Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Compute $\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$ Compute the given line integral:
*
*$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$
Let $P=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},Q=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}, R=\fr... | Consider
$$f(x,y,z)=\sqrt{x^2+y^2+z^2}+C$$
then $\vec \nabla f=F$ and
$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}=f(3,4,5)-f(0,0,0)=5\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Binomial coefficient and series expansion I had to expand $y=\sqrt{1+x^2}$ into the Maclaurin series, which (prior to simplification) would give me: \begin{align} \sqrt{1+x^2}= 1+\frac{x^2}{2}+ \frac{1/2(1/2-1)x^4}{2}+\frac{1/2(1/2-1)(1/2-2)x^6}{3!}+...+\frac{1/2(1/2-1)...(1/2-n+1)x^{2n}}{n!}+...\end{align}
What is the... | A convenient generalisation of the binomial coefficient is given for real (or even complex) values $\alpha$ by
\begin{align*}
\binom{\alpha}{n}:=
\begin{cases}
\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}&n\geq 0\tag{1}\\
0&n<0
\end{cases}
\end{align*}
Using (1) we can write
\begin{align*}
\sqrt{1+x^2}&= 1+\frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equation $\tan x = \tan 2x \tan 4x \tan 8x$ In the question we have the equality
$$\tan 6^{\circ} \tan 42^{\circ} = \tan 12^{\circ} \tan 24^{\circ}$$
which is equivalent to
$$ \tan 6^{\circ} = \tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ}$$
This means that the equation
$$\tan x = \tan 2x \tan 4x \tan 8x$$
has th... | $$\tan x=\tan 2x\tan 4x\tan 8x$$
$$\frac{\sin x}{\cos x}=\frac{2 \sin x \cos x}{\cos 2x}\left(\frac{2\sin 4x \sin 8x}{2 \cos 4x \cos 8x}\right) $$
If $\sin x\neq0$ then,
$$\frac{\cos 2x}{2\cos^2 x}=\frac{\cos 4x - \cos 12x}{\cos 4x + \cos 12x}$$
Now,using componendo-dividendo,
$$\frac{\cos 2x+ 2\cos^2x}{2\cos^2 x-\cos ... | {
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"url": "https://math.stackexchange.com/questions/4149132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve... | Another way to tackle this problem:
\begin{align}\frac 1{x+1} - \frac 1{x}=-\frac 1{x(x+1)} &\le \frac 1{x-1} - \frac 1{x-2}= -\frac 1{(x-2)(x-1)}\\
\iff \frac{1}{(x-2)(x-1)}&\le \frac{1}{x(x+1)}
\end{align}
Now observe that:
*
*outside of $\,(-1,0)\cup(1,2)$, both denominators are positive, so
$$\frac{1}{(x-2)(x-1... | {
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Show that $a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}$ is decreasing I need to show that the sequence $a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}}$ is decreasing.
I tried showing $a_{n+1}\le a_n$ but it was too messy.
I did manage to do it by showing that $f'(x)$ will be negative at some p... | $$a_n=\frac{\sqrt{n^{2}+2021n+420}}{\sqrt{n^{3}+2022n+420}} \implies a_n^2=\frac {n^{2}+2021n+420}{n^{3}+2022n+420 }$$
Long division or Taylor series
$$a_n^2 =\frac{1}{n}+\frac{2021}{n^2}-\frac{1602}{n^3}+O\left(\frac{1}{n^4}\right)$$ Doing the same
$$a_{n+1}^2=\frac{1}{n}+\frac{2020}{n^2}-\frac{5643}{n^3}+O\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Lagrange multiplier does not find global minimum I want to minimize the function
$$f(r_1,r_2,r_3)=\frac{1}{6}(r_1^2+r_2^2+r_3^2)-(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3})$$
subject to the conditions:
$$g(r_1,r_2,r_3)=\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3})=0$$
$$r_1,r_2... | Calling
$$
L(r_k,\lambda) = \frac{1}{6}(r_1^2+r_2^2+r_3^2)-\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)-\lambda\left(\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)\right)
$$
the stationary points for this lagrangian are the solutions for
$$
0=\nabla L = \cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151249",
"timestamp": "2023-03-29T00:00:00",
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Show power series is solution to differential equation I am currently studying for my analysis 2 course, and I've run into a couple of questions from old exams regarding power series as solutions for differential equations which I have a trouble completing.
We have a specific power series i.e.
$$f(x)=\sum_{n=0}^{\infty... | I will get you started. The goal is to get an expression
$$
x^2y''-3xy'+(3-4x^2)y=\sum_{n=0}^\infty a_n x^n
$$
where $a_n$ is a function of $n$ alone. You sort of got that in your last equation, but you have $x^4$ in your expression for the coefficient of $x^{4n+3}$. If all goes well, you should have $a_n=0$ for all $n... | {
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"timestamp": "2023-03-29T00:00:00",
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Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy$. Q: Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \, dx\,dy$.
I imagined that $x$ should varie between $\dfrac{1}{9}$ and $\sqrt{\dfrac{2}{3}}$ and $y$ varie betwee... | The 'x' limit $y^2$ means $x=y^2$, so we have $y=\sqrt{x}$
The 'x' limit $\sqrt{y}$ means $x=\sqrt{y}$, so we have $y=x^2$
Notice you have exchanged lower and upper 'x' limits
For the 'y' limit $y=\frac{1}{3}$ we must get the $x=y^2 = \frac{1}{9}$ function because $\left(\frac{1}{3} \right)^2 < \sqrt{\frac{1}{3}}$.
Sam... | {
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"timestamp": "2023-03-29T00:00:00",
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if $u = \arccos \left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right)$ then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ Edit: this is a repost
if possible i need some sort of shorcut, 80 questions, 90 min to solve these type of questions i have only 1 min time solt for each question
Context: i've looked at P... | I got :
$$\tag{1} \boxed{ x \frac{\partial u}{\partial x} = -~ \frac{2x \sqrt{y} + x \sqrt{x} -y \sqrt{x}}{2~\sin(u) ~(\sqrt{x} + \sqrt{y})^2}} $$
$$\tag{2} \boxed{ y \frac{\partial u}{\partial y} = -~ \frac{2y \sqrt{x} + y\sqrt{y} -x \sqrt{y}}{ 2~\sin(u)~ (\sqrt{x} + \sqrt{y})^2}} $$
$$x \frac{\partial u}{\partial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find $\lim\limits_{x\to 0} \int_0^x \frac{\sin^2(u)}{\sin(x^3)}du $ How to find
$$\lim\limits_{x\to 0} \int_0^x \frac{\sin^2(u)}{\sin(x^3)}du ?
$$
Where should I even start with this one? Can I pull out $\frac{1}{\sin(x^3)}$ outside the integral as $x$ is an constant and we're finding integral with respect to to... | $$\lim_{x \to 0} \left(\frac{1}{\sin(x^3)} \int_0^x \sin^2 u \ du \right)$$
$$= \lim_{x \to 0} \left( \frac{1}{\sin(x^3)} \left[\frac{1}{2}u - \frac{1}{4} \sin2u \right]_0^x \right)$$
and when $x^3$ is small, $\sin x \approx x - \frac{x^3}{3!}$ and $\sin x^3 \approx x^3$, so:
$$= \lim_{x \to 0} \left(\frac{1}{x^3} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153951",
"timestamp": "2023-03-29T00:00:00",
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Question from pathfinder for Olympiad mathematics 2 If $p$, $q$, $r$ are the real roots of equation $x^3-6x^2+3x+1=0$, determine the possible value of
$p^2q+q^2r+r^2p$.
My Attempt:
$p+q+r=6 (1)$
$pq+qr+pr=3 (2)$
$pqr=-1 (3)$
Multiplying (1) and (2) and substituting (3) in it...
I came close but still unable t... | Since $v_1=p^2q+q^2r+r^2p$ is not a symmetric polynomial, there might be multiple values. Another possible value is $v_2=p^2r+q^2p+r^2q.$
Now $v_1+v_2$ and $v_1v_2$ are symmetric. So we can express those values in terms of $p+q+r,$ $pq+pr+qr,$ and $pqr.$
Then solve for the two values, $v_1,v_2.$
I’ll do the harder case... | {
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What values of $n$ is $A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$ an integer?
Suppose $a_i\in \mathbb{N} $ such that $a_i\ne a_j$ and $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ for $i \ne j$ $($take $a_{n+1}=a_1)$ and $\displaystyle A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+... | $$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{15}{6}+\frac{24}{15}+\frac{10}{24}+\frac{2}{10}=10$$
$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{30}{7}+\frac{14}{30}+\frac{24}{14}+\frac{2}{24}=13$$
$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{36}{9}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Q is to prove that integer just above ($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers. Q is to prove that integer just above($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers.
