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Sove the recurrence relation $a_n = 5a_{n-1}-8a_{n-2}+4a_{n-3}$ and $a_0 = 0$, $a_1 = 0$, $a_2 = 1$
Consider the recurrence relation $a_n = 5a_{n-1}-8a_{n-2}+4a_{n-3}$ and $a_0 = 0$, $a_1 = 0$, $a_2 = 1$. Find a closed form expression for $a_n$.
I have computed the OGF $A(z)$ of $(a_n)_n$ as
$$A(z) = \frac{z^2}{1-5z+8z^2-4z^3}.$$
According to Wolfram Alpha this has the following partial fraction decomposition:
$$= \frac{-3}{2(1-2z)} + \frac{1}{2(1-2z)^2} + \frac{1}{1-z}$$
and using the Generalised Binomial Theorem yields:
$$= -\frac{3}{2} \sum_{n = 0}^\infty 2^nz^n + \frac{1}{2} \sum_{n=0}^\infty \binom{-2}{n}2^nz^n + \sum_{n=0}^\infty z^n.$$
Thus we should have
$$a_n = [z^n]A(z) = -\frac{3}{2}2^n + \frac{1}{2} \binom{-2}{n}2^n +1 = -3 \cdot 2^{n-1} + \binom{-2}{n}2^{n-1} +1 $$
However, this does not seem to be true. Could you please tell me what I am doing wrong?
| Starting from
$$\frac{z^2}{1-5z+8z^2-4z^3}$$
then
\begin{align}
\sum_{n=0}^{\infty} a_{n} \, x^n &= \frac{x^2}{1-5x+8x^2-4x^3} = \frac{x^2}{(1-x)(1-2 x)^2} \\
&= \frac{1}{1 - x} - \frac{3}{2 (1 - 2 x)} + \frac{1}{2 \, (1 - 2 x)^2} \\
&= \sum_{n} x^n - \frac{3}{2} \, \sum_{n} 2^n \, x^n + \frac{1}{2} \, \sum_{n} 2^n (n+1) \, x^n \\
&= \sum_{n=0}^{\infty} \left(1 - 3 \cdot 2^{n-1} + 2^{n-1} \, (n+1) \right) \, x^n \\
&= \sum_{n=0}^{\infty} \left(1 + 2^{n-1} \, (n-2) \right) \, x^n
\end{align}
which gives
$$ a_{n} = 1 + 2^{n-1} \, (n-2). $$
As a check:
\begin{align}
\phi_{n} &= 5 \, a_{n-1} - 8 \, a_{n-2} + 4 \, a_{n-3} \\
&= (5 - 8 + 4) + 2^{n-4} \, (5 \cdot 4 \, (n-3) - 8 \cdot 2 \, (n-4) + 4 \, (n-5)) \\
&= 1 + 2^{n-1} \, (n-2) = a_{n}
\end{align}
which is the expected result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Attempt to solve $\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$ I tried to solve this limit. Is it correct? There exists a more straightforward way?
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$$
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2-1+1}{1+x}\right)} -\exp\left(x^2\ln\left(1+\frac1x\right)\right)$$
$$\lim_{x\to +\infty} \exp(x-1)\cdot\exp{\left(\frac{1}{1+x}\right)} -\exp\left(x^2\left(\frac1x-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right)\right)$$
$$\lim_{x\to +\infty} \exp(x-1)\cdot{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)} -\exp\left(x-\frac{1}{2}+o(1)\right)$$
$$\lim_{x\to +\infty} \exp\left(-\frac12\right)\cdot\exp\left(x-\frac12\right)\cdot{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)} -\exp\left(x-\frac{1}{2}+o(1)\right)$$
$$\lim_{x\to +\infty}\underbrace{\exp\left(x-\frac12\right)}_{\to+\infty}\left[\exp\left(-\frac12\right)\cdot{\underbrace{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)}_{\to0}} -\underbrace{\exp\left(o\left(1\right)\right)}_{\to 1}\right] $$
$$= +\infty\cdot\left[0-1\right] = -\infty$$
| You are correct. A slightly shorter way with a more general result: let $a$ be a real number then, as $x\to +\infty$,
$$\begin{align}
\exp{\left(\frac{x^2}{a+x}\right)}& - \left(1+\frac1x\right)^{x^2}=
\exp{\left(\frac{x}{1+\frac{a}{x}}\right)}-\exp{\left(x^2\ln\left(1+\frac{1}{x}\right)\right)}\\
&=\exp{\left(x\left(1-\frac{a}{x}+o\Big(\frac{1}{x}\Big)\right)\right)}-\exp{\left(x^2\left(\frac{1}{x}-\frac{1}{2x^2}+o\Big(\frac{1}{x^2}\Big)\right)\right)}\\
&=e^x\left(e^{-a+o(1)}-e^{-1/2+o(1)}\right)\to
\begin{cases}
+\infty & \text{if $a<1/2$}\\
-\infty & \text{if $a>1/2$}
\end{cases}
\end{align}
$$
For the case $a=1/2$, we need a more precise expansion and the result is $-\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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In any triangle $\triangle ABC$, show that $4R\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})=r$ This is a problem problem I found in a JEE examination prep textbook, it was a "starred" question which I believe implies that it is more challenging than usual. It goes as follows:
In any triangle $\triangle ABC$, show that $$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
Hint: $$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=\Delta$$
Here is my attempt at it. I want to know if this is correct and if there any better alternative approaches to achieve the same result, please do share them!
We know that:
$$\Delta=rs$$
Using the given hint:
$$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=rs$$
$$16R^2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\frac{a+b+c}{2}\right)$$
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)\right)$$
Now, focusing on the equation of the right hand side for a bit, we know that:
$$A+B+C=\pi$$
$$\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$$
$$\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$
$$2\cos\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)$$
$$4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
Now substituting this back into the original problem:
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=\left(4r\right)\left[\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\right]$$
And that gives us:
$$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
| By using the area formula
$$\Delta=\sqrt{s(s-a)bc}\sin{\frac{A}{2}},$$
at each vertex of the triangle, we can have
$$\Delta^6=s^3(s-a)(s-b)(s-c)(abc)^2\sin^2{\frac{A}{2}}\sin^2{\frac{B}{2}}\sin^2{\frac{C}{2}}.$$
After simplifying by Heron's formula,
$$\Delta^4=s^2(abc)^2\sin^2{\frac{A}{2}}\sin^2{\frac{B}{2}}\sin^2{\frac{C}{2}}.\tag{1}$$
On the other hand,
$$\Delta^4=\Delta^2\Delta^2=(sr)^2\left(\frac{abc}{4R}\right)^2.\tag{2}$$
From $(1)$ and $(2)$, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find all positive integers that are the cube of the sum of their digits
(HMMT 2000 Guts #5). Find all positive integers that are the cube of the sum of their digits.
I've found a (very) tedious way to resolve the question described below. The tedious part is highlighted in a similar way to the question at the top.
Suppose n is such a number. Note that the sum of the digits of a number is congruent to its remainder modulo 9. So $n^3 \equiv n \mod 9,$ which implies $n\equiv 0,1,-1\mod 9.$ If n has a single digit, then $n=n^3,$ so $n=1$ as $n>0.$ $ n$ is a perfect cube, and the only two digit perfect cubes are $27$ and $64,$ neither of which work, so there are no two-digit solutions. The only three digit perfect cubes are $125,216,343,512,729.$ But $(1+2+5)^3 > 125, (2+1+6)^3 > 216, (3+4+3)^3 > 343, (5+1+2)^3 = 512, (7+2+9)^3 > 729.$ Hence the only three-digit solution is $512.$
Now my question is, is there a better way to solve this question than to literally check all 4 to 6 digit cubes?
Note that there are no solutions with $7$ or more digits, as the largest such number would be $(9\cdot 7)^3 = 250047 < 10^6.$
| Let $n=k^3.$ If $n$ has $6$ digits, the largest it could be is $(6\cdot 9)^3 = 157464$, which shows that one digit must be $1$. So in fact the largest it could be is $(1+5\cdot 9)^3 = 97336$ which is too small. So there are no $6$ digit (or larger) solutions. (I think this is the gist of @dezdichado's comment.)
Because the cube root of $100,000 = 46.4\ldots$, we know that $k\leq 46.$ Using your observation that $k\equiv -1, 0, 1 \pmod{9}$ leaves us with only these $16$ possibilities for $k$:
$$1, 8, 9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46.$$
We can just check these or we can play a bit more:
Since $(5\cdot 9)^3 = 91125$, we see that a $5$-digit solution must have at least one non-$9$ digit. Since $(4\cdot 9+8)^3 = 85184$, the largest the first two digits can be is $79$, so at least one digit is at most $7$. Since $4\cdot 9 +7 = 43$ we can cross off the last three numbers from the list.
Of these, $1, 8, 17, 18, 26, 27$ work.
A couple of observations: It surprises me that the last two pairs are consecutive, and both are of the shape $-1, 0 \pmod{9}$ so I feel like I've missed something. Since $1$ is a solution, it doesn't seem like modular considerations would eliminate $1 \pmod{9}$.
| {
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Is there another way of solving this binomial problem? The problem is as follows: in $$(1+x)^n$$ find $n$ such that the fifth coefficient is $70$ and the seventh coefficient equals $28$.
I put $$ \binom{n}{4}=70 $$ $$\binom{n}{6}=28 $$
I ended up with two polynomial equations:
$$ n^4-6n^3+11n^2-6n-1680=0 $$
$$ n^6-15n^5+85n^4-225n^3+274n^2-120n-20160=0 $$
and the solution to this binomial is $n=8$.
Is there another (simpler) way to solve this problem?
| Divide $\binom{n}{4}$ by $\binom{n}{6}$, it will give us
$$
\frac{70}{28} = \frac{\binom{n}{4}}{\binom{n}{6}} = \frac{\frac{n!}{4!\left(n-4\right)!}}{\frac{n!}{6!\left(n-6\right)!}} = \frac{6!\left(n-6\right)!}{4!\left(n-4\right)!} = \frac{6\cdot5\cdot4!\left(n-6\right)!}{4!\left(n-4\right)\cdot\left(n-5\right)\cdot\left(n-6\right)!} = \frac{30}{\left(n-4\right)\left(n-5\right)}
$$
hence
$$
\left(n-4\right)\left(n-5\right) = 12 \Longrightarrow n^{2} -9n+8 = 0
$$
the solutions are $n=8$ and $n=1$, since we want a solution greater than $6$, we must have $n=8$.
| {
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"url": "https://math.stackexchange.com/questions/4601559",
"timestamp": "2023-03-29T00:00:00",
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Deflation of eigenvalue Could you explain to me the deflation?
For example if we have the matrix $A=\begin{pmatrix}1 & -0.5 & -1.5\\ -15 & -2.5 & 4.5\\ -15 & -4.5 & 2.5\end{pmatrix}$ how do we apply the deflation of the eigenvalue $\lambda_1=4$ ?
$$$$
EDIT :
I have done the following :
We have the eigenvalue $\lambda_1=4$. The corresponding eigenvector is : $v_1=\begin{pmatrix}-1\\ 3 \\ 1\end{pmatrix}$, right?
Since $v_1$ does not have $1$ as the component of largest modulus, we multiply $v_1$ by a permutation matrix $P$ which interchanges the largest element and the first element: We interchange rows $1$ and $2$ of $A$ : $v_1'=\begin{pmatrix}3 \\-1\\ 1\end{pmatrix}$
$$\begin{pmatrix}-15 & -2.5 & 4.5\\1 & -0.5 & -1.5\\ -15 & -4.5 & 2.5\end{pmatrix}$$
We want to transform the vector into the vector $e_1$, so we divide the first row by $3$, add it to the second row and subtract it from the third row : $v_1''=\begin{pmatrix}1 \\0\\ 0\end{pmatrix}$
$$\begin{pmatrix}-5 & -5/6 & 1.5\\-4 & -4/3 & 0\\ -10 & -11/3 & 1\end{pmatrix}$$ Since we have interchanged the rows $1$ and $2$, now we have to interchange the columns $1$ and $2$ and so we get the matrix $$\begin{pmatrix} -5/6 & -5 &1.5\\-4/3 & -4 & 0\\ -11/3 & -10 &1\end{pmatrix}$$ Then $\lambda_1=4$ should be an eigenvalue of this matrix with $e_1$ the corresponding eigenvector, but this is not like that. What am I doing wrong?
Is the deflated matrix equal to $$\begin{pmatrix} -4 & 0\\ -10 &1\end{pmatrix}$$ ?
| We have $v_1=\begin{pmatrix}-1\\3\\1\end{pmatrix}$.
Let $P$ be the permutation matrix that swaps the first 2 elements.
Then $Pv_1=v_1'=\begin{pmatrix}3\\-1\\1\end{pmatrix}$ has the element with the largest magnitude first.
Let $R$ be the matrix of row operations that transforms $v_1'$ into $e_1$. That is $Rv_1'=e_1$.
Then $R=\begin{pmatrix}1/3\\1/3 & 1\\-1/3 &&1\end{pmatrix}$.
Let $B=RPAP^{-1}R^{-1}$. Then $B$ is similar to $A$ and:
$$Be_1=RPAP^{-1}R^{-1}e_1=RPAP^{-1}v_1'=RPAv_1=RP\lambda_1 v_1=\lambda_1Rv_1'=\lambda_1e_1$$
So $B$ has $\lambda_1$ as eigenvalue with eigenvector $e_1$.
You have found $RPAP^{-1}=\begin{pmatrix} -5/6 & -5 &1.5\\-4/3 & -4 & 0\\ -11/3 & -10 &1\end{pmatrix}$. We can verify that if we multiply it with $v_1'$ we find $\lambda_1 e_1$.
If we multiply it on the right with $R^{-1}=\begin{pmatrix}3\\-1&1\\1&&1\end{pmatrix}$ then we find $B=\begin{pmatrix}4&-5&3/2\\&-4&0\\&-10&1\end{pmatrix}$.
So $B$ has indeed $e_1$ as an eigenvector for $\lambda_1=4$.
The resulting deflated matrix is $B_1=\begin{pmatrix}-4&0\\-10&1\end{pmatrix}$, which we already had in $RPAP^{-1}$ since $R^{-1}$ affected only the leftmost column. This matches your findings.
We can verify that $B_1$ has indeed the same remaining eigenvalues $-4$ and $1$ as $A$, which are on its diagonal since $B_1$ is a triangular matrix.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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On generalizing $\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$? Given $\color{blue}{t = 2\pi/p}$ for the appropriate prime $p=4m+1$.
I. Sine
$$\begin{align}
\frac{5+\sqrt{5}}8 &=\sin^2(t)\\
\frac{13+\sqrt{13}}8 &=\sin^2(t)+\sin^2(3t)+\sin^2(4t)\\
\frac{17+\sqrt{17}}8 &=\sin^2(3t)+\sin^2(5t)+\sin^2(6t)+\sin^2(7t)\\
\frac{29+\sqrt{29}}8 &=\sum_{k=1}^7\sin^2(a_k\, t)
\end{align}$$
with the seven $a_k = 1,4,5,6,7,9,13.$ And so on for other prime $p=4m+1.$ For the opposite sign, one uses the remaining integers $b_k \leq \frac{p-1}2.$ For example,
$$\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$$
where the $a_k$ are simply $2^n$. (The next Fermat prime $p=257$ isn't so nice since it has $256/4 = 64$ sine terms.)
II. Cosine
This uses the same set of multipliers $a_k$.
$$\begin{align}
\frac{3-\sqrt{5}}8 &=\cos^2(t)\\
\frac{11-\sqrt{13}}8 &=\cos^2(t)+\cos^2(3t)+\cos^2(4t)\\
\frac{15-\sqrt{17}}8 &=\cos^2(3t)+\cos^2(5t)+\cos^2(6t)+\cos^2(7t)\\
\frac{27-\sqrt{29}}8 &=\sum_{k=1}^7\cos^2(a_k\, t)
\end{align}$$
with the same seven $a_k = 1,4,5,6,7,9,13.$ For the opposite sign,
$$\frac{15+\sqrt{17}}8 =\cos^2(t)+\cos^2(2t)+\cos^2(4t)+\cos^2(8t)$$
III. Conclusion
Given prime $p=4m+1$ and $t = 2\pi/p.$ The pattern clearly is,
$$\frac{p\pm\sqrt{p}}8 = \sum_{k=1}^m \sin^2(a_k\, t)$$
$$\frac{(p-2)\mp\sqrt{p}}8 = \sum_{k=1}^m \cos^2(a_k\, t)$$
Question: I used Mathematica's integer relations to find the above examples. But, for any prime $p=4m+1$, what is a clever and faster algorithm to derive the correct set of $a_k$?
| I found an algorithm to derive the correct set $H$ of $a_k$ for any prime $\ p=4m+1 \ $ (or its complement).
$H=\{ 1\}$.
For $k$ from $2$ to $2m$:
*
*Compute $\alpha$ such that $\ \ k^2 \equiv \alpha \pmod{4m+1}\ \ $ and $\ \ 1\leqslant \alpha \leqslant 4m \ \ $
*if $\ 1\leqslant \alpha \leqslant 2m \ $ then $\ \beta =\alpha \ $ else $\ \beta = 4m+1-\alpha $
*$H = H\cup \{\beta \}$
We can use "Quadratic Gauss sums" to show the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of Quadrilateral $ABCD$. A puzzle for 10th graders. As the title suggests, the problem in this post was meant to be a puzzle for 10th graders, so claims the person who posted this on a language exchange platform: The problem is as follows:
Given a Quadrilateral $ABCD$ with internal point $P$, where $AP=1$, $BP=2$ and $CP=3$, and unknown sides $k$ and $2k$, compute the area of this Quadrilateral.
I first tried to inscribe this quadrilateral into a square but that approach did not turn out successful, I was thinking if there are any other ways to solve it, perhaps via setting up a coordinate system, or via a trigonometric method. I will share my own successful approach below as an answer!
| The problem is described as
As the title suggests, the problem in this post was meant to be a puzzle for 10th graders,
but I imagine for very mathematically talented students. I
simplify the original problem into an equivalent form.
Suppose a square $ABCE$ is such that the side lengths of $AB$
and $CE$ are equal to $b+c$ and the side lengths of $AE$ and $BC$
are equal to $a+d$.
Suppose that there is a point $P$ in the square such that the projection of $AP$ and $BP$ onto $AE$ and $BC$ has length $a$ and,
the projection of $BP$ and $CP$ onto $AB$ and $CE$ has length $c$.
Of course, since $ABCE$ is a square, its side length is $b+c=a+d$.
Finally, suppose we are given that the ratios of the segments $AP:BP:CP:EP = 1:2:3:x$ where $x$ is unknown.
We don't need to know $x$, but using the Pythagorean theorem
$$ \frac{a^2+b^2}{1^2} =
\frac{a^2+c^2}{2^2} =
\frac{c^2+d^2}{3^2} =
\frac{b^2+d^2}{x^2} $$
but $$ (a^2+b^2)+(c^2+d^2)=(a^2+c^2)+(b^2+d^2)$$
which implies that $\, 1^2+3^2 = 2^2+x^2 $ and $\,x=\sqrt{6}.$
Now, to solve for $\,a,b,c,d\,$ let
$$ b=a+u,\quad c=a+v,\quad d=a+u+v.$$
We know that $$ p_1 := 2^2(a^2+b^2) - 1^2(a^2+c^2) = 0, $$
and $$ p_2 := 3^2(a^2+b^2) - 1^2(c^2+d^2) = 0. $$
Compute the polynomial resultant
$$ \text{res}(p_1,p_2,v) = 8(a^2-2u^2)(2a^2+2au+u^2). $$
Solve for the positive real root of the second factor and
let $r:=u$. Then
$$a=\sqrt{2}r,\quad b=(1+\sqrt{2})r,\quad c=(4+\sqrt{2})r,
\quad d = (5+\sqrt{2})r.$$
In the original problem, we are given that $a^2+b^2=1^2,$ but
$a^2+b^2 = (5+2\sqrt{2})r^2$ which implies $r^2=1/(5+2\sqrt{2}).$
The area of the square is $(a+d)^2=(5+2\sqrt{2})^2r^2=5+2\sqrt{2}.$
In the original problem, point $D$ is the midpoint of $AE$,
and the area of the quadrilateral $ABCD$ is $3/4$ times the area of
the square $ABCE$ and so the final answer is $\frac34(5+2\sqrt{2}).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of $a<\frac{1}{a}I am currently working on Velleman's How to Prove It section 3.2, question nine (question eight in the second edition).
Why does he hint in the solution section to go through the steps to first prove $a<1$ and $a<0$, rather than just using a proof by contradiction by assuming $a\ge-1$ to conclude the goal in one step? Can someone point out where I have gone wrong in this proof?
Suppose $a$ and $b$ are nonzero real numbers. Suppose $a<\frac{1}{a}<b<\frac{1}{b}$ and $a\ge-1$. Because $a$ is nonzero, either $a=-1$ or $a>0$. Clearly it is not the case that $a>0$ because this contradicts $a<\frac{1}{a}$. It cannot be the case that $a=-1$, as this also contradicts $a<\frac{1}{a}$. Therefore we conclude $a<-1$ as required.
EDIT: Right I can see the problem with my proof. The solution manual has this: "Suppose that $a\ge-1$. Dividing by the positive
number $−a$, we conclude that $−1\ge\frac{1}{a}$. Combining $a\ge-1$ and $−1\ge\frac{1}{a}$ we get $a\ge\frac{1}{a}$,
contradicting the fact that $a<\frac{1}{a}$. Therefore $a<−1$." Why can he divide by the "positive number" $-a$?
| Suppose $a$ and $b$ are nonzero real numbers such that $a < \frac{1}{a} < b < \frac{1}{b}$. We want to show that $a < -1$.
To this end, suppose towards a contradiction that $a \geq -1$. Then there are three cases to consider:
Case 1: Suppose that $-1 \leq a < 0$. Then since $a$ is negative, we know that $-a$ is positive, so we can divide it from both sides of the inequality $-1 \leq a$ to obtain $\frac{1}{a} \leq -1$. But since $a < \frac{1}{a}$, it follows that $-1 \leq a < \frac{1}{a} \leq -1$ so that $-1 < -1$, a contradiction.
Case 2: Suppose that $a \geq 1$. Then since $a$ is positive, we can divide it from both sides of the inequality $1 \leq a$ to obtain $\frac{1}{a} \leq 1$. But since $a < \frac{1}{a}$, it follows that $1 \leq a < \frac{1}{a} \leq 1$ so that $1 < 1$, a contradiction.
Case 3: Suppose that $0 < a < 1$. Then since $a$ is positive, we can divide it from both sides of the inequality $a < 1$ to obtain $1 < \frac{1}{a}$. But since $\frac{1}{a} < b$, it follows that $1 < b$. Since $b$ is positive, we can divide it from both sides of this inequality to obtain $\frac{1}{b} < 1$. But since $1 < b < \frac{1}{b} < 1$, we conclude that $1 < 1$, a contradiction.
Thus, we conclude that $a < -1$, as desired. $~~\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Where did I go wrong with this integration? $$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2-3x^2}}$$
$$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2(1-\frac32x^2)}}$$
$$\frac 1{\sqrt 2}\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {1-\frac32x^2}}$$
$$\bigg(\frac 1{\sqrt 2}\sin^{-1}{\sqrt {\frac 32 x}}\bigg)\bigg|_0^\frac 1{\sqrt 3}$$
$$\frac 1{\sqrt 2} \times \frac {\pi}{4}$$
However, the answer is $\frac 1{\sqrt 3} \times \frac {\pi}{4}$. Where did I go wrong?
| We use $f'(x)=\frac{1}{\sqrt{1-x^2}}$ for $f(x)=\sin^{-1}(x)$ and $x\in(-1,1)$. Using the chain rule gives $f'(x)=\frac{a}{\sqrt{1-(ax)^2}}$ for $f(x)=\sin^{-1}(ax)$. Hence, we have
\begin{align*}
\int_{0}^{1/\sqrt{3}}\frac{1}{\sqrt{2-3x^2}}\mathrm dx
&=\sqrt{\frac{2}{2\cdot 3}}\int_{0}^{1/\sqrt{3}}\frac{\sqrt{\frac{3}{2}}}{\sqrt{1-\left(\sqrt{\frac{3}{2}}x\right)^2}}\mathrm dx
=\sqrt{\frac{1}{3}}\left[\sin^{-1}\left(\sqrt{\frac{3}{2}}x\right)\right]_0^{1/\sqrt{3}}\\
&=\sqrt{\frac{1}{3}}\left(\frac{\pi}{4}-0\right)
=\frac{\pi}{4\sqrt{3}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$\mathbb{Q}(\sqrt{-23}) \subseteq \mathbb{Q}(\zeta_{23})$ I want to prove $\mathbb{Q}(\sqrt{-23}) \subseteq \mathbb{Q}(\zeta_{23})$ using the hint from "Introduction to Cyclotomic Fields" Exercise 2.1. $\zeta_{23}$ is the primitive 23-th root of unity.
