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$\langle 3 \rangle$ 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 obscures the fact that $1 + 2 \omega = \sqrt{-3}$, and since $(\sqrt{-3})^2 = -3$, it follows that $\langle 3 \rangle = \langle \sqrt{-3} \rangle^2$. Meaning that $\langle 3 \rangle$ ramifies, it doesn't split. This even though $$\left( \frac{3 - \sqrt{-3}}{2} \right) \left( \frac{3 + \sqrt{-3}}{2} \right) = 3.$$
That's because $$\left( \frac{-1 + \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{-\sqrt{-3} - 3}{2},$$ or $\omega(1 + 2 \omega) = \omega + 2 \omega^2$. Since $\omega$ is a unit, the ideals $\langle 1 + 2 \omega \rangle$ and $\langle \omega + 2 \omega^2 \rangle$ are in fact the same.
Are these calculations correct? Have I drawn the right conclusion?
| 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
$$\frac{\frac{3 - \sqrt{-3}}{2}}{\frac{3 + \sqrt{-3}}{2}}=\frac{(3 - \sqrt{-3})^2}{(3 + \sqrt{-3})(3 - \sqrt{-3})}=\frac{6-6\sqrt{-3}}{12}=\frac12-\frac{\sqrt{-3}}2,$$
is a unit since it has norm $1$.
| {
"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
} |
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",
"answer_count": 1,
"answer_id": 0
} |
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 recurrence relation but failed to get its solution. The relation is $R_{n+1} = R_{n}+\sum _{k=2}^{n-1}(k-1)(n-k); R_{4} = 1$.
I tried to use generating function to solve it.
|
The relation is $\;R_{n+1} = R_{n}+\sum _{k=2}^{n-1}(k-1)(n-k); \;R_{4} = 1\,$.
First off:
$$
\begin{align}
\sum _{k=2}^{n-1}(k-1)(n-k) = \sum _{k=1}^{n-2} k(n-k-1) &= (n-1) \cdot \sum _{k=1}^{n-2}k - \sum _{k=1}^{n-2} k^2 \\
&= (n-1) \cdot \frac{(n-2)(n-1)}{2} - \frac{(n-2)(n-1)(2n-3)}{6} \\
&= \frac{n(n-1)(n-2)}{6} \\
&= \binom{n}{3}
\end{align}
$$
(The combinatorial interpretation of this is fairly straightforward: when adding the $\,(n+1)^{th}\,$ vertex to the previous $n$-gon, the new intersections of diagonals are precisely those where one of the diagonals originates at the newly added vertex.)
Then, telescoping:
$$
\begin{align}
R_n &= R_{n-1} + \binom{n-1}{3} \\
&= R_{n-2} + \binom{n-2}{3} + \binom{n-1}{3} \\
& \cdots \\
&= R_4 + \binom{4}{3}+\binom{5}{3}+\ldots + \binom{n-2}{3} + \binom{n-1}{3} \\
&= \binom{3}{3} + \binom{4}{3}+\binom{5}{3}+\ldots + \binom{n-2}{3} + \binom{n-1}{3} \\
&= \binom{n}{4}
\end{align}
$$
The last step used the hockey-stick identity $\displaystyle {\binom {n+1}{k+1}}=\sum_{j=k}^{n} \binom{j}{k}\,$.
| {
"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-tres-quadrados (in portuguese).
Any clue?
| 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, $c^2$ is divisible by $4$, so the division will come out even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
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?
| $$
I=\int \frac 1 {1+\tan x} \, dx=\int \frac {\cos x}{\cos x +\sin x} \, dx\\
J=\int \frac {\sin x}{\cos x +\sin x} \, dx\\
I+J=\int \frac {\cos x+\sin x}{\cos x +\sin x} \, dx = x+C_1\\ \\
I-J=\int \frac {\cos x-\sin x}{\cos x +\sin x} \,dx=\int \frac { d\left( \sin x +\cos x\right) }{ \cos x +\sin x} =\ln \left| \sin x +\cos x \right| + C_2 \\
2I = x+\ln \left| \sin x +\cos x\right| + C_3 \\
I=\frac 1 2 \left[ x+\ln \left| \sin x +\cos x \right| + C_3 \right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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,
$$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$
I then took $\bigl(2^x\bigl)$ common and wrote it as,
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
After further simplification I got,
$$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$
Taking $-2^{99}$ common I got,
$$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$
Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
So,
$$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
| 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 $$2^x=\frac{3\cdot 2}4\cdot\frac{1-2^{-100}}{1-4^{-100}}=\frac{3}{2(1+2^{-100})}\\ x=\frac{\ln 3-\ln(1+2^{-100})}{\ln 2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
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 result I get was:
$$ T(n) = 4^n T(\frac{n}{2^k}+3k)+[\sum_{i=0}^{k-1}\frac{4^in}{2^{i+1}}+4^i*3*(i-1)]$$
All of this is new to me, all the examples i saw online are way more different than this excercise
Am I getting closer to an answer?
Note: No base case was given
Thanks in advance!
| One way might be$$A=\sum _{ i=0 }^{ k-1 } 4^{ i }(i-1)=-1+{ 4 }^{ 2 }+{ 4 }^{ 3 }\cdot 2+{ 4 }^{ 4 }\cdot 3+...+{ 4 }^{ k-1 }\left( k-2 \right) \\ A+1={ 4 }^{ 2 }+{ 4 }^{ 3 }\cdot 2+{ 4 }^{ 4 }\cdot 3+...+{ 4 }^{ k-1 }\left( k-2 \right) \\ 4\left( A+1 \right) ={ 4 }^{ 3 }+{ 4 }^{ 4 }\cdot 2+{ 4 }^{ 5 }\cdot 3+...+{ 4 }^{ k }\left( k-2 \right) \\ \left( A+1 \right) -4\left( A+1 \right) ={ 4 }^{ 2 }+\left( { 4 }^{ 3 }\cdot 2-{ 4 }^{ 3 } \right) +\left( { 4 }^{ 4 }\cdot 3-{ 4 }^{ 4 }\cdot 2 \right) +...+\left( { 4 }^{ k-1 }\left( k-2 \right) -{ 4 }^{ k-1 }\left( k-3 \right) \right) -{ 4 }^{ k }\left( k-2 \right) \\ -3\left( A+1 \right) ={ 4 }^{ 2 }+{ 4 }^{ 3 }+...+{ 4 }^{ k-1 }-{ 4 }^{ k }\left( k-2 \right) \\ -3\left( A+1 \right) =\frac { { 4 }^{ 2 }\left( 1-{ 4 }^{ k } \right) }{ 1-4 } -{ 4 }^{ k }\left( k-2 \right) ={ 4 }^{ k }\left( \frac { 22 }{ 3 } -k \right) \\ A=\frac { { 4 }^{ k } }{ -3 } \left( \frac { 22 }{ 3 } -k \right) -1\\ $$
or finding a derivative of $\sum _{ i=0 }^{ n }{ { x }^{ n } } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 add -1(third row) to the 2nd row->
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 2 & 0 & 4
\end{bmatrix}
add -2(2nd row) to the third row ->
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 0 & 10 & 0
\end{bmatrix}
But then this seems to have no solution because $10z = 0$.... ugh
EDIT
As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?
| I don't understand your way to obtain the RREF, we can proceed as follow
$$\begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2
\end{bmatrix}\stackrel{R3-R2}\to \begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\0 & -1 & 5 & -2
\end{bmatrix}\stackrel{R2-2\cdot R1}\to \begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & -1 & 5 & -2
\end{bmatrix}\stackrel{3\cdot R3+R2}\to \begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 0 & 10 & 0
\end{bmatrix}$$
and since the matrix is full rank (we have three pivots) we have an unique solution that is
*
*from the third row: $z=0$
*from the second row: $y=2$
*from the first row: $x=1$
| {
"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 $a \sec\theta - c \tan\theta =d$ and $b \sec\theta+ d \tan\theta= c$
Squaring and adding them,we get
$a^2\sec^2\theta+c^2\tan^2\theta-2ac\sec\theta\tan\theta+b^2\sec^2\theta+d^2\tan^2\theta+2bd\sec\theta\tan\theta=d^2+c^2$
$a^2\sec^2\theta+c^2\tan^2\theta+b^2\sec^2\theta+d^2\tan^2\theta-d^2-c^2=2ac\sec\theta\tan\theta-2bd\sec\theta\tan\theta$
I am stuck here.
| 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 forward, the step was $(a+b)^3 - 41\cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.
| 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) $0<3s^2+6s-\sqrt\Delta<2s(6+3s)$
L:$3s^2+6s>\sqrt\Delta$
$(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$
$4(6+3s)(s^3-697s^2)>0$
$0<s<697$
R:$-3s^2-6s<\sqrt\Delta$ It is always true
(2) $0<3s^2+6s+\sqrt\Delta<2s(6+3s)$
L: always true
R: $3s^2+6s>\sqrt\Delta$
It is true for $0<s<697$
Overall $a+b=s=1,2,...695$ or $696$
$sum=242556$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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} \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{1\cdot2\cdot3\cdots(n-1)} -\frac{1}{2(1\cdot2\cdot3\cdots n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{(n-1)!} - \frac{1}{2}\sum_ {n=2}^\infty \frac{1}{n!}\right) \\[10pt]
& = e - \frac{1}{2} (e- 1)= \frac{1}{2}(e+1)
\end{align}
Is it correct ???
if not correct then any hints/solution will be appreciated..
thanks in advance
| 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?
| {
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"url": "https://math.stackexchange.com/questions/2872093",
"timestamp": "2023-03-29T00:00:00",
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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) $
$$\int_{-\infty}^{\infty}\left( \frac{\cos{\left (x \right )}}{x^{4} + 1} \right)dx=\frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right) $$
The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $\gamma_{R}$, assuming $R > 1$ define
$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R $$
$$\gamma_{R}^{2}(t) = Re^{it} \, \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
One can call these two curves taken together $\gamma_{R}$ or $\gamma$, after picking our $\gamma_{R}$ one can consider $$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz.$$
It's trivial that,
$$\oint_{\gamma_{R}}g(z)~ dz = \oint_{\gamma_{R}^{1}} g(z) dz+ \oint_{\gamma_{R}^{2}}g(z) ~dz.$$
It's imperative that
$$\displaystyle \oint_{\gamma_{R}^{1}} g(z) dz \rightarrow \lim_{R \rightarrow \infty }\int_{-R}^{R} \frac{e^{ix}}{1+x^{4}} \, \, \text{as}\, \, R \rightarrow \infty $$
Using the Estimation Lemma one be relived to see
$$\bigg |\oint_{\gamma_{R}^{2}} g(z) dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|g(z)|\leq \pi \cdot \frac{1}{R^{4} - 1}. $$
Now it's safe to say that
$$\lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}\frac{1}{1+z^{4}} dz \bigg| \rightarrow 0. $$
It's easy to note after all our struggle that
$$ \int_{-\infty}^{\infty} \frac{cos(x)}{1+x^{4}} = Re \int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^{4}} = Re\bigg( \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right)\bigg) = \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right) $$
The finial leg of our conquest is to consider that
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i \sum_{j=1,2,3,4} Ind_{\gamma} \cdot \operatorname{Res_{f}(P_{j})}$$
It's easy to calculate that
$$\operatorname{Res_{z = \sqrt[4]{-1}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}}$$
$$\operatorname{Res_{z = \sqrt[4]{-1}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}}$$
$$\operatorname{Res_{z = -(-1)^{3/4}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}$$
$$\operatorname{Res_{z = (-1)^{3/4}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}$$
Putting the pieces together we have that
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i + \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}= \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right).$$
| From the previous attempt the last line,
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i + \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}= \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right).$$
Is incorrect and one should sum the poles inside $\gamma_{R}$ hence,
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i +\operatorname{Res}\limits\limits_{ z=\omega}\frac{e^{iz}}{z^4+1} = \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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 $k=1,2,\dots$
$$\left(\frac{d^2}{dr^2}\right)\left(\frac 1r\frac{d}{dr}\right)^{k-1}\left(r^{2k-1}\phi(r)\right)=\left(\frac 1r\frac{d}{dr}\right)^{k}\left(r^{2k}\phi'(r)\right)$$
For $k=1$, both sides equal $2\phi'+r\phi''$; for $k=2$, they equal $8\phi'+7r\phi''+r^2\phi'''$. Abbreviating $\frac{d}{dr}$ with $d$, I think that the following identity will be useful in a proof by induction
\begin{aligned} d \frac 1r d &= \frac 1r d^{-1}d^2\\
&=\left(\frac 1r d + \frac 1{r^2}\right)d^{-1}d^2 \\
&=\frac 1r d^2 + \frac 1{r^2}d \end{aligned}
From this, it also follows that $d^2=rd\frac 1r d-\frac 1r d$. However, I tried to insert these identities in multiple places without success.
| 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 $r\frac{d}{dr}\phi$ ,we get$$\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k} \frac{d \phi}{d r}\right)=r\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\left(r\frac{d}{d r}\right)\phi$$
notice that$ \frac{d}{dr}r\varphi=\left(r\frac{d}{d r}+1\right)\varphi$,so
$$
\begin{align*}
\left(\frac{d}{dr}\right)^2\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k-1} \phi\right)&=\left(\frac{d}{dr}\right)^2r\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi\\
&=\left(\frac{d}{dr}\right)\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi
\end{align*}
$$
and$$\begin{align*}
\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k} \frac{d \phi}{d r}\right)&=\frac{1}{r}\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\left(r\frac{d}{d r}\right)\phi\\
&=\frac{1}{r}\left(r\frac{d}{d r}\right)\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi\\
&=\left(\frac{d}{d r}\right)\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi
\end{align*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874973",
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"source": "stackexchange",
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"answer_count": 3,
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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 than one with the same side length. The smallest of these where $A^2+B^2=C^2\land A^2+D^2=E^2$ is $A=15$ such as in $(15,8,17)$.
$$m_{min}=\lceil\sqrt{15}\space\rceil=\lceil 3.872983346\rceil=4\text{ and }m_{max}=\frac{A+1}{2}=\frac{16}{2}=8$$
Testing for $m=4\text{ to }8$, we find integers when $(m,n)=(4,1)\text{ and }(8,7)$
$$\text{For }(4,1)\quad A=4^2-1^1=15\quad B=2*4*1=8\quad C=4^2+1^2=17$$
$$\text{For }(8,7)\quad A=8^2-7^2=15\quad D=2*8*7=112\quad E=8^2+7^2=113$$
Sometimes there is only one match such as the smallest $(3,4,5)$ where $m_{min}=m_{max}=2.$ At other times, there are many matches such as for $A=105$ where $m_{min}=11$ and $m_{max}=53.$ and we find $(11,4), (13,8), (19,16),\text{ and }(53,52).$ For these respective values of $(m,n)$ we have:
$$A,B,C=105,88,137\quad D,E=208,233\quad F,G=608,617\quad H,I=5512,5513$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 1
} |
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
$$J_n=\frac{\sin x}{(a+b \cos x)^n}-nb \int \frac{\sin^2 x\:dx}{{(a+b \cos x)^{n+1}}}$$
Can we proceed here?
| 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 what comes next, but I get:$$(1-n)I_{n-1}=(1-2n)aI_n+\frac{b\sin x}{ (a+b\cos x)^n}-nb^2(1-\frac{a^2}{b^2})I_{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
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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
$$2^{20}-2^{10}+1 = \frac{1+4 \times (2^7)^4}{1+4 \times (2^2)^4} =
\frac{1-2^8+2^{15}}{1+2^3+2^5} \times \frac{1+2^8+2^{15}}{1-2^3+2^5}=
\frac{32513}{41} \times \frac{33025}{25}=793 \times 1321.$$
It's easy to see that $793 = 13 \times 61$ and, as the solution is a product of three primes, we are done. So the largest prime divisor is 1321.
| {
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"source": "stackexchange",
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"answer_count": 1,
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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 $x-\frac{1}{2}=t$
$$f((t+1)^2+\frac{3}{4})=f(t^2+\frac{3}{4})$$
Let $g(t)$ be a polynomial defined as
$$g(t)=t^2+\frac{3}{4}$$
Can I conclude anything from this?
| $$f((t+1)^2+\frac{3}{4})=f(t^2+\frac{3}{4})$$
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
$$g(x+1)=g(x)$$ Thus period of your function $g(x)=f(x^2-x+1) $ is 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886867",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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}\left(x\right)-S_{n}\left(x\right)\right|\nrightarrow0
\end{equation}
for some $x\in[0,\;\frac{\pi}{2})$.
