have shown it holds for all cases.
Note: We used the fact that $\frac{2x}{5+x}$ and $\frac{2x}{3+2x}$ are increasing functions for $x \ge 0$.
To prove this, let $0 \le x < y$. Then
$$10 x < 10 y$$
$$10 x + 2xy < 10y + 2xy$$
$$2x(5+y) < 2y(5+x)$$
$$\frac{2x}{5+x} < \frac{2y}{5+y}$$
which shows $\frac{2x}{5+x}$ is increasing. The proof for $\frac{2x}{3+2x}$ is analogous.
| {
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"timestamp": "2023-03-29T00:00:00",
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Contour integral over semicircle Evaluate the integral $$\int_{2}^{i} \frac{dz}{1-z^2}$$
where the path is the semicircle $|z-1|=1$ in $\mbox{Im}(z)\leq0$ and the imaginary axis.
My attempted solution:
The primitive function to the definite integral is $$\frac{1}{2}(\log(z+1)-\log(1-z))$$
I need to find a suitable branch. If we draw a line segment from $0,0$ to $i-1$ we have a line that doesn't cross the path. The branch would then be $(\frac{-5\pi}{4}, \frac{3\pi}{4})$. Is this thought process correct?
| We have that
$$\frac{d}{dz}\left(\frac{1}{2}\log\left(\frac{z+1}{z-1}\right)\right)=\frac{1}{1-z^2}$$
where the $\log$ has a branch cut along $[0,+\infty)$. Hence
$$\int_{\gamma} \frac{dz}{1-z^2}=\left[\frac{1}{2}\log\left(\frac{z+1}{z-1}\right)\right]_2^i=\left[\frac{1}{2}\log\left(w\right)\right]_3^{-i}=
\frac{1}{2}\left(\frac{3\pi}{2}i-\ln(3)\right)=-\frac12 \log{3} + i \frac{3\pi}{4}.$$
Note that $\gamma$ is transformed by $\frac{z+1}{z-1}$ into a the upper half-circle from $3$ to $-1$ union the lower quarter-circle from $-1$ to $-i$.
This curve does not intersect $[0,+\infty)$.
P.S. Direct calculation.
Let $\gamma_1$ be the lower half circle from $2$ to $0$:
$$\int_{\gamma_1} \frac{dz}{1-z^2}
=\int_{t=0}^{-\pi} \frac{d(1+e^{it})}{1-(1+e^{it})^2}=
-i\int_{t=0}^{-\pi} \frac{dt}{2+e^{it}}=-\frac{1}{2}\log 3+\frac{i\pi}{2}.$$
(confirmed by WA).
Let $\gamma_2$ be the segment from $0$ to $i$:
$$\int_{\gamma_2} \frac{dz}{1-z^2}=\int_{t=0}^1 \frac{d(it)}{1-(it)^2}=
i\int_{t=0}^1 \frac{dt}{1+t^2}=i[\arctan(t)]_0^1=\frac{i\pi}{4}.$$
Hence,
$$\int_{\gamma} \frac{dz}{1-z^2}=\int_{\gamma_1} \frac{dz}{1-z^2}+\int_{\gamma_2} \frac{dz}{1-z^2}=-\frac12 \log{3} + i \frac{3\pi}{4}.$$
| {
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How to prove that $|z^2| = |z|^2$ where $z = a+bi$? I just started my topic on complex numbers and I'm stuck on this question.
What I have managed to get (I go wrong here, I don't know where though):
$z^2 = (a+bi)^2 = a^2 + b^2$, so $|z^2| = \sqrt{(a^2 + b^2)^2 + 0^2} = \sqrt{a^4+2a^2b^2+b^4}$
and
$|z| = a^2+b^2$ so $|z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$
Suggestions?
| If you use $z=r\cdot e^{i\theta}$, the proof is easy.
$|z^2|=|r^2\cdot e^{2i\theta}|=r^2$ and $|z|^2=r^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove "For every positive integer n, $1^{n}$ + $2^{n}$ + ... + $n^{n}$ < $(n+1)^{n}$." by using induction? How can I prove the following theorem:
For every positive integer $n$, $1^n + 2^n + ... + n^n \lt (n+1)^n$
by using induction?
