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How to find the value of $4\cos(\frac{\pi}{26})+\tan(\frac{2\pi}{13})$ I have found in wolfram alpha that $\displaystyle 4\cos\left(\frac{\pi}{26}\right)+\tan\left(\frac{2\pi}{13}\right)=\sqrt{13+2\sqrt{13}}$.
How to prove this identity ? Thank you.
| This is old problem can see: How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$
show that: The follow nice trigonometry
$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate the limit $\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$ I need to evaluate the limit without using l'Hopital's rule.
$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$
| If we are allowed to use Taylor series:
Since $\sin x = x + O(x^3)$, we have $1+x\sin x = 1 + x^2 + O(x^4)$.
Thus, $\sqrt{1+x\sin x} = 1+\dfrac{1}{2}x^2 + O(x^4)$.
Since $\cos x = 1 - \dfrac{1}{2}x^2 + O(x^4)$, we have $\cos 2x = 1 - 2x^2 + O(x^4)$.
Thus, $\sqrt{\cos 2x} = 1-x^2 + O(x^4)$.
Since $\tan x = x + O(x^3)$, we have $\tan \dfrac{x}{2} = \dfrac{1}{2}x + O(x^3)$.
Thus, $\tan^2 \dfrac{x}{2} = \dfrac{1}{4}x^2 + O(x^4)$.
Therefore, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} = \dfrac{\left(1+\tfrac{1}{2}x^2 + O(x^4)\right) - \left(1-x^2+O(x^4)\right)}{\tfrac{1}{4}x^2 + O(x^4)} = \dfrac{\tfrac{3}{2}x^2+O(x^4)}{\tfrac{1}{4}x^2+O(x^4)} = \dfrac{\tfrac{3}{2}+O(x^2)}{\tfrac{1}{4}+O(x^2)}$.
Hence, as $x \to 0$, $\dfrac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\tan^2\tfrac{x}{2}} \to \dfrac{\tfrac{3}{2}}{\tfrac{1}{4}} = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Proving $n^2(n^2+16)$ is divisible by 720 Given that $n+1$ and $n-1$ are prime, we need to show that $n^2(n^2+16)$ is divisible by 720 for $n>6$.
My attempt:
We know that neither $n-1$ nor $n+1$ is divisible by $2$ or by $3$, therefore $n$ must be divisible by both $2$ and $3$ which means it must be divisible by $6$.
So, $n = 6k$ and since $n>6$ we must have $k>3$. So the expression becomes $36k^2(36k^2+16) = 144k^2(9k^2+4)$
...
But then I get stuck.
Could someone please guide me towards a solution.Thanks.
| Hint $\ $ Specialize $\ c,a,b = 4,3,2\ $ below
Lemma $\ \ 5\nmid n\pm 1,\,\ ab\mid n,\,\ \color{#c00}{b^2\!\mid 5c\!-\!4}\ \Rightarrow\ {\rm lcm}(5,a^2,b^4)\mid n^2(n^2\!+5c\!-\!4)$
$\begin{eqnarray}{\bf Proof}\quad\! &&a\mid n\,\Rightarrow\ a^2\!\mid n^2,\ \ \ {\rm and}\ \ \ \ b\mid n\,\Rightarrow\, b^2\mid n^2,\ n^2\!+\color{#c00}{5c\!-\!4}\\\ \\
&&5\nmid n\pm1\,\Rightarrow\, 5\mid n\ \ {\rm or}\ \ 5\mid \color{#0a0}{n\pm 2}\,\Rightarrow\,5\mid n^2\ \ {\rm or}\ \ 5\mid \!\!\!\!\underbrace{(n^2-4)}_{\large\color{#0a0}{(n-2)(n+2)}}\!\!\!\! + 5c\end{eqnarray}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove the identity... $$\frac{\cos{2x}-\sin{4x}-\cos{6x}}{\cos{2x}+\sin{4x}-\cos{6x}}=\tan{(x-15^{\circ})}cot{(x+15^{\circ})}$$
So, here's what I've done so far, but don't know what do do next:
$$\frac{\cos{2x}-2\sin{2x}\cos{2x}-\cos{6x}}{\cos{2x}+2\sin{2x}\cos{2x}-\cos{6x}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}=...$$
I have no idea what to do next. I have a big feeling that I'm going in a totally wrong direction. Is there anything I can do with the expression in the beginning?
| Well, don't expand to $\sin(x)$ or $\cos(x)$...
Consider complex numbers...
Let $z = \exp( 2 \textbf{i} x )$
Then
$
\begin{eqnarray}
2 \cos(2x) &=& z + \bar{z}\\
2 \sin(2x) &=& - \textbf{i} \Big( z - \bar{z} \Big)\\
2 \sin(4x) &=& - \textbf{i} \Big( z^2 - \bar{z}^2 \Big)\\
&=& - \Big( z + \bar{z} \Big) \textbf{i} \Big( z - \bar{z} \Big)\\
2 \cos(6x) &=& z^3 + \bar{z}^3\\
&=& \Big( z + \bar{z} \Big) \Big( z^2 - z\bar{z} + \bar{z}^2 \Big)
\end{eqnarray}
$
Then we can write
$
\begin{eqnarray}
2\cos(2x) + 2\sin(4x) - 2\cos(6x)
&=& \Big(z + \bar{z}\Big)
\left[ 1 - \textbf{i} \Big( z - \bar{z} \Big) - z^2 + z\bar{z} - \bar{z}^2 \right]\\
&=& \Big(z + \bar{z}\Big)
\left[ - \textbf{i} \Big( z - \bar{z} \Big) - z^2 + 2 z\bar{z} - \bar{z}^2 \right]\\
&=& \Big(z + \bar{z}\Big)
\left[ \textbf{i}^2 \Big( z - \bar{z} \Big)^2 - \textbf{i} \Big( z - \bar{z} \Big) \right]\\
&=& - \textbf{i} \Big( z - \bar{z} \Big) \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) + 1 \right]\\
\end{eqnarray}
$
and
$
\begin{eqnarray}
2\cos(2x) - 2\sin(4x) - 2\cos(6x)
&=& - \textbf{i} \Big( z - \bar{z} \Big) \Big(z + \bar{z}\Big) \left[ - \textbf{i} \Big( z - \bar{z} \Big) - 1 \right]\\
\end{eqnarray}
$
So
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \frac{ - \textbf{i} \Big( z - \bar{z} \Big) - 1 }{ - \textbf{i} \Big( z - \bar{z} \Big) + 1 }\\
\end{eqnarray}
$
But $\sin(2x) = - \textbf{i} \Big( z - \bar{z} \Big)$, whence
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \frac{ 2 \sin(2x) - 1 }{ 2 \sin(2x) + 1 }\\
\end{eqnarray}
$
We can write
$
\begin{eqnarray}
\frac{ 2 \sin(2x) - 1 }{ 2 \sin(2x) + 1 }
&=& \frac{ 2 \sin(2x) - 2 \sin(2y) }{ 2 \sin(2x) + 2 \sin(2y) }\\
\end{eqnarray}
$
where $2 \sin(2y) = 1$, thus $\sin(2 y) = \frac{1}{2}$, so $2 y = 30^o$,
whence $y = 15^o$.
But
$
\begin{eqnarray}
\sin(2x) \pm 2 \sin(2y) &=&
2 \sin\big( x \pm y \big) \cos\big( x \mp y \big)
\end{eqnarray}
$
so
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \frac{ \sin\big( x - 30^o \big) \cos\big( x + 30^o \big) }{ \sin\big( x - 30^o \big) \cos\big( x - 30^o \big) }\\
\end{eqnarray}
$
So we obtain
$
\begin{eqnarray}
\frac{\cos(2x) - \sin(4x) - \cos(6x)}{\cos(2x) + \sin(4x) - \cos(6x)}
&=& \tan\big( x - 30^o \big) \cot\big( x + 30^o\big)
\end{eqnarray}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Maclaurin series of (1+x)^(1/x) how can i find the Maclaurin series of $f(x)=(1+x)^{1 \over x}$?
$f(0)$ is not even defined, or should I define it as $f(0)=e$?
I stopped at the first derivative as it gets terribly messy.
thank you
| Hint
Start with $$\log(f)=\frac{1}{x} \log(1+x)=\frac{1}{x}\Big(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)\Big)= 1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ Now $$f=e^{\log(f)}=e^{1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)}=e ~~ e^{-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)}$$ Now, set $y=-\frac{x}{2}+\frac{x^2}{3}$ and use $e^y=1+y+\frac{y^2}{2}+...$ Replace $y$ by its expression as a function of $x$ and develop. You will end with
$$f=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/840451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the limit $\lim_{x\rightarrow 0}\left[ \frac{\ln(\cos x)}{x\sqrt{1+x}-x} \right]$ Evaluate the limit:
$$\lim_{x\rightarrow 0}\left[ \frac{\ln(\cos x)}{x\sqrt{1+x}-x} \right]$$
I actually was able to find the limit is $-1$ after applying L'Hôpital's rule twice.
I wonder if that was the intention of this exercise or there's an "easier" way.
Thanks.
| We need to proceed as follows $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log(\cos x)}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{\cos x - 1}\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}1\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0} \frac{\cos x - 1}{x^{2}}\cdot\frac{x^{2}}{x\sqrt{1 + x} - x}\\
&=\lim_{x \to 0} \frac{-1}{2}\cdot\frac{\sqrt{1 + x} + 1}{1} = -1\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$
So,the last two digits are $8 \text{ and } 9$.
$$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one?
| A variant to minimise the computation time of the order of $9\bmod100$: by the Chinese remainder theorem,
$$\mathbf Z/100\mathbf Z\simeq\mathbf Z/4\mathbf Z\times \mathbf Z/25\mathbf Z. $$
Now $\varphi(25)=20$, and $9\equiv 1\mod4$. Hence $9^{20}\equiv 1\mod100$, so the order of $9$ is a divisor of 20.
Fast exponentiation algorithm shows it has order $10$. Thus
$$9^{9^9}\equiv9^{9^9\bmod 10}\equiv 9^{(-1)^9\bmod 10}\equiv 9^{-1}\mod100. $$
Bézout's identity $\;100_11\cdot 9=1\;$ then shows $\; 9^{-1}\equiv -11\equiv 89\mod100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proving $x+\sin x-2\ln{(1+x)}\geqslant0$
Question: Let $x>-1$, show that
$$x+\sin x-2\ln{(1+x)}\geqslant 0.$$
This is true. See http://www.wolframalpha.com/input/?i=x%2Bsinx-2ln%281%2Bx%29
My try: For
$$f(x)=x+\sin x-2\ln{(1+x)},\\
f'(x)=1+\cos{x}-\dfrac{2}{1+x}=\dfrac{x-1}{1+x}+\cos{x}=0\Longrightarrow\cos{x}=\dfrac{1-x}{1+x}.$$
So
$$\sin x=\pm\sqrt{1-{\cos^2{x}}}=\pm \dfrac{2\sqrt{x}}{1+x}$$
If $\sin x=+\dfrac{2\sqrt{x}}{1+x}$, I can prove it. But if $\sin x=-\dfrac{2\sqrt{x}}{1+x}$, I cannot. See also http://www.wolframalpha.com/input/?i=%28x-1%29%2F%28x%2B1%29%2Bcosx
This inequality seems nice, but it is not easy to prove.
Thank you.
| Let $f(x)=x+\sin{x}-2\ln(1+x).$
Thus, $$f'(x)=1+\cos{x}-\frac{2}{1+x}=2-\frac{2}{1+x}-(1-\cos{x})=\frac{2x}{1+x}-2\sin^2\frac{x}{2}\leq0$$
for all $-1<x\leq0,$ which says that for $-1<x\leq0$ we have
$$f(x)\geq f(0)=0.$$
Let $x\geq5.$
Thus, since $$\left(x-2\ln(1+x)\right)'=1-\frac{2}{1+x}=\frac{x-1}{x+1}>0,$$ we obtain:
$$f(x)\geq x-1-2\ln(1+x)>4-2\ln6>0.$$
Id est, it's enough to prove our inequality for $0\leq x\leq5.$
We see that $f'(x)=0,$ when $\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}$ or $\sin\frac{x}{2}=-\sqrt{\frac{x}{x+1}}.$
Also, $\sin\frac{x}{2}+\sqrt{\frac{x}{x+1}}\geq\sqrt{\frac{x}{x+1}}\geq0$ for all $0\leq x\leq5$.
Thus, all critical points of $f$ on $[0.5]$ gives the following equation. $$\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}.$$
We see that $\sin\frac{x}{2}$ decreases on $[\pi,5]$ and $\sqrt{\frac{x}{x+1}}$ increases on $[\pi,5]$.
Also, we have $$\sin\frac{\pi}{2}>\sqrt{\frac{\pi}{\pi+1}}$$ and $$\sin\frac{5}{2}<\sqrt{\frac{5}{5+1}},$$
which says that the equation $$\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}$$ has an unique root $x_1\in[\pi,5].$
Also, we see that $f'(x)<0$ for all $x\in\left(\pi,x_1\right)$ and $f'(x)>0$ for all $x\in(x_1,5),$ which says $x_1$ gives a minimum point and $x_{min}=4.06268...$.
Now, we'll prove that the equation
$$\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}$$ has exactly two roots on $[0,\pi]$.
Indeed, on this set it's equivalent to
$$\sin^2\frac{x}{2}=\frac{x}{x+1}$$ or
$$\frac{1}{\sin^2\frac{x}{2}}=1+\frac{1}{x}$$ or $\tan\frac{x}{2}=\sqrt{x}$.
But $\tan$ is a convex function and $\sqrt{x}$ is a concave function, which says that the equation $\tan\frac{x}{2}=\sqrt{x}$ has two roots maximum.
$0$ is one of them and $$\sqrt{\frac{\pi}{2}}-\tan\frac{\pi}{4}>0$$ and $$\sqrt{\frac{3\pi}{4}}-\tan\frac{3\pi}{8}<0,$$
which says that our equation has last root $x_2\in\left(\frac{\pi}{2},\frac{3\pi}{4}\right)$ and $x_2=1.88176...$.
Also, we saw that $f'(x)>0$ for all $x\in(0,x_2)$ and $f'(x)<0$ for all $x\in(x_2,x_1)$, which gives
$$\min_{[0,5]}f=\min\left\{f(0),f\left(x_{1}\right)\right\}=\min\{0,0.0226...\}=0$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Ellipse cutting orthogonally If the curves $ax^2+by^2=1$ and $a'x^2+b'y^2=1$ cut orthogonally, then :
A)$\displaystyle \frac{1}{b}+\frac{1}{b'}=\frac{1}{a}+\frac{1}{a'}$
B)$\displaystyle \frac{1}{b}-\frac{1}{b'}=\frac{1}{a}-\frac{1}{a'}$
C)$\displaystyle \frac{1}{b'}-\frac{1}{b}=\frac{1}{a}-\frac{1}{a'}$
D)$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{a'}+\frac{1}{b'}=0$
I tried solving it but failed. My doubt is "Can two ellipses having common centre ever cut orthogonally"?
| HINT:
Let the two curves intersect at $(h,k)$
Solve for $(h,k)$ in terms $a,b,a',b'$
Now, the gradient $(m_1)$ of $ax^2+by^2=1$ is $-\dfrac{ah}{bk}$
As the two curves cut orthogonally, $m_1m_2=-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the equation Prove that
$$\int_0^{\infty}\exp\left(-\left(x^2+\dfrac{a^2}{x^2}\right)\right)\text{d}x=\frac{e^{-2a}\sqrt{\pi}}{2}$$
Assume that the equation is true for $a=0.$
| In general
$$
\begin{align}
\int_{x=0}^\infty \exp\left(-ax^2-\frac{b}{x^2}\right)\,dx&=\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\
&=\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx.
$$
Let $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then
$$
I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.
$$
Let $t=x\;\rightarrow\;dt=dx$, then
$$
I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Adding the two $I_t$s yields
$$
2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}},
$$
where $I$ is a Gaussian integral. Thus
$$
\begin{align}
\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx
&=\large\color{blue}{\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}}.
\end{align}
$$
In our case, put $a=1$ and $b=a^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Mathematical Induction Matrix Example I'm a little rusty and I've never done a mathematical induction problem with matrices so I'm needing a little help in setting this problem up.
Show that $$\begin{bmatrix}1&1\\1&1\end{bmatrix}^{n} = \begin{bmatrix}2^{(n-1)}&2^{(n-1)}\\2^{(n-1)}&2^{(n-1)}\end{bmatrix}$$ for every $n\ge 1$.
| The case $n=1$ is clear since $2^0 = 1$. So suppose that
$$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^n = \begin{pmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{pmatrix} \quad \quad *$$
for some $n \geq 1$ and let us prove that
$$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n+1} = \begin{pmatrix} 2^{n}&2^{n} \\ 2^{n} & 2^{n}\end{pmatrix}. $$
We have
$$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n+1} = \begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n} \begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix} \overset{*}{=} \begin{pmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{pmatrix}\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2\cdot 2^{n-1}&2\cdot 2^{n-1} \\ 2\cdot 2^{n-1} & 2\cdot 2^{n-1}\end{pmatrix} =\begin{pmatrix} 2^{n}&2^{n} \\ 2^{n} & 2^{n}\end{pmatrix}. $$
And thus the relation is true for every $n \in \mathbb{N}$
| {
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"answer_id": 0
} |
Integrate $1/(x^5+1) $from $0$ to $\infty$? How can I calculate the integral $\displaystyle{\int_{0}^{\infty} \frac{1}{x^5+1} dx}$?
| The change of variable $\frac{1}{1+x^{5}}=u$ gives $dx=-\frac{u^{-1-\frac{1}{5}}(1-u)^{-\frac{4}{5}}}{5}du$
Hence our integral reduces to $$\frac{1}{5}\int_{0}^{1}{u^{\frac{4}{5}-1}(1-u)^{\frac{1}{5}-1}du}=\frac{1}{5}\beta(1-\frac{1}{5},\frac{1}{5})=\frac{1}{5}\Gamma(1-\frac{1}{5})\Gamma(\frac{1}{5})=\frac{1}{5}\frac{\pi}{\sin(\frac{\pi}{5})}$$ we can actually simplify things a bit more by considering the following identity, deduced by De'Moivres formula $$\sin(5x)=16\sin^{5}(x)-20\sin^{3}(x)+5\sin(x)$$ pluggin in $x=\frac{\pi}{5}$ combined with the fact that $\sin(\frac{\pi}{5})>0$ We get $$16\sin^{4}(\frac{\pi}{5})-20\sin^{2}(\frac{\pi}{5})+5=0$$ which we can begin by solving for $\sin^{2}(\frac{\pi}{5})$ and then taking the positive root. We will end up with $\sin(\frac{\pi}{5})=\sqrt{5-\sqrt{5}}$ hence your integral will finally evaluate at $$\frac{1}{5}\frac{\pi}{\sqrt{5-\sqrt{5}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/847679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answer. :/
| Hint
First of all, you could start factoring both numerator and denominator $$\frac{3x^2+3x-6}{2x^2+6x+4}=\frac{3(x^2+x-2)}{2(x^2+3x+2)}$$ In order to factor them, you can then compute the roots of the quadratics : for the numerator, the roots are obviously $1$ and $-2$ and for the denominator $-1$ and $-2$ (you can solve for them or just find by inspection).So, $$\frac{3x^2+3x-6}{2x^2+6x+4}=\frac{3(x^2+x-2)}{2(x^2+3x+2)}=\frac{3(x-1)(x+2)}{2(x+1)(x+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?
| Note that $\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^3}}$, with a similar expression for the other one. Make the substitution $t=1/x$. Our expression becomes
$$\frac{2-\sqrt[3]{1+t+t^3}-\sqrt[3]{1-t+t^3}}{t},$$
and we want to find the limit of this as $t$ approaches $0$ from the right. Note that the conditions for using L'Hospital's Rule are met. So our limit is
$$\lim_{t\to 0^+} -\frac{1}{3}(1+3t^2)(1+t+t^3)^{-2/3}-\frac{1}{3}(-1+3t^2)(1-t+t^3)^{-2/3}.$$
This one is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Is this procedure for $5^{300} \bmod 11$ correct? I'm new to modular exponentiation. Is this procecdure correct?
$$5^{300} \bmod 11$$
$$5^{1} \bmod 11 = 5\\
5^{2} \bmod 11 = 3\\
5^{4} \bmod 11 = 3^2 \bmod 11 = 9\\
5^{8} \bmod 11 = 9^2\bmod 11 = 4\\
5^{16} \bmod 11 = 4^2 \bmod 11 = 5\\
5^{32} \bmod 11 = 5^2 \bmod 11 = 3$$
$$5^{300} = 3 + 3 + 3 + 3 +3 +3 + 3 + 3 +3 +4 + 9$$
| You're almost there. Instead of your last line, you want:
$$5^{300} \equiv 5^{4}5^{8}5^{32}5^{256} \pmod{11}$$
Now you replace each of those factors with the modular equivalent you found before (e.g. $5^4\implies 9$)
Let me know if you need more help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
I need help solving these limits I've been struggling to solve the following limits for about an hour. I've tried using conjugates as well as common factors, but it gets me nowhere. Wolfram Alpha does not provide steps for these limits. I could really use some help.
(x greater than 1) (SOLVED)
(SOLVED)
Thanks for taking your time!
| Addressing the third limit here.