Any hint?
Is this correct?
\begin{align}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right| & =\left|\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(n+1\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\right|\\
& =\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(n+1\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \geq\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(2n\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \geq\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(2n\right)^{2}}+\cdots+\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \geq\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}+\cdots+\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& =\frac{n\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber
\end{align}
then let $x=\frac{\pi}{2}-\frac{1}{n}$, so we have
\begin{align}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right| & \geq\frac{n\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\\
& =\frac{n\left(n+1\right)\left(\sin\left(\frac{\pi}{2}-\frac{1}{n}\right)\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \rightarrow0.25,\:not\:0\nonumber
\end{align}
| 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 $\lim_{M\to \infty}\sup_{M<m<m'}\|S_m-S_{m'}\|=0$ because $\|S_m-S_{m'}\|\leq \|S_m-S\|+\|S-S_{m'}\|.$
Given $M\in \Bbb N,$ take $m>M$ and let $m'=2m+1.$ Then take $x\in [0,\pi/2)$ where $x$ is close enough to $\pi /2$ that $(\sin x)^{2m+1}>1/2.$ Then $(\sin x)^j>1/2$ for $m+1\leq j\leq 2m+1.$ And we have $$\|S_m-S_{m'}\|\geq |S_m(x)-S_{m'}(x)|=$$ $$=\sum_{j=m+1}^{2m+1}\frac {j(\sin x)^j}{j^2+2}>
\sum_{j=m+1}^{2m+1}\frac {j/2}{j^2+2}>$$ $$> \sum_{j=m+1}^{2m+1}\frac {j/2}{2j^2}=\sum_{j=m+1}^{2m+1} \frac {1}{4j}>$$ $$> \frac {1}{4}\sum_{j=m+1}^{2m+1}\int_j^{j+1}(\frac {1}{x})dx=$$ $$=\frac {1}{4}\int_{m+1}^{2m+2}(\frac {1}{x})dx=\frac {1}{4}\ln 2.$$
So convergence of $S_m$ cannot be uniform.
| {
"language": "en",
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"question_score": "4",
"answer_count": 2,
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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 using Taylor approximations:
$$\sin \frac{1}{k} \approx \frac{1}{k} - \frac{1}{3!k^3} + O(\frac{1}{k^5})$$
$$\arctan \frac{1}{k} \approx \frac{1}{k} - \frac{1}{3k^3} + O(\frac{1}{k^5})$$
$$\sin \frac{1}{k} - \arctan \frac{1}{k} \approx \frac{1}{k} - \frac{1}{3!k^3} + O(\frac{1}{k^5}) - \left(\frac{1}{k} - \frac{1}{3k^3} + O(\frac{1}{k^5}) \right) \approx \frac{1}{6k^3} + O(\frac{1}{k^5})$$
$$\sum_{k=1}^{\infty} (\sin \frac{1}{k} - \arctan\frac{1}{k}) \approx \sum_{k=1}^{\infty} (\frac{1}{6k^3} + O(\frac{1}{k^5})) < \infty$$ So the series converges. $\square$
My questions are:
*
*Is my answer correct?
*Is there anything wrong with using Taylor approximations to determine convergence?
*Is there a rule of thumb for how to tackle questions which ask you to determine the convergence or divergence of series? If there is, please share - I could really use the help.
| 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.
| {
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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 $1^k + 2^k + \ldots + n^k$ for $k=3,4$.
It is easy to see that $|A| = 1^2 + 2^2 + \ldots + n^2$, since from the sum rule we have $$|A| = \sum_{i=0}^{n+1} |\{(x,y,i): 1\leq x,y,z \leq n+1, i> \max\{x,y\}\}| = \sum_{i=0}^{n+1} i^2$$ (as we can choose $x$ and $y$ in $i \times i$ ways for each $i$).
However I can't see why is $|A|$ equals $\dfrac{n(n+1)(2n+1)}{6}$.
| 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 $\sum_{i=1}^n i^2=|A|=2\binom{n+1}{3}+\binom{n+1}{2}=\frac{n(n+1)(2n+1)}{6}$
| {
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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:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$
Dividing numerator and denominator by x:
$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$
$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$
$$= \frac{7}{\sqrt{4+0}-2}$$
$$=\frac{7}{2-2}$$
$$=\infty$$
Conclusion: Limit does not exist.
Why is my solution wrong?
Correct answer: $\frac{-7}{4}$
| Because, when you divide the denominator by $x$, you forgot $x$ is supposed to be negative, so that
$$x=-\sqrt{x^2}.$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
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 have
$$\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)\le \frac{1}{n}-\frac{1}{n}+\frac{1}{2n^2}=\frac{1}{2n^2}$$
and therefore the given series converges by comparison test with $\sum \frac 1{2n^2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 $c=-2y^2-7$. Thus discriminant equals to $9y^2+16y^2+56=25y^2+56$.
Now $x = \frac{3y±\sqrt{25y^2+56}}{4}$ , $x$ to be an integer $\sqrt{25y^2+56}$ has to be an integer too. I substituted $y$ as nonnegative integer first, because the answer wouldn't differ anyway as it is squared and found that when $y=1$, $\sqrt{25y^2+56}=9$, so we get $x = \frac{3±9}{4}$ and when we have plus sign we get $x = \frac{3+9}{4}=3$. So there we have it, $x=3, y=1$ is on of the solution. But $y=-1$ will also work because $y$ is squared, again, we get $x = \frac{-3±9}{4}$, if we have minus sign we have $x=-3$. Thus another solution, leaving us with:
$$ x=3, y=1$$
$$ x=-3, y=-1$$
I checked other integers for $y$ but none of them lead to solution where $x$ is also an integer. But here's problem, how do I know for sure that these two solutions are the only solutions, I can't obviously keep substituting $y$ as integers, as there are infinitely many integers. So that's why I came here for help.
| 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 $y<0, u-5y\ge1+5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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
$$
\begin{align}
\sum_k(-1)^k\frac{\binom{n}{k}}{\binom{r+k}{m}}
&=\sum_k(-1)^k\frac{\binom{n-1}{k}+\binom{n-1}{k-1}}{\binom{r+k}{m}}\tag1\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac1{\binom{r+k}{m}}-\frac1{\binom{r+k+1}{m}}\right)\tag2\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac{\frac{r+k+1}{m+1}}{\binom{r+k+1}{m+1}}-\frac{\frac{r+k+1-m}{m+1}}{\binom{r+k+1}{m+1}}\right)\tag3\\
&=\frac{m}{m+1}\sum_k(-1)^k\frac{\binom{n-1}{k}}{\binom{r+k+1}{m+1}}\tag4\\
&=\frac{m}{m+n}\frac1{\binom{r+n}{m+n}}\tag5
\end{align}
$$
Explanation:
$(1)$: Pascal's Rule
$(2)$: substitute $k\mapsto k+1$ in $(-1)^k\left.\binom{n-1}{k-1}\middle/\binom{r+k}{m}\right.$
$(3)$: put things over a common denominator
$(4)$: subtract and pull out the factor of $\frac{m}{m+1}$
$(5)$: repeat $n$ times
Application to the Question
$$
\begin{align}
\sum_{k=0}^{n-r}(-1)^kn\frac{\binom{n-1}{r-1}\binom{n-r}k}{r+k}
&=n\binom{n-1}{r-1}\sum_{k=0}^{n-r}(-1)^k\frac{\binom{n-r}k}{\binom{r+k}{1}}\\
&=n\binom{n-1}{r-1}\frac1{n-r+1}\frac1{\binom{n}{n-r+1}}\\
&=n\binom{n-1}{r-1}\frac1n\frac1{\binom{n-1}{n-r}}\\[6pt]
&=1
\end{align}
$$
| {
"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 with $b_n$ the last digit of $a_n$, I want to show that
$$
b_n =
\begin{cases}
3, & \text{if $n$ is odd}, \\
7, & \text{if $n$ is even}.
\end{cases}
$$
There is a hint to prove that for each $n \in \mathbb{N}$, if $a_n \equiv 3 \pmod{5}$ then $a_{n+1} \equiv 2 \pmod{5}$ and if $a_n \equiv 2 \pmod{5}$ then $a_{n+1} \equiv 3 \pmod{5}$.
First of all, if we take $a_n \equiv 3 \pmod{5}$, then $a_{n+1}=3^3\pmod{5} \equiv 2 \pmod{5}$.
If $a_n \equiv 2 \pmod{5}$, then $a_{n+1}=2^2 \pmod{5}=4$. Or am I doing something wrong?
And also how does it follow, if we have shown the hint, that $b_n$ is $3$ when $n$ is odd, and $7$ if $n$ is even?
| 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 prime factorization of $a_n$, but the only prime dividing $a_n$ is 3, so this is not possible.
In the same way, from $a_n\equiv 2\textrm{ (mod 5)}$ you get $a_n \equiv 7\textrm{ (mod 10)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $$\ln(1+x) = \sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^nx^{n+1}}{n+1}$$
My first thought was to use a series expansion, $$\ln\left(1+3x+2x^2\right) = \sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^n(3x+2x^2)^{n+1}}{n+1}$$ and compare this sum with the sum on the right hand side.
$$\sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^n\left(3x+2x^2\right)^{n+1}}{n+1} = \sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^{n-1}\left(2^n+1\right)x^{n}}{n}$$
From here, I would try and manipulate the expression within the sum to give the expression on the right hand side.
I know I can turn the sum on the left into
$$\sum\limits_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}\left(3x+2x^2\right)^{n}}{n}$$
which is slightly closer to the sum on the right. However, when comparing both sides, the expression I get seems quite incorrect since the highest degree on the left side is $x^{2n}$.
$$\left(3x+2x^2\right)^n = \left(2^n+1\right)x^n$$
Where have I gone wrong? How I can better approach this problem?
| Hint: $1+3x+2x^2=(1+x)(1+2x)$, so you just add two Taylor series.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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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
$$g(x)-g(\frac{x}{2^{n}})=x^2+\frac{x^2}{4}+\frac{x^2}{16}+...+\frac{x^2}{4^{n-1}}=x^2 \frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}$$
Since $f$ is continuous at $x=0$, so is $g$. Taking the limit by $n$ you get
$$g(x)-g(0)=\frac{4}{3} x^2$$
Moreover, $f(x)-f(\frac{x}{2})=g(x)$ implies $g(0)=0$.
Now, we need to solve
$$f(x)-f(\frac{x}{2})=\frac{4}{3}x^2$$
which is exactly the equation above, multiplies by $\frac{4}{3}$. Therefore, solving exactly as above we get
$$f(x)-f(0)=\frac{4}{3}\frac{4}{3}x^2=\frac{16}{9}x^2$$
This shows that
$$f(x)=\frac{16}{9}x^2 +c$$
where $c=f(0)$
are all the solutions.
P.S. To make this more clear, the equation can be reduced via (*) to two equations of the type
$$h(x)-h(\frac{x}{2})=r(x)$$
with $h, r$ continuous. This equation can be solved as above: By induction
$$h(x)-h(\frac{x}{2^n})=\sum_{k=0}^{n-1}r(\frac{x}{2^n})$$
Using the continuity of $h$ at $x=0$ (we don't even need $h$ to be continuous at other points) we get
$$h(x)-h(0)=\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$
(Note here that, if the series is divergent it implies that there is no solution which is defined at that $x$).
So, by calculating the series, you get the solution
$$h(x)=c+\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$
where $c=h(0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 perpendicular to curved suface at (a,b,c) would be
$ \begin{bmatrix} {2a} \\
{4b}\\
{-2c} \\
\end{bmatrix} $.
Distance is $\frac{10}{2\sqrt{a^2+4b^2+c^2}}$ .
Then $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix} \cdot \begin{bmatrix} {2a} \\
{4b}\\
{-2c} \\
\end{bmatrix} = 2a^2+4b^2-2c^2 = \sqrt{a^2+b^2+c^2}\cdot 2\sqrt{a^2+4b^2+c^2}\cdot \cos{0} =10$.
and $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix} \times \begin{bmatrix} {2a} \\
{4b}\\
{-2c} \\
\end{bmatrix} = \begin{bmatrix} {-2bc-4bc} \\
{2ac+2ac}\\
{4ab-2ab} \\
\end{bmatrix} = \begin{bmatrix} {-6bc} \\
{4ac}\\
{2ab} \\
\end{bmatrix}= \sqrt{()} \cdot \sqrt{()} \cdot \sin{0}$ = 0
$\quad \to$ For this to be true, two of a,b, and c have to be zero.
So the answer is $\frac{10}{2\sqrt{10}}=\frac{\sqrt{10}}{2}$.
Is this correct? and is there a better way?
| 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$$\left\{\begin{array}{l}a=\lambda a\\b=2\lambda b\\c=-2\lambda c\\a^2+2b^2-c^2=5.\end{array}\right.$$This system has $4$ solutions: $\pm\left(0,\sqrt{\frac52},0\right)$ and $\pm\left(\sqrt5,0,0\right)$. Therefore, the minimal distance is $\sqrt{\frac52}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+2n+1)+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+12n+6+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+13n+7$
$\Rightarrow$ $10^n \cdot 10^1 \gt 6n^2+13n+7$
$\Rightarrow$ $(6n^2+n)\cdot10 \gt 6n^2+13n+7$
$\Rightarrow$ $60n^2+10n \gt 6n^2+13n+7$
I am stuck at this point. What techniques or tricks are there to solve the rest?
It is of interest to me, because I am currently practicing a lot of exercises related to convergences, inequalities and mathematical induction.
Any hints guiding me to the right direction I much appreciate.
| 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 $n \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
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,
$$
\frac{125^{6}\times 25^{-3}}{(5^{2})^{-3}\times25^{7}}
$$
But, I get very confused and I end up getting stuck in calculations with negative exponents (i.e. $b^{-a}$) and I do not know how to solve them. So I'd like to know. How can I resolve powers with negative exponents?
| \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.
\end{align}
| {
"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
} |
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
$$ \begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix} = \begin{pmatrix}
2 \\
1 \\
2 \\
\end{pmatrix}, \text{where }A = \begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Here, notice that in order to find solution of this equation system, we need to multiply both sides of the equation by $A^{-1}$ from the left. However, applying Gauss-Jordan elimination,
$$\begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} \rightarrow \begin{pmatrix}
1 & 1 & 1 \\
0 & 0 & -1 \\
0 & 0 & 1 \\
\end{pmatrix} \rightarrow \begin{pmatrix}
1 & 1 & 0 \\
0 & 0 & -1 \\
0 & 0 & 0 \\
\end{pmatrix} \rightarrow \begin{pmatrix}
1 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{pmatrix}$$
so as you can see, we cannot have identity matrix by using row elementary operations. When it can be done, i.e. $A$ is invertible, we can say $[A\ \ 0]$ can be written as $[I\ \ 0]$ since they are row equivalent.
| {
"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 the limit of $\cos (2 \pi \sqrt{n^2 + n})$? I know that $\cos (n\pi) = (-1)^n$, if $n$ is a positive integer but then what?
| 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
$$\sqrt{n^2 + n}=n+\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$
$$2\pi\sqrt{n^2 + n}=(2n+1)\pi -\frac{\pi}{4 n}+O\left(\frac{1}{n^2}\right)$$
$$\cos(2 \pi \sqrt{n^2 + n})\sim -\cos \left(\frac{\pi }{4 n}\right)=-1+\frac{\pi ^2}{32 n^2}+O\left(\frac{1}{n^4}\right)$$
$$\sin ^2 (\pi \sqrt{n^2 + n}) =1-\frac{\pi ^2}{64 n^2}+O\left(\frac{1}{n^4}\right)$$ then the limit and how it is approached.
