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How to integrate $\int_0^{\pi/2}\frac{1}{\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)}\,\mathrm{d}x$ I got stuck at this definite integral, any idea?
$$\int_0^{\pi/2}\frac{1}{\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)}\,\mathrm{d}x$$
Thank you.
| Under $2x\to x$ and $t=\tan(\frac x2)$, one has
\begin{eqnarray}
&&\int_0^{\pi/2}\frac{1}{\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)}\,\mathrm{d}x\\
&=&\int_0^{\pi/2}\frac{1}{1-\frac12\sin(2x)}\,\mathrm{d}x=\int_0^{\pi}\frac{1}{2-\sin(x)}\,\mathrm{d}x=\int_0^{\pi}\frac{1}{2-\frac{2\tan(\frac x2)}{1+\tan^2(\frac x2)}}\,\mathrm{d}x\\
&=&\frac12\int_0^\pi\frac{\tan^2(\frac x2)}{\tan^2(\frac x2)-\tan(\frac x2)+1}dx=\frac12\int_0^\infty\frac{t^2}{t^2-t+1}\frac{2}{t^2+1}dt\\
&=&\int_0^\infty\left(\frac{t}{t^2-t+1}-\frac{t}{t^2+1}\right)dt\\
&=&\frac12\ln\left(\frac{t^2-t+1}{t^2+1}\right)+\frac{\sqrt3}{3}\arctan\left(\frac{2t-1}{\sqrt3}\right)\bigg|_0^\infty\\
&=&\frac{2\sqrt3\pi}{9}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the range of $f(x)= \frac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\frac 56, \frac 32\right]$ then what is $c$?
If the range of the function $f(x)= \dfrac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\dfrac 56, \dfrac 32\right]$ then $c$ is equal to?
Attempt:
$y= \dfrac{x^2+x+c}{x^2+2x+c}$
For real values of $x$, $\Delta \ge 0$
$\implies 4y^2(1-4c) +1-4y(1-2c) - 4c \ge 0$
What do I do next? I am really unable to understand the concept to be followed after this. Could someone explain that?
The answer is:
$c= 4$
| Write $f(x)=1-\frac x{x^2+2x+c}$ As $x \to \pm \infty$ this goes to $1$. If the denominator has a real root it will go off to $\pm \infty$ so $c \gt 1$. The denominator is then always positive, so the cases where $f(x) \gt 1$ are where $x \lt 0$ and the cases where $f(x) \lt 1$ are where $x \gt 0$. Take the derivative, set to zero, and find the $x$ value of the local maximum as a function of $c$. Plug that $x$ value into $f(x)$ and set it equal to $\frac 32$. You will get an equation for $c$. You can then check that the minimum has $f(x)=\frac 56$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ using the Taylor series of $e^{x+2 }$.
Answer:
$$
e^{x+2}=1+(x+2)+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\ldots
$$
Integrating, we get
$$
\int e^{x+2} dx=x+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\ldots
$$
Again integrating, we get
$$
\int e^{x+2} dx=\frac{x^2}{2}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\frac{(x+2)^5}{5!}+\ldots\\
\Rightarrow \int e^{x+2} dx=\frac{x^2}{2}+(x+2)^3 \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}\\
\Rightarrow \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}=\frac{\int e^{x+2} dx-\frac{x^2}{2}}{(x+2)^3}
$$
Is this the sum of the series ?
| There is no need for integrals.
Multiply the sum by $(x+2)^3$ to let the exponents match the denominators and add the missing initial terms to obtain an exponential:
$$1+(x+2)+\frac{(x+2)^2}2+(x+2)^3S=\sum_{n=-3}^\infty\frac{(x+2)^{n+3}}{(n+3)!}=e^{x+2}$$
from which you draw $S$ (which is not what you found).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{k=0}^{n}{n\choose 3k}\le\frac{1}{3}(2^n+2) \text{such that } n>0$ and $k$ is the largest number for which $3k\le n$ I tried it using induction
For $n=1,\displaystyle{1\choose 0}\le\dfrac{1}{3}(2^1+2)$
Let for $m\le n $ it is true i.e $\displaystyle \sum_{k=0}^{n}{m\choose 3k}\le\dfrac{1}{3}(2^m+2)$
Now we need to prove for $m+1 \le n,\displaystyle \sum_{k=0}^{n}{m+1\choose 3k}\le\dfrac{1}{3}(2^{m+1}+2)$
Now $\displaystyle \sum_{k=0}^{n}{m+1\choose 3k}=\sum_{k=0}^{n}{m\choose 3k}+\sum_{k=1}^n {m\choose {3k-1}}$
Now the in the RHS, $\displaystyle \sum_{k=0}^{n}{m\choose 3k}+\sum_{k=1}^n {m\choose {3k-1}}\le \dfrac{1}{3}(2^m+2)+\underbrace{\sum_{k=1}^n {m\choose {3k-1}}}_{\text{Lets call this X}}=\dfrac{1}{3}(2^m+2)+X$
But how do I show $X \le \dfrac{1}{3}(2^m)$ in order to prove $\displaystyle \sum_{k=0}^{n}{m\choose 3k}+\sum_{k=1}^n {m\choose {3k-1}} \le \dfrac{1}{3}(2^{m+1}+2)$
| It's not hard to notice that the left-hand side is equal to $\frac 13 \left((1+\zeta_3)^n + (1+\zeta_3^2)^n + (1+\zeta_3^3)^n\right)$, where $\zeta_3$ is the third root of unity. Now the inequality is equivalent to:
$$(1+\zeta_3)^n + (1+\zeta_3^2)^n \le 2$$
However now we have that $1+\zeta_3 = \zeta_6$ and $1+\zeta_3^2 = \zeta_6^5$ and we have that as the left-hand side is real:
$$ (1+\zeta_3)^n + (1+\zeta_3^2)^n = \left| (1+\zeta_3)^n + (1+\zeta_3^2)^n \right| = \left|\zeta_6^{5n} + \zeta_6^{n}\right| \le \left|\zeta_6^{5n}\right| + \left|\zeta_6^{n}\right| = \left|\zeta_6\right|^{5n} + \left|\zeta_6\right|^{n}=2$$
Hence the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Certain types of tangent values I was messing around with values of the tangent function and came across something interesting. For example, we have
$$\tan^2(\frac{\pi}{4\cdot2}) = \dfrac{\sqrt{4} - \sqrt{2}}{\sqrt{4} + \sqrt{2}}, \tan^2(\frac{\pi}{3\cdot4}) = \dfrac{\sqrt{4} - \sqrt{3}}{\sqrt{4} + \sqrt{3}}, \tan^2(\frac{\pi}{1\cdot1}) = \frac{\sqrt{1} - \sqrt{1}}{\sqrt{1} + \sqrt{1}}.$$
All of these values satisfy
$$\tan^2(\frac{\pi}{m\cdot n}) = \dfrac{\sqrt{m} - \sqrt{n}}{\sqrt{m} + \sqrt{n}}.$$
How could I go about to find all pairs $(m,n)$ which satisfy the equation?
|
Incomplete answer: I can't proceed through the last step but I think for those with higher level math knowledge, it might be proceedable.
Consider$$\tan^2\left(\frac{\pi}{m\cdot n}\right) =\frac{\sin^2\left(\frac{\pi}{m\cdot n}\right)}{\cos^2\left(\frac{\pi}{m\cdot n}\right)}=\frac{1-\cos^2\left(\frac{\pi}{m\cdot n}\right)}{\cos^2\left(\frac{\pi}{m\cdot n}\right)}$$
Let$$x={\cos^2\left(\frac{\pi}{m\cdot n}\right)}$$
Then
$$ \frac{1-x}{x}=\dfrac{\sqrt{m} - \sqrt{n}}{\sqrt{m} + \sqrt{n}}$$
Simplifying
$$2x=\frac{\sqrt{m} + \sqrt{n}}{\sqrt m}$$
Note that
$$\cos\left(\frac{2\pi}{m\cdot n}\right)+1=2x$$
Hence,
$$\cos\left(\frac{2\pi}{m\cdot n}\right)=\frac{\sqrt{m} + \sqrt{n}}{\sqrt m}-1=\sqrt{\frac nm}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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find all the value of x,y z? Let $A$ be the complex $3 × 3$ matrix
$A = \begin{bmatrix}2&0&0 \\ x&2 & 0 \\y&z&-1\end{bmatrix}$
Find all triples $(x, y, z)$ for which the characteristic and minimal polynomials of $A $ are different.
My attempts : the Characteristic polynomial of $A = (\lambda +1)(\lambda -2)^2$
Case $1$: if $x=y= z \neq 0$ ,then minimial polynomial of $A = (\lambda +1)(\lambda -2)^2$
Case $2$ : if $x=y= z =0$,then minimial polynomial of $A = (\lambda +1)(\lambda -2)$
Here im confused that how can i find all triples $(x, y, z)$ for which the characteristic and minimal polynomials of $A $ are different.
| The minimal polynomial is either $(\lambda+1)(\lambda-2)$ or $(\lambda+1)(\lambda-2)^2.$
We can try to compute $(A+I)(A-2I)$ and see whether we obtain $0$.
$$A+I=\begin{bmatrix}3&0&0 \\ x&3 & 0 \\y&z&0\end{bmatrix}, A-2I=\begin{bmatrix}0&0&0 \\ x&0 & 0 \\y&z&-3\end{bmatrix} $$
$$(A+I)(A-2I)=\begin{bmatrix} 0 & 0 & 0 \\ 3x & 0 & 0\\ xz & 0 & 0\end{bmatrix} $$
Can read off the required condition from the matrix above?
Edit:
We obtain $3x=0$ and $xz=0$.
which means $x=0$ and ($x=0$ or $z=0$)
which is equivalent to $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Why is $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$ defined at -1, 0 and not at 1 Why does the function:
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$
Is defined at $x = -1, x = 0$ (we will have $1/0$ in the fraction then) and not at $x = 1$ according to the following graph?
Live example
| $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$
$f(x)=\frac{(x-1)(x+1)(x)}{(x-1)(x+1)(x)} \cdot \frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}} $
$f(x)=\frac{(x+1)(x)+(x-1)(x)}{2(x-1)(x)+(x-1)(x+1)}$
$f(x)=\frac{x^2+x+x^2-x}{2x^2-2x+x^2-1}$
$f(x)=\frac{2x^2}{3x^2-2x-1}$
$f(x)=\frac{2x^2}{(3x+1)(x-1)}$
So the vertical asymptotes at $x=-\frac{1}{3}$ and $x=1$
while there only holes at $x=0$ and $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Express the sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ in terms of ${x_0, x_1, n}$ Given a sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ how can one express it in terms of ${x_0, x_1, n}$. Can this be generalized for ${x_n = \alpha x_{n-1} + \beta x_{n-2}}$
I've tried to use the following approach:
$$\eqalign{
& 1x_0 = x_01 \\
& zx_1 = x_1z \\
& z^2x_2 = (2x_1 + 3x_0)z^2 \\
& z^3x_3 = (2x_2 + 3x_1)z^3 \\
& z^4x_4 = (2x_3 + 3x_2)z^4 \\
& ... \\
& z^nx_n = (2x_{n-1} + 3x_{n-2})z^n
}$$
Then sum LHS with RHS which will produce:
$${
x_0 + \sum\limits_{k = 1 }^na_nz^n = x_0 +zx_1 + 2\sum\limits_{k = 2}^nx_{n-1}z^n + 3 \sum\limits_{k = 2}^n x_{n-2}z^n
}$$
Let
$${
G(z) = x_0 + \sum\limits_{k = 1 }^na_nz^n
}$$
Then RHS may be expressed in terms of ${G(z)}$. For example
$${
2\sum\limits_{k = 2}^na_{n-1}z^n = 2z\sum\limits_{k = 2}^na_{n-1}z^{n-1} = 2z(\sum\limits_{k = 1}^na_{n}z^{n} + x_0 - x_0) = 2z(G(z) - x_0)
}$$
Applying those transformations I eventually got ${G(z)}$ expressed in terms of z and ${x_1, x_0}$. But at this point I got stuck.
I got a sum of fractions:
$${
\frac{3x_0 - x_1}{4(1+z)}+\frac{x_0+x_1}{4(1-3z)}
}$$
I guess i could expand the fractions into series and find their sum, but i am not supposed to know about such expansions at the point of the book i took the problem from.
All of the above feels like a wrong approach. So the question is whether this can be done in a more elegant way.
| Let $f(x)$ be the generating function for some sequence $(a_{n})_{n\in\mathbb{N}}$ such that
\begin{align*}
f(x)=\frac{c}{bx+d}=\sum^{\infty}_{n=0}{a_{n}x^{n}},\ c,b,d\in\mathbb{Z}.\tag{1}
\end{align*}
Then
\begin{alignat*}{3}
f(x)=\frac{c}{bx+d}&\implies f^{\prime}(x) &&=-1!\left(\frac{b}{c}\right)^{1}f(x)^{2}\\
&\implies f^{\prime\prime}(x) &&=+2!\left(\frac{b}{c}\right)^{2}f(x)^{3}\\
&\implies f^{\prime\prime\prime}(x) &&=-3!\left(\frac{b}{c}\right)^{3}f(x)^{4}\\
&\qquad &&\vdots\\
&\implies f^{n}(x) &&=(-1)^{n}n!\left(\frac{b}{c}\right)^{n}\big(f(x)\big)^{n+1},
\end{alignat*}
where $f^{n}(x)$ denotes the $n^{\text{th}}$ derivative of $f$ with respect to $x.$ If we evaluate $f^{n}(0)$, the Maclaurin series expansion of $f(x)$ is then given by
\begin{alignat*}{4}
&\qquad \qquad \qquad \qquad \quad f^{n}(0)&&=(-1)^{n}n!\left(\frac{b}{c}\right)^{n}f(0)^{n+1}\\
& &&=(-1)^{n}n!\left(\frac{b}{c}\right)^{n}\left(\frac{c}{d}\right)^{n+1}\\
&\qquad \qquad \qquad \quad \implies f(x) &&=\sum_{n=0}^{\infty }c\ (-1)^{n}\left(\frac{1}{d}\right)^{n+1}{b^{n}x^{n}}.\tag{2}\\
\end{alignat*}
It follows from (1) and (2) that the explicit form of the sequence $(a_{n})_{n\in\mathbb{N}}$ is given by
\begin{align*}a_{n}=c\ (-1)^{n}\left(\frac{1}{d}\right)^{n+1} b^{n}.\tag{3}\\ \end{align*}
If we let $c=x_{0}+x_{1},\ b=-12,$ and $d=4$, as in (1) then
\begin{alignat*}{3}
g(x)=\frac{x_{0}+x_{1}}{4-12z}&\overset{(3)}{\implies} g_{n}&&=(-1)^{n}(x_{0}+x_{1})(-12)^{n}\left(\frac{1}{4}\right)^{n+1}\\
&\qquad &&=(x_{0}+x_{1})(-1)^{n}(-1)^{n}(3)^{n}(4)^{n}\left(\frac{1}{4}\right)^{n}\left(\frac{1}{4}\right)\\
&\ \implies g_{n} &&=\left(\frac{1}{4}\right)3^{n}(x_{0}+x_{1}).\tag{4}
\end{alignat*}
Similarly
\begin{alignat*}{3}
h(x)=\frac{3x_{0}-x_{1}}{4+4z}&\overset{3}{\implies} h_{n}&&=(-1)^{n}(3x_{0}-x_{1})(4)^{n}\left(\frac{1}{4}\right)^{n+1}\\
&\qquad &&=(3x_{0}-x_{1})(-1)^{n}(4)^{n}\left(\frac{1}{4}\right)^{n}\left(\frac{1}{4}\right)\\
&\ \implies h_{n} &&=(-1)^{n}\left(\frac{1}{4}\right)(3x_{0}-x_{1}).\tag{5}\\
\end{alignat*}
It follows from the linearity of differentiation, (superposition principle), that we may combine (4) and (5) to obtain the desired formula for $x_{n}$. Whence,
\begin{align}x_{n}=\left(\frac{1}{4}\right)3^{n}(x_{0}+x_{1})+ (-1)^{n}\left(\frac{1}{4}\right)(3x_{0}-x_{1})\tag*{$\square$}\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\frac{0.5}{ \cos^2(30)} = \frac{\tan(30)}{\cos(30)} $, but $\frac{0.5}{ \cos^2(13) } \neq \frac{\tan(13)}{\cos(13)} $? Is just it a coincidence that
$$\frac{0.5}{ \cos^2(30)} = \frac{\tan(30)}{\cos(30)} $$
However
$$\frac{0.5}{ \cos^2(13) } \neq \frac{\tan(13)}{\cos(13)} $$
is not equal ? And if not does anyone know a reason why they just so happen to be equatable?
| Rewrite the right side:
$$
\begin{align}
\frac{0.5}{\cos^2(30^\circ)}&=\frac{\tan(30^\circ)}{\cos(30^\circ)}\\
&=\frac{\sin(30^\circ)}{\cos^2(30^\circ)}\\
&=\frac{0.5}{\cos^2(30^\circ)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving inverse function problems The following is my (likely completely incorrect) attempt.
$$f(x)= 3x + 5, \qquad \qquad g(x)= 1 - 2x$$
*
*Show that $f{^{-1}}(2) < g(-3)$
$$\frac{1}{3\cdot 2+5} < (1-2\cdot(-3))\implies\frac{1}{11} < 7$$
*Find the value of $a$ such that $f(2a) = g{^{-1}}(a)-2$
$$6a+5 = \frac{1}{-1-2a}$$
$$6a = \frac{1}{4-2a}$$
$$8a = \frac{1}{4}\implies a = \frac{1}{32}$$
*Solve $\frac{f(x)}{2} + \frac{g(x)}{3} = -23$
$$\frac{3x+5}{2} + \frac{1-2x}{3} = -23$$
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23$$
$$2x+17 = \frac{-23}{6}\implies x = -\frac{125}{12}$$
| I think $$f^{-1}(x)=\frac{1}{3}(x-5)$$ so $$f^{-1}(2)=\frac{2}{3}-\frac{5}{3}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Example 14, Chapter 1, Higher Algebra by Henry Sinclair. I stumbled across the question 14 in Higher Algebra by Henry Sinclair and S.R Knight:
If $$(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2$$
Show $$x:a=y:b=c:z$$
I started by expanding the trinomials on both sides:
$$a^2x^2+b^2y^2+c^2z^2 + (a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = a^2x^2 + b^2y^2 + c^2z^2 + (axby + axcz) + (byax + bycz) + (czax +czby)$$
That simplifies to $$(a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = (axby + axcz) + (byax + bycz) + (czax +czby)$$
Adding like terms, I got
$$(a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = 2(axby + 2axcz + 2bycz)$$
Rearranging, I finally obtained:
$$(ay-bx)^2 + (az-cx)^2 + (bz-cy)^2 = 0$$
Now, I realized that if $\frac{x}{a} = \frac{y}{b}$ then $ay = bx$. It then follows that $ay - bx = 0$. This also applies for the other fractions.
However, in the context of this problem, we do not know that $x:a = y:b = z:c$. So, I'm not sure how I can proceed to prove it with just the equation above.
Any help would be appreciated.
| If $p^2+q^2+r^2=0$ where $p,q,r\in \mathbb{R}$, then we know that $p=q=r=0$.
Hence we do know that $ay-bx=az-cx=bz-cy=0$.
Remark: This question is basically asking when does equality holds for Cauchy-Schwarz inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I show that a complex number $z$ is a root of $z^6-1$? This problem would be less confusing to me if one was not subtracted from $z^6$.
Here is the problem:
If $z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$, show that $z$ is a root of $z^6-1$.
My approach to this problem involved using de Moivre's theorem to convert $z$ to it's complex polar form. I think converting it to complex polar form first makes it easy to identify the root:
$$z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$$
$$a=cos(\frac{\pi}{3}), b=sin(\frac{\pi}{3})$$
$$r=\sqrt{cos(\frac{\pi}{3})^2+sin(\frac{\pi}{3})^2}$$
$$r=1$$
$$\theta=Tan^1(\frac{b}{a})$$
$$\theta=Tan^-1(\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})})$$
$$\theta=\frac{\pi}{6}$$
$$\therefore z=1\angle{\frac{\pi}{6}}$$
This part is the confusing bit. Because $1$ is subtracted from $z^6$, I decided to expand the whole thing and then show that $z$ is a root of $z^6-1$.
$$(cos(\frac{\pi}{3})+jsin(\frac{\pi}{3}))^6-1$$
$$(\frac{1}{2}+\frac{\sqrt{3}}{2}j)^6-1$$
Noting that $j^6=-1$...
$$(\frac{1}{2}-\frac{\sqrt{3}}{2})^6-1$$
The equation becomes...
$$(-\frac{13}{32})-1$$
Which equates to...
$$-\frac{45}{32}$$
Convert to complex polar form...
$$z^6=-\frac{45}{32}\angle{0rad}$$
$$z^6=(-\frac{45}{32}\angle{0rad})^6$$
$$z^6=(-\frac{45}{32})^6\angle{(0)}^6$$
$$z^6=(-\frac{45}{32})^6\angle{6(0+2\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(12\pi-11\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(\pi n)}$$
Noting that $z$ is raised to the first power, make $n=1$
$$z^6=(-\frac{45}{32})^6\angle{\pi (1)}$$
I get...
$$z^6=(-\frac{45}{32})^6\angle{\pi}$$
Clearly the complex polar form I got is not equal to that of $z$. Can someone lead me to the right path?
| HINT
Note that according to JuliusL33t and rtybase suggestions by De Moivre's formula
*
*$z=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})=e^{i\frac{\pi}{3}}\implies z^6=\cos(\frac{6\pi}{3})+i\sin(\frac{6\pi}{3})$
or also by exponential form
*
*$z=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})=e^{i\frac{\pi}{3}}\implies z^6=e^{6\cdot i\frac{\pi}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that: $\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)dx=\pi?$ I just wonder if this integral is correct.
How can we show that?
$$\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)\mathrm dx=\color{blue}\pi$$
Let $u=\tan x$
$$\int_{0}^{\infty}\left(\frac{1}{\ln u}+\frac{2-u^{4/3}}{1-u}\right)\frac{\mathrm du}{1+u^2}$$
Let $u=e^{y}$
$$\int_{1}^{\infty}\left(\frac{1}{y}+\frac{2-e^{4y/3}}{1-e^{y}}\right)\frac{e^y\mathrm du}{1+e^{2y}}$$
$$\frac{1}{2}\int_{-\infty}^{\infty}\left(\frac{1}{y}+\frac{2-e^{4y/3}}{1-e^{y}}\right)\frac{\mathrm dy}{\cosh y}$$
I can't continue...
| Let $I$ be given by the integral
$$I=\int_0^{\pi/2}\left(\frac{1}{\log(\tan(x))}+\frac{2-\sqrt[3]{\tan^4(x)}}{1-\tan(x)}\right)\,dx\tag1$$
Enforce the substitution $x\mapsto \arctan(x)$ in $(1)$ to reveal
$$\begin{align}
I&=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{2-x^{4/3}}{x-1}\right)\,\frac{1}{1+x^2}\,dx\\\\
&=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{1}{x-1}\right)\,\frac{1}{1+x^2}\,dx+\int_0^\infty \left(\frac{1-x^{4/3}}{1-x}\right)\,\frac{1}{1+x^2}\,dx\tag2
\end{align}$$
The first integral on the right-hand side of $(2)$ is easy to evaluate by symmetry. Let $J$ be given by
$$J=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{1}{x-1}\right)\,\frac{1}{1+x^2}\,dx\tag3$$
and enforce the substitution $x\mapsto 1/x$. Then, we find that
$$J=\int_0^\infty \left(-\frac{1}{\log(x)}-\frac{x}{1-x}\right)\,\frac{1}{1+x^2}\,dx\tag4$$
Adding $(3)$ and $(4)$, we find that
$$2J=\int_0^\infty \frac{1}{1+x^2}\,dx=\frac\pi2$$
Hence, $J=\frac\pi4$.
