Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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how do you solve $a^2+b^2+c^2=d^3$ let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$
| It is a theorem that one can identically solve,
$$x_1^2+x_2^2+\dots+x_n^2 = (y_1^2+y_2^2+\dots+y_n^2)^k$$
for any positive integer n and k. Thus the kth power of n squares is itself the sum of n squares. For example, for $n =3$, we have,
$k=2:$
$$(a^2-b^2-c^2)^2+(2ab)^2+(2ac)^2 = (a^2 + b^2 + c^2)^2$$
$k=3:$
$$a^2(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find z: $|\frac{z-1}{z-i}|=1$ and $|\frac{z-3i}{z+i}|=1$ Give a complex number z:
$|\frac{z-1}{z-i}|=1$ and $|\frac{z-3i}{z+i}|=1$
Find z ?
Could someone help me through this problem ?
| ${\left| {z - 1} \right|^2} = {\left| {z - i} \right|^2}$ and ${\left| {z - 3i} \right|^2} = {\left| {z + i} \right|^2}$, for $z = x + iy$ this gives you ${\left( {x - 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2}$ and ${x^2} + {\left( {y - 3} \right)^2} = {x^2} + {\left( {y + 1} \right)^2}$, since ${\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\Big(\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}\Big)$ $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}$
Converging or Diverging? I guess I have to lower the fraction so that the roots will get away and I will have $\frac{1} {n}$ that diverges. But... | Recall that $(1+\varepsilon)^p\approx 1+p\varepsilon$; thus
$$ \frac{\sqrt[3]{(n+1)^2} - \sqrt[3]{n^2}}{n}
= \frac{n^{2/3} \sqrt[3]{(1+\frac1n)^2} - \sqrt[3]{n^2}}{n}
\approx \frac{n^{2/3}\left(1+\frac2{3n}\right) - n^{2/3}}{n}
= \frac{2}{3n^{4/3}}
$$
So it should converge.
To justify the conclusion more formally, we c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove inequality using AM-GM inequality. $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ I know it is trivial to prove straight forward but I need to prove it using AM-GM inequality.
| \begin{align*}
\sqrt a + \sqrt b
&= \frac{(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}) +
(\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab})}{2} \\
&\ge \sqrt{\big(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}\big)
\big(\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab}\big)} \\
&= \sqrt{\big(\sqrt a + \sqrt b\big)^2 - 2 \sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exact closed form expression of $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$ Exact closed form of this expression $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$
I assume this means there is just one $2^0$ and one $2^{2n}$ and a double of all the terms in between?
| Since all terms are divisible by $2^0+\cdots+2^n$, you can split off that common factor:
$$
(2^0+\cdots+2^n)\times(2^0+\cdots+2^n)=(2^{n+1}-1)^2=2^{2n+2}-2^{n+2}+1.
$$
In fact, the above misreads the question to be about $(2^0+\cdots+2^n)+(2^1+\cdots+2^{n+1})+\cdots+(2^n+\cdots+2^{2n})$ rather than about $(2^0+\cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ Finding the closed form of:
$$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$
where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$
It appears when we try to determine the summation $\displaystyle \sum_{n=1}^\infty\frac{... | HINT: Consider $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{2^nn^2}+\operatorname{Li}_4\left(\frac{1}{2}\right)$ and then express the remaining sum as a double integral. After some work, you get
$$\int_0^1 \frac{\displaystyle\log(x)\operatorname{Li}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 1
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Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ How can I prove that
$$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$$
| $$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}= \sum_{n=0}^{\infty }\frac{(2n+1)!}{(n+1)!n!}2^{-(3n+4)} = \dfrac{1}{16}\sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$$
with $x = \dfrac{1}{8}$.
We are going to compute $f(x) = \sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$ using the integral presentation of binomial coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two eq... | Put $A(x)=(x+a)(x+b)(x+c)=x^3+ux^2+vx+w$. Then
$$\frac{A^{\prime}(x)}{A(x)}=\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}$$ Hence we get $A^{\prime}(\omega)=2\omega^2 A(\omega)$ and $A^{\prime}(\omega^2)-2\omega A(\omega^2)=0$, or if $Q(x)=xA^{\prime}(x)-2A(x)$, that $Q(\omega)=Q(\omega^2)=0$. Hence $Q(x)$ is divisible by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How does the recursion relation work in the solution to this differential equation (using series)? Sorry for the vague title but it would not let me post the first step and last step of this equation (too many characters!).
How does $$\dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} = \... | From what I understand, the product I denote as $P$
$$(3n+1)(3n)(3n−2)(3n−3)⋯10⋅9⋅7⋅6⋅4⋅3$$
is really (by writing the product pairwise)
$$P = (3\cdot 4)(6\cdot 7)(9\cdot 10)\cdots(3n\cdot(3n+1))$$
the pattern is multiplying a pair, one of which is a multiple of three and the other, the increment of that. First, notice ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Help finding the residue of $1/(z^8+1)$ Help finding the residue of $1/(z^8+1)$
I'm integrating over $\{ Re^{it} | 0 \leq t \leq \pi \}$, and I found 4 simple poles at $z_0=e^{in\pi/8}$ where $n = 0,...,3$ and I'm trying to calculate $res(1/(z^8+1),z_0)$
calculating this: $$\lim_{z\to z_0} (z-z_0)f = \lim_{z\to z_0}\f... | $$z^8+1=(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})$$
$$Res_{z=\sqrt[8]i}= \lim_{x\to\sqrt[8]i}\dfrac{(z-\sqrt[8]i)}{(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Finding the limit of a function with ArcTan I've found difficulties finding this limit ( without using Taylor series approximation, as it's intended for the secondary-school ):
$$
\lim_{x\ \to\ \infty}\left[\,
{x^{3} \over \left(\,x^{2} + 1\,\right)\arctan\left(\,x\,\right)} - {2x \over \pi}
\,\right]
$$
Thanks.
| i think the answer is $4/\pi^2.$
i will make a change of variable $u = 1/x$ and will use the fact $\tan 1/u = \pi/2 - \tan u$ for $u > 0.$
$\begin{eqnarray}
lim_{x \to \infty} \frac{x^3}{(1+x^2) \tan x} - \frac{2x}{\pi} =
\lim_{u \to {0+}} \frac{1}{u(1+u^2)\tan(1/u)} - \frac{2}{\pi u} \\
= \frac{2}{u(1+u^2)(\pi-2\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$ How to find the sum of the following series:
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
Any hints.
| According to Maple it's $1 - 5^{1/3}/2$. It looks like this comes from the
Maclaurin series for $(1-t)^{-1/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that
$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$
How can I show this?
| Hint: $a^3 + b^3 > a^2b + ab^2$,
and use AM-GM inequality for $6$ terms for the other one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to integrate $\frac{\sqrt{z+1}}{z}$ How to integrate $\frac{\sqrt{z+1}}{z}$
Anyone could help me? Thanks
| $u=\sqrt{z+1}$, $z=u^2-1$. Then
$$\begin{gathered}
\int {\frac{{\sqrt {z + 1} }}
{z}dz} = \int {\frac{{2{u^2}}}
{{{u^2} - 1}}du} = 2\int {\left( {1 + \frac{1}
{{{u^2} - 1}}} \right)du} = 2\int {\left[ {1 + \frac{1}
{2}\left( {\frac{1}
{{u - 1}} - \frac{1}
{{u + 1}}} \right)} \right]du} \hfill \\
= 2\left( {u +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Feyman's Triangle? How do you find the area of the inner triangle if the outside triangle is equilateral If triangle $ABC$ is equilateral,$BD/BC=1/3, CE/CA=1/3,$ and $ AF/AB=1/3$. What is the ratio of the area of triangle? I have problems analyzing this triangle I tried to use phythagorean, heron's formula, the formula... | There are many proofs of this result (see here for a generalized version), but perhaps the easiest is using the theorem of Menelaus, along with some area ratios.