Using $n=1000$, the "exact" value would be $$\sin ^2\left(10 \sqrt{10010} \pi \right)\approx 0.99999984594$$ while the above approximation would give
$$1-\frac{\pi ^2}{64000000}\approx 0.99999984578$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 said square root.
Please help.
| 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 and list those terms as the continued fraction.
So, carrying out this procedure gives $$x_0=\sqrt{119},\quad a_0=\lfloor\sqrt{119}\rfloor=10,$$ $$x_1=\frac{1}{\sqrt{119}-10}=\frac{\sqrt{119}+10}{19},\quad a_1=\lfloor\frac{\sqrt{119}+10}{19}\rfloor=1,$$ $$x_2=\frac{1}{\frac{\sqrt{119}+10}{19}-1}=\frac{1}{\frac{\sqrt{119}-9}{19}}=\frac{19}{\sqrt{119}-9}=\frac{19(\sqrt{119}+9)}{38}=\frac{\sqrt{119}+9}{2},\quad a_2=\lfloor \frac{\sqrt{119}+9}{2}\rfloor=9,$$ $$x_3=\frac{1}{\frac{\sqrt{119}+9}{2}-9}=\frac{\sqrt{119}+9}{19},\quad a_3=\lfloor\frac{\sqrt{119}+9}{19}\rfloor=1,$$ $$x_4=\frac{1}{\frac{\sqrt{119}+9}{19}-1}=10+\sqrt{119},\quad a_4=\lfloor10+\sqrt{119}\rfloor=20,$$ $$x_5=\frac{1}{10+\sqrt{119}-20}=\frac{1}{\sqrt{119}-10}=x_1.$$ Since we have reached a loop in terms, we stop. Thus $$\sqrt{119}=[10;\overline{1,9,1,20}].$$ From here, we note that the period of this continued fraction is $4$ and so, the fundamental solution arises in the third partial convergent. We compute it as follows $$[10;1,9,1]=10+\frac{1}{1+\frac{1}{9+\frac{1}{1+1}}}=\frac{120}{11}.$$ So the pair $(120,11)$ is the smallest solution. Indeed $$120^2-119\times 11^2=1.$$ Finally, to generate all the solutions to this equation, we take the form $$(120+11\sqrt{119})^n,$$ where $n\in\Bbb N$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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 share). I'm not sure if I should start off with this equation or go a different way.
| 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
$$x_1^2+x_2^2+x_3^2=\sigma_1^2-2\sigma_2=s_2.$$
Thus, what you need is
$$A=(s_2-x_1^2)(s_2-x_2^2)(s_2-x_3^2)=s_2^3-(x_1^2+x_2^2+x_3^2)s_2^2+(x_1^2x_2^2+x_1^2x_3^2+x_1^2x_3^2)s_2-x_1^2x_2^2x_3^2.$$
The first two terms are cancelled. Then we determine
$$x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2=\sigma_2^2-2\sigma_1\sigma_3.$$
Hence, we get
$$A=(\sigma_2^2-2\sigma_1\sigma_3)s_2-\sigma_3^2.$$
If you plug in all the numbers $\sigma_1=-24/7$, $\sigma_2=2016/7=288$, $\sigma_3=-i/7$, I believe it is a monster. WolframAlpha says it should be $$A=-\frac{2293235711}{49} + i\frac{2654208}{2401}.$$
| {
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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.
$$ |\frac{x-iy}{(x+iy)(x-iy)}| <3 $$
$$ |\frac{x-iy}{x\cdot x + x(-iy) + iy \cdot x + iy (-iy)}| <3$$
$$ |\frac{x-iy}{x^2-xyi+xyi-(iy)^2}|<3 $$
Utilize the fact that $i^2=-1$
$$ |\frac{x-iy}{x^2+y^2}|<3 $$
$$ |\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}|<3 $$
$$ \sqrt{(\frac{x}{x^2+y^2})^2+(-\frac{iy}{x^2+y^2})^2}<3 $$
$$ (\frac{x}{x^2+y^2})^2+(-\frac{iy}{x^2+y^2})^2<3^2 $$
$$ \frac{x^2}{x^4+2x^2y^2+y^4}+\frac{i^2y^2}{x^4+2x^2y^2+y^4}<3^2 $$
$$ $$
I tried to plot the inequality with wolfram alpha and i got area outside of circle in $(0,0)
Not quite sure if this is correct.
| |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
} |
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 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \frac { 2 - x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { 2 } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 3 } } $
Why can we just add $ 2 - x ^ { 2 } $ in the numerator?
Step 1 to step 2, as well as step 3 to step 4 is clear to me.
| 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
} |
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}=$$\frac{\sqrt{2014x-x^2}}{2x-2014}-... \\ =\frac{\sqrt{2014/x-1}}{2-2014/x}-...
$
I got stuck so I tried substituting $u=2014-x$,
thus $\int_{0}^{2014}{\frac{u}{\sqrt{2014-u}+u}}du=... ?$
I found the value of 1007 using the value of integrand at x=0, 1007 and 2014.
But cannot solve integral. How can it be solved?
| If you substitute $x \mapsto 2014-x$, you get
$$ 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. $$
Hence
\begin{align}
1007 &= \frac{1}{2}\int_0^{2014} dx = \frac{1}{2}\int_0^{2014}\frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} + \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx\\
&= \frac{1}{2}(I+I) = I.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$. Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 \ x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 \ -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x \ -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
| $$ \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( - x - 1 \right) } + \left( 2 \right) $$
$$ \left( - x + 1 \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( - x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - x^{2} - x + 1 \right) }{ \left( - x - 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + 1 }{ 2 } \right) }{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } $$
$$ \left( x^{3} + 1 \right) \left( \frac{ - x - 1 }{ 2 } \right) - \left( x^{2} + 1 \right) \left( \frac{ - x^{2} - x + 1 }{ 2 } \right) = \left( -1 \right) $$
..................................
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Please show detailed steps of double integration of absolute difference Please show detailed steps of integration
$$\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 - y\int_y^1dx\}dy$$
$$=\frac{1}{2}\int_0^2\frac{1}{2}y^2dy + \frac{1}{2}\int_0^2\frac{1}{2} -y + \frac{1}{2}y^2dy$$
$$
=2/3 + 1/6 = 5/6
$$
I can't get the $\frac{2}{3}$ solution.
| The region of integration, $0\le y\le 2$ and $0\le x\le 1$, is a rectangle in the xy-plane. y> x, so |y- x|= y- x above the line form the lower left corner, (0, 0), to the upper right corner, (1, 2) and y< x, so |y- x|= x- y below it. That line is given by y= 2x so the integral is $\int_{x= 0}^1 \int_{y= 0}^{2x} (x- y) dydx+ \int_{x= 0}^1\int_{y= 2x}^2 (y- x) dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\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}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\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^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
| All you had expected, was solved by other solutions. I'd like to note some points. Maybe it has another taste. We know that for every natural number $n ≥ 1$, if $P(x)$ is a polynomial of degree $n$ with leading coefficient $a$, then $$\lim_{x\to\infty}P(x)=\infty (-\infty)$$ whether $a$ is positive (negative). This means equally that, the limit of such polynomial at infinity, can be evaluated by the term which identify the order of $P(x)$. So at infinity, $$n^3-3\approx n^3,~~~2n^2+n−1\approx 2n^2$$ Now do the limit with the two result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
} |
An Olympiad question about equality How many $(x, y)$ positive integer pairs have $y^2-x^2=2y+7x+4$ equality?
I can't solve this Olympiad question.
$\textbf{Solution:}$ If above equality is regulated,we obtain $(2x+2y-5)(2x-2y+9)=29$. And, $29$ is prime we can easily find solution.
*
*$\textbf{1-)}$ How can we regulate it to obtain $(2x+2y-5)(2x-2y+9)=29$?
*$\textbf{2-)}$ Are there the general method for this kind of questions?
Thank you for help...
| It's just completing the square,
$$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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The $m$-order submatrixs respectively from two $m\times n$ matrixs
Suppose that $A,B\in Mat_{m\times n}(\mathbb{C}),m\leq n.\det(A\overline{A^{'}})\ne 0,\det(B\overline{B^{'}})\ne 0.$
Show that
$$ \exists C\in Mat_{m\times m}(\mathbb{C}),s.t. B=CA. \Longleftrightarrow $$$$\exists \lambda\in \mathbb{C},s.t. \forall 1\leq i_{1}< i_{2}<\cdots<i_{m}\leq n,$$
$$ \det (B\begin{pmatrix}
1& 2& \cdots& m\\
i_{1}& i_{2}& \cdots& i_{m}
\end{pmatrix})=\lambda\cdot\det( A \begin{pmatrix}
1& 2& \cdots& m\\
i_{1}& i_{2}& \cdots& i_{m}
\end{pmatrix}).$$
$A\in Mat_{m\times n}(\mathbb{C})$ means $A$ is an $m$-by-$n$ complex matrix.
$\overline{A^{'}}$ means the conjugate transpose of $A$.
$A\begin{pmatrix}
1& 2& \cdots& m\\
i_{1}& i_{2}& \cdots& i_{m}
\end{pmatrix} $ is a submatrix of $A$ ,in which the set of row indices is $\{1,2,\cdots ,m\} $ and the set of column indices is$\{i_{1},i_{2},\cdots ,i_{m}\}$.
$\Rightarrow_{.}$ It is easy to verify.
$\Leftarrow_{.}$ In order to simplify this question, I let $m=2,n=3.$$A=(a_{ij})_{2\times 3},B=(b_{ij})_{2\times 3}.$
Then
$$\det\begin{pmatrix}
b_{11}& b_{12} \\
b_{21}& b_{22}
\end{pmatrix}=\lambda\cdot \det\begin{pmatrix}
a_{11}& a_{12} \\
a_{21}& a_{22}
\end{pmatrix},$$
$$\det\begin{pmatrix}
b_{11}& b_{13} \\
b_{21}& b_{23}
\end{pmatrix}=\lambda\cdot \det\begin{pmatrix}
a_{11}& a_{13} \\
a_{21}& a_{23}
\end{pmatrix},$$
$$\det\begin{pmatrix}
b_{12}& b_{13} \\
b_{22}& b_{23}
\end{pmatrix}=\lambda\cdot \det\begin{pmatrix}
a_{12}& a_{13} \\
a_{22}& a_{23}
\end{pmatrix}.$$
How can I find a matrix $\begin{pmatrix}
c_{11}&c_{12} \\
c_{21}&c_{22}
\end{pmatrix},$
satisfying the following equations:
$$
\begin{cases}
\qquad\qquad\qquad\quad\lambda=\det\begin{pmatrix}
c_{11}&c_{12} \\
c_{21}&c_{22}
\end{pmatrix}; & \\
\begin{pmatrix}
b_{11}& b_{12}& b_{13}\\
b_{21}& b_{22}& b_{23}
\end{pmatrix}=\begin{pmatrix}
c_{11}&c_{12} \\
c_{21}&c_{22}
\end{pmatrix}\begin{pmatrix}
a_{11}& a_{12}& a_{13}\\
a_{21}& a_{22}& a_{23}
\end{pmatrix}.&
\end{cases}
?$$
| Let $A = [A_1\;A_2]$ and $B = [B_1\;B_2]$ and assume that $A_1$ and $B_1$ are invertible. If this is not the case, rearrange the columns. Set $D := \det(A_1)$. Then $\det(B_1) = \lambda D$. Define $C := B_1A_1^{-1}$. Then $CA_1 = B_1$.
Now, let $a$ be an arbitrary column in $A_2$ and $b$ the corresponding one in $B_2$. We have to prove that $Ca = b$. Let $x := A_1^{-1}a$ and $y := B_1^{-1}b$. Let $j\in\{1,\ldots,m\}$ and let $A_1'$ be the matrix $A_1$ with missing $j$-th column. Define $B_1'$ accordingly. Then (denoting by $A_1^k$ the $k$-th column of $A_1$)
\begin{align*}
\det(A_1'|a)
&= \det(A_1'|A_1x) = \det(A_1'|\sum_{k=1}^mx_kA_1^k) = \sum_{k=1}^mx_k\det(A_1'|A_1^k)\\
&= x_j\det(A_1'|A_1^j) = (-1)^{m-j} x_j\det(A_1) = (-1)^{m-j}Dx_j.
\end{align*}
Analogously,
$$
\det(B_1'|b) = (-1)^{m-j}y_j\det(B_1) = (-1)^{m-j}\lambda Dy_j.
$$
Therefore,
$$
(-1)^{m-j}\lambda Dy_j = \det(B_1'|b) = \lambda\det(A_1'|a) = (-1)^{m-j}\lambda Dx_j.
$$
This implies $y = x$ and thus $Ca = CA_1x = B_1 x = B_1y = b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the limit by using the definition of derivative. Here is the problem.
Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of
$$ 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_{x\to 1}
\frac{1}{x-1}\frac{\frac{\arctan{\sqrt{f(x)}}-\arctan{\sqrt{f(1)}}}{x-1}-\frac{\arctan{{f(x)}}-\arctan{{f(1)}}}{x-1}}
{\left [\frac{\arcsin{\sqrt{f(x)}}-\arcsin{\sqrt{f(1)}}}{x-1}-\frac{\arcsin{{f(x)}}-\arcsin{{f(1)}}}{x-1} \right ]^2}
$$
and reduced that big chunks by using $f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ which I got
$$
\begin{split}
L&=\lim_{a\to 1}
\frac{1}{a-1}\frac{\Big [ \arctan\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arctan{f(x)} \Big ]'_{x=a}}
{\Big [ \arcsin\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arcsin{f(x)} \Big ]'_{x=a}}\\[2em]
&=\lim_{a\to 1}
\frac{1}{a-1} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{1+(f(a))^2}f'(a)}
{\left [ \frac{1}{\sqrt{1-f(a)}}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{\sqrt{1-(f(a))^2}}f'(a) \right ]^2}\\[2em]
&=\lim_{a\to 1}
\frac{1}{f'(a)\frac{a-1}{1-f(a)}} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}\\[2em]
&=\lim_{a\to 1}{-\frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}}\\[2em]
&=\boxed{(\sqrt{2}+1)^2}
\end{split}
$$
but the answer keys tell me that the answer of this problem is $L=\left( \frac{\sqrt{2}+1}{2}\right)^2$.
So, Can someone please explain to me what did I do wrong?
| There are several steps that are not clearly allowed.
The first step is okay:
$$
\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) - \arcsin 1) \right)^2 } \\
= \frac{1}{x-1} \frac{ \frac{\arctan \sqrt{f(x)} - \arctan 1}{x-1} - \frac{\arctan f(x) - \arctan 1}{x-1} }{ \left( \frac{\arcsin \sqrt{f(x)} - \arcsin 1}{x-1} - \frac{\arcsin f(x) - \arcsin 1}{x-1} \right)^2 } \\
$$
Then you assume not only that the limit can be distributed over the internal terms, which would give derivatives at $x=1$, e.g. $\left[ \arctan\sqrt{f(x)} \right]'_{x=1},$ but even that you can take the limits of the derivatives, $\lim_{a \to 0} \left[ \arctan\sqrt{f(x)} \right]'_{x=a}.$
Also, as gimusi has already pointed out, $\arcsin \sqrt{f(x)}$ and $\arcsin f(x)$ are not differentiable when $f(x)=1.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus) $$\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$$
Since $$\binom{1}{0} = \binom{1}{1} = 1$$
it's true.
Now suppose that is true with $n$ if is true with $n + 1$
Then, multiply both sides by two
$$2(2^n) = 2(\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n})$$
$$2^{n+1} = 2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n}$$
$$2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n} = \binom{n}{0} + \binom{n}{0} + \binom{n}{1} + \binom{n}{1} + ... + \binom{n}{n} + \binom{n}{n} $$
The first term have two equal term, then, you sum the last one with the first one of the next term, and you'll get this
$$\binom{n}{0} + \binom{n}{1} +... + \binom{n}{n-1} + \binom{n}{n}$$
If we use this equation (Already proved)
$$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $$
Of course, we'll have two term without sum, one $\binom{n}{0}$ and $\binom{n}{n}$
We can write these two term like this
$$\binom{n}{0} = \binom{n}{n} = \binom{n+1}{0} = \binom{n+1}{n+1}$$
Then, we get
$$2^{n+1} = \binom{n+1}{0} + \binom{n+1}{1} + ... + \binom {n+1}{n+1}$$
And it's already proved. Note: Just if we take $0! = 1$
I have to prove these too.
