Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why is this nested sum formula true I've been trying to get this sum:
$\sum_{i}^{n} \sum_{j=0}^{n-i}j$ into a closed formula but couldn't really understand how to "unpack" that nested sum.
It occured to me that the answer is:
$$\sum_{i=1}^n \left(\sum_{j=0}^{n-i} j\right) = \frac16 n (n^2-1)$$
But I can't figure out w... | From sum of first n natural numbers, we get
$$\sum_{i = 1}^{n}\sum_{j = 0}^{n - i}j = \sum_{i = 1}^{n}\frac{(n - i)(n - i + 1)}{2}$$
$$=\frac{1}{2}\sum_{i = 1}^{n}n^2 - (2n + 1)i +i^2+n$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - (2n + 1)\frac{1}{2}\sum_{i = 1}^{n}i + \frac{1}{2}\sum_{i = 1}^{n}i^2$$
Using sum of first n natu... | {
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"timestamp": "2023-03-29T00:00:00",
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area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.
|AB| = a, P is midpoint of segment AB
| As achille hui says in the coments, $$(1-r)^2 = |AF|^2 = \frac{1}{2^2} + |PF|^2 = \frac{1}{2^2} + |DF|^2 - \frac{1}{4^2} = \frac{1}{2^2} + \left(r + \frac14\right)^2 - \frac{1}{4^2}.$$ Solve for $r$ and you get $r=\dfrac{3}{10}.$
Since area of intersection between the two larger circles is $\dfrac{2\pi}{3}-\dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Simple Trigonometric Problem Here is a Trig problem, and I am missing some understanding of some basic algebraic rule:
$if \sin\theta= \frac{m^2 +2mn}{m^2+2mn+2n^2}$
then prove that $\tan\theta = \frac {m^2 +2mn}{2mn+2n^2},$ In the picture below can anyone explain me how did they achieve the step from 1 to 2 highlighte... | \begin{align*}
\sqrt{1-\left(\frac{m^2+2mn}{m^2+2mn+2n^2}\right)^2}
&= \sqrt{\frac{(m^2+2mn+2n^2)^2 - (m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4(m^2+2mn)2n^2 + 4n^4}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4n^2(m^2+2mn+n^2)}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4n^2(m+n)^2}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I prove $\pi ^2=\sum_{n=0}^{\infty }\frac{1}{(2n+1+\frac{a}{3})^2}+\frac{1}{(2n+1-\frac{a}{3})^2}$ Proving this formula
$$
\pi^{2}
=\sum_{n\ =\ 0}^{\infty}\left[\,{1 \over \left(\,2n + 1 + a/3\,\right)^{2}}
+{1 \over \left(\, 2n + 1 - a/3\,\right)^{2}}\,\right]
$$
if $a$ an even integer number so that
$$
a \ge... | Start with the well known? expansion of $\cot z$ and differentiate,
$$\cot z = \sum_{n=-\infty}^\infty \frac{1}{z - n\pi}
\implies \frac{1}{\sin(z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z - n\pi)^2}
\implies \frac{\pi^2}{\sin(\pi z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z - n)^2}
$$
Substitute $z$ by $-\frac{1+\alpha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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What is $\sin x \sin 2x \sin 4x$ converted to a sum? Converting production to sum. I tried to find the formulas but i could find the formulas only for two identities not for three that is why I couldn't solve. Anybody help?
| $$ \begin{align*} \sin x \sin 2x \sin 4x &= \frac{1}{2}(\cos x - \cos 3x) \sin 4x \\&= \frac{1}{2}\cos x \sin 4x - \frac{1}{2}\cos 3x \sin 4x \\&= \frac{1}{4}(\sin 5x + \sin 3x) - \frac{1}{4}(\sin 7x + \sin x) \\&= \frac{1}{4} (\sin 5x + \sin 3x - \sin 7x - \sin x) \end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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which answer is correct and why? $\arcsin \left(\sin\frac{11\pi}{4}\right)$ Find the value of $\arcsin\left(\sin\dfrac{11\pi}4\right)$
I'm confused if the answer should be $\dfrac{\pi}4$ or $\dfrac{3\pi}4$.
my calculator says it's $\dfrac{\pi}4$ however I don't understand why that is correct? Is that because of the ra... | The range of the arcsine function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Thus,
$$\arcsin\left[\sin\left(\frac{11\pi}{4}\right)\right] \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$
Since
\begin{align*}
\sin\left(\frac{11\pi}{4}\right) & = \sin\left(2\pi + \frac{3\pi}{4}\right)\\
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that if a rectangle's sides are all odd, then it's diagonal is irrational? In trying to write an alternate and simple proof that at least one leg of a right triangle is a multiple of 4 using Dickson's method of generating triples, I came across quite an interesting observation that ifall the sides of a rectangle ... | If the sides of the rectangle are integers, then the square of the length of the diagonal is also an integer since
$d^2 = l^2 + w^2$
and the integers are closed under multiplication and addition.
The only integers that have rational square square roots are perfect squares. Since
$d = \sqrt{l^2 + w^2}$
the length ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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How do I prove the maximum of this function I have the function
$$y = x - \sqrt{x^2 - 1}$$
which must have a maximum of $1$ at $x = 1$, as after that you're taking $x$ and subtracting something slightly smaller than $x$, tending to $0$ as $x$ tends to infinity, however its derivative of
$$1 - \frac{x}{\sqrt{x^2 - 1}}$... | To get maximum of this function, clearly, $x \ge 0$. So we have:
$$x-\sqrt{x^2-1} = x-\sqrt{x^2-1} \times \frac{x+\sqrt{x^2-1} }{x+\sqrt{x^2-1} } = \frac{1 }{x+\sqrt{x^2-1} } \ge \frac{1}{1+\sqrt{1^2-0}}= 1,$$
Note that $x+\sqrt{x^2-1}$ is an increasing function at $[1, +\infty]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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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... | Clearly, the two of the three roots of $$\frac1{a+x}+\frac1{b+x}+\frac1{c+x}=\frac2x$$ are $\omega,\omega^2$
On simplification, $$2x^3+2(a+b+c)x^2+2(ab+bc+ca)x+2abc=3x^3+2x^2(a+b+c)+x(ab+bc+ca)$$
$$\iff x^3+x^2(0)-(ab+bc+ca)x-2abc=0$$
If $a$ is the third root, using Veita's formula $$a+\omega^2+\omega=-\dfrac01$$
Hope ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating $\displaystyle\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x} \frac{t^3\ln(1-t)}{t^4 + 4}\,dt$
Evaluate the following limit: $$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt$$
Any advice on how to tackle this problem ?
| Using Fundamental Theorem of Calculus
$$\frac{d}{dx}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt=
\frac{x^3\ln(1-x)}{x^4 + 4}$$
$$\begin{align}
\lim_{x\space\to\space0} \dfrac{1}{x^5}\int_0^{x}\dfrac{t^3\ln(1-t)}{t^4 + 4}\,dt&=\lim_{x\space\to\space0} \frac{1}{5x^4}\cdot
\frac{x^3\ln(1-x)}{x^4 + 4}\tag{1}\\
&=\lim_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $E(X)$ and $Var(X)$ In a box there are $30$ balls, $20$ are black and $10$ are red.