We now proceed to evaluate the second integral on the right-hand side of $(2)$. To that end, let $K$ be the integral given by
$$\begin{align}
K&=\int_0^\infty \left(\frac{1-x^{4/3}}{1-x}\right)\,\frac{1}{1+x^2}\,dx\\\\
&\overbrace{=}^{x\mapsto x^3}3\int_0^\infty \frac{1-x^4}{1-x^3}\frac{x^2}{1+x^6}\,dx\\\\
&=3\int_0^\infty \frac{x^2(1+x+x^2+x^3)}{(1+x+x^2)(1+x^6)}\,dx\\\\
&\overbrace{=}^{x\mapsto 1/x}3\int_0^\infty \frac{x(1+x+x^2+x^3)}{(1+x+x^2)(1+x^6)}\,dx\\\\
&=3\int_0^\infty \frac{x(x+1)}{(x^2+x+1)(x^4-x^2+1)}\tag5\\\\
&=\frac{3\pi}{4}
\end{align}$$
where $(5)$ can be evaluated using partial fraction expansion.
Putting it together yields the coveted equality
$$\int_0^{\pi/2}\left(\frac{1}{\log(\tan(x))}+\frac{2-\sqrt[3]{\tan^4(x)}}{1-\tan(x)}\right)\,dx=\pi$$
| {
"language": "en",
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For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes. Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$
By induction: $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$. Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note: This argument was taken from Wikipedia.
(2) $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$
If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$
(3) For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$
From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$
(4) Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:
${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.
| A combinatorial argument
Divide $3n$ distinctive elements into $n$ groups, each of which contains 3 elements. To pick $n$ elments out of them, there are many ways, including the following categories:
1) pick one element out of each group, which gives $3^{n}$ different ways
2) pick one element out of each group for $n-k$ group, then pick $k$ elments out of remaining $k$ groups, so that every group of the remaining $k$ groups contributes either $0$ elements or at least $2$ elements. This gives ${ n \choose {n-k}}3^{n-k} A_k$ ways, where $A_k$ is the way of picking $k$ elments out of remaining $k$ groups, so that every group of the remaining $k$ groups contributes either $0$ elements or at least $2$ elements. Obviously $A_k \ge 1$, except for $k=1$we have $A_1=0$
Remark that the above ways are exlcusive for different $k$.
With ${ n \choose {n-2}}3^{n-2} A_2 = { n \choose {n-2}}3^{n-2} 6 = n(n-1)3^{n-1}\ge 3^{n-1} n + \frac{n(n-1)}{2}3^{n-2}$ (the last inequality is equivalent to $5n-11\ge 0$)
we have
${ {3n} \choose {n}} \ge 3^n + \sum_{k=2}^n { n \choose {n-k}}3^{n-k} A_k\ge \sum_{k=0}^n { n \choose {n-k}}3^{n-k} =(3+1)^n =4^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to solve the limit:$ \lim_{x \to 0,y \to 0} \frac{x^2y^2}{x^3+y^3} $ I confronted a problem, that is
$$
\lim_{x \to 0,y \to 0} \frac{x^2y^2}{x^3+y^3}
$$
I tried to use $x=r\cos(\theta) , y=r\sin(\theta)$ to solve the problem, and I got
$$\lim_{r \to 0}r(\frac{\cos^2(\theta)\sin^2(\theta)}{\cos^3(\theta)+\sin^3(\theta)})$$
And no matter what value $\theta$ takes, we all have the limit is zero.
My question is:
(1) Am I right? Or
(2) If we can change the limit into such "$rf(\theta)$" form, when r goes to zero, will the limit go to zero?
| As an alternative note that
*
*$x=0 \,,y\to 0\implies \frac{x^2y^2}{x^3+y^3}=0$
*$x=t \,, y=t^2-t\,,t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^2(t^2-t)^2}{t^3+(t^2-t)^3}=\frac{t^6-2t^5+t^4}{t^3+t^6-3t^5+3t^4-t^3}=\frac{t^2-2t+1}{t^2-3t+3}\to \frac13$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How do I go about simplifying $1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$ to $\sin x\cos x$
This is a tricky trig problem I'm stuck with. The problem is asking me to simplify $$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$$
to $\sin x\cos x$.
What I've been doing so far is trying to remove those $tan$ functions.
$$1-\left(\frac{\sin^2x\tan x-\cos^2}{\tan x+1}\right)$$
$$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\tan x+1}\right)$$
$$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\frac{\sin x}{\cos x}+1}\right)$$
I made this complicated. Is there a simple way to do this problem?
| We have
$$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1} =1-\frac{\sin^2x\tan x+\cos^2 x}{\tan x+1}=1-\frac{\sin^3x+\cos^3 x}{\sin x + \cos x}=\\=\frac{\sin x(1-\sin^2x)+\cos x(1-\cos^2 x)}{\sin x + \cos x}=\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\sin x \cos x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding radius of capsule defined by total volume and height of inner cylinder I have a capsule shape that is comprised of a cylinder and two half-sphere end caps, and I want to dynamically resize it. As I stretch the capsule by increasing/decreasing the height of the cylinder, the total volume should remain constant by adjusting the radius. This means, I need to define the radius in terms of height and volume.
My starting equation looks like this:
Total volume = Sphere volume + Cylinder volume
$$W_v = \frac{4}{3}\pi r^3 + h \pi r^2$$
Given this equation, I don't know how to solve for radius (r).
Am I approaching this from the right direction? It seems that, when it comes to weird circle problems, or pi in general, it often comes down to integrating/derivating(??) a solution. I know basic Calc, but I can't see how to use it here. Also, since this is for a computer game and needs to run in realtime, I'm really hoping the solution isn't too ... problematic. ;) I know I can fudge an okay representation, but if the real solution is elegant, I'd rather have that. Plus after circling poor attempts to factor and simplify for a while, I'm genuinely curious to know what the answer is now.
| Non-Complex Solution
The math involved for solving for $r$ goes a bit beyond basic calculus, I can spare you the long explanations and tell you that $V = \frac{4}{3}\pi r^3+h\pi r^2$ is solved for $r$ as:
$$\frac{1}{4} \left(-\frac{\sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}{\sqrt[3]{\pi }}-\frac{\sqrt[3]{\pi } h^2}{\sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}-h\right)$$
That's the only solution not involving complex numbers.
Complex Soltution
The solution(s) involving complex numbers are:
$$r = \frac{\left(1 \mp i \sqrt{3}\right) \sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}{8 \sqrt[3]{\pi }}+\frac{\sqrt[3]{\pi } \left(1 \pm i \sqrt{3}\right) h^2}{8 \sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}-\frac{h}{4}$$
Note that this is two separate solutions due to the flipping of the signs. Also notice the formatting of the "minus-plus sign" and the "plus-minus sign". This means that when $\mp$ is negative, $\pm$ is positive. When $\mp$ is positive, $\pm$ is negative.
Graphical Representation of the Functions
I thought it'd be best to represent the data graphically.
The three functions are labeled in these representations as:
*
*$f(h,V) = \frac{1}{4} \left(-\frac{\sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}{\sqrt[3]{\pi }}-\frac{\sqrt[3]{\pi } h^2}{\sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}-h\right)$
*$f\text{complex1}(h,V) = \frac{\left(1-i \sqrt{3}\right) \sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}{8 \sqrt[3]{\pi }}+\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\pi } h^2}{8 \sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}-\frac{h}{4}$
*$f\text{complex2}(h,V) = \frac{\left(1+i \sqrt{3}\right) \sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}{8 \sqrt[3]{\pi }}+\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\pi } h^2}{8 \sqrt[3]{4 \sqrt{3} \sqrt{12 V^2-\pi h^3 V}+\pi h^3-24 V}}-\frac{h}{4}$
All three functions, transposed onto a single $3D$ graph looks like so:
To further represent the data, we can visualize the functions transforming when we put them into the form $f\left(x \cdot h, \; y \cdot V\right)$:
Note that in this gif, $h$ and $V$ are both running from $-10$ to $10$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the probability of getting at least two different numbers from first roll when three six-sided dice are rolled a second time? I roll three six sided fair dice. Three different numbers show. What is the probability that at least two different numbers show in the second roll that occurred in the first roll? For example, first roll (1,2,3), second roll (3,1,4)?
Attempt
\begin{align*}
\Pr(\text{of at least two the same}) & = 1 - \Pr(\text{exactly $0$ the same}) - \Pr(\text{exactly $1$ the same})\\
& = 1 - \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} - \frac{1}{3} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot 3\\
& = 0.625
\end{align*}
In the second term, I am multiplying by $3$ to account for the number of placements the number can be in. Is this approach correct? Thanks!
| Let's count all desired possibilities out the $6^3$ total potential results:
*
*${3 \choose 3} \times 3!=6$ ways of getting all three original values in some order
*${3 \choose 2} \times {6-3 \choose 1} \times 3!= 54$ ways of getting two different original values and one non-original value in some order
*${3 \choose 1} \times {2 \choose 1} \times \frac{3!}{2!} = 18$ ways of getting a double original value and a single original value in some order
making the probability $\dfrac{6+54+18}{6^3} = \dfrac{78}{216}=\dfrac{13}{36}\approx 0.361$, the same as drhab found another way
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing the coefficient of the term of a certain degree in a polynomial Given the polynomial
${1\over8}((1+z)^9 + 3(1-z)^4(1+z)^5 + (1-z)^6(1+z)^3)$
(which is the weight enumerator of a code)
how do I find out the coefficient of $z^2$?
The solution given is ${1 \over 8}(36-12+0) = 3$.
I got $36$ for the $z^2$ coefficient of $(1+z)^9$ using the Binomial Theorem, but I don't know how to get $-12$ for the $z^2$ coefficient of $3(1-z)^4(1+z)^5$. By using the Binomial Theorem separately on $(1-z)^4$ and $(1+z)^5$ I get the following two polynomials, repsectively:
$z^4-4z^3+6z^2-4z+1$
$z^5+5z^4+10z^3+10z^2+5z+1$
I am unsure what to do next, or even if this is going in the right direction.
| It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{[z^2]}&\color{blue}{\frac{1}{8}\left((1+z)^9+3(1-z)^4(1+z)^5+(1-z)^6(1+z)^3\right)}\\
&=\frac{1}{8}\left([z^2](1+z)^9+3[z^2](1-z)^4(1+z)^5+[z^2](1-z)^6(1+z)^3\right)\tag{2}\\
&=\frac{1}{8}\left\{\binom{9}{2}+3\left(\binom{4}{0}[z^2]-\binom{4}{1}[z^1]+\binom{4}{2}[z^0]\right)(1+z)^5\right.\\
&\qquad \qquad\qquad \left.+\left(\binom{6}{0}[z^2]-\binom{6}{1}[z^1]+\binom{6}{2}[z^0]\right)(1+z)^3\right\}\tag{3}\\
&=\frac{1}{8}\left\{\binom{9}{2}+3\left(\binom{4}{0}\binom{5}{2}-\binom{4}{1}\binom{5}{1}+\binom{4}{2}\binom{5}{0}\right)\right.\\
&\qquad\qquad\qquad\left.+\left(\binom{6}{0}\binom{3}{2}-\binom{6}{1}\binom{3}{1}+\binom{6}{2}\binom{3}{0}\right)\right\}\tag{4}\\
&=\frac{1}{8}\left(36+3\left(10-20+6\right)+\left(3-18+15\right)\right)\\
&\,\,\color{blue}{=\frac{1}{8}\left(36-12+0\right)}\\
&\,\,\color{blue}{=3}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we use the linearity of the coefficient of operator.
*In (3) we select the coefficient of $z^2$ in $(1+z)^9$ and we again use the linearity of the coefficient of operator together with the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we select the coefficient of $z^k, k=0,1,2$ accordingly.
| {
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If $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$ , then $\triangle ABC$ is a right triangle
Show that the triangle whose angles satisfy the equality
$$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$$
is right-angled.
I've tried many times, but was unsuccessful.
| COMMENT.- First note that if $C=90^{\circ}$ then you have $\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\circ}$ you can consider that $\sin C=\sin (180^{\circ}-(A+B))=\sin(A+B)$ and $\cos C= \cos (180^{\circ}-(A+B))=-\cos(A+B)$ so you get
$$\sin^2 A+\sin^2 B+\sin^2 (A+B)=2(\cos^2A+\cos ^2 B+\cos^2(A+B))$$
You have to know simple formulas of trigonometry to finish.
EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\sin^2 A+\sin^2 B+\sin^2 (A+B) = 2\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\dfrac{\pi}{2}$ (the "if" is obvious and we need the "only if"). So put $A + B = \dfrac{\pi}{2}\pm h$ with $h\ne 0$ and look at the functions
$$\sin^2(x)+\sin^2(\dfrac{\pi}{2}\pm h-x)+\sin^2(\dfrac{\pi}{2}\pm h),\quad0\lt x\lt\dfrac{\pi}{2}$$ which becomes for $h$ positive and $h$ negative respectively $$f_1(x)=\sin^2(x)+\cos^2(x-h)+\cos^2(h)\qquad (1)\\f_2(x)=\sin^2(x)+\cos^2(x+h)+\cos^2(h)\qquad (2)$$ Now you can prove that $$\begin{equation}f_1(x)\begin{cases}\lt2\text{ when } 0\lt x\lt h\\=2 \text { when }x=h \space\text {discarded because } B\ne\dfrac{\pi}{2} \\\gt 2\text { when }h\lt x\lt\dfrac{\pi}{2}\\ \end{cases}\end{equation}$$
and that $f_2(x)$ is smaller than $2$ on $0\lt x\lt \dfrac{\pi}{2}$.
This requires elementary calculus that I guess you know how to handle.
| {
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A triangle inscribed in a circle of radius $2$ has angles $45^\circ$ and $60^\circ$. What's its area? Points $A$, $B$ and $C$ are on the circumference of a circle with radius 2 such that $\angle BAC = 45$ and $\angle ACB = 60$ . Find the area of $\triangle ABC$ .
I tried using the Law of Sines on the three separate triangles formed find the side lengths of $AB$, $BC$, and $AC$, and tried to continue with Herons Formula to end up with $(3 + \sqrt{3})/3$, but I got the wrong answer. Any thoughts on a different way to solve this?
| The circumcircle of a triangle is centered at the intersection of the perpendicular bisector of its sides
Call this center $O$, and draw $OA, OB, OC$ and effectively you have divided the triangle into three isosceles triangles(because the sides are the radii of the circle).
Now, we know that the base angles of an isosceles triangle, are equal. So we have this system of equations, (assume angles $A$ is divided into angles $x,y$, $B\to x,z$ and $C\to z,y$): $$x+y=45^\circ\\x+z=60^\circ\\y+z=75^\circ$$
Which gives you: $$x\to 15^\circ, y\to 30^\circ, z\to45^\circ$$
From here you can workout the interior angles: $$\angle BOA \to 150^\circ, \angle AOC \to 120^\circ, \angle COB\to 90^\circ$$
The area of the triangle is defined by: $$A_\triangle=\frac12 a b \sin C$$
Looking at $\triangle BOC$, you know that $BC=a=2\sqrt2$
Using the law of sines, you get $b$ as: $$\frac{\sin 120^\circ}{b}=\frac{\sin 30^\circ}2\iff b=\frac{2 \sin 120^\circ}{\sin 30^\circ}$$
Now, we can solve for the area as: $$\begin{align} A_{\triangle ABC}&=\frac12(2\sqrt2)(\frac{2 \sin 120^\circ}{\sin 30^\circ})\sin75\\ &=\frac{\left(2 \sqrt{2}\right) \left(2 \sqrt{3}\right) \left(\sqrt{3}+1\right)}{\frac{2}{2} 2 \left(2 \sqrt{2}\right)} \\&=\sqrt{3} \left(\sqrt{3}+1\right) \end{align}$$
| {
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"question_score": "1",
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Ask for the rational roots of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$ I consider the problem: ask the rational roots of $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$
I try to use the theory of ellipic curves (If we have two rational roots, we could have the third collinear one) and calculte the genus of $\Sigma_{p}:=\{[a,b,c]: p(a,b,c)=0\}$, where $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c).(*)$$
How to get the genus? I know I should use:
The Riemann-Hurwitz theorem: $$2g(\Sigma)-2=B_{p}(f)-2\deg(f),$$
where $g(\Sigma)$ is the genus of Riemann surface $\Sigma$ and $B_{p}(f)$ branch numbers of $f$ at $p$.
I try to bring $(*)$ into the form $y^2=x(x-1)(x-\lambda), \lambda \ne0,1,$ since I know the genus is one.
Maybe help?
$p(a,b,c)=a^3+b^3+c^3-3a^2b-3ab^2-3a^2c-3ac^2-3cb^2-3bc^2.$
| There is one idea. To search for the solution of the equation.
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=q$$
If we know any solution $(a,b,c)$ of this equation. Then it is possible to find another $(a_2, b_2, c_2)$. Make such a change.
$$y=(a+b+2c)(q(a+b)-c)-(a+b)^2-(a+c)(b+c)$$
$$z=(a+2b+c)(2b-qa-(q-1)c)+(b+2a+c)(2a-qb-(q-1)c)$$
Then the following solution can be found by the formula.
$$a_2=((5-4q)c-(q-2)(3b+a))y^3+((5-4q)b+4(1-q)c)zy^2+$$
$$+(3c+(q-1)(a-b))yz^2-az^3$$
$$b_2=((5-4q)c-(q-2)(3a+b))y^3+((5-4q)a+4(1-q)c)zy^2+$$
$$+(3c+(q-1)(b-a))yz^2-bz^3$$
$$c_2=2(q-2)cy^3+3(2-q)(a+b)zy^2+$$
$$+((5-4q)(a+b)+2(1-q)c)yz^2+(2c-(q-1)(a+b))z^3$$
| {
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"url": "https://math.stackexchange.com/questions/2779545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational.
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?
| By hypothesis there exist $a$ and $b$ such that $x$ is a root of the polynomials $f(X)=X^2-2X+a$ and $g(x)=X^3-5X+b$. Since $x$ is irrational, $[\mathbb Q(x):\mathbb Q]=2$ and the minimal polynomial of $x$ is $f(X)$. This implies that $g(X)$ is a multiple of $f(X)$, so
$$g(X)=(X-c)f(X)$$
and equalising the coefficients we find that $a=-1$, $b=-2$ and $c=-2$. Hence $x$ is a root of $X^2-2X-1$, and $x^3-5x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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} |
Probability box of cards I have to solve next task: There is box with 12 red cards, 8 blue cards and 7 green cards. Four cards are taken from it, and must find:
a) probability to has 2 red, 1 blue and 1 green card
b) all 4 cards to be the same color:
My logic: a)
$$\frac{C\frac{2}{12}.C\frac{1}{8}.C\frac{1}{7}}{C\frac{4}{27}} = \frac{66.8.7}{17550} = \frac{3696}{17550} = 0.21$$
b) $$\frac{ C\frac{4}{12} }{ C\frac{4}{27} } + \frac{C\frac{4}{8}}{\frac{4}{27}} + \frac{C\frac{4}{7}}{C\frac{4}{27}} = \frac{11880}{421200} + \frac{1680}{421200} + \frac{840}{421200} = \frac{14400}{421200} = 0.034$$
Are my calculations correct?
| There are some mistakes in combinatorics notation, but the final answers are OK. In question b), you shouldn't use the combinatorics notation in the second step and on, just in the first expression.
I mean:
a) $\frac{\binom{12}2\binom{8}1\binom{7}1}{\binom{27}4}=\frac{3969}{17550}=0.2105$
b) $\frac{\binom{12}4+\binom{8}4+\binom{7}4}{\binom{27}4}=\frac{495+70+35}{17550}=\frac{600}{17550}=0.034$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| Just to suggest another method.
$$\int \frac{1}{(x^2+1)^2} \, dx =\int \frac{x^2+1-x^2}{(x^2+1)^2} \, dx =\int \frac{1}{x^2+1} \, dx -\int \frac{x^2}{(x^2+1)^2} \, dx.$$ You can compute the first antiderivative very easily and for the second one, by parts $$\int \frac{x^2}{(x^2+1)^2} \, dx= \int x \cdot \frac{x}{(x^2+1)^2} \, dx=-\frac{1}{2}\int x\cdot \frac{\partial}{\partial x}\left(\frac{1}{x^2+1}\right) \, dx.$$ I think you can now easily finish the exercice.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this integral $\int_0^1 \frac{x\log ^2(\sqrt{x^2+1}+1)}{\sqrt{1-x^2}} \, dx$ How to evaluate $$A=\int_0^1 \frac{x \log ^2\left(\sqrt{x^2+1}+1\right)}{\sqrt{1-x^2}} \, dx$$See the details here from a similar question: Evaluating $\int_0^1 \frac{z \log ^2\left(\sqrt{z^2+1}-1\right)}{\sqrt{1-z^2}} \, dz$.
By applying the same process, I find:
$$A=\int_0^1\ln^2\left(\sqrt{2-y^2}+1\right)dy=\int_1^{\sqrt2}\frac {x\log^2 (x+1)}{\sqrt{2-x^2}}dx=\sqrt2 \int_{0}^{\pi/4}\cos\varphi\ln^2\left(\sqrt2 \cos\varphi+1\right)d\varphi.$$
$$A=\ln^22+32(1+\sqrt2)\int_{0}^{\sqrt2-1}\frac{t^2}{\left(1+t^2\right)^2(t+1+\sqrt2)(-t+1+\sqrt2)}\ln\left(\sqrt2\, \frac{1-t^2}{1+t^2}+1\right)dt.$$
Now,put $$A=\int_0^1 \frac{x \log ^2\left(\sqrt{x^2+1}+1\right)}{\sqrt{1-x^2}} \, dx, B=\int_0^1 \frac{x \log ^2\left(\sqrt{x^2+1}-1\right)}{\sqrt{1-x^2}} \, dx.$$
We have:
$$A-B=4\int_0^{1}\frac {x\log x\log(\sqrt{1+x^2}+1)}{\sqrt{1-x^2}}dx-4\int_0^{1}\frac {x\log^2 x}{\sqrt{1-x^2}}dx$$
$$A+B=-2\int_0^{1}\frac {x\log (\sqrt{1+x^2}-1)\log(\sqrt{1+x^2}+1)}{\sqrt{1-x^2}}dx+4\int_0^{1}\frac {x\log^2 x}{\sqrt{1-x^2}}dx$$
$$\int_0^{1}\frac {x\log^2 x}{\sqrt{1-x^2}}dx=\ln^22-\frac{\pi^2}{12}-2\ln2+2.$$
But how to deduce these integrals.(using Mathematica?)
| Substitute $x^2=\sin 2\theta$ to get $\sqrt{1\pm x^2}= \cos\theta\pm\sin\theta$
\begin{align}
I=& \int_0^1 \frac{x \log ^2(\sqrt{1+x^2}+1)}{\sqrt{1-x^2}} \, dx\\
=& \ \frac12\int_0^{\pi/2}\ln^2(\cos\theta+\sin\theta+1)(\cos\theta+\sin\theta)\ d\theta\\
=& \int_0^1 \ln^2\frac{2(1+t)}{1+t^2}\ d\left( \frac{t-1}{1+t^2}\right)\>\>\>\>\>\>\>t=\tan\frac{\theta}2\\
\overset{ibp}=&\ \ln^22 + \int_0^1 \ln\frac{2(1+t)}{1+t^2}\left(\frac1{1+t}+\frac{2}{1+t^2}+\frac{4(t^2-1)}{(1+t^2)^2}\right)dt
\end{align}
where the remaining integrals are readily manageable, given by
\begin{align}
& \int_0^1 \ln\frac{2(1+t)}{1+t^2}\ \frac1{1+t}\ dt
=\frac34\ln^22+\frac{\pi^2}{48}\\
& \int_0^1 \ln\frac{2(1+t)}{1+t^2}\ \frac{1}{1+t^2}\ dt
= G-\frac\pi{8}\ln2\\
& \int_0^1 \ln\frac{2(1+t)}{1+t^2}\ \frac{t^2-1}{(1+t^2)^2}\ dt
= -\frac\pi8-\frac34\ln2+\frac12
\end{align}
Substitute above results into $I$ to obtain
$$I=\frac74\ln^22+\frac{\pi^2}{48}+ 2G-\frac\pi4\ln2-\frac\pi2-3\ln2+2
$$
| {
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"url": "https://math.stackexchange.com/questions/2786924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A specific question about a factor theorem proof I know there are tons of answers to questions about the factor theorem, however, I couldn't find what I was looking for so apologies if someone has already answered this before.
