Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is $x=\tan\theta$ the only way for this integral? Evaluate $$\int\frac{dx}{(1+x^2)^4}$$ Now I did solve it, but I used the mentioned substitution and after a lot of converting into double angles, I did it. But, it doesn't look like a good approach. Is there a better way? Thanks.
| If you let $x=\tan\theta$ then
$$\begin{align}\int\frac{dx}{(x^2+1)^4}&=\int\cos^6\theta\,d\theta\\
&=\frac1{32}\int\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)d\theta\\
&=\frac1{32}\left(\frac16\sin6\theta+\frac32\sin4\theta+\frac{15}2\sin2\theta+10\theta\right)+C\\
&=\left(\frac16\left(32\sin^5\theta-32\sin^3\theta+6\sin\theta\right)\cos\theta\right.\\
&\left.+\frac32\left(4\sin\theta-8\sin^3\theta\right)\cos\theta+\frac{15}2\left(2\sin\theta\right)\cos\theta+10\theta\right)+C\\
&=\frac1{32}\left[\left(\frac{16}3\frac{x^5}{(x^2+1)^{5/2}}-\frac{52}3\frac{x^3}{(x^2+1)^{3/2}}+22\frac x{(x^2+1)^{1/2}}\right)\frac1{(x^2+1)^{1/2}}+10\tan^{-1}x\right]+C\\
&=\frac16\frac{x^5}{(x^2+1)^3}-\frac{13}{24}\frac{x^3}{(x^2+1)^2}+\frac{11}{16}\frac x{(x^2+1)}+\frac5{16}\tan^{-1}x+C\end{align}$$
In the above I expanded $\cos^6\theta=\left(\frac{e^{i\theta}+e^{-i\theta}}2\right)^6$ by the binomial theorem and then used ran the Chebyshev polynomials of the second kind out to
$$\begin{array}{rl}\sin(2\theta)=&2\sin\theta\cos\theta\\
\sin(3\theta)=&3\sin\theta-4\sin^3\theta\\
\sin(4\theta)=&4\sin\theta\cos\theta-8\sin^3\theta\cos\theta\\
\sin(5\theta)=&16\sin^5\theta-20\sin^3\theta+5\sin\theta\\
\sin(6\theta)=&32\sin^5\theta\cos\theta-32\sin^3\theta\cos\theta+6\sin\theta\cos\theta\end{array}$$
Then since $x=\tan\theta$, $\cos\theta=\frac1{\sqrt{x^2+1}}$, and $\sin\theta=\frac x{\sqrt{x^2+1}}$. Since the result was fairly complicated, I verified it with numerical quadrature.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| Hint. One may apply
$$
\left( \frac1{f}\right)'=-\frac{f'}{f^2} \tag1
$$ with
$$
f(x)=x^2+3^3.
$$
Find $f'(x)$ then use $(1)$ and conclude with $\dfrac2{x^2+3^3}=2 \times \dfrac1{x^2+3^3}=2 \times \dfrac1{f}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
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Large Prime numbers Say we have $2^{690} + 345^4$ and we want to figure out whether this is a prime number.
I feel that we could break down the numbers into their respective prime factors (prime factorization) and use modular arithmetic to calculate whether these numbers are prime numbers or not.
$690 = 2\cdot 3 \cdot 5 \cdot 23$ and $345 = 3\cdot 5\cdot 23$
How should I proceed?
| The factor lulu means is $17$. To show $$2^{690}+345^4\equiv 0\mod 17$$ you can reduce the exponents modulo $\phi(17)=16$ and the bases modulo $17$. This gives $2^2+5^4$. Since $5^2\equiv 8\mod 17$ and $5^4\equiv 8^2\equiv 13\equiv-4\mod 17$, we get $$2^2+5^4\equiv 4+(-4)=0\mod 17$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the integral $\int \frac{x^2+x}{(e^x+x+1)^2}dx$
Evaluate $$\int \frac{x^2+x}{(e^x+x+1)^2}dx$$
I tried converting in the form of Quotient rule(seeing the square in the denominator), neither am I able to make the denominators' derivative in the numerator. Some hints would be great. Thanks.
| One may observe that
$$
\begin{align}
\left(\frac{x+1}{e^x+x+1}\right)'&=\frac{1 \times(e^x+x+1)-(x+1)(e^x+1)}{(e^x+x+1)^2}
\\\\&=\frac{(e^x+x+1)-(x+1)(e^x+x+1)+x(x+1)}{(e^x+x+1)^2}
\\\\&=\frac{-x(e^x+x+1)}{(e^x+x+1)^2}+\frac{x^2+x}{(e^x+x+1)^2}
\\\\&=\frac{\left((e^x+x+1)'-(e^x+x+1)\right)(e^x+x+1)}{(e^x+x+1)^2}+\frac{x^2+x}{(e^x+x+1)^2}
\\\\&=\frac{(e^x+x+1)'}{(e^x+x+1)}-1+\frac{x^2+x}{(e^x+x+1)^2}
\end{align}
$$ giving
$$
\int \frac{x^2+x}{(e^x+x+1)^2}dx=x-\ln\left|e^x+x+1\right|+\frac{x+1}{e^x+x+1}+C
$$
for any constant $C$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the least positive integer $n$ for which $n!$ is divisible by $3^8$? What is the least positive integer $n$ for which $n!$ is divisible by $3^8$?
I am not sure how to tackle this problem without just putting in random low digits as $n$.
| An additional $3$ in the prime factorization can only occur, if the number is divisble by $3$. If it is divisible by $9$, $3^2$ occurs additionally, if it is divisible by $27$, $3^3$ occurs aditionally and so on.
$3!$ is divisble by $3$
$6!$ is divisble by $3^2$
$9!$ is divisible by $3^4$
$12!$ is divisible by $3^5$
$15!$ is divisible by $3^6$
$18!$ is divisible by $3^8$
The answer is therefore $18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding and b in an equation with a floor function \begin{align}
a^2 + ba + c &= 0 & \text{{No real roots.}}\\
\lfloor a^2 \rfloor + ba + c &= 0 & \text{{At least one real roots.}}
\end{align}
Are there any values of b and c that will make the given number of roots correct?
The first thing I thought about doing was finding the discriminant .In order for the first equation to have no real roots, we must have $b^2 – 4c < 0$. That means that $b^2 < 4c$.
I know that the discriminant of the second one must be 0 but I am not sure how to express it because a floor function is involved. What should I do?
Will this approach get me anywhere or are there any better methods?
| \begin{align}
x^2 + bx + c &= 0 & \text{{No real roots.}}\\
\lfloor x^2 \rfloor + bx + c &= 0 & \text{{At least one real root.}}
\end{align}
Let's start by noting that the discriminant of $x^2 + bx + c$ is
$b^2 - 4c$. Hence $x^2 + bx + c$ has no real roots if and only if $b^2 - 4c < 0$, which is true if and only if $c - \dfrac{b^2}{4} > 0$. For convenience, let's define $$\Delta = c - \dfrac{b^2}{4}$$
and from here on, it will turn out to be necessary to assume that $0 < \Delta < 1$.
Let $f(x) = \lfloor x^2 \rfloor + bx + c$
If $x$ is an integer, then $f(x) = x^2 + bx + c$ which we assume has no real roots. So the roots of $f(x)$ cannot be an integer.
Since $x^2 - 1 < \lfloor x^2 \rfloor \le x^2$, then
$x^2 + bx + (c-1) < f(x) < x^2 + bx + c$
So, if $f(x) = 0$,
\begin{array}{c}
x^2 + bx + (c-1) < 0 < x^2 + bx + c \\
-x^2 - bx - c < 0 < -x^2 - bx + (1-c) \\
0 < x^2 + bx + c < 1 \\
0 < \left(x + \dfrac b2 \right)^2 + \Delta < 1 \\
0 \le \left(x + \dfrac b2 \right)^2 < 1 - \Delta \\
\left|x + \dfrac b2 \right| \lt \sqrt{1 - \Delta}
\end{array}
So all possible solutions to $f(x) = 0$ are in the interval
$\left( -\dfrac b2 - \sqrt{1 - \Delta}, -\dfrac b2 + \sqrt{1 - \Delta} \right)$.
This range can be partitioned into intervals of the form $(-\sqrt{n+1}, -\sqrt n]$ or $[\sqrt n, \sqrt{n+1})$ and each interval will contain $0$ or $1$ solutions as follows.
If $-\dfrac{n+c}{b}\in (\sqrt{n+1}, \sqrt n]$ then it is a root of $f(x)$.
If $-\dfrac{n+c}{b}\in [-\sqrt n, -\sqrt{n-1})$ then it is a root of $f(x)$.
Example
Let $f(x) = \lfloor x^2 \rfloor + 3x + 3$
Then $0 < \Delta = \dfrac 34 < 1$ and $\sqrt{1 - \Delta} = \dfrac 12$
So all possible solutions to $f(x) = 0$ are in the interval
$\left( -\dfrac 32 - \dfrac 12, -\dfrac 32 + \dfrac 12 \right) = ( -2, -1)$.
\begin{array}{|c|c|c|c|}
\hline
\text{n} & \text{interval} & a=-\dfrac{n+c}{b} & \lfloor a^2 \rfloor + ba + c \\
\hline
4 & (-\sqrt 5, -\sqrt 4] & -\dfrac 73 & \text{not in interval} \\
3 & (-\sqrt 4, -\sqrt 3] & -2 & \text{not in interval} \\
2 & (-\sqrt 3, -\sqrt 2] & -\dfrac 53 & 2 - 5 + 3 = 0 \\
1 & (-\sqrt 2, -\sqrt 1] & -\dfrac 43 & 1 - 4 + 3 = 0 \\
0 & (-\sqrt 1, -\sqrt 0] & -1 & \text{not in interval} \\
\hline
\end{array}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$. All of its vertices satisfy $x^2+y^2+z^2=1$
Find the remaining two vertices with respect to $a$ and $b$.
| Let $c=\dfrac{\sqrt{2}}{3}$, $d=\dfrac{\sqrt{2}}{\sqrt{3}}$
With these notations, the solution is:
$A_1(0,0,1)$ (given), $A_2(2c,0,-1/3)$, $A_3(-c,d,-1/3)$ and $A_4(-c,-d,-1/3)$
The way I have done it:
1) As the center of mass is necessarily the origin, knowing that a regular tetrahedron has its center of mass at the $1/4$ of its height, points $A_2, A_3, A_4$ constituting the (horizontal) base of the tetrahedron must all have their $z$ coordinate equal to -1/3.
In particular $b=-1/3$.
2) Obtaining $a$: because of the sphere constraint $a^2+0^2+(1/3)^2=1$, we have $a=2c$.
3) Obtaining edge length (squared) : Knowing now the coordinates $A_1$ and $A_2$, the value of the square of the length of any edge is $(A_1A_2)^2=8/3$.
4) Let $A_3=(e,f,-1/3)$ and $A_4(g,h,-1/3)$. Then "sphere constraints" give
$e^2+f^2+1/9=1 \ \ (1)$ and $g^2+h^2+1/9=1 \ \ (2)$.
5) Edge lengths constraints (as seen before) i.e., $(A_1A_3)^2=8/3$ and $(A_1A_4)^2=8/3$ give equations $(e-1)^2+f^2+1/9=8/3 \ \ (3)$ and $(g-1)^2+h^2+1/9=8/3 \ \ (4)$.
6) Using equations (1), (2), (3), (4) the four unknowns $e,f,g,h$ are easily obtained.
Remark: if you know rotation matrices, after items 1) and 2) one can proceed to a computation that applies the following rotation matrix (axis $0z$, angle $2 \pi/3$) twice to vector $\vec{OA_2}=\begin{bmatrix}2\sqrt{2}/3\\0\\-1/3\end{bmatrix}$:
$$R=\begin{bmatrix}-1/2& -\sqrt{3}/2& 0\\\sqrt{3}/2& -1/2& 0\\0& 0& 1\end{bmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$ Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$.
This is the last part of a problem that I am working on. So far, we have shown that the cosine series for $x^2$ is
$x^2=\frac{\pi^2}{3}+4\sum_1^\infty (-1)^n\frac{\cos(nx)}{n^2}$
The sine series for $x^2$ is
$x^2=2\pi\sum_1^\infty(-1)^{n+1}\frac{\sin nx}{n}-\frac{8}{\pi}\sum_1^\infty\frac{\sin ((2n-1)x)}{(2n-1)^3}$
The sine series for x is
$x=2(\sin x -\frac{\sin 2x}{2}+ \frac{\sin 3x}{3}-\dots)$
and using the above we got that
$\frac{\pi^3}{32}=1 - \frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots$.
Not sure if any of this matters, I just don't know how to use this information to solve the last part. I don't really know if this information is needed at all, but I assume it is.
| Let $\sum_{n=1}^\infty \frac{1}{(2n-1)^3}=s$, then we have
\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{n^3} &=& \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{1}{(2n-1)^3} \frac{1}{2^{3m}} \\
&=& \sum_{m=0}^\infty \frac{1}{8^m} \sum_{n=1}^\infty \frac{1}{(2n-1)^3} \\
&=& \frac{1}{1-\frac{1}{8}} s \\
&=& \frac{8}{7} s
\end{eqnarray*}
Where we first count all odd numbers, then the numbers divisble by $2$ but not $4$, then divisble by $4$ not $8$ et cetera. This gives us the sum over $m$ in the first line.
And we may switch order of summation as we sum only positive numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem solving limit infinity/infinity. I cannot solve this limit:
$$\lim_{x\rightarrow\infty}
\frac {(3x^2-4) \left( \sqrt [3]{2x^2+1}+1
\right)^2}{ (2x-1) \left( 4-\sqrt {8x^3-2}
\right)x^{3/2}}$$
I make this:
$${\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}} =3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}-4\,{\frac {
\left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1 \right) \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+6\,{\frac {\sqrt {x}\sqrt [3]
{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) }}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1
\right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+3\,{\frac
{\sqrt {x}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) }}-4\,{\frac {1}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{
x}^{3}-2} \right) {x}^{3/2}}}$$
then
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac { \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1
\right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}6\,{\frac {\sqrt {x}\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right)
\left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{
3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac {1}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}}
$$
but I cannot solve this
NOTE: I cannot use L'hopital for finding this limit.
| In general, if $\lim_{x \to \infty} f(x) = a$ and $\lim_{x \to \infty} g(x) = b$ then continuity of multiplication shows that $\lim_{x \to \infty} f(x)g(x) = ab$.
Write $${1 \over 2x-1} = {1 \over x(2-{1 \over x})} ={1 \over x} {1 \over 2 - {1 \over x}},$$ then since
$$\lim_{x \to \infty} {1 \over 2 - {1 \over x}} = {1 \over 2} \text{ and } \lim_{x \to \infty} {1 \over x} = 0$$ we see that
$$\lim_{x \to \infty} {1 \over 2x-1} = 0.$$
Write $${1 \over 4 -\sqrt{8 x^3 - 2} } = {1 \over x^{3 \over 2}( {4 \over x^{3 \over 2}}-\sqrt{8 - {2 \over x^3}} )} = {1 \over x^{3 \over 2}} { 1 \over {4 \over x^{3 \over 2}}-\sqrt{8 - {2 \over x^3}} )} .$$
Then $$\lim_{x \to \infty} {1 \over x^{3 \over 2}} = 0$$ and
$$\lim_{x \to \infty} {1 \over {4 \over x^{3 \over 2}}-\sqrt{8 - {2 \over x^3}} )} = -{ 1 \over \sqrt{8}},$$ and so
$$\lim_{x \to \infty} {1 \over 4 -\sqrt{8 x^3 - 2} } = 0.$$
You can see that $\lim_{x \to \infty} { 1 \over x^{3 \over 2}} = 0$.
It follows that the limit is $0\cdot 0 \cdot 0 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If sum of n natural number is 20 then what is their max. product? If sum of n natural number is 20 then what is their max. product ?
| Let $s=20$ denote the given sum.
As almagest noted, $k(k+2)\lt(k+1)^2$ shows that the factors can differ by at most $1$. Thus there must be $j$ factors of $\left\lceil\frac sn\right\rceil$ and $n-j$ factors of $\left\lfloor\frac sn\right\rfloor$, with $j$ determined by
$$j\left\lceil\frac sn\right\rceil+(n-j)\left\lfloor\frac sn\right\rfloor=s$$
and thus
\begin{align}
j&=\frac{s-n\left\lfloor\frac sn\right\rfloor}{\left\lceil\frac sn\right\rceil-\left\lfloor\frac sn\right\rfloor}=s-n\left\lfloor\frac sn\right\rfloor=n\left(\frac sn-\left\lfloor\frac sn\right\rfloor\right)=n\left\{\frac sn\right\}=s\bmod n\;,
\end{align}
where $\{x\}$ is the fractional part of $x$. Thus, the solution has $s\bmod n$ factors of $\left\lceil\frac sn\right\rceil$ and $n-(s\bmod n)$ factors of $\left\lfloor\frac sn\right\rfloor$.
As shown here (and apparently also in almagest's answer, though I don't understand the argument there), if the intent was to maximise this over all $n$, the optimal case has only factors of $2$ and $3$, so it has either $\left\lfloor\frac sn\right\rfloor=2$, or $\frac sn=3$, and since $3^2\gt2^3$, among these cases it's the one with lowest $n$, which is $n=\left\lceil\frac s3\right\rceil$, and the product is
$$
3^{s\bmod\left\lceil\frac s3\right\rceil}2^{\left\lceil\frac s3\right\rceil-s\bmod\left\lceil\frac s3\right\rceil}=2^{\left\lceil\frac s3\right\rceil}\left(\frac32\right)^{s\bmod\left\lceil\frac s3\right\rceil}=3^{\left\lceil\frac s3\right\rceil}\left(\frac23\right)^{(-s)\bmod 3}\;.
$$
In the present case, for $s=20$, the optimal $n$ is $7$ and the maximal product is
$$3^{\left\lceil\frac{20}3\right\rceil}\left(\frac23\right)^{(-20)\bmod 3}=2\cdot3^6=1458\;.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Presentations of rational numbers into product Prove or disprove:
For every rational number $q$ there exist rational numbers $q_1$, $q_2$, $q_3$ and $q_4$, such that:
*
*$q_1\times q_2\times q_3\times q_4 = q$
*$q_1+q_2+q_3+q_4 = 0$
My work so far:
I was able to solve it for $q_1$, $q_2$, $q_3$, $q_4$ and $q_5$:
*
*$-q\times\frac{q}{2}\times\frac{q}{2}\times\frac{2}{q}\times\left(-\frac{2}{q}\right)=q$
*$-q+\frac{q}{2}+\frac{q}{2}+\frac{2}{q}+\left(-\frac{2}{q}\right)=0$
| It is a terrible solution (I obtained it by solving some diophantine equations):
$$q_1=\frac{(q-2^{10}\cdot3^{4})^2}{2^7\cdot3^4\cdot(q+2^9\cdot3^4)};$$
$$q_2=-4-\frac{q}{2^7\cdot3^4};$$
$$q_3 = \frac{2^3\cdot3\cdot(q+2^9\cdot3^4)}{q-2^{10}\cdot3^4};$$
$$q_4 = -\frac{2^{11}\cdot3^7\cdot q}{(q-2^{10}\cdot3^4)(q+2^{9}\cdot3^4)},$$
but it is a solution.