In Q , by integer just above means that:
For an example , which is the in... | Part of the problem is that the image was a bit obscured and so the transcription was a bit off. Here is a more accurate transcription with tags added for reference:
$$\require{cancel}
\left(\sqrt3+1\right)^{2n}=\left(4+2\sqrt3\right)^n=2^n\left(2+\sqrt3\right)^n\tag1
$$
$$
\!\!\!\!I+f=2^n\left(2+\sqrt3\right)^n=2^n\l... | {
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"url": "https://math.stackexchange.com/questions/4157841",
"timestamp": "2023-03-29T00:00:00",
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How to compute this determinant
I think adding some rows or column to each other will get a triangular matrix which is easy to compute but i cant see how
| \begin{align}
\Delta_n&= \left|\begin{matrix}
1 & 2 & 3 & \cdots & n-1 & n\\
-1 & x & 0 & \cdots & 0 & 0 \\
0 & -1 & x & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & x & 0 \\
0 & 0 & 0 & \cdots & -1 & x
\end{matrix}\right|
\tag{1}\label{1}
\end{align}
By expanding the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many possible Jordan decompasitions for an Endomorphism? Let V be an vector space and $\phi \in End_\mathbb{C}(V)$ a linear map with the characteristic polynomial $P_\phi(X) = (x-2)^3(x-5)^2$. How many possible Jordan normal form are there for $\phi$?
| Hint: You should also provide the minimal polynomial that could be one of
*
*$(x-2)(x-5)$,
*$(x-2)(x-5)^2$,
*$(x-2)^2(x-5)$,
*$(x-2)^2(x-5)^2$,
*$(x-2)^3(x-5)$,
*$(x-2)^3(x-5)^2$.
With that in mind the posibles respectively are
$\left[\begin{array}{ccccc}
2&0&0&0&0\\
0&2&0&0&0\\
0&0&2&0&0\\
0&0&0&5&0\\
0&0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinatorial proof of ${{n+1}\choose 3}-{{n-1}\choose 3}=(n-1)^2.$
Prove that ${{n+1}\choose 3}-{{n-1}\choose 3}=(n-1)^2.$
I found the algebraic proof of the above statement.
So we have to show that $$ \frac{(n+1)(n)(n-1)}{3\times 2}-\frac{(n-1)(n-2)(n-3)}{3\times 2} \stackrel{?}{=} (n-1)(n-1)$$ $$\implies \frac{(n+... | I would use Pascal's identity to get
$$\binom{n+1}{3}-\binom{n-1}{3} = \left[\binom{n+1}{3} - \binom{n}{3}\right] + \left[\binom{n}{3}-\binom{n-1}{3}\right] = \binom{n}{2} + \binom{n-1}{2}.$$
This can be solved algebraically:
$$\binom{n}{2} + \binom{n-1}{2} = \frac{n(n-1)}{2} + \frac{(n-1)(n-2)}{2} = \frac{(n-1)[(n-2)+... | {
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Show that $2 ^{n(n+1)} \mid 32 · φ(2^{2^n} − 1) $
Show that for every $n \in \mathbb{N}$
$$2 ^{n(n+1)}\mid 32 · \varphi(2^{2^n}− 1)$$
where $\varphi$ denotes the Euler-Phi function.
So the right hand side can be broken down into:
$$(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-3}}+1)\cdots(2^1 + 1)(2-1)$$
Since $φ$ is mult... | We could use your idea of breaking $2^{2^n} - 1$ up into
$$(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-3}}+1) \ldots (2^1 + 1)(2-1)$$
Next, we could determine appropriate lower limits of each $\varphi\left(2^{2^k} + 1\right)$ and then add them together.
However, another similar (but I think a bit easier) way, with your idea... | {
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Calculate $\displaystyle \iiint_V x^2 \,dx\, dy\, dz$ where $V=\{(x, y, z)|z^2\le \frac{x^2}{4}+\frac{y^2}{9}\le 2z\}$ Calculate $\displaystyle \iiint_V x^2 \,dx\, dy \,dz$, where $V=\{(x, y, z)\mid z^2\le \frac{x^2} {4} + \frac{y^2}{9}\le 2z\}$.
I made the substitution $x=2r\cos\theta$, $y=3r\sin \theta$ and $z=z$. Fr... | Your working is absolutely correct. The region is bound below by the paraboloid and above by the cone and we could have alternatively set up the integral wrt $z$ first. The surfaces intersect at $z = 2$ and at intersection, $r \leq 2$.
We have $z^2 \leq r^2 \leq 2z \implies \dfrac{r^2}{2} \leq z \leq r$. So the integr... | {
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If $x+2y+3z=10$ find the maximum value of $x^2y^2z^6$. If $x+2y+3z=10$ where x,y,z are postive real numbers, find the maximum value of $x^2y^2z^6$.
I started by trying to find some integer solutions. The first thing I found was $(x,y,z)=(1,3,1)$. But the value of $x^2y^2z^6$ was way too small (it was a multiple choice ... | I presume $x,y,z$ are non-negative integers. Otherwise, if they are allowed to be negative, we may choose their magnitude to be arbitrarily large.
Write : \begin{align*}
10 &= x + 2y + 3z\\
&= \frac{x}{2} + \frac{x}{2} + y + y + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2}\\
&\geqsl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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what is the residue of $\sin(z^2)/z^7$ what is the residue of $$\frac{\sin(z^2)}{z^7} \text{ at } z=0.$$ I'm finding trouble at the limit part at the end.
I am using the formula:
$$\operatorname{res}(f,z_0)=\lim_{z\to z_0}\frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} ((z-z_0)^n f(z))$$
| Recall that
$$\sin(z)= \sum_{n=0}^\infty \frac{(-1)^{n} z^{2n+1}}{(2n+1)!}=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$$
Which also means that
$$\sin(z^2)=z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}+\cdots$$
And thus
\begin{align}
\frac{1}{z^7}\sin{(z^2)}&=\frac{1}{z^7}\left( z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}+\cdots\right)\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve this ominous integral? Let $ n\ge 1 $ be a positive integer. How do I prove the generalization: $$ \int_0^1\frac{\arctan(x)\log^{2n}(x)}{1+x} \, dx =\frac{\pi}{4}\left(1-2^{-2 n}\right) \zeta(2 n+1)(2 n)!+\frac{1}{2} \beta(2 n+2)(2 n) !-\frac{\pi}{16} \lim _{s \rightarrow 0}\left(\frac{d^{2 n}}{d s^{2 n}... | The general "closed-form" is:
$$\int _0^1\frac{\arctan \left(x\right)\ln ^{2n}\left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{4}\left(1-2^{-2n}\right)\zeta \left(2n+1\right)\left(2n\right)!+\frac{1}{2}\beta \left(2n+2\right)\left(2n\right)!-\frac{\pi }{16}$$
$$\lim _{s\to 0}\left(\frac{d^{2n}}{ds^{2n}}\left(\csc \left(\frac{... | {
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"url": "https://math.stackexchange.com/questions/4184005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y}$? I want to understand this multivariate limit.
WolframAlpha says $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} = 0$
But what if I take a curve $y = -x$, then the limit doesn't exist, right? Wouldn't this make the limit inexistent?
I tried to prove using sandwic... | $$
\begin{aligned}
\lim\limits_{(x, y) \rightarrow (0, 0)}\frac{1-\cos(x)}{x + y} &=
\left|
\begin{array}{c}
\text{changing the variables:} \\
x = r\cos(\theta) \\
y = r\sin(\theta) \\
(r \in (0, +\infty), \text{ }\theta \in [0, 2\pi]) \\
\Downarrow \\
(x, y) \rightarrow (0, 0) \Leftrightarrow r \rightarrow 0
\end{arr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Prove $3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx -\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx =\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3})$
Prove that
$$
I
=3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx
-\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx
=\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3}).
$$
whe... | Integrate by parts and then substitute $x=\tan \frac t2$
\begin{align}
I=& \>\int_{0}^{1}\frac{x\arctan x}{x^2+\frac13}\,\mathrm dx
-\int_{0}^{1}\frac{x\arctan x}{x^2+3}dx
=\frac14 \int_0^{\frac\pi2} \ln\frac{1+\frac12 \cos t}{1-\frac12 \cos t}dt
\end{align}
Let $J(a) = \int_0^{\frac\pi2}\ln(1+\cos a\cos t)dt$. Then
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$ I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.
Prove the following inequality without using calculator:
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3... | Your solution is fine. If you want a non-calculus approach, note that
\begin{align*}
f(x) &= (x+1)^{1/3}-x^{1/3}
\\
&= \dfrac{\left((x+1)^{1/3}-x^{1/3}\right)\left((x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}\right)}{(x+1)^{2/3}+2(x+1)^{1/3}x^{1/3}+x^{2/3}}
\\
&= \dfrac{(x+1)-x}{(x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}}
\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + ... | $\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}+\sqrt{10}}\!=\!\dfrac{\sqrt{\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)^2}}{2\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)}\!=\!\dfrac{\sqrt{\dfrac12}+\sqrt{\dfrac52}}{2\left(\sqrt{\dfrac12}+\sqrt{\dfrac52}\right)}\!=\!\dfrac12.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
The easiest way to solve the following integral : $\int_{\pi/6}^{\pi/3} \frac{\sin x}{\sin x+\cos x} \mathrm{d}x$ Today I passed the test to access to medicine, and I found some really cool questions on the mathematics part, anyways I just went to see the difficulty of the test, and sorry for being far from my main que... | HINT
Make the change of variable $y = \pi/2 - x$. Then we get that
\begin{align*}
I & =
\int_{\pi/6}^{\pi/3}\frac{\sin(x)}{\sin(x) + \cos(x)}\mathrm{d}x = \int_{\pi/6}^{\pi/3}\frac{\cos(y)}{\cos(y) + \sin(y)}\mathrm{d}y
\end{align*}
Then add both expressions together and see what happens.