Since
$\frac{x^k - y^k}{x-y} = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$
${x^k - y^k} = (x-y)(x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1})$
hence ${x^k - y^k} = (x-y)q_k(x)$
As usual, we write our polynomial $ p$ as
$p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 $
$p(x)-p(y) = $
$a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 $
$ -(a_ny^n + a_{n-1}y^{n-1} + ... + a_1y + a_0)$
$= a_n(x^n-y^n) + a_{n-1}(x^{n-1}-y{n-1}^) + ... + a_1(x-y)$
The terms have the form $a_k(x^k-y^k)$. But $x^k-y^k = (x-y)q_k(x)$, and substitute this in:
$p(x)-p(y)=$
$a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)$
$(x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1)$
$= (x-y)q(x)$ where $q(x) = a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1$
If $p(a) = 0$ and $y = a$ then
$p(x)-p(a) = (x-a)q(x) $ Q.E.D
My question is: why we don't write $p(x) - p(y)$ as
$a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)q_1(x)$
$(x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1q_1(x))$.
NOTE! That I'm asking whether or not $a_1q_1(x) = a_1$
Can anyone please explain this to me thanks!
| It's exactly what's happening.
Observe that $q_1$ is just the constant $1$ polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
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find the value of $ S = \sum _{k=1}^{\infty }\left(\frac{\left(-1\right)^{k+1}}{(k)(2k+1)}\right)$ Find the value of $\frac{1}{3}-\frac{1}{10}+\frac{1}{21}-\frac{1}{36}+...$
I have no idea how to solve this infinite sum, I appreciate any help, thanks in advance.
| Observe you have
\begin{align}
\operatorname{Log}(1+i)=\sum^\infty_{n=1} (-1)^n\frac{i^n}{n}=\sum^\infty_{k=1}\frac{(-1)^k}{2k}+i\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k-1}
\end{align}
where $\operatorname{Log}z$ is the principal branch.
Hence it follows
\begin{align}
\operatorname{Log}(1+i)=\log 2+ i\arg (1+i) = \log 2 + i\frac{\pi}{4}
\end{align}
which means
\begin{align}
\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k+1}=1-\frac{\pi}{4}.
\end{align}
For the other problem, we see that
\begin{align}
\sum^\infty_{k=1}(-1)^{k+1} \frac{1}{2k(2k+1)} = \sum^\infty_{k=1}(-1)^{k+1}\left( \frac{1}{2k}-\frac{1}{2k+1}\right)=-\frac{1}{2}\sum^\infty_{k=1}\frac{(-1)^k}{k}-\frac{\pi}{4} = \frac{1}{2}\log 2+\frac{\pi}{4}-1.
\end{align}
| {
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Computing the surface integral of the octant of a sphere with polar coordinate substitution Let me first describe where I start:
$$\iint_Sz^2\,dS$$
We want to compute the surface integral of the octant of a sphere $S$.
The radius = 1.
The sphere is centered at the origin.
$$S=x^2+y^2+z^2=1.$$
$$z=f(x,y)=\sqrt{1-x^2-y^2}$$
$R$ is the projection of $S$ on the $xy$-plane.
Now we compute the normalization factor used to project the integral on the $xy$-plane. First we compute the derivative $\frac{\partial f}{\partial x}$:
If
$$m=1-x^2-y^2$$
and
$$n=\sqrt{m}\,,$$
then the derivative of $n$ is
$$n'=\frac12m^{-\frac12},$$
and the derivative of $m$ (with regard to $x$) equal to:
$$m'=-2x.$$
Now we can compute $\frac{\partial f}{\partial x}$ using the chain rule:
$$\frac{\partial f}{\partial x}=n'\cdot m'=\frac12m^{-\frac12}\cdot m'=\frac12(1-x^2-y^2)^{-\frac12}\cdot-2x=-{\frac{x}{\sqrt{(1-x^2-y^2)}}}.$$
Since $\sqrt{(1-x^2-y^2)}=z$:
$$\frac{\partial f}{\partial x}=-{\frac xz}$$
And in the same manner (using the derivative of $m$ with regard to $y$) we can calculate $\frac{\partial f}{\partial y}$:
$$\frac{\partial f}{\partial y}=-{\frac yz}$$
Now we can calculate the normalization factor:
$$\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}=\sqrt{1+\frac{x^2}{z^2}+\frac{y^2}{z^2}}=\frac1z\sqrt{x^2+y^2+z^2}.$$
Since $x^2+y^2+z^2=1$:
$$\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}=\frac1z$$
Now we can calculate the projection $R$ of the sphere $S$ on the $xy$-plane:
$$\iint_Sz^2\,dS=\iint_R z^2\frac1z \,dx\,dy=\iint_R z\;dx\,dy.$$
Substituting for $z$ finishes the conversion of the surface integral (remember $z=f(x,y)=\sqrt{1-x^2-y^2}$):
$$\iint_Sz^2dS=\iint_R \sqrt{1-x^2-y^2}\;dx\,dy.$$
This is where my problem starts:
The book I'm reading says if we convert this to polar coordinates, the integration should be trivial.
So we convert to polar coordinates:
$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$z=f(x,y)=f(r\cos\theta,r\sin\theta)$$
Calculate the Jacobian determinant:
$$\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\\ \end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r\,(\cos^2\theta + \sin^2\theta)=r$$
Substitute:
$$\iint_Sz^2dS=\iint_R \sqrt{1-x^2-y^2}\;dx\,dy=\iint_T\sqrt{1-r^2\cos^2\theta-r^2\sin^2\theta}\cdot r \cdot dr\,d\theta$$
$$\iint_T\sqrt{1-r^2(\cos^2\theta+\sin^2\theta)}\cdot r \cdot dr\,d\theta = \iint_T\sqrt{1-r^2}\cdot r \cdot dr\,d\theta=\iint_T\sqrt{r^2-r^4}\cdot dr\,d\theta$$
Since the radius $r=1$ it is easy to see that:
$$\int^1_0\sqrt{r^2-r^4}\cdot dr=\int^1_0(r^2-r^4)^{\frac12}\cdot dr = \int^1_0 \frac{(r^2-r^4)^{\frac32}}{\frac32}=0$$
According to the book the result of the calculation of the surface of the sphere in the first octant should be $\pi/6$.
That won't happen if $\int^1_0\sqrt{r^2-r^4}\cdot dr=0$.
The domain of $\theta$ is:
$$0\le\theta\le\frac12\pi$$
So where am I going wrong?
| The surface integral is given by
$$\iint_S\,dS=\iint_R \frac1{\sqrt{1-x^2-y^2}}\;dx\,dy=$$
that is by polar coordinates
$$=\int_0^{\pi/2}\,d\theta\int_0^1 \frac{r}{\sqrt{1-r^2}}\,dr =\frac{\pi}2\left[-\sqrt{1-r^2}\right]_0^1=\frac{\pi}2$$
that is correct since the surface of the whole sphere is equal to $4\pi$.
The given integral is
$$\iint_Sz^2\,dS=\iint_R {\sqrt{1-x^2-y^2}}\;dx\,dy=$$
that is by polar coordinates
$$=\int_0^{\pi/2}\,d\theta\int_0^1 r{\sqrt{1-r^2}}\, dr=\frac{\pi}2\left[-\frac13(1-r^2)^\frac32\right]_0^1=\frac{\pi}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
| Note that
$$
d/dx((x - 1/2) \sin^{-1}(\sqrt x)) = \frac{x - 1/2}{2 \sqrt{x(1 - x)} } + \sin^{-1}(\sqrt x )
$$
So you can easily integrate
$$
\int\sin^{-1}\sqrt{x}dx \\
= (x - 1/2) \sin^{-1}(\sqrt x) - \int \frac{x - 1/2}{2 \sqrt{x(1 - x)} } dx \\
= (x - 1/2) \sin^{-1}(\sqrt x) + \frac{ \sqrt{x(1 - x)}}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Does the following multivariable limit exist? What should be the value of $A$ for $\lim_{(x,y)\to (0,0) } f(x,y) $ to exist?
$$
f(x,y) = \begin{cases}
\frac{x^4+y^4}{y(x^2+y^2 ) } , \quad y \neq 0 \\
A , \quad y=0
\end{cases}
$$
Thanks in advance!
It seems that $\frac{x^4+y^4}{(x^2+y^2 ) } \to 0$ , but $\frac{1}{y} \to \infty$ does not help me that much. In addition, substituting polar coordinates does not give me anything useful, because of the expression $\cot(\theta) \cos^3 (\theta) $ that is not bounded. Any help will be appreciated
| We have that
*
*for $x=y=t\to 0 \implies \large{\frac{x^4+y^4}{y(x^2+y^2 ) } =\frac{2t^4}{2t^3}=t\to 0}$
*for $x=t$ and $y=t^2$ with $t\to 0 \implies \large{\frac{x^4+y^4}{y(x^2+y^2 ) } =\frac{t^4+t^8}{t^4+t^6}=\frac{1+t^4}{1+t^2}\to 1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does this always converge? So, I was researching the whole quadrature thing recently and decided to define right triangles area(with legs a and b) as a sum of the inscribed square and two smaller triangles.
Square inscribed in right triangle with legs a and b
To find its area I have defined following recursive expression:
$$S\bigl(a, b\bigr) = \bigl(\frac{ab}{a+b}\bigr)^2 + S\bigl(\frac{a^2}{a+b}, \frac{b}{a+b}\bigr) + S\bigl(\frac{ab}{a+b}, \frac{b^2}{a+b}\bigr)$$
And, as I was trying to find S(a, b) in finite terms(and assuming that I don't know that it is $\frac{ab}{2}$), expanding S on the right side, I came up to following:
$$S\bigl(a, b\bigr) = \bigl(\frac{ab}{a+b}\bigr)^2\sum_{i=0}^\infty \biggl(\frac{a^2}{(a+b)^2}+\frac{b^2}{(a+b)^2}\biggr)^i$$
And thus, $S\bigl(a, b\bigr) = \frac{ab}{2}$
But what if we don't limit this problem to just positive a and b?
Does it converge?
And to what it converges?
Edit: Typo in series
| For starters, for $S$ to be well defined we need $a+b \neq 0$. For each $(a,b) \in \mathbb{R}$, $\bar{S}(a,b) := S(a,b)(\frac{a+b}{ab})^2$ is nothing more than a geometric series. It will converge if and only if
$$
\big|\frac{a^2+b^2}{(a+b)^2}\big| = \frac{a^2+b^2}{(a+b)^2} < 1
$$
This will occur if $a^2 + b^2 < (a+b)^2 = a^2 +2ab + b^2$, that is, when $2ab >0$. Therefore, both $a$ and $b$ have to share the same sign and be nonzero. As for the value of convergence, since we're dealing with a geometric series,
$$
\bar{S}(a,b) = \frac{1}{1-\frac{a^2+b^2}{(a+b)^2}} = \frac{1}{\frac{2ab}{(a+b)^2}} = \frac{(a+b)^2}{2ab}
$$
Now,
$$
S(a,b) = \big(\frac{ab}{a+b}\big)^2\bar{S}(a,b) = \frac{a^2b^2}{2ab} = \frac{ab}{2}
$$
like predicted. To sum up, the biggest domain we can consider is $D := \{(a,b) : ab > 0\}$, and here $S(a,b) = \frac{ab}{2}$, coinciding with the values known for positive $a$ and $b$.
| {
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Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
My Attempt
Using Partial fractions
$$
\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\frac{1}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)}+\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=I_1+I_2\\
I_1=\bigg[\frac{1}{b^2}.\frac{1}{ab}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0=\bigg[\frac{1}{ab^3}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0
$$
$$
I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=\frac{b^2-a^2}{b^2}\bigg[\frac{x}{(a^2+b^2x^2)^2}-\int\frac{-2.2x.x}{(a^2+b^2x^2)^3}dx\bigg]
$$
How do I evaluate the integral $I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$ ?
| $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}+\int_0^\infty\frac{x^2dx}{(a^2+b^2x^2)^2}=I_1+I_2$
Without bounds and constant of integration we have putting $x=\dfrac {at}{b}$ with integration by parts
$$I_1=\frac ab\left(\frac{t}{(a^2+b^2t^2)^2}-\int\frac{-4a^2t}{(a^2+a^2t^2)^3}tdt\right)$$ $$I_1=\frac ab\left(\frac{t}{(a^2+b^2t^2)^2}+\frac{1}{2a^4}(\arctan(t)-\frac 14\sin(4\arctan(t))\right)$$
It follows coming back to $x$,
$$I_1=\frac{x}{a^2+b^2x^2)^2}+\frac{1}{2ba^3}\arctan\left(\frac{bx}{a}\right)-\frac{1}{2ba^3}\frac14\sin(4\arctan\left(\frac{bx}{a}\right)$$
$I_2$ is easier and we have
$$I_2=\frac{1}{2ab^3}\arctan\left(\frac{bx}{a}\right)-\frac{1}{4ab^3}\sin\left(2\arctan(\frac{bx}{a}\right)$$
I leave the calculations with the bounds $0$ and $\infty$ to finish.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}\neq 0$ According to solution, $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}=\frac{1}{2}$. Why it is so, when power of polynomial in denominator is greater than in numerator?
| $$ \left( x^{4} + 2 x + 1 \right) $$
$$ \left( x^{3} - 2 x - 1 \right) $$
$$ \left( x^{4} + 2 x + 1 \right) = \left( x^{3} - 2 x - 1 \right) \cdot \color{magenta}{ \left( x \right) } + \left( 2 x^{2} + 3 x + 1 \right) $$
$$ \left( x^{3} - 2 x - 1 \right) = \left( 2 x^{2} + 3 x + 1 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 3 }{ 4 } \right) } + \left( \frac{ - x - 1 }{ 4 } \right) $$
$$ \left( 2 x^{2} + 3 x + 1 \right) = \left( \frac{ - x - 1 }{ 4 } \right) \cdot \color{magenta}{ \left( - 8 x - 4 \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 3 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x^{2} - 3 x + 4 }{ 4 } \right) }{ \left( \frac{ 2 x - 3 }{ 4 } \right) } $$
$$ \color{magenta}{ \left( - 8 x - 4 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - 4 x^{3} + 4 x^{2} - 4 x - 4 \right) }{ \left( - 4 x^{2} + 4 x + 4 \right) } $$
$$ \left( x^{3} - x^{2} + x + 1 \right) \left( 2 x - 3 \right) - \left( x^{2} - x - 1 \right) \left( 2 x^{2} - 3 x + 4 \right) = \left( 1 \right) $$
$$ \left( x^{4} + 2 x + 1 \right) = \left( x^{3} - x^{2} + x + 1 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$
$$ \left( x^{3} - 2 x - 1 \right) = \left( x^{2} - x - 1 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x + 1 \right) } $$
$$ \left( x^{4} + 2 x + 1 \right) \left( 2 x - 3 \right) - \left( x^{3} - 2 x - 1 \right) \left( 2 x^{2} - 3 x + 4 \right) = \left( x + 1 \right) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the DE: $y''+1/(2y^3)=0$. \begin{align}
y''+\frac{1}{2y^3}=0\\
\end{align}
This is a second order differential equation which doesn't contain $x$ explicitly.
Let $y'=p(y)$, so that $y''=p(y)*p'(y)$.
\begin{align}
pp'+\frac{1}{2y^3}=0\\
\int{pdp}=\int{-\frac{1}{2y^3}dy}\\
\frac{p^2}{2}=-\frac{1}{2}*\frac{-1y^{-2}}{2}+C_1\\
p^2=\frac{1}{2y^2}+C_1\\
p=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
$p$ was equal to $y'$
\begin{align}
y'=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
I've tried to continue with trigonometric substitution but that didn't work for me.
Can you please help me out?
Thank you in advance!
Edit: adding final solution
Because of the answers below, I found the solution:
\begin{align}
\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}=\pm{x}dx\\
\int{\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}}=\int{\pm{x}dx}\\
\frac{1}{2C_1}\int{\frac{d({1+C_1y^2})}{\sqrt{1+C_1y^2}}}=\pm\frac{1}{\sqrt2}x\\
2\sqrt{1+C_1y^2}=\pm{\frac{2C_1}{\sqrt{2}}}(x+C_2)\\
\sqrt{1+C_1y^2}=\pm{\frac{C_1}{\sqrt{2}}}(x+C_2)\\
2(1+C_1y^2)=C_1^2(x+C_2)\\
\end{align}
| Try to find $x$ as a function of $y^2$ instead, for taking it next from there, in terms of $y^2$:
$$ dx= \frac{\sqrt2y dy}{\sqrt{1+2 C y^2}} = \frac{d y^2}{\sqrt{2(1+2 C y^2})} $$
leading to inverse circular/hyperbolic function solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit problem with summation: $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \dots + \frac{n}{n^2 +n}$ $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} + \dots + \frac{n}{n^2 +n}$
question is when we take limit we can seperate things right ? So we can write $\lim_{n\to\infty} \frac{1}{n^2 + n}$ + $\lim_{n\to\infty} \frac{2}{n^2 + n}$ + .... $\lim_{n\to\infty} \frac{n}{n^2 + n}$
if we take limits one by one we get zeroes.
we get sum = 0
but if we do sum first than take limit
$\lim_{n\to\infty} \frac{\frac{n.(n+1)}{2}}{n^2 + n}$ with simplification we get 1/2
so did my first question wrong ? can't we take limits first than do the sum ?
| By Stolz-Cesaro
$$\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} + \frac{n}{n^2 +n}=\lim_{n\to\infty} \frac{\sum_{k=1}^n k}{n^2+n}=\lim_{n\to\infty} \frac{\sum_{k=1}^{n+1} k-\sum_{k=1}^n k}{(n+1)^2+(n+1)-n^2-n}=\lim_{n\to\infty} \frac{n+1}{2n+2}=\frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $2^{x-z} + 2^{y-z} = 1$ Suppose we have an equation $2^{x-z} + 2^{y-z} = 1$,where $x, y, z$ are integers from $0$ to $9$. $x, y, z$ can have same values. I guess $x$ and $y$ should have $9$ possible values like $x=y=0$ and $z=1$ and so on . Or should it have more values?
| The only possible case is $x-z = -1 = y-z\implies x = y = z-1$. Thus $z = 1,2,3,4,5,6,7,8,9$, and $x = y = 0,1,2,3,4,5,6,7,8$
| {
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Find real $a$ for which $x^2 +|x-a|-3\lt 0$ is satisfied for negative $x$. *Have problem solution given at: https://math.stackexchange.com/a/460988/424260, which am unable to understand. *
This problem's graphical soln. takes $|x-a|\lt 3-x^2$, & has right arm of the graph for $|x-a|$ first touching the curve for negative $a$, & then the left arm for positive $a$, with range between these two values of $a$ being asked for.
Am unable to get the stated values as given at the above link's answer :$$x^2+x-3\lt a\lt -x^2+x+3$$ has both l.h.s. ($x^2+x-3$) & r.h.s. ($-x^2+x+3$) having the roots given by same values $x= \frac{-1\pm \sqrt{13}}{2}$.
So, the second para. is incomprehensible to me.
Update The approach chosen by the solution's analytical approach (also the one desired by me) is to not consider the two graphs separately, but to consider the points for which the modulus changes from $(x-a) \lt 0 $ to $\ge 0$. The range gives the value of $a$, & need take minima of the left side $(x^2+x-3)$ that represents the intersection of the right arm of the term $|x-a|$ with the parabola, as shown below:
$x-a\ge 0\implies x^2+x-3\lt a$
So, the minima represents the minimum positive value of $a$ that would yield the intersection of the right arm of $|x-a|$ with the parabola. This is minima as it is the smallest value of $x$ for the 'combined' curve, where any intersection is possible.
This minima can be found by taking derivative of the equality : $x^2+x-3=a$ wrt $x$, i.e. $\frac{\delta a}{\delta x} = 2x +1$, and taking second dervative to verify minima, get : $\frac{\delta^2 a}{\delta^2 x} = 2$; a positive value; hence minima. This minima is given at : $x = -\frac 12$. For this value of $x$, the value of $x^2 +x-3 = \frac14-\frac 12-3= -\frac {13}{4}$.
Next, need find maxima , as the left arm would intersect the curve; leading to 'cumulative' curve having a maximum value of $x$.
This maxima can be found by taking derivative of the equality : $-x^2+x+3=a$ wrt $x$, i.e. $\frac{\delta a}{\delta x} = -2x +1$, and taking second dervative to verify maxima, get : $\frac{\delta^2 a}{\delta^2 x} = -2$; a negative value; hence maxima. This maxima is given at : $x = \frac 12$. For this value of $x$, the value of $-x^2 +x-3 = -\frac14+\frac 12+3= \frac14 +3 = \frac{13}{4}$.
Have two issues here:
(i) as the solution (at the given link) takes the same value of minima $x=-\frac12$ to get the maxima point for intersection of the left arm for positive value of $x$, by substituting in $-x^2+x+3$, i.e. $-\frac14-\frac12+3\implies -\frac34 +3\implies \frac94$.
(ii) the value obtained by me for maxima is invalid as it should be less than $\sqrt{3}$; but it is greater than that.
However, the values obtained by me for maxima ($\frac{13}4 = 3.25$) and minima (-$\frac{13}4 = -3.25$) values of $a$ in terms of $x$ are correct graphically, as shown at: https://www.desmos.com/calculator/h0fuoqe7jc
Update 2 The link has given a very nice idea to just take the minima instead of finding roots of the concerned two quadratic equations; but it seemingly falters in finding maxima & it uses the symmetric properties of both hyperbola & $|x-a|$ to unnecessarily find the intersection of left arm too at the same $x$ value.
The parabola is symmetrical around the y-axis, hence it is logical that $a$ values (for maxima & minima) too are just a sign change only. Similarly, the function $|x-a|$ too is. In fact symmetric property is used by solution (at the link given) to derive the left arm 's value for $a$ for given $x=\frac12$.
In fact the value $a=\frac94$ is not apparent as to where it should belong to, as it gives two intersections of the left arm with parabola. Am very confused at its graphical interpretation. The only possible interpretation is that the second intersection point is having coordinates $x=-\frac12, y = \frac{11}4$, i.e. the chosen value of minima ($x=-\frac12$). So, it was a wrong idea for the solution (at given link) to put the minima value for left arm equation to find maxima.
Update 3 Need find not maxima, but the value of $a$ for which negative $x$ values are there. So, from minima $-\frac{13}4$ to $x\lt 3$ for which the left arm intersects parabola at y-axis. Will arrive at the value $a=3$ as below:
The $y$ coordinate is same for both parabola and the left arm intersection at y axis. It is a unique point. The value $-x^2+x+3= a$ gives the left arm equation. But, $x=0$, so $a=3$.
So, the value range for $a$ is from minima to less than $3$: $[-\frac{13}4, 3)$.
| Preparation:
*
*If we encounter an inequality $|x|<\beta$. If $\beta \le 0$, then no such $x$ exists. Otherwise, it is equivalent to $-\beta < x< \beta$.
Now, when we encounter the inequality $x^2+|x-a|-3<0$, this is equivalent to $$|x-a| < 3-x^2.$$
$$|a-x|< 3-x^2$$
Hence to have a solution, we need the condition that $3-x^2>0$, that is we are interested in the solution in $(-\sqrt3,0)$. Also, from the preparation material, we let $\beta=3-x^2$, and we have
$$-(3-x^2)<a-x < 3-x^2$$
which is just $$x^2+x-3<a<-x^2+x+3$$
Hence $a$ take values from $(-\frac{13}4, 3).$
Remark: It is puzzling why you write $|x-a|<x^2-3$.
Edit $1$:
*
*The question of interest is find the value of $a$ when a negative $x$ exists that satisfies the inequality. Hence, we just have to look at negative part. In fact, we just have to look at $(-\sqrt3, 0)$. In fact, it is within this range that we have $x<0$ and also $x^2+x-3 < -x^2+x+3$ which is equivalent to $2(x^2-3)<0$.
*The curve $x^2+x-3=\left( x+\frac12\right)^2-3-\frac14=\left( x+\frac12\right)^2-\frac{13}4$ attains the minimal value at point $\left(-\frac12, -\frac{13}4 \right)$. Note that we have $-\sqrt{3} < -\frac12$.
*The curve $-x^2+x+3=-(x^2-x)+3=-\left(x-\frac12 \right)^2+3+\frac14$ which attains maximal value at point $\left(\frac12, \frac{13}4 \right)$. The maximal is attained at point $\left( \frac12, \frac{13}4\right)$. However, recall that we are interested in $-\sqrt{3}<x<0$, hence the supremum is $-0^2+0+3=3$.
*Hence $-\frac{13}4<a<3$
*Here is a desmos link, we can see when does $y=a$ lies in the region of interest.