I know that $p=23$ is the only prime ramified in $\mathbb{Q}(\zeta_{23})$ hence in $\mathbb{Q}(\sqrt{d})$, thus we have $d=\pm 23$. And I'm wondering how to show $d=-23$ instead of $d=23$. The book says when $p \equiv 1 \pmod 4$ we choose $+$, other we choose $-$. I don't know how to prove it.
Thanks for helps!
| also, if $x^{23} = 1$ but $x \neq 1,$ take
$$z = x + x^2 + x^3 + x^4 + x^6 + x^8 +x^9 + x^{12} + x^{13} + x^{16} + x^{18} $$
Then
$$ z^2 + z + 6 = 0 $$
because gathering the same exponents mod 23 comes out to
$$z =6 \left( 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 +x^7 + x^8 +x^9 + x^{10} + x^{11} +x^{12} + x^{13} + x^{14} + x^{15} + x^{16} + x^{17} + x^{18} + x^{19} + x^{20} + x^{21} +x^{22}
\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Area of isosceles triangle I need to solve following problem:
The vertex of the isosceles triangle $ABC$ is the point $A(-1, 0)$, and the vertices $B$ and $C$ belong to the parabola $y^2 = 4x$. If the point $(0, 0)$ is the orthocenter of triangle $ABC$, then its area is equal to ?
I've tried to use fact that points C and B lies on parabola to express their coordinates $B(x,2\sqrt x)$ and $C(x,-2\sqrt x)$
Also i can express height and area of triangle as:
Area = $2\sqrt{x}(x+1)$.
Height = $x+1$
Can you help me to figure it out ?
| Given that the line connecting $B$ to the origin is $y = mx$, $y^2 = 4x$ gives $m^2 x^2 - 4x = 0$ $ \implies x(m^2 x - 4) = 0$, so the other intersections apart from $(0, 0)$ are when $x = \frac{4}{m^2}$, and $y = \pm \frac{4}{m}$.
Now finding the equation of the perpendicular which must pass through $(-1, 0)$, we have:
$$y - 0 = -\frac{1}{m} (x - -1)$$
$$-\frac{4}{m} = -\frac{1}{m} (\frac{4}{m^2} + 1)$$
$$-4m^2 = -4 - m^2$$
$$4 = 3m^2, m^2 = 4/3$$
and thus $x = 4/(4/3) = 3$ leading to an area of $2 \sqrt{3} (3 + 1) = \boxed{8 \sqrt3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
kernel of matrix in given basis Linear transformation $f$ is given in basis ${v_1, v_2, v_3}$ by matrix $M$:
$$\begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix}.$$
Find kernel and image of matrix in terms of basis ${v_1, v_2, v_3}$
Okay so in order to find kernel I transform matrix using elementary row operations and I find out that $v=(1,2,4)$ is a vector such that if I apply matrix on this vector ($Mv$) i get $0$ vector. And every sclar multiple of $v$ has same property of course, exluding them none of other vectors have this property so the kernel is one dimensional.
My concern is about what is kernel in terms of ${v_1, v_2, v_3}$. Is it $span((1,2,4))$ or is it $span((v_1+2v_2+4v_3))$ or maybe its something else.
I know if the matrix was given in standard basis kernel would be $span((1,2,4))$.
Answering my question please explain why.
| Consider the matrix representation of linear transformation $f$ in the ordered basis $ \beta=\{v_{1},v_{2},v_{3}\}$ given by $$[f]_{\beta}=\begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix}$$
By definition, the image of $[f]_{\beta}$ can be find as
$${\rm im}([f]_{\beta})=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\in {\bf R}^3: \exists \begin{pmatrix} a\\b\\c\end{pmatrix}\in {\bf R}^{3}: \begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix} \right\}$$
that is, solving the linear system we have
$${\rm im}([f]_{\beta})=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\in {\bf R}^{3}: x-2y+z=0 \right\}={\rm span}\left\{\begin{pmatrix}2\\1\\0\end{pmatrix},\begin{pmatrix} -1\\0\\1\end{pmatrix}\right\}$$
Since $$[f(v)]_{\beta}=[f]_{\beta}\cdot [v]_{\beta}, \quad (*)$$ then using $(*)$
$${\rm im}(f)={\rm span}\left\{2\cdot v_{1}+1\cdot v_{2}+0\cdot v_{3},-1\cdot v_{1}+0\cdot v_{2}+1\cdot v_{3} \right\}$$
$$\boxed{{\rm im}(f)={\rm span}\left\{2v_{1}+v_{2},-v_{1}+v_{3} \right\}}$$
By definition, the kernel of $[f]_{\beta}$ can be find as
$$\ker([f]_{\beta})=\left\{\begin{pmatrix}a\\b\\c\end{pmatrix}\in {\bf R}^{3}: \begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \right\}$$
that is, solving the linear system we have
$$\ker([f]_{\beta})=\left\{\begin{pmatrix}a\\b\\c\end{pmatrix}\in {\bf R}^{3}: y=2x, z=4x, x\in {\bf R} \right\}={\rm span}\left\{\begin{pmatrix}1\\2\\4\end{pmatrix} \right\}$$
Using $(*)$ we have
$${\rm ker}(f)={\rm span}\left\{1\cdot v_{1}+2\cdot v_{2}+4\cdot v_{3} \right\}$$
$$\boxed{\ker(f)={\rm span}\left\{ v_{1}+2v_{2}+4v_{3}\right\}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An $n\times n$ matrix $A$ that contains each of $1,2,\cdots, n$ exactly $n$ times and $A^2\equiv 0\pmod n$ Construct a matrix $A \in \{1,\,2, ...,\,n\}^{n \times n}$ such that
(1) Each integer between $[1, n]$ occurs in $A$ exactly $n$ times,
(2) For all integers $i,\,j \in [1, n]$, $(A^2)_{ij} \equiv 0\,(\text{mod}\,n)$,
or report such $A$ does not exist. Here $A^2 := A*A$ is matrix multiplication.
This problem is not proposed by me. It is proposed on the Codeforces forum: https://codeforces.com/blog/entry/111297. For odd $n$ we can construct $A$ easily: Just fill every row of $A$ with $1,\,2, ...\,n$. But what about even $n$?
| We have the following interesting family of examples. If $n$ is divisible by $4$, then take $B$ to be the matrix whose columns are
$$
c_1 = \left(1,3,\dots,\frac n2 - 1, 0\dots,0,\frac n2 + 1,\frac n2 + 3,\dots, n-1\right),\\
c_2 = \left(0,0,\dots,0, 1,\dots,1,0,0,\dots, 0\right),
$$
and take $C$ to be the matrix whose rows are
$$
r_1 = (1,\dots,1), \quad r_2 = (n,\dots,4,2,2,4,\dots,n)
$$
Then the matrix $A = BC$ contains each element of $\{1,\dots,n\}$ $n$ times and satisfies $A^2 = 0$. For $n = 4,8,$ this produces the examples
$$
\left[\begin{matrix}1 & 1 & 1 & 1\\4 & 2 & 2 & 4\\4 & 2 & 2 & 4\\3 & 3 & 3 & 3\end{matrix}\right], \quad \left[\begin{matrix}1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\3 & 3 & 3 & 3 & 3 & 3 & 3 & 3\\8 & 6 & 4 & 2 & 2 & 4 & 6 & 8\\8 & 6 & 4 & 2 & 2 & 4 & 6 & 8\\8 & 6 & 4 & 2 & 2 & 4 & 6 & 8\\8 & 6 & 4 & 2 & 2 & 4 & 6 & 8\\5 & 5 & 5 & 5 & 5 & 5 & 5 & 5\\7 & 7 & 7 & 7 & 7 & 7 & 7 & 7\end{matrix}\right],
$$
and the pattern of the entries continues in this fashion for larger multiples of $4$.
Proof: We want to show that $A^2 \equiv 0 \pmod{n}$, i.e. $BCBC \equiv 0 \pmod n$. Replacing entries of a matrix with equivalent values modulo $n$ will not affect this result, so we rewrite the first column of $B$ as
$$
\left(1,3,\dots,\frac n2 - 1, 0,\dots,0,-(\frac n2 + 1),-(\frac n2 + 3),\dots, -1\right).
$$
With this, we find that the dot-product of the first column of $B$ and either row of $C$ is equal to $0$. Thus, the product $CB$ has the form
$$
CB \equiv \pmatrix{0&m_1\\0&m_2} \pmod{n}.
$$
The upper-right entry of the product $CB$ is given by
$$
\overbrace{1 + 1 + \cdots + 1}^{n/2} = \frac n2.
$$
The lower-right entry of the product $CB$ is given by
$$
2 \cdot (2 + 4 + \cdots + n/2) = 4 \cdot \frac{(n/4)(n/4 + 1)}{2} = \left( \frac n4 + 1\right)\frac n2.
$$
Thus, $m_1$ and $m_2$ are multiples of $n/2$. On the other hand, the entries of $r_2$ are all equal to a multiple of $2$.
Now, the product $CBC$ has rows $m_1 r_2$ and $m_2 r_2$. Because we can write
$$
m_i r_2 = \left(\frac n2\right)\left(\frac{m_i}{n/2}\right) \cdot 2 \left( \frac {r_2}2\right) = n\, \left(\frac{m_i}{n/2}\right)\left( \frac {r_2}2\right),
$$
we can conclude that $CBC$ has entries that are multiples of $n$. It follows that the same is true for $A^2 = B(CBC)$. That is, $A^2 \equiv 0 \pmod n$, which is what we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding $\cot(\beta)$, knowing $\sin(\alpha+\beta)=4/5$ and $\sin(\alpha-\beta)=3/5$ Let's assume that for $0<\beta<\alpha<\frac{\pi}{2}$, $\sin(\alpha+\beta) = \frac{4}{5}$, and $\sin(\alpha-\beta) = \frac{3}{5}$. Then, how could we find $\cot(\beta)$?
$$\sin(\alpha+\beta)+\sin(\alpha-\beta) = 2\sin(\alpha)\cos(\beta) = \frac{7}{5}$$
$$\sin(\alpha+\beta)-\sin(\alpha-\beta) = 2\sin(\beta)\cos(\alpha) = \frac{1}{5}$$
$$\tan(\alpha)\cot(\beta) = 7$$
But I am not sure where this would lead us.
| $$
\begin{aligned}
& \sin ^2(\alpha+\beta)+\sin ^2(\alpha-\beta)=\frac{16}{25}+\frac{9}{25}=1 \\
\Rightarrow \quad & \sin ^2(\alpha+\beta)=1-\sin ^2(\alpha-\beta)=\cos ^2(\alpha-\beta)\\ \Rightarrow \quad &
(\sin \alpha \cos \beta+\sin \beta \cos \alpha)^2=(\cos \alpha \cos \beta+\sin \alpha \sin \beta)^2 \\\Rightarrow \quad &\left(\cos ^2 \beta-\sin ^{2} \beta\right)\left(\cos ^2 \alpha-\sin ^{2} \alpha\right)=0 \\ \Rightarrow \quad
&\cos ^2 \beta=\sin ^2 \beta \quad \textrm{ or }\cos ^2 \alpha=\sin ^2 \alpha \\\Rightarrow \quad & \cot\beta =1 \textrm{ or } \tan \alpha =1 \\ \Rightarrow \quad & \cot \beta=1 \quad \textrm{ or } \quad 7 \quad \textrm{ (By }\tan \alpha \cot \beta = 7)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Maximizing/minimizing $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$
Assume that $f : [0, 2\pi]\rightarrow \mathbb{R}$ is a function such that $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$. Then, how can we maximize/minimize $f$?
We can re-parametrize our function $f$ by defining another function $g : [-1, 1]\rightarrow \mathbb{R}$ function such that for every $t\in [-1, 1]$,
$$g(t) = \sqrt{2}\sqrt{1-t^2}-4t$$
$$\frac{dg}{dt} = \frac{d}{dt}\left(\sqrt{2}\sqrt{1-t^2}-4t\right) = \frac{\sqrt{2}t}{\sqrt{1-t^2}} + 4 = 0$$
From which we conclude that $g$ attains its maximum/minimum at $\left(-\frac{2\sqrt{2}}{3}, g\left(-\frac{2\sqrt{2}}{3}\right)\right), (1, g(1))\in \mathbb{R}^2$ respectively.
| Use Cauchy-Schwarz inequality: $f^2(\theta) = \left(\sqrt{2}\cos\theta - 4\sin\theta\right)^2\le ((\sqrt{2})^2+4^2)(\cos^2\theta+\sin^2\theta)=18\implies -3\sqrt{2} \le f(\theta) \le 3\sqrt{2}$. Can you conclude the min and max for $f$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate this double Integral in polar coordinates. $$ \iint y^2(a^2-x^2)^{0.5}dxdy $$ over $x^2+y^2\le a^2$
I have evaluated this in $x-y$ plane and got $32a^5/45$. Please help in evaluating the same in polar coordinates.
Ive tried putting $x=r\cos\theta$ and $y=r\sin\theta$ and After considering the Jacobian I am getting this expression
$$ \int\limits_0^a\int\limits_0^{2π}r^3\sin^2θ(a^2-r^2cos^2θ)^{0.5}\,drdθ $$
Please help me evaluate this.
| Let $D$ be the disk of radius $a$ centered at the origin. We can parameterize the region in polar coordinates by letting
$$\begin{cases}x = a r \cos(\theta) \\ y = a r \sin(\theta) \end{cases}$$
with $(r,\theta) \in [0,1] \times [0,2\pi]$. The Jacobian in this case is almost identical to the typical one you get when converting to standard polar coordinates (sans $a$):
$$J = \begin{bmatrix}a \cos(\theta) & -ar \sin(\theta) \\ a\sin(\theta) & ar\cos(\theta)\end{bmatrix} \implies |\det J| = a^2r$$
Now the integral is
$$\begin{align*}
I &= \iint_D y^2 \sqrt{a^2 - x^2} \, dx \, dy \\[1ex]
&= \int_0^{2\pi} \int_0^1 a^2r^2\sin^2(\theta) \sqrt{a^2 - a^2 r^2 \cos^2(\theta)} \cdot a^2r \, dr \, d\theta \\[1ex]
&= 2a^5 \int_0^\pi \int_0^1 r^3 \sin^2(\theta) \sqrt{1 - r^2 \cos^2(\theta)} \, dr \, d\theta \tag{1} \\[1ex]
&= -\frac{2a^5}3 \int_0^\pi \left[\frac{\sin^5(\theta)}{\cos^2(\theta)} - 2 \tan^2(\theta) \int_0^1 r \left(1-r^2\cos^2(\theta)\right)^{3/2} \, dr\right] \, d\theta \tag{2} \\[1ex]
&= -\frac{2a^5}3 \int_0^\pi \left[\frac{\sin(\theta)(1-\cos^2(\theta))^2}{\cos^2(\theta)} - \frac{\sin^2(\theta)}{\cos^4(\theta)} \int_{\sin^2(\theta)}^1 s^{3/2} \, ds \right] \, d\theta \tag{3} \\[1ex]
&= -\frac{2a^5}3 \cdot -\frac{16}{15} \tag{4} \\[1ex]
&= \frac{32a^5}{45}
\end{align*}$$
*
*$(1)$ : exploit symmetry about $\theta=\pi$; that is, replace $\theta\mapsto\pi-\theta$ to get an integral of an even function (with respect to $\theta$) over $[-\pi,\pi]$. We end up with an integral over $[0,\pi]$, and on this interval we have $\sin(\theta)\ge0$ so we can pick positive square roots whenever they show up.
*$(2)$ : integrate by parts with respect to $r$, letting $u=r^2$ :
$$\begin{align*}
\int_0^1 r^3 \sqrt{1 - r^2 \cos^2(\theta)} \, dr &= uv\big|_0^1 - \int_0^1 v\,du \\[1ex]
&= -\frac{\sin^3(\theta)}{3\cos^2(\theta)} + \frac2{3\cos^2(\theta)} \int_0^1 r \left(1-r^2\cos^2(\theta)\right)^{3/2} \, dr
\end{align*}$$
*$(3)$ : substitute $s = 1 - r^2 \cos^2(\theta)$
*$(4)$ : partial fractions and evaluate; in particular, we rewrite to the more readily integrable form,
$$\sec^2(\theta) - \frac{3\sin(\theta)}{\cos^2(\theta)} - \sec^2(\theta)(1+\tan^2(\theta)) + \frac{\sin(\theta)}{\cos^4(\theta)} + 3\sin(\theta) - \sin(\theta)\cos^2(\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplifying $\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$ I need help to simplify this expression.
$$\frac{x^4(y-z)+y^4(z-x)+z^4(x-y)}{(y+z)^2+(z+x)^2+(x+y)^2}$$
I only know to start, I extracted the $(x-y)$ from top one, and then I don't know what's next. Extracting $(x-y)$ from the numerator:
$$(x^4y - y^4x) - (zy^4-zx^4)+z^4(x-y)$$
$$ = xy(x^3-y^3)- z(x^4-y^4)+z^4(x-y)$$
$$ = xy(x-y)(x^2+xy+y^2)-z(x-y)(x+y)(x^2+y^2)+z^4(x-y)$$
$$ = (x-y)\left[ xy(x^2+xy+y^2) - z(x+y)(x^2+y^2) +z^4 \right]$$
$$ = (x-y)(x^3y+x^2y^2+xy^3-zx^3-zy^3+z^4-zxy^2-zx^2y)$$
| I like Pavel's answer, but here's a slightly different way.
Starting with $(x-y)(x^3y+x^2y^2+xy^3−zx^3−zy^3+z^4−zxy^2−zx^2y)$, rearrange the longer expression to:
$$(x^3y-zx^3) + (xy^3-zxy^2) + (x^2y^2 - zx^2y) + (z^4 - zy^3)$$
$$ = (y-z)(x^3 + xy^2 + x^2y) - z(z^3-y^3)$$
$$ = (y-z)(x^2y + xy^2 + x^2y -z(y^2+yz+z^2)) $$
Giving us $(x-y)(y-z)(x^2y+xy^2+x^2y-y^2z-yz^2-z^3)$ as the numerator. Again, rearrange the new long expression as:
$$(xy^2-zy^2) + (x^3-z^3) + (x^2y-yz^2)$$
$$ = (x-z)y^2 + (x-z)(x^2+xz+z^2)+y(x-z)(x+z)$$
And we can factor out the $(x-z)$. This gives the numerator as:
$$(x-y)(y-z)(x-z)(x^2+y^2+z^2+xy+xz+yz)$$
But if we expand the denominator, we get:
$$x^2+2xy+y^2 + x^2+2xz+z^2 + y^2+2yz+z^2 = 2(x^2+y^2+x^2+xy+xz+yz)$$
And we recognize that as having the same expression in parentheses as the last expression in the numerator! So when we divide we get:
$$\frac{(x-y)(y-z)(x-z)}{2} = - \frac{(x-y)(y-z)(z-x)}{2}$$
Which is the same as Pavel's answer. Taking the negative out just makes the numerator look symmetrical. If you expand the numerator, you get another symmetrical expression:
$$\frac12 \big(xy(x-y) + yz(y-z) + zx(z-x) \big)$$
which is longer but, I think, just a bit nicer-looking.
| {
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"url": "https://math.stackexchange.com/questions/4622054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show $k \cdot a + c \cdot b \ge \frac{1}{2} \cdot (k+c) \cdot (a+b)$? Assume k, c, a, b $\gt$ 0 and that $k \ge c$ and $a \ge b$. Prove $k \cdot a$ + $c \cdot b$ $\ge$ $\frac{1}{2}\cdot (k+c) \cdot (a+b)$. This is what I have so far:
$2\cdot k\cdot a +2\cdot c \cdot b$ = $(k\cdot a + k\cdot a)$ + $(c\cdot b + c\cdot b)=(k\cdot a + c\cdot b)$ + $(k\cdot a + c\cdot b)\ge(k\cdot a + c\cdot b)$ + $(k\cdot b + c\cdot b)$
| We have $k≥c$ and $a≥b$. And we have to prove that $ka+bc≥(k+c)(a+b)/2$. Now by algebraic expansion we can write it out like these $$ka+bc≥(k+c)(a+b)/2$$ $$or, ka+bc≥kb+ac$$. Since $k≥c$ and $a≥b$ by Appling rearrangement inequality, the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
} |
If $a,b>0$ and $a+b=2$ , prove that $a^{2b}+b^{2a}+(\frac{a-b}2)^2\leqslant2$ If $a,b>0$ and $a+b=2$ , prove that
$$
a^{2b}+b^{2a}+(\frac{a-b}2)^2\leqslant2
$$
The equality occurs if and only if $(a,b)\sim(1,1)$ or $(a,b)\sim(2,0)$ or its cyclic permutations.
My attempt:
By symmetry and the constraint, we may let $x=a-1=1-b$ , the inequality converts to
$$
(1+x)^{2(1-x)}+(1-x)^{2(1+x)}+x^2\leqslant2
$$
I tried to let $f(x):=(1+x)^{2(1-x)}+(1-x)^{2(1+x)}+x^2$ , but its derrivative is too complicated. I also tried to write $(1+x)^{1-x}$ as $e^{2(1-x)\ln(1+x)}$ and apply inequalities like $\ln x\leqslant x-1$ , $e^x\leqslant\dfrac{1}{1-x}$ and such, but it would be either too complicated or too crude.
How to solve it?
| Variant @Macavity .
We use a form of the Young's inequality wich is a somewhere a generalization of the Bernoulli's inequality :
Let $a,b>0$ and $0<v<1$ then we have :
$$av+b(1-v)\geq a^vb^{1-v}$$
Taking account of this theorem and putting :
$a=(1+x)$$\quad$$b=1$$\quad$$v=(1-x)$ we get $0< x<1$:
$$(1+x)^{2(1-x)}\leq ((1+x)(1-x)+1-(1-x))^2$$
Again :
$a=(1-x)^2,b=1,v=x$
$$(1-x)^{2x}\leq (1-x)^{2}x+1-x$$
Or :
$$(1-x)^{2x+2}\leq (1-x)^2((1-x)^{2}x+1-x)$$
Adding the inequalities we got :
$$(1-x)^3x^2\geq 0$$
We have also :
$$f\left(x\right)=\frac{2^{\left(1-x^{2}\right)}\left(1+x^{\frac{1}{3}}\right)}{2}x^{2+\frac{1}{\ln2}-\ln12}+1-x^{2},h(x)=f(1-x),g\left(x\right)=(1+x)^{2(1-x)}$$
Then for $x\in(0,1)$ :
$$h(x)>g(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
What is the probability f(p) of A winning the match?
Here's my approach:
Let A denote the event "A wins the game" and B denote the event "B wins the game"
I am supposed to calculate the probability f(p)
There are two possible cases:-
i) A wins within 4 games
Possible cases: AAA,ABAA,BAAA,AABA
ii) A doesn't win within 4 games
Possible cases: AABB(A and B occur 2 times each) ABABABAB...(A and B occur k times each) AA
AABB can be arranged in 6 ways
ABABAB.... can be arranged in $(2!)^k$ ways (A and B are adjacent to one another)
Now :
f(p) = P(case i) + P(case ii)
P(case ii) = $(6 (p^2) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k. p^2 $=$(6 (p^4) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k$
= $(6. 2 . p^5 . (1-p)^3)/(1-2p+2p^2)$ (The denominator seems correct)
P(case i ) = $p^3 +3p^3(1-p)$
f(p) = $p^3 +3p^3(1-p) + \frac{(6. 2 . p^5 . (1-p)^3)}{(1-2p+2p^2)}$
The expected answer is not produced by solving the above equation.Where did I go wrong?
Edit:-
I finally figured out the mistake that I made in the above solution.
In P(case ii) = $(6 (p^4) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k $, I replaced $\sum_{k=1}^∞$ with $\sum_{k=0}^∞$ . There has to be one case when B does not occur after the tie, i.e. A occurs only two times.
P(case ii) = $(6 (p^4) (1-p)^2 ) \sum_{k=0}^∞ 2^k . p^k . (1-p)^k $
= $(6 (p^4) (1-p)^2 ) .[1+2p(1-p)^1+2^2.p^2.(1-p)^2+...]$
=$(6 (p^4) (1-p)^2 ) .\frac{1}{1-2p(1-p)}$
f(p) = P(case i) + P(case ii)
=$p^3 +3p^3(1-p)+(6 (p^4) (1-p)^2 ) .\frac{1}{1-2p(1-p)}$
=$p^3[1+3(1-p)+\frac{6p(1-p)^2}{1-2p(1-p)}]$
=$p^3[1+(3-3p)+\frac{6p-12p^2+6p^3}{1-2p+2p^2)}]$
=$p^3[\frac{(1-2p+2p^2)+(3-6p+6p^2-3p+6p^2-6p^3)+(6p-12p^2+6p^3)}{1-2p+2p^2)}]$
=$p^3\frac{(4-5p+2p^2)}{1-2p+2p^2}$
| I would treat the $2$-$2$ case separately. Calculate the probability that $B$ has won by the end of Game $4$. Then you know the probability of arriving at a $2$-$2$ tie.
$A$ wins within $4$ games with probability $p^3+3p^3(1-p)=4p^3-3p^4=p^3(4-3p)$.