If $q = -2^{9}\cdot3^4$ or $q= 2^{10}\cdot3^4$ we shell obtain the decomposution for $\frac{q}{2^{12}\cdot 3^{12}}$ and after this multiplay $q_i$ by $2^{3}\cdot 3^{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786071",
"timestamp": "2023-03-29T00:00:00",
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Shortest distance between two lines in 3-dimensional space Can someone explain to me how to solve this question?
Find the shortest distance between the lines
$L_1 = \left\{t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} : t \in \mathbb{R}\right\}$ and $L_2 = \left\{s \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}: s \in \mathbb{R}\right\}$
Thanks
| The difference vector is
$$
d = L_2(s) - L_1(t)
= s (1,2,3) + (1,0,0) - t (1,1,1)
= (s - t + 1, 2s - t, 3s - t)
$$
Then
$$
q = \lVert d \rVert^2 = d^2 = (s-t+1)^2 + (2s - t)^2 + (3s - t)^2
$$
Then the gradient is
$$
q_s =2(s-t+1)+ 4 (2s-t) + 6(3s-t) = 28s-12t + 2 \\
q_t = -2(s-t+1)-2(2s-t) -2(3s-t) = -12s+6t-2
$$
It vanishes for
$$
\left(
\begin{array}{rr|r}
28 & -12 & -2 \\
-12 & 6 & 2
\end{array}
\right)
\to
\left(
\begin{array}{rr|r}
16 & -6 & 0 \\
-12 & 6 & 2
\end{array}
\right)
\to
\left(
\begin{array}{rr|r}
16 & -6 & 0 \\
4 & 0 & 2
\end{array}
\right)
\to \\
\left(
\begin{array}{rr|r}
16 & -6 & 0 \\
1 & 0 & 1/2
\end{array}
\right)
\to
\left(
\begin{array}{rr|r}
0 & -6 & -8 \\
1 & 0 & 1/2
\end{array}
\right)
\to
\left(
\begin{array}{rr|r}
1 & 0 & 1/2 \\
0 & 1 & 4/3
\end{array}
\right)
$$
So $s = 1/2$ und $t = 4/3$. This gives
$q = d^2 = (1/6)^2 + (-1/3)^2 + (1/6)^2 = 1/18 + 1/9 = 3/18 = 1/6$
and $d = 1/\sqrt{6}$.
(Large version)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$. Prerequisite
This problem is found in "Trigonometry" by I. M. Gelfand [in English].
It is asked in the section "Double the angle". So, assume that I know the sin/cos angle additions [i.e.: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, etc.] as well as everything learned prior.
I've check other sources and they say to use Morrie's Law, however I have not actually learned it in the book.
Problem
Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$.
Hint: If the value of the given expression is $M$, find $M \cos 10^\circ$.
| Here's the strategy that I would use.
Note that
$$\sin 10^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ}{\sin 30^\circ}$$
And that
$$\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 20^\circ\sin 30^\circ\sin 40^\circ \sin 50^\circ\sin 60^\circ\sin 70^\circ\sin 80^\circ}{\sin 20^\circ\sin 40^\circ\sin 60^\circ\sin 80^\circ}$$
See where this is going?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Simplifying a definite integral expression
If $a_1,a_2,a_3$ are the three values of a which satisfy the equation $$\int_{0}^ {\pi/2}(\sin x+a\cos x)^3dx-\frac{4a}{\pi-2}\int_{0}^{\pi/2}x\cos x dx=2$$ Find the value of $a_1+a_2+a_3$.
Now I did find the value of the second integral which comes out to be $\frac{\pi-2}{2}$. But I am unable to find the value ofhe first integral. Please help, thanks.
| To find the value of the first integral,
using that
$$\sin^3\alpha=\frac{-\sin(3\alpha)+3\sin\alpha}{4}\quad\text{and}\quad\cos^3\alpha=\frac{3\cos\alpha+\cos(3\alpha)}{4}$$
gives
$$\begin{align}&\int_{0}^{\pi/2}(\sin x+a\cos x)^3dx\\&=\int_{0}^{\pi/2}(\sin^3x+3a\sin^2x\cos x+3a^2\sin x\cos^2x+a^3\cos^3x)dx\\&=\left[\frac{\cos(3x)-9\cos x}{12}+a\sin^3x-a^2\cos^3x+\frac{9\sin x+\sin(3x)}{12}a^3\right]_{0}^{\pi/2}\\&=\left(a+\frac{8}{12}a^3\right)-\left(\frac{-8}{12}-a^2\right)\\&=\frac 23a^3+a^2+a+\frac 23\end{align}$$
Finally, use Vieta's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of two consecutive squares equal square $N^2 + (N+1)^2 = K^2$, find all solutions for $N<200$
I have done some factoring and also realized that $ K=[n\sqrt{2}]+1$ in eventual solutions, where $[x]$ denotes the greatest integer less than.
The problem is from Nordic math competition 1994.
| We want solutions to $k^2=n^2+(n+1)^2$. So $(k,n,n+1)$ is a Pythagorean triple. Moreover it is a primitive triple since $n,n+1$ are coprime. So we must be able to write it as $k=a^2+b^2,n=2ab,n+1=a^2-b^2$ for $n$ even or $k=a^2+b^2,n=a^2-b^2,n+1=2ab$ for $n$ odd.
If $n$ is odd, then $a^2-b^2+1=2ab$ and $a>b$, so we have $(a+b)^2-2a^2=1$. If $n$ is even, we have $2ab+1=a^2-b^2$, so $(a+b)^2-2a^2=-1$.
So we need solutions to the Pell equation $r^2-2s^2=\pm1$. The smallest such solution is $(r_1,s_1)=(1,1)$. We generate the others by $r_{n+1}=r_n+2s_n,s_{n+1}=r_n+s_n$. So we get $(3,2),(7,5),(17,12),(41,29)$.
These give the solutions $(k,n)=(5,3),(29,20),(169,119)$, and the next is obviously outside the range.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of $p(x)\operatorname{csch}(x)$ I'd like to calculate the following integral
$$\int_{-\infty}^{+\infty}\frac{x^4 \left(\frac 1 {a^2+x^2} +\frac 1 {b^2+x^2}\right)}{\sinh^2(x\pi /c)} \, dx$$
where $a$, $b$ and $c$ are positive constants. Any suggestions? I probably have to use contour integrals, but I'm not sure of which would be the most convenient contour, nor if there's an easy way (I know that the solution has something like the Trigamma function on it).
Thanks a lot! You've been so helpful with my previous questions!
| Obviously two of the parameters are redundant, so let us just study:
$$\begin{eqnarray*} I(a) &=& \int_{0}^{+\infty}\frac{x^4}{\sinh^2(\pi x)}\frac{dx}{(x^2+a^2)}\\&=&-a^2\int_{0}^{+\infty}\frac{x^2\,dx}{(x^2+a^2)\sinh^2(\pi x)}+\int_{0}^{+\infty}\frac{x^2\,dx}{\sinh^2(\pi x)}\\&=&\frac{1}{6\pi}-a^2\int_{0}^{+\infty}\frac{x^2}{\sinh^2(\pi x)}\frac{dx}{x^2+a^2}\\&=&\frac{1}{6\pi}+\frac{a}{2\pi}+a^4\int_{0}^{+\infty}\left(\frac{1}{\sinh^2(\pi x)}-\frac{1}{\pi^2 x^2}\right)\frac{dx}{x^2+a^2}\tag{1}\end{eqnarray*}$$
We may recall that:
$$ \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{2} $$
so, by considering $-\frac{d^2}{dz^2}\log(\cdot)$ of both sides,
$$ \frac{1}{z^2}-\frac{\pi^2}{\sinh^2(\pi z)} = \sum_{n\geq 1}\left(\frac{1}{(n+iz)^2}+\frac{1}{(n-iz)^2}\right)\tag{3}$$
and now we may compute $(1)$ through $(3)$, since:
$$ \int_{0}^{+\infty}\frac{dz}{(z^2+a^2)}\left(\frac{1}{(n+iz)^2}+\frac{1}{(n-iz)^2}\right)=\frac{\pi}{a(n+a)^2}.\tag{4}$$
$(4)$ gives that $(1)$ just depends on $\psi'(a)$, the already mentioned trigamma function:
$$ \psi'(a) = \sum_{n\geq 0}\frac{1}{(n+a)^2}.\tag{5}$$
By $(1),(4)$ and $(5)$ we get:
$$ I(a)=\int_{0}^{+\infty}\frac{x^4\,dx}{(x^2+a^2)\sinh^2(\pi x)}=\color{red}{\frac{1}{6\pi}+\frac{a}{2\pi}-\frac{a^3}{\pi}\psi'(a+1)}.\tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Where did I go wrong in finding maximum? The question in my book is given as:
If $x^2+y^2+z^2=1$ for $x,y,z$ belongs to all real numbers ($x,y,z$ are independent), then find the maximum of $x^3+y^3+z^3-3xyz$.
What I tried:
As all variables are independent I took the derivative of first equation and I got $x+y+z=0$.
Then for maximum of second equation I took the derivative and equated to zero
$$3x^2+3y^2+3z^2-3(xy+yz+zx)=0$$
and now for maximum
$$xy+yz+zx=1$$
(this is because $x^2+y^2+z^2=1$)
We know that $(x+y+z)^3=x^3+y^3+z^3+3[(x+y+z)(xy+yz+zx)]-3xyz$ and here left hand side is zero since $x+y+z=0$ so we get
$x^3+y^3+z^3-3xyz=0$ but this is wrong!
I think I made error by taking the derivative of $x^2+y^2+z^2$ but i don't know why it is wrong.
Any help will be appreciated!
| I'm not sure why you are taking derivatives of $x^2+y^2+z^2=1$ and setting them to zeroes. $x^2+y^2+z^2=1$ is your constraint, and $x^3+y^3+z^3-3xyz$ is your function.
For this problem you can use Lagrange Multipliers.
Set $g(x)=x^2+y^2+z^2$ and $f(x)=x^3+y^3+z^3-3xyz$
$g_x=2x$ $f_x=3x^2-3yz$
$g_y=2y$ $f_y=3y^2-3xz$
$g_z=2z$ $f_z=3z^2-3xy$
$$<f_x,f_y,f_z>=\lambda<g_x,g_y,g_z>$$
$$3x^2-3yz=\lambda(2x)$$
$$3y^2-3xz=\lambda(2y)$$
$$3z^2-3zy=\lambda(2z)$$
$$\lambda=\frac{3x^2-3yz}{2x}=\frac{3y^2-3xz}{2y}=\frac{3z^2-3zy}{2z}$$
Along with the condition $x^2+y^2+z^2=1$, after calculations we can get $x=\pm\frac{\sqrt3}{3}$, $y=\pm\frac{\sqrt3}{3}$, $z=\pm\frac{\sqrt3}{3}$
So in order to get the maximum value of $x^3+y^3+z^3-3xyz$, x,y,z must all be positive numbers, so $x=y=z=\frac{\sqrt3}{3}$
Plug it into the function and find the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the properties of the sum $\sum_{k=0}^n (-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k}$ I have to show that $$\displaystyle\sum\limits_{k=0}^n (-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k} = \begin{cases} 1\ \text{if}\ n=0 \\ 0\ \text{if}\ n>0 \end{cases}$$
My try: I have tried to use snake oil method
$$\sum_{k=0}^n \binom{m+1}{k}\binom{m+n-k}{n-k}$$ $$=\sum_m{\sum_{k=0}^n \binom{m+1}{k}\binom{m+n-k}{n-k}}x^m$$ $$={\sum_{k=0}^n(-1)^k \binom{m+1}{k}\sum_m\binom{m+n-k}{n-k}}x^m$$ We know that the second sum $\frac{1}{(1-x)^{1+n-k}}=\sum_m\binom{m+n-k}{n-k}x^m$
$$=\sum_{k=0}^n(-1)^k\binom{m+1}{k}\frac{1}{(1-x)^{1+n-k}}$$$$=\frac{1}{(1-x)^{1+n}}\sum_{k=0}^n(-1)^k\binom{m+1}{k}(1+x)^k$$ How should I proceed after that?
Edit: My method might be absolutely wrong so if anyone might show me some other method I would really appreciate it
| Note: Here we show that the snake oil method which was applied by OP does also work.
Let $S_m(n)$ denote the series
\begin{align*}
S_m(n)=\sum_{k=0}^n(-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k}\qquad\qquad m,n\geq 0
\end{align*}
The following is valid for $m\geq 0$
\begin{align*}
S_m(n)=
\begin{cases}
1\qquad&\qquad n=0\\
0\qquad &\qquad n>0
\end{cases}
\end{align*}
We obtain
\begin{align*}
\sum_{n=0}^\infty S_m(n)x^n&=\sum_{n=0}^\infty \sum_{k=0}^n(-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k}x^n\\
&=\sum_{k=0}^\infty(-1)^k\binom{m+1}{k}\sum_{n=k}^\infty\binom{m+n-k}{n-k}x^n\tag{1}\\
&=\left(\sum_{k=0}^\infty(-1)^k\binom{m+1}{k}x^k\right)\left(\sum_{n=0}^\infty\binom{m+n}{n}x^n\right)\tag{2}\\
&=(1-x)^{m+1}\sum_{n=0}^\infty\binom{-(m+1)}{n}(-1)^nx^n\tag{3}\\
&=\frac{(1-x)^{m+1}}{(1-x)^{m+1}}\tag{4}\\
&=1
\end{align*}
and the claim follows.
Comment:
*
*In (1) we exchange the sums and factor out which does not depend on the inner sum.
*In (2) we shift the index $n$ of the inner sum so that it starts from $0$. We factor out $x^k$ (from $x^{n+k}$) and observe the sums are now separated.
*In (3) we apply the binomial theorem to the left-hand sum and use the identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}
*In (4) we apply the binomial series expansion and the result is $1+0x+0x^2+\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$ Infinite product
$F_{n}:=[1,1,2,3,5,8,\cdots]$ and
$L_{n}:=[1,3,4,7,\cdots]$
for $n=1,2,3,\cdots$ respectively.
$\frac{1+\sqrt5}{2}=\phi$
Show that,
$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$$
We took the idea from this site
Expand the product
(1)
$$\left(1+\frac{1}{F_{3}L_{3}}\right)\cdot\left(1+\frac{1}{F_{5}L_{5}}\right)\cdot\left(1+\frac{1}{F_{9}L_{9}}\right)\cdots=\frac{3}{\phi^2}$$
$F_{n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt5}$
$L_{n}=\phi^n+(-\phi)^{-n}$
$F_{n}L_{n}=\frac{\phi^{2n}-(-\phi)^{-2n}}{\sqrt5}=F_{2n}$
Rewrite (1)
$$\left(1+\frac{1}{F_{6}}\right)\cdot\left(1+\frac{1}{F_{10}}\right)\cdot\left(1+\frac{1}{F_{18}}\right)\cdots=\frac{3}{\phi^2}$$
Still doesn't help much to get from LHS to RHS.
We have $F_{2n}=F^2_{n+1}-F^2_{n-1}$
I have substituted in, but the formula seem too messy and more complicated.
Can anybody please give a hand?
| Hint. One may use the following result,
$$
\prod_{n=1}^{\infty}\left(1+\frac{F_b}{F_{2^na+b}}\right)=\frac{1-(-1)^b\phi^{-2a-2b}}{1-\phi^{-2a}} \tag1
$$
which is proved by J. Sondow here (see proposition 1 on page 3), where $\phi=\dfrac{1+\sqrt5}{2}$ and where $F_{n}:=[1,1,2,3,5,8,\cdots]$ are the Fibonacci numbers.
Applying $(1)$ with $a=2$, $b=2$ using $F_{2^n+1}L_{2^n+1}=F_{2^{n+1}+2}$ gives the announced infinite product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find the spectrum of graphs for adjacency matrix Find the spectrum of graphs for adjacency matrix $(A)$ below :
$$ \left[
\begin{matrix}
0 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 0 \\
\end{matrix}
\right] $$
My attempt :
The spectrum of a graph is the list of eigenvalues of the adjacency matrix with the multiplicities. I do the calculation by $det (xI - A)$, I find the $(xI - A)$ as below :
$$ \left[
\begin{matrix}
x & -1 & -1 & 0 & 0 \\
-1 & x & -1 & 0 & 0 \\
-1 & -1 & x & -1 & -1 \\
0 & 0 & -1 & x & -1 \\
0 & 0 & -1 & -1 & x \\
\end{matrix}
\right] $$
$det (xI - A) = x^5-6x^3-4x^2+5x+4$
Then I stack to find the eigenvalues of the adjacency matrix with the multiplicities.
| $(x-1)(x+1)^2(x-\frac{1-\sqrt{17}}{2})(x-\frac{1+\sqrt{17}}{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Computing the residue of a rational function The real integral I am trying to compute with residues/contour integration is
$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^3} \,dx$ For $a$ positive and by using the complex integral
$$\int_{C_R}\frac{z^2}{(z^2+a^2)^3} \,dz$$
For some arc bounding only the positive pole, $a i$. I am not sure how to evaluate this pole of order 3.
