Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Computation of the $2n$-th order determinant Compute the determinant of the $2n-\text{th}$ order.
$$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\\2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\end{vmatrix}$$
My attempt:
I noticed the two following blocks:
$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\end{vmatrix}\;\&\;\begin{vmatrix}2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\end{vmatrix}$
I switched the blocks because I was dealing with the determinant of the even-order:
$$\begin{vmatrix}2&0&\ldots&0&0&0&0&\ldots&0&3\\0&2&\ldots&0&0&0&0&\ldots&3&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\\0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&3&\ldots&0&0&0&0&\ldots&2&0\\3&0&\ldots&0&0&0&0&\ldots&0&2\end{vmatrix}$$
Then I saw we can subtract $j-\text{th column}$ multiplied by $-\frac{3}{2}$ from the $(n-j+1)-\text{column}\;\forall j\in\{1,\ldots,2n\}$
Then I got a $\text{lower-triangular}$ matrix with entries $-\frac{5}{2}$ on the main diagonal.
My final result is: $$D_{2n}=\left(-\frac{5}{2}\right)^{2n}=\left(\frac{5}{2}\right)^{2n}$$
Is this correct?
| *
*When you switch the two blocks you switch $n$ pairs of rows therefore the determinant comes multiplied by $(-1)^n$.
*When you perform the column additions only half of the diagonal entries of the matrix are altered.
The final result should be $(-1)^n2^n(-\frac{5}{2})^n=(-1)^{n+1}2^n(\frac{5}{2})^n=(-1)^{n+1}5^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Diagonalizability of a matrix
Show that $$ A :=\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 2 \\ -2 & 0 & 3 \end{pmatrix}$$ is diagonalizable.
What I did:
First, I determined the characteristic polynomial $$\chi_A(X) = \det(X \cdot E_3-A)=(X-3)(X-1)(X-1)=X^3-5X^2+7X-3,$$
so the eigenvalues are $3$ and $1$.
I then determined the eigenspaces of each eigenvalue:
$$X=3: \left(\begin{array}{@{}ccc|c@{}}
2 & 0 & 0 & 0 \\
-2 & 2 & 2 & 0 \\
-2 & 0 & 0 & 0 \\
\end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}}
2 & 0 & 0 & 0 \\
0 & 2 & 2 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right),$$ so $x_1=0$, $x_2=-x_3$, $x_3=x_3$ and thus $V_3(C) = \left< \begin{pmatrix} 0\\-1\\1 \end{pmatrix} \right>$.
Analogous:
$$X=1: \left(\begin{array}{@{}ccc|c@{}}
0 & 0 & 0 & 0 \\
-2 & 0 & 2 & 0 \\
-2 & 0 & -2 & 0 \\
\end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}}
-2 & 0 & 2 & 0 \\
0 & 0 & -4 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right),$$ so $x_1=x_3=0$, $x_2=x_2$ and thus $V_1(A) = \left< \begin{pmatrix} 0\\1\\0 \end{pmatrix} \right>$.
It now follows that $\dim(V_3(C)) + \dim(V_1(A)) = 1+1=2 \lt 3 = \dim(A)$ and because of the $\lt$, A shouldn't be diagonalizable, but it is.
So where's the mistake? Thanks in advance!
| Your error lies in the computation of the eigenspace corresponding to the eigenvalue $1$, which is $2$-dimensional; it is spanned by $(1,0,1)$ and by $(0,1,0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generalized version of calculating expected minimum value rolled until $5$ is obtained from a fair die The following is a modified interview question.
Given an $n$-sided fair die where $n\geq 1.$
You roll a die until you get a $m$ where $1\leq m\leq n$.
Calculate the expected value of the minimum rolled.
The original interview question takes $n=6$ (standard fair die) and $m=5$.
I mange to solve the problem and I reproduce my attempt below.
The expected minimum value rolled is $\frac{137}{60}$ because if $X$ is the minimum value rolled up to and including $5,$ then
$$P(X=x) = \frac{1}{x(x+1)} \quad \text{for }x=1,2,3,4 \quad \text{and} \quad P(X=5) = \frac{1}{5}.$$
So,
$$E(X) = \sum_{x=1}^5 xP(X=x) = \frac{137}{60}.$$
I am trying to solve the generalized version of the problem.
By the same spirit, let $Y$ be the minimum value rolled up to and including $m.$ Then
$$P(Y=y) = \frac{(y-1)!}{(y+1)!} = \frac{1}{y(y+1)} \quad \text{for }y =1,2,...,m-1 \quad \text{and}\quad P(Y=m) = \frac{1}{m}.$$
Therefore,
$$E(Y) = \sum_{y=1}^m y P(Y=y) = 1 + \sum_{y=1}^{m-1} \frac{1}{y+1} = 1 + \sum_{y=2}^{m} \frac{1}{y}.$$
| A possibly simpler derivation of the fact that $P(Y \geq y) = \frac1y,$
where $Y$ is the smallest number rolled among all the rolls up to and including the first roll of $m.$
Consider the set of numbers $\{1, 2, \ldots, y-1, m\}.$
Let $Z$ be the first number rolled from among the numbers in that set.
(We know at least one number is rolled, since we know that eventually we roll $m$.)
Each number is equally likely to be the first rolled. With probability $\frac1y,$ that number is $m.$
If $m$ is the first number rolled from that set, then $Y \geq y.$
If $m$ is not the first number rolled from that set, then $Y < y.$
Hence $P(Y \geq y) = P(\text{$m$ is rolled before any number less than $y$}) = \frac1y.$
Note that $n$ is irrelevant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Products of Orthogonal Vectors Let $a,b,c$ be orthonormal vectors in $\mathbb{R}^n$. Then is $\sum_{k=1}^n a_{k}b_{k}c_{k} = 0$?
I tried two examples and it was true, but I don't immediately see why.
| Counterexample:
$$a = \begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt 3}\\\frac{1}{\sqrt 6}\end{pmatrix},~~b = \begin{pmatrix}\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt 3}\\\frac{1}{\sqrt 6}\end{pmatrix},~~ c = \begin{pmatrix}0\\\frac{1}{\sqrt 3}\\\frac{-2}{\sqrt 6}\end{pmatrix}.~~$$
Then
$$\sum_k a_kb_kc_k = \frac{2\sqrt 3 - \sqrt 6}{18} \neq 0.$$
Rational counterexample:
$$a = \frac{1}{3} \begin{pmatrix}1\\2\\-2\end{pmatrix},~~b = \frac{1}{3} \begin{pmatrix}2\\1\\2\end{pmatrix},~~c = \frac{1}{3} \begin{pmatrix}-2\\2\\1\end{pmatrix}.$$
In this case, $$\sum_k a_kb_kc_k = \frac{-4}{27} \neq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the sum of the series: $ \sum_{n=1}^\infty (-1)^{n-1}\frac{\cos nx}{n(n+1)} $
Find the sum of the series:
$$
\sum_{n=1}^\infty (-1)^{n-1}\frac{\cos nx}{n(n+1)}
$$
My attempt:
$$
\begin{aligned}
&\sum_{n=1}^\infty (-1)^{n-1}\frac{\cos nx +i\sin nx}{n(n+1)}=\\
&=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{(\cos x +i\sin x)^n}{n}-\frac{(\cos x +i\sin x)^n}{n+1}\right)=\\
&=[\cos x + i\sin x = t]=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{t^n}{n}-\frac{t^n}{n+1}\right)=\dots=\\
&=\ln(1+t)+\frac{1}{t}\left(-\ln|1+t|+t\right)
\end{aligned}
$$
But I don't know how to end the solution. How to ged rid of $i$ inside the logarithms?
P.S. The answer my textbook gives me is
$
(1+\cos x)\ln\left(2\cos\frac{x}{2}\right)+\frac{1}{2}x\sin x - 1
$.
| We have that
$$\cos((n+1)x)+\cos(nx)=2\cos(x/2)\cos((n+1/2)x)$$
Therefore, we split the given series into a telescopic sum and another convergent series,
$$S(x):=\sum_{n=1}^\infty (-1)^{n-1}\frac{\cos(nx)}{n(n+1)}=
\sum_{n=1}^\infty \left((-1)^{n-1}\frac{\cos(nx)}{n}-(-1)^{n}\frac{\cos(n+1)x}{n+1}\right)\\
+2\cos(x/2)\text{Re}\left(\sum_{n=1}^\infty\frac{(-1)^{n} (e^{ix})^{n+1/2}}{n+1}\right).$$
Finally, after recalling that $\ln(1+z)=\sum_{n=1}^\infty (-1)^{n-1}\frac{z^n}{n}$ for $|z|\leq 1$ and $z\not=-1$, we find
$$\begin{align}
S(x)&=\cos(x)+2\cos(x/2)\text{Re}\left(e^{-ix/2}(\ln(1+e^{ix})-e^{ix})\right)\\
&=\cos(x)+\cos(x/2)(\cos(x/2)\ln(2+2\cos(x))+x\sin(x/2))
-2\cos^2(x/2)\\
&=(1+\cos(x))\ln\left(2\cos(x/2)\right)+\frac{x\sin(x)}{2} - 1.
