Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\cos \left(\frac{2\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{4\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{6\pi }{7}\right)^{\frac{1}{3}} =?$ I have been trying to solve the following question for a long time:
Find $a,b,c,d$ such that: ($a,b,c,d$ are primes)
$\cos \left(\frac{2\pi }{7}\right)^{\frac{1}{3}}+\cos... | Try Ramanujan's cubic polynomial, this polynomial don't need to find 3 roots. If coefficients are satisfied the condition, then sum of cuberoots of zeroes can be evaluated directly from its coefficients.
EDIT: With your results, you established the equation: $8t^3+4t^2-4t-1=0$ which is equivalent $t^3+\frac{1}{2}t^2-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is... | You want to divide by $x^3+x = x(x^2+1)$, hence by the Chinese remainder theorem it is enough to check the remainders $\pmod{x}$ (which is obviously zero) and $\pmod{x^2+1}$. This remainder is simply given by setting $x^2\equiv -1$, such that
$$ x^{10}+x^{11}+\ldots+x^{19}+x^{20} \equiv (-1-x+1+x)+(-1-x+1+x)+(-1-x+1+x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
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Evaluating $ \frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx$ I recently came across a problem on definite integration and couldn't solve it despite my efforts. It goes as
$$
\frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx
$$
The $2-x^2$ term and the upper limit along... | After the first step from @ClaudeLeibovici’s answer, you can do a crazy u-sub: $$u=\frac{x^2-1}{x^2+1}$$ so $$x=\sqrt{\frac{1+u}{1-u}}$$ By chain rule, $$\implies dx=\frac{1}{2\sqrt{\frac{1+u}{1-u}}}\cdot\frac{2}{(1-u)^2}du$$$$=\frac{du}{(1+u)^\tfrac12(1-u)^\tfrac32}$$ Also, $$\frac{1}{\sqrt{1+x^4}}= \frac{1}{\sqrt{1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Determine the power series representation of the function $f(x)=\sqrt{(4+x)^3}$ and indicate the radius of convergence I want to find a representation of the function mentioned above, so I took into account that:
$$f(x)=\sqrt{(4+x)^3}=8\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$$ and developing the binomial series for $\... | Starting from
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\,x^n \tag{1}$$
which is important and should be treated as a fundamental "building block", in my opinion
one gets by termwise integration
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(1-2n)}\,x^n \tag{2}$$
and by integrating again
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\int_0^{2\pi}\frac{\cos(\phi)-R}{1-2R\cos(\phi)+R^2}\cos(n\phi)~d\phi$? I wish to calculate
$$I(R)=\int_0^{2\pi}\frac{\cos(\phi)-R}{1-2R\cos(\phi)+R^2}~\cos(n\phi)~d\phi,$$
where $n\in\mathbb{N}$, $R\in[0,1)$.
Based on trial and error from plugging numbers into Wolfram alpha I think the answer is
$$I(... | Define: $z=e^{i\phi}$
$$\cos(\phi)=\frac{1}2\left(z+\frac{1}z \right),~~~~\cos(n\phi)=\frac{1}2\left(z^n+\frac{1}{z^n} \right),~~~~d\phi=\frac{1}{iz}dz$$
If $n=0$,
$$\begin{align}
I&=\frac{1}{2i}\oint \frac{z^2-2Rz+1}{z(z-R)(1-Rz)}dz=\frac{1}{2i}\cdot 2\pi i\cdot \left(Res[z=0]+Res[z=R]\right)\\
\\
&=\pi\cdot\left( -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .
Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^... | A partial answer
Let
$$F=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}$$
So we can write $F=G+H$, where
$$G=\sum_{j=0}^{n+1}{2n+2\choose 2j}{j \choose n}, \quad H=\sum_{j=1}^{n+1} {2n+2 \choose 2j-1}{j-1/2 \choose n}$$
Only 2 terms in $G$ are nonzero, when $j=n,n+1$.
Hence $$G={2n+2 \choose 2n} {n \choose n}+{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Interesting integral $\int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$? Noting that
$\displaystyle d(x \sin x+\cos x)=x \cos xdx,\tag*{} $
we have
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x=-\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos x} d\left(\frac{1}{x \sin x+\cos x}\... | Well, we are trying to solve:
$$\mathcal{I}\left(x\right):=\int\underbrace{\left(\frac{x}{x\sin\left(x\right)+\cos\left(x\right)}\right)^2}_{:=\space\mathscr{I}\left(x\right)}\space\text{d}x\tag1$$
Rewrite the integrand using $\sin^2\left(x\right)+\cos^2\left(x\right)=1$:
$$\mathscr{I}\left(x\right)=\left(\frac{x}{x\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the differential equation: $6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$ I have the following differential equation before me:
$6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$
I tried solving it by reducing to Bernoulli form of first order differential equation.
I divided both sides of the equa... | I am writing $\frac{dy}{dx}$ as $y'$
Our equation is $$6\operatorname{cos}^2(x)y'=\operatorname{sin}(x)y-2\operatorname{sin}^3(x)y^4$$ $=$
$$y'-\frac{\operatorname{sin}(x)y}{6\operatorname{cos}^2(x)}=-\frac{\operatorname{sin}^3(x)y^4}{3\operatorname{cos}^2(x)}$$ Dividing by $y^4$,
$$\frac{y'}{y^4}-\frac{\operatorname{s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then
*
*$ \ \displaystyle{ \alpha<-2 }$ ;
*$ \ \displaystyle{ 3<\alpha<7 }$ ;
*it is impossible... | If $r$ and $s$ are the roots of $x^2+2(\alpha-1)x-\alpha+7$, then$$\left\{\begin{array}{l}r+s=2-2\alpha\\rs=7-\alpha.\end{array}\right.$$Therefore$$\left\{\begin{array}{l}2-2\alpha<0\\7-\alpha>0;\end{array}\right.$$in other words, $\alpha\in(1,7)$. But, in fact, we cannot have $\alpha\in(1,3]$, because $(r-s)^2>0$, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}$ where $x,y,z>0$ and $x+y+z=3$.
Prove $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x},$$ where $x,y,z>0$ and $x+y+z=3$.
Maybe we can show $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\g... | Let $z=\max\{x,y,z\}.$
Thus, $$\sum_{cyc}\left(\frac{x}{y^2}-\frac{x^2}{y}\right)=$$
$$=\left(\tfrac{x}{y^2}+\tfrac{y}{z^2}+\tfrac{z}{x^2}-\tfrac{1}{x}-\tfrac{1}{y}-\tfrac{1}{z}\right)-\left(\tfrac{x^2}{y}+\tfrac{y^2}{z}+\tfrac{z^2}{x}-x-y-z\right)+\sum_{cyc}\left(\tfrac{1}{x}-1\right)=$$
$$=\left(\tfrac{x}{y^2}+\tfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically? I'm trying to solve the said equation in the thread title algebraically.
$$(x^3+1)=2\sqrt[3]{2x-1}$$
Cubing both sides and simplifying:
$$x^9+3x^6+3x^3-16x+9 = 0$$
Not sure if this can be solved algebraically?
Edit: WA gives $3$ solutions $x=1,\frac{1}{2}(-1-\sqrt{5}),\fra... | Rearrange as $$\underbrace{\frac{x^3+1}{2}}_{f(x)}=\underbrace{\sqrt[3]{2x-1}}_{g(x)}$$
Since $f(x)$ is a bijective function on $\mathbb R$, it must have an inverse. But note that the inverse of $f(x)$ is $g(x)$. Thus, if the two curves intersect, they must intersect ON the line $y=x$ (because $f$ and $g$ are mirror im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Bounding a simple ratio of bivariate functions For any $x,y \ge 0$, define $f(x,y) := x^2+y^2 + 2c xy$, where $c := \sqrt{2/\pi}$. It is clear that
$$
\sqrt{2/\pi}\cdot (x+y)^2 \le f(x,y) \le (x+y)^2.
\tag{1}
$$
This is an immediate consequence of the fact that $c \le 1 \le 1/c$.