In class we have discussed how we can write a function $f(z)$ with a pole of order $n$ at $z_0$ multiplicatively as
$$f(z)=\frac{g(z)}{(z-z_0)^n}$$ or additively as
$$f(z)=\frac{a_{-n}}{(z-z_0)^n}+\cdots+\frac{a_{-1}}{z-z_0}+h(z)$$
For holomorphic functions on a disc around the pole $h,g$, which I think is equivalent to finding a Laurent series
This yields the formula for the residue
$$\text{Res}(f,z_0)=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[
(z-z_0)^nf(z)]$$
Which gets messy. Is there a more elegant way?
| Computing the residue as @joriki suggested
\begin{align*}
\text{Res}(f,z_0)&=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[
(z-z_0)^nf(z)]\\
&\Rightarrow \text{Res}(f,ai)=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2}
[(z-ai)^{3}\frac{z^2}{(z^2+a^2)^3}]=
\frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2}
[(z-ai)^{3}\frac{z^2}{(z-ai)^3(z+ai)^3}]\\
&=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2}
[\frac{z^2}{(z+ai)^3}]=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d}{dz}
[\frac{z(2ai-z)}{(z+ai)^4}]\\
&=\frac{1}{2}\lim_{z\rightarrow ai}\frac{2(ai-z)}{(z+ai)^4}
-4\frac{z(2ai-z)}{(z+ai)^5}=
-4\frac{1}{2}\lim_{z\rightarrow ai}\frac{z(2ai-z)}{(z+ai)^5}\\
&=-2\frac{(ai)^2}{(2ai)^5}=\frac{1}{16ia^3}
\end{align*}
yielding then for our integral:
\begin{align*}
\int_{C_R}\frac{z^2}{(z^2+a^2)^3}dz&=2\pi i \text{Res}(f,ai)=\frac{2\pi i}{16ia^3}
=\frac{\pi}{8a^3}
\end{align*}
Now, to send the arc legnth to zero, we parametrize the arc as usual $\gamma_R(t)=Re^{it}$, $t\in [0,\pi]$,
and bounding our integrand using the max and path length
\begin{align*}
|\int_{\gamma_R}\frac{z^2}{(z^2+a^2)^3}&|\leq
\max_{|z|=R}|\frac{z^2}{(z^2+a^2)^3}|*||\gamma_R(t)||=
\frac{R^2}{(R^2+a^2)^3}\pi R\sim cR^{-3}\rightarrow 0\;\text{as}\;
R\rightarrow \infty
\end{align*}
Yielding our answer:
\begin{align*}
\lim_{R\rightarrow \infty}\int_{-R}^{R}\frac{z^2}{(z^2+a^2)^3}dz=\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)^3}dz=\frac{\pi}{8a^3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Can anyone help me with this finite sum? I have to calculate the sum
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1}$
We can re-write the sum as follows
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k-1+1}{(3^{k+1}-1)(3^k-1)}$
And then we obtain
$\displaystyle\sum_{k=1}^n \displaystyle\frac{1}{3^{k+1}-1}+\displaystyle\sum_{k=1}^n \displaystyle\frac{1}{(3^{k+1}-1)(3^k-1)}$
But I don't know what to do with the last two sums. Can anyone help me with them. Thanks
| Telescopic!
$$\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1} = \frac{1}{2}\left(\frac{1}{3^k-1}-\frac{1}{3^{k+1}-1}\right) $$
hence:
$$ \sum_{k=1}^{n}\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1} = \color{red}{\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3^{n+1}-1}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convolution of Gaussian and error function I am trying to evaluate the following integral:
$$
\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}\Phi(x-t)dx
$$
where
$$
\Phi(y) = \frac{1}{2} + \frac{1}{2}erf\left(\frac{y}{\sqrt{2}}\right)
$$
I have tried integration by parts, a substitution, but nothing useful came out of this
| The key is to express as integral of a gaussian.
$$
I(t) = \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2} \right)~\Phi(x-t)~dx\\
= \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2} \right)~\left[
1-\Phi(t-x)
\right]~dx\\
=\sqrt{2\pi}-\int_{-\infty}^\infty{
\exp\left(-\frac{x^2}{2} \right)~
\Phi(t-x)
~dx
}
$$
Plug in
$$
\Phi(t-x)=\int_{\infty}^{t}{
\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{(y-x)^2}{2} \right)
~dy
}
$$
to get
$$
I(t)=\sqrt{2\pi}-\int_{-\infty}^\infty{
\exp\left(-\frac{x^2}{2} \right)~
\int_{\infty}^{t}{
\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{(y-x)^2}{2} \right)
~dy
}
~dx
}\\
=\sqrt{2\pi}-
\int_{\infty}^{t}{
\frac{1}{\sqrt{2}}\exp \left(-\frac{y^2}{4} \right)
~dy
}\\
=\sqrt{2\pi}-
\sqrt{2\pi}\Phi\left(\frac{t}{\sqrt{2}} \right)\\
=\sqrt{2\pi}\Phi\left( -\frac{t}{\sqrt{2}} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does one integrate $x^2 \frac{e^x}{(e^x+1)^2}$? How can I show this? $$ \int_{-\infty}^{\infty} x^2 \frac{e^x}{(e^x+1)^2} dx = \pi^2/3$$
I tried applying residuals, but the pole is of infinite(?) order.
| We have
$$\frac d{dx}\frac1{e^x+1}=\frac{-e^x}{(e^x+1)^2}$$
Also
$$\frac{e^x}{(e^x+1)^2}=\frac{e^{-x}}{(1+e^{-x})^2}$$
So
$$\begin{align}\int_{-\infty}^{\infty}x^2\frac{e^x}{(e^x+1)^2}dx&=2\int_0^{\infty}x^2\frac{e^x}{(e^x+1)^2}dx=-2\int_0^{\infty}x^2\frac d{dx}\frac1{e^x+1}dx\\
&=\left.-2x^2\frac1{e^x+1}\right|_0^{\infty}+4\int_0^{\infty}\frac x{e^x+1}dx\\
&=0+4\int_0^{\infty}\frac{xe^{-x}}{e^{-x}+1}dx=4\sum_{k=0}^{\infty}(-1)^k\int_0^{\infty}xe^{-(k+1)x}dx\\
&=4\sum_{k=0}^{\infty}(-1)^k
\frac{\Gamma(2)}{(k+1)^2}=4\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}\\
&=4\left(1-\frac24\right)\zeta(2)=2\frac{\pi^2}6=\frac{\pi^2}3\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
| Induction is fundamental to the structure of mathematics. The $``\cdots"$ in the expression $1+ 5+ 9 + \cdots +(4n-3)$ cannot be explained rigorously without using some form of induction. All of the picture proofs here have an implied $``\cdots"$ in them. Similarly, $\text{$\Sigma$-notation}$ and telescoping sums cannot be defined without some form of induction.
Putting that aside, I have always been curious about the process of converting sums to integrals, so I offer the following flawed solution.
Note that
$\int_\limits k^{k+1}(4x-5)\,dx
= \left. (2x^2-5x) \right|_k^{k+1}
= (2k^2+4k+2-5k-5)-(2k^2-5k)
= 4k-3$.
So
\begin{align}
\sum\limits_{k = 1}^n (4k - 3)
&= \sum\limits_{k = 1}^n \left( \int_\limits k^{k+1}(4x-5)\,dx \right) \\
&= \int_\limits 1^{n+1}(4x-5)\,dx
&\text{(This step requires induction.)} \\
&= \left. (2x^2-5x) \right|_1^{n+1} \\
&= (2n^2 + 4n+2 -5n - 5) - (2 - 5) \\
&= 2n^2-n
\end{align}
Addendum
I got the integrand by solving
$4k - 3 = \int_\limits k^{k+1}(2ax+b)\,dx = 2ak + (a+b)$
The implication is that
\begin{align}
1 &= u_2 - u_1 \\
5 &= u_3 - u_2 \\
9 &= u_4 - u_3 \\
&\vdots \\
4k-3 &= u_{k+1} - u_k \\
&\vdots
\end{align}
where $u_k = 2k^2 - 5k$.
Which converts the sum to the sum of a collapsing series.
This implies Lynn's graphical proof shown below and, though it still requires induction, I thought it shows an interesting way to compute sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 13
} |
limit $\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}$ without Hospital Is it possible to find $$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}$$ without l'Hospital's rule?
I have $\lim_{x\to 0}\frac{\tan x}{x}=1$ proved without H. but it doesn't help me with this complicated limit (however I'm sure I have to use it somehow).
I know the answer is $\frac{1}{3}$, so I tried to estimate: $0<\frac{\tan x-x}{x^2\tan x}\le\frac{1}{3}\cdot\frac{\tan x}{x}+g(x)$ with various $g$ and prove that $g(x)\to 0$, but with no result.
| $$L=\lim_{x\to 0}\frac{x-\tan x}{x^2 \tan x}=\lim_{x\to 0}\frac{\cos x-\frac{\sin x}{x}}{x^2}\cdot\frac{x}{\sin x}=\lim_{x\to 0}\frac{1-2\sin^2\frac{x}{2}-\cos\frac{x}{2}\frac{\sin(x/2)}{(x/2)}}{x^2} $$
gives $L=A+B$ where:
$$ A = -\frac{1}{2}+\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2} = -\frac{1}{2}+\lim_{x\to 0}\frac{2\sin^2\frac{x}{4}}{x^2} = -\frac{1}{2}+\frac{1}{8}=-\frac{3}{8}$$
and
$$ B = \lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{\sin(x/2)}{x/2}\right)=\frac{1}{4}\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{\sin x}{x}\right)$$
(assuming such a limit exists) fulfills:
$$ 3B = 4B-B = \lim_{x\to 0}\left(\frac{\sin(x/2)}{x/2}-\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{2\sin(x/2)}{\sin (x)}-1\right)$$
so that:
$$ B = \frac{1}{3}\lim_{x\to 0}\frac{1}{x^2}\left(\frac{1}{\cos\frac{x}{2}}-1\right)=\frac{1}{3}\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2}=\frac{1}{3}\lim_{x\to 0}\frac{2\sin^2\frac{x}{4}}{x^2}=\frac{1}{24}$$
and $L=A+B=-\frac{3}{8}+\frac{1}{24}=\color{red}{\large-\frac{1}{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$. I came across this problem and i could not solve it despite many tries.
Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$.
I think the question can be solved more easily..Can anybody suggest any other method.
| Using $a^2+b^2 =(a+b)^2-2ab$, we have
$3^{2008} +4^{2009}=(3^{1004})^2+(2^{2009})^2=(3^{1004}+2^{2009})^2 - 2 \cdot 2^{2009} \cdot 3^{1004}$
So we have $3^{2008} +4^{2009}=(3^{1004}+2^{2009})^2- 2^{2010} \cdot 3^{1004}$
Now, using $a^2-b^2=(a-b)(a+b)$, we have
$3^{2008} +4^{2009}=(3^{1004}+2^{2009}+2^{1005} \cdot 3^{502})(3^{1004}+2^{2009}-2^{1005} \cdot 3^{502})$
Finally, it is not hard to see that $(3^{1004}+2^{2009}-2^{1005} \cdot 3^{502}) > 2009^{182}$, which is left as an exercise to the reader.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Solve first order partial diferential equation Consider the Cauchy problem:
$$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
It is easy to show that the characteristic equations are given by:
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}{2qu-2yq}=dt$$
By the Lagrange-Charpit's method, we need a first integral $\Psi$ different to $\Phi=x^2p+y^2q-u^2$.
How I can find $\Psi$? Maybe we should get directly the characteristic curves of the previous system?
Many thanks!
| $$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
FIRST METHOD :
The characteristics equations are :
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{u^2}$$
First characteristic curve, from $\frac{dx}{x^2}=\frac{dy}{y^2} \quad\to\quad \frac{1}{x}-\frac{1}{y}=c_1$
Second characteristic curve, from $\frac{dy}{y^2}=\frac{du}{u^2} \quad\to\quad \frac{1}{y}-\frac{1}{u}=c_2$
General solution expressed on implicit form :
$$\Phi\left(c_1\:,\:c_2\right)= \Phi\left(\frac{1}{x}-\frac{1}{y}\:,\:\frac{1}{y}-\frac{1}{u}\right)$$
where $\Phi$ is any differentiable function of two variables.
Solving the implicit equation for its second variable leads to :
$$\frac{1}{y}-\frac{1}{u}=F\left(\frac{1}{x}-\frac{1}{y} \right)$$
where $F$ is any differentiable function.
$$u(x,y)=\frac{1}{\frac{1}{y}-F\left(\frac{1}{x}-\frac{1}{y} \right)}$$
With condition $u(x,2x)=1 \quad\to\quad \frac{1}{\frac{1}{2x}-F\left(\frac{1}{x}-\frac{1}{2x} \right)}=1 \quad\to\quad F\left(\frac{1}{2x} \right)= \frac{1}{2x}-1$
This determines the function $F(Z)=Z-1$ and so, $F\left(\frac{1}{x}-\frac{1}{y} \right)= (\frac{1}{x}-\frac{1}{y})-1$
$$u(x,y)=\frac{1}{\frac{1}{y}-\left(\frac{1}{x}-\frac{1}{y}-1 \right)}= \frac{1}{\frac{2}{y}-\frac{1}{x}+1}= $$
$$u(x,y)=\frac{xy}{2x-y+xy} $$
SECOND METHOD :
Change : $\begin{cases} x=\frac{1}{X}\\ y=\frac{1}{Y}\\ u=\frac{1}{U}\end{cases}$ transforms the PDE : $\frac{x^2}{u^2}\frac{\partial u}{\partial x}+\frac{y^2}{u^2}\frac{\partial u}{\partial y}=1$ to :
$$\quad U_X+U_Y=1$$
Solving this elementary linear PDE leads to :
$$U=Y+f(X-Y)$$
any differentiable function $f$.
The condition $u(x,2x)=1$ is transformed to $\frac{1}{U(\frac{1}{x},\frac{1}{2x})}=\frac{1}{U(X,\frac{X}{2})}=1 \quad\to\quad U(X,\frac{X}{2} )=1$
$1=\frac{X}{2}+f(X-\frac{X}{2}) \quad\to\quad f(\frac{X}{2})=1-\frac{X}{2} $ which determines $f(Z)=1-Z$
$f(X-Y)=1-(X-Y)=1-X+Y $
$U=Y+f(X-Y)=Y+(1-X+Y)=1-X+2Y$
$\frac{1}{u}=1-\frac{1}{x}+2\frac{1}{y} \quad\to\quad u=\frac{xy}{2x-y+xy}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| $$2^{3^m}+1=h\cdot 3^{m}\implies 2^{3^{m+1}}+1=(h\cdot 3^m-1)^3+1=\\=(h\cdot 3^m)((h\cdot 3^{m}-1)^2-(h\cdot 3^m-1)+1)=\\=(h\cdot 3^m)(h^2\cdot 3^{2m}-3h\cdot 3^m+3)$$ if $m\ge 1$, then $3$ divides $h^2\cdot3^{2m}-3h\cdot 3^m+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that $z = \ln (x^2+y^2) +2\tan^{-1}(y/x)$ satisfies the laplaces's equation.
I can get to the first partial derivative of $\partial z / \partial x$ of course, but after that I get this, which I don't know how to differentiate further... I am talking about c part of the question. Attaching my working...
| Differentiating once:
$$\frac{\partial z}{\partial x}\left( \ln(x^2+y^2)+2\tan^{-1}\left( \frac{y}{x}\right )\right )=\frac{2x}{x^2+y^2}+\frac{-2\frac{y}{x^2}}{\frac{x^2+y^2}{x^2}}=\frac{2x}{x^2+y^2}-\frac{2y}{x^2+y^2}=\frac{2x-2y}{x^2+y^2}$$
Differentiating again:
$$\frac{\partial z}{\partial x}\left ( \frac{\partial z}{\partial{x}}\right )=\frac{\partial}{\partial x}\left( \frac{2x-2y}{x^2+y^2}\right )=\frac{2(x^2+y^2)-2x(2x-2y)}{(x^2+y^2)^2} \\ =\boxed{\frac{-2x^2+2y^2+2xy}{(x^2+y^2)^2}}$$
Now let us show $\frac{\partial^2 z}{\partial y^2}$:
$$\frac{\partial}{\partial y}\left( \ln(x^2+y^2)+2\tan^{-1}\left( \frac{y}{x}\right )\right )=\frac{2y}{x^2+y^2}+\frac{\frac{1}{x}}{1+(\frac{y}{x})^2}=\frac{2y+x}{x^2+y^2}$$
$$\frac{\partial }{\partial y}\left( \frac{2y+x}{x^2+y^2}\right )=\frac{2(x^2+y^2)-(2y)(2y+x)}{(x+y)^2}=\boxed{\frac{2x^2-2y^2-2xy}{(x^2+y^2)^2}}$$
Therefore,
$$\frac{\partial^2 z}{\partial x^2}+\frac{\partial ^2 z}{\partial y^2}=\frac{-2x^2+2y^2+2xy}{(x^2+y^2)^2}+\frac{2x^2-2y^2-2xy}{(x^2+y^2)^2}=0 \checkmark$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Systems of equations with unknown constant How do I solve this system? It says I must row reduce it to solve it (depending on parameter $a$).
$(8−a)x_1 + 2x_2 + 3x_3 + ax_4 = 2$
$x_1 + (9−a)x_2 + 4x_3 + ax_4 = 1$
$x_1 + 2x_2 + (10−a)x_3 + ax_4 = 2$
$x_1 + 2x_2 + 3x_3 + ax_4 = 2$
I have reduced it to the form
$$\left[
\begin{array}{cccc|c}
1 & 2 & 3 & a & 2\\
7-a & 0 & 0 & 0 & 0\\
0 & 7-a & 1 & 0 & -1\\
0 & 0 & 7-a & 0 & 0\\
\end{array}
\right]$$
How do I proceed from here? If there were no $a$ then I would be able to put it in row reduced echelon form, but now because of the $a$ I don't know how to continue.
| Permuting the columns,
$$\det \left[
\begin{array}{ccc|c}
1 & 2 & 3 & a\\
\hline
7-a & 0 & 0 & 0\\
0 & 7-a & 1 & 0\\
0 & 0 & 7-a & 0\\
\end{array}
\right] = \pm \det \left[
\begin{array}{c|ccc}
a & 1 & 2 & 3\\
\hline
0 & 7-a & 0 & 0\\
0 & 0 & 7-a & 1\\
0 & 0 & 0 & 7-a\\
\end{array}
\right] = \pm a (7-a)^3$$
Thus, if $a = 0$ or $a = 7$, then the matrices above are singular and the linear system being studied either has infinitely many solutions or it has none. We have three cases to consider.
$a \neq 0$ and $a \neq 7$
The linear system has the unique solution
$$\begin{bmatrix} 0\\ \dfrac{1}{a-7}\\ 0\\ \dfrac{2a-16}{a (a-7)}\end{bmatrix}$$
$a = 0$
The 4th column is zero, i.e., $x_4$ is free. However, the system is then inconsistent, as $-\frac{2}{7} \neq 2$.
$a = 7$
The southwest $3 \times 3$ block contains only zeros on the main diagonal. There are two degrees of freedom. The solution set is the $2$-dimensional affine space
$$\begin{bmatrix} 0\\ 0\\ -1\\ \frac{5}{7}\end{bmatrix} + \gamma_1 \begin{bmatrix} 1\\ 0\\ 0\\ -\frac{1}{7}\end{bmatrix} + \gamma_2 \begin{bmatrix} 0\\ 1\\ 0\\ -\frac{2}{7}\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this?
| $a=(x^2-1)=(x+1)(x-1)$
$b=(x^2+2x+1)=(x+1)^2=(x+1)(x+1)$
So, we have:
$$\frac{(x+3)}{a}-\frac{(x-2)}{b}$$
$$\frac{b(x+3)-a(x-2)}{ab}$$
$ab=(x+1)^3(x-1)$
$b(x+3)=(x+3)(x+1)(x+1)$
$a(x-2)=(x-2)(x+1)(x-1)$
Then,
$$\frac{(x+3)(x+1){(x+1)}-(x-2){(x+1)}(x-1)}{{(x+1)^3}(x-1)}$$
$$\frac{{(x+1)}[(x+3)(x+1)-(x-2)(x-1)]}{{(x+1)^{3}}(x-1)}$$
$$\frac{(x+3)(x+1)-(x-2)(x-1)}{(x+1)^2(x-1)}$$
Expanding:
$$\frac{7x+1}{x^3+x^2-x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Help with a limit using big O? $\lim_{x\to 0} \frac{4sin^{2}(\frac{x}{2})-x^{2}cos(\frac{x}{2})}{4x^{2}sin^{2}(\frac{x}{2})}$ is equal to $\frac{1}{24}$ apparently but I can't work it out.