\end{align}$$
where
$$\ln(1+e^{ix})=\ln(|1+e^{ix}|)+i\text{Arg}(1+e^{ix})
=\frac{1}{2}\ln(2+2\cos(x))+\frac{ix}{2}.$$
P.S. Your textbook answer is valid for $x\in (-\pi,\pi)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $9\mid a^6+a^{6^2}+\ldots+ a^{6^{27}},\;a\in\mathbb N$
Prove $$9\mid a^6+a^{6^2}+\ldots +a^{6^{27}},a\in\mathbb N$$
My attempt:
\begin{aligned}&a \equiv 2\pmod{9}\implies a^{6} \equiv 2^{6} \equiv 1\pmod{9}\\&\begin{array}{l}{a \equiv 3\pmod{9}\implies a^{6} \equiv 3^{6} \equiv 0\pmod{9}} \\{a \equiv 4\pmod{9}\implies a^{6} \equiv\left(2^{6}\right)^{2} \equiv 1\pmod{9}}\\a\equiv 5\equiv -4\pmod{9}\implies a^6\equiv(-4)^6=4^6\equiv 1\pmod{9}\end{array}\end{aligned}
$$a^{6^i}=\left(a^6\right)^{6^{i-1}}\equiv [0,1]\pmod{9}\implies a^{6^i}=9k\;\lor\;a^{6^i}=9k+1,\;k\in\mathbb N_0$$
$$a^6+a^{6^2}+\ldots +a^{6^{27}}=\sum_{i=1}^{27}\left(a^6\right)^{6^{i-1}}=\sum_{i=1}^{27}(9k_i+[0,1])=\underbrace{\sum_{i=1}^{27}9k_i+[0,27]}_{\text{multiple of}\; 9}$$
//edited mistake [27,54]//
Is this correct and is there a more efficient method?
| Observe that $$a^m(a^n-1)\equiv0\pmod9$$ for any integer $a$ if $m\ge2$ and $6\mid n$
So, for integer $k\ge1,$ $$a^{6^{k+1}}-a^6=a^6(a^{6^k}-1)\equiv0\pmod9$$
$$\implies\sum_{k=1}^na^{6^k}\equiv na^6\pmod9$$
Here $n=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.
| If $x, y, z > 0$, then we see that
$$
\begin{align}
\frac{x}{y} + \frac{ y }{z+x } + \frac{z }{x } - 2 &= \frac{ x^2(z+x) + xy^2 + yz(z+x) - 2xy(z+x) }{ xy(z+x) } \\
&= \frac{ yz^2 + x(x - y)z + x\left(x^2 - y^2 \right) }{ xy(z+x) }.
\end{align}
$$
Now consider the numerator, which is a quadratic in $z$; it has the discriminant
$$
\begin{align}
x^2(x- y)^2 - 4 xy \left(x^2 - y^2 \right) &= x^4 - 2x^3y + x^2y^2 - 4x^3y + 4xy^3 \\
&= x^4 - 6x^3y + x^2y^2 + 4xy^3 \\
&=
\end{align}
$$
The idea is to see if the discriminant is negative for all values of $x, y > 0$, or not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$ The Function is $$f(x) = \log_3\sqrt{(x+3)}−\log_3(9−x^2)$$
and I need to figure out arguments for which $$ f(x) < 0 $$
So I calculated the domain of function which is $ D: (-3;3)$
However I am still unable to solve $f(x) < 0$
I simplified $ \log_3\sqrt{(x+3)}−\log_3(9−x^2) < 0$ to $\log_3\frac{\sqrt{(x+3)}}{(9-x^2)} < 0$
which gets me to $$ \frac{\sqrt{(x+3)}}{9-x^2} < 1$$
But the solutions of the above equation are complex numbers and I definitely should get them as my result. So, what I am doing wrong here?
Would apprecite every answer
| Hint $
$\begin{align*}
f(x) &= \log_3\sqrt{(x+3)}−\log_3(9−x^2) \\
&= \frac{1}{2}\log_3(x+3) - \log_3((3-x)(3+x)) \\
&=\frac{1}{2}\log_3(x+3) - (\log_3(3-x)+\log_3(3+x))
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of the series $\sum_{k=1}^\infty \frac{\cos 2kx}{k^2}$
Find $S_1(x)$:
$$
S_1(x)=\sum_{k=1}^\infty \frac{\cos 2kx}{k^2}
$$
My attempt:
$$
\begin{aligned}
&\text{Let } S_2=\sum_{k=1}^\infty \frac{\sin 2kx}{k^2}\\
&S_1+i\cdot S_2=\sum_{k=1}^\infty\frac{(\cos x+i\sin x)^{2k}}{k^2}=[z=\cos x+i\sin x]=\\
&=\sum_{k=1}^\infty\frac{z^{2k}}{k^2}=S\\
&\left(\frac{z^{2k}}{k^2}\right)'_z=\frac{2z^{2k-1}}{k}\\
&\sum_{k=1}^\infty\frac{2z^{2k-1}}{k}=\frac{2}{z}\sum_{k=1}^\infty
\frac{(z^2)^k}{k}=-\frac{2\ln(1-z^2)}{z}=T\ \ \ \ \ \ (|z|<1)
\end{aligned}
$$
Then I tried to integrate $T$ (from $0$ to $z$) in order to get sum $S$. But this integral seems to get me back to the polylogarithm I was trying to avoid. So, I don't see how I can write $S_1$ with the finite number of terms (without the polylogarithm function).
| By "guessing" the appropriate function, we can find this sum with Fourier series.
First guess:
$$x^2 \sim \frac{\pi^2}{3} +\sum_{k=1}^\infty (-1)^k \frac{\cos k x}{k^2}.$$
After guessing that the function is a parabola (quadratic in $x$), found the cosine series for:
$$f(x)=(x-a)^2, \textrm{ on the interval } \left(0, \frac{\pi}{2}\right).$$
It became apparent that $a=\frac{\pi}{2}$ works.
$$f(x) \sim \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos 2 k x$$
$$a_0 = \frac{4}{\pi} \int_0^{\frac{\pi}{2}} \left(x-\frac{\pi}{2}\right)^2 \, dx = \frac{\pi^2}{6}$$
$$a_k = \frac{4}{\pi} \int_0^{\frac{\pi}{2}} \left(x-\frac{\pi}{2}\right)^2 \cos 2kx \, dx = \frac{1}{k^2}$$
$$\left(x-\frac{\pi}{2}\right)^2 \sim \frac{\pi^2}{12} + \sum_{k=1}^\infty \frac{\cos 2 k x}{k^2}.$$
So
$$ \sum_{k=1}^\infty \frac{\cos 2 k x}{k^2} \sim \left(x-\frac{\pi}{2}\right)^2 - \frac{\pi^2}{12} .$$
The series is equal to the function on the right hand side in the interval $x \in (0,\pi)$, and its periodic extension, with period $\pi$.
Of course, this is the answer @Gary came up with in the comments.
| {
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"timestamp": "2023-03-29T00:00:00",
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Drawing red and blue balls from a box by $2$ people, A box contains five blue and eight red balls. Jim and Jack start drawing balls from the box, respectively, one at a time, at random, and without replacement until a blue ball is drawn. What is the probability that Jack draws the blue ball?
The books says the answer is:
$8\cdot5\cdot11!+8\cdot7\cdot6\cdot5\cdot9!+8\cdot7\cdot6\cdot5\cdot4\cdot5\cdot7!+ 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 5 \cdot 5! = 2, 399, 846, 400$
Therefore, the answer is $\frac{2, 399, 846, 400}{13!} = 0.385$.
I do not understand the solution the book provided!