We use this figure. Consider collinear points $A, K, E$ on the sides of $\triangle DBG$. Menelaus's Theorem tells us:
$$\begin{align}
\frac{DA}{AB} \frac{BK}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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All solutions of the recurrence relation Find all solutions of the recurrence relation $$ a_n = 2a_{n-1}+15a_{n-219}-64a_{n-3}+k $$
| Plug $a_n=b^n$ into the above equation, ignoring the RHS, and get that $b$ satisfies
$$b^3-2 b^2-15 b+36=0 $$
Solutions are $b=3$ (double root) and $b=-4$. The general homogeneous solution is then
$$a_n = (A+B n) 3^n + C (-4)^n $$
For the particular solution, plug in $a_n = D \cdot 2^n$ and get that
$$D \left (1 - 1 - ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving messy integral with modulus and trigonometry. If $$a\in \mathbb R,\int_{a-\pi}^{3\pi+a}|x-a-\pi|\sin(x/2)dx=-16$$ then a can be?
I tried substituting $x-a=u$ and then breaking into two integrals removing modulus then used $\int \sin x=-\cos x,\int x\sin x=\sin x-x\cos x$
| The evaluation of the integral on the left hand side of the integral equation
\begin{equation*}
\int_{a-\pi }^{3\pi +a}\left\vert x-a-\pi \right\vert \sin \left( x/2\right)\tag{1}
\,dx=-16
\end{equation*}
can be carried out starting with the substitution suggested by GFauxPas $y=x-a-\pi $ in a comment, [edit] splitting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
| \begin{align}
\int_{0}^{1}\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2+b^2}}&=\int_{0}^{\large k_1}\frac{b\sec^2t}{(b^2\tan^2t+a^2)\sqrt{b^2\tan^2t+b^2}}\mathrm dt\tag1\\
&=\int_{0}^{\large k_1}\frac{\cos t}{b^2\sin^2t+a^2\cos^2t}\mathrm dt\tag2\\
&=\int_{0}^{\large k_1}\frac{\cos t}{\left(b^2-a^2\right)\sin^2t+a^2}\mathrm dt\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ M... | We could also start the induction step from the RHS. This way we need only to cope with a square of a binom and not with a cube of a binom.
Induction step $n \rightarrow n+1$:
\begin{align*}
\left(\sum_{i=1}^{n+1}i\right)^2&=\left(\sum_{i=1}^{n}i+(n+1)\right)^2\\
&=\left(\sum_{i=1}^{n}i\right)^2 + 2(n+1)\sum_{i=1}^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}... | I assume the constant $a$ is positive. This is because if $a\lt 0$ there is no minimum.
We equivalently want to minimize $\frac{(x+a)^4}{x^2-a^2}$ with $x\gt a$, or equivalently minimize
$$\frac{(x+a)^3}{x-a}$$
with $x\gt a$. The derivative is
$$\frac{3(x+a)^2(x-a)-(x+a)^3}{(x-a)^2},$$
difficult to get wrong. The top i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Evaluation of $2$ limit problems Evaluation of $(a)\;\;\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
and $(b)\;\; \displaystyle \lim_{x\rightarrow 0^{-}}\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+...............+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+... | Not a full answer, just expanding my comment about $(a)$.
We can rewrite the function (keeping in mind that $x \to \infty$) like this:
\begin{align}
f(x)&:=\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} - \sqrt{x^4 + 2x^3 -cx^2 + 3x - d} =\\
&=x^2\left(1 + \frac a{2x} + \frac 3{2x^2} + \frac b{2x^3} + \frac 1{x^4} - 1 - \frac 1x + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $\int\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\,dx$? I'm trying to solve $$\int_0^1\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ To calculate it I first tried to calculate the primitive function. So let $$\int\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ first I added $+1-1$ to numerator then split it and tr... | Mathematica 11.1 gives that this integral is equal to
$$
\frac{2}{11} \left(i 3^{17/12} \, _2F_1\left(1,\frac{11}{6};\frac{23}{12};\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)-i \sqrt{3} \, _2F_1\left(1,\frac{11}{6};\frac{23}{12};\frac{1}{6} \left(3+i
\sqrt{3}\right)\right)-3+ 3^{23/12}\right)
$$
the function of interes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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show that the series converges. Using the definition, show that the series $\sum_{n=1}^\infty \frac{1}{n(n+3)}$ converges. Determine the limit.
This is what I've managed to do so far:
$\frac{1}{n(n+3)}$ = $\frac{1}{3n}$ -$\frac{1}{3(n+3)}$
$\sum_{n=1}^\infty \frac{1}{n(n+3)}$ = $\sum_{n=1}^\infty \frac{1}{3n}$ -$\frac... | prove that for the partial sum is hold
$$-1/3\, \left( m+1 \right) ^{-1}-1/3\, \left( m+2 \right) ^{-1}-1/3\,
\left( m+3 \right) ^{-1}+{\frac {11}{18}}
$$
and take then the Limit for $m$ tends to infinity
| {
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"url": "https://math.stackexchange.com/questions/1059601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
| After checking that $S(n)=1+2^n+3^3+4^n\equiv0$ mod $10$ for $n=1$, $2$, and $3$ (i.e., computing $S(1)=10$, $S(2)=30$ and $S(3)=100$), induction does the rest, thanks to the difference
$$\begin{align}
S(n+4)-S(n)&=2^n(2^4-1)+3^n(3^4-1)+4^n(4^4-1)\\
&=15\cdot2^n+80\cdot3^n+255\cdot4^4\\
&\equiv5\cdot2^n+0\cdot3^n+5\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
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In triangle ABC, find the range of sinAsinC. In $\triangle$ABC, $\angle$B=$\frac{\pi}{3}$. Find the range of sinAsinC.
I used C=120-A to simplify the equation in sinC and then applied $-1<sinC<1$ but answer did not matched.
| From given, $A+C = \frac{2\pi}3$.
$$\sin A\sin C = \frac{\cos(A-C)-\cos(A+C)}{2} = \frac{\cos(\frac{2\pi}3-2C)-\cos\frac{2\pi}3}{2}$$
Find the maximum and the minimum of the first $\cos$ term.
For maximum, choose $C = \frac\pi3$, which gives $\cos(\frac{2\pi}3-2C) = 1$.
For minimum, choose $C = 0$ or $\frac{2\pi}{3}$,... | {
"language": "en",
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"question_score": "2",
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$4 $ balls selected at random from $ 15$ $15$ balls $ 4$ selected
$1$ blue, $2 $ green, $3$ red, $4$ white, $5$ yellow
What is the probability that $2$ are red and at least $1$ is white?