Can you take it from there?
ED... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$ Question: Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$
Thoughts: the "standard" method of starting by squaring (twice) to get rid of the square roots, because I don't have a nice way of showing the resulting pol... | Since the comment has answered the OP's stated question, it is open season.
Alternative approach
If $a$ is rational and $b$ is irrational, then $(a+b)$ is irrational.
Further, if $a$ is irrational, then $\sqrt{a}$ is also irrational.
The proof to the 2nd assertion above is that if $\sqrt{a}$ can be rationally expressed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Comparing sets $\{8^n-7n-1: n \in \Bbb{N}\}$ and $\{49(n-1): n \in \Bbb{N}\}$
If
$$\begin{align}
X&=\{8^n-7n-1: n \in \Bbb{N}\} \\
Y&=\{49(n-1): n \in \Bbb{N}\}
\end{align}$$
then,
a) $X\subset Y \qquad$ b) $Y\subset X\qquad$ c) $X=Y\qquad$d) none of these
I know this question can be solved by taking $X=8^n-7n-1$ an... | $8^n - 7n - 1 = (8^n - 1) - 7n = (8-1)(8^{n-1} + ..... + 8 + 1) - 7n = 7[(8^{n-1} + .... + 8 + 1) - n]$.
So $7|8^n - 7n - 1$.
And $8^n \equiv 1^n \equiv 1 \pmod 7$ so $(8^{n-1} + .... + 8 + 1) - n\equiv (1+1+1+....+1) -n \equiv n - n \equiv 8 \pmod 7$
So $7|(8^{n-1} + .... + 8 + 1) - n$ and $49|8^n - 7n - 1=7[(8^{n-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find locus of mid-point of chord of circle I am working through a pure maths book as a hobby and am struggling with the very last part of the question below. I give the whole of the question just in case it proves relevant to the answer.
A circle passes through the points A, B and C which have coordinates $(0,3),(\sqrt... | Equation of circle is $x^2+y^2-2y-3=0$
Equation of line is $y = mx - 3$
Substituting $y$ from equation of line into equation of circle, we get
$x^2 + (mx-3)^2 - 2 (mx-3) - 3= 0$
Simplifying, $(1+m^2) x^2 - 8 mx + 12 = 0$
As the line intersects the circle at two points, the sum of roots of the above equation is the sum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:
*
*For every $n \geq 2$, equation $(1)$ always has a unique positive s... | For you first question, a picture may help clarify what is happening. This one comes from Wolfram Alpha when $n=7$
In effect your function, let's call it $f_n(x)$, is the sum of $n$ hyperbolae with $n$ vertical asymptotes at $-1,-2,\ldots, -n$, a locally decreasing function of $x$ where it exists, and with $f_n(x) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Ambiguity in Integration done by substitution. I was solving this integral $$I= \int x\sqrt{1-x^2} \mathrm dx$$ My solution is:
Let $x=\sin y$, then $|\cos y|=\sqrt{1-x^2}$ I now use method of substitution for the cases $\cos y\geq 0$ and $\cos y<0$. The correct answer is $-\frac{(1-x^2)^{3/2}}{3} +C$. I am able to obt... | Suppose $~~I=\int x(\sqrt{1-x^2}) dx $
Let $(1-x^2)=u ~~\Longrightarrow -2xdx=du$
$\Longrightarrow dx=\frac{-1}{2x}du $
Now we have,
$$I=\int x(\sqrt{1-x^2}) dx = \int \frac{-1}{2x}x \sqrt{u}du = \frac{-1}{2} \int\sqrt{u} du $$
$$= \frac{-1}{2} \left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right) +c= \frac{-u^\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequa... | We want to show :
$$2(ab+bc+ca)>a^2+b^2+c^2$$
With :
$$a+b-c\geq0$$
$$a+c-b\geq 0$$
$$c+b-a\geq 0$$
As the inequalities are homogenous we need to show $0<x<1$ and $0<y<1$:
$$(x^2+y^2+1)<2(xy+x+y)$$
With :
$$x+y-1\geq0$$
$$x+1-y\geq 0$$
$$y+1-x\geq 0$$
Or :
$$(x^2+y^2+x+y)<2(xy+x+y)$$
Or :
$$2(x+y)<2(xy+x+y)$$
Done !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{2}{3}\int_0^1\frac{\ln x}{1+x^2}dx=\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx$ Proof the following
$$\frac{2}{3}\int_0^1\frac{\ln x}{1+x^2}dx=\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx$$
I've tried to use x=tan(u) substitution, but don't know how to proceed with this
$$\frac{2}{3}\int_0^{π/4}\ln(\tan(u))du=\i... | Utilize the integral
$$\int_0^1 \frac{2(y^2-1)\cos 2t}{y^4+1-2y^2\cos 4t}dy =\ln(\tan t)
$$
to prove
\begin{align}
&\int_0^{2-\sqrt3}\frac{\ln x}{1+x^2}dx \\= & \int_0^\frac\pi{12}\ln(\tan t)~dt
= \int_0^1 \int_0^{\frac\pi{12} }
\frac{2(y^2-1)\cos 2t}{y^4+1-2y^2\cos 4t}dt \ dy\\= &\ -\frac12\int_0^1\frac1y \tan^{-1}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ diverges First of all, I noticed that $\sin^2(\pi(x+\frac{1}{x}))$ "approaches" $\sin^2(\pi x)$ as $x \to \infty$. In other words, it seems that the area under one period of $\sin^2(\pi(x+\frac{1}{x}))$ should converge to the value of the area under one period of $... | What's wrong with this less formal approach?
Replace the original integrand with sin^2 x , since the factor of pi goes away with a simple change of variable, and since the x term dominates the 1/x term over the interval. Then every occurrence of the new integrand as x increases is non-negative and identical in magnitud... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Necessary condition on $|x-1|$ to assure $|x^2 -1| < \frac{1}{2}$. Is my solution correct? How to get the tightest answer? Question: Determine a condition on $|x-1|$ that will assure that:-
$$|x^2 -1| < \frac{1}{2}$$
My solution:-
Let $f(x) = x^2$
$$\lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} x^2 = 1$$
To fi... | I like how you are trying to use the concept of continuity here. But the problem is much simpler. One can determine the necessary and sufficient condition on $|x-1|$
by simply solving the inequality directly.
First, an algebraic solution:
$$|x^2-1|<\frac{1}{2}\iff -\frac{1}{2}<x^2-1<\frac{1}{2}
\iff \frac{1}{2}<x^2<\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Hard and non-trivial inequality with three variable reals For all reals $a$, $b$, $c$, show that $$a^2+b^2+c^2 \geq a\sqrt[\leftroot{-1}\uproot{1}4]{\frac{b^4+c^4}{2}} + b\sqrt[\leftroot{-1}\uproot{1}4]{\frac{c^4+a^4}{2}} + c\sqrt[\leftroot{-1}\uproot{1}4]{\frac{a^4+b^4}{2}}.$$
I tried to use Holder inequality:
$$\left... | Your second attempt gives another solution!
Indeed, for non-negative variables we need to prove that:
$$2(a^2+b^2+c^2)^4\geq3(a+b+c)^2\sum_{cyc}(a^4b^2+a^4c^2)$$ or
$$2(a^2+b^2+c^2)^4\geq3(a+b+c)^2\left(\sum_{cyc}a^2\sum_{cyc}a^2b^2-3a^2b^2c^2\right).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Is it just a coincidence that the solution to $y=x$, $y=mx+b$ ($m<1$) is the sum of an infinite geometric series with first term $b$ and ratio $m$? The solution for the system of linear equations of $y = x$ and $y = mx + b$ is
$$x=y= \frac{b}{1-m}.$$
I noticed that this is also the sum of an infinite geometric series, ... | It follows by induction.
As $x = mx+b$ we can replace $x$ with $mx+b$ indefinitely.
$y = x \\= mx + b=m(mx+b) + b \\= m^2x + mb + b\\=m^2(mx+b)+mb+b \\= m^3x + m^2b + m b + b\\\dots\\b + mb + m^2b + m^3b + m^4b + ......$
This shouldn't be a surprise. If $|m| < 1$ then $b + mb + m^2b + m^3b + m^4b + ...... \\= b(1 + m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=... | I don't know if this is faster for you, but here's how I did it.
The first thing you need to notice is that higher powers of roots of the equation can be reduced to linear terms only. If $x_i$ is a root of $x = 5 - x^2$, then:
$$\begin{align*}x_i^2 & = 5 - x_i \\ x_i^3 & = 5x_i - x_i^2 \\ & = 6x_i - 5\end{align*}$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Challenging geometry problem from AMC 2013 Question 29 from the senior division of the 2013 Australian mathematics competition.