This is how I approached the problem:
There are $13!$ factorial ways of picking out all the balls. But there are $\frac{13!}{5!\cdot8!}$ distinguishable ways of picking out all the balls. So I don't see how we got different denominators. Second, I was hoping to calculate the probability that the first $8$ balls are red, the complement of which would give me an answer that at least one blue ball would be picked in the first $8$ picks, and divide that by two for the probability of Jack. There is $1$ way of picking $8$ red balls in a row out of $\frac{13!}{5!\cdot8!}$.
$P[$Jack picking blue$] =\frac{1-\frac{1}{13\cdot 11\cdot 9}}{2}\approx 0.499$
How did I go wrong? And what was the thinking of the author?
| It is absolutely irrelevant to take $13!$ or $\frac{13!}{5!8!}$ events. You can either consider all permutations of balls of the same color as different or not. This does not effect on the ratio since both coefficients $5!8!$ in denominator and numerator are cancelled.
And what is relevant is how you calculates number of events where Jack picks blue ball. How the probability that first $8$ balls are red relates to this? Yes, if the first $8$ balls are red, then Jack cannot win. But this is not the unique case when he lost. If the first drawn ball is blue, he also lost. If the first two balls are red and the third is blue, Jack also lost. And so on.
First Jim picks ball. He must pick red ball ($8$ variants). Next Jack can pick blue or red. If he picked blue ($5$ variants) then the game is over and the other permutations of $5+8-1-1=11$ balls are irrelevant. If we will not distinguish the balls of the same color, the number of elementary events which correspond to the case when Jack wins by his first step is
$$
\frac{\color{red}8\cdot \color{blue}5\cdot 11!}{5!8!}.
$$
If Jack picked red ($7$ variants), then Jim should also pick red ($6$ variants) and then Jack can pick blue or red. If he picked blue ($5$ variants), then the game is over and the potential $5+8-4=9$ balls can permute in any order, so if we will not distinguish the balls of the same color, the number of elementary events which correspond to the case when Jack wins by his second step is
$$
\frac{\color{red}8\cdot \color{red}7\cdot \color{red}6\cdot \color{blue}5\cdot 9!}{5!8!}.
$$
Next, there are
$$
\frac{\color{red}8\cdot \color{red}7\cdot \color{red}6\cdot \color{red}5 \cdot \color{red}4 \cdot \color{blue}5\cdot 7!}{5!8!}
$$
variants where Jack wins by its third step. And there are the
$$
\frac{\color{red}8\cdot \color{red}7\cdot \color{red}6\cdot \color{red}5 \cdot \color{red}4 \cdot \color{red} 3 \cdot \color{red}2 \cdot \color{blue}5\cdot 5!}{5!8!}
$$
variants to win by fourth step of Jack.
All the other variants are impossible since there is not enough amount of red balls.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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maximum value of $ac+bd$
If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is
what i try
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$
$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$
but answer is $3$
How do i get right answer help me please
| Let $a = r_1 \cos p, b = r_1 \sin p, c = r_2 \cos q$, and $ d = r_2 \sin q$, where $0 < r_1 ≤ \sqrt{2}$ and $0 < r_2 ≤ 2$. This satisfies the inequalities mentioned in the problem.
Then:
$$ac+bd = r_1 r_2\cos p \cos q + r_1 r_2 \sin p \sin q = r_1 r_2 ( \cos p \cos q + \sin p \sin q) = r_1 r_2 \cos(p-q)$$
Since $\cos x ≤ 1$ for all real $x$, the maximum value is $\sqrt{2} \cdot 2 \cdot 1 = 2 \sqrt{2}$.
Sidenote: This is only one choice of variables. Another choice of variables such as $a = r_1 \sin p, b = r_1 \cos p, c = r_2 \cos q, d = r_2 \sin q$ leads to the same result. Try it yourself!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the shortest distance from the origin to the surface of $x=yz+10$ we can use the distance formula, but for the sake of simplicity we can use $f(x,y,z)=x^2+y^2+z^2$ and let $\phi(x,y,z)=x-yz=10$. Via Lagrange multiplier,
\begin{align}\partial_xf+\partial_x\phi\cdot\lambda=0\\
2x+\lambda=0\tag{i}\\
\partial_yf+\partial_y\phi\cdot\lambda=0\\
2y-z\lambda=0\tag{ii}\\
\partial_zf+\partial_z\phi\cdot\lambda=0\\
2z-\lambda y=0\tag{iii}
\end{align}
from equating (ii) and (iii) we find that $$2z=\frac{\lambda^2z}{2}\implies \lambda=\pm 2$$
for $\lambda=2,y=z,$ and $x=-1$, which means
$$-1-yz=10$$
$$-yz=11$$
but since $y=z$
$$y^2=-11$$
Therefore $\lambda=-2\implies y=-z,x=1$
\begin{align}
-y^2=-9\\
y=3\\
z=-3
\end{align}
Therefore the shortest distance from the origin to the plane is $\sqrt{9+9+1}=\sqrt{19}$
I believe this should be correct, but I have no idea how to verify it, could anyone help me confirm this, or point out where I made a mistake? Thanks!
| We substitute $x = 10 - yz$ into squared distance $d^{2} = x^{2}+y^{2}+z^{2}$ to obtain $ d^{2}=100+20yz+(yz)^{2}+y^{2}+z^{2}$. Rewrite the equation as $d^{2}=19+(y+z)^{2}+(yz+9)^{2}$. This squared distance is minimized if $y+z=0$ and $yz+9=0$, the minimum squared distance is then $19$ and the minimum distance is $\sqrt{19}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553356",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Roots of the equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ The equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ has
(A) all its solution real but not all positive
(B) only two of its solution real
(C) two of its solution positive and two negative
(D) none of its solution real
My approach is as follow $(x^2+3x+4)^2+3(x^2+3x+4)+4-x=y$
$f(x)=y=x^4+6x^3+20x^2+32x+32$
$f(-x)=y=x^4-6x^3+20x^2-32x+32$
Using Descartes rule no positive roots but the possible ways we can have either 4,2,0 negative roots.
$f'(x)=4x^3+18x^2+40x+32$
$f''(x)=12x^2+36x+40$ which is imaginary
From here I am not able to approach.
| $\displaystyle f(x)=(x^2+3x+4)^2+3(x^2+3x+4)+4-x$
$\displaystyle f(x)=\bigg[\bigg(x+\frac{3}{2}\bigg)^2+\frac{7}{4}\bigg]^2+3\bigg[\bigg(x+\frac{3}{2}\bigg)^2+\frac{7}{4}\bigg]+4-x$
put $\displaystyle x+\frac{3}{2}=t\in \mathbb{R}$
$\displaystyle f(t)=\bigg(t^2+\frac{7}{4}\bigg)^2+3\bigg(t^2+\frac{7}{4}\bigg)+4-t+\frac{3}{2}$
$\displaystyle f(t)=t^4+\frac{49}{16}+\frac{7}{2}t^2+3t^2+\frac{21}{4}+4+\frac{3}{2}-t$
$\displaystyle f(t)=t^4+\frac{13}{2}\bigg[t^2-\frac{2}{13}t+\frac{1}{13^2}+\frac{43}{26}-\frac{1}{13^2}\bigg]$
$\displaystyle f(t)=t^4+\frac{13}{2}\bigg[\bigg(t-\frac{1}{13}\bigg)^2+\frac{43\cdot 13-2}{26\cdot 13}\bigg]>0$
| {
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"answer_count": 4,
"answer_id": 3
} |
How to solve a differential equation $(x^2y+y^5)dx+(x^3-xy^4)dy=0$?
Solve the following differential equation:
$$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
I noticed that it is not exact, since:
$$
(x^2y+y^5)'_y=x^2+5y^4\ne3x^2-y^4=(x^3-xy^4)'_x
$$
Then I tried to express $y'$:
$$
y'=\frac{x^2y+y^5}{xy^4-x^3}
$$
And now I don't understand what type of differential equation it is and how to solve it.
Could someone help me?
| Probably, I can help a bit.