My attempt:
$\lim_{x\to 0} \frac{4(\frac{x}{2}+O(x^{3}))^{2}-x^{2}(1-\frac{(\frac{x}{2})^{2}}{2}+O(x^{4}))}{4x^{2}(\frac{x}{2}+O(x^{3}))^{2}} $
$= \lim_{x\to 0}\frac{\frac{x^{4}}{8}+O(x^{4})}{x^{4}+O(x^{6})}=\frac{1}{8}$
Thanks in advance.
| Using, almost as you did (but using one more term for the first expansion), $$\sin \left(\frac{x}{2}\right)=\frac{x}{2}-\frac{x^3}{48}+O\left(x^4\right)$$ $$\sin^2 \left(\frac{x}{2}\right)=\frac{x^2}{4}-\frac{x^4}{48}+O\left(x^5\right)$$ $$\cos \left(\frac{x}{2}\right)=1-\frac{x^2}{8}+O\left(x^4\right)$$ $$4\sin^2 \left(\frac{x}{2}\right)-x^2\cos \left(\frac{x}{2}\right)=4\left(\frac{x^2}{4}-\frac{x^4}{48}+O\left(x^5\right)\right)-x^2\left(1-\frac{x^2}{8}+O\left(x^4\right)\right)=\frac{x^4}{24}+O\left(x^5\right)$$ I am sure that you can take it from here.
Edit
Let me explain "better" why we had to include the first two terms for the expansion of $\sin \left(\frac{x}{2}\right)$ (at least the one in numerator).
It is clear that the denominator will be something like $x^4$. So, for the limit, the numerator must be at least $x^4$ too. We have this for $x^2\cos \left(\frac{x}{2}\right)$ using a single term for $\cos \left(\frac{x}{2}\right)$ (just as you did). A term in $x^4$ will appear when squaring $\sin \left(\frac{x}{2}\right)$ since $$\left(\frac{x}{2}-\frac{x^3}{48}\right)^2=\frac{x^2}{4}-\frac{x^4}{48}+\frac{x^6}{2304}$$ the $x^4$ term coming from the croos product in $(a-b)^2=a^2-2ab+b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let R be a Boolean ring with unity. Show that $0$ is the only nilpotent element and $1$ is the only unit. A ring is said to be Boolean if $x^2=x$ for all $x \in R$ .
I tried it this way: let $x \neq 0$ ,
$x^n=0$ and
$x^2=x$ so $x^{2n}=x^{n}=(-x)^n$ How do I show that $x=0$?
For unit part:
Let x be a unit so there exits $y \neq 0$ and and $xy=1$ then $xy=yx$ as we know that the Boolean ring is commutative and then $(xy)^2=xy=1$ and we have $x^2=x$ and $y^2=y$ so $(xy)^2=xy=x^2y^2$ then??
Thanks for any help!
| For the first part, $x^2=x$ implies that $x^3=x^2x=x^2=x$ and similarly $x^n=x$ for all $n\geq 1$. Therefore if $x^n=0$ for some $n$ then $x=0$.
For the second part, since $x^2=x$ it follows that $x(1-x)=0$. Therefore if $xy=1$ then multiplying by $1-x$ we get $1-x=0$, so $x=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a simpler proof?
| Notice that
$$ \left( \sqrt[3]{a+b}+\sqrt[3]{a-b} \right)^3= 2a + 3\cdot \sqrt[3]{a^2-b^2}\left(\sqrt[3]{a-b}+\sqrt[3]{a+b} \right)$$
Now with your question
Consider
$$x= \sqrt[3]{a+\sqrt {a^2+b^3}} +\sqrt[3]{a-\sqrt {a^2+b^3}} $$
Let $\sqrt{a^2+b^3} = c$
$$ x= \sqrt[3]{a+c} +\sqrt[3]{a-c} $$
$$ \Leftrightarrow x^3 = 2a + 3 \cdot \sqrt[3]{a^2-c^2}\left(\sqrt[3]{a-c}+\sqrt[3]{a+c} \right) $$
$$ x^3 = 2a + 3\cdot \sqrt[3]{a^2-c^2}(x)$$
I think you can take it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the sum of the $81$ products in the $9 \times 9$ multiplication grid? What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.
Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?
| Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \ =\ \ 6\times 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 5
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A variation of Ahmed's integral $\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $ Given that the closed form exist, evaluate the following Integral:
$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$
| Since:
$$ \frac{z+4}{z^2-12 z+16} = \frac{\bar{\varphi}}{z-(2\bar\varphi)^2}+\frac{\varphi}{z-(2\varphi)^2}\tag{1}$$
the integral can be computed by exploiting the relations between the dilogarithm and the golden ratio, namely Landen's identity. A first step of partial fraction decomposition followed by integration by parts leads to:
$$ I = \frac{\pi\log 5}{8}+\frac{1}{4}\int_{0}^{\pi/2}\log\left(\frac{4-2\sin(t)-\sin^2(t)}{4+2\sin(t)-\sin^2(t)}\right)\,dt \tag{2}$$
and if we replace $t$ with $\frac{\pi}{2}-2\arctan u$ we get:
$$ I = \frac{\pi\log 5}{8}+\frac{1}{2}\int_{0}^{1}\log\left(\frac{1+10v^2+5v^4}{5+10v^2+v^4}\right)\frac{dv}{1+v^2}\tag{3}$$
where the last integral is a dilogarithmic integral, namely the imaginary part of
$$\text{Li}_2\left[i\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)\right]+\text{Li}_2\left[i\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)\right]\tag{4}$$
that is a special value of the Clausen function $\text{Cl}_2$. It is also useful to notice that:
$$ \frac{1+10v^2+5v^4}{5+10v^2+v^4}=\frac{v}{\tanh(5\text{arctanh}(v))}. \tag{5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the solution of the Cauchy problem for the pde Find the solution of the Cauchy problem for the pde
$x\dfrac{\partial z }{\partial x}+y\dfrac{\partial z}{\partial y}=z$
on $D=\{(x,y,z):x^2+y^2\neq 0,z>0\}$ with the initial condition $x^2+y^2=1,z=1$.
By Lagrange's Equations
$\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{dz}{z}$
If I solve the above I get ;$\dfrac{x}{y}=c_1,\dfrac{y}{z}=c_2,\dfrac{x}{z}=c_3$
But I am unable to use the boundary conditions.How to do this?
| The set of characteristic equations is :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$$
With the condition $z=1$ on the curve $x^2+y^2=1$
So, it should be easy to satisfy this condition if this curve would be a characteristic curve. So, it is of interest to make appear the terme $x^2+y^2$ in a combination of the characteristic equations :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{xdx}{x^2}=\frac{ydy}{y^2}=\frac{xdx+ydy}{x^2+y^2}$$
This uses the well known identity : $\frac{A}{B}=\frac{C}{D}=\frac{A+C}{B+D}$
$$\frac{xdx+ydy}{x^2+y^2}=\frac{dz}{z}$$
$$\frac{1}{2}\frac{d(x^2+y^2)}{x^2+y^2}=\frac{dz}{z}$$
$$z=C\:\sqrt{x^2+y^2}$$
with the condition $1=C\sqrt{1}\quad\to\quad C=1\quad$. So the solution is :
$$z=\sqrt{x^2+y^2}$$
NOTE:
A more extended application of the method of characteristics should be to express the general solution of the PDE.
As seen above, a first characteristic equation is :
$$\frac{z}{\sqrt{x^2+y^2}}=C_1$$
As already found by learnmore, another characteristic curve is $\frac{x}{y}=C_2$
Thus, an implicit form of the general solution is the equation :
$$\Phi\left(\frac{z}{\sqrt{x^2+y^2}} \:,\: \frac{x}{y}\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Solving it for the the first variable gives the explicit form of the general solution :
$$\frac{z}{\sqrt{x^2+y^2}} =F\left( \frac{x}{y}\right)$$
where $F$ is any differentiable function.
The condition $z=1 \:,\: x^2+y^2=1 \quad\to\quad \frac{1}{\sqrt{1}} =F\left( \frac{x}{y}\right)$
This implies that $F$ is a constant function $F=1 \quad\to\quad \frac{z}{\sqrt{x^2+y^2}} =1$
$$z=\sqrt{x^2+y^2}$$
Alternatively, one could easily solve the PDE with the change to polar coordinates.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of k when $f(x) = x^3 -4x^2 + 6x + k$ If $ \ \ a ,b \ \ $and $\ \ c \ \ $ are roots of $ \ \ f(x) = x^3 -4x^2 + 6x + k \ \ $
and $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ \ \ $then find the value of $ \ k $
I have to tell you honestly that I cannot solve this question or find the way to get value of $ \ k$
I started with $(a + b +c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) \ \ $ $\Rightarrow a^2 + b^2 + c^2 = 4$
then $a^3 +b^3 + c^3 = (a+b+c)(a^2 + b^2 +c^2) - (ab+bc+ac)(a+b+c) + 3abc$
$ \ \ \ \ \ \ \ \ a^3 +b^3 + c^3 = -8 + 3k$
apply to $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c)$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 -3k = 3(a+b)(b+c)(a+c)$
finally , I don't know how relation between $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ $ and $(a+b)(b+c)(a+c)$
Please tell me the way to solve this question correctly.
thank you in advance.
| We know: $a+b+c=4, ab+bc+ca=6, abc=-k$
Using the condition and $a^2+b^2+c^2=4$ you can get that:
$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}=1=\frac{1}{4-c^2}+\frac{1}{4-a^2}+\frac{1}{4-b^2}\Leftrightarrow -a^2b^2c^2+4(a^2b^2+b^2c^2+c^2a^2)-16(a^2+b^2+c^2)+64=a^2b^2+b^2c^2+c^2a^2-8(a^2+b^2+c^2)+48 \Leftrightarrow a^2b^2c^2-3(a^2b^2+b^2c^2+c^2a^2)+8(a^2+b^2+c^2)-16=0$
Now: $a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=36+8k$
$k^2-3(36+8k)+32-16=0$
$k^2-24k-92=0$
$k_{1,2}=12\pm2\sqrt{59}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:
Suppose there is some point at which the line touches the curve, $a$, which will obviously obey the equation $\left(a,\frac{a^2}{a-1}\right)$. The derivative is calculated as
\begin{align*}
f(x) = \frac{x^2}{x-1} = x^2(x-1)^{-1} \Rightarrow f'(x) = 2x(x-1)^{-1} -x^2(x-1)^{-2} = \frac{x^2-2}{(x-1)^2} \Rightarrow f'(a) = \frac{a^2-2}{(a-1)^2}.
\end{align*}
Filling this into the equation we get
\begin{align*}
y - \left(\frac{a^2}{a-1}\right) = \frac{a^2-2}{(a-1)^2}\left(x-a\right).
\end{align*}
Obviously this equation will contain the point $(2,0)$, so we fill it in and obtain the answer:
\begin{align*}
\frac{a^2}{a-1} = \frac{a^2-2}{(a-1)^2}\left(a-2\right) \Rightarrow a^2 +2a-4=0 \Rightarrow (a+1)^2=5 \Rightarrow a = -1\pm\sqrt 5.
\end{align*}
The equations are getting so complicated I'm pretty sure I'm doing it wrong, but I'm not quite sure what I'm doing wrong so I'm wondering if StackExchange could offer some input on my method?
| First, you can simply $f(x)$ a little and write
$$ f(x) = \frac{x^2}{x-1} = \frac{x^2-1+1}{x-1} = (x + 1) + \frac{1}{x-1}. $$
Then we can calculate the derivative to obtain:
$$ f'(x) = 1 - \frac{1}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}. $$
If the line passes through $(2,0)$, it has the form $y = m(x - 2)$. If it also passes through $(a,f(a))$ we must have
$$ f(a) = m(a - 2) \implies m = \frac{f(a)}{(a-2)}. $$
Finally, we must have
$$ f'(a) = m = \frac{f(a)}{(a-2)} \implies \frac{a(a-2)^2}{(a-1)^2} = \frac{a^2}{a - 1} \iff a(a-2)^2 = a^2(a-1). $$
One solution is obtained by taking $a = 0$. For other solutions, we must solve
$$ (a-2)^2 = a^2 - 4a + 4 = a(a-1) = a^2 - a \implies \\
4 - 3a = 0 \implies a = \frac{4}{3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ then can we find $\lim_{n\to \infty} x_n$? Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence of real numbers defined as follows: $x_1=1$ and $x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ then can we find $\lim_{n\to \infty} x_n$ ?
| To see that the limit exist, consider
$\epsilon_{n+1}=\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_{n}} = \frac{\frac{3+2\frac{ (3+2 x_n)}{3+x_n}}{3+\frac{3+2 x_n}{3+x_n}}-\frac{3+2 x_n}{3+x_n}}{\frac{3+2 x_n}{3+x_n}-x_n}=\frac{3}{12+5x_n}$
Since $x_n>0$ from its definition, $0<\epsilon_{n+1} < 1 $, which mean it converges.
And then the limit $x=\lim_{n\to \infty}x_n$ is found by
$ x= \frac{3+2x}{3+x}$, $x= \frac{1}{2}(\sqrt{13}-1)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1)} + \frac{C}{(x-1)} = \frac{-1}{x}+\frac{1/2}{(x+1)} + \frac{1/2}{(x-1)} $$
Back to the integral:
$$2(-\int_2^\infty\frac{1}{x}+\frac{1}{2}\int_2^\infty\frac{1}{(x+1)} + \frac{1}{2}\int_2^\infty\frac{1}{(x-1)})dx$$
$$\lim_{a\to2..{b\to\infty}}2[(-\ln(x)+\frac{\ln(x+1)}{2} + \frac{\ln(x-1)}{2}]_a^b$$
$$2[(-\ln(b)+\frac{\ln(b+1)}{2} + \frac{\ln(b-1)}{2}]-2[(-\ln(2)+\frac{\ln(3)}{2} + \frac{\ln(1)}{2}] $$
$$(-\infty+\infty+\infty) + (\ln(3/4)) $$
So, the integral does not converge? I am unsure about the last part, if anyone can confirm me the answer or explain why it's wrong... thank you guys!
| I know you're asking if it converges, but unless I'm interpreting your answer incorrectly I think it's wrong. As others have pointed out before me you can see it converges via a limit comparison test to see $x\rightarrow\infty$. So here's how I get $\ln(4/3)$:
$$\int\dfrac{2}{x(x+1)(x-1)}dx$$
$$=2\int\dfrac{1}{x(x+1)(x-1)}dx$$
$$=2\int-\frac{1}{x}+\dfrac{1}{2(x+1)}+\dfrac{1}{2(x-1)}dx$$
$$=2\left(-\ln|x|+\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|\right)+C$$
Now the bounds:
$$2\left(-\ln|2|+\frac{1}{2}\ln|2+1|+\frac{1}{2}\ln|2-1|\right)\Rightarrow-\ln\left(\frac{4}{3}\right)$$
$$2\left(-\ln|\infty|+\frac{1}{2}\ln|\infty+1|+\frac{1}{2}\ln|\infty-1|\right)\Rightarrow0$$
$$=0-\left(-\ln\left(\frac{4}{3}\right)\right)$$
$$=\ln\left(\frac{4}{3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac {\sin x}{1-\sin x}-\frac {\sin x}{1+\sin x}\equiv 2\tan^2 x$ I need assistance with proving the following identity:
$$\frac {\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x} \equiv 2\tan^2 x$$
What I have done so far is expanded them:
$$\frac {\sin x\;(1+\sin x)}{(1-\sin x)(1+\sin x)}-\frac {\sin x\;(1-\sin x)}{(1+\sin x)(1-\sin x)}$$
So therefore:
$$\frac {\sin x+\sin^2 x}{1-\sin^2x}-\frac {\sin x-\sin^2 x}{1-\sin^2 x}$$
I'm completely stuck on what to do next. Any pointers would be appreciated. Thanks for your time!
| You have
$$\frac{\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x}=\frac {\sin x+\sin^2x}{1-\sin^2x} -\frac {\sin x-\sin^2x}{1-\sin^2x}$$
Then, use that $\frac{B}{A}-\frac CA=\frac{B-C}{A}$ to get $$\frac {\sin x+\sin^2x-(\sin x-\sin^2x)}{1-\sin^2x}=\frac{2\sin^2x}{1-\sin^2x}$$
Now use $1-\sin^2x=\cos^2x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Convert $16x^2+56x-80=0$ to the form of $(\text{something})^2=D$ Sorry about this very basic question, I want to convert the
equation $16x^2+56x-80=0$ to the form $(\text{something})^2=D$, I know that the answer is $(4x+7)^2-129$, but how can I convert this without usnig calculator
| Here is an expansion which had helped me when I had come to know about completing the squares for the first time:
$ax^2+bx+c=0$
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + c = 0 $
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + (\frac{b}{2\sqrt{a}})^2 + c - (\frac{b}{2\sqrt{a}})^2 = 0$
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + (\frac{b}{2\sqrt{a}})^2 = (\frac{b}{2\sqrt{a}})^2 - c$
$(\sqrt{a}x+ \frac{b}{2\sqrt{a}})^2 = \frac{b^2}{4a} - c$
Clearly, $a \neq 0$ as this is a polynomial of second degree i.e. a quadratic equation. Hope it helps.
Cheers
| {
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Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$
Could someone please explain to my why this is the case?
Thanks a lot.
| $\lim_{x\to\infty} x(\sqrt{x^2+1}-x)=\lim_{x\to\infty} \sqrt{x^4+x^2}-x^2$
$=\lim_{x\to\infty}\frac{(\sqrt{x^4+x^2}-x^2)(\sqrt{x^4+x^2}+x^2)}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{x^4+x^2-x^4}{\sqrt{x^4+x^2}+x^2}$
$=\lim_{x\to\infty}\frac{x^2}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}\rightarrow \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Which number is greater, $11^{11}$ or $9^{12}$?
Which number is greater than $11^{11}$ or $9^{12}$?
My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
| This is a variant on Servaes's answer. Note that $3^5=243=2\cdot11^2+1$. Using a binomial expansion and some extremely crude upper bounds, we find
$$\begin{align}
3\cdot9^{12}
&=3^{25}\\
&=(2\cdot11^2+1)^5\\
&=32\cdot11^{10}+80\cdot11^8+80\cdot11^6+40\cdot11^4+10\cdot11^2+1\\
&\lt32\cdot11^{10}+80\cdot11^8+11^8+11^8+11^8+11^8\\
&\lt32\cdot11^{10}+121\cdot11^8\\
&=32\cdot11^{10}+11^{10}\\
&=3\cdot11^{11}
\end{align}$$
and thus $9^{12}\lt11^{11}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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How can I develop $f(x) = (x^2+1)\cdot \arctan(x)$ into a Taylor series about $x_{0} = 0$? How can I develop $f(x) = (x^2+1)\cdot \arctan(x)$ into a Taylor series about $x_{0} = 0$ ?
I see that $f(x) = \frac{\arctan(x)}{\arctan'(x)}$ and I also know what the Taylor expansion of $\arctan(x)$ looks like about zero. But I don't know how to develope what they wanted. I get a series that is not a power series...
| The key here is that you want the same power of $x$ for all the terms of the sum.
\begin{align*}
(x^2+1)\arctan(x) & = (x^2+1) \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=1}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}+x\\
& =\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}-\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+3}+x\\
& =x+\sum_{k=0}^{+\infty}(-1)^k \left(\frac{1}{2k+1}-\frac{1}{2k+3}\right)x^{2k+3}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Bounding a series: $\frac{\pi}{2} < \sum_{n=0}^\infty \frac{1}{n^2 + 1} < \frac{3\pi}{2} $ I have the following statement -
$$\frac{\pi}{2} < \sum_{n=0}^\infty \dfrac{1}{n^2 + 1} < \frac{3\pi}{2} $$
So I tried to prove this statement using the integral test and successfully proved the lower bound. But when I tried to calculate the upper bound I was required to calculate the integral from -1 - $\int_{-1}^{\infty} \frac{1}{x^2 +1}\,dx$.