Question. What is a good upper-bound fo... | It is natural to guess that the maximum is achieved when $x=y$, that is:
\begin{equation}
\frac{(x+y)^2}{x^2+y^2+2cxy}\le \frac{2}{1+c},
\end{equation}
which is equivalent to
\begin{equation}
2(x^2+y^2+2cxy) \ge (1+c)(x+y)^2,
\end{equation}
which is true because $LHS-RHS = (1-c)(x-y)^2 \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th ter... | This reduces the problem to the evaluation of a second derivative.
$$
1\cdot 2 + 2\cdot 3+ 3\cdot 4+ \cdots + n\cdot (n+1)
= \sum_{k=1}^{n} k(k+1) \\
= \left.\frac{d^2}{dx^2}(1+x+x^2+x^3+\cdots+x^{n+1})\right|_{x=1} \\
= \left.\frac{d^2}{dx^2}\frac{x^{n+2}-1}{x-1}\right|_{x=1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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I can't come up with the intended solution to $\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18$ I was going trough some easy algebra problems when I encountered
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18.
$$
As you can see the problem is easily solvable with AM > GM
I fairly quickly came up with thi... | Here is the intended solution (this is from a Junior Balkan Mathematical Olympiad TST of Bulgaria).
By repeated use of AM-GM,
$$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab \geq \frac{2ab}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Arc length of curve stuck with integration Data from exercise
$$y=\frac{4}{3}x^2+2\\
x\in[-1,1]$$
Formula for length of curve
$$L=\int_a^{b}\sqrt{1+(f(x)')^2}\ dx$$
So far i have
$$y'=\frac{8}{3}x$$
$$\int_{-1}^{1}\sqrt{1+\frac{64}{9}x^2}\ dx$$
Substition
$$t^2=\frac{64}{9}x^2$$
$$t=\frac{8}{3}x$$
$$\frac{3}{8}dt=dx$$
... | Call the indefinite integral $I$
$$I\equiv\int\sqrt{1+t^2}\,\mathrm dt$$
Through integration by parts with $u=\sqrt{1+t^2}$ and $v=x$, then
$$I=t\sqrt{1+t^2}-\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt$$
The integrand can also be rewritten as
\begin{align*}
I & =\int\frac {1+t^2}{\sqrt{1+t^2}}\,\mathrm dt\\ & =\int\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches)
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.
I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\mi... | Go with Trigonometry.
Let $x=a\sin \alpha ,\, y=a\cos \alpha$, then we have:
$$\begin{aligned}&3a\sin \alpha+4a\cos \alpha=15\\
\implies &15\leq \sqrt {9a^2+16a^2}=5|a|\\
\implies &|a|\ge 3.\end{aligned}$$
$$\sqrt {x^2+y^2}=|a|\ge 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1... | As an alternative by half-angle formula $\tan \frac{\theta}2=\frac{1-\cos \theta}{\sin \theta}$ we have (note that $\cos \alpha \neq \sin \alpha$):
$$\tan\left(\dfrac{3\pi}{4}+\alpha\right)
=\frac{1-\cos \left(\frac{3\pi}{2}+2\alpha\right)}{\sin \left(\frac{3\pi}{2}+2\alpha\right) }
=\frac{1-\sin(2\alpha)}{-\cos(2\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | I've tried to answer in a way which skips over some of the tedious calculation so as to make it easier to understand what this question means.
$$f(a)=4x^2-2ax+a^2-5a+4>0$$
Consider, for the moment, $x$ to be a constant. Then we have a quadratic expression in $a$ with roots at (details omitted):
$$a_1=- \sqrt{-3x^2+5x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 8
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The number of four-digit numbers that have distinct digits and are divisible by $99$ We try to find the number of four-digit numbers that have distinct digits and are divisible by $99$.
Let a number be $N = abcd$, then we have $9| N$ and $11|N$.
Thus $9| a+b+c+d$ and $a+c \equiv b+d \mod 11$.
I listed all the four-dig... | Note that:
For a number to be divisible by $99$, the sum of doublets should be divisible by $99$.
Proof:
$abcdef=ab(10^4)+cd(10^2)+ef$
and $10^{2n} \mod 99 =1$
So, $abcdef \mod 99 = ab + cd + ef$
So, $ab+cd+ef$ should be divisible by $99$.
Now let's solve your problem:
Now for a $4$-digit number ($abcd$) to be divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving Rate of Convergence I am investigating the following coupled sequence:
\begin{align*}
y_0 &= 1\\
x_{n+1} &= \sqrt{1 + \frac{1}{y_n}}\\
y_{n+1} &= \sqrt{1 - \frac{1}{x_{n+1}}}\\
\end{align*}
I am trying to show
\begin{align*}
\lim_{n \to \infty} \frac{\left |x_{n+1} - \varphi\right |}{\left |x_n - \varphi\right... | After the edit, it should be simple. You have:
$$x_{n+1} = f(x_n): \quad f(x) = \sqrt{1 + \dfrac{1}{\sqrt{1-\frac 1x}}},\,\, x_1 =\sqrt{2}.$$
Some, but not too terrible, calculus will show that $1 <x<f(x)$ if $x\in (1,\varphi).$ This tells us that $x_n$ is increasing. In fact, after removing square roots $f(x)>x$ is eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Compute the area of Quadrilateral $ABCD$ As title suggests, the question is to solve for the area of the given convex quadrilateral, with two equal sides, a side length of 2 units and some angles:
I have solved the problem with a synthetic geometric approach involving some angle chasing. However, I believe my solution... | Alternative approach:
Area$(\triangle ADC) = \dfrac{1}{2} \times \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)].$
Area$(\triangle ABC) = \dfrac{1}{2} \times \overline{BC} \times [ ~\overline{AC} \sin(75^\circ)].$
Therefore, $~\text{Area}(\triangle ADC) ~=~ \text{Area}(\triangle ABC).$
Therefore
$$\text{Area(qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
From a question I asked before this, I have trouble actually with the numbers manipulating part.
Using trigo identity, $\sin^2 \frac{\pi}{12} ... | You can also use $$\cos(x)+\sin(x)=\sqrt{2}\sin(x+\frac{\pi}4)$$
In this instance with $x=\frac{\pi}{12}$ you get to calculate $\sin(\frac{\pi}3)$ which is known.
The advantage is that you don't get nested root, nor have to rationalize $\sqrt{3}-1$ on denominator, just fraction addition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
SO(2)(K) isomorphic to a $K$-split Torus and its rank Let $K$ be a field and $SO(2)(K):= \{Q \in GL_2(K) \ \vert \ Q^T Q= I \text{ and } \det(Q)=1 \}$ the special orthogonal group.
Is there any characterisation known to decide for which fields $K$ the special orthogonal group $SO(2)(K)$ is toric, ie $SO(2)(K) \cong (K^... | First as an algebraic variety, $SO_2(K)\simeq V(x^2+y^2=1)$. Indeed, let $\begin{pmatrix} x & a \\ y & b\end{pmatrix}\in SO_2(K)$, we have the two conditions $xa+yb=0$ and $xb-ya=1$ have completely determined $a,b$ since $\det\begin{pmatrix} x & y \\ -y & x\end{pmatrix}=1\not=0$.
On the other hand, $K^{\times}\simeq V(... | {
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"url": "https://math.stackexchange.com/questions/4544642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus
*
*Verify the following limits
(iii) $\lim\limits_{n\to \infty} \left
[\sqrt[8]{n^2... | I want to present you a very general method that does not involve any trick. Basically this is using the taylor expansion of $(1+x)^\alpha$ around $0$ to get en equivalent of the sequence :
\begin{aligned}
(1+x)^{\alpha} &=1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^{2}+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
My Attempt:
On rearranging, I get, $(x^2+x+1)(3x+7)+2=0$
Or, $3x^3+10x^2+10x+9=0$
Derivative of the cubic is $9x^2+20x+10$
It is z... | $x^2+x+1 + \dfrac{2}{x^2+x+1} \ge 2\sqrt 2$
Largest value of $x^2 - 2x-6$ in the given interval is $2$ at $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Closed Form Formula for Nonlinear Recurrence $a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$ I'm trying to find a closed form solution to the sequence
$a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$
I tried using a generating function approach in the following way:
Let $$f(x) = \sum_{n=1}^\infty a_n x^n$$
Then multiplying the ... | Here is an intuition: Rewrite the recurrence relation s
$$ a_{n+1} = \frac{1}{2}\left(a_n + \frac{\ell^2}{a_n}\right), \qquad \ell = \sqrt{10}. $$
This is precisely what we get when we apply the Newton-Raphson method to $x^2 - \ell^2$. So,
*
*for any positive initial value, $(a_n)$ converges to $\ell$, and
*for any ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Showing that $(x,y)=\left(\frac{c\sin t+d}{a+b\cos t},\frac{e\cos t+f}{a+b\cos t}\right)$ parameterizes an ellipse I want to show that
$$\mathbf{P}(t) = (X, Y) = \left( \dfrac{ c \sin t + d}{a + b \cos t} , \dfrac{ e \cos t + f }{a + b \cos t } \right)$$
is actually an ellipse, given that $ | b | \lt | a | $
How can I... | Try to solve $\cos t$ and $\sin t$ in terms of $x$ and $y$:
\begin{align}
\cos t &= \frac{f-ay}{by-e} \\
\sin t &= \frac{(ae-bf)x+bdy-de}{c(e-by)} \\
\cos^2 t+\sin^2 t &=
\left(
\frac{f-ay}{by-e}
\right)^2+
\left[
\frac{(ae-bf)x+bdy-de}{c(e-by)}
\right]^2 \\
c^2(by-e)^2 &=c^2(f-ay)^2+[(ae-bf)x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine the values of c and d so that E[Y] = c and Var(Y ) = 1 Question
Supppose $X$ is a random PDF $f(x)=\frac{2x}{c^{2}d^{2}}, \quad 0<x<cd$ where $c > 0, \; d > 0$.
