Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\langle 3 \rangle$ in $\textbf{Z}[\omega]$ ramifies, not splits, right? Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.
As you know, $$\omega = \frac{-1 + \sqrt{-3}}{2}$$ is a complex cubit root of unity, that is, $\omega^3 = 1$. This slightly obs... | Yes your conclusion is correct, and the factorisation
$$\left(\frac{3 - \sqrt{-3}}{2}\right) \left(\frac{3 + \sqrt{-3}}{2}\right) = 3$$
is no problem since the factors on the left side are associate. Indeed
$$\frac{\frac{3 - \sqrt{-3}}{2}}{\frac{3 + \sqrt{-3}}{2}}=\frac{(3 - \sqrt{-3})^2}{(3 + \sqrt{-3})(3 - \sqrt{-3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving that $\cot^220^\circ + \cot^240^\circ + \cot^280^\circ = 9$.
Prove that $\cot^220^\circ + \cot^240^\circ + \cot^280^\circ = 9$.
I tried bringing them all to $\cot^220^\circ$ but it didn't work. How do I proceed?
| $$\cot3x=\frac{\cot^3 x-3\cot x}{3\cot^2 x-1}$$
so
$$\cot^23x=\frac{\cot^6 x-6\cot^4 x+9\cot^2x}{9\cot^4x-6\cot^2 x+1}.$$
Then for $a\in\{\pi/9,2\pi/9,4\pi/9\}$, $\cot^2a=1/3$. So these
$\cot^2a$ are the roots of
$$\frac{y^3-6y^2+9y}{9y^2-6y+1}=\frac13.$$
But this re-arranges to $y^3-9y^2+\cdots=0$. etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Recurrence relationer of intersection points formed by the diagonals of a convex polygon. Derive a recurrence relation to represent the number of intersection points formed by the diagonals of a convex polygon with n vertices. Show that the solution of the recurrence relation is $\binom n4$.
I have derived the recurren... |
The relation is $\;R_{n+1} = R_{n}+\sum _{k=2}^{n-1}(k-1)(n-k); \;R_{4} = 1\,$.
First off:
$$
\begin{align}
\sum _{k=2}^{n-1}(k-1)(n-k) = \sum _{k=1}^{n-2} k(n-k-1) &= (n-1) \cdot \sum _{k=1}^{n-2}k - \sum _{k=1}^{n-2} k^2 \\
&= (n-1) \cdot \frac{(n-2)(n-1)}{2} - \frac{(n-2)(n-1)(2n-3)}{6} \\
&= \frac{n(n-1)(n-2)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there a perfect square that is the sum of $3$ perfect squares? This is part of a bigger question, but it boils down to:
Is there a square number that is equal to the sum of three different square numbers?
I could only find a special case where two of the three are equal? https://pir2.forumeiros.com/t86615-soma-de-tr... | We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd. Let $c^2=2n+1, n=\frac {c^2-1}2$ and
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=\frac {c^2-4}4$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Integrate $\int \frac{1}{1+ \tan x}dx$ Does this integral have a closed form?
$$\int \frac{1}{1+ \tan x}\,dx$$
My attempt:
$$\int \frac{1}{1+ \tan x}\,dx=\ln (\sin x + \cos x) +\int \frac{\tan x}{1+ \tan x}\,dx$$
What is next?
| $$
I=\int \frac 1 {1+\tan x} \, dx=\int \frac {\cos x}{\cos x +\sin x} \, dx\\
J=\int \frac {\sin x}{\cos x +\sin x} \, dx\\
I+J=\int \frac {\cos x+\sin x}{\cos x +\sin x} \, dx = x+C_1\\ \\
I-J=\int \frac {\cos x-\sin x}{\cos x +\sin x} \,dx=\int \frac { d\left( \sin x +\cos x\right) }{ \cos x +\sin x} =\ln \left| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1... | From scratch, the quantity you are looking at is $$\sum_{k=0}^{99} 2^{2(x-k)}-2^{x-k}=4^x\sum_{k=0}^{99}4^{-k}-2^x\sum_{k=0}^{99}2^{-k}=4^x\cdot\frac{4^{-100}-1}{-\frac34}-2^x\frac{2^{-100}-1}{-\frac12}=\\=2^x\left(2^x\cdot \frac{4-4^{-99}}{3}-2+2^{-99}\right)$$
That quantity is $0$ if and ony if $$2^x=\frac{3\cdot 2}4... | {
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"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Evaluating $\sum_{i=0}^{k-1} 4^i(i-1)$ for recurrence relation exercise I need help to solve the following sum:
$$\sum_{i=0}^{k-1} 4^i(i-1)$$
I'm doing some exercises about recurrence relations in algorithms and this sum came up.
The exercise stands like:
$$T(n) = \frac {1}{2}n + 4T(\frac{n}{2} + 3)$$
And the resul... | One way might be$$A=\sum _{ i=0 }^{ k-1 } 4^{ i }(i-1)=-1+{ 4 }^{ 2 }+{ 4 }^{ 3 }\cdot 2+{ 4 }^{ 4 }\cdot 3+...+{ 4 }^{ k-1 }\left( k-2 \right) \\ A+1={ 4 }^{ 2 }+{ 4 }^{ 3 }\cdot 2+{ 4 }^{ 4 }\cdot 3+...+{ 4 }^{ k-1 }\left( k-2 \right) \\ 4\left( A+1 \right) ={ 4 }^{ 3 }+{ 4 }^{ 4 }\cdot 2+{ 4 }^{ 5 }\cdot 3+...+{ 4 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Gauss elimination. Where did I go wrong?
Gaussian elimination with back sub:
So my starting matrix:
\begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2
\end{bmatrix}
multiply the 2nd and 3rd row by -1 * (first row):
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 2 & 0 & 4
\end{bmatrix}
then ad... | I don't understand your way to obtain the RREF, we can proceed as follow
$$\begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2
\end{bmatrix}\stackrel{R3-R2}\to \begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\0 & -1 & 5 & -2
\end{bmatrix}\stackrel{R2-2\cdot R1}\to \begin{bmatrix}
1 & -1 & 1 & -1
\\0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $\frac {(a^2+b^2+c^2+d^2)(\sin 20°)}{(bd-ac)}.$ Given that $a,b,c,d \in R$, if $$ a \sec(200°) - c \tan(200°) =d$$ and $$b \sec(200°) + d \tan(200°) = c$$ then find the value of $$\dfrac {(a^2+b^2+c^2+d^2)(\sin 20°)}{(bd-ac)}.$$
My attempt:Let $\theta=200^o$ then our equations become $a \sec\theta - ... | Hint:
Method$\#1:$
Use $\sec(180^\circ+y)=-\sec y,\tan(180^\circ+y)=+\tan y$
Solve for $\sec20^\circ,\tan20^\circ$
Finally $\sin t=\dfrac{\tan t}{\sec t}\ \ \ \ (1)$
Here $t=20^\circ$
Method$\#2:$
Solve for $\sec200^\circ,\tan200^\circ$
Use $(1),$ here $t=200^\circ$
Finally $\sin(180^\circ+y)=-\sin y $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$
If $a+b$ is an positive integer and $a\ge b$ and
$a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.