If someone can explain why it will be great , thanks!
| Note that
$$ \sum_{n=0}^\infty \dfrac{1}{n^2 + 1}= \sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{n^2+1}dx>\sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{x^2+1}dx=\int_{0}^\infty\frac{1}{x^2+1}dx=\frac{\pi}{2}. $$
Also note
$$ \sum_{n=0}^\infty \dfrac{1}{n^2+1}=1+\sum_{n=1}^\infty \dfrac{1}{n^2 + 1}<1+\sum_{n=1}^\infty \dfrac{1}{n^2}=1+\frac{\pi^2}{6}<\frac{3\pi}{2}. $$
Thus
$$ \frac{\pi}{2}<\sum_{n=0}^\infty \dfrac{1}{n^2+1}<\frac{3\pi}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Antiderrivative of ${d^2 y \over dx^2} = 1-x^2$
At any point $(x,y)$ on a curve, ${d^2 y \over dx^2} = 1-x^2$, and an equation of the tangent line to the curve at the point $(1,1)$ is $y=2-x$. Find an equation of the curve.
This is what I've done
$${d^2 y \over dx^2} = 1-x^2
\\ \int dy' = \int (1-x^2)dx
\\y' = x- {x^3 \over 3} +C
\\ {dy\over dx}=x- {x^3 \over 3} +C
\\\int dy =\int (x- {x^3 \over 3} +C)dx
\\y= {x^2 \over 2}-{1\over 3} \cdot {x^4 \over 4}+c_1x + c_2
\\ y= {x^2 \over 2}-{x^4 \over 12}+c_1x +c_2$$
Do I now substitute (1,1) in this? I don't this is right. Someone help me. Thank you!
| Since the equation of the tangent line to the curve at $(1,1)$ is $y=-x+2$, we know that $dy/dx|_{(x,y)=(1,1)}=-1=1-\frac{1}{3}+C \Rightarrow C=-\frac{5}{3}$
Thus, $y=\frac{x^2}{2}-\frac{x^4}{12}-\frac{5}{3}x+D$ and we know that
$1=\frac{1}{2}-\frac{1}{12}-\frac{5}{3}+D\Rightarrow D=-\frac{9}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Convergence and Limit of a Recursive Sequence from the multiples of $a_n$ I'm having trouble with this recursive sequence problem. I'm supposed to find the limit, assuming that is it convergent, but I can't seem to get the answer.
$a_1 = 1, a_{n+1}= \frac {2a_n}{7+a_n} $
| We may just find the explicit form of $a_n$. By setting $a_n=\frac{p_n}{q_n}$ we have:
$$\begin{pmatrix} p_{n+1} \\ q_{n+1}\end{pmatrix}=\begin{pmatrix} 2 & 0 \\ 1 & 7\end{pmatrix}\begin{pmatrix} p_{n} \\ q_{n}\end{pmatrix}$$
where the characteristic polynomial of the involved matrix is $(x-2)(x-7)$.
By the Cayley-Hamilton theorem it follows that:
$$ p_n = A\cdot 7^n + B\cdot 2^n,\qquad q_n = C\cdot 7^n + D\cdot 2^n $$
and since $p_1=q_1=1$ and $p_2=2,q_2=8$ we have:
$$ p_n = 2^{n-1},\qquad q_n = \frac{6}{35}\cdot 7^n-\frac{1}{10}\cdot 2^n $$
so:
$$ a_n = \color{red}{\frac{5}{6\cdot \left(\frac{7}{2}\right)^{n-1}-1}}$$
and $\lim_{n\to +\infty} a_n = \color{red}{0}$ trivially follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$
Prove that
$$
\begin{vmatrix}
1 & a^2 + bc & a^3 \\
1 & b^2 + ac & b^3 \\
1 & c^2 + ab & c^3 \\
\end{vmatrix}
=(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$$
My work
I am getting - sign extra , please tell me why I am getting this
| It is faster to exploit Gaussian elimination:
$$\begin{eqnarray*} \det\begin{pmatrix}1 & a^2+bc & a^3 \\ 1 &b^2 +ac & b^3 \\ 1 & c^2+ab & c^3\end{pmatrix}&=&\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &(b-a)(a+b-c) & (b-a)(b^2+ba+a^2) \\ 0 & (c-a)(a+c-b) & (c-a)(c^2+ca+a^2)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &a+b-c & b^2+ba+a^2 \\ 0 & a+c-b & c^2+ca+a^2\end{pmatrix} \\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ a+c-b & c^2+ca+a^2\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ 2(c-b) & (c-b)(a+b+c)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)(c-b)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ 2 & a+b+c\end{pmatrix}\end{eqnarray*}$$
and:
$$ (a+b-c)(a+b+c)-2(a^2+ab+b^2) = (a+b)^2-c^2-2(a^2+ab+b^2)=-(a^2+b^2+c^2).$$
An even faster approach is to notice that the determinant is clearly zero if $a=b,a=c$ or $b=c$. So the determinant is $(a-b)(b-c)(c-a)$ times a symmetric second-degree polynomial, that is not hard to find.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Inequality with $\arctan$ I try to show that $x \cdot \arctan\left( \frac{1}{x^2} \right)$ is monotonically decreasing, but I can't solve this inequality with $\arctan$. Can somebody show me how to do this?
$$
x \in [1, \infty)
$$
$$
\arctan\left( \dfrac{1}{x^2} \right) \leq \dfrac{2x^2}{x^4+1}
$$
| Putting $x=1/y$, it is enough to show that, for $0 < y \leq 1$
\begin{equation*}
\arctan(y^2) \leq \frac{2y^2}{1+y^4}
\end{equation*}
If we let
\begin{equation*}
f(y) = \frac{2y^2}{1+y^4} - \arctan(y^2)
\end{equation*}
then
\begin{equation*}
f'(y) = \frac{2y(1-3y^4)}{(1+y^4)^2}
\end{equation*}
$f'(y)=0$ when $y = 0$ or $y = 3^{-1/4}$. At $y=0$ $f(y) = 0$ and at $y = 3^{-1/4}, f(y) > 0$. Also $f(1) = 1-\frac{\pi}{4} > 0$. Thus it follows that $f(y) > 0$ for $0 < y \leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}$ What's the value of $$\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}?$$
I've tried the AM-GM inequality with no luck.
Also tried to right the inequality as:
$$\left(\left(2-\frac{1}{2}\right)\left(2-\frac{1}{3}\right)\dots\left(2-\frac{1}{n+1}\right)\right)^\frac{1}{n}.$$
Would be happy to hear any hints/proofs.
| Hint
$$\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\frac{(2n+1)!}{{{2}^{n}}n!(n+1)!}$$
Note
$$\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\left( \frac{2\times 4\times 6\times \cdots \times 2n}{{{2}^{n}}(1\times 2\times 3\times \cdots \times n)} \right)\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\frac{(2n+1)!}{{{2}^{n}}n!(n+1)!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.
So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb + \binom{n}{n}x^n$$
Now on integrating both sides and putting $x=1$.
I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is $$\frac{2^{n+1}-1}{n+1}.$$
Where does that -1 term in the numerator come from?
| Remember when you integrate both sides, there will be the constant of integration. You need to first find this constant of integration. That's where the $-1$ comes from. In actual fact, it is $\frac{-1}{n+1}$ but looks like $-1$ because it's been absorbed into the numerator.
You may like to try $x=0$ to find this $+C$ value...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_ 1^2\frac{1}{x}dx$ with a Riemann Sum I'm trying to solve
$$ \int_ 1^2 \frac{1}{x} \ dx $$
using Riemann sums, however I'm having trouble solving it.
| $$
\begin{align}
\int_1^2\frac1x\,\mathrm{d}x
&=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac nk\cdot\frac1n\\
&=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac2{2k}\right)\\
&=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k\tag{1}
\end{align}
$$
By the alternating series test, the limit in $(1)$ exists. Call it $\alpha$.
In the derivation of $(1)$, we have the equation
$$
\begin{align}
\int_1^2\frac1x\,\mathrm{d}x
&=\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+k}\\
&=\alpha\tag{2}
\end{align}
$$
In this answer, It is shown that $\left(1+\frac1n\right)^n$ increases and $\left(1+\frac1n\right)^{n+1}$ decreases to $e$. Thus,
$$
\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}\le\dots\le\left(1+\frac1{n+k}\right)^{n+k}\tag{3}
$$
and
$$
\left(1+\frac1n\right)^{n+1}\ge\left(1+\frac1{n+1}\right)^{n+2}\ge\dots\ge\left(1+\frac1{n+k}\right)^{n+k+1}\tag{4}
$$
Putting $(3)$ and $(4)$ together yields, for $0\le k\le n$,
$$
\left(1+\frac1{n+k}\right)^{\frac{n}{n+1}}\le\left(1+\frac1{n+k}\right)^{\frac{n}{n+k}\frac{n+k+1}{n+1}}\le\left(1+\frac1n\right)^\frac{n}{n+k}\le\left(1+\frac1{n+k}\right)\tag{5}
$$
Therefore,
$$
\begin{align}
e^\alpha
&=\lim_{n\to\infty}\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{6}
\end{align}
$$
and by $(5)$
$$
\begin{align}
\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
&\le\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\dots\left(1+\frac1{2n}\right)\\
&=\frac{2n+1}{n+1}\tag{7}
\end{align}
$$
Using the other direction of $(5)$, we get
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{8}
$$
By the Squeeze Theorem, $(6)$, $(7)$, and $(8)$ say
$$
e^\alpha=2\tag{9}
$$
Therefore,
$$
\int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{10}
$$
| {
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"url": "https://math.stackexchange.com/questions/1847818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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As $ b$ varies range of $m(b)$ is? Let $f(x)=(1+b^2)x^2+2bx+1$ and let $m(b)$ be the minimum value of $f(x)$.As $b$ varies range of $m(b)$ is ?
I can find the minima by putting $f'(x)=0$ and hence find $m(b)$.But after that i'm not able to find range of $m(b)$.
| Finding minimum :
$$f(x)=(1+b^2)x^2+2bx+1$$
$$f(x)=(1+b^2)\left[x^2+\frac{2b}{1+b^2}x+\frac{1}{1+b^2}\right]$$
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{(1+b^2)^2}+\frac{1}{1+b^2}\right]$$
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{(1+b^2)^2}+\frac{1+b^2}{(1+b^2)^2}\right]$$
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2+\frac{b^2}{(1+b^2)^2}\right]$$
Thus $$f(x)=(1+b^2)\left(x+\frac{b}{1+b^2}\right)^2+\frac{1}{1+b^2}$$
Since $$\left(x+\frac{b}{1+b^2}\right)^2 \geq 0$$
$$(1+b^2)\left(x+\frac{b}{1+b^2}\right)^2 \geq 0$$
$$(1+b^2)\left(x+\frac{b}{1+b^2}\right)^2+\frac{1}{1+b^2} \geq \frac{1}{1+b^2}$$
$$f(x) \geq \frac{1}{1+b^2}$$
Therefore $$f(x)_{min}=m(b)=\frac{1}{1+b^2}$$
Finding the range of the minimum :
Now take $y=m(b)$ , for all real $b$ ,
$$y=\frac{1}{1+b^2} \Rightarrow y+yb^2=1$$
$$yb^2+y-1=0$$
Since $b$ is a variable , this is a quadratic equation of $b$.
Thus for all real $b$ , its discriminant must be greater than or equal to zero.
That is , $$0^2-4y(y-1) \geq 0$$
$$(y-1)y \leq 0$$
Thus $$0 < y \leq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\det{\begin{bmatrix} A & B \\ C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC) $? I found this identity in Wikipedia,
When $D$ is a $1 \times 1$ matrix, $B$ is a column vector, and $C$ is a row vector, then
$$\det{\begin{bmatrix} A & B \\ C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC)= (D+1)\det(A) -\det(A+BC)$$
I wonder how to prove that. Thanks!
| Observe that
\begin{equation*}
\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
= \begin{bmatrix}
A & B + 0\\
C & (D-1) + 1
\end{bmatrix}.
\end{equation*}
Therefore, using multilinearity of determinants
\begin{equation*}
\det\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
= \det \begin{bmatrix}
A & 0\\
C & D - 1
\end{bmatrix} +
\det \begin{bmatrix}
A & B\\
C & 1
\end{bmatrix}.
\end{equation*}
Now, the first determinant on the RHS is equal to $\det(A) \det(D -1) = (D - 1) \det(A)$, and the second determinant is $\det([1])\det(A - B[1]^{-1}C) = \det(A - BC)$, from which the first form of the result follows. The second form is obtained similarly by writing $D = (D + 1) - 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved?
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$
This is my attempt. It is too big, I am sure there is a better way of calculating this, but my knowledge is not very good (the questions are at the end of the explanation):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor = \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$
by Bertrand's postulate $\lfloor \sqrt{p_{i+1}} \rfloor \lt \lfloor \sqrt{2 \cdot p_i} \rfloor$ so it is possible to replace in a first step $\lfloor \sqrt{3} \rfloor$ by $\lfloor \sqrt{2 \cdot 2} \rfloor$:
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2 \cdot 2} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$
then recursively doing the same with the next prime $\lfloor \sqrt{5} \rfloor \lt \lfloor \sqrt{2\cdot 3} \rfloor \lt \lfloor \sqrt{2\cdot (2 \cdot 2)} \rfloor$, etc. the result is:
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
$=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$
$\sqrt{2} \lt 2$:
$\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}} \lt 2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} $
Finally, $\sum_{i=1}^n 2^i = (2^{i+1}-1)-2^0 = 2^{i+1}-2$, thus:
$2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} \lt 2 \cdot (2^{(\frac{n}{2}+1)}-2)=2^{(\frac{n}{2}+2)}-4$
So finally the expression would be:
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P \lt 2^{(\frac{n}{2}+2)}-4$
I would like to ask the following questions:
*
*Are the calculations correct or there are gaps / errors?
*Is there a better approach? Thank you!
UPDATE 2016/7/17: Thank you very much for the answers, for the sake of completeness I have sketched the solutions everybody kindly provided, compared with $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$. The reason is that this sequence is expected to grow slower than the sequence of the sum of the first $n$ elements, so it is a good expression to compare with. The only solution that was too big was mine, kindly improved by mathlove, that is why I wanted to ask for better values really under the value of the sum of the first $n$ natural numbers.
Blue: $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$
Red (mathlove) = $\sum_{i=1}^{n}\sqrt{p_i}\lt \sum_{i=1}^{n}2^{\frac i2}=\frac{2^{\frac{n+1}{2}}-\sqrt 2}{\sqrt 2-1}$
Yellow (Marco Cantarini) = $n\sqrt{n\log\left(n\right)+n\log\left(\log\left(n\right)\right)}-\frac{2\left(n-1\right)\sqrt{n-1}}{3\sqrt{\log\left(2\right)}}+1$
Green (LeGrandDODOM) (leading term only): $\displaystyle\frac 23 n^{\frac 32}\sqrt{\ln n}$
| Some asymptotics that should lead to a better bound with some additional work
From $p_n\sim n\ln n$ you can derive $\lfloor \sqrt{p_n} \rfloor\sim \sqrt{ n\ln n}$
Since $\sum_i \lfloor \sqrt{p_i} \rfloor$ diverges, the asymptotics above imply
$$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor \sim \sum_{i=1}^n\sqrt{ i\ln i} $$
This last sum can be estimated via integrals, yielding
$$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor \sim \int_1^n \sqrt{ t\ln t} dt$$
Integrating by parts, $\displaystyle \int_1^n \sqrt{ t\ln t} dt = \frac 23 n^{\frac 32}\sqrt{\ln n}-\int_1^n \sqrt{ \frac{t}{\ln t}} dt$
Since $\displaystyle \sqrt{ \frac{t}{\ln t}}dt = o\left( \sqrt{ t\ln t} \right) $ and $\int_1^\infty \sqrt{ t\ln t} dt = \infty$,
$$\displaystyle \int_1^n \sqrt{ \frac{t}{\ln t}} dt = o\left(\int_1^n \sqrt{ t\ln t} dt \right)$$
Therefore, $\displaystyle \int_1^n\sqrt{ \frac{t}{\ln t}}dt \sim \frac 23 n^{\frac 32}\sqrt{\ln n}$
and $$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor \sim\frac 23 n^{\frac 32}\sqrt{\ln n}$$
As a result, you should expect a sharp upper bound to have $\displaystyle\frac 23 n^{\frac 32}\sqrt{\ln n}$ as its leading term.
| {
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"url": "https://math.stackexchange.com/questions/1850686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
| We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation.
Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd then, $$x^2+y^2+z^2\equiv(0+0+1)\mod4 ; 2xyz \equiv 0 \mod4$$ So, the only possibility is that all are even.
Let x = 2X , y = 2Y ,z = 2Z. Then, $$4X^2 +4Y^2 +4Z^2 = 16XYZ $$ $$\implies X^2 + Y^2 + Z^2 = 4XYZ$$ The same argument goes show that X,Y,Z are even. The process can be continued indefinitely.
This is possible only when x = y = z = 0. So ,there exist no other solution over $\mathbb{N} $
| {
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"url": "https://math.stackexchange.com/questions/1851751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$ what is the value of $P\times Q\times R$ Problem
The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$
And we are asked to find $P\times Q\times R$ by using the identities:
$P(x)=Q(x)\times D(x)+R(x)$
where $P(x)$ is the equations and $Q(x)$ is the quotient and $D(x)$ is the divisor and $R(x)$ is the remainder.
My thought process
so I had set $$ax^2+bx+c=Q(x)\times(x-6)\times(x-P)+0$$
and in the next equation:
$$cx^2+bx+a=Q(x)\times(x-P)(x-Q)+0$$
I really don't know where to go from here
| $$ax^{ 2 }+bx+c=0\\ 6p=\frac { c }{ a } \\ cx^{ 2 }+bx+a=0\\ \\ qr=\frac { a }{ c } \\ 6pqr=1\Rightarrow \quad pqr=\frac { 1 }{ 6 } $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$? The primary question was:
What is the minimum value for $(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
$\color{red}{\text{But sorry guys! I messed it up! my question is:}}$ (Edited)
What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
I feel like this needs to deal with AM-GM inequality but I don't know how to apply that onto this case. It is easy but like I still don't know how to do it! Can someone help me with it?
| Since $a+b+c=1$, Re-write all $1$s to $a+b+c$, then
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\left(\frac{a+b+c}{a}-1\right)\left(\frac{a+b+c}{b}-1\right)\left(\frac{a+b+c}{c}-1\right)=\left(\frac{b+c}{a}\right)\left(\frac{a+c}{b}\right)\left(\frac{a+b}{c}\right)$$
Using AM-GM inequality,
$$b+c\geq2\sqrt{bc}$$
$$a+c\geq2\sqrt{ac}$$
$$a+b\geq2\sqrt{ab}$$
For these equations, equality holds when $a=b=c=1/3$.