It seems to me, you can try substitution
$$
f = - \frac{1}{xy},\quad g = - \frac{y^3}{x^3}.
$$
Since
$$
df = \frac{dx}{x^2 y} + \frac{dy}{x y^2},\quad
dg = \frac{3 y^3 dx}{x^4} - \frac{3 y^2 dy}{x^3},
$$
you can write
$$
df + \frac{1}{3} dg = \frac{\left(x^2y + y^5\right)dx+ \left(x^3-x y^4\right)dy}{x^4 y^2}.
$$
Now solution should be straightforward.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do I get the inverse of this equation? The equation is the following:
$$y = \frac{x}{x-2}$$
What I did is:
$$y = \frac{x}{x-2}$$
$$y \times (x-2) = x$$
$$\frac{x-2}{x} = \frac{1}{y}$$
$$\frac{x}{x}-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{x}{2} = y$$
I've got a lot of doubts about that last step. Besides that, the result according to the textbook is the following:
$$ f^{−1}(x)=-\frac{2x}{x−1} $$
| A different approach is to note that
$$
y = \frac{x}{x-2} = \frac{(x-2)+2}{x-2} = 1+\frac{2}{x-2}
$$
Hence, $(x-2)(y-1) = 2$ and $x = \frac{2}{y-1}+2 = \frac{2y}{y-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Inequality $a^{2b}+b^{2a}\leq \cos(ab)^{(a-b)^2}$ It's well know that we have (inequality of Vasile Cirtoaje) :
Let $a,b>0$ such that $a+b=1$ then :
$$a^{2b}+b^{2a}\leq 1$$
The following is a add of mine :
Let $a,b>0$ such that $a+b=1$ then :
$$a^{2b}+b^{2a}\leq \Big(\cos(ab)\Big)^{(a-b)^2}$$
I try the following inequalities :
$$a^{2b}\leq \beta \Big(\cos(ab)\Big)^{(a-b)^2}$$
$$b^{2a}\leq (1-\beta) \Big(\cos(ab)\Big)^{(a-b)^2}$$
And take log on both side but I'm stuck on this way.
I try furthermore to use convexity but I gave up quickly.
Finally I try Bernoulli's inequality but it's too weak .
Any helps are very appreciated .
Thanks a lot for your time and patience .
| Partial answer
Due to symmetry, assume that $a \ge b$.
By using $a = 1- b$ and $b\in (0, \frac{1}{2}]$, the desired inequality is written as
$$(1-b)^{2b} + b^{2 - 2b} \le (\cos (b - b^2))^{(1-2b)^2}.$$
Let us first prove the case when $b\in (0, \frac{1}{3}]$.
We give the following auxiliary results (Facts 1 through 3). Their proof is not difficult and hence omitted.
Fact 1: $(1-x)^r \le 1 - rx + \frac{1}{2}(r^2-r)x^2$ for $x\in (0, \frac{1}{2}]$ and $0 < r \le 1$.
Fact 2: $\cos y \ge 1 - \frac{1}{2}y^2$ for $y \in (0, +\infty)$.
Fact 3: From Bernoulli inequality, $(1-x)^r \ge 1 - rx$ for $r \ge 1$ and $0 < x < 1$.
From Facts 2 and 3, we have for $b \in (0, \frac{1}{2}]$,
\begin{align}
&(\cos (b - b^2))^{(1-2b)^2} \\
\ge\ & [1 - \tfrac{1}{2}(b - b^2)^2]^{(1-2b)^2}\\
= \ & [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}[1 - \tfrac{1}{2}(b - b^2)^2]^{1+(1-2b)^2}\\
\ge \ & [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}
\Big(1 - \tfrac{1}{2}(b - b^2)^2[1 + (1-2b)^2] \Big).
\end{align}
From Fact 1, by letting $x = b, r = 2b$, we have for $b \in (0, \frac{1}{2}]$,
$$(1-b)^{2b} \le (1-b)(1+b - b^2 - 2b^3).$$
Thus, it suffices to prove that for $b\in (0, \frac{1}{3}]$,
$$(1-b)(1+b - b^2 - 2b^3) + b^{2-2b} \le [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}
\Big(1 - \tfrac{1}{2}(b - b^2)^2[1 + (1-2b)^2] \Big)$$
or
$$\frac{b^2(2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3)}{-b^4+2 b^3-b^2+2}-b^{2-2b}\ge 0$$
or
$$\frac{2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3}{-b^4+2 b^3-b^2+2}-b^{-2b}\ge 0$$
or
$$\ln \frac{2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3}{-b^4+2 b^3-b^2+2} + 2b \ln b \ge 0.$$
It is not difficult with computer. Indeed, denote LHS by $f(b)$. One can prove that $f''(b) > 0 $ for $b\in (0, \frac{1}{3}]$.
Then one can prove that $f'(b) < 0$ for $b\in (0, \frac{1}{3}]$. Omitted.
| {
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Determine whether $\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}$ diverges or not
Determine whether $\{a_n\}$ is convergent or not, where
$$a_n:=\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}.$$
At least, we can obtain$$\sqrt[n]{1+\sqrt[n]{2+\sqrt[n]{3+\cdots+\sqrt[n]{n}}}}\le\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt[n]{n}}}}<2,$$which imlies $\{a_n\}$ is bounded. But is it monotonic?
| We have
$$
\sqrt[n]{n} \le n,
$$
$$
\sqrt[n]{{n - 1 + \sqrt[n]{n}}} \le \sqrt[n]{{n + \sqrt[n]{n}}} \le \sqrt[n]{{2n}} \le 2n,
$$
$$
\sqrt[n]{{n - 2 + \sqrt[n]{{n - 1 + \sqrt[n]{n}}}}} \le \sqrt[n]{{n + \sqrt[n]{{n - 1 + \sqrt[n]{n}}}}} \le \sqrt[n]{{n + 2n}} \le 3n,
$$
etc. Hence $$
\sqrt[n]{{1 + \sqrt[n]{{2 + \cdots + \sqrt[n]{n}}}}} \le \sqrt[n]{{1 + (n - 1)n}} \to 1.
$$
On the other hand $a_n >1$. Hence, the limit is $1$.
| {
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"url": "https://math.stackexchange.com/questions/3559511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find $a, b \in \mathbb{R}$ such that the polynomial $f(x)=x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$ has a triple root. I am given the following polynomial:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$$
with $a, b \in \mathbb{R}$.
I have to find $a$ and $b$ such that the given polynomial has a triple root.
I know that if a polynomial has a triple root $\alpha$ then we have:
$$f(\alpha) = 0$$
$$f'(\alpha) = 0$$
$$f''(\alpha) = 0$$
And in previous exercises (where the degree of the polynomial was $4$, not $5$) I could use $f''(\alpha) = 0$ to find $2$ $\alpha$'s and find $a$ and $b$ for each $\alpha$. I could do that because the second derivative of a $5$th degree polynomial is a quadratic so I could find the $\alpha$'s. Here, it's different. We have:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b$$
$$f'(x) = 5x^4+28x^3+57x^2+52x+a$$
$$f''(x) = 20x^3+84x^2+114x+52$$
And when I try to solve $f''(\alpha) = 0$ I get:
$$20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$$
And I don't know how to find the $\alpha$'s. So, what approach should I use to solve this exercise?
| $20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$
$10\alpha^3 + 42\alpha^2 + 57\alpha + 26=0$
Rational roots theorem says if there is are rational root it is $\pm\frac {1,2,13}{1,2,5}$ and it's clearly negative.
And we can try by brute force $\alpha =-2$.
If we don' want to try them all we can note such large coefficients can be reduced by letting $\alpha = b-1$ and
$10(b-1)^3 + 42(b-1)^2 + 57(b-1) + 26=0$
$10b^3 +(-30+42)b^2 + (30-2*42+57)b +(-10+42-57+26) =0$
$10b^3 +12b^2 +3b +1=0$ and as $10+3 = 12+1$ we can see $b=-1$ is a solution.
Or we could let $b=c-1$ and get
$10(c-1)^3 + 12(c-1)^2 + 3(c-1) + 1 =0$
$10c^3 + (-30 + 12)c^2 + (30-24 + 3)c + (-10+12-3+) = 0$
$10c^3 -18c^2 +9c =0$ which has $c=0$ as solutions as well as $\frac {18\pm\sqrt{18^2-4*9*10}}{20}$ which are not real solutions.
but at any rate $\alpha = -2$ is a solution.
So $f'(-2) = 5*16-28*8+57*4-52*2+a=0$
$a = -80 + 224 - 228 + 104= 20$
And $f(-2) = -32 + 7*16 - 19*8 + 26*4 -2*20 + b$
so $b=32-112 +152-104 + 40 = 8$.
| {
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"source": "stackexchange",
"question_score": "2",
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Evaluate the challenging $\int_0^{\infty } \frac{\sin (x)}{\sqrt{x} \left(\cos ^2(x)+1\right)} \, dx$ in terms of Legendre Chi function How to prove $$\int_0^{\infty } \frac{\sin (x)}{\sqrt{x} \left(\cos ^2(x)+1\right)} \, dx=
\sqrt{2\pi} \sum _{k=1}^{\infty } \frac{(-1)^{k-1} \left(\sqrt{2}-1\right)^{2 k-1}}{\sqrt{2 k-1}}$$
Any help will be appreciated.