Edit $2$:
The set of interest is \begin{align}A&=\{a: \exists x \in (-\sqrt{3}, 0), y_1(x) < a <y_2(x)\} \\
&= \{a: \exists x \in (-\sqrt{3}, 0), y_1(x) < y <y_2(x), y=a\} \\\end{align}
Let $D=\{(x,y): -\sqrt{3}<x<0, y_1(x) < y<y_2(x) \}$, we are interested in knowing for which value of $a$ does the $y=a$ intersect with $D$.
Also notice that $D$ doesn't include it's boundary. Hence $a\ne -\frac{13}4$.
Here is another desmos link that shaded the region in $D$. notice that $D$ doesn't include any boundary point. We can see that $A=(\inf_{(x,y) \in D}y, \sup_{(x,y)\in D}y)=\left( - \frac{13}4, 3 \right).$
Remark: The original solution in the previous posts is more rigorous. My approach uses a graph to help visualization.
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Maximum value of $p+q+r$ where $p=q(4-q), q=r(4-r), r=p(4-p)$ and $p,q,r\in R$ I would be grateful if somebody could please help me out.
Let $p,q,r\in R$. If $p=q(4-q), q=r(4-r), r=p(4-p)$, find the maximum value of $$p+q+r$$
My attempt:
$$p+q+r$$
$$=q(4-q)+(4-r)+(4-p)$$
$$=-(p^2+q^2+r^2-4p-4q-4r)$$
$$=-[(p-2)^2+(q-2)^2+(r-2)^2-12]$$
$$=12-[(p-2)^2+(q-2)^2+(r-2)^2]$$
$$$$
$$\Rightarrow p+q+r=12-[(p-2)^2+(q-2)^2+(r-2)^2]$$
$$$$
Thus it remains to minimize $[(p-2)^2+(q-2)^2+(r-2)^2]$
This clearly attains its minimum value at $p=2,q=2,r=2$. Hence, it seems that the minimum value of $$p+q+r=12-[0]=12$$
This does not match with the given answer.
$$$$
Additionally, I seem to be arriving at a contradiction since if $q=2$, $p=q(4-q)=2(4-2)=4$ (which clearly doesn't match with $p=2$) and so on for $r$ and $p$.
| Since (Cauchy-Shwartz inequality) $$ {1\over 3}(p+q+r)^2 \leq p^2+q^2+r^2 $$
and $$ p^2+q^2+r^2= 3(p+q+r)$$
we have $$ {1\over 3}(p+q+r)^2 \leq 3(p+q+r)$$
we get $p+q+r\leq 9$.
Value $9$ is achieved if $p=q=r=3$. So $\max\{p+q+r\}=9$
| {
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Integrate $\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx$
Integrate $\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx$
$\newcommand{\intd}[1]{\,\mathrm{d}#1}$
My Attempt
Put $t=\tan x\implies\intd t=\sec^2x\intd x$
\begin{align*}
\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx&=\int_0^{\pi/2}\frac{\tan x\cdot\sec^2x}{(1+m^2\tan^2 x)\sec^2x}\intd x\\
&=\int_0^{\infty}\frac{t}{(1+m^2t^2)(1+t^2)}\intd t
\end{align*}
Put $t^2=y\implies 2t\intd t=\intd y$
\begin{align*}
\frac{1}{2}\int_0^\infty\frac{dy}{(1+m^2y)(1+y)}&=\frac{1}{2}\int_0^\infty\bigg[\frac{1}{1-m^2}\cdot\frac{1}{1+y}+\frac{m^2}{m^2-1}\cdot\frac{1}{1+m^2y}\bigg]dy\\
&=\frac{1}{2(1-m^2)}\int_0^\infty\frac{dy}{1+y}-\frac{m^2}{2(1-m^2)}\int_0^\infty\frac{dy}{1+m^2y}\\
&=\bigg[\frac{1}{2(1-m^2)}\log|1+y|-\frac{1}{2(1-m^2)}\log|1+m^2y|\bigg]_0^\infty\\
&=\bigg[\frac{1}{2(1-m^2)}\log|\frac{1+y}{1+m^2y}|\bigg]_0^\infty=\color{red}{\frac{1}{2(1-m^2)}\bigg[\frac{\infty}{\infty}-0\bigg]}\end{align*}
I think I am getting stuck here as the substitutions does not seem to give the solution ?
Doubt
$$
\lim_{y\to 0}\log|\frac{1+y}{1+m^2y}|=\log 1=0\\
\lim_{y\to \infty}\log|\frac{1+y}{1+m^2y}|=\lim_{y\to \infty}\log|\frac{\frac{1}{y}+1}{\frac{1}{y}+m^2}|=\log\frac{1}{m^2}
$$
So what really we are doing with definite integrals ?. Does this mean that we are actually computing the upper and lower limits and taking the difference to find the definite integral ?
Note: Expanding in terms of $\sin x$ and $\cos x$ gives the solution, no doubt about limit going to infinity, not looking for that.
| You have one error $$\frac{m^2}{1-m^2}\int\frac{1}{1+m^2y}\,\mathrm{d}y=\frac{1}{1-m^2}\log|1+m^2y|+C$$
So your final answer is
\begin{align*}
\bigg[\frac{1}{2(1-m^2)}\log|1+y|-\frac{1}{2(1-m^2)}\log|1+m^2y|\bigg]_0^\infty&=\frac{1}{2(1-m^2)}\left[\log|1+y|-\log|1+m^2y|\right]_0^\infty\\
&=\frac{1}{2(1-m^2)}\log\left|\frac{1+y}{1+m^2y}\right|\bigg\rvert_0^\infty\\
&=\frac{1}{2(1-m^2)}\log\left(\frac1{m^2}\right)\\
&=\boxed{\frac{\log m}{m^2-1}}
\end{align*}
| {
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Gamma Identities from Inverse Transform I have been working on a transform (the notes are a little rough and the hosting website has some fraction formatting issues recently) but the idea is there:
Basically, the transform $\mathcal{I}_x[f(x)](s)$, will extract the coefficients of the scaled inverse function, $f^{-1}(x)/x$, in a manner analogous to the Mellin transform.
I have found a pattern that seems to generate Gamma function identities that Mathematica cannot simplify or verify. This was done by acting on simple polynomial expressions for which I previously found that the principal inverse (via series reversion) was a generalised hypergeometric function which led to the following results:
\begin{align}
\mathcal{I}[x+x^2](s) = \frac{\Gamma(1-\frac{2s}{1})\Gamma(\frac{s}{1}) }{\Gamma(2-s)}\\
\mathcal{I}[x+x^3](s) = \frac{\Gamma(1-\frac{3s}{2})\Gamma(\frac{s}{2})}{2 \Gamma(2-s)} \\
\mathcal{I}[x+x^4](s) = \frac{8 \cdot 3^{s-\frac{5}{2}}\pi \Gamma(-4s/3)\Gamma(s/3)}{\Gamma(2/3-s/3)\Gamma(4/3-s/3)\Gamma(-s/3)} \stackrel{?}{=}\frac{\Gamma(1-\frac{4s}{3})\Gamma(\frac{s}{3})}{3 \Gamma(2-s)}
\end{align}
the last one appears to be true, but Mathematica won't simplfy it.
In short the generalised conjecture is that the Mellin transform of this generalised hypergeometric function has a very simple result
$$
\mathcal{M}_x\left[\;_{(m-1)}F_{(m-2)}\left(\left\{\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}{m}\right\};\left\{\frac{2}{m-1},\cdots,\frac{m-2}{m-1},\frac{m}{m-1}\right\};-\frac{m^mx^{m-1}}{(m-1)^{m-1}}\right)\right](s) = \frac{\Gamma\left(1-\frac{m s}{m-1}\right)\Gamma\left(\frac{s}{m-1}\right)}{(m-1)\Gamma(2-s)} = \mathcal{I}_x[x+x^m](s)
$$
which only makes sense for integer $m$. Any way to prove this general statement? How do we simplify the identity that Mathematica couldn't?
| For the Gamma function identity, we use the multiplication formula for the Gamma function, choosing $n=3, z=\frac{2}{3}-\frac{s}{3}$:
\begin{align}
\Gamma\left(nz\right)&=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right)\\
\Gamma\left(2-s\right)&=\frac{1}{2\pi}3^{3/2-s}\Gamma\left( \frac{2}{3}-\frac{s}{3} \right)\Gamma\left(1-\frac{s}{3} \right)\Gamma\left( \frac{4}{3}-\frac{s}{3} \right)
\end{align}
Now, with the functional relation $ \Gamma\left( 1+u \right)=u\Gamma(-u) $ applied with $u=-s/3$ and $u=-4s/3$
\begin{equation}
\frac{\Gamma(1-\frac{4s}{3})\Gamma(\frac{s}{3})}{3 \Gamma(2-s)}=\frac{8\pi\Gamma(-\frac{4s}{3})\Gamma(\frac{s}{3})}{3^{5/2-s}\Gamma\left( \frac{2}{3}-\frac{s}{3} \right)\Gamma\left(-\frac{s}{3} \right)\Gamma\left( \frac{4}{3}-\frac{s}{3} \right)}
\end{equation}
as expected.
For the Mellin transform question, we evaluate the inverse transform of the function
\begin{equation}
f(s)=\frac{\Gamma(1-\frac{ms}{m-1})\Gamma(\frac{s}{m-1})}{ \Gamma(2-s)}
\end{equation}
assuming that the domain of analicity is $0<\Re (s) <\frac{m-1}{m}$. It can be noticed that the poles in the left complex half-plane are $s_p=-p(m-1)$, with $p=0,1,2\ldots$ Their residues are
\begin{equation}
r_p=\frac{(m-1)\Gamma(mp+1)}{\Gamma\left(2+p(m-1) \right)}\frac{(-1)^p}{p!}
\end{equation}
In the following we use the multiplication formula above with $z=1/n$,
\begin{equation}
\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)=(2\pi)^{(n-1)/2}n^{-1/2}
\end{equation}
to write
\begin{align}
\Gamma\left(np\right)&=(2\pi)^{(1-n)/2}n^{np-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(p+\frac{k}{n}\right)\\
&=(2\pi)^{(1-n)/2}n^{np-(1/2)}\Gamma(p)\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)\left( \frac{k}{n} \right)_p\\
&=n^{np-1}\Gamma(p)\prod_{k=1}^{n-1}\left( \frac{k}{n} \right)_p
\end{align}
Using the functional relation this residue can be converted into a ratio of Gamma functions:
\begin{align*}
r_p&=\frac{mp(m-1)}{p(m-1)\left( 1+p(m-1) \right)}\frac{\Gamma(mp)}{\Gamma\left(p(m-1) \right)}\frac{(-1)^p}{p!}\\
&=\frac{m}{\left( m-1 \right)\left(p+\frac{1}{m-1} \right)}
\frac{m^{mp-1}\Gamma(p)\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{(m-1)^{(m-1)p-1}\Gamma(p)\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\frac{(-1)^p}{p!}\\
&=\frac{\Gamma\left(p+\frac{1}{m-1} \right)}{\Gamma\left(p+1+\frac{1}{m-1} \right)}
\frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}\\
&=\frac{\left(\frac{1}{m-1} \right)_p\Gamma\left( \frac{1}{m-1} \right)}{\left(1+\frac{1}{m-1} \right)_p\Gamma\left(1+ \frac{1}{m-1} \right)}
\frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}\\
&=\left( m-1 \right)\frac{\left(\frac{1}{m-1} \right)_p}{\left(1+\frac{1}{m-1} \right)_p}
\frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}
\end{align*}
The inverse transform can thus be expressed as a hypergeometric series:
\begin{align*}
\mathcal{M}_x^{-1}\left[ \frac{1}{m-1}f(s)\right]&=\frac{1}{m-1}\sum_{p=0}^\infty r_px^{p(m-1)}\\
&=\;_{(m-1)}F_{(m-2)}\left(\left\{\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}{m}\right\};
\left\{\frac{2}{m-1},\cdots,\frac{m-2}{m-1},\frac{m}{m-1}\right\};-\frac{m^mx^{m-1}}{(m-1)^{m-1}}\right)\\
\end{align*}
which is the proposed identity.
| {
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"url": "https://math.stackexchange.com/questions/2808784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof an Inequality of a sequence $\frac{1}{2^1-1}+\frac{1}{2^2-1}+\cdots+\frac{1}{2^n-1}\lt \frac{5}{3}$ I have tried to amplify it like this
$$\frac{1}{2^1-1}+\frac{1}{2^2-1}+\cdots+\frac{1}{2^n-1}\lt 1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}} $$
which obviously too big. Any proper ways to amplify it ?
| Let $$ T = \sum_{n=1}^{\infty} \frac{1}{2^n-1}.$$
$$T < \frac{1}{1} + \frac{1}{3} + \sum_{n=3}^{\infty} \frac{1}{2^n-2} = \frac{4}{3} + (\frac{T}{2} - \frac{1}{2}). $$
Hence $ T < \frac{5}{3}$.
Based on what you were trying to do, I'm guessing you wanted to use induction by strengthening the induction hypothesis. If so, this is how you can proceed to solve the problem in a logical manner.
The goal is to prove a statement of the form
$$\sum_{i=1}^n \frac{1}{2^i-1} < \frac{5}{3} - f(n)$$
In order to apply induction easily, the induction step requires $\frac{1}{2^{n+1}-1} < f(n) - f(n+1) $.
Letting $f(n) = \frac{1}{2^n-2}$ satisfies that constraint. (You can determine this form by testing a few fractions. $\frac{1}{2^n}$ doesn't work.)
It remains to find a base case for which the statement is true. Note that it is not true for $n = 1$.
Note: The final 2 paragraphs motivate/explain the 2 "magic steps" in the first solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculation of $ \int\limits_{-\pi}^\pi \cos^4(x)dx$ using Fourier series of $\cos^2(x)$ The Fourier series of $\cos^2(x)={1\over2}+{1\over2}\cos(2x)$ is $a_0=1, a_2={1\over2}$ and the other terms are zero since $$Ff(x)={a_0\over2}+\sum\limits_{k=1}^{\infty}[a_k\cos(kx)+b_k\sin(kx)]$$
So in order to calculate $\int\limits_{-\pi}^\pi \cos^4(x)dx $ we use Parseval's identity :
$$\|f\|^2={1\over T}\int\limits_{-T/2}^{T/2}f^2(x)dx=a_0^2+{1\over2}\sum\limits_{k=1}^{\infty}a_k^2+b_k^2$$
But what I get is false:
$${1\over 2\pi}\int\limits_{-\pi}^\pi (\cos^2(x))^2dx=1+{1\over2}\Big({1\over2}\Big)^2={9\over 8}$$ $$\int\limits_{-\pi}^\pi \cos^4(x)dx=2\pi\cdot{9\over8}={9\pi\over4}$$
But it should be ${3\over4}\pi$
What am I doing wrong?
| It should be that
$$ \frac{1}{L}\int_{-L}^L f^2(x)\, dx = \frac{a_0^2}{2} + \sum_{n=1}^\infty \left(a_n^2 + b_n^2\right). $$
Hence,
$$ \frac{1}{\pi}\int_{-\pi}^\pi \cos^4(x)\, dx = \frac{a_0^2}{2} + a_1^2 = \frac{1}{2} + \left(\frac{1}{2}\right)^2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} $$
and the desired answer follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction? Can you provide the steps and corresponding explanations to prove the following predicate by induction?
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.
Thank you.
Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.
| Hint: For the left hand side inequality we have
$$ 1^3 + 2^3 +\ldots +n^3 < \frac{n^4}4 + n^3 < \frac{n^4}4 + \frac14 (4n^3 + 6n^2 + 4n + 1) = \frac{(n+1)^4}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime.
Answer:
We have to prove $\gcd(a^2+b^2,a^2b^2)=1$, $\gcd(a,b)=1$. Then $\gcd(a^2,b^2)=1$. Now we have $\gcd(a^2,a^2+b^2)=1$ and $\gcd(b^2,a^2+b^2)=1$ by $\gcd(a,b)=\gcd(p\pm q)$.
Then I am lost.
| Let $p$ be a common prime factor of $a^2 + b^2$ and $a^2 b^2$, and recall that for any integer $n$, $p | n \iff p | n^2$. Thus $p | a^2 b^2$ implies $p | ab$, so $p$ must divide both of $a^2 + b^2 \pm 2ab = (a\pm b)^2$; i.e., both $a+b$ and $a-b$ must be multiples of $p$. Thus, both $a$ and $b$ are multiples of $p/2$. If $p$ is odd, then $a$ and $b$ have $p$ as a common factor, so we're done. If $p = 2$, then $p | a^2 b^2$ forces at least one of $a$ and $b$ to be even; as $a^2 + b^2$ is even, both $a$ and $b$ must be even. QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the maximum value of $2x^2 - 3xy - 2y^2$ subject to the condition that $25x^2 - 20xy + 40y^2 = 36$
Let $x$ and $y$ be real numbers. Find the maximum value of $2x^2 - 3xy - 2y^2$ subject to the condition that $25x^2 - 20xy + 40y^2 = 36$.
So first I factored out $2x^2 - 3xy - 2y^2 = (x-2y)(2x+y)$, and because "maximum" kinda reminded me of the AM-GM, I went ahead to use the formula and got $(\frac{3x-y}{2})^2 \ge 2x^2 - 3xy - 2y^2.$ But that didn't seem to be helping me much so I moved $36$ to the LHS of the equation given so that $25x^2 - 20xy + 40y^2 -36 = 0$, which can also be written as $25x^2 - 20xy + 4(10y^2-9) = 0$ or $25x^2 - 20xy + 4(\sqrt{10}y + 3)(\sqrt{10}y - 3)=0$. The latter isn't quite nice because it's almost there but not quite — I don't think I can factor it unless $20xy$ were $20\sqrt{10}xy$, which isn't the case.
So at this point, I'd really like some help :)
| For a calculus-free solution, start with the factorization $2x^2 - 3xy - 2y^2 = (x-2y)(2x+y)$ and make the substitution $u = x-2y$, $v = 2x+y$. Then the condition $$25x^2 - 20xy + 40y^2 = 36$$ turns into $$9u^2 + 4v^2 = 36$$ so we are maximizing $uv$ subject to this condition.
Now we can apply AM-GM: $9u^2 + 4v^2 \ge 2 \sqrt{(9u^2)(4v^2)} = 12|uv|$, or to rearrange this a bit, $$uv \le \frac{9u^2+4v^2}{12} = \frac{36}{12} = 3.$$ This upper bound is achieved when $9u^2 = 4v^2$, the equality condition of AM-GM (and when $uv = |uv|$, because $uv \le |uv|$ was the other inequality we used). If we wanted to, we could solve this to find the values of $x$ and $y$ that give us the maximum value of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve Differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ with substitution I am trying to solve this differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ using substitution, with the initial value $y(1)=0$ This is what I've done so far:
$$
\begin{align*}
y'(t) &= \frac{t^2+3y(t)^2}{2ty(t)} = \frac{t^2}{2ty(t)} + \frac{3y(t)^2}{2ty(t)} \\
&= \frac{t}{2y(t)} + \frac{3y(t)}{2t} \\
&= \frac{1}{\frac{2y(t)}{t}} + \frac{3y(t)}{2t} \\
\\
\text{Substitution:}\\
f\Big(\frac{y(t)}{t}\Big) &= y'(t) \ \ \text{with} \ \ f(z(t)) = \frac{1}{2z(t)} + \frac{3}{2}z(t) \\
z(t)&= y(t) \ \ \text{and} \ \ z_0 = z(t_0) = z(1) = 0 \\
\\
z'(t) &= \frac{1}{t} \Big( f(z(t) - z(t)\Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{3z(t)}{2} - \frac{2z(t)}{2} \Big) \\
&= \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)}{2} \Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)^2}{2z(t)} \Big) \\
&= \frac{1}{t} \frac{1+z(t)^2}{2z(t)} \\
\\
\text{Seperation of variables:}\\
g(t) &= \frac{1}{t} \ ,\ \ h(z(t)) = \frac{1+z(t)^2}{2z(t)} \\
\\
\int_{t_0=1}^t \frac{1}{t} d\tau &= \Big[\ln(\tau)\Big]_1^t = ln(t) - 0 = ln(t) \\
\\
\int_{z_0=0}^{z(t)} \frac{1+x}{2x} dx &= \int_{0}^{z(t)} \frac{1}{2x} dx + \frac{x}{2x} dx = \int_{0}^{z(t)} \frac{1}{2x} dx + \int_{0}^{z(t)} \frac{1}{2} dx \\
&= \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{1}{2} \int_{0}^{z(t)} 1 \ dx = \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} \\
&= \frac{1}{2} \int_{\alpha}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} = \frac{1}{2} \Big[ \ln(x) \Big]_\alpha^{z(t)} + \frac{z(t)}{2} \ \ \ \Big( \alpha \in \big(0,z(t)\big)\Big) \\
&= \frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} \\
\\
\lim_{\alpha \rightarrow 0} \ &\frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} = \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2} \\
\\
\text{Now solve for z(t):} \\
ln(t) &= \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2}
\end{align*}
$$
This is the point where I don't know how to continue.
| Hint
$$2t\ y(t)\ y'(t)=t^2+3y(t)^2$$
It's a Bernouilli's equation.
You can solve it this way
$$t(y^2)'=t^2+3y^2$$
It becomes then a first order linear equation ($v=y^2$)
$$tv'-3v=t^2$$
That's easy to solve with integrating factor
| {
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"url": "https://math.stackexchange.com/questions/2821466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$?
Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$?
By letting $g(x)=x^3+3x^2+3x+5$, it is easy to see that $g'(-1) = g''(-1) = 0$, and so $f'(-1) = f''(-1) = f'''(-1) = 0$.
However, I have no idea how to compute $f^{(n)}(-1)$ for any positive integer $n$, and in particular for $n=99$.
Any idea would be appreciated.
| Note that in a neighbourhood of $x=-1$,
$$f(x)=\frac{1}{x^3+3x^2+3x+5}=\frac{1/4}{1+\frac{(x+1)^3}{4}}=\sum_{n=0}^{\infty}\frac{f^{(n)}(-1)}{n!}(x+1)^n.$$
Now recall that for $|t|<1$,
$$\frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^n t^n.$$
Can you take it form here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric integral with cosinus I cannot solve the equation below:
I know what I typed is wrong byt I can't understand where it went sour.
$$\int \frac{1}{\cos(5x)} dx$$
$$\left[
\begin{align*}
5x=s\\
x=\frac{s}{5};
\frac{dx}{ds}=\frac{1}5
\end{align*}
\right]
$$
$$
\frac{1}{5}\int \frac{1}{\cos (s)}\frac{\cos (s)}{\cos (s)}\,ds=
\frac{1}{5}\int \frac{\cos (s)}{\cos^2(s)}\,ds=\frac{1}{5}\int \frac{\cos (s)}{1-\sin^2(s)}\,ds
$$
$$\left[
\begin{align*}
\sin(s)=y\\
\frac{dy}{ds}=-\cos(s)
\end{align*}
\right]
$$
$$
\frac{1}{5} \cdot -\frac{1}{2}\int -\frac{1}{1-y}-\frac{1}{1+y} dy
$$
$$
\left[ -(-\ln(1-y) - \ln(1+y)\right]
=
\left[ \ln\frac{(1-y)}{(1+y)}\right]
=
\left[ \ln\frac{(1-\sin(s))}{(1+\sin(s))}\right]
=
\left[ \ln\frac{(1-\sin(5x))}{(1+\sin(5x))}\right]
$$
| Hint. Note that taking the derivative of
$$\ln\left(\frac{1-\sin(5x)}{1+\sin(5x)}\right)=\ln(\underbrace{1-\sin(5x)}_{>0})-\ln(\underbrace{1+\sin(5x)}_{>0})$$
we get
$$\frac{-5\cos(5x)}{1-\sin(5x)}-\frac{5\cos(5x)}{1+\sin(5x)}=
-\frac{10\cos(5x)}{1-\sin^2(5x)}=-\frac{10}{\cos(5x)}.$$
So you are not so far from the correct answer: you just forgot to consider the constant $\frac{1}{5} \cdot -\frac{1}{2}$ in the last line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
| Let $I(a)=\int_0^1 \frac{\ln(a x+\sqrt{1-x^2})}{x}dx$. Then, $I(0) = \int_0^1 \frac{\ln\sqrt{1-x^2}}{x}dx
\overset{x^2\to x}
=-\frac{\pi^2}{24}$
$$I’(a)=\int_0^1 \frac{dx}{a x+\sqrt{1-x^2}}
=\frac1{1+a^2}\left(\frac\pi2+a\ln a \right)$$
and
$$\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx
=I(1)= I(0)+\int_0^1I’(a)da \\
\hspace{20mm}= -\frac{\pi^2}{24}+ \frac\pi2\int_0^1 \frac{da}{1+a^2}
+ \int_0^1 \frac{a\ln a}{1+a^2}da= \frac{\pi^2}{16}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Pull constant out of summation In the solution of my homework there's this step that I don't understand:
$$\begin{align}\mathsf {Var}[X] & = \sum_{k=1}^{n}\dfrac{n(k-1)}{(n-k+1)^2}\\ &= n \sum_{k=1}^{n}\dfrac{(n-k)}{k^2}\\&= n^2 \sum_{k=1}^{n}\dfrac{1}{k^2}-n \sum_{k=1}^{^n}\dfrac{1}{k}\end{align}$$
I would have thought, that when I pull $n$ out of the $\sum$ that I'd just get:
$n\sum_{k=1}^n\frac{(k-1)}{(n-k+1)^2}$.