Let $X$ be the number of red in a selection of two balls drawn without replacement then $$X=I_1 + I_2$$ where $I_1 = 1$ if red is drawn first, else 0 the same thing applies to $I_2$ in the second draw. Find $$E(I_1),\,\, E(I_2),\,\... | The variables $I_1, I_2$ are indicator variables and therefore you can use that $$E[I_i]=P(I_i=1)$$
for $i=1,2$. In more detail, you have that
$$E[I_1]=1\cdot P(I_1=1)+0\cdot P(I_1=0)=1\cdot\frac{10}{30}=\frac{1}{3}$$
and by conditiong on $I_1$ also that
$$\begin{align*}E[I_2]&=E[E[I_2|I_1]]=E[I_2|I_1=1]P(I_1=1)+E[I_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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elementary properties of cyclotomic polynomials How can one rewrite $1+x^2+x^4+x^8+\cdots x^{2^n}$ as a product of cyclotomic polynomials? more general how can we express $1+x^p+\cdots+x^{p^n}$, where $p$ is a prime, in term of product of cyclotomic polynomials?
| Let $p(x)=1+x^2+x^4+x^6...+x^{2n}$. Assume that $n>1$, otherwise when $n=1$, $p(x)=x^2+1$, which is the only irreducible polynomial in this case. The case that $n$ is even, $p(x)=(1+x+x^2+x^3+x^n)(1-x+x^2-x^3+x^n)$. When $n$ is odd, the best possible factorization is $p(x)=(x^2+1)(1+x^4+x^8+x^{4k})$ where $2k+1=n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"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.
| I am applying Robert Israel's suggestion -
I have learned that this is generally a good thing to do.
The general term in
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot$
seems to be
$\begin{array}\\
a_n
&=(-1)^n\dfrac{\prod_{k=1}^n (3k-2)}{5^n n!}\\
&=\dfrac{\prod_{k=1}^n (2-3k)}{5^n n!}\\
&=\dfrac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Two squares and angle can someone answer this please?
I have two equal squares (picture). We know the lengths of three lines inside them. One of them is upside down. Find angle alpha. The picture is not perfect.
Thanks
| Following is one boring way of solving the problem using coordinate geometry.
Even though I suspect there are more geometry intuitive and clever solutions,
a solution is a solution and here we go...
Choose a coordinate system such that the lower left corner of the upper square is the origin.
Let $\ell$ be the side and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Integration giving different answers (trig substitution) Integrating $\sin^3x\cos^5x$, i get 2 different answers, using techniques that should both be valid.
| HINT:
Establish that $$\frac{\cos^8x-\sin^8x}8-\frac{\cos^6x-\sin^6x}6-\frac{\sin^4x}4$$ is a constant
If $u_n=\cos^nx-\sin^nx,$
$$(\cos^nx-\sin^nx)(\cos^2x+\sin^2x)=\cdots$$
$$\implies u_n=u_{n+2}+\cos^2x\sin^2x\cdot u_{n-2}$$
$$\implies u_{n+2}=u_n-\cos^2x\sin^2x\cdot u_{n-2}$$
Now, $u_0=0,u_2=\cos^2x-\sin^2x$
$$n=2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Maximum and minimum of $f(x,y)=xy$ when $x^2 + y^2 + xy =1$ It is asked to find the maximum and minimum points of the function
$$f(x,y)=xy$$
when $x^2 + y^2 + xy=1$
I've tried Lagrange and obtained
$$\lambda = \frac{y}{2x+y}=\frac{x}{2y+x}$$
but what should I do with this? Any other suggestion?
Thanks!
| An elementary approach: using $2xy\leq x^2+y^2$, you can write
$$
xy=\frac{1}{3}(2xy+xy)\le\frac{1}{3}(x^2+y^2+xy)=\frac{1}{3}1=\frac{1}{3}
$$
with equality iff $x=y=\frac{1}{\sqrt{3}}$. For minimum, note that
$$
xy+1=xy+(x^2+y^2+xy)=(x+y)^2\geq 0\implies xy\geq-1
$$
where equality realizes when $(x,y)=(1,-1)$ or $(x,y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\lim_{x \to 0}$ $\sqrt{3x^2 + 4} = 2$ using the definition We have $\lim_{x \to 0}$ $\sqrt{3x^2 + 4} = 2$
Proof: $\vert \sqrt{3x^2+4} - 2 \vert = \cdots$ I got $\left \vert \frac{3x^2}{\sqrt{3x^2+4}+2} \right \vert = \frac{\vert x \vert \vert 3x \vert}{\sqrt{3x^2+4}+2}$
Then what I do next?
$\vert 3x \vert$ $\lt... | You found
$$\vert \sqrt{3x^2+4} - 2 \vert = \left| \frac{3x^2}{\sqrt{3x^2+4}+2} \right|. $$
Hence,
$$\vert \sqrt{3x^2+4} - 2 \vert = \left| \frac{3x^2}{\sqrt{3x^2+4}+2} \right| \leqslant \frac{3}{4}|x|^2$$
and $\vert \sqrt{3x^2+4} - 2 \vert < \epsilon$ if $|x| < \delta = 2\sqrt{\epsilon/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Limit, having trouble applying L'Hospital rule I have been asked to solve for this limit:
$$\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
I try to find the derivative of the numerator and denominator:
$$\frac{6\sin(3x)\cos(3x)}{2\sin(2x)}$$ but that still gives me a $\frac{0}{0}$. How do I solve this
|
Using L'Hospital Again
$$\begin{align}\lim_{x \to 0} \frac{6\sin(3x)\cos(3x)}{2\sin(2x)}
&=\lim_{x \to 0} \frac{18 \cos ^2(3 x)-18 \sin ^2(3 x)}{4\cos(2x)}\\
&=\lim_{x \to 0} \frac{18 \cos ^2(3 x)}{4\cos(2x)}-\frac{18 \sin ^2(3 x)}{4\cos(2x)}\\
&=\frac{18}{4}+0\\
&=\frac{9}{2}\\
\end{align}$$
You can do it without ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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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
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
| For real $a,b\neq 0$ Mathematica gives
$$
\Phi(a,b) = \frac{1}{a\sqrt{b^2-a^2}}\cdot\arctan\left(\frac{\sqrt{b^2-a^2}}{a\sqrt{1+b^2}}\right)
$$
| {
"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": 3
} |
Triple Integral to Find Volume Question: Use a triple integral to find the volume of the solid enclosed by the parabaloids $y=x^2+z^2$ and $y=8-x^2-z^2$.
My attempt: The best I can figure, this object looks kind of like a football oriented along the $y$-axis from $y=0$ to $y=8$ and is symmetric about the $y$-axis and t... | The volume will be represented by the triple integral:
$\int \int \int _{\Omega} dV = \Gamma$ where $ \Omega$ is the solid enclosed by the two paraboloids. The paraboloids intersect in $y = 4$ So:
$\Gamma = \int \int_{\Omega'} \int_{x^{2}+z^{2}}^{8-(x^{2}+z^{2})}dV$. Where $\Omega'= $$\{(x,z) \lvert x^{2} + z^{2} \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\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}$$ M... | Induction $n-1\to n$ makes it more compact.
$$\left(\sum_1^{n-1}i+n\right)^2=\left(\sum_1^{n-1}i\right)^2+2n\sum_1^{n-1}i+n^2=\left(\sum_1^{n-1}i\right)^3+n^2(n-1)+n^2=\left(\sum_1^{n-1}i\right)^3+n^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
When is a particular sum $\Theta(n)$? Define
$$S_n = \prod_{x=1}^{\lceil\frac{n}{\ln{n} }\rceil} \left(\frac{1}{\sqrt{n}} + \frac{2x}{n}\left(z_n-\frac{1}{\sqrt{n}} \right)\right) .$$
I am trying to work out necessary and sufficient conditions for a simply stated, positive,
non-increasing function $z_n$ with $1/\sq... | Note that
$$
\prod_{x=1}^{\lceil n/\ln n \rceil} \max\bigg\{ \frac{1}{\sqrt{n}}, \frac{2x}{n}\bigg(z_n-\frac{1}{\sqrt{n}} \bigg)\bigg\} \le S_n \le \prod_{x=1}^{\lceil n/\ln n \rceil} 2\max\bigg\{ \frac{1}{\sqrt{n}}, \frac{2x}{n}\bigg(z_n-\frac{1}{\sqrt{n}} \bigg)\bigg\},
$$
and so
\begin{align*}
-\ln S_n &= \sum_{x=... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
| There are already 7 nice solutions. I want to generalize the result further, by rationalization and substitutions, of the given integral to $$
I(m ,n):=\int \frac{\sec ^m x}{(\sec x+\tan x)^{n}} d x= \int \frac{\sec ^{m} x(\sec x-\tan x)^{n}}{\left(\sec ^{2} x-\tan ^{2} x\right)^{n}} d x=\int \frac{(1-\sin x)^{n}}{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c$ are positive real numbers, then $\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$
If $a$, $b$, and $c$ are positive real numbers such that $abc=1$, then prove that
$$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$$
Progress
I think the relevant concept would be the application o... | Writing $a=x/y$, $b=y/z$, and $c=z/x$, we simplify the LHS:
$$
\frac{xz}{xy+yz}+\frac{xy}{xz+yz}+\frac{yz}{xy+xz}=\frac{r}{s+t}+\frac{s}{r+t}+\frac{t}{s+r}
$$
where $r=xz$, $s=xy$, and $t=yz$. It remains to invoke Nesbitt's inequality.