I got every question except this one and have no clue how to solve it.
| Applying Menelaus Theorem a few times:
$$\frac{RY}{RQ}\times \frac{QP}{PT}\times\frac{TX}{XY}=1$$
$$\frac{3}{6}\times \frac{QP}{PT}\times\frac{TX}{XY}=1$$
$$\frac{TX}{XY}=\frac{2PT}{QP}...[1]$$
$$\frac{SQ}{ST}\times \frac{TX}{XY}\times\frac{YZ}{ZQ}=1$$
$$\frac{11}{11+QT}\times \frac{TX}{XY}\times\frac{2}{1}=1$$
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Near-Pythagorean triplets: What are the general solutions to $a^2+b^2=c^2-1$? Obtaining the most general solution to a quadratic Diophantine equation in three variables is often easier if the equation is homogeneous. For example, by focusing on "primitive" solutions, it is easy to show that all Pythagorean triplets ca... | This is only a partial answer. I managed to find several families of solutions:
Family 1
$$
\begin{align}
a &= 2\left[(4k^2+1)n \pm k\right],\\
b &= 2\left[(4k^2+1)n^2 \pm 2kn - k^2\right],\\
c &= 2\left[(4k^2+1)n^2 \pm 2kn - k^2\right] + 4k^2 + 1.
\end{align}
$$
Family 2
$$
\begin{align}
a &= 2\left[\left(2k(k+1)+1 \v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
How to find characteristic polynomial of a matrix $A$ whose elements are of the form $a_{ij} = a_ia_j$ for all $1\leq i,j \leq n$ For $n > 1$,
$$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix}$$
The characteristic polynomial of m... | $$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix} = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\
a_2 & a_2 & ... a_2 \\.... \\
a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Algebra in Geometric Proof of Quadratic Reciprocity I am trying to understand the proof of quadratic reciprocity form the George Andrews textbook on number theory (this proof follows geometry and Eisenstein's thinking). I understand what is happening conceptually, but the finer points of inequality algebra are not obvi... | After lots of questions and lots of attempts, here is the definitive algebra justifying the claim from this proof:
Consider the algebraic justification for $(m,n)=\left(\frac{p+1}{2}-m,\frac{q+1}{2}-n\right)$. Let $(m,n)$ be some lattice point within $H$, and let $(u,v)=\left(\frac{p+1}{2}-m,\frac{q+1}{2}-n\right)$ be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$3\times 2^m + 4 = n^2$ I have been trying to solve this problem but I couldn't find any way to find the answers. Find all natural numbers m and n such that:
$3\times\ 2^m + 4 = n^2$
This is a question from the Moroccan Maths Olympiad (2019.)
I tried bringing $4$ to the other side and factorising to get: $3\times 2^m=(... | $$3\cdot2^{m} + 4 = n^2$$
We take the three cases $m=3a, m=3a+1,$ and $m=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet m=3a$
Let $X = 3\cdot2^a, Y=3n$, then we get $Y^2 =X^3 + 36.$
According to LMFDB, this elliptic curve has $(X,Y)=(-3,\pm 3)$, $(0,\pm 6)$, $(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Finding $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$
What is the limit of $\,\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ ?
My thought was $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right) = \lim\limits_{x... | $$\lim_{x\to0}\left(\csc^2(x)-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\right)$$
Both numerator and denominator tend to zero, and they are continuous and differentiable, so L'Hopital's rule applies. The derivative of the numerator is: $2x-2\sin(x)\cos(x)=2x-\sin(2x)$. The derivative of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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If $x^3+y^2$ and $x^2+y^3$ are integers, show that $x$, $y$ are integers
Given rational numbers $x$, $y$ such that $x^3+y^2$ and $x^2+y^3$ are integers, prove that $x$, $y$ are integers.
For this problem I don't even know how to start tackling it. I tried various ways:
*
*Letting $x = \frac{a}{b}$, $y = \frac{c}{d}... | Suppose $x=a/d$ and $y=b/d$ with $d\gt0$ and $\gcd(a,b,d)=1$ (i.e., write $x$ and $y$ with the smallest possible common denominator). Then $x^2+y^3=m$ implies $b^3=(md^2-a^2)d$, which implies $d\mid b^3$. Likewise, $x^3+y^2=n$ implies $d\mid a^3$. Thus
$$0\lt d=\gcd(a^3,b^3,d)\le\gcd(a^3,b^3,d^3)=(\gcd(a,b,d))^3=1^3=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 0
} |
Artist needing to determine geometric angle for sculpture based on platonic solid Dear Mathematicians I need your help for a new sculpture!
I will attach images but first imagine 2 hexagons - where one is rotated 30 deg. They are separated by 12 equilateral triangles. I need to confirm the angle between the 2 faces a... | You are requesting a dihedral angle of a (regular) hexagonal antiprism. This is calculated as follows. Consider a hexagon with vertices $v_0, \ldots, v_5$, where $$v_k = \left(\cos \frac{\pi}{3} k, \sin \frac{\pi}{3} k, d \right),$$ and another hexagon with vertices $w_0, \ldots, w_5$ where
$$w_k = \left(\cos \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I prove that all quadratic equations are not injective? I was trying to prove that any quadratic formula ($ax^2 + bx + c$) will not be injective, but I have a little problem.
I started by assuming $f(x) = f(y)$.
We can put x and y into the general form of quadratic function, and we get
$ax^2 + bx + c = ay^2 + b... | This is good stuff so far. You now know that you get $x = y$ if you have $a(x + y) + b \neq 0$, in which case, there is no contradiction. So, let's focus on when $a(x + y) + b = 0$. Can you find two values of $x$ and $y$, that are not equal to each other, but for which $a(x + y) + b = 0$? Or, equivalently, $x + y = -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Sufficient condition for infinite product being $>0$? I'm trying to solve the following problem:
When is $\prod_{k=1}^{\infty}\frac{\left(\frac{1}{2}-\frac{1}{2}k^{-\alpha}\right)^{1/2}+\left(\frac{1}{2}+\frac{1}{2}k^{-\alpha}\right)^{1/2}}{2^{1/2}}>0$, for $\alpha >0$?
Since $\frac{1}{2}-\frac{1}{2}k^{-\alpha}$ con... | Taking the logarithm, you see that the product is $>0$ if and only if
$$ \sum_{k=1}^{+\infty}\log\left(\frac{\sqrt{1-k^{-\alpha}}+\sqrt{1+k^{-\alpha}}}{2}\right) $$
converges.
As $k\rightarrow +\infty$,
$$ \begin{aligned} \log\left(\frac{\sqrt{1-k^{-\alpha}}+\sqrt{1+k^{-\alpha}}}{2}\right) &= \log\left(1+\frac{1}{2k^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and... | If $a + b > 4$, then $a > 2$ or $b > 2$. If, for instance, $a > 2$, then $a^2 +b^2 \geq a^2 > 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Proving Fibonacci sum with induction Let the Fibonacci sequence be defined as $f_1 = f_2 = 1$ and $f_n = f_{n-1} + f_{n-2}$
Prove that $f_1f_2+f_2f_3+f_3f_4+...+f_{2n-1}f_{2n}+f_{2n}f_{2n+1} = f^2_{2n+1} - 1$
I was able to verify that the claim holds for $n=1$, since $(1)(1)+(1)(2) = 2^2 - 1$
Now, assuming that the cla... | Hint.
If the statement holds for $n$, then
$f_1f_2 + \dots + f_{2n}f_{2n+1}+ f_{2n+1}f_{2n+2}+f_{2n+2}f_{2n+3}= (\boxed{f_{2n+1}^2}-1) +\boxed{f_{2n+1}f_{2n+2}}+f_{2n+2}f_{2n+3}= f_{2n+1}\underbrace{(f_{2n+1} + f_{2n+2})}_{=\text{what is it?}} + f_{2n+2}f_{2n+3} -1 = \dots \text{(similar step)} = f_{2n+3}^2-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplifying $(1+i)^n-(1-i)^n$ I'm trying to simplify this expression because there is a simpler way of writing this. But the furthest I've got is $$(1+i)^n-(1-i)^n=\sum_{k=0}^{n} \binom nk i^{k}-\sum_{k=0}^{n} \binom nk (-1)^{k}i^{k}.$$
| In the case of you being not used to complex numbers, here is an answer with basic algebra.