$B$ wins within $4$ games with probability $(1-p)^3+3p(1-p)^3=(1-p)^3(1+3p).$
A $2$-$2$ tie occurs with probability $y=1-[p^3(4-3p)+(1-p)^3(1+3p)]$.
In a $2$-$2$ tie, the probability that $A$ wins immediately is $p^2$, the probability that $B$ wins immediately is $(1-p)^2$, and the probability that we reach another tie is $2p-2p^2$. This, if you reach the tied state, the probability $x$ that $A$ eventually wins the game satisfies $x=p^2+(2p-2p^2)x$, so $x=\dfrac{p^2}{2p^2-2p+1}$.
Thus, the probability that $A$ wins is $p^3(4-3p)+xy$.
I haven't worked through the algebra to confirm this results in the expected answer, nor have I worked through your algebra to see whether the mistake is there or in the way you've set up the problem.
A tweak to this approach is to observe that if the match is tied $1$-$1$, the probability of an $A$ win is $x$, calculated above, because we're already in the position where the first player to lead by $2$ wins.
$A$ will lead $2$-$0$ with probability $p^2$ and once there, will win without going to "overtime" with probability $p(2-p)$.
$B$ will lead $2$-$0$ with probability $(1-p)^2$ and once there will win without going to "overtime" with probability $1-p^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$
If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?
My approach is as follows
Given $g\left( x \right) = ax + c$ & $f\left( x \right) = {x^b} + 3 \Rightarrow {\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = x$
$\frac{{g\left( x \right) - c}}{a} = x$
${\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = \frac{{g\left( x \right) - c}}{a}$
$g\left( x \right) = g\left( x \right) \Rightarrow {g^{ - 1}}\left( {g\left( x \right)} \right) = x$
I am not able to proceed.
| Hint: Recall that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$. You've already compute each inverse separately, so now we just compose them to get:
$$
(g \circ f)^{-1}(x) = f^{-1}(g^{-1}(x))
= f^{-1}\left(\frac{x - c}{a}\right)
= \left( \left(\frac{x - c}{a}\right) - 3 \right)^{\frac{1}{b}}
= \left( \frac{x - 3a - c}{a}\right)^{\frac{1}{b}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do i solve the integral of $x*\ln(x-1)$? I've been having some trouble trying to solve this integral. I feel like i'm going in circles with this. What i did first is use integration by parts in this manner.
$$\int x\cdot \ln(x-1)\;dx = \ln(x-1)\cdot \frac{1}{2}x^2- \int \frac{1}{x-1}\cdot \frac{1}{2}x^2 \; dx$$
$$= \ln(x-1)\cdot \frac{1}{2}x^2-\frac{1}{2}\int \frac{1}{x-1}\cdot x^2\; dx=$$
So i get $\int \frac{1}{x-1}\cdot x^2 \; dx$ on which i should use integration by parts again i assume.
$$\int \frac{1}{x-1}\cdot x^2\; dx = \frac{1}{1-x} \cdot \frac{1}{2}x^2-\int \frac{-1}{(x-1)^2} \cdot\frac{1}{2}x^2 \; dx$$
And again.
$$\frac{1}{2}\int \frac{-1}{(x-1)^2}\cdot x^2\; dx=x^2 \cdot \frac{-1}{(x-1)^2} - \int x^2 \cdot \frac{1}{x-1}\;dx.$$
$\int x^2 \cdot \frac{1}{x-1}\;dx$ yields $x^2 \cdot \ln(x-1) - \int 2x\cdot \ln(x-1)\;dx$
I feel like I'm messing something up, however i can't tell what. I've also tried to use $f(x)=x \space, g'(x) = \ln(x-1)$ in the first step. Any suggestion is greatly appreciated.
| When you're integrating a rational expression, step one is to make sure it's a proper rational expression. By long division:
$$\frac{x^2}{x-1} = x^2+1+\frac{1}{x-1},$$
which is much easier to integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Unable to crack $\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx=\frac{\pi^2}{30}$ I am unable to solve the integral
$$\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx=\frac{\pi^2}{30} $$
after a number of attempts, except with a few observations below.
1). Despite the appearance, I do not believe this integral is related to the Ahmed integral or any variations of known solution.
2). The integral is related to a more complex one posted here. However, the answer offered is non conventional, which I could not fully appreciate. I would prefer an integration approach.
3). Numerically, I could establish that
$$\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx
=2 \int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt5 x}{1+x^2}dx
- \int_0^{\sqrt3} \frac{\cot^{-1}\sqrt5 x}{1+x^2}dx $$
where the RHS can be worked out with an elaborate procedure.
So, I would like to see either a direct solution to the integral, or, at least, establish the integral relationship above analytically.
| Perform the substitution $t=\frac{1}{\sqrt{2-x^2}}$ to check that,
$$
\tag{1}\int_{0}^{\frac{1}{\sqrt{3} } } \frac{\operatorname{arccot} \left ( \sqrt{2-x^2} \right ) }{
1+x^2}\text{d}x=
\int_{\frac{\sqrt{2}}{2} }^{\sqrt{\frac{3}{5}}}
\frac{\arctan\left ( t \right ) }{\sqrt{2t^2-1}(3t^2-1) }\text{d}t.
$$
The second integral can be done by evaluating
$$
\tag{2}\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y
=\frac{\pi^2}{30}.
$$
To prove $(2)$, use the identity
$$
\int_{0}^{\infty} {e^{-t^2x^2}}\text{d}x
=\frac{\sqrt{\pi} }{2\sqrt{t} }.
$$
Then I can get(classical result $\int_{0}^{1} \frac{e^{-t^2x^2}}{1+t^2}\text{d}x
=\frac{\pi}{4}e^{t^2}\left ( 1-\operatorname{erf}(t)^2 \right )$ is used)
$$
\begin{align*}
&\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y\\
=& \frac{2}{\sqrt{\pi} } \int_{0}^{1} \int_{0}^{1}
\frac{1}{(1+x^2)(1+y^2)} \left (\int_{0}^{\infty}e^{-t^2(3+x^2+y^2)}\text{d}t\right ) \text{d}x\text{d}y\\
\overset{{\scriptsize\text{Fubini}}}{=}&
\frac{2}{\sqrt{\pi} } \int_{0}^{\infty}e^{-3t^2}\left ( \int_{0}^{1} \frac{e^{-t^2x^2}}{1+x^2}\text{d} x\right )^2\text{d}t\\
=&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16} \int_{0}^{\infty}
e^{-x^2}\left(1-\operatorname{erf}(x)^2\right)^2\text{d}x\\
=&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16} \int_{0}^{\infty}
e^{-x^2}\left(1-2\operatorname{erf}(x)^2+\operatorname{erf}(x)^4\right)\text{d}x\\
=&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16}\left(\int_{0}^{\infty}e^{-x^2}\text{d}x
-2\int_{0}^{\infty}e^{-x^2}\operatorname{erf}(x)^2\text{d}x
+\int_{0}^{\infty}e^{-x^2}\operatorname{erf}(x)^4\text{d}x\right)\\
=&\frac{\pi^2}{30}.
\end{align*}
$$
The last equality follows from their primitives
$$
\frac{\mathrm{d} }{\mathrm{d} x}
\left ( \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)^{n+1} \right )
=(n+1)e^{-t^2}\operatorname{erf}(x)^{n}.
$$
And
$$
\begin{align*}
&\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y\\
=&\int_{0}^{1} \frac{\arctan\left ( \sqrt{\frac{2+x^2}{4+x^2} } \right ) }{
\left ( 1+x^2 \right )\sqrt{2+x^2} }\text{d}x
\\
\overset{{\scriptsize t=\sqrt{\frac{2+x^2}{4+x^2} }}}{=}&\int_{\frac{\sqrt{2}}{2} }^{\sqrt{\frac{3}{5}}}
\frac{\arctan\left ( t \right ) }{\sqrt{2t^2-1}(3t^2-1) }\text{d}t\\
=&\int_{0}^{\frac{1}{\sqrt{3} } } \frac{\operatorname{arccot} \left ( \sqrt{2-x^2} \right ) }{
1+x^2}\text{d}x=\frac{\pi^2}{30}.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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How to solve the given differential equation?
Given $T$ is a function in $t$ $(x>0)$. And $T(@t=0)=40$. And which satisfies,
$$ \frac{dt}{4}= \frac{3\cdot dT}{2} + \frac{T\cdot dt}{400+t}$$
Then find $T$
I tried putting $400+t=u$ and then,
$$ \frac{du}{4}= \frac{3\cdot dT}{2} + \frac{T\cdot du}{u}$$
And multiplying by $4u$ both sides gives,
$$u\cdot du= 6\cdot u\cdot dT+ 4\cdot T \cdot du$$
And by integrating by parts bay taking common $4$ gives,
$$ \frac{u^2}{2}=4\cdot u \cdot T+ \int 2\cdot u \cdot dT+c$$
But now I am stuck here. Can someone help me solving this?
| Under your substitution, the ODE becomes homogeneous
$$\frac32 \, dT = \left(\frac14 - \frac T{400+t}\right) \, dt \implies \frac{dT}{du} = \frac16 \left(1-\frac{4T}u\right)$$
Further substituting $U=\frac Tu$ and $\frac{dU}{du}=\frac{u\frac{dT}{du} - T}{u^2}$ gives the separable ODE,
$$u\,\frac{dU}{du} + U = \frac{1-4U}6 \implies \frac{dU}{1-10U} = \frac{du}{6u}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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For odd $k$, $\sum_{j=0}^{N-1}\cos^k(\pi j/N) = 1$. Is there a nicer proof? Is this well known? In the course of answering this question, I stumbled into the following result: for any integer $N > 1$ and any odd $k$,
$$
\sum_{j=0}^{N-1}\cos^k(\pi j/N) = 1.
$$
My proof, which I suspect is much more complicated than it needed to be, was as follows:
\begin{align}
\sum_{j=0}^{N-1}\cos^k\left(\frac{\pi j}{N}\right) &=
\sum_{j=0}^{N-1} 2^{-k}(\omega^{j} + \omega^{-j})^k
\\ & =
2^{-k} \sum_{j=0}^{N-1} \sum_{\ell = 0}^k \binom k{\ell}\omega^{(k - 2\ell)j}
\\ & =
2^{-k} \sum_{\ell = 0}^k \binom{k}{\ell} \sum_{j=0}^{N-1}\omega^{(k - 2\ell)j}
\\ & =
2^{-k} \sum_{\ell = 0}^k \binom{k}{\ell}
\cdot \begin{cases}
\frac {1 - \omega^{(k - 2\ell)N}}{1 - \omega^{(k - 2\ell)}}
& 2N \nmid (k - 2\ell)\\
N & 2\ell = k
\end{cases}
\\ & =
2^{-k} \sum_{\ell = 0}^k \binom{k}{\ell}
\cdot \begin{cases}
\frac {1 - (-1)^{(k - 2\ell)}}{1 - \omega^{(k - 2\ell)}}
& 2N \nmid (k - 2\ell)\\
N & 2\ell = k
\end{cases}
\\ & =
2^{-k} \sum_{\ell = 0}^k \binom{k}{\ell} \frac {1 - (-1)}{1 - \omega^{k - 2\ell}}
\\ & =
2^{-k} \sum_{\ell = 0}^k \binom{k}{\ell} \frac 2{1 - \omega^{k - 2\ell}}
\\ &=
2^{-k} \sum_{m = 0}^{(k-1)/2} \binom{k}{m} \left[\frac 2{1 - \omega^{k - 2m}} + \frac 2{1 - \omega^{-(k - 2m)}}\right]
\\ & =
2^{-k} \sum_{m = 0}^{(k-1)/2} \binom{k}{m} 4\operatorname{Re}\left[\frac 1{1 - \omega^{k - 2m}}\right]
\\ & =
2^{-k} \sum_{m = 0}^{(k-1)/2} \binom{k}{m} 4\operatorname{Re}\left[\frac {1 - \omega^{-(k - 2m)}}{|1 - \omega^{k - 2m}|^2}\right]
\\ & =
2^{-k} \sum_{m = 0}^{(k-1)/2} \binom{k}{m} 4\operatorname{Re}\left[\frac {(1 - \cos((k - 2m)\pi/N)) + i\sin((k - 2m)\pi/N)
}{(1 - \cos((k - 2m)\pi/N))^2 + \sin^2((k - 2m)\pi/N)}\right]
\\ & =
2^{-k} \sum_{m = 0}^{(k-1)/2} \binom{k}{m} 4
\frac{1 - \cos((k - 2m)\pi/N)}{2(1 - \cos((k - 2m)\pi/N))}
\\ & =
2^{-k} \sum_{m = 0}^{(k-1)/2} 2\binom{k}{m}
=
2^{-k} \sum_{m = 0}^{k} \binom{k}{m} = 1
\end{align}
Is there a nicer proof? Is this fact well known?
| The result follows since $\cos \pi j/N=-\cos \pi (N-j)/N$ for $j=1,..N-1$, while for $j=0$ we get $1$. Now $a=-b$ implies $a^k=-b^k$ for any odd $k$. So all but the first term cancel in pairs (and the middle term is zero for $N$ even)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of possibilities for creating words of length four A random generator is generating words of length $4$ out of the alphabet $\{0,1,2,3,4,5,6,7,8,9\}$, for example $1234$.
What is the probability to get
*
*all different numbers?
*exactly one pair of same numbers?
*exactly two pairs of the same numbers?
As to (1), my answer is
$$
\frac{10\cdot 9\cdot 8\cdot 7}{10^4}=0.504
$$
As to (2), my answer is
$$
\frac{10\cdot\binom{4}{2}\cdot 9\cdot 2\cdot 8}{10^4}=0.864
$$
Because first we have $10$ options to choose one element of the alphabet, and $\binom{4}{2}$ possibilities to choose two positions to put it. Then we have $9$ options to choose an alement different from the first one and two options to put it. Finally, we have $8$ options to choose the third element and exactly one possibility to put it down.
As to (3) my answer is
$$
\frac{10\cdot\binom{4}{2}\cdot 9}{10^4}=0.054
$$
| The first answer is correct. The second and third are incorrect since it does not matter which of the values is selected first when you pick two numbers that appear the same number of times.
Exactly one pair of same numbers: There are $10$ ways to select the number which appears twice, $\binom{4}{2}$ ways to select two of the four positions for that number, $\binom{9}{2}$ ways to select the two numbers which each appear once, and $2!$ ways to arrange those numbers in the remaining two positions. Hence, the probability that the word contains exactly one pair of numbers is
$$\frac{\dbinom{10}{1}\dbinom{4}{2}\dbinom{9}{2}2!}{10^4}$$
Exactly two pairs of the same numbers: There are $\binom{10}{2}$ ways to select the two values which will each appear twice and $\binom{4}{2}$ ways to choose two of the four positions for the larger of those values. Hence, the probability of obtaining two numbers which each appear twice in the string should be
$$\frac{\dbinom{10}{2}\dbinom{4}{2}}{10^4}$$
In the comments, you asked about the cases that one number appears three times and another number appears once and the case that all four numbers are the same. Both of your proposed answers are correct.
Sanity Check: The number $4$ can be partitioned in five ways:
\begin{align*}
4 & = 4\\
& = 3 + 1\\
& = 2 + 2\\
& = 2 + 1 + 1\\
& = 1 + 1 + 1 + 1
\end{align*}
Hence, there are five possible cases:
*
*all four numbers are the same, which you correctly found can occur in $\binom{10}{1}$ ways
*one number appears three times and another number appears once, which you correctly found can occur in $\binom{10}{1}\binom{4}{3}\binom{9}{1}$ ways
*two numbers each appear twice, which can occur in $\binom{10}{2}\binom{4}{2}$ ways
*one number appears twice and two other numbers each appear once, which can occur in $\binom{10}{1}\binom{4}{2}\binom{9}{2}2!$ ways
*four numbers each appear once, which can occur in $\binom{10}{4}4!$ ways
Notice that the sum of these probabilities is
$$\frac{\dbinom{10}{1} + \dbinom{10}{1}\dbinom{4}{3}\dbinom{9}{1} + \dbinom{10}{2}\dbinom{4}{2} + \dbinom{10}{1}\dbinom{4}{2}\dbinom{9}{2}2! + \dbinom{10}{4}4!}{10^4} = 1$$
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the $n$th power of a 3-by-3 circulant matrix Consider the matrix given as $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}$$
Write down a formual for $A^n$ for $n\in\mathbb{N}$.
$$$$
My attempt: The first that comes to mind is to diagonalize it and hence find the formual for $A^n$, but that is very messy, so I tried to do something else it goes as:
bserve that $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}=a_0\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+a_1\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}+a_2\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix}$$
Let $U=\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}$ then we will have that
$$\begin{matrix}U^2=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix} & \text{ and } & U^3=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\end{matrix}$$
Hence we have $A=a_0I+a_1U+a_2U^2 = a_0U^3+a_1U+a_2U^2=(a_0U^2+a_1I+a_2U)U$
This got me thinking that there might be an easy way to solve the above problem but, I was not able to make any further progress.
Please Help and thanks in advance.
| I think you've reached a decent result and by just using the multinomial theorem you can obtain a generalized solution.
$A=a_0I+ a_1U + a_2 U^2 \\
A^n = (a_0I+ a_1U + a_2 U^2)^n\\$
Then the coeffecients of $U^3$, $U^2$ and $U^1$ are determined using the multinomial theorem and just plugged back into the result matrix.
$$
\begin{matrix}
\sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\
\sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\
\sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\
\end{matrix}
$$
The answer may seem obvious but unfortunately the multinomial expansion can't be simplified further (as per my knowledge). Even if it could it would just be a huge mess.
| {
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"url": "https://math.stackexchange.com/questions/4639166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Prove that $\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge (n^2-1)^n$ Let $n$ be a positive integer and $x_1,...,x_n$ positive reals such that $x_1+...+x_n=1$. Prove that $$\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge (n^2-1)^n$$
Now notice that if we apply Jensen's Inequality to $$f(x)=\ln\left(\frac{1}{x_i^2}-1\right)$$ We would get the exact inequality...except that $f$ is not convex on $(0,1)$. But I do know this,
Claim: If for all $k\le n$ we have $0< a_k\le 1/2$. Then, $$\prod_{i=1}^n\left(\frac{1}{a_i}-1\right)\ge \left(\frac{n}{a_1+...+a_n}-1\right)^n$$
The proof is not too hard. Just consider $$g(x)=\ln\left(\frac{1}{x}-1\right)$$
$g$ is convex on $(0,1/2)$ and you can finish with Jensen. Now what if we let $a_i=x_i^2$? Well, we would get $$\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge \left(\frac{n}{x_1^2+...+x_n^2}-1\right)^n\ge (n^2-1)^n$$
Because $x_1+...+x_n=1$ implies, $$\sum_{i=1}^n x_i^2\ge \frac{1}{n}\left(\sum_{i=1}^n x_i\right)^2=\frac{1}{n}$$
And we're done...Or are we? The argument is only valid if $x_i^2\le 1/2$ for all $i$. Well if there exists $x_i^2>1/2$ Can we still get the inequality?
By the way there is a way to solve the inequality without any use of Jensen. But I want to know if we can solve it like that.
| Lemma:$$\left(\dfrac{1}{x}-x\right)\left(\dfrac{1}{y}-y\right)\ge 4^3\left(\dfrac{z^3}{3^9\cdot x^2y^2}\right)^{\frac{1}{4}}~~~~~(n=3)$$
$$
LHS=\dfrac{1}{xy}\left(\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}+xy\right)\left(x+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)\left(y+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)\ge RHS$$
then
$$\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{1}{y^2}-1\right)\ge 4^3\left(\dfrac{z^3}{3^9\cdot x^2y^2}\right)^{\frac{1}{4}}\cdot\dfrac{1}{xy}$$
for any postive $n$ we have use AM-GM same methods
$$\left(\dfrac{1}{x^2_{1}}-1\right)\left(\dfrac{1}{x^2_{2}}-1\right)
\ge \left(n-\dfrac{1}{n}\right)^2\left(\dfrac{1}{n}\right)^{\dfrac{n^3-3n^2+2n-2}{(n-1)^2(n+1)}}\cdot(x_{1}x_{2}\cdots x_{n})^{\frac{n}{(n-1)^2}}\left(\dfrac{1}{x_{1}x_{2}}\right)^{\frac{n^2+1}{n^2-1}}\cdot\dfrac{1}{x_{1}x_{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ Prove $(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ for $a,b,c>0$ and $abc\geq1$.
My try: $(a+1)x^2-4xy+(b+1)y^2+(a+1)x^2-4xz+(c+1)z^2+(b+1)y^2-4yz+(c+1)z^2\geq0$
Then I had to prove that $(a+1)x^2-4xy+(b+1)y^2\geq2x^2-4xy+2y^2$, but I realized that there's no way of using $abc\geq1$.
Any help is appreciated!
| We can assume at least two of ${x}$, ${y}$, ${z}$ non negative, otherwise we take their additive inverse. Let ${y}$, $z\ge0$.
Consider $x{}$ the only variable. The inequality can be regarded as a quadratic function of $x$. Since the coefficient of $x^2$is positive, we only need $\frac\Delta4\le0$, which requires
\[(a+1)\left((b+1)y^2+(c+1)z^2\right)\ge(y+z)^2+2(a+1)yz.\]Or $(ab+a+b)y^2+(ac+a+c)z^2\ge2(a+2)yz$. Using AM-GM, it suffices to show
\[(ab+a+b)(ac+a+c)\ge a+2.\]Expand the left side, since we have $ab\cdot ac\ge a$, $ab\cdot c$, $ac\cdot b\ge1$ and the rest $\ge0$, we're done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$.
How do I find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$?
I have tried plugging in values, tried manipulating and inequality by means but that didn't worked. How do I approach this question?
| Capitalizing on writing the expression as a sum of two positive numbers, indicates the value of a:
$$\begin{align}E(a,b,c)&=\underbrace{(b^2-a^2)}_{\ge 0}\underbrace{(c-b)}_{\ge 0}+\underbrace{c^2(1-c)}_{\ge 0}, a,b,c \lt 1\\ &\le (b^2-0)(c-b)+c^2(1-c), \boxed{a= 0}\end{align}$$
The first term $b^2(c-b)$ could be analyzed using $f(x)=x^2(c-x), x\lt c$. This cubic has a maximum at $x=\frac 23 c$, therefore $\boxed{b=\frac 23 c}$
Replacing $b=\frac 23 c$, the original expression after considering $a=0$ becomes:
$$E(a,b,c)\lt \frac {4}{27}c^3+c^2(1-c), E(a,b,c)\lt -\frac{23}{27}c^3+c^2$$
The function $g(x)=-\frac{23}{27}x^3+x^2$ has a maximum at $x=\frac{18}{23}$. Therefore $\boxed{c=\frac{18}{23}}$.
Replacing $a,b,c$ finds the maximum of the original expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Fraction of prime number between consecutive powers of two A $n$-bit integer is an integer $x$ such that $2^{n-1} \le x < 2^n$. In [1] it is claimed without proof that a corollary of the prime number theorem is that:
For any $n > 1$, the fraction of $n$-bit integers that are prime is at least $\frac{1}{3n}$.
Can somebody point me to a proof of this?
It is (perhaps) worth noticing that a direct application of the prime number theorem allows to prove an asymptotic version of the above (with a slightly better constant).
$$
\begin{align*}
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \pi(2^n-1) - \pi(2^{n-1})\bigg) &=
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \pi(2^n) - \pi(2^{n-1}) \bigg) \\&=
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \frac{2^n}{n \ln 2} - \frac{2^{n-1}}{(n-1)\log2} \bigg)\\ & =
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \frac{2^n}{2n \ln 2 } \cdot\frac{n-2}{n-1} \bigg) = 1.
\end{align*}
$$
Therefore $\frac{\pi(2^n-1) - \pi(2^{n-1})}{2^{n-1}} \sim \frac{1}{n \ln 2}$, and for every $\varepsilon >0$ there is some $n_0$ such that, for all $n \ge n_0$, the fraction of $n$-bit integers that are prime is at least $\frac{1-\varepsilon}{n \ln 2}$.
A not-so-elegant way to prove the the claim would be that of choosing $\varepsilon =1-\frac{\ln 2}{3} \approx 0.7689$ and finding an upper bound on $n_0$ that lies in the range for which $\pi(\cdot)$ has been computed. Are such upper bounds known?
[1] Jonathan Katz, Yehuda Lindell. Introduction to Modern Cryptography (3rd edition). CRC Press. ISBN 9781351133012.
| This is just a more detailed write-up of the strategy proposed by Marty Cohen's answer.
We can prove the stronger claim: for $n \ge 1$ there are at least $\frac{2^{n-2}}{2n}$ $n$-bit prime numbers (i.e., the fraction of $n$-bit prime numbers is at least $\frac{1}{2n}$).
According to [1], for $x \ge 55$,
$\dfrac{x}{\log x+2}
\lt \pi(x)
\lt \dfrac{ n}{\log x-4}
$.