So we get
$$\left(\frac{b+c}{a}\right)\left(\frac{a+c}{b}\right)\left(\frac{a+b}{c}\right)\geq\frac{2\sqrt{bc}}{a}\frac{2\sqrt{ac}}{b}\frac{2\sqrt{ab}}{c}=8\frac{abc}{abc}=8$$
Thus the minimum is $8$ when $a=b=c=\frac{1}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{1}{\cos^2\frac{\pi}{7}}+ \frac {1}{\cos^2\frac {2\pi}{7}}+\frac {1}{\cos^2\frac {3\pi}{7}} = 24$ Someone gave me the following problem, and using a calculator I managed to find the answer to be $24$.
Calculate $$\frac {1}{\cos^2\frac{\pi}{7}}+ \frac{1}{\cos^2\frac{2\pi}{7}}+\frac {1}{\cos^2\frac{3\pi}{7}}\,.$$
The only question left is, Why? I've tried using Euler's Identity, using a heptagon with Law of Cosine w/ Ptolemy's, etc. but the fact that the cosine values are all squared and in the denominator keeps getting me stuck.
If $\zeta=e^{\frac{2\pi i}{7}}$, then the required expression is
$$4\left(\frac{\zeta^2}{(\zeta+1)^2}+\frac{\zeta^4}{(\zeta^2+1)^2}+\frac{\zeta^6}{(\zeta^3+1)^2}\right).$$
How do we simplify this result further?
| Incomplete solution
The idea behind this solution is using Vieta's cubic formulas to rewrite the equality in a simpler, more manageable way. Letting $\cos^2\frac{\pi}{7}=\alpha$, $\cos^2\frac{2\pi}{7}=\beta$, and $\cos^2\frac{3\pi}{7}=\gamma$, we have this:
$$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=24$$
$$\frac{\alpha\beta+\alpha\gamma+\beta\gamma}{\alpha\beta\gamma}=24$$
By letting $\alpha,\beta,$ and $\gamma$ be roots to a cubic, we use Vieta's formula's to see that $-\frac{c}{d}=24$
Now, consider the three denominators as roots to a cubic polynomial. Let's make a third degree equation using them:
$$(x-\cos^2\frac{\pi}{7})(x-\cos^2\frac{2\pi}{7})(x-\cos\frac{3\pi}{7})$$
Let's use substitutions to avoid some nasty simplified expression. Let $u=\cos\frac{\pi}{7}$. Using some angle addition trig identities, we transform our equation into this:
$$(x-u^2)(x-4u^2+4u-1)(x-16u^6+24u^4-9u^2)$$
Multiply them out:
$$x^3+(-16u^6+24u^4-14u^2+4u-1)x^2+(80u^8-64u^7-104u^6+96u^5+25u^4-40u^3+10u^2)x+(-64u^{10}+64u^9+80u^8-96u^7-12u^6+36u^5-9u^4)$$
Consider the form $ax^3+bx^2+cx+d$. Per Vietas' formulas, $-c/d=24$.
$$\frac{80u^6-64u^5-104u^4+96u^3+25u^2-40u+10}{64u^8-64u^7-80u^6+96u^5+12u^4-36u^3+9u^2}=24$$
I will continue to work on proving this last part, but I figured at least simplifying your problem to this might be of some asssistance, and someone else may be able to validate the last equation before I do. Basically, if you prove the last equality you prove the question posted.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $E(X\mid X+Y)$ if $(X,Y)$ is centered normal with known covariance matrix
The random variable $(X,Y)$ has a two dimensional normal distribution with mean $(0,0)$ and covariance matrix $\begin{pmatrix}
4&2 \\
2&2
\end{pmatrix}$. Find $E(X\mid X+Y)$.
I am completely lost with this question.
| Let's pose $Z = X+Y$. The random variable $(X,Z)$ has a two dimensional normal distribution with mean $$(E(X),E(Z)) ) (E(X), E(X)+E(Y)) = (0,0)$$ and covariance matrix $$\pmatrix{E(X^2) & E(XZ)\\E(ZX) & E(Z^2)} = \pmatrix{E(X^2) & E(X^2)+E(XY) \\E(X^2) + E(YX) & E(X^2)+E(Y^2)+E(2XY)} = \\
\pmatrix{4 & 4+2 \\4 + 2 & 4 + 2 + 2\cdot2} = \pmatrix{4 & 6\\6 & 10}.$$
The correlation coefficient of $X$ and $Z$ is:
$$\rho = \frac{E(XZ)}{\sqrt{E(X^2)E(Z^2)}} = \frac{6}{\sqrt{4 \cdot 10}} = 3\frac{\sqrt{10}}{10}.$$
The conditional expected value is:
$$E(X\mid Z) = \mu_X + \rho \sqrt{\frac{E(X^2)}{E(Z^2)}}(Z-\mu_Z) = \\
= 0 + 3\frac{\sqrt{10}}{10}\sqrt{\frac{4}{10}}(Z-0) = \frac{3}{5}Z.$$
For the last step, take a look here.
| {
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"url": "https://math.stackexchange.com/questions/1862551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Last digit on $3^{100}$ How to find the last digit on $3^{100}$?
Is there any proper method to solve such questions without calculator of course?
| $3^{100}=(3^4)^{25 }=81^{25}=(80 + 1)^{25}=a*80^{25}+b*80^{24}+.....+x*80^2+y*80 + 1$ where a,b,c,....,y are integer coeffients obtained by expanding $(80 +1)^{25} $. We can calculate these coefficients if we wanted to, but it's enough to know they are integers.
Notice, every term $k*80^j $ is divisible by 10 so
$3^{100}=10M + 1$ where $M= (a*80^{25}+.....+y*80)/10$. $M $ is a whole number, so the last digit is 1.
This is more detail than we need.
It's a basic fact that if we use the notation $a \equiv b \mod n $ to mean that for integers $a,b$ and $n $ that $a$ and $b $ will have the same remainder when divided by $n $ that if $a \equiv 1 \mod n $ then $ak \equiv k \mod n $.
(Because $a \equiv 1 \mod n$ means $a=jn +1$ so $ak =(jk)n+k $ so $ak \equiv k \mod n$.)
Therefore if $a^k \equiv 1 \mod n $ than $a^{mk}=a^k*a^k*....a^k\equiv 1*1*...*1 \mod n $.
So as $3^4=81 \equiv 1 \mod 10$, it follows that $3^{4m} \equiv 1 \mod 10$ and will have 1 as a last digit.
More interesting is what are the last 2 digits of $3^{100} $. $3^4=81 \mod 100$ so $3^8=6400+160+1\equiv 61 \mod 100$. $3^{12} \equiv 81*61 \equiv (6+8)*10+1\equiv 41 \mod 100$. $3^{16}\equiv 61^2\equiv (6+6)10+1\equiv 21\mod 100$. $3^{20}\equiv 21*81\equiv (8+2)10+1 \equiv 1701 \equiv 1\mod 100$. So $3^{20m}\equiv 1 \mod 100$ so the last two digits of $3^{100}$ are 01.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral of a rational function with linear denominator: $ \int \frac{ x^7}{(x+1)}{dx} $ $$ \int \frac{ x^7}{(x+1)}{dx} $$
$$ \int \frac{ \left(x^7 + x^6 - x^6 - x^5 + x^5 + x^4 -x^4 - x^3 + x^3 + x^2 - x^2 -x^1 + x^1 +1 -1\right ) }{\left(x+1\right)}{dx}$$
$$ \int { \left(x^6 - x^5 + x^4 - x^3 + x^2 -x +1 - \frac{1}{x+1}\right)}{dx}$$
$$\frac{x^7}{7} - \frac{x^6}{6} + \frac{x^5}{5} -\frac{x^4}{4}+ \frac{x^3}{3} - \frac{x^2}{2} + \frac{x^1}{1} -\frac{\log(x+1)}{1} +C $$
Is there any other good way to do this ?
| Same as your method : But try to use reduction formulas for higher degrees :
Let
$$I_n= \int \frac{x^n}{(x+a)}dx$$
$$I_n= \int \frac{(x+a)x^{n-1}-ax^{n-1}}{(x+a)}dx$$
$$I_n= \int x^{n-1}dx-a\int \frac{x^{n-1}}{(x+a)}dx$$
So $$I_n+aI_{n-1}=\frac{x^n}{n}$$
Now you have to find $I_7$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the Inequality $\frac{1}{1-x}-\frac{x(3-x)(2-x)(13x^4-50x^3+89x^2-84x+36)}{4(1-x)(2x(1-x))^2}<1$ Can anyone suggest any hints to prove the following inequality:
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
for all $x \in (0,1)$?
| One may see that for the initial inequality to hold true it is sufficient to prove that
$$
1-\frac{(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)}{16 x(1-x)^2 }<0,\quad x \in (0,1), \tag1
$$ then setting
$$
\begin{align}
&f(x)=16 x(1-x)^2-(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)
\end{align}
$$ one gets$$
\begin{align}
&f'(x)=700-2044 x+2535 x^2-1668 x^3+575 x^4-78 x^5 \in [20,700] \tag2
\\\\&f'(x)>0\implies f \nearrow, \quad x \in (0,1), \quad f(0)=-216, \quad f(1)=-8,
\end{align}
$$ giving $$ f(x)<0, \quad x \in (0,1), \tag3$$
which yields $(1)$ then yielding the initial inequality.
Remark. By setting $t=1-x$ in $f'(x)$ above, one gets a polynomial with positive coefficients:
$$
20+68 t+201 t^2+148 t^3+185 t^4+78 t^5 >0,\quad t \in (0,1),
$$
proving $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$ So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated
| Assuming the right expression to be $$\dfrac{1+\sqrt{\dfrac{4-x}{4+x}}}{1-\sqrt{\dfrac{4-x}{4+x}}}$$
$$\tan^{-1}\dfrac{1+\sqrt{\dfrac{4-x}{4+x}}}{1-\sqrt{\dfrac{4-x}{4+x}}}=\dfrac\pi4+\tan^{-1}\sqrt{\dfrac{4-x}{4+x}}$$
Let $\tan^{-1}\sqrt{\dfrac{4-x}{4+x}}=y\ge0\implies0\le y<\dfrac\pi2$
$$\dfrac{4-x}{4+x}=\tan^2y,\dfrac x4=\cdots=\cos2y\implies2y=\arccos\dfrac x4$$
| {
"language": "en",
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if $x \neq 0$ and $x = \sqrt{4xy - 4y^2}$, then how does expressing $x$ in terms of $y$ mean $x = 2y$? I have the equation $x=\sqrt{4xy - 4y^2}$, and I know that $x=2y$ when expressed in terms of $y$, but I'm not sure of the process to get there.
I know that
\begin{align}\sqrt{4xy - 4y^2} &= \sqrt{4y(x-y)}\\
&= 2 \sqrt{y(x-y)} \\
&= 2(xy-y^2)^{\frac12}
\end{align}
but pretty stumped.
| $x =\sqrt{4xy + 4y^2}$
$x^2 = 4xy - 4y^2$
$x^2 - 4xy + 4y^2 = 0$
$x = \frac{4y \pm \sqrt{16y^2 - 16y^2}}{2}$
$x = \frac{4y \pm \sqrt{0}}{2}$
$x = 2y$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
|
$${ \left( x-1 \right) }^{ 2 }-{ \left( x-5 \right) }^{ 2 }=\left( x-1-x+5 \right) \left( x-1+x-5 \right) =4\left( 2x-6 \right) =8\left( x-3 \right) $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that this inequality doesn't hold Given $(a,b,c) \in \mathbb R^3_+$ show that atleast one of the real numbers $a(1-b)$, $ b(1-c)$ and $c(1-a)$ is less than or equal to 1\4.
I tried to show it by contradiction i.e Suppose that $$a(1-b) > \frac{1}{4} $$
$$b(1-c) > \frac{1}{4} $$$$c(1-a) > \frac{1}{4} $$
So as to end with a contradiction but it ended up in a mess. Please help.
| Suppose that for all $a,b,c$ $$a(1-b) > \frac{1}{4}\\b(1-c) > \frac{1}{4}\\c(1-a) > \frac{1}{4}$$ We then have $$a+b+c\gt \frac 34+(ab+ac+bc)=\frac 34+\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}$$ It follows that if $a=b=c$ we have $$12a\gt 15a^2+3\Rightarrow(15a-6)^2+45\lt 0$$ This is absurde.
| {
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Solving $2^x+2^y+1=3^z$ in integers I have reasons to believe that there should be an elementary, relatively simple way to find all solutions of the equation in the title in positive integers $x>y$ and $z$. Any ideas?
| I think I was eventually able to recover (?) the elementary solution mentioned in the literature. I will show below that the only solutions in positive integers $x,y,z$ are $x=4,y=6,z=4$, $x=6,y=4,z=4$, and $x=y=z=2$.
Reducing $2^x+2^y+1=3^z$ modulo $3$, we conclude that both $x$ and $y$ are even; say, $x=2a$ and $y=2b$, leading to $4^a+4^b+1=3^z$. Reducing now modulo $4$ we see that $z$ is even, too, and the substitution $z=2c$ gives $4^a+4^b+1=9^c$.
Assume without loss of generality that $a\le b$. If $a=1$, then $4^b+5=9^c$ whence $b=1$ and therefore also $c=1$, as otherwise reducing modulo $16$ we would get $9^c\equiv 5\pmod{16}$, which is impossible (the order of $9$ modulo $16$ is $2$). Thus $a\ge 2$, and reducing $4^a+4^b+1=9^c$ modulo $16$ we obtain $9^c\equiv 1\pmod{16}$, whence $c$ is even; we write $c=2u$ to get $4^a+4^b+1=81^u$.
We know that $a\ge 2$. If, indeed, $a=2$, then $4^b+17=81^u$, whence $(9^u-2^b)(9^u+2^b)=17$, leading to $u=1,b=3$; that is, $x=z=4$, $y=6$.
We are thus left with the case where $b\ge a\ge 3$. Reducing $4^a+4^b+1=81^u$ modulo $64$ and observing that the order of $81$ modulo $64$ is $4$, we conclude that $u$ is divisible by $4$, whence $4^a+4^b+1=3^{16v}$, $u=4v$. Reducing then modulo $85$ (which is a divisor of $3^{16}-1$), we get $4^a+4^b\equiv 0\pmod{85}$. Dividing through by $4^a$ gives $4^{b-a}\equiv -1\pmod{85}$. However, there is no power of $4$ congruent to $-1$ modulo $85$, as one can easily verify (the order of $4$ modulo $85$ is $4$). This shows that there are no solutions with $a,b\ge 3$.
| {
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How to solve $\int \frac{1}{x^2+4x+7} dx$? How to solve $\int \frac{1}{x^2+4x+7} dx$?
I think the first step is to write it in the following form: $$\int \frac{1}{(x+2)^2+3} dx$$
| $$\int \frac { 1 }{ (x+2)^{ 2 }+3 } dx=\int { \frac { dx }{ 3\left( 1+{ \left( \frac { x+2 }{ \sqrt { 3 } } \right) }^{ 2 } \right) } } =\frac { 1 }{ \sqrt { 3 } } \int { \frac { d\left( \frac { x+2 }{ \sqrt { 3 } } \right) }{ 1+{ \left( \frac { x+2 }{ \sqrt { 3 } } \right) }^{ 2 } } } =\\=\frac { 1 }{ \sqrt { 3 } } \arctan { { \left( \frac { x+2 }{ \sqrt { 3 } } \right) } } +C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Another way to solve $| x^2 - 4x + 3 | + | x^2 - 6x + 8 | =\sqrt{4|x|^2 - 20x + 25}$ Let $| x^2 - 4x + 3 | + | x^2 - 6x + 8 | =\sqrt{4|x|^2 - 20x + 25}$ then number of integral values of $|x|$ is
My work
$=
\begin{cases}
2x^2 - 10x + 11= -2x + 5, & x\ge 1\\
-2x+5 = -2x+5 , & 1 \lt x \le 2
\\
-2x^2 + 10x - 11= -2x + 5 ,& 2 \lt x \le \frac{5 }{2}\\
-2x^2 + 10x - 11= 2x - 5 ,& \frac{5}{2} \lt x \le 3 \\
2x - 5=2x - 5, & 3 \lt x \le 4
\\
2x^2 - 10x + 11= 2x - 5 ,& x \gt4
\end{cases}$
I want to know that how this question can be done in another way .
| Note that $|x|^2 = x^2$ for all $x\in \mathbb R$. Using @juantheron's comment, notice that
$$|x^2-4x+3-(x^2-6x+8)| = |2x-5| = \sqrt{4x^2-20x+25}.$$
Finally, note that
$$(x^2-4x+3)(x^2-6x+8) = (x-1)(x-2)(x-3)(x-4).$$
So the set of solutions is $[1,2]\cup[3,4]$.
| {
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Getting the wrong result when solving a limit I correctly solved to following limit like this:
$$\lim_{x\to\infty} \left(\frac{1}{x^2}\right)^{\frac{2x}{x+1}}$$
$$ = \lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2x}{x(1+\frac{1}{x})}}$$
$$ = \lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2}{(1+\frac{1}{x})}}$$
$$ = 0^2 = 0$$
However, I usually use the following formula for these kind of limits: $\lim_{x\to\infty} (1+\frac{1}{v})^v = e$ This is simple one that doesn't really need this extra complexity, but anyway. The problem is that I get a different and wrong result:
$$\lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2x}{x+1}}$$
$$= \lim_{x\to\infty} (1 + \frac{1}{x^2} - 1)^{\frac{2x}{x+1}}$$
$$= \lim_{x\to\infty} (1 + \frac{1-x^2}{x^2})^{{\frac{2x}{x+1} \frac{1-x^2}{x^2} \frac{x^2}{1-x^2}}}$$
$$= e^{ \lim_{x\to\infty} {{\frac{2x}{x+1} \frac{1-x^2}{x^2}}}}$$
$$= e^{ \lim_{x\to\infty} {{\frac{2x-2x^3}{x^3+x^2}}}} = e^{ \lim_{x\to\infty} {{\frac{-2x^3}{x^3}}}} = e^{-2}$$
| $$\frac{1-x^2}{x^2}$$ does not approach zero as $x$ approaches infinity, so you cannot simply write that $$\lim_{x\rightarrow\infty}\left(1+\frac{1-x^2}{x^2}\right)^{\frac{x^2}{1-x^2}}=e$$
| {
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Integration of root of quadratic over quadratic I need help to get a generalized solution for this type equations:
$$ \Gamma(v) = \int\frac{(1-v^2)^{\frac{1}{2}}}{a^{2}-b^{2}v^{2} }dv $$
NOTE: This is not a homework question. I am in high school and I was reading up some extra material and landed on this from another equation.
Sorry for the bad formatting.