Substitution yields:
$$\frac{\sqrt[]{1-(-8)}-3}{2+(-8)^{2/3}}=\frac{3-3}{2-2}=\frac{0}{0}$$
which is an indeterminate form. Therefore we can apply L'Hospital's rule to obtain
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-\frac{1}{2\sqrt[]{1-x}}}{\frac{1}{3x^{2/3}}}}=-\frac{3}{2}\lim_{x\rightarrow-8}{\frac{x^{2/3}}{\sqrt[]{1-x}}}=-\frac{3}{2}\cdot\frac{4}{3}=-2$$
$\textbf{Edit:}$
Another solution without L'Hospital's rule:
First multiply by the conjugate of the numerator over itself:
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}\frac{\sqrt[]{1-x}+3}{\sqrt[]{1-x}+3}=\lim_{x\rightarrow-8}{\frac{-(x+8)}{(2+\sqrt[3]{x})(\sqrt[]{1-x}+3)}}$$
Now to avoid the indeterminate form we want to find a way to get a $x+8$ in the denominator, to cancel the one in the numerator. Multiplying the top and bottom by $4-2\sqrt[3]{x}+x^{2/3}$ will do the trick. That's because of the sum of cubes formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$.
So from above
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-(x+8)}{(2+\sqrt[3]{x})(\sqrt[]{1-x}+3)}}\cdot\frac{4-2\sqrt[3]{x}+x^{2/3}}{4-2\sqrt[3]{x}+x^{2/3}}=\lim_{x\rightarrow-8}{\frac{-(x+8)(4-2\sqrt[3]{x}+x^{2/3})}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+x^{2/3})(\sqrt[]{1-x}+3)}}=\lim_{x\rightarrow-8}{\frac{-(x+8)(4-2\sqrt[3]{x}+x^{2/3})}{(x+8)(\sqrt[]{1-x}+3)}}$$
Canceling the $x+8$ gives
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-(4-2\sqrt[3]{x}+x^{2/3})}{\sqrt[]{1-x}+3}}=-\frac{4+4+4}{3+3}=-\frac{12}{6}=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?
| I could not get any ideas of getting the answer analytically.
So I just wrote the matlab program as below
format rat
Initial=1
i=1;
count=0;
while i<100,
Initial = Initial*(i+1)/i;
i=i+2;
count=count+1;
a(count+1)= Initial;
end;
Got the answer 12 after flooring the end result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. The problem asks to show that $$f(x,y) = \left\{ \begin{align} \frac{x^3y^2}{x^4+y^4}, & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{align} \right.$$ is continuous at the origin, however it has resisted my bravest efforts. I have attempted using $x^4+y^4 \geq y^4$ and therefore $$\left\vert \frac{x^3y^2}{x^4+y^4} \right\vert \leq \left\vert \frac{x^3}{y^2} \right\vert$$ and similar strategies but they have failed. Trying to disprove it and see if it's discontinuous has only strengthened the belief that it's continuous.
| To expand on my comments, let $\epsilon > 0 $ be given, then our limit, in polar coordinates is of the form
$$f(r,\theta)={r^5\cos^3\theta\sin^2\theta\over r^4(\cos^4\theta+\sin^4\theta)}$$
this satisfies
$$|f(r,\theta)-0|\le |r|\left|{\cos^3\theta\sin^2\theta\over \cos^4\theta+\sin^4\theta}\right|\le {r\over m_\theta}={\delta\over m_\theta}$$
with $m_\theta$ the minimum value of $\cos^4\theta+\sin^4\theta$ for $0\le\theta<2\pi$. What is this minimum?
Taking derivatives, we see that it is possible when $-4\cos^3\theta\sin\theta+4\sin^3\theta\cos\theta=0\quad (*)$
three cases
(i) $\theta=0,\pi$ then the function's value is $1$.
(ii) $\theta=\pi/2, 3\pi/2$ again, value $1$.
(iii) $\cos\theta\sin\theta\ne 0$ then we have $\sin^2\theta=\cos^2\theta$ by manipulating $(*)$. This only happens when $\theta$ is on the line $y=\pm x$, whence both $\sin^4\theta$ and $\cos^4\theta$ are positive, so the minimum is a positive number, in fact it's easily seen to be $m_\theta={1\over 2}$
Hence we choose $\delta={\epsilon\over 2}$ and we are guaranteed convergence to $0$, as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \operatorname{span}\left\{ \begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right\}$
and the vector $v = \begin{pmatrix}9 \\0 \\0\end{pmatrix}$
How can I find the orthogonal projection of $v$ on $V$?
This is what I did so far:
\begin{align}&P_v(v)=\langle v,v_1\rangle v_1+\langle v,v_2\rangle v_2 =\\=& \left\langle\begin{pmatrix}9 \\0 \\0\end{pmatrix},\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}\right\rangle\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}+\left<\begin{pmatrix}9 \\0 \\0\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right>\begin{pmatrix}1 \\3 \\4\end{pmatrix} = \begin{pmatrix}10 \\29 \\38\end{pmatrix}\end{align}
Is this the right method to compute this?
| Hint
You have to construct by the Gram Schmidt procedure an orthonormal basis $(e_1,e_2)$ from the given basis of $V$ and then
$$P_v(V)=\langle v,e_1\rangle e_1+\langle v,e_2\rangle e_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/857942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to prove $ \int _0 ^1 \left ( \sqrt{1-x^2}\right )^n dx = \prod_{k=1} ^n \frac {2k}{2k+1} $ How to prove this integral identity?
$$ \int _0 ^1 \left ( \sqrt{1-x^2}\right )^n dx = \prod_{k=1} ^n \frac {2k}{2k+1} $$
↑ This identity is false. It should be corrected to $ \int _0 ^1 \left ( {1-x^2}\right )^n dx = \prod_{k=1} ^n \frac {2k}{2k+1} $
I'm sorry for causing confusion
| Hint:
$$I_{n+1}:=\int _0 ^1 \left ( \sqrt{1-x^2}\right )^n dx =\int _{-\frac{\pi}{2}} ^0 \left ( \sqrt{1-\sin^2 t}\right )^n \cos t dt=\int _{-\frac{\pi}{2}} ^0 \cos^{n+1} t dt=$$
integrating by parts
$$\left[\sin t \cos^{n}t\right]_{-\frac{\pi}{2}} ^0+n\int _{-\frac{\pi}{2}} ^0 \sin t\cos^{n-1} t \sin t dt=n\int _{-\frac{\pi}{2}} ^0 \cos^{n-1} t (1-\cos^2 t) dt$$
$$=n\int _{-\frac{\pi}{2}} ^0 \cos^{n-1} t dt-n\int _{-\frac{\pi}{2}} ^0 \cos^{n+1} t dt=nI_{n-1}-nI_{n+1}$$
Hence: $$I_{n+1}=\frac{n}{n+1}I_{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Finding the derivative of $\sqrt{x+\sqrt{x^2+5}}$ How to derive $y=\sqrt{x+\sqrt{x^2+5}}$ at $x=2$.I used logarithmic differentiation and chain rule over and over again but I can't get the right answer
| \begin{align*}
\frac{d}{dx}y & = \frac{d}{dx}\sqrt{x+\sqrt{x^2+5}}\\
& = \frac{1}{2}\frac{1}{\sqrt{x+\sqrt{x^2+5}}}\cdot \frac{d}{dx}(x+\sqrt{x^2+5})\\
& = \frac{1}{2}\frac{1}{\sqrt{x+\sqrt{x^2+5}}} \,\left(1+\frac{x}{\sqrt{x^2+5}}\right)\\
\end{align*}
Now simplify as best as you can.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/859254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Optimize $x^2 + y^2 +2z^2 +z(x^2-y^2)$ subject to $x+y=2$ $$x^{2}+y^{2}+2z^{2}+zx^{2}-zy^{2}\overset{\left(x=2-y\right)}{\longrightarrow}4-4y+2y^{2}+2z^{2}+4z-4yz\rightarrow FOC: \; \begin{cases}
-4+4y-4z=0\\
4z+4-4y=0
\end{cases}\rightarrow y=1+z\rightarrow f\left(x,y,z\right)=2$$
at any point that satisfies
$$\begin{cases}
x=1-z\\
y=1+z
\end{cases}$$
Hessian $4\begin{pmatrix}1 & -1\\
-1 & 1
\end{pmatrix}$ is positive semidefinite, so I cannot claim that these points are points of strict local minimum.
Question: how to show that these points are points of (not only local but) global minimum? As far as I understand I need to prove $f(x,y,z)\geq 2$, but I cannot imagine how to deal with inequality of 3 variables.
| Hint : $4-4y+2y^{2}+2z^{2}+4z-4yz$ can be rewritten as $2(1+(z-y+1)^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this P(x) if $ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $ Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that
$$ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $$
This problem is IMO Shortlist 2013
let $$P(x)=\sum_{i=0}^{n}a_{i}x^i,a_{i}\in R$$
then
$$\sum_{i=0}^{n}[a_{i}x^3(x+1)^i-mx^2(x+1)^i+(x+1)^i]+\sum_{i=0}^{n}[x^3(x-1)^i+mx^2(x-1)^i+(x-1)^i]=2\sum_{i=0}^{n}[x^{3+i}-mx^{i+1}+x^i]$$
then I can't
| Momentarily setting aside the trivial solution, one option is to rearrange in the form of a finite difference equation, like so:
$$(x^3+1)(P(x+1)-2P(x)+P(x-1))-mx^2(P(x+1)-P(x-1))+2mxP(x)=0$$
Adjusting the definition of a polynomial $P(x)$ might make sense in this case, where it appears that finite differences play a significant role. In particular, using the binomial form of polynomial terms would result in a definition of an $n$th degree polynomial as
$$P(x)=\sum_{i=0}^na_i{x\choose i}$$
and then the values of $P(x+1)-2P(x)+P(x-1)$ and $P(x+1)-P(x-1)$ are easily determined as
$$P(x+1)-2P(x)+P(x-1)=\sum_{i=0}^na_i\left({x+1\choose i}-2{x\choose i}+{x-1\choose i}\right)
=\sum_{i=2}^na_i{x-1\choose i-2}$$
$$P(x+1)-P(x-1)=\sum_{i=0}^na_i\left({x+1\choose i}-{x-1\choose i}\right)
=\sum_{i=1}^na_i\left({x\choose i-1}+{x-1\choose i-1}\right)$$
and then the full difference equation looks like
$$(x^3+1)\sum_{i=0}^{n-2}a_{i+2}{x-1\choose i}-mx^2\sum_{i=0}^{n-1}a_{i+1}\left({x\choose i}+{x-1\choose i}\right)+2mx\sum_{i=0}^na_i{x\choose i}=0$$
Reorganizing, we have extra terms $Q(x)=-a_nmx^2\left({x\choose n-1}+{x-1\choose n-1}\right)+2a_nmx{x\choose n}+2a_{n-1}mx{x\choose n-1}$ which rearranges to
$$Q(x)=a_nmx^2\left(\frac2n-1\right){x-1\choose n-1}+mx(2a_{n-1}-a_nx){x\choose n-1}$$
and main sum
$$Q(x)+\sum_{i=0}^{n-2}\left[\left(a_{i+2}(x^3+1)-a_{i+1}mx^2\right){x-1\choose i}+\left(2a_imx-a_{i+1}mx^2\right){x\choose i}\right]=0$$
Clearly, the low-exponent monomials in $x$ will have many terms to deal with, but it appears that the high-exponent monomials in $x$ will only occur a few times each, so it would be best to analyze this sum starting from the high-exponent terms first. So starting with all terms where $x^{n+1}$ appears, we have
$$a_nx^3{x-1\choose n-2}-a_nmx^2\left({x\choose n-1}+{x-1\choose n-1}\right)+2a_nmx{x\choose n}=0$$
So either $a_n = 0$ or is irrelevant for the solution. We can further divide through by $x$ to get
$$x^2{x-1\choose n-2}-mx\left({x\choose n-1}+{x-1\choose n-1}\right)+2m{x\choose n}=0$$
$$x^2\frac{(x-1)!}{(n-2)!(x+1-n)!}-mx\frac{x!}{(n-1)!(x+1-n)!}-mx\frac{(x-1)!}{(n-1)!(x-n)!}+2m\frac{x!}{n!(x-n)!}=0$$
$$n(n-1)x^2(x-1)!-mnx^2(x-1)!-nmx(x+1-n)(x-1)!+2mx(x+1-n)(x-1)!=0\\
n(n-1)x^2-mnx^2-nmx(x+1-n)+2mx(x+1-n)=0\\
n(n-1)x-mnx-mn(x+1-n)+2m(x+1-n)=0$$
In order for all $x$-dependent terms to cancel, $n$ must be either $0$ or $1$. If $n=0$ then $P(x)=c$ for some constant $c$, but this fails a simple sanity check with the original equation ($0=-2mxc,m\neq 0$) unless $c=0$. So we have either $P(x)=0$ which means that $m$ is unlimited except for not being $0$, or $P(x)=a_1x+a_0$ which means $n=1$. In that case, our main sum is $0$ already, and since $Q(x)$ has only a single term with $a_{n-1}$ present and that term has a lower exponent of $x$ than all the others, we must have $a_0=0$, so our solution set is $P(x)=ax$.
Given this solution set, what is the limitation on $m$ when $n=1$? All terms in $Q(x)$ cancel, so we have to go back to the original again, which works out as
$$(x^3 - mx^2 +1 )a(x+1) + (x^3+mx^2+1)a(x-1) =2(x^3 - mx +1 )ax\\
ax^4+ax^3-amx^3-amx^2+ax+a+ax^4-ax^3+amx^3-amx^2+ax-a=2ax^4-2amx^2+2ax\\
0=0$$
From this, it is apparent that $m$ can take on any non-zero integer value.
Sources: my own work over the past $20$+ years, as supplemented by mathworld and wikipedia.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum related to zeta function I was trying to evaluate the following sum:
$$\sum_{k=0}^{\infty} \frac{1}{(3k+1)^3}$$
W|A gives a nice closed form but I have zero idea about the steps involved to evaluate the sum. How to approach such sums?
Following is the result given by W|A:
$$\frac{13\zeta(3)}{27}+\frac{2\pi^3}{81\sqrt{3}}$$
Any help is appreciated. Thanks!
| Notice that
$$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{1}{27} \sum_{n=0}^{\infty} \frac{1}{(k+\frac{1}{3})^{3}} = - \frac{1}{54} \psi_{2}\left(\frac{1}{3} \right) $$
where $\psi_{2}(x)$ is the second derivative of the digamma function.
Differentiating the multiplication formula for the digamma function twice and letting $q=3$,
$$\psi_{2}(x) + \psi_{2} \left( x+ \frac{1}{3} \right) + \psi_{2} \left(x+ \frac{2}{3} \right) = 27 \psi_{2}(3x) .$$
Therefore, $$ \begin{align} \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) &= 27 \psi_{2}(1) - \psi_{2}(1) \\ &=27 \left( -2 \zeta(3) \right) + 2 \zeta(3) = -52 \zeta(3). \tag{1}\end{align} $$
And differentiating the reflection formula for the digamma function twice,
$$ \psi_{2} (x) - \psi_{2}(1-x) = - 2\pi^{3} \cot(\pi z) \csc^{2}(\pi z) .$$
Therefore, $$\psi_{2} \left(\frac{1}{3} \right) - \psi_{2} \left( \frac{2}{3}\right) = - \frac{8 \pi^{3}}{3 \sqrt{3}} . \tag{2}$$
Adding $(1)$ and $(2)$,
$$ \psi_{2} \left( \frac{1}{3}\right) = -26 \zeta(3) - \frac{4 \pi^{3}}{3 \sqrt{3}} .$$
So
$$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^{3}} = \frac{13 \zeta(3)}{27} + \frac{2 \pi^{3}}{81 \sqrt{3}} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/869589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove by induction that $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$ is decreasing I want to prove that the following sequence is monotonously decreasing:
$A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$
I think it should be pretty easy to do so by induction:
*
*$A_1 = \binom{3}{2}\cdot{(\frac{60}{100})}^2\cdot{(\frac{40}{100})}^1+\binom{3}{3}\cdot{(\frac{60}{100})}^3\cdot{(\frac{40}{100})}^0 = 0.648$
*$A_2 = \binom{6}{4}\cdot{(\frac{60}{100})}^4\cdot{(\frac{40}{100})}^2+\binom{6}{5}\cdot{(\frac{60}{100})}^5\cdot{(\frac{40}{100})}^1+\binom{6}{6}\cdot{(\frac{60}{100})}^6\cdot{(\frac{40}{100})}^0 = 0.54432$
*Let's assume that $A_{k} > A_{k-1}$
*Let's prove that $A_{k+1} > A_{k}$
So how do I go about with that? Or is there another way?
| It happens that
$$ A_k = \left(\frac{18}{125}\right)^k\binom{3k}{2k}\phantom{}_2 F_1\left(1,-k;1+2k;-\frac{3}{2}\right),$$
and a clever idea is to use the Stirling approximation together with the Gauss continued fraction for the hypergeometric function in order to give tight bounds for $A_k$, then prove $A_{k+1}<A_{k}$ for any $k$ big enough.
$$\phantom{}_2 F_1\left(1,-k;1+2k;-\frac{3}{2}\right)=1+\frac{\frac{3k}{2+4k}}{1-\frac{\frac{3k-3}{4+4k}}{1+\frac{3k-3}{4+4k}+\frac{\frac{3(2-k)}{2(3+2k)}}{1+\ldots}}}\in\left[4-\frac{60}{k+20},4-\frac{60}{k+53}\right],$$
so the asymptotic behaviour of $A_k$ is:
$$A_k=\left(\frac{243}{250}\right)^k\sqrt{\frac{12}{\pi k}}\left(1+\Theta\left(\frac{1}{k}\right)\right)$$
and $\{A_k\}_{k\in\mathbb{N}}$ decreases for sure from a certain point on. If we compute the constants hidden in the $\Theta$-notation, we can also notice that
$$\left(\frac{250}{243}\right)^k A_k$$
is a decreasing function, hence $A_k$ is a decreasing function, too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/870063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
| When $n\equiv 1\pmod{5}$, $n^4-1\equiv 1^4-1\equiv 0$.
When $n\equiv 2\pmod{5}$, $n^4-1\equiv 2^4-1=16-1=15\equiv 0$.
When $n\equiv 3\pmod{5}$, $n^4-1\equiv 3^4-1=81-1=80\equiv 0$.
When $n\equiv 4\pmod{5}$, $n^4-1\equiv 4^4-1=256-1=255\equiv 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Evaluate $\int_0^1 \sqrt{2x-1} - \sqrt{x}$ $dx$ I'm trying to calculate the area between the curves $y = \sqrt{x}$ and $y= \sqrt{2x-1}$
Here's the graph:
I've already tried calculating the area with respect to $y$, i.e.
$\int_0^1 (\frac{y^2+1}{2} - y^2)$ $ dx$
[since $y^2=x$ for the first curve and $\frac{y^2+1}{2}=x$ for the second curve]
And the result for that integral is $1/3$ which should be the same for the result of the integral in question
$\int_0^1 (\sqrt{2x-1} - \sqrt{x})$ $dx$
But, $\int_0^1 (\sqrt{2x-1} - \sqrt{x})$ $dx$ = $\int_0^1 (2x-1)^{1/2}$ $dx$ - $\int_0^1 {x}^{1/2} $ $dx$
[let $2x-1 = u$ and $2 dx = du$]
$\Rightarrow$ $\frac 12 \int_0^1 (u)^{1/2}$ $du$ - $\frac{2x^{3/2}}{3}$
$=\frac{u^{3/2}}{3}$ - $\frac{2x^{3/2}}{3}$
$=\frac{(2x-1)^{3/2}}{3}$ - $\frac{2x^{3/2}}{3}$
Evaluated at $x=1$ and $x=0$ and subtracting:
$(\frac{-1}{3})-(\frac{-1}{3})$
Why?
| I would say that you can integrate as you have suggested, however, despite the variable y:
$\int_0^1 (\frac{y^2+1}{2} - y^2)\color{red}{dy}=\frac{1}{2}[y-\frac{y^3}{3}]_0^1=\frac{1}{3}$
There is no reason to go back to the variable x.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a recurrence (with the form of a convolution) involving binomial coefficients While dealing with a problem related to intersection of hyperplanes I have come across the following recurrence to obtain the values of $K_{j}$
\begin{array}{cccccccccc}
1 & = & K_{1}\tbinom{n+1}{0}\\
1 & = & K_{1}\tbinom{n+2}{1} & +K_{2}\tbinom{n+2}{0}\\
1 & = & K_{1}\tbinom{n+3}{2} & +K_{2}\tbinom{n+3}{1} & +K_{3}\tbinom{n+3}{0}\\
1 & = & K_{1}\tbinom{n+4}{3} & +K_{2}\tbinom{n+4}{2} & +K_{3}\tbinom{n+4}{1} & +K_{4}\tbinom{n+4}{0}\\
& \vdots\\
1 & = & K_{1}\tbinom{n+j}{j-1} & +K_{2}\tbinom{n+j}{j-2} & +K_{3}\tbinom{n+j}{j-3} & +K_{4}\tbinom{n+j}{j-4} & +\cdots & +K_{j-2}\tbinom{n+j}{2} & +K_{j-1}\tbinom{n+j}{1} & +K_{j}\tbinom{n+j}{0}
\end{array}
That is, the convolution
$$1=\sum_{i=1}^{j}K_{i}\binom{n+j}{j-i}$$
or said otherwise, if we consider the sequence of coefficients, $K_{1},K_{2},K_{3},\ldots$
the j
th term is given by the recurrence
$$K_{j}=1-\sum_{i=1}^{j-1}K_{i}\binom{n+j}{j-i}$$
A direct calculation gives the following solutions
\begin{array}{cccc}
K_{1}= & +1 & = & +1\\
K_{2}= & -{n+1 \choose 1} & = & -\left({n+1 \choose 1}\right)\\
K_{3}= & +{n+2 \choose 2} & = & +\left({n+1 \choose 2}\right)\\
K_{4}= & -{n+3 \choose 3} & = & -\left({n+1 \choose 3}\right)\\
K_{5}= & +{n+4 \choose 4} & = & +\left({n+1 \choose 4}\right)\\
\vdots & \vdots & & \vdots\\
K_{j}= & (-1)^{j-1}{n+j-1 \choose j-1} & = & (-1)^{j-1}\left({n+1 \choose j-1}\right)
\end{array}
where ${n \choose i}=\frac{n(n-1)\cdots(n-i+1)}{i!}$
and $\left({N \choose i}\right)=\frac{n(n+1)\cdots(n+i-1)}{i!}$
However, this proves nothing.