Determine the values of $c$ and $d$ so that $\mathbb{E}[X]= c $ and $\mathrm{Var}(x) = 1$
My approach
My approach has been to try and isolate the var... | I am sorry, I am having a hard time following what you were trying to do. You are given that the pdf of $X$ is $f(x)=\frac{2x}{c^{2}d^{2}}$ with $c,d>0$ and $0<x<cd$.
The first sanity check is that your pdf has to be non-negative everywhere and integrate to $1$. You have
$$
\int_{-\infty}^\infty f(x) dx
= \int_0^{cd} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.
I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exap... | There doesn't seem to be any general factored form, but on the few polynomial factors you mention in the question, we can answer exactly when they appear. From symmetry and homogeneity, it is enough to consider the simpler $$p(x)=(1+x)^n-(1+x^n)$$ as with $x = b/a$, we have $a^n p(x)$ is the expression you want to fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\sqrt{x^2+x+1}+\sqrt{x^2+\frac{3x}{4}}=\sqrt{4x^2+3x}$ Solve the equation $$\sqrt{x^2+x+1}+\sqrt{x^2+\dfrac{3x}{4}}=\sqrt{4x^2+3x}$$
The domain is $$x^2+\dfrac{3x}{4}\ge0,4x^2+3x\ge0$$ as $x^2+x+1>0$ for every $x$. Let's raise both sides to the power of 2: $$x^2+x+1+x^2+\dfrac{3x}{4}+2\sqrt{(x^2+x+1... | following your original approach you should have gotten in line 4:
$$2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}-1$$
and, squaring both sides and collecting like terms, yields
$32x^3+151x^2+88x-16=0$
which can be factored as
$(x+4)(32x^2+23x-4)=0$
from which you get $x=-4$ and two extraneous irr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove PQ is parallel to BC with angle chase? I've been working on this geometry problem for a long time, but I haven't been able to make any progress. I have tried angle chase but that didn't get me anywhere.
Acute triangle $ABC$ has $AB<AC.$ Let $D, E, F$ be the foots of perpendiculars from $A, B, C \text{ and } H $ ... | Let D be the origin and DA be the $y$-axis. Suppose $A(0,a)$, $B(b,0)$, $C(c,0)$, with $a\not=0$ and $b\not=c$.
$AB$: $x/b+y/a=1$
$CH$: $y=b/a(x-c)$
$F$: $(\frac{b(a^2+bc)}{a^2+b^2}, \frac{ab(b-c)}{a^2+b^2})$
$AC$: $x/c+y/a=1$
$BH$: $y=c/a(x-b)$
$E$: $(\frac{c(a^2+bc)}{a^2+c^2}, \frac{ac(c-b)}{a^2+c^2})$
The $y$-coordi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I can't find the solution of $\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$ I can't find the solution of
$$\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$$
Computing for $x$ goes to $0$ it gives a $1^\infty$ type of indeterminate form. I tried to solve... | $$\begin{align*}
\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}} &= \lim_{x\to0} \exp\left(\ln\left(\left(1+\frac x{(x-1)^2}\right)^{\frac1{\sqrt{x+1}-1}}\right)\right) \tag{1} \\[1ex]
&= \exp\left(\lim_{x\to0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}{\sqrt{x+1}-1}\right) \tag{2} \\[1ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Isolating $θ$ in $a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$. I have an equation that follows this form:
$a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$ where $a,b,c,d,e,f \in \mathbb{R}$ and $a\neq 0, d\neq 0$. I am trying to isolate $θ.$
How can this... | You can convert this expression to a 4-th order polynomial using a wassestrass substitution
$$ \cos \theta = \frac{1-t^2}{1+t^2} $$
$$ \sin \theta = \frac{2 t}{1+t^2} $$
which converts the trig expression into the following polynomial
$$ a \frac{4 t^2}{(1+t^2)^2} +b \frac{2 t}{1+t^2} + c +d \frac{(1-t^2)^2}{(1+t^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577312",
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"source": "stackexchange",
"question_score": "1",
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Which number is larger? Using Binomial Theorem Which expression is larger,
$$
99^{50}+100^{50}\quad\textrm{ or }\quad 101^{50}?
$$
Idea is to use the Binomial Theorem:
The right hand side then becomes
$$
101^{50}=(100+1)^{50}=\sum_{k=0}^{50}\binom{50}{k}1^{50-k}100^k=100^{50}+\sum_{k=0}^{49}\binom{50}{k}100^k
$$
The le... | It could be amazing to compare
$$(2x)^x +(2 x-1)^x \qquad \text{and} \qquad (2 x+1)^x$$ or, better, their logarithms.
So, consider that you look for the zero of function
$$f(x)=\log \left((2x)^x +(2 x-1)^x\right)-\log \left((2 x+1)^x\right)$$ which has a trivial solution $x=2$.
So
$$x\gt 2 \qquad \implies\qquad (2x)^x ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the solution for $x+22=-6\sqrt{2x+9}$ So, I want to solve for x in the radical equation:
$x + 22 = -6\sqrt{2x+9}$
By Squaring each expression we get:
$(x + 22)^2 = (-6\sqrt{2x+9})^2$
$ x^2 + 44x + 484 = 36\cdot(2x+ 9) $
$ x^2 + 44x + 484 = 72x + 324 $
Now by solving the quadratic equation:
$ x^2 - 28x + 160 = ... | Just because we can write a question doesn't mean the question has any solutions. For example I can with a perfectly straight face ask you to solve $|x-3|=-7$ (which is impossible as $|x-3| \ge 0 > -7$) and if you solve by squaring both sides to get $(x-3)^2 = (-7)^2$ and you do the work ($x^2-6x + 9 = 49; x^2-6x-40=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is wrong in the proof of max value of $|iz+z_0|$
If $|z-i|\leq 2,z_0 = 5+3i$ then find max value of $|iz+z_0|.$
Now ,
$|iz+z_0| \leq |iz|+|z_0| \cdots \text{(By Triangle Inequality.)}$
Now consider the diagram in Argand Plane -
Observe that $|z| \leq 3.$
$\Rightarrow |iz+z_0| \leq \boxed{3 + \sqrt{34}}$
But the... | Where you were incorrect is not accounting for the change of $z \to iz$. While its true that $|z| = |iz|$, this does not always apply when adding complex numbers, $|z + z_0| ≠ |iz + z_0|$.
In your method, when you substituted $|iz| = 3$, you assimilated that similar substitution also applies to $|iz + z_0|$ towards, as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Inequality with parameters Let $2x(ac+bd)+2y(ad-bc)+(a^2+b^2)(1+z)+(c^2+d^2)(1-z) ≥ 0$ for any $a, b, c , d \in \mathbb{R}$. Prove that $x^2+y^2+z^2 ≤ 1$.
My attempt:
Take $2$ vectors: ${\vec{u}}=(a;b),{\vec{v}}=(c;d)$.
Then $a c+b d=\left(\vec{u},\vec{v}\right)\nonumber=\,u v\cos\varphi$ is a dot product of these ve... | Letting $c = 1, d = 0$, the hypothesis yields
$$2ax - 2by + (a^2 + b^2)(1+z) + 1 - z \ge 0, \quad \forall a, b\in \mathbb{R}. \tag{1}$$
From (1), clearly we have $z + 1 \ge 0$.