Now the way this question is worded makes it seem as though you should account for the possibility of the case being $2$ red balls ... | $\frac{\binom{3}{2}\binom{4}{1}\binom{8}{1}}{\binom{15}{4}}+\frac{\binom{3}{2}\binom{4}{2}}{\binom{15}{4}}$
So you are right
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}... | Since
$$\left(\sin{\theta}+\dfrac{1}{\sin{\theta}}\right)^2+\left(\cos{\theta}+\dfrac{1}{\cos{\theta}}\right)^2=5+\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}$$
Use Cauchy-Schwarz inequality we have
$$\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}\ge\dfrac{(1+1)^2}{\sin^2{\theta}+\cos^2{\theta}}=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving uniform continuity and uniform discontinuity Could someone please explain to me how to show uniform continuity and not uniformly continuous for the following:
$f(x) = \frac{1}{x^2}$ for $A = [1, \infty)$ show uniform continuity
$f(x) = \frac{1}{x^2}$ for $B = [0, \infty)$ show that $f$ is not uniformly continu... | *
*$|\frac{1}{x^2}-\frac{1}{y^2}|\leq |\frac{y^2-x^2}{x^2y^2}|=|\frac{(y-x)(y+x)}{x^2y^2}|\leq |\frac{x+y}{x^2y^2}||y-x|\leq (|\frac{1}{xy^2}|+|\frac{1}{yx^2}|)|x-y|\leq 2|x-y|$
2.$\frac{1}{x^2}$ is not uniformly continuous. $|\frac{1}{x^2}-\frac{1}{y^2}|\leq |\frac{y^2-x^2}{x^2y^2}|$ Near $0$ quantity is very large.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find upper and lower bound without using formula? I am studying discrete math for tomorrow's exam and got stuck in the below question. I tried to google it and couldn't find anything usefull.
Prove the following sum is theta(n^2) (we have to find O(n^2) and omega (n^2))
1) $P(n)= 1+2+3+4+\cdots +n$
2) $P(n) =n+(... | For 1) if $n = 2k$, then $P(n) = P(2k) = 1+2+3+\cdots + 2k = (1+2k) +(2+ (2k-1)) + (3+(2k-2)) + \cdots + (k+(2k-(k-1))) = (2k+1) +(2k+1) + \cdots + (2k+1) = k(2k+1) = \dfrac{n}{2}\cdot (n+1) = \dfrac{n(n+1)}{2}$.
If $n = 2k+1$, then:
$1 + 2 + 3 +\cdots + (2k+1) = (1+2+3+\cdots + 2k) + (2k+1) = (1+2+\cdots + +(n-1)) + n... | {
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"source": "stackexchange",
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How can I understand solving the equation? $$\begin{align}
&\left[(\sqrt[4]{p}-\sqrt[4]{q})^{-2} + (\sqrt[4]{p}+\sqrt[4]{q})^{-2}\right] : \frac{\sqrt{p} + \sqrt{q}}{p-q} \\
&= \left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right) \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqr... | $$(A+B)^2+(A-B)^2=(A^2+2AB+B^2)+(A^2-2AB+B^2)=2(A^2+B^2)$$ By using this identity, You can simplify the numerator. $$\dfrac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}=\dfrac{2(\sqrt{p}+\sqrt{q})}{(\sqrt{p} - \sqrt{q})^{2}}$$ Rest will be obvious. Good luck.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How did Sir Isaac Newton develop and formulate the famous binomial theorem? After completing combination, I have started to read Binomial Theorem. My book only mentioned about Pascal's Triangle. And the formula was then given straightforward. But how did Sir Issac Newton actually develop this theorem?? How did he formu... | About the Binomial Theorem.
When expanding the powers of a binomial by hand and grouping the terms by identical powers, it is not very hard to observe the pattern:
$$(x+y)^0=1$$
$$(x+y)^1=x+y$$
$$(x+y)^2=x^2+2xy+y^2$$
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
$$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
$$\dots$$
*
*the expansion ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I prove $\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$ Can the residue theorem prove this?
$$\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$$
| Hint
$$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$
and $$(a-b)^3=(a^3-b^3)-3ab(a-b)$$
so
\begin{align*}\dfrac{1}{n^3(n+1)^3}&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-3\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\left(\dfrac{1}{n}-\dfrac{1}{(n+1)}\right)\\
&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-\dfrac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to transform the product to sum? I just wonder that how to prove that
$$
\prod_{m=1}^{n}\Big(x-2\cos\frac{m\pi}{n+1}\Big)=\sum_{k=0}^{[n/2]}(-1)^{k}\binom{n-k}{k}x^{n-2k}.
$$
Similarly, how to transform the product
$$
\prod_{m=1}^{n}\Big(x+2\cos\frac{2m\pi}{n}\Big)\overset{?}{=}\sum_{k}b_k x^k?
$$
| Let, $\displaystyle P(x) = \sum\limits_{k=0}^{[n/2]} (-1)^k\binom{n-k}{k}x^{n-2k}$
We show, $\displaystyle P(\omega + \omega^{-1}) = 0$ where, $\displaystyle\omega = \exp{\frac{2\pi i}{2n+2}}$ is a $(2n+2)^{th}$ roots of unity.
$\begin{align} P(\omega+ \omega^{-1}) &= \sum\limits_{k=0}^{[n/2]} (-1)^k\binom{n-k}{k}(\ome... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration without complex analysis on rational-improper integral Evaluate:
$$\int_{0}^{\infty} \frac{1}{x^6 + 1} \,\mathrm dx$$
Without the use of complex-analysis.
With complex analysis it is a very simple problem, how can this be done WITHOUT complex analysis?
| Setting $x=\dfrac1y,$
$$I=\int_0^\infty\frac1{1+x^6}dx=\cdots=\int_0^\infty\frac{x^4}{1+x^6}dx$$
$$2I=\int_0^\infty\frac{1+x^4}{1+x^6}dx=\int_0^\infty\frac{(1+x^2)^2-2x^2}{1+x^6}dx$$
$$=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx-\frac23\int_0^\infty\frac{3x^2}{1+x^6}dx$$
$$I_1=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx=\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x... | By definition, if $r$ is a nonnegative real number, then $\sqrt{r}$ denotes the unique nonnegative real number whose square is $r$. Note that $\sqrt{r}$ and $-\sqrt{r}$ both have the property that their square is $r$, but there is only one nonnegative square root.
If $x$ is any real number, then $x^2$ is nonnegative, s... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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For a Diophantine equation $x^2+py^2=z^2$ show that $z$ is necessarily odd. For a Diophantine equation $x^2+py^2=z^2$ where $p$ is a prime of the form $p\equiv 1(mod4)$ and $(x,y,z)=1$.
Show that $z$ is necessarily odd.
| Suppose to the contrary that $z$ is even.
If $y$ is even, then $x^2$ is even, so $x$ is even, contradicting the fact that $\gcd(x,y,z)=1$. Similarly, if $x$ is even, then $py^2$ is even, and since $p$ is odd $y^2$ is even, and therefore $y$ is even, and again $\gcd(x,y,z)\gt 1$.