Is it possible to derive a solution from the initial recurrence (convolution)?
| We start from the convolution $$1=\sum_{i=1}^j K_i \binom{n+j}{j-i}$$ then multiply both sides by $x^{j-1}$ and sum from $j=1$ to $\infty$. For the RHS we obtain
$$\sum_{j=1}^\infty \sum_{i=1}^j K_i \binom{n+j}{j-i}x^{j-1}=\sum_{i=1}^\infty \sum_{j=i}^\infty K_i \binom{n+j}{j-i}x^{j-1}$$ where we have reversed the order of summation.
The inner sum over $j$ may then be rewritten as
$$\sum_{j=i} \binom{n+j}{j-i}x^{j-1}=\sum_{j=0}^\infty\binom{n+i+j}{j}x^{j+i-1}
=\frac{x^{i-1}}{(1-x)^{n+i+1}}$$ where we have recognized the negative exponent binomial series $(1-x)^{-n-1}=\sum_{k=0}^\infty\binom{n+k}{k} x^k$. Completing the outer summation over $i$ then gives the RHS as
$$ \sum_{i=1}^\infty K_i \frac{x^{i-1}}{(1-x)^{n+i+1}}=\frac{1}{x (1-x)^{n+1}}\sum_{i=1}^\infty K_i \left(\frac{x}{1-x}\right)^i.$$ But the LHS is simply the geometric series $$\sum_{j=1}^\infty x^{j-1}=1+x+x^2+\cdots=\frac{1}{1-x},$$ and so we may write $$\sum_{i=1}^\infty K_i \left(\frac{x}{1-x}\right)^i=x(1-x)^n.$$ The substitution $x\mapsto x(1+x)^{-1}$ then brings this to the form
$$\sum_{i=0}^\infty K_{i+1} x^{i+1}=\frac{x}{(1+x)^{n+1}}=\sum_{i=0}^\infty \binom{n+i}{i} (-1)^i x^{i+1}$$ using the same binomial series as earlier. Identifying coefficients then finally yields $$K_i = (-1)^{i-1}\binom{n+i-1}{i-1}=(-1)^{i-1}\frac{(n+1)(n+2)\cdots(n+i-1)}{(i-1)!}$$ which is equivalent to the formula deduced in the question statement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Given system of equations $a+b = 2, ab=4$ solve $a^2+b^2=?$ and $a^3=?$ I am trying to solve $a^2+b^2$ and $a^3$ given $a+b = 2, ab=4$. I have the key with the answers $a^2+b^2=-4$ and $a^3=-8$ but am wondering which steps to take to get to that answer.
My understanding of the problem so far is that from this system of equations it is not possible to get integer solutions for $a$ and $b$ (which is not asked for). I have not really gotten further in solving it than this realization.
| HINT:
$a^2+b^2=(a+b)^2-2(ab)$
and
$a^3+b^3=(a+b)^3-3ab(a+b)$
EDIT: To further elongate the second equation
$a^3+b^3=-16$
Now, substitute b=4/a and solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/873583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $26$ is the one and only number between square and cube $x^2 + 1 = z = y^3 - 1$
Why $z = 26 $ and only $26$ ? Is there an elementary proof of that ?
| Instead write $$x^2+2=y^3$$ so that $x^2+2=(x+\sqrt{-2})(x-\sqrt{-2})$ is a norm from the integer ring $\Bbb Z[\sqrt{-2}]$ which is Euclidean. So then it is clear that $\gcd(x+\sqrt{-2},x-\sqrt{-2})\mid\sqrt{-2}$, so that if $\sqrt{-2}\nmid x$ they are coprime. Since $x$ is an integer, this means their $\gcd$ is $\sqrt{-2}$ iff $x$ is even and 1 otherwise.
Case 1: $\sqrt{-2}\mid x$ whence $x=2m$ so that we have $x^2+2=4m^2+2=2(m^2+1)$ so $2\mid y$ and $m=2k+1$ necessarily. But then $x^2+2=4(k^2+k+1)$ and $k^2+k+1$ is always odd $\bmod 4$, a contradiction since $8\mid y^3$. Hence $x$ is odd which puts us in
Case 2: $\gcd(x+\sqrt{-2},x-\sqrt{-2})=1$
Then $x+\sqrt{-2}=(a+b\sqrt{-2})^3$ and here's the kicker:
$$(a+b\sqrt{-2})^3=a^3+3a^2b\sqrt{-2}-6ab^2-2b^3\sqrt{-2} = (a^3-6ab^2)+(3a^2b-2b^3)\sqrt{-2}.$$
Now since $3a^2b-2b^3=b(3a^2-2b^2)=1$ it must be that $b=\pm 1$ and similarly $3a^2-2b^2=\pm 1$. Since $b^2=1$ this means $3a^2=2\pm 1$ clearly $3a^2=1$ is impossible, so $a=\pm 1$ gives us our only solutions.
So $x=a^3-6a=a(-5)$ the only positive possibility is $x=5$, which immediately gives $y=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
What is remainder when $5^6 - 3^6$ is divided by $2^3$ (method) I want to know the method through which I can determine the answers of questions like above mentioned one.
PS : The numbers are just for example. There may be the same question for BIG numbers.
Thnx.
| Another method is to simply reduce mod $8$. We know that $5^2 = 25 \equiv 1 \bmod 8$ and $3^2 = 9 \equiv 1 \bmod 8$.
Thus:
$5^6 - 3^6 = (5^2)^3 - (3^2)^3 \equiv 1^3 - 1^3 \equiv 0 \bmod 8$.
So $8$ divides $5^6 - 3^6$ with no remainder.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A problem on nested radicals Find the value of $x$ for all $a>b^2$ if:
$$\large x=\sqrt{a-b\sqrt{a+b\sqrt{a-b{\sqrt{a+b.......}}}}}$$
My attempt
$$\large x=\sqrt{a-b\sqrt{(a+b)x}}$$
$$\large x^4=(a-b)^2(a+b)x$$
$$\large x=((a-b)^2(a+b))^{1/3}$$ (real root)
Question: Is my solution correct??
| Your approach is not right unless your original equation was: $$x=\sqrt{(a-b)\sqrt{(a+b)\sqrt{(a-b)\sqrt{(a+b)\cdots}}}}$$
However, it's almost there.
$\displaystyle\begin{align}
x & = \sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+b\cdots}}}}
\\ x & =\sqrt{a-b\sqrt{a+bx}}
\\x^2 & = a-b\sqrt{a+bx}
\\(a-x^2)^2 & = b^2(a+bx)
\\ a^2-2ax^2+x^4 & = ab^2 + b^3x
\\ 0 & = x^4-2ax^2-b^3 x +a^2-ab^2
\\ 0 & = (x^2 +bx-a+b^2)(x^2-bx-a)
\\ \therefore x & \in \left\{\frac {-b-\sqrt{4a-3b^2}}2, \frac{-b+\sqrt{4a-3b^2}}2, \frac{b-\sqrt{4a+b^2}}2, \frac{b+\sqrt{4a+b^2}}2 \right\}
\end{align}$
Now apply the criteria $a>b^2$ to select among them for the satisfactory answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/874672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ordered partitions of an integer (with a twist) I would like to know how to prove (preferably algebraically) that $P_1(2,n)=F_{2n+1}$, where $P_1(2,n)$ is what I define to be the number of ordered partitions of an integer, where the number $1$ has 2 possible colours. For example,
\begin{align}
2
&=2\\
&={\color\red{1}}+{\color\red{1}}\\
&={\color\green{1}}+{\color\green{1}}\\
&={\color\green{1}}+{\color\red{1}}\\
&={\color\red{1}}+{\color\green{1}}\\
\end{align}
so in this case, $P_1(2,2)=5$.
This is my (failed) attempt to derive a formula for $P_1(k,n)$. Consider a case where the number $n$ is expressed as a sum of $m$ natural numbers, out of which there are $j$ number of $1$s. The number of possible ways to do so is given by
$$\binom{m}{j}[x^n]x^j(x^2+x^3+...)^{m-j}$$
Now we look at how many partitions we can get from this particular case. Since there are $j$ number of $1$s, we would have to multiply by $k^j$, and since $0\le j \le m$ and $1 \le m \le n$ we have
\begin{align}
P_1(k,n)
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}[x^n]x^j(x^2+x^3+...)^{m-j}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}[x^{n-2m+j}](1-x)^{-(m-j)}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{m-j+n-2m+j-1}{n-2m+j}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
&=2^{n-1}+\sum^{n-1}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
\end{align}
I have tried using the "snake-oil" method to compute this sum, that is,
$$P_1(2,n)=2^{n-1}+[x^{n-1}]\sum_{n\ge 1}\sum^{n-1}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}x^{n-1}$$
but I, embarrassingly, have trouble with changing the order of summation and determining the limits of summation.
This is where I am stuck, and I would like to seek your help in finding a closed form for this sum, and in particular, proving that
\begin{align}
P_1(2,n)
&=2^{n-1}+\sum^{n-1}_{m=1}\sum^{m}_{j=0}2^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
&=F_{2n+1}
\end{align}
Help will be greatly appreciated. Thank you.
| Thanks to @paw88789's insightful explanation, I was able to derive a closed form for $P_1(2,n)$ without using any prior knowledge of the answer. The recurrence relation is
\begin{align}
a_n
&=2a_{n-1}+a_{n-2}+...+a_1+1\\
&=1+a_{n-1}+\sum^{n-1}_{m=1}a_m
\end{align}
Applying the forward difference operator $\Delta_n$ to both sides yields
\begin{align}
a_{n+1}-a_n&=a_n-a_{n-1}+a_n\\
a_{n+2}&=3a_{n+1}-a_n
\end{align}
Define the generating function
$$A(x)=\sum_{n\ge 1}a_nx^n$$
In terms of $A(x)$, the recurrence relation is
$$\frac{A(x)-2x-5x^2}{x^2}=\frac{3A(x)-6x}{x}-A(x)$$
Solving for $A(x)$ gives
\begin{align}
A(x)
&=\frac{2x-x^2}{1-3x+x^2}\\
&=\frac{1-x}{1-3x+x^2}-1\\
\end{align}
Extract the coefficient
\begin{align}
a_n
&=[x^n]\left(\frac{1-x}{1-3x+x^2}\right)\\
&=\frac{1}{\sqrt{5}}[x^n]\left(\frac{\tau}{x-(\tau+1)}-\frac{\phi}{x-(\phi+1)}\right)\\
&=\frac{1}{\sqrt{5}}\left(\phi\left(\frac{1}{\phi +1}\right)^{n+1}-\tau\left(\frac{1}{\tau +1}\right)^{n+1}\right)\\
&=\frac{1}{\sqrt{5}}\left(\phi(1+\tau)^{n+1}-\tau(1+\phi)^{n+1}\right)\\
&=\frac{1}{\sqrt{5}}\left(\phi^{2n+1}-\tau^{2n+1}\right)\\
\end{align}
where $\phi=\dfrac{1+\sqrt 5}{2}$ and $\tau=\dfrac{1-\sqrt 5}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Show that: $ (\csc\theta - \sin\theta)(\sec\theta - \cos\theta) \equiv \frac{1}{\tan\theta + \cot\theta}$ I am having hard time solving this question, I start with the L.H.S and my answer always boils down to
$\sin\theta\cos\theta$
And I don't know what to do after that.
| $$ (\csc\theta - \sin\theta)(\sec\theta - \cos\theta) =\frac{(1-\sin^2\theta)(1-\cos^2\theta)}{\sin\theta \cos\theta}= \frac{(\cos^2\theta)(\sin^2\theta)}{\sin\theta \cos\theta}=(\cos\theta)(\sin\theta)=\frac{(\cos\theta)(\sin\theta)}{\sin^2\theta+ \cos^2\theta}=\frac{1}{\tan\theta + \cot\theta}$$
The last inequality comes from dividing numerator and denominator by the same number $(\cos\theta)(\sin\theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/878118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of this double integral? Let $C$ be the subset of the plane given by
$$ C \colon= \{ \ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x^2 + y^2 \leq 1 \}.$$
Then what is the value of the double integral
$$ \int_{C} \int (x^2 + y^2) \ dx \ dy?$$
My work:
In $C$, we have $-1 \leq x \leq 1$ and $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$. So we have
$$ \int_C \int (x^2 + y^2) \ dx \ dy = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \ (x^2 + y^2) \ dy \ dx = 2 \int_{-1}^1 (x^2 \sqrt{1-x^2} + \frac{ (\sqrt{1-x^2})^3}{3} ) \ dx = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx. $$ Now let $x= \sin t$. Then we obtain
$$ \int_C \int (x^2 + y^2) \ dx \ dy = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx = \frac{4}{3} \int_{t = 0}^{\pi/2} (2\sin^2 t + 1) \cos^2 t \ dt \ \ = \frac{1}{3} \int_{t=0}^{\pi/2} (2\sin^2 2t + 4 \cos^2 t) \ dt = \frac{4}{3} \int_{t=0}^{\pi/2} ( 1 - \cos 4t + 2(1+ \cos 2t) ) \ dt = \frac{1}{3} \int_{t=0}^{\pi/2} (3 + 2 \cos 2t - \cos 4t) \ dt = \frac{\pi}{2}. $$ Am I right?
| If you use polar coordinates you will get the following
$$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$
| {
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"url": "https://math.stackexchange.com/questions/879489",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
| Let $a+b+c=3$.
Hence, we need to prove that:
$$\sum_{cyc}\frac{a}{(3-a)^2}\geq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{a}{(3-a)^2}-\frac{1}{4}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)(9-a)}{(3-a)^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(a-1)(9-a)}{(3-a)^2}-2(a-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\geq0$$
and since $\{a,b,c\}\subset(0,3)$, we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Finding (or rather expanding) the product $(5-xy)(3+xy)$ Given the product $(5-xy)(3+xy)$
I tried the following,
As we know, $(x+a)(x+b)=x^2+(a+b)x+ab$
Here $x$ is $xy$. But $xy$ has two signs$-$ and $+$.
How do I solve this.
| $$ \begin {align*} (5 - xy)(3 + xy) &= 5 \cdot 3 + 3 \cdot (-xy) + 5 \cdot (xy) + (-xy) \cdot (xy) \\&= 15 - 3xy + 5xy - (xy)^2 \\&= 15 + 2xy - x^2y^2. \end {align*} $$Hope that makes sense!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Why does $\sum_{n=1}^\infty \sqrt{n+1}-\sqrt{n}$ diverge? Why does $\sum_{n=1}^\infty \sqrt{n+1}-\sqrt{n}$ diverge?
Using the ratio test I get the following. First of all since $u_n=\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$
Then $\left| \frac{u_{n+1}}{u_n} \right|=\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}}\lt 1$.
By the ratio test, the series should converge, but it does not. What am I doing wrong?
| The ratio test actually doesn't conclude anything. Note that
$$
\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}} = \frac{\sqrt{1 + 1/n} + 1}{\sqrt{1+2/n}+\sqrt{1+1/n}} \to \frac{1+1}{1+1} = 1
$$
Now notice that for large $n$ we have
$$
n > 2\sqrt{n+1} > \sqrt{n+1} + \sqrt{n} \implies \frac{1}{n} < \frac{1}{\sqrt{n+1}+\sqrt{n}}
$$
So we get that the series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $a+b+c \geq ab+bc+ca$, given an additional constraint
If $a,b,c$ are positive real numbers satisfying
$$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1$$
then I'd like to prove that
$$a+b+c \geq ab+bc+ca\,.$$
Additional info:
We should only use Cauchy-Schwarz (preferred to be used at least once) and AM-GM. We are not allowed to use induction.
Things I have tried so far: Using the Cauchy-Schwarz on the given constraint inequality, I can show
$$\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)\big((a+b+1)+(b+c+1)+(c+a+1)\big)\geq (1+1+1)^2$$
$$\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)(2(a+b+c)+3)\geq9$$
$$2(a+b+c)+3\geq 9$$
$$a+b+c\geq 3$$
So my idea is right now to show $3\geq ab+bc+ca$.
And trying Cauchy-Schwarz on problem statment:$$(a+b+c)(a^2b+b^2c+c^2a)\geq(ab+bc+ca)^2$$
| Hint:Use Cauchy-Schwarz inequality
$$(a+b+c^2)(a+b+1)\ge (a+b+c)^2$$
so
$$\dfrac{1}{a+b+1}\le\dfrac{a+b+c^2}{(a+b+c)^2}$$
so
$$1\le \sum_{cyc}\dfrac{1}{a+b+1}\le\sum_{cyc}\dfrac{a+b+c^2}{(a+b+c)^2}=\dfrac{2(a+b+c)+a^2+b^2+c^2}{(a+b+c)^2}$$
so
$$a+b+c\ge ab+bc+ac$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, ...
What is the missing number?
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
| $$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}$$
The above is the same as $\displaystyle\frac1{16},\frac2{16},\frac3{16},\frac4{16},\frac5{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Limit of a definite integral We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$
Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5+u^6}$$
Then by partial fractions, which I did manually and chcecked with WolframAlpha afterwards, it becomes $$\begin {align}
2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\left(\frac{1}{u^5}-\frac{1}{u^4}+\frac{1}{u^3}-\frac{1}{u^2}+\frac{1}{u}-\frac{1}{1+u}\right) du =\\
\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}2\left(\log{u}-\log{(1+u)}+\frac{1}{u}-\frac{1}{2u^2}+\frac{1}{3u^3}-\frac{1}{4u^4}\right) du =\\
\lim_{x \to 0}\int_{\sin x}^{x}\left(\log{t}-2\log{(1+\sqrt{t})}+\frac{2}{\sqrt{t}}-\frac{1}{t}+\frac{2}{3t^{3/2}}-\frac{1}{2t^2}\right) dt
\\\end{align}$$
Fianlly we obtain the following limit:
$$\lim_{x \to 0}\left(\log {x}-\log {\sin x}+2\log {(1+\sqrt{x})}-2\log {(1+\sqrt{\sin x})}+\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}-\frac{1}{x}+\frac{1}{\sin x}+\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}-\frac{1}{2x^2}+\frac{1}{\sin^2 x}\right)$$
Here's where I stuck. It gets messy when I try to calculate $\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}$ and $\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}$. The rest is rather doable - de l'Hospital's rule is useful with $-\frac{1}{x}+\frac{1}{\sin x}$ which is $0$ in limit, so as logarithm expressions (obviously) and Taylor expansion helps with $-\frac{1}{2x^2}+\frac{1}{\sin^2 x}$ which, in turn, equals $1/6$ when $x$ approaches $0$.
Did I make any mistakes? I hope not, but even so I'm not certain what to do with this horrible limit. I'd be glad if anyone could point out what to do.
| When you substitute a new variable in, you must not forget to change the limits of your integral.
So,
instead of $$x, \sin x$$
We have
$$\sqrt{x}, \sqrt{\sin x}$$
Everything else looks correct.
EDIT
To solve the limit:
$$\lim_{x\to 0} \frac{2}{\sqrt{x}} + \frac{2}{\sqrt{\sin x}}$$
$$ = \lim_{x\to 0} \frac{2(\sqrt{\sin{x}} - \sqrt{x})}{\sqrt{x}\sqrt{\sin x}}$$
Use l'hopsitals
Can you get it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Find the following limit Find $\displaystyle \lim_{x \rightarrow 1} \frac{x\log x}{x-x^4}$.
My approach was that canceling out both $\displaystyle x$, then I have $\displaystyle \frac{\log{x}}{1-x^3}$.
Since $\displaystyle 1-x^3 = (1-x)(x^2 - x + \frac{1}{2})$, so that $\displaystyle \frac{\log{x}}{1-x^3}$ is same as $\displaystyle \frac{\log{x}}{(1-x)(x^2 -x +\frac{1}{2})}$
Then I don't know how to carry on from here.