The linked wikipedia article contains a standard Cauchy-Schwarz proof:
$$
[(s+t)+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the region that is enclosed by the cardioid $r=2+2\sin(\theta)$ We just learned polar integration, so I know that's how we're supposed to do it. I have a problem though: I'm getting a negative answer.
What I did:
Using the graph, which is:
I figured out that every $2\pi$ it repeated, so I did
$$\int_... | Your integral is correct, but you must have made a mistake in integration.
\begin{array} \\
\frac{1}{2} \int_0^{2\pi} (2 + 2\sin \theta)^2 \, \mathrm{d}\theta &= \frac{1}{2} \int_0^{2\pi} (4 + 8 \sin \theta + 4 \sin^2 \theta) \, \mathrm{d}\theta \\
&= \frac{1}{2} \int_0^{2\pi} \left(4 + 8 \sin \theta + 4 \left(\frac{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the first two non vanishing maclaurin terms Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!... | The limit in question is more easily solved by a little bit of algebra. Note that $$\frac{\sin(x+x^3)-x}{x^3}=\frac{\sin (x+x^3)-(x+x^3)}{(x+x^3)^3}\cdot (1+x^2)^3+1$$ Putting $t=x+x^3$ we can see that $t\to 0$ and hence the first factor $(\sin t - t) /t^3$ in first term tends to $-1/6$ and the desired limit is thus $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find an integral with fractions How to find the integral $$\int_0^\infty \frac{e^{-x^2}}{(x^2+1/2)^2}dx?$$
I find it is difficult to do if I integrate by parts...What's the trick?
| $$\int\dfrac{e^{-x^2}}{\left(x^2+\dfrac{1}{2}\right)^2}dx=-\int\dfrac{e^{-x^2}}{2x}d\left(\dfrac{1}{x^2+\dfrac{1}{2}}\right)=-\dfrac{e^{-x^2}}{2x\left(x^2+\dfrac{1}{2}\right)}-\int\dfrac{e^{-x^2}}{x^2}$$
and
$$-\int\dfrac{e^{-x^2}}{x^2}=\int e^{x^2}d\dfrac{1}{x}=\dfrac{e^{-x^2}}{x}+2\int e^{-x^2}dx$$
so
$$I=\int_{0}^{+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
| HINT : You can consider each right-most digit.
$$1:1,1,1,1,1,\cdots$$
$$2^n:2,4,8,6,2,\cdots$$
$$3^n:3,9,7,1,3,\cdots$$
$$4^n:4,6,4,6,4,\cdots$$
So, we have
$$1+2+3+4=1\color{red}{0},1+4+9+6=2\color{red}{0},1+8+7+4=2\color{red}{0},1+6+1+6=14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}... | $\begin{align}(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 & =5+\sec^2 x+\csc^2 x\\&=5+1+\tan^2 x+1+\cot^2x\\&=7+\tan^2x+\cot^2x\\&\geq9\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Improper integral: $\int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}$ I've tried many ways, but it seems that it didn't work:
$$
\int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}
= \int_0^1\frac{x^{n-1}}{(x+1)(2x+1)\cdots(nx+1)}\mathrm dx
= \cdots
$$
Any help would be appreciated!
| We have the partial fraction decomposition
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac{a_0}{x}+\frac{a_1}{x+1}+\frac{a_2}{x+2}+\ldots+\frac{a_n}{x+n}$$
Now as $x\to 0$, the part with $a_0$ dominates, and we have
$$a_0=\lim_{x\to0}\frac{x}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}$$
Similarily, as $x\to -1$, the part with $a_1$ d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Easy way to calculate the determinant of a big matrix? Given this matrix:
\begin{matrix}
2 & 3 & 0 & 9 & 0 & 1 & 0 & 1 & 1 & 2 & 1 \\
1 & 1 & 0 & 3 & 0 & 0 & 0 & 9 & 2 & 3 & 1 \\
1 & 4 & 0 & 2 & 8 & 5 & 0 & 3 & 6 & 1 & 9 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 \\
2 & 2 & 4 & 1 & 1 & 2 & 1 & 6 & 9 & 0 & 7 \\
0 & 0 ... | It is possible to notice that the third column has only one non-zero value, $4$. Hence we cancel out the third column and the fifth row. Now the (original) fourth row has only one non-zero value, $5$, hence we cancel out the fourth row and the sixth column and continue this way. After six steps, i.e. after using the el... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
graphing $\frac{x^3-x+1}{x^2}$ I want to graph:
$$f(x) = \frac{x^3-x+1}{x^2}$$
so I took the first derivative:
$$f'(x) = \frac{x^3+x-2}{x^3}$$
but this function is hard to find the signals. In other words, it's hard to find where the function is increasing or decreasing. Also, the second derivative is even worse. I'm a... | Study of sign of $f':$
Write the numerator as
$$x^3+x-2=(x-1)(x^2+x+2).$$ Now, $(x-1)(x^2+x+2)=0\iff x=1.$ So,
$$\begin{array}{ccc} x & (-\infty,1) & (1,\infty) \\ \mathrm{sign}(x^3+x-2) & - & +\end{array}$$
It is clear that $x^3=0\iff x=0.$ Now
$$\begin{array}{ccc} x & (-\infty,0) & (0,\infty) \\ \mathrm{sign}(x^3) &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find Minimum value of $P=\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{3-2xy}{5-x^2-y^2}$ Given: $x,y\in (-\sqrt2;\sqrt2)$ and $x^4+y^4+4=\dfrac{6}{xy}$
Find Minimum value Of $$P=\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{3-2xy}{5-x^2-y^2}$$
Could someone help me ?
| You could try this approach, although I think there is a simpler way.
Firstly, apply the change of variables
$$
v=x^4+y^4, \ w = \frac{1}{xy}
$$
in order to simplify the domain $D$ to
$$
D = \{\ v(w) = 6w-4\ : w \in (-\infty,-1/2) \cup (1/ 2,\infty)\ \}.