Also I don't understand the second line. Could someone please tell me what steps are missing and how they solved this? Thanks!
| Alternatively, note that the summation is reversed:
$$\begin{align}\color{red}{\mathsf {Var}[X]} & \color{red}{= \sum_{k=1}^{n}\dfrac{n(k-1)}{(n-k+1)^2}}\\
&=n\left(\frac{1-1}{n^2}+\frac{2-1}{(n-1)^2}+\frac{3-1}{(n-2)^2}+\cdots+\frac{n-2-1}{3^2}+\frac{n-1-1}{2^2}+\frac{n-1}{1^2}\right) \\
&=n\left(\frac{n-1}{1^2}+\frac{n-2}{2^2}+\frac{n-3}{3^2}+\cdots+\frac{2}{(n-2)^2}+\frac{1}{(n-1)^2}+\frac{0}{n^2}\right) \\
&\color{red}{=n \sum_{k=1}^{n}\dfrac{(n-k)}{k^2}}\\
&=n \sum_{k=1}^{n}\left(\dfrac{n}{k^2}-\frac{k}{k^2}\right)\\
&\color{red}{= n^2 \sum_{k=1}^{n}\dfrac{1}{k^2}-n \sum_{k=1}^{^n}\dfrac{1}{k}}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829571",
"timestamp": "2023-03-29T00:00:00",
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How to calculate the integral $\int_{0.5}^{1} \frac{1}{\sqrt{2x-x^2}}$? I have the following integration problem:
$$ \int_{0.5}^{1} \frac{1}{\sqrt{2x-x^2}} $$
And I can see I should probably be completing the square here. I may be missing something extremely obvious, but wouldn't this mean I have to take the negative outside the expression to do this? I need the $x^2$ term by itself, don't I?
Any help would be greatly appreciated!
| There's no hyperbolic function here.
Set
$$
t=\sqrt{\frac{x}{2-x}}
$$
so $(2-x)t^2=x$ and
$$
x=\frac{2t^2}{1+t^2}=\frac{2t^2+2-2}{1+t^2}=2-\frac{2}{1+t^2}
$$
whence
$$
dx=\frac{4t}{(1+t^2)^2}
$$
and the integral becomes
\begin{align}
\int_{1/2}^1\frac{1}{2x-x^2}\,dx
&=\int_{1/2}^1\frac{1}{x}\sqrt{\frac{x}{2-x}}\,dx \\[4px]
&=\int_{1/\sqrt{3}}^{1}\frac{1+t^2}{2t^2}t\frac{4t}{(1+t^2)^2}\,dt \\[4px]
&=2\int_{1/\sqrt{3}}^1\frac{1}{1+t^2}\,dt \\[4px]
&=2\Bigl[\arctan t\Bigr]_{1/\sqrt{3}}^1 \\[4px]
&=2\frac{\pi}{4}-2\frac{\pi}{6} \\[4px]
&=\frac{\pi}{6}
\end{align}
| {
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For any $n \in \mathbb{N}$, show that: $\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} < \frac{5}{6}$. For any $n \in \mathbb{N}$, show that:
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} < \frac{5}{6}$$
I wrote the sum as $H_{2n} - H_{n}$, where $H_{k} = \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{k}$ (the kth harmonic number). After that, I was searching for inequalities with harmonic numbers, but I didn't find anything useful.
Can you, please, give me a hint? I don't want the full proof. Thank you!
| We need to prove that
$$\frac{1}{n+1}-\frac{1}{n}+\frac{1}{n+2}-\frac{1}{n}+...+\frac{1}{n+n}-\frac{1}{n}<\frac{5}{6}-1$$ or
$$\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{n}{n+n}>\frac{n}{6}.$$
Now, by C-S
$$\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{n}{n+n}=$$
$$=\frac{1^2}{1(n+1)}+\frac{2^2}{2(n+2)}+...+\frac{n^2}{n(n+n)}\geq$$
$$\geq\frac{(1+2+...+n)^2}{1(n+1)+2(n+2)+...+n(n+n)}=$$
$$=\frac{\frac{n^2(n+1)^2}{4}}{n(1+2+...+n)+1^2+2^2+...+n^2}=$$
$$=\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=\frac{3n(n+1)}{2(5n+1)}>\frac{n}{6}.$$
| {
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Calculating a flux integral
Let $$F=(xe^{xy}-2xz+2xy\cos^2 z, y^2\sin^2 z-y e^{xy}+y, x^2+y^2+z^2)$$ and $V$ be the solid in space bounded by $z=9-x^2-y^2$ and $z=0$. I am trying to compute the flux integral $\iint_{\partial V}F\cdot n \ dS$, $n$ being the outward unit normal.
Setting $r(x,y)=(x,y,9-x^2-y^2)$, I found that $r_x\times r_v=(2x,2y,1)$ and $$\iint_{\partial V}F\cdot n \ dS=\iint_D F\cdot r_x\times r_y \ dA$$ where $$F\cdot r_x\times r_y=2e^{xy}(x^2-y^2)-36x^2+4x^4+4x^2y^2+4x^2y\cos^2(9-x^2-y^2)+2y^3\sin^2(9-x^2-y^2)+2y^2;$$ after the substitution $x=r\cos t, \ y=r\sin t$ I get $$f(r,t)=r(F\cdot r_x\times r)=2r^3e^{r^2\sin t \cos t}(\cos^2 t-\sin^2 t)-36r^3\cos^2 4r^5\cos^2 t + \\4r^4\cos^2 t\sin t \cos^2{(9-r^2)}+2r^4\sin^3 t \sin^2(9-r^2)+2r^3\sin^2 t $$ and I need to compute $$\int_0^{2\pi}\int_0^3 f(r,t)drdt$$
I wonder whether I can further simplify $f(r,t)$? The current expression looks to cumbersome and it seems like a hassle to compute the integral if no simplifications can be made.
| Why aren't you using the divergence theorem? Compute $$\begin{align} {\rm div}\,F(x,y,z) &= \partial_x(xe^{xy}-2xz+2xy\cos^2 z)+\partial_y(y^2\sin^2 z-y e^{xy}+y)+\partial_z(x^2+y^2+z^2) \\ &= \require{cancel} \cancel{e^{xy}} + \cancel{xye^{xy}} -
\cancel{2z} + 2y\cos^2z + 2y\sin^2 z - \cancel{e^{xy}} - \cancel{xye^{xy}}+1+\cancel{2z} \\ &= 2y + 1.\end{align}$$Then $$\begin{align} \iint_{\partial V} F\cdot n\,{\rm d}S &= \iiint_V {\rm div}\,F(x,y,z)\,{\rm d}x\,{\rm d}y\,{\rm d}z \\ &= \iiint_V 1+2y \,{\rm d}x\,{\rm d}y\,{\rm d}z \\ &= \int_0^{2\pi}\int_0^3\int_0^{9 - r^2} (1+2r\sin \theta)r\,{\rm d}z\,{\rm d}r\,{\rm d}\theta, \end{align}$$which is easy. Can you finish?
| {
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} |
Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$ I found in this link the reduction formula
$$
\int_0^{\pi/2} \sin^m x \cos^n x \ dx =
\begin{cases}
\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2} (\frac{\pi}{2}) & m, n \text{ even}\\[5pt]
\frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)} & \text{otherwise}
\end{cases}
$$
How can I prove it?
| Let $I_{m,n}=\int_0^{\pi/2} \sin^m x \cos^n x \ dx$, integrating by parts we find that
$$
I_{m,n}=\frac{n-1}{m+1} I_{m+2,n-2} \qquad (1) \\
I_{m,n}=\frac{m-1}{n+1} I_{m-2,n+2} \qquad (2)
$$
Using (1) when $n$ is odd,
$$
I_{m,n}=\frac{(n-1)(n-3) \cdots 2}{(m+n-2) \cdots (m+1)}I_{m+n-1,1}=\frac{(n-1)(n-3) \cdots 2}{(m+n)(m+n-2) \cdots (m+1)} \quad (3)
$$
Interchaging $m$ and $n$ in (3) we find $I_{m,n}$ when $m$ is odd.
Then
$$
I_{m,n} = \frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)}
$$
when either $m$ or $n$ are odd.
If $n$ and $m$ are even
$$
I_{m,n}=\frac{(n-1)(n-3) \cdots 1}{(m+n-1) \cdots (m+1)}I_{m+n,0}
$$
On the other hand, let $J_k=\int_0^{\pi/2} \sin^k x \ dx$. Again,integrating by parts we find:
$$
J_k=\frac{k-1}{k}J_{k-2}
$$
In particular when $k$ is even
$$
J_k=\frac{(k-1)(k-3) \cdots 1}{k(k-2) \cdots 2}\left(\frac{\pi}{2}\right)
$$
Then
$$
\begin{align}
I_{m,n}&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n-1)(m+n-3) \cdots 1} \left[\frac{(m+n-1)(m+n-3) \cdots 1}{(m+n)(m+n-2) \cdots 2} \left(\frac{\pi}{2}\right)\right]\\
&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2}\left(\frac{\pi}{2}\right)
\end{align}
$$
| {
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} |
$n,m$ are natural positive integers. F$(n,1)$=F$(1,n)=1$ F$(n,m+1)$+F$(n+1,m)=$F$(n+1,m+1)$ Write F($2,n$) and F($3,n$) as a function of $n$ How do I solve this problem? I do not know where to begin nor does anyone close to me. I am lost and I need help. Thank you so much for whoever helps.
| Picture a quarter-plane rectangular grid in $n$ and $m$. Then the equations
$$
F(1,n) = F(n,1) = 1
$$
place the value $1$ along the edges of that grid. Now look at the equation
$$
F(n+1,m+1) = F(n+1,m) + F(n, m+1)
$$
Whenever we have an empty square with filled in values below and to the immediate left, we can now fill in that square using said equation. Fo for example
$$F(2,2) = F(2,1) + F(1,2) = 1+1 = 2\\
F(3,2) = F(3,1) + F(2,2) = 1+2 = 3 \\ \vdots \\
F(n,2) = F(n,1) + F(n-1,2) = 1 + n-1 = n
$$
And once you have filled in that row,
$$F(2,3) = F(2,2) + F(1,3) = 2+1 = 3\\
F(3,3) = F(3,2) + F(2,3) = 3+3 = 6 \\
F(4,3) = F(4,2) + F(3,3) = 6+6 = 10 \\ \vdots \\
F(n,3) = F(n,2) + F(n-1,3) = n + \frac{n(n-1)}{2} = \frac{(n+1)n}{2}
$$
| {
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Find the volume common to sphere $x^2+y^2+z^2$ and the cylinder $x^2+y^2Find the volume common to sphere $x^2+y^2+z^2<1$ and the cylinder $x^2+y^2<ax$
I set up the following integral :
$$I=2\cdot \iiint_{z=0}^{\sqrt{a^2-x^2-y^2}} dz\,dy\,dz = 2\cdot\iint_E\sqrt{a^2-x^2-y^2} \, dy\,dx$$ where $E:x^2+y^2=ax$
Now under polar coordinates
$E:r=a\cos(\theta)$
and so $0\leq r \leq a\cos(\theta)$ and $-\pi/2\leq \theta \leq \pi/2$
$$I= 2\cdot\int_{\theta=-\pi/2}^{\pi/2}\,\int_{r=0}^{a\cos(\theta)} \sqrt{a^2-r^2}\,r\,dr\,d\theta=2\cdot (1/2)\cdot(2/3)\int_{\theta=-\pi/2}^{\pi/2}a^3\cdot(1-\sin^3(\theta))d\theta = \frac{2a^3}{3}\cdot\{\int_{\theta=-\pi/2}^{\pi/2} d\theta - \int_{\theta=-\pi/2}^{\pi/2}\sin^3(\theta)d\theta\} = \frac{2\pi a^3}{3}$$
However answer given to me is $$\frac{2a^3}{3}\cdot(\pi - \frac43)$$
Where am I making the mistake? Or is it the case that the answer given to me is incorrect?
| Assuming $a<1/2$, here's the proper integral:
$$\int\limits_{x=0}^a \int\limits_{y=0}^\sqrt{x(a - x)} \int\limits_{z=0}^\sqrt{1 - x^2 - y^2} 4\ dx\ dy\ dz$$
| {
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Proving $ 2\arcsin\left(\sinh(x)\right) + \arccos\left( 2 - \cosh(2x) \right) = 0 $
$$ 2\arcsin\big( \sinh(x) \big) + \arccos\big( 2 - \cosh(2x) \big) = 0 $$
If this is true, then for what values of $x$? (Clearly, the $\arcsin(\cdot)$ and $\arccos(\cdot)$ are not valid for every real argument.)
| Let
$$
L(x) = 2\arcsin(\sinh(x)) \\
R(x) = -\arccos(2-2\cosh(x))
$$
I'll show that these functions are equal for $x \ge 0$ when $|\sinh(x)| \le 1$.
Step 1: check that $L(0) = R(0)$. Yes.
Step 2: Check that $L'(x) = R'(x)$ for all $x$.
Start from the following: for $0 \le u \le 1$, we have
$$
\sqrt{4u^2 - 4u^4} = 2u \sqrt{1-u^2},
$$
because $u > 0$, so we don't need the absolute value sign on the right (although including it would probably deal with the case where $x < 0$ later, but I leave that to OP).
Then
\begin{align}
\sqrt{1 - (1 - 4u^2 + 4u^4)} &= 2u \sqrt{1-u^2} \\
\sqrt{1 - (1 - 2u^2)^2} &= 2u \sqrt{1-u^2}.
\end{align}
If we replace $u$ with $\sinh(x)$, we get
\begin{align}
\sqrt{1 - (1 - 2\sinh^2(x))^2} &= 2\sinh(x) \sqrt{1-\sinh^2(x)}\\
\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\sinh(x) \sqrt{1-\sinh^2(x)}\\
\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\cosh(x)\sinh(x) \sqrt{1-\sinh^2(x)}\\
\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= \sinh(2x) \sqrt{1-\sinh^2(x)}\\
2\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\sinh(2x) \sqrt{1-\sinh^2(x)}\\
\frac{2\cosh(x)}{\sqrt{1-\sinh^2(x)}} &= \frac{2\sinh(2x)}{\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2}}\\
\frac{2\cosh(x)}{\sqrt{1-\sinh^2(x)}} &= \frac{2\sinh(2x)}{\sqrt{1 - (2 - \cosh(2x))^2}}\\
\end{align}
The left hand side of this is just $L'(x)$; the right hand side is $R'(x)$. Because they're equal, and $L(0) = R(0)$, we have that $L(x) = R(x)$ within the specified domain (by the FTC, if you want to get fancy).
In short: the OP's statement, with a sign-fix, is correct for positive $x$; a similar proof, without the sign-change, presumably demonstrates the original claim for negative $x$. The whole thing is an exercise in algebra and the FTC.
| {
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Help on evaluating $\int_{0}^{\infty}\frac{\cos\frac2x}{(x^2+4)^2}dx$ I try to integrate $I=\int_{0}^{\infty}\frac{\sin^2\frac1x}{(x^2+4)^2}dx$.
Using identity $2\sin^2x=1-\cos(2x)$
$$I=\frac{1}{2}\int_{0}^{\infty}\frac{dx}{(x^2+4)^2}-\frac{1}{2}\int_{0}^{\infty}\frac{\cos\frac2x}{(x^2+4)^2}dx=\frac{1}{2}(I_1-I_2)$$
Using the reduction formula for
$I_1=\frac{1}{8}\int_{0}^{\infty}\frac{dx}{x^2+4}=\frac{\pi}{32}$. But, I am not sure how to evaluate
$$I_2=\int_{0}^{\infty}\frac{\cos\frac2x}{(x^2+4)^2}dx$$
Let $u=2x^{-1}$
$$I_2=\frac{1}{2}\int_{0}^{\infty}\cos(u)\cdot \frac{du}{(1+u^2)^2}=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\int_{0}^{\infty}\frac{u^{2n}}{(1+u^2)^2}du$$
Applying the reduction formula to $$\int_{0}^{\infty}\frac{u^{2n}}{(1+u^2)^2}du=0$$
Did I evaluate $I_2$ correctly?
| I see two problematic points in your calculation of $I_2$.
When carrying out the substitution carefully I get:
*
*$u = \frac{2}{x} \Rightarrow \frac{du}{dx} =-\frac{2}{x^2}=-\frac{u^2}{2}$
$$I_2=\int_{0}^{\infty}\cos(2/x)\cdot \frac{dx}{(x^2+4)^2}=\int_{\infty}^{0} \frac{\cos(u)}{(\frac{4}{u^2} + 4)^2} \frac{-2du}{u^2}=\frac{1}{8}\int_{0}^{\infty} \frac{u^2\cos(u)}{(u^2 + 1)^2} du$$
Besides this you write "$\int_{0}^{\infty}\frac{u^{2n}}{(1+u^2)^2}du \color{red}{=} 0$", which cannot be, because the integrand is strictly positive.
Another possible way to show that $\color{blue}{I_2=0}$ is to use residues:
*
*Consider $\int_{0}^{\infty} \frac{u^2\cos(u)}{(u^2 + 1)^2} du = \frac{1}{2}\int_{-\infty}^{\infty} \frac{u^2\cos(u)}{(u^2 + 1)^2} du$
*$\int_{-\infty}^{\infty} \frac{u^2\cos(u)}{(u^2 + 1)^2} du = \Re \left(2\pi i Res_i \left( \frac{z^2}{(z^2 + 1)^2}e^{iz}\right) \right) = \Re \left(2\pi i Res_i \left( \frac{z^2}{(z + i)^2(z-i)^2}e^{iz}\right) \right)$
$$\color{blue}{Res_i} \left( \frac{z^2}{(z + i)^2(z-i)^2}e^{iz}\right) = \left. \left( \frac{2z(z+i)^2-z^2\cdot 2(z+i)}{(z+i)^4}e^{iz} + \frac{z^2}{(z+i)^2}\cdot ie^{iz}\right) \right|_{z=i} = \left( -\frac{i}{4} + \frac{i}{4} \right)e^{-1} \color{blue}{ = 0}$$
| {
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How to show in $R^2$, the $f(x,y)=(x^4+y^4)^{1/4}$ defines a norm on $R^2$. How to show in $R^2$, the $f(x,y)=(x^4+y^4)^{1/4}$ defines a norm on $R^2$, i.e., how to prove the inequality:
$$((x_1+x_2)^4+(y_1+y_2)^4)^{1/4}\le (x_1^4+y_1^4)^{1/4}+(x_2^4+y_2^4)^{1/4}$$
| There is a nice proof of this Minkowski by Holder:
Let $x_1=a$, $y_1=b$, $x_2=c$ and $y_2=d$.
Thus, we need to prove that
$$\sqrt[4]{a^4+b^4}+\sqrt[4]{c^4+d^4}\geq\sqrt[4]{(a+b)^4+(c+d)^4}.$$
Indeed, by Holder we obtain:
$$\left(\sqrt[4]{a^4+b^4}+\sqrt[4]{c^4+d^4}\right)^4=$$
$$=a^4+b^4+4\sqrt[4]{(a^4+b^4)^3(c^4+d^4)}+6\sqrt{(a^4+b^4)(c^4+d^4)}+$$
$$+4\sqrt[4]{(a^4+b^4)(c^4+d^4)^3}+c^4+d^4\geq$$
$$\geq a^4+b^4+4(a^3c+b^3d)+6(a^2c^2+b^2d^2)+$$
$$+4(ac^3+bd^3)+c^4+d^4=(a+c)^4+(b+d)^4.$$
| {
"language": "en",
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"source": "stackexchange",
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Find a real matrix $B$ such that $B^3 = A$
Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$
I think I am doing something wrong here, so let me describe my attempt:
1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambda_1 = -8$ and $\lambda_2 = 1$ and the corresponding eigenvectors $ \vec v_1 = \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix} = x + y = 0 \Rightarrow -x = y \Rightarrow \begin{bmatrix}1\\-1\end{bmatrix}$ and $ \vec v_2 = \begin{bmatrix}1 & -\frac{1}{2}\\0 & 0\end{bmatrix} = x - \frac{1}{2}y = 0 \Rightarrow 2x = y \Rightarrow \begin{bmatrix}1\\2\end{bmatrix}$
2) With that being done I proceeded with computing $D = \begin{bmatrix}-8 & 0\\0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$ and check everything with $D = PAP^{-1}$
3) Now I thought I will simple find a diagonal matrix $M = PBP^{-1}$ and $M^3 = D$ and the easiest solution I came up with was $M = \begin{bmatrix}\sqrt[3]{-8} & 0\\0& 1\end{bmatrix}$ so basically $M = D^{\frac{1}{3}}.$ So that $B = PMP^{-1}$. But now come the tricky part, if I compute $B$ it results in a complex matrix not a real. //It is real!
Have I perhaps overlooked something here or miscalculated the solution for $B$?
Edit:
As Cameron pointed out my calculator and I totally failed as it was in complex mode and computed one of the non-real cube roots instead of -2. So $M = \begin{bmatrix}-2 & 0\\0 & 1\end{bmatrix}$ and consequentially $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$
| The characteristic polynomial of $A$ is
\begin{align}
p(\lambda)&= (-5-\lambda)(-2-\lambda)-18\\
&= \lambda^2+7\lambda-8 \\
&= (\lambda +8)(\lambda-1).
\end{align}
So $(A+8I)(A-I)=0$. That means
\begin{align}
A(A+8I)&=(A+8I) \\
A(A-I) &= -8(A-I) \\
I &= \frac{1}{9}((A+8I)-(A-I)) \\
A^{1/3}&=\frac{1}{9}((A+8I)+2(A-I)) \\
&= \frac{1}{9}(3A+6I) = \frac{1}{3}(A+2I) \\
&= \frac{1}{3}\begin{pmatrix}-3 & 3 \\ 6 & 0\end{pmatrix}
= \begin{pmatrix}-1 & 1 \\ 2 & 0 \end{pmatrix}.
\end{align}
| {
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Evaluating $\int_0^1\frac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$
$$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$$
Attempt:
If we write: $f(x)= x^4 + x^3+ x^2$, we get:
$$I = \displaystyle\int_0^1 \dfrac{3f(x)+x^3}{(f'(x)+1)^2}\, dx$$
I have no idea how to proceed. Integration by parts/ substitution can't help.
I don't need the full solution. Just a guiding hint would suffice.
| Since this integral is very closely linked to the Quotient Rule for differentiation (due to the square in the denominator), we will try to write the numerator as an expression of the form $$P'(x)\cdot(4x^3+3x^2+2x+1)-P(x)\cdot(4x^3+3x^2+2x+1)'$$ so $$3x^4+4x^3+3x^2=P'(x)\cdot(4x^3+3x^2+2x+1)-2P(x)\cdot(6x^2+3x+1).$$
We see that $P$ must be at least a quadratic, so $P(x)=ax^2+bx+c$ for some real numbers $a,b,c$. Then $$\begin{align}3x^4+4x^3+3x^2&=(2ax+b)\cdot(4x^3+3x^2+2x+1)-2(ax^2+bx+c)(6x^2+3x+1)\\&=8ax^4+(6a+4b)x^3+(4a+3b)x^2+(2a+b)x+b-12ax^4-(6a+12b)x^3\\&\,\,\,\,\,\,-(2a+6b+12c)x^2-(2b+6c)x-2c\\&=-4ax^4-8bx^3+(2a-3b-9c)x^2+(2a-b-6c)x+b-2c\end{align}$$ Hence $$a=-\frac34,\quad b=-\frac12,\quad c=-\frac14$$ This means that $$P(x)=-\frac14(3x^2+2x+1)$$ so $$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx=-\frac14\left[\frac{3x^2+2x+1}{4x^3 + 3x^2 + 2x+ 1}\right]_0^1=\frac1{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Find all $z$ for which $z^2+2z+2$ is real positive. Find all $z$ for which $z^2+2z+2$ is real positive.