Let $1+i = \alpha$, $1-i = \beta$, then $\alpha-\beta = 2i$, $\alpha\beta=2$, $\alpha/\beta = i$ and $\beta^2 = -2i$.
\begin{align*}
z_n:= \alpha^n - \beta^n &= (\alpha- \beta)(\alpha^{n-1} + \alpha^{n-2}\beta + \dots + \alpha \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Determining the transformation matrix for Jordan normal form. Let $$A=
\begin{pmatrix}
2 & 0 & -4 & -4 \\
0 &4&2&3\\
2&0&8&4\\
-1&0&-2&2\\
\end{pmatrix}$$
I want to find Jordan normal form of $A$ and the transformation matrix $P$. $(P^{-1}AP=J).$
I could find the Jordan form of $A$ is $$\begin{pmatrix}
4&0&0&0\\
0&4&1&... | You need to be careful with your choice of $p_2$; not just any vector will do. Note that
$$p_2 = (A - 4I)p_3 = (A - 4I)^2 p_4,$$
which means we need $p_2$ to be in the columnspace (or range, if you like) of $(A - 4I)^2$. It also needs to be an eigenvector, i.e. an element of $\operatorname{null} (A - 4I)$ so we ought t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac... | We have
$$
\begin{array}{l}
S = \frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{5 \cdot 6 \cdot 7}} + \frac{1}{{9 \cdot 10 \cdot 11}} + \cdots = \\
= \sum\limits_{0 \le k} {\frac{1}{{\left( {4k + 1} \right)^{\,\overline {\;3\,} } }}} = \sum\limits_{0 \le k} {\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
Expected value of sequence of rv I'm stuck with this problem.
Let $X_n$ be a sequence of random variables with $X_1$ the random number chosen at $(0,1)$. $X_2$ be the random number chosen at $(1-X_1,1)$ and so on. This is, $X_n$ is the random number chosen at $(1 - X_{n-1}, 1)$. What is $E[X_n]$?
I have used condition... | First of all compute the conditional expectation:
$$\begin{aligned}E[X_n|X_{n-1}]&=\frac{1}{1-(1-X_{n-1})}\int_{(X_{n-1}-1,1)}xdx=\\
&=\frac{1}{X_{n-1}}\frac{1}{2}(1-(X_{n-1}-1)^2)=\\
&=\frac{1-X_{n-1}^2-1+2X_{n-1}}{2X_{n-1}}=\\
&=\frac{2-X_{n-1}}{2}
\end{aligned}$$
We can recast the problem as a sequential problem:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Identity $\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}$ In the book, Control Systems Engineering - frequency design, the author used the equality $$\phi_{max}=\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}$$
Is this some famous identity?
Am I seriously missing out sin... | Another way to show two functions are equal.
Let:
$$f(x)=\arctan{\frac{1-x}{2\sqrt{x}}}$$
$$g(x)=\arcsin{\frac{1-x}{1+x}}$$
Both are defined and $C^1$ for $x>0$.
Now,
$$\arctan'(x)=\frac{1}{1+x^2}$$
$$\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$$
Therefore
$$f'(x)=\left(\frac{1-x}{2\sqrt{x}}\right)'\frac{1}{1+\left(\dfrac{1-x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Refinement about :$\left(\left(1-x\right)^{-\left(2x\right)}-1\right)\left(\left(x\right)^{-\left(2\left(1-x\right)\right)}-1\right)\geq 1$ Claim :
Let $0.5\leq x<1$ then it seems we have :
$$\left(\left(1-x\right)^{-2x}-\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^... | Sketch of a proof:
Denote
$$A = (1 - x)^{2(1 - x)}, \quad B = (1 - x)^{2x - 1},
\quad C = x^{2x}.$$
The desired inequality is written as
$$\left(\frac{A}{(1 - x)^2} - \frac{B}{2^4x(1 - x)^2C}\right)(x^{-2}C - 1) \ge 1.$$
Denote $a = \ln 2$. Let
\begin{align*}
A_1 &= \frac{p_1x^2 + p_2x + p_3}{(8a^3 - 24a^2
+ 48a - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Area enclosed between $2$ curves
Question: The area enclosed by the curves $$ y^2 = x \ \text{and} \ y^2 = \ 3x - 1, \ \ \ \ \ \text{where} \ \ 0\leq \ x\leq \frac{1}{2} \ $$
My work:
To begin with I re-arranged the curves in terms of $y$ to the power $1$
$$y_{1} = \sqrt{x}$$
and
$$y_{2} = \sqrt{3x-1}$$
and then p... | Say $A_1$ is area between parabola $y^2 = x$ and line $x = \frac{1}{2}$, which you write as $f(x_1)$ and $A_2$ is the area between parabola $y^2 = 3x-1$ and line $x = \frac{1}{2}$
You need to multiple $A_1$ by $2$ as the answer you have is only upper / lower half of the area.
Coming to $A_2$, your mistake is that you a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
Find all functions: $f:\mathbb R\to\mathbb R$ that satisfy all the following three conditions:
*
*$f(-x)=-f(x)$
*$f(x+1)=f(x)+1$
*$f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
I assume $f(... |
$f:\Bbb{R} \to \Bbb{R}(\text{odd}), f(x+1)=f(x)+1, f(\frac{1}{x})=\frac{f(x)}{x^2}.$
\begin{align}
& f\left(1+\dfrac{1}{x}\right)=f\left(\dfrac{1}{x}\right)+1 = \dfrac{f(x)}{x^2}+1. \tag{1}\label{1} \\
& f\left(\dfrac{x+1}{x}\right)=\dfrac{f\left(\dfrac{x}{x+1}\right)}{\left(\dfrac{x}{x+1}\right)^2}\\
& = \dfrac{(x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Evaluate $\int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}dx$ using elementary methods I found this problem while doing some integration from my problem practice book (unkown name). It said to evaluate it using elementary methods.
Please help me evaluating the following integral using elementary methods
$$\int \frac{(x^2-1)(x... | Presented below is a full solution via elementary integration.
The polynomial $x^4-2x^3-6x-1$ admits two real roots, $a=$ -0.1650 and $b=$ 2.8068, computed with the standard quartic-root formulae below
$$a,b=\frac12\left(1+\sqrt r \pm \sqrt{3-r+{14}{r^{-\frac12}}}\right)\tag1
$$
with $ r= 1+ \left(\frac43\right)^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\... | If you're confused about why your own method seemingly missed that the derivative is discontinuous, see here. The magic happens in line $4$. We have
$$\begin{split}
f'(x)
&= \frac{1}{\sqrt{1-\left(\frac{x+\sqrt{1-x^2}}{\sqrt 2}\right)^2}}\cdot \frac{1}{\sqrt 2}\left(1-\frac{x}{\sqrt{1-x^2}}\right) \\
&= [\ldots] \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$?
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?
I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequali... | Another way.
By C-S
$$10a^2+10b^2+c^2=$$
$$=\sqrt{\left(8a^2+\frac{c^2}{2}+8b^2+\frac{c^2}{2}+2a^2+2b^2\right)\left(\frac{c^2}{2}+8a^2+\frac{c^2}{2}+8b^2+2b^2+2a^2\right)}\geq\sqrt{(2ac+2ac+2bc+2bc+2ab+2ab)^2}=\sqrt{16}=4.$$
The equality occurs, when
$$\left(2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\frac{c}{\sqrt2},\sqrt2a,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Finding equation of tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
Find the equation of the tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
This is in the context of learning implicit differentiation.
First, I apply $\frac{dy}{... | As
$$
\sin\left(\sin^{-1}x+\sin^{-1}y\right)=\sin\left(\frac{\pi}{6}\right)
$$
gives
$$
\sqrt{1-x^2} y+x \sqrt{1-y^2}=\frac{1}{2}
$$
calling $f(x,y) = \sqrt{1-x^2} y+x \sqrt{1-y^2}$ we have the tangent at $p_0=\left(0,\frac 12\right)$ as
$$
(p-p_0)\cdot\nabla f(p_0)=0
$$
with $p = (x,y),\ \ \nabla f(x,y) = \left\{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using logarithmic differentiation or otherwise, differentiate $y = (x −1) (x − 2) (x − 3$)? And show that $y' = 3x^2-12x+11$ So far this is my method:
\begin{align*}
y = (x-1)(x-2)(x-3) & \Longleftrightarrow \ln(y) = \ln(x-1) + \ln(x-2) + \ln(x-3)\\\\
& \Longleftrightarrow \frac{y'}{y} = \frac{1}{x-1} + \frac{1}{x-2} +... | The product rule is straightforward.
$$ y= (x-1)(x-2)(x-3)$$
$$ y' = (x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)=$$
$$ x^2-5x+6 + x^2-4x+3+x^2-3x+2 =$$
$$ 3x^2-12x+11$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}... | $$\sum\limits_{i=1}^n \frac{1}{3^i} = \frac{3n-1}{2 \cdot 3^n}$$
So $$\lim\limits_{n \to \infty} \frac{3n-1}{2 \cdot 3^n} = \frac{1}{2}$$
So for finite $k$:
$$\lim\limits_{n \to k} \sum\limits_{i=1}^n \frac{1}{3^i} < \frac{1}{2}$$
or for all finite $n$:
$$\sum\limits_{i=1}^n \frac{1}{3^i} + \underbrace{\sum\limits_{i=n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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If $a_n=b_n = \frac{(-1)^n}{n+1}$, how to deduce that the Cauchy product series converges? If $c_n=\sum_{k=0}^{n}a_kb_{n-k}$, we have to prove that,
$$c_n=(-1)^n\frac{2}{j+2}\left(1+\frac{1}{2}+\cdots+\frac{1}{k+1}\right)$$
But what I get is:
$$c_n=(-1)^n\sum_{k=0}^{n}\frac{1}{(k+1)((n-k)+1)}=(-1)^n\left(\frac{1}{1(n-(... | We have
$$\tag{*}\sum_{n=0}^\infty c_n = \sum_{n=0}^\infty (-1)^n\underbrace{\sum_{k=0}^n \frac{1}{(k+1)(n-k+1)}}_{d_n},$$
where $d_n$ can be expressed in terms of the harmonic sum $H_{n+1} = \sum_{k=1}^{n+1} \frac{1}{k}$, viz.
$$d_n =\sum_{k=0}^n \frac{1}{(k+1)(n-k+1)}= \sum_{k=0}^n\frac{1}{n+2} \left(\frac{1}{k+1} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Show that for all except perfect powers of 2, it's true.