If $z + 1 = 0$, (1) yields
$$2ax - 2by + 2 \ge 0, \quad \forall a, b \in \mathbb{R}$$
which results in $x = y = 0$.
Thus, $x^2 + y^2 + z^2 \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(x^2−1) \bmod 8$ $\in \{ 0 , 3 , 7 \}, \forall x \in \mathbb{Z}$. It must be verified that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$. First some definitions. Using the following theorem a definition for $\bmod$ is provided:
Theorem. For all $a \in \mathbb{Z}$ and $d \in \m... | Since $x^2-1\bmod8$ only depends on $x\bmod8,$ its possible values are obtained by taking 8 consective values for $x,$ e.g. $x\in\{-3,-2,-1,0,1,2,3,4\}.$
$(\pm3)^2-1\equiv0\bmod8.$
$(\pm2)^2-1\equiv3\bmod8.$
$(\pm1)^2-1\equiv0\bmod8.$
$0^2-1\equiv7\bmod8.$
$4^2-1\equiv7\bmod8.$
Your observation that $(x+4)^2\equiv x^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584496",
"timestamp": "2023-03-29T00:00:00",
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Prove that $a_n$ is a Cauchy sequence, if $|a_{n+1} - a_n| < \frac{1}{n(n+1)}$ Obviously, $\frac{1}{n(n+1)}$ converges to $0$, as $n \to \infty$
But it doesn't say anything to me, since, for example, $a_n = \sqrt{n}$ diverges, but $|a_{n+1} - a_n| \to 0$
So I need to look at this problem from another point of view.
Seq... | For $n<m$, write
$$
a_n-a_m = \sum_{k=n}^{m-1}a_k-a_{k+1},
$$
so that the triangle inequality yields
$$
|a_n-a_m| \leqslant \sum_{k=n}^{m-1}|a_k-a_{k+1}|.
$$
From what you have done, you should find
$$
|a_n-a_m| \leqslant \frac{1}{n}-\frac{1}{m} \leqslant \frac{1}{n},
$$
and the result should come up easily now.
This i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find y from $x^2 y^3 d x+x\left(1+y^2\right) d y=0$ Suppose $\alpha (x,y)=\frac{1}{xy^3}$ is integral factor of equation $$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$
Check $\alpha (x,y)$:
$x^2 y^3 d x+x\left(1+y^2\right) d y=0 \mid \cdot \frac{1}{x y^3} \quad x \neq 0, y \neq 0$
$x d x+y^3\left(1+y^2\right) d y=0 \qua... | $$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$
Separable ODE :
$$xdx=-\left(\frac{1}{y^3}+\frac{1}{y} \right)dy$$
Integrate :
$$\frac12 x^2=\frac{1}{2y^2}+\ln|y|+\text{constant}$$
Solution on the form of inverse function :
$$x(y)=\pm\sqrt{\frac{1}{y^2}+2\ln|y|+c}$$
Explicit solution :
$$\ln|y|=\frac12 x^2+\frac{1}{2y^2}+\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4587878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum square sum of inscripted triangle Problem: In a right triangle $ABC$, the hypotenuse is long $4$ and the angle in $B$ is $30°$. Calling $N$ the midpoint on the side $AB$ (the hypotenuse), and $M$ the middle point on the side $CB$, consider a random point on the side $AC$, lets call it $P$ and $AP = x$. Find the... | You can apply the law of cosines to $\triangle{APN}$ to have
$$PN^2=x^2+2^2-4x\cos 60^\circ=x^2-2x+4$$
Also, note that $CB=2\color{red}{\sqrt 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving $\,\sin^272^\circ - \sin^260^\circ = \frac18\left(\sqrt5 - 1\right)$ I'm fairly new into proving trigonometric equations, but I believe I'm getting better at it day after day. Yet, I have stumbled upon one such which left me at an impasse. The problem is presented as follows:
$$\sin^272^\circ - \sin^260^\circ =... | Now
$$\sin^2 72°=\cos^2(90°-72°)=\cos^2(18°)=\frac{1+\cos36°}{2}=\frac{1+\frac{\sqrt 5+1}{4}}{2}$$
$$=\frac{5+\sqrt 5}{8}$$
Thus
$$\sin^2 60°=\frac 34$$
Hence
$$\color{orange}{\sin^272^\circ - \sin^260^\circ} = \frac{5+\sqrt 5}{8}-\frac 34=\frac{5+\sqrt 5-6}{8}=\color{red}{\frac18\left(\sqrt5 - 1\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Proving $\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$ I encountered the following binomial equality:
$$\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$$
which I know it's true, but I don't know how to prove it directly. I entered the left-hand-side to Mathemat... | We seek to show that
$$\sum_{q=0}^K (-1)^q {2n+1-q\choose q}
{2n-2q\choose K-q} = \frac{1}{2} (1+(-1)^K).$$
The LHS is
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\sum_{q=0}^K
(-1)^q \frac{1}{z^{q+1}} (1+z)^{2n+1-q}
\frac{1}{w^{K-q+1}} (1+w)^{2n-2q}
\; dz \; dw
\\ = \frac{1}{2\pi i} \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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To find the sum of this series I need to find the sum of this series $$
\frac{n}{n^2}+\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{n}{n^2+(n-1)^2}
$$
Here general term $=\frac{n}{n^2+r^2}, r=0,1,2, \ldots$
$$
\begin{aligned}
& =\frac{1}{n}\left[\frac{1}{1+\left(\frac{r}{n}\right)^2}\right] \\
& =\frac{d x}{1+x^2}\... | Upper limit= $\lim_{n\to\infty} \frac{n}{n}$,
Lower limit= $\lim_{n\to\infty} \frac{0}{n}$.
That gives $1$ and $0$.
In general if this type of sum comes up with the lower limit being $a_n$ and upper being $ b_n$. Then Upper limit of integral = $\lim_{n\to\infty} \frac{b_n}{n}$,
Lower limit of integral = $\lim_{n\to\inf... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of ${a_n^2}+{b_n^2}+{c_n^2}-a_nb_n-b_nc_n-c_na_n$
Let $$a_n=\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots$$
$$b_n=\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\cdots$$
$$c_n=\binom{n}{2}+\binom{n}{5}+\binom{n}{8}+\cdots$$
Then find the value of $${a_n^2}+{b_n^2}+{c_n^2}-a_nb_n-b_nc_n-c_na_n$$
I wrote all exp... | I saw this question or a similar in this site but it is not easy to find it. With the hints in the comments:
By binomial expansion, we have
$$\begin{array}
.(1+1)^n&=&\binom{n}{0}&+\binom{n}{1}&+\binom{n}{2}&+\binom{n}{3}\cdots\\
(1+w)^n&=&\binom{n}{0}&+\binom{n}{1}w&+\binom{n}{2}w^2&+\binom{n}{3}w^3+\cdots\\
(1+w^2)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Does $\sum\limits_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$ converge? My Attempt
First, check the limit
$$\lim_{n \to \infty} \frac{3^n + 4^n}{2^n + 5^n} =
\lim_{n \to \infty} \frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} = 0.$$
So, we cannot conclude anything.
I us... | I prefer the Ratio Test. In fact,
$$ \lim_{n\to\infty}\frac{\frac{3^n+4^n}{2^n+5^n}}{(\frac{4}{5})^n}=\lim_{n\to\infty}\frac{3^n+4^n}{4^n}\cdot\frac{5^n}{2^n+5^n}=1>0 $$
which implies that $\sum_{n=1}^\infty\frac{3^n+4^n}{2^n+5^n}$ converges since $\sum_{n=1}^\infty(\frac{4}{5})^n$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Why is $2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$ for $x \in \mathbb{N}$? I have found following formula stands:
$2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$
($x \in \mathbb{N}$)
I have wondered why this is true. I have checked its rightness on the graph drawing tools, but I want to know the mathematical proof of this.
Although I ... | You can set $f(x)=2\sqrt{x+1}-2-\frac 12\sqrt{x}$
The derivative $f'(x)=\frac 1{\sqrt{x+1}}-\frac 1{4\sqrt{x}}$
It is zero when $(4\sqrt{x})^2=(\sqrt{x+1})^2\iff 16x=x+1\iff x=\frac 1{15}$
The derivative is $-$ then $+$ so $f$ is $\searrow$ then $\nearrow$ with a minimum in $\frac 1{15}$.