Thanks
| Substituting $v=\sin\theta$ gives $\sqrt{1-v^2}=\cos\theta$, $dv=\cos\theta\,d\theta$ so we have
\begin{eqnarray}
\int\dfrac{\cos^2\theta}{a^2-b^2\sin^2\theta}\,d\theta&=\int\dfrac{1-\sin^2\theta}{a^2-b^2\sin^2\theta}\,d\theta\\
&=\dfrac{1}{b^2}\int\left[1-\dfrac{a^2-b^2}{2a}\left(\frac{1}{a-b\sin\theta}+\frac{1}{a+b\sin\theta} \right) \right]\,d\theta
\end{eqnarray}
The rational functions of $\sin\theta$ can be completed by the substitution
\begin{equation}
t=\tan\left(\frac{\theta}{2} \right),\, d\theta=\frac{2dt}{1+t^2},\,\sin\theta=\frac{2t}{1+t^2}
\end{equation}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of given limit when $f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\frac{1}{\tan 2^nx}$
Question Statement:-
If $\displaystyle f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\dfrac{1}{\tan 2^nx}$, then find the value of
$$\lim_{x\to 0} \bigg( (\cos x)^{g(x)}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}} \bigg)$$
I am not able to find value of $g(x)$. Could someone help me as how to calculate value of $g(x)?$
| Basically, what you have to show (lab bhattacharjee gave you the hint while I was typing) is that $$f(x)=\cot (x)-\cot \left(2^n x\right)$$ which makes $$g(x)=\cot (x)$$ Now consider $$A=\cos(x)^{\cot(x)}\qquad , \qquad B=\bigg(\frac{1}{\cos (x)} \bigg)^{\frac{1}{\sin (x)}} $$ So $$\log(A)=\cot(x) \log(\cos(x))\qquad , \qquad \log(B)=-\frac{1}{\sin (x)}\log(\cos(x))$$ and use Taylor series to get $$\log(A)=-\frac{x}{2}+\frac{x^3}{12}+O\left(x^4\right)$$ $$\log(B)=\frac{x}{2}+\frac{x^3}{6}+O\left(x^4\right)$$ Now, using $y=e^{\log(y)}$ and Taylor again $$A=1-\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{16}+O\left(x^4\right)$$ $$B=1+\frac{x}{2}+\frac{x^2}{8}+\frac{3 x^3}{16}+O\left(x^4\right)$$ $$A+B=2+\frac{x^2}{4}+\frac{x^3}{4}+O\left(x^4\right)$$ which shows the limit and how it is approached.
For illustration puroposes, let us compute the value of the expression for $x=\frac \pi 4$. The exact value is $$\frac{1}{\sqrt{2}}+2^{\frac{1}{\sqrt{2}}}\approx 2.33963$$ while the above formula gives $$2+\frac{\pi ^2}{64}+\frac{\pi ^3}{256}\approx 2.27533$$ and we are far away from $x=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.
Hence show that $(217)^\frac{1}{3} \simeq 6.0092$
My attempt :
$$\frac{1+x+x^2}{(1-x)^2}=\frac{(1+x+x^2)(1-x)}{(1-x)^3}$$
$$\frac{1+x+x^2}{(1-x)^2}=\frac{1-x^3}{(1-x)^3}$$
$$=\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
$$=\frac{1}{\left(1+(-x)\right)^3}+\frac{1}{\left(1+\left(-\frac{1}{x}\right)\right)^3}$$
$$=(1+(-x))^{-3}+\left(1+\left(-\frac{1}{x}\right)\right)^{-3}$$
How can I proceed after this ? Is there another method ? Is my method correct ?
| In another approach, as inquired in the OP, we write the term of interest as
$$\begin{align}
\frac{1+x+x^2}{(1-x)^2}&=1-\frac{3}{1-x}+\frac{3}{(1-x)^2}\\\\
&=1-3\sum_{n=0}^\infty x^n+3\frac{d}{dx}\sum_{n=0}^\infty x^n\\\\
&=1-3\sum_{n=0}^\infty x^n+3\sum_{n=1}^\infty nx^{n-1}\\\\
&=1-3\sum_{n=0}^\infty x^n+3\sum_{n=0}^\infty (n+1)x^n\\\\
&=1+3\sum_{n=0}^\infty nx^n
\end{align}$$
Hence, the coefficient on the $n$'th term is indeed $3n$ as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a$ and $b$ are integers then $a^2 -4b-2$ does not equal zero. I was just wondering if the rough outline of this proof was logically rigorous.
If a and b are integers then $a^2 -4b-2$ does not equal zero.
Proof:
Suppose for the sake of contradiction that $a^2-4b-2=0$.
Then we have $a^2 = 2(2b+1)$. $a^2$ is even which implies that $a$ is even.
If $a$ is even then $a = 2k$ for some integer $k$.
$(2k)^2 = 2(2b+1)$
$2(2k^2) = 2(2b+1)$
$2k^2 = 2b+1$
Let $w = 2k^2$ where since$ k^2 $is an integer, $w$ is even.
Let $v = 2b+1$ where since $b$ is an integer, $v$ is odd.
Then we have, $w = v$, which is a contradiction.
QED
Thanks for the feedback!
| I think the best way to think about this is to work modulo 4:
$0^2 \equiv 0$ mod 4
$1^2 \equiv 1$ mod 4
$2^2 \equiv 0$ mod 4
$3^2 \equiv 1$ mod 4
Clearly $4b + 2 \equiv 2$ mod 4. So indeed it is not possible that $a^2$ = 4b + 2. Similarly you can show $a^2$ - 4b - 3 is not equal to zero either.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the fraction $\frac{n^3 + 2n}{n^4 + 3n^2 + 1}$ is irreducible for $n \in \mathbb{N}$
Prove that the fraction $\dfrac{n^3 + 2n}{n^4 + 3n^2 + 1}$ is in lowest terms for every positive integer $n$.
I just don't know how to solve this. I tried polynomial division, expressing the gdc of the two terms as a linear combination, and factorizing the sum of the two terms but nothing really leads anywhere. I'd really appreciate some help.
| You can do it with linear combinations:
$$
(n^4+3n^2+1)-n(n^3+2n)=n^2+1 \tag{1}
$$
Then
$$
(n^3+2n)-n(n^2+1)=n \tag{2}
$$
and
$$
(n^2+1)-n\cdot n=1 \tag{3}
$$
Just work backwards if you need the linear combination, but this is already sufficient.
\begin{align}
1
&=\color{red}{(n^2+1)}-n\cdot\color{red}{n}
&&\text{by (3)}\\
&=\color{red}{(n^2+1)}-
n\cdot\bigl(\color{green}{(n^3+2n)}-n\color{red}{(n^2+1)}\bigr)
&&\text{by (2)} \\
&=(-n)\color{green}{(n^3+2n)}+(n^2+1)\color{red}{(n^2+1)}
&&\text{reorder}\\
&=(-n)\color{green}{(n^3+2n)}+
(n^2+1)\bigl(\color{green}{(n^4+3n^2+1)}-n\color{green}{(n^3+2n)}\bigr)
&&\text{by (1)}\\
&=(n^2+1)\color{green}{(n^4+3n^2+1)}+(-n^3-2n)\color{green}{(n^3+2n)}
&&\text{reorder}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$
Integrate $$\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$$.
My Try:
$$\int \frac{x\ln (x)}{x^3(1-1/x^2)^{3/2}} dx=\int \frac{\frac{1}{x^2}\ln (x)}{(1-\frac{1}{x^2})^{3/2}} dx.$$
Putting $x=\frac{1}{t}$,
I get
$$\int \frac{t^2\ln (\frac{1}{t})}{(1-t^2)^{3/2}(-t^2)} dt=\int \frac{\ln ({t})}{(1-t^2)^{3/2}} dt.$$
After this what to do?
| Check that
$$\int\frac1{t^{3/2}}dx=-2t^{-1/2}+C\implies \int\frac{f'(x)\,dx}{f(x)^{3/2}} =-2f(x)^{-1/2}+C$$
and now just observe that $\;2x=(x^2-1)'\;$ , so by parts:$${}$$
$$\begin{cases}&u=\log x,&u'=\cfrac1x\\{}\\&v'=\cfrac x{(x-1)^{3/2}},&v=-\cfrac1{\sqrt{x^2-1}}\end{cases}\;\;\implies$$$${}$$
$$\int\frac{x\log x}{(x^2-1)^{3/2}}dx=-\frac{\log x}{(x^2-1)^{3/2}}+\int\frac{dx}{x\sqrt{x^2-1}}$$
and remember (or prove...)
$$\int\frac{dx}{x\sqrt{x^2-1}}=-\arctan\frac1{\sqrt{x^2-1}}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real function with complex antiderivative $\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}$? Consider this indefinite integral (I'm interested in the interval $x>0$):
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx$$
By substitution:
$$u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}, \qquad x=\frac{1}{2u \sqrt{u^2-1}}$$
We get a closed form antiderivative:
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx=\sqrt{2} \tanh^{-1} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}} +C$$
Now the inverse hyperbolic tangent is real only for the argument in $(-1,1)$. But in our case the argument for all real $x$ is $ > 1$.
How can the function with real values on $x>0$ have a complex antiderivative?
Edit
To be clear, this is the correct antiderivative. I.e. by differentiating it we get the function under the integral.
How would you explain this without appealing to complex analysis, i.e. branches? This is a real valued function for positive $x$, so this can be given as an assignment to a first year calculus student for example
Edit 2
The derivation of the antiderivative after the substitution:
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx=\sqrt{2} \int \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}} \frac{dx}{\sqrt{x^2+1}}=\sqrt{2} \int u \frac{dx}{\sqrt{x^2+1}}$$
$$\sqrt{x^2+1}=x(2u^2-1)$$
$$dx=-x\frac{2u^2-1}{u(u^2-1)}du$$
$$\frac{dx}{\sqrt{1+x^2}}=-\frac{du}{u(u^2-1)}=\frac{du}{u(1-u^2)}$$
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx=\sqrt{2} \int \frac{du}{1-u^2}=\sqrt{2} \tanh^{-1} u$$
| I got the answer
$$\text{arctanh} = \frac12 \ln \frac{1+x}{1-x}$$
consider $\ln(x)$ where $x < 0 $
$$\ln(x) = \ln(|x| (-1))=$$
$$ = \ln(x) + \ln(-1)= \qquad \text{(properties of logs)} $$
$$= \ln(x) + \ln(e^{i \pi}) $$
$$= \ln(x) + i \pi$$
so in your integration, the log of a negative number generates a constant $i \pi$ (or $ki \pi$ where $k$ is a constant)- this just disappears into the constant of integration, similar to other constants that can be generated and absorbed in integrations involving inverse hyperbolic functions
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Interesting Logarithmic Integral: $\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx $ Other than numerical approximation, how can we calculate the closed form of this integral?
$$\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx $$
| OK as I suggested in a comment here is an approach.Let us begin with the Taylor series of $\ln^2(1+x)$. It is known to be:
$$\ln^2 (1+x) = 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n$$
Then we have successively:
\begin{align*}
\int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x &= 2 \int_{0}^{1} \frac{\ln^2 x}{x} \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n \, {\rm d}x \\
&=2 \int_{0}^{1} \ln^2 x \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^{n-1} \, {\rm d}x\\
&= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} \int_{0}^{1} x^{n-1} \ln^2 x \, {\rm d}x\\
&= 2 \sum_{k=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n^4}\\
&=2 \sum_{k=2}^{\infty} \frac{(-1)^n \left [ \mathcal{H}_n - \frac{1}{n} \right ]}{n^4} \\
&= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} - 2 \sum_{n=2}^{\infty} \frac{(-1)^n}{n^5} \\
&= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} - 2 - \frac{15 \zeta(5)}{16}
\end{align*}
Well for the Euler sum first of all we have the generating function:
\begin{align}
\sum^\infty_{n=1}\frac{H_n}{n^3}z^n
=&2{\rm Li}_4(z)+{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right)\\
&+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\\
&+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90}
\end{align}
Integrating once and plugging $z=-1$ we get the value of the sum. I am not presenting the full calculatios but it suffices to say that:
$$\sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} = \frac{\pi^2 \zeta(3)}{4} -\frac{43 \zeta(5)}{32}+1$$
Thus:
$$\int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x = \frac{\pi^2 \zeta(3)}{3} - \frac{29 \zeta(5)}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does the limit exist for $\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}$? $$\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}\le \lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3}}=\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2}x^{\frac{3}{2}}}=\frac{\pi}{\sqrt{2}\cdot2}$$
So the limit of $\displaystyle\frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}$ is finite?
| $$\displaystyle\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}=\lim_{x\to \infty} \frac{\arctan(x)}{\sqrt{2+4\frac{1}{x}+\frac{1}{x^2}}}=\frac{\pi}{\sqrt{2}\cdot2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882386",
"timestamp": "2023-03-29T00:00:00",
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In ℕ⁺, can the sum of three squares equal the sum of two squares? Are there any examples where:
$a² + b² + c² = p² + q²\qquad {a, b, c, p, q ∈ ℕ⁺}\tag{1}$
If not, can $(1)$ be disproven?
| One need only to pick out 3 distinct naturals, with the only constraint that they aren't all even:
Suppose you pick $(p,q,r) \in \mathbb N^3, \ \ \ $ let $r \gt q \ \land \ r\gt p ,\ \ $ then there exists a natural number $N \not \equiv 2 \pmod 4 $, which can be converted to a difference of two squares in at least one way, such that:
$$p^2+r^2=q^2+N \qquad | \qquad N=d_1 \cdot d_2, \quad N \not \equiv 2 \pmod 4 $$
And then converting $N$ to a difference of two squares using the identity below:
$N=\alpha^2-\beta^2 \quad | \quad (\alpha,\beta) \in \mathbb N^2\implies$
$$p^2+r^2+\beta^2=q^2+\alpha^2$$
Note that, to allow that $N \not \equiv 2 \pmod 4, \ \ $if you chose only one even, it needs to go on the L.H.S., but if you choose two they can be placed in any position.
Ex: Pick $(1,2,5). \ \ $ Then we have $2^2+5^2=1^2+N \quad | \quad N=28=14\cdot 2=8^2-6^2 \ \implies$
$$2^2+5^2+6^2=1^2+8^2$$
Suppose you don't want to pick three numbers repeatedly, then you can pick one well-chosen pell equation:
For nonsquare integers $C$ and $D$, if you take any pell-type equation of the form:
$$x^2-Dy^2=C \qquad | \qquad C \not \equiv 2 \pmod 4 \quad \land \quad D \not \equiv 2 \pmod 4$$
then it can be converted into a sum of three squares equaling a sum of two squares. You only need to rewrite $C$ and $D$ as a difference of two squares:
$$C=\beta^2-\alpha^2, \quad D=\gamma^2-\delta^2 \quad \implies$$
$$\alpha^2+x^2+(\delta y)^2=\beta^2+(\gamma y)^2$$
That you can always find $\alpha, \beta, \gamma, \delta$ that works is evident from:
*
*The identity $$N=d_1\cdot d_2=\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2$$
*The constraint that $C$ and $D$ are not $ 2 \pmod 4$. Note that if they are, you need only to multiply the pell type equation by $4$, producing $(2x)^2+Ey^2=F$, and using the above identity, and
*All natural numbers have at least 2 divisors. And all natural numbers that are not $ 2 \pmod 4$ have at least 2 divisors of the same parity, so that their sum or difference is even and you can divide by 2.
Thus for each solvable pell-type equation of the above form, there are an infinity of associated "five-tuplets" which meet the question.
Example: $x^2-13y^2=3 \quad \to \quad x^2-(7^2-6^2)y^2=2^2-1^2 \quad \implies$
$$1^2+x^2+(6y)^2=2^2+(7y)^2$$
With the first two solutions $(4,1)$ and , well $(256,71)$ (!) we have
$$1^2+4^2+6^2=2^2+7^2$$
$$1^2+256^2+426^2=2^2+497^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Counting multiple of $3$ in a triangular arrangement of the integers. I write
:$$\mathbb{N}_\triangle = \begin{matrix}
&&&&&21&\ldots \\
&&&&15&20&\ldots \\
&&&10&14&19&\ldots \\
&&6&9&13&18&\ldots \\
&3&5&8&12&17&\ldots \\
1&2&4&7&11&16&\ldots
\end{matrix}$$
Can we determine how many numbers in any given column of $\mathbb{N}_\triangle$ are divisible by $3$. I need help proving the following claim:
If $f_3(n)$ counts the numbers in the $n^{th}$ column of
$\mathbb{N}_\triangle$ that are divisible by $3$ then $$f_3(n)
=\Bigg\lfloor{n+2 \above 1.5pt 3}\Bigg\rfloor$$
For example $$f_3(6) = \Bigg\lfloor{6+2 \above 1.5pt 3}\Bigg\rfloor=\Bigg\lfloor{8 \above 1.5pt 3}\Bigg\rfloor=\Bigg\lfloor 2.66666\Bigg\rfloor =2$$ and surely there are $2$ numbers in column six that are divisible by $3$. I suspect there are no multiples of $3$ on the base of the triangle since possibly $$\Bigg(3, 1+ {n(n-1) \above 1.5pt 2}\Bigg)=1$$ In fact if $(3,T_n)=1$ then I suspect that there are no numbers in the $n^{th}$ row that are divisible by $3$. Surely there are consecutive triangular numbers divisible by $3$.
| The $n$ th column is the sequence of $n$ consecutive numbers starting at $1+\frac{n (n-1)}{2}$. Let's call $b$ this base value.
The sequence length is $n$. $n$ starts at 1 like the height of its column. $k$ is an integer.
Each 3 consecutive numbers sequence contains exactly one multiple of 3.
When starting with a base number like $3k+2$ , $n$ covers $\Bigg\lfloor\frac{n+1}{3}\Bigg\rfloor$ 3-multiples.
If starting with $3k$, $n$ covers $\Bigg\lfloor\frac{n+2}{3}\Bigg\rfloor$ 3-multiples and with $3k+1$ , $\Bigg\lfloor\frac{n}{3}\Bigg\rfloor$ .
It appears that $1+\frac{n (n-1)}{2}$ may be 1 or 2 modulo 3 and is never 0 :
if $n=0$ modulo 3 , we have $1+\frac{n (n-1)}{2} = 1$ modulo 3 , then $f_3(n) = \Bigg\lfloor\frac{n}{3}\Bigg\rfloor$
if $n=1$ modulo 3 , we have
$1+\frac{n (n-1)}{2} = 1$ modulo 3 , then $f_3(n) = \Bigg\lfloor\frac{n}{3}\Bigg\rfloor$
if $n=2$ modulo 3 , we have
$1+\frac{n (n-1)}{2} = 2$ modulo 3 , then $f_3(n) = \Bigg\lfloor\frac{n+1}{3}\Bigg\rfloor$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to turn the recurrence $a_n=2a_{n−1}−a_{n−2}$, with $a_0=−1$ and $a_1=1$ into a quadratic, to find the closed form? I am having trouble understanding how to turn the recurrence into a quadratic equation. What principles are involved, rules etc.?
| You can write the recurrence with matices:
$$
\begin{pmatrix}
a_{n-1} \\
a_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-1 & 2 \\
\end{pmatrix}
\begin{pmatrix}
a_{n-2} \\
a_{n-1} \\
\end{pmatrix}
$$
Then,
$$
\begin{pmatrix}
a_{n-1} \\
a_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-1 & 2 \\
\end{pmatrix}^{n-1}
\begin{pmatrix}
-1 \\
1 \\
\end{pmatrix}
$$
You can Jordan-decompose the matrix in the power as $M = S J S^{-1}$
$$
\begin{pmatrix}
0 & 1 \\
-1 & 2 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
-1 & 1 \\
\end{pmatrix}
$$
To do that manually, you can use the characteristic polynomial, which probably is what you meant with quadratic equation.
$$
\det(M - \lambda \mathbb{1})
=
\begin{vmatrix}
-\lambda & 1 \\
-1 & 2-\lambda \\
\end{vmatrix}
= \lambda^2 -2 \lambda + 1
= (\lambda^2 - 1)^2
= 0
$$
That means that there is only the eigenvalue $\lambda = 1$. Then you find eigenvectors and etcetera. Probably there is some answer which already explains how to do the decomposition in detail.