Update: Using @uniquesailor's hint the problem is solved. Indeed, set $b=3-2 \sqrt{2}$ and make use of $\frac{1}{\cos ^2(x)+1}=\frac{2}{3 \left(\frac{1}{3} \cos (2 x)+1\right)}$, one may broke the integrand into Fourier series based on his Poisson type formula. Then, upon using Fresnel's result $\int_0^{\infty } \frac{\sin (x)}{\sqrt{x}} \, dx=\sqrt{\frac{\pi }{2}}$ and the trigonometric identity $2 \sin (x) \cos (2 n x)=\sin ((2 n+1) x)-\sin ((2 n-1) x)$, the integral is transformed to RHS after rearranging. According to Benidict RHS is also equivalent to $-i \sqrt{2\pi} \chi_{\frac{1}{2}}(i (\sqrt{2}-1))$.
| Not quite complete, but this looks promising....
$$
\Phi(z,s,a) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-at}}{1-ze^{-t}}\;dt
$$
$$
\chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^\nu} = 2^{-\nu} z \Phi(z^2,\nu,\frac{1}{2})
$$
For your sum we clearly have something like $\nu = \frac{1}{2}, z = (\sqrt{2}-1)$
$$
\chi_{\frac{1}{2}}((\sqrt{2}-1)) = \sum_{k=0}^\infty \frac{(\sqrt{2}-1)^{2k+1}}{\sqrt{2k+1}} =\frac{2^{-1/2}(\sqrt{2}-1)}{\Gamma(\frac{1}{2})}\int_0^\infty \frac{x^{-\frac{1}{2}}e^{-x/2}}{1-(\sqrt{2}-1)^2e^{-x}}\;dx
$$
$$
\chi_{\frac{1}{2}}((\sqrt{2}-1)) = \sum_{k=1}^\infty \frac{(\sqrt{2}-1)^{2k-2}}{\sqrt{2k-2}} =\frac{2^{-1/2}(\sqrt{2}-1)}{\sqrt{\pi}}\int_0^\infty \frac{1}{\sqrt{x}}\frac{e^{-x/2}}{1-(\sqrt{2}-1)^2e^{-x}}\;dx
$$
Might have to rearrange the exp/trig terms? There is a missing factor of $(-1)^{k-1}$, perhaps change $z$ slightly. I haven't checked it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Exponential equals rational fraction $2^{-3x^3+5x^2-x}=\frac{x^2+1}{x}$ I'm trying to solve this equation:
$$2^{-3x^3+5x^2-x}=\frac{x^2+1}{x}$$
I can see that $x=1$ is a solution and I'm struggling find. others.
The exponential on the left side is always positive, so right must also be >0.
$\frac{x^2+1}{x}>0$
$x^2 >0 \implies x^2+1>0$
so $x>0$. I don't know what else to do.
| From your work, because $x > 0$, we can apply AM-GM on the right hand side:
$$\frac{x^2+1}{x}\geq \frac{2x}{x}=2$$
Therefore, since the exponential is increasing over $\mathbb{R}$:
$$2^{-3x^3+5x^2-x}\geq 2^1\Rightarrow -3x^3+5x^2-x \geq 1\Rightarrow (3x+1)(x-1)^2\leq 0$$
The unique solution is thus $x=1$.
| {
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Given $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ prove $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$ Given a sequence $(a_n)_{n\geq 1}$ such that $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ where $ n\geq1$
prove that $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$
I tried to prove that $a_n$ is increasing but it I just can't get around to it. Thank you in advance!
| We first need to show $\{a_n\}$ is an increasing sequence, and the argument of this part could adopt the proof above by Prof. Blatter. Once this is overcome, then $a_n > 0$ for all $n \ge 1$. Next , $a_{n} = (a_n - a_{n-1})+ (a_{n-1} - a_{n-2}) +...=(a_2 -a_1) + a_1 = (1+\dfrac{a_{n-1}}{a_n})+ (1 + \dfrac{a_{n-2}}{a_{n-1}}) + (1 + \dfrac{a_1}{a_2}) + a_1 > (n-1)+1 = n$. Thus we have: $ 1< \dfrac{a_{n+1}}{a_n} = 1 + \dfrac{1}{a_n} + \dfrac{1}{a_{n+1}} < 1+\dfrac{1}{n}+\dfrac{1}{n+1}$ and this shows $\dfrac{a_{n+1}}{a_n} \to 1$ as $n \to \infty$. This implies $a_{n+1} - a_{n} = 1 + \dfrac{a_n}{a_{n+1}}\to 1+1 = 2$ as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $ I_1=\int_{-\infty}^{+\infty}\frac{\cos 3x}{5+6x+9x^2}dx $ using Fourier transform.
Evaluate integral:
$$
I_1=\int\limits_{-\infty}^{+\infty}\frac{\cos 3x}{5+6x+9x^2}dx
$$
I used Fourier transform in the following form:
$$
\hat{f}(y)=F[f(x)]=\int\limits_{-\infty}^{+\infty}f(x)e^{ixy}dx
$$
Therefore, I have the following properties and formulas:
$$
\begin{aligned}
&F[f(kx)]=\frac{1}{|k|}\hat{f}\left(\frac{y}{k}\right),\ \ k\ne 0\\
&F[f(x-a)]=e^{iay}\hat{f}(y),\ \ a\in\mathbb{R}\\
&F\left[\frac{1}{1+x^2}\right]=\pi\cdot e^{-|y|}
\end{aligned}
$$
So, here is my attempt:
Suppose $I_2=i\cdot\int\limits_{-\infty}^{+\infty}\frac{\sin 3x}{5+6x+9x^2}dx$. Therefore:
$$
\begin{aligned}
&I_1+I_2=\int\limits_{-\infty}^{+\infty}\frac{e^{3ix}}{5+6x+9x^2}dx=F\left[\frac{1}{5+6x+9x^2}\right]\Bigm|_{y=3}\\
&\frac{1}{5+6x+9x^2}=\frac{1}{4}\cdot\frac{1}{\left(\frac{3x}{2}+\frac{1}{2}\right)^2+1}=g(x)\\
&F\left[\frac{1}{\left(\frac{3x}{2}\right)^2 +1}\right]=\frac{2}{3}\pi e^{-\frac{2}{3}|y|}\ \ \Rightarrow\ \
F\left[\frac{1}{\left(\frac{3x}{2}-\left(-\frac{1}{2}\right)\right)^2 +1}\right]=e^{-\frac{1}{2}iy}\cdot\frac{2}{3}\pi e^{-\frac{2}{3}|y|}\\
&F[g(x)]=\frac{\pi}{6}e^{-\frac{1}{2}iy}e^{-\frac{2}{3}|y|}\Rightarrow I_1+I_2=\frac{\pi}{6}e^{-\frac{3}{2}i}\cdot e^{-2}\\
&I_1=\text{Re}(I_1+I_2)=\frac{\pi}{6e^2}\cos\frac{3}{2}
\end{aligned}
$$
However, the correct answer is this: $$I_1=\frac{\pi}{6e^2}\cos 1$$
I've tried different approaches, and yet I don't realize what is wrong with my solution. Any help will be appreciated!
| Mistake when changing variables?
$$F\left[\frac{1}{\left(\frac{3x}{2}-\left(-\frac{1}{2}\right)\right)^2 +1}\right]=e^{-\frac{1}{3}iy}\cdot\frac{2}{3}\pi e^{-\frac{2}{3}|y|}$$
| {
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Prove that $a^2+b^2=p$ has a unique solution $(a,b)\in\mathbb{Z}_{\ge0}^2$ with $a\le b$, where $p\equiv 1\pmod4$ is prime. I am trying to prove the following:
Consider $p$, a prime natural number with $p\equiv 1\pmod4$. Show
that the equation $a^2 + b^2 = p$ has a unique solution $a, b \in \mathbb{Z}_{≥0}$, up to swapping $a$ and $b$.
I know if $p\equiv 1(\bmod4)$ then $p = (a + bi)(a − bi)$ is a prime factorisation, which I think gets me part of the way there, but I don't know how to prove it.
| The uniqueness part follows from the identity
$$a^2c^2-b^2d^2=(a^2+b^2)c^2-b^2(c^2+d^2)\,.$$
Thus, if $a^2+b^2=p$ and $c^2+d^2=p$, then $a^2c^2\equiv b^2d^2\pmod{p}$, or $ac\equiv \pm bd\pmod{p}$.