We could calculate $f(\frac 1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assuming x is small, expand $\frac{\sqrt{1-x}}{\sqrt{1+2x}}$ up to and including the term in $x^{2}$ I have tried this many times but can't quite land on the correct answer.
The correct answer:
$1-\frac{3x}{2}+\frac{15x^{2}}{8}$
These are the steps I took:
*
*Re wrote it as: $\left ( 1-x \right )^{\frac{1}{2}}\left (... | The first terms of the Taylor expansion of $\sqrt{1+2x}$ near $0$ are $1+x-\frac{x^2}{\color{red}2}$; therefore, the first terms of the Taylor expansion of $\frac1{\sqrt{1+2x}}$ near $0$ are $1-x+\frac{3x^2}2$. And if you multiply $1-\frac x2-\frac{x^2}8$ with $1-x+\frac{3x^2}2$ and then you eliminate those terms whose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Verifying $\sqrt{\frac{2}{3}}\cos\left(\frac{1}{3}\arccos\sqrt{\frac{3^3}{2^5}}\,\right)=\cos\frac\pi5$ I can verify the equation using Excel. But is there another derivation?
$$\sqrt{\frac{2}{3}}\cos\left[\frac{1}{3}\arccos\left(\!\!\sqrt{\frac{3^3}{2^5}}\right)\right]=\cos\left(\frac{\pi}{5}\right)$$
Thank you for he... | Let $u=\sqrt{\frac{2}{3}}\cos\left(\theta\right)$, where $\theta=\frac{1}{3}\arccos\left(\sqrt{\frac{3^3}{2^5}}\right)$.
By the triple angle formula, we have
$$4\cos^3(\theta) -3\cos(\theta)-\frac{3}{4}\sqrt{\frac{3}{2}} = 0.$$
i.e. $u$ is a root of the cubic $$q(x) = 6\sqrt{\frac{3}{2}}x^3-3\sqrt\frac{3}{2}x-\frac{3}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Summation of $\sum_{r=1}^n \sin(2r-1)\theta$ In a CIE A Level Further Mathematics question paper, the following question appeared:
By considering $$\sum_{r=1}^n z^{2r-1}$$ where $z=\cos\theta +i\sin\theta$, show that if $\sin ≠ 0$,
$$\sum_{r=1}^n \sin(2r-1)\theta =\frac{\sin^2n\theta}{\sin\theta}$$
My attempt at solv... | The evaluation should be
$$
\eqalign{
\Im\left(\frac{\cos\left(2n\theta\right)+i\sin\left(2n\theta\right)-1}{2i\sin\left(\theta\right)}\right) &= \Im\left(\frac{\cos\left(2n\theta\right)}{2i\sin\left(\theta\right)}+\frac{\sin\left(2n\theta\right)}{2\sin\left(\theta\right)}-\frac{1}{2i\sin\left(\theta\right)}\right) \cr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after... | Let us begin with noticing that
\begin{align*}
f(x) & = x^{6} + x^{5} + 4x^{4} - 12x^{3} + 4x^{2} + x + 1\\\\
& = x^{3}\left[\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12\right] \tag{$x\neq 0$}
\end{align*}
Hence, if we make the change of variable $u = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
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Evaluating $\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)$. (WolframAlpha says it doesn't exist; I get $0$.) I have tried to solve limit, which wolfram says that DNE, but according to my calculations it is equal to 0. Limit is given below
$$\begin{align}
\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)
&=\sin(\sqrt{n^2+1}\pi-n\pi+n... | The step $\sin(\sqrt{n^2+1}\pi) = (-1)^n\sin(\sqrt{n^2+1}\pi-n\pi)$ holds if $n$ is an integer but does not hold if $n$ is not an integer.
Let's look at this another way.
$\sqrt{n^2+1}$ is unbounded and increasing. When $n$ is large $\sqrt{n^2+1}\approx n.$
Suppose $n\in \mathbb R$
For large values of $n$ there will b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so... | One way is as follows. It can be proved that $f(x) = \sqrt{x}$ is an increasing function. So we have:
$$x_1\lt x_2 \implies \sqrt{x_1}\lt \sqrt{x_2} \implies \sqrt{x_2} - \sqrt{x_1} \gt 0$$This implies that $x,y\gt 0$. Also $g(x) = x^2$ is an increasing function for $x\ge 0$, then:
$$x_1\lt x_2 \implies x_1^2\lt x_2^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving $\left\lfloor(\frac{1+\sqrt{5}}{2})^{4n+2}\right\rfloor-1$ is a perfect square for $n=0,1,2,\ldots$ Let $$S_n = \left \lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1$$ ($n=0, 1, 2, \ldots$).
Prove that $S_n$ is a perfect square.
In Art of Problem Solving website, there is a hint
$$
\begin{align}
... | By considering the binomial expansions of $\left(\frac{1+\sqrt5}2\right)^{4n+2}$ and $\left(\frac{1-\sqrt5}2\right)^{4n+2}$ it can be all non-integer terms cancel out when the two are added together, giving an integer result. Therefore we can say that $$\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Solving $y''(x) + \epsilon y'(x) + 1 = 0$ using power series We are given
\begin{align}
\begin{cases}
y''(x) + \epsilon y'(x) + 1 =0, \ 0 < \epsilon <<1\\
y(0)=0, \ y'(0)=1
\end{cases}
\end{align}
and asked to solve this using the solution form $y(x) = \sum_{n=0}^{\infty}\epsilon^{n}y_n(x)$.
Doing the known method for ... | You methodology is basically correct but there are some slips in the algebra. I feel more comfortable writing $y_n(x)=a_nx^n.$ Then $y=a_0+\epsilon a_1x+\epsilon^2a_2x^2+...$ Then $y(0)=a_0=0$ and$y^{\prime}(0)=\epsilon a_1$, so $a_1=1/\epsilon$. To find $\epsilon y^{\prime}$, differentiate $y$ termwise, multiply by $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determining the sum of $\frac{a_{n+1}}{a_n}$ where $a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$
Let $a_0=1, a_1=\frac{1}{2}, a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$, then find $\lim_{n\to\infty} \sum_{k=0}^{n}\frac{a_{k+1}}{a_k}=\dots$
We have $(a_n)$ is strictly decreasing as $a_{n+1}-a_n=\frac{-a_n(a_n+1)}{1+(n+1)a_n}<0, a_n>0$ ... | Consider the sequence $b_n=na_n$ and notice that $b_{n}-b_{n+1}=\frac{b_n}{1+\frac{n+1}{n}b_n}\in(0,b_n)$ for $n>0$, so $b_n$ is positive and decreasing, and thereby convergent. Let $b=\lim_{n\rightarrow\infty}b_n$. Combining the convergence with the above we have $0=\lim_{n\rightarrow\infty}(b_{n+1}-b_n)=\frac{b}{1+b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How fast does $a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}$ decay as $n \to \infty$? Let
\begin{equation}
a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}
\end{equation}
for $n \in \mathbb{N}$.
Then, $a_n$ is clearly a monotone decreasing sequence of positive numbers... | Alternative solution:
With the substitution $x = \sqrt y$, we have
$$a_n = \int_{1/2}^1 \frac{1}{(1/2 + y)^{n+1/2}}\frac{1}{2\sqrt y}\,\mathrm{d}y.$$
It is easy to prove that, for all $y \in [1/2, 1]$,
$$\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2}(y - 1/2) \le \frac{1}{2\sqrt y} \le \frac{\sqrt 2}{2}.$$
We have
$$\int_{1/2}^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Nth Derivative of $\frac{x}{x-1}$ So starting off, rewrite $x*\dfrac1{x-1}$. $d/dx=\dfrac{d}{dx}(x)*\dfrac{1}{x-1}+\dfrac{d}{dx}\dfrac{1}{x-1}*x$.
$\dfrac{d}{dx}(x)=1$. $x^{n}=nx^{n-1}$.
$\dfrac{1}{x-1}=(x-1)^{-1}$.
$\dfrac{d}{dx}(\dfrac{1}{x-1})=(-1)(x-1)^{-2}=\dfrac{-1}{(x-1)^{2}}$.