I tried a lot to solve it, i came to a step after which i was not able to proceed for... | Note: I solve it base on that a,b are positive numbers(not integers)
suppose $s=a+b$
Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$
Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$
$\Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$
$0<s<=2794$
for one solution, $0<a<s$
(1) $0<3s^2+6s-\sqrt\Delta<2s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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compute the summation $\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$ compute the summation
$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$
My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$
Now
\begin{align}
& = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}... | For $n\ge1,$$$\dfrac{2n-1}{2^n n!}=\dfrac1{2^{n-1}(n-1)!}-\dfrac1{2^nn!}=f(n-1)-f(n)$$
where $f(m)=\dfrac1{2^mm!}$
Can you recognize the Telescoping nature and the surviving term(s) of the summation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Calculating $\int_{-\infty}^{\infty}\left( \frac{\cos{\left (x \right )}}{x^{4} + 1} \right)dx$ via the Residue Theorem? In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$\text{Proposition} \, \, \, (1) $
$$\int_{-\infty}^{\... | From the previous attempt the last line,
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i + \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}= \frac{\sqrt{2}}{2 e^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Identity for Euler-Darboux-Poisson (Evans 2.4 lemma 2) The following identity is used in the treatment of the multivariate wave equation (Evans on p. 74)
Lemma 2 (Some useful identities). Let $\phi:\Bbb R\rightarrow \Bbb R$ be $C^{k+1}$. Then for $k=1,2,\dots$
$$\left(\frac{d^2}{dr^2}\right)\left(\frac 1r\frac{d}{dr}\... | consider$$ \left(\frac{1}{r} \frac{d}{d r}\right)(t^m\phi)=r^{n-2}\left(r\frac{d}{d r}+m\right)\phi$$
so$$\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k-1} \phi\right)=r\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi$$
change $\phi$ into $r\frac{d}{dr}\phi$... | {
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"url": "https://math.stackexchange.com/questions/2874973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there any 2 primitive pythagorean triples who share a common leg? So is it possible for:
$\gcd(a,b,c)=1$
$a^2+b^2=c^2$
and
$\gcd(a,d,e)=1$
$a^2+d^2=e^2$
?
| We can find different triples with the same odd legs, if they exist, using this function of $(m,A)$:
$$\text{We can let }n=\sqrt{m^2-A}\text{ where m varies from }\lceil\sqrt{A}\space\rceil\text{ to }\frac{A+1}{2}$$
There is always a Pythagorean triple with side $A=(2m-1),m\in\mathbb{N}\land m\ge2$ and sometimes more t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Reduction Formula for $I_n=\int \frac{dx}{(a+b \cos x)^n}$ Reduction Formula for $$I_n=\int \frac{dx}{(a+b \cos x)^n}$$
I considered $$I_{n-1}=\int \frac{(a+b \cos x)dx}{(a+b \cos x)^n}=aI_n+b\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$
Let $$J_n=\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$ using parts for $J_n$ we get
$$J_n=\fr... | If I've got my coefficients right, $$\sin^2 x =1-\cos^2 x=-\frac{1}{b^2}(a+b\cos x)^2+\frac{2a}{b^2}(a+b\cos x)+1-\frac{a^2}{b^2}.$$
Thus $$\frac{\sin x (a+b\cos x)^{-n}-J_n}{nb}=\int\frac{1-\cos^2 x}{(a+b\cos x)^{n+1}}dx=(1-\frac{a^2}{b^2})I_{n+1}+\frac{2a}{b^2}I_n-\frac{1}{b^2}I_{n-1}.$$You'll want to double-check wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the largest of the three prime divisors of the number $13^4 + 16^5 - 172^2$ I was able to factor out only the prime 13,thus $13^4 + 16^5 - 172^2=13\cdot 80581$
What should be done to solve it? (Maybe some clever factorization, modulo, or anything else?)
| Since $13^4−172^2 = (13^2-172)(13^2+172) = -3 \times 341 = -1023 = 1-2^{10}$ we see that
$$13^4+16^5−172^2 = 2^{20}-2^{10}+1.$$
Next, since
$$\frac{1+x^{30}}{1+x^{10}} = x^{20}-x^{10}+1,$$
then, using the Sophie Germain identity on both numerator and denominator, we have that
$$2^{20}-2^{10}+1 = \frac{1+4 \times (2^7)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886489",
"timestamp": "2023-03-29T00:00:00",
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Given that $f(x^2+x+1)=f(x^2-x+1)$ for all $x$, is $f(x)$ periodic?
Given that $f(x)$ is a function defined on $\mathbb{R}$, satisfying
$$f(x^2+x+1)=f(x^2-x+1)\;\;\; \forall\;\;\;x\in\mathbb{R}$$ Is $f(x)$
periodic?
My Attempt:
$$f((x+\frac{1}{2})^2+\frac{3}{4})=f((x-\frac{1}{2})^2+\frac{3}{4})$$
Let $x-\frac{1}{... | $$f((t+1)^2+\frac{3}{4})=f(t^2+\frac{3}{4})$$
Definition of a periodic function $f(x+a)=f(x)$
As you can see in your equation $a$ is 1
or you can do it like this $$f(x^2+x+1)=f(x^2-x+1)\;\;\; \forall\;\;\;x\in\mathbb{R}$$
Now put $x+1$ in the second equation in place of $x$ they both become equal which means
$$g(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$ is not uniformly convergent on $[0,\;\frac{\pi}{2})$? Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$
is not uniformly convergent on $[0,\;\frac{\pi}{2})$?
I was thinking that we need to show partial sums
\begin{equation}
\left|S_{2n}\... | For any functions $f,g$ from $[0,\pi/2)$ to $\Bbb R$ let $\|f-g\|=\sup \{|f(x)-g(x)|: x\in [0,\pi/2)\}\in [0,\infty].$
For $m\in \Bbb N$ let $S_m(x)=\sum_{n=1}^m(n(\sin x)^n)/(2+n^2).$
Suppose that $S_m\to S$ uniformly on $[0,\pi/2).$ Then $\|S-S_m\|\to 0.$ Then $\lim_{M\to \infty}\sup_{M<m<m'}\|S_m-S_{m'}\|=0$ becau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Does the series $\sum_{k=1}^{\infty} (\sin \frac{1}{k} - \arctan\frac{1}{k})$ converge? I am shockingly terrible at determining whether or not infinite series converge or not... I'm stuck on the problem:
Does the series $\sum_{k=1}^{\infty} (\sin \frac{1}{k} - \arctan\frac{1}{k})$ converge?
I attempted solving it usi... | Your strategy does work. Why? Because the terms beyond $1/k^3$ are an arbitrarily small fraction of that term for sufficiently large $k$, so their contribution to large-$k$ terms are bounded on both sides by multiples of the $1/k^3$ part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Derive the following identity $1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.
Count the elements of the following set
$$A=\{(x,y,z): 1\leq x,y,z \leq n+1, z>\max\{x,y\}\}.
$$
From this derive the following identity:
$$1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
In the same manner find the formula for... | On the one hand, $|A|=\sum_{i=1}^n i^2$.
On the other hand, we can also count the elements of $A$ by grouping them according to the order of $x$ and $y$.
*
*when $x\neq y$ means that $x<y<z$ or $y<x<z$, so it contributes $2\binom{n+1}{3}$.
*when $x=y$, the amount of $(x,x,z)$ is $\binom{n+1}{2}$.
So $\sum_{i=1}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\s... | Because, when you divide the denominator by $x$, you forgot $x$ is supposed to be negative, so that
$$x=-\sqrt{x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Nature of infinite series $ \sum\limits_{n\geq 1}\left[\frac{1}{n} - \log(1 + \frac{1}{n})\right] $ $$\sum\limits_{n\geq 1}\left[\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)\right]$$
Is it convergent or divergent?
Wolfram suggests to use comparison test but I can't find an auxiliary series.
| We have that
$$\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)= \frac{1}{n}-\frac{1}{n}+\frac{1}{2n^2}+O\left(\frac1{n^3}\right)=\frac{1}{2n^2}+O\left(\frac1{n^3}\right)$$
therefore the given series converges by limit comparison test with $\sum \frac 1{n^2}$.
As an alternative, since $\log (1+x)\ge x-\frac12 x^2$ we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. T... | Hint:
Clearly, if $x,y$ is a solution, $-x,-y$ will also be.
Let $25y^2+56=u^2\iff(u-5y)(u+5y)=56$ where $u\ge0$
As $5y-u,5y+u$ have the same parity, both must be even
$\implies\dfrac{u-5y}2\cdot\dfrac{u+5y}2=14$
Now $14=\pm1\cdot\pm14,\pm2\cdot\pm7$
If $y>0, u+5y\ge1+5\implies\dfrac{u+5y}2\ge3$ i.e., $=7$ or $14$
If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Combinatorics sum to 1 using Identity I want to prove that the following equals 1:
$W=\sum_{i=0}^{n-r}(-1)^i n \frac{C_{n-1}^{r-1}C_{n-r}^i}{(r+i)}$. I tried mathematical induction and succeeded. Is there any known identity of combinatorics that could be used to prove it directly?
| A Slightly More General Identity
$$
\begin{align}
\sum_k(-1)^k\frac{\binom{n}{k}}{\binom{r+k}{m}}
&=\sum_k(-1)^k\frac{\binom{n-1}{k}+\binom{n-1}{k-1}}{\binom{r+k}{m}}\tag1\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac1{\binom{r+k}{m}}-\frac1{\binom{r+k+1}{m}}\right)\tag2\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac{\frac{r+k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Last digit of sequence of numbers
We define the sequence of natural numbers
$$
a_1 = 3 \quad \text{and} \quad a_{n+1}=a_n^{a_n},
\quad
\text{ for $n \geq 1$}.
$$
I want to show that the last digit of the numbers of the sequence $a_n$ alternates between the numbers $3$ and $7$. Specifically, if we symbolize wit... | It follows directly from the hint:
The last digit of $a_n$ is the residue class of $a_n$ mod 10.