If $ac\equiv +bd\pmod{p}$, then note that
$$p^2=(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2\,.$$
Show that this means $ac=bd$, and so $(a,b)=(d,c)$. If $ac\equiv -bd\pmod{p}$, then note that
$$p^2=(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\,.$$
Show that this means $ad=bc$, and so $(a,b)=(c,d)$.
The existence part is done by first observing that $x^2\equiv -1\pmod{p}$ has a solution $x\in\mathbb{Z}$. Now, consider the set $$S:=\big\{0,1,2,\ldots,\lfloor\sqrt{p}\rfloor\big\}\times\big\{0,1,2,\ldots,\lfloor\sqrt{p}\rfloor\big\}\,.$$ Note that $$|S|=\big(\lfloor\sqrt{p}\rfloor+1\big)^2>\big(\sqrt{p}\big)^2=p\,.$$ Ergo, by the Pigeonhole Principle, there are two distinct pairs $(a_1,b_1)$ and $(a_2,b_2)$ in $S$ such that $$a_1x-b_1\equiv a_2x-b_2\pmod{p}\,.$$ Let $a:=|a_1-a_2|$ and $b:=|b_1-b_2|$. We have that
$$a^2+b^2= (a_1-a_2)^2+(b_1-b_2)^2\equiv (b_1-b_2)^2\left(x^2+1\right)\equiv 0\pmod{p},$$
and $0\leq a,b\leq \lfloor\sqrt{p}\rfloor<\sqrt{p}$. Since $a$ and $b$ cannot be both zero (otherwise $(a_1,b_1)=(a_2,b_2)$), we have $$0<a^2+b^2<2p\,.$$
As $a^2+b^2$ is divisible by $p$, $a^2+b^2=p$ must hold.
| {
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"url": "https://math.stackexchange.com/questions/3574873",
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"source": "stackexchange",
"question_score": "1",
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Combinatorics: Number of ways to distribute 12 balls in group, with given conditions
The number of ways in which 12 identical balls can be grouped in four marked non empty sets A,B,C,D such that n(A) < n(B) is ?
The answer given is 70.
This is how I solved it:-
Now to distribute 12 identical balls to 4 different group such that no group is empty, we have $a+b+c+d = 12$, $$^{12-1}C_{4-1}$$ ways but we cannot have $a=b$ (as $a$ and $b$ are variables and interchangeable n(A) < n(b) and n(A) > n(b) mean the same), which also means that all $a, b, c,d$ should not be equal (as $a,b,c,d$ are variables and interchangeable) i.e. we need to exclude cases of $a=b=c=d$, but there is only one case that is possible which is $a=b=c=d=3$,
thus total ways $=$ $$^{12-1}C_{4-1}-1$$ but that equals 164 but answer given is 70. What am I doing wrong? How will you solve the question?
| We have that, concerning the number of solutions, we can re-state the problem as
$$
\eqalign{
& \left\{ \matrix{
1 \le a,b,c,d \hfill \cr
a < b \hfill \cr
a + b + c + d = 12 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{
0 \le x_{\,1} ,x_{\,2} ,x_{\,3} ,x_{\,4} \hfill \cr
x_{\,1} < x_{\,2} \hfill \cr
x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 8 \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
0 \le x_{\,1} ,y_{\,2} ,x_{\,3} ,x_{\,4} \hfill \cr
x_{\,2} = x_{\,1} + y_{\,2} + 1 \hfill \cr
2x_{\,1} + y_{\,2} + x_{\,3} + x_{\,4} = 7 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{
0 \le x_{\,1} ,x_{\,2} ,x_{\,3} ,x_{\,4} \hfill \cr
x_{\,2} + x_{\,3} + x_{\,4} = 7 - 2x_{\,1} \hfill \cr} \right. \cr}
$$
Therefore the number of solutions will be the number of weak compositions of $7-2k \; | \,0 \le k \le 3$ into $3$ parts, i.e.
$$
\eqalign{
& N = \sum\limits_{k = 0}^3 {
\left( \matrix{ 7 - 2k + 3 - 1 \cr 7 - 2k \cr} \right)}
= \sum\limits_{k = 0}^3 {\left( \matrix{ 9 - 2k \cr 2 \cr} \right)} = \cr
& = \left( \matrix{ 9 \cr 2 \cr} \right) + \left( \matrix{ 7 \cr 2 \cr} \right)
+ \left( \matrix{ 5 \cr 2 \cr} \right) + \left( \matrix{ 3 \cr 2 \cr} \right)
= {{9 \cdot 8 + 7 \cdot 6 + 5 \cdot 4 + 3 \cdot 2} \over 2} = \cr
& = 70 \cr}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there a formula for expanding these kind of things (what kind of expression is it, by the way?)?
This is my solution:
$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$
$= \displaystyle\frac{[(\sqrt5+2)^3+(\sqrt5-2)^3][(\sqrt5+2)^3-(\sqrt5-2)^3]}{8\sqrt5}$
$=\displaystyle\frac{(\sqrt5+2+\sqrt5-2)[(\sqrt5+2)^2\color{red}{+}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2](\sqrt5+2-\sqrt5+2)[(\sqrt5+2)^2\color{red}{-}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2]}{8\sqrt5}$
$=\displaystyle\frac{[2\sqrt5(5+4\sqrt5+4+\color{red}{5-4}+5-4\sqrt5+4][4(5+4\sqrt5+4\color{red}{-(5-4)}+(5-4\sqrt5+4)]}{8\sqrt5}$
$=\displaystyle\frac{2584\sqrt5}{8\sqrt5}$
$=323$
Because of the multiplication, I still got the same answer as given in the book. However, is the book or I correct in terms of the positive/negative signs(in red)?
| The book is correct. Notice the signs in the identities:
$$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$
$$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$
Let $a = (\sqrt{5}+2)^2$ and $b = (\sqrt{5}-2)^2$ and plug in to the second formula to recover your equation.
Your arithmetic happened to work by the lucky circumstance of $(18+1)(18-1)$ equalling $(18-1)(18+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find center of circle, which tangent to $y=2$ at $(3,2)$ and $y=-x\sqrt 3 +2$ Using $$r=\frac{|Ax + By + c|}{\sqrt{A^2+B^2}}$$
I get $$b=2+a\sqrt3$$ and stuck,because the options are on numbers. What do i do next?
|
Tangent lines meet at $B=(0,2)$,
and the minor angle at $B$ is $60^\circ$,
so we can construct an equilateral $\triangle ABC$,
for which the minor and major circles tangent to $y=2$ at $T=(3,2)$
are the inscribed and escribed circles, respectively.
The side length of the equilateral $\triangle ABC$
$|BC|=|CA|=|AB|=a=b=c=|AB|=2\cdot|BT|=6$.
So, the radius of the inscribed circle is
\begin{align}
r&=\tfrac13\cdot a\cdot\tfrac{\sqrt3}2
=\sqrt3
.
\end{align}
And the radius of the escribed circle is
\begin{align}
r_c&
=\frac{a+b+c}{a+b-c}\cdot r
=\frac{3\,a}{a}\cdot r
=3\,r
,
\end{align}
so the coordinates of the centers are
\begin{align}
I&=(3,2-\sqrt3)
\approx(3,0.268)
,\\
O_c&=(3,2+3\sqrt3)
\approx(3,7.196)
.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
solving $|2x+1|-|5x-2|\geq1$ Solve:
$|2x+1|-|5x-2|\geq1$
$\Rightarrow |2x+1|\geq1+|5x-2|$
Then I squared the inequality
$\Rightarrow 4x^2-4x+1\geq1+2|5x-2|+25x^2-20x+4$
$\Rightarrow-21x^2+16x-2|5x-2|-4\geq0$
Then I separated into two cases
The first case
$$\begin{cases}-21x^2+16x-10x+4-4\geq0...(a)\\x\geq0...(b) \end{cases}
$$
From (a) we get that
$x\in[0,\frac{2}{7}]$
$(a)\cap(b)\\
\Rightarrow x\in[0,\frac{2}{7}]$
The second case
$$\begin{cases}-21x^2+16x+10-4-4\geq0...(c)\\x<0...(d)\end{cases}$$
From (c) we get that
$x\in[\frac{4}{7},\frac{2}{3}]$
$(c)\cap(d)\\
\Rightarrow x\in\emptyset$
So the solutions to the initial inequality should be $x\in [0,\frac{2}{7}]$
But wolfram said the solutions were $[\frac{2}{7},\frac{2}{3}]$
Where did I make the mistake???
| Your first case is
a) $-21x^2 + 16x - 10x + 4 - 4 \ge 0$
b) $5x-2\ge 0$
so by a) $(21x-6)x \le 0$ so either $x \le 0$ and $x \ge \frac 27$ (impossible) or $x \ge 0$ and $x \le 27$ so $[0, \frac 27]$.