Combining them both we get $\dfrac... | $\frac {x}{1-x} = \frac {x}{1-x}+1-1 = \frac {1}{1-x} - 1\\
\frac {d}{dx} (1-x)^{-1}-1= (1-x)^{-2}\\
\frac {d^n}{dx^n} (1-x)^{-1}-1= n!(1-x)^{-(n+1)}\\
$
With what you have above, you do change the sign of your expression in the first step. I don't know if that is intentional.
at $\frac {d}{dx} x\frac {1}{x-1} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Integration $\int_0^{\pi/2} \frac{dx}{(3 + 5 \cos x)^2}$ I had tried to solve this integral; using the substitution $\tan(x/2) =t$, and $\cos x= \frac{1-t^2}{1+t^2}$. But after making terms in $t$, I am not able to integrate further as numerator contains quadratic and denominator contains biquadratic.
$\int\limits_0^{\... | An alternative approach is to integrate both sides of
$$\left(\frac{5\sin x}{3 + 5 \cos x}\right)’
= \frac{16}{(3 + 5 \cos x)^2}+\frac3{3 + 5 \cos x}$$
to simply the integral
\begin{align}
\int_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}dx
= \frac{5}{48}
-\frac1{16}\int_0^{\pi/2} \frac3{3 + 5 \cos x}dx
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding angle with geometric approach I would like to solve this problem just with an elementary geometric approach. I already solved with trigonometry, e.g. using the Bretschneider formula, finding that the angle $ x = 15° $. Any idea?
I edited showing how I computed the $ x $ value using the Bretschneider formula for... | Here I report the original figure, to which I added the segments $DH$ bisecting the angle $\widehat{FDG}$ and $EK$ orthogonal to $AD$.
Let's define
$$
d = AB, \qquad e = EF.
$$
We have, by Pythagorean theorem,
$$
AD = \sqrt{ AC^2 + CD^2 } = \sqrt{ (a+d)^2 + a^2 }.
$$
where $a$ is the known length of the side of the sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4623615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How do I justify $\cot 60$ being $\frac{\sqrt{3}}{3}$? Apologies if this is a basic question, my trigonometry skills simply isn't nowhere near up to scratch.
I have a real problem grappling with how $\cot$ functions. At least in my limited understanding, the standard values found in most table simply doesn't reflect ho... | $\ \sqrt{3}\ $ is defined to be the (unique*) positive real number whose square is $\ 3.\ $ We therefore have:
$$ \frac{1}{\sqrt{3}} = 1\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\left( \sqrt{3} \right)^2} = \frac{\sqrt{3}}{3}. $$
$$$$
Alternatively,
$$\frac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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All natural number solutions for the equation $a^2+b^2=2c^2$ $a$, $b$ and $c$ of all Pythagorean triplets can be written in the form
$$
\begin{split}
a &= 2mn\\
b &= m^2-n^2 \\
c &= m^2+n^2
\end{split}
$$
where $m$ and $n$ are natural numbers. For any natural number $m$ and $n$, this set of equations will give a Pythag... | Here is a proof on why natural solutions exist for $a, b, c$, as a matter of fact, infinite solutions.
$$a^2 + b^2 \equiv 0 \mod 2 \to (a + b)^2 - 2ab \equiv 0 \mod 2 \\ \quad \\ \implies a + b \equiv 0 \mod 2 $$
More importantly, this suggests $$a \equiv b \mod 2 \implies \text{ both $a - b$ and $a + b$ must be even}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Probability of a sum being divisible by $4$ Roll $n$ six-sided dice.
What is the probability that the sum of the results is divisible by $4$?
This is an 11th grade problem, and I really don't know -after a lot of searching-what tools do I need to solve this problem using only high school knowledge.
| I'm not quite sure that generating functions and some complex numbers are included in 11th grade (not sure what it corresponds to in Sweden), but you can get a closed solution for each value of $n$ if you know about these. Even though I suspect this is too complicated.
Let the generating function be
\begin{equation}
f_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continuity of function of two variable by Epsilon-Delta technique. Given a function
$f(x,y)= \frac{x^3y^2+x}{x^2+y^2}$ when $(x,y) \neq (0,0)$. Show that the function $f(x,y)$ is continuous at $(1,1)$ by Epsilon-Delta definition.
I can show it easily by the fraction of two continuous functions is continuous. But I need... | For any $\epsilon > 0$ there exists a $\delta>0$ such that
$\sqrt {(x-1)^2 + (y-1)^2} < \delta \implies \left | \frac {x^3y^2 + x - x^2 - y^2}{x^2+y^2}\right | <\epsilon$
The numerator can be simplified and factored.
$(x^3-1)y^2 - x(x-1) = (x-1)((x^2+x+1)(y^2)-x)$
$(x-1) < \delta$
Let $\delta\le 1$
$|x|< 2, |y| < 2$
$|... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Triple summation having $i,j,k$ as variables
Finding value of
$(1)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk$
$(2)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i\leq j\leq k\leq n}ijk$
For $(1) $ what I have done as
For without any restriction as
$ (S)=\displaystyle \mathop{\sum\sum\sum}_{1\leq i... | Your approach is smart. It just needs a little revision. Based on Faulhaber's formula
we use
\begin{align*}
S_n&=\sum_{1\leq i,j,k\leq n}ijk=\left(\sum_{i=1}^ni\right)^3=\frac{1}{8}n^3(n+1)^3\\
S^{\prime}_n&=\sum_{1\leq i=j,k\leq n}ijk=\left(\sum_{j=1}^{n}j^2\right)\left(\sum_{k=1}^nk\right)\\
&=\frac{1}{6}n(n+1)(2n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Sums with nested radicals related to $1/\pi^2$ Context:
I evaluated some sums related to $1/\pi^{2}$ that for a reason I don't know I can't find in the literature.
$$(1)\hspace{.5cm}{1\over 2\sqrt{2}}+{1\over 2^{3}\sqrt{2+\sqrt{2}}}+{1\over 2^{5}\sqrt{2+\sqrt{2+\sqrt{2}}}}+\cdots=1-{8\over \pi^{2}},$$
$$(2)\hspace{.5cm... | I will consider here only the third series with $\varphi$ the golden ratio.
I will establish it as an application of the following identity, justifying the presence of a $\frac{1}{\pi^2}$ term :
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
Its proof by different means is given in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{al... | Without derivatives:
Remark: I used this trick in my answer before.
We want to prove that $2\cos \frac{50\pi}{180} + \sqrt 3 > 3$.
Let $u = \cos \frac{50\pi}{180}$ and $v = \frac{3 - \sqrt 3}{2}$. We need to prove that $u > v$.
We have $u \ge \cos \frac{60\pi}{180} = 1/2$ and $v > \frac{3 - 2}{2} = 1/2$.
Thus, $4u^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Find the roots of the equation $(1+\tan^2x)\sin x-\tan^2x+1=0$ which satisfy the inequality $\tan x<0$ Find the roots of the equation $$(1+\tan^2x)\sin x-\tan^2x+1=0$$ which satisfy the inequality $$\tan x<0$$
Shold I solve the equation first and then try to find which of the roots satisfy the inequality? Should I use ... | $tanx < 0 $ essentially means $ x \in \frac{(2n-1)\pi}{2} ,n\pi $ and the by taking intersection of this and the solution that you obtained, the final answer is $x = 2n\pi -\frac{\pi}{6}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
| Given integral $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Let's factor out $x^2 $ from the denominator.
$\Rightarrow \int \frac{1}{x}\sqrt{\frac{1+x^2}{1-x^2}}dx$
Now multiply and divide the numerator and the denominator by $\sqrt{1+x^2}$
$\Rightarrow \int \frac{1+x^2}{x\sqrt{1-x^4}}dx$
Now we can split the integral into 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Trying to do integration using residue theorem prove using residue theorem $$\int_{0}^{2\pi}\frac{\cos2\theta}{5+4\cos\theta}d\theta={\pi}/6$$
I tried by using $$z=e^{i\theta}$$
now $$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+z^{-1}}{2}$$
$$dz=izd\theta$$
$$\int_{0}^{2\pi}\frac{\cos2\theta}{5+4\cos\theta}d\... | $\displaystyle \int_{|z|=1}\frac{(z^4+1)dz}{2iz^2(2z+1)(z+2)}$, should be converted to $\displaystyle \frac{1}{i}\int_{|z|=1} \frac{1+z^4}{4z^2(z+2)(z+1/2)}$. From there, you can see you have poles at $z=-2$, $z=0$ (order $2$), and $z=-\dfrac{1}{2}$. Since we only care about the poles inside the unit circle, we can eva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the volume between sphere and hyperboloid I am trying to find the volume between the sphere: $x^2+y^2+z^2=9$ and the hyperboloid $x^2+y^2-z^2=1$. I set the integral as: $$\int_{0}^{2\pi}\int_{-3}^{3}\int_{\sqrt{1+z^2}}^{\sqrt{9-z^2}}rdrdzd\theta$$ but it does not give me the correct answer. Can somebody tell me... | Your parametrization of the solid is wrong. Notice that $\sqrt{z^2+1}<\sqrt{9-z^2}$ when $-2<z< 2$, but $\sqrt{z^2+1}>\sqrt{9-z^2}$ when $|z|>2$.