$$
The minimal value is now given by invoking the chain rule
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How can I understand solving the equation? $$\begin{align}
&\left[(\sqrt[4]{p}-\sqrt[4]{q})^{-2} + (\sqrt[4]{p}+\sqrt[4]{q})^{-2}\right] : \frac{\sqrt{p} + \sqrt{q}}{p-q} \\
&= \left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right) \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqr... | If it helps, try expressing it in another way, recall that we can also express $\sqrt[n]{{a}^m}$ as $a^{m/n}$, so we have
$$\begin{align}\left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right)&=\left(\frac{1}{(p^{1/4}-q^{1/4})^2}+\frac{1}{(p^{1/4}+q^{1/4})^2}\right)\\
\end{align}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factorial identity $\left(\tfrac{1}{2}\right)!$ to get Waallis I asked the wrong question here, my fault :(
How does one see, using $n! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^n \frac{k}{k+n}$, that
$$\left(\frac{1}{2}\right)! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfra... | Looks like you need to show that
$$\prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right) = \frac{\pi}{2}.$$
This Wikipedia article uses Euler's infinite product for the sine function:
$$\frac{\sin x}{x} = \prod_{k=1}^{\infty}\left(1 - \frac{x^2}{k^2 \pi^2}\right).$$
The result follows quickly by applying $x = \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$ Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution
$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$
$$x^2+1=\frac{u^2-2}{u^2-1}$$
$$2xdx=\frac{-2u}{(u^2-1)^2}$$
$$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}dx}$$
$$\int{\frac{... | $$u=\frac{\sqrt{x^2-1}}x$$
$$\frac{du}{dx}=-\frac{\sqrt{x^2-1}}{x^2}+\frac{2x}{x\cdot2\sqrt{x^2-1}}$$
$$=\frac{-(x^2-1)+x^2}{x^2\sqrt{x^2-1}}$$
$$\implies\int\frac{dx}{(1+x^2)\sqrt{x^2-1}}=\int\frac{x^2}{1+x^2} \frac{dx}{x^2\sqrt{x^2-1}}$$
Now $u^2=\dfrac{x^2-1}{x^2}\implies\dfrac1{x^2}=1-u^2$
$\implies\dfrac{x^2}{1+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Integral evaluation (step-by-step) I'm trying to evaluate the integral by exponent. Could you help me with following steps?
Integral: $$\int \frac{1}{4+\sin(x)} dx$$
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\int \frac{1}{4+sin(x)} dx = \int \frac{1}{4+\frac{e^{ix}-e^{-ix}}{2i}} dx = \int \frac{2i}{8i+e^{ix}-e^{-ix}} dx$$... | Let be $t = \tan(\frac{x}{2})$ so that $$\sin x=\frac{2t}{1+t^2},\quad\cos x=\frac{1-t^2}{1+t^2},\quad\operatorname{d}\!x=\frac{2 \operatorname{d}\!t}{1 + t^2}$$
and the integral becomes
$$
\int \frac{\operatorname{d}\!t}{2t^2+t+2}=\frac{1}{2}\int \frac{\operatorname{d}\!t}{\left(t+\frac{1}{4}\right)^2+\frac{15}{16}}=8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
A problem from KVS 2014 I was doing the the following problem-
Prove that $$\frac { \sqrt { a+b+c } +\sqrt { a } }{ b+c } +\frac { \sqrt { a+b+c } +\sqrt { b } }{ c+a } +\frac { \sqrt { a+b+c } +\sqrt { c } }{ a+b } \ge \frac { 9+3\sqrt { 3 } }{ 2\sqrt { a+b+c } }. $$
I normalized this to $a+b+c=1$ and simplified t... | we consider $f(x)=\frac{1}{1-\sqrt{x}}$ for $0<x<1$ and the tangent line at the point $x=\frac{1}{3}$ we get $y=\frac{3}{4}\sqrt{3}(2+\sqrt{3})x+\frac{3}{4}$. We have for $0<x<1$ $$f(x)\geq \frac{3}{4}\sqrt{3}(2+\sqrt{3})x+\frac{3}{4}$$. Thus we get $$\frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\geq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x... | You have $x^2 = y^2 \implies x = \pm y$. However, if $x = y$, then $x^2 = y^2$.
Also note that $\sqrt{x^2} = |x|$, since we adopt the convention that $\sqrt{x}$ is the nonnegative square root of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| the trouble with your approach stems from $0 \times \infty$ being indeterminate.
here is an explanation.
(a) what you are doing is approximating $\sqrt{x^2+1} = x + \cdots$
(b) $\sqrt{x^2 + 1} - x = 0 + \cdots$
(c) $x(\sqrt{x^2+1} - x) = x(0 + \cdots) = 0?$
you think the answer is 0. but, perhaps $x \times \cdots \ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Why do we have $u_n=\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}}=O(\frac{1}{n^3})$? Why do we have
*
*$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=O\left(\dfrac{1}{n^3}\right)$
*$u_n=e-\left(1+\frac{1}{n}\right)^n\sim \dfrac{e}{2n}$
any help would be appreciated
| $$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=$$
$$\dfrac{\sqrt{n^2+1}-\sqrt{n^2-1}}{\sqrt{n^2-1}\sqrt{n^2+1}}=$$
$$\dfrac{(\sqrt{n^2+1}-\sqrt{n^2-1})(\sqrt{n^2+1}+\sqrt{n^2-1})}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{n^2+1-n^2+1}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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.
| $\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}}$
$\frac{\sqrt{2}}{2}=\frac{2^{\frac{1}{2}}}{2^1}=2^{\frac{1}{2}-1}=2^{-\frac{1}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
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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
| We have
$$
8^a+1+1\geq3\sqrt[3]{8^a}=3\times 2^a,
$$
$$
8^b+1+1\geq3\sqrt[3]{8^b}=3\times 2^b,
$$
$$
8^c+1+1\geq3\sqrt[3]{8^c}=3\times 2^c,
$$
$$
2^a+2^b+2^c\geq 3\sqrt[3]{2^{a+b+c}}=3.
$$
It follows that
\begin{eqnarray}
8^a+8^b+8^c&\geq&3(2^a+2^b+2^c)-6\\
&=&(2^a+2^b+2^c)+2(2^a+2^b+2^c-3)\\
&\geq&2^a+2^b+2^c.
\end{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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?
| $\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}=\frac{\sqrt{2}+ \sqrt{3} - \sqrt5}{2\sqrt{6}}$ I think you can do the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Asymptotics of $\int_{0}^{+\infty}\!\!\frac{dx}{\sinh^2(\epsilon \sqrt{x^2+1}) } $ for $\epsilon$ near $0$ How to find an asymptotic expansion, for $\epsilon$ near $0$, of the following integral
$$
I(\epsilon):=\int_{0}^{+\infty}\frac 1{\sinh^2 (\epsilon \sqrt{x^2+1}) } {\rm d}x.
$$
As $\epsilon \rightarrow 0$, I have ... | My calculation shows that
$$I(\epsilon)
:= \int_{0}^{\infty} \frac{dx}{\sinh^{2}(\epsilon\sqrt{x^{2}+1})}
= \frac{\pi}{2\epsilon^{2}} - \frac{1}{\epsilon} + \pi \epsilon \sum_{n=1}^{\infty} \frac{1}{(\pi^{2}n^{2} + \epsilon^{2})^{3/2}} \tag{1}. $$
In particular, if we expand the infinite sum on the RHS, we get
$$ I(\... | {
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"url": "https://math.stackexchange.com/questions/1084044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Evaluate: $\int e^{x^4}(x+x^3+2x^5)e^{x^2} dx$ Evaluate:
$$\int e^{x^4}(x+x^3+2x^5)e^{x^2} dx$$
I know the answer of this integral but got stuck at how to solve this. It seems to be the form like $ \int e^x(f(x)+f''(x))dx = e^x f(x)+C$
| $$\begin{aligned}
\int e^{x^4+x^2}\left ( x+x^3+2x^5 \right )\,dx &=\frac{1}{2}\int e^{x^4+x^2}\left ( 2x+2x^3+4x^5 \right )\,dx \\
&= \frac{1}{2}\int \left ( 2xe^{x^4+x^2}+x^2e^{x^4+x^2}\left ( 4x^3+2x \right ) \right )\,dx\\
&= \frac{1}{2}\int \left [ \left ( x^2 \right )'e^{x^2+x^4}+x^2 \left ( e^{x^4+x^2} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation of the telescoping series $\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}$ $(i)$ Verify that $$\frac{1}{1+(n-1)x} - \frac{1}{1+nx} = \frac{x}{(1+(n-1)x)(1+nx)}$$
$(ii)$ Hence show that for $x \ne 0$, $$\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}=\frac{N}{1+Nx}$$
Deduce that the infinite series $\frac{1}{1.\frac{3}{2}} ... | From the LHS (The series), you can see that for:
At n=1, we have $\frac{x}{1+x}$
At n=2, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}=\frac{2}{1+2x}$ (By simplifying)
At n=3, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}=\frac{3}{1+3x}$ (Again by simplifying)
At n=4, we have $\frac{x}{1+x}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 the inequality between harmonic and arithmetic means we have that
$$
\frac{1}{\sqrt{1+xy}}+\frac{1}{\sqrt{1+yz}}+\frac{1}{\sqrt{1+xz}}\ge \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}}.