Let $T_n$ denotes the least natural such that
$$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$$
Find all naturals $m$ such that $m\ge T_m$.
I think it’s true for any $m$ except perfect powers of $2.$
For perfect powers of $2.$
Let $m=2^k$. We must have $2^{k+1} \m... | As Adam Rubinson points out in a comment, you also have to show that $n\ge T_n$ when $n$ is not a power of $2$. So given such a positive integer $n$, we want to find $m\le n$ such that $n\mid T_m=\frac12 m(m+1)$, i.e. $2n\mid m(m+1)$.
So write $n$ as $n=a\cdot2^b$ where $a$ is odd. Then we want $a\cdot 2^{b+1}\mid m(m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Number of terms in product of two monomials with common terms I am trying to find the number of terms in the expression
$$(x+y+z)^{20}(w+x+y+z)^2$$.
I understand that the number of terms in $(x+y+z)^{20} = \binom{22}{2}$ and the number of terms in $(w+x+y+z)^2 = \binom{5}{3}$. If the variables were distinct, then you w... | This is a variation continuing OPs approach. OP already observed the number of different terms in
\begin{align*}
(x+y+z)^{20}\quad\text{is}\quad\binom{22}{2}\text{.}
\end{align*}
Since
\begin{align*}
(x+y+z)^{20}&(w+\left(x+y+z\right))^2\\
&=(x+y+z)^{20}(w^2+2w\left(x+y+z\right)+\left(x+y+z\right))\\
&=w^2(x+y+z)^{20}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $\int\frac{x+1}{x^2+x+1}dx$ Find
$\int\frac{x+1}{x^2+x+1}dx$
$\int \frac{x+1dx}{x^2+x+1}=\int \frac{x+1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
From here I don't know what to do.Write $(x+1)$ = $t$?
This does not work.Use partial integration?I don't think it will work here.
And I should complete square then find.
| You will have to make the numerator as a sum of two functions like this
$\int \frac{x+\frac{1}{2}}{(x+\frac{1}{2})^2+\frac{3}{4}}dx+\int \frac{\frac{1}{2}}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
The first integral will result in $ln$ and the second in $atan$.
$\frac 1 2 \ln (x^2+x+1)+ \frac 1 {\sqrt{3}} \arctan{\frac {(2x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question regarding the largest proper divisor of $x^n+1$ Let $x$ and $n$ be positive integers, with $n$ odd, and let $d$ be the largest proper divisor of $x^n+1$.
QUESTION: Is there a way to characterize the $x$ and $n$ satisfying $d > x^{(n+1)/2}$?
Example: $17^3+1=4914$ has largest [proper] divisor $d=2457 > 17^2 =... | Lemma: Let $x,n$ be positive integers with $n\ge 3$. Let $d$ be the largest proper divisor of $x^n+1$. Then $d<x^{(n+1)/2}$ if and only if one of the following is true
*
*$x^n+1$ is prime.
*$x^n+1=pq$ for primes $x^{(n+1)/2}>q>p>x^{(n-1)/2}$
Proof: Assume that $d<x^{(n+1)/2}$. If $x^n+1$ is prime, we are done. Other... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Summation notation $ 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \ldots + \frac{1}{1+2+3 + \ldots +n} + \ldots $ How to write the sum
$$
1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \ldots + \frac{1}{1+2+3 + \ldots +n} + \ldots
$$
in summation (∑) notation.There are 2 series here, one the entire 1 + (1/1+2)... series and the oth... | $$\sum_{n=1}^{\infty}\frac{2}{n(n+1)}$$ which is equivalent to
$$\sum_{n=1}^{\infty}\frac{1}{T_n}$$
$T_n$ being the $n^{th}$ triangle number.
Clearly, it converges to $2$. This is because the sequence is equivalent to
$$\sum_{n=1}^{\infty}\frac{2}{n}-\frac{2}{n+1}$$ which is telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Alternative approach:
From the constraints,
$(a + b)^2 = (c + d)^2 \implies$
[since $a^2 + b^2 = c^2 + d^2$]
$2ab = 2cd \implies ab = cd \implies$
$a(c + d - a) = cd$.
This allows a quadratic equation in $a$ to be formed, that may be solved in terms of $c,d$.
$a^2 - (c+d)a + cd = 0.$
Therefore,
$\displaystyle a = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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How to multiply polynomials in $\mathbb Z[x]/(x^n+1)$? Consider the ring
$$R_q=\mathbb Z_q[X]/(X^n + 1),$$
where $q\equiv 1 \bmod 2n$ and $n$ is a power of $2$.
This is the quotient ring where the cosets are represented by polynomials up to $n-1$ in order.
I'd like to compute $c = a · b \bmod (X^n + 1)$, where $a$ and ... | The command you want in the Wolfram language is PolynomialMod. See documentation here. PolynomialMod[(1x^0+2x^1+3x^2+4x^3+5x^4+6x^5+7x^6+8x^7) * (2x^0+5x^1+8x^2+11x^3+14x^4+17x^5+20x^6+23x^7) , {x^8+1,2}]
will return x^2 + x^4 + x^5 after reducing first mod $x^8+1$, and then reducing coefficients mod $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\int_{-1}^1 \frac{1}{|x-y|^{2/3}} \frac{1}{|x|^{1/3}} dx \approx -2\log y$ as $y\to 0$ I want to show that
$$\int_{-1}^1 \frac{1}{|x-y|^{2/3}} \frac{1}{|x|^{1/3}} dx \approx -2\log y$$
as $y\to 0$. Here, $f(y) \approx g(y)$ means that $\lim_{y\to 0} \frac{f(y)}{g(y)}=1$.
By computing the integral and finding a... | With $x = ty$ and $s=-t$, we find
\begin{align*}
&\int_{ - 1/y}^{1/y} {\frac{1}{{\left| {t - 1} \right|^{2/3} }}\frac{1}{{\left| t \right|^{1/3} }}dt} = \int_1^{1/y} {\frac{1}{{(t - 1)^{2/3} }}\frac{1}{{t^{1/3} }}dt} + \int_{ - 1/y}^0 {\frac{1}{{(1 - t)^{2/3} }}\frac{1}{{( - t)^{1/3} }}dt} \\ & \quad + \int_0^1 {\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $\left(x^y + y^x \right)\left( \frac{1}{x} + \frac{1}{y} \right)\ge 4$ For arbitrary $x, y > 0$, prove
$$
\left(x^y + y^x \right) \left( \frac{1}{x} + \frac{1}{y} \right) \ge 4
$$
By plotting, it seems true and somewhat tight, but I cannot find a proof for it.
If $x\ge 1,\ y \ge 1$ or $x\le 1,\ y\le 1$, we have
... | Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$(x^y + y^x)(1/y + 1/x)
\ge \left(\sqrt{\frac{x^y}{y}} + \sqrt{\frac{y^x}{x}}\right)^2.$$
It suffices to prove that
$$\sqrt{\frac{x^y}{y}} + \sqrt{\frac{y^x}{x}} \ge 2. \tag{1}$$
This inequality has been proved
in show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Can trigonometric substitution be used to solve this integral? The integral in question is $$\int \frac{x+1}{9x^2+6x+5}dx.$$ I first completed the square in the denominator giving $(3x+1)^2+4$ and proceeded to perform a $u$-substitution with $u = 3x+1$, $du=3~dx$, and $x=(u-1)/3$. After simplifying, I was left with $$\... | There is a simpler way to determine this integral: rewrite first the numerator:
$$x+1=\frac 1{18}(18x+6)+\frac 23,$$
split the integral in two:
\begin{align}
\int \frac{x+1}{9x^2+6x+5}\,\mathrm dx&=\frac 1{18}\int \frac{18x+6}{9x^2+6x+5}\,\mathrm dx+\frac23\int\frac{\mathrm dx}{(3x+1)^2+4}\\
&=\frac 1{18}\ln(9x^2+6x+5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate sum, if the recurrence is given? Let $a_n=4a_{n-1}-a_{n-2}$, for $n\geq 3$. Prove that $\sum\limits_{n=1}^\infty \cot^{-1}{a_n^2}=\frac{\pi}{12}$, if $a_1=2$, $a_2=8$.
This recurrent formula is easy to solve. Characteristic equation $x^2-4x+1=0$ has two roots, $2\pm \sqrt{3}$. Therefore general soluti... | Note: This is one part of a solution, the other part is in cineel's answer.