$\sum_{k}^n \binom{n}{m} = 2^{n-1}$ If $m$ is even. And
$\sum_{j}^n \binom{n}{j} = 2^{n-1}$ If $j$ is odd. Then, I just said that
If
$$\sum_{m}^n \binom{n}{m} + \sum_{j}^n \binom{n}{j} = \sum_{k=0}^n \binom{n}{k} $$
Then
$$2^{n - 1} + 2^{n - 1} = 2^n$$
Which is true, then, I already prove this. And I have a last one.
$$\sum_{i=0}^n (-1)^i\binom{n}{i} = 0$$
if n is odd. Then
$$\binom{n}{0} - \binom{n}{1} + ... + \binom{n}{n-1} - \binom{n}{n} = 0$$
And that's can be solve knowing that $$\binom{n}{k} = \binom{n}{n - k}$$
And if n is even
$$\binom{n}{0} - \binom{n}{1} + ... - \binom{n}{n-1} + \binom{n}{n} = 0$$
That means that every negative term if when n is odd, then, we can use our two last prove to prove it
If
$$\sum_{m}^n \binom{n}{m} - \sum_{j}^n \binom{n}{j} = 0$$
Then
$$2^{n-1} - 2^{n-1} = 0$$
Which is true.
And that's it, I want to know if my proves are fine and are rigorous too and what is the meaning of every combinatorics prove .
I want to know too better approaches to prove these (Or forms more intuitive)
| These are corollaries of the Binomial Theorem, which states that
$$(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}$$
which means that the number of subsets of a set with $n$ elements is $2^n$.
If we set $x = 1$ and $y = -1$, we obtain
$$0^n = [1 + (-1)]^n = \sum_{k = 0}^{n} 1^{n - k}(-1)^{k} = \sum_{k = 0}^{n} (-1)^k\binom{n}{k}$$
Notice that each term in which $k$ is even is positive and each term in which $k$ is odd is negative. Hence,
$$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} - \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} = 0 \tag{1}$$
which means the number of subsets with an even number of elements is equal to the number of subsets with an odd number of elements.
Since every subset has an even number of elements or an odd number of elements,
$$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} + \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} = \sum_{k = 0}^{n} \binom{n}{k} = 2^n \tag{2}$$
Adding equations 1 and 2 yields
$$2\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = 2^n \implies \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = 2^{n - 1}$$
which means the number of subsets with an even number of subsets is $2^{n - 1}$.
Since the number of subsets with an odd number of elements is equal to the number of subsets with an even number of elements, the number of subsets with an odd number of elements is
$$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} = 2^{n - 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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]$
But I get confused in the quadratic case! Any insight would be great!
| 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 succeed when $\vert F \vert < \infty$), what can be done?
Well, it is worth noting that certain well-known, ordinary procedures such as completing the square and its cousin the quadratic formula may be implemented as long as (1) applies. Indeed, with
$q(x) = 0 \tag 3$
we may write
$x^2 + ax = -b, \tag 4$
and then add $a^2/4$ to each side; by (1), this is a well-defined element of $F$; we obtain
$\left (x + \dfrac{a}{2} \right )^2 = a^2 + ax + \dfrac{a^2}{4} = \dfrac{a^2}{4} - b; \tag 5$
if then there exists $c \in F$ with
$c^2 = \dfrac{a^2}{4} - b, \tag 6$
(4) becomes
$\left (x + \dfrac{a}{2} \right )^2 = c^2, \tag 7$
whence
$x + \dfrac{a}{2} = \pm c, \tag 8$
or
$x = -\dfrac{a}{2} \pm c. \tag 9$
So the problem of finding solutions to (3) is tantamount to discovering if $a^2/4 - b$ is a perfect square in $F$; indeed, we may even write (9) in the form
$x = -\dfrac{a}{2} \pm \sqrt{\dfrac{a^2}{4} - b} = \dfrac{-a \pm \sqrt{a^2 -4b}}{2}, \tag{10}$
familiar to us as the famous "quadratic formula", assuming of course such a $c \in F$ as in (6) exists.
In the present case of $F = \Bbb Z_7$, the non-zero squares are
$1 = 1^2 = 6^2, \; 2 = 3^2 = 4^2, \; 4 = 2^2 = 5^2; \tag{11}$
if
$q(x) = x^2 + x + 1, \tag{12}$
we have
$a = b = 1, \tag{13}$
whence
$c^2 = \dfrac{a^2}{4} - b = \dfrac{1}{4} - 1 = 2 - 1 = 1; \tag{14}$
thus
$c = 1, 6, \tag{15}$
and
$x = -\dfrac{1}{2} \pm 1 = 3 \pm 1 = 2, 4; \tag{16}$
both these results are easily checked.
Of course, if we are searching for roots by systematic evaluation of $q(x)$ at different elements of $F$, once we have discovered one of them, such as $2$, we may find the other via dividing $q(x)$ by $x - 2$, or (much less work!) recalling that $-a$ must be their sum, whence the second zero is $-1 - 2 = -3 = 4$ in $\Bbb Z_7$.
If we encounter a quadratic for which $a^2/4 - b$ is not a perfect square, then such a $q(x)$ will not split over $F$, and we must consider the extension field $F[x]/(q(x))$ to split $q(x)$; but that is another chapter in a long, long story.
| {
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"url": "https://math.stackexchange.com/questions/2919776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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?
Thanks!
| 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$$
Differentiating, which we do term by term does not change the radius of convergence, giving us
$$\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \ldots \quad |x| < 1.$$
Differentiating a second time gives us
$$\frac{2}{(1 - x)^3} = (2 \cdot 1) + (3 \cdot 2)x + (4 \cdot 3)x^2 + \ldots \quad |x| < 1$$
Recall that $\binom{n}{2} = \frac{1}{2}n(n - 1)$, so we have
$$\frac{1}{(1 - x)^3} = \binom{2}{2} + \binom{3}{2}x + \binom{4}{2}x^2 + \ldots \quad |x| < 1.$$
Dividing by $16$, and substituting $x = \frac{1}{4}$, we get
$$\frac{4}{27} = \frac{1}{16} \cdot \frac{1}{\left(1 - \frac{1}{4}\right)^3} = \frac{\binom{2}{2}}{4^2} + \frac{\binom{3}{2}}{4^3} + \frac{\binom{4}{2}}{4^4} + \ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
| Since $7\cdot 2^k+1$ is odd we have that
$$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)$$
and since one among $n$ and $n+1$ must be odd therefore amd since $7$ is prime, one of them must be equal to $7$ and then the unique solution is $2^{k-2}=8 \implies k=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the integral $\leq 1/\sqrt{5} \leq 1$.
Correct?
Unsure how to progress from here
| Slightly Sharper Bounds
For $x\geq 0$, we have by the AM-GM Inequality that
$$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}{\sqrt{x^3+4}}\leq \frac{1}{\sqrt{3x+2}}$$
for every $x\in[0,2]$. Consequently,
$$\frac{1}{\sqrt{2}}\,\ln\left(\sqrt{2}+1\right)=\int_0^2\,\frac{1}{\sqrt{2}x+2}\,\text{d}x\leq \int_0^2\,\frac{1}{\sqrt{x^3+4}}\,\text{d}x\leq \int_0^2\,\frac{1}{\sqrt{3x+2}}\,\text{d}x=\frac{2\sqrt{2}}{3}\,.$$
Hence, we have a slight improvement to the original bounds:
$$\frac1{\sqrt{3}}<0.623< \int_0^2\,\frac{1}{\sqrt{x^3+4}}\,\text{d}x<0.943<1\,.$$
The improvement is not significant, nonetheless, and the actual value of $\displaystyle\int_0^2\,\frac{1}{\sqrt{x^3+4}}\,\text{d}x$ is roughly $0.854$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$ $$
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}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$
I would like to find the minimum of
$$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(\cos ^2(x)+1\right) \sec ^2(x)\right)^{n-1} \left(2 \left(\cos ^2(x)+1\right)\tan (x) \sec ^2(x)-2 \tan (x)\right)+n \left(\left(\sin ^2(x)+1\right) \csc^2(x)\right)^{n-1} \left(2 \cot (x)-2 \left(\sin ^2(x)+1\right) \cot (x) \csc^2(x)\right).$$
I think this is not the best way.
| You're probably right. But there is a way without using calculus. You should only use Cauchy's inequality:
$$ \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\left(\sqrt{\left(1+\frac{1}{\sin^2x}\right)\left(1+\frac{1}{\cos^2x}\right)}\right)^n=2\left(\sqrt{1+\frac{1}{\sin^2x}+\frac{1}{\cos^2x}+\frac{1}{\sin^2x\cos^2x}}\right)^n= \\ = 2\left(\sqrt{1+\frac{2}{\sin^2x\cos^2x}}\right)^n=2\left(\sqrt{1+\frac{8}{\sin^22x}}\right)^n\geq 2\left(\sqrt{1+8}\right)^n=2\cdot3^n$$
We must now prove there is $x_1$ such that $f(x_1)=2\cdot3^n$.
To that end, notice that $f(x)=2\cdot3^n$ is equivalent to following system
\begin{cases}
\sin^2x=\cos^2x, \\
\sin^22x=1,
\end{cases}
which has $x_1=\pi/4$ as solution. Now note that
$$f\left(\frac{\pi}{4}\right)=\left(\frac{1+1/2}{1/2}\right)^n+\left(\frac{1+1/2}{1/2}\right)^n=2\cdot3^n.$$
Thereby, the minimum value of $f(x)$ is
$$f_{min}=2\cdot3^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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About Spliting Field What is splitting field of polynomial $X^3+X+\bar 1$ in $\mathbb{F_5}$.
Attempt: To show this we first want to check irreducibility which is clear in case of finite field since there are only 5 elements so put values and check it has no root in $\mathbb{F_5}$. After that how to proceed?
| My answer to this question with software is in the following form.
First, I find a positive integer $n$ such that in the decomposition of $x^n-1$ in modulo $5$ we have the factor $x^3+x+1$. In your question we get $n=62$. In fact, we get
$$
(x^{62}-1) \pmod{5}=
\left( {x}^{3}+2\,x+4 \right) \left( {x}^{3}+2\,x+1 \right) \left(
{x}^{3}+{x}^{2}+3\,x+4 \right) \left( {x}^{3}+4\,{x}^{2}+x+1 \right)
\left( x+1 \right) \left( {x}^{3}+3\,{x}^{2}+4 \right) \left( {x}^{
3}+{x}^{2}+3\,x+1 \right) \left( {x}^{3}+2\,{x}^{2}+x+4 \right)
\left( {x}^{3}+{x}^{2}+1 \right) \left( {x}^{3}+3\,{x}^{2}+4\,x+1
\right) \left( {x}^{3}+x+4 \right) \left( {x}^{3}+4\,{x}^{2}+3\,x+4
\right) \left( {x}^{3}+3\,{x}^{2}+x+1 \right) \left( {x}^{3}+x+1
\right) \left( {x}^{3}+{x}^{2}+4\,x+1 \right) \left( {x}^{3}+4\,{x}
^{2}+4 \right) \left( {x}^{3}+{x}^{2}+x+4 \right) \left( x+4
\right) \left( {x}^{3}+4\,{x}^{2}+3\,x+1 \right) \left( {x}^{3}+2\,
{x}^{2}+1 \right) \left( {x}^{3}+2\,{x}^{2}+4\,x+4 \right) \left( {x
}^{3}+4\,{x}^{2}+4\,x+4 \right)
$$
Then, we obtain a positive integer $k$ such that $62\mid 5^k-1$ which is $k=3$. Then, we construct $\mathbb{F}_{5^3}$ with any irreducible polynomial such as $\bf f$.
Finally, we choose elements of the field $\mathbb{F}_{5^3}$ that their order is $62$ and test which of these elements satisfy $x^3+x+1=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability of at least one king in a 13-card hand? What is the probability of drawing 13 cards and having at least one king?
Here is what I came up with:
$$1 - \frac{\begin{pmatrix}
48\\
13
\end{pmatrix}}{\begin{pmatrix}
52\\
13
\end{pmatrix}}$$
Is this correct?
Thanks.
| $\begin{pmatrix}52 \\ 13\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 52 card deck. $\begin{pmatrix}52 \\ 48\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 48 card deck- with the kings missing. So your $\frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48\end{pmatrix}}$ is the probability that a 13 card hand randomly dealt from a 52 card deck has no kings and, yes, $1- \frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48\end{pmatrix}}$ is the probability that such a hand does contain at least one king.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929495",
"timestamp": "2023-03-29T00:00:00",
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Solve for $u$ the PDE $(x − y − 1)u_x + (y − x − u + 1)u_y = u$ if $u=1$ on $x^2+(y+1)^2=1.$
Solve the Cauchy problem $$(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)=0$$
so $g_1(x,y,z)=x+y+z=c_1.$ I have not managed to find the second relation $g_2(x,y,z)=0$, needed in order to get $F(g_1,g_2)=0$ for some $F$. (As far as I am concerned, there are no standard procedures in these cases, one has to work on trial-and -error to find the exact expressions).
Thanks in advance.
| $$(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}$$
With $v=x-y-1$
$$\frac{du}{u}=\frac{dv}{2v+u}$$
is a separable ODE which solution is $v=-u+c_2u^2$ . Thus $x-y-1=-u+c_2u^2$ . The second characteristic equation is :
$$\frac{x-y-1+u}{u^2}=c_2$$
The general solution of the PDE can be expressed on the form of implicit equation :
$$\frac{x-y-1+u}{u^2}=F(x+y+u) \tag 2$$
where $F$ is an arbitrary function.
Boundary condition in order to determine the function $F$ :
$u=1$ on $x^2+(y+1)^2=1$ , so $\frac{x-y-1+1}{1^2}=F(x+y+1)$
$$F(x+y+1)=x-y$$
Let $X=x+y+1=\pm\sqrt{1-(y+1)^2}+y+1$
$(X-y-1)^2=1-(y+1)^2.\quad$ To be solved for $y$ which leads to
$y=\frac{X-2\pm\sqrt{2-X^2}}{2}\quad;\quad x=\frac{X\mp\sqrt{2-X^2}}{2}\quad;\quad x-y=\mp\sqrt{2-X^2}+1$
$$F(X)=1\mp\sqrt{2-X^2}$$
So, $F(X)$ is determined. We put it into the above general solution $(2)$ where $X=x+y+u$.
$$\frac{x-y-1+u}{u^2}=1\mp\sqrt{2-(x+y+u)^2}$$
The solution which complies to the PDE and the boundary condition is expressed on the form of an implicit equation :
$$x-y-1+u-u^2\pm u^2\sqrt{2-(x+y+u)^2}=0 \tag 3$$
To express $u(x,y)$ on explicit form we have to solve a polynomial equation of sixth degree. Thus there is no closed form for $u(x,y)$. The final answer is the above implicit form $(3)$.
| {
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Show that for any integer $n$, there will always exist integers $x,y$ such that $n^3 = x^2-y^2$ Problem: Show that for any integer $n$, there will always exist integers $x,y$ such that $n^3 = x^2-y^2$
My attempt:
If $n^3 = x^2-y^2$, then $x = y \mod(n)$, and at the same time $x = -y \mod(n)$,
So that $2x = 0 \mod(n)$. I could then split the problem into two cases:
If $2^{-1}$ exists, then $x=kn$ for some integer $k$, and so the equation reduces to: $n^2(n-k^2)=-y^2$, so that $n<k^2$. This is about how far I have gotten, not sure how to proceed. Hints appreciated.
If $2^{-1}$ does not exist...
Edit: I made a mistake above, but now that the answer is given I will just leave it up for now.
| One way to systematically arrive at an answer is to assume that
$$
\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
$$
Next, comparing coefficient of constant forces $a_0=\pm b_0$. Having $a_0 = b_0$ would force $a_1 = b_1$ when comparing coefficient of $n^1$. This in turn forces coefficient of $n^3$ in $x^2 - y^2$ to be zero which is not possible, so $a_0 = -b_0$.