Now if you have $a_n \equiv 3 \textrm{ (mod 5)}$ it follows $a_n \equiv 3 \textrm{ (mod 10)}$ or $a_n \equiv 8 \textrm{ (mod 10)}$. But $a_n \equiv 8 \textrm{ (mod 10)}$ would mean that $a_n$ is even, so $2$ comes in the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $\ln\left(1+3x + 2x^2\right) = 3x - \frac{5x^2}{2} + \frac{9x^3}{3} -\cdots + \left(-1\right)^{n-1}\frac{2^n+1}{n}x^n+\cdots$ Show that $$\ln\left(1+3x + 2x^2\right) = 3x - \frac{5x^2}{2} + \frac{9x^3}{3} -\cdots + \left(-1\right)^{n-1}\frac{2^n+1}{n}x^n+\cdots$$
I know that $$\ln(1+x) = \sum\limits_{n=0}^{\i... | Hint: $1+3x+2x^2=(1+x)(1+2x)$, so you just add two Taylor series.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$
Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.
| OK, since the hint seems not to be enough, here is the solution.
First,
Setting $g(x)=f(x)-f(\frac{x}{2})$ we have
$$g(x)-g(\frac{x}{2})=x^2 \\
f(x)-f(\frac{x}{2})=g(x) \tag{*}$$
We solve the first equation.
By induction
$$g(x)-g(\frac{x}{2^{n}})=x^2+\frac{x^2}{4}+\frac{x^2}{16}+...+\frac{x^2}{4^{n-1}}=x^2 \frac{1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the shortest distance from a point to curved surface
Find the shortest distance from a point $(0,0,0)$ to curved surface $x^2+2y^2-z^2=5.$
What I have done is,
Let a point in the curved surface be $(a,b,c)$, and $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix}$ be vector v.
A vector w perpendicu... | No, this is not correct. The vector $w$ is indeed orthogonal to the surface. Therefore, what you want to know is when $v$ and $w$ are collinear. That is, when is there a number $\lambda$ such that $v=\lambda w$. What this means is that you must solve the system$$\left\{\begin{array}{l}a=\lambda a\\b=2\lambda b\\c=-2\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+... | Left to show:
$54n^2-3n -7 >0$, $n \ge 1.$
$54n^2-3n-7 \ge 54 n^2 -3n^2-7=$
$51n^2 - 7 >0$.
Above inequality true for $n \ge 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Solve power with negative exponent $\frac{125^{6}\times 25^{-3}}{(5^{2})^{-3}\times25^{7}}$ I am currently studying about exponents and powers for college calculus discipline. In the meantime I came across negative exponents, like this $25^{-3}$ and $(5^{2})^{-3}$.
I have this calculation to solve,
$$
\frac{125^{6}\tim... | \begin{align}
\frac{125^6\times 25^{-3}}{(5^2)^{-3}\times 25^7}=
\frac{\left(\dfrac{125^6}{25^3}\right)}{\left(\dfrac{25^7}{(5^2)^3}\right)}=
\dfrac{125^6}{25^3}\dfrac{(5^2)^3}{25^7}=
\dfrac{(5^3)^6}{(5^2)^3}\dfrac{(5^2)^3}{(5^2)^7}=
\dfrac{5^{3\times 6 + 2\times 3}}{5^{2\times 3 + 2\times 7}}=
5^{24-20}=
5^4=
625.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Gauss Jordan elimination reduces to row-echelon form always? I am reading this text:
and I'm wondering if gauss-jordan elimination always leads to an identity matrix on the left? If so, that helps me understand this passage:
I'm trying to figure out why [A 0] can be rewritten as [I 0]. Why is this?
| Gauss-Jordan elimination leads to identity matrix if the equation system has a unique solution. Book gives an example when this is the case so here is an example when this is not the case:
Example: Take the equation system
$$x+y+z = 2$$
$$x+y = 1$$
$$z = 2$$
which has no solution and can be represented as
$$ \begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I find the limit of the following sequence $\sin ^2 (\pi \sqrt{n^2 + n})$? How can I find the limit of the following sequence:
$$\sin ^2 (\pi \sqrt{n^2 + n})$$
I feel that I will use the identity $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n})), $$
But then what? how can I deal with... | Assuming that you want to know more than the limit itself.
Using $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n}))$$ use Taylor series
$$\sqrt{n^2 + n}=n+\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$
$$2\pi\sqrt{n^2 + n}=(2n+1)\pi -\frac{\pi}{4 n}+O\left(\frac{1}{n^2}\right)$$
$$\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Pell's equation (or a special case of a second order diophantine equation) Question Find integers $x,y$ such that $$x^2-119y^2=1.$$
So far I've tried computing the continued fraction of $\sqrt{119}$ to find the minimal solution, but either I messed up or I don't know where to stop computing a rough approximation of sai... | We will compute the continued fraction by the following algorithm (may or may not be the most efficient way in doing this, but it's the way I was taught!):
We set the following $$x:=\sqrt{119},\quad x_0=x,\quad a_i=\lfloor x_i\rfloor,\quad x_{i+1}=(x_i-a_i)^{-1},$$
and then we stop once we reach a loop in $x_i$ terms a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Zeroes of a polynomial. Evaluate an expression Let $x_1,x_2,x_3$ be the zeros of the polynomial $7x^3+24x^2+2016x+i$. Evaluate $(x_1^2+x_2^2)(x_2^2+x_3^2)(x_3^2+x_1^2)$.
My thoughts: I've tried $7(x-x_1)(x-x_2)(x-x_3)=0$ and expanded it out to match the polynomial given and got an ugly system of equations (which I can ... | The general method is to express the symmetric polynomial $A=(x_1^2+x_2^2)(x_2^2+x_3^2)(x_3^2+x_1^2)$ as a polynomial of elementary symmetric polynomials: $\sigma_1=x_1+x_2+x_3$, $\sigma_2=x_1x_2+x_2x_3+x_3x_1$, $\sigma_3=x_1x_2x_3$. Then you apply Vieta's formula.
Let's do the first step. Observe that
$$x_1^2+x_2^2+x_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$ Problem
Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$
Attempt to solve
if i set $z=x+iy$ when $(x,y) \in \mathbb{R}, z \in \mathbb{C}$
$$ |\frac{1}{x+iy}|<1 $$
Trying to multiply denominator and nominator with complex conjugate.
$$ |\frac{x-iy}... | |z| > 1
that means outside the circle of radius 1 , centered at (0,0)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } ... | Is because for any $y>0$
$$
\sqrt{y}=\frac{y}{\sqrt{y}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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integral of $\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx} $
Solve $\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx} $.
Answer is 1007.