The first elements of the sequence are
$$
a_n = 2, 8, 30, 112, 418, 1560, 5822, 21728, 81090, \ldots
$$
Using the identity
$$
\cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1} \left( \frac{xy-1}{x+y}\right)
$$
we compute the first few partial sums:
$$
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}... | Your partial fraction is incomplete. At least you can further decompose $\frac{1}{n(n+2)}$. Anyway the correct decomposition is $\frac{1}{2}(\frac{1}{n} + \frac{1}{n+2}) - \frac{1}{n+1}$.
Now for any $3 \le n \le k$, $\frac{1}{n}$ will appear twice with coefficient $\frac{1}{2}$ and once with coefficient $-1$, therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2}$ Edit: A friend of mine used a contour integral to solve this problem and his final answer matches with mine. It seems that Wolphram´s is incorrect
I saw the following integral here and went to proof it.
$$... | It can be shown by the Abel-Plana formula, hoping for convergence, that:
$$\sum_{x\ge0} f(x)=\frac{f(0)}2 +\int_0^\infty f(x)dx+i\int_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}dx$$
Note that when $f(x)=-\frac i2 x \ln(1-x^2)\implies f(ix)-f(-ix)=x\ln(x^2+1)$:
Let’s plug in to the formula:
$$\sum_{x\ge0} -\frac i2 x \ln(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Error of $\left(1+\frac{1}{n}\right)^n$ For a positive $\varepsilon$, how do you find $n$ that satisfies $e-\left(1+ \frac{1}{n}\right)^n<\varepsilon$?
Below is the context.
Definition
$$e=\lim_{n\rightarrow\infty}\left(1+ \frac{1}{n}\right)^n$$
Attempt
It is proved that $e=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$. On the ... | This is an alternate way to derive an $n$ that works, reasonable close to the optimal one given in Claude Leibovici's answer, without using any special function.
Notice for $x > 0$,
$$\left(\frac1x\right)' = -\frac1{x^2} < 0\quad\text{ and }\quad \left(\frac1x\right)'' = \frac{2}{x^3} > 0$$
$\frac1x$ is strictly decrea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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common ratio of a geometric sequence formed from three terms of an arithmetic sequence The problem is to prove that if we have an arithmetic sequence and we choose $3$ numbers from it which form a geometric sequence as well, and also in the new geometric sequence these $3$ numbers are after each other with no gap, the ... | Let $(a_k)$ be an arithmetic sequence with initial term $a_1$ and common difference $d$. Then its $k$th term is given by $a_k = a_1 + d(k - 1)$. Hence,
\begin{align*}
a_m & = a_1 + d(m - 1)\\
a_n & = a_1 + d(n - 1)\\
a_p & = a_1 + d(p - 1)
\end{align*}
Since $a_m, a_n, a_p$ are successive terms of a geometric sequenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How could I use telescoping to find $f(1) + f(2) + f(3) + \cdots + f(100) $? How could I use telescoping to find $f(1) + f(2) + f(3) + \cdots + f(100) $?
Let $f(x)$ be a function defined by $f(x) = x^6 - 3x^5 + 5x^4 - 5x^3 + 3x^2 - x$. Compute the sum of the base-ten digits of the sum $f(1) + f(2) + f(3) + \cdots + f(1... | The key observation is that $$7f(x) = 7x^6 - 21x^5 + 35x^4 - 35x^3 + 21x^2 - 7x$$ has coefficients that look suspiciously like terms in Pascal's triangle; specifically, $\binom{7}{k}$ for $k \in \{1, 2, \ldots, 6\}$. So in fact $$7f(x) = x^7 - (x-1)^7 - 1,$$ and we no longer need to define $g$ because $$\sum_{x=1}^{10... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*... | $$
\begin{aligned}
I&=\int_{0}^{\pi / 2} \frac{x}{\sin (x)} d x\\&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)} d x \\
&=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)} \frac{1}{\frac{\cos \left(\frac{x}{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
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Proof recursion is a subset of Lucas numbers I need to prove that the recursion $a_n=\frac{a_{n-1}^2+5}{a_{n-2}}$ for $a_0=2,a_1=3$ are the Lucas numbers with even index. I would like to use induction, but I got a fraction that I'm not sure how to simplify into the clean recursion for the Lucas numbers. Is induction th... | We know that
$$L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n \quad \forall\ n \in \mathbb{N}$$
you can see it here for example. To simplify the calculations, let
$$x=\frac{1+\sqrt{5}}{2}\quad and\quad y=\frac{1-\sqrt{5}}{2}$$
where $x^2+y^2=3$ and $xy=-1$. Then, (with $n\geq 2$)
\begin{ali... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Triangular prism in a sphere A sphere of unit radius touches all edges of some triangular prism. What can be the volume of this prism?
The prism is regular because it is inscribed in the sphere. We express the radius by the Pythagorean theorem $1=\sqrt{(0,5 a)^{2}+\left(\frac{a}{\sqrt{3}}\right)^{2}}$. So, $a_{1}=-\fra... | If the edge length is $2 a $, then the circumcircle of the the triangular face (which is an equilateral triangle) is
$ R = \sqrt{ 1^2 - a^2 }$
but $ R = \dfrac{2a}{\sqrt{3}} $
Hence, $1 - a^2 = \dfrac{4 a^2}{3} $
making $a^2 = \dfrac{3}{7} $
The volume is $(2 a) \dfrac{\sqrt{3}}{4} (4 a^2) = (6) (\dfrac{1}{\sqrt{7}}) \... | {
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"url": "https://math.stackexchange.com/questions/4285408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\cos^{-1}(-x)=\pi-\cos^{-1}x$ without geometry. Let
$$
\cos^{-1}x=a
\implies x=\cos a
$$
and
$$
\cos^{-1}(-x)=b
\implies -x=\cos b
$$
Hence we have
$$
\cos a+\cos b=0
$$
Using
$$
\cos(A+B)+\cos(A-B)=2\cos A\cos B
$$
with
$$
A=\frac{a+b}{2} \\
\text{ and } \\
B=\frac{a-b}{2}
$$
we get
$$
\cos a+\cos b=2\cos\lef... | Let $f(x) = \cos^{-1}(-x) + \cos^{-1}x$. Note that
$$
f^{\prime}(x) = -\dfrac{1}{\sqrt{1 - (-x)^2}}\cdot (-1) - \dfrac{1}{\sqrt{1 - x^2}} = 0 \quad \Rightarrow \quad f(x) = k, \quad k \in \mathbb{R}
$$
For $x = 0$, we have $k = \cos^{-1}(-0) + \cos^{-1}0 = \pi/2 + \pi/2 = \pi$. Thus,
$\cos^{-1}(-x) = \pi - \cos^{-1} x$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find the general solution to $y''+2y'+4y=\sin x \cos x + 3x^2$ Given this equation $y''+2y'+4y=\sin x \cos x + 3x^2$, how do I find the general solution? It is obvious that this is a differential equation, so that's a start. However, this is a second order differential equation. And it doesn't have any elementary solut... | First, solve the homogeneous equation. Here, you got $\lambda^2+2\lambda+4=0$, which gives $\lambda_1 = -1+ i\sqrt3, \lambda_2 = -1-i\sqrt3,$ so the homogeneous solution is $$y_h(x)=c_1 e^{-x}\cos(\sqrt3 x) + c_2 e^{-x}\sin(\sqrt3 x).$$
Now to get the particular solution, note that we have $f(x)=\frac{1}{2} \sin(2x) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Fourth Degree Pell Equation I got stuck at the following of my research problem:
Prove that only solution to equation $4b^2-3a^4=1$ for odd positive integers $a$, $b$ is $(1,1)$.
I made factorization -->$3a^4 = (2b-1)(2b+1)$
from here since 2b-1 and 2b+1 are coprime,
$$2b-1 = 3x^4, 2b+1 = y^4$$
or
$$2b-1 = x^4, 2b+1 ... | Comment: May be this idea helps:
$4b^2-3a^4=1 \Rightarrow 3(b-a^2)(b+a^2)=(1-b)(1+b)$
We consider following cases:
Case 1:
$$\begin{cases} 3b-3a^2=1-b\\b+a^2=1+b\rightarrow
a^2=1\rightarrow a=\pm 1\end {cases}$$
which gives $b= 1$ for $a=+1$
Case 2:
$$\begin{cases} 3b+3a^2=1+b\\b-a^2=1-b\end {cases}$$
which gives no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove that: $(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\ge16abc$
Let $a,b,c\ge0$, prove that: $$(a^2+1)(a^2+b^2+1)(a^2+b^2+c^2+1)\ge16abc$$
I tried AM-GM as below:
$$a^2+1\ge2a; a^2+b^2\ge2ab; a^2+b^2+c^2\ge ab+bc+ca$$ The rest is proving the inequality which is not true for all non-negative real numbers: $a(2ab+1)(ab+bc+ca+1... | If you write it as a quadratic in $c$ and find its discriminant, then it suffices to prove:
$$(a^2+1)^2(a^2+b^2+1)^3\geq 64a^2b^2.$$
This can be done with AM-GM since:
$$(a^2+1)^2 = \left(a^2+\frac 13+\frac 13+\frac 13\right)^2\geq (4\cdot3^{- 3/4}a^{ 1/2})^2=\dfrac{16}{3\sqrt{3}}a$$
and so
$$(a^2+b^2+1)^3 = \left(b^2+... | {
"language": "en",
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Finding all primes $p$ satisfying $p=a^2+b^2$ and $p$ divides $a^3+b^3-4$ .