Now comparing coefficient of $n^1$ gives $b_1 = - a_1$. Comparing coefficients $n^3$ and $n^2$, we arrive at
$$
4a_2a_1 = 1, \quad 4a_2a_0 = 0
$$
Hence $a_0 = 0$. This leads to
$$
\begin{align}
x &= a_2n^2 + n/(4a_2)\\
y &= a_2n^2 - n/(4a_2)
\end{align}
$$
Only $a_2 = 1/2$ gives integral values of $x,y$ so we get
$$
\begin{align}
x &= n(n+1)/2\\
y &= n(n-1)/2
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| As an alternative by
$$\sin^3 x+ \cos ^3 x=1 \iff \sin x\cdot \sin^2+\cos x\cdot \cos^2 x=1$$
since $\sin^2 x+ \cos ^2 x=1$, the given equality is a weighted mean of $\sin x$ and $\cos x$ which holds if and only if
*
*$\sin x=1,\,\cos x=0$
or
*
*$\cos x=1,\,\sin x=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$ I want to prove that $$\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}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}} $$
So,
$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}}=\frac{\theta^{2}}{4}-\frac{\pi^{2}}{12}.$$
I'm looking for a $\theta$ such that $(-1)^{n}\cos(n\theta)=(-1)^{n+1}\Rightarrow \cos(n\theta)=-1$ for all $n\in\mathbb{N}$. How can I choose that $\theta$?
| Split it!
$$\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{even}}\frac{1}{n^{2}} - \sum_{n ~\text{even}}\frac{1}{n^{2}} = \\
=\sum_{n=1}^{\infty}\frac{1}{n^2}-2\sum_{n ~\text{even}}\frac{1}{n^{2}} = \frac{\pi^2}{6} - 2\sum_{n ~\text{even}}\frac{1}{n^{2}}.$$
Now, notice that:
$$\sum_{n ~\text{even}}\frac{1}{n^{2}}=\sum_{i =1}^{\infty}\frac{1}{(2i)^{2}} = \frac{1}{4}\sum_{i =1}^{\infty}\frac{1}{i^{2}} = \frac{1}{4}\frac{\pi^2}{6}.$$
Therefore:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}} = \frac{\pi^2}{6} - 2\frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{12}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Rolles theorem used for solving equation $ax^3+bx^2+cx+d=0$ If a,b,c,d are Real number such that
$\frac{3a+2b}{c+d}+\frac{3}{2}=0$. Then the equation $ax^3+bx^2+cx+d=0$ has
(1) at least one root in [-2,0]
(2) at least one root in [0,2]
(3) at least two root in [-2,2]
(4) no root in [-2,2]
I am doing hit and trial method by using $f'(x)=0$, put x=1, we get $3a+2b+c=0$, putting $3a+2b=-c$ in
$\frac{3a+2b}{c+d}+\frac{3}{2}=0$, i get relation between c & d, but not able to proceed.
| Let $$f(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx.$$
Then $f0)=0$ and $f(2)=\frac{16a}{4}+\frac{8b}{3}+\frac{4c}{2}+2d=\frac{48a+32b+24c+24d}{12}=\frac{16(3a+2b)+24(c+d)}{12}=0$ (from the condition given).
Now apply Rolle's theorem. There exists a point in $[0,2]$, where the derivative of $f$ must be zero.
| {
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Finding value of $ \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$
Finding value of $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$
Try: $$\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}\cdot \prod^{n}_{k=1}\frac{2k}{2k+1}$$
$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4 \cdot 6\cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}\times \frac{(2\cdot 4 \cdot 6 \cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}$$
$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4\cdot 6\cdot 2n)^4}{(1\cdot 2\cdot 3\cdots 2n)^2} = \lim_{n\rightarrow \infty}2^{4n}\cdot \frac{(n!)^4}{(2n!)^2}$$
Did not find any clue how to solve from that point
Could some help me to solve it, Thanks
| With Stirling's approximation $n!\sim\sqrt{2\pi n}(n/e)^n$ we get:
\begin{align}
\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}
&=\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\\
&=\prod^{n}_{k=1}\frac{(2k)^2}{(2k-1)(2k)}\cdot\frac{(2k)^2}{(2k)(2k+1)}\\
&=4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k-1)(2k)}\cdot 4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k)(2k+1)}\\
&=2^{4n}\cdot\frac{(n!)^2}{(2n)!}\cdot\frac{(n!)^2}{(2n+1)!}\\
&=2^{4n}\cdot\frac{(n!)^4}{((2n)!)^2(2n+1)}\\
&\sim 2^{4n}\cdot\frac{(\sqrt{2\pi n}(n/e)^n)^4}{(\sqrt{2\pi 2n}(2n/e)^{2n})^2(2n+1)}\\
&\sim 2^{4n}\cdot\frac{4\pi^2n^2(n/e)^{4n}}{4\pi n(2n/e)^{4n}(2n)}\\
&\sim 2^{4n}\cdot\frac{\pi n^2}{2n^2 2^{4n}}\\
&\xrightarrow{n\to\infty}\frac\pi 2\\
\end{align}
| {
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} |
Prove that $1/6 < \int_0^1 \frac{1-x^2}{3+\cos(x)}dx < 2/9$
Prove that $$\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{1-t^2}{3+\cos(t)}dt - \frac{1}{6}$$ and then arrive at something using monotony, but still no answer.
I even tried to compute the integral using the substitution $\displaystyle t = \tan \left(\frac{x}{2} \right),$ but then I arrive at an integral of the form $ \displaystyle \int \frac{(\arctan(x))^2}{x^2+a}dx, a \in \mathbb{R}, $ which I do not know to compute.
| For any $x\in(0,1)$
$$ \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] $$
hence the given integral is bounded between $\frac{1}{6}=\int_{0}^{1}\frac{1-x^2}{4}\,dx$ and $\frac{1}{6}+\frac{1}{144}$.
| {
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Find $A$ if solution to $Ax=b$ is given
Solution to $Ax=b$ is $x=\begin{bmatrix}
1\\
0\\
1\\
0
\end{bmatrix}+\alpha_{1}\begin{bmatrix}
1\\
1\\
-1\\
0
\end{bmatrix}+\alpha_{2}\begin{bmatrix}
1\\
0\\
1\\
1\end{bmatrix}$. For $b=(1,2,1)^T$, find $A$.
Somehow, Gilbert Strang says that it is obvious that first and third column should add up to $(1,2,1)^T$, and then he says that second column is third minus first and fourth is -(first + third). I don't see how he concluded all of this, so any clarification is very welcome.
| You have that
$$
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 \end{bmatrix}=b
$$
for every $\alpha_1$ and $\alpha_2$.
In particular you have
$$
\begin{cases}
a_1+a_3=b & (\alpha_1=0, \alpha_2=0) \\[4px]
2a_1+a_2+a_3=b & (\alpha_1=1, \alpha_2=0) \\[4px]
2a_1+a_3+a_4=b & (\alpha_1=0, \alpha_2=1)
\end{cases}
$$
This is quite similar to a linear system, isn't it?
$$
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
2 & 1 & 1 & 0 & b \\
2 & 0 & 1 & 1 & b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
0 & 1 & -1 & 0 & -b \\
0 & 0 & -1 & 1 & -b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
0 & 1 & -1 & 0 & -b \\
0 & 0 & -1 & 1 & -b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
0 & 1 & -1 & 0 & -b \\
0 & 0 & 1 & -1 & b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 \\
0 & 0 & 1 & -1 & b \\
\end{bmatrix}
$$
Thus $a_1=-a_4$, $a_2=a_4$, $a_3=a_4+b$. The fourth column can be anything; so the matrix $A$ is
$$
\begin{bmatrix} -x & x & x+b & x \end{bmatrix}
$$
where $x$ is an arbitrary $3\times 1$ column.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof Check: Let $f_{0}(x)=\frac{1}{1-x},$ and $f_{n}(x)=f_{0}(f_{n-1}(x))$ for any positive integer $n$. Find $f_{2018}(2018)$ Let us denote by $F$ the set of real-valued functions of a real variable such that $$F=\left\{\frac{1}{1-x},\frac{x-1}{x},x\right\}.$$ We shall prove that this set forms an abelian group under the compositon of functions. We first notice that the set has closure and is commutative. Indeed,
\begin{alignat*}{2}
\left(x\right)\circ\left(\frac{1}{1-x}\right) &= \left(\frac{1}{1-x}\right)\in F &&= \left(\frac{1}{1-x}\right)\circ \left(x\right)\\
\left(x\right)\circ\left(\frac{x-1}{x}\right) &= \left(\frac{x-1}{x}\right)\in F &&= \left(\frac{x-1}{x}\right)\circ \left(x\right)\tag{1}\\
\left(\frac{1}{1-x}\right)\circ\left(\frac{x-1}{x}\right) &= \ \quad\left(x\right)\in F &&= \left(\frac{x-1}{x}\right)\circ \left(\frac{x-1}{x}\right).
\end{alignat*}
Now, we clain that $x$ is the identity element of the set. Indeed,
\begin{alignat*}{2}
\left(x\right)\circ\left(\frac{1}{1-x}\right) &= \left(\frac{1}{1-x}\right)\circ \left(x\right) &&= \frac{1}{1-x}\\
\left(x\right)\circ\left(\frac{x-1}{x}\right) &= \left(\frac{x-1}{x}\right)\circ \left(x\right) &&= \frac{x-1}{x}\tag{2}
\end{alignat*}
$$ \left(x\right)\circ\left(x\right) = x.$$
Moreover, we see from (1) and (2) that every element is invertible, hence the set forms an abelian group under the composition operator. We would now like to show that $F$ is a cyclic group:
Let us define the $n^{\text{th}}$ power $f^{n}$ of an element of $f\text{ of } F$ as the following
$$f^n =\begin{cases}
e & : n = 0 \\
f^{n-1} \circ f & : n > 0 \\
\left({f^{-n}}\right)^{-1} & : n < 0
\end{cases}$$
where $e$ is the identity element of $F.$ We have to prove that every element in $F$ can be expressed as the power of a single element from $F$. We claim this generator is $\frac{1}{1-x}$. Indeed,
$$\frac{1}{1-x}^{0}=x,\qquad \frac{1}{1-x}^{1}=\frac{1}{1-x},\qquad \frac{1}{1-x}^{2}=\frac{x-1}{x},\\$$
hence $F$ is the cyclic group of order 2 generated by $\frac{1}{1-x}$. Namely,
$$ F=\left\{\frac{1}{1-x}^{0},\frac{1}{1-x}^{1},\frac{1}{1-x}^{2} \right\}=\left<\frac{1}{1-x}\right>.\\$$
All cyclic groups of equal order being isomorphic to each other, consider the bijection $\phi\colon\mathbb{Z}/3\mathbb{Z}\to{F}\colon\phi(2^{n})=\frac{1}{1-x}^{n}$. We would like to show this is a homomorphism from $\mathbb{Z}/3\mathbb{Z}$ to $F$.
Note that $\phi(2^{n})=\frac{1}{1-x}^{n}$ holds for all $n$, for let $n$ be an integer such that $n = q k + r$ for $0 \le r < k$. Now
$$
2^n = 2^r,\qquad \text{and}\qquad \frac{1}{1-x}^r = \frac{1}{1-x}^n,\\
$$
thus
$$
\phi (2^{n}) = \phi (2^{r}) = \frac{1}{1-x}^r = \frac{1}{1-x}^n.
$$
Now let $x,y$ be in $\mathbb{Z}/3\mathbb{Z}$. Since $\mathbb{Z}/3\mathbb{Z}=<2>,$ it follows that there exist integers $s,t$ such that
$$x=2^{s} ,\qquad \text{and}\qquad y=2^{r}, $$
thus,
\begin{align*}
\phi(xy)&=\phi(2^{s}2^{t})\\
\qquad &=\phi(2^{s+t})\\
\qquad &=\frac{1}{1-x}^{s+t}\\
\qquad &=\frac{1}{1-x}^{s}\frac{1}{1-x}^{t}\\
\qquad &=\phi(2^{s})\phi(2^{t})\\
\qquad &=\phi(x)\phi(y)
\end{align*}
so $\phi$ is a homomorphism. As $\phi$ is a bijection, then it is also an isomorphism from $\mathbb{Z}/3\mathbb{Z}$ to $F$, hence $\mathbb{Z}/3\mathbb{Z} \cong F$, as desired. Lastly, let $h\colon \mathbb{N}\to \mathbb{Z}/3\mathbb{Z}$ be a function mapping a positive integer into its equivalence class modulo 3. Then
\begin{alignat*}{4}
&\qquad \qquad \qquad \qquad \quad f_{2018}(x)&&=(\phi\circ h)(x)\\
& &&=\phi(h(2018))(x)\\
& &&=\phi(2)(x)\\
& &&=\phi(2)(x)=\frac{1}{1-x}^2\\
& &&=x\\
&\qquad \qquad \qquad \quad \implies f_{2018}(2018) &&=2018
\end{alignat*}
| 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
} |
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 $-\infty.$
Proof
$\forall M>0$,$\exists X=\max(1,M+1)>0, \forall|x|>X$：
\begin{align*}
\left|\frac{x^3+1}{x^2+1}\right|&=\left|x-\frac{x-1}{x^2+1}\right|\\&\geq |x|-\left|\frac{x-1}{x^2+1}\right|\\&\geq |x|-\frac{|x|+1}{x^2+1}\\&\geq |x|-\frac{x^2+1}{x^2+1}\\&=|x|-1\\&>X-1\\&\geq M.
\end{align*}
Please verify the proof above.
| 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",
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"source": "stackexchange",
"question_score": "4",
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"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 $sin(0)$). Any tips?
$$\frac{y^4\sin(x)}{x^2+y^4}$$
| 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 $v=y^2$
$$\frac{y^4x}{x^2+y^4}=\frac{uv^2}{u^2+v^2}=r\cos \theta\sin^2 \theta \to 0$$
| {
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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 $y \to 0$. Thus:
$$\lim_{y \to 0} \frac{(y+4)^4 - 4^{y+4}}{y} = \\ = \lim_{y \to 0}\frac{(y+4)^4}{y} - \lim_{y \to 0} \frac{4^{(y+4)}}{y} $$
And this step is not allowed from the get go, as I am deducting infinities, which is indeterminate. What I should have done though:
$$4^4 \lim_{y \to 0} \frac{(1+y/4)^4-1+(4^y-1)}{y} = \\ 4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \frac{4^y-1}{y} \right) = \\
=4^4\left(\frac{1}{4} \cdot 4 - \ln 4 \right) = 256(1-\ln 4)$$
| 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.
| {
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"source": "stackexchange",
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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 now and keep messing up somewhere. It would much be appreciated if you could point out my mistake and lead me in the right direction.
Attempt: $ $ $ y = 4 - x$, $ $ $y = \sqrt{5-x}+1$,$ $ a= 1 $ $ b= 4
$\pi\int_1^4(\sqrt{5-x}+1)^2-(4-x)^2dx$ = $\pi\int_1^4-x^2+7x+2\sqrt{5-x}-10 dx$
$\pi(\frac{-x^3}3 +\frac{7x^2}2 +\frac{4/(5-x)^{3/2}}3-10x) |^4_1$ = $\pi(\frac{-64}3+\frac{112}2+\frac{4}3-40+\frac{1}3-\frac{7}2-\frac{32}3+10)$
= -24.60914245312
| 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.
$V=\pi\int_1^5 (\sqrt{5-x}+1)^2 dx$-$\pi\int_4^5 (-\sqrt{5-x}+1)^2 dx - \pi r^2h/3$ where r is the maximum radius of the cone mentioned above, i.e. 3, and h is it's height, also 3.
$$\int_1^5 (5-x)+1+2\sqrt{5-x}=\int_1^5 6-x+2\sqrt{5-x}dx=6x-x^2/2-\frac{4}{3}(5-x)^\frac{3}{2}|_1^5$$
$$=30-\frac{25}{2}-6+\frac{1}{2}+\frac{32}{3}=142\frac{2}{3}$$
$$\int_4^5 (1-\sqrt{5-x})^2dx=\int_4^5 6-x-2\sqrt{5-x}dx=6x-x^2/2+\frac{4}{3}\sqrt{5-x}|_4^5$$=$$30-25/2-24+8-\frac{4}{3}=\frac{1}{6}$$
$$V=\pi22\frac{2}{3}-\pi\frac{1}{6}-9\pi=(13\frac{1}{2})\pi$$
| {
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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!}{5!}$$4^5$ ways for this case.