So $x$ and $y$ are both odd. Thus $x^2\... | {
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Why do we have $u_n=\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}}=O(\frac{1}{n^3})$? Why do we have
*
*$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=O\left(\dfrac{1}{n^3}\right)$
*$u_n=e-\left(1+\frac{1}{n}\right)^n\sim \dfrac{e}{2n}$
any help would be appreciated
| Hint: $\left(\dfrac{1}{n^2-1}\right)^{1/2} = \left(n^2-1\right)^{-1/2} = \dfrac{1}{n}\left(1-\frac{1}{n^2}\right)^{-1/2} = \dfrac{1}{n} + \mathcal{O}\left(\frac{1}{n^3}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stochastic matrix A^n=const. Let $A\in M(n,n)$ be a doubly stochastic matrix such that
$$
A^k=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n\\
\end{pmatrix}
$$
for some $k>1$. Does it follow that
$$
A=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n\... | No, it does not follow that
$$
A=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n
\end{pmatrix}.
$$
Here is a (minimal) counterexample. Take
$$
J_3=\begin{pmatrix}
1/3 & 1/3 & 1/3\\
1/3 & 1/3 & 1/3\\
1/3 & 1/3 & 1/3
\end{pmatrix},\qquad N=\begin{pmatrix}
1 & 1 & -2\\
-1 & -1 & 2\\
0 & ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
| The number $\frac{1}{\sqrt{2}}$ is defined to be the number such that, when you multiply it by $\sqrt{2}$, you get $1$. Symbolically, it is the solution to $x\cdot \sqrt{2}=1$. That's how division works - it's the inverse of multiplication. What is $\frac{\sqrt{2}}2\cdot \sqrt{2}$? Why, it's $\frac{2}2=1$ - so it must ... | {
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"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$ $a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$
I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next
| Let $x=2^a,y=2^b,z=2^c$. Then $x,y,z>0$ and
$$ xyz=2^{a+b+c}=1.$$
By Holder's inequality
$$(x^3+y^3+z^3)(1+1+1)(1+1+1)\ge (x+y+z)^3.$$
Therefore,
$$x^3+y^3+z^3\ge\dfrac{(x+y+z)^3}{9}\ge (x+y+z)$$
because use AM-GM inequality
$$(x+y+z)^2\ge (3\sqrt[3]{xyz})^2=9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Removing the root squares from this expression? I would like to understand how to remove the root squares from this expression:
$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$
How to do it?
| Here's a push
You can try rationalising it by multiplying the expression by its conjugate.
$$\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}\times\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$
and you can continue it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$ Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$
| By AM-HM or C-S we have
$$\frac{\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}}{3} \geq \frac{3}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$
Therefore
$$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}} \geq \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$
To prove your claim, you need to ... | {
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"source": "stackexchange",
"question_score": "5",
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Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there... | Reorganize the term like $$(x^4+4+4x^2)+(3x^3+6x)-4x^2=(x^2+2)^2+3x(x^2+2)-4x^2$$ than it is easy to come up with $$(x^2+4x+2)(x^2-x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 4,
"answer_id": 0
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Simple limit problem without L'Hospital's rule $$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}$$
We are not supposed to use any derivatives yet, but I can't find any formula that helps here. It's a $\frac{0}{0}$ indeterminate form, and all I think of doing is
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3... | HINT:
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} =\frac{x^4+1-2}{x-1}\cdot\frac{\sqrt[3]{x^2}+\sqrt[3]x+1}{\sqrt{x^4 + 1} + \sqrt{2}}$$
Now as $x\to1,x\ne1,x-1\ne0,$ so we can safely cancel out $x-1$
| {
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"source": "stackexchange",
"question_score": "6",
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Let $a_n=2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt{k}}$. Show $a_n$ converges and $1<\lim_{n\to\infty}a_n<2$. I am able to show that $\lim_{n\to\infty}a_n\le 2$, but I can't think of a way to show strict inequality.
Here's my work so far:
By Riemann sums,
$$\int_1^{n+1}\frac{1}{\sqrt x}\,dx<\sum_{k=1}^n\frac{1}{\sqrt k}<\int... | You just have to make your initial estimate more quantitative:
\begin{eqnarray*}
2\sqrt{n+1}-2-\sum_{m=1}^{n}\frac{1}{\sqrt{m}} & = & \int_{1}^{n+1}\frac{1}{\sqrt{x}}\,{\rm d}x-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}\\
& = & \sum_{m=1}^{n}\int_{m}^{m+1}\underbrace{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{m}}}_{\leq0}\,{\rm d}x\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Finding the limit of $\lim_{n\rightarrow\infty}\frac{1}{n}[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n-1}{n})^{2}] $ Here's my try:
$\lim_{n\rightarrow\infty}\frac{1}{n}[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n-1}{n})^{2}] = $$\lim_{n\rightarrow\infty}[\frac{1}{n}(a^{2}+\frac{2a}{n}+\frac{1}{... | Did you use that
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{\color{red}{6}}\quad?$$
Remark: You can simply find the result using the Riemann sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $ \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ and probability I begin with this problem:
Calculate the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ when $n\rightarrow \infty$.
It's natural to think of recurrence relation. Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=I_n$. By integration by pa... | Here is a possibly simpler way for proving your theorem. Since $\sin x$ is strictly increasing in the range $[0,\pi/2]$, for each $\epsilon > 0$ we can bound
$$
\begin{align*}
\int_0^{\pi/2} \sin^n x \, dx &= \int_0^{\pi/2-\epsilon} \sin^n x \, dx + \int_{\pi/2-\epsilon}^{\pi/2} \sin^n x \, dx \\ &\leq \frac{\pi}{2} \s... | {
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"url": "https://math.stackexchange.com/questions/1095308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire... | Let's examine a simple identity:
$$9+10+11+12=13+14+15$$
Notice that there are $4$ terms on the left and $3$ on the right - so let's "use up" one of the terms on the left. We can split $9=3+3+3$ and distribute these additional $+3$ terms to the other $3$ summands on that side to get:
$$(10+\color{red} 3)+(11+\color{red... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{... | **Hint: **$$\sum_{n=1}^\infty \frac{1}{(n+1)!} = \sum_{n=2}^\infty \frac{1}{n!} \\ = \sum_{n=0}^\infty \frac{1}{n!}-\frac{1}{1!}-\frac{1}{0!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
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series convergence. Dirichlet test $$\sum(-1)^{\lfloor\frac{n^3+n+1}{3n^2-1}\rfloor}\frac{\ln(n)}{n}$$
I thought about using Dirichlet test. $\ln(n)/n$ is a decreasing sequence that tends to 0 but I have problem with proving it. I also can
t tell anything about $(-1)^{\lfloor\frac{n^3+n+1}{3n^2-1}\rfloor}$.
| To prove that $\frac{\log(n)}{n}$ is decreasing, notice that:
$$\frac{\log(n+1)}{n+1}=\frac{\log n}{n+1}+\frac{\log\left(1+\frac{1}{n}\right)}{n+1}\leq \frac{\log n}{n+1}+\frac{1}{n(n+1)}$$
hence if $n> e,$
$$\frac{\log(n+1)}{n+1}< \log n\left(\frac{1}{n+1}+\frac{1}{n(n+1)}\right)=\frac{\log n}{n}.\tag{1}$$
To prove th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| *
*Struggle to find a real root. By trying small positive and negative integers, you find that $P(-2)=0$. So $x+2$ is a factor.
*Apply the change of variable $y=x+2$:
$$P(x)=P(y-2)=(y-2)^3+4(y-2)^2+(y-2)+6=y^3-2y^2-3y.$$
Now you easily factor as
$$P(y-2)=y(y^2-2y-3),$$
and revert to the original
$$P(x)=(x+2)\left((x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to decompose this matrix exponential? I would some help with the steps to decompose the below matrix exponential.
$\exp\left[ \zeta \left ( \begin{matrix}
-\cos(x) & i \sin(x) \\
-i \sin(x) & \cos(x)
\end{matrix} \right ) \right] = \left ( \begin{matrix}
\cosh(\zeta)-\sinh(\zeta)\cos(x) & i \sinh(\zeta)\sin(x) \\... | First, check your equality when $\sin(x)=0$.