You guys are amazing, Thank you so much :) I have received so many useful ideas to solve one limit problem. Learnt so much today.
| You can also do using Taylor series built at $x=1$. So, $$\log(x)=(x-1)-\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$ $$\frac{1}{1-x^3}=-\frac{1}{3 (x-1)}+\frac{1}{3}-\frac{2 (x-1)}{9}+\frac{1}{9} (x-1)^2+O\left((x-1)^3\right)$$ Multiplying the first by the second leads to $$\displaystyle \frac{\log{x}}{1-x^3}=-\frac{1}{3}+\frac{x-1}{2}-\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$
Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$
My attempt at the question:
Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x:
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$
Therefore
$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}} $$
Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$
But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2.
we get: $$f'(1) = \frac13 $$
Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.
| Differentiate both sides to get $f'(x-\frac{2}{x})(1+\frac{2}{x^2})=\frac{1}{2\sqrt{x-1}}$. Then use $x=2$ to evaluate, $f'(1)=\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Recurrence of the form $2f(n) = f(n+1)+f(n-1)+3$ Can anyone suggest a shortcut to solving recurrences of the form, for example:
$2f(n) = f(n+1)+f(n-1)+3$, with $f(1)=f(-1)=0$
Sure, the homogenous solution can be solved by looking at the characteristic polynomial $r^2-2x+1$, so that in general a solution for the homogenous equation is of the form $f^h(n) = c_1+c_2n$. But how does one deal with the constant 3 in this case?
| $$2f(n) = f(n+1)+f(n-1)+3$$
$$f(n+1) = 2f(n) - f(n-1) - 3$$
$$
\begin{bmatrix} f(n+1) \\ f(n) \end{bmatrix}
=
\begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix} f(n) \\ f(n - 1) \end{bmatrix}
+
\begin{bmatrix} -3 \\ 0 \end{bmatrix}
$$
$$
\begin{bmatrix} f(n+1) \\ f(n) \\ 1 \end{bmatrix}
=
\begin{bmatrix} 2 & -1 & -3 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} f(n) \\ f(n - 1) \\ 1 \end{bmatrix}
$$
$$
\begin{bmatrix} f(n+1) \\ f(n) \\ 1 \end{bmatrix}
=
\begin{bmatrix} 2 & -1 & -3 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^n
\begin{bmatrix} f(1) \\ f(0) \\ 1 \end{bmatrix}
$$
And if that form isn't closed enough for you, Jordan Decomposition:
$$\begin{align}
\begin{bmatrix} f(n+1) \\ f(n) \\ 1 \end{bmatrix}
&=
\left(\begin{bmatrix} -3 & -3 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} -3 & -3 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}^{-1}\right)^n
\begin{bmatrix} f(1) \\ f(0) \\ 1 \end{bmatrix}
\\ & =
\begin{bmatrix} -3 & -3 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}^n
\begin{bmatrix} -3 & -3 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}^{-1}
\begin{bmatrix} f(1) \\ f(0) \\ 1 \end{bmatrix}
\\ & =
\begin{bmatrix} -3 & -3 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & n & \frac{n(n-1)}2 \\ 0 & 1 & n \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} -3 & -3 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}^{-1}
\begin{bmatrix} f(1) \\ f(0) \\ 1 \end{bmatrix}
\\ &=
\begin{bmatrix} n+1 & -n & \frac{-3n^2 - 3n}2 \\ n & 1 - n & \frac{-3n^2 + 3n}2 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} f(1) \\ f(0) \\ 1 \end{bmatrix}
\end{align}$$
$$f(n) = n\,f(1) + (1-n)\,f(0)+ \frac{-3n^2 + 3n}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $\tan 54^\circ=\frac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$ How to prove this identity without using the actual values of $\tan54^\circ$ and $\sin24^\circ$
$$\tan 54^\circ=\dfrac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$$
Edit: I still don't get it, I am stuck at:
$$\dfrac{\cos 24^\circ+\sqrt{3}\sin 24^\circ}{\sqrt{3}\cos 24^\circ-\sin 24^\circ}$$
P.S. : Is this coincidence or a more general form can be obtained?
| Finally,,,after much effort I have a proof.
Note that $2\cdot 36^{\circ} =180^{\circ}-3\cdot 36^{\circ}$ therefore
$$\sin 2\cdot 36^{\circ}\cdot =\sin 3\cdot 36^{\circ}$$ and thus
$$2\sin36^{\circ}\cos 36^{\circ}=3\sin 36^{\circ}-4\sin^336^{\circ}$$
and therefore
$$2\cos 36^{\circ}=3-4\sin^2 36^{\circ}$$
so
$$2\cos 36^{\circ}=3\cos^2 36^{\circ}-\sin^2 36^{\circ}=(\sqrt{3}\cos36^{\circ}-\sin 36^{\circ})(\sqrt{3}\cos36^{\circ}+\sin36^{\circ})$$
Now
$$\sin24^{\circ}=\sin(60^{\circ}-36^{\circ})=\frac{\sqrt{3}}{2}\cos36^{\circ}-\frac{1}{2}\sin36^{\circ}$$
and therefore with the last equation we have
$$\cos 36^{\circ}=\sin24^{\circ}(\sqrt{3}\cos36^{\circ}+\sin36^{\circ})$$
dividing by $\sin36^{\circ}$ we get
$$\cot 36^{\circ}=\sin24^{\circ}(\sqrt{3}\cot36^{\circ}+1)$$
Note that $\tan54^{\circ}=\tan(90^{\circ}-36^{\circ})=\cot36^{\circ}$
$$\tan54^{\circ}=\sin24^{\circ}(\sqrt{3}\tan54^{\circ}+1)$$ from which
$$\tan54^{\circ}=\frac{\sin24^{\circ}}{1-\sqrt{3}\sin24^{\circ}}$$
follows.
The moral is that $2\cdot 18^{\circ}+3\cdot 18^{\circ}=90^{\circ}$ yields some nice trig relations that should be taken note of.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/889971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A nice trignometric identity How to prove that:
$$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=\dfrac{\sqrt{13}-1}{4} $$
I have a solution but its quite lengthy, I would like to see some elegant solutions. Thanks!
| As @Tunk-Fey said in the comment, we only need to show $4x^2 + 2x - 3 =0$ with $x = \cos\frac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13} $ since we know $x$ should be positive.
Take $\theta = \frac{2\pi}{13}$, $w = \exp(i\theta)$ and $$A = w + w^3 + w^4$$
$$B = w^{-1} + w^{-3} + w^{-4}$$
we have $A + B = 2x$, thus we only need to show $(A+B)^2 + A+B - 3 = 0$
We will need the following facts:
since we have $w^{13} - 1 = (w-1)\sum_{k=0}^{12}w^k=0$ thus $\sum_{k=0}^{12}w^k = 0$, dividing by $w^6$($\neq 0$) gives
$$\sum_{k=1}^6(w^k + w^{-k}) + 1 =0$$
and we have also $$w^5 + w^{-5} = w^8 + w^{-8}$$ $$w^6 + w^{-6} = w^7 + w^{-7}$$
Simple computation gives
\begin{align}
(A+B)^2 =&w^2 + w^{-2} + w^{6} + w^{-6} + w^8 + w^{-8} + 6\\
&+2(w + w^{-1} + w^2 + w^{-2} + w^3 + w^{-3} + w^4 + w^{-4} + w^5 + w^{-5} + w^{7}+w^{-7}) \\
\end{align}
replacing $w^{7}+w^{-7}$ by $w^{6}+w^{-6}$ gives
\begin{align}
(A+B)^2 = &w^2 + w^{-2} + w^{6} + w^{-6} + w^8 + w^{-8} + 6+2(-1) \\
=& w^2 + w^{-2} + w^5 + w^{-5} + w^6 + w^{-6} + 4
\end{align}
Now adding together $(A+B)^2$, $A+B$ and $-3$ allows to conclude.
| {
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"url": "https://math.stackexchange.com/questions/890052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$ Question:
$$\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$$
Prove that L.H.S.=R.H.S.
My Efforts:
L.H.S.$$=\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}\times\frac{\csc\theta+1}{\csc\theta+1}$$
$$=\frac{\tan^2\theta(1-\cos\theta)(\csc^2\theta-1)}{(1-\cos^2\theta)(\csc\theta+1)}$$
| Hint:
$$\tan^2\theta (\csc^2\theta-1) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1}{\sin^2\theta}-1\right) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1}{\sin^2\theta}-\frac{\sin^2\theta}{\sin^2\theta}\right) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1-\sin^2\theta}{\sin^2\theta}\right) = \frac{\sin^2\theta}{\cos^2\theta}\frac{\cos^2\theta}{\sin^2\theta} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find the minimum of the $n$ such $99^n+100^n<101^n$ Question:
Find the smallest positive integer $n$ such that
$$99^n+100^n<101^n$$
My idea: This is equivalent to
$$\left(\dfrac{99}{101}\right)^n+\left(\dfrac{100}{101}\right)^n<1$$
so
$$\left(1-\dfrac{2}{101}\right)^n+\left(1-\dfrac{1}{101}\right)^n<1$$
Use this:
$$(1+x)^n\ge 1+nx,x>-1,n\ge 1$$
so
$$1-\dfrac{2}{101}n+1-\dfrac{1}{101}n<1\Longrightarrow n>\dfrac{101}{3}=33.666$$
But I found $n=34$ is not sufficient. I used a computer to find $n\ge 49$.
I want to see how to find it by hand. Thank you.
| First of all, $n \ge 34$ is a necessary (but not sufficient) condition for that inequality to be true. Remember that $1+nx \le c$ does not imply that $(1+x)^n \le c$.
We can get a better bound by manipulating the inequality as follows:
$99^n + 100^n < 101^n$
$101^n - 99^n > 100^n$
$\left(\dfrac{101}{100}\right)^n - \left(\dfrac{99}{100}\right)^n > 1$
$\left(1+\dfrac{1}{100}\right)^n - \left(1-\dfrac{1}{100}\right)^n > 1$
Using the binomial theorem, this becomes:
$2\dbinom{n}{1}\dfrac{1}{100} + 2\dbinom{n}{3}\dfrac{1}{100^3} + 2\dbinom{n}{5}\dfrac{1}{100^5} + \cdots > 1$
All the terms on the left side are positive and first term is $\dfrac{n}{50}$, so the inequality holds for $n \ge 50$.
This is a sufficient condition, but not a necessary one. For $n = 49$, the first two terms exceed $1$, so the inequality holds for $n = 49$ as well.
Now, all that remains is to show that the inequality does not hold for $n \le 48$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove that $ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$? Let $ a,b,c>0$ be such that $ a+b+c=1$. How prove that
$ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$?
| Let $a+b=2u$ and $ab=v^2$.
Hence, we need to prove that $3(a^3+b^3)+1-3(1-a-b)\geq\frac{a^2+b^2+1-4(1-a-b)}{a+b}$ or
$$6u(4u^2-3v^2)+1-3(1-2u)\geq\frac{4u^2-2v^2+1-4(1-2u)}{2u},$$
which is a linear inequality of $v^2$, which says that it's enough to prove our inequality for an extremal value of $v^2$, which happens in the following cases.
*
*$v^2\rightarrow0^+$.
Let $b\rightarrow0^+$. We obtain $3a^3+3a-2\geq\frac{a^2+4a-3}{a}$ or
$$3a^4+2a^2-6a+3\geq0,$$
which is true by AM-GM: $$3a^4+2a^2+3=3a^4+2a^2+4\cdot\frac{3}{4}\geq6\sqrt[6]{3\cdot2\cdot\left(\frac{3}{4}\right)^4}a>6a$$
2. $b=a$, which gives $$6a^3+6a-2\geq\frac{2a^2+8a-3}{2a}$$ or
$$12a^4+10a^2-12a+3\geq0$$ or
$$\left(a^2-\frac{1}{4}\right)^2+\frac{4}{3}\left(a-\frac{3}{8}\right)^2\geq0,$$
which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$ I would like to know how to evaluate the integral
$$\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$$
I tried expanding the integrand as a series but made little progress as I do not know how to evaluate the resulting sum.
\begin{align}
\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx
&=\int^1_0\sum_{n \ge 1}\frac{x^n}{n^2}\sum_{k \ge 0}x^k\ln{x}dx\\
&=\sum_{n \ge 1}\frac{1}{n^2}\sum_{k \ge 0}\int^1_0x^{n+k}\ln{x}dx\\
&=-\sum_{n \ge 1}\frac{1}{n^2}\sum_{k \ge 0}\frac{1}{(n+k+1)^2}
\end{align}
I am aware that a similar question has been answered here, however, I find that the answers are not detailed enough for someone who has a shallow understanding on Euler sums, such as myself, to fully comprehend. Hence, I would like to seek your help on the techniques that can be used to evaluate this integral. Thank you.
| \begin{align}\text{J}&=\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[-\ln(1-x)\ln x\mathrm{Li}_2(x)\Big]_0^1}_{=0}+\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(x)}{x}dx}_{=-\frac{1}{2}\text{Li}^2_2(1)}-\underbrace{\int_0^1 \frac{\ln^2(1-x)\ln x}{x}dx}_{\text{IBP}}\\
&=-\frac{\pi^4}{72}-\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx\\
&\overset{\text{IBP}}=-\frac{\pi^4}{72}-\left[\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1\frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1-\\&\int_0^1 \frac{1}{1-x}\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\
&=-\frac{\pi^4}{72}-\int_0^1 \frac{1}{1-x}\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\
&=-\frac{\pi^4}{72}-\int_0^1 \int_0^1 \left(\frac{x\ln^2(tx)}{(1-tx)(1-x)}-\frac{\ln^2 t}{(1-x)(1-t)}\right)dtdx\\
&=-\frac{\pi^4}{72}+\int_0^1\int_0^1\left(\frac{\ln^2(tx)}{(1-t)(1-tx)}+\frac{\ln^2 t}{(1-t)(1-x)}-\frac{\ln^2(tx)}{(1-t)(1-x)}\right)dtdx\\
&=-\frac{\pi^4}{72}+\int_0^1\int_0^1\left(\frac{\ln^2(tx)}{(1-t)(1-tx)}-\frac{\ln^2 x}{(1-t)(1-x)}\right)dtdx-\\&2\underbrace{\left(\int_0^1 \frac{\ln t}{1-t}dt\right)}_{=-\frac{\pi^2}{6}}\left(\int_0^1 \frac{\ln x}{1-x}dx\right)\\
&=-\frac{5\pi^4}{72}+\int_0^1 \left(\frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)-\frac{1}{1-t}\left(\int_0^1\frac{\ln^2 x}{1-x}dx\right)\right)dt\\
&=-\frac{5\pi^4}{72}+\int_0^1 \frac{1}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^2 u}{1-u}du\right)dt\\
&\overset{\text{IBP}}=-\frac{5\pi^4}{72}-\underbrace{\int_0^1 \frac{\ln^3 t}{1-t}dt}_{=-6\zeta(4)=-\frac{\pi^4}{15}}+\underbrace{\int_0^1 \frac{\ln(1-t)\ln^2 t}{1-t}dt}_{-\frac{\pi^4}{72}-\text{J}}\\
&=-\frac{\pi^4}{60}-\text{J}\\
&=\boxed{-\dfrac{\pi^4}{120}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
} |
Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$. The exercise is:
Calculate:$$\int\frac{dx}{x\sqrt{x^2-2}}$$
My first approach was:
Let $z:=\sqrt{x^2-2}$ then $dx = dz \frac{\sqrt{x^2-2}}{x}$ and $x^2=z^2+2$ $$\int\frac{dx}{x\sqrt{x^2-2}} = \int\frac{1}{x^2}dz = \int\frac{1}{z^2+2}dz = \frac{1}{2}\int\frac{1}{(\frac{z}{\sqrt{2}})^2 + 1}dz = \frac{1}{2} \arctan(\frac{z}{\sqrt{2}})$$
$$ = \frac{1}{2}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$
Yet when e.g. Maple derives the expression, it turns out my solution is false. I also cannot find the mistake, and I'd be glad if someone could point it out to me.
I am interested in your solutions, but in another approach I substituted $z:=\frac{\sqrt{2}}{x}$ and multiplied the integrand with $\frac{\sqrt{2}x}{\sqrt{2}x}$ and it worked out just fine.
| In the step$$\frac{1}{2}\int\frac{1}{(\frac{z}{\sqrt{2}})^2 + 1}dz=\frac{1}{2} \arctan(\frac{z}{\sqrt{2}})$$ you made a small mistake. The integral $$\int\frac{1}{\left(\frac{x}{a}\right)^2+1}dx=a\cdot \arctan\left(\frac{x}{a}\right)$$ not $\arctan\left(\frac{x}{a}\right)$
So if you multiply by the factor of $\sqrt{2}$ that you forgot, your answer is right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Definite integral into indefinitie series Convert $\displaystyle \int_0^1 e^{x^2}\, dx$ to an infinite series.
| As $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$
$$\int_0^1e^{x^2}dx\\\eqalign{
&=\int_0^1\left(1+x^2+\frac{(x^2)^2}{2!}+\frac{(x^2)^3}{3!}+\cdots\right) dx\\
&=\int_0^1\left(1+x^2+\frac{(x^4)}{2!}+\frac{(x^6)}{3!}+\cdots\right)dx\\
&=\left(x+\frac{x^3}{3.1!}\frac{x^5}{(5.2)}+\frac{x^7}{(7.3!)}+\cdots\right)_0^1\\
&=\frac1{1.0!}+\frac1{3.1!}+\frac1{(5.2!)}+\frac1{(7.3!)}+...\\
&=\sum_{k=0}^{\infty}\frac1{k!(2k+1)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Could someone show me what has been done and explain why?
| Both solutions are correct provided your $d$ satisfies the condition $y(1) = 1$. Note that in your solution $y(1) = \frac{A}{3} + \frac{B}{2} + c + d = 1$ so setting $d = 1 - (\frac{A}{3} + \frac{B}{2} + c)$ would be the way to go.
Note also that
$$\frac{Ax^3}{3} + \frac{Bx^2}{2} + cx + \underbrace{1-\left(\frac{A}{3} + \frac{B}{2}+c\right)}_{d} =\\ = \frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to evaluate $\sum_{k=1}^n\ln\left(2\cos\left(\frac{2\pi\cdot3^k}{3^n+1}\right)+1\right)$ By using wolfram alpha, it seems like that
$$\sum_{k=1}^n\ln\left(2\cos\left(\frac{2\pi\cdot3^k}{3^n+1}\right)+1\right)=0 \text{ for all }n\in\mathbb{N}.$$
But I don't know how to prove this identity. Thank you very much.
| Taking exponential of both sides, you want
$$ \prod_{k=1}^n \left(2 \cos\left(\dfrac{2\pi\cdot 3^k}{3^n+1}\right)+1\right) = 1 $$
Now note that $1 + 2 \cos(x) = \sin(3x/2)/\sin(x/2)$
and the product telescopes to become
$$\dfrac{\sin \left( \dfrac{3^{n+1}\pi}{3^n+1}\right)}{\sin \left( \dfrac{3 \pi}{3^n+1}\right)} = \dfrac{\sin \left( 3\pi - \dfrac{3\pi}{3^n+1}\right)}{\sin \left( \dfrac{3 \pi}{3^n+1}\right)} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Problem getting the real roots of this complex expression I'm trying to get the real roots of this expression:
$$\dfrac{1}{z-i}+\dfrac{2+i}{1+i} = \sqrt{2}$$
Where $i^2=-1$ and $z=x+iy$.
I tried to simplify that with Algebra, and then separate the real and imaginary parts in both sides of the expression to obtain an equation system, so I would solve it to obtain the roots for both $x$ and $y$. But all I get is a mess!
Any help would be appreciated, thank you! :)
P.S. It comes again from a Russian book, it says the answer is: there aren't real solutions. And with the procedure I said, I got real solutions!
P.P.S. I'd write down what I did, but I don't have the written steps anymore, sorry :(
| This can be done by "getting the unknown alone" by simplifying the fraction and undoing each operation on the left side.
We get
$$\begin{gathered}
\frac{1}{{z - i}} + \frac{{2 + i}}{{1 + i}} = \sqrt 2 \\
\frac{1}{{z - i}} + \frac{3}{2} - \frac{1}{2}i = \sqrt 2 \\
\frac{1}{{z - i}} = \sqrt 2 - \frac{3}{2} + \frac{1}{2}i \\
z - i = - \frac{1}{3} + \left( { - \frac{2}{3}\sqrt 2 - 1} \right)i \\
z = - \frac{1}{3} - \frac{2}{3}\sqrt 2 i \\
\end{gathered} $$
As you see, there is no real solution: just one complex one. I left out the hairy division and reciprocal steps: let me know if you need them.
Here are more details on the first division:
$$\begin{gathered}
\frac{{2 + i}}{{1 + i}} \\
= \frac{{(2 + i)(1 - i)}}{{(1 + i)(1 - i)}} \\
= \frac{{2 - 2i + i - {i^2}}}{{1 - {i^2}}} \\
= \frac{{2 - 2i + i + 1}}{{1 + 1}} \\
= \frac{{3 - i}}{2} \\
= \frac{3}{2} - \frac{1}{2}i \\
\end{gathered} $$
And here is the reciprocal:
$$\begin{gathered}
\frac{1}{{\sqrt 2 - \frac{3}{2} + \frac{1}{2}i}} \\
= \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{\left( {\sqrt 2 - \frac{3}{2} + \frac{1}{2}i} \right)\left( {\sqrt 2 - \frac{3}{2} - \frac{1}{2}i} \right)}} \\
= \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{{{\left( {\sqrt 2 - \frac{3}{2}} \right)}^2} - {{\left( {\frac{1}{2}i} \right)}^2}}} \\
= \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{\left( {2 - 3\sqrt 2 + \frac{9}{4}} \right) + \frac{1}{4}}} \\
= \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{\frac{9}{2} - 3\sqrt 2 }} \\
= \frac{{\left( {\sqrt 2 - \frac{3}{2} - \frac{1}{2}i} \right)\left( {\frac{9}{2} + 3\sqrt 2 } \right)}}{{\left( {\frac{9}{2} - 3\sqrt 2 } \right)\left( {\frac{9}{2} + 3\sqrt 2 } \right)}} \\
= \frac{{\frac{9}{2}\sqrt 2 + 3 \cdot 2 - \frac{{27}}{4} - \frac{9}{2}\sqrt 2 - \frac{9}{4}i - \frac{3}{2}\sqrt 2 i}}{{\frac{{81}}{4} - 9 \cdot 2}} \\
= \frac{{ - \frac{3}{4} + \left( { - \frac{3}{2}\sqrt 2 - \frac{9}{4}} \right)i}}{{\frac{9}{4}}} \\
= - \frac{1}{3} + \left( { - \frac{2}{3}\sqrt 2 - 1} \right)i \\
\end{gathered} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Sum of the $11^\mathrm{th}$ power of the roots of the equation $x^5+5x+1=0$
Find the sum of the $11^\mathrm{th}$ power of the all roots of the equation
$$
x^5+5x+1=0
$$
My Attempt:
Let $R=\{\alpha,\beta,\gamma,\delta,\mu\}$ be the set of all roots of the equation ${x^5+5x+1=0}$, and let $x\in R$. Then we have
$$
\begin{align}
x^5 &= -\left(5x+1\right)\\
x^{10}&=25x^2+1+10x\\
\tag1 x^{11}&=25x^3+10x^2+x
\end{align}
$$
Taking the sum of $(1)$ on all elements $x\in R$ gives us
$$
\sum_{x\in R}x^{11}=25\cdot\sum_{x\in R}x^3+10\cdot\sum_{x\in R}x^2+\sum_{x\in R}x
$$
How can I solve the problem from this point?
| You can use the technique of Newton's Sums which @user157227's comment hinted at
SOLUTION:
Basically imagine if you factored the polynomial into its 5 roots
$$(x - r_1)(x - r_2)..(x - r_5)$$
And then expanded that out:
it quickly will become evident you get an answer of the form
$$a_0 + a_1x + a_2x^2 ... a_4x^4 + x^5$$
such that
$$ \begin{matrix} a_4 = r_1 + r_2 ... r_5 \\ a_3 = r_1r_2 + r_1r_3 + ... + r_1r_5 + r_2 r_3 + ... + r_2 r_5 + ... r_4 r_5 \\ a_2 = r_1 r_2 r_3 + r_1 r_2 r_4 + ... r_3 r_4 r_5 \\ a_1 = r_1r_2r_3r_4 + ... +r_2 r_3r_4r_5 \\ a_0 = r_1r_2r_3r_4r_5 \end{matrix} $$
So to solve the case for sum of all 11th powers of the roots we need to only solve for the third, 2nd and single powers based on your formula. We start with:
the sum of all first powers is merely $a_4: \ (r_1 + r_2 .. r_5)$
To find the square of all the powers note that we can compute
$$(r_1 + r_2 + \ ... \ + r_5)^2 = \sum(r_i^2) + 2\sum(r_ir_j) = \sum(r_i^2) + 2a_3 \rightarrow (a_4)^2 - 2a_3 = \sum(r_i^2) = \Omega_1 $$
To find the cube of all the powers
$$(r_1 + r_2 + \ ... \ r_5)^3 = \sum(r_i)^3 + 3\sum(r_i^2r_j) + 6\sum(r_ir_jr_k) = 6(r_1r_2r_3 + ... r_3r_4r_5) + 3(r_1^2 + ... + r_5^2)(r_1 + ... + r_5) - 2\sum(r_i)^3 \rightarrow \sum(r_i)^3 = -\frac{1}{2} \left( a_4^3 - 6(a_2) + 3(\Omega_1)a_4 \right) = \Omega_2$$
Now since $a_4,a_3,a_2$ are equal to 0 it follows that
$$25\sum(\Omega_2) + 10\sum(\Omega_1) + a_4 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{Wrong}$$
Try 2:
Put
$z= x+\sqrt{1+x^2}$
$$\implies x-z
=\sqrt{1+x^2}\implies x^2+z^2-2xz
=1+x^2\implies x
=\frac{z^2-1}{2z}$$
$$\mathrm dz
=\left(1+\frac{x}{\sqrt{1+x^2}}\right)\mathrm dx
=\frac{z\mathrm dx}{x-z}=\frac{-2z^2\mathrm dx}{1+z^2}$$
$$I
=\int\frac{(z^2-1)\ln z}{2z}.\frac{(1+z^2)\mathrm dz}{-2z^2}$$
$$=\int\frac{(z^4-1)\mathrm dz}{4z^3}
=\frac14\int\left(z-\frac1{z^3}\right)\mathrm dz
=z^2/2+2/z^2+C\tag{Wrong}$$
Try 3:
Put
$z
=\sqrt{1+x^2},\mathrm dx
=x/\sqrt{1+x^2}\mathrm dx$
$$I
=\int \ln(x+z)\mathrm dz
=\int \ln(z+\sqrt{z^2-1})\mathrm dz$$
Don't know how to solve this integral.
[Note that if I take $u=z+\sqrt{z^2-1}$, it is $=\sqrt{1+x^2}+\sqrt{1+x^2-1}=x+\sqrt{1+x^2}$; same as first try.]
What's wrong in try 1 & 2, how to further solve try 3 and the best method to solve this question?
Update: Sorry, I don't know hyperbolic/inverse hyperbolic trigonometry.
| Continuation of Try 1:
$z=\ln(x+\sqrt{1+x^2})\implies e^{z}=x+\sqrt{1+x^2}\implies(e^z-x)^2=1+x^2\implies$
$\;\;\;e^{2z}-2xe^z=1\implies2xe^z=e^{2z}-1\implies x=\frac{1}{2}(e^z-e^{-z}),$
so $I=\int\frac{1}{2}z(e^z-e^{-z})\;dz=\frac{1}{2}[\int ze^z \;dz-\int ze^{-z}\;dz$]. $\;\;$ Now use integration by parts.
[Notice that we could have used $\frac{1}{2}(e^z-e^{-z})=\sinh z$].
Continuation of Try 2:
$z=x+\sqrt{1+x^2}\implies z-x=\sqrt{1+x^2}\implies z^2-2xz+x^2=1+x^2\implies 2xz=z^2-1\implies$
$x=\frac{1}{2}(z-\frac{1}{z})\implies dx=\frac{1}{2}(1+\frac{1}{z^2})dz$.
Then $\displaystyle I=\int\frac{\frac{1}{2}(z-\frac{1}{z})\ln z}{z-\frac{1}{2}(z-\frac{1}{z})}\frac{1}{2}\left(1+\frac{1}{z^2}\right)dz=\frac{1}{2}\int\frac{(z^2-1)\ln z}{z^2+1}\cdot\frac{z^2+1}{z^2}\;dz$
$\;\;\;\displaystyle=\frac{1}{2}\int\left(1-\frac{1}{z^2}\right)\ln z\;dz.$ $\;\;$Now use integration by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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Writing the complex number $z = 1 - \sin{\alpha} + i\cos{\alpha}$ in trigonometric form Now I can't finish this problem:
Express the complex number $z = 1 - \sin{\alpha} + i\cos{\alpha}$ in trigonometric form, where $0 < \alpha < \frac{\pi}{2}$.
So the goal is to determine both $r$ and $\theta$ for the expression: $z = r(\cos{\theta} + i\sin{\theta})$
I've done this so far:
*
*First of all I obtained $r = \sqrt{(1-\sin{\alpha})^2 + \cos^2{\alpha}} = \sqrt{1 + 2 \sin{\alpha} + \sin^2{\alpha} + \cos^2{\alpha}} = \sqrt{2(1 - \sin{\alpha})}$ (possible thanks to the condition over $\alpha$).
*Now I tried to get $\theta = \arctan{\left(\frac{\cos{\alpha}}{1-\sin{\alpha}}\right)}$
And here it is where I get stuck... how to determine $\theta$ with such an expression?
I already know $0 < 1-\sin{\alpha} < 1$ and $0 < \cos{\alpha} < 1$ under the given conditions.
Any help will be appreciated. Thank you :)
P.S. I think (according to my search results here) there are no questions about this problem. I hope you won't mind if it is a duplicate.
| $$z=1-\sin \alpha +i \cos \alpha\\=\{(\cos^2 \dfrac{ \alpha}{2}+\sin^2 \dfrac{ \alpha}{2})-2\sin \dfrac{ \alpha}{2}\cos \dfrac{ \alpha}{2}\}+i(\cos^2 \dfrac{ \alpha}{2}-\sin^2 \dfrac{ \alpha}{2})\\=
(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})^2+i(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})(\cos \dfrac{ \alpha}{2}+\sin \dfrac{ \alpha}{2})\\=(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})\{(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})+i(\cos \dfrac{ \alpha}{2}+\sin \dfrac{ \alpha}{2}) \}\\=\sqrt2(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})\{\cos(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} )+i\sin(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} ) \}$$
Hence this complex number has modulus $$r=\sqrt2(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})=2\cos(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} )$$ and the principle argument $$\theta=(\dfrac{ \alpha}{2}+\dfrac{ \pi}{4} ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \sqrt{\frac{2}{3}}\right) =\frac{4}{3 \sqrt{3}}$$
If my work is not much right then please rectify it
| This is a slight simplification of Seyed's solution.
We have $$f(x) = \sin (x)\sin (2x) = 2(\cos (x) - {\cos ^3}(x))=2g(\cos(x))$$
for an extreme we should have
$$0=g'(y) = 1-3 y^2,y=\cos(x)$$
so finally
$$ y = \frac{1}{\sqrt 3 } \to g(y) = \frac{4}{{3\sqrt 3 }}=f(x)$$
$$ y = - \frac{1}{\sqrt 3 } \to g(y) = - \frac{4}{{3\sqrt 3 }}=f(x)$$
which clearly shows the maximum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(x-y)(y-z)(x-z)=16$$
find this follow minimum of the value
$$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$
My idea: since
$$\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}=\dfrac{x-z}{(x-y)(y-z)}+\dfrac{1}{x-z}$$
so
$$I=\dfrac{(x-z)^2}{16}+\dfrac{1}{x-z}=\dfrac{(x-z)^2}{16}+\dfrac{1}{2(x-z)}+\dfrac{1}{2(x-z)}\ge\dfrac{3}{4}$$
if and only if $(x-z)=2$,so $(x-y)(y-z)=8$
But we know $$(x-z)^2=[[(x-y)+(y-z)]^2\ge 4(x-y)(y-z)$$
so this is wrong,
Now I let $x-z=t$ it is clear $t\ge 4$,so
$$\dfrac{t^2}{16}+\frac{1}{t}=f(t)\Longrightarrow f'(t)\ge 0,t\ge 4$$
so
$$f(t)\ge f(4)=\dfrac{5}{4}$$
My Question: I fell my methods is ugly,I think this problem have other simple methods.Thank you
| First, let $ a=x-y $, $b=y-z$, and $c=x-z$. Then, note that $ a + b - c = 0 $; i.e. $c=a+b$ and $a,b,c>0$. Finally, $abc=16$; i.e. $ab(a+b)=16$. Then, $$ I = \frac {1}{a} + \frac {1}{b} + \frac {1}{a+b}. $$You can either finish like lab bhattacharjee did, or you can use the method of Lagrange Multipliers. I will post the Lagrange solution. Note that $$ \nabla I = \left< - \frac {1}{a^2} - \frac {1}{(a+b)^2}, - \frac {1}{b^2} - \frac {1}{(a+b)^2} \right> $$and $$ \nabla \left( \text{constraint} \right) = \nabla \left( ab(a+b) \right) = \left< b(2a+b), a(a+2b) \right>. $$Now, these two gradients must be proportional, so we get $$ \frac {b(2a+b)}{\frac{1}{a^2} + \frac {1}{(a+b)^2}} = \frac {a(a+2b)}{\frac{1}{b^2} + \frac {1}{(a+b)^2}}. $$Doing some algebra yields $ a = b $ when $a,b>0$. Hence, $c=a+b=2a$, so $$ abc = a \cdot a \cdot \left( 2a \right) = 16 \iff a = 2, $$ giving $ b = 2 $ and $ c = 4 $, and the minimum value of $I$ is, therefore, $$ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {1}{2} + \frac {1}{2} + \frac {1}{4} = \boxed {\dfrac {5}{4}}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the number of solutions to an equation under bounds of $x$ I need to find the number of solutions to this equation under the following circumstances.
$$x_1 + x_2 + x_3 = 20$$ where $x_1, x_2, x_3 \in \Bbb Z$ and $1\le x_1 \le 4$, $ 2\le x_2 \le 10$ and $3\le x_3 \le 12$
I'm not 100% sure how to do it but have completed this (i'm not sure if it's anywhere close to being correct):
Attempted Solution
Total # possible solutions of $x_1 + x_2 + x_3 = 20$ is ${22 \choose 20} = 231$
Rearranging the initial equation/s:
i) $x_1 \le 4$ and $x_1 \ge 2$
ii) $x_2 \le 10$ and $x_2 \ge 3$
iii) $x_3 \le 12$ and $x_3 \ge 4$
Find (i) for "$\le$"
$${(20 - 4) + 3 - 1 \choose (20 - 4)} $$
$${16 + 3 -1 \choose 16}$$
$${18 \choose 16} = 153 $$(solutions that have $x_1 \ge 5$).
Therefore the # solutions where $x_1 \le 4$ $$= 231 - 153$$
$$ = 78$$
This is repeated for the $x_2$ and $x_3$ to get 165 and 186 respectively.
Now to find for "$\ge$":
Same process as above except not "$-$ing" this number away from 231.
This gives,
i) ${20 \choose 18} = 190$
ii) ${19 \choose 17} = 171$
iii) ${18 \choose 16} = 153$
Then to find the total number of solutions for each equation(i,ii,iii) I take the first number calculated away from the second.
i) $190 - 78 = 112$
ii) $171 - 165 = 6$
iii) $153 - 186 = -33$
Then to find the total number of solutions for the whole initial equation:
$$ sol = 231 - 112 - 6 - (-)33 $$
$$ sol = 146 $$
Is this correct? If not, where am I going wrong?
| What you have to find is just:
$$\begin{eqnarray*}&&[x^{20}](x+x^2+x^3+x^4)(x^2+\ldots+x^{10})(x^3+\ldots+x^{12})\\&=&[x^{14}](1+\ldots+x^3)(1+\ldots+x^8)(1+\ldots+x^9)\\&=&[x^{14}]\frac{(1-x^4)(1-x^9)(1-x^{10})}{(1-x)^3}\\&=&[x^{14}](1-x^4-x^9-x^{10}+x^{13}+x^{14})\sum_{j=0}^{+\infty}\binom{j+2}{2}x^j\\&=&\binom{16}{2}-\binom{12}{2}-\binom{7}{2}-\binom{6}{2}+\binom{3}{2}+1\\&=&\color{red}{22}.\end{eqnarray*}$$
As an alternative approach, you can check that we have $4$ solutions with $x_1=1$, $5$ solutions with $x_1=2$, $6$ solutions with $x_1=3$ and $6$ solutions with $x_1=4$, and:
$$ 4+5+6+7 = \color{red}{22}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate
$$1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + 4\cdot 10^4+\dots + n\cdot 10^n$$
as a expression, with a proof so I could actually understand it if possible?
| Let $$S_n = 1*10^1+2*10^2+3*10^3 + \cdots n*10^n$$
$$10S_n = 1*10^2+2*10^3+3*10^4 + \cdots n*10^{n+1}$$
$$S_n - 10S_n = 1*10^1 + 1*10^2 + 1*10^3 +\cdots1*10^n-n*10^{n+1}$$
$$-9S_n = 10*\frac{1-10^n}{1-10} - n*10^{n+1}$$
Simplifying, $$S = \frac{10}{81}*(9*10^n n-10^n+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the most distant points on a curve Hy, I have to ask someone for help with this problem.
I have a curve with this implicit equation:
$$\left ( x^2 + y^2 \right )^2 = x^3 + y^3$$
I have to find the most distant coordinates from the center of coordinate system, and the most distant coordinates from y-axis.
The graph looks like this Wolfram graph plot
Do I have to use Lagrange method to solve this and how to set the equation up if so ?
Thank you very much !
| Convert to polar coordinates:
$$\begin{align}(r^2)^2&=r^3(\cos^3\theta + \sin^3\theta)\\
r&=\cos^3\theta + \sin^3\theta\qquad \because r\neq0\\
\frac {dr}{d\theta}&=-3\cos^2\theta \sin\theta+3\sin^3\theta\cos\theta=0\qquad
\quad \text{at max/min}\\
\sin\theta\cos\theta(\sin\theta-\cos\theta)&=0\\
&\Rightarrow \begin{cases}
\sin\theta&=0\qquad \qquad \qquad \quad \Rightarrow \theta=n\pi\\
\cos\theta&=0\qquad \qquad \qquad \quad \Rightarrow \theta=(n+\frac 12)\pi\\
\sin\theta&=\cos\theta\Rightarrow \tan\theta=1\Rightarrow \theta=(n+\frac 14)\pi
\end{cases}
\end{align}$$
Checking each case with using $\frac {d^2r}{d\theta^2}$ will confirm that the maximum value for $r$ occurs at
$$\theta=2n\pi, (2n+\frac12)\pi$$
where
$$r=1$$
i.e. the most distant coordinates from the origin are, in $(x,y)$ terms,
$$(1,0), (0,1)\qquad \blacksquare$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic equation - solving for x Question:
Solve the equation $$(x+2)(x+3)(x+8)(x+12) = 4x^2$$
I tried to solve the equation by expanding the LHS and then equating it to the RHS, but that just doesn't seem to be feasible. I am probably missing a key point here, which would make the question a lot easier. Help please!
| I've just got an easier way. Setting $A=x^2+24$ gives you
$$\begin{align}(x+2)(x+3)(x+8)(x+12)=4x^2&\iff (x+2)(x+12)(x+3)(x+8)=4x^2\\&\iff (x^2+14x+24)(x^2+11x+24)=4x^2\\&\iff (A+14x)(A+11x)=4x^2\\&\iff A^2+25xA+150x^2=0\\&\iff (A+10x)(A+15x)=0.\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this $\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$ let $n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}},c\in (0,2),c=2\cos{x}$, show that
$$\dfrac{\sin{(nx)}}{\sin{x}}\ge\dfrac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$
where $0<x<\dfrac{\pi}{2}$
My idea: let $$a_{n}=\dfrac{\sin{(nx)}}{\sin{x}}$$
then for any $k\in[1,n]$, then we have
$$a^2_{k}-a^2_{k-1}=\dfrac{1}{\sin^2{x}}[\sin^2{kx}-\sin^2{(k-1)x}]=\dfrac{\sin{(2k-1)x}\sin{x}}{\sin^2{x}}=\dfrac{\sin{(2k-1)x}}{\sin{x}}$$
since
$$n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}}=\dfrac{\pi}{2x}\Longrightarrow 0<kx\le nx<\dfrac{\pi}{2}$$
then
$2kx\le 2n<\pi$, so $x\le (2k-1)x<\pi-x$
so $\sin{(2k-1)x}\ge \sin{x}>0$,so
$a^2_{k}-a^2_{k-1}\ge 1$
$$a^2_{n}=\sum_{k=2}^{n}(a^2_{k}-a^2_{k-1})+a^2_{2}\ge n\Longrightarrow a_{n}\ge \sqrt{n}$$
then I can't it,
I'm sorry, I just to eat
,and I'm come back
| Next time, I will to wait until the OP stops changing his/her question.
Note that
$$\frac{\sin nx}{\sin x}=U_n(\cos x)=2^{n-1}\prod_{k=1}^{n-1}\left(\cos x-\cos\left(\frac{k\pi}{n}\right)\right)$$
where $U_n$ is the Chebyshev polynomial of the second kind.
Now, for $0<x<\frac{\pi}{2n}$ (which is equivalent to the inequality with the $``c"$ thing,)
we have
$$\cos x-\cos\left(\frac{k\pi}{n}\right)\geq \cos\left( \frac{\pi}{2n}\right)-\cos\left(\frac{k\pi}{n}\right)>0$$
for $k=1,\ldots,n-1$. Thus
$$\frac{\sin nx}{\sin x}\geq2^{n-1}\prod_{k=1}^{n-1}\left(\cos \left(\frac{\pi}{2n}\right)-\cos\left(\frac{k\pi}{n}\right)\right)=\frac{\sin n(\frac{\pi}{2n})}{\sin (\frac{\pi}{2n})}=\frac{1}{ \sin (\frac{\pi}{2n})}.$$
This inequality is clearly better than the proposed one. In fact,
using $\sin (\frac{\pi}{2n})<\frac{\pi}{2n}$. The proposed inequality follows if we have
$$
\frac{2n}{\pi}\geq \frac{1}{\sqrt{3}}(2n-1)^{3/4}\iff n(2n)^3\geq \frac{\pi^4}{18}(2n-1)^3
$$
and this is clearly true if $n\geq \frac{\pi^4}{18}\approx 5.4$. So, the proposed inequality is true for $n\geq 6$. For $n\leq 5$ we can check directly that
$$\frac{1}{\sin(\frac{2n}{\pi})}\geq \frac{1}{\sqrt{3}}(2n-1)^{3/4}.$$
So, the proposed inequality is valid for every $n$. Note
that the inequality that we obtained is asymptotically much stronger.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series
\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}
Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.
| Another possible approach is to use the discrete Fourier transform.
Let $\omega=\exp\frac{2\pi i}{3}$. Then:
$$f(n)=\frac{1}{3}\left(1+\omega^n+\omega^{2n}\right)=\mathbb{1}_{n\equiv 0\!\pmod{3}}(n),$$
hence:
$$\color{red}{\sum_{n=0}^{+\infty}\frac{1}{(3n)!}}=\sum_{n=0}^{+\infty}\frac{f(n)}{n!}=\frac{1}{3}\left(\exp(1)+\exp(\omega)+\exp(\omega^2)\right)=\color{red}{\frac{e}{3}+\frac{2}{3\sqrt{e}}\cos\frac{\sqrt{3}}{2}.}$$
The other two series can be computed with the same technique, by noticing that:
$$f_1(n)=\frac{1}{3}\left(1+\omega^2\cdot\omega^n+\omega\cdot\omega^{2n}\right)=\mathbb{1}_{n\equiv 1\!\pmod{3}}(n),$$
$$f_2(n)=\frac{1}{3}\left(1+\omega\cdot\omega^n+\omega^2\cdot\omega^{2n}\right)=\mathbb{1}_{n\equiv 2\!\pmod{3}}(n).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Addition of fractions repetition and convergence Is this a new mathematical concept?
$$
\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} \cdots = \frac{1}{n-1}
$$
If it isn't then what is this called?
I haven't been able to find anything like this anywhere.
| $$
if\\n\neq 1 \\ s=\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+...\\multiply\\ s \\by \\n \\\frac{1}{n}s=\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+\frac{1}{n^5}...\\now\\s-\frac{1}{n}s =\frac{1}{n}+(\frac{1}{n^2}-\frac{1}{n^2})+(\frac{1}{n^3}-\frac{1}{n^3})+(\frac{1}{n^4}-\frac{1}{n^4})+(\frac{1}{n^5}-\frac{1}{n^5})+...\\so \\s-\frac{1}{n}s =\frac{1}{n}
$$
$$
s(\frac{n-1}{n})=\frac{1}{n}
$$
$$
s=\frac{1}{n-1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove $\sqrt{x}+\sqrt{y}+\sqrt{\frac{x+y+2}{xy-1}}\ge 2\left( \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\sqrt{\frac{xy-1}{x+y+2}}\right)$ How prove
$\sqrt{x}+\sqrt{y}+\sqrt{\frac{x+y+2}{xy-1}}\ge 2\left( \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\sqrt{\frac{xy-1}{x+y+2}}\right)$
for $8x\ge13, 8y \ge 13$?
| For easy calculation,let $a=\sqrt{x},b=\sqrt{y} \implies a+b+\sqrt{\dfrac{a^2b^2-1}{a^2+b^2+2}} \ge 2\left(\dfrac{1}{a}+\dfrac{1}{b}+\sqrt{\dfrac{a^2+b^2+2}{a^2b^2-1}}\right) \iff (a+b)\dfrac{ab-2}{ab}+\dfrac{a^2+b^2+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge 0$
we have two cases:
1) $ab\ge 2 $
$\dfrac{a^2+b^2+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge \dfrac{2ab+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}}=\dfrac{-2(ab-2)(ab+1)}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \iff (ab-2)\left( \dfrac{a+b}{ab}-\dfrac{2(ab+1)}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}}\right) \ge 0 \iff \\ \dfrac{a+b}{ab}-\dfrac{2(ab+1)}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge 0 \iff \dfrac{2}{\sqrt{ab}}-\dfrac{2(ab+1)}{\sqrt{(a^2b^2-1)(2ab+2)}}\ge0 \iff \\ \dfrac{1}{\sqrt{ab}} \ge \dfrac{1}{\sqrt{2(ab-1)}} \iff ab\ge2 $
2)$ab < 2$
$\dfrac{a^2+b^2+4-2a^2b^2}{\sqrt{(a^2b^2-1)(a^2+b^2+2)}} \ge (a+b)\dfrac{2-ab}{ab}$
$t=ab<2,p=a^2+b^2 \ge 2t \iff \dfrac{p+4-2t^2}{\sqrt{(t^2-1)(p+2)}}\ge \dfrac{(2-t)\sqrt{p+2}}{t} \iff \\A(t)p^2+B(t)p+C(t) \ge 0 \\ A(t)=(-t^4+4t^3-2t^2-4t+4)=t^2(2-t)+(t-1)^2(2t+2) > 0\\ B(t)=(-2t^5+2t^4+2t^3-6t^2+8) \\ C(t)=4t^6-4t^5-12t^3+16t$
$4tA(t)+B(t)=-6t^5+18t^4-6t^3-22t^2+16t+8 >0 \implies A(t)p^2+B(t)p+C(t) \ge A(t)(2t)^2+B(t)(2t)+C(t)=4t(t+1)^2(2-t)^3 >0$
So the "=" will hold then $a=b, ab=2 \implies x=y=2$
there is no limitation for $x,y$ except $x>0,y>0,xy>1$
QED
| {
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Taking Calculus in a few days and I still don't know how to factorize quadratics Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or something like that..
Can someone explain to me simply how I would step by step factorize something like $4x^2 + 16x - 19$ ?
| First of all, to find the roots of $4x^2+16x-19$ we have to calculate the discriminant:
If the second degree polynomial is of the form $$ax^2+bx+c=0$$ the discriminant is given from the formula: $$\Delta=b^2-4 \cdot a \cdot c$$
So the discriminant in this case is the following:
$4x^2+16x-19=0 \Rightarrow \Delta=16^2-4 \cdot 4 \cdot (-19)=256+304 \Rightarrow \Delta=560$
Then the solutions are given from the formula:
$$x_{1.2}=\frac{-b \pm \sqrt{\Delta}}{2 \cdot a}$$
Therefore we have the following solutions:
$x_{1,2}=\frac{-16 \pm \sqrt{560}}{2\cdot 4}=\frac{-16 \pm \sqrt{560}}{8}$
$x_1=\frac{-16-4 \sqrt{35}}{8}=-2-\frac{\sqrt{35}}{2} \text{ and } x_2=\frac{-16 +4\sqrt{35}}{8}=-2+\frac{\sqrt{35}}{2}$
Knowing that $$ax^2+bx+c=0 \Rightarrow a(x-x_1)(x-x_2)=0$$
we have the following:
$$4x^2+16x-19=4 \left ( x- \left (-2-\frac{\sqrt{35}}{2} \right ) \right ) \left ( x- \left (-2+\frac{\sqrt{35}}{2} \right ) \right )$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I understand Pythagorean theorem 1) I understand the formula
$$\frac{BC}{AB}=\frac{BH}{BC},\ \frac{AC}{AB}=\frac{AH}{AC}$$
But I can't understand the formula is obtained
$$BC^2=AB\times BH \ \text{and}\ AC^2=AB\times AH$$
Why if somebody multiplies $AB * BH = AC^2$?
2) How is the formula obtained?
$$\frac{\sqrt{AB^2-AC^2}}{AC}=\sqrt{1-\left(\frac{AC}{AB}\right)^2}$$
I understand first part, but I can't understand second part
| Each equation of the second pair is obtained from an equation in the first pair by "cross-multiplying," or equivalently getting rid of the denominators.
For example, start from $\frac{BC}{AB}=\frac{BH}{BC}$. Multiply both sides by $(AB)(BC)$. We get $(BC)(BC)=(AB)(BH)$.
Similarly, from $\frac{AC}{AB}=\frac{AH}{AC}$, by multiplying both sides by $(AB)(AC)$ we get $(AC)(AC)=(AB)(AH)$.
To finish, take the two equations $BC^2=(AB)(BH)$ and $AC^2=(AB)(AH)$ and add. We get
$$BC^2+AC^2=AB(BH+AH)=(AB)(AB)=AB^2.$$
Remark: As to your question 2, the quoted formula is wrong. For example, let $AB=5$ and $AC=4$. For these values, the the left-hand side of your expression is equal to $\frac{3}{4}$. The right-hand side is equal to $\frac{3}{5}$.
Added: It turns out that OP wants to prove the correct
$$\frac{\sqrt{AB^2-AC^2}}{AB}=\sqrt{1-\left(\frac{AC}{AB}\right)^2}.\tag{1}$$
Rewrite the left side of (1) as $\sqrt{\frac{AB^2-AC^2}{AB^2}}$. Now divide term by term. We have $\frac{AB^2}{AB^2}=1$ and $\frac{AC^2}{AB^2}=\left(\frac{AC}{AB}\right)^2$, so we obtain the right-hand side of (1)/
The context is that if $\alpha$ is the angle at $A$, then
$\sin\alpha=\frac{BC}{AB}= \frac{\sqrt{AB^2-AC^2}}{AB}$. So on the left-hand side of (1) we have $\sin\alpha$, and on the right-hand side we have $1-\cos^2\alpha$.
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Maximizing $ \frac {y + z + yz}{\left( 1 + y + z \right)^2} $ I was solving a WOOT Problem of the Day and I boiled the problem down to finding the maximum value of $$ \frac {y+z+yz}{\left( 1 + y + z \right)^2}. $$Intuitively, this seems to occur when $y=z$, since $y$ is symmetric, and after that, it is easy to find that $$ \frac {y+z+yz}{\left( 1 + y + z \right)^2} \le \frac {1}{3}. $$This seems "intuitive" but is not a proof. Any ideas for a formal proof?
| If you compute the derivatives of $$F=\frac {y+z+yz}{\left( 1 + y + z \right)^2}$$ you have, after simplifications $$F'_y=\frac{-y (z+1)+z^2+1}{(y+z+1)^3}$$ $$F'_z=-\frac{y^2-2 y (2 z+1)+(z-2) z+3}{(y+z+1)^4}$$ Using $F'_y=0$ gives $y=\frac{z^2+1}{z+1}$ and the numerator of $F'_z$ simplifies to $$(1+z+z^2)(z^2+2 z-1)=0$$ So the solutions are $$y=2 \left(-1-\sqrt{2}\right),z=-1-\sqrt{2},F=\frac{1}{18} \left(1-3 \sqrt{2}\right)$$ $$y=2 \left(-1+\sqrt{2}\right),z=-1+\sqrt{2},F=\frac{1}{18} \left(1+3 \sqrt{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to show $\int_{0}^{\pi / 2} \ln\left(\tan x - \sqrt{2 \tan x} + 1\right){d}x = 0 $ Problem: Show that
$$\int_{0}^{\pi / 2} \ln\left(\tan x - \sqrt{2 \tan x} + 1\right)\,\mathrm{d}x = 0 $$
If possible, I would like to use regular single-variable calculus methods, with only substitutions, IBP, partial fractions and so on, which does not involve series manipulation.
Thanks.
| Substitute $t^2=\tan x$, together with $t^4+1=(t^2-\sqrt2 t+1)(t^2+\sqrt2 t+1) $
\begin{align}
I=&\int_{0}^{\pi / 2} \ln\left(\tan x - \sqrt{2 \tan x} + 1\right)\,dx \\
=&\int_0^\infty \frac{2t\ln(t^2-\sqrt2 t+1)}{t^4+1}dt\\
=&\int_0^\infty \frac{t\ln(t^4+1)}{t^4+1}\overset{y=t^2}{dt}
+ \frac12\int_{-\infty}^\infty \underset{=J}{\frac{t\ln\frac{t^2-\sqrt2 t+1}{t^2+\sqrt2 t+1}}{t^4+1}}dt\tag1\\
\end{align}
Further decompose the integral $J$ as follows
\begin{align}
J=& \ \frac1{\sqrt2}\int_{-\infty}^\infty \overset{y=\sqrt2t-1}{\frac{\ln(t^2-\sqrt2t+1)}{t^2-\sqrt2t+1}}+ \overset{y=\sqrt2t+1}{ \frac{\ln(t^2+\sqrt2t+1)}{t^2+\sqrt2t+1}}-\overset{y=\frac{t-1/t}{\sqrt2}}{\frac{(1+t^2)\ln(t^4+1)}{t^4+1}}\ dt\\
=& \ 2\int_{-\infty}^\infty \frac{\ln \frac{y^2+1}2}{y^2+1}dy
- \int_{-\infty}^\infty \frac{\ln (2(y^2+1))}{y^2+1}dy
=2 \int_{0}^\infty \frac{\ln \frac{y^2+1}8}{y^2+1}dy
\end{align}
Plug $J$ into (1) to arrive at
\begin{align}
I =\frac12 \int_0^\infty \frac{\ln(y^2+1)}{y^2+1}dy
+ \int_{0}^\infty \frac{\ln \frac{y^2+1}8}{y^2+1}dy
=\ \frac32 \int_0^\infty \frac{\ln \frac{y^2+1}4}{y^2+1}dy=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/919970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Selecting at least one ball of each color An urn contains five red, six white and seven blue balls. Five balls are selected without replacement. Find the probability that at least one ball of each color is selected.
Answer (attempt):
Getting red balls = $5 \choose 1$
Getting white balls = $6 \choose 1$
Getting blue balls = $7 \choose 1$
Remaining two balls = $15 \choose 2$
Total selections = $18 \choose 5$
P = $\frac{{5 \choose 1} {6 \choose 1}{7 \choose 1}{15 \choose 2}}{18 \choose 5}$
The answer seems to be wrong.
| For the split up of $5$ there are $2$ possibilities:
$5=1+2+2$
$5=1+1+3$
Leading to $$\binom{5}{1}\binom{6}{2}\binom{7}{2}+\binom{5}{2}\binom{6}{1}\binom{7}{2}+\binom{5}{2}\binom{6}{2}\binom{7}{1}+\binom{5}{1}\binom{6}{1}\binom{7}{3}+\binom{5}{1}\binom{6}{3}\binom{7}{1}+\binom{5}{3}\binom{6}{1}\binom{7}{1}$$
possibilities under the condition that from each color at least one ball is selected.
In total there are: $$\binom{18}{5}$$ possibilities.
These facts enable you to find the probability.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
exponent y=x^a sequences. While analyzing square and cube functions, i found the following:
for y=x^2
x=1, y=1
+3
x=2, y=4 +2
+5
x=3, y=9 +2
+7
x=4, y=16 +2
+9
x=5, y=25
increase of increase (well, how else should i say this) is +2.
What does this signify?
Same pattern for y=x^3, only number in question being +6 and appearing after another round of measuring increase:
x=1, y=1
+7
x=2, y=8 +12
+19 +6
x=3, y=27 +18
+37 +6
x=4, y=64 +24
+61
x=5, y=125
| I think that you are just showing that, whatever $x$ could be, the $n^{th}$ derivative of $x^n$ is a constant $$\frac{d^2}{dx^2}(x^2)=2$$ $$\frac{d^3}{dx^3}(x^3)=6$$ This corresponds to the number of steps required to arrive to your constant term. With $x^4$, one more round would give you $24$ as constant. In fact $$\frac{d^n}{dx^n}(x^n)=n!$$ what you would get after $n$ rounds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Third point of a triangle in the complex plane I have an equilateral triangle with two points equal to $(2+2i)$ and $(5+i)$. I want to find the third point(s) (there are $2$ of these). I have that the side length of the triangle is $\sqrt{10}$.
| (I'm purposely not trying
to be clever here
and doing all calculations in my head.)
The distance
$d$
between the two known points
satisfies
$d^2
=(2-5)^2+(2-1)^2
=10
$.
If the point is
$(x, y)$,
then
$(x-2)^2+(y-2)^2
=(x-5)^2+(y-1)^2
=10
$,
or
$x^2-4x+4+y^2-4y+4
=x^2-10x+25+y^2-2y+1
=10
$,
or
$x^2-4x+y^2-4y+8
=x^2-10x+y^2-2y+26
=10
$.
Subtracting the first two,
$6x-2y-18 = 0$,
or
$y = 3x-9$.
Substituting this
in the first equation,
$10
=x^2-4x+(3x-9)^2-4(3x-9)+8
=x^2-4x+9x^2-54x+81-12x+36+8
=10x^2-70x+125
$
or
$10x^2-70x+115=0$
or
$2x^2-14x+23=0$
Using the good old
quadratic equation formuls,
$D^2
= b^2-4ac
= 14^2-4\cdot 2 \cdot 23
=196-184
=12
$
so
$D = 2\sqrt{3}$.
The roots are therefore
$x
=\dfrac{-b \pm D}{2 a }
=\dfrac{14\pm 2\sqrt{3}}{2\cdot 2}
=\dfrac{7\pm \sqrt{3}}{2}
$
and
$y
=3x-9
=\dfrac{21\pm 3\sqrt{3}-18}{2}
=\dfrac{3\pm 3\sqrt{3}}{2}
$.
Note:
I find it somewhat
embarrassing and dishearting
how many errors
I made
while doing this
all in my head.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$? I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\dfrac{\sqrt{16x^2-9}}x\,\mathrm{d}x$.
$$v=4x\implies\mathrm{d}v=4\,\mathrm{d}x$$
$$\int\frac{\sqrt{v^2-9}}{v}\,\mathrm{d}v\implies a=3,\quad v=3\sec\theta,\quad\mathrm{d}v=3\sec\theta\tan\theta\,\mathrm{d}\theta.$$
$$v^2-9=9\sec^2\theta-9=9\left(\sec^2\theta-1\right)=9\tan^2\theta.$$
$$\int\frac{3\tan\theta\cdot3\sec\theta\tan\theta}{3\sec\theta}\,\mathrm{d}\theta=3\int\tan^2\theta\,\mathrm{d}\theta.$$
$$3\int\left(\sec^2\theta-1\right)\mathrm{d}\theta=3\left(\tan\theta-\theta\right).$$
And this where I feel I am going wrong. So from my understanding since I used $\sec\theta$ for the substitution the triangle I am using to use this integral has a hypotenuse of $v$ or $4x$ since I initially assigned that value to b and the adjacent side of $\theta$ is $a$ or $3$, and for the missing side I got $\sqrt{v^2-9}$ or $\sqrt{16x^2-9}$. So for $\tan\theta$ I got $\dfrac{\sqrt{16x^2-9}}3$ and for $\theta$ I got $\sec^{-1}\left(\dfrac{v}3\right)$ or $\cos\left(\dfrac{4x}3\right)$. So my final answer is $$\sqrt{16x^2-9}-3\cos\left(\frac{4x}3\right)$$
but this is wrong so if someone could tell me where I am going wrong it'd be greatly appreciated. Thanks in advance.
| You're almost there...
Since $x = \frac {3}{4} \sec \theta$, $\sec \theta = \frac {4x}{3}$; thus $\cos \theta = \frac {3}{4x}$ and $\theta = \arccos \frac {3}{4x}$.
By the Pythagorean theorem, $\tan \theta = \frac {\sqrt {16x^2-9}} {3}$. With a little bit of cancellation, your final answer will be...
$$\sqrt {16x^2-9} - 3 \arccos \frac {3}{4x} + C$$
NOTE: I used $\arccos \theta$ versus $\cos^{-1}\theta$ for clarity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seems to make it worse. Any help would be appreciated.
| You have another solution using Taylor series. Rewrite $$x\Big(\sqrt {x^2+a} - \sqrt {x^2+b}\Big)=x^2\Big(\sqrt{1+a/x^2}-\sqrt{1+b/x^2}\Big)$$ and remember that, when $y$ is small $\sqrt{1+y}\simeq 1+y/2$. Replace successively $y$ by $a/x^2$ and $b/x^2$ and you are done.
If you continue in the same spirit you coulod show a beautiful thing $$x\Big(\sqrt {x^2+a} - \sqrt {x^2+b}\Big)=\frac{a-b}{2}-\frac{a^2-b^2}{8 x^2}+\frac{a^3-b^3}{16 x^4}-\frac{5
\left(a^4-b^4\right)}{128 x^6}+O\left(\left(\frac{1}{x}\right)^7\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$ Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$
I choose to use the formula $\displaystyle f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$
Which after some work I found to be
$\frac{3-\sqrt{2x+1}}{3\sqrt{2x+1}(x-4)}$
Which is basically back to where I started. Is there any way to get the $x-4$ out of the denominator?
| Hint: Try multiplying by the conjugate:
$$
\frac{3 + \sqrt{2x+1}}{3 + \sqrt{2x+1}}
$$
Then the numerator becomes:
$$
(3)^2 - (\sqrt{2x + 1})^2 = 9 - (2x + 1) = -2x + 8 = -2(x - 4)
$$
so that you can cancel the $(x - 4)$ term.
At the request of the comments, here's all the work:
\begin{align*}
\left. \frac{d}{dx}\right|_{x=4} \frac{1}{\sqrt{2x+1}}
&= \lim_{x\to 4} \frac{\frac{1}{\sqrt{2x+1}} - \frac{1}{\sqrt{2(4)+1}}}{x - 4} \\
&= \lim_{x\to 4} \frac{\frac{1}{\sqrt{2x+1}} - \frac{1}{3}}{x - 4} \\
&= \lim_{x\to 4} \frac{3-\sqrt{2x+1}}{3\sqrt{2x+1}(x-4)} \\
&= \lim_{x\to 4} \frac{-2(x - 4)}{3\sqrt{2x+1}(x-4)(3 + \sqrt{2x+1})} &\text{using my hint}\\
&= \lim_{x\to 4} \frac{-2}{3\sqrt{2x+1}(3 + \sqrt{2x+1})} \\
&= \frac{-2}{3\sqrt{2(4)+1}(3 + \sqrt{2(4)+1})} \\
&= \frac{-1}{27}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple differentiation from first principles problem I know this is really basic, but how do I differentiate this equation from first principles to find $\frac{dy}{dx}$:
$$
y = \frac{1}{x}
$$
I tried this:
$$\begin{align}
f'(x) = \frac{dy}{dx} & = \lim_{\delta x\to 0} \left[ \frac{f(x + \delta x) - f(x)}{\delta x} \right] \\
& = \lim_{\delta x\to 0} \left[ \frac{(x+\delta x)^{-1} - x^{-1}}{\delta x} \right] \\
& = \lim_{\delta x\to 0} \left[ \frac{1}{\delta x(x + \delta x )} - \frac{1}{x(\delta x)} \right] \\
& = \lim_{\delta x\to 0} \left[ \frac{1}{x(\delta x)} + \frac{1}{(\delta x)^2} - \frac{1}{x(\delta x)} \right] \\
& = 1
\end{align}$$
which is obviously wrong, since $f'(x) = - \frac{1}{x^2}$. Where am I going wrong?
| $$\begin{align} f(x+h)-f(x) &= \dfrac{1}{x+h} - \dfrac{1}{x} \\ &= \dfrac{x-h-x}{x(x+h)} \\ &= \dfrac{-h}{x(x+h)} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/929832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$ Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$
$ $
This appears to be an easy problem, but it is consuming a lot of time, I am wondering if an easy way is possible.
WHAT I DID :
Wrote this as $\int \dfrac{1}{4}(\sin^22x)\cos^2x$
And then I wrote $\cos^2x$ in terms of $\cos(2x)$, I get an Integral which is sum of two known integrals, I did them, I got answer $\dfrac{\pi}{16}$
| First look at the antiderivative $$I=\int\sin^2(x)\cos^4(x)\hspace{1mm}dx$$ and now use $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$ so $$\cos^4(x)=\Big(\frac{\cos(2x)+1}{2}\Big)^2=\frac{1}{4}\Big(\cos^2(2x)+2 \cos(2x)+1\Big)$$ $$\cos^4(x)=\frac{1}{4}\Big(\frac{\cos(4x)+1}{2}+2 \cos(2x)+1\Big)$$ $$\cos^4(x)=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x)+\frac{3}{8})$$ So, now $$sin^2(x)\cos^4(x)=\frac{1-\cos(2x)}{2}\Big(\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x)+\frac{3}{8}\Big)$$ Develop and as a result you will have simple cosines and products of cosines; now remember and use the fact that $$\cos(a)\cos(b)=\frac{1}{2}\Big(\cos(a+b)+\cos(a-b)\Big)$$ Normally, you will end with $$\sin^2(x)\cos^4(x)=\frac{1}{32} \cos (2 x)-\frac{1}{16} \cos (4 x)-\frac{1}{32} \cos (6 x)+\frac{1}{16}$$ and so $$I=\frac{x}{16}+\frac{1}{64} \sin (2 x)-\frac{1}{64} \sin (4 x)-\frac{1}{192} \sin (6 x)$$ For the given integration bounds only the first term has to be used and the result is $\frac{\pi}{16}$.
In my opinion, it is easier to use Euler formula $$\cos x=\frac{e^{ix}+e^{-ix}}2,\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ so $$\sin^2(x)=-\frac{1}{4} e^{-2 i x}-\frac{1}{4} e^{2 i x}+\frac{1}{2}$$ $$\cos^4(x)=\frac{1}{4} e^{-2 i x}+\frac{1}{4} e^{2 i x}+\frac{1}{16} e^{-4 i x}+\frac{1}{16}
e^{4 i x}+\frac{3}{8}$$ $$\sin^2(x)\cos^4(x)=\frac{1}{64} e^{-2 i x}+\frac{1}{64} e^{2 i x}-\frac{1}{32} e^{-4 i x}-\frac{1}{32}
e^{4 i x}-\frac{1}{64} e^{-6 i x}-\frac{1}{64} e^{6 i x}+\frac{1}{16}$$ $$\sin^2(x)\cos^4(x)=\frac{\cos(2x)}{32}-\frac{\cos(4x)}{16}-\frac{\cos(6x)}{32}+\frac{1}{16}$$ and the integration become extremely simple
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Number of Lattice Points in a Triangle
Problem
Let the co-ordinates of the vertices of the $\triangle OAB$ be $O(1,1)$, $A(\frac{a+1}{2},1)$ and $B(\frac{a+1}{2},\frac{b+1}{2})$ where $a$ and $b$ are mutually prime odd integers, each greater than $1$. Then find the number of lattice points inside $\triangle OAB$, i.e., not on the borders of $\triangle OAB$. How does the answer change if the restriction $\operatorname{gcd}(a,b)=1$ is removed?
Solution
Let $L(a,b)$ be the number of lattice points inside $\triangle OAB$. The linear transformation $T(x,y)=(x-1,y-1)$ on the triangle $OAB$ do not have any effect on $L(a,b)$. Hence $L(a,b)$ is equal to the number of lattice points inside the triangle $O'A'B'$, where $O'=(0,0)$, ${\textstyle A'=(\frac{a-1}{2},0)}$ and ${\textstyle B'=(\frac{a-1}{2},\frac{b-1}{2})}$.
Set ${C'=(0,\frac{b-1}{2})}$. By symmetry the number of lattice points inside the rectangle $O'B'C'$ is $L(a,b)$. The number of lattice point inside the rectangle $O'A'B'C'$ is ${\textstyle \frac{a-3}{2} \cdot \frac{b-3}{2}}$. Consequently
$$2L(a,b) = \frac{a-3}{2} \cdot \frac{b-3}{2} - K$$
where $K$ are the number of lattice points on the straight line between $O'$ and $B'$. This line is given by $y= \frac{b-1}{a-1}x$, yielding
$$K = |\{ {\textstyle 0 < x < \frac{a-1}{2} \mid \frac{b-1}{a-1}x \in \mathbb{N} \}| }$$
By letting $d=\text{gcd}(a-1,b-1)$, we obtain $a-1=rd$ and $b-1=sd$ for two positive coprime integers $r$ and $s$. Therefore
$$\frac{b-1}{a-1}x = \frac{sx}{r} \in \mathbb{N} \;\; \Leftrightarrow \;\; \frac{x}{r} \in \mathbb{N}$$
Now ${\textstyle 0 < x < \frac{a-1}{2} = \frac{rd}{2}}$, implying ${\textstyle 0 < \frac{x}{r} = \frac{d}{2}}$. Since $d$ is even (since $d=\gcd(a-1,b-1)$ and $a$ and $b$ are both odd positive integers) give us ${\textstyle K = \frac{d}{2}-1}$. Thus by (1)
$${\textstyle L(a,b) = \dfrac{(a-3)(b-3)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}}$$
But I think that by Pick's Theorem the answer should be,
$${\textstyle L(a,b) = \dfrac{(a-1)(b-1)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}}$$
Which one is correct?
| The area of $\Delta OAB$ is $K = \dfrac{1}{2} \cdot \dfrac{a-1}{2} \cdot \dfrac{b-1}{2} = \dfrac{(a-1)(b-1)}{8}$.
The number of points on the boundary (line segments $OA$, $AB$, and $BO$) is
$\underbrace{\dfrac{a-1}{2}}_{OA}+\underbrace{\dfrac{b-1}{2}}_{OB}+\underbrace{\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\right)}_{BO}$ (you missed the first two terms of this).
So, by Pick's Theorem, the area is
$K = I+\dfrac{1}{2}B-1 = I+\dfrac{1}{2}\left[\dfrac{a-1}{2}+\dfrac{b-1}{2}+\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\right)\right]-1$
$= I + \dfrac{a-1}{4} + \dfrac{b-1}{4}+\dfrac{\text{gcd}(a-1,b-1)}{4}-1$.
Hence, the number of points in the interior is
$I = \dfrac{(a-1)(b-1)}{8} - \dfrac{a-1}{4} - \dfrac{b-1}{4} - \dfrac{\text{gcd}(a-1,b-1)}{4} + 1$
$= \dfrac{(a-3)(b-3)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}$ (after a bit of simplification).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Decomposition of shear matrix into rotation & scaling How can I decompose the affine transformation:
$$ \begin{bmatrix}1&\text{shear}_x\\\text{shear}_y&1\end{bmatrix}$$
into rotation and scaling primitives?
$$ \begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$
\begin{bmatrix}\text{scale}_x&0\\0&\text{scale}_y\end{bmatrix}
| You can only do this if $\textrm{shear}_x = -\textrm{shear}_y$, in which case
$$
\begin{align}
\theta &= \textrm{atan}(\textrm{shear}_y)\\
\textrm{scale}_y&=\sqrt{\textrm{shear}_y^2 + 1}\\
\textrm{scale}_x&=-\sqrt{\textrm{shear}_y^2 + 1}\\
\end{align}
$$
To see this we can identify both matrices:
$$
\begin{bmatrix}1& \text{shear}_x \\ \text{shear}_y&1\end{bmatrix} =
\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}
\begin{bmatrix}\text{scale}_x&0\\0&\text{scale}_y\end{bmatrix}
$$
$$
\begin{bmatrix}1& \text{shear}_x \\ \text{shear}_y&1\end{bmatrix} =
\begin{bmatrix}\text{scale}_x~\cos\theta&-\text{scale}_y~\sin\theta\\ \text{scale}_x~\sin\theta&\text{scale}_y~\cos\theta\end{bmatrix}
$$
Which yields the following equation system:
$$
\begin{eqnarray}
1 &= \text{scale}_x~\cos\theta\\
1 &= \text{scale}_y~\cos\theta\\
\text{shear}_x &= -\text{scale}_y~\sin\theta\\
\text{shear}_y &= \text{scale}_x~\sin\theta\\
\end{eqnarray}
$$
. This can only work if neither $\text{scale}_x$, $\text{scale}_y$, or $\cos\theta$ are zero. If that is the case, then we can immediately see that the scaling is isotropic ($\text{scale}_x=\text{scale}_y=\textrm{scale}$) and that
$$\textrm{scale} = \frac{1}{\cos\theta}$$
. We can use this to simplify the other two equations
$$
\begin{align}
\text{shear}_x &= - \frac{1}{\cos\theta}~\sin\theta = - \tan\theta\\
\text{shear}_y &= \frac{1}{\cos\theta}~\sin\theta = \tan\theta\\
\end{align}
$$
. Both can only be true at the same time, if $\text{shear}_x = -\text{shear}_y$, yielding above constraint. From here, we can find values for both $\theta$ and $\textrm{scale}$:
$$
\begin{align}
\theta &= \textrm{atan}~\textrm{shear}_y\\
\textrm{scale} &= \frac{1}{\cos{\theta}} = \frac{1}{\cos(\textrm{atan}~\textrm{shear}_y)} = \sqrt{\textrm{shear}_y^2 + 1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/931090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Taking limits of powers containing 'x'? If we have $lim_{x\rightarrow 0}\sqrt{x^2+sinx-tanx}$ we can write this as
$\sqrt{lim_{x\rightarrow \:0}\left(x^2+sinx-tanx\right)}$
OR If we have $lim_{x\rightarrow \:0}\left(x^3+secx-cosecx\right)^n$. We can write this as
$\left(lim_{x\rightarrow \:\:0}\left(x^3+secx-cosecx\right)\right)^n$
So if we have a limit as follows:
$lim_{x\rightarrow \:\:\:0}\left(\frac{sinx}{x}\right)^{\frac{1}{x^2}}$
We expand the limit as :
$lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right)^{\frac{1}{x^2}}=lim_{x\rightarrow 0}\left(\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}.....}{x}\right)^{\frac{1}{x^2}}=lim_{x\rightarrow 0}\left(1+\left[-\frac{x^2}{3!}+\frac{x^4}{5!}....\right]\right)^{\frac{1}{x^2}}=lim_{x\rightarrow \:0}\left(\left(1+\left[-\frac{x^2}{3!}+\frac{x^4}{5!}....\right]\right)^{\frac{1}{-\frac{x^2}{3!}+\frac{x^4}{5!}....}}\right)^{\frac{-\frac{x^2}{3!}+\frac{x^4}{5!}....}{x^2}}$$=e^{\left(lim_{x\rightarrow 0}\frac{-\frac{x^2}{3!}+\frac{x^4}{5!}....}{x^2}\right)}=e^{\left(lim_{x\rightarrow \:0}-\frac{1}{3!}+\frac{x^2}{5!}....\right)=}e^{-\frac{1}{6}}$
So why can't we take the limit as we did in the initial case as:
$lim_{x\rightarrow \:\:\:0}\left(\frac{sinx}{x}\right)^{\frac{1}{x^2}}$$=lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right).lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right).lim_{x\rightarrow \:0}\left(\frac{sinx}{x}\right)......\left(\frac{1}{x^2}times\right)=\left[lim_{x\rightarrow 0}\left(\frac{sinx}{x}\right)\right]^{lim_{x\rightarrow 0}\left(\frac{1}{x^2}\right)}=\left(1\right)^{lim_{x\rightarrow \:\:0}\frac{1}{x^2}}=1$
which is wrong. I know I am wrong somewhere, please point it out.
| The fallacy in your reasoning is that you took:
$$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}}=1$$
which is of the form ${(\sim1)}^{(\sim\infty)}$, which is indeterminate.
$$\lim_{x\to0}\sqrt{x^2+\sin x-\tan x}=\lim_{x\to0}\sqrt{x^2+\color{red}{(x-x^3/6+O(x^5))}-\color{blue}{(x+x^3/3+O(x^5))}}\\=\lim_{x\to0}\sqrt{x^2-x^3/2+O(x^5)}=0$$
Actually:
$$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}}=\lim_{x\to0}\exp\left(\frac1{x^2}\ln\left(\frac{\sin x}x\right)\right)=\lim_{x\to0}\exp\left(\frac1{x^2}\left(\frac{\sin x}x-1\right)\right)=\lim_{x\to0}\exp\left(\frac1{x^2}\left(\color{red}{(1-x^2/6+O(x^4))}-1\right)\right)=e^{-1/6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating an indefinite integral $\int\sqrt {x^2 + a^2} dx$ indefinite integral $$\int\sqrt {x^2 + a^2} dx$$
After some transformations and different substitution, I got stuck at this
$$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$
I am not sure I am getting the first step correct. Tried substituting $ x=a\tan \theta$ but that doesn't help either.
| Consider the integral
\begin{align}
I = \int \sqrt{x^{2}+ a^{2}} \, dx.
\end{align}
Make the substitution $x = a \sinh(t)$, $dx = a \cosh(t) dt$, it is seen that
\begin{align}
I &= a \int \sqrt{ a^{2} (1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\
&= a^{2} \int \sqrt{\cosh^{2}(t)} \cosh(t) \, dt \\
&= a^{2} \int \cosh^{2}(t) \, dt \\
&= \frac{a^{2}}{2} \int (1 + \cosh(2t)) dt \\
&= \frac{a^{2}}{2} \left[ t + \frac{1}{2} \sinh(2t) \right] \\
&= \frac{a^{2}}{2} \left[ t + \sinh(t) \cosh(t) \right].
\end{align}
Now back substitute to obtain
\begin{align}
\int \sqrt{x^{2}+ a^{2}} \, dx &= \frac{a^{2}}{2} \left[ \sinh^{-1}(x/a) + (x/a) \cosh(\sinh^{-1}(x/a)) \right] \\
&= \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \sinh^{-1}\left( \frac{x}{a} \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
x^8+366120 x^6+87829920 x^4+13038019200
x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
| Second non-zero coeff of long poly seems to be $3(n-1)(n^2+2n-12)$.
Could try subtracting $(x^2+3(n^2+2n-12))^{n-1}$ from each long poly to eliminate first two terms and see if the rest looks manageable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
How find this limit $\lim_{x\to 0}\frac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\frac{x^3}{3}+\frac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$ Find this limit
$$I=\lim_{x\to 0}\dfrac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$$
I think
$$I=6\lim_{x\to 0}\dfrac{\int_{0}^{x}(\sin{t}\ln{(1+t)}-t^2+\dfrac{t^3}{2})dt}{x^5}$$
Iuse $\sin{x}=x-\dfrac{x^3}{6}+o(x^3)$
and
$$\ln{(1+x)}=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+o(x^3)$$
so
$$\sin{x}\ln{(1+x)}=x^2-\dfrac{x^3}{2}+\dfrac{1}{6}x^4+o(x^4)$$
so
$$I=\dfrac{1}{5}$$
Have other methods?
| We can apply LHR as follows. First we note that $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\text{ (via LHR)},\text{ and }\lim_{x \to 0}\frac{e^{x^{2}} - 1}{x^{2}} = 1$$ and then we can write $$\begin{aligned}L &=\lim_{x\to 0}\dfrac{{\displaystyle \int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}}{(x-\sin{x})(e^{x^2}-1)}\\
&= \lim_{x \to 0}\dfrac{{\displaystyle \int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}}{\dfrac{x - \sin x}{x^{3}}\cdot x^{3}\cdot\dfrac{e^{x^{2}} - 1}{x^{2}}\cdot x^{2}}\\
&= 6\lim_{x \to 0}\dfrac{{\displaystyle \int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}}{x^{5}}\\
&= 6\lim_{x \to 0}\dfrac{\sin x\log(1 + x) - x^{2} + \dfrac{x^{3}}{2}}{5x^{4}}\text{ (Using L'Hospital's Rule)}\\
&= \frac{3}{5}\cdot\lim_{x \to 0}\dfrac{2\sin x\log(1 + x) - 2x^{2} + x^{3}}{x^{4}}\\
&= \frac{3}{5}\cdot\lim_{x \to 0}\dfrac{2(\sin x - x)\log(1 + x) + 2x\log(1 + x) - 2x^{2} + x^{3}}{x^{4}}\\
&= \frac{3}{5}\cdot\left\{\lim_{x \to 0}\dfrac{2(\sin x - x)}{x^{3}}\cdot\dfrac{\log(1 + x)}{x} + \lim_{x \to 0}\dfrac{2\log(1 + x) - 2x + x^{2}}{x^{3}}\right\}\\
&= \frac{3}{5}\cdot\left(-\frac{1}{3} + \lim_{x \to 0}\dfrac{2\log(1 + x) - 2x + x^{2}}{x^{3}}\right)\\
&= \frac{3}{5}\cdot\left(-\frac{1}{3} + \lim_{x \to 0}\dfrac{\dfrac{2}{1 + x} - 2 + 2x}{3x^{2}}\right)\text{ (Using L'Hospital's Rule)}\\
&= \frac{3}{5}\cdot\left(-\frac{1}{3} + \frac{2}{3}\lim_{x \to 0}\dfrac{1 - (1 - x)(1 + x)}{x^{2}(1 + x)}\right)\\
&= \frac{3}{5}\cdot\left(-\frac{1}{3} + \frac{2}{3}\lim_{x \to 0}\dfrac{1}{1 + x}\right)\\
&= \frac{3}{5}\cdot\frac{1}{3} = \frac{1}{5}\end{aligned}$$ Note that L'Hospital's rule should be used together with various manipulations to simplify limit expressions. Just applying L'Hospital repeatedly without any simplification does not help in most cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving a recurrence equation that yields polynomials I am trying to solve the following recurrence equation:
$$
T(n) = kT(n - 1) + nd
$$
I have expanded the first 4 values ($n = 1$ was given):
$$\begin{align}
T(1) & = 1 \\
T(2) & = kT(2-1) + 2d = k + 2d \\
T(3) & = kT(3-1) + 3d = k(k + 2d) + 3d = k^2 + 2kd + 3d \\
T(4) & = kT(4-1) + 4d = k(k^2 + 2kd + 3d) + 4d = k^3 + 2k^2d + 3dk + 4d\\
\end{align}$$
I was able to convert the above into the following summation, in hopes of finding a closed-form solution:
$$
T(n) = k^{n - 1} + d \sum_{i=1}^{n-1} (i + 1) (k^{n-i-1})
$$
But after this point I am stuck. Was creating the summation the correct thing to do, or is there something better I could try?
| You can convert your recursion into a more linear recursion using the following technique:
$$T(n) = k~T(n-1) + n~d$$
$$\frac 1d ~ T(n) - \frac 1d~k~T(n-1) = n \tag{A}$$
$$\frac 1d ~ T(n - 1) - \frac 1d~k~T(n-2) = n - 1 \tag{B}$$
$$\frac 1d ~ T(n) + \left(-k~\frac 1d - \frac 1d\right)~ T(n - 1) + \frac 1d~k~T(n-2) = 1 \tag{C}$$
$$T(n) = (k + 1)~ T(n - 1) - k~T(n-2) + d \tag{D}$$
(A) is just a rewrite, (B) is the equation for $n-1$, (C) = (A) - (B), (D) is just a rewrite.
(D) is pretty standard form, my preferred approach to finishing it is matrices:
$$
\begin{bmatrix}T(n + 2) \\ T(n + 1) \\ 1\end{bmatrix} =
\begin{bmatrix}
k+1 & -k & d \\
1 & 0 & 0 \\
0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} T(n + 1) \\ T(n) \\ 1\end{bmatrix}$$
$$
\begin{bmatrix}T(n + 2) \\ T(n + 1) \\ 1\end{bmatrix} =
\begin{bmatrix}
k+1 & -k & d \\
1 & 0 & 0 \\
0 & 0 & 1 \end{bmatrix} ^n
\begin{bmatrix} T(1) \\ T(0) \\ 1\end{bmatrix}$$
Jordan Decomp:
$$P = \begin{bmatrix}
1 & 1 & 1 \\
\frac 1k & 1 & 0 \\
0 & 0 & \frac{1-k}d
\end{bmatrix}, D = \begin{bmatrix}
k & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{bmatrix}$$
$$
\begin{bmatrix}T(n + 2) \\ T(n + 1) \\ 1\end{bmatrix} = \left(P~D~P^{-1}\right)^n
\begin{bmatrix} T(1) \\ T(0) \\ 1\end{bmatrix} = \left(P~D^n~P^{-1}\right)
\begin{bmatrix} T(1) \\ T(0) \\ 1\end{bmatrix}$$
Since $D^n = \begin{bmatrix}
k^n & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \end{bmatrix}$, you get:
$$
\begin{bmatrix}T(n + 2) \\ T(n + 1) \\ 1\end{bmatrix} =
\begin{bmatrix}\frac{{k}^{n+1}-1}{k-1} & -\frac{k\,\left( {k}^{n}-1\right) }{k-1} & \frac{d\,\left( {k}^{n+1}-2\,k+1\right) }{{\left( k-1\right) }^{2}} \\ \frac{{k}^{n}-1}{k-1} & -\frac{{k}^{n}-k}{k-1} & \frac{d\,\left( {k}^{n}-k\right) }{{\left( k-1\right) }^{2}} \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} T(1) \\ T(0) \\ 1\end{bmatrix}$$
Finally:
$$T(n+1) = \frac{{k}^{n}-1}{k-1} ~T(1) -\frac{{k}^{n}-k}{k-1} ~T(0) + \frac{d\,\left( {k}^{n}-k\right) }{{\left( k-1\right) }^{2}}$$
I know this isn't the approach that you started with, but it is a very general approach to solving even non-linear recurrences. Hope it helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x}
=\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x
=\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2}
$$
I noticed something interesting, namely that
$$
\begin{align*}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
& = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\
& = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
= \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
\end{align*}
$$
The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$).
But I had problems proving the first equality. Does anyone have some quick hints?
| Split the integral into two terms with limit $\left[0,\frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2},\pi\right]$
\begin{align}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\
\end{align}
Using identity
\begin{align}
\int_a^{b} f(x) \,\mathrm{d}x = \int_a^{b} f(a+b-x) \,\mathrm{d}x
\end{align}
Also the facts that $\cos\left(\frac{\pi}{2}-x\right)=\sin x$, $\sin\left(\frac{\pi}{2}-x\right)=\cos x$, $\cos\left(\frac{3\pi}{2}-x\right)=-\sin x$, and $\sin\left(\frac{3\pi}{2}-x\right)=-\cos x$, we get
\begin{align}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x &=\int_0^{\pi/2} \frac{(\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(-\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x \\
&=\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\tag{1}
\end{align}
Let
\begin{align}
I=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
\end{align}
Since
\begin{align}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
\end{align}
then
\begin{align}
2I&=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\\
&=\int_0^{\pi} \frac{(\cos x)^2 + (\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\\
I&=\frac{1}{2}\int_0^{\pi} \frac{1}{1 + \cos x \sin x} \,\mathrm{d}x\tag{2}
\end{align}
Using $(2)$, we get
\begin{align}
I&=\int_0^{\pi} \frac{1}{2 + 2\cos x \sin x} \,\mathrm{d}x\\
&=\int_0^{\pi} \frac{1}{2 + \sin (2x)} \,\mathrm{d}x\qquad\Rightarrow\qquad x\mapsto2x\\
&=\frac{1}{2}\int_0^{2\pi} \frac{1}{2 + \sin x} \,\mathrm{d}x\tag{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution?
Any help would be appreciated!
| The numbers are small, so general techniques are not necessary. However, we describe, in this particular case, the method of Hensel Lifting.
It is clear that the solutions modulo $7$ are $x\equiv \pm 3\pmod{7}$. We lift the solution $x\equiv 3\pmod{7}$ to a solution of $x^2\equiv 2\pmod{7^2}$.
Any solution to $x^2\equiv 2\pmod{7^2}$ that is congruent to $3$ modulo $7$ has shape $x=3+7t$. Square. Modulo $7^2$, the result is congruent to $9+(2)(3)(7)t\pmod{7^2}$. We want this to be congruent to $2$ modulo $7^2$, so we want
$$9+(2)(3)(7)t\equiv 2\pmod{7^2}.$$
A little manipulation turns this to
$$(2)(3)t\equiv -1\pmod{7}.$$
Thus $t=1$ works, and we have
$$x\equiv 10\pmod{7^2}.$$
Now lift this solution to a solution of $x^2\equiv 2\pmod{7^3}$. We look for a solution of the shape $x\equiv 10+7^2t\pmod{7^3}$.
Squaring, we get
$$100+(2)(10)(7^2)t\equiv 2\pmod{7^3},$$
which simplifies to
$$(2)(10)t\equiv -2\pmod{7}.$$
A solution is $t\equiv 2\pmod{7}$. That gives solution $x\equiv 10+2(49)\pmod{7^3}$.
By general theory, the only solutions are therefore $x\equiv \pm 108\pmod{7^2}$.
Remark: If we wished to, we could continue, and lift to a solution modulo $7^4$, $7^5$, and so on.
Hensel lifting is an important general technique for solving polynomial congruences modulo prime powers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Unclear step in half-angle formula derivation (trigonometric identities) In deriving the half-angle formulas, my textbook first says: "Let's take the following identities:"
$$\cos^2\left(\frac a2\right)+\sin^2\left(\frac a2\right)=1;$$
$$\cos^2\left(\frac a2\right)-\sin^2\left(\frac a2\right)=\cos(a);$$
These identities I know. But then the texbook says "through addition and subtraction, we respectively arrive at:"
$$2\cos^2\left(\frac a2\right)=1+\cos(a)$$
$$2\sin^2\left(\frac a2\right)=1-\cos(a)$$
I failed to catch what exactly is added and what is substracted to arrive from the first two formulas to the second pair. Give me a hint, please.
| Let $a, b$ be real numbers. Recall that
$$
e^{ia}=\cos a + i \sin a
$$ where $i^2=1.$
Then
$$
e^{i(a+b)}=\cos (a+b) + i \sin (a+b)
$$
But
$$\begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\\\\&=(\cos a + i \sin a)(\cos b + i \sin b)
\\\\&=(\cos a\cos b-\sin a\sin b) + i (\sin a\cos b+\sin b\cos a).
\end{align}
$$
Consequently
$$
\begin{align}
\cos (a+b)&=\cos a\cos b-\sin a\sin b\\
\sin (a+b)&=\sin a\cos b+\sin b\cos a.
\end{align}
$$
Putting $a=b$ gives
$$
\begin{align}
\cos (2a)&=\cos^2 a-\sin^2 a \tag1\\
\cos (2a) &=2\cos^2 a-1 \tag2
\end{align}
$$ since
$$ \cos^2 a+\sin^2 a=1. \tag3$$
From $(2)$ we easily get $$\cos^2 a=\frac{1+\cos (2a)}{2} \tag4
$$
From $(1)$ you also have
$$\cos (2a) =1-2 \sin^2 a \tag5 $$
giving
$$\sin^2 a=\frac{1-\cos (2a)}{2}.\tag6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate limit of ratio of these definite integrals How do I evaluate the following limit?
$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}$
| $\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}=\lim_{n\to\infty}\frac{\int_{0}^1\left(2+x-x^2\right)^n dx}{\int_{0}^1\left(2+2x-4x^2\right)^n dx}$
Let $I_{n}=\int_{0}^1\left(2+x-x^2\right)^n dx$
$(2+x-x^2)^n\geq 2^n$ for $0\leq x\leq 1$.Hence,$I_{n}\geq 2^n$
Let $J_{n}=\int_{0}^1\left(2+2x-4x^2\right)^n dx$
Now,$J_{n}=\frac{1}{2}\int_{0}^2(2+x-x^2)^n dx=\frac{1}{2}\int_{0}^1(2+x-x^2)^n dx+\frac{1}{2}\int_{1}^2(2+x-x^2)^n dx$
Let $K_{n}=\int_{1}^2(2+x-x^2)^n dx$
Now,$x^2-2x+1\geq 0$.Therefore, $2+x-x^2\leq 3-x$.Thus,
$K_{n}=\int_{1}^2(2+x-x^2)^n dx\leq\int_{1}^2(3-x)^n dx=\frac{2^{n+1}-1}{n+1}$
$K_{n}=\int_{1}^2(2+x-x^2)^n dx\leq\frac{2^{n+1}-1}{n+1}$
$0\leq \frac{K_{n}}{I_{n}}\leq \frac{2^{n+1}-1}{2^n(n+1)}$
$\lim_{n\to \infty} \frac{K_{n}}{I_{n}}=0$
$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}=\lim_{n\to \infty}\frac{I_{n}}{J_{n}}=\lim_{n\to \infty}\frac{2I_{n}}{I_{n}+K_{n}}=\lim_{n\to \infty}\frac{2}{1+\frac{K_{n}}{I_N}}=2$
Is there any other straight forward way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Integral $\int_{0}^{1} \frac{x^{2n +1}}{\sqrt{1 - x^2}} dx$ with n Find a general expression for $$\int_{0}^{1} \frac{x^{2n +1}}{\sqrt{1 - x^2}} dx$$ for every natural n.
Is there any common algorythm for such integrals?
| Let $x^2\mapsto x$, then
\begin{align}
\int^1_0\frac{x^{2n+1}}{\sqrt{1-x^2}}{\rm d}x
&=\frac{1}{2}\int^1_0x^n(1-x)^{-1/2}\ {\rm d}x\\
&=\frac{1}{2}\beta\left(n+1, \small{\frac{1}{2}}\right)\\
&=\frac{1}{2}\frac{n!\sqrt{\pi}}{(n+\frac{1}{2})\frac{(2n)!}{2^{2n}n!}\sqrt{\pi}}\\
&=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}\\
\end{align}
if $n\in\mathbb{N}$. Similarly,
\begin{align}
\int^1_0\frac{x^{2n}}{\sqrt{1-x^2}}{\rm d}x
&=\frac{1}{2}\beta\left(n+\small{\frac{1}{2}}, \small{\frac{1}{2}}\right)\\
&=\frac{1}{2}\frac{\frac{(2n)!}{2^{2n}n!}\pi}{n!}\\
&=\frac{\pi}{2^{2n+1}}\binom{2n}{n}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Solving $ \frac{1}{ a} = \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $ with additional conditions How to solve this equation
$$ \frac{1}{a}= \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $$
where $$ b = \sqrt{ (x-a/2)^2 + y^2 + z^2 )}$$
& $$ c = \sqrt{ (x+a/2)^2 + y^2 + z^2 )}$$
We have to get an equation in x y z and a!
I have tried to rationalize and do squaring but it becomes cumbersome.
Answer given is $y^2+z^2=15/4$
I get this by putting x=0.
| With the given info, the solution $y^2+z^2=\frac{15}{4}$ (or $\frac{15}{4}a^2$) is incorrect when $x=0$. One can verify the last statement by Mathematica. Here is another way to see it.
By the AM-HM inequality, we have
$$
\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq\frac{4}{\sqrt{b}+\sqrt{c}}.
$$
Now, use $x+y\leq\sqrt{2(x^2+y^2)}$ twice:
$$
\sqrt{b}+\sqrt{c}\leq\sqrt{2(b+c)}\leq\sqrt{2}(2b^2+2c^2)^{1/4}\leq\sqrt{2}[(a^2+4u+4x^2)]^{1/4}.
$$
Here $u=y^2+z^2\geq 0$. Putting this together, we get
$$
\frac{1}{a}\geq\frac{4}{\sqrt{2}[a^2+4u+4x^2]^{1/4}}\implies a^2+4u+4x^2\geq 64a^4.
$$
When $x=0$, all the inequalities above reduce to equalities, giving us
$$
a^2+4u=64a^4\implies u=\frac{1}{4}(-a^2+64a^4).
$$
Thus, when $x=0$, $u$ as given above for any $a$ (sufficiently large to ensure $u\geq 0$) will solve $\frac{1}{a}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$.
| {
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"url": "https://math.stackexchange.com/questions/960246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tricky integration/functions problem For $x>0$, let $f(x) = \displaystyle \int_1^x\frac{\ln t}{1+t}dt$. Find the function $f(x) + f(1/x)$ and show that $f(e) + f(1/e) = 1/2$.
Any help would be thoroughly appreciated.
| Let $g(x)=f(x)+f(\frac{1}{x})$. Then $ g^{\prime}(x)=\frac{\ln x}{1+x}+\frac{\ln\frac{1}{x}}{1+\frac{1}{x}}\cdot\left(\frac{-1}{x^2}\right)=\frac{\ln x}{1+x}+\frac{-\ln x}{x+1}\cdot\left(\frac{-x}{x^2}\right)=\frac{x\ln x+\ln x}{x(x+1)}=\frac{(\ln x)(x+1)}{x(x+1)}=\frac{\ln x}{x}$,
so $g(x)=\frac{1}{2}(\ln x)^2+C$, and $C=g(1)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
Progress
We have $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$.
Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$.
How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
| Let $d = gcd(a+b,a-b)$. So $d |(a-b)$, and $d|(a+b)$. Thus:
$d |(a-b)+(a+b) = 2a$, and $d |(a+b) - (a-b) = 2b$. So $d | gcd(2a,2b) = 2gcd(a,b) = 2$.
Thus $d = 1 $ or $2$.
For the other one, observe that $2(a^2 + b^2) = (a+b)^2 + (a-b)^2$, and $a^2 - b^2 = (a-b)(a+b)$. Thus if $ d = gcd(a^2+b^2, a^2-b^2)$, then $d |(a^2+b^2)+(a^2-b^2) = 2a^2$, and $d|(a^2+b^2)-(a^2-b^2) = 2b^2$. Thus: $d |gcd(2a^2,2b^2) = 2gcd(a^2,b^2) = 2(gcd(a,b)^2) = 2\cdot 1 = 2$. Thus $d = 1$ or $2$.
Note: $gcd(a^2,b^2) = (gcd(a,b))^2$ is easy to prove.
| {
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"url": "https://math.stackexchange.com/questions/964203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Does the series $\sum_{n=1}^{\infty}\sin\left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$ converge? Let $\alpha$ be such that $0\leq \alpha \leq 1$. Since $\sin n$ has no limit as $n$ tends to $\infty$, I'm having trouble with finding if the series $$\sum_{n=1}^{\infty}\sin \left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$$ is convergent? Thanks.
| For simplicity, let $a_n = \alpha^2 \sin n + (-1)^n$. Notice that since $\sin$ has period $2\pi$,
$$\sin \left( 2\pi \sqrt{n^2 + a_n} \right) = \sin \left( 2\pi \sqrt{n^2 + a_n} - 2\pi n \right) = \sin \frac {4\pi^2 (\sqrt{n^2 + a_n})^2 - 4\pi^2 n^2} {2\pi \sqrt{n^2 + a_n} + 2\pi n} = \\
\sin \frac {4\pi^2 a_n} {2\pi \sqrt{n^2 + a_n} + 2\pi n} ,$$
therefore your series becomes
$$\sum \limits _{n \ge 1} \sin \frac {4\pi^2 a_n} {2\pi \sqrt{n^2 + a_n} + 2\pi n} .$$
Let us study the fraction inside the sine. Since
$$| a_n | = | \alpha^2 \sin n + (-1)^n | \le \alpha^2 + 1$$
and $\alpha$ is a constant, we deduce that the numerator is bounded. The denominator, on the other hand, tends to $\infty$, so the whole fraction will tend to $0$. In particular, there exist $n_0$ such that for $n \ge n_0$ the fraction will stay in $[0, \frac \pi 2]$, so its sine will be positive.
This allows us to use the limit comparison test to "drop" the sine, i.e. we deduce that the series has the same behaviour as
$$\sum \limits _{n \ge 1} \frac {4\pi^2 a_n} {2\pi \sqrt{n^2 + a_n} + 2\pi n} $$
(we use that if $x_n \to 0$, then $\dfrac {\sin x_n} {x_n} \to 1$).
Dropping those annoying constants and keeping in mind that $a_n$ is bounded, we can use the same test again to further simplify the denominator and obtain
$$\sum \limits _{n \ge 1} \frac {a_n} n .$$
Expanding this, we get
$$\alpha^2 \sum \limits _{n \ge 1} \frac {\sin n} n + \sum \limits _{n \ge 1} \frac {(-1)^n} n .$$
The first series is convergent by Abel-Dirichlet's test (this has been shown several times on MSE), while the second one is clearly convergent by Leibniz's test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.
$$
Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that:
$$
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.
$$
Finding $\cos\alpha$ and $\sin\beta$:
$$\begin{align}
\cos\alpha & = \frac{\sqrt{4-x^2}}{2}, \\[0.1in]
\sin\beta & = \sqrt{1-x^2}.
\end{align}$$
So plugging everything in:
$$\begin{align}
\sin(2\arcsin x + \arccos x) & = \frac{x}{2} \cdot x + \frac{\sqrt{4-x^2}}{2} \cdot \sqrt{1-x^2}\\[0.1in]
& = \frac{x^2 + \sqrt{4-x^2}\sqrt{1-x^2}}{2}.
\end{align}$$
But this doesn't seem to lead to the right side. Is my method incorrect?
| $f(x)=\sin(2\arcsin x + \arccos x) - \sqrt{(1-x^2)}=\sin(\arcsin x + \frac\pi 2) - \sqrt{(1-x^2)}=\cos(\arcsin x)-\sqrt{(1-x^2)}$.
$f'(x)=0$, hence $f$ is constant.
In particular $f(x)=f(0)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the sum $\sum_{n=1}^\infty \frac{n}{(1+x)^{2n+1}}$ Find the sum
$$\sum_{n=1}^\infty \frac{n}{(1+x)^{2n+1}}.$$
Indicating the interval of convergence for $x$.
My attempt:
Let $ t=\frac{1}{x+1}$. Then, applying the root test,
$$\lim_{n\to \infty} \{n t^{2n+1}\}^{1/n} = |t|^2 < 1 \iff |t| <1.$$
Then, we have that $|x-1| >1 \iff x < -2 \text{ or } x >0$.
Now, consider the series
$$\sum_{n=1}^\infty n t^n = \frac{t}{(t-1)^2}, \quad |t|<1.$$
So, since $|t|< 1 \Rightarrow |t^2| <1$,
$$\sum_{n=1}^\infty n t^{2n }= \frac{t^2}{(t^2-1)^2}, \quad |t|<1.$$
Multiplying by $t,$
$$\sum_{n=1}^\infty n t^{2n +1 }= \frac{t^3}{(t^2-1)^2}, \quad |t|<1.$$
If we substitute back, we have what we want.
I want to know if my steps are correct. I have doubts about the interval of convergence part.
Thanks for your effort!
| Starting with the series
\begin{align}
\sum_{n=1}^{\infty} n \, t^{n} = \frac{t}{(1-t)^{2}}
\end{align}
then it is seen that
\begin{align}
\sum_{n=1}^{\infty} \frac{n}{(1+x)^{2n+1}} = \frac{1}{(1+x)^{3}} \cdot \frac{(1+x)^{4}}{[(1+x)^{2}-1]^{2}} = \frac{1+x}{x^{2} \, (2+x)^{2}}.
\end{align}
The series does not converge for $x \in\{0, -2\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\;\int x^3\sqrt{x^2 + 2}\,dx$ Integrate the following: $$\int x^3\sqrt{x^2 + 2}\,dx$$
I understand how to do basic integration by parts but I don't know what to do with
$\sqrt{x^2+2}$.
Do I divide the $\sqrt{x^2+2}$ by 2 first to make it becomes to $\sqrt{2}\sqrt{\frac{x^2}{2}+1}$ ? If so, how do I keep going?
Thank you for any help!
| If you let $u = x^2 + 2\implies du = 2x\,dx \iff \dfrac {du}{2} = x\,dx$,
And note that $x^3 = x\cdot x^2,$ where $x^2 = u-2$,
Then your integral is equivalent to $$\int x^3\sqrt{x^2 +2}\,dx = \int x^2 \sqrt{x^2 + 2}(x\,dx)$$ $$ = \frac 12 \int (u-2)\sqrt u \,du = \frac 12\int \left(u^{3/2} -2u^{1/2}\right)\,du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with a differential equation?
I'm confused about what the question is asking. I solved the following equations:
$$y'' + 4y = 0 \implies y = c_1\cos2x + c_2\sin2x$$
$$y'' + 4y = \sin x \implies y= c_3\cos2x + c_4\sin2x + \frac{\sin x}{3}$$
Using the given initial values, the coefficients for the second solution (the one that's the solution for the equation with $\sin x$ on the RHS) can be solved for. I don't understand how or why you would be able to solve for the coefficients for the first solution though ($c_1$ and $c_2$). The given value is outside of the domain of the first function, hence it doesn't say anything about it so it can't be used to solve for the coefficients.
How can $c_1$ and $c_2$ be solved for?
| You are asked to find a solution where $y$ and $y^\prime$ are continuous at $\frac{\pi}{2}$. Now in the region up to $\frac{\pi}{2}$ the solution is dictated by the initial conditions (at $x=0$) giving $c_3 = 1$ and $2 c_4 + \frac{1}{3} = 2$ (that is, $c_4 = \frac{5}{6}$.
$$
y = 1 \cdot \cos (2x) + \frac{5}{6} \sin (2x) + \frac{1}{3} \sin x
$$
Now we can find the value and first derivative at $\frac{\pi}{2}$:
$$
y(\frac{\pi}{2}) = -1 + \frac{5}{6} (0) + \frac{1}{3} = -\frac{2}{3}
$$
$$
y^\prime = 0 - \frac{5}{3} + 0 = - \frac{5}{3}
$$
So $c_1$ and $c_2$ are determined by the need for the limit of the solution having a value of $-\frac{2}{3}$ and a first derivative of $-\frac{5}{3}$ as $x\rightarrow \frac{\pi}{2}^+$. Thre is nothing fancy about the limit, it is just the value in this case. So,
$$
-c_1 + 0 = -\frac{2}{3}
$$ and
$$
0 - 2 c_2 = -\frac{5}{3}
$$
Thus $c_1 = \frac{2}{3}$ and $c_2 = \frac{5}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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