I tried multiplying $\sqrt{x}-\sqrt{2014-x}\;$,
which results in $\frac{\sqrt{2014-x}(\sqrt{x}-\sqrt{2014-x})}{2x-2014}=$$\frac{\sqrt{2014x-x^2}}{... | If you substitute $x \mapsto 2014-x$, you get
$$ I = \int_0^{2014} \frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} \, dx = \int_{0}^{2014} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx. $$
Hence
\begin{align}
1007 &= \frac{1}{2}\int_0^{2014} dx = \frac{1}{2}\int_0^{2014}\frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$. Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous... | $$ \left( x^{3} + 1 \right) $$
$$ \left( x^{2} + 1 \right) $$
$$ \left( x^{3} + 1 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x \right) } + \left( - x + 1 \right) $$
$$ \left( x^{2} + 1 \right) = \left( - x + 1 \right) \cdot \color{magenta}{ \left( - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Please show detailed steps of double integration of absolute difference Please show detailed steps of integration
$$\int_0^2 \int_0^1 0.5|x-y| dxdy$$
$$=\frac{1}{2}\int_0^2\int_0^y(y-x)dxdy + \frac{1}{2}\int_0^2\int_y^1(x-y)dxdy$$
$$=\frac{1}{2}\int_0^2\{y\int_0^ydx - \int_0^yxdx\}dy + \frac{1}{2}\int_0^2\{\int_y^1xdx ... | The region of integration, $0\le y\le 2$ and $0\le x\le 1$, is a rectangle in the xy-plane. y> x, so |y- x|= y- x above the line form the lower left corner, (0, 0), to the upper right corner, (1, 2) and y< x, so |y- x|= x- y below it. That line is given by y= 2x so the integral is $\int_{x= 0}^1 \int_{y= 0}^{2x} (x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911118",
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"source": "stackexchange",
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The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\... | All you had expected, was solved by other solutions. I'd like to note some points. Maybe it has another taste. We know that for every natural number $n ≥ 1$, if $P(x)$ is a polynomial of degree $n$ with leading coefficient $a$, then $$\lim_{x\to\infty}P(x)=\infty (-\infty)$$ whether $a$ is positive (negative). This mea... | {
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"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
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An Olympiad question about equality How many $(x, y)$ positive integer pairs have $y^2-x^2=2y+7x+4$ equality?
I can't solve this Olympiad question.
$\textbf{Solution:}$ If above equality is regulated,we obtain $(2x+2y-5)(2x-2y+9)=29$. And, $29$ is prime we can easily find solution.
*
*$\textbf{1-)}$ How can we regu... | It's just completing the square,
$$y^2-2y-(x^2+7x)=4$$
$$y^2-2y+1-1-\left(x^2+7x + \left(\frac72\right)^2 \right)+\left(\frac72\right)^2=4$$
$$(y-1)^2-1-\left(x+ \frac72 \right)^2+\left(\frac72\right)^2=4$$
Multiply everything by $4$.
$$(2y-2)^2-4-\left(2x+ 7 \right)^2+49=16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The $m$-order submatrixs respectively from two $m\times n$ matrixs
Suppose that $A,B\in Mat_{m\times n}(\mathbb{C}),m\leq n.\det(A\overline{A^{'}})\ne 0,\det(B\overline{B^{'}})\ne 0.$
Show that
$$ \exists C\in Mat_{m\times m}(\mathbb{C}),s.t. B=CA. \Longleftrightarrow $$$$\exists \lambda\in \mathbb{C},s.t. \forall 1\... | Let $A = [A_1\;A_2]$ and $B = [B_1\;B_2]$ and assume that $A_1$ and $B_1$ are invertible. If this is not the case, rearrange the columns. Set $D := \det(A_1)$. Then $\det(B_1) = \lambda D$. Define $C := B_1A_1^{-1}$. Then $CA_1 = B_1$.
Now, let $a$ be an arbitrary column in $A_2$ and $b$ the corresponding one in $B_2$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the limit by using the definition of derivative. Here is the problem.
Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of
$$ L = \lim_{x \to 1} {\frac{\arctan{\sqrt{f(x)}-\arctan{f(x)}}}{ \left (\arcsin{\sqrt{f(x)}}-\arcsin{f(x)}\right)^2}} $$
I have tried using
$$
L=\lim... | There are several steps that are not clearly allowed.
The first step is okay:
$$
\frac{ \arctan \sqrt{f(x)} - \arctan f(x) }{ \left( \arcsin \sqrt{f(x)} - \arcsin f(x) \right)^2 }
= \frac{ (\arctan \sqrt{f(x)} - \arctan 1) - (\arctan f(x) - \arctan 1) }{ \left( (\arcsin \sqrt{f(x)} - \arcsin 1) - (\arcsin f(x) - \arcsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus) $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ I'll use induction to solve prove this. Then
$$\sum_{k=0}^n \binom{n}{k} = 2^n = \binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n}$$
First prove with n = 1
$$\binom{1}{0} + \binom{1}{1} = 2^1... | These are corollaries of the Binomial Theorem, which states that
$$(x + y)^n = \sum_{k = 0}^{n} \binom{n}{k} x^{n - k}y^k$$
If we set $x = y = 1$, we obtain
$$2^n = (1 + 1)^n = \sum_{k = 0}^{n} 1^{n - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$
which means that the number of subsets of a set with $n$ elements is $2^n$.
I... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Solving roots of a polynomial on $\mathbb{Z}_p$ This is probably a simple question. and I would like to work out an example.
How do we solve $x^2 + x + [1] = 0$ over the field $\mathbb{Z}_7$?
I tried a simple case, for example: $[4] x - [3] = 0$, where I find $x = [3][4]^{-1}$ where we find $[4]^{-1} = [2]$ so $x = [6]... | Suppose $F$ is any field with
$\text{char}(F) \ne 2, \tag 1$
and
$q(x) = x^2 + ax + b \in F[x] \tag 2$
is a quadratic polynomial with coefficients in $F$; we wish to find the zeroes of $q(x)$; short of simply evaluating $q(x)$ for succesive elements $x \in F$ (which in any event will in principle only be guaranteed to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Generating Function Sum and Combinotorics Problem:
The sum $\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$ has a finite value. Determine that value.
I am quite stuck on how to do this. Can somebody give me <only> a hint or hints to get going?
Th... | Hint: start repeatedly differentiating the generating function for $\frac{1}{1 - x}$.
Answer:
We have the generating function/Maclaurin series for $\frac{1}{1 - x}$:
$$\frac{1}{1 - x} = 1 + x + x^2 + \ldots \quad |x| < 1$$
Differentiating, which we do term by term does not change the radius of convergence, giving ... | {
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Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
| Since $7\cdot 2^k+1$ is odd we have that
$$7\cdot 2^k+1=(2n+1)^2=4n^2+4n+1 \iff7\cdot 2^k=2^2n(n+1) \iff 7\cdot 2^{k-2}=n(n+1)$$
and since one among $n$ and $n+1$ must be odd therefore amd since $7$ is prime, one of them must be equal to $7$ and then the unique solution is $2^{k-2}=8 \implies k=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the ... | Slightly Sharper Bounds
For $x\geq 0$, we have by the AM-GM Inequality that
$$x^3+2\cdot1\geq 3\sqrt[3]{x^3\cdot1^2}=3x\,.$$
That is,
$$x^3+4\geq 3x+2\,.$$
When $0\leq x\leq 2$, we also get
$$x^3+4\leq \left(x^{\frac{3}{2}}+2\right)^2\leq \big(\sqrt{2}x+2\big)^2\,.$$
Therefore,
$$\frac{1}{\sqrt{2}x+2}\leq \frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$ $$
y =\ 2\ {x} - {x}^2
$$
$$
y =\ -{x}
$$
According to me , the area
$$
\int_{0}^{2}{2x\ -\ { x} ^2}\, dx \ + \int_{2}^{3}{\ {x} ^2\ -\ 2{x} }\, dx \\
$$
Which gives the area $ \frac{8}{3}$
But the answer is $ \frac{9}{2}$
| You should go around the path in clockwise fashion:
$$
\int_0^3(2x-x^2)\,dx+\int_3^0(-x)\,dx=
\int_0^3(2x-x^2+x)\,dx=\Bigl[\frac{3}{2}x^2-\frac{1}{3}x^3\Bigr]_0^3=\frac{27}{2}-9=\frac{9}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$
I would like to find the minimum of
$$f(x)=\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n,$$
where $n$ is a natural number.
I know there is possible by derivate, but
$$f'(x)=n \left(\left(... | You're probably right. But there is a way without using calculus. You should only use Cauchy's inequality:
$$ \frac{a_1+a_2}{2}\geq \sqrt{a_1a_2}.$$
$$ \left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n=\left(1+\frac{1}{\sin^2x}\right)^n+\left(1+\frac{1}{\cos^2x}\right)^n \geq \\ \geq 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
About Spliting Field What is splitting field of polynomial $X^3+X+\bar 1$ in $\mathbb{F_5}$.
Attempt: To show this we first want to check irreducibility which is clear in case of finite field since there are only 5 elements so put values and check it has no root in $\mathbb{F_5}$. After that how to proceed?
| My answer to this question with software is in the following form.