Defining $L(n) = \dfrac{2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}$ we have:
$
\pi(2^n -1 )-\pi(2^{n-1}) = \pi(2^n)-\pi(2^{n-1}) \ge L(n).
$
As a hint that we are going towards the right direction, we notice that $L(n)$ is roughly:
$$
\dfrac{ 2^n}{n\log 2}-\dfrac{ 2^{n-1}}{(n-1)\log 2} \approx
\dfrac{2^n}{n\log 2}-\dfrac{ 2^{n-1}}{n\log 2}
= \dfrac{2^{n-1}}{n\log 2} > \dfrac{2^{n-1}}{n}.
$$
More formally, let $\ell(n) = \dfrac{2n \log 2}{n \log 2 + 2} - \dfrac{n \log 2}{(n-1) \log 2 - 4}$:
$$
L(n) = \dfrac{2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}
= \dfrac{2^{n-1}}{n \log2} \left( \dfrac{2n \log 2}{n \log 2 + 2} - \dfrac{n \log 2}{(n-1) \log 2 - 4} \right) = \dfrac{2^{n-1}}{n \log2} \cdot \ell(n).
$$
Taking the derivative of $\ell(n)$ we see that:
$$
\dfrac{d \ell(n)}{dn} = \dfrac{4 \log 2}{(2 + n \log 2)^2} + \frac{(4 + \log 2) \log 2}{(n \log 2 - 4 - \log 2)^2} \ge 0,
$$
therefore $\ell(n)$ is defined for all $n \ge 10$ and monotonically increasing. Evaluating $\ell(10)$ we see that $\ell(10) \ge \frac{1}{2}$. Therefore:
$$
L(n) \ge \dfrac{2^{n-1}}{n \log2} \cdot \dfrac{1}{2} \ge \frac{2^{n-1}}{2n}.
$$
This proves the claim for $n \ge 10$. For $ 2 \le n < 10$ we can exhaustively inspect $\pi(2^n -1 )-\pi(2^{n-1})$.
[1] "Explicit Bounds for Some Functions of Prime Numbers",
Barkley Rosser
American Journal of Mathematics
Vol. 63, No. 1 (Jan., 1941), pp. 211-232
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to evaluate $\int x \sqrt{x^2 - x}\ dx$
Problem is to integrate $ x \sqrt{x^2 - x}$.
My attempt:
I made it ready for a substitution $u = x^2 - x$
$$\begin{aligned} \int x \sqrt{x^2 -x}\ dx &= \int (2x-1)\sqrt{x^2 - x} \ dx - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \int \sqrt u\ du - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \frac23(x^2 -x)^{3/2} + C_1 - \int(x-1) \sqrt{x^2 -x}\ dx\end{aligned}$$
I don't know how to continue from here.
Alternatively I tried this:
$$\begin{aligned} \int x \sqrt{x^2 - x}\ dx &= \int x^2 \sqrt{1- \frac{1}{x}}dx\\ & \overset{1- \frac1x = t^2}{=} \int \frac{2t^2}{(1-t^2)^4}\ dt\\& \overset{t =\sin(\theta)}{=} \int \frac{2\sin^2(\theta) \cos(\theta)\ d\theta}{\cos^4(\theta)}\\& = \int 2 \tan(\theta)\tan(\theta) \sec(\theta) \ d\theta\\ & = \int 2\sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& \overset{\sec(\theta) = u}{=} \int 2 \sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& = \int 2 \sqrt{u^2 - 1}\ du\\& = u \sqrt{u^2 - 1} - \ln|u + \sqrt{u^2- 1}| + C\\& = \sqrt{x^2 - x} - \ln|x + \sqrt{x^2 - x}| + C\end{aligned}$$
This method is very tedious. Is there any easy way to do the original integral?
| Hyperbolic function for $\int \sqrt{u^2-u}du$, using integration by parts yields
$$
\begin{aligned}
K & =u \sqrt{u^2-1}-\int \frac{u^2-1+1}{\sqrt{u^2-1}} d u \\
& =u \sqrt{u^2-1}-K-\int \frac{d u}{\sqrt{u^2-1}}
\end{aligned}
$$
Putting $y=\cosh u$ yields
$$
K=\frac{1}{2}\left(u \sqrt{u^2-1}-\operatorname{arccosh} u\right)+C
$$
Hence $$
I=\frac{1}{3}\left(x^2-x\right)^{\frac{3}{2}}+\frac{1}{16}\left[2(2 x-1) \sqrt{x^2-x}-\operatorname{arccosh}(2 x-1)\right] +C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Combinatorics: Amount of options for majority Say we have $2n$ people. Then the amount of options to form a majority (e.g. in a commission) are
$ \binom{2n}{n+1} + \binom{2n}{n+2} + \cdots + \binom{2n}{2n}$
I want to prove, that this is equal to $\frac{1}{2} \left[ 2^{2n} - \binom{2n}{n} \right]$
But I'm stuck. I have the this formula: $ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n $
So I can say $ \binom{2n}{n+1} + \binom{2n}{n+2} + \cdots + \binom{2n}{n} = 2^{2n} - \left[ \binom{2n}{n} + \binom{2n}{n-1} + \cdots + \binom{2n}{n-n} \right] = 2^{2n} - \left[ \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n} \right] $
But I can't see where to go from there, even when writing the binomial coefficients as $ \frac{n!}{k!(n-k)!} $
Can anyone help me with it?
| By the property of binomial coefficients $$\binom{n}{r}=\binom{n}{n-r}$$
$\implies \binom{2n}{n+1}=\binom{2n}{n-1}$
$\implies\binom{2n}{n+2}=\binom{2n}{n-2}$ and so on
Let $$A=\binom{2n}{n+1}+\cdots+\binom{2n}{2n}$$ Also
$$A=\binom{2n}{0}+\cdots+\binom{2n}{n-1}$$
So $$2A=2^{2n}-\binom{2n}{n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving Generating Function holds a specific recurrence. Consider the generating function $$\dfrac{1}{1 − 2x − x^2} = \sum_{n=0}^{\infty}a_nx^n$$
Prove that for each integer $n \ge 0$,
$$a_n^2+a_{n+1}^2 = a_{2n+2}$$
Hint: Find a $2 \times 2$ matrix $A$ such that
$$A^{n+2} =\begin{bmatrix}a_n &a_{n+1}\\
a_{n+1} &a_{n+2}\end{bmatrix}$$
and consider the top left entry of the matrix product $A^{n+2}A^{n+2}$.
Looking at the hint, I think about how we used matrices and eigenvalues to find the closed form expression of recurrence relations, but I can only really do that for stuff of the form $a_n=a_{n-1}+a_{n-2}$, and I'm not sure how the matrix product fits in there.
Instead, I tried to find the generating function for the recurrence listed above as:
$$\begin{align*}
a_n^2+a_{n+1}^2 &= a_{2n+2}\\
\left(x^n\right)^2a_n^2+\left(x^n\right)^2a_{n+1}^2 &= \left(x^n\right)^2a_{2n+2}\\
\sum_{n=0}^{\infty}\left(x^n\right)^2a_n^2+\sum_{n=0}^{\infty}\left(x^n\right)^2a_{n+1}^2 &= \sum_{n=0}^{\infty}\left(x^n\right)^2a_{2n+2}\\
A(x)^2+\dfrac{A(x)^2}{x^2} &= \sum_{n=0}^{\infty}x^{2n}a_{2n+2}\\
A(x)^2+\dfrac{A(x)^2}{x^2} &= \dfrac{A(x)}{x^2}\\
x^2A(x)^2+A(x)^2&=A(x)\\
A(x)=\dfrac{1}{x^2+1}
\end{align*}$$
But that's clearly not what we wanted. What mistakes did I make, and how does the hint fit into all of this? Thanks!
| Define $f(x)$ by
$$f(x)= \frac{1}{1-2x-x^2} = \sum_{n=0}^{\infty} a_n x^n$$
so
$$f(x) -2x f(x) -x^2 f(x) = 1$$
From this equation we deduce $a_0=1$, $a_1=2$, and $a_{n+2} -2a_{n+1} -a_n = 0$ for $n \ge 0$. It follows by induction on $n$ that if we let
$$A = \begin{pmatrix}
0 &1 \\
1 &2 \end{pmatrix}$$
then
$$A^{n+2} =\begin{pmatrix}
a_n &a_{n+1} \\
a_{n+1} &a_{n+2} \end{pmatrix}$$ for $n \ge 0$.
Now we can compute
$$A^{n+2} \cdot A^{n+2} = \begin{pmatrix}
a_n^2+a_{n+1}^2 & a_n a_{n+1} + a_{n+1} a_{n+2} \\
a_n a_{n+1} + a_{n+1} a_{n+2} & a_{n+1}^2 a_{n+2}^2
\end{pmatrix}$$
On the other hand,
$$A^{n+2} \cdot A^{n+2} = A^{2n+2+2} = \begin{pmatrix}
a_{2n+2} & a_{2n+3} \\
a_{2n+3} & a_{2n+4} \end{pmatrix}$$
Comparing the upper-left elements in these two matrices, we see
$$a_n^2+a_{n+1}^2 = a_{2n+2}$$
| {
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Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$. Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$.
Let $[x]$ denote $\lceil{x-\frac{1}{2}}\rceil$.
Using Desmos, I got $2.46736022133$ and WolframAlpha does not give me a solution. My intuition tells me that it might be possible to find an exact value using Trapezoidal Reimann Sums but I am not really sure how to go about doing it. After my attempt, I got stuck but I was at a point where I could plug it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. Why did it come out so nicely?
My attempt:
Where $A_n$ denotes the area of the $nth$ trapezoid from the right:
$$A=\frac{h}{2}(a+b)$$
$$A_n=\frac{\frac{2}{2n-1}-\frac{2}{2n+1}}{2}(\frac{2n}{2n-1}+\frac{2n}{2n+1})$$
$$A_n=\frac{\frac{4n+2}{4n^{2}-1}-\frac{4n-2}{4n^{2}-1}}{2}\left(\frac{4n^{2}+2n}{4n^{2}-1}+\frac{4n^{2}-2n}{4n^{2}-1}\right)$$
$$A_n=\frac{2}{4n^{2}-1}\left(\frac{8n^{2}}{4n^{2}-1}\right)$$
$$A_n=\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
Then:
$$\int_{0}^{2}x[\frac{1}{x}]dx=\sum_{n=1}^{\infty}A_n=\sum_{n=1}^{\infty}\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
I do not know how to solve this infinite summation so I plugged it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. How did it get to this conclusion? Is there a more efficient way to solve this?
| $$\frac{16 n^2}{\left(4 n^2-1\right)^2}=-\frac{1}{2 n+1}+\frac{1}{(2 n+1)^2}+\frac{1}{2 n-1}+\frac{1}{(2 n-1)^2}$$
$$\sum_{n=1}^p\frac{16 n^2}{\left(4 n^2-1\right)^2}=\frac{1}{2} \left(\psi ^{(0)}\left(\frac{3}{2}\right)-\psi
^{(0)}\left(p+\frac{3}{2}\right)\right)+$$
$$\frac{1}{8} \left(-2 \psi ^{(1)}\left(p+\frac{3}{2}\right)+\pi
^2-8\right)+$$ $$\frac{1}{2} \left(\psi ^{(0)}\left(p+\frac{1}{2}\right)-\psi
^{(0)}\left(\frac{1}{2}\right)\right)+
\frac{1}{8} \left(\pi ^2-2 \psi
^{(1)}\left(p+\frac{1}{2}\right)\right)$$
Using the asymptotics
$$\sum_{n=1}^p\frac{16 n^2}{\left(4 n^2-1\right)^2}=\frac{\pi ^2}{4}-\frac{1}{p}+\frac{1}{2
p^2}+O\left(\frac{1}{p^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proving that $0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)$ I need help proving that
$$
0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)
$$
For $n > 2$ and positive $k$.
I've tried all sorts of identities, and nothing have worked. Any help would be greatly appreciated.
The things I have tried are the following:
Using a identity for multiples angle formula:
$$
k\sin \left(\frac{2 \pi}{n}\right) >\sin\left(k\cdot\frac{2 \pi }{n}\right)=2^{k -1}\prod_{i=0}^{k-1}\sin \left(\frac{\pi i}{k}+\frac{2 \pi }{n}\right)
$$
$$
k \geq 2^{k -1}\prod_{i=1}^{k-1}\sin \left(\frac{\pi i}{k}+\frac{2 \pi }{n}\right)
$$
Which I could not make sense of.
I also tried rewriting the inequality
$$
k\sin \left(\frac{2 \pi}{n}\right) +\sin \left(\frac{2 \pi (n -k)}{n}\right)>0
$$
Which got me here:
$$\begin{split}
&=(k-1)\sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi (n -k)}{n}\right)\\
&=(k-1)\sin \left(\frac{2 \pi}{n}\right)+2\sin \left(\frac12\left[\frac{2 \pi}{n}+\frac{2 \pi (n -k)}{n}\right]\right)\sin \left(\frac12\left[\frac{2 \pi}{n}-\frac{2 \pi (n -k)}{n}\right]\right)
\end{split}$$
But I could not make that work either.
| I am assuming that $k,n\in\mathbb{Z}$.
Suppose that for some $k$, we have
$$
\sin\left(\frac{2\pi k}{n}\right)\le k\sin\left(\frac{2\pi}{n}\right)\tag1
$$
Note that $(1)$ is trivially true for $k=1$.
For $k\ge2$, we have
$$
k\,\overbrace{\sin\left(\frac{2\pi}4\right)}^1\gt k\,\overbrace{\sin\left(\frac{2\pi}3\right)}^{\sqrt3/2}\gt1\ge\overbrace{\sin\left(\frac{2\pi k}{n}\right)}^{\le1}\tag2
$$
Thus, $(1)$ is true for $n=3$ and $n=4$. Therefore, assume, $n\ge5$.
$$
\begin{align}
\sin\left(\frac{2\pi(k+1)}{n}\right)
&=\sin\left(\frac{2\pi k}{n}\right)\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{2\pi k}{n}\right)\sin\left(\frac{2\pi}{n}\right)\tag{3a}\\
&\le k\sin\left(\frac{2\pi}{n}\right)\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{2\pi k}{n}\right)\sin\left(\frac{2\pi}{n}\right)\tag{3b}\\
&=\left(k\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{2\pi k}{n}\right)\right)\sin\left(\frac{2\pi}{n}\right)\tag{3c}\\
&\le(k+1)\sin\left(\frac{2\pi}{n}\right)\tag{3d}
\end{align}
$$
$\text{(3a):}$ sine of a sum identity
$\text{(3b):}$ apply $(1)$ with $\cos\left(\frac{2\pi}n\right)\ge\cos\left(\frac{2\pi}5\right)=\frac{-1+\sqrt5}4$
$\text{(3c):}$ distribute a product over a sum
$\text{(3d):}$ $\cos(x)\le1$ and $\sin\left(\frac{2\pi}n\right)\ge0$ for $n\ge2$
Inequality $(1)$ is true for $k=1$. Then $(3)$ and induction prove $(1)$ for any $k\ge1$.
| {
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| $$\sum_{i=0}^ni-\sum_{i=0}^{n-1}i=S_1(n)-S_1(n-1)=n,$$
so that $S_1(n)$ must be a polynomial of the second degree in $n$.
By the method of undeterminate coefficients, noting that there is no constant term as $S_1(0)=0$:
$$S_1(n)-S_1(n-1)=n=(an^2+bn)-(a(n-1)^2+b(n-1))=\\=2an+b-a,$$
and by identification
$$a=b=\frac12.$$
$$S_1(n)=\frac{n^2+n}2.$$
Let us generalize to the sum of squares,
$$S_2(n)-S_2(n-1)=n^2=(an^3+bn^2+cn)-(a(n-1)^3+b(n-1)^2+c(n-1))=\\=3an^2+(-3a+2b)n+a-b+c,$$
giving
$$a=\frac13,b=\frac12,c=\frac16.$$
$$S_2(n)=\frac{2n^3+3n^2+n}6.$$
For any power, you get a triangular system of equations where you recognize a part of Pascal's triangle, with alternating signs, as in the sum of cubes:
$$S_3(n)-S_3(n-1)=n^3=4an^3+(-6a+3b)n^2+(4a-3b+2c)n+(-a+b-c+d),$$
$$\color{blue}{4}a=1\\-\color{blue}{6}a+\color{blue}{3}b=0\\\color{blue}{4}a-\color{blue}{3}b+\color{blue}{2}c=0\\-\color{blue}{1}a\ +\color{blue}{1}b-\color{blue}{1}c+\color{blue}{1}d=0,$$
giving
$$a=\frac14,b=\frac12,c=\frac14,d=0.$$
$$S_3(n)=\frac{n^4+2n^3+n^2}4.$$
| {
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"question_score": "136",
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"answer_id": 17
} |
Explain why calculating this series could cause paradox? $$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots
= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$
$$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$
thanks.
| In Matt's answer I began discussing with Matt E over several comments, which I think should be written out as an answer.
As Matt pointed out, this is a rearrangement of this conditionally convergent series which is why you have this sort of paradox.
However it was unclear about how this is exactly a rearrangement, as the equities seems perfectly legal - even for a conditionally convergent series.
*
*$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = 1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2})$ is the first step, which is legal as you simply replace the negative terms by pairs of a positive and negative terms, but you don't change the order of summation from the original series which makes this exchange legit.
*$1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2}) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots - 1 - \frac{1}{2} - \ldots$ this is where things break, you've taken a conditionally convergent series and changed the order, basically we've performed infinitely many commutations in order to rearrange the series into this order, and that is what breaks the summation.
The rearrangement wasn't very obvious, but it was hiding there with its big sharp pointy teeth... and when you stepped too close to its cave - it jumped out at you and bit your head off.
The series in the question is closely reminding me of the one my calculus teacher used when he first showed us what changing conditionally convergent series can do, although his was even less obvious.
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to prove an identity in radicals? (4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
How to prove it is right equality?
I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.
Any ideas?
| Trivial. Lets simplify the left part:
multiply first fraction to the sum of expressions in its denominator, and get whole part in second fraction and do with second fraction the same that we did with first:
$(\frac{4(3+\sqrt{5})}{9-5})^2 - (\frac{25-19 - \sqrt{6}}{5-\sqrt{6}}) = (3+\sqrt{5})^2 - (5-\frac{19(5+\sqrt{6})}{25-6})^2 = (3+\sqrt{5})^2 - 6 = 8 + 6\sqrt{5}$
Now lets prove this:
$8+6\sqrt{5}=2\sqrt{61+24\sqrt{5}}$
divide by two and square:
$4+3\sqrt{5} = \sqrt{61+24\sqrt{5}}$
$16+45+24\sqrt{5}=61+24\sqrt{5}$
Thats all.
| {
"language": "en",
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} |
Partial sum of the series $\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$ An Exercise from Apostol's Introduction to Analytic Number Theory which I am not able to solve.
Let $\mathsf{S_{n}}$ denote the $n$-th partial sum of the series: $$\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$$
Prove that for every integer $k>1$, there exists integers $m$ and $n$, such that $$s_{m}-s_{n} =\frac{1}{k}$$
| (EDIT: My original post took $p$ to be the smallest prime that divides $k$. This is unnecessary. You can take $p$ to be any number other than 1 that divides $k$. Doing so gives you not just one solution but $d(k)-1$ solutions, one for each of the divisors of $k$ other than 1.)
(SECOND EDIT: See the answer to this question for all solutions in $m$ and $n$.)
First,
$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}.$$
Then $$s_n = \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1}.$$
Thus, for any $m,n \geq 1$, $$s_m - s_n = \frac{1}{n+1} - \frac{1}{m+1} = \frac{m-n}{(m+1)(n+1)}.$$
We want to find $m,n$ such that this last expression is equal to $\frac{1}{k}$. Let $p$ be any positive integer other than 1 that divides $k$. Take $m = (p-1)k-1$. Clearly $m$ is a positive integer.
Then we want to show that the resulting $n$ that solves $$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k}$$ is also an integer.
We have
$$\frac{m-n}{(m+1)(n+1)} = \frac{1}{k} \Rightarrow (m-n)k = (m+1)(n+1) $$
$$\Rightarrow ((p-1)k-1 - n)k = (p-1)k(n+1) \Rightarrow (p-1)k-1 - n = (p-1)n + p-1 $$
$$\Rightarrow pn = (p-1)k - p \Rightarrow n = \frac{(p-1)k}{p} - 1,$$
which means that $n$ is an integer because $p|k$.
Thus we have a family of solutions $$m = (p-1)k-1, n = \frac{(p-1)k}{p} - 1,$$
where $p$ is any positive integer other than 1 that divides $k$.
This works even in the case $k = 2$ because then we just get $m = 1$, $n = 0$.
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges? Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
Thought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys
| Another way is
If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$
We have that
$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$
$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} < 2$
Thus $S_n < 2$
thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
} |
An inequality like Riemann sum involving $\sqrt{1-x^2}$ How can I prove that for every positive integer $n$ we have
\begin{equation*}
\frac{n\pi}{4}−\frac{1}{\sqrt{8n}}<\frac{1}{2}+\sum_{k=1}^{n−1}\sqrt{1−\frac{k^2}{n^2}}?
\end{equation*}
| Write the inequality as
$$\frac{\pi}{4} < \frac{1}{2n} + \frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-\left(\frac{k}{n}\right)^2} + \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$
The left-hand side $\pi/4$ is the area of the part of the unit circle that
lies in the first quadrant (below the curve $y=f(x)=\sqrt{1-x^2}$).
We want to interpret the right-hand side as the area of a region $D$ which
covers that quarter circle.
Note that $f$ is concave, so that its graph lies below any tangent line.
Thus the trapezoid bounded by the lines $x=a-\epsilon$ and $x=a+\epsilon$
and by the $x$ axis and the tangent line through $(a,f(a))$ will cover the
corresponding part of the circle:
$$\int_{a-\epsilon}^{a+\epsilon} f(x) dx < 2\epsilon f(a).$$
Thus, taking $D$ to be the union of the following pieces does the trick:
*
*A rectangle of height 1 between $x=0$ and $x=1/2n$.
*Trapezoids as above, of width $\frac{1}{n}$ and centered at $x=k/n$ for $k=1,\ldots,n-1$.
*A trapezoid as above, of width $\frac{1}{2n}$ and centered at $x=1-1/4n$. This last one has area
$$\frac{1}{2n} f(1-1/4n) = \frac{1}{2n} \sqrt{\frac{1}{2n} - \frac{1}{16n^2}} < \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$
| {
"language": "en",
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"source": "stackexchange",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Here's mine. I'm answering late, I know that, but I am still answering it.
We'll use the expansion of $\tanh^{-1}$:
$$\frac{1}{2}\log\frac{1+y}{1-y}=\sum_{n\geq0}\frac{y^{2n+1}}{2n+1},\quad|y|<1$$
We start with this inequality:
$$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx=\int_{-1}^{1}\frac{1}{1+2xy+y^2}dx\,dy$$
The LHS of this equality gives:
$$\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx=\int_{-1}^{1}\frac{\arctan \frac{x+y}{\sqrt{1-x^2}}}{\sqrt{1-x^2}}dx\Biggr|_{y=-1}^{y=1}\\
\quad\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad=\int_{-1}^{1}\frac{\pi}{2\sqrt{1-x^2}}dx=\frac{\pi^2}{2}$$
The RHS of the former equality yields:
\begin{align}
\int_{-1}^{1}\int_{-1}^{1}\frac{1}{1+2xy+y^2}dy\,dx&=\int_{-1}^{1}\frac{\log(1+2xy+y^2)}{2y}dy\Biggr|_{x=-1}^{x=1}\\
&=\int_{-1}^{1}\frac{\log\frac{1+y}{1-y}}{y}dy\\
&=2\int_{-1}^{1}\sum_{n\geq0}\frac{y^{2n}}{2n+1}dy\\
&=4\sum_{n\geq0}\frac{1}{(2n+1)^2}
\end{align}
Hence,
$$\sum_{r\geq0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
Now
$$\frac{3}{4}\zeta(2)=\zeta(2)-\frac{1}{4}\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{m\geq1}\frac{1}{(2m)^2}=\sum_{r\geq0}\frac{1}{(2r+1)^2}=\frac{\pi^2}{8}$$
Solving this we get
$$\zeta(2)=\frac{\pi^2}{6}$$
as desired.
Source:https://www.emis.de/journals/GM/vol16nr4/ivan/ivan.pdf
Here are more proofs.
| {
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} |
Evaluation of Gaussian integral $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx$ How to prove
$$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$
| Here's a proof that only requires elementary, but clever, calculus manipulations.