$\textbf{Case 2:}$ Two 1s and One 2 OR NO 2s --> 1x1x5x4x4x4 = $5x4^3$. This can be arranged in $\dfrac{6!}{2!3!}$ ways. So there are $(6x4^3)$$\dfrac{6!}{2!3!}$ ways for this case.
$\textbf{Case 3:}$ Three 1s and Two, One or NO 2s --> 1x1x1x5x5x4 = 4x$5^2$. This can be arranged in $\dfrac{6!}{2!3!}$ ways. So there are $(4x5^2)$$\dfrac{6!}{2!3!}$ ways for this case.
$\textbf{Case 4:}$ Four 1s and Two, One or NO 2s --> 1x1x1x1x5x5 = $5^2$. This can be arranged in $\dfrac{6!}{4!2!}$ ways. So there are $(5^2)$$\dfrac{6!}{4!2!}$ ways for this case.
$\textbf{Case 5:}$ Five 1s and One or NO 2s --> 1x1x1x1x1x5 = 5. This can be arranged in $\dfrac{6!}{5!}$ ways = 6 ways. So there are $5^3$ ways for this case.
$\textbf{Case 6:}$ Six 1s and NO 2s --> 1x1x1x1x1x1 = 1. There is only one way to arrange this so there is only 1 way for this case.
With this logic...I would add the number of ways from each case to get my answer.
| 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 really $1 \cdot 1 \cdot (4+1) \cdot 4^3$ where the $4+1$ is four ways to not get a $2$ and one way to get one. If you don't get one there are $\frac {6!}{2!4!}={6 \choose 2}=15$ ways to arrange them.
| {
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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 $f(\frac{1}{2})=\frac{3\sqrt{3}}{4}$
The motivation behind this function comes from maximising the area of an inscribed triangle in the unit circle, for anyone that is curious.
My Attempt
$$f(x)=(1+x)\sqrt{1-x^2}=\sqrt{(1-x^2)(1+x)^2}=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}}$$
By the AM-GM Inequality, $\sqrt{ab}\leq \frac{a+b}{2}$, with equality iff $a=b$
This means that
$$\sqrt 3 \sqrt{ab} \leq \frac{\sqrt 3}{2}(a+b)$$
Substituting $a=1-x^2, b=\frac{(1+x)^2}{3}$,
$$f(x)=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}} \leq \frac{\sqrt 3}{2} \left((1-x^2)+\frac{(1+x)^2}{3}\right)$$
$$=\frac{\sqrt 3}{2} \left(\frac{4}{3} -\frac{2}{3} x^2 + \frac{2}{3} x\right)$$
$$=-\frac{\sqrt 3}{2}\frac{2}{3}(x^2-x-2)$$
$$=-\frac{\sqrt 3}{3}\left(\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\right)$$
$$\leq -\frac{\sqrt 3}{3}\left(-\frac{9}{4}\right)=\frac{3\sqrt 3}{4}$$
Both inequalities have equality when $x=\frac{1}{2}$
Hence, $f(x)$ is maximum at $\frac{3\sqrt 3}{4}$ when $x=\frac{1}{2}$
However, this solution is (rather obviously I think) heavily reverse-engineered, with the two inequalities carefully manipulated to give identical equality conditions of $x=\frac{1}{2}$. Is there some better or more "natural" way to find the minimum point, perhaps with better uses of AM-GM or other inequalities like Jensen's inequality?
| $$\dfrac{1+x+1+x+1+x+3(1-x)}{3+1}\ge\sqrt[4]{(1+x)^33(1-x)}$$
Square both sides
| {
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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 can think of is by adding up the fractions and expanding, but took too long and I eventually got stuck. Is there any other method I can use to help me in this question?
| 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...
| {
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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} - \underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}}_{-\ln(2)}
\end{align}
We know, then
$$\ln(1 + x) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k} = x - \frac{x^2}{2} + \frac{x^2}{3} - \frac{x^2}{4} + \cdots$$
and for $x = 1$ we have
$$\ln(1 + 1) = \ln(2) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{1^k}{k} = 1 \underbrace{- \frac{1^2}{2} + \frac{1^2}{3} - \frac{1^2}{4} + \cdots}_{s_2} = 1- s_2$$
Which implies
$$s_2 = \ln(2) -1$$
Therefore
\begin{align}
S_1 &= 4(s_2) - (-\ln(2))\\
&= 4(\ln(2) -1) - (-\ln(2))\\
&= 4\ln(2) -4 + \ln(2)\\
&= 5\ln(2) -4\\
&= \ln(32) -4\\
\end{align}
See:
the sum: $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$ using Riemann Integral and other methods
| {
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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 $\sqrt[3]{2}+i$.
| 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 $\Bbb Q(\sqrt[3]2, i) = \Bbb Q(\sqrt[3]2 + i)$:
$\alpha = \sqrt[3]2 + i; \tag 1$
$\alpha - i = \sqrt[3]2; \tag 2$
$(\alpha - i)^3 = 2; \tag 3$
$\alpha^3 - 3 \alpha^2 i - 3 \alpha + i = 2; \tag 4$
$i(1 - 3\alpha^2) = 2 + 3\alpha - \alpha^3; \tag 5$
$i = \dfrac{\alpha^3 - 3\alpha - 2}{3\alpha^2 - 1} = \dfrac{2 + 3\alpha - \alpha^3}{1 - 3\alpha^2} \in \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i); \tag 6$
$\sqrt[3]2 = \alpha - i \in \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i); \tag 7$
from (6), (7) and the fact that $\Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i) \subset \Bbb Q(\sqrt[3]2, i)$ we affirm
$\Bbb Q(\sqrt[3]2, i) \subset \Bbb Q(\alpha) \Longrightarrow \Bbb Q(\sqrt[3]2, i) = \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i). \tag 8$
N.B. We may in fact derive from (6) a sixth degree polynomial satisfied by $\alpha$:
$\alpha^3 - 3\alpha - 2 = i(3\alpha^2 - 1), \tag 9$
$(\alpha^3 - 3\alpha - 2)^2 = -(3\alpha^2 - 1)^2, \tag{10}$
$\alpha^6 + 9\alpha^2 + 4 - 6\alpha^4 - 4\alpha^3 + 6\alpha = -9\alpha^4 + 6\alpha^2 - 1, \tag{11}$
$\alpha^6 + 3\alpha^4 - 4\alpha^3 + 3\alpha^2 + 6\alpha + 1 = 0, \tag{12}$
consistent with the results of Parcly Taxel.
| {
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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 $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?
| It seems right and it is bounded, as a minor observation we can improve the upper bound
$$x_n < \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \le\sqrt 2$$
| {
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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
$$e>\frac{n}{\sqrt[n]{n!}}.\tag1$$
Moreover, we can find that, for $k=0,1,\cdots,n-1.$ $$\frac{n^k}{k!}< \frac{n^n}{n!}.$$
Thus
\begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\
&< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\
&=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\
&=(2n+1)\cdot \frac{n^n}{n!},
\end{align*}
which shows that
$$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.\tag2$$
Combining $(1)$ and $(2)$, we have
$$e>\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}\to e(n \to \infty).$$
By the squeeze theorem,
$$\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
Please correct me if I'm wrong. Many thanks.
| 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 have
$$|\log f(n)-1|\le \frac{C\log n}{n}$$ Then $\log f(n)-1$ tends to $0$ so $f(n)$ tends to $e$ as you have show.
| {
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"source": "stackexchange",
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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; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)}=a+b+c+d-p-q-r-s$
My Working
I was able to calculate partial fractions as follows,
$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)(x-a)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)(x-b)}+..$
But I cannot proceed to deduction part. Highly appreciated if someone can give me a hint to work this out. Thank you!
| This is $$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}+\frac{D}{x-d}$$
The result should be this here
$$1+{\frac {{c}^{4}-{c}^{3}p-{c}^{3}q-{c}^{3}r-{c}^{3}s+{c}^{2}pq+{c}^{2
}pr+{c}^{2}ps+{c}^{2}qr+{c}^{2}qs+{c}^{2}rs-cpqr-cpqs-cprs-cqrs+pqrs}{
\left( -c+a \right) \left( -c+b \right) \left( -d+c \right)
\left( x-c \right) }}+{\frac {-{d}^{4}+{d}^{3}p+{d}^{3}q+{d}^{3}r+{d}
^{3}s-{d}^{2}pq-{d}^{2}pr-{d}^{2}ps-{d}^{2}qr-{d}^{2}qs-{d}^{2}rs+dpqr
+dpqs+dprs+dqrs-pqrs}{ \left( x-d \right) \left( -d+a \right)
\left( -d+b \right) \left( -d+c \right) }}+{\frac {-{b}^{4}+{b}^{3}p
+{b}^{3}q+{b}^{3}r+{b}^{3}s-{b}^{2}pq-{b}^{2}pr-{b}^{2}ps-{b}^{2}qr-{b
}^{2}qs-{b}^{2}rs+bpqr+bpqs+bprs+bqrs-pqrs}{ \left( x-b \right)
\left( -b+a \right) \left( -c+b \right) \left( -d+b \right) }}+{
\frac {{a}^{4}-{a}^{3}p-{a}^{3}q-{a}^{3}r-{a}^{3}s+{a}^{2}pq+{a}^{2}pr
+{a}^{2}ps+{a}^{2}qr+{a}^{2}qs+{a}^{2}rs-apqr-apqs-aprs-aqrs+pqrs}{
\left( -b+a \right) \left( -c+a \right) \left( -d+a \right)
\left( x-a \right) }}
$$
| {
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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 given was
$$ S_n = \frac{1}{10}n(n+1)(n+2)(n+3)(2n+3). $$
My approach
First, I get $$ U_n=n(n+2)(n+1)^2. $$
Then I tried to make $U_n = V_n - V_{n-1}$
in order to get $S_n = V_n - V_0$.
But I really don't know how can I figure this out.
| 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.
| {
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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
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+b^2)\cos^2x-2ac\cos x+(c^2-b^2)=0\\
\implies\cos^2x-\frac{2ac}{a^2+b^2}\cos x+\frac{c^2-b^2}{a^2+b^2}=0
$$
$$
a\cos\alpha+b\sin\alpha=c\implies a\cos^2\alpha\cos\beta+b\sin\alpha\cos\alpha\cos\beta=c\cos\alpha\cos\beta\\
a\cos\beta+b\sin\beta=c\implies a\sin\alpha\sin\beta\cos\beta+b\sin\alpha\sin^2\beta=c\sin\alpha\sin\beta\\
c\cos(\alpha+\beta)=a\cos\beta+a\sin\alpha\cos\beta.(\sin\beta-\sin\alpha)+b\sin\alpha+b\sin\alpha\cos\beta(\cos\alpha-\cos\beta)\\
$$
I think its getting complicated to solve now. What is the simplest way to solve this kind of problems?
| 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
$$\cos(\alpha+\beta)+i\sin(\alpha+\beta)=e^{i\alpha}e^{i\beta}=\frac{a^2-b^2}{a^2+b^2}+i\frac{2ab}{a^2+b^2}.$$
| {
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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 examples with squares or cubes, dealing with radicals can twist into strange paths.
The above expression ($a=\sqrt{(y^2+x^2)^3}$) can turn into a monster just as you add another simple term:
$$a=\sqrt{(y^2+x^2)^3} + \sqrt{(y^2-x^2)^3}$$
I can not see now how squaring the two sides could help in y isolation...
| 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,
$$y=\pm \sqrt{a^\frac{2}{3}-(a+x)^2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $y$-axis lies along this line. There are many ways to do this, one of which is the transformation matrix
$$\mathbf{T} := \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix}$$
Geometrically, this matrix is the rotation of the coordinate system around the origin counterclockwise by 45 degrees, thus turning the line $x = y$ counterclockwise just the right amount to align with the $y$-axis (as it lies at a 45-degree angle) multiplied by a scaling by $\sqrt{2}$ to make the entries integer and so easy to work with. Under this transformation, a coordinate vector $(x, y)$ transforms as
$$(x, y) \mapsto (x - y, x + y)$$
That is, define $u := x - y$ and $v := x + y$ as your new coordinates. Then the denominator becomes $u$, and the numerator is solved as follows: note that $v^2 = (x^2 + 2xy + y^2)$ so
$$x^2 + y^2 + 5xy = (x^2 + 2xy + y^2) + 3xy = v^2 + 3xy$$
and now, by recognizing the sum-and-difference-of-the-same-thing pattern, that you can take $u + v = 2x$ and $u - v = -2y$ so $(u + v)(u - v) = -4xy$ and $\frac{3}{4} (u + v)(u - v) = 3xy$. Thus
$$\begin{align}\frac{x^2 + y^2 + 5xy}{x - y} &= \frac{v^2 - \frac{3}{4}(u + v)(u - v)}{u} \\
&= \frac{v^2 - \frac{3}{4}(u^2 - v^2)}{u} \\
&= \frac{v^2 - \frac{3}{4}u^2 + \frac{3}{4}v^2}{u}\\
&= \frac{-\frac{3}{4}u^2 + \frac{7}{4}v^2}{u}\\
&= -\frac{3}{4} u + \frac{7}{4} \frac{v^2}{u} \\
&= \frac{7}{4} \frac{v^2}{u} - \frac{3}{4} u
\end{align}$$
and now when you take the limit - note that the origin is unchanged, so the limit is still to $(0, 0)$ but now in $(u, v)$ coordinates instead of $(x, y)$ coordinates - the second term is zero, but the first has no limit: if $v$ goes to 0 first, then the limit as $u$ goes to 0 is 0, but if $u$ goes to 0 first, the limit becomes $\pm \infty$, and is such for every value of $v$ except 0 for such an approach. That is, the function is unboundedly large in an arbitrarily small circle around the origin, and thus the limit at that origin does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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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-1}{2n}\frac{2n-3}{2n-2}I_{n-1}=\cdots\\
&=\frac{(2n-1)!!}{(2n)!!}I_1=4^{-n}\binom{2n}{n}\frac \pi 2
\end{align*}
By Stirling's formula,
$$\int_0^1\frac{dx}{(1+x^2)^{n+1}}=I_{n+1}+O(\tfrac 1 n)\sim \frac{1}{\sqrt{\pi n}}\cdot\frac\pi 2=\frac{\sqrt\pi}{2\sqrt n}$$
Thus the original limit equals
$$\sqrt n\int_0^1\frac{dx}{(1+x^2)^n}\sim\frac{\sqrt{\pi n}}{2\sqrt{n-1}}\to\frac{\sqrt \pi}{2}=0.8862\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
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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 $x,y$, such that $2a=x+2y-1, 2b=y(y-1)$, but using this idea, $x,y$ won’t be always positive.
| 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 $B=(0,2b+1)$ and having its vertex in the second quarter plane.
Starting from the origin, we can perform a sequence of Viéte jumps:
$$
(x,y)
\mapsto \big(x,2b+1-2x-y\big)
\mapsto \big(2a+1-2(2b+1-2x-y)-x,2b+1-2x-y\big)
\mapsto \ldots
$$
All these lattice points lie on the parabola. We continue this sequence until we get onto the arc $AB$.