I have proved that "for every real number $x > 0$ and every non-negative integer $n$, $x^{n} + n \cdot x^{n-1} \le (x+1)^n$".
It might be useful. Thank you.
| This looks like the same proof as GAVD, but without sigma notation. I didn't use your lemma explicitly either.
Let $P(n)$ be the statement that $1^n + 2^n + 3^n + \dots + n^n < (n+1)^n$. Then $P(1)$ is $1 < 2$, which is definitely true.
Suppose $P(k)$ is true for some $k$. That is, suppose
$$
1^k + 2^k + \dots + k^k < (k+1)^k
$$
We want to show $P(k+1)$ is true. Now
\begin{align*}
1^{k+1} + 2^{k+1} + 3^{k+1} + \dots + k^{k+1} + (k+1)^{k+1}
&\leq(k+1)1^k + (k+1)2^k + \dots + (k+1)k^k \\&\quad\quad+ (k+1)(k+1)^k \\
&= (k+1)\left(1^k + 2^k + \dots + k^k + (k+1)^k\right) \\
&\stackrel{(*)}{\leq} (k+1)\left((k+1)^k + (k+1)^k\right)
= 2(k+1)^{k+1}
\end{align*}
The point marked $(*)$ is where we used the inductive hypothesis.
Since $2 < k+1$, we have
$$
1^{k+1} + 2^{k+1} + 3^{k+1} + \dots + k^{k+1} + (k+1)^{k+1}
\leq (k+1)^{k+2}
$$
which establishes that $P(k+1)$ is true.
Therefore, by induction, $P(n)$ is true for all $n$.
| {
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1967 HSC 4 unit Mathematics Question 2 Screenshot from the examination paper
[...asking about partial fraction decomposition of $$\frac{1 - abx^2}{(1-ax)(1-bx)} $$ and related formulas...]
This question is taken from the New South Wales Higher School Certificate 4 unit (highest level possible) paper of 1967. I wasn't sure how to approach this problem. In the end, I decided to let x = 1 and this gave me u$_r$ = -1/n and R$_n$(x) = (1 - ab)x when I compared terms on left and right but I'm not sure if this is the correct approach. Any help appreciated on this one.
| Let us do things formally without worrying about convergent issues. As mentioned by Jack Lam, the partial fraction decomposition of the function is given by
\begin{align}
\frac{1-abx^2}{(1-ax)(1-b)} = \frac{1}{1-bx}+\frac{1}{1-ax}-1.
\end{align}
Formally, we will us the geometric series representation for the two fractions on the right-hand side, i.e. we have
\begin{align}
\frac{1}{1-bx} + \frac{1}{1-ax} =&\ \sum^\infty_{r=0} b^rx^r + \sum^\infty_{r=0}a^rx^r = \sum^\infty_{r=0}(b^r+a^r)x^r \\
=&\ \sum^n_{r=0}(b^r+a^r)x^r + \sum^\infty_{r=n+1} b^rx^r+\sum^\infty_{r=n+1}a^rx^r\\
=&\ \sum^n_{r=0}(b^r+a^r)x^r + b^{n+1}x^{n+1}\sum^\infty_{r=0}b^rx^r+ a^{n+1}x^{n+1}\sum^\infty_{r=0}a^rx^r\\
=&\ \sum^n_{r=0}(b^r+a^r)x^r + \frac{b^{n+1}x^{n+1}}{1-bx}+ \frac{a^{n+1}x^{n+1}}{1-ax}.
\end{align}
Now, going back to the original function, we have
\begin{align}
\frac{1-abx^2}{(1-ax)(1-bx)} =&\ \sum^n_{r=0}(b^r+a^r)x^r + \frac{b^{n+1}x^{n+1}}{1-bx}+ \frac{a^{n+1}x^{n+1}}{1-ax}- 1\\
=&\ \sum^n_{r=1}(b^r+a^r)x^r + \frac{b^{n+1}x^{n+1}}{1-bx}+ \frac{a^{n+1}x^{n+1}}{1-ax}+1.