Second, notice that $\zeta\begin{pmatrix}-\cos(x)& i\sin(x)\\-i\sin(x)& \cos(x)\end{pmatrix}=\zeta\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}-\cos(x)& \sin(x)\\\sin(x)& \cos(x)\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}$
and $\begin{pmatrix}i& 0\\ 0& 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Reverse engineering a Taylor expansion We have the sum: $$S(x) = \frac{x^4}{3(0!)} + \frac{x^5}{4(1!)} + \text{ }...$$ And we are told to sum the series to obtain a finite expression. My guess was to reverse engineer the expression in order to find a function which has $S(x)$ or some form of it as a taylor expansion. $... | $$\begin{eqnarray*}S(x)&=&x^4\sum_{n\geq 0}\frac{x^j}{j!}\cdot\frac{1}{j+3}=x^4\sum_{n\geq 0}\frac{x^j}{j!}\int_{0}^{1}y^{j+2}\,dy=x^4\int_{0}^{1}y^2\sum_{n\geq 0}\frac{(xy)^j}{j!}\,dy\\&=&x^4\int_{0}^{1}y^2 e^{xy}\,dy=x\int_{0}^{x}z^2 e^z\,dz=x\cdot\left[\left(z^2-2z+2\right)e^{z}\right]_{0}^{x}\\&=&x\left(-2+(x^2-2x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding Cartesian coordinates of remaining vertices of triangle, given a vertex and angle from y-axis I have an isosceles triangle $ABC$, where the height $h$ and angle at vertex $A$ are known. The Cartesian coordinates of vertex $A$ are also known to be $\left(x,y\right)$.
If the angle between the y-axis and the line ... | Absolutely.
The angle between the $y$-axis and $AB$ is $\theta - \tfrac{A}{2}$, and the angle between the $y$-axis and $AC$ is $\theta + \tfrac{A}{2}$.
So you will get
\begin{align*}
B &= (x - r \sin(\theta - \tfrac{A}{2}), y + r \cos(\theta - \tfrac{A}{2}) \\
C &= (x - r \sin(\theta + \tfrac{A}{2}), y + r \cos(\theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Complex Solution I was looking at a simple complex problem and I came by this:
Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\fra... | When you write $z=x+yi$ you are looking for real $x$ and $y$.
There's really nothing bad in considering also the other two possibilities, but they will give just the same numbers.
Indeed, if you take $x=i$, you obtain, from $xy=-2$, that $y=-2/i=2i$ and so
$$
z=x+yi=i+2i^2=-2+i
$$
which has already been obtained. The o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$
$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}... | Hint: We are finding
$$\sum_1^\infty \frac{1}{2}\left(\left(\frac{1}{n}-\frac{1}{n+1}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right) \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn... | I will address both parts of your question:
Part 1: Proving that $a_n>0$.
From your final form of $a_n$
$$a_n=\frac{2n-2\sqrt{(n-1)(n+1)}}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}$$
we can notice that the denominator is always positive (when $n\ge1$), so it remains to show that $2n>2\sqrt{(n-1)(n+1)}=2\sqrt{n^2-1}$. This is q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Given $x,y,z>0$: $\frac{2}{3x+2y+z+1}+\frac{2}{3x+2z+y+1}=(x+y)(x+z)$. Find Minimum Value Of: $P=\frac{2(x+3)^2+y^2+z^2-16}{2x^2+y^2+z^2}$
Given $x,y,z>0$: $\frac{2}{3x+2y+z+1}+\frac{2}{3x+2z+y+1}=(x+y)(x+z)$ $(1)$
Find Minimum Value Of:
$P=\frac{2(x+3)^2+y^2+z^2-16}{2x^2+y^2+z^2}$
I found $2x+y+z\geq 2$ from (1) but... | I don't know why I have this reslut:
since Use Cauchy-Schwarz inequality,we have
$$\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}$$
so
$$\dfrac{2}{3x+2y+z+1}+\dfrac{2}{3x+2y+z+1}\ge \dfrac{8}{6x+3y+3z+2}\ge \dfrac{8}{6x+3\sqrt{2(y^2+z)}+2}$$
other hand
$$(x+y)(x+z)\le\dfrac{(2x+y+z)^2}{4}\le\dfrac{(2x+\sqrt{2(y^2+z^2)})^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Analyze the continuity of the following function Here in my book I have such an exercise with the explanation given below, but still there is something the authors didn't add, but simply put "...after some operations...". Here is such an exercise:
Analyze the continuity of the following function:
$$f:\Bbb R^2\to\Bbb R... | if $|y|\leq 1$, $$\sqrt{x^4+y^2}\geq \sqrt{x^4+y^4}$$
and thus
$$|f(x,y)|=\left|\frac{3x^2y}{\sqrt{x^4+y^2}}\right|\leq\underbrace{\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|}_{\leq 1}|3y|\leq |3y|\underset{(x,y)\to(0,0)}{\longrightarrow 0}$$
therefore $f$ is continuous on $\mathbb R^2$.
Justification for $\left|\frac{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a}$ Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a},a \geq 0.$
One way to do it is using the formula
$$
a^3+2 - a^2-2 \sqrt{a}=(\sqrt{a}-1)^2(1+(a+1)(\sqrt{a}+1)^2) \geq 0.
$$
But I hope there is a better way.
| Substitute $a=b^2$, then $a^3-a^2-2\sqrt a+2=b^6-b^4-2b+2$, which has $1$ as a root, so we can factor it out:
$$b^6-b^4-2b+2=(b-1)(b^5+b^4-2)$$
The polynomial $b^5+b^4-2$ has still one as a root, so factor more:
$$b^5+b^4-2=(b-1)(b^4+2b^3+2b^2+2b+2)$$
So $a^3-a^2-2=(b-1)^2(b^4+2b^3+2b^2+2b+2)$, which is nonnegative for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ where $f(n+1) = f(f(n))+1$ Consider checking function $\mathbb{N}\to \mathbb{N}$ relationship $f(n+1) = f(f(n))+1$, for any positive integer $n$.
Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ for any positive inte... | i am going to try. the defining relation for $f$ gives us the inequality $$nf(n) = n\{f(f(n-1)) + 1\} =n + nf(f(n-1)) \ge n $$ because the range of $f$ consists of positive integers.
we can use $jf(j) \ge j$ again and again to get
$$ nf(n) \ge n + nf(f(n-1)) \ge n + nf(n-1) \ge n + n(n-1) = n^2$$
therefore
$$1f(1) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$ $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$
Attempt: $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = \dfrac {2n+2-1}{n^2(n+1)^2}$
$S_1=2 \sum_{n=1}^\infty [\dfrac {n+1}{n^2(n+1)^2} ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
$=2 \sum_{n=1}^\infty [\dfra... | First note that $$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]$$
By telescoping the series, we can see that
$$
\lim\limits_{z\to\infty}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\dots +\frac{1}{z^2}-\frac{1}{(z+1)^2}\right]
$$
$$
= \lim\limits_{z\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Closed form as sum and combinatorial of Fibonacci numbers How can I prove that the Fibonacci numbers that are defined as $F_n=F_{n-1}+F_{n-2}, \; n \geq 2$ and $F_0=0,\ F_1=1,\ F_2=1$ have the form:
$$F_n=\sum_{k=0}^{n-1} \binom{n-1-k}{k}, \; n\ge 2 $$
I am aware of the gen. function that is:
$$\frac{x}{1-x-x^2} =\sum_... | This one can also be done using complex variables.