Let $I_A = \int_0^A e^{-x^2}dx$.
$$\begin{split} I_A^2 &= \int_0^A \int_0^A e^{-(x^2+y^2)}dxdy\\
&= \int_0^A \int_0^A \sum_{n\geq 0} \frac{(-1)^n}{n!}(x^2+y^2)^ndxdy\\
&=\sum_{n\geq 0} \frac{(-1)^n}{n!}\int_0^A \int_0^A\sum_{k=0}^n{n \choose k}x^{2k}y^{2n-2k}dxdy\\
&=\sum_{n\geq 0} \frac{(-1)^n}{n!}\sum_{k=0}^n{n \choose k}\frac{A^{2k+1}}{2k+1}\frac{A^{2n-2k+1}}{2n-2k+1}\\
&= \sum_{n\geq 0} \frac{(-1)^n}{n!}A^{2n+2}\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}\frac{1}{2n-2k+1}
\end{split}$$
Now, note that
$$\begin{split}
\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}\frac{1}{2n-2k+1} &= \frac 1 {2n+2}\sum_{k=0}^n{n \choose k}\left (\frac{1}{2k+1}+\frac{1}{2n-2k+1}\right)\\
& =\frac 1 {n+1}\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}
\end{split}
$$
Thus,
$$\begin{split} I_A^2 &=
\sum_{n\geq 0} \frac{(-1)^n}{(n+1)!}A^{2n+2}\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}\\
&= \sum_{n\geq 0} \frac{(-1)^n}{(n+1)!}A^{2n+2}\sum_{k=0}^n{n \choose k}\int_0^1x^{2k}dx\\
&= \sum_{n\geq 0} \frac{(-1)^n}{(n+1)!}A^{2n+2}\int_0^1(1+x^2)^ndx\\
&= \int_0^1\frac 1 {1+x^2}\sum_{n\geq 0}\frac{(-1)^n}{(n+1)!}A^{2n+2}(1+x^2)^{n+1}dx\\
&= \int_0^1\frac{1-e^{-A^2(1+x^2)}}{1+x^2}dx\\
&= \frac{\pi}4 + \mathcal O\left(e^{-A^2}\right)
\end{split}$$
Taking the limit as $A\rightarrow+\infty$ yields the result.
| {
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Prove by induction $T(n) = 2T(\frac{n}{2}) + 2$ I'm stuck with this induction proof:
So far, given:
$\begin{align*}
T(1) & = 2 \\
T(n) & = 2T(n/2)+2 \\
& = 2(2T(n/[2^2])+2) + 2 \\
& = [2^2]T(n/[2^2]) + [2^2] + 2 \\
& = [2^2](2T(n/[2^2])+2) + [2^2] + 2 \\
& = [2^3]T(n/[2^3]) + [2^3] + [2^2] + 2 \\
& = [2^3]T(n/[2^3]) + 2\{[2^2] + [2^1] + 1\} \\
& \vdots \\
& = [2^k]T(n/[2^k]) + 2\{2^{k} - 1\}
\end{align*}$
How then do I show this to be correct (the proof). So far I have:
Let $(n/[2^k]) = 1$
$\Rightarrow n = 2^k$
So, $T(n) = nT(1) + 2(n - 1)$
$T(n) = 4n - 2$ //This is where I'm stuck.
Proof (by induction):
When $n = 1$, $T(1) = 2$.
Assume $T(k)$ is true [$T(n) = 4n - 2$] //This is where I am stuck.
| Here is how I attempt this from the point you left off:
We know that $\rm T(1) = 2$. We are trying to prove that $\rm T(n) = 4n-2$.
This is trivially true for $n = 1$:
$ \rm \begin{eqnarray*}
T(1) &=& 4(1) - 2\\
&=& 4 - 2\\
&=& 2 \\
\end{eqnarray*}
$
Assume
$\rm T(k) = 4k - 2 $
From the original definition:
$\rm T(k+1) = 2T( [k+1] / 2 ) + 2 $
//since we assumed up to $T(k)$ is correct and $(k+1)/2$ is less than $T(k)$, we substitute:
So, We have
$ \rm \begin{eqnarray*}
&2&( 4( (k+1) /2) - 2 ) + 2 \\
&=& 4(k+1)- 4 + 2\\
&=& 4(k+1) - 2
\end{eqnarray*}
$
Proven.
| {
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Inequality of three variable polynomial I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that
$$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
How this can be done? Is it possible to prove stronger inequalities like
$$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$
or even
$$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
without computer?
| You want to apply AM-GM to some of the terms on the LHS to get some of the terms on the RHS. A naive way to do this such as
$$b^6 + c^6 + 2a^2 b^3 c \ge 4 a b^3 c^2$$
uses too many of the $11$ terms relative to the $40$ terms and only proves the weaker inequality
$$a^6 + b^6 + c^6 + abc(ab^2 + bc^2 + ca^2) \ge 2abc(a^2 b + b^2 c + c^2 a)$$
after cyclic summation. A slightly better combination of terms is
$$5a^6 + c^6 + 12a^2 b^3 c \ge 18 a^3 b^2 c$$
but this still uses too many of the $11$ terms relative to the $40$ terms and proves the inequality
$$a^6 + b^6 + c^6 + 2abc(ab^2 + bc^2 + ca^2) \ge 3abc(a^2 b + b^2 c + c^2 a)$$
after cyclic summation. You get a strong enough result by mixing some of the $40$ terms with each other to get
$$4a^6 + 12a^2 b^3 c + 4a^3 b c^2 \ge 20 a^3 b^2 c$$
which proves
$$a^6 + b^6 + c^6 + 4abc(ab^2 + bc^2 + ca^2) \ge 5abc(a^2 b + b^2 c + c^2 a).$$
Now multiply by $10$ and apply AM-GM to the remaining terms to conclude. The question of whether one can do better than this using AM-GM is a geometric question about using convex linear combinations of the vectors $(6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 3, 1), (3, 1, 2), (1, 2, 3)$ to obtain the vectors $(3, 2, 1), (2, 1, 3), (1, 3, 2)$ and one might be able to prove that the above result is optimal.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An inequality on Cevians Let $\displaystyle AD$, $\displaystyle BE$, $\displaystyle CF$ be three cevians concurrent at $\displaystyle P$ inside the $\displaystyle \Delta ABC$.
Prove or disprove that:
$$\displaystyle \dfrac{AD}{AP} + \dfrac{BE}{BP} + \dfrac{CF}{CP} \ge \dfrac{9}{2}$$
| The inequality is true!
It can be shown that (see proof at the end of the answer)
$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$
Note that this implies that
$$\displaystyle \frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF} = 2$$
as $\displaystyle 1 - \frac{PD}{AD} = \frac{AP}{AD}$ etc.
Now we have the inequality (easily shown using $\text{AM} \ge \text{GM}$) that
$$\displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9$$
This shows that
$$\displaystyle (\frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF})(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$
and so
$$\displaystyle 2(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$
i.e.
$$\displaystyle \frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP} \ge \frac{9}{2}$$
Note that the equality occurs only when $\displaystyle \frac{AP}{AD} = \frac{BP}{BE} = \frac{CP}{CF} = \frac{2}{3}$, which implies that $\displaystyle P$ is the centroid.
Proof
Let us try showing that
$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$
Consider the figure (repeated from above for convenience).
Note, if you are worried about acute triangle vs obtuse etc, a simple affine transformation will do to transform the triangle into an equilateral triangle.
Let X be the foot of perpendicular from A to BC and Y be the foot of the perpendicular from P to BC.
$\displaystyle \triangle AXD$ and $\triangle PYD$ are similar and thus
$\displaystyle \frac{PY}{AX} = \frac{PD}{AD}$.
Now $\displaystyle \frac{PY}{AX} = \frac{|\triangle PBC|}{|\triangle ABC|}$
where $\displaystyle |\triangle MNO|$ is the area of $\displaystyle \triangle MNO$.
Thus $\displaystyle \frac{PD}{AD} = \frac{|\triangle PBC|}{|\triangle ABC|}$
Similarly
$\displaystyle \frac{PE}{BE} = \frac{|\triangle PAC|}{|\triangle ABC|}$
$\displaystyle \frac{PF}{CF} = \frac{|\triangle PAB|}{|\triangle ABC|}$
Adding gives us
$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = \frac{|\triangle PAB| + |\triangle PAC| + |\triangle PBC|}{|\triangle ABC|} = \frac{|\triangle ABC|}{|\triangle ABC|} = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find the derivative of $y = \sin(\arctan x) + \tan(\arcsin x)$? My question is, how can I solve the following derivative question?
$$y = \sin(\arctan x) + \tan(\arcsin x)$$
Thanks in advance
| First you want to simplify $\tan (\arcsin x)$ and $\sin(\arctan x)$. Let me show
you the latter one.
Suppose $\theta=\arctan x$. Then by definition $-\pi/2<\theta<\pi/2$ and $\tan\theta=x$.
Since $\sec \theta>0$ (because $-\pi/2<\theta<\pi/2$), and $$\sec^2\theta=1+\tan^2 \theta =1+x^2,$$ you get
$\sec \theta=\sqrt{1+x^2}$, so $\cos \theta =1/\sqrt{1+x^2},$ and
$$\sin\theta=\tan\theta \cos\theta =\frac{x}{\sqrt{1+x^2}},$$
i.e. $\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}$.
Similarly, you can prove $\tan (\arcsin x)=\frac{x}{\sqrt{1-x^2}}.$
Using quotient rule, you can prove that the derivative of $\frac{x}{\sqrt{1+x^2}}$
and $\frac{x}{\sqrt{1-x^2}}$ are
$$\frac{1}{(1+x^2)^{3/2}} \mbox{ and }\frac{1}{(1-x^2)^{3/2}} $$
respectively. Hence the derivative of your function is $$\frac{1}{(1+x^2)^{3/2}}+\frac{1}{(1-x^2)^{3/2}}. $$
${\bf Edit.}$ Or you can use chain rule to get:
$$\cos(\arctan x)\frac{1}{1+x^2}+\sec^2(\arcsin x)\frac{1}{\sqrt{1-x^2}}$$
Now simplify it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.
Please give an explanation of how the answer
| $\rm\ 0 = a\ x^3 + b\ x^2 + 25 - (x^2 - 4\ x + 5)\ (a\ x + 5) = (4\ a + b - 5)\ x^2 + (20 - 5\ a)\ x\ \Rightarrow \ a,\: b = \ldots$
Note: $\rm\ a\ x + 5\ $ comes from comparing leading and constant coefficients.
| {
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Solve $\cos(\theta) + \sin(\theta) = x$ for known $x$, unknown $\theta$? After looking at the list of trigonometric identities, I can't seem to find a way to solve this. Is it solvable?
$$\cos(\theta) + \sin(\theta) = x.$$
What if I added another equation to the problem:
$$-\sin(\theta) + \cos(\theta) = y,$$
where $\theta$ is the same and $y$ is also known?
Thanks.
EDIT:
OK, so using the linear combinations I was able to whip out:
$$a \sin(\theta) + b \cos(\theta) = x = \sqrt{a^2 + b^2} \sin(\theta + \phi),$$
where $\phi = \arcsin \left( \frac{b}{\sqrt{a^2 + b^2}} \right) = \frac{\pi}{4}$ (as long as $a\geq 0$)
Giving me:
$$x = \sin(\theta + \frac{\pi}{4}) \text{ and } \arcsin(x) - \frac{\pi}{4} = \theta.$$
All set! Thanks!
| Another method goes by noting that $\cos^2\theta+\sin^2\theta=1$. We have $\cos\theta+\sin\theta=x$, so $$\cos^2\theta+2\cos\theta\sin\theta+\sin^2\theta=x^2,$$ or $2\cos\theta\sin\theta=x^2-1$. But $2\cos\theta\sin\theta=\sin(2\theta)$, so $2\theta=\sin^{-1}(x^2-1)$, or $$ \theta=\frac12\sin^{-1}(x^2-1).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluate $\int \cos^3 x\;\sin^2 xdx$ Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.
$$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx
= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$
let $u = \sin x$, $du = \cos xdx$
$$\int(1-u^2)u^2du = \int(u^2-u^4)du
= \frac{u^3}{3} - \frac{u^5}{5} +C$$
Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C
| \begin{align*}
\frac{\sin^3 (x)}{3} - \frac{\sin^5 (x)}{5} &= \sin^3 (x) (\frac{1}{3} - \frac{\sin^2 (x)}{5});\\
\cos(2x) &= 1 - 2 \sin^2(x)\\
\sin^2(x) &= \frac{1- \cos(2x)}{2}.
\end{align*}
Hence, we get,
\begin{align*}
\sin^3 (x) \left(\frac{1}{3} - \frac{\sin^2 (x)}{5}\right) &= \sin^3 (x) \left(\frac{1}{3} - \frac{1 - \cos(2x)}{10}\right)\\
& = \sin^3(x) \left(\frac{10 - 3 + 3 \cos(2x)}{30}\right)\\
& = \frac{\sin^3(x)}{30} (3 \cos(2x)+7)
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/21589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
Find lowest integer value with integer x of given function I need to find integer x with which function's y gets lowest integer values
$$f(x)=\frac{x^2-x-17}{x-2}$$
I tried to find derivative, but it never equals 0.
Other steps was to change expression
$$f(x)=\frac{x^2-x-17}{x-2}=1+\frac{(x-5)(x+3)}{x-2}=x-2+\frac{3(x-7)}{x-2}$$
But didn't notice any solutions
| For $n < -13$ we have $n+1 < f(n) < n+2$ and so $f(n)$ is not an integer for integer $n$.
For $n > 17$ we have $n < f(n) < n+1$ amd so $f(n)$ is not an integer for integer $n$.
So you just have to check the 31 integers remaining. The integer values are $f(-13)=-11$, $f(-3)=1$, $f(-1)=5$, $f(1)=17$, $f(3)=-11$, $f(5)=1$, $f(7)=5$, $f(17)=17$. So the integer minima are $-11$ when $n=-13$ or $n=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
| Consider the contour integral
$$
\oint_C\frac{\pi\cot\pi z}{z^4}\ dz
$$
where $C$ is the counter-clockwise square contour centered at origin with vertices $\left(N+\frac12\right)(\pm1\pm i)$.
Lemma
Suppose that the function $\phi(z)$ is holomorphic at $z=n\in\mathbb{Z}$ with $\phi(n)\neq0$, then $\pi\phi(z)\cot\pi z$ has a simple pole at $n$ with residue $\phi(n)$.
Proof
Note that $\tan\pi z$ have simple zeros at $z=n$, hence $\pi\phi(z)\cot\pi z$ have simple poles there and
$$
\text{Res}\left[\pi\phi(z)\cot\pi z\ ;\ n\right]=\text{Res}\left[\frac{\pi\phi(z)}{\tan\pi z}\ ;\ n\right]=\frac{\pi\phi(n)}{\pi\sec^2\pi n}=\phi(n).
$$
Thus, by residue theorem for $z\neq0$ we obtain
$$
\sum_{n=-N}^N\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=n\right]=\sum_{n=-N,\ n\neq0}^N\frac1{n^4}=2\sum_{n=1}^N\frac1{n^4}.\tag1
$$
From the Taylor series of $\cot\pi z$ at $z=0$ we obtain
\begin{align}
\frac{\pi\cot\pi z}{z^4}&=\frac\pi{z^4}\cos\pi z\csc\pi z\\
&=\frac\pi{z^4}\left(1-\frac{(\pi z)^2}{2!}+\frac{(\pi z)^4}{4!}-\frac{(\pi z)^6}{6!}+\cdots\right)\left(\frac1{\pi z}+\frac{\pi z}{6}+\frac{7(\pi z)^3}{360}+\cdots\right)\\
&=\frac1{z^5}\left(1-\frac{(\pi z)^2}{2!}+\frac{(\pi z)^4}{4!}-\frac{(\pi z)^6}{6!}+\cdots\right)\left(1+\frac{(\pi z)^2}{6}+\frac{7(\pi z)^4}{360}+\cdots\right)\\
\end{align}
Expanding the series above, we see that
$$
\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=0\right]=-\frac{\pi^4}{2!\cdot6}+\frac{\pi^4}{4!}+\frac{7\pi^4}{360}=-\frac{\pi^4}{45}.\tag2
$$
Observe that at any point on the boundary, we have
$$
\left|\frac{\pi\cot\pi z}{z^4}\right|\le\frac{\pi\coth\frac\pi2}{\left(N+\frac12\right)^4}.\tag3
$$
Proof
Putting $z=x+iy$ and using the trigonometric sum formulas and basic identities, we have
$$
|\cot\pi z|^2=\left|\frac{\cos\pi z}{\sin\pi z}\right|=\frac{\sinh^2\pi y+\cos^2\pi x}{\cosh^2\pi y-\cos^2\pi x}.
$$
On the vertices sides of contour $C$, we have $x=\pm\left(N+\frac12\right)$ giving $\cos\left(N+\frac12\right)\pi=0$, hence
$$
|\cot\pi z|=|\tanh\pi y|\le1.
$$
On the horizontal sides we have $0\le\cos^2\pi x\le1$, hence
$$
|\cot\pi z|^2\le\frac{\sinh^2\pi y+1}{\cosh^2\pi y-1}=\frac{\cosh^2\pi y}{\sinh^2\pi}=\coth^2\pi y.
$$
Therefore
$$
|\cot\pi z|\le\coth\pi y=\coth\left(N+\frac12\right)\pi\le\coth\frac\pi2
$$
Thus, on the boundary of contour $C$ we have
$$
|\cot\pi z|\le\max\left[1,\coth\frac\pi2\right]=\coth\frac\pi2
$$
From $(3)$ and the property
$$
\left|\int_C f(z)\ dz\right|\le ML,
$$
where $M$ is $\max|f(z)|$ on C and $L$ is the length of $C$, we obtain
$$
\oint_C\frac{\pi\cot\pi z}{z^4}\ dz\le\frac{\pi\coth\frac\pi2}{\left(N+\frac12\right)^4}\cdot8\left(N+\frac12\right)=\frac{8\pi\coth\frac\pi2}{\left(N+\frac12\right)^3}\to0
$$
as $N\to\infty$. Thus, using $(1)$ and $(2)$ also using residue theorem we obtain
\begin{align}
\lim_{N\to\infty}\frac{1}{2\pi i}\oint_C\frac{\pi\cot\pi z}{z^4}\ dz&=0\\
\sum_{n=-\infty}^\infty\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=n\right]+\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=0\right]&=0\\
2\sum_{n=1}^\infty\frac1{n^4}-\frac{\pi^4}{45}&=0\\
\large\color{blue}{\sum_{n=1}^\infty\frac1{n^4}}&\large\color{blue}{=\frac{\pi^4}{90}}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
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Can this integral $\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$ be written in the form of a elliptic integral I am trying to find the magnetic field due to an elliptic loop of wire. How to do integrals of the type $$\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$$ Where a and b are the axes of the ellipse. From dimensional considerations in the problem, I have deduced that the value of the integral must be must reduced to $$\frac{ L}{(ab)^2}$$ where $L$ is the circumference of the ellipse. How do I show it.
OK. So now I know from wikipedia that the circumference of ellipse involves something called elliptic integrals, so thankfully, I wouldnt have to be looking for a closed form. But does this reduce to the elliptic integral? Or have I done something wrong?
| (I assume that $a > b$ in this answer.)
Hans's answer covers the OP's needs, but I'll try out the pedestrian approach here.
First, a trivial simplification:
$$\int_0^{2\pi} \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}=\int_0^\pi \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}+\int_\pi^{2\pi} \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}$$
$$=\int_0^\pi \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}+\int_0^{\pi} \frac{\mathrm d\theta}{(a^2 \cos^2(\theta+\pi)+b^2\sin^2(\theta+\pi))^{3/2}}$$
$$=2\int_0^\pi \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}$$
Another simplification can be done as follows:
$$=2\int_{-\pi/2}^{\pi/2} \frac{\mathrm d\theta}{(a^2 \cos^2(\theta+\pi/2)+b^2\sin^2(\theta+\pi/2))^{3/2}}$$
$$=2\int_{-\pi/2}^{\pi/2} \frac{\mathrm d\theta}{(a^2 \sin^2\theta +b^2\cos^2\theta)^{3/2}}$$
or, since the integrand is an even function,
$$=4\int_{-\pi/2}^0 \frac{\mathrm d\theta}{(a^2 \sin^2\theta +b^2\cos^2\theta)^{3/2}}$$
$$=4\int_0^{\pi/2} \frac{\mathrm d\theta}{(a^2 \cos^2\theta +b^2\sin^2\theta)^{3/2}}$$
Now, to get the integral into something recognizable, we use the Pythagorean identity and then try to factor out constants, like so:
$$=4\int_0^{\pi/2} \frac{\mathrm d\theta}{(a^2(1-\sin^2\theta)+b^2\sin^2\theta)^{3/2}}$$
$$=4\int_0^{\pi/2} \frac{\mathrm d\theta}{(a^2-(a^2-b^2)\sin^2\theta)^{3/2}}$$
$$=\frac4{a^3}\int_0^{\pi/2} \frac{\mathrm d\theta}{(1-m\sin^2\theta)^{3/2}}$$
where we let $m=1-\frac{b^2}{a^2}$, and then introduce the substitution $\theta=\mathrm{am}(u|m)$, $\mathrm d\theta=\mathrm{dn}(u|m)\mathrm du$ (where $\mathrm{am}(u|m)$ is the Jacobian amplitude and $\mathrm{dn}(u|m)$ is a Jacobian elliptic function):
$$=\frac4{a^3}\int_0^{K(m)} \frac{\mathrm{dn}(u|m)\mathrm du}{(1-m\mathrm{sn}^2(u|m))^{3/2}}$$
(since $\sin(\mathrm{am}(u|m))=\mathrm{sn}(u|m)$). Using the identity $\mathrm{dn}^2(u|m)+m\mathrm{sn}^2(u|m)=1$, the integral turns into
$$=\frac4{a^3}\int_0^{K(m)} \frac{\mathrm du}{\mathrm{dn}^2(u|m)}=\frac4{a^3}\int_0^{K(m)} \mathrm{nd}^2(u|m)\mathrm du$$
Using formula 22.16.20 in the DLMF, the integral evaluates to
$$\frac4{a^3}\frac{E(m)}{1-m}=\frac4{ab^2}E\left(1-\frac{b^2}{a^2}\right)$$
where $E(m)$ is the complete elliptic integral of the second kind, and since $L=4aE\left(1-\frac{b^2}{a^2}\right)$, your supposition is correct.
| {
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More Computing Integrals This particular problem has been giving me trouble, and while the math dept tutors did help a great deal, the resulting answer hasn't been accepted by the online homework submission website. Find the definite integral of
$$\int\frac{x(x+1)}{2x^{3}+3x^{2}-13}$$
The work done so far with the help of the tutors is
$$\int\frac{x(x+1)}{2x^{3}+3x^{2}-13} = \int\frac{x^{2}+x}{2x^{3}+3x^{2}-13}$$
Let u=$2x^{3}+3x^{2}-13$
and ${u}'= 6x^{2}+6x = 6(x^{2}+x)$
$$\int\frac{1}{6}\cdot\frac{{u}'}{u}= \frac{1}{6}\int\frac{{u}'}{u}$$
$$\frac{1}{6}\ln~u= \frac{1}{6}\ln(2x^{3}+3x^{2}-13)+C$$
What is missing from this solution?
| What's missing is what you're integrating with respect to. You need to put a $ dx $ in the first 3 integrals of your post (and a $ du $ in any integrals involving $ u $).
Our integral is $ \int \frac{x^2 + x}{2x^3 + 3x^2 - 13} \ dx $. We make the substitution $ u = 2x^3 + 3x^2 - 13 $. This gives us $ \frac{du}{dx} = 6x^2 + 6x = 6(x^2 + x) $, and so $ dx = \frac{du}{6(x^2+x)} $. When we substitute this into the integral, the $ x^2 + x $ cancels to give $ \int \frac{1}{6} \frac{1}{u} \ du $. This evaluates to $ \frac{1}{6} \ln{|u|} + C $. We then substitute back in for $ x $, to leave $ \frac{1}{6}\ln{|2x^3 + 3x^2 - 13|} + C.$
Note that a substitution wasn't necessary. In general, $ \int \frac{f'(x)}{f(x)} \ dx = \ln{|f(x)|} + C $. In this case, we have something of the form $ \int \frac{ k f'(x)}{f(x)} \ dx $, where $ k $ is a constant. So we can integrate directly to give $ \int \frac{ k f'(x)}{f(x)} \ dx = k \ln{|f(x)|} + D $. You just need to spot that the numerator of the integrand is a constant multiple of the derivative of the denominator.
| {
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A differential equation (nonlinear First-Order) how to solve this equation:
$(Px-y)(Py+x)=h^2P$
that $P=\frac{dy}{dx}$
and $h$ is a constant.
| Assume $h\neq0$ for the key case:
Let $u=x^2+y^2$ ,
Then $\dfrac{du}{dx}=2x+2y\dfrac{dy}{dx}$
$\therefore\left(\dfrac{x}{2y}\left(\dfrac{du}{dx}-2x\right)-y\right)\dfrac{1}{2}\dfrac{du}{dx}=\dfrac{h^2}{2y}\left(\dfrac{du}{dx}-2x\right)$
$\left(x\left(\dfrac{du}{dx}-2x\right)-2y^2\right)\dfrac{du}{dx}=2h^2\left(\dfrac{du}{dx}-2x\right)$
$x\dfrac{du}{dx}-2x^2-2y^2=2h^2\left(1-\dfrac{2x}{\dfrac{du}{dx}}\right)$
$x\dfrac{du}{dx}-2u=2h^2-\dfrac{4h^2x}{\dfrac{du}{dx}}$
$2u+2h^2=x\dfrac{du}{dx}+\dfrac{4h^2x}{\dfrac{du}{dx}}$
$u+h^2=\dfrac{x}{2}\dfrac{du}{dx}+\dfrac{2h^2x}{\dfrac{du}{dx}}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore u+h^2=x^2\dfrac{du}{dv}+\dfrac{h^2}{\dfrac{du}{dv}}$
$u+h^2=v\dfrac{du}{dv}+\dfrac{h^2}{\dfrac{du}{dv}}$
Let $s=u+h^2$ ,
Then $\dfrac{ds}{dv}=\dfrac{du}{dv}$
$\therefore s=v\dfrac{ds}{dv}+\dfrac{h^2}{\dfrac{ds}{dv}}$
$s\dfrac{dv}{ds}=v+\dfrac{h^2}{\left(\dfrac{ds}{dv}\right)^2}$
$v=s\dfrac{dv}{ds}-h^2\left(\dfrac{dv}{ds}\right)^2$
Which reduces to Clairaut's ODE.