\end{align}
Multiply both sides of the above equation by $(1-ax)(1-bx)$ yields
\begin{align}
1-abx^2 =&\ (1-ax)(1-bx)\left[\sum^n_{r=1}(b^r+a^r)x^r+1 \right] + [(1-ax)b^{n+1} +(1-bx)a^{n+1}]\ x^{n+1}\\
=&\ (1-ax)(1-bx)\sum^n_{r=0}u_rx^r + x^{n+1} R_n(x)
\end{align}
where
\begin{align}
R_n(x) = (1-ax)b^{n+1} +(1-bx)a^{n+1}
\end{align}
and
\begin{align}
u_r =
\begin{cases}
1 & \text{ if } r =0,\\
a^r+b^r & \text{ if } r\geq 1
\end{cases}.
\end{align}
| {
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If $a$, $b$ and $c$ are three positive real numbers, prove that $\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$ If $a$, $b$ and $c$ are three positive real numbers, prove that $$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$$
I think we are supposed to use the AM-GM inequality here; or maybe this could even be solved using the $T_2$'s Lemma, although I'm a bit skeptical about that.
| Let us define $t=a+b+c$. First we note that
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b}=
\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+
\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)$$
so by Titu's lemma:
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{t}{2}$$
and
$$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geq\frac{3^2}{2(a+b+c)}=\frac{9}{2t}.$$
Therefore, we have to check that
$$\frac{t}{2}+\frac{9}{2t}\geq3$$
for $t>0$. But this follows from the fact that the polynomial $p(t)=t^2-6t+9=(t-3)^2\geq0$.
| {
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Find the limit of the sequence $x_{n+1}=\sqrt{3x_n}$
Let $x_{n+1}=\sqrt{3x_n}$ and $x_1=1$. Prove $x_n=3^{1-(\frac{1}{2^{n-1}})}$ for all $n$ and find the limit of $\{x_n\}$.
Notes: The first few terms of the sequence are $1,\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}} ...$ I do not know from this how to find the equation for the $x_n$ term in terms of $3$. The second part can be answered as follows, please tell me if my thinking is correct.
\begin{align*}
x_n&=3^{1-(\frac{1}{2^{n-1}})} \\
\lim_{n\rightarrow \infty}x_n&=\lim_{n\rightarrow \infty}3^{1-(\frac{1}{2^{n-1}})} \\
&=3^{1-\lim_{n\rightarrow \infty}(\frac{1}{2^{n-1}})} \\
&=3^{1-(\frac{1}{\lim_{n\rightarrow \infty}2^{n-1}})} \\
&= 3
\end{align*}
Also I have looked at Show that the sequence $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$ converges and find its limit. But there is no discussion on how to find something similar to $x_n=3^{1-(\frac{1}{2^{n-1}})}$ .
| Hint
Considering the sequence $$x_{n+1}=\sqrt{3x_n}$$ take logarithms of both sides $$\log(x_{n+1})=\frac{\log(3)}2+\frac 12 \log(x_n)$$ Now define $y_n=\log(x_n)$ which makes the new sequence to be $$y_{n+1}=\frac 12 y_n+\frac{\log(3)}2$$ which is much easier to manipulate.
For the general sequence $$y_{n+1}=a y_n+b\qquad (y_1=0)$$ the general solution is simply $$y_n=\frac{1-a^{n-1}}{1-a} b$$ So, if $a<1$, the limit is just $\frac b {1-a}$; then the limit for the original sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$ Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$
My Attempt:
$LHS=\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{(\sin\frac{x}{2}-\cos\frac{x}{2})(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\sin\frac{x}{2}\cos\frac{x}{2})}{2+2\sin \frac{x}{2}\cos\frac{x}{2}}=\frac{\sin\frac{x}{2}-\cos\frac{x}{2}}{2}$
$RHS=\frac{\cos x}{3}=\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{3}$
But my LHS and RHS are not equal.Please help me.
| $$\frac{\sin^3(0/2)-\cos^3(0/2)}{2+\sin0}=-\frac{1}{2}\neq\frac{1}{3}=\frac{\cos0}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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