More formally, let $P_n=(x_n,y_n)$ with
$$
x_0=y_0=0, \quad
x_{n+1}=2a-4b-1+3x_n+2y_n,
\quad y_{n+1}=2b+1-2x_n-y_n.
$$
Notice that $x_{n+1}+y_{n+1}=x_n+y_n+2(a-b)$, so $x_n+y_n=2(a-b)n$. Hence, we must take the unique index $n$ with $2b+1<2(a-b)n<2a+1$ in order to find $P_n$ inside arc $AB$. Solving the equations $f(x,y)=0$, $x+y=2(a-b)n$, the coordinates are
$$
x_n = \big(2(a-b)n-(2b+1)\big)n, \quad
y_n = \big((2a+1)-2(a-b)n\big)n.
$$
(The picture shows the case $a=8$, $b=6$.)
| {
"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
} |
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:
$$
A = E_n \dots E_2 \cdot E_1 \cdot B
$$
therefore $AM = 0$ becomes
$$
AM = (E_n \dots E_2 \cdot E_1 \cdot B) \cdot M = E_n \dots E_2 \cdot E_1 \cdot (B \cdot M) = 0$$
This allows me to find a matrix M that, multiplied left by B, will give me the zero matrix. I got that
$$ B = \begin{bmatrix} 1 & 0 & -1 & 0 &0\\
0 & 1 & 2 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} $$
However, the only column I can think of that would give me a zero is in the form:
$$ x = \begin{bmatrix}
c\\
-2c\\
c\\
0\\
0\\
\end{bmatrix} $$
Setting that as one of the columns of M and filling the other entries with zeroes would only give me a matrix of rank 1.
Am I missing something or is my logic totally flawed?
| We have that
$$A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
-1 & 1 & 3 & -1 & 0\\
-2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix}\to \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
0 & 1 & 2 & 1 & 1\\
0 & 1 & 2 & 3 & 5\\ 0 & -1 & -2 & -5 & -9\\ \end{bmatrix}\to$$
$$\begin{bmatrix}
1 & 0 & -1 & 2 &1\\
0 & 1 & 2 & 1 & 1\\
0 & 0 & 0 & 2 & 4\\
0 & 0 & 0 & -2 & -4\\ \end{bmatrix}\to
\begin{bmatrix}
1 & 0 & -1 & 2 &1\\
0 & 1 & 2 & 1 & 1\\
0 & 0 & 0 & 2 & 4\\
0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$
therefore since the rank of the matrix is equal to $r=3$ the null space has dimension $\dim(N(A))= n-r=5-3=2\,$ and we can find $x_1$ and $x_2$ linearly independent such that
$$Ax_1=Ax_2=0$$
| {
"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\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
| 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 $$(s^3 + s^2 + s - 1) p(s) + 4$$ for some cubic polynomial $p$. But we're given that $s^3 + s^2 + s = 1$, so the first term is zero, and thus $$\color{#df0000}{\boxed{c^6 - 4 c^4 + 8 c^2 = 4}} .$$
| {
"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 can be computed with $\vec{y}=A \vec{x}+b$.
Since it's the unit circle I tried to map it:
$$T((0,0)^T)=A(0,0)^T+b=(0,0)+b$$
but I don't get the transformation.
In which way can it be done?
| 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
$$x^2+y^2+xy-x-y=a\cdot(x+y+b)^2+c\cdot(x-y+d)^2-ab^2-cd^2.$$
Matching coefficients you get the conditions
\begin{align}a+c=1\\
2a-2c=1\\
2ab+2cd=-1\\
2ab-2cd=-1
\end{align}
solving these you get everything you need. Indeed you get
$$a=\frac34,\ c=\frac14,\ b=-\frac{2}3,\ d=0$$
So you can rewrite the equation as
\begin{align}x^2+y^2+xy-x-y&=\frac34\cdot(x+y-2/3)^2+\frac14\cdot(x-y)^2-1/3\\ &=\left(\frac{\sqrt{3}}{2}(x+y)-\frac{\sqrt{3}}3\right)^2+\left(\frac{x-y}{2}\right)^2-1/3.\end{align}
Now we go from
$$\left(\frac{\sqrt{3}}{2}(x+y)-\frac{\sqrt{3}}3\right)^2+\left(\frac{x-y}{2}\right)^2-1/3\leq 1$$
to
$$\frac34\left(\frac{\sqrt{3}}{2}(x+y)-\frac{\sqrt{3}}3\right)^2+\frac34\left(\frac{x-y}{2}\right)^2\leq 1$$
which is equal to
$$\left(\frac{3}{4}(x+y)-\frac12\right)^2+\left(\frac{\sqrt{3}}{4}(x-y)\right)^2\leq 1.$$
So the affine transform is
$$\begin{pmatrix}\frac{3}{4} & \frac{3}{4}\\ \frac{\sqrt{3}}{4} & -\frac{\sqrt{3}}{4}\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}+\begin{pmatrix}-\frac12 \\ 0 \end{pmatrix}$$
| {
"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
} |
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
$n+1+(p-1)n>p\sqrt[p]{n^{p-1}(n+1)}$
$ \Longleftrightarrow $ $pn+1>p\sqrt[p]{n^{p-1}(n+1)}$
$ \Longleftrightarrow $ $1>p\sqrt[p]{n^{p-1}}(\sqrt[p]{n+1}-\sqrt[p]{n})$
$ \Longleftrightarrow $ $\frac{1}{\sqrt[p]{n^{p-1}}}>p(\sqrt[p]{n+1}-\sqrt[p]{n})$
$ \Longrightarrow $ $ 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}}}$
For left hand side of inequality
$n-1+(p-1)n>p\sqrt[p]{(n-1)n^{p-1}}$
$ \Longleftrightarrow $ $pn-1>p\sqrt[p]{(n-1)n^{p-1}}$
$ \Longleftrightarrow $ $\frac{1}{\sqrt[p]{n^{p-1}}}<p(\sqrt[p]{n}-\sqrt[p]{n-1})$
$ \Longrightarrow $ $\frac{1}{\sqrt[p]{1}}+\frac{1}{\sqrt[p]{2^{p-1}}}+...+\frac{1}{\sqrt[p]{n^{p-1}}}< p\sqrt[p]{n}$
Hope that someone will have nicer solution.
| Note that $f(x)=\dfrac{1}{\sqrt[p]{x^{p-1}}}=x^{(1/p-1)}$ is a decreasing function
$$\frac{1}{\sqrt[p]{(k+1)^{p}}}
<\int_{k}^{k+1} \frac{dx}{\sqrt[p]{x^{p-1}}}
<\frac{1}{\sqrt[p]{k^{p}}}$$
$$\sum_{k=0}^{n-1} \frac{1}{\sqrt[p]{(k+1)^{p}}}
<\int_{0}^{n} \frac{dx}{\sqrt[p]{x^{p-1}}}
\implies
\frac{1}{\sqrt[p]{1^{p}}}+\ldots+\frac{1}{\sqrt[p]{n^{p}}} < p\sqrt[p]{n}$$
$$\int_{1}^{n} \frac{dx}{\sqrt[p]{x^{p-1}}}
< \sum_{k=1}^{n} \frac{1}{\sqrt[p]{k^{p}}}
\implies
p\left( \sqrt[p]{n+1}-1 \right)<\frac{1}{\sqrt[p]{1^{p}}}+\ldots+\frac{1}{\sqrt[p]{n^{p}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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: $$\mathrm{d}v=\mathrm{d}x\Rightarrow v=x\\u=\frac1{(ax^2+b)^n}\Rightarrow \mathrm{d}u=-2an(ax^2+b)^{-n-1}x\mathrm{d}x$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2n\int\frac{ax^2\mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2n\int\frac{ax^2+b-b}{(ax^2+b)^{n+1}}\mathrm{d}x$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2n\int\frac{ax^2+b}{(ax^2+b)^{n+1}}\mathrm{d}x-2nb\int\frac{\mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2n\int\frac{\mathrm{d}x}{(ax^2+b)^n}-2nb\int\frac{\mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x^2}{(ax^2+b)^n}+2nI_n-2nbI_{n+1}$$
Now we solve for $I_{n+1}$ in terms of $I_n$:
$$I_n=\frac{x^2}{(ax^2+b)^n}+2nI_n-2nbI_{n+1}$$
$$2bI_{n+1}=\frac{x^2}{(ax^2+b)^n}+2nI_n-I_n$$
$$I_{n+1}=\frac{x^2}{2b(ax^2+b)^n}+\frac{2n-1}{2b}I_n$$
Cleverly replacing $n$ with $n-1$ gives
$$I_{n-1+1}=\frac{x^2}{2b(ax^2+b)^{n-1}}+\frac{2(n-1)-1}{2b}I_{n-1}$$
$$I_{n}=\frac{x^2}{2b(ax^2+b)^{n-1}}+\frac{2n-3}{2b}I_{n-1}$$
And there's your reduction formula. Plugging in $a=1,\quad b=1,\quad n=2$ will give your integral without the bounds plugged in. Note that this formula only works when $b\neq0$, $n$ is an integer, and $n\geq2$.
| {
"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
} |
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't find any way to solve it, plz help me.. Advance thanks to u
| 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
$$f(z)=\frac{1+z^2+z^4}{(1+z^2)(1+z^4)}=\frac{1}{2\,(1+z^2)}+\frac{1}{4\,\Biggl(\frac{1}{2}+\left(z+\frac{1}{\sqrt{2}}\right)^2\Biggr)}+\frac{1}{4\,\Biggl(\frac{1}{2}+\left(z-\frac{1}{\sqrt{2}}\right)^2\Biggr)}$$
for all $z\in\mathbb{C}$ with $|z|<1$. That is,
$$\int\,f(x)\,\text{d}x=\frac{1}{2}\,\arctan(x)+\frac{\arctan(\sqrt{2}\,x+1)+\arctan(\sqrt{2}\,x-1)}{2\sqrt{2}}+\text{constant}\,.$$
That is,
$$\int_0^1\,f(x)\,\text{d}x=\frac{1}{2}\,\left(\frac{\pi}{4}\right)+\frac{\left(\frac{3\pi}{8}-\frac{\pi}{4}\right)+\left(\frac{\pi}{8}+\frac{\pi}{4}\right)}{2\sqrt{2}}=\frac{\pi}{8}\,\big(1+\sqrt{2}\big)\,.$$
However, since the series representation of $f(z)$ compactly converges for $z\in\mathbb{C}$ such that $|z|<1$, we can then integrate $f(z)=1-z^6+z^8-z^{14}+\ldots$ term-by-term to get
$$\int_0^1\,f(x)\,\text{d}x=1-\frac{1}{7}+\frac{1}{8}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\ldots\,.$$
This shows that
$$1-\frac{1}{7}+\frac{1}{8}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\ldots=\frac{\pi}{8}\,\big(1+\sqrt{2}\big)\approx 0.948059449\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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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 example if $x = 2, y = 2$, we have $4$ rectangles of size $(1,1)$, $2$ rectangles of size $(1, 2)$, $2$ rectangles of size $(2, 1)$ and $1$ rectangle of size $(4,4)$. So the total answer is $4\cdot 1+ 2\cdot2 + 2\cdot2+1\cdot4 = 16$
Is there easy way to solve this for different $x$ and $y$?
| 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:
$$\begin{align}A_{1\times y}&=1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1=\\
&=\sum_{i=1}^yi\cdot (y-i+1)=\\
&=(y+1)\sum_{i=1}^y i-\sum_{i=1}^yi^2=\\
&=(y+1)\cdot \frac{y(y+1)}{2}-\frac{y(y+1)(2y+1)}{6}=\\
&=\frac{y(y+1)[3y+3-2y-1]}{6}=\\
&=\frac{y(y+1)(y+2)}{6}.\end{align}$$
Consider $2\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{2\times y}&=\color{red}{2}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\
&+\color{red}{1}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]=\\
&=\color{red}{2}\times [A_{1\times y}]+\color{red}{1}\times 2\times [A_{1\times y}]\\
&=(\color{red}2+\color{red}1\times 2)\times [A_{1\times y}]\\
&=4\cdot \frac{y(y+1)(y+2)}{6}.\end{align}$$
Consider $3\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{3\times y}&=\color{red}{3}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\
&+\color{red}{2}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]+\\
&+\color{red}{1}\times [3\times y+6\times (y-1)+9\times (y-3)+\cdots +3y\times 1]=\\
&=(\color{red}{3}+\color{red}2\times 2+\color{red}1\times 3)\times [A_{1\times y}]=\\
&=10\cdot \frac{y(y+1)(y+2)}{6}.\end{align}$$
And now consider $x\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{x\times y}&=\ \ \ \ \ \ \ \ \ \ \color{red}{x}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\
&+\color{red}{(x-1)}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]+\\
&+\color{red}{(x-2)}\times [3\times y+6\times (y-1)+9\times (y-3)+\cdots +3y\times 1]=\\
& \ \ \vdots\\
&+\ \ \ \ \ \ \ \ \ \ \ \color{red}{1}\times [x\times y+2x\times (y-1)+3x\times (y-3)+\cdots +xy\times 1]=\\
&=(\color{red}{x}+\color{red}{(x-1)}\times 2+\color{red}{(x-2)}\times 3+\cdots +\color{red}1\times x)\times [A_{1\times y}]=\\
&=[A_{x\times 1}]\cdot [A_{1\times y}]=\\
&=\frac{x(x+1)(x+2)}{6}\cdot \frac{y(y+1)(y+2)}{6},\end{align}$$
which complies with the brilliant answer given by Todor Markov (+1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 tried bringing in $\sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?
| 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 $z+\dfrac1z=2\cos\dfrac{2k\pi}5, \left(z+\dfrac1z\right)^2=\cdots=2\left(1+\cos\dfrac{4k\pi}5\right)$
$z^2+\dfrac1{z^2}=2\cos\dfrac{4k\pi}5,\left(z^2+\dfrac1{z^2}\right)^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $$S'\leq \frac{1}{3}S$$
We can assume that side of a triangle $ABC$ is $1$. Further let $CM =x$ and $\angle KCM =\gamma$. Rotate $M$ around $C$ for $-60^{\circ}$ in to $F$. Then the area of triangle $AMF$ is the one we are looking for and it's area is area of $AMCF$ minus area of equilateral triangle $CFM$, so $$4S' = -x^2\sqrt{3}+2x\sin (60^{\circ}+\gamma)$$
and this should be easy to calculate that is less than ${\sqrt{3}\over 3}$.
If we see $S'$ quadratic function on $x$ we get:
$$ 4S'\leq {1\over \sqrt{3}}\sin (60^{\circ}+\gamma)\leq {1\over \sqrt{3}}$$
From here we can see that equality is achieved iff $\gamma = 30^{\circ}$ and $x= {\sqrt{3}\over 3} = {2\over 3}v$ where $v$ is altitude of triangle $ABC$. That is, equality is achieved iff $M$ is gravity centre of $ABC$.
I'm interested in different solutions (for example without trigonometry).
| 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 $a = AM$, $b = BM$, $c = CM$ and $S'$ be the area of a triangle with sides $a,b,c$. In this coordinate system, $S = \frac{3\sqrt{3}}{4}$, we want to show $S' \le \frac{\sqrt{3}}{4}$. Using Heron's formula, this is equivalent to
$$16S'^2 = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) \stackrel{?}{\le} 3$$
Identify euclidean plane with the complex plane. The vertices $A,B,C$ corresponds to $1, \omega, \omega^2 \in \mathbb{C}$ where $\omega = e^{\frac{2\pi}{3}i}$ is the cubic root of unity. Let $z$ be the complex number corresponds to $M$ and $\rho = |z|$, we have
$$
\begin{cases}
a^2 = |z-1|^2 = \rho^2 + 1 - (z + \bar{z})\\
b^2 = |z-\omega|^2 = \rho^2 + 1 - (z\omega + \bar{z}\omega^2)\\
c^2 = |z-\omega^2|^2 = \rho^2 + 1 - (z\omega^2 + \bar{z}\omega)
\end{cases}
\implies
a^2 + b^2 + c^2 = 3(\rho^2 + 1)
$$
Thanks to the identity $\omega^2 + \omega + 1 = 0$, all cross terms involving $\omega$ explicitly get canceled out.
Doing the same thing to $a^4 + b^4 + c^4$, we get
$$\begin{align}a^4 + b^4 + c^4
&= \sum_{k=0}^2 (\rho^2 + 1 + (z\omega^k + \bar{z}\omega^{-k}))^2\\
&= \sum_{k=0}^2\left[ (\rho^2 + 1)^2 + (z\omega^k + \bar{z}\omega^{-k})^2\right]\\
&= 3(\rho^2 + 1)^2 + 6\rho^2\end{align}$$
Combine these, we obtain
$$16S'^2 = 3(\rho^2+1)^2 - 12\rho^2 = 3(1 - \rho^2)^2$$
Since $M$ is inside triangle $ABC$, we have $\rho^2 \le 1$. As a result,
$$S' = \frac{\sqrt{3}}{4}(1-\rho^2) \le \frac{\sqrt{3}}{4} = \frac13 S$$
Solution 2 - using circle inversion.