From b) $x \ge \frac 25$ and $a\cap b = \emptyset$.
Second case
c) $-21x^2 + 26x -8\ge 0$ and d) $5x - 2< 0$
For c) $(7x -4)(3x -2) \le 0$ and so $\frac 47\le x \le \frac 23$.
And d) yields $x < \frac 25$. $c\cap d= c$
So solution is $[\frac 47, \frac 23]$.
....
Your error was takeing b) and d) as $x\ge 0$ and $x < 0$ and I honestly have no idea why you did that. The term we had in modulus was $|5x-2|$ so you need to take those as $\ge, < 0$.
| {
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"source": "stackexchange",
"question_score": "1",
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Verifying a continued fraction related to $\logφ$. The continued fraction is the following,
$${1+\cfrac{1\cdot 2}{3φ+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7φ+\cfrac{3\cdot 4}{9+\ddots}}}}}=\frac{2}{3\logφ}\tag{1}$$
Where,
$$φ=\frac{1+\sqrt{5}}{2}$$
Something with I found in an backside of my previous notes where I kept my recreational math works. So can the closed form be verified in any sort (most preferably by using already established identities)?
| This can be deduced from Gauss's continued fraction for $_2F_1$, written as $$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}.$$ With $a=b=1$ and $c=3/2$, computing the values of $_2F_1$, we obtain $$1+\cfrac{1\cdot 2}{3/z+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7/z+\cfrac{3\cdot 4}{9+\ddots}}}}=1+\frac{2z}{3}\frac{_2F_1(2,1;5/2;-z)}{_2F_1(1,1;3/2;-z)}=\frac{\sqrt{z(1+z)}}{\sinh^{-1}\sqrt{z}}.$$ Now we put $z=1/\varphi=\varphi-1$ and note that $$\sinh^{-1}\frac{1}{\sqrt\varphi}=\log\left(\frac{1}{\sqrt\varphi}+\sqrt{\frac{1}{\varphi}+1}\right)=\log\left[\sqrt\varphi\left(\frac{1}{\varphi}+1\right)\right]=\frac32\log\varphi.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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What is the sum of the following infinite series? $$
\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \cdots
$$
So basically I separated it into two series
where:
one of them is $\left(\frac{1}{3}\right)^n$
And I use geometric series formula to find that this series equals $\frac{1}{2}$.
But I can't figure out the series of the other one.
Apparently the answer for the series combined is: $\frac{5}{8}$
What is the other series?
| It seems you have
$$\begin{equation}\begin{aligned}
& \frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \ldots \\
& = \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + \frac{1}{729} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \ldots\right) \\
& = \sum_{i=1}^{\infty}\left(\frac{1}{3}\right)^i + \sum_{i=1}^{\infty}\left(\frac{1}{9}\right)^i \\
& = \frac{\frac{1}{3}}{1 - \frac{1}{3}} + \frac{\frac{1}{9}}{1 - \frac{1}{9}} \\
& = \frac{\frac{1}{3}}{\frac{2}{3}} + \frac{\frac{1}{9}}{\frac{8}{9}} \\
& = \frac{1}{2} + \frac{1}{8} \\
& = \frac{5}{8}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I was able to split the sum into $2$ parts in the second line due to the series being absolutely convergent, with details about this in the Rearrangements and unconditional convergence section. Also note I used, such as described in Geometric series, that for $|r| \lt 1$, you have
$$\sum_{i=0}^{\infty}ar^i = \frac{a}{1 - r} \tag{2}\label{eq2A}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Moscow papyrus area of hemisphere I found this interesting text on wikipedia that explains how egyptians calculated the area of (supposedly) hemisphere, referred to as "basket". It is contained in the Moscow Papyrus. Here is the text:
Example of calculating a basket. You are given a basket with a mouth of $4 \frac{1}{2}$. What is its surface? Take $\frac{1}{9}$ of $9$ (since) the basket is half an egg-shell. You get $1$. Calculate the remainder which is $8$. Calculate $\frac{1}{9}$ of $8$. You get $\frac{2}{3} + \frac{1}{6} + \frac{1}{18}$. Find the remainder of this $8$ after subtracting $\frac{2}{3} + \frac{1}{6} + \frac{1}{18}$. You get $7 + \frac{1}{9}$. Multiply $7 + \frac{1}{9}$ by $4 + \frac{1}{2}$. You get $32$. Behold this is its area. You have found it correctly.
Now I am trying to understand what all these numbers represent by imagining this "basket" and I have been unsuccessful. Maybe someone understands the logic behind all of this to explain in more detail?
EDIT
So to be more specific. The algorithm obviously works ignoring the small error compared to our current formula. They were very practical and didn't have sophisticated mathematical knowledge. So this is why I essentially want that whoever answers this forgets about our modern mathematics and thinks exclusively in the context of this algorithm and its operations to somehow make sense of this.
| If the diameter of the hemisphere is $d=4.5$ then
*
*Take $\frac19$ of $2d$ to give $\frac{2}{9}d$ leaving $\frac89$ of $2d$, i.e. $2d - \frac{2}{9}d = \frac{16}{9}d$
*Take $\frac19$ of $\frac{16}{9}d$ to give $\frac{16}{81}d$ leaving $\frac89$ of $\frac{16}{9}d$, i.e. $\frac{16}{9}d - \frac{16}{81}d = \frac{128}{81}d$
*Multiply $\frac{128}{81}d$ by $d$ to give $\frac{128}{81}d^2$
The suggestion is that the curved surface area of a hemisphere is about $\frac{128}{81}$ times the square of the diameter
We know the surface area of a sphere is $4\pi r^2 = \pi d^2$, implying the curved surface area of a hemisphere is $2\pi r^2 = \frac{\pi}{2} d^2$
So this suggests that $\frac{\pi}{2} \approx \frac{8}{9} \times \frac{8}{9} \times 2 = \frac{128}{81}$ and thus $\pi \approx \frac{256}{81}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$
Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$.
My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies \left(\frac{\sin^4x}{a}+\frac{\cos^4x}{b}\right)^2=\frac{1}{(a+b)^2} \\ \implies \frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}-\sin^2x\cos^2x\left(\frac{\sin^2x}{a}-\frac{\cos^2x}{b}\right)^2=\frac{1}{(a+b)^2}.$$
Therefore, if we can prove that $$\frac{\sin^2x}{a}-\frac{\cos^2x}{b}=0,$$ then we are done. But, I am not able to prove the same.
| [This solution assumes that $x$ is real. It does not work if you allow for complex $x$.]
First, if $ a > 0 > b$ and (WLOG) $ a + b > 0$, then we have
$\frac{1}{a} < \frac{1}{a+b} = \frac{\sin^4}{a} + \frac{ \cos^4 }{b} \leq \frac{1}{a} $.
Hence, we must have (WLOG) $ a, b > 0 $.
Now, we may apply Titu's lemma, which gives us
$\frac{ \sin^4 x}{ a} + \frac{ \cos^4 x } { b} \geq \frac{ ( \sin^2 x + \cos^2 x)^2 } { a + b } = \frac{1}{a+b}$.
The condition tells us that equality holds, thus we have $\frac{ \sin^2 x } { a} = \frac{ \cos^2 x } { b}$.
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Proving a polynomial divides other polynomials Given $\mathbb{F}_p[u]$ over $\mathbb{F}_p$ with $p$ prime, how can I prove that $u^2-cu+d$ irreducible is a factor of both $u^p+u-c$ and $u^{p+1}-d$?
| When $p=2,$ the claim is trivial, so assume $p$ is an odd prime. In the finite field $\mathbb F_{p^2}=\mathbb F_p[u]/(u^2-cu+d)$ (it is a field by the irreducibility of $u^2-cu+d$,) we want to show $u^p+u-c=0$ and $u^{p+1}-d=0.$
Assuming the first equation, we immediately have that $u^{p+1}-d=u^{p+1}+u^2-cu=u(u^p+u-c)=0,$ so all that is left to do is show $u^p+u-c=0.$
Now, in $\mathbb F_p[u]/(u^2-cu+d),$ we can express $u^p=au+b,$ for some $a,b\in\mathbb F_p.$ Then, taking both sides to the $p$-th power, we have $u^{p^2}=u=(au+b)^p=au^p+b=a(au+b)+b=a^2u+(a+1)b.$ Thus, $a^2=1$ and $(a+1)b=0.$
If $b=0,$ we would have $u^p=au,$ or $u^{p-1}=a.$ Then, $(u^{p-1})^p=u^{p^2-p}=u^{-p}=a,$ so $u=a^{-2},$ a contradiction.