$\dfrac{dv}{ds}=s\dfrac{d^2v}{ds^2}+\dfrac{dv}{ds}-2h^2\dfrac{dv}{ds}\dfrac{d^2v}{ds^2}$
$\dfrac{d^2v}{ds^2}\left(2h^2\dfrac{dv}{ds}-s\right)=0$
$\therefore\begin{cases}\dfrac{d^2v}{ds^2}=0\\2h^2\dfrac{dv}{ds}-s=0\end{cases}$
$\begin{cases}v=as+b\\v=\dfrac{s^2}{4h^2}+c\end{cases}$
$\therefore\begin{cases}as+b=as-h^2a^2\\\dfrac{s^2}{4h^2}+c=\dfrac{s^2}{2h^2}-\dfrac{s^2}{4h^2}\end{cases}$
$\begin{cases}b=-h^2a^2\\c=0\end{cases}$
$\therefore\begin{cases}v=as-h^2a^2\\v=\dfrac{s^2}{4h^2}\end{cases}$
$\begin{cases}x^2=au+h^2a-h^2a^2\\x^2=\dfrac{(u+h^2)^2}{4h^2}\end{cases}$
$\begin{cases}x^2=ax^2+ay^2+h^2a-h^2a^2\\x^2=\dfrac{(x^2+y^2+h^2)^2}{4h^2}\end{cases}$
$\begin{cases}(a-1)x^2+ay^2=h^2a^2-h^2a\\x^2=\dfrac{(x^2+y^2+h^2)^2}{4h^2}\end{cases}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence relation (linear, second-order, constant coefficients) Q1. Find the general solution to the difference equation
$$ a_{n} - 4a_{n-1} + 3a_{n-2} = 6 $$
Q2. Solve the difference equation
$$ a_{n} - a_{n-1} - 2a_{n-2} = 0, a_0 = 2, a_1 = 4 $$
I am completely lost in solving recurrence relation questions. Can anyone guide me in steps to solve the following 2 questions?
| Simple, general way of disposing such equations is using generating functions.
Define:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
Write the recurrence so it hasn't subtractions in indices:
$$
a_{n + 2} - 4 a_{n + 1} + 3 a_n = 6
$$
Multiply by $z^n$, sum over $n \ge 0$ an recognize the resulting terms:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} - 4 \frac{A(z) - a_0}{z} + 3 A(z) = 6 \frac{1}{1 - z}
$$
Substituting your initial values, and solving for $A(z)$, writing that as partial fractions:
$$
A(z) = \frac{3}{(1 - z)^3} + \frac{5}{2 (1 - z)} + \frac{5}{2 (1 - 3 z)}
$$
Using the generalized binomial theorem you can read off the coefficients:
\begin{align}
a_n &= 3 \binom{-3}{n} (-1)^n + \frac{5}{2} + \frac{5}{2} \cdot 3^n \\
&= 3 \binom{n + 3 - 1}{3 - 1} + \frac{5}{2} + \frac{5}{2} \cdot 3^n \\
&= \frac{3 (n + 2) (n + 1)}{2} + \frac{5}{2} + \frac{5}{2} \cdot 3^n \\
&= \frac{5 \cdot 3^n + 3 n^2 + 9 n + 11}{2}
\end{align}
The other one I leave as a exercise for the gentle reader.
| {
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Where are these additional solutions coming from? Solve for $x$: $2\sin(2x)-\sqrt{2} = 0$ in interval $[0,2\pi)$
Step $1$: Add $\sqrt{2}$ and divide by $2$ to get $\sin(2x) = \dfrac{\sqrt{2}}{2}$
Step $2$: Set $2x$ equal to the angles where $\sin(x) = \dfrac{\sqrt{2}}{2}$:
$2x = \dfrac{\pi}{4}$ and $2x = \dfrac{3\pi}{4}$
Step $3$: Solve for $x$ by dividing by $2$: $x = \dfrac{\pi}{8}$ and $x = \dfrac{3\pi}{8}$
My textbook also lists $\dfrac{9\pi}{8}$ and $\dfrac{11\pi}{8}$ as additional solutions, anyone know where they may have came from? thanks
| $x\in[0,2\pi)$ means $2x\in[0,4\pi)$, so you have find the solutions in this larger interval.
| {
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} |
Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
| The problem is:
$$x+5 = (x - 1)^2$$
doesn't imply
$$\sqrt{x + 5} = x - 1$$
but
$$\sqrt{x + 5} = \vert x - 1\vert$$
| {
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} |
Confused by textbook solution to trig problem The following is part of a question and the solution from my textbook:
Question
Given that $2 \sin{2\theta} = \cos{2\theta}$, show that $\tan{2\theta} = 0.5$
Solution
$2 \sin{2\theta} = \cos{2\theta}$
$\Rightarrow \frac{2 \sin{2\theta}}{\cos{2\theta}} = 1$
$\Rightarrow 2 \tan{2\theta} = 1~~~\tan{2\theta} = \frac{\sin{2\theta}}{\cos{2\theta}}$
I am unsure about this part: Since it is possible that $\cos(2\theta) = 0$, isn't it bad form to divide by $\cos(2\theta)$?
| If $\cos(2\theta) = 0$, then solving for $\theta$ we get one value of $\theta =\frac{\pi}{4}$, which says that the value of $$2 \sin{2\theta} = 2 \sin\frac{\pi}{2} = 2 \neq 0$$
which doesn't satisfy the given hypothesis $2 \sin{2\theta} = \cos{2 \theta}$.
| {
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Limit of eigenvectors versus eigenvectors of limit Consider the matrix
$$
A=\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & g
\end{pmatrix},
$$
where $g$ is a real parameter. If I set $g=0$ and calculate the normalized eigenvectors of $A|_{g=0}$ with Mathematica, I find that they are
$$
v_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\
v_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},\
v_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.
$$
If instead I calculate the eigenvectors of $A$ leaving $g$ as an unknown and then take their limit as $g\to 0$, I find
$$
u_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\
u_2 = \frac{1}{\sqrt{6}}\begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix},\
u_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.
$$
My question is, why are these two sets of eigenvectors different?
| Both $(-1,1,0)^T$ and $(-1,0,1)^T$ are eigenvectors of $-1$; so is $(-1,-1,2)^T$, as
$$(-1,-1,2) = -(-1,1,0) + 2(-1,0,1).$$
You are just taking a different basis for the eigenspace corresponding to $-1$.
It's likely just an artifact of how Mathematica finds a basis for the eigenspace; the eigenvalues of the matrix are $-1$,
$$\frac{1+g+\sqrt{g^2-2g+9}}{2},$$
and
$$\frac{1+g-\sqrt{g^2-2g+9}}{2}$$
so that there is, up to sign, only one normal eigenvector for each eigenvalue when $g\neq 0$ (note the quadratic in the square root is always positive, so those two eigenvalues never coincide, and neither is equal to $-1$ unless $g=0$). But at the limit you end up with a matrix that has a repeated eigenvalue (corresponding to $\lambda=-1$) and in that case you have many different ways of obtaining a basis.
| {
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Transforming power series Definition:
Let $$a_0 = a_1 = 1, \; a_{n+2} = a_{n+1} + (n+1) \cdot a_n, \; n \geq 0$$
Exercise:
Prove that $$\sum_{n\geq 0} \frac{a_n}{n!} x^n = \exp \left( x + \frac{1}{2} x^2 \right)$$
I don't know how to start with this. I do know that $$e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n, \; e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ but how do I get this wrapped up?
Thank you in advance!
| Let $$S(x)= \sum_{n\ge 0} \frac{a^n}{n!}x^n$$
and note that
$$S'(x)= \sum_{n\ge 1} \frac{a^n}{(n-1)!}x^{n-1}$$
Replacing in the first equation the value of $a_n$ given by the recursion, $a_{n}=(a_{n+2}-a_{n+1})/(n+1)$ we get
$$S(x) = S_2(x) - S_1(x)$$
where
$$S_1(x) = \sum_{n\ge 0} \frac{a^{n+1}}{(n+1)!}x^n =\frac{1}{x} [ S(x)-a_0 ]$$
and
$$S_2(x) = \sum_{n\ge 0} \frac{a^{n+2}}{(n+1)!}x^n =\frac{1}{x} [ S'(x)-a_1 ]$$
What gives
$$S(x) \; \left( x+1 \right) = S'(x) + a_0 - a_1 = S'(x)$$
This first order differential equation is straightforward:
$$\frac{S'(x)}{S(x)} = \log'{S(x)} = x+1 \Rightarrow \log{S(x)} = \frac{1}{2}x^2+x +C$$
which, together with the initial condition $S(0)=1$, gives $S(x)=\exp(x+x^2/2)$.
| {
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How to find remainder modulo $n$, when $n$ is a large number I am doing RSA questions and I really could use help! Can someone show me a simple way to find $25^9 \pmod{33}$?
| In the case of $25^9 \pmod{33}$ it is pretty straightforward to use exponentiation by squaring because $9 = 8 + 1 = 2^3 + 2^0$.
$25 \equiv 25 \pmod{33}$
$25^2 \equiv 31 \pmod{33}$
$25^4 \equiv 31^2 \equiv 4 \pmod{33}$
$25^8 \equiv 4^2 \equiv 16 \pmod{33}$
$25^9 \equiv 25^8 \times 25 \equiv 16 \times 25 \equiv 4 \pmod{33}$
| {
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
| Firstly, calculate the sum $S=\sum_{k=1}^n k(k+1)$ which is:
$S=1\cdot 2+2\cdot 3+ \cdots +n(n+1)$,
multiplying $S$ by 3
we get:
$3S=1\cdot 2\cdot 3+2\cdot 3\cdot (4-1)+3\cdot 4\cdot (5-2)+ \cdots +n\cdot (n+1)\cdot (n+2-(n-1))$
$3S=1 · 2 · 3 + 2 · 3 · 4 − 1 · 2 · 3 + 3 · 4 · 5 − 2 · 3 · 4 + · · ·
+ n(n + 1)(n + 2) − (n − 1)n(n + 1)$
This telescoping series collapses to yield:
$$3S=n(n+1)(n+2)$$
$$S=\frac{n(n+1)(n+2)}{3}$$
On the other side we have:
\begin{alignat*}{2}
&\sum_{k=1}^n k(k+1)&&=\sum_{k=1}^n k^2+k \\
& &&=\sum_{k=1}^n k^2+\sum_{k=1}^n k \\
&\frac{n(n+1)(n+2)}{3}&&=\sum_{k=1}^n k^2+\frac{n(n+1)}{2} \\
&\sum_{k=1}^n k^2&&=\frac{n(n+1)(n+2)}{3}-\frac{n(n+1)}{2} \\
&\sum_{k=1}^n k^2&&=\frac{n(n+1)(2n+1)}{6}
\end{alignat*}
| {
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| My favorite technique for proving symmetric inequalities of positive numbers (particularly if you have a computer algebra package) is to note that if the inequality is symmetric, then w.l.o.g. we can assume the variables are in sorted order, then rewrite the inequality using the smallest variable and the consecutive differences, expand everything algebraically and note that all the coefficients are positive.
Using the example at hand
$(x + y + z)^3 - 27 x y z \ge 0$
assume w.l.o.g. $x\le y \le z$ and let $y=x+a$ and $z = x + a + b$, so
$\begin{align*}(x + y + z)^3 - 27 x y z &= (3x + 2a + b)^3 - 27 x (x+a)(x+a+b) \\ &= 9 a b x + 6 a b^2 + 9 x a^2 + 9 x b^2 + 12 b a^2 + b^3 + 8a^3\end{align*}$
which is greater than or equal to $0$ as all of $x$, $a$, and $b$ are.
This trick does not always work, but it works surprisingly often.
| {
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Logarithm rules What can I do with these expression: $2^{\log _{\frac{4}{3}}n}$ and $2^{\log _{4}n}$ if I don't want to have $n$ in the exponent?
I tried nothing because I didn't have any good ideas.
Thanks.
| For example, $2^{\log_4n}=2^{2\cdot\frac{1}{2}\cdot\log_4n}=((2^{2})^{\log_4n})^{\frac{1}{2}}=(4^{\log_4n})^{\frac{1}{2}}=n^{\frac{1}{2}}=\sqrt{n}.$ In this case, I am using the fact that $2^2=4$ and the property that says that $a^{\log_ab}=b.$
For the other expression, we would need to use that $2^{\log_2{\frac{4}{3}}}=\frac{4}{3}.$ Let us represent $\log_2{\frac{4}{3}}$ by $c$. Of course, $c\neq 0$. Then $2^{\log_{\frac{4}{3}}n}=2^{c\cdot\frac{1}{c}\cdot\log_{\frac{4}{3}}n}=((2^{c})^{\log_{\frac{4}{3}}n})^{\frac{1}{c}}=(\frac{4}{3})^{\log_{\frac{4}{3}}n})^{\frac{1}{c}}=n^{\frac{1}{c}}=n^{\frac{1}{\log_2{\frac{4}{3}}}}.$
| {
"language": "en",
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Number of Ways of Writing $n$ as Sum I have a general question about numbers:
How many ways can we write $n$ as the sum of the numbers $1$,$2$ and $3$?
I know that we start with the following functions: $$\frac{1}{1-z} = 1+z+z^2+ \dots$$ $$\frac{1}{1-z^2} = 1+z^2+ z^4+ z^6 + \dots$$ $$\frac{1}{1-z^3} = 1+ z^3+ z^{6}+ z^{9} + \dots$$
Multiplying these together we get $$\sum C(n) z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$$ where $C(n)$ is our answer. In other words, by multiplying out the polynomials and looking at the coefficients, I can verify that I get the answer. But how do I know that this gives me the answer? How would I get an explicit formula for $C(n)$? I think I have to do something with partial fractions. So $$\frac{1}{(1-z)(1-z^2)(1-z^3)} = \frac{A}{1-z} + \frac{B}{1-z^2}+ \frac{C}{1-z^3}$$ and we have to solve for $A$, $B$ and $C$?
| Denote by $N_3=\{1,2,3\}\,$ then your question can be reformulated:
Find the number of partitions of natural number n in parts over set $N_3$ that number we denote
$p(n,N_3)$, $n=6k+l$, where $ k\in \{0,1,2,...\}$,and $ l\in \{0,1,2,3,4,5\}\,$
Using recurrences (no generating functions) I find that exact solution is.
[1] $p(6k,N_3)=3k^2+3k+1$
[2] $p(6k+1,N_3)=3k^2+4k+1$
[3] $p(6k+2,N_3)=3k^2+5k+2$
[4] $p(6k+3,N_3)=3k^2+6k+3$
[5] $p(6k+4,N_3)=3k^2+7k+4$
[6] $p(6k+5,N_3)=3k^2+8k+5$
To explain each formula is needed a lot of introduction
example: $p(10,N_3)=14$ because
[1,1,1,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,1,2]
[1,1,1,1,1,1,1,3]
[1,1,1,1,1,1,2,2]
[1,1,1,1,1,2,3]
[1,1,1,1,2,2,2]
[1,1,2,2,2,2]
[1,1,1,1,3,3]
[1,1,1,2,2,3]
[1,1,2,2,2,2]
[1,2,2,2,3]
[2,2,2,2,2]
[1,3,3,3]
[2,2,3,3]
From our formulas we have
$p(10,N_3)=p(6\times 1+4)=3\times 1^2+7\times 1+4=14$ there $k=1$
| {
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Get number of elements of a square matrix given a vector that has upper right elements of that matrix Suppose I have some vectors:
*
*$[1,2,1]$ its length is $3$ wich represents a matrix like $\begin{pmatrix}
1 & 2\\
-1 & 1
\end{pmatrix}$ the complete matrix would have $4$ elements
*$[1,2,1,3,4,1]$ its length is $6$ wich represents $\begin{pmatrix}
1& 2& 3\\
-1& 1 & 4\\
-1& -1 &1
\end{pmatrix}$ the complete matrix would have $9$ elements
*$[1,2,1,3,4,1,5,6,7,1]$ its length is $10$ wich represents $\begin{pmatrix}
1 & 2& 3& 5\\
-1 &1 & 4& 6\\
-1 &-1 &1 & 7\\
-1 &-1 &-1 &1
\end{pmatrix}$ the complete matrix would have $16$ elements
*$[1,2,1,3,4,1,5,6,7,1,8,9,10,11,1]$ its length is $15$ wich represents
$\begin{pmatrix}
1 &2 &3 &5 &8 \\
-1 &1 &4 &6 &9 \\
-1 &-1 &1 &7 &10 \\
-1 &-1 &-1 &1 &11 \\
-1 &-1 &-1 &-1 &-1
\end{pmatrix}$ the complete matrix would have $25$ elements
How do I find a general formula to get all elements of a square matrix from the number of elements in a vector?
So
*
*$3\to 4$
*$6\to 9$
*$10\to 16$
*$15\to 25$
*etc...
also: How would you know the matrix size $2\times 2$, $3\times 3$,...$n\times n$?
| As Billy explains, you can find an explicit formula by solving a quadratic, the result of which appears in Fabian's answer.
An alternative is to note that $$\frac{m^2}{2}\lt\frac{m(m+1)}{2}\lt\frac{(m+1)^2}{2}.$$ You are given $n=\frac{m(m+1)}{2}$, and you want $m^2$. The inequalities above yield $$m^2<2n<(m+1)^2,$$ so $$m<\sqrt{2n}<m+1.$$ This implies that $m=\lfloor\sqrt{2n}\rfloor$ (the greatest integer not greater than $\sqrt{2n}$). Finally, if the vector has $n$ entries, then the matrix has $\lfloor\sqrt{2n}\rfloor^2$ entries.
| {
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Integrate square of the log-sine integral: $\int_0^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$ $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx=-\frac{\pi}{2}\ln(2)$ is an integral that is common.
But, how can we show $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx=\frac{{\pi}^{3}}{24}+\frac{\pi}{2}\ln^{2}(2)$?.
Does anyone have any ideas on how to approach $\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$?.
Thank you very much.
| Alternative solution:
\begin{align}
A&=\int_0^{\frac{\pi}{2}}x\ln^2(\sin x)dx\\
&\overset{y=\sin x}=\int_0^1 \frac{\ln^2 y\arcsin y}{\sqrt{1-y^2}}dy\\
&\overset{\text{IBP}}=\left[\left(\int_0^y \frac{\ln^2 t}{\sqrt{1-t^2}}dt\right)\arcsin y\right]_0^1-\int_0^1 \frac{1}{\sqrt{1-y^2}}\left(\int_0^y \frac{\ln^2 t}{\sqrt{1-t^2}}dt\right)dy \\
&=\frac{\pi}{2}\int_0^1 \frac{\ln^2 t}{\sqrt{1-t^2}}dt-\int_0^1 \int_0^1 \frac{y\ln^2(ty)}{\sqrt{1-y^2}\sqrt{1-t^2y^2}}dtdy\\
&=\frac{\pi}{2}\int_0^1 \frac{\ln^2 t}{\sqrt{1-t^2}}dt+\int_0^1 \int_0^1 \frac{y\ln^2 y}{\sqrt{1-y^2}\sqrt{1-t^2y^2}}dtdy-\\&\int_0^1 \int_0^1 \frac{y\ln^2 t}{\sqrt{1-y^2}\sqrt{1-t^2y^2}}dtdy-
2\underbrace{\int_0^1 \int_0^1 \frac{y\ln y\ln(ty)}{\sqrt{1-y^2}\sqrt{1-t^2y^2}}dtdy}_{u=ty}\\
&=\frac{\pi}{2}\int_0^1 \frac{\ln^2 t}{\sqrt{1-t^2}}dt+\int_0^1 \left[\frac{\arcsin(ty)}{\sqrt{1-y^2}}\right]_{t=0}^{t=1}\ln^2 ydy+\\&\int_0^1 \frac{1}{t}\left[\text{arcsinh}\left(\frac{t\sqrt{1-y^2}}{\sqrt{1-t^2}}\right)\right]_{y=0}^{y=1}\ln^2 tdt-2\int_0^1 \frac{\ln y}{\sqrt{1-y^2}}\left(\int_0^y \frac{\ln t}{\sqrt{1-t^2}}dt\right)dy\\
&=\frac{\pi}{2}\int_0^1 \frac{\ln^2 t}{\sqrt{1-t^2}}dt+A-\underbrace{\int_0^1 \frac{\text{arcsinh}\left(\frac{t}{\sqrt{1-t^2}}\right)\ln^2 t}{t}dt}_{\text{IBP}}-\left(\int_0^1 \frac{\ln y}{\sqrt{1-y^2}}dy\right)^2\\
&=\frac{\pi}{2}\int_0^1 \frac{\ln^2 t}{\sqrt{1-t^2}}dt+A+\frac{1}{3}\int_0^1 \frac{\ln^3 t}{1-t^2}dt-\left(\int_0^1 \frac{\ln y}{\sqrt{1-y^2}}dy\right)^2\\
\end{align}
Moreover,
\begin{align}
\int_0^1 \frac{\ln y}{\sqrt{1-y^2}}dy&=\int_0^{\frac{\pi}{2}}\ln(\sin x)dx=-\frac{\pi}{2}\ln 2\\
\int_0^1 \frac{\ln^3 t}{1-t^2}dt&=-\frac{45}{8}\zeta(4)=-\frac{\pi^4}{16}
\end{align}
Therefore,
\begin{align}
\boxed{\displaystyle \int_0^1 \dfrac{\ln^2 t}{\sqrt{1-t^2}}dt=\int_0^{\frac{\pi}{2}}\ln^2(\sin x)dx=\dfrac{\pi\ln^2 2}{2}+\dfrac{\pi^3}{24}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 6,
"answer_id": 4
} |
Solve by induction: $n!>(n/e)^n$ To Prove : $n! > (n/e)^n$
The question seems easy but it ain't; anyone up for it ?
| I had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
$$\begin{align}
a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\
b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n}
\end{align}$$
and
$$\underset{n \rightarrow \infty}{\lim} a_n ~~~=~~~ \underset{n \rightarrow \infty}{\lim} b_n ~~~=~~~ e \\ ~ \\$$
While both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import rcParams
rcParams.update({'figure.autolayout': True})
pts = np.arange(0, 20, 1)
a_n = lambda n: (1+1/n)**n
b_n = lambda n: (1-1/n)**(-n)
plt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = "bx", markersize = "5", markeredgewidth = "2", label = "$a_n$")
plt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = "rx", markersize = "5", markeredgewidth = "2", label = "$b_n$")
plt.plot(pts, [np.exp(1)]*len(pts), color = "black", linewidth = 2, label = "$e$")
plt.xlim(1.5, 14.5)
plt.ylim(2.0, 3.5)
plt.legend(loc = "best")
plt.setp(plt.gca().get_legend().get_texts(), fontsize = "22")
plt.show()
So we're going to use the following inequality:
$$\forall n \in \mathbb{N} ~ : ~~~~~ \left ( 1 + \frac{1}{n} \right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \tag*{$\circledast$} \\ ~ \\$$
Thesis
$$\forall n \in \mathbb{N}, ~ n \geq 2 ~ : ~~~~~ e \cdot \left ( \frac{n}{e} \right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \cdot e \cdot \left ( \frac{n}{e} \right )^n \\ ~ \\$$
Proof By Induction
Base Case
We begin with $n = 2$ and get
$$\begin{align}
& ~ && e \cdot \left ( \frac{2}{e} \right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \cdot e \cdot \left ( \frac{2}{e} \right )^2 \\ ~ \\
& \Leftrightarrow && e \cdot \frac{4}{e^2} ~~~~&&<~~~~ 1 \cdot 2 ~~~~&&<~~~~ 2 \cdot e \cdot \frac{4}{e^2} \\ ~ \\
& \Leftrightarrow && \frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \frac{8}{e} \\ ~ \\
&\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\
\end{align}
$$
Which is a true statement.