My Attempt :
Let $z=x+iy, (x,y\in R)$:
$$z^2+2z+2= (x^2-y^2+2ixy) +2(x+iy)+2$$
$$=(x^2-y^2+2x+2)+i(2xy+2y)$$
| You want $2xy+2y = 2y(x+1)=0$ and $x^2-y^2+2x+2 = (x+1)^2-y^2+1>0$. From the first equation, $y=0$ or $x=-1$; if $y=0$, you want $(x+1)^2+1>0$, which is always true; if $x=-1$, you want $-y^2+1>0$, which is true when $|y|<1$. So the solutions are
\begin{equation*}
\{x+iy\,|\,y=0\text{ or }(x=-1\text{ and }|y|<1)\}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| $\begin{align}
(\sin^2x+\cos^2x)^2-2\sin^2(2x) & = \sin^4x+\cos^4x + 2\sin^2x\cos^2x-2\sin^2(2x) \\& =
\sin^4x+\cos^4x + 2\sin^2x\cos^2x-2(2\sin x \cos x)^2
\\& =
\sin^4x+\cos^4x -6\sin^2x\cos^2x
\\& \neq
\sin^4x+\cos^4x
\end{align} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Converting cartesian rectangular equation to it's corresponding polar equation I try to find a manner to convert cartesian rectangular equation like $y = f(x)$ to it's polar equation $r = r(\theta)$.
Given the function $f(x) = y = 5x^4$, lets try to get the polar equation:
$$x = r\cdot \cos(\theta), \quad y = r\cdot \sin(\theta)$$
So:
$$r\cdot \sin(\theta) = 5\cdot r^4 \cdot \cos^4(\theta)$$
$$ \Rightarrow r^3= \frac{\sin(\theta)}{5\cdot \cos^4(\theta)} $$
$$ \Rightarrow r(\theta) = \sqrt[3] {\frac {\sin(\theta)}{5\cdot \cos^4(\theta)}}$$
My question is, That's really the correct answer? if so, how we can convert it back to rectangular equation? I tried to convert the polar equation we received to a rectangular equation using the following equations
$$\theta = \arctan\Bigr(\frac yx\Bigr), \quad r = \sqrt{x^2+y^2}$$
The result I got was everything except $y = 5x^4$.
Can someone point me the right direction to get the correct solution?
Thanks for Help!!
| Using the inverse relations, the Cartesian equation reads
$$\sqrt{x^2+y^2}=\sqrt[3]{\frac{\sin\left(\arctan\dfrac yx\right)}{5\cos^4\left(\arctan\dfrac yx\right)}}=\sqrt[3]{\frac{\tan\left(\arctan\dfrac yx\right)}{5\cos^3\left(\arctan\dfrac yx\right)}}.$$
Now notice that $$\frac1{\cos^2\theta}=\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta}=\tan^2\theta+1$$ and
$$\frac1{\cos^3\theta}=(\tan^2\theta+1)^{3/2}.$$
Substituting, we have
$$\sqrt{x^2+y^2}=\sqrt[3]{\frac{\tan\left(\arctan\dfrac yx\right)}{5\cos^3\left(\arctan\dfrac yx\right)}}
=\sqrt[3]{\dfrac y{5x}\left(\tan^2\left(\arctan\dfrac yx\right)+1\right)^{3/2}}
=\sqrt[3]{\dfrac y{5x}\left(\dfrac{y^2}{x^2}+1\right)^{3/2}}=\sqrt[3]{\dfrac y{5x^4}}\sqrt{x^2+y^2}.$$
A final simplification yields
$$\frac y{5x^4}=1.$$
Easy as that :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I prove $\tan15^\circ=2-\sqrt3$ using the triple-angle formulas?
Prove $\tan15^\circ=2-\sqrt3\,\,$ using following results $$\sin3x=3\sin{x}-4\sin^3x$$
$$\cos3x=4\cos^3x-3\cos{x}$$
I know that if I substitute $x=15^\circ\,$I can write$$\sin45=3\sin15-4\sin^315$$
$$\cos45=4\cos^315-3\cos15$$ What can I do next? any hints?
| Solution
For convenience, denote $\sin 15^o=x, ~~\cos 15^o=y.$ Obviously, $0<x<y$ and $x^2+y^2=1$.
Thus, we obtain $$3x-4x^3=\frac{\sqrt{2}}{2},\tag1$$ and $$4y^3-3y=\frac{\sqrt{2}}{2}. \tag2$$
By $(1)-(2)$, we have $$3x+3y-4x^3-4y^3=(x+y)[3-4(x^2-xy+y^2)]=0.\tag3$$
Thus, $$x^2-xy+y^2=\frac{3}{4}.\tag4$$
Hence $$xy=\frac{1}{4}.\tag5$$
Further,$$(x+y)^2=x^2+y^2+2xy=1+2\cdot \frac{1}{4}=\frac{3}{2}.\tag6$$
Hence,$$x+y=\frac{\sqrt{6}}{2}.\tag7$$
From $(5),(7)$, we may see that $x, y$ are the roots for the quadratic equation $$t^2-\frac{\sqrt{6}}{2}t+\frac{1}{4}=0.\tag8$$
By solving it, we have $$x=\frac{\sqrt{6}-\sqrt{2}}{4},~~y=\frac{\sqrt{6}+\sqrt{2}}{4}.$$
As a result, $$\tan 15^o=\frac{x}{y}=2-\sqrt{3}.$$
Note
Now,you must have seen that the algebraic method is too complicated. Indeed, there exists an elegant geometrical solution without much calculation and even without a word. Can you get the same result from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$ Show that $ f(x)=x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$
I tried to simplify it by putting $x^5=y$
It simplifies the polynomial but I cannot put it in the case of the divisor.
So I assumed that $x^2+1$ is a divisor of $f(x)$
Then examine the assumption is correct then $x^2+1=(x-i)(x+i)$
Now for $x=\pm i$ using synthetic division it leaves remainder 0
Am I at the right direction, please tell
| Now, even if we don't have any insight we can always resort to long division:
$(x^2 + 1)| \begin{cases}x^{18} & & - x^{16} &&+ x^{14} &... & & \\
x^{20} & & & && x^{15} &&&&& x^{10}&&&&& x^5\\
x^{20} & & x^{18}\\
--&--&--&--&--&--\\
&& -x^{18}\\
&& -x^{18} && - x^{16}&\\
--&--&--&--&--&--\\
&&&&x^{16}&+x^{15}\\
...\\
\end{cases}$
We'll call that plan B.
(Which will give us $(x^2 + 1)(x^{18}-x^{16} + x^{14} + x^{13} - x^{12} - x^{11} + x^{10} + x^{9} -x^7 + x^5)$ but it's long and tedious and not very insightful. And wouldn't be practical for proving something like $x^2 + 1$ divides $x^{100} + x^{75} + x^{50} + x^{25}$ which ... I think it does.)
....
But if we try to be insightful:
The way I see it there are two things to realize.
1) $a^3 + a^2 + a + 1 =$
$(a^3 + a^2) + (a+1)=$
$a^2(a+1) + (a+1) = $
$(a^2 + 1)(a+1)$.
To make this more general (although really hard to read:)
$a^{nk - 1} + a^{nk - 2} + .... + a + 1= $
$(a^{nk-1} + .... + a^{(n-1)k}) + (a^{(n-1)k -1} + ..... + a^{(n-2)k}) + ..... + (a^{k-1} + ... + 1) =$
$a^{(n-1)k}(a^{k-1} + .... + 1) + a^{(n-2)k}(a^{k-1} + .... + 1) + ... + a^k(a^{k-1} + ..... + 1) + (a^{k-1} + .... + 1) = $
$(a^{(n-1)k} + a^{(n-2)k} + .... 1)(a^{k-1} + a^{k-2} + ... + 1)$.
... Yeesh ...
but that means
$x^{20} + x^{15} + x^{10} + x^5 = $
$x^5(x^{15} + x^{10} + x^5 + 1) = $
$x^5(x^{10} + 1)(x^5 + 1)$.
... and factoring the $(x^{10} + 1)$ looks like it has potential.
2) $(a -1)(a^{k-1} + a^{k-2} + .... + 1) = a^k -1$
is well known. But we can "flip signs" depending on parity:
$(a+1)(a^{k-1} - a^{k-2} + a^{k-3} - ... + a - 1) = a^k - 1$ if $k$ is even
$(a+1) (a^{k-1} - a^{k-2} + a^{k-3} - ... -a + 1)= a^k + 1$ if $k$ is odd.
So
$(x^{10} + 1) = ((x^2)^5 + 1) = (x^2 + 1)(x^8 - x^6 + x^4 - x^2 + 1)$.
....
And that's it
$x^{20} + x^{15} + x^{10} + x^5 = $
$x^5(x^{10} + 1)(x^5 + 1)=$
$x^5(x^2 + 1)(x^8 - x^6 + x^4 -x^2 + 1)(x^5 + 1)$
Which has $x^2 + 1$ as a factor.
...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
How to find the general term of the following sequence? Consider the following recurrence problem:
\begin{align}
d_{i-1} &= 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1} - F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \\
\varphi_{i-1} &= -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) + F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, ,
\end{align}
where $d_i, i \in \{2, \cdots , N+1\}$ represent displacements, $\varphi_i$ inclinations, and $F$ is a known force acting at the nodes $N$ and $N+1$.
We require by the system symmetry that $d_{N+1}=d_N = d_\mathrm{C}$ and $\varphi_N = -\varphi_{N+1}= \varphi_\mathrm{C}$, where $d_\mathrm{C}$ and $\varphi_\mathrm{C}$ are still to be determined from the boundary conditions:
$d_1 = 0$ (zero displacement) and $2\varphi_1+\varphi_2 = 3d_2$ (zero torque)
In order to proceed, i have tried to first determine $d_{N-1}$ and $\varphi_{N-1}$ from the system above, and then $d_{N-2}$ and $\varphi_{N-2}$, etc... in a recursive way and then try to find out the general term of the resulting sequences.
For the term $N-1$, we obtain
\begin{align}
d_{N-1} &= d_\mathrm{C}+2\varphi_\mathrm{C}-F \, \\
\varphi_{N-1} &= -9\varphi_\mathrm{C}+3F \, .
\end{align}
Analogously, we get for the term $N-2$
\begin{align}
d_{N-2} &= d_\mathrm{C}-18\varphi_\mathrm{C}+4F \, \\
\varphi_{N-2} &= 89\varphi_\mathrm{C}-24F \, .
\end{align}
I was wondering whether there is a particular way to figure out the general term of such sequence.
Any help or suggestions are most welcome.
| Thanks again to Yuri for the helpful insights. I can say that I have now found the solution to my problem.
First of all, let us define $D_i = d_i-d_{i-1}$, in the same way as Yuri did, and let us get a recurrence equation that involves $\varphi_i$ only.
Actually, original problem (which is equivalent to the one stated above) was
\begin{align}
d_{i+1}-2d_i+d_{i-1} - \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1}\right) + F \left( \delta_{i, N} + \delta_{i, N+1} \right) &=0\, , \\
d_{i+1}-d_{i-1} - \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) &= 0 \, .
\end{align}
Accordingly, it can easily be shown that for $0 <i <N$,
\begin{align}
D_{i+1}-D_i &= \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1} \right) \, , \\
D_{i+1}+D_i &= \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) \, .
\end{align}
This leads to
\begin{align}
D_i = \frac{7}{24} \varphi_i + \frac{1}{24} \varphi_{i-2} + \frac{2}{3} \varphi_{i-1} =
\frac{1}{24} \varphi_{i+1}+\frac{7}{24} \varphi_{i-1} + \frac{2}{3} \varphi_i \, , \tag{1} \label{1}
\end{align}
or
$$
\frac{1}{24} \left( \varphi_{i+1}-\varphi_{i-2} \right)
+\frac{3}{8} \left( \varphi_i-\varphi_{i-1} \right) =0\, .
$$
Using Yuri's powerful approach, and posing $\varphi_i = A/p^i$ yields
$$
p^3+9p^2-9p-1=0 \, ,
$$
whose solutions are $p=1$ and $p=-5\pm 2 \sqrt{6}$.
Finally,
$$
\varphi_i = C_1 + C_2 \left(-5-2\sqrt{6}\right)^i + C_3 \left(-5+2\sqrt{6}\right)^i \, .
$$
(again, by noting that $\left(-5-2\sqrt{6}\right)\left(-5+2\sqrt{6}\right)=1$.)
Therefore,
$$
d_i
=D_i + D_{i-1} + \cdots + D_2 =
\sum_{n=2}^{n=i} D_n \, ,
$$
where the expression of $D_n$ follows readily from Eq. \eqref{1}, and using the fact that $d_1=0.$
The final expression read
$$
D_i = C_1 +
C_2 \left( -1+\frac{\sqrt{6}}{2} \right) \left( -5-2\sqrt{6} \right)^i +
C_3 \left( -1-\frac{\sqrt{6}}{2} \right) \left( -5+2\sqrt{6} \right)^i \, ,
$$
and
$$
d_i = (i-1)C_1 +
C_2 \left( -1+\frac{5\sqrt{6}}{12} \right)
\left[ 49+20\sqrt{6}-\left( -5-2\sqrt{6} \right)^{i+1} \right] +
C_3 \left( 1+\frac{5\sqrt{6}}{12} \right)
\left[ -49+20\sqrt{6}+\left( -5+2\sqrt{6} \right)^{i+1} \right] \, .
$$
Clearly, $d_1=0$.
Finally, at this point, the 3 unknown coefficients can readily be determined from the 3 boundary conditions.
I have checked that the solution is very well behaved and is in full agreement with the numerical solution.
| {
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"url": "https://math.stackexchange.com/questions/2849643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
$\sum_{N=6}^{\infty}\frac{6}{N}\binom{N-1}{5}p^6(1-p)^{N-6}$
Find the sum: $\sum\limits_{N=6}^{\infty}\frac{6}{N}\binom{N-1}{5}p^6(1-p)^{N-6}$
I could not find a way to manipulate $\binom{N-1}{5}$ to get any suitable form here. Note that $\binom{N-1}{5}p^6(1-p)^{N-6}$ are probability masses of negative binomial distribution.
| What we want to find:
$$\sum_{k=0}^\infty \frac{6}{k+6}\binom{k+5}{k}p^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$\sum_{k=0}^\infty \binom{k+5}{k}z^k =\frac{1}{(1-z)^6} $$
Some manipulation, and integration:
$$\int_0^{1-x}\sum_{k=0}^\infty \binom{k+5}{k}z^{k+5} dz =\sum_{k=0}^\infty \frac{1}{k+6}\binom{k+5}{k}(1-x)^{k+6} = \int_0^{1-x} \frac{z^5}{(1-z)^6}dz = \frac{300x^4-300x^3+200x^2-75x+12}{60x^5}+\ln(x)-\frac{137}{60} $$
After substitution, and some multiplying by constants:
$$S(p) =\frac{300p^5-300p^4+200p^3-75p^2+12p}{10(1-p)^6}+\frac{6p^6\ln(p)}{(1-p)^6}-\frac{137p^6}{10(1-p)^6} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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'Completing the square' in higher degrees Let $p_2=ax^2+bxy+cy^2$, $a,b,c \in \mathbb{R}$ with $a \neq 0$;
$p_2$ is homogeneous of degree $2$.
By 'completing the square trick' we obtain:
$p_2=a(x^2+\frac{b}{a}xy+\frac{c}{a}y^2)=
a(x^2+2x\frac{b}{2a}y+\frac{b^2}{4a^2}y^2-\frac{b^2}{4a^2}y^2+\frac{c}{a}y^2)=
a((x+\frac{b}{2a}y)^2+(\frac{c}{a}-\frac{b^2}{4a^2})y^2)=
a(x+\frac{b}{2a}y)^2+a\frac{4ac-b^2}{4a^2}y^2$
Now consider $p_3=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d \in \mathbb{R}$ with $a \neq 0$; $p_3$ is homogeneous of degree $3$.
Similarly to the above trick we obtain:
$p_3=a(x^3+\frac{b}{a}x^2y+\frac{c}{a}xy^2+\frac{d}{a}y^3)=
a(x^3+3x^2\frac{b}{3a}y+3x(\frac{b}{3a}y)^2+(\frac{b}{3a}y)^3
-3x(\frac{b}{3a}y)^2+\frac{c}{a}xy^2-(\frac{b}{3a}y)^3+\frac{d}{a}y^3)=
a(x^3+3x^2\frac{b}{3a}y+3x(\frac{b}{3a}y)^2+(\frac{b}{3a}y)^3)+
a(-3x\frac{b^2}{9a^2}y^2+\frac{c}{a}xy^2-\frac{b^3}{27a^3}y^3+\frac{d}{a}y^3)=
a(x+\frac{b}{3a}y)^3+
a(\frac{c}{a}-3\frac{b^2}{9a^2})xy^2+a(\frac{d}{a}-\frac{b^3}{27a^3})y^3=
a(x+\frac{b}{3a}y)^3+
\frac{9ac-3b^2}{9a}xy^2+\frac{27a^2d-b^3}{27a^2}y^3=
a(x+\frac{b}{3a}y)^3+
\epsilon xy^2+\delta y^3$
where $\epsilon:=\frac{9ac-3b^2}{9a}=\frac{3ac-b^2}{3a}$ and $\delta:=\frac{27a^2d-b^3}{27a^2}$.
Now let $G: (x,y) \mapsto (x-\frac{b}{3a}y,y)$; $G$ is an (affine) automorphism of $\mathbb{R}$. We get that: $G(p_3)=G(a(x+\frac{b}{3a}y)^3+
\epsilon xy^2+\delta y^3)= aG(x+\frac{b}{3a}y)^3+
\epsilon G(xy^2)+\delta G(y^3)=ax^3+
\epsilon (x-\frac{b}{3a}y)y^2+\delta y^3=
ax^3+
\epsilon xy^2+ (\delta-\epsilon\frac{b}{3a})y^3$
Therefore, $\frac{\partial (G(p_3))}{\partial x}=3ax^2+\epsilon y^2$,
in which every monomial has even $x$ degree and even $y$ degree.
Is there a similar trick for higher odd degrees $2n+1 \in \{5,7,\ldots\}$? Namely, can one find an affine automorphism $G$ of $\mathbb{R}[x,y]$ such that every monomial in $\frac{\partial G(p_{2n+1})}{\partial x}$ has even $x$ degree and even $y$ degree. ?
The problem is that, for example, for degree $5$ the above trick only yields the existence of an automorphism $H$ such that:
$H(p_5)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$, and then
$\frac{\partial (Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5)}{\partial x}
= 5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains $2Dxy^3$.
Related questions are: i (former question of mine), ii (geometric interpretation) and iii (which is not exactly what I have asked; I did not ask for a sum of squares).
Thank you very much!
| By dehomogenizing the equation, this is equivalent to finding a substitution $x\mapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $\Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $x\mapsto x+c$ is a group action by $\Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_{2n+1}x^{2n+1}+a_{2n-1}x^{2n-1}+\cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to rewrite this optimization problem in standard form? Consider the following problem
\begin{eqnarray*}
\underset{y}{\max} & f(y)\\
s.t. & y_{1}A_{1}+y_{2}A_{2}+y_{3}A_{3}+S_{1}=C_{1},\\
& y_{4}A_{4}+y_{5}A_{5}+y_{6}A_{6}+y_{7}A_{7}+S_{2}=C_{2},\\
& y_{4}A_{8}+y_{5}A_{9}+y_{6}A_{10}+y_{7}A_{11}+S_{3}=C_{3},\\
& y_{4}A_{12}+y_{5}A_{13}+y_{6}A_{14}+y_{7}A_{15}+y_{2}A_{16}+y_{3}A_{17}+S_{4}=C_{4},\\
& S_{1},S_{2},S_{3},S_{4}\succeq0,
\end{eqnarray*}
where $y_{i}$ are scalar variables. How can we rewrite this optimzation
problem in the following form
\begin{eqnarray*}
\underset{x}{\max} & f(x)\\
s.t. & \sum_{i=1}^{i=m}x_{i}B_{i}+T=D,\\
& T\succeq0,
\end{eqnarray*}
where $x_{i}$ are scalar variables please? Thanks.
| We have
$$
{
\begin{array}{rl}
\underset{y}{\max} & f(y)\\
y_1
\left[\begin{array}{cccc}
A_1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0
\end{array}\right]
+
y_2
\left[\begin{array}{cccc}
A_2&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&A_{16}
\end{array}\right]
+
y_3
\left[\begin{array}{cccc}
A_3&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&A_{17}
\end{array}\right]
+&
\\
+y_4
\left[\begin{array}{cccc}
0&0&0&0\\0&A_4&0&0\\0&0&A_8&0\\0&0&0&A_{12}
\end{array}\right]
+y_5
\left[\begin{array}{cccc}
0&0&0&0\\0&A_5&0&0\\0&0&A_{9}&0\\0&0&0&A_{13}
\end{array}\right]
+y_6
\left[\begin{array}{cccc}
0&0&0&0\\0&A_6&0&0\\0&0&A_{10}&0\\0&0&0&A_{14}
\end{array}\right]
+
&\\
+y_7
\left[\begin{array}{cccc}
0&0&0&0\\0&A_7&0&0\\0&0&A_{11}&0\\0&0&0&A_{15}
\end{array}\right]
+
\left[\begin{array}{ccc}
S_1&0&0&0\\0&S_2&0&0\\0&0&S_3&0\\0&0&0&S_4
\end{array}\right]
&=
\left[\begin{array}{cccc}
C_1&0&0&0\\0&C_2&0&0\\0&0&C_3&0\\0&0&0&C_4
\end{array}\right]
\\
\left[\begin{array}{ccc}
S_1&0&0&0\\0&S_2&0&0\\0&0&S_3&0\\0&0&0&S_4
\end{array}\right]
&\succeq 0
\end{array}
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating a common integral for $\arcsin(x)$ using a different method
The common method of evaluating the integral below is by using the substitution $x = a \sin u$ which gives:
$$\int\frac{dx}{\sqrt{a^2-x^2}}= \arcsin\Bigl(\frac xa\Bigr) + C.$$
But, using this substitution led me down a path that seems to give a different answer (or, at least, a different form). Did I make a mistake, or are the two forms somehow equivalent?
$$x = ai\tan u $$
$$x^2=-a^2\tan^2(u)du$$
$$dx = ai\sec^2(u)du$$
$$1+\tan^2(u)=\sec^2(u)$$
\begin{align}
\int\frac{dx}{\sqrt{a^2-x^2}} & = \int\frac{1}{\sqrt{a^2+a^2\tan^2(u)}}\cdot ai\sec^2(u)du \\
& = ai\int \frac{\sec^2(u)}{\sqrt{a^2(1+\tan^2(u))}}du \\
& = \frac{ai}{a} \int \frac {\sec^2(u)}{\sec(u)}du \\
& = i \int \sec(u)du \\
& =i \ln \lvert \sec(u)+\tan(u)\rvert + C \\
& = i \ln \bigl| \frac {\sqrt{x^2-a^2}}{ai} + \frac {x}{ai} \bigr| + C \\
& = i \ln \bigl| \frac {-i\sqrt{x^2-a^2}}{a} - \frac {ix}{a} \bigr| + C \\
& = i \ln \bigl| -i\frac {\sqrt{x^2-a^2}+x}{a} \bigr| + C\\
& = i \Big[ \ln \lvert -i \rvert + \ln \bigl|\frac {\sqrt{x^2-a^2}+x}{a} \bigr| \Big] + C\\
& = i \ln \bigl|\frac {\sqrt{x^2-a^2}+x}{a} \bigr| + C
\end{align}
| Hint:
Note that $$y = \sin(x) = \frac{e^{ix}-e^{-ix}}{2} = \frac{e^{i2x}-1}{2e^{ix}}.$$
Consequently, $$(e^{ix})^2+2y\cdot e^{ix}-1=0 \implies e^{ix} = -y\pm\sqrt{y^2+1} \implies x = ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$[1+(\frac{1+i}{2})][1+(\frac{1+i}{2})^2][1+(\frac{1+i}{2})^{2^2}]...[1+(\frac{1+i}{2})^{2^n}]=(1-\frac{1}{2^{2^n}})(1+i)$,where $n\ge 2$
Show that $$\!\!\!\left[1+\left(\frac{1+i}{2}\right)\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^2\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^{2^2}\right]\cdots\left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]\!\!\!=\left(1-\frac{1}{2^{2^n}}\right)(1+i)$$ for $n\ge 2$.