First, I find a positive integer $n$ such that in the decomposition of $x^n-1$ in modulo $5$ we have the factor $x^3+x+1$. In your question we get $n=62$. In fact, we get
$$
(x^{62}-1) \pmod{5}=
\left( {x}^{3}+2\,x+4 \right) \left( {x}^{3}+2\,x+1 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of at least one king in a 13-card hand? What is the probability of drawing 13 cards and having at least one king?
Here is what I came up with:
$$1 - \frac{\begin{pmatrix}
48\\
13
\end{pmatrix}}{\begin{pmatrix}
52\\
13
\end{pmatrix}}$$
Is this correct?
Thanks.
| $\begin{pmatrix}52 \\ 13\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 52 card deck. $\begin{pmatrix}52 \\ 48\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 48 card deck- with the kings missing. So your $\frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve for $u$ the PDE $(x − y − 1)u_x + (y − x − u + 1)u_y = u$ if $u=1$ on $x^2+(y+1)^2=1.$
Solve the Cauchy problem $$(x − y − 1)u_x + (y − x − u + 1)u_y = u,$$
if $u=1$ on $x^2+(y+1)^2=1.$
Attempt. $$\frac{dx}{x-y-1}=\frac{dy}{y-x-z+1}=\frac{dz}{z}$$
so $$\frac{dx+dy}{(x-y-1)+(y-x-z+1)}=\frac{dz}{z}\iff d(x+y+z)... | $$(x − y − 1)u_x + (y − x − u + 1)u_y = u \tag 1$$
$$\frac{dx}{x-y-1}=\frac{dy}{y-x-u+1}=\frac{du}{u}\quad\text{is correct}$$
You rightly found a first characteristic equation :
$$x+y+u=c_1$$
A second characteristic equation comes from
$$\frac{dx-dy}{(x-y-1)-(y-x-u+1)}=\frac{du}{u}=\frac{d(x-y-1)}{2(x-y-1)+u}$$
With $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that for any integer $n$, there will always exist integers $x,y$ such that $n^3 = x^2-y^2$ Problem: Show that for any integer $n$, there will always exist integers $x,y$ such that $n^3 = x^2-y^2$
My attempt:
If $n^3 = x^2-y^2$, then $x = y \mod(n)$, and at the same time $x = -y \mod(n)$,
So that $2x = 0 \mod(n)$.... | One way to systematically arrive at an answer is to assume that
$$
\begin{align}
x &= a_2 n^2 + a_1 n + a_0\\
y &= b_2 n^2 + b_1 n + b_0
\end{align}
$$
We may assume $a_2,b_2>0$, changing signs if necessary. Then comparing coefficient of $n^4$ in
$$
n^3 = x^2 - y^2
$$
forces
$$
a_2 = b_2
$$
Next, comparing coefficient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| As an alternative by
$$\sin^3 x+ \cos ^3 x=1 \iff \sin x\cdot \sin^2+\cos x\cdot \cos^2 x=1$$
since $\sin^2 x+ \cos ^2 x=1$, the given equality is a weighted mean of $\sin x$ and $\cos x$ which holds if and only if
*
*$\sin x=1,\,\cos x=0$
or
*
*$\cos x=1,\,\sin x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Proving that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$ I want to prove that $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}, $$
using the Fourier Series for the $2\pi$-periodic function $f(\theta)=\theta^{2},\quad (-\pi<0<\pi)$, that is,
$$\theta^{2}=\frac{\pi^{2}}{3}+4\sum_{n=1... | Split it!
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\sum_{n ~\text{odd}}\frac{1}{n^{2}} - \sum_{n ~\text{even}}\frac{1}{n^{2}}.$$
Add and subtract the "even" part:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\left(\sum_{n ~\text{odd}}\frac{1}{n^{2}} + \sum_{n ~\text{even}}\frac{1}{n^{2}}\right) - \sum_{n ~\text{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Rolles theorem used for solving equation $ax^3+bx^2+cx+d=0$ If a,b,c,d are Real number such that
$\frac{3a+2b}{c+d}+\frac{3}{2}=0$. Then the equation $ax^3+bx^2+cx+d=0$ has
(1) at least one root in [-2,0]
(2) at least one root in [0,2]
(3) at least two root in [-2,2]
(4) no root in [-2,2]
I am doing hit and trial m... | Let $$f(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx.$$
Then $f0)=0$ and $f(2)=\frac{16a}{4}+\frac{8b}{3}+\frac{4c}{2}+2d=\frac{48a+32b+24c+24d}{12}=\frac{16(3a+2b)+24(c+d)}{12}=0$ (from the condition given).
Now apply Rolle's theorem. There exists a point in $[0,2]$, where the derivative of $f$ must be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding value of $ \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$
Finding value of $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$
Try: $$\lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \frac{2k}{2k+1} = \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1... | With Stirling's approximation $n!\sim\sqrt{2\pi n}(n/e)^n$ we get:
\begin{align}
\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}
&=\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\\
&=\prod^{n}_{k=1}\frac{(2k)^2}{(2k-1)(2k)}\cdot\frac{(2k)^2}{(2k)(2k+1)}\\
&=4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k-1)(2k)}\cdot 4^n\cdot\prod^{n}_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove that $1/6 < \int_0^1 \frac{1-x^2}{3+\cos(x)}dx < 2/9$
Prove that $$\frac{1}{6}<\int_0^1 \frac{1-x^2}{3+\cos(x)}dx < \frac{2}{9}. $$
I tried using known integral inequalities (Cauchy-Schwarz, Chebyshev) but I did not arrive at anything. Then I also tried considering functions of the form $$f(x) = \int_0^x \frac{... | For any $x\in(0,1)$
$$ \frac{1-x^2}{3+\cos x}-\frac{1-x^2}{4} = \frac{(1-x^2)\sin^2\frac{x}{2}}{2(3+\cos x)}\in\left[0,\frac{x^2(1-x)}{12}\right] $$
hence the given integral is bounded between $\frac{1}{6}=\int_{0}^{1}\frac{1-x^2}{4}\,dx$ and $\frac{1}{6}+\frac{1}{144}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find $A$ if solution to $Ax=b$ is given
Solution to $Ax=b$ is $x=\begin{bmatrix}
1\\
0\\
1\\
0
\end{bmatrix}+\alpha_{1}\begin{bmatrix}
1\\
1\\
-1\\
0
\end{bmatrix}+\alpha_{2}\begin{bmatrix}
1\\
0\\
1\\
1\end{bmatrix}$. For $b=(1,2,1)^T$, find $A$.
Somehow, Gilbert Strang says that it is obvious that first an... | You have that
$$
A
\begin{bmatrix}
1+\alpha_1+\alpha_2 \\
\alpha_1 \\
1-\alpha_1+\alpha_2 \\
\alpha_2
\end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}
$$
If $A=[a_1\ a_2\ a_3\ a_4]$, this means that
$$
(1+\alpha_1+\alpha_2)a_1
+\alpha_1a_2
+(1-\alpha_1+\alpha_2)a_3
+\alpha_2 a_4
=\begin{bmatrix} 1 \\ 2 \\ 1 \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Proof Check: Let $f_{0}(x)=\frac{1}{1-x},$ and $f_{n}(x)=f_{0}(f_{n-1}(x))$ for any positive integer $n$. Find $f_{2018}(2018)$ Let us denote by $F$ the set of real-valued functions of a real variable such that $$F=\left\{\frac{1}{1-x},\frac{x-1}{x},x\right\}.$$ We shall prove that this set forms an abelian group under... | Another method:
Check that : $$f_0(x)=\frac {1}{1-x}$$
$$f_1(x)=1-\frac 1x$$
$$f_2(x)= x$$
And now hereafter it would be quite obvious that $$f_n(x) =
\begin{cases}
\frac{1}{1-x}, & \text{if $n=3k$} \\
1-\frac 1x, & \text{if $n=3k+1$} \\
x, & \text{if $n=3k+2$}
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition. Problem
Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition.
Note:
The problem asks us to prove that, no matter $x \to +\infty$ or $x \to -\infty$, the limit is $\infty$,which may be $+\infty$ or $-\infty.... | Your proof looks ok to me.
An option:
$x^3+1=(x+1)(x^2-x+1);$
$x^2+1<x^2+x =x(x+1)$, for $x >1$.