Let $a = AM, b = BM, c = CM$ again. Let $\Delta(u,v,w)$ be the area of a triangle with sides $u,v,w$. In particular, $S = \Delta(1,1,1)$ and $S' = \Delta(a,b,c)$. We will use the fact $\Delta(u,v,w)$ is homogeneous in $u,v,w$ with degree $2$.
Consider the circle inversion with respect to a unit circle centered at $A$.
Under such an inversion, $B,C$ get mapped to itself while $M$ mapped to a point $M'$ with $$AM' = \frac{1}{a}, BM' = \frac{b}{a}, CM' = \frac{c}{a}$$
We can decompose the quadrilateral $ABM'C$ in two manners. $\triangle ABC + \triangle BM'C$ and $\triangle ABM' + \triangle AM'C$. This leads to
$$\begin{align}
&\verb/Area/(ABC) + \verb/Area/(BM'C) = \verb/Area/(ABM') + \verb/Area/(AM'C)\\
\iff &
S + \Delta(1,\frac{b}{a},\frac{c}{a}) = \Delta(1,\frac{b}{a},\frac{1}{a}) + \Delta(1,\frac{c}{a},\frac{1}{a})\\
\iff &
Sa^2 + S' = \Delta(1,a,b) + \Delta(1,a,c)
\end{align}
$$
By a similar argument, we have
$$
Sb^2 + S' = \Delta(1,b,a) + \Delta(1,b,c)\quad\text{ and }\quad
Sc^2 + S' = \Delta(1,c,a) + \Delta(1,c,b)
$$
Summing these three equalities together and notice
$$\Delta(1,a,b) + \Delta(1,b,c) + \Delta(1,c,a)
= \verb/Area/(ABM) + \verb/Area/(BCM) + \verb/Area/(CAM)
= S$$
We obtain
$$3S' = 2S - S(a^2+b^2+c^2)$$
For any $\triangle XYZ$ and point $P$ in the plane, we know the expression $XP^2 + YP^2 + ZP^2$ is minimized when $P$ is the centroid of $\triangle XYZ$. For an equilateral triangle of side $1$, the centroid is at a distance $\frac{1}{\sqrt{3}}$ from the vertices. This implies $a^2 + b^2 + c^2 \ge 1$.
As a result, $3S' \le S$ and we are done.
| {
"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 equality cases. I'm mostly interested if there is a way to derive this reasonably, other than just magically guessing the solutions and then doing degree-counting. This was left as a "good exercise" in Mildorf that has bothered me for years.
| Too long for comment
Using $\sin(\pi-A)=+\sin A,\cos(\pi-B)=-\cos B,\sin2C=2\sin C\cos C$
$$\dfrac{\sin^2\dfrac{2\pi}7}{\sin^2\dfrac{\pi}7}=4\cos^2\dfrac\pi7=2+2\cos\dfrac{2\pi}7$$
$$\dfrac{\sin^2\dfrac{\pi}7}{\sin^2\dfrac{3\pi}7}=\dfrac{\sin^2\dfrac{6\pi}7}{\sin^2\dfrac{3\pi}7}=4\cos^2\dfrac{3\pi}7=2+2\cos\dfrac{6\pi}7$$
$$\dfrac{\sin^2\dfrac{3\pi}7}{\sin^2\dfrac{2\pi}7}=\dfrac{\sin^2\dfrac{4\pi}7}{\sin^2\dfrac{2\pi}7}=4\cos^2\dfrac{2\pi}7=2+2\cos\dfrac{4\pi}7$$
From How to solve $8t^3-4t^2-4t+1=0$ and factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$,
the roots of $$x^3+x^2-2x-1=0$$ are $2\cos\dfrac{2\pi}7,2\cos\dfrac{4\pi}7,2\cos\dfrac{6\pi}7$
Replace $x$ with $y-2$ to find $$y^3-5y^2+6y-1=0$$
| {
"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
} |
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:
$$
\frac{f(x)+f(y)+f(z)}{3} + f(\frac{x+y+z}{3})-\frac{2}{3}[f(\frac{x+y}{2}) +f(\frac{y+z}{2}) +f(\frac{z+x}{2})] \ge 0
$$
I tried to prove this statement with this assumption that f is convex. First, I used the definition of convexity for each terms of the right hand side:
$$f(\frac{x+y}{2})\le \frac{1}{2}f(x)+\frac{1}{2}f(y)$$
$$f(\frac{y+z}{2})\le \frac{1}{2}f(y)+\frac{1}{2}f(z)$$
$$f(\frac{z+x}{2})\le \frac{1}{2}f(z)+\frac{1}{2}f(x)$$
$$\Longrightarrow f(\frac{x+y}{2}) +f(\frac{y+z}{2}) +f(\frac{z+x}{2}) \le f(x)+f(y)+f(z)
$$
Then after using definition for $f(x)+f(y)+f(z)/3$ and replacing above statement in the main inequality, I obtained:
$$
\frac{f(x)+f(y)+f(z)}{3} + f(\frac{x+y+z}{3})-\frac{2}{3}[f(\frac{x+y}{2}) +f(\frac{y+z}{2}) +f(\frac{z+x}{2})] \ge f(\frac{x+y+z}{3})-\frac{f(x)+f(y)+f(z)}{3}
$$
According to the convexity of $f$, the later expression must be nonpositive but our purpose is to prove it is nonnegative. It seems we encounter with a contradiction! I tried other types of convex combinations, but I just obtained either of parts are greater or smaller than a same term, for example this one:
first part:$$\frac{f(x)+f(y)+f(z)}{3} + f(\frac{x+y+z}{3}) \ge 2f(\frac{x+y+z}{3})$$
second part:$$2[\frac{1}{3}f(\frac{x+y}{2}) +\frac{1}{3}f(\frac{y+z}{2}) +\frac{1}{3}f(\frac{z+x}{2})] \ge 2f(\frac{x+y+z}{3})$$
I think it needs to obtain a tighter bound or maybe some new information about continuous convex functions.
| 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
} |
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:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \cos(\frac{\pi}{3})$
$(\frac{2}{2}) \cdot \frac{3}{5} + \frac{1}{2} \cdot (\frac{5}{5})$
$\frac{6}{10} + \frac{5}{10}$
$\frac{11}{10}$
| 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
} |
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 assuming it's on the tangent line.
How do you find the equation of the tangent line? Explanation needed.
| 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 $AC$ is $x=7$ and the tangent line $AB$ is $y=\frac5{12}x+\frac{13}{12}.$
| {
"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, $S=2\pi\int_0^{2\pi}b\sin\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta$
About $y-$axis is, $S=2\pi\int_0^{2\pi}a\cos\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta$from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.Thanks in advance.
| 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 the numerator must vanish at $x=2$, we can use l'Hopital to obtain
$$\lim_{x\to 2}\frac{x^2-cx+d}{x^2-4}=\lim_{x\to 2}\frac{2x-c}{2x}=3$$
from which we can find $c$ and then $d$.
| {
"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 use the Picard-Iteration, i get ($ s:= \begin{pmatrix}
x(t) \\
y(t)\end{pmatrix} $ )
$$ s_1 = \begin{pmatrix}
2 \\
-2t\end{pmatrix} $$
$$ s_2 = \begin{pmatrix}
2-t^2 \\
-2t\end{pmatrix} $$
$$ s_3 = \begin{pmatrix}
2-t^2 \\
\frac{1}{3} t^3-2t\end{pmatrix} $$
I assume that $ s_{\infty} = \begin{pmatrix}
2cos t \\
-2sint \end{pmatrix} $
But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 = \begin{pmatrix}
2-t^2 \\
\frac{1}{3} t^3-2t\end{pmatrix} $
How do i get this right?
| 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:
$$
\frac{1}{x^p}=\frac{1}{\Gamma(p)}\int_{0}^{\infty}e^{-xt}t^{p-1}dt\tag{2}
$$
By plugging $(2)$ into $(1)$ I get the following double integral:
$$
\frac{1}{\Gamma(p)}\int_{0}^{\infty}\int_{0}^{\infty}e^{-xt}t^{p-1}\cos xdtdx\tag{3}
$$
However, I unable to proceed any further as I am unclear as to what method should I use in order to compute this integral. I thought that an appropriate change of variables could transform it into a product of two gamma functions but I cannot see how that would work. Any help would be greatly appreciated.
| 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 be rewritten as
$$\int_0^\infty \frac{\cos x}{x^p} \, dx = \frac{1}{\Gamma (p)} \int_0^\infty \int_0^\infty e^{-xt} \cos x t^{p - 1} \, dt \, dx,$$
or
$$\int_0^\infty \frac{\cos x}{x^p} \, dx = \frac{1}{\Gamma (p)} \int_0^\infty t^{p - 1} \int_0^\infty e^{-xt} \cos x \, dx \, dt,$$
after changing the order of integration.
The inner $x$-integral can be readily found. Either using integration by parts twice, or recognising the integral as the Laplace transform for the function $\cos x$, as
$$\int_0^\infty e^{-xt} \cos x \, dx = \frac{t}{1 + t^2},$$
we have
$$\int_0^\infty \frac{\cos x}{x^p} \, dx = \frac{1}{\Gamma (p)} \int_0^\infty \frac{t^p}{1 + t^2} \, dt.$$
Enforcing a substitution of $t \mapsto \sqrt{t}$ leads to
\begin{align}
\int_0^\infty \frac{\cos x}{x^p} \, dx &= \frac{1}{2 \Gamma (p)} \int_0^\infty \frac{t^{\frac{p}{2} - \frac{1}{2}}}{1 + t} \, dt = \frac{1}{2 \Gamma (p)} \int_0^\infty \frac{t^{\frac{p + 1}{2} - 1}}{(1 + t)^{\frac{p + 1}{2} + \frac{1 - p}{2}}}.
\end{align}
As this is exactly of the form of the Beta function (see the second of the integral representations in the link) we have
\begin{align}
\int_0^\infty \frac{\cos x}{x^p} \, dx &= \frac{1}{2 \Gamma (p)} \text{B} \left (\frac{p + 1}{2}, \frac{1 - p}{2} \right )\\
&= \frac{1}{2 \Gamma (p)} \Gamma \left (\frac{p}{2} + \frac{1}{2} \right ) \Gamma \left (\frac{1}{2} - \frac{p}{2} \right ) \\
&= \frac{1}{2 \Gamma (p)} \Gamma \left (\frac{p}{2} + \frac{1}{2} \right ) \Gamma \left [1 - \left (\frac{p}{2} + \frac{1}{2} \right ) \right ] \\
&= \frac{1}{2 \Gamma (p)} \frac{\pi}{\sin (p + 1)\pi/2} \qquad (*)\\
&= \frac{\pi}{2 \Gamma (p) \cos \left (\frac{\pi p}{2} \right )},
\end{align}
as required. Note Euler's reflection formula was used in ($*$).
| {
"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:
$\displaystyle \frac{2}{y^{2}} - \frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$\displaystyle y = -\frac{5}{2} \pm \frac{\sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $\displaystyle y = \frac{58}{4} \pm \frac{10\sqrt{33}}{4}$
But this isn't the correct solution.
Also for $\#10$ I do the same thing:
Let $y = \sqrt n$ then $y^2 = n$
So I have $\displaystyle \frac{3}{y^2} - \frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $\displaystyle y = -\frac{3}{2}, \frac{1}{3}$
Then $n = \frac{9}{4}, \frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1)\ (x-7)^2 -13(x-7) +36=0 \qquad \qquad 4)\ 3(w/6)^2 -8(w/6) +4=0 \\
2)\ (1-3x)^2 -13(1-3x) +36=0 \qquad \qquad 5)\ 3(w^2-2)^2 -8(w^2-2) +4=0 \\
3)\ x^4 -13x^2 +36=0 \qquad \qquad 6)\ \frac{3}{p^2} -\frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
| 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 solutions of the original equation are solutions of the new equation. But the new equation might have more solutions than the original!
Look at a simpler example first: If you want to solve $\sqrt{x} = x-2$ for real $x \ge 0$, squaring yields $x = (x-2)^2$ with solutions $x = 1$ or $x = 4$. But $1 = (x-2)^2$ does not imply that $1 = x-2$, so you should see why you cannot conclude that the original equation had the same solutions. It is also incorrect to just exclude the solutions based on 'domain', as this example shows clearly.
Symbolically, ( $x \in \mathbb{R}_{\ge0}$ and $\frac2x - \frac5{\sqrt{x}} = 1$ and $y = \sqrt{x}$ ) imply ( $x \in \mathbb{R}_{\ge0}$ and $\frac2x - \frac5{\sqrt{x}} = 1$ and $y^2 = x$ ) and hence imply ( $y \in \mathbb{R}$ and $\frac2{y^2} - \frac5y = 1$ and $y^2 = x$ ), but the reverse implications do not necessarily hold. The implications nevertheless allow you to conclude that every $x,y$ satisfying ( $x \in \mathbb{R}_{\ge0}$ and $\frac2x - \frac5{\sqrt{x}} = 1$ and $y = \sqrt{x}$ ) must also satisfy ( $y \in \mathbb{R}$ and $\frac2{y^2} - \frac5y = 1$ and $y^2 = x$ ), so you can limit your search to just the solutions of the latter, which are easier to find.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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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 $x^2+y^2=9$
$$\cases{x=3\cos t \\ y=3\sin t} \text{ for } 0\le t\le2\pi$$
Now how do I calculate $\int_CF\cdot dr$?
Can anyone explain how to solve this?
| 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
$\vec r = \begin{pmatrix} 3 \cos t \\ 3\sin t \end{pmatrix}, \tag 5$
$d \vec r = \dfrac{d \vec r}{dt} dt = \begin{pmatrix} -3\sin t \\ 3\cos t \end{pmatrix} dt, \tag 5$
and
$F = \begin{pmatrix} 3\sin t \\ -3\cos t \end{pmatrix}; \tag 6$
then
$F \cdot d \vec r = \begin{pmatrix} 3\sin t \\ -3\cos t \end{pmatrix} \cdot \begin{pmatrix} -3\sin t \\ 3\cos t \end{pmatrix} dt$
$= (-9 \sin^2 t - 9 \cos^2 t)dt = -9(\sin^2 t + \cos^2 t)dt = -9(1)dt = -9dt; \tag 7$
finally,
$\displaystyle \int_C F \cdot d \vec r = \int_C -9 dt = -9 \int_C dt = -9(2\pi) = -18\pi. \tag 8$
| {
"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 ∞} $$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)}$ - ${1\over 2}$
$\left|{(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)} - {1\over 2} \right|$ < ε
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)\over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)\over(8n^7-14n^2+10)}$ < ε
$ {(1)\over(4n^4)}$ = $ {(n^3)\over(4n^7)}$ < $ {(3n^3)\over(4n^7)}$ = $ {(6n^3)\over(8n^7)}$ < $ {(6n^3)\over(8n^7-14n^2+10)}$ < ε+7
$ {(1)\over(ε+7)}$ < $4n^4$
$ {1\over4ε+28^{1\over4}}$ < n = N
Now I have my N so I can writ my proof:
Let ε, N > 0 and N < n. Suppose N = $ {1\over4ε+28^{1\over4}}$ . Then it must follow that:
$ {1\over4ε+28^{1\over4}}$ < n
$ {1\over4ε+28}$ < $n^4$
$ {1\overε+7}$ < $4n^4$
$ {1\over 4n^4}$ < $ε+7$
from there on I dont know how to use my inequalities to get back to what I started with.
| I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$\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\mathbb{N_{>0}}$$
Thus $$\lim_{n\to\infty}\Biggl(\frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}\Biggr)=\lim_{n\to\infty}\Biggl(\frac{2+3\frac{1}{n^2}+4\frac{1}{n^6}}{4-7\frac{1}{n^5}+5\frac{1}{n^7}}\Biggr)=\frac{2+0+0}{4-0+0}=\frac{1}{2}$$
| {
"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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