I took $\frac{1+i}{2}=\frac{1}{\sqrt2}e^{i\frac{\pi}{4}}$ and tried solving but i could not reach the RHS.Please help.
| Notice that, for any $x \neq 1,$
\begin{align*}
\prod_{k=0}^n(1+x^{2^k})&=(1+x)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})\\&=\frac{(1-x)(1+x)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})}{1-x}\\&=\frac{(1-x^2)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})}{1-x}\\&\vdots\\&=\frac{1-x^{2^{n+1}}}{1-x}.
\end{align*}
Thus, \begin{align*}\prod_{k=0}^n\left[1+\left(\frac{1+i}{2}\right)^{2^k}\right]&=\frac{1-\left(\dfrac{1+i}{2}\right)^{2^{n+1}}}{1-\dfrac{1+i}{2}}=\frac{1-\left[\left(\dfrac{1+i}{2}\right)^2\right]^{2^{n}}}{\dfrac{1}{1+i}}\\&=\left[1-\left(\frac{i}{2}\right)^{2^n}\right](1+i)\\&=\left(1-\frac{1}{2^{2^n}}\right)(1+i).\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Double summation with a variable stopping point I am interested in calculating the following double summation:
$\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\frac{1}{4}^k \frac{1}{2}^{n-k-2}$
I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.
The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.
| \begin{aligned}
\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\left(\frac{1}{4}\right)^k \left(\frac{1}{2}\right)^{n-k-2} &= \sum_{n=2}^ \infty \sum_{k =0}^{n-2}\left(\frac{1}{2}\right)^{2k} \left(\frac{1}{2}\right)^{n-k-2} \\
&= \sum_{n=2}^ \infty \sum_{k =0}^{n-2}\left(\frac{1}{2}\right)^{n+k-2} \\
&= \sum_{n=2}^ \infty \left(\frac{1}{2}\right)^{n-2}\sum_{k =0}^{n-2}\left(\frac{1}{2}\right)^{k} \\
&= \sum_{n=2}^ \infty \left(\frac{1}{2}\right)^{n-2} \frac{1-\left(\frac 12\right)^{n-1}}{1-\frac 12} \\
&= \sum_{n=2}^ \infty\left(\frac{1}{2}\right)^{n-3} - \left(\frac{1}{2}\right)^{2n-4} \\
&= 2 \sum_{n=0}^ \infty\left(\frac{1}{2}\right)^n -\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n \\
&= 2 \frac{1}{1-\frac 12}-\frac{1}{1-\frac 1 4} \\
&= \frac 8 3
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is there any way to find minimum using AM, GM?
| Hint: We have
$$\frac{x^4+5x^2+7}{x^2+3}\geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21\geq 7x^2+21$$
$$x^2(3x^2+8)\geq 0$$ the equal sign holds it $$x=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Evaluating $\int\sqrt{\frac{2-x}{x-3}} dx$
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
Need help in spotting my mistake:
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
$x-2 = t^2$
$\implies dx = 2t dt$
$$2 \int \sqrt{\dfrac{1}{1-t^2}} t^2dt$$
$t= \sin \theta $
$dt = \cos\theta d\theta$
$$\int (1- \cos 2\theta )d\theta $$
$=\theta - \dfrac{\sin 2\theta}{2}$
$$= \arcsin(t)- t\sqrt{1-t^2}$$
$= \arcsin (\sqrt{x-2})- \sqrt{(x-2)(3-x)}+C$
But answer given is: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$
| Find the derivatives of both answers. You will find that yours is the correct one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Closed form for $\sin(n\arctan(x))$ Is there a closed form for the function $\sin(n\arctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?
From playing around with a few values, it seems that
$$\sin\arctan(x)=\frac{x}{\sqrt{1+x^2}},~\sin(2\arctan x)=\frac{2x}{1+x^2},~\sin(3\arctan x)=\frac{3x-x^3}{(1+x^2)^{3/2}},$$
I can see that the denominator is $(1+x^2)^{\tfrac12n}$ but can't quite see the form of the numerator.
Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $\sin\theta$ but $\sin n\theta$ for some integer $n.$
| We have that
$$\tan^{-1}(x)=\frac{i}{2}\log\left(\frac{1-ix}{1+ix}\right)$$
And
$$\sin(x)=\frac{i}{2}e^{-ix}-\frac{i}{2}e^{ix}$$
So
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\exp\left(\frac{i}{2}n\log\left(\frac{1-ix}{1+ix}\right)\right)-\frac{i}{2}\exp\left(\frac{i}{2}n\log\left(\frac{1+x}{1-ix}\right)\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{(1-ix)^\frac{n}{2}}{(1+ix)^\frac{n}{2}}-\frac{(1+ix)^\frac{n}{2}}{(1-ix)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{(1-ix)^n-(1+ix)^n}{(x^2+1)^\frac{n}{2}}\right)$$
Now we can use the Binomial theorem:
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}(-ix)^k-\sum_{k=0}^{n} \binom{n}{k}(ix)^k}{(x^2+1)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{i}{2}\left(\frac{\sum_{k=0}^{n} \binom{n}{k}x^k((-i)^k-i^k)}{(x^2+1)^\frac{n}{2}}\right)$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}}\sum_{k=0}^{n} \binom{n}{k}x^k\frac{i((-i)^k-i^k)}{2}$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \left(\binom{n}{1}x^1-\binom{n}{3}x^3+\binom{n}{5}x^5+\dots\right)$$
$$\sin(n\tan^{-1}(x))=\frac{1}{(x^2+1)^\frac{n}{2}} \sum_{k=0}^{n}\binom{n}{k}\cos\left((n-1)\frac{\pi}{2}\right)x^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the least value of the expression $x^2+2xy+2y^2+4y+7$ Find the least value of the expression
$x^2+2xy+2y^2+4y+7$
I am not able to solve this equation though i am able to differentiate it.
| For $x^2+axy+bx+cy-d=0,$
$$x^2+x(ay+b)+cy-d=0$$
As $x$ is real, the discriminant must be $\ge0$
i.e., $$(ay+b)^2\ge4(cy-d)\iff4d\ge4cy-(a^2y^2+2aby+b^2)$$
$$=-b^2-a^2y^2-2y(ab-2c)$$
$$=\left(\dfrac{ab-2c}a\right)^2-b^2-\left(ay+\dfrac{ab-2c}a\right)^2$$
$$\ge\left(\dfrac{ab-2c}a\right)^2-b^2$$
the equality occurs if $ay+\dfrac{ab-2c}a=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$ Prove that if $a,b,c$ are the lengths of the edges of a given triangle, then the following inequality holds:
$\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$.
I have tried so far using the Ravi substitution, but the expression got even uglier. I have also tried making the substitutions: $x = a + b$, $y = b + c$, $z = a + c$, and after using $3(a^2 + b^2 + c^2) \geq (a + b + c)^2$, this resulted in showing that: $x + y + z \geq \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}$, where we also know (from triangle inequality) that: $z \geq \frac{x+y}{3}$ and the other 2 analogues, but at this point I got stuck.
| Computer assisted proof.
The sides of the triangle can be cyclically rearranged, so that $a$ is the smallest side. Then the other two sides are of the shape $a+s$, $a+s+t$ with $0\le s$ and $0\le t<a$. (The last inequality, $t< a$, insures that the biggest side, $a+s+t$, is $< a+(a+s)$.
We have then to consider the two cases, where $a+s$, and $a+s+t$ are either $b,c$, or $c,b$.
We compute the difference of the quantities in the OP:
sage: var('a,b,c,s,t');
sage: f(a,b,c) = (a+b)^2 / (b+c)
sage: E(a,b,c) = 6*(a^2+b^2+c^2)/(a+b+c) - f(a,b,c) - f(b,c,a) - f(c,a,b)
sage: G(a,b,c) = E(a,b,c).numerator()
sage: G(a,a+s,a+s+t).expand()
8*a^3*s^2 + 18*a^2*s^3 + 11*a*s^4 + 2*s^5 + 8*a^3*s*t
+ 36*a^2*s^2*t + 34*a*s^3*t + 9*s^4*t + 8*a^3*t^2 + 30*a^2*s*t^2
+ 33*a*s^2*t^2 + 10*s^3*t^2 + 6*a^2*t^3 + 10*a*s*t^3 + 3*s^2*t^3
- a*t^4 - 2*s*t^4 - t^5
sage: G(a,a+s+t,a+s).expand()
8*a^3*s^2 + 18*a^2*s^3 + 11*a*s^4 + 2*s^5 + 8*a^3*s*t
+ 18*a^2*s^2*t + 10*a*s^3*t + s^4*t + 8*a^3*t^2 + 12*a^2*s*t^2
- 3*a*s^2*t^2 - 6*s^3*t^2
+ 6*a^2*t^3
- 2*a*s*t^3 - 7*s^2*t^3 - a*t^4 - 4*s*t^4 - t^5
(Sage code was inserted. Slightly rearranged to fit in the window.)
So we have to dominate the minus terms by using the plus terms.
In the first case things are simple, we have for instance $8a^3t^2 -at^4-t^5>0$, and $8a^3st-2st^4>0$.
In the second case we have more to type, there is no intelligence involved more than matching the degrees of $s$ while covering the minus terms with the plus terms:
$$
\begin{aligned}
8 \, a^{3} s^{2} + 18 \, a^{2} s^{2} t - 3 \, a s^{2} t^{2} - 7 \, s^{2} t^{3} &\ge 0\ ,\\
18 \, a^{2} s^{3} - 6 \, s^{3} t^{2} &\ge 0\ ,\\
8 \, a^{3} s t - 2 \, a s t^{3} - 4 \, s t^{4} &\ge 0\ ,\\
8 \, a^{3} t^{2} + 6 \, a^{2} t^{3} - a t^{4} - t^{5} &>0\ .
\end{aligned}
$$
$\square$
Personal note: The given proof misses by intention any beauty, since the author of the problem also tried desperately to offer an inequality without symmetry, without a visual possibility of a Chebyshev rearrangement, and with the $\ge$ sign on the side that does not allow an application of a Cauchy inequality, so that things may become symmetric after some steps. I hope (s)he may enjoy this proof, at least as i enjoyed typing it.
| {
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"url": "https://math.stackexchange.com/questions/2874232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $a$, $b$ and $c$ are positive then $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$
If $a$, $b$ and $c$ are positive then $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$.
I tried to solve this problem by C-S. But I can't sovle it.
Things I have done so far:
$\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq \frac{(\sum\limits_{cyc}a)^2}{2.\sum\limits_{cyc}a^2+\sum\limits_{cyc}ab}\geq \frac{(\sum\limits_{cyc}a)^2}{3.\sum\limits_{cyc}a^2}=\frac{\sum\limits_{cyc}a^2+2.\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}=\frac{1}{3}+\frac{2\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}\geq ?$
| Very strightforward solution:
Given inequality is equivalent to:
$$\frac{a^{2}}{b^{2}+c^{2}+bc}+\frac{b^{2}}{c^{2}+a^{2}+ca}+\frac{c^{2}}{a^{2}+b^{2}+ab}-1\geq 0$$
or:
$$a^6+a^5 b+a^5 c-a^3 b^3-a^3 b^2 c-a^3 b c^2-a^3 c^3-a^2 b^3 c-a^2 b c^3+a b^5-a b^3 c^2-a b^2 c^3+a c^5+b^6+b^5 c-b^3 c^3+b c^5+c^6 \ge 0$$
If you introduce $T[p,q,r]=\sum_{sym}a^pb^qc^r$ you get:
$$\frac12T[6,0,0]+T[5,1,0]-T[3,2,1]-\frac12 T[3,3,0]\ge0$$
...which is obviously true because:
$$T[6,0,0]\ge T[3,3,0],\quad T[5,1,0]\ge T[3,2,1]$$
(Muirhead)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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First Order Separable differential Equation
Problem:
Solve the following differential equation:
\begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}
Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, dx - \int \frac{dy}{y} &=& c_1 \\
2 \ln{|x^3+1|} - \ln{|y|} &=& c_1 \\
\ln{(x^3+1)^2} - \ln{|y|} &=& c_1 \\
\ln{ \Big( \frac{(x^3+1)^2}{|y|} \Big) } &=& c_1 \\
(x^3+1)^2 &=& c|y| \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
(x^3+1)^2 &=& |cy| \\
\end{eqnarray*}
Is my answer different from the book's answer? I believe it is. What am I missing?Any idea of how to proceed?
Thanks,
Bob
| Solve the equation by separate $x,y$ :
$$\int\frac{6x^2}{x^3+1}dx=\int\frac{dy}{y}\\
\Rightarrow2\ln|x^3+1|+C_1=\ln|y|+C_2\\
\Rightarrow e^{2\ln|x^3+1|+C_1}=e^{\ln|y|+C_2}\\
\Rightarrow |x^3+1|^2=\frac{e^{C_2}}{e^{C_1}}|y|$$
Because $c=\frac{e^{C_2}}{e^{C_1}}>0$, $|cy|=c|y|$ if $c>0$ is specified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proof verification of an exercise involving a functional equation Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function and $a \in \mathbb{R}$ such that
$$f(m+n) = f(m) + f(n) + a$$
$$f(2) = 10, f(20) = 118$$
Find $a$ and $f$.
I found this exercise at the beginning of a Real Analysis textbook. I've never solved a functional equation before, but here's my solution (attempt):
i) Using induction it's easy to verify that for $m, n \in \mathbb{N}$ we have $f(m \cdot n) = m (f(n) + a)$, since
$$f((m+1)n) = f(mn + n) = f(mn) + f(n) + a$$
ii) Then $118 = f(20) = f(10 \cdot 2) = 10 (f(2) + a) = 10 (10 + a) \Rightarrow a = \frac{9}{5}$
iii) Then $f(0) = f(0 + 0) = f(0) + f(0) + a \Rightarrow f(0) = -a = -\frac{9}{5}$.
We also get $10 = f(2) = f(1+1) = f(1) + f(1) + \frac{9}{5} = 2 f(1) + \frac{9}{5} \Rightarrow f(1) = \frac{41}{10}$
iv) Then finally we can define $f$ recursively by $f(0) = -\frac{9}{5}$ and $$f(n+1) = f(n) + f(1) + \frac{9}{5} = f(n) + \frac{41}{10} + \frac{9}{5} = f(n) + \frac{59}{10}$$
EDIT
Then thanks to the user lulu, the real pattern should be $f(m \cdot n) = m f(n) + (m-1) a$ instead. Using this in ii) then gives $a = 2$. Then we get in iii) that $f(0) = -2$ and $f(1) = 4$, so $f$ is defined by $f(n+1) = f(n) + 6$. And now it's pretty obivous that $f(n) = 6n - 2$ solves the equation.
| HINT:
We have $$f(20)=2f(10)+a=10f(2)+9a\implies a=2$$
Now consider $$f(m-n)=f(m)+f(-n)+2\\f(m+n)=f(m)+f(n)+2$$ and adding gives $$f(m-n)+f(m+n)=2f(m)+f(-n)+f(n)+4$$ or $$f(2m)-2=2f(m)+f(0)-2+4$$ Now what do we know about $f(0)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$? How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$?
My try:
All possible $4$-digit numbers $= 7 \cdot 7 \cdot 7 \cdot 6$
$4$-digits numbers NOT containing $3, 5 = 5 \cdot 5 \cdot 5 \cdot 4$
Answer= $7 \cdot 7 \cdot 7 \cdot 6-5 \cdot 5 \cdot 5 \cdot 4 =1558$
Is that OK ? If not, please explain my mistake.
Thank you.
| Your answer is incorrect, because the negation of the statement that the positive integers includes the digits $3$ and $5$ is that it does not contain the digit $3$ or does not contain the digit $5$. What you subtracted is the number of $4$-digit positive integers that contain neither the digit $3$ nor the digit $5$. However, numbers such as $3460$ and $2145$ are also inadmissible.
How many $4$-digit positive integers can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ with repetition contain the digits $3$ and $5$?
If there were no restrictions, the thousands place could be filled in six ways (since $0$ is excluded) and each of the three remaining places could be filled in seven ways (assuming repetition of digits is permitted). Hence, there are a total of $$6 \cdot 7 \cdot 7 \cdot 7$$ four-digit positive integers that can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ with repetition.
From these, we must subtract those that do not contain the digits $3$ and $5$. Numbers that do not contain the digits $3$ and $5$ do not contain the digit $3$ or do not contain the digit $5$.
How many four-digit postive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition do not contain the digit $3$?
There are five ways to fill the thousands place (since neither $0$ nor $3$ is permitted) and six ways to fill each of the remaining places (since $3$ is not permitted). Hence, there are $$5 \cdot 6 \cdot 6 \cdot 6$$ such numbers.
By symmetry, there are also
$$5 \cdot 6 \cdot 6 \cdot 6$$
four-digit positive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition that do not contain the digit $5$.
However, if we subtract those numbers that do not contain the digit $3$ and those numbers that do not contain the digit $5$ from the total, we will have subtracted those numbers that contain neither the digit $3$ nor the digit $5$ twice. We only want to subtract them once, so we must add them to the total.
How many four-digit positive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition contain neither the digit $3$ nor the digit $5$?
There are four ways to fill the thousands place (since neither $0$ nor $3$ nor $5$ is permitted) and five ways to fill each of the remaining three places (since neither $3$ nor $5$ is permitted). Hence, there are
$$4 \cdot 5 \cdot 5 \cdot 5$$ such numbers.
By the Inclusion-Exclusion Principle, the number of positive four-digit positive integers formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition that contain the digits $3$ and $5$ is
$$6 \cdot 7 \cdot 7 \cdot 7 - 2 \cdot 5 \cdot 6 \cdot 6 \cdot + 4 \cdot 5 \cdot 5 \cdot 5$$
More formally, let's follow the suggestion of Nicolas FRANCOIS made in the comments. Let the universal set, $U$, be the set of four-digit positive integers that may be formed from the digits $0, 1, 2, 3, 4, 5, 6$ with repetition; let $A$ be the event that such a four-digit positive integer includes the digit $3$; let $B$ be the event that such a four-digit positive integer includes the digit $5$. What we wish to calculate is $$|A \cap B| = |U| - |(A \cap B)^C| = |U| - |A^C \cup B^C| = |U| - (|A^C| + |B^C| - |A^C \cap B^C|)$$
What we showed above is that
\begin{align*}
|U| & = 6 \cdot 7 \cdot 7 \cdot 7\\
|A^C| & = 5 \cdot 6 \cdot 6 \cdot 6\\
|B^C| & = 5 \cdot 6 \cdot 6 \cdot 6\\
|A^C \cap B^C| & = 4 \cdot 5 \cdot 5 \cdot 5
\end{align*}
which gives
\begin{align*}
|A \cap B| & = |U| - |(A \cap B)^C|\\
& = |U| - |A^C \cup B^C|\\
& = |U| - (|A^C| + |B^C| - |A^C \cap B^C|)\\
& = |U| - |A^C| - |B^C| + |A^C \cap B^C|\\
& = 6 \cdot 7 \cdot 7 \cdot 7 - 5 \cdot 6 \cdot 6 \cdot 6 - 5 \cdot 6 \cdot 6 \cdot 6 + 4 \cdot 5 \cdot 5 \cdot 5\\
& = 6 \cdot 7 \cdot 7 \cdot 7 - 2 \cdot 5 \cdot 6 \cdot 6\cdot 6 + 4 \cdot 5 \cdot 5 \cdot 5
\end{align*}
How many $4$-digit positive integers can be formed using the digits $0, 1, 2, 3, 4, 5, 6$ without repetition contain the digits $3$ and $5$?
In this problem, either $3$ or $5$ is the leading digit or neither is.
$3$ or $5$ is the leading digit:
*
*Choose which of them is the leading digit.
*Choose the place occupied by whichever number in the set $\{3, 5\}$ that has not been used as the leading digit.
*Now that $3$ and $5$ have been placed, choose which of the remaining numbers occupies the first open place.
*Choose which of the remaining numbers fills the final open place.
There are $$2 \cdot 3 \cdot 5 \cdot 4$$ such numbers.
Neither $3$ nor $5$ is the leading digit:
*
*Choose which of the other non-zero digits occupies the thousands place.
*Choose which of the remaining places is occupied by the digit $3$.
*Choose which of the remaining places is occupied by the digit $5$.
*Choose which of the remaining digits fills the remaining place.
There are $$4 \cdot 3 \cdot 2 \cdot 4$$ such numbers.
Since these cases are mutually exclusive and exhaustive, the answer is found by adding the results for the two cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
N-th power of a matrix I have to find the n-th power for the following matrix $$A=\begin{pmatrix} 1+\sqrt3 & 1-\sqrt3\\ \sqrt3 - 1 & \sqrt3 +1\end{pmatrix}
$$ My thoughts is that here could be the same situation as for $B=\begin{pmatrix} \sin x & - \cos x\\ \cos x & \sin x\end{pmatrix}
$ which give $$B^n=\begin{pmatrix} \sin (nx) & - \cos (nx) \\ \cos (nx) & \sin (nx) \end{pmatrix}.
$$ So I think I have to make a connection between A and B, however in A I dont have only $1$ term per place so I dont know how to proceed. $$\frac12 A=\begin{pmatrix} \sin(\pi /6) +\cos(\pi /6) &\sin(\pi /6) - \cos(\pi /6) \\ \cos(\pi /6) - \sin(\pi /6) &\sin(\pi /6) +\cos(\pi /6) \end{pmatrix}
$$ Couls you help me?
| As can be easily verified
$$
\left(\begin{array}{cc}
a & -b\\
b & a
\end{array}\right)=\left(\begin{array}{cc}
\sin\theta & -\cos\theta\\
\cos\theta & \sin\theta
\end{array}\right)\left(\begin{array}{cc}
b\cos\theta+a\sin\theta & a\cos\theta-b\sin\theta\\
-a\cos\theta+b\sin\theta & b\cos\theta+a\sin\theta
\end{array}\right)
$$
but
$$
b\cos\theta+a\sin\theta=\sqrt{a^{2}+b^{2}}\left(\frac{b}{\sqrt{a^{2}+b^{2}}}\cos\theta+\frac{a}{\sqrt{a^{2}+b^{2}}}\sin\theta\right)=\rho\cos\left(\theta_{0}-\theta\right)
$$
and then
$$
\left(\begin{array}{cc}
b\cos\theta+a\sin\theta & a\cos\theta-b\sin\theta\\
-a\cos\theta+b\sin\theta & b\cos\theta+a\sin\theta
\end{array}\right)=\rho\left(\begin{array}{cc}
\cos\left(\theta_{0}-\theta\right) & \sin\left(\theta_{0}-\theta\right)\\
-\sin\left(\theta_{0}-\theta\right) & \cos\left(\theta_{0}-\theta\right)
\end{array}\right)
$$
so
$$
\left(\begin{array}{cc}
a & -b\\
b & a
\end{array}\right)=\rho\left(\begin{array}{cc}
\sin\theta & -\cos\theta\\
\cos\theta & \sin\theta
\end{array}\right)\left(\begin{array}{cc}
\cos\left(\theta_{0}-\theta\right) & \sin\left(\theta_{0}-\theta\right)\\
-\sin\left(\theta_{0}-\theta\right) & \cos\left(\theta_{0}-\theta\right)
\end{array}\right)=\rho\left(\begin{array}{cc}
\sin\theta_{0} & -\cos\theta_{0}\\
\cos\theta_{0} & \sin\theta_{0}
\end{array}\right)
$$
hence
$$
\left(\begin{array}{cc}
a & -b\\
b & a
\end{array}\right)^{n}=\rho^{n}\left(\begin{array}{cc}
\sin\left(n\theta_{0}\right) & -\cos\left(n\theta_{0}\right)\\
\cos\left(n\theta_{0}\right) & \sin\left(n\theta_{0}\right)
\end{array}\right)
$$
with
$$
\theta_0 = \arctan\left(\frac ab\right)\\
\rho = \sqrt{a^2+b^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int _0^1\frac{dx}{(e^x-1)^{1/3}}$ Study the convergence of the following integral for $\alpha\in\mathbb{R}$ and evaluate it for $\alpha = \frac{1}{3}$ if it converges for that value.