Inductive Hypothesis
Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step
$$\begin{align}
& ~ && e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
& = && (n+1) \cdot \frac{1}{e} \cdot e \cdot \left ( \frac{n+1}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( \frac{n+1}{n} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( 1 + \frac{1}{n} \right )^n\\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot e\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n!\\ ~ \\
& = && (n+1)!\\ ~ \\
& = && (n+1) \cdot n!\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n \cdot e \cdot \left ( \frac{n}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n}{e} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( \frac{n}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{-(n+1)} \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
\end{align}
$$
Conclusion
Therefore the statement holds $\forall n \in \mathbb{N}, ~ n \geq 2$.
$$\tag*{$\square$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Determinant of a special skew-symmetric matrix Simple calculation show that:
$$
\begin{align}
\det(A_2)=\begin{vmatrix}
0& 1 \\
-1& 0
\end{vmatrix}&=1\\
\det(A_4)=\begin{vmatrix}
0& 1 &1 &1 \\
-1& 0 &1&1\\
-1& -1& 0&1\\
-1& -1& -1&0
\end{vmatrix}&=1
\end{align}
$$
Here is my question:
Is it true that $\det(A_{2n})=1$ for all $n\in{\mathbb Z_+}$?
With MAPLE, I tried some large $n$. And I guess it is true. But temporarily I have no idea how to show it.
| Let $A_{2n}$ the $2n\times 2n$ matrix defined by $(A_{2n})_{i,j} = \begin{cases}
1&\mbox{if }i<j\\
0&\mbox{if }i=j\\
-1&\mbox{if } i>j
\end{cases}$.
We can show the result ($\det(A_{2n}) = 1$) by induction. The case $n=1$ and $n=2$ has been solved. Now, we can write $A_{2n+2}=\begin{pmatrix}A_{2n}&B\\-B^t &C\end{pmatrix}$ where $B$ is a $2n\times 2$ matrix whose entries are all $1$ and $C=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Consider the line $l_{2n+2}$: we can transform it, computing $l_{2n+2}\leftarrow l_{2n+2}-l_{2n+1}$. It becomes $(0,\ldots,0,-1,-1)^t$ (the $2n$ first components are $0$). After that we make $c_{2n+2}\leftarrow c_{2n+2}-c_{2n+1}$: we get $(0,\ldots,0,1,0)$. Now we conclude, expanding by the last column and after the last row (the first expansion gives $(-1)^{2n-1+2n}=-1$ and the second $(-1)^{2n-1+2n-1}\cdot (-1)$), hence we get $\det A_{2n+2}=(-1)\cdot(-1)\cdot\det A_{2n} =\det A_{2n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 = (x^2 + 3x + 2)^3 $have?
How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 =
(x^2 + 3x + 2)^3 $have?
If I didn't make any mistake that equation could be reduced to the form:$$(2+6x^2)(x+2)^3=(8–2x^2 )^3$$
However,I don't know how to approach next as more simplification and then forming up the factors again would be more time consuming,I suppose this problem could be solved in a more easy way,any ideas?
| Using the idea in Aryabhata's comment, since
$$\left( 8-2x^{2}\right) ^{3}=2^{3}\left( 4-x^{2}\right) ^{3}=8
(2-x)^{3}(2+x)^{3},$$
your second equation
$$\left( 2+6x^{2}\right) \left( x+2\right) ^{3}=\left( 8-2x^{2}\right) ^{3}\tag{1}$$
is equivalent to
$$\begin{eqnarray*}
\left( 2+6x^{2}\right) \left( x+2\right) ^{3} &=&8( 2-x)^{3}(2+x)
^{3}.\tag{2}
\end{eqnarray*}$$
So the factor $(x+2)^3$ in both sides yields the real triple root $x=-2$. For $x\ne -2$ this equation $(2)$ is equivalent to
$$\begin{eqnarray*}
\left( 2+6x^{2}\right) &=&-8\left( x-2\right) ^{3}\\
\left( 2+6x^{2}\right) +8\left( x-2\right) ^{3} &=&0 \\
\left( 1+3x^{2}\right) +4\left( x-2\right) ^{3} &=&0\text{.}
\end{eqnarray*}$$
By inspection we see that $x=1$ is a root and by long division or Ruffini's rule, we find:
$$
\begin{eqnarray*}
\left( 1+3x^{2}\right) +4\left( x-2\right) ^{3} =4x^{3}-21x^{2}+48x-31=\left( x-1\right) \left( 4x^{2}-17x+31\right).
\end{eqnarray*}$$
Alternatively we could have applied the rational zero theorem: all the rational roots of the equation $$d_{n}x^{n}+d_{n-1}x^{n-1}+\ldots +d_{0}=0,$$ where all the coefficients are integers, are of the form $\frac{p}{q}=\frac{\text{a factor
of }d_{0}}{\text{a factor of }d_{n}}$.
Hence, for $x\ne -2$, the equation $(1)$ is equivalent to
$$\left( x-1\right) \left( 4x^{2}-17x+31\right) =0.\tag{3}$$
Since the discriminant of the quadratic term is negative $$\Delta
=17^{2}-4\times 4\times 31=-207<0,$$ the original equation has the real
solution $x=1$ due to the factor $(x-1)$ and the triple real solution $x=-2$ due to the common factor $(x+2)^3$. In the total, two distinct real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/60054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?
The answer says $1$, but I am not sure how to derive it.
| let $a+\frac{1}{b} = b+\frac{1}{c} = c+\frac{1}{a} = k$
$ab + 1 = bk, bc + 1 = ck, ca + 1 = ak \Rightarrow ab=bk-1, bc=ck-1, ca=ak-1$
$abc + c = bck = (ck-1)k = ck^2-k \Rightarrow abc + k=c(k^2-1)$
$abc + a = ack = (ak-1)k = ak^2-k \Rightarrow abc+k =a(k^2-1)$
$abc+b=abk=(bk-1)k=bk^2-k \Rightarrow abc+k=b(k^2-1)$
$c(k^2-1)=a(k^2-1)=b(k^2-1)=abc + k$
$a\neq b \neq c \neq 0 \Rightarrow k^2-1=0 \Rightarrow k^2=1 \Rightarrow k= \pm1$
$abc+k=c(k^2-1)=0 \Rightarrow abc=-k=\pm1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/61054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How could we find the sum of reciprocals of the roots of this equation?
A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How
could we find the sum of reciprocals of the roots of this equation?
I need some hints for solving this problem.
| Also, just for the fun of it, since you have a root with multiplicity two:
$$
ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2
$$
Then the root is $x^* = \pm\sqrt{\frac{c}{a}}$ and from $a = 2c$, we get $x^* = \pm\sqrt{\frac{1}{2}}$. Hence, $$
\frac{1}{x_1}+\frac{1}{x_2} = \frac{2}{x^*}=\pm 2{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/61690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Show that every prime $p>3$ is either of the form $6n+1$ or of the form $6n+5$ Show that every prime $p>3$ is either of the form $6n+1$ or of the form $6n+5$, where $n=0,1,2, \dots$
| Every integer is of the form $6n$ or $6n+1$ or $6n+2$ or $6n+3$ or $6n+4$ or $6n+5$ for some integer $n$. This is because when we divide an integer $m$ by $6$, we get a remainder of $0$, $1$, $2$, $3$, $4$, or $5$.
If an integer $m>2$ is of the form $6n$ or $6n+2$ or $6n+4$, then $m$ is even and greater than $2$, and therefore $m$ is not prime.
If an integer $m>3$ is of the form $6n+3$, then $m$ is divisible by $3$ and greater than $3$, and therefore $m$ is not prime.
We have shown that an integer $m>3$ of the form $6n$ or $6n+2$ or $6n+3$ or $6n+4$ cannot be prime. That leaves as the only candidates for primality greater than $3$ integers of the form $6n+1$ and $6n+5$.
Comment: In fact, it turns out that there are infinitely many primes of the form $6n+1$, and infinitely many primes of the form $6n+5$. Showing that there are infinitely many of the form $6n+5$ is quite easy, it is a small variant of the "Euclid" proof that there are infinitely many primes. Showing that there are infinitely many primes of the form $6n+1$ requires more machinery. But your question did not ask for such a proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/64414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| for $a,b,c>=0$
we know $(a-b-c)^2>=0$
i.e $a^2 + b^2 + c^2 - 2 (ab + bc + ca) \ge 0$
i.e $a^2 + b^2 + c^2 \ge ab + bc + ca$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/64868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 12,
"answer_id": 7
} |
Variance in a Discrete Probability Distribution (Homework) For my Homework I have four values: $N_1, N_2, A, C$.
Two values are chosen at random without replacement.
X denotes how many N values are chosen.
$$P(x=0) = \frac16 \qquad
P(x=1) = \frac23 \qquad
P(x=2) = \frac16 $$
find the mean and variance.
I have taken the mean to be $\frac{(0+1+2)}{3} = \frac33 = 1$ However, I think this is wrong, and that the distribution must be taken into account.
The Variance is to be determined by the equation $ \text {Var}(x)=E(x^2)-E(x)^2$
However I am not certain what value to assign $x$.
| For any function $f$, the expectation of $f(X)$ is given by: $\newcommand{\E}{\mathrm{E}}$
$$ \E(f(X)) = \sum_{i=0}^2 \ \Pr[X = i] \cdot f(i) = \sum_{i=0}^2 \ p_i f(i) .$$
The expectation of $X$ is
$$
\E(X) = 0 \cdot p_0 + 1 \cdot p_1 + 2 \cdot p_2 = \frac{1}{6} \cdot 0 + \frac{2}{3} \cdot 1 + \frac{1}{6} \cdot 2 = 1.
$$
$$
\E(X^2) = 0 \cdot p_0 + 1^2 \cdot p_1 + 2^2 \cdot p_2 = \frac{1}{6} \cdot 0^2 + \frac{2}{3} \cdot 1^2 + \frac{1}{6} \cdot 2^2 = \frac{4}{3}.
$$
The variance of $X$ is
$$
\mathrm{Var}(X) = \E(X^2) - \E(X)^2 = \frac{4}{3} - 1^2 = \frac{1}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $ I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$
$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$
The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless...
Can you help me find a useful substitution?
$$ x=\cos(2t) $$
$$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$
$$ u=\tan(t/2) $$
$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$
But now it looks even more complicated... ?
| Just to give a complete answer using the method of the square. Using the easily proved equation
$$\left ( \sqrt{1+x} + \sqrt{1-x} \right )^2 = 2\left ( 1+ \sqrt{1-x^2} \right )$$
one can get
\begin{align*}
\int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{1+x} + \sqrt{1-x} + 2} &= \int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{2\left (1+\sqrt{1-x^2} \right )}+2} \\
&\!\!\!\!\!\overset{x = \sin u}{=\! =\! =\! =\! =\!} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{2\left ( 1+\cos u \right )}+2}\, \mathrm{d}u\\
&=\int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{4 \cos^2 \frac{u}{2}}+2} \, \mathrm{d}u \\
&= \int_{-\pi/2}^{\pi/2} \frac{\cos u}{2 \cos \frac{u}{2} + 2} \, \mathrm{d} u \\
&= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{1+ \cos \frac{u}{2}} \, \mathrm{d}u \\ &= \frac{1}{4} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\cos^2 \frac{u}{4}} \, \mathrm{d}u \\ &= {\require{cancel}\cancelto{0}{\frac{1}{4} \left [ 4 \tan \frac{u}{4} \cos u \right ]_{-\pi/2}^{\pi/2}}} + \int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \end{align*}
The last integral can be dealt as follows:
\begin{align*}
\int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \; &\overset{y=u/4}{=\! =\! =\! =\!} \; 4 \int_{-\pi/8}^{\pi/8} \tan y \sin 4y \, \mathrm{d}y\\
&=4 \int_{-\pi/8}^{\pi/8} \tan y \left ( 4 \sin y \cos^3 y - 4 \sin^3 y \cos y \right )\, \mathrm{d}y \\
&=4 \int_{-\pi/8}^{\pi/8} \left ( 4 \sin^2 y \cos^2 y - 4 \sin^4 y \right )\, \mathrm{d}y \\
&= 16 \int_{-\pi/8}^{\pi/8} \left ( \sin y \cos y \right )^2 \, \mathrm{d} y - 16 \int_{-\pi/8}^{\pi/8} \sin^4 y \, \mathrm{d}y \\
&= \left (\frac{\pi}{2} - 1 \right ) + \left (-1 + 4 \sqrt{2} - \frac{3\pi}{2} \right ) \\
&= 4\sqrt{2} - \pi - 2
\end{align*}
Basically as in @RE60K's answer. The results arise in view of the identities:
$$\sin^4 x = \frac{1}{8} \left ( 3 + \cos 4x - 4 \cos 2x \right )$$
and
\begin{align*}
\int_{-\pi/8}^{\pi/8} \sin^2 x \cos^2 x \, \mathrm{d}x &= \int_{-\pi/8}^{\pi/8} \left ( \sin x \cos x \right )^2 \, \mathrm{d} x\\
&=\int_{-\pi/8}^{\pi/8} \left ( \frac{\sin 2x}{2} \right )^2 \, \mathrm{d}x\\
&= \frac{1}{4} \int_{-\pi/8}^{\pi/8} \sin^2 2x \, \mathrm{d}x\\
&=\frac{1}{8}\int_{-\pi/8}^{\pi/8} \left ( 1- \cos 4x \right ) \, \mathrm{d}x \\
&= \frac{1}{8} \left ( \frac{\pi}{4} - \frac{1}{2} \right ) \\
&= \frac{\pi}{32} - \frac{1}{16}
\end{align*}
I know I'm 9 years late but I wanted to present the method of squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/67172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Probability question about married couples
If four married couples are arranged in a row, what is the probability
that no husband sits next to his wife?
Would it be
$1- \frac{2(4!)}{8!}$?
| Let's count the complement. If at least one couple stands together, you have $\binom{4}{1} = 4$ ways of choosing the couple, $2$ ways of arranging them, and $7$ positions in which to place them (the first person in the couple cannot stand in the last place of the line), and the remaining 6 positions can be filled in $6!$ ways; this gives a total of $4\times 2\times 7\times 6! = 8!=40320$ ways.
However, this overcounts by counting arrangements in which more than one couple sits together twice. If at least two couples stand together, we have $4\times 3$ ways of choosing the first and second couple, $4$ ways of arranging them, 15 ways to stand them in the line in that order, and $4!$ of filling in the remaining 4 places, for a total of $4\times 3\times 4\times 15\times 4! = 17280$ ways.
If at least 3 couples stand together, we have $4\times 3\times 2$ ways of choosing the couples (in order), $8$ ways of arranging them amongst themselves, 10 ways of arranging them in the line, and $2$ ways of filling the remaining seats, for a total of $4\times 3\times 2\times 8\times 10\times 2= 3840$ ways.
If all four couples sit together, we have $4\times 3\times 2\times 1$ way of arranging the couples, and $2^4 = 16$ ways of arranging each couple, for a total of $384$ ways.
So, let's see: when we count "at least one couple stands together", we count the arrangements with exactly one couple together once, the arrangements with exactly two couples together twice; the arrangements with exactly three couples three times, and the arrangements with all four couples together four times.
When we could "at least two couples together", we count the arrangements with exactly two couples together once, the arrangements with exactly three couples together $\binom{3}{2}=3$ times; and the arrangements with all four couples together $\binom{4}{2}=6$ times.
When we could "at least three couples together", we count the arrangements with exactly three couples together once, and the arrangements with all four couples together $\binom{4}{3}=4$ times.
So if we take:
$$(\text{at least one}) - (\text{at least two}) + (\text{at least three}) - (\text{all four})$$
we get the count exactly right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/68541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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separation of variables Solve and prove Using separation of variables Solve
$$\frac{dy}{dx} = 3y +1.$$
My final answer I got was:
$$y = \pm C \frac{e^x}{3} - \frac{1}{3}.$$
But I don't how to take the derivative of that to get it back to my orig problem
| First, note that a constant divided by $3$ is just a constant, so you could rewrite what you have without the fraction. Also, plus or minus a constant is just a constant. So you could rewrite your solution as
$$y = Ce^x - \frac{1}{3}.$$
To check, you just take the derivative: $C$ is a constant, so
$$\frac{dy}{dx} = \frac{d}{dx}\left(Ce^x - \frac{1}{3}\right) = C\frac{d}{dx}e^x - \frac{d}{dx}\frac{1}{3} = Ce^x.$$
Then you compute $3y + 1 = 3Ce^x - 1 + 1 = 3Ce^x$.
This is not the same as $\frac{dy}{dx}$, so this is not a correct answer.
And, no, it wasn't my "simplifications" that did you in:
$$\frac{d}{dx}\left( \pm \frac{Ce^x}{3} - \frac{1}{3}\right) = \pm\frac{C}{3}\frac{d}{dx}e^x = \pm\frac{C}{3}e^x,$$
and
$$3\left(\pm\frac{Ce^x}{3}-\frac{1}{3}\right) + 1 = \pm Ce^x\neq \pm\frac{Ce^x}{3}.$$
So in fact, you did a mistake in your derivation.
From
$$\frac{dy}{dx} = 3y+1,$$
we must consider the possibility that $3y+1=0$; this will occur if $y=-\frac{1}{3}$; that is one possible solution.
If $3y+1\neq 0$, then separating variables we get
$$\frac{dy}{3y+1} = dx,$$
and integrating both sides, we have
$$\int\frac{dy}{3y+1} = \int\,dx.$$
To do the integral on the left, we let $u=3y+1$; then $du = 3dy$, so $\frac{1}{3}du = dy$. Hence
$$\int\frac{dy}{3y+1} = \int\frac{\frac{1}{3}du}{u} = \frac{1}{3}\int\frac{du}{u} = \frac{1}{3}\ln|u|+C = \frac{1}{3}\ln|3y+1|+C;$$
so, consolidating constants, we have:
$$\begin{align*}
\int\frac{dy}{3y+1} &= \int\,dx\\
\frac{1}{3}\ln|3y+1| &= x+C&&C\text{ an arbitrary constant}\\
\ln|3y+1| &= 3x + D&&D\text{ an arbitrary constant}\\
e^{\ln|3y+1|} &= e^{3x+D}\\
|3y+1| &= e^De^{3x}\\
|3y+1| &= Ae^{3x} &&A\text{ a positive constant}\\
3y+1 &=\pm Ae^{3x} &&A\text{ a positive constant}\\
3y+1 &= Be^{3x} &&B\text{ a non-zero constant}\\
3y &= Be^{3x} - 1\\
y &= \frac{B}{3}e^{3x} - \frac{1}{3}\\
y &= Ke^{3x} - \frac{1}{3} &&K\text{ a non-zero constant}
\end{align*}$$
I set $D=3C$, still a constant, then $A=e^D$, a positive constant (because $e^D$ is always positive); then $B=\pm A$, a nonzero constant (because $A$ is never zero, so $B$ is never zero); and finally $K=\frac{B}{3}$, a nonzero constant (because $B$ is never zero).
So putting it together with the special case considered before, we have that the solutions are:
$$\text{Either }y=-\frac{1}{3}\quad\text{or}\quad y = Ke^{3x}-\frac{1}{3}\text{ with }K\text{ non-zero.}$$
Then we may realize that if we allow $K=0$, then we get the special case, so we can summarize the two cases by saying
$$ y = Ke^{3x} - \frac{1}{3},\quad K\text{ an arbitrary constant.}$$
Now take derivatives and plug in to verify this is correct.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove those "curious identities"? How to prove
$$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$
and
$$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
| Define $\zeta_n = e^{2 \pi i/n}$.
Proposition
For odd integer $n \geq 1$,
\begin{align}
\prod_{k = 1}^{n-1}(\zeta_n^{k} - \zeta_n^{-k}) = n.
\end{align}
and
\begin{align}
\prod_{k = 1}^{n-1} \sin( \tfrac{2 \pi k }{n} ) = \tfrac{n}{(2 i)^{n-1}}.
\end{align}
Proof: The claimed identities follow from the identity
\begin{align}
z^n - 1 = \prod_{ k =0}^{n-1} (z - \zeta_n^{k}) = \prod_{ k =0}^{n-1} (z - \zeta_n^{-2k}).
\end{align}
Writing $z = x/y$, we have
\begin{align}
x^n - y^n = \prod_{k = 0}^{n-1} ( \zeta_n^{k} x - \zeta_n^{-k} y).
\end{align}
Thus,
\begin{align}
n y^{n-1} = \lim_{x \to y} \frac{x^n - y^n}{x - y} = \lim_{x \to y} \ \ \prod_{k = 1}^{n-1} ( \zeta_n^{k} x - \zeta_n^{-k} y) = y^{n-1} \ \prod_{k = 1}^{n-1} ( \zeta_n^{k} - \zeta_n^{-k} ).
\end{align}
For the second identity, let $x =e^{\pi i z}$ and $y = e^{- \pi i z}$ and recall the complex exponential representation of the sine function. This yields
\begin{align}
n = \lim_{z \to 0} \frac{\sin n \pi z}{\sin z } = (2 i)^{n-1} \lim_{z \to 0} \ \ \prod_{k = 1}^{n-1} \sin( \pi z + \tfrac{2 \pi k }{n} ) = (2 i)^{n-1} \prod_{k = 1}^{n-1} \sin( \tfrac{2 \pi k }{n} ).
\end{align}
Similar reasoning works to prove the identities that you mention.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a problem involving continuity of a piecewise function I'm trying to solve the following problem:
Determine which $n\in\mathbb{N}$ make the following function
$f:\mathbb{R}^3\rightarrow \mathbb{R}$ continuous at the origin:
$ f(x,y,z)=\left\{
\begin{array}{cr}
\frac{(\cos^2\left(|x|+|y|\right)-1)\sin(y^2+z^2)}{(x^2+y^2+z^2)^{n/2}}
& \text{if } (x,y,z) \neq 0\\
0 & \text{if } (x,y,z) = 0\\
\end{array} \right. $
It's obvious that $f$ is continuous at the origin if $n\leq 2$, since:
$\left|\frac{(\cos^2\left(|x|+|y|\right)-1)\sin(y^2+z^2)}{(x^2+y^2+z^2)^{n/2}}\right|\leq\frac{|\cos^2\left(|x|+|y|\right)-1|\,|y^2+z^2|}{|x^2+y^2+z^2|}\leq\frac{|\cos^2\left(|x|+|y|\right)-1|\,|y^2+z^2|}{|y^2+z^2|}=|\cos^2\left(|x|+|y|\right)-1|\rightarrow 0$
I'm almost sure the function isn't continuous if $n> 2$ but I can't seem to find a way to prove it.
| The question is for $n$ integer, but we can take $a\in\mathbb R$ instead of $n$. We denote $f_a$ the corresponding function. We have for $(x,y,z)\neq (0,0,0)$
$$f_a(x,y),z)=-\frac{\sin^2(|x|+|y|)\sin (y^2+z^2)}{(x^2+y^2+z^2)^{a/2}}.$$
Now, fix a real number $m$. We have
$$f_a(x,mx,mx)=-\frac{\sin(2|mx|)\sin(2m^2x^2)}{(1+2m^2)^{a/2}|x|^a}\overset{0}{\sim}-\frac{4|m^3||x|^{3-a}}{(1+2m^2)^{a/2}},$$
hence $f_a$ cannot be continuous if $3-a\leq 0$ hence $a\geq 3$. Conversely, if $a<3$ we have thanks to then inequality $|\sin t|\leq |t|$:
\begin{align*}
|f_a(x,y,z)|&\leq\frac{(|x|+|y|)(y^2+z^2)}{(x^2+y^2+z^2)^{a/2}}\\
&\leq \sqrt 2\frac{\sqrt{x^2+y^2}(y^2+z^2)}{(x^2+y^2+z^2)^{a/2}}\\
&\leq \sqrt 2(x^2+y^2+z^2)^{\frac{3-a}2},
\end{align*}
and we can conclude since the exponent is positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/70721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this Pythagorean Triple? How do I prove that $(2mn, m^2 - n^2, m^2 + n^2)$ is true for $m>n>0$?
Since $m^2 + n^2$ is the hypotenuse, I applied the Pythagoren theorem:
$(2mn)^2 + (m^2 - n^2)^2 = (m^2 +n^2)^2$ and simplified it so that I would get $(m^2 + n^2)^2 = (m^2 + n^2)^2$ but I wasn't able to prove anything. How do I continue from here?
Thanks!
| Realizing that I made a silly mistake, I reevaluated the equation and:
Given the Pythagorean Triple $(2mn, m^2-n^2,m^2+n^2)$ and $m>n>0$, we can apply the Pythagorean theorem to see if they are equal:
$(2mn)^2+(m^2-n^2)^2 = (m^2+n^2)^2$
$4m^2n^2 + (m^4-2m^2n^2+n^4) = (m^2+n^2)^2$
$m^4+2m^2n^2+n^4=(m^2+n^2)^2$
$(m^2+n^2)^2 = (m^2 + n^2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/70808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from "The Cauchy-Schwarz Master Class") Exercise 1.4 from a great book The Cauchy-Schwarz Master Class asks to prove the following:
For all positive $x$, $y$ and $z$, one has
$$x+y+z \leq 2 \left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right).$$
Introduction to the exercise says:
There are many situations where Cauchy's inequality conspires with symmetry to provide results that are visually stunning.