Determine all primes $p$ such that there exist integers $a, b$ satisfying $p=a^2+b^2$ and $a^3+b^3-4$ is divisible by $p$.
So, $p | ab(a+b)+4$ after writing $a^3+b^3-4=(a+b)(p-ab)-4$ and $p=2$ is a solution if $p|4$. If $p$ doesn't divide $4$... | Assume $p>2$, and let $s=a+b$. Then
$$ab=\frac{(a+b)^2-(a^2+b^2)}2\equiv \frac{s^2}2\pmod p.$$
Now,
$$a^3+b^3=(a+b)(a^2-ab+b^2)\equiv (a+b)(-ab)\equiv -\frac{s^3}2\pmod p.$$
This implies that $s^3\equiv -8$ modulo $p$. This means that either $p\mid s+2$ or $p\mid s^2-2s+4$.
In the first case,
$$s\geq p-2\implies p=a^2+... | {
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"answer_count": 2,
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} |
Find domain and range of $\log_2\left(\sin(2x)+\cos(2x)\right)$ I'm having trouble finding domain and range of this function.
Can somebody give me a HINT please? Thanks.
| \begin{align}
\cos(2x)+\sin(2x)
&=\cos(2x)\,r\cos \phi + \sin(2x)\,r\sin \phi \\
&= r\cos(2x-\phi)
\end{align}
\begin{cases}
r\cos \phi =1\\
r\sin \phi =1
\end{cases}
Finding $r$:
\begin{align}
r^2 (\sin^2\phi +\cos^2 \phi) &= 1^2+1^2\\
r &=\sqrt{2}
\end{align}
Finding $\phi$:
\begin{align}
\frac{r\sin \phi}{r\cos \phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the measure of the largest angle of a triangle formed by the reciprocals of the altitudes of a given triangle For reference:
The measures of the sides of a triangle
are $3$, $5$ and $7$. Calculate the measure of the largest angle
of a triangle whose sides are the inverses of the heights of the first triangle (Ans... | Elaborating @Stinking Bishop's hint, consider triangle $ABC$ with sides $a$, $b$ and $c$ (in standard notation). Let altitudes from each vertex to the opposite side be $h_a$, $h_b$ and $h_c$. Now, if the area of the triangle is $A$, we get, $$\frac1{h_a}=\frac{a}{2A}$$ $$\frac1{h_b}=\frac{b}{2A}$$ $$\frac1{h_c}=\frac{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\fra... | Another way.
Let $x^2+y^2+z^2=t(xy+xz+yz).$
Thus, $t\geq1$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}\left(\frac{x}{\sqrt{1+x^2}}+\frac{1}{x^2}\right)=\sum_{cyc}\frac{2x}{2\sqrt{(x+y)(x+z)}}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x^2y^2}{x^2y^2z^2}\geq$$
$$\geq\sum_{cyc}\frac{2x}{2x+y+z}+\frac{\sum\limits_{cyc}xy\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the sum of arcus cotangents? Say I wish to find the sum of three arcus cotangents. I wanted to start with a smaller example, say
$$\mathrm{arccot}{x} + \mathrm{arccot}{y}.$$
I know that
$$
\mathrm{arccot}{x} =
\begin{cases}
\arctan{\frac{1}{x}} & \text{if $x>0$} \\
\pi + \arctan{\frac{1}{x}} & \text{if $x<0$}... |
I know that $$ \mathrm{arccot}(x) = \begin{cases}
\arctan{\frac{1}{x}} & \text{if $x>0$} \\ \pi + \arctan{\frac{1}{x}} &
\text{if $x<0$} \end{cases} $$
And a small addition: $$=\frac\pi2\quad\text{if }x=0.$$
Based on this definition (there are two common definitions of arccot), we have that on $\mathbb R,$ $$\arct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
High dimensional generalizations of $\int_{\sqrt{2} }^{\sqrt{3} } \frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } dy =\frac{5\pi^2}{96}$ Let $Q=n_1+n_2+n_3+1$,$\mathbf{s}=(n_1,n_2,n_3)$.
Define
$$
A(\mathbf{s})
=\int_{D}\prod_{n=1}^{Q-1}\frac{1}{1+x_n^2}\int_{1}^{\infty}\left(Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y... | I've found a direct way.
Let me give an example,
$$
I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2} }\text{d}x\text{d}y\text{d}z.
$$
Use the fact that
$$
\frac{1}{\sqrt{x} } =\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-xt^2}\text{d}t
$$,
and rewrite the integral into
$$I = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
In the figure, determine $x$ as a function of $R$. For reference: In the figure, determine $x$ as a function of $R$.
My progress: I don't know and I can affirm that $\angle FGD$ is $90^o$ ...I think you would need to demonstrate ...but I can't
$\triangle FDB: (R+r)^2 = R^2+(2R-2r)^2\\R^2+2Rr+r^2=R^2+4R^2-4Rr+4r^2\\6Rr... | If the radius of the other half circle is $r$, you already showed that $ \small \displaystyle r = \frac{2R}{3}$.
Now applying Descartes' theorem,
$ \small \displaystyle \left(- \frac{1}{2R} + \frac{1}{R} + \frac{1}{2R/3} + \frac{1}{x} \right)^2 = 2 \left(\frac{1}{4R^2} + \frac{1}{R^2} + \frac{1}{(2R/3)^2} + \frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$x^4+x^2=\frac{11}{5}$ then what is the value of $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$?
$x^4+x^2=\frac{11}{5}$ then what is the value of $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}{x+1}}$ ?
My approaches which didn't yield anything
I tried manipulating $\sqrt[3]{\frac{x+1}{x-1}}+\sqrt[3]{\frac{x-1}... | Let $a = ...$ "which again is not useful", wrong here, it is.
Let $a = x^2$
$a^2 + a - \frac{11}{5} = 0$
$a = \frac{-1 \pm \sqrt{1 - 4 * 1 * \frac{-11}{5}}}{2}$
$a = \frac{-1 \pm \sqrt{\frac{49}{5}}}{2}$
$a = \frac{-1 \pm \frac{7\sqrt{5}}{5}}{2}$
$a = \frac{-5\pm {7\sqrt{5}}}{10}$
$x^2 = \frac{-5\pm {7\sqrt{5}}}{10}$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and ... | Check out my analysis of your question.
Equation is 2xcosα−3ysinα=0. This equation passes through origin and points A and B are on x axis. Assume slope of equation be tan $\phi$.
I have illustrated the diagram here
now product of length of two perpendiculars will be 5 sin$^2$ $\phi$. From equation of line slope of li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logar... | \begin{gather*}
\lim _{x\rightarrow \infty }\frac{1}{\ln( 1+x) -\ln x} -x\\
=\lim _{x\rightarrow \infty }\frac{1}{\ln\left( 1+\frac{1}{x}\right)} -x\\
=\lim _{x\rightarrow \infty }\frac{1-x\cdotp \ln\left( 1+\frac{1}{x}\right)}{\ln\left( 1+\frac{1}{x}\right)}\\
Let\ us\ make\ a\ substitution\ h=\frac{1}{x}\\
Then,\ h\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Showing the formal identity $(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$
How to show this formal identity (or you can assume $|x|<1$)?
$$(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$$
I can show that the latter is $$=1-x+x^3-x^4+x^6-x^7+\cdots$$ but how to show this is equal to the infinite ... | Notice that $(1+x+x^2)(1-x+x^2)=(x^2+1)^2-x^2=1+x^2+x^4$. So: multiply both sides by $1+x+x^2$:
$$\begin{array}{rcl}(1+x+x^2)\cdot(\text{LHS})&\equiv&(1+x+x^2)(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots\\&=&(1+x^2+x^4)(1-x^2+x^4)(1-x^4+x^8)\cdots\\&=&(1+x^4+x^8)(1-x^4+x^8)\cdots\\&&\vdots\end{array}$$
The second row shows th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then does $a | 2^m-1$ or $a| 2^m+1$
Determine whether for any integer $a$, if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then $a | 2^m-1$ or $a| 2^m+1$.
Suppose $a\not |2^m-1$ and $a\not| 2^m+1$. Then there exist primes $p_1$ and $p_2$ so t... | Let $a$, $k$ and $m$ be positive integers with $k\geq2$ and
$$a^k=(2^m-1)(2^m+1).$$
The two factors on the right hand side are coprime, because they are both odd and their difference is $2$. Then by unique factorization, both are $k$-th powers. That is, there exist positive integers $b$ and $c$ such that
$$2^m-1=b^k\qq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.