Once you have the decomposition, it's easy to see that
$$
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}^{n-1}
=
\begin{pmatrix}
1 & n-1 \\
0 & 1 \\
\end{pmatrix}
$$
Therefore,
$$
\begin{pmatrix}
0 & 1 \\
-1 & 2 \\
\end{pmatrix}^{n-1}
=
\begin{pmatrix}
1 & -1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 & n-1 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
-1 & 1 \\
\end{pmatrix}
$$
And finally,
$$
\begin{pmatrix}
a_{n-1} \\
a_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 & n-1 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
-1 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-1 \\
1 \\
\end{pmatrix}
=
\begin{pmatrix}
2n-3 \\
2n-1 \\
\end{pmatrix}
$$
The general term is
$$a_n = 2n-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
solution of absolute value equation Question: If $x + |x| + y = 7$ and $x + |y| - y = 6$, then $x + y =?$
My solution:
I considered each cases of $x$ and $y$ i.e $x$ positive and $y$ positive, $x$ positive and $y$ negative, $x$ negative and $y$ positive, and finally $x$ negative and $y$ negative. After solving I found $x= 4$ and $y=-1$ is the solution. So, their sum is $4-1= 3$.
Is there any smarter way to solve this type of equation?
| You should consider one case:
If $y\ge0$ then $|y|=y$ , and
$$x+|y|-y=x=6$$
and
$$\underbrace{x+|x|}_{12}+y=7\implies y=-5$$
It is contradiction, hence $y<0.$ Similarly $x\le 0$ then $y=7$ whereas $x+\underbrace{|y|-y}_{0}=6$. So $x>0$ and $y<0$, we have
\begin{cases}
2x+y=7\\
x-2y=6
\end{cases}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find the sum upto 20 terms $$\frac{1}{4}+\frac{1\times 2}{4\times 5}+\frac{1\times2\times 3}{4\times5\times6}+\frac{1\times2\times3\times4}{4\times 5\times 6\times 7}+\cdots(\operatorname{upto 20 terms})=?$$
I am trying like this...
$$\frac{1}{4}+\frac{1}{10}+\frac{1}{20}+\frac1{35}+\frac1{56}+\cdots $$
Here some terms can be written like $(\frac1{n-1}-\frac1{n})$, but not all of them.
Also I am unable to find any pattern relating to combinations. Please show the right approach.
| $$\begin{eqnarray*}\sum_{n=1}^{N}\frac{n!}{(n+3)!/6} &=& 6\sum_{n=1}^{N}\left(\frac{1}{(n+1)(n+2)(n+3)}\right)\\&=&3\sum_{n=1}^{N}\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right)\\&=&\color{red}{\frac{1}{2}-\frac{3}{(N+2)(N+3)}}.\end{eqnarray*} $$
An equivalent approach is to use Euler's beta function. We have:
$$ \sum_{n=1}^{N}\frac{6n!}{(n+3)!}=3\sum_{n=1}^{N}\frac{\Gamma(3)\Gamma(n+1)}{\Gamma(n+4)}=3\sum_{n=1}^{N}B(3,n+1) $$
hence:
$$ \sum_{n=1}^{N}\frac{6n!}{(n+3)!}=3\int_{0}^{1}\sum_{n=1}^{N}x^n(1-x)^2\,dx =3\int_{0}^{1}\left(x-x^{N+1}\right)(1-x)\,dx$$
and:
$$ \sum_{n=1}^{N}\frac{6n!}{(n+3)!}=\frac{1}{2}-3\, B(2,N+2)=\color{red}{\frac{1}{2}-\frac{3}{(N+2)(N+3)}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| HINT:
$$\sum_{n=0}^\infty a^n\sum_{m=0}^n b^m=\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}$$
Can you finish?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $\binom{x+2}{5} = 126$ for $x$ How can we solve for $x$ in the equation $\binom{x+2}{5} = 126$? By expanding, we get $(x+2)(x+1)x(x-1)(x-2) = 126(5!)$, and by trial and error, we can see $x = 7$, but how can we solve for this without trial and error? Also, how do we know only one solution exists?
|
Use:
*
*$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$
*$$a!=\prod_{p=1}^{a}p=a\times(a-1)\times(a-2)\times\dots\times3\times2\times1$$
So, we get:
$$\binom{x+2}{5}=126\Longleftrightarrow\frac{(x+2)!}{5!((x+2)-5)!}=126\Longleftrightarrow$$
$$\frac{(x+2)!}{120(x-3)!}=126\Longleftrightarrow\frac{(x+2)!}{(x-3)!}=15120$$
Expanding $\frac{(x+2)!}{(x-3)!}$:
$$\frac{(x+2)!}{(x-3)!}=x(x-2)(x-1)(x+2)(x+2)=x(x^4-5x^2+4)$$
So, we get:
$$x(x^4-5x^2+4)=15120\Longleftrightarrow(x-7)(x^4+7x^3+44x^2+308x+2160)=0$$
So, we will get $7$ as a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer
Let $a$ and $b$ be integers. Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer.
I determined that since $2a^2-1 \equiv 1,7 \pmod{8}$ we must have $b^2+2 \equiv 3 \pmod{8}$ in order for the fraction to be an integer. I didn't see how to find a contradiction from here.
| Hint: Show that $2a^2-1$ is divisible only by primes $\equiv 1,7\pmod 8$, whereas $b^2+2$ must have a prime divisor not of that form. You already did the latter part.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find three consecutive positive integers a,b and c such that a+b is a perfect square and b+c is a perfect cube. Find three consecutive positive integers a,b and c such that a+b is a perfect square and b+c is a perfect cube.
Any ideas on how to begin?
| For all $x,y \in \mathbb{N}$, $x^3-y^2=2$ according to the given question, where $x=a+b$ and $y=b+c$. However, $x^3 \equiv 0,\pm1 \pmod4$ and $y^2 \equiv 0,1 \pmod4$.
So to satisfy the given equation, the following must hold: $$x^3-y^2\equiv 2 \pmod4$$
For this means that the required $x,y$ are such that $x^3 \equiv -1\equiv 3 \pmod4$ and $y^2 \equiv 1 \pmod4$. That is, the required forms are $$x^3=4n+3$$ $$y^2=4n+1$$
See if this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Find $x+y+z$ From Equations Including Radicals Let $(x,y,z)$ be the real solution to the system of equations
\begin{align*}
x+y &= \sqrt{4z-1} ,\\
y+z &= \sqrt{4x- 1} , \\
z+x &= \sqrt{4y-1} .
\end{align*}
Find $x+y+z.$
I could add all the equations up but that doesn't do any good. Thanks in advance!
| $$x+y=\sqrt{4z-1}\tag{1}$$
$$y+z=\sqrt{4x-1}\tag{2}$$
$$z+x=\sqrt{4y-1}\tag{3}$$
Adding $(1),(2)$ and $(3)$, we get
$$x+y+z=\sqrt{x-\frac14}+\sqrt{y-\frac14}+\sqrt{z-\frac14}$$
Let $a=\sqrt{x-\frac14}$,$b=\sqrt{y-\frac14}$ and $c=\sqrt{z-\frac14}$
Using these substitutions, we get
$$\left(a^2+\frac14\right)+\left(b^2+\frac14\right)+\left(c^2+\frac14\right)=a+b+c$$
$$ \Longleftrightarrow \left(a-\frac12\right)^2+\left(b-\frac12\right)^2+\left(c-\frac12\right)^2=0$$
Thus, the only possible solution is $\color{blue}{\boxed{\color{red}{(x,y,z)=\left(\frac12,\frac12,\frac12\right)}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Limit of the difference between two exponential functions
Find the limit $$\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).$$
I did nothing by now.
| We can proceed as follows
\begin{align}
L &= \lim_{x \to \infty}(x + 3)^{1 + 1/x} - x^{1 + 1/(x + 3)}\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 1)\log(x + 3)}{x}\right) - \exp\left(\frac{(x + 4)\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\left\{\exp\left(\frac{(x + 1)\log(x + 3)}{x} - \frac{(x + 4)\log x}{x + 3}\right) - 1\right\}\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\cdot\frac{e^{t} - 1}{t}\cdot t \text{ (note that }t \to 0\text{ as }x \to \infty)\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\cdot t\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\left(\frac{(x + 1)\log(x + 3)}{x} - \frac{(x + 4)\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}x\cdot x^{1/(x + 3)}\left(\log\frac{x + 3}{x} + \frac{\log(x + 3)}{x} - \frac{\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}x\left(\log\frac{x + 3}{x} + \frac{\log(x + 3)}{x} - \frac{\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\left(x\log\frac{x + 3}{x} + \log(x + 3) - \frac{x\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\left(3\cdot\dfrac{\log(1 + 3/x)}{3/x} + \log\frac{x + 3}{x} + 3\cdot\frac{\log x}{x + 3}\right)\notag\\
&= 3\cdot 1 + \log 1 + 3\cdot 0\notag\\
&= 3\notag
\end{align}
Here we have used the standard limits $$\lim_{h \to 0}\frac{\log(1 + h)}{h} = 1,\,\lim_{x \to \infty}\frac{\log x}{x} = 0$$ and because of these limits the expression $$x^{1/(x + 3)} = \exp\left(\frac{\log x}{x + 3}\right) \to 1$$ as $x \to \infty$ and the variable $$t = \frac{(x + 1)\log(x + 3)}{x} - \frac{(x + 4)\log x}{x + 3} = \log\frac{x + 3}{x} + \frac{\log(x + 3)}{x} - \frac{\log x}{x + 3}$$ tends to $0$ as $x \to \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.
| The function
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
is defined for every real $x$, because $|2x|<1+x^2$. We also have
$$
f'(x)=\frac{1}{\sqrt{1-\left(\dfrac{2x}{1+x^2}\right)^2}}
\cdot 2\frac{(1+x^2)-x\cdot 2x}{(1+x^2)^2}=
2\frac{1+x^2}{|1-x^2|}\frac{1-x^2}{(1+x^2)^2}
$$
so
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & \text{if $|x|<1$} \\[6px]
-\dfrac{2}{1+x^2} & \text{if $|x|>1$}
\end{cases}
$$
Thus we can write
$$
f(x)=\begin{cases}
a+2\arctan x & \text{if $|x|<1$} \\[6px]
b-2\arctan x & \text{if $|x|>1$}
\end{cases}
$$
Since $f(0)=0$, we have $a=0$. Since $\lim_{x\to\infty}f(x)=0$ we have $b=\pi$. Finally
$$
f(x)=\begin{cases}
2\arctan x & \text{if $|x|\le 1$} \\[6px]
\pi-2\arctan x & \text{if $|x|>1$}
\end{cases}
$$
The function is not differentiable at $-1$ and $1$.
You can now write your complicated function as
$$
g(x)=\arcsin\frac{2^{x+1}\cdot 3^x}{1+36^x}=f(6^x)
$$
by noticing that $2^{x+1}\cdot 3^x=2\cdot 6^x$ and $36^x=(6^x)^2$.
Thus, by the chain rule,
$$
g'(x)=f'(6^x)\cdot 6^x\ln 6
$$
provided $6^x\ne1$, that is, $x\ne0$. Explicitly,
$$
g'(x)=\begin{cases}
\dfrac{2\cdot 6^x\cdot \ln 6}{1+36^x} & \text{if $x>0$} \\[6px]
-\dfrac{2\cdot 6^x\cdot \ln 6}{1+36^x} & \text{if $x<0$} \\[6px]
\end{cases}
$$
The function is not differentiable at $0$, as it can be seen in the diagram (and proved directly).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$
If $ a, b, c \in R^{+}$ and $\frac{1}{a + 1}$ + $\frac{1}{b + 1}$ + $\frac{1}{c + 1}$ = 2, then $\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$.
I don't know how to start.
| The condition gives $\sum\limits_{cyc}\left(\frac{1}{1+a}-1\right)=-1$ or $\sum\limits_{cyc}\frac{a}{1+a}=1$.
Thus, by C-S $1=\sum\limits_{cyc}\frac{a}{1+a}\geq\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3+a+b+c}$, which gives what you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
homework - probability problem A dice is thrown 3 times and the sum of the 3 numbers thrown is 15. Then, what is the probability that the first throw was a four?
My Attempt:
The sample space for the above problem, $S = 6^3$.
Combinations of thrown numbers such that their sum is $15$ and the first digit is $4 = [\{4, 5, 6\}, \{4, 6, 5\}]$
Thus, the total number of possible sets are 2. Thus, the probability for the required event (say $A$),
$$P(A) = \frac{2}{216}$$
$$\therefore P(A) = \frac{1}{108}$$
In my book, the answer given is $\frac{1}{5}$. How's is my answer?
| You have to consider all the possible ways in which you can obtain $15$.
Those possibilities are:
$\{3, 6, 6 \}$, $\{6, 3, 6 \}$, $\{6, 6, 3 \}$
$\{5, 5, 5 \}$
$\{4, 5, 6 \}$, $\{4, 6, 5 \}$, $\{5, 4, 6 \}$, $\{5, 6, 4 \}$, $\{6, 4, 5 \}$, $\{, 5, 4 \}$
Total = $10$
There are only $2$ cases in which $4$ is the first element, so
$$P = \frac{2}{10} = \frac{1}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit with little-o It's a while I'm trying to verify this limit:
$$\lim_{x\rightarrow 0^+} \frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0,$$
this is my attempt:
Since $x \rightarrow 0$ I can do the substitution:
$$\sqrt{1+x^2}=1+o(1)$$
$$\sin{x}=x(1+o(1))$$
and I obtain:
$$\frac{\log(x+1+o(1))-x(1+o(1))}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0$$
Now I do these substitutions for $y \rightarrow 0$
$$\log (y+1)= y(1+o(1))$$
$$a^y-1 = y(\log a + o(1))$$
$$\sin y = y(1 + o(1))$$
and I obtain:
$$\frac{(x+o(1))(1+o(1))-x(1+o(1))}{x^2(\log 2+o(1))x^2(1+o(1))}=\frac{x+o(1)-x+o(x)}{(x^2 \log 2 + o(x^2))(x^2+o(x^2))}=\frac{o(1)}{x^4 \log 2 + o(x^4)}.$$
But the limit with $x \rightarrow 0$ of this makes infinite, not $0$ as it should.
Where is my error?
Thank you.
| You have
\begin{align}
\sqrt{1+x^2}&=1+\frac{x^2}2-\frac{x^4}8+o(x^6)\\ \ \\
\log(1+x)&=x-\frac{x^2}2+\frac{x^3}3+o(x^4)\\ \ \\
\sin x&=x-x^3/6+o(x^5)\\ \ \\
2^x&=1+x\log2+o(x^2)\\ \ \\
e^x&=1+x+o(x^2)
\end{align}
Then
\begin{align}
\frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))}
&=\frac{\log(1+x+x^2/2-x^4/8+o(x^6))-(x-x^3/6+o(x^5))}{(x^2\log2+o(x^4))(\sin(x^2+o(x^4)))}\\ \ \\
&=\frac{\log(x+\sqrt{1+x^2})-\sin x}{(x^2\log2+o(x^4))(x^2+o(x^4))}\\ \ \\
&=\frac{\log(x+\sqrt{1+x^2})-\sin x}{o(x^4)}.
\end{align}
So the denominator is clearly $o(x^4)$. Let us work on the numerator: using Wolphram Alpha,
\begin{align}
\log(x+\sqrt{1+x^2})&=\log(1+x+\frac{x^2}2-\frac{x^4}8+o(x^6))\\ \ \\
&=x+\frac{x^2}2-\frac{x^4}8-\frac{(x+\frac{x^2}2-\frac{x^4}8)^2}2
+\frac{(x+\frac{x^2}2-\frac{x^4}8)^3}3+o(x^6)\\ \ \\
&=x-\frac{x^3}6+o(x^5).
\end{align}
Then
$$
\log(x+\sqrt{1+x^2})-\sin x=x-\frac{x^3}6+o(x^5)-\left(x-\frac{x^3}6+o(x^5)\right)=o(x^5).
$$
Thus
$$
\frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))}=\frac{o(x^5)}{o(x^4)}=o(x).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $A+B+C+D=2\pi$, prove that: If $A+B+C+D=2\pi$, prove that: $$\cos A+\cos B+\cos C+\cos D=4\cos\frac {A+B}{2}\cdot\cos\frac {A+C}{2}\cdot\cos\frac {B+C}{2}$$.
My Approach:
Here,
$$A+B+C+D=2\pi$$
$$A+B=2\pi - (C+D)$$
$$ \sin(A+B)=\sin(2\pi-(C+D))$$
$$\sin(A+B)=-\sin(C+D)$$
Again,
$$\cos(A+B)=\cos(2\pi-(C+D))$$
$$\cos(A+B)=\cos(C+D)$$
Now,
$$L.H.S=\cos A+\cos B+\cos C+\cos D$$
$$=2 \cos\frac {A+B}{2}\cdot\cos\frac {A-B}{2} + 2 \cos\frac {C+D}{2}\cdot \cos\frac {C-D}{2}$$.
I got stuck at here. Please help me to complete the proof.
| As $D=2\pi-(A+B+C),\cos D=\cos(A+B+C)$
$$\implies\cos A+\cos B+\cos C+\cos(A+B+C)$$
$$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+2\cos\dfrac{A+B+C-C}2\cos\dfrac{A+B+C+C}2$$ (using Prosthaphaeresis Formulas)
$$=2\cos\dfrac{A+B}2\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B+2cc}2\right)$$
Apply Prosthaphaeresis Formula on $$\cos\dfrac{A-B}2+\cos\dfrac{A+B+2C}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the largest positive integer $n$ such that $2^n$ divides $3^{4096} −1$. I dont know how to start with this questions.It is an olympiad sum.
| HINT:
Observe that the highest power of $2$ that divides $3^2-1$ is $3$
Now $3^{2n}=(1+8)^n\equiv1\pmod8\implies3^{2n}+1\equiv2\pmod8\equiv2\pmod4$
So, the highest power of $2$ that divides $3^{2n}+1$ is $1$ for integer $n\ge0$
So, the highest power of $2$ that divides $3^{(2^2)}-1=(3^2-1)(3^2+1)$ will be $3+1$
So, the highest power of $2$ that divides $3^{(2^3)}-1=(3^2-1)(3^2+1)(3^4+1)$ will be $3+1+1$
Observe that $4096=2^{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.