$$
Now if we prove that
$$
\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{10}
$$
we are done. but by Cauchy-Schwarz we have
$$
\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there... | In general any quartic equation can be solved. If the polynomial happens to be the product of two irreducible polynomials over $\mathbb Z$, the following method allows to find the coefficients easily. Reducing the coefficients mod $2$, one has $$f(x):=x^4+3x^3+6x+4=x^4+x^3=x^2(x^2+x).$$ Lifting back to ${\mathbb Z}$-co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 4,
"answer_id": 1
} |
Simple limit problem without L'Hospital's rule $$\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... | Let $\sqrt[3]x-1=y\implies x=(1+y)^3,x^4=[(1+y)^3]^4=(1+y)^{12}$
$$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}=\lim_{y\to0}\frac{\sqrt{(1+y)^{12}+1}-\sqrt2}y$$
$$=\lim_{y\to0}\frac{(1+y)^{12}+1-2}{(\sqrt{(1+y)^{12}+1}+\sqrt2)y}$$
$$=\lim_{y\to0}\frac{1+1+12y+O(y^2)-2}y\cdot\frac1{\lim_{y\to0}(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.
Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$,
then $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$ y^2+ e^{2x}-2(e^x)y= 1$$
So if ... | Let $y=\ln(x+\sqrt{x^2+1})$, then:
$$e^{y}=x+\sqrt{x^2+1}$$
$$e^{y}-x=\sqrt{x^2+1}$$
$$(e^{y}-x)^2=(\sqrt{x^2+1})^2$$
$$e^{2y}-2xe^{y}+x^2=x^2+1$$
$$e^{2y}-2xe^{y}=1$$
$$e^{2y}-1=2xe^{y}$$
$$x=\frac{e^{y}-e^{-y}}{2}$$
So $\displaystyle f^{-1}(y)=\frac{e^{y}-e^{-y}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Solving $\tan x-\tan(2x)=2\sqrt{3}$ $$\tan x-\tan(2x)=2\sqrt{3}$$
TRY #1
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3}&\implies\tan x=2\sqrt{3}+\tan{2x}\\
&\implies \tan^2x=\tan^2(2 x)+4 \sqrt{3} \tan(2 x)+12\\
&\implies\tan^2x=(\frac{2\tan x}{1-\tan^2 x})^2+4\sqrt{3}\frac{2\tan x}{1-\tan^2x}+12
\end{align*}$$
but this w... | How about
$$\tan x-\frac{2\tan x}{1-\tan^2 x}=2\sqrt 3$$
Removing the fraction will give you a cubic equation. At least that is easier than the quartic!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I show that $\lim_{n\to\infty} 2^n \left( \frac{n}{n+1} \right ) ^{n^2} = 0$? How to calculate limit of the following expression:
$$2^n \left( \frac{n}{n+1} \right ) ^{n^2} $$
I know that limit of this sequence is equal to zero, but how to show that?
| $(\frac{n}{n+1})^{n^{2}} = ( 1 -\frac{1}{n+1})^{n^{2}}$. For sufficiently large $n$,
$(1 - \frac{1}{n+1})^{n+1}$ is very close to $e^{-1}$, so is less than $\frac{2}{5}$ as $e >2.5.$ Then $( 1 -\frac{1}{n+1})^{n^{2}} < ( 1 -\frac{1}{n+1})^{n^{2}-1} < (\frac{2}{5})^{n-1}$. So $2^{n}(\frac{n}{n+1})^{n^{2}} < \frac{2^{2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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Permutation in discrete math Is the permutation
$$\begin{pmatrix} 1& 2 &3 &4 &5 &6&7 \\ 7 & 4 & 2 & 1 & 3 & 6 & 5 \end{pmatrix}$$
even or odd?
The product of disjoint cycles is $$\begin{pmatrix}1& 7&5&3&2&4\end{pmatrix}\begin{pmatrix}6\end{pmatrix}$$
and the transposition are $$\begin{pmatrix}1 & 7 \end{pmatrix}\begin... | This is as you said $$(1,7,5,3,2,4),$$
and then it is equal to
$$(1,7)(7,5)(5,3)(3,2)(2,4),$$
and hence it is odd.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{... | $$
\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{\left( {n - 1} \right)!\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}} = \sum\limits_{n = 0}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 6
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| $x=-2$ is a solution of the equation $$x^3+4x^2+x-6$$ thus you can divide $$x^3+4x^2+x-6$$ by $$x+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Working on sequence, possibly recursive I am working on this problem which asks to find if the sequence converges or not and if so the value it converges to. I am not sure how to deal with this type of question, but I feel like it may be a recursive relation. It is $a_n=\dfrac{1^2}{n^3}+\dfrac{2^2}{n^3}+ \cdots +\dfrac... | First, note that $1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}$. Then:
\begin{align}
\frac{1^2}{n^3} + \cdots + \frac{n^2}{n^3} &= \frac{\frac{n(n+1)(2n+1)}{6}}{n^3}\\
&\underset{n\to\infty}{=} \frac{2n^3}{6n^3}\\
&= \frac{1}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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irreducibility in $\mathbb Q[X]$ Are these polynomials irreducible in $\mathbb Q[X]$:
1) $x^4+3x^3+x^2-2x+1$
What I did: I reduced it modulo $3$, then saw that it has no roots, so then I checked all $9$ monic polynomials of $\mathbb F_3$ and got $3$ of them, then checked their values under multiplication (as I calculat... | (3) To "see" it better I would change $x=x+y$ and $y=x-y$ to get
$$P(x,y)=2x^3-2yx^2-2(y^2-2)x+2y^3+1$$
So, a non-trivial factorization would have to be of the form $2(x+p(y))(x^2+q(y)x+r(y))$.
From this $p(y)r(y)=2y^3+1$. But $2y^3+1$ is irreducible over $\mathbb{Q}$. So, either $p(y)$ or $r(y)$ is a constant.
If $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it i... | The number you seek is
\begin{equation}
L
= \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k^2}
= \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\cdot\frac{1}{n}\tag{1}
\end{equation}
Now, letting
\begin{align*}
\Delta x &= \frac{1}{n} & x_k &= \frac{k}{n}
\end{align*}
allows us to rewrite (1) as
$$
L = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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What is the remainder when polynomial $f(x)$ is divided by $(x+1)(x-3)$ when $f(-1) = -4$ and $f(3) = 2$? A polynomial $f(x)$ gives remainder $2$ when divided by $(x-3)$ and gives a remainder $-4$ when divided by $(x+1)$. What is the remainder when $f(x)$ is divided by $(x^2 - 2x - 3)$?
I have shortened the question by... | You are on the right track.
Hint: For some polynomial $s(x),$ and for some $a,b \in \mathbb{R},$
$f(x) = (x^2 - 2x - 3) \cdot s(x)+(ax+b) = (x+1)(x-3) \cdot s(x)+(ax+b).$
Can you see why?
How then can you use this and everything else you know about $f$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Implicit Differentiation: $(x/y)+(y/x) =1$ Hi can anyone please tell me where I goes wrong with this question:
Find $ \frac{dy}{dx} $ for the curves defines by this equation:
\begin{align}
\frac{x}{y} + \frac{y}{x} = 1
\end{align}
Here is what I did:
\begin{align}
&\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\... | if you going to do implicit differentiation you might as well multiply by $xy$ to get $x^2 + y^2 = xy$ before differencing. now differencing gives you $2xdx + 2ydy = xdy + ydx.$ this can also be written as
$$\dfrac{dy}{dx} = \dfrac{y - 2x}{2y-x} = \dfrac{y(y-2x)}{y(2y-x)}=
\dfrac{y(y-2x)}{2y^2 - xy} = \dfrac{y(y-2x)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\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}... | Your diagram is perfect. All these cancel.. It's a telescoping sum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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... | We have, $$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1} =\frac{1}{\sqrt{n} + \sqrt{n-1}} - \frac{1}{\sqrt{n+1} + \sqrt{n}} \ge 0$$
and $\displaystyle \sum a_n$ telescoping!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find sum of the roots of quadratic polynomials
The zeroes of a quadratic polynomial $x^2+ax+b$ are $c$ and $d$ and the zeroes of a quadratic polynomial $x^2+cx+d$ are $a$ and $b$. Find the value of $a+b+c+d$.
The thing doesn't make sense how to use the two equations to get the sum?
Should I use the Vieta's formulas t... | Hint You know that
$$x^2 + ax + b = (x-c)(x-d) = x^2 - (c+d)x + cd\\
x^2 + cx + d = (x-a)(x-b) = x^2 - (a+b)x + ab$$
Now compare the coefficients to get four equations.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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indefinite integration the result is -1 instead of $1 \over 2$ Below is from a book,
When 0 $\le$ x < 1, F(x) = $\int_0^x$ t dt = $x^2 \over 2 $;
When 1 $\le$ x < 2, F(x) = $\int_0^1$ t dt + $\int_1^x$(2-t) dt
= -$x^2 \over 2$ + $2x$ $\color{blue}{-1}$;
However, I think the last part of above in blue, the -1 ... | The problem is in your last integral:
$$
\int_1^x (2-t)dt = 2t - \frac{t^2}{2} \bigg|_1^x = \left( 2x-\frac{x^2}{2} \right) -\left( 2-\frac{1}{2} \right) = -\frac{x^2}{2} + 2x -\frac{3}{2}.
$$
Then, we add the term $\int_0^1 t dt = \frac{1}{2}$, which yields that
$$
F(x) = \int_0^1 t dt + \int_1^x (2-t)dt = \frac{1}{2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a}$ Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a},a \geq 0.$
One way to do it is using the formula
$$
a^3+2 - a^2-2 \sqrt{a}=(\sqrt{a}-1)^2(1+(a+1)(\sqrt{a}+1)^2) \geq 0.
$$
But I hope there is a better way.
| You can do it with the AM-GM inequality:
$$
\frac{a^3+a^3+1}{3}\ge \left(a^3\cdot a^3\cdot 1\right)^{\frac{1}{3}}=a^2
$$
$$
2\cdot\frac{a^3+1+1+1+1+1}{6}\ge 2\cdot\left(a^3\cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \right)^{\frac{1}{6}}=2\sqrt{a}
$$
Adding these inequalities yields the desired result.
| {
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"url": "https://math.stackexchange.com/questions/1112733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$ $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$
Attempt: $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = \dfrac {2n+2-1}{n^2(n+1)^2}$
$S_1=2 \sum_{n=1}^\infty [\dfrac {n+1}{n^2(n+1)^2} ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
$=2 \sum_{n=1}^\infty [\dfra... | $$\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]$$
$$=\bigl[\frac{\pi^{2}}{6}-\bigl(\frac{\pi^{2}}{6}-1\bigr)\bigr]=1.$$
using the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.
Otherwise:
$$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)... | {
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"url": "https://math.stackexchange.com/questions/1113954",
"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$ How would I go about solving this following limit?
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
My attempts:
Direct substitution yields the limit to be undefined, also ruling out the possibility of using L'Hospital's Rule.
I don't see any clever sub... | i will use the maclaurin series $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots, \ \tan x = x + \frac{x^3}{3} + \cdots.$
now we can expand
$\begin{align}
\ln(1 + \tan x) &= \tan x - \frac{\tan^2 x}{2} +\frac{\tan^3 x}{3} + \cdots \\
&= x + \frac{x^3}{3}+\cdots -\frac{x^2}{2}+\cdots + \frac{x^3}{3} + \cdots\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"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.
| We have $y=x-\sqrt{4-x^2}$. Function domain
\begin{equation*}
4-x^2\ge 0,~~|x|\le 2
\end{equation*}
Now find the value of function at the ends of the segment
\begin{equation*}
y(-2)=-2, ~~y(2)=2
\end{equation*}
Find the extrema
\begin{equation*}
y^{\prime} = 1-\frac{-2x}{2\sqrt{4-x^2}}=0
\end{equation*}
\begin{equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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CHKMO 2015 and cubic equations Let $a,b,c$ be distinct real numbers. If the equations $E_1: ax^3+bx+c=0, E_2: bx^3+cx+a=0$ and $E_3: cx^3+ax+b=0$ have a common root, prove that at least one of these equations has three real roots(not necessarily distinct).
Suppose $r$ is a common root, then adding the three equations i... | If $abc = 0$, WLOG $a = 0$, and so the common root must b$ x = - \frac{c}{b}$. However, it is easy to check that $ 0 = b \frac{-c^3}{b^2} + c \frac{-c}{b} = \frac{ -c^3-c^2b } { b^2} = - \frac{ c^2 (c+b) } {b^2}$, hence we must have $ c = -b$, which gives us $x=1$. Then, $ 0 = bx^3 + cx + a = bx^3 - bx = bx(x-1)(x+1)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 terms in your sum are $p_nx^n$. If you solve the recurrence relation, you can find that $p_n$ grows as $r^n$ for some $r$, the larger root of the characteristic equation. You need the terms to decease faster than $\frac 1n$, so need $|x| \lt \frac 1r$
| {
"language": "en",
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"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 $$M=\{z \in \mathbb{N} | \exists x, y \in \mathbb{Z}: x^4+y^4=z^2, x \cdot y \neq 0 \} \subseteq \math... | Since the discussion in the comments is getting complicated, let's do this directly.
Let $d=\gcd(z-x^2, z+x^2)$. Clearly $2|d$ since $2|z-x^2$ and $2|z+x^2$, seeing as how each of $z,x^2$ are odd, so we may write $d=2k$.
Now, for any divisor $f|(z-x^2)$ and $f|(z+x^2)$ then $f$ divides sums and differences of these thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Generating Functions and Polynomial Expansions Give a formula similar to:
$\frac{1-x^{m+1}}{1-x} = 1 + x + x^2 + ... + x^m$
For the following
(a) $1 + x^4 + x^8 + ... + x^{24}$
(b) $x^{20} + x^{40} + ... + x^{180}$
Workings
a. $1 + x^4 + x^8 + ... + x^{24}$
$g(x) = \frac{1-x^{25}}{1-x^4}$
b. $x^{20} + x^{40} + ... + x^... | You almost got it! But are just a little bit off. One way to check each of these is to introduce $y = x^{4}$ for a) and $z = x^{20}$ for b). This means $$1 + x^4 + x^8 + ... + x^{24} = 1+y+y^2+\ldots +y^6 \\ = \frac{1-y^{7}}{1-y} \\ = \frac{1-x^{28}}{1-x^4}$$ and $$x^{20} + x^{40} + ... + x^{180} = z+z^2+\ldots +z^8 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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?
| You may write
$$
x^2 - \sqrt{x^4 - x^2 + 1}=\frac{(x^2)^2 - (x^4 - x^2 + 1)}{x^2 + \sqrt{x^4 - x^2 + 1}}=\frac{1 - \frac{1}{x^2}}{1 + \sqrt{1 - \frac{1}{x^2} + \frac{1}{x^4}}} \to \frac12
$$ as $x \to +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Parametrization of solutions of diophantine equation $x^2 + y^2 = z^2 + w^2$ I need integer solutions of $x^2 + y^2 = z^2 + w^2$ parametrized. Can it be done? Thanks.
| [Partial solution.]
Look for rational solutions to $$x_1^2+y_1^2-z_1^2=1\tag{1}$$ first.