Suppose we seek to show that
$$F_n = \sum_{k=0}^{n-1} {n-1-k \choose k}.$$
Introduce the integral repesentation
$${n-1-k\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1-k}}{z^{k+1}} \; dz.$$
This gives for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1... | $$\begin{eqnarray*}\sum_{n=1}^{N}\left(2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\right)&=&\color{red}{\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n-1}\right)}+\color{blue}{\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n+1}\right)}\\&=&\color{red}{\sqrt{N}}+\color{blue}{1-\sqrt{N+1}}\\&=&1-\frac{1}{\sqrt{N}+\sqrt{N+1}},\tag{1}\end{eqnarray*}$$
so, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find the range of $ x-\sqrt{4-x^2}$ $Y=x-\sqrt{4-x^2}$. How to find these types of functions' range?
I just know that the answer is $R=\{y\in\mathbb{R}\mid-2\sqrt{2}\leq y\leq 2\}$, but I have no idea how to find it step by step.
| Hint
Let
$x=2\cos{t}$,then
$$x-\sqrt{4-x^2}=2\cos{t}\pm 2\sin{t}$$
case 1
if $t\in[0,\pi]$,then
$$x-\sqrt{4-x^2}=2\cos{t}-2\sin{t}=2\sqrt{2}\cos{\left(t+\dfrac{\pi}{4}\right)}\in [-2\sqrt{2},2] $$
if case 2,$t\in[\pi,2\pi]\Longrightarrow \dfrac{5}{4}\pi\le t+\dfrac{\pi}{4}\le\dfrac{9\pi}{4}$
then
$$x-\sqrt{4-x^2}=2\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Range of trigonometric functions I would like to know if there is a simple approach to find the range of functions in the form:
$$\sin x\sin2x$$
$$\cos x\cos3x$$
$$\sin 2x\cos 4x$$
For example, finding the range of a function in the form:
$$a\cos\theta + b\sin\theta$$ is simple (the minimum value is $-\sqrt{a^2 + b^2}... | We have always $ a^2\leq a^2+b^2\ \Rightarrow\ |a|\leq\sqrt{a^2+b^2} \Rightarrow \frac{|a|}{\sqrt{a^2+b^2}}\leq 1\Rightarrow-1\leq\frac{a}{\sqrt{a^2+b^2}}\leq1$ so there exists a $\alpha $ that $ \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. Beside We know that $\sin\alpha=\pm\sqrt{1-\cos^2\alpha}=\pm\sqrt{1-(\frac{a}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\in... | There are a couple of ways to see this. Firstly, draw a right triangle, call it $ABC$ (with $C$ being the right angle), with side lengths $a$, $b$ and $c$ with the usual convention. Then $\arcsin(\frac{b}{c})$ is the measure of the angle $CBA$. Additionally, $\arccos(\frac{b}{c})$ is the angle of the angle of the oppos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
So first we have two potentially problematic points which are $1,\infty$
We split the integral to $$\int_1^2 \frac{x\ln x}{x^4-1} dx + \int_2^\infty \frac{x\ln x}{x^4-1} dx$$
Now first I t... | For showing, that $I_{2}$ converges, you might use $\ln(x)\leq x$, which is valid for $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ an... | An easy approach is to simply construct an ellipse with its upper half satisfying the above conditions.
An ellipse is defined via $2$ numbers $a$ and $b$ which are each the half of the major and minor axis of the ellipse.
Then all points $(x,y)$ which suffice the following equation are on the ellipse:
$$ \frac{x^2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
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Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$. Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$.
I came up with the formula $\displaystyle\sum_{r=1}^{2n} (-1)^{r+1}(2r-1)^2=-8n^2$ but I got stuck proving it by ind... | First, show that this is true for $n=1$:
$\sum\limits_{r=1}^{2}(-1)^{r+1}\cdot(2r-1)^2=-8$
Second, assume that this is true for $n$:
$\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2=-8n^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{r=1}^{2(n+1)}(-1)^{r+1}\cdot(2r-1)^2=$
$\color{red}{\sum\limits_{r=1}^{2n}(-1... | {
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"url": "https://math.stackexchange.com/questions/1123189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating function for Pell numbers Problem: The Pell numbers $p_n$ are defined by the recurrence relation \begin{align*} p_{n+1} = 2p_n + p_{n-1} \end{align*} for $n \geq 1$. The initial conditions are $p_0 = 0$ and $p_1 = 1$.
a) Determine the generating function \begin{align*} P(x) = \sum_{n=0}^{\infty} p_n x_n \en... |
The radius of convergence is the distance to the nearest singularity
For $(b)$ you can advance as (based on your calculations) using partial fraction
$$P(x)= \frac{x}{(1-2x-x^2)}= \frac{A}{x-a} + \frac{B}{x-b} $$
where $a,b$ are the roots of $1-2x-x^2$ and $A$ and $B$ need to be determined. The calculations gives: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The equation $x^4+y^4=z^2$ has no integer solution The equation $$x^4+y^4=z^2$$ has no integer solution for $(x, y, z), x \cdot y \neq 0 , z >0$.
We suppose that there is a solution $(x, y, z)$.
We consider the set $$M=\{z \in \mathbb{N} | \exists x, y \in \mathbb{Z}: x^4+y^4=z^2, x \cdot y \neq 0 \} \subseteq \math... | Since $x,z$ are odd, certainly $z\pm x^2$ are even, hence the gcd is at least $2$. Any common divisor of $z+x^2$ and $z-x^2$ also divides their sum $2z$ and their difference $2x^2$ and their product $y^4$, i.e., $\gcd(z-x^2,z+x^2)\mid \gcd(2x^2,y^4,2z)\mid 2\gcd(x^2,y^4,z)\mid 2\gcd(x^4,y^4,z^4)=2\gcd(x,y,z)^4=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta
I am having difficulty finding the $\delta$ value.
Here is what I have done so far:
What I want to show:
$$\forall \epsilon > 0, \exists \delta > 0 \; such \; that \; 0<\mid x -2 \mid < \d... |
$$\text{Write the polynomial as a sum of powers of $x-2$.}$$
$$x^4-2x^3+x+3=(x-2)^4+6(x-2)^3+12(x-2)^2+9(x-2)+5$$
The most efficient way to do this is to apply Ruffini-Horner repeatedly with the polynomial with the value $x=2$.
Then $$|x^4-2x^3+x+3-5|=|(x-2)^4+6(x-2)^3+12(x-2)^2+9(x-2)|\leq(1+6+12+9)|x-2|$$
when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$ $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1}) = ?$
I don't know how to solve the indetermination there... is it possible to rearrange the expression in brackets in order to use L'Hospital or Taylor Series?
| Hint: $$x^2-\sqrt{x^4-x^2+1}=\left(x^2-\sqrt{x^4-x^2+1}\right)\cdot\frac{x^2+\sqrt{x^4-x^2+1}}{x^2+\sqrt{x^4-x^2+1}}=\frac{(x^2)^2-\left(\sqrt{x^4-x^2+1}\right)^2}{x^2+\sqrt{x^4-x^2+1}}$$
Once you have that simplified, multiply by $\dfrac{1/x^2}{1/x^2}$ and recognize that $x^2=\sqrt{x^4}$ to distribute the $1/x^2$ 'int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplification of Complex Number. I would appreciate any hints for the following problem:
Given that
$z=\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\cos^2\theta}$
show that $\vert z\vert=2\sin\theta$ and arg $z=\theta$
Update:
Using John's suggestion I now have
$\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\co... | I'm assuming $0 \le \theta < \pi/2$. Since \begin{align}(1 - \cos 4\theta) + i\sin 4\theta &= 2\sin^2 2\theta + i(2\sin 2\theta \cos 2\theta)\\
& = 2\sin 2\theta (\sin 2\theta + i \cos 2\theta)\\
& = 2\sin 2\theta e^{i(2\theta - \pi/2)}
\end{align}
and
$$\sin 2\theta + 2i\cos^2 \theta = 2\sin \theta \cos \theta + 2i\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2 + p^2 = b^2$ then $2(a+p+1)$ is a perfect square We are given $$ a^2 + p^2 = b^2 $$ where $a,b\in\mathbb{Z}$ and $p$ is prime. We are to show that $$2(a+p+1)$$ is a perfect square. Is there any elegant ways to go about this problem? Struggling to find a proof myself. Thanks in advance.
| We have $a^2 + p^2 = b^2$ so $p^2 = b^2-a^2 = (b+a)(b-a)$.