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\alpha}} x$$
The integral converges for $\alpha \lt 1$, if I am not wrong, so I can evaluate:
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}}$$
whose primitive, using some razionalizations and change of variables, is: $\frac{1}{2}\log|4(e^x-1)^\frac{2}{3}-4(e^x-1)^\frac{1}{3}+4|+\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\left(2(e^x-1)^\frac{1}{3}-1\right)\right)-\log|(e^x-1)^\frac{1}{3}+1|+C$
Anyone knows if there is a less intricate way in order of obtain it than almost 3 sheet full of calculations? Thank you
| Take
$$t = (e^x - 1)^{1/3}$$
or
$$x = \ln( t^3 + 1 )$$
Then
$$dt = \frac{1}{3}e^x(e^x-1)^{-2/3} \ dx$$
Hence
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}}
=
3{\Large\int} _{0}^{(e-1)^{1/3}}\frac{1 }{t e^x(e^x-1)^{-2/3}}\ dt
=
3{\Large\int} _{0}^{(e-1)^{1/3}}\frac{t}{(t^3 + 1)}\ dt
$$
The last integral could be written as
$$
{\displaystyle\int} _{0}^{(e-1)^{1/3}}\dfrac{t}{\left(t+1\right)\left(t^2-t+1\right)}\,\mathrm{d}t$$
Perform Partial Fraction Decomposition
$${\displaystyle\int} _{0}^{(e-1)^{1/3}}\left(\dfrac{t+1}{3\left(t^2-t+1\right)}-\dfrac{1}{3\left(t+1\right)}\right)\mathrm{d}t$$
Take the first integral above, i.e. ${\displaystyle\int}\dfrac{t+1}{t^2-t+1}\,\mathrm{d}t$ could be written as
$$\class{steps-node}{\cssId{steps-node-7}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{2t-1}{t^2-t+1}\,\mathrm{d}t+\class{steps-node}{\cssId{steps-node-8}{\dfrac{3}{2}}}{\displaystyle\int}\dfrac{1}{t^2-t+1}\,\mathrm{d}t$$
which is equal to (with some straight forward math)
$$\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)$$
Now, we need $\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)$
which is easily shown to be
$$\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)==\ln\left(t+1\right)$$
Finalizing we get
$$3\left(-\dfrac{\ln\left(\left|t+1\right|\right)}{3}+\dfrac{\ln\left(t^2-t+1\right)}{6}+\dfrac{\arctan\left(\frac{2t-1}{\sqrt{3}}\right)}{\sqrt{3}}\right)\Big\vert_0^{(e-1)^{1/3}}$$
which is
$$3\left(\dfrac{\ln\left(\left(\mathrm{e}-1\right)^\frac{2}{3}-\sqrt[3]{\mathrm{e}-1}+1\right)}{6}+\dfrac{\arctan\left(\frac{2\cdot\sqrt{3}\sqrt[3]{\mathrm{e}-1}-\sqrt{3}}{3}\right)}{\sqrt{3}}-\dfrac{\ln\left(\sqrt[3]{\mathrm{e}-1}+1\right)}{3}+\dfrac{{\pi}}{2{\cdot}3^\frac{3}{2}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Sum of Geometric Series Formula I just need the formula for the sum of geometric series when each element in the series has the value $1/2^{j+1}$, where $j = 0, 1, 2, \ldots, n$. Please help.
Someone told me it is:
$$S = 2 - \frac{1}{2^n}$$
I am not sure if its right because he has given me no proof and I couldn't prove it when I calculate it manually. Say for example:
$$S = 1/2 + 1/4 + 1/8 = .875$$
But when using the formula given above, with $n=3$ (since there are $3$ elements):
$$S = 2 - 1/8 = 1.875$$
The answers are not the same. Please enlighten me with this issue.
| Consider the $n$th partial sum
$$S_n = \sum_{j = 0}^{n} r^j = 1 + r + r^2 + \cdots + r^n$$
of the geometric series
$$\sum_{j = 0}^{\infty} r^j$$
with common ratio $r$.
If we multiply $S_n$ by $1 - r$, we obtain
\begin{align*}
(1 - r)S_n & = (1 - r)(1 + r + r^2 + \cdots + r^n)\\
& = 1 + r + r^2 + \cdots + r^n - (r + r^2 + r^3 + \cdots + r^{n + 1})\\
& = 1 - r^{n + 1}
\end{align*}
If $r \neq 1$, we may divide by $1 - r$ to obtain
$$S_n = \frac{1 - r^{n + 1}}{1 - r}$$
In particular, if $r = 1/2$, we obtain
\begin{align*}
S_n & = \sum_{r = 0}^{n} \left(\frac{1}{2}\right)^j\\
& = \frac{1 - \left(\frac{1}{2}\right)^{n + 1}}{1 - \frac{1}{2}}\\
& = \frac{1 - \left(\frac{1}{2}\right)^{n + 1}}{\frac{1}{2}}\\
& = 2\left[1 - \left(\frac{1}{2}\right)^{n + 1}\right]\\
& = 2\left(1 - \frac{1}{2^{n + 1}}\right)\\
& = 2 - \frac{1}{2^n}
\end{align*}
which is the formula you were given.
However, you want
\begin{align*}
\sum_{j = 0}^{n + 1} \frac{1}{2^{j + 1}} & = \frac{1}{2} \sum_{j = 0}^{n} \frac{1}{2^j}\\
& = \frac{1}{2} \sum_{j = 0}^{n} \left(\frac{1}{2}\right)^n\\
& = \frac{1}{2}\left[2 - \frac{1}{2^n}\right]\\
& = 1 - \frac{1}{2^{n + 1}}
\end{align*}
As a check, observe that when $n = 2$
$$\sum_{j = 0}^{2} \frac{1}{2^{j + 1}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} = 0.875$$
and
$$1 - \frac{1}{2^{2 + 1}} = 1 - \frac{1}{2^3} = 1 - \frac{1}{8} = \frac{7}{8} = 0.875$$
In your calculation, you used $n = 3$ because you did not take into account the fact that if the index starts with $0$, the third term is $n = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $x$, given: $x^2 + \frac{9x^2}{(x+3)^2} = 16$ Here's an equation:
$$x^2 + \frac{9x^2}{(x+3)^2} = 16$$
First, I subtracted 16 from both sides and factored $x^2$ so I would get a quadratic equation, but with no success. Also, I can see that the equation can be rewritten as:
$$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$
But I can't see how can I use that information.
What should I do?
| $$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$
multiply by $(x+3)^2$
$$(x+4)(x-4)(x+3)^2 + 9x^2 = 0$$
$$x^4+6x^3+2x^2-96x-144=0$$
$$(x^2-2x-6)(x^2+8x+24)=0$$
real solutions : $x=1\pm \sqrt{7}$
complex solutions : $x=-4\pm\iota 2\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $
Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4 +y^2 $
$x^2 = y^4 +y^2 - 41$
$x^2 -49 = y^4 +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.
| Here is yet another way in. Note that $y$ must be a factor of $30$ as must $x+y$ (both from the second equation).
You can rewrite the first as $(y^2+x)(y^2-x)=xy+11=$ and use the second $=41-y^2$
Now note also that changing the signs of both $x$ and $y$ still gives a solution, so we can assume $y\gt 0$.
If both $x$ and $y$ are positive $xy+11\gt 0$ and $y\le 6$
If $y$ is positive and $x$ is negative, then $x+y\gt 0$ from the second equation so $y\gt -x$, and since $y^2 \ge y$ we have also that $y^2\gt -x$, whence $(y^2+x)(y^2-x)\gt 0$ and $y\le 6$
So positive $y$ is restricted to $y\le 6$ and the possibilities are easily tested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Let $a$, $b$, $c$, $d$ be the roots of $x^4 + x + 1 = 0$. Find $a^4 + b^4 + c^4 + d^4$.
Let $p(x) = x^4 + x + 1 = 0$, and let $a$, $b$, $c$, $d$ be its roots.
Find $a^4 + b^4 + c^4 + d^4$.
I have no idea how to start solving this problem.
| For any of the roots, $r^4=-r-1$ so that the requested sum is $-a-b-c-d-4$, and by the Vieta's formula, the sum of the roots is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values for $a,b,c,d$
Given $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$ for any
real number $x$ and $y$ , find the value of $a,b,c,d$
| Starting from the equation
\begin{align*}
x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2
\end{align*}
we could go on as follows.
First step: Simplify and put each terms to the left-hand side. We obtain
\begin{align*}
4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy=0\tag{1}
\end{align*}
by noting that the terms $x^3$ and $y^2$ cancel out.
We have in (1) a bivariate polynomial in $x$ and $y$ at the left-hand side and the zero polynomial at the right-hand side. We have equality with the zero-polynomial iff the coefficients of $x^ky^l$ are zero for each $k,l$ at the left-hand side.
The next step is to either arrange the left-hand side by increasing powers of $x$ or by increasing powers of $y$. Since there is a somewhat tricky unknown $c$, which is a power of $x$ and which makes arrangement by increasing powers of $x$ cumbersome, we decide to collect according to increasing powers of $y$.
Second step: By arranging according to increasing powers of $y$ we obtain
\begin{align*}
(4x^2+3x-bx^c+dx)\color{blue}{y}+(ax+7x)\color{blue}{y^2}=0
\end{align*}
Now we are ready to harvest. We can now easily determine the unknowns $a,b,c,d$ by comparing corresponding powers of $x$.
Third step: Compare corresponding powers of $x$.
We start with the easy one:
\begin{align*}
ax+7x=0
\end{align*}
Since the coefficient of $x$ has to be zero we see $\color{blue}{a=-7}$.
The other expression is
\begin{align*}
4x^2+3x-bx^c+dx=0
\end{align*}
We have a linear term in $x$, namely $3x+dx=0$ and conclude $\color{blue}{d=-3}$. We see that setting $\color{blue}{c=2}$ gives a second quadratic term $4x^2-bx^2=0$ and we finally conclude $\color{blue}{b=4}$.
| {
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"url": "https://math.stackexchange.com/questions/2883282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$ Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$
$\alpha>0$ & $\beta<\frac{\pi}{2}$
I get the following step after some substitution
$\cos2\alpha + \cos2\beta+\cos2(\alpha+\beta)=-\frac{3}{2}$ from here not able to proceed.
| One has $$\cos \alpha\cos\beta(\cos\alpha\cos\beta - \sin\alpha\sin\beta) = -\frac{1}{8}$$
$$1 - \tan\alpha\tan\beta = -\frac{1}{8}(1+\tan^2\alpha)(1+\tan^2\beta)$$
$$8-8\tan\alpha\tan\beta = -1-\tan^2\alpha-\tan^2\beta - \tan^2\alpha\tan^2\beta$$
$$(\tan\alpha\tan\beta-3)^2 + (\tan\alpha-\tan\beta)^2 = 0$$
One then has
$$\tan\alpha=\tan\beta=\sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$
WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$
Then Im stuck here, help me solve it.
| Next to the usage of Lagrange multipliers, one can also solve it directly function-wise by assuming $c(a,b)=6-a-b$, which leads to the following derivatives to find the extrema:
$$\left\{\begin{align}
\frac{\partial A}{\partial a} &= (ab-2)(12-3a-2b)&=0 \\
\frac{\partial A}{\partial b} &= (a^2-4)(6-a-2b)&=0 \\
\end{align}\right.$$
This gives us various solutions:
*
*${\partial A}/{\partial b}=0$ gives us $a=2$ leading to:
*
*$(a,b,c)=(2,1,3)$ with A = 36
*$(a,b,c)=(2,3,1)$ with A = 36
*${\partial A}/{\partial b}=0$ gives us $2b=6-a$ resembling $b=c$, leading to:
*
*$6-2a=0$ or $(a,b,c)=(3,3/2,3/2)$ with A = 38.25
*$ab=2$ which creates a quadratic eqution having : $(a,b,c)=(3\pm\sqrt5,(3\mp\sqrt 5)/2,(3\mp\sqrt 5)/2)$ with A = 32
From all these five points, the Hessian matrix shows that they are saddle points, with two exceptions, $(a,b,c)=(3-\sqrt5,(3+\sqrt 5)/2,(3+\sqrt 5)/2)$ which is a local minimum and $(a,b,c)=(3,3/2,3/2)$ which is a local maximum.
As the domain of investigation is $a>0,b>0,c>0$ we should also investigate these various planes for extrema:
*
*$a=0$ gives that $A=72-24b +4b^2$ which has a local minimum in $b=3$ with A = 36
*$b=0$ gives that $A=72-24a+3a^2$ which has a local minimum in $a=4$ with A = 24
Note: these points are not true extrema of the function $A(a,b,c(a,b))$
The real maximal value, however, is reached when $(a,b,c)$ tends to $(0,0,6)$ with A=72.
Note: this point is the highest value reached, which lays just outside of the domain and is a supremum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show equivalence with Legendre symbol Let $p,q$ be two odd primes for which $p=2q+3$. I want to show that $\left( \frac{q}{p}\right)=1 \iff q=\pm 1 \pmod{12}$.
I have thought the following.
$$\left( \frac{q}{p}\right)=\left( \frac{p}{q}\right) (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}=\left( \frac{p}{q}\right) (-1)^{\frac{(q+1)(q-1)}{2} }=\left( \frac{p}{q}\right)$$
So we know that if there is a $x$ such that $x^2 \equiv q \pmod{p}$, then there is also a $y$ such that $y^2 \equiv p \mod{q}$.
But how can we get the desired equality for $q$ ?
| Assume $(\frac{q}{p}) = 1.$ Then we have $(\frac{p}{q}) = (-1)^{\frac{(q + 1)(q - 1)}{2}} = 1.$ Note that $q \equiv 1, 5, 7, 11 \pmod{12}.$ ($q \neq 3$ because then $p = 9$) We have $(\frac{p}{q}) = (\frac{3}{q}) = 1$ implies that $(\frac{q}{3}) = (-1)^{\frac{q - 1}{2}}.$ So if $q \equiv 5 \pmod{12}, (\frac{q}{3}) = (\frac{2}{3}) = -1 = (-1)^{2 + 6k} = 1.$ Contradiction. If $q \equiv 7 \pmod{12}$ then $(\frac{1}{3}) = 1 = (-1)^{3 + 6k} = -1.$ Contradiction.
To show the reverse direction, we want to show that $(\frac{q}{p}) = (\frac{p}{q}) = 1.$ Thus, we want to show that $(\frac{p}{q}) = (\frac{3}{q}) = 1.$ We have that $(\frac{3}{q})(\frac{q}{3}) = (-1)^{\frac{q - 1}{2}}.$ If $q \equiv 1 \pmod{12}$ then $(\frac{q}{3}) = 1.$ And thus, $(\frac{3}{q}) = (-1)^{6k} = 1.$ If $q \equiv -1 \pmod{12}$ then $(\frac{q}{3}) = -1$ and $-(\frac{3}{q}) = (-1)^{-1 + 6k} = -1.$ Thus, $(\frac{p}{q}) = 1 = (\frac{q}{p}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\frac{3}{\sqrt{-5x^2-4\sqrt{5}x+2}}$
Integrate $\frac{3}{\sqrt{-5x^2-4\sqrt{5}x+2}}$
For this question I first factored out the 3, seeing as its a constant
$$3\int\frac{1}{\sqrt{-5x^2-4\sqrt{5}x+2}}dx$$
I then noticed that this integral has a similar form to the integral of $\arcsin(x)$
$$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin(\frac{x}{a})+C, \vert x \vert \lt a$$
The similarities between these two integrals is clear, but I dont know how to get from $$-5x^2-4\sqrt{5}x+2$$ to $$a^2-x^2$$
Any help or ideas would be highly appreciated!
| A rather messy completing the square:
$-5x^2-4\sqrt{5}x+2$
$-5\left(x^2+\frac{4\sqrt{5}x}{5}\right)+2$
$2-5\left(x+\frac{2\sqrt{5}}{5} \right)^2+4$
$(\sqrt{6})^2-\left(\sqrt{5}\left(x+\frac{2\sqrt{5}}{5} \right)\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given that $z=\cos\theta+i\sin\theta$, show that $Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$
Given that $z=\cos\theta+i\sin\theta$, show that
$Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$
For this question I had to show that the real part of $\frac{z-1}{z+1}=0$
To find that I first substituted $z$ with $\cos\theta+i\sin\theta$ to get
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}$$
I then multiplied by the conjugate of the denominator
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}\cdot\frac{\cos\theta-i\sin\theta+1}{\cos\theta-i\sin\theta+1}$$
Which, when expanded, gives me
$$\frac{\cos^2\theta+\sin^2\theta+2i\sin\theta-1}{\cos^2\theta+\sin^2\theta+2i\cos\theta+1}$$
After that,
$$\frac{2i\sin\theta}{2\cos\theta+2}=\frac{i\sin\theta}{\cos\theta+1}$$
How do I proceed?
Edit: Just realized (after being told by a commenter) that $\frac{i\sin\theta}{\cos\theta+1}$ is imaginary. I won't delete the question. Ínstead I'll leave it here, as a testament to my stupidity, for everyone to see. May God have mercy on my soul during my exam.
| You are done indeed
$$\frac{i\sin\theta}{\cos\theta+1}=i\cdot \frac{\sin\theta}{\cos\theta+1}$$
is purely imaginary.
As an alternative since $z\bar z=|z|^2=\cos^2 \theta+\sin^2 \theta=1$ we have that
$$\frac{z-1}{z+1}=\frac{z-1}{z+1}\frac{\bar z+1}{\bar z+1}=\frac{z\bar z+z-\bar z-1}{z\bar z+z+\bar z+1}=i\cdot \frac{\,\Im(z)}{\Re(z)+1}$$
which is again purely imaginary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Coefficient of $x^{27}$ in $(1-x^{10})^6/(1-x)^6$. It's been a while since I have been playing with these so excuse me if it is too obvious.
How can I represent
$$ f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6} $$
as sum of powers of $x$
I am especially interested in the coefficient in front of $x^{27} $ in that sum.
The book I am reading gives this coefficient as obviously being
$${32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$
but I don't know where this comes from.
Many thanks in advance.
| As an alternative recall that
$$(1-x^{10} )=(1-x)(1+x+x^2+\ldots+x^9)$$
therefore
$$f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6}=(1+x+x^2+\ldots+x^9)^6$$
then we can evaluate the coefficient for $x^{27}$ by the stars and bars method by the equivalent problem, discussed here, of the numbers of way to distribute $b=27$ balls in $c=6$ container with no more than $n=9$ balls in any container that is
$$N(b,c,n)=\sum_i(-1)^i\binom{c}i\binom{b+c-1-i(n+1)}{c-1}=$$
$$=\binom{27+6-1}{5}-\binom{6}{1}\binom{27+6-1-(9+1)}{5}+\binom{6}{2}\binom{27+6-1-2(9+1)}{5}=$$
$$={32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Evaluate $\lim_{n\to \infty} \sum_{r=1}^n {\frac{r^4}{4r^2-1}}$ $$\lim_{n\to \infty} \sum_{r=1}^n {\frac{r^4}{4r^2-1}}$$
I am a class 12th student and this question is part of a previously given assignment.My solution is as follows...
$$\lim_{n\to \infty} \sum_{r=1}^n {\frac{r^4}{4r^2-1}}$$
$$=\frac{1}{16}\lim_{n\to \infty}
\sum_{r=1}^n\left({4r^2+1}+ \frac{1}{4r^2-1}\right)$$
$$=\lim_{n\to \infty}\frac{n(n+1)(2n+1)}{24} \;+\;\lim_{n\to \infty}{(n/16)}+\;\;\lim_{n\to \infty}\sum_{r=1}^n\frac{1}{16(4r^2-1)}$$
what should i do after that? ,
$\mathbf {i\,think\,answer\,is \, \infty} $
| You could write
$$\frac{r^4}{4 r^2-1}=\frac{1}{16}+\frac{r^2}{4}+\frac{1}{32}\left(\frac{1}{2 r-1}-\frac{1}{2 r+1}\right)$$ making, as you almost did,
$$\sum_{r=1}^n {\frac{r^4}{4r^2-1}}=\frac{n}{16}+\frac{ n (n+1) (2 n+1)}{24} +\frac{1}{32}\sum_{r=1}^n\left(\frac{1}{2 r-1}-\frac{1}{2 r+1}\right)$$ and notice a telescoping sum
$$\sum_{r=1}^n\left(\frac{1}{2 r-1}-\frac{1}{2 r+1}\right)=1-\frac{1}{2 n+1}$$
Adding all terms and simplifying, this would give
$$S_n=\sum_{r=1}^n {\frac{r^4}{4r^2-1}}=\frac{n^4+2 n^3+2 n^2+n}{6 (2 n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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convergence of sequence defined recursively $\frac{2}{a_{n+2}}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$ How to prove that the sequence ${a_n}$ defined by $\frac{2}{a_{n+2}}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$ for $n\geq1$ and $0\lt{a_1}\lt{a_2}$ converges? How to find its limit? I truly do not know how to proceed further. I tried to use A.M>G.M inequality but with no success.
| As it was suggested in the comments use $b_n=\frac{1}{a_n}$ and then solve $$2b_{n+2}=b_{n+1}+b_n$$
using characteristic polynomials technique (or look at this similar question), leading to the following polynomial
$$2x^2-x-1=0$$
with solutions $x_1=1$, $x_2=-\frac{1}{2}$ and general solution for the recurrence
$$b_n=Ax_1^n+Bx_2^n=A+B\left(-\frac{1}{2}\right)^n \Rightarrow
a_n=\frac{1}{A+B\left(-\frac{1}{2}\right)^n}$$
and $\lim\limits_{n\rightarrow\infty} a_n = \frac{1}{A}$. Given $a_1=\frac{1}{A-\frac{B}{2}}$ and $a_2=\frac{1}{A+\frac{B}{4}}$, it's not difficult to find $A$:
$$A-\frac{B}{2}=\frac{1}{a_1} , 2A+\frac{B}{2}=\frac{2}{a_2} \Rightarrow
A=\frac{1}{3a_1}+\frac{2}{3a_2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conversion of $\int_0^1\frac{\log( 1+x)}{x}dx$ to summation We know,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^nf\left(\frac rn\right)=\int_0^1f(x)dx$$
Now,we have $\int_0^1\frac{\log( 1+x)}{x}dx$
So,$f(x)=\frac{\log (1+x)}{x}$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.
Converting to summation,
$$\lim_{n\to\infty}\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}$$
Now this should evaluate to $1-\frac{1}{2^2}+\frac1{3^3}-\cdots$.How do I carry on the simplification?
Thanks for any help!
| HINT
Following the previous suggestion by TheSimpliFire
$$\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}=1-\frac{\sum r}{2n^2}+\frac{\sum r^2}{3n^3}-\cdots\to1-\frac1{2\cdot 2}+\frac1{3\cdot 3}\ldots=$$$$=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^2}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12}$$
indeed from the well known results
*
*Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)
*Sum of Reciprocals of Squares of Odd Integers
we have
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\frac14\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}8-\frac{\pi^2}{24}=\frac{\pi^2}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate the integral $\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$ $$I=\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$$
My attempt:$$I=\int_{0}^{1}\log x d(\log(x+\sqrt{1+x^2}))$$
$$=\log x\log(x+\sqrt{1+x^2})|_0^1-\int_{0}^{1}\frac{\log(x+\sqrt{1+x^2})}{x}dx$$
I don't know how to proceed below, please help me.
That is different to me.
| \begin{align}
\int_0^1 x^\alpha\ dx &= \dfrac{x^{\alpha+1}}{\alpha+1}\\
\dfrac{d}{d\alpha}\int_0^1 x^\alpha\ dx &=\dfrac{d}{d\alpha}\dfrac{x^{\alpha+1}}{\alpha+1}\\
\int_0^1 x^\alpha\ln x\ dx &=\dfrac{x^{\alpha+1}((\alpha+1)\ln x-1)}{(\alpha+1)^2}\Big|_0^1=\dfrac{-1}{(\alpha+1)^2}
\end{align}
\begin{align}
I
&= \int_{0}^{1} \frac{\ln x}{\sqrt {1+x^2}}dx \\
&= \int_0^1\ln x\left(1-\dfrac12x^2+\dfrac38x^4+\cdots\right)\ dx\\
&= \int_{0}^{1} \frac{\ln x}{\sqrt {1+x^2}}dx \\
&= \int_0^1\left(\ln x-\dfrac12x^2\ln x+\dfrac38x^4\ln x+\cdots\right)\ dx \\
&= -1+\dfrac{1}{18}-\dfrac{3}{200}+\cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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.