We know that $p_0=(-1,0,0)$ is a solution. Let $(a,b,c)$ be any set of integers. Then solve $p_0+t(a,b,c)=(x_1,y_1,z_1)$. $(-1+at)^2+(bt)^2-(ct)^2 =1$, or $1-2at+a^2t^2+b^2t^2-c^2t^2=1$ or $2at = a^2t^2+b^2t^2-c^2t^2$. Now, $t=0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2} $... | This one can also be done using complex variables.
Starting from the generating function
$$\sum_{n\ge 0} P_n(x) t^n
= \frac{1}{\sqrt{1-2xt+t^2}}$$
Call $P_n(0) = Q_n$ so that
$$\sum_{n\ge 0} Q_n t^n
= \frac{1}{\sqrt{1+t^2}}.$$
Now using Lagrange Inversion we get
$$Q_{2n} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing a sequence defined recursively is convergent Given the recursively defined sequence
$$ a_1 = 0, a_{n+1} = \frac 1{2+a^n} $$
Show it converges. I'm working with Cauchy sequences, and proved in a previous question that any sequence of real numbers $(a_n)$ satisfying:
$$|a_n - a_{n+1}| \leq \frac1{2^n} $$
is conv... | Continuing as Thomas Andrews suggested,
$a_{n+1}-a_{n+2}
=\frac1{2+a_n}-\frac1{2+a_{n+1}}
=\frac{(2+a_{n+1})-(2+a_n)}{(2+a_n)(2+a_{n+1})}
=\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}
$
so
$|a_{n+1}-a_{n+2}|
=\big|\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}\big|
<|\frac{a_{n+1}-a_n}{4}|
$.
From this,
$|a_{n+k}-a_{n+k+1}|
<|\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Differentiation using the Chain Rule $$y=\frac { \cos(1+x) }{ 1+\cos(x) } $$
Steps I took:
$$y'=\frac { (-\sin(1+x))(1+\cos(x))-(\cos(1+x))(-\sin(x)) }{ (1+\cos(x))^2 } $$
$$y'=\frac { (\cos(1+x))(\sin(x)) }{ (1+\cos(x))^2 } -\frac { (\sin(1+x))(1+\cos(x)) }{ (1+\cos(x))(1+\cos(x)) } $$
$$y'=\frac { (\cos(1+x))(\sin(x)... | you can do this with without chain rule. let us see.
$\begin{align}
\dfrac{\cos (1+x)}{1+\cos x} &= \dfrac{\cos 1 \cos x - \sin 1 \sin x}{1+\cos x}\\
&= \dfrac{\cos 1 (1+\cos x) -\cos 1 - \sin 1 \sin x}{1+\cos x}\\
&= \cos 1 - \dfrac{\cos 1}{1 + \cos x} -\sin 1\tan(x/2)
\end{align}$
we can now take the derivative of
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the equation $4x=y^2+z^2+1$ has no integer solution
Show that the equation
$$4x=y^2+z^2+1$$
has no integer solution.
I divided throughout by $4$ to get
$$x=\frac{y^2}{4}+\frac{z^2}{4}+\frac{1}{4}$$
but not sure if that is correct
| $$4x=y^2+z^2+1$$4x is even ,so other side must be even ,so one of y or z is odd and ohter is even
suppose y is odd and z is even $$y=2k+1 \rightarrow y^2=4k^2+4k+1\\$$$$z=2q \rightarrow z^2=4q^2$$put them in furmula $$4x=(4k^2+4k+1)+(4q^2)+1\\4x=4(k^2+k+q^2)+2 $$now divide both side by 2 $$2x=2(k^2+k+q^2)+1$$ and it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve this system without WolframAlpha How can I solve this system without using WolframAlpha or any other program?
$$\begin{equation}
\begin{cases}
2x_1+\lambda x_2+3x_3+4x_4=1\\x_1-x_2+9x_3+7x_4=3\\3x_1+5x_2+\lambda x_3+5x_4=1\\x_1+2x_2+x_4=0
\end{cases}
\end{equation}$$
| i will use $k$ for $\lambda.$ look at the augmented matrix
$\begin{align}\pmatrix{2&k&3&4&1&|a\\1&-1&9&7&3&|b\\3&5&k&5&1&|c\\1&2&0&1&1&|d}
&\to \pmatrix{1&2&0&1&1&|d\\1&-1&9&7&3&|b\\3&5&k&5&1&|c\\2&k&3&4&1&|a} \\
&\to \pmatrix{1&2&0&1&1&|d\\0&-3&9&6&-2&|b-d\\0&-1&k&3&-2&|c-3d\\0&k-4&3&2&-1&|a-2d} \\
&\to \pmatrix{1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expressio... | According to Maple, there are no real solutions. There are complex solutions.
They have $a$ as a root of the irreducible quartic $x^4+4 x^2-3 x+7$.
But as for the value of $(a+5)(b+5)(c+5)(d+5)$: note that
$(a+x)(b+x)(c+x)(d+x)$ is a monic quartic polynomial in $x$. Find a
monic quartic with values $15,\;45,\;133,\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 4
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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{... | When you take a factor out of the square-root, it still gets square-rooted, but you have been squaring them. For example, $\sqrt{n^2+1}=\sqrt{n^2(1+1/n^2)}=\sqrt{n^2}\sqrt{1+1/n^2}=n\sqrt{1+1/n^2}$
| {
"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": 2
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For each of the following rules, either prove that it is true in every group $G$, or give a counterexample to show that it is false in some groups. Let $J$ be a group that consists of six matrices:
$I = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}, A = \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}, B = \begin{pmatrix} 0 &... | The second to last statement is false. Consider for example the integers with multiplication.
The last statement is true. Let $z=x^{-1}y.$
The other answers look good.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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summation of a trigonometric series How to evaluate $\tan^2(1) + \tan^2(3) + \tan^2(5) + \tan^2(7) + \ldots + \tan^2(89)$?
Angles are given in degrees.
I tried converting $\tan(89)$ as $\cot(1)$ and then tried combining $\tan(1)$ and $\cot(1)$, but later got stuck.
| My answer is based on a rather beautiful answer given to a related question, that can be found here.
Expressing the summation in radians, we wish to evaluate
$$S=\sum_{m=0}^{45-1}\tan^2\left\{\frac{(2m+1)\pi}{2\cdot90}\right\}=\sum_{m=0}^{45-1}\tan^2\left\{\frac{(m+\frac{1}{2})\pi}{2\cdot45}\right\}$$
Using the well-kn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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 $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$
| $0\leq x^2+y^2<2+xy$. Squaring it, you get $x^4+y^4+2x^2y^2<4+4xy+x^2y^2\Rightarrow x^4+y^4<4+4xy-x^2y^2=8-(4+x^2y^2-4xy)=8-(2-xy)^2\leq 8$
| {
"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": 2
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Probability of drawing a yellow ball after 1 or 2 have been drawn. The question is as follows:
An opaque urn contains ten balls, five red, four yellow, and one green. Three balls are drawn from the urn in sequence but at random, without replacement.
What is the conditional probability of drawing a yellow ball gi... | Let $T$ be the event yellow on third, and let $A$ be the event at least one yellow in the first two trials. We want $\Pr(T|A)$. By the definition of conditional probability, we have
$$\Pr(T|A)=\frac{\Pr(A\cap T)}{\Pr(A)}.\tag{1}$$
We compute the two probabilities on the right-hand side of (1).
The event $A$ can happen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
| COMMENT.-In attention to the note by Community above I show the following:
$$LHS=\frac{\sin(\frac{15\pi}{18})}{\sin(\frac{11\pi}{18})\sin(\frac{4\pi}{18})}\\$$ Hence one has $$\sin(\frac{15\pi}{18})=4\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})\sin(\frac{\pi}{18})$$ Note that $\dfrac{15\pi}{18}=150^{\circ}=180^{\circ}-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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... | No need to transform into polar form.
Put, $x^2+y^2=z$. Then , $z\to 0$ as $(x,y)\to (0,0)$.