Hence, we have that $b\neq 0,$ so since $(a+1)b=0,$ we have that $a=-1.$ Now, since $u^2-cu+d=0,$ $(u^2-cu+d)^p=cu^{2p}-cu^p+d=(2c-2b)u+(b^2-cb)=0,$ showing $b=c.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587273",
"timestamp": "2023-03-29T00:00:00",
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Norm of an integral operator in $L^2$ Let T an operator in $L^2((-1,1))$ such that $Tf(x) = \int_{-1}^{1} (xy + y^2)f(y)dy$.
Calculate the norm of $T$.
I tried calculating $\left\lVert Tf(x) \right\rVert$ first, but I got stucked with the square of the inside integral. Any help?
| Observe
\begin{align}
Tf(x) = x\int^1_{-1} dy\ yf(y)+\int^1_{-1}dy\ y^2f(y)=: \mu x +v
\end{align}
then we see that
\begin{align}
\|Tf\|_{2}^2 = \int^1_{-1} dx\ |\mu x+v|^2 = \int^1_{-1}dx\ \mu^2x^2+\mu v x + v^2= \frac{2\mu^2}{3}+2v^2.
\end{align}
Next, observe that for any $f \in L^2(-1, 1)$, we can write
\begin{align}
f(x) = \sum^\infty_{n=0} c_n P_n(x) = \frac{c_0}{\sqrt{2}}+ c_1\sqrt{\frac{3}{2}}x+c_2\sqrt{\frac{5}{8}}(3x^2-1)+\cdots
\end{align}
where $P_n$ is the normalized $n$th degree Legendre polynomial. Hence it follows
\begin{align}
\|f\|_2^2= \sum^\infty_{n=0}|c_n|^2
\end{align}
with
\begin{align}
\mu =\int^1_{-1} dy\ yf(y) =\sqrt{\frac{2}{3}}c_1
\end{align}
and
\begin{align}
v=\int^1_{-1} dy\ y^2f(y) = \frac{\sqrt{2}}{3}c_0+\frac{2}{3}\sqrt{\frac{2}{5}}c_2.
\end{align}
Finally, we see that
\begin{align}
\|Tf\|_2^2=&\ \frac{2}{3}\left(\sqrt{\frac{2}{3}}c_1 \right)^2+2\left(\frac{\sqrt{2}}{3}c_0+\frac{2}{3}\sqrt{\frac{2}{5}}c_2 \right)^2\\
\le&\ \frac{4}{9}c_1^2+2\left(\frac{2}{9}+\frac{8}{45}\right)(c_0^2+c_2^2)\\
\leq&\ 2\left(\frac{2}{9}+\frac{8}{45}\right)(c_0^2+c_1^2+c_2^2)= \frac{4}{5}(c_0^2+c_1^2+c_2^2)
\end{align}
which means
\begin{align}
\|T\|_\text{op} \leq \sqrt{\frac{4}{5}}\approx 0.894.
\end{align}
Consider $f(x) = P_0(x)+cP_2(x)$, then we see that
\begin{align}
\|Tf\|_2^2=2\left(\frac{\sqrt{2}}{3}+\frac{2}{3}\sqrt{\frac{2}{5}}c \right)^2=\frac{4}{5}(1+c^2) = \frac{4}{5}\|f\|_2^2
\end{align}
provided $c=\frac{2}{\sqrt{5}}$. Hence $\|T\|_\text{op} = \sqrt{4/5}$.
| {
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Evaluate $ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx$
Evaluate
$$ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx,$$
My attempt : $u=\tan\frac{x}{2} \rightarrow x=2\arctan(u) \rightarrow \frac{2}{1+u^2}du=dx$
$$\sin x= \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{2u}{1+u^2}$$
$$\cos x= \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}= \frac{1-u^2}{1+u^2}$$
I was looking to see if this is the right way to go and also i am not sure how to evaluate the boundaries . Could you help me out ?
| Just convert into exponentials
$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$
The extrema won't change.
Substitute, arrange and you get
$$\int_{\pi/4}^{\pi/2} \frac{e^{ix}(2 + i) + e^{-ix(i-2)}}{e^{ix}(2i + 1) + e^{-ix}(2i - 1)}\ \text{d}x$$
You can split the integral in two pieces, then collect $e^{ix}$ up and down in the first piece and $e^{-ix}$ up and down in the second piece. This leads you to
$$\int_{\pi/4}^{\pi/2} \frac{2+i}{2i + 1 + e^{-2ix}(2i-1)}\ \text{d}x + \int_{\pi/4}^{\pi/2} \frac{i-2}{2i - 1 + e^{2ix}(2i+1)}\ \text{d}x$$
The integrals are rather easy for it's just a denominator exponential integral (you can always call with other letters the complex constants).
Hence we have:
$$I_1 = \left(\frac{1}{10}-\frac{3 i}{40}\right) \left(\pi +2 i \log \left(\frac{9}{2}\right)\right)$$
$$I_2 = \left(\frac{1}{10}+\frac{3 i}{40}\right) \left(\pi -2 i \log \left(\frac{9}{2}\right)\right)$$
Eventually
$$I = I_1 + I_2 = \frac{1}{10} \left(2 \pi +\log \left(\frac{729}{8}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$ Evaluate the following integral :
$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$
I was tried use change variable ,
If I use $x=y^2$ integral becomes :
$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$
From here I have one idea the derivative under sing integral but I got I difficult integration :
$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$
I already to see you hints or solution!
| In my post, I had found a formula for the integral
$$I(a)=\int_{0}^{\infty} \frac{\ln \left(x^{4}+2 x^{2} \cos 2 a+1\right)}{1+x^{2}} d x=2 \pi \ln \left(2 \cos \frac{a}{2}\right)$$
Factorise $x^8+1= \left(x^4+\sqrt{2} x^2+1\right)\left(x^4-\sqrt{2} x^2+1\right)$ split our integral into two integrals with quartic polynomials.
$$
\begin{aligned}
I&=2 \int_0^{\infty} \frac{\ln \left[\left(x^4+\sqrt{2} x^2+1\right)\left(x^4-\sqrt{2} x^2+1\right)\right]}{1+x^2} d x \\
&=2\left[\int_0^{\infty} \frac{\ln \left(x^4+\sqrt{2} x^2+1\right)}{1+x^2} d x+\int_0^{\infty} \frac{\ln \left(x^4-\sqrt{2} x^2+1\right)}{1+x^2} d x\right] \\
&=4 \pi\left\{\ln \left[2 \cos \left(\frac{1}{4} \cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)\right]+\ln \left[2 \cos \left(\frac{1}{4} \cos ^{-1}\left(\frac{-\sqrt{2}}{2}\right)\right)\right]\right\} \\
&=4 \pi\left\{\ln \left(2 \cos \frac{\pi}{16}\right)+\ln \left(2 \cos \frac{3 \pi}{16}\right)\right\} \\
&=4 \pi \ln (\sqrt{2}+\sqrt{2+\sqrt{2}})
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $z=\cos \theta + i \sin \theta$, express $\displaystyle \frac {1}{1-z \cos \theta}$ in the form $a+i\cdot b$. In order to get rid of the $z$, should I substitute $z=\cos\theta + i\sin \theta$ into the complex number, or what conjugate should I multiply the complex number by?
I have tried substituting $z=\cos\theta + i\sin \theta$ into the complex number, but only got this far:
$\displaystyle \frac {1}{1-z \cos \theta}$
$= \displaystyle\frac{1}{1-(\cos\theta + i\sin\theta)(\cos\theta)}$
$=\displaystyle\frac{1}{1-\cos^2\theta-i\cos\theta\sin\theta}$
As for multiplying the complex number by a conjugate, I have used $\big(\displaystyle\frac {1}{z}-\frac{1}{\cos\theta}\big)$, $\big(\displaystyle z-\frac{1}{z}\big)$ and $(1+z\cos\theta)$ but to no avail.
I have only learnt de Moivre's theorem, and I haven't learnt $\cosθ+i\sinