Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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
} |
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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"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 $6$.
... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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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:
$$\frac{\cos{2x}-2\sin{2x}\cos{2x}-\cos{6x}}{\cos{2x}+2\sin{2x}\cos{2x}-\cos{6x}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-... | Well, don't expand to $\sin(x)$ or $\cos(x)$...
Consider complex numbers...
Let $z = \exp( 2 \textbf{i} x )$
Then
$
\begin{eqnarray}
2 \cos(2x) &=& z + \bar{z}\\
2 \sin(2x) &=& - \textbf{i} \Big( z - \bar{z} \Big)\\
2 \sin(4x) &=& - \textbf{i} \Big( z^2 - \bar{z}^2 \Big)\\
&=& - \Big( z + \bar{z} \Big) \textbf{i} \Big(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Maclaurin series of (1+x)^(1/x) how can i find the Maclaurin series of $f(x)=(1+x)^{1 \over x}$?
$f(0)$ is not even defined, or should I define it as $f(0)=e$?
I stopped at the first derivative as it gets terribly messy.
thank you
| Hint
Start with $$\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)}$$ Now, set $y=-\frac{x}{2}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate the limit $\lim_{x\rightarrow 0}\left[ \frac{\ln(\cos x)}{x\sqrt{1+x}-x} \right]$ Evaluate the limit:
$$\lim_{x\rightarrow 0}\left[ \frac{\ln(\cos x)}{x\sqrt{1+x}-x} \right]$$
I actually was able to find the limit is $-1$ after applying L'Hôpital's rule twice.
I wonder if that was the intention of this exercis... | We need to proceed as follows $$\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... | {
"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
} |
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:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \c... | A variant to minimise the computation time of the order of $9\bmod100$: by the Chinese remainder theorem,
$$\mathbf Z/100\mathbf Z\simeq\mathbf Z/4\mathbf Z\times \mathbf Z/25\mathbf Z. $$
Now $\varphi(25)=20$, and $9\equiv 1\mod4$. Hence $9^{20}\equiv 1\mod100$, so the order of $9$ is a divisor of 20.
Fast exponentia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Proving $x+\sin x-2\ln{(1+x)}\geqslant0$
Question: Let $x>-1$, show that
$$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}=\dfr... | 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 obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
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)$\displaysty... | 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, $m_1m_2=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/843720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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
$$
\begin{align}
\int_{x=0}^\infty \exp\left(-ax^2-\frac{b}{x^2}\right)\,dx&=\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Mathematical Induction Matrix Example I'm a little rusty and I've never done a mathematical induction problem with matrices so I'm needing a little help in setting this problem up.
Show that $$\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}$$ for every $... | The case $n=1$ is clear since $2^0 = 1$. So suppose that
$$\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} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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 $$\frac{1}{5}\int_{0}^{1}{u^{\frac{4}{5}-1}(1-u)^{\frac{1}{5}-1}du}=\frac{1}{5}\beta(1-\frac{1}{5},\frac{1}{5})=\frac{1}{5}\Gamma(1-\frac{1}{5})\Gamma(\frac{1}{5})=\frac{1}{5}\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answ... | Hint
First of all, you could start factoring both numerator and denominator $$\frac{3x^2+3x-6}{2x^2+6x+4}=\frac{3(x^2+x-2)}{2(x^2+3x+2)}$$ In order to factor them, you can then compute the roots of the quadratics : for the numerator, the roots are obviously $1$ and $-2$ and for the denominator $-1$ and $-2$ (you can so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B ... | Note that $\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^3}}$, with a similar expression for the other one. Make the substitution $t=1/x$. Our expression becomes
$$\frac{2-\sqrt[3]{1+t+t^3}-\sqrt[3]{1-t+t^3}}{t},$$
and we want to find the limit of this as $t$ approaches $0$ from the right. Note that the condi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Is this procedure for $5^{300} \bmod 11$ correct? I'm new to modular exponentiation. Is this procecdure correct?
$$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 = ... | 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
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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 ta... | 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
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-\frac{1}{2\sqrt[]{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
M... | I could not get any ideas of getting the answer analytically.
So I just wrote the matlab program as below
format rat
Initial=1
i=1;
count=0;
while i<100,
Initial = Initial*(i+1)/i;
i=i+2;
count=count+1;
a(count+1)= Initial;
end;
Got the answer 12 after flooring the end result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. The problem asks to show that $$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.$$ is continuous at the origin, however it has resisted my bravest efforts. I have attempted using $x^4+y^4 \geq y^4$ and ... | To expand on my comments, let $\epsilon > 0 $ be given, then our limit, in polar coordinates is of the form
$$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}... | {
"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
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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 orthog... | 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
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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 = \pro... | 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
$$\left[\sin t \cos^{n}t\right]_{-\frac{\pi}{2}} ^0+n\int _{-\frac{\pi}{2}} ^0 \sin t\cos^{n-1} t \sin t dt=n\int _{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the derivative of $\sqrt{x+\sqrt{x^2+5}}$ How to derive $y=\sqrt{x+\sqrt{x^2+5}}$ at $x=2$.I used logarithmic differentiation and chain rule over and over again but I can't get the right answer
| \begin{align*}
\frac{d}{dx}y & = \frac{d}{dx}\sqrt{x+\sqrt{x^2+5}}\\
& = \frac{1}{2}\frac{1}{\sqrt{x+\sqrt{x^2+5}}}\cdot \frac{d}{dx}(x+\sqrt{x^2+5})\\
& = \frac{1}{2}\frac{1}{\sqrt{x+\sqrt{x^2+5}}} \,\left(1+\frac{x}{\sqrt{x^2+5}}\right)\\
\end{align*}
Now simplify as best as you can.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/859254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Optimize $x^2 + y^2 +2z^2 +z(x^2-y^2)$ subject to $x+y=2$ $$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{... | 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",
"answer_count": 1,
"answer_id": 0
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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 $$P(x)=\sum_{i=0}... | Momentarily setting aside the trivial solution, one option is to rearrange in the form of a finite difference equation, like so:
$$(x^3+1)(P(x+1)-2P(x)+P(x-1))-mx^2(P(x+1)-P(x-1))+2mxP(x)=0$$
Adjusting the definition of a polynomial $P(x)$ might make sense in this case, where it appears that finite differences play a s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/863142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Sum related to zeta function I was trying to evaluate the following sum:
$$\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\s... | 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 ... | {
"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... | 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 $A_{k+1}<A_{k}$... | {
"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}... | 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",
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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... | 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 ord... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 o... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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}... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 c... | {
"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}... | If you use polar coordinates you will get the following
$$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$
| {
"language": "en",
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"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 rare... | 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
$$\sum_{cyc}\frac{(a-1)^2(9-2a)}{(3-a)^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Finding (or rather expanding) the product $(5-xy)(3+xy)$ Given the product $(5-xy)(3+xy)$
I tried the following,
As we know, $(x+a)(x+b)=x^2+(a+b)x+ab$
Here $x$ is $xy$. But $xy$ has two signs$-$ and $+$.
How do I solve this.
| $$ \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*} $$Hope that makes sense!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\sum_{n=1}^\infty \sqrt{n+1}-\sqrt{n}$ diverge? Why does $\sum_{n=1}^\infty \sqrt{n+1}-\sqrt{n}$ diverge?
Using the ratio test I get the following. First of all since $u_n=\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$
Then $\left| \frac... | The ratio test actually doesn't conclude anything. Note that
$$
\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}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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) ... | 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
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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 $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\s... | When you substitute a new variable in, you must not forget to change the limits of your integral.
So,
instead of $$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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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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 $... | 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 $$\displaystyle \frac{\log{x}}{1-x^3}=-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$
Find $f'(1)$ if $$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}} $... | 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",
"answer_count": 4,
"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 homogen... | $$2f(n) = f(n+1)+f(n-1)+3$$
$$f(n+1) = 2f(n) - f(n-1) - 3$$
$$
\begin{bmatrix} f(n+1) \\ f(n) \end{bmatrix}
=
\begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix} f(n) \\ f(n - 1) \end{bmatrix}
+
\begin{bmatrix} -3 \\ 0 \end{bmatrix}
$$
$$
\begin{bmatrix} f(n+1) \\ f(n) \\ 1 \end{bmatrix}
=
\begin{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Prove $\tan 54^\circ=\frac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$ How to prove this identity without using the actual values of $\tan54^\circ$ and $\sin24^\circ$
$$\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}{\sq... | 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
$$2\cos 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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A nice trignometric identity How to prove that:
$$\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}$$
we have $A + B = 2x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$ Question:
$$\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\f... | Hint:
$$\tan^2\theta (\csc^2\theta-1) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1}{\sin^2\theta}-1\right) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1}{\sin^2\theta}-\frac{\sin^2\theta}{\sin^2\theta}\right) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1-\sin^2\theta}{\sin^2\theta}\right) = \frac{\sin^2\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How find the minimum of the $n$ such $99^n+100^n<101^n$ Question:
Find the smallest positive integer $n$ such that
$$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$$
Use thi... | First of all, $n \ge 34$ is a necessary (but not sufficient) condition for that inequality to be true. Remember that $1+nx \le c$ does not imply that $(1+x)^n \le c$.
We can get a better bound by manipulating the inequality as follows:
$99^n + 100^n < 101^n$
$101^n - 99^n > 100^n$
$\left(\dfrac{101}{100}\right)^n - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How prove that $ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$? Let $ a,b,c>0$ be such that $ a+b+c=1$. How prove that
$ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$?
| Let $a+b=2u$ and $ab=v^2$.
Hence, we need to prove that $3(a^3+b^3)+1-3(1-a-b)\geq\frac{a^2+b^2+1-4(1-a-b)}{a+b}$ or
$$6u(4u^2-3v^2)+1-3(1-2u)\geq\frac{4u^2-2v^2+1-4(1-2u)}{2u},$$
which is a linear inequality of $v^2$, which says that it's enough to prove our inequality for an extremal value of $v^2$, which happens in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integral $\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$ I would like to know how to evaluate the integral
$$\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$$
I tried expanding the integrand as a series but made little progress as I do not know how to evaluate the resulting sum.
\begin{align}
\int^1_0\frac{\ln{x} \ ... | \begin{align}\text{J}&=\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[-\ln(1-x)\ln x\mathrm{Li}_2(x)\Big]_0^1}_{=0}+\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(x)}{x}dx}_{=-\frac{1}{2}\text{Li}^2_2(1)}-\underbrace{\int_0^1 \frac{\ln^2(1-x)\ln x}{x}dx}_{\text{IBP}}\\
&=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$. The exercise is:
Calculate:$$\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}{\sq... | 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 $\sqrt{2}$... | {
"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 $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$
$$\int_0^1e^{x^2}dx\\\eqalign{
&=\int_0^1\left(1+x^2+\frac{(x^2)^2}{2!}+\frac{(x^2)^3}{3!}+\cdots\right) dx\\
&=\int_0^1\left(1+x^2+\frac{(x^4)}{2!}+\frac{(x^6)}{3!}+\cdots\right)dx\\
&=\left(x+\frac{x^3}{3.1!}\frac{x^5}{(5.2)}+\frac{x^7}{(7.3!)}+\cdots\right)_0^1\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$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.$$
Co... | Both solutions are correct provided your $d$ satisfies the condition $y(1) = 1$. Note that in your solution $y(1) = \frac{A}{3} + \frac{B}{2} + c + d = 1$ so setting $d = 1 - (\frac{A}{3} + \frac{B}{2} + c)$ would be the way to go.
Note also that
$$\frac{Ax^3}{3} + \frac{Bx^2}{2} + cx + \underbrace{1-\left(\frac{A}{3} ... | {
"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
$$\dfrac{\sin \left( \dfrac{3^{n+1}\pi}{3^n+1}\right)}{\sin \left( \dfrac{3 \pi}{3^n+1}\right)} = \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Problem getting the real roots of this complex expression I'm trying to get the real roots of this expression:
$$\dfrac{1}{z-i}+\dfrac{2+i}{1+i} = \sqrt{2}$$
Where $i^2=-1$ and $z=x+iy$.
I tried to simplify that with Algebra, and then separate the real and imaginary parts in both sides of the expression to obtain an eq... | This can be done by "getting the unknown alone" by simplifying the fraction and undoing each operation on the left side.
We get
$$\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} + \fr... | {
"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
} |
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
$$
\beg... | You can use the technique of Newton's Sums which @user157227's comment hinted at
SOLUTION:
Basically imagine if you factored the polynomial into its 5 roots
$$(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$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$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{... | 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 th... | {
"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
} |
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 =... | $$z=1-\sin \alpha +i \cos \alpha\\=\{(\cos^2 \dfrac{ \alpha}{2}+\sin^2 \dfrac{ \alpha}{2})-2\sin \dfrac{ \alpha}{2}\cos \dfrac{ \alpha}{2}\}+i(\cos^2 \dfrac{ \alpha}{2}-\sin^2 \dfrac{ \alpha}{2})\\=
(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})^2+i(\cos \dfrac{ \alpha}{2}-\sin \dfrac{ \alpha}{2})(\cos \dfrac{ \alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \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 \... | 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
$$ y = \frac{1}{\sqrt 3 } \to g(y) = \frac{4}{{3\sqrt 3 }}=f(x)$$
$$ y = - \frac{1}{\sqrt 3 } \to g(y) = - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(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)}... | 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 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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... | What you have to find is just:
$$\begin{eqnarray*}&&[x^{20}](x+x^2+x^3+x^4)(x^2+\ldots+x^{10})(x^3+\ldots+x^{12})\\&=&[x^{14}](1+\ldots+x^3)(1+\ldots+x^8)(1+\ldots+x^9)\\&=&[x^{14}]\frac{(1-x^4)(1-x^9)(1-x^{10})}{(1-x)^3}\\&=&[x^{14}](1-x^4-x^9-x^{10}+x^{13}+x^{14})\sum_{j=0}^{+\infty}\binom{j+2}{2}x^j\\&=&\binom{16}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate
$$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)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the most distant points on a curve Hy, I have to ask someone for help with this problem.
I have a curve with this implicit equation:
$$\left ( x^2 + y^2 \right )^2 = x^3 + y^3$$
I have to find the most distant coordinates from the center of coordinate system, and the most distant coordinates from y-axis.
The grap... | Convert to polar coordinates:
$$\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{ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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",
"url": "https://math.stackexchange.com/questions/909670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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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 $$a_{n}=\dfrac{\sin{(nx)}}{\sin{x}}$$
the... | 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 ineq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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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 Wolfra... | 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:
$$\color{red}{\sum_{n=0}^{+\infty}\frac{1}{(3n)!}}=\sum_{n=0}^{+\infty}\frac{f(n)}{n!}=\frac{1}{3}\left(\exp(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Addition of fractions repetition and convergence Is this a new mathematical concept?
$$
\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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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 so... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 8
} |
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?
$$\frac{\sqr... | Each equation of the second pair is obtained from an equation in the first pair by "cross-multiplying," or equivalently getting rid of the denominators.
For example, start from $\frac{BC}{AB}=\frac{BH}{BC}$. Multiply both sides by $(AB)(BC)$. We get $(BC)(BC)=(AB)(BH)$.
Similarly, from $\frac{AC}{AB}=\frac{AH}{AC}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximizing $ \frac {y + z + yz}{\left( 1 + y + z \right)^2} $ I was solving a WOOT Problem of the Day and I boiled the problem down to finding the maximum value of $$ \frac {y+z+yz}{\left( 1 + y + z \right)^2}. $$Intuitively, this seems to occur when $y=z$, since $y$ is symmetric, and after that, it is easy to find tha... | If you compute the derivatives of $$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... | {
"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 s... | 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 \u... | {
"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 bal... | For the split up of $5$ there are $2$ possibilities:
$5=1+2+2$
$5=1+1+3$
Leading to $$\binom{5}{1}\binom{6}{2}\binom{7}{2}+\binom{5}{2}\binom{6}{1}\binom{7}{2}+\binom{5}{2}\binom{6}{2}\binom{7}{1}+\binom{5}{1}\binom{6}{1}\binom{7}{3}+\binom{5}{1}\binom{6}{3}\binom{7}{1}+\binom{5}{3}\binom{6}{1}\binom{7}{1}$$
possibilit... | {
"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 pa... | 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 $$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Third point of a triangle in the complex plane I have an equilateral triangle with two points equal to $(2+2i)$ and $(5+i)$. I want to find the third point(s) (there are $2$ of these). I have that the side length of the triangle is $\sqrt{10}$.
| (I'm purposely not trying
to be clever here
and doing all calculations in my head.)
The distance
$d$
between the two known points
satisfies
$d^2
=(2-5)^2+(2-1)^2
=10
$.
If the point is
$(x, y)$,
then
$(x-2)^2+(y-2)^2
=(x-5)^2+(y-1)^2
=10
$,
or
$x^2-4x+4+y^2-4y+4
=x^2-10x+25+y^2-2y+1
=10
$,
or
$x^2-4x+y^2-4y+8
=x^2-10x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$? I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\dfrac{\sqrt{16x^2-9}}x\,\mathrm{d}x$.
$$v=4x\implies\mathrm{d}v=4\,\mathrm{d}x$$
$$\int\frac{\sqrt{v^2-9}}{v}\,\mathrm{d}v\implies a=3,... | 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...
$$\sqrt {16x^2-9}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seem... | You have another solution using Taylor series. Rewrite $$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)$$ and remember that, when $y$ is small $\sqrt{1+y}\simeq 1+y/2$. Replace successively $y$ by $a/x^2$ and $b/x^2$ and you are done.
If you continue in the same spirit you coulod sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$ Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$
I choose to use the formula $\displaystyle f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$
Which after some work I found to be
$\frac{3-\sqrt{2x+1}}{3\sqrt{2x+1}(x-4)}$
Which is basically back to where I started. Is... | Hint: Try multiplying by the conjugate:
$$
\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)
$$
so that you can cancel the $(x - 4)$ term.
At the request of the comments, here's all the work:
\begin{align*}
\left. \frac{d}{dx}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple differentiation from first principles problem I know this is really basic, but how do I differentiate this equation from first principles to find $\frac{dy}{dx}$:
$$
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} \rig... | $$\begin{align} f(x+h)-f(x) &= \dfrac{1}{x+h} - \dfrac{1}{x} \\ &= \dfrac{x-h-x}{x(x+h)} \\ &= \dfrac{-h}{x(x+h)} \end{align}$$
| {
"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
} |
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 $\c... | 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 $$\cos^4(x)=\Big(\frac{\cos(2x)+1}{2}\Big)^2=\frac{1}{4}\Big(\cos^2(2x)+2 \cos(2x)+1\Big)$$ $$\cos^4(x)=\frac{1}{4}\Big(\frac{\cos(4x)+1}{2}+2 \cos(2x)+1\Big)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Number of Lattice Points in a Triangle
Problem
Let the co-ordinates of the vertices of the $\triangle OAB$ be $O(1,1)$, $A(\frac{a+1}{2},1)$ and $B(\frac{a+1}{2},\frac{b+1}{2})$ where $a$ and $b$ are mutually prime odd integers, each greater than $1$. Then find the number of lattice points inside $\triangle OAB$, i.e.... | The area of $\Delta OAB$ is $K = \dfrac{1}{2} \cdot \dfrac{a-1}{2} \cdot \dfrac{b-1}{2} = \dfrac{(a-1)(b-1)}{8}$.
The number of points on the boundary (line segments $OA$, $AB$, and $BO$) is
$\underbrace{\dfrac{a-1}{2}}_{OA}+\underbrace{\dfrac{b-1}{2}}_{OB}+\underbrace{\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Decomposition of shear matrix into rotation & scaling How can I decompose the affine transformation:
$$ \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}$$
\begin{bmatrix}\text{scale}_x&0\\0... | You can only do this if $\textrm{shear}_x = -\textrm{shear}_y$, in which case
$$
\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& ... | {
"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\rig... | 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.
$$\lim_{x\to0}\sqrt{x^2+\sin x-\tan x}=\lim_{x\to0}\sqrt{x^2+\color{red}{(x-x^3/6+O(x^5))}-\color{blue}{(x+x^3/3+O(x^5))}}\\=\lim_{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating an indefinite integral $\int\sqrt {x^2 + a^2} dx$ indefinite integral $$\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$$
I am not sure I am getting the first step correct. Tried substituting $ x=... | Consider the integral
\begin{align}
I = \int \sqrt{x^{2}+ a^{2}} \, dx.
\end{align}
Make the substitution $x = a \sinh(t)$, $dx = a \cosh(t) dt$, it is seen that
\begin{align}
I &= a \int \sqrt{ a^{2} (1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\
&= a^{2} \int \sqrt{\cosh^{2}(t)} \cosh(t) \, dt \\
&= a^{2} \int \cosh^{2}(t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left... | 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
$$I=6\lim_{x\to 0}\dfrac{\int_{0}^{x}(\sin{t}\ln{(1+t)}-t... | We can apply LHR as follows. First we note that $$\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}\... | {
"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):
$$\begin{align}
T(1) & = 1 \\
T(2) & = kT(2-1) + 2d = k + 2d \\
T(3) & = kT(3-1) + 3d = k(k + 2d) + 3d = k^2 + 2kd + 3d \\
T(... | You can convert your recursion into a more linear recursion using the following technique:
$$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 \... | {
"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
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 +... | Split the integral into two terms with limit $\left[0,\frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2},\pi\right]$
\begin{align}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution?
Any help would be appreciated!
| The numbers are small, so general techniques are not necessary. However, we describe, in this particular case, the method of Hensel Lifting.
It is clear that the solutions modulo $7$ are $x\equiv \pm 3\pmod{7}$. We lift the solution $x\equiv 3\pmod{7}$ to a solution of $x^2\equiv 2\pmod{7^2}$.
Any solution to $x^2\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Unclear step in half-angle formula derivation (trigonometric identities) In deriving the half-angle formulas, my textbook first says: "Let's take the following identities:"
$$\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);$$
These identitie... | Let $a, b$ be real numbers. Recall that
$$
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}
$$
Consequently
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculate limit of ratio of these definite integrals How do I evaluate the following limit?
$\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}$
| $\lim_{n\to\infty}\frac{\int_{0}^1\left(x^2-x-2\right)^n dx}{\int_{0}^1\left(4x^2-2x-2\right)^n dx}=\lim_{n\to\infty}\frac{\int_{0}^1\left(2+x-x^2\right)^n dx}{\int_{0}^1\left(2+2x-4x^2\right)^n dx}$
Let $I_{n}=\int_{0}^1\left(2+x-x^2\right)^n dx$
$(2+x-x^2)^n\geq 2^n$ for $0\leq x\leq 1$.Hence,$I_{n}\geq 2^n$
Let $J_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_{0}^{1} \frac{x^{2n +1}}{\sqrt{1 - x^2}} dx$ with n Find a general expression for $$\int_{0}^{1} \frac{x^{2n +1}}{\sqrt{1 - x^2}} dx$$ for every natural n.
Is there any common algorythm for such integrals?
| Let $x^2\mapsto x$, then
\begin{align}
\int^1_0\frac{x^{2n+1}}{\sqrt{1-x^2}}{\rm d}x
&=\frac{1}{2}\int^1_0x^n(1-x)^{-1/2}\ {\rm d}x\\
&=\frac{1}{2}\beta\left(n+1, \small{\frac{1}{2}}\right)\\
&=\frac{1}{2}\frac{n!\sqrt{\pi}}{(n+\frac{1}{2})\frac{(2n)!}{2^{2n}n!}\sqrt{\pi}}\\
&=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}\\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $ \frac{1}{ a} = \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $ with additional conditions How to solve this equation
$$ \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 )}$$
We have to get an equ... | With the given info, the solution $y^2+z^2=\frac{15}{4}$ (or $\frac{15}{4}a^2$) is incorrect when $x=0$. One can verify the last statement by Mathematica. Here is another way to see it.
By the AM-HM inequality, we have
$$
\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq\frac{4}{\sqrt{b}+\sqrt{c}}.
$$
Now, use $x+y\leq\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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$, an... | {
"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 \exis... | 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... | {
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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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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}\r... | For simplicity, let $a_n = \alpha^2 \sin n + (-1)^n$. Notice that since $\sin$ has period $2\pi$,
$$\sin \left( 2\pi \sqrt{n^2 + a_n} \right) = \sin \left( 2\pi \sqrt{n^2 + a_n} - 2\pi n \right) = \sin \frac {4\pi^2 (\sqrt{n^2 + a_n})^2 - 4\pi^2 n^2} {2\pi \sqrt{n^2 + a_n} + 2\pi n} = \\
\sin \frac {4\pi^2 a_n} {2\pi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\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.
$... | $f(x)=\sin(2\arcsin x + \arccos x) - \sqrt{(1-x^2)}=\sin(\arcsin x + \frac\pi 2) - \sqrt{(1-x^2)}=\cos(\arcsin x)-\sqrt{(1-x^2)}$.
$f'(x)=0$, 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",
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"answer_id": 2
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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... | 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... | {
"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
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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 ke... | 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",
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"answer_id": 1
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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 (t... | 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 (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/974107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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