How to prove that inequality? And how does one benefit from the "symmetry"? What is the general idea behind this "conspiracy"?
| $(x+y+z)^2 = \left(x\sqrt{\frac{y+z}{y+z}} +y\sqrt{\frac{x+z}{x+z}}+z\sqrt{\frac{x+y}{x+y}}\right)^2\leq\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right)(y+z+x+z+x+y)$. You will get it. I didn't read that book, but I believe Srivatsan is right.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Divisibility question Let $r$ be an integer greater than $2$.
Is there a simple way of showing that $2^r$ divides
$\left(\begin{array}{c}
{2}^{r-2} \\
k
\end{array}\right) 2^{2k}$
but it does not divide
$\left(\begin{array}{c}
2^{r-l} \\
k
\end{array}\right) 2^{2k}$ for $l>2$ and for all $k$ greater than
$0$ and less or equal than $2^{r-2}$?
Equivalently, is there a simple way of showing that the element $5$ has order
$2^{r-2}$ in the group of units of the ring $\mathbb{Z}_{2^r}$?
The two questions are connected by writing $5 = 4 + 1$ and using the binomial expansion formula to study the order of $5$.
| I don't know off-hand about the binomial coefficients, but here's a proof of the order of $5$.
For $r=3$, the order of $5$ is $2$, which agrees with what we want. So assume that $r\gt 3$.
Assume that the order of $5$ has order $2^{r-2}$ modulo $2^{r}$. The order of $5$ modulo $2^{r+1}$ is a multiple of $2^{r-2}$. It cannot be of order $2^r$ (the order of the group), because both $2^{r}-1$ and $2^r+1$ have order $2$ and they are not congruent modulo $2^{r+1}$ so the group of units is not cyclic. So the order of $5$ is either $2^{r-2}$ or $2^{r-1}$.
Since the order of $5$ modulo $2^{r}$ is $2^{r-2}$, and the order modulo $2^{r-1}$ is $2^{r-3}$, then $5^{2^{r-3}}\equiv 1 \pmod{2^{r-1}}$, so $5^{2^{r-3}}\equiv 1 + k2^{r-1}\pmod{2^{r}}$, with $k=0,1$. Can't be $k=0$, because then the order of $5$ modulo $2^r$ would divide $2^{r-3}$, so $5^{2^{r-3}}\equiv 1 + 2^{r-1}\pmod{2^r}$. Hence $5^{2^{r-3}}\equiv 1 + 2^{r-1} + k2^r\pmod{2^{r+1}}$, with $k=0$ or $1$. Squaring both sides, we have:
$$\begin{align*}
5^{2^{r-2}} &\equiv (1 + 2^{r-1} + k2^r)^2\pmod{2^{r+1}}\\
&\equiv 1 + 2^{2r-2} + k2^{2r} + 2^r + k2^{r+1} + k2^{2r}\pmod{2^{r+1}}\\
&\equiv 1+2^r\pmod{2^{r+1}}.
\end{align*}$$
So the order of $5$ is $2^{r-1}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/74061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Interscholastic Mathematics League Senior B #12 Compute the product of the nonreal roots of the equation $x^4+4x^3+6x^2+1004x+1001=0$.
So here is what I have done so far. I got two of the roots to be zero and 4 since $x^3(x+4)+(6x^2+1004x+1001)=0$. But for the other two roots by using the quadratic formula I get $$\frac{-1004 \pm \sqrt{1004^2-24024}}{12}.$$ This contest was already completed but I couldn't solve that quadratic equation during the competition. Can someone please post a easier way to solve this problem?
| $4$ is not a root by the rational root theorem and $0$ is not a root since the polynomial is $1001$ at $x=0$. We have $(x+1)^4+1000(x+1)=0$, so one real root is $-1$. Taking that out we have a sum of cubes, $(x+1)^3+10^3=0$, which factors as $(x+11)((x+1)^2-10(x+1)+100)=0$. Eliminating the real root $-11$ we are left with $(x+1)^2-10(x+1)+100=x^2-8x+91=0$. The product of the roots of a monic quadratic is the constant term, so the product of the complex roots is $91$. To see that the product of roots is the constant term, expand $(x-a)(x-b)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Formula for completing the square? My math teacher said that this was the formula for completing the square.
Original function: $$ax^2 + bx + c$$
Completed square: $$a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$
However, using this formula I'm not getting the same answers that I would get just by determining the stuff myself. Is this correct?
| I usually find that completing the square by hand for each example is better than just using the formula.
For your example,
$$4x^2 + 4x + 5 = 4\left(x^2 + x + \frac{5}{4}\right).$$
This is of the form $x^2 + x + \mathrm{const}$, so you to find a number $a$ such that $(x+a)^2 = x^2 + x + \mathrm{const}$. The solution is $a=\frac{1}{2}$, giving
$$4x^2 + 4x + 5 = 4\left(x+\frac{1}{2}\right)^2 + \mathrm{const}$$
Expanding the RHS, you find that the constant is $4$, so the whole expression is
$$4x^2 + 4x + 5 = 4\left(x + \frac{1}{2} \right)^2 + 4 = (2x + 1)^2 + 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/76772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Irreducible polynomial which is reducible modulo every prime
How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?
For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$ gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$?
| When $p=2$ then just note $x^4+1=(x+1)^4$.
Now if $p$ is odd then $8\mid p^2-1 \implies x^4+1\mid x^{p^2-1}-1\mid x^{p^2}-x$. Let $a$ be a root of $x^4+1$ in some extension of $\mathbb F_p$. So, $[\mathbb F_p(a):\mathbb F_p]=4$ if we let $x^4+1$ is irreducible over $\mathbb F_p$. But from $x^4+1\mid x^{p^2}-x$ we can say $a\in\mathbb F_{p^2} \implies [\mathbb F_p(a):\mathbb F_p]\leq 2$, a contradiction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$ I have to prove:
$$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$
Using the following:
$$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$
My work is that first I substitute $n^3$ for $6 {n \choose 3} + 6 {n \choose 2} + n$. Then I go and invoke the sum over that (I am assuming this is how it works). That is,
$$\sum\limits^n_{i=1} \bigg(6 {i \choose 3} + 6 {i \choose 2} + i \bigg)$$
$$6 \sum\limits^n_{i=1} {i \choose 3} + 6 \sum\limits^n_{i=1} {i \choose 2} + \sum\limits^n_{i=1} {i \choose 1}$$
The summation identity is:
$$\sum\limits^n_{i=0} {i \choose k} = {n+1 \choose k+1}$$
invoking it over the sums yields:
$$6{n+1 \choose 4} + 6{n+1 \choose 3} + {n+1 \choose 2}$$
I think these are the right steps, but simplification seems a bit difficult to get the ending result.
| It’s just algebra $-$ tedious, but straightforward:
$$\begin{align*}
6\binom{n+1}4&+6\binom{n+1}3+\binom{n+1}2\\
\strut\\
&=\frac{6(n+1)n(n-1)(n-2)}{4!}+\frac{6(n+1)n(n-1)}{3!}+\frac{(n+1)n}2\\
&=n(n+1)\left(\frac{(n-1)(n-2)}{4}+(n-1)+\frac12\right)\\
&=\frac{n(n+1)}4\big((n-1)(n-2)+4(n-1)+2\big)\\
&=\frac{n(n+1)}4\big((n-1)(n+2)+2\big)\\
&=\frac{n(n+1)}4(n^2+n)\\
&=\frac{n^2(n+1)^2}4\\
&=\left(\frac{n(n+1)}2\right)^2
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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$1+1/(2+3)+1/(4+5+6)...$ What is the value of the sum of the series
$$\frac{1}{1}+\frac{1}{2+3}+\frac{1}{4+5+6}+\dotso\;?$$
And this:
$$\frac{1}{1}+\frac{1}{2\cdot3}+\frac{1}{4\cdot5\cdot6}+\dotso\;?$$
| Note $a+(a+1)+\cdots+(b-1)+b= (b-a+1)(b+a)/2$. Set $b=n(n+1)/2,a=n(n-1)/2+1$:
$$(\circ)=\sum_{n=1}^\infty \left((n(n+1)/2-[n(n-1)/2+1]+1)\cdot\frac{n(n+1)/2+[n(n-1)/2+1]}{2}\right)^{-1}$$
$$=\sum_{n=1}^\infty\frac{2}{n(n^2+1)}=\sum_{n=1}^\infty\left(\frac{2}{n}-\frac{1}{n+i}-\frac{1}{n-i}\right)=\sum_{n=1}^\infty\int_0^12x^{n-1}-x^{n+i-1}-x^{n-i-1}dx$$
$$=\int_0^1\left(\sum_{n=1}^\infty x^{n-1}\right)(2-x^i-x^{-i})dx=\int_0^1\frac{1-x^i}{1-x}dx+\int_0^1\frac{1-x^{-i}}{1-x}dx$$
$$=\psi(1+i)+\psi(1-i)+2\gamma.$$
(Where $\psi$ is the digamma function.) That's the first series. Second one doesn't look so easy.
| {
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Converting base 10 fractions into other bases How do you convert
$14\frac{8}{13}$
into base 3?
I was able to convert $\frac{3}{7}$ into base 3 by constantly multiplying by 3 and dividing the numerator by denominator until I finally got a repetition, but this method doesn't seem to work for $14\frac{8}{13}$.
The correct answer is supposed to be $112.\overline{121}_{3}$
| Just like you convert a fraction to decimal. $8\cdot 3=24=13(1)+11,$ so the first digit (ternit?) is $1. \ \ 11\cdot 3=33=13(2)+7,$ so the second digit is $2. \ \ 3\cdot 7=21=13(1)+8$ and so on. I get $112.\overline{121}_{3}$
Alternately you can notice that $\frac{8}{13}=\frac{16}{3^3-1}$, so the repeat is $3$ digits long and is $16_{10}=121_3$
| {
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Need help finding smallest value of $x^2 + y^2$ I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula.
| For fun, we give a couple of solutions that are not the intended ones. The solutions are very similar, but the first is expressed algebraically, while the
second brings in the geometry.
$1$) Note that
$$(2x+3y)^2+(3x-2y)^2=13(x^2+y^2).$$
Thus, given that $2x+3y=6$,
$$13(x^2+y^2)= 36+(3x-2y)^2.$$
If we can manage to make $3x-2y=0$, then $13(x^2+y^2)$ will be as small as possible. But the system of two linear equations $2x+3y=6$, $3x-2y=0$ has a solution. There,
$$13(x^2+y^2)=36,$$
so the smallest possible value of $x^2+y^2$ is $36/13$.
$2$) Look at the problem geometrically. We want to find the smallest radius $r$ such that the circle $x^2+y^2=r^2$ meets the line $2x+3y=6$. If we draw a picture, we can see that for this smallest $r$, the line $2x+3y=6$ will be tangent to the circle. Let the point of tangency be $T(a,b)$. The line from the origin to $T$ is perpendicular to the tangent line.
The line $2x+3y=6$ has slope $-2/3$. So the line from the origin to $T$ has slope the negative of the reciprocal of $-2/3$. Thus
$$\frac{b}{a}=\frac{3}{2}.$$
This equation simplifies to $3a-2b=0$. We also have $2a+3b=6$. Now we can solve for $a$ and $b$. But let's not bother. Use the fact that
$$(3a-2b)^2+(2a+3b)^2=13(a^2+b^2)$$
to conclude that $r^2=a^2+b^2=36/13$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the integral of $x(1-x)^8$? I want to find the integral of $x (1-x)^8$. How do I go about this? For example, which rule do I use from http://integral-table.com ? Thanks!
| $$x(1-x)^8 = x-8 x^2+28 x^3-56 x^4+70 x^5-56 x^6+28 x^7-8 x^8+x^9$$
so
$$\begin{array}{cl}
&& \int x(1-x)^8 \,dx \\
&=& \int x-8 x^2+28 x^3-56 x^4+70 x^5-56 x^6+28 x^7-8 x^8+x^9 \,dx \\\\
&=& \int x \,dx - 8 \int x^2 \,dx +28 \int x^3 \,dx - 56 \int x^4 \,dx + 70 \int x^5 \,dx - 56 \int x^6 \,dx + 28 \int x^7 \,dx - 8 \int x^8 \,dx + \int x^9 \,dx \\
&=& \tfrac{1}{2} x^2 - \tfrac{8}{3} x^3 + \tfrac{28}{4} x^4 - \tfrac{56}{5}x^5 + \tfrac{70}{6} x^6 - \tfrac{56}{7} x^7 + \tfrac{28}{8} x^8 - 8 \tfrac{1}{9} x^9 + \tfrac{1}{10} x^{10} + C
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Square Roots of Complex Number $3-4i$
What I did
$z^2=3-4i$
$(a+bi)^2 = 3-4i$
$a^2-b^2+2abi = 3-4i$
Then got 2 simultaneous equations
$a^2-b^2=3$ and $2ab=-4$
Solve for $a^2$ in 1st equation: $a^2=3+b^2$
Subbed into 2nd equation to power of 2
$(3+b^2)b^2=4$
$b^4+3b^2-4=0$
Is there a better way than below? Solving power 4 equation then cubic?
Then solved solved power 4 equation ...
$(b-1)(Ab^2+Bb^2+Cb+D)=b^4+3b^2-4$
$(b-1)(b^3+b^2+4b+4)=b^4+3b^2-4$
then solved for cubic equation, getting ...
$(b-1)(b+1)(b^2+4)=0$
So
$b=\pm 1 \text{ or } 2i$
What did the book do now?
I did: (subbing into $a^2=3+b^2$)
When $b=1, a^2=3+1^2 \Longrightarrow a = 2$
When $b=-1, a^2=3+(-1)^2 \Longrightarrow a=2$
When $b=2i, a^2=3+(2i)^2=3+4(-1)=-1, a=i$
So I will have 3 equations
*
*$2+i$ // why is this not in the book?
*$2-i$
*$i-2$
UPDATE (In response to @David Mitra)
Ok. I let $x=a^2$ and $b=b^2$ giving the quadratic equation: $x=3+y$ and $xy=4$. Then (after subbing) $(y-1)(y+4)=0$
Then when $y=1, x=4$. $y=-4, x=-1$
Then $b=\pm 1 or \pm 2i$ and $a=\pm 2i or \pm i$
Finally testing equations:
$(2+i)^2 = 3+4i$ (rej)
$(2-i)^2=5$ what to do?
$(i-2)^2=3$
$(-2-i)^2=3$
... 8 equations (but 4 unique)
| The high school way:
If $(x+iy)^2=3-4i$, expanding the l.h.s. and identifying the real and imaginary parts, you obtain the system of equations
$\;\begin{cases}
x^2-y^2=3,\\2xy=-4.
\end{cases}$
Now, you can also write that $|x+iy|^2=|3-4i|$, i.e. $\;x^2+y^2=5$. Therefore you obtain the linear system in $x^2$ and $y^2$:
$$\begin{cases}
x^2-y^2=3,\\x^2+y^2=5
\end{cases}\iff \begin{cases}
x^2=4,\\y^2=1
\end{cases}$$
Thus, $x=\pm 2,\;y=\pm 1$. Furthermore, the unused equation $xy=-2$ says that $x$ and $y$ have opposite signs. Consequently the solutions are
$$x+iy=\pm(2-i).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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alternative proof to induction, binomial sum identity
$\displaystyle \sum_{q=0}^{k} \begin{pmatrix}n-1+q\\ n-1 \end{pmatrix} = \begin{pmatrix}n+k\\n \end{pmatrix} $
induction
$\begin{pmatrix}n \\ k \end{pmatrix} := \frac{n!}{(n-k)!k!}$
Beginning of induction : $k=0 \rightarrow \begin{pmatrix} n-1 \\ n-1 \end{pmatrix} = \begin{pmatrix} n \\ n \end{pmatrix} = 1 $
Induction step : $k\rightarrow k+1: \sum_{q=0}^{k+1} \begin{pmatrix} n-1+q\\n-1 \end{pmatrix} = (\sum_{q=0}^{k} \begin{pmatrix} n-1+q\\n-1 \end{pmatrix}) + \pmatrix{n-1+k+1\\n-1} $
$\displaystyle = \begin{pmatrix}n+k \\ n \end{pmatrix} + \begin{pmatrix} n+k \\ n-1 \end{pmatrix} = \frac{(n+k)!}{k!(n)!}+ \frac{(n+k)!}{(k+1)!(n-1)!} =$
$\displaystyle =\frac{(n+k)!(k!n!+(k+1)k!(n-1)!}{k!n!(k+1)!(n-1)!} = \frac{(n+k)!(n+k+1)}{n!(k+1)!} = \begin{pmatrix}n+k+1 \\ n \end{pmatrix}$
Do you know any other way to show this? Please do tell.
| $\binom{n+k}{n}$ is the number of ways of choosing $n$ objects from $n+k$ possibilities.
In order to select $n$ objects out of $n+k$ possibilities, we can also proceed as follows: number the objects from $1$ throught $n+k$. First, pick the object with the largest number that you will select; it must be numbered somewhere between $n$ and $n+k$ (otherwise, you cannot pick $n$ objects total); say it is numbered $n+q$, with $q\in\{0,\ldots,k\}$. Then you still have to pick $n-1$ objects from the first $n+q-1$ objects; this can be done in precisely $\binom{n+q-1}{n-1}$ ways.
Since you can do this for each value of $q\in\{0,\ldots,k\}$, adding up these possibilities will result in a selection of $n$ objects from $n+k$ possibilities; that is,
$$\sum_{q=0}^{k}\binom{n+q-1}{n-1} = \binom{n+k}{n}.$$
| {
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Solve $8^{x+1} = 32 \cdot\sqrt2$ without $\log$ I need help solving the equation $8^{x+1} = 32 \cdot\sqrt2$. The obvious answer is to use log, but that is reserved for the next section. The example given for this section of questions is:
$4^x = 8$
$(2^2)^x = 2^3$
$2x = 3$
$x = \dfrac{3}{2}$
The example looks obvious and easy to solve, but I do not understand where you'd use this for my question.
| The idea is that everything in sight is a power of $2$.
$$
\begin{eqnarray*}
8^{x+1} &=& 32 \sqrt{2} \\
(2^3)^{x+1} &=& 2^5\cdot 2^{1/2} \\
2^{3(x+1)} &=& 2^{5 + 1/2}\\
\end{eqnarray*}
$$
Since both sides are expressed as a power of the same base, we can set the exponents equal:
$$
\begin{eqnarray*}
3(x+1) &=& \frac{11}{2} \\
x+1 &=& \frac{11}{6}\\
x &=& \frac{11}{6} - 1 \\
&=& \frac{5}{6}
\end{eqnarray*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to solve $\pm y \equiv 2x+1 \pmod {13}$ with Chinese remainder theorem or iterative method? how to solve $\pm y \equiv 2x+1 \pmod {13}$ with Chinese remainder theorem or iterative method?
It comes from solving $x^2+x+1 \equiv 0 \pmod {13}$ (* ) and background is following:
13 is prime. (* ) holds under Euclidean lemma if and only if $4(x^2+x+1) \equiv \pmod {13}$
or if and only if $(2x+1)^2 \equiv -3 \pmod {13}$. So if $p=13$, so by Euler's criterion
$[ \frac{-3}{13} ] \equiv (-3)^{\frac{13-1}{2}} = (-3)^6 = 9^3 \equiv (-4)^3 =-64 \equiv 1 \pmod{13} $. Hence equation $y^2 \equiv -3 \pmod{13}$ has two incongruent solution( lemma 4.1.3) $\pm y$ so solutions of the equations $\pm y \equiv 2x+1 \pmod{13}$ are solutions of the equation (* )
So my most important question is how you change equation $\pm y \equiv 2x+1 \pmod{13}$ to the form $ax\equiv b \pmod{13} $ in other words to the form where you can use either Chinese remainder theorem or iterative method to solve $\pm y \equiv 2x+1 \pmod{13}$ and finally (* )?
Finally just because of curiosity. Is $[\frac{-3}{7}]\equiv (-3)^3 = -27 \equiv -1 \pmod{7}$? So is mod(-27,7)=1 or -1?
| Either, $\pm y-1$ is divisible by 2, so you divide by 2, or $\pm y+12$ is divisible by 2, so divide this by 2.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this proof about Mersenne numbers acceptable? I want to prove following property of Mersenne numbers :
If $p > 3$ then $M_p\equiv 1 \pmod {6\cdot p}$
So, according to Fermat's Little Theorem we may write :
$2^p\equiv 2 \pmod p \Rightarrow 2^p-2=a\cdot p \Rightarrow 2^p-1=a\cdot p +1 \Rightarrow$
$\Rightarrow 2^p-1\equiv 1 \pmod p \Rightarrow M_p\equiv 1 \pmod p$
Now, since $M_p\equiv 7 \pmod {24} $ it follows that :
$M_p=24\cdot a +7 =24\cdot a+6+1=6\cdot(4a+1)+1 \Rightarrow M_p\equiv 1 \pmod 6$
Therefore we may conclude :
$(6 \mid (M_p-1) \land p \mid (M_p-1))\Rightarrow 6\cdot p \mid(M_p-1) \Rightarrow M_p \equiv 1 \pmod{6\cdot p}$
Am I correct ?
| Note that $M_p= 2^p -1 = (2-1)(2^{p-1} + 2^{p-2} + \dots + 2^2 + 2^1 + 1)$. If $p>3$ then modding out by $2^3=8$ leaves $2^2+2^1+1 =7$ behind. Also $M_p = 2^p-1$ and $p$ is odd, so $2^p \equiv 2 \mod 3$, which means $M_p \equiv 1\mod 3$. The Chinese Remainder Theorem then applies and shows that $M_p \equiv 7 \mod (3*8)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit of $\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$ $$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$
Can't find a means to resolve. The answer is $10$ by graphing.
| We give two solutions. The first uses a standard trick that you have probably already seen. The second uses the definition of the derivative.
You are familiar with the identity
$$(\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v, \qquad \qquad (\ast)$$
at least in its incarnation $(a-b)(a+b)=a^2-b^2$. We will use $(\ast)$ twice. First, we multiply the top and bottom of your expression by
$$(x +\sqrt{x^2+5x+2})(x+\sqrt{x^2+0.5x + 1}).$$
We get
$$\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\frac{(x - \sqrt{x^2+5x+2})(x +\sqrt{x^2+5x+2})(x+\sqrt{x^2+0.5x + 1})}{(x-\sqrt{x^2+0.5x + 1})(x+\sqrt{x^2+0.5x + 1})(x +\sqrt{x^2+5x+2})}.$$
The product of the first two terms on top is $-(5x+2)$, and the product of the first two terms at the bottom is $-(0.5x+1)$. It follows that
$$\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\frac{(5x+2)(x+\sqrt{x^2+0.5x + 1})}{ (0.5x+1)(x +\sqrt{x^2+5x+2})}.\qquad\qquad(\ast\ast)$$
The behaviour of the right-hand side of $(\ast\ast)$ for large $x$ is almost obvious. To do the calculation formally, divide top and bottom by $x^2$, as follows:
$$\frac{(5x+2)(x+\sqrt{x^2+0.5x + 1})}{ (0.5x+1)(x +\sqrt{x^2+5x+2})}=
\frac{(5+\frac{2}{x})(1+\sqrt{1+ \frac{0.5}{x} + \frac{1}{x^2}})}{ (0.5+\frac{1}{x})(1 +\sqrt{1+\frac{5}{x}+\frac{2}{x^2}})}.\qquad\qquad(\ast\ast\ast)$$
If we look at the right-hand side of $(\ast\ast\ast)$, it should be clear that as $x \to\infty$, the top approaches $10$ and the bottom approaches $1$.
Another way: Divide top and bottom by $x$. We get
$$\frac{1 - \sqrt{ 1+\frac{5}{x}+\frac{2}{x^2}}} {1-\sqrt{ 1+\frac{0.5}{x} + \frac{1}{x^2} }}.$$
Make the substitution $t=\frac{1}{x}$, and change the sign of top and bottom. We are now interested in the limit as $t$ approaches $0$ from the right of
$$\frac{\sqrt{1+5t+2t^2}-1} {\sqrt{1+0.5t + t^2 }-1}.$$
Looks nice than before! We could now use a trick much like in the first solution. But we will use another idea. Multiply top and bottom by $t-0$. We get
$$\frac{\sqrt{ 1+5t+2t^2}-1}{t-0}\frac{t-0}{\sqrt{ 1+0.5t + t^2 }-1}.$$
What is
$$\lim_{t\to 0}\frac{\sqrt{ 1+5t+2t^2}-1}{t-0}?$$
If we look back on the definition of the derivative, we will recognize
this limit as the derivative of $f(t)=\sqrt{1+5t+2t^2}$ at $t=0$. Calculate $f'(0)$. It is $5/2$. The limit of the second term is $1/g'(0)$, where $g(t)=1+0.5t +t^2$. Calculate. We get $1/(0.5/2)$. The product of the two limits is $10$.
Comment: The technique (in another answer) that uses the Taylor expansion is more sophisticated, but no harder. In the long run it is much more useful.
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Subsets and Splits
Fractions in Questions and Answers
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