$\dfrac{(x+1)(x^2-x+1)}{x^2+1}>$
$\dfrac{(x+1)(x^2-x+1)}{x(x+1)}=$
$x-1+1/x >x-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Limit of $\frac{y^4\sin(x)}{x^2+y^4}$ when $(x,y) \to (0,0)$ I have to find the following limit when $x,y$ tend towards $0$. I think the limit doesn't exist (thanks to Wolfram Alpha), but all I can find no matter what path I use (I've tried $y=x², x=y^2, x=0, y=0$) is that the limit equals $0$ (because of the $sin(0)$)... | We have that
$$\frac{y^4\sin x}{x^2+y^4}=\frac{\sin x}{x}\frac{y^4x}{x^2+y^4}$$
then recall that $\frac{\sin x}{x}\to 1$.
To show that $\frac{y^4x}{x^2+y^4} \to 0$ we can use inequalities or as an alternative by $u=x$ and $v=y^2$
$$\frac{y^4x}{x^2+y^4}=\frac{uv^2}{u^2+v^2}=r\cos \theta\sin^2 \theta \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim_{x \to 4} \frac{x^4-4^x}{x-4}$, where is my mistake? Once again, I am not interested in the answer. But rather, where is/are my mistake(s)? Perhaps the solution route is hopeless:
Question is: evaluate $\lim_{x \to 4} \frac{x^4 -4^x}{x-4}$.
My workings are:
Let $y=x-4$. Then when $x \to 4$, we have that ... | The step where you rewrite the limit as the difference of two other limits
((i) and (ii)) is not legitimate.
You can only equate a limit to a sum or difference of two limits
if both those limits converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the volume of the solid obtained by rotating the region bounded by: Q: "Find the volume of the solid obtained by rotating the region bounded by:
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$x + y = 4$, $ $ $ $$x = 5 - (y-1)^2$
$ $ $ $ $ $ $ $ $ $ $ $ $ $ About the $x-axis$."
I have attempted this problem for a while no... | The equation $x+y=4$ cuts out a cone upon rotation from the solid bounded by the parabola. This occurs for $x\in[1,4]$.For $x>4$, $-\sqrt(5-x)+1>0$ and so also contributes to the integral. You can reduce the work you do by using geometry and subtraction.
$V=\pi\int_1^5 (\sqrt{5-x}+1)^2 dx$-$\pi\int_4^5 (-\sqrt{5-x}+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957308",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Combinatorics rolling a die six times How many ways are there to roll a die six times such that there are more ones than twos?
I broke this up into six cases:
$\textbf{EDITED!!!!!}$
$\textbf{Case 1:}$ One 1 and NO 2s --> 1x4x4x4x4x4 = $4^5$. This can be arranged in six ways: $\dfrac{6!}{5!}$. So there are $\dfrac{6!}{... | Case 1 is wrong because there are $4$ numbers that are neither $1$ nor $2$, so there are $6 \cdot 4^5$ ways to roll one $1$ and no $2$s. That error repeats.
Added: Case 2 is wrong because the number of ways to arrange the numbers is dependent on whether there is a $2$ or not. The $1 \cdot 1 \cdot 5 \cdot 4^3$ is r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Maximise $(x+1)\sqrt{1-x^2}$ without calculus Problem
Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$
With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking that $f''(\frac{1}{2})<0$ to obtain the final answer of $f(\frac{1}{2})=\fra... | $$\dfrac{1+x+1+x+1+x+3(1-x)}{3+1}\ge\sqrt[4]{(1+x)^33(1-x)}$$
Square both sides
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$
Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $$\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$$
I have attempted multiple times in this question and the only method that I ... | Hint $${ab+4\over a+2} = {2ab+8\over 2a+4} = {2ab+abc\over 2a+4} = {ab\over 2}\cdot{2+c\over a+2}$$
Similary for the rest of terms. Now use AM-GM for 3 terms...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How do you show that $\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$? how to show that $$\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$$? Can I use the Alternating Series test and how?
| $$S_1= \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$$
$$\dfrac{3n-1}{n^2 + n} = \dfrac{3-\frac{1}{n}}{n + 1} = \left(\frac{4}{n+1} - \frac{1}{n}\right)$$
So
\begin{align}
S_1 &= \sum_{n=1}^{\infty}(-1)^{n}\left(\frac{4}{n+1} - \frac{1}{n}\right)\\
&= 4\underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n+1}}_{s_2} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How can I prove that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}+i)$? It is easy to show that $\mathbb{Q}(\sqrt[3]{2}+i)\subseteq \mathbb{Q}(\sqrt[3]{2},i)$.
But how I can show that $\mathbb{Q}(\sqrt[3]{2},i)\subseteq\mathbb{Q}(\sqrt[3]{2}+i)$?
I can't find a way to express $\sqrt[3]{2}$ in terms of $\sqrt[3]{2}... | Follows a sequence of maneuvers which shows how to express $i$ and $\sqrt[3]2$ in terms of $\sqrt[3]2 + i$, and hence that $\Bbb Q(\sqrt[3]2, i) = \Bbb Q(\sqrt[3]2 + i)$:
$\alpha = \sqrt[3]2 + i; \tag 1$
$\alpha - i = \sqrt[3]2; \tag 2$
$(\alpha - i)^3 = 2; \tag 3$
$\alpha^3 - 3 \alpha^2 i - 3 \alpha + i = 2; \tag 4$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{... | It seems right and it is bounded, as a minor observation we can improve the upper bound
$$x_n < \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \le\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$. Solution
Notice that
$$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$
Thus, we obtain
$$e>\frac{n}{... | You are right with your limit $e$. Using Stirling's formula you have putting $f(n)=\dfrac{n}{\sqrt[n]{n!}}$,
$$\log f(n)=\log n-\frac{n\log n-n+O(\log n)}{n}\Rightarrow \log f(n)-1=\frac{O(\log n)}{n}$$ By definition of Big $O$ notation there is some positive constant $C$ and some $n_0$ such thar for all $n\gt n_0$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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How can partial fractions be used for deductions?
Find partial fractions of the expression,$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}$
. Hence deduce that; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(... | This is $$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}+\frac{D}{x-d}$$
The result should be this here
$$1+{\frac {{c}^{4}-{c}^{3}p-{c}^{3}q-{c}^{3}r-{c}^{3}s+{c}^{2}pq+{c}^{2
}pr+{c}^{2}ps+{c}^{2}qr+{c}^{2}qs+{c}^{2}rs-cpqr-cpqs-cprs-cqrs+pqrs}{
\left( -c+a \right) \left( -c+b \right) \left( -d+c \right)
\left( x-c \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sum the first $n$ terms of the series $1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots$ The question
Sum the first $n$ terms of the series:
$$ 1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots. $$
This was asked under the heading using method of difference and the answer ... | We have:
$$\sum_{r=2}^n{r^2(r+1)(r-1)}=\sum_{r=2}^n{r^2(r^2-1)}=\sum_{r=2}^n{r^4}-\sum_{r=2}^n{r^2}$$
Just use (or show) the well-known formulae to evaluate this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+... | Setting $z=e^{ix}$, the equation can be rewritten in a quadratic form
$$a\frac{z+z^{-1}}2+b\frac{z-z^{-1}}{2i}=c,$$
$$(a-ib)z^2-2cz+a+ib=0$$
and by Vieta, the product of the roots (in $z$) is
$$\frac{a+ib}{a-ib},$$ giving the identity
$$\cos(\alpha+\beta)+i\sin(\alpha+\beta)=e^{i\alpha}e^{i\beta}=\frac{a^2-b^2}{a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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How Isolate y from $a=\sqrt{(y^2+(a+x)^2)^3}$ Just that!
I'm asking for a method to isolate $y$ from such expressions:
$$a=\sqrt{(y^2+(a+x)^2)^3}$$
or even easier:
$$a=\sqrt{(y^2+x^2)^3}.$$
EDIT:
I'm just trying to revisit some of the basics in algebraic equation manipulation.