Now, $$\lim_{(x,y)\to (0,0)}\frac{\tan (x^2+y^2)}{\arctan \left(\frac{1}{x^2+y^2}\right)}$$
$$=\lim_{z\to 0}\frac{\tan z}{\arctan (1/z)}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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... | This is best explained with a picture:
The length you have on the LHS is the length of the dotted red path in the unit square.
Now use some reflections in order to "straighten" that path. You will get:
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \sqrt{(1+a)^2+(2-a)^2}$$
and now it is trivial that t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Let p be prime. Prove that: Let p be prime:
$p^2\choose p$ is congruent to p (mod $p^2$) and $2p\choose p$ is congruent to 2(mod$p^2)$
I know that when p is prime p|$p\choose k$ where $p\choose k$ can be defined as $p!/k!(p-k)!$ and have tried playing around from there but cannot seem to make it work. Any help would be... | Doing the case for $p=2$ can be done by hand, so I assume $p$ is odd.
$\binom{p^2}{p}=\frac{1\cdot 2 \cdot 3\dots p^2}{(1\cdot 2 \cdot 3 \cdot p)(1\cdot 2\cdot 3\dots \cdot p^2-p)}=\frac{(p^2-p+1)(p^2-p+2)\dots p^2}{1\cdot 2\cdot \dots p}$ If we cancel all of the factors of $p$ we get a complete residue class mod $p^2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have
Base case let $p(2)$
\begin{align*}
\frac{1}{2}+ \frac{1}{2+2} + \... | Consider $P(n+1) - P(n)$. This difference has only three terms, all the middle terms cancel. And looking at those three terms you can easily show $P(n+1)-P(n)>0$. That proves that if $P(n)>7/12$ then $P(n+1)$ must also be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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 = ... | hint: add the two equations to get: $(a_n+b_n) = (a_{n-1} + b_{n-1}) + 1\to a_n+b_n = n+1$. Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $f(2)$ if $f$ satisfies $2f(x)-3f(\frac1x)=x^2$ The following expression is given, and we are asked to find $f(2)$.
\begin{equation}
2f(x)-3f\left(\frac{1}{x}\right) =x^2
\end{equation}
Does a unique and well defined answer exist? Why? and what is it?
We have that $f(1)=f(-1)=-1$.
| Hint: Note that
$$
\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}
\begin{bmatrix}
f(x)\\
f(1/x)
\end{bmatrix}
=
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}
$$
and
$$
\det\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}=-5\ne0
$$
Completed Answer
Now that over a year has passed, I think it is appropriate to finish the answer.
$$
\be... | {
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} |
the total differential equation $(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$ In solving the total differential equation $$(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$$
Substituting $x =uz$ and $y=vz$ and reached at a position where, i face the following
$$(v^2 - v^3 +u^2v)du - (u^2+... | Your equation $(v^2 - v^3 +u^2v)du - (u^2+u^3-uv^2)dv=0$ is OK. To continue, see below :
Instead of such a complicated calculus, a little bit of observation would have given the integrating factor $\frac{1}{u^2v^2}$ for $(v^2 - v^3 +u^2v)du - (u^2+u^3-uv^2)dv=0$
$$\left(\frac{1}{u^2}-\frac{v}{u^2}+\frac{1}{v}\right)du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help with conditional probability Assume a box contains 11 balls: 5 red, 4 blue, and 2 yellow. A ball is drawn and its color noted. If the ball is yellow, it is replaced; otherwise, it is not. A second ball is then drawn and its color is noted.
What is the probability that the first ball was yellow, given that the seco... | There are $5$ cases where $\color{red}{\text{the second ball is red}}$, out of which, in $3$ cases $\color{green}{\text{the first ball is yellow}}$:
*
*$\color{red }{P( RR)= \frac{5}{11}\cdot\frac{4}{10}}$
*$\color{black}{P( RB)= \frac{5}{11}\cdot\frac{4}{10}}$
*$\color{black}{P( R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$.
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$
My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so... | You already have a great answer from Kim. Another way would be to use Cauchy-Schwarz inequality to get the same bounds on $xy$:
$$(1+xy)^2 = (x^2+2y^2)(2y^2+x^2) \ge 8x^2y^2$$
$$\implies -1-xy \le 2\sqrt2 xy \le 1+xy$$
$$\implies -\frac1{2\sqrt2+1} \le xy \le \frac1{2\sqrt2-1}$$
$$\implies 1-\frac1{2\sqrt2+1}\le 1+xy=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $a^2+2b+c$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares
Let $a,b,c$ (with $a,b,c>1$) be postive integers,and such that $\color{#0a0}{\text{$a^2+2b+c$}}$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares.
Show that:
$$a+b+c=276$$
We note that
$$a^2+2b+c\ge (a+1)^2,b^2+2c+a\ge (b+1)^2,c^2+2a+b\ge (c+1)... | Without loss of generality let $a=\min(a,b,c)$. Then there are two cases: $a \leq b \leq c$ and $a \leq c \leq b$.
*
*Suppose $a \leq b \leq c$. Then $c^2 < c^2+2a+b \leq c^2 + 3c < (c+2)^2$, hence $c^2+2a+b=(c+1)^2$ yielding $2a+b=2c+1$. We also have $b^2 < b^2+2c+a = b^2+2a+b-1+a \leq b^2+4b - 1 < (b+2)^2$, hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1170097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sum\limits_{n}\frac{1}{z+n}$ diverges
How to bound the sequence $\sum\limits_{n\ge0}\frac{1}{z+n}$ from below ?
Actually the problem is from complex analysis, and it is given that $z\in U=\mathbb C\setminus\mathbb R_{\le0}$, but I think it can be treated as in the real case.
So for $z>0, n\in \mathbb N$, I have to... | You've only shown that $\sum_{n\ge 0}^\infty \frac{1}{z + n}$ diverges for $z > 0$. However, you missed the case when $z = a + bi$ with $b\neq 0$. Write
$$\frac{1}{z + n} = \frac{1}{(a + n) + bi} = \frac{a + n}{(a + n)^2 + b^2} - \frac{b}{(a + n)^2 + b^2}i.\tag{*} $$
The first term on the right-most side of $(*)$ domin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the slope of the tangent line to the graph of $f^{-1}$ Given function $f$, find the slope of the line tangent to the graph $f^{-1}$ at the point on the graph $f^{-1}$. $f(x)=\sqrt{5x}$; $(4,\frac{16}{5})$?
Here is what I have thus far:
$f'(x)= \frac{\sqrt{5}}{2\sqrt{x}}$
can it also be $f'(x)= \frac{2.5}{\sqrt{5x... | look at what the function $$y=f(x) = (5x)^{1/2} \text{ does at } x = 16/5, y = 4.$$ the derivative of $f$ is $$f'(x) = \frac12 (5x)^{-1/2}\times 5,\quad
f'(16/5) = \frac 52 \times \frac 14= \frac58 $$
so far we have at $$x = {16}/{5}, y=f(16/5) = 4, f'(16/5) = \frac58.$$ therefore
$$ y = 4, f^{-1}(4) = 16/5, \left(f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_\limits{x\to 0}\left({\tan x\over x}\right)^{1\over 1-\cos x}$. Find $\lim_\limits{x\to 0}\left({\tan x\over x}\right)^{1\over 1-\cos x}$. Is there a way to do it without differentiating so many times? That is exhausting and confusing and will probably cause errors. I would really appreciate your help with t... | Using basic Taylor series as $x \to 0$,
$$
\frac{1}{1-\cos x} = \frac{1}{1-(1-\frac{x^2}{2} + o(x^2))} \sim \frac{2}{x^2}
$$
and
$$
\log\left(\frac{\tan x}{x}\right) = \log\left(1 + \frac{x^2}{3} + o(x^2)\right) \sim \frac{x^2}{3}.
$$
Therefore
$$
\frac{\log\left(\frac{\tan x}{x}\right)}{1-\cos x} \sim \frac{2}{x^2}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
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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.