Therefore $b-a=1$ and $p^2 = b+a=2a+1$
Therefore $2(a+p+1) = p^2 + 2p + 1 = (p+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Inclusion and exclusion in combinatorics You have 15 identical balls and must divide them into 4 drawers stacked on top of each other with the following limitations:
*
*You have at least 2 balls in each drawer
*There will be no more than 5 balls in the top drawer and there will be no more than 5 balls in the bottom... | Generating Function Approach
Inclusion-Exclusion is what the question asks for, and there are some good answers using inclusion-exclusion, but here is another approach using generating functions that gives the same answer as inclusion-exclusion.
By taking the products of the generating functions for the number of ways... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$... | If $\cos4x=0,4x=(2n+1)\dfrac\pi2$ where $n$ is any integer
$x=(2n+1)\dfrac\pi8$ where $n\equiv0,1,2,3\pmod4$
Now $\cos4x=2\cos^22x-1=\cdots=8\cos^4x-8\cos^2x+1$
So, the roots of $8c^4-8c^2+1=0$ are $\cos(2n+1)\dfrac\pi8$ where $n\equiv0,1,2,3\pmod4$
So, the equation whose roots are $1+\cos(2n+1)\dfrac\pi8=y\iff\cos(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
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Legendre polynomial to show identity, can't spot mistake Using Legendre polynomial generating function
\begin{equation}
\sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}}
\end{equation}
Or $$ P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n] $$
Show$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2} $... | \begin{equation}
\begin{split}
P_n(x) &= \frac{1}{2^nn!}[(x+1)^n\cdot(x-1)^n]^{(n)}
\\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)}
\\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k
\\ &= \frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I ca... | let $y = \dfrac{x(x+1)^{x+1}}{(x+2)^{x+2}},$ then
$\begin{align} \ln y &= \ln x + (x+1)\ln(x+1)-(x+2)\ln(x+2)\\
&=\ln x +(x+1)[\ln x + \ln(1 + 1/x)] -(x+2)[\ln x+ \ln(1 + 2/x)]\\
&=\left(1+x+1-x-2 \right)\ln x +(x+1)\left(1/x + \cdots\right)-(x+2)\left(2/x +\cdots\right)\\
&=-1 + \cdots
\end{align}$
so $$\lim_{x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
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The set $\{1,2,3,\ldots,n\}$, where $n \geq 5$, can be divided into two subsets so that the sum of the first is equal to the product of the second A peer of mine showed me earlier today this problem, taken from a 7th grade math contest :
Let $A=\{1,2,3,\ldots,n\}$; (where $n \geq 5$) prove that $A$ can be divided int... | The sum of all the numbers is
$$
\sum_{k=1}^nk=\frac{n(n+1)}2\tag{1}
$$
We will find $a$ and $b$ so that
$$
\overbrace{\left(\sum_{k=1}^nk\right)-a-b-1}^{\text{sum of all but $a$, $b$, and $1$}}=\overbrace{\vphantom{\left(\sum_1^n\right)}1ab}^{\text{product}}\tag{2}
$$
That is, we need to find $a$ and $b$ so that
$$
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 0
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Problem with infinite series sum I'm trying to calculate the infinite sum of the series
$$\sum_{i=1}^{n}\frac{(i-1)^2}{n^3}. $$
Computing it with WolframAlpha results in $$\frac{2n^2-3n+1}{6n^2}.$$
I'm trying to solve it by considering it as an arithmetic series and using the formula for the nth $$\sum_{i=1}^{n}a_i=\... | $$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\\0^2+1^2+2^2+...+(n-1)^2=\frac{(n-1)(n-1+1)(2(n-1)+1)}{6}=\\\frac{n(n-1)(2n-1)}{6}$$so $$ \sum_{i=1}^{n}\frac{(i-1)^2}{n^3}=\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2=\\\frac{\frac{n(n-1)(2n-1)}{6}}{n^3}=\\\frac{(n-1)(2n-1)}{6n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{... | Hint:
$$
\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}=\lim _{n\to \infty}\sqrt{\dfrac{n+\sqrt{n^2+1}}{3n+1}}.
$$
So you only need to determine the limit of the term inside the square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int... | $\displaystyle\int\sqrt{\frac{x}{x+1}}dx$
$\displaystyle x=y^{2}-1\Rightarrow dx=2ydy$
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\int\frac{\sqrt{y^{2}-1}}{y}2ydy=2\int\sqrt{y^{2}-1}dy$
$\displaystyle y=ch(z)\Rightarrow dy=sh(z)dz$
$\displaystyle sh(z)=\sqrt{ch^{2}(z)-1}=\sqrt{y^{2}-1} $
$\displaystyle\int\sqrt{\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
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From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
| you can do $\dfrac{1-\cos x}{\sin x} = \dfrac{\cos 0 - \cos x}{\sin 0 + \sin x}=\dfrac{2\cos x/2 \cos x/2}{2 \sin x/2 \cos x/2} = \tan(x/2).$
we used $\cos a -\cos b = 2\cos (a-b)/2 \cos(a+b)/2$ and $\sin a + \sin b = 2 \sin(a+b)/2 \cos (a-b)/2$ with $a = 0$ and $b = x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How can I prove that $2/9$x$ and $y$ are real numbers.
Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?
Then can I use that to prove that for any natural number $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$
| $x^2 - xy + y^2 = 1$ is an ellipse. to find out the minor and major axis, we change variables $$x = \xi \cos t - \eta \sin t, y = \xi \sin t + \eta \cos t$$ where $t$ will be determined later.
the ellipse in $\xi, \eta$ coordinates is $$\xi^2(\cos^2 t -\sin t \cos t+\sin^2 t) + \xi \eta(-2\cos t \sin t+\sin^2 t - \cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$ $$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$
Can someone help me to solve it?