Even it's sooo easy to find the school ex... | Taking the square of both sides of $a=\sqrt{(y^2+(a+x)^2)^3}$, one may get
$$ a^2=(y^2+(a+x)^2)^3$$
Now we apply the cube root,
$$ a^\frac{2}{3}=y^2+(a+x)^2$$
Then, we have
$$ y^2=a^\frac{2}{3}-(a+x)^2$$
Finally, we take the square root of both sides,
$$y=\pm \sqrt{a^\frac{2}{3}-(a+x)^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2+5xy}{x-y}$ does not exist $$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2+5xy}{x-y}$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic.
Hints? Thank you.
| A coordinate transformation can work - but you just have to get a little bit more ingenious. The other answers have shown that the trick is the line $x = y$, so we should choose a transformation which makes analyzing it a bit more palatable. Consider the transformation to a tilted coordinate system in which the $y$-axi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding $\lim_{n\to \infty}\sqrt n \int_0^1 \frac{\,dx}{(1+x^2)^n}$ $$
\lim_{n\to\infty} n^{1/2}
\int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0
$$
Is my answer correct?
But I am not sure of method by which I have done.
| Let
\begin{align*}
I_n&=\int_0^\infty\frac{dx}{(1+x^2)^n}=\left.\frac{x}{(1+x^2)^n}\right|_{x=0}^\infty+\int_0^\infty\frac{2nx^2\,dx}{(1+x^2)^{n+1}}\\
&=2n\int_0^\infty\frac{dx}{(1+x^2)^n}-2n\int_0^\infty\frac{dx}{(1+x^2)^{n+1}}=2nI_n-2nI_{n+1}
\end{align*}
So we get
\begin{align*}I_{n+1}&=\frac{2n-1}{2n}I_n=\frac{2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
} |
On existence of positive integer solution of $\binom{x+y}{2}=ax+by$ How can I prove this?
Prove that for any two positive integers $a,b$ there are two positive integers $x,y$ satisfying the following equation:
$$\binom{x+y}{2}=ax+by$$
My idea was that $\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$ and choose... | Consider the equation $f(x,y)=x^2+2xy+y^2-(2a+1)x-(2b+1)y=0$.
If $a=b$ then this is satisfied along the parallel lines $x+y=0$ and $x=y+2a+1$, so we can choose say $x=1$, $y=2a$.
From now on we assume $a>b$. Then the equation is satisfied along a parabola, passing through points $O=(0,0)$, $A=(2a+1,0)$ and $B=(0,2b+1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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What is the remainder when $4^{10}+6^{10}$ is divided by $25$? Without using calculator, how to decide? Must go with last two digits of $4^{10}+6^{10}$, can tell the last digit is $2$. How to tell the tenth digit of the sum?
Thanks!
| $$4^4=256 \equiv 6 \mod (25)$$
$$ 4^8 \equiv 36 \equiv 11 \mod (25)$$
$$ 4^{10} \equiv 16\times 11= 176 \equiv 1 \mod (25)$$
$$6^2 =36 \equiv 11 \mod (25)$$
$$6^4 \equiv 121 \equiv -4 \mod (25)$$
$$6^8 \equiv 16 \mod (25)$$
$$6^{10} \equiv 176 \equiv 1 \mod (25)$$
$$4^{10} + 6^{10} \equiv 2 \mod (25)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding a matrix M of rank 2 such that AM = 0
Let
$$ A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
-1 & 1 & 3 & -1 & 0\\
-2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix} $$
Find a $ 5 \times 5$ matrix $M$ with rank 2 such that $AM =0_{4\times5}$
My logic was to row reduce $A$ into a matrix $B$ such that:
$$
A =... | We have that
$$A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
-1 & 1 & 3 & -1 & 0\\
-2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix}\to \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
0 & 1 & 2 & 1 & 1\\
0 & 1 & 2 & 3 & 5\\ 0 & -1 & -2 & -5 & -9\\ \end{bmatrix}\to$$
$$\begin{bmatrix}
1 & 0 & -1 & 2 &1\\
0 & 1 & 2 & 1 & 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\bi... | Write $s := \sin x, c := \cos x$.
Hint Using $s^2 + c^2 = 1$, we can write our expression as
$$c^6 - 4 c^4 + 8 c^2 = (1 - s^2)^3 - 4 (1 - s^2)^2 + 8 (1 - s^2) = -s^6 - s^4 - 3 s^2 + 5 .$$
Now, perform polynomial long division by $s^3 + s^2 + s - 1$.
Carrying out long division gives that our expression is $$(s^3 + s^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Finding the affine transformation that turns the unit circle into $x^2+y^2+xy-x-y =1$
How to find an affine transformation $T: \mathbb{R^2} \to \mathbb{R^2}$ such that
$$T(\lbrace (x,y) \in \mathbb{R^2}:x^2+y^2 \leq 1\rbrace)= \lbrace (x,y) \in \mathbb{R^2}:x^2+y^2+xy-x-y \leq 1 \rbrace)$$
The affine transformation c... | Let's first picture the set on the right:
Now that we know what we are dealing with an ellipse that is tilted by $45^\circ$ it is natural to try to rewrite the latter using the coordinates $t=x+y$ and $u=x-y$. Now since it's an ellipse it makes sense to try rewriting
$$x^2+y^2+xy-x-y=a\cdot(x+y+b)^2+c\cdot(x-y+d)^2-ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inequality with fraction and n-th root Prove that $$ p(\sqrt[p]{n+1}-1)< \frac{1}{\sqrt[p]{1}}+\frac{1}{\sqrt[p]{2^{p-1}}}+...+\frac{1}{\sqrt[p]{n^{p-1}}}< p\sqrt[p]{n} \quad p\in \mathbb{N},p\ge 2 $$
I used AM-GM to prove it.
For right hand side of inequality
From AM-GM we get
$n+1+(p-1)n>p\sqrt[p]{n^{p-1}(n+1)}$
... | Note that $f(x)=\dfrac{1}{\sqrt[p]{x^{p-1}}}=x^{(1/p-1)}$ is a decreasing function
$$\frac{1}{\sqrt[p]{(k+1)^{p}}}
<\int_{k}^{k+1} \frac{dx}{\sqrt[p]{x^{p-1}}}
<\frac{1}{\sqrt[p]{k^{p}}}$$
$$\sum_{k=0}^{n-1} \frac{1}{\sqrt[p]{(k+1)^{p}}}
<\int_{0}^{n} \frac{dx}{\sqrt[p]{x^{p-1}}}
\implies
\frac{1}{\sqrt[p]{1^{p}}}+\ldo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integrate squared trigonometric function I'm trying to integrate $\int_a^b \left( \frac{1}{1+x^2} \right)^2 dx$
I know that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$, but how can I integrate with the squared part?
I've tried substitution with no success.
| Consider the integral
$$I_n=\int\frac{\mathrm{d}x}{(ax^2+b)^n}$$
Integration by parts: $$\mathrm{d}v=\mathrm{d}x\Rightarrow v=x\\u=\frac1{(ax^2+b)^n}\Rightarrow \mathrm{d}u=-2an(ax^2+b)^{-n-1}x\mathrm{d}x$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2n\int\frac{ax^2\mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2n\int\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Find the sum of $1-\frac17+\frac19-\frac1{15}+\frac1{17}-\frac1{23}+\frac1{25}-\dots$
Find the sum of $$1-\frac17+\frac19-\frac1{15}+\frac1{17}-\frac1{23}+\frac1{25}-\dots$$
a) $\dfrac{\pi}8(\sqrt2-1)$
b) $\dfrac{\pi}4(\sqrt2-1)$
c) $\dfrac{\pi}8(\sqrt2+1)$
d) $\dfrac{\pi}4(\sqrt2+1)$
I have tried a lot.. But i can'... | Here is an alternative solution in the case you do not know the cotangent series given in the amazing solution by user10354138. Let $$f(z):=1-z^6+z^8-z^{14}+z^{16}-z^{22}+\ldots\text{ for }z\in\mathbb{C}\text{ such that }|z|<1\,.$$
Then, $(1-z^8)\,f(z)=1-z^6$, so
$$f(z)=\frac{1+z^2+z^4}{(1+z^2)(1+z^4)}=\frac{1}{2\,(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Given big rectangle of size $x, y$, count sum of areas of smaller rectangles. Let's say we have two integers $x$ and $y$ that describe one rectangle, if this rectangle is splitten in exactly $x\cdot y$ squares, each of size $1\cdot 1$, count the sum of areas of all rectangles that can be formed from those squares.