When $-2\leq z\leq 2$, the part of the solid looks like a pipe and it is parametrized as $-2\leq z\leq 2$, $0\leq r\leq\sqrt{z^2+1}$, $0\leq\theta\leq 2\pi$. We can find the ... | {
"language": "en",
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"source": "stackexchange",
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About an integral of the MIT Integration Bee Finals (2023) I would like to solve the first problem of the MIT Integration Bee Finals, which is the following integral :
$$\int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx$$
I tried substitution $u=\tan(x)$, King Property, but nothing leads me to th... | I want to generalize the integral as
$$
I(a,n)=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(\sin x+\cos x)^{2 n}}=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(1+\sin 2 x)^n} d x
$$
where $0<a<1$ and $n\in N$.
Letting $t=\tan x$ gives
$$
I(a,n)=\int_0^{\infty} \frac{t^{a}}{\left(1+\frac{2 t}{1+t^2}\right)^n} \cdot \frac{d t}{1+t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Hint with floors and ceiling proof Prove that $$\forall n \in \mathbb{Z},\left \lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$$
I decided to approach it by extending floor and ceiling definition and got
$$ \frac{n-1}{2} \leq \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \... | Let $n = 2m + 1$ for some $m \in \mathbb{Z}$. In this case, $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = m$. Similarly, $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor m + \frac{1}{2} \right\rfloor = m$. Consequently, $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find integral of $\int\frac{x^2+3}{\sqrt{x^2+6}}dx$ from Cambridge IGCSE Additional Mathematics $$\int\frac{x^2+3}{\sqrt{x^2+6}}dx$$
I tried to $$\int\frac{x^2+6-3}{\sqrt{x^2+6}}dx=\int\frac{x^2+6}{\sqrt{x^2+6}}dx-\int\frac{3}{\sqrt{x^2+6}}dx= \int\sqrt{x^2+6}dx - 3\int\frac{1}{\sqrt{x^2+6}}dx$$
Then got stuck
In exam-... | Since your work $$\int \frac{x^{2}+3}{\sqrt{x^{2}+6}}dx=\int \frac{x^{2}+6-3}{\sqrt{x^{2}+6}}dx=\int \sqrt{x^{2}+6}dx-3\int \frac{1}{\sqrt{x^{2}+6}}dx,$$ which it is correct. Now we are looking for primitives of $G(\sqrt{a^{2}+x^{2}})$, we can try with the substitution $x=a\tan t$.
*
*$\int \sqrt{x^{2}+6}dx\underset{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Difficulties with estimation of epsilon-delta limit proof I have to proof $\lim_{x\to 5}\frac{4x-9}{3x-16}=-11$. I have hard time to evaluate $\frac{1}{|3x-16|}$.
So I start that let $|x-5|<$. I need to show that $|\frac{4x-9}{3x-16}-(-11)|<ε$.
$|\frac{4x-9}{3x-16}-(-11)|$ = $|\frac{37x-185}{3x-16}|$ = $\frac{5|x-5|}{|... | If $|x-5|<\frac16$, then $5-\frac16<x<5+\frac16$, and therefore $-\frac32<3x-16<-\frac12$. So, $|3x-16|>\frac12$.
So, if you take $\delta\leqslant\frac16$ and if $|x-5|<\delta$, you have$$\left|\frac{4x-9}{3x-16}-(-11)\right|=\frac{37|x-5|}{|3x-16|}<74|x-5|.$$So, you can take $\delta=\min\left\{\frac\varepsilon{74},\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4651858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,... | $\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.
$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$
$= ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisi... | Take base cases 1, 2, and 3, and then deduce case $n+3$ from case $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Finding a Function That Approaches Another Function One of my math professors gave me the following challenge. It isn't graded, it's just for fun.
Consider the function:
\begin{equation*}
f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1).
\end{equation*}
I want to find which of the following functions $f_n$ is ... | Yes looks like it must be b)
Consider the following:
There is only one $x$ factor, so this eliminates choices c) and d).
The function is odd, i.e $f(-x) = -f(-x)$, so this eliminates a).
Thus b) must be the answer.
To find the answer by hand you can do the following:
Start with
$$f(x) = x + x^3 + x^5 + \cdots = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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partitions with even number of even parts - partitions with odd number of even parts Let $e(n)$ be the number of partitions of $n$ with even number of even parts and let $o(n)$ denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that $\sum_... | Now,
$$ \frac{1}{(1-x)(1+x^2)(1-x^3)\dots} $$
$$ = (1+x+x^2+\dots)(1-x^2+x^4-\dots)(1+x^3+x^6+\dots)(1-x^4+x^8-\dots)\dots $$
Consider a term which provides a negative coefficient to $x^n$
If we pick $x^{2n_2}$ from the second term, $x^{4n_4}$ from the fourth and so on,
The coefficient of $-1$ is $$(-1)^{n_2 + n_4 + \d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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differential equations I am trying to work through an example from Jordan and Smiths book Nonlinear Ordinary Differential Equations. It's example $6.1$ on page $195$. The question reads:
Obtain an approximate solution of the family of equations $x^{\prime\prime} + x = ex^3$
with $x(e, 0) = 1$, $x^{\prime}(e,0) = 0... | Well, as you can see in the book, proposing a regular perturbation
$$
x(t) = x_0(t) + \varepsilon x_1(t) + \varepsilon^2 x_2(t) + ...
$$
we have
$$
x_0'' + \varepsilon x_1 '' + \varepsilon^2 x_2 '' + ... + x_0 + \varepsilon x_1 + \varepsilon^2 x_2 + ... = \varepsilon (x_0 + \varepsilon x_1 + ... )^3
$$
and
$$
(x_0 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/6594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
| Let $\omega$ be a primitive fifth root of unity. The quadratic residues in $\mathbb{Z}/(5\mathbb{Z})^*$ are $1$ and $4$, hence by setting $q=\omega^1-\omega^2-\omega^3+\omega^4$ such Gauss sum fulfills $q^2=5$. On the other hand
$$\omega^1-\omega^2-\omega^3+\omega^4= 2\cos\frac{2\pi}{5}-2\cos\frac{4\pi}{5} \tag{A}$$
is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 11,
"answer_id": 8
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Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $ I am stuck while proving this identity I verified it using induction but like the other two (1) (2), I am seeking a more of a general way (algebraic will be much appreciated)
$$\frac {1}{(n-1)!... | HINT:
$$(1+x)^{n} - (1-x)^{n} = 2 \Biggl[ {n \choose 1} x + {n \choose 3} x^{3} + \cdots \Biggr]$$
| {
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How to calculate this efficiently? If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?
I solved it in the following way :
Expanding we get, the coefficient of $ x $ as $ (m-n) = 3 \cdots (1)$ and coefficient of $ x^2 $ as $ \frac{n(n-1)}{2} + ... | For the coefficient of $x$ to be 3, $m-n=3$ as you said, so $m>n$ and $m-n>0$ and we can factor the original product:
$$\begin{align}
(1+x)^m(1-x)^n&=(1+x)^{m-n+n}(1-x)^n
\\
&=(1+x)^{m-n}(1+x)^n(1-x)^n
\\
&=(1+x)^{m-n}\left((1+x)(1-x)\right)^n
\\
&=(1+x)^{m-n}(1-x^2)^n
\\
&=(1+x)^3(1-x^2)^n
\end{align}$$
Now, the coeff... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | Just as a curiosity, a one-line-real-analytic-proof I found by combining different ideas from this thread and this question:
$$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
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Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$ In this post, David Speyer, actually, gave an expression for $\displaystyle \frac{t}{e^{t}-1}$.
The question is can we sum the given series, using that expression, if not how does one sum this series. $$\sum\limits_{n=1... | What you require here are the Eisenstein series. In particular the evaluation of
$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$
at $\tau = i. $ Rearrange to get
$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$
See Lambert s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However... | This may be an easier method $$\int\frac{1}{\sin{x} \cdot \cos{x}} \ dx = \int\frac{\sec^{2}{x}}{\tan{x}} \ dx$$ by multiplying the numerator and denominator by $\sec^{2}{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/9075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Height of a tetrahedron How do I calculate the height of a regular tetrahedron having side length $1$ ?
Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table?
| You can also use trig based on the dihedral angle between two faces of the tetrahedron.
Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right ang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Suggest a tricky method for this problem Find the value of:
$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots $$
| This can also be done using Wallis' product.
For instance see a similar problem: Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$
and
Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$
Use
$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \sin^{2n} x \ \text{dx} =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integral of $\arcsin(\sin(x))$ I'm having trouble with this integral
$$\int\arcsin(\sin x)\,\mathrm dx$$
The problem is with the intervals of definition for each function :/ if someone could dumb it down for me.
*
*$\arcsin\colon [-1, 1] \to [-\pi/2, \pi/2]$.
*$\sin:\mathbb{R}\to [-1, 1]$.
right?
But what a... | Let's have $f(x)=\arcsin(\sin(x))$ and calculate $\displaystyle F(x)=\int_0^x f(t)dt$
Since $f$ is $2\pi-$periodic and $F(2\pi)=0$ by symetry, we have that $F$ is $2\pi-$periodic too.
The idea to introduce $s(x)=\operatorname{sgn}(\cos(x))$ will be useful to find a closed form.
$F(x)=x\arcsin(\sin(x))-\frac{s(x)}2x^2+... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?
| If you want to know if $a^n + b^n$ is divisible by $a+b$ (or by $a-b$, perhaps), you can always try long division, whether explicitly or in your head. I can't figure out a way to do the LaTeX or ASCII art for it here to do it explicitly, so I'll show you how one would do it "in one's head".
For example, for $a^3+b^3$, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*}
\lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}
&=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\... | An argument why it should be negative.
When $x>0$, we have $\sqrt{4+x} > 2$ and hence $\frac{1}{\sqrt{4+x}} < \frac{1}{2}$ and hence
$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$
Similarly, When $x<0$, we have $\sqrt{4+x} < 2$ and hence $\frac{1}{\sqrt{4+x}} > \frac{1}{2}$ and hence
$$\frac{\frac{1}{\sqrt{4+x}} ... | {
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"url": "https://math.stackexchange.com/questions/22704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now
| Observe that $(5 - 2\sqrt{6}) = \frac{1}{5+2\sqrt{6}}$.
So, if we set $a = 5+ 2 \sqrt{6}$, we have
\begin{equation}
a^x +a^{-x} +2 = 100.
\end{equation}
The above expression is symmetric in $x$, so if $x = k$ is a solution, $x = -k$ is also a solution. Consider $f(x) = a^x +a^{-x} +2$ for $x>0$. You can show that this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/28157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
All positive integers $n$ such that $n \mid 2^{n-1}+1$ How can one find all positive integers $n$ for which $n \mid 2^{n-1}+1$?
| The idea behind this solution is that if $n>1$ then $n$ must be odd, so $2^{n-1}+1$ is a sum of coprime squares, and we get $n\equiv 1 \bmod{4}$, so that $2^{n-1}+1$ is a sum of coprime fourth powers and $n\equiv 1 \bmod{8}$, etc...
We claim that if $p$ is an odd prime such that $p\mid x^{2^k}+1$ for some natural $k$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/31892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Characterization of two-step 2x2 stochastic matrices Show that:
A 2 x 2 stochastic matrix is two-step transition matrix of a Markov chain if and only if the sum of its principal diagonal terms is greater than or equal to $1$.
| Encode the Markov matrix $\begin{pmatrix}1-a & a\\ b & 1-b\end{pmatrix}$ by the couple $(a,b)$ with $a$ and $b$ between $0$ and $1$. The square of this matrix is encoded by the couple
$$
(a(2-a-b),b(2-a-b)),
$$
hence the question is to determine for which $a$ and $b$ between $0$ and $1$, there exists $x$ and $y$ betwe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/32000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving indetermination in limit I'm trying to solve this limit, but I can't get it out of a $\frac{0}{0}$ indetermination:
$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$
Maybe there is something I'm missing. Thanks a lot in advanced.
| $$\frac{x-4}{5 - \sqrt{x^2 + 9}} = \frac{x-4}{25 - (x^2 + 9)} \times (5 + \sqrt{x^2 + 9}) = \frac{x-4}{16 - x^2} \times (5 + \sqrt{x^2 + 9}) = \frac{5 + \sqrt{x^2 + 9}}{-(x+4)}$$
Hence, $$\lim_{x \rightarrow 4} \frac{x-4}{5 - \sqrt{x^2 + 9}} = \lim_{x \rightarrow 4} \frac{5 + \sqrt{x^2 + 9}}{-(x+4)} = \frac{5+5}{-8} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/37948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Linear Algebra Question My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system
$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$
is consistent for any values of $a,b$, and $c$ that satisfy that equation.
Thanks,
| Perform Gaussian elimination on the system. For each row with 3 zeros on the left, the corresponding equation on the right is a restriction on $a$, $b$, $c$.
Here are the details:
$$
\begin{array}{rrrl}
2 & 1 & -1 & a \\
1 & -2 & -3 & b \\
-3 & -1 & -2 & c
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/39844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Multiplication of coefficients in Dirichlet series This appears to be a relationship:
$\sum\limits_{p\;\text{prime}} \frac{1}{p^s} = \log\zeta (s) - \sum\limits_{n=1}^{\infty}\frac{\sqrt{a_{n}b_{n}}}{n^{s}}$
where $a_{n}$ is a sequence of fractions beginning $\frac{0}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{2}, \frac{1}... | After a quick glance at your sequences, it appears that $$a_n =\frac{\Lambda(n)}{\log n}$$ and $$\sqrt{a_n b_n}=\frac{\Lambda(n)}{\log n} -\chi_p (n)$$ where $\chi_p(n)$ is the indicator function for the primes.
The sequence $b_n$ doesn't really matter too much, except the fact that it is $$\frac{\Lambda(n)}{\log n} -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/42840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute the expression $(a^2 + 4b - 1)(b^2 + 4a - 1)$ without calculating the roots of $x^2 - x - 5 = 0$ Let $a$ and $b$ be the roots of this equation:
$$x^2 - x - 5 = 0$$
Find the value of
$$(a^2 + 4b - 1)(b^2 + 4a - 1)$$
Without calculating the values of a and b.
I saw this on a problems site and tried it but I got 1... | This is an exercise in elementary symmetric polynomials. Write $s_1=a+b$ and $s_2=ab$. From the equation
$$
x^2-x-5=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-s_1x+s_2
$$
we can read that $s_1=1$ and $s_2=-5$.
The number that you wanted to know is
$$
(a^2+4b-1)(b^2+4a-1)=a^2b^2+4(a^3+b^3)+16ab-4(a+b)-(a^2+b^2)+1.
$$
We need to e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/42996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Finding $A^n$ for a matrix I have a matrix $$
A =
\left[ {\begin{array}{cc}
1 & c \\
0 & d \\
\end{array} } \right]
$$
with $c$ and $d$ constant. I need to find $A^n$ ($n$ positive) and then need to prove that formula using induction.
I would like to check that the formula I derived is correct:
$$
A^n =
\... | Mathematical induction is the way to go, but first you want to have a "target." I'm not sure what you did, but I think you confused $c$ and $d$ somewhere along the line in your calculations. Let's see a first few values:
$$\begin{align*}
A&= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\\
A^2 &= \left(\begin{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/44368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Factoring Quantities Question I was doing an exercise and ran into a problem with their use of factoring. Here is the problem specific to where the issue occurs:
$$
\frac{(x^2 + 1)^{1/2} - x^2 (x^2 + 1)^{-1/2}}{x^2 + 1} = \frac{(x^2 + 1)^{-1/2}\left( x^2 + 1 - x^2\right)}{x^2 + 1}
$$
If $(x^2+1)^{-1/2}$ was factored ... | When you factor out a $(x^2+1)^{-\frac{1}{2}}$, the remaining quantity is multiplied by $(x^2+1)^{\frac{1}{2}}$. That is, if we have some expression
$$(x^2+1)^a \cdot f(x)$$
where $f(x)$ is some polynomial and $a$ is any number, when we factor out a $(x^2+1)^{-\frac{1}{2}}$ we get
$$(x^2+1)^{-\frac{1}{2}}\left((x^2+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/44613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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.