result of online calculator: 3/5
| Rewrite $$A=\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{\sqrt{x}\sqrt{1+1/x}-\sqrt{x}\sqrt{1-2/x}}{\sqrt{x}\sqrt{1+2/x}-\sqrt{x}\sqrt{1-3/x}}=\frac{\sqrt{1+1/x}-\sqrt{1-2/x}}{\sqrt{1+2/x}-\sqrt{1-3/x}}$$ Replace $\frac 1x$ by $y$; so $$A=\frac{\sqrt{1+y}-\sqrt{1-2y}}{\sqrt{1+2y}-\sqrt{1-3y}}$$ Now, use th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Finding $\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})} $ I'm having trouble understanding how the $\displaystyle\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})}$. I used the product law to set it up as $\displaystyle\frac{\lim_{(x,y)\rightarrow (0,0)} \ta... | $$\lim\limits_{(x,y)\to (0,0)} \frac{\tan\left(x^2+y^2\right)}{\arctan\left(\frac{1}{x^2+y^2}\right)}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\arctan\left(\frac{1}{r^2\cos^2\phi+r^2\sin^2\phi}\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\tan\left(r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make ... | First, show that this is true for $n=1$:
$2^{4+2}+3^{1+2}=13\cdot7$
Second, assume that this is true for $n$:
$2^{4n+2}+3^{n+2}=13k$
Third, prove that this is true for $n+1$:
$2^{4(n+1)+2}+3^{(n+1)+2}=$
$2^4\cdot2^{4n+2}+3^1\cdot3^{n+2}=$
$16\cdot2^{4n+2}+3\cdot3^{n+2}=$
$(13+3)\cdot2^{4n+2}+3\cdot3^{n+2}=$
$13\cdot2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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How to prove $\sum_{k=0}^{\infty}k^2x^{k} = \frac{x(1+x)}{(1-x)^3}\text{, }|x| < 1$? How do I prove that the summation $$\sum_{k=0}^{\infty}k^2x^{k} = \dfrac{x(1+x)}{(1-x)^3}\text{, }|x| < 1\text{?}$$
| $$\sum_{k=0}^{\infty}x^k = \frac{1}{1-x}$$
Taking derivatives both sides with respect to $x$, the first and the second derivatives yield respectively:
$$\sum_{k=1}^{\infty}kx^{k-1} = \frac{1}{(1 - x)^2}$$
$$\sum_{k=2}^{\infty}k(k-1)x^{k-2} = \frac{2}{(1-x)^3}$$
Now:
$$\sum_{k=0}^{\infty}k^2x^k = \sum_{k=1}^{\infty}k^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$
Prove that the following inequality
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
holds for arbitrary real numbers $a$, $b$ and $c.$
Someone says, "It's very easy problem. It can also... | By Minkowski's inequality,
\begin{align}&\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - a)^2} \\
&= \|(a,1-b)\|_2 + \|(b,1-c)\|_2 + \|(c,1-a)\|_2\\
&= \frac{\|(a,1-b)\|_2 + \|(1 - c,b)\|_2}{2} + \frac{\|(b,1-c)\|_2 + \|(1-a,c)\|_2}{2}\\
& + \frac{\|(c,1-a)\|_2 + \|(1-b, a)\|_2}{2}\\
&\ge \frac{\|(a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Convolutions and the Gaussian distribution Suppose $X_1$ and $X_2$ are independent random variables each with the standard Gaussian distribution. Compute, using convolutions, the density of the distribution of $X_1 + X_2$ and show $X_1 + X_2 = \sqrt{2}X$ where X has standard Gaussian distribution.
I wrote down the dens... | For $X, Y\sim N(0,1)$, we have
$${f_{X + Y}}\left( x \right) = \int_{ - \infty }^\infty {{f_X}\left( {x - y} \right){f_Y}\left( y \right)dy} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - \frac{1}{2}{{\left( {x - y} \right)}^2}}}{e^{ - \frac{1}
{2}{y^2}}}dy} = \frac{1}{{2\pi }}\int_{ - \infty }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use proof by induction to prove $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$ Use proof by induction to prove that that $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$
Inductive hypothesis: Assume there exists $k\in \mathbb{N}$ s.t.
$$ \frac{1}{k... | Rewrite as$$n!\ge2^n.$$
Then
$$4!\ge2^4$$
and
$$n!\ge2^n\land n+1\ge2\implies(n+1)!\ge2^{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Solving quadratic diophantine equations in two variables I've looked at the recommended questions, but none of them seem to match my question.
Consider the equation $2015 = \frac{(x+y)(x+y-1)}{2} - y + 1$.
This can trivially be simplified to $4030 = x^2 + 2xy - x + y^2 - 3y +2$.
According to Wolfram Alpha, the integer ... | The equation is
$$x^2 + 2xy + y^2 - x -3y - 4028 = 0.$$
This can be modified to
$$(x+y)^2 - (x+y) - (2y+4028) = 0.$$
Let's denote $u = x+y$ so we have
$$u^2 - u = 2(y+2014)$$
Solving for $y$, we have
$$y = \frac{u^2-u}{2}-2014.$$
Then, remembering that $u=x+y$, we can solve for $x$:
$$x = u-y = u-\frac{u^2-u}{2}+2014 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Adding distances/weights to absorbing markov chain in presence of an absorbing state, I want to calculate mean/expected 'distance' from any state to that absorbing state. What I mean by distance is that I want to give different lengths from one particular state to another.
For example, below i wrote a simple coin toss,... | One way to approach this is to create another Markov chain where the states are the transitions that have positive probability in the original Markov chain, plus one state for the initial state. We only need to include transitions from transient states since that's all we're interested in. So we exclude the transition ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
System of Recurrence Relations Solve the following System of Recurrence Relation:
$$a_n = 2a_{n-1} - b_{n-1} + 2, a_0 = 0$$
$$b_n = -a_{n-1} + 2b_{n-1} - 1, b_0 = 1$$
Workings:
$b_n - 2b_{n-1} = -a_{n-1} - 1$
$a_n = 2a_{n-1} - b_{n-1} + 2$
$a_{n+1} = 2a_n - b_n + 2$
$-2a_n = -4a_{n-1} + 2b_{n-1} - 4$
$a_{n+1} + 2a_n = ... | One way is to define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and $B(z) = \sum_{n \ge 0} b_n z^n$, write your recurrences with indices shifted so that there are no subtractions in indices:
$\begin{align}
a_{n + 1} &= 2 a_n - b_n + 2 \\
b_{n + 1} &= - a_n + 2 b_n - 1
\end{align}$
Multiply both recurrences by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving the following set of real numbers is a field Show that the following set $A$ of real numbers under addition and multipication is a field:
$A = {a + b\sqrt{2} : a,b \ \text{rational}}$
I am not sure if I am right but here is what I have thus far:
Closure: Let $a_{1} + b_{1}\sqrt{2}$, $a_{2} + b_{2}\sqrt{2}\in \m... | Please can we use the fact that it is closed under addition, subtraction, multiplication and division to prove that it is a field.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Permutation's decomposition into transpositions
*
*Transposition is a cycle with 2 elements.
*Any permutation can be decomposed into a product of transpositions.
For example, for permutation
$\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 1 & 4
\end{pmatrix}$
decomposition is
$\begin{pmatrix}
1 & 3
\end{pmatrix}
\begin{... | A cycle of length 2 is called a transposition, of course it switches (transposes) two elements.
$$
(1 3) =
\left(
\begin{matrix}
1 & 2 & 3 & 4 \\
3 & 2 & 1 & 4
\end{matrix}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? How can I prove that
$$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$
I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
| A simple proof by induction starts by noting that
$$ \tag 1 \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{2}{k(k+2)},$$
$$ \tag 2 \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \frac{1}{(k+2)(k+3)} = \frac{3}{k(k+3)}.$$
By induction we then see that
$$ \tag 3 \sum_{n=0}^N \frac{1}{(k+n)(k+n+1)} = \frac{N+1}{k(k+N+1)}. $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
| An idea : you have
$$\frac{1}{x}+\frac{1}{y} = \frac{x+y}{xy}$$
It imply that $x+y | xy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
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.