For ... | It is useful to move from simpler cases to more complex ones!
Consider $1\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{1\times y}&=1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1=\\
&=\sum_{i=1}^yi\cdot (y-i+1)=\\
&=(y+1)\sum_{i=1}^y i-\sum_{i=1}^yi^2=\\
&=(y+1)\cdot \frac{y(y+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $z = cis(2k\pi/5)$, $z \neq 1$, then what is $(z+1/z)^2+(z^2 + 1/z^2)^2=$? question 20, part c in the picture:
I substituted the first time as $4 \cos^2(2k \pi/5)$ and the second term as $4 \cos^2(4k \pi/5)$, and then tried writing one term in terms of the other using the identity $\cos 2a = 2 \cos^2 a- 1$. I even t... | Hint:
If $5t=2k\pi,5\nmid k,\cos t\ne1$
$\cos3t=\cdots==\cos2t$
The roots of
$0=\dfrac{4\cos^3t-2\cos^2t-3\cos t+1}{\cos t-1}=4\cos^2t+2\cos t-1=0$ will be $$t=2k\pi,k\equiv\pm1,\pm2\pmod5$$
Now $z+\dfrac1z=2\cos\dfrac{2k\pi}5, \left(z+\dfrac1z\right)^2=\cdots=2\left(1+\cos\dfrac{4k\pi}5\right)$
$z^2+\dfrac1{z^2}=2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$M$ is a point in an equalateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$.
$M$ is a point in an equilateral triangle $ABC$ with the area $S$. Prove that $MA, MB, MC$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$
... | Solution using complex numbers. Copied from a deleted question per request.
( update - I have added another solution using circle inversion at end)
Solution 1 - using complex numbers.
Choose a coordinate system so that triangle $ABC$ is lying on the unit circle centered at origin and $A$ on the $x$-axis. Let $a = AM$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
System of equations with three variables Characterize all triples $(a,b,c)$ of positive real numbers such that
$$ a^2-ab+bc = b^2-bc+ca = c^2-ca+ab. $$
This is the equality case of the so-called Vasc inequality. I think the answer is that $a=b=c$ or $a:b:c = \sin^2(4\pi/7) : \sin^2(2\pi/7) : \sin^2(\pi/7)$ and cyclic e... | Too long for comment
Using $\sin(\pi-A)=+\sin A,\cos(\pi-B)=-\cos B,\sin2C=2\sin C\cos C$
$$\dfrac{\sin^2\dfrac{2\pi}7}{\sin^2\dfrac{\pi}7}=4\cos^2\dfrac\pi7=2+2\cos\dfrac{2\pi}7$$
$$\dfrac{\sin^2\dfrac{\pi}7}{\sin^2\dfrac{3\pi}7}=\dfrac{\sin^2\dfrac{6\pi}7}{\sin^2\dfrac{3\pi}7}=4\cos^2\dfrac{3\pi}7=2+2\cos\dfrac{6\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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proving inequality through convexity and continuity Suppose $f$ is a continuous function. Prove $f$ is convex iff this inequality holds:
$$
\frac{f(x)+f(y)+f(z)}{3} + f(\frac{x+y+z}{3}) \ge \frac{2}{3}[f(\frac{x+y}{2}) +f(\frac{y+z}{2}) +f(\frac{z+x}{2})]
$$
if we move all the terms to the left hand side, we'll have:
$... | This is not true for functions $f:\mathbb R^n \to \mathbb R$ for $n\ge2$:
Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \co... | Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Finding the equation of the line tangent to $x^2+y^2-6x+4y-3=0$ passing through $(7,4)$. How to proceed, not knowing point of tangency?
Find the equation of the tangent line to the circle $x^2+y^2-6x+4y-3=0$ which passes through the point $(7,4)$.
Graphing the circle, $(7,4)$ is a point not on the circle. So, I am as... | Without calculus, refer to the graph:
$\hspace{2cm}$
To find the line $AB$, let $B(x,y)$, then:
$$\begin{cases}AB=AC=6 \\ OB=4\end{cases} \Rightarrow \begin{cases} (7-x)^2+(4-y)^2=6^2 \\ (3-x)^2+(-2-y)^2=4^2 \end{cases} \Rightarrow \\
(x,y)=C(7,-2); B\left(\frac{19}{13}, \frac{22}{13}\right)$$
Hence, the tangent line $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the area of the surface formed by revolving the given curve about $(i)x$-axis and $(i)y$-axis
Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=a\cos\theta ,y=b\sin\theta,0\le\theta\le2\pi$$
About $x-$axis is, $S=2\pi\int_0^{2\pi}b\sin\theta \sqrt{a^2(\sin\t... | Hint: For the first one let $\cos\theta=u$
$$
S=2\pi\int_0^{\pi}b\sin\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta
=2\pi\int_{-1}^1 b\sqrt{a^2+(b^2-a^2)u^2}\ du
$$
and then let substitution $\sqrt{b^2-a^2}u=a\tan\phi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding function given limit $$\lim_{x\to2} \frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.
I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?
| We have that
$$\lim_{x\to 2}\frac{x^2-cx+d}{x^2-4}=3$$
requires that for $x=2$
$$x^2-cx+d=0 \implies 4-2c+d=0 \implies d=-4+2c$$
then
$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$
then
$$\lim_{x\to 2}\frac{x^2-cx+d}{x^2-4}=\lim_{x\to 2}\frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$
As an alternative once we recognize that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Picard Iteration/ index i have the following System of Differential Equations
$$ \begin{pmatrix}
x'(t) \\
y'(t)
\end{pmatrix} = \begin{pmatrix}
0 && 1\\
-1 && 0
\end{pmatrix} \begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix} \ and \ \begin{pmatrix}
x(0) \\
y(0)
\end{pmatrix}= \begin{pmatrix}
2 \\
0
\end{pmatrix} $$
When i ... | There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate this nonelementary integral? Let $x>0$. I have to prove that
$$
\int_{0}^{\infty}\frac{\cos x}{x^p}dx=\frac{\pi}{2\Gamma(p)\cos(p\frac{\pi}{2})}\tag{1}
$$
by converting the integral on the left side to a double integral using the expression below:
$$
\frac{1}{x^p}=\frac{1}{\Gamma(p)}\int_{0}^{\infty}e^{... | So let us follow your initial line of thought and convert the integral to a double integral. As you correctly observe, as
$$\frac{1}{x^p} = \frac{1}{\Gamma (p)} \int_0^\infty e^{-xt} t^{p - 1} \, dt,$$
which, by the way, is just the Laplace transform for the function $x^{p -1}$, as a double integral your integral can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \fr... | The other answers did not point out the logical gaps in your reasoning. It is unstated in your question but presumably you want to find non-negative real $x$ satisfying the given equation, so for any such $x$ you can let $y = \sqrt{x}$, and then you know that $y^2 = x$. Thus if you substitute $x$ by $y^2$ then solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
How to calculate the line integral with respect to the circle in counterclockwise direction Consider the vector field $F=<y,-x>$.
Compute the line integral $$\int_CF\cdot dr$$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.
My Try:
The circle is $x^2+y^2=9$
$$\cases{x=3\cos t \\ y=3\sin t... | The radial vector is
$\vec r = \begin{pmatrix} x \\ y \end{pmatrix}; \tag 1$
also,
$F = \begin{pmatrix} y \\ - x \end{pmatrix}; \tag 2$
with
$x = 3 \cos t, \tag 3$
$y = 3 \sin t, \tag 4$
we have
$\vec r = \begin{pmatrix} 3 \cos t \\ 3\sin t \end{pmatrix}, \tag 5$
$d \vec r = \dfrac{d \vec r}{dt} dt = \begin{pmatrix} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove the limit lim n→∞ $ {(2n^7 + 3n^5 + 4n) \over (4n^7 − 7n^2 + 5)} $ = 1/2 Im trying to use the formal definition of a limit to prove
$\lim \limits_{x \to ∞} $$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n... | I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$\frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=\frac{2+3\frac{1}{n^2}+4\frac{1}{n^6}}{4-7\frac{1}{n^5}+5\frac{1}{n^7}}$$
Note that $$\lim_{n\to\infty}\Bigl(\frac{1}{n^k}\Bigr)=0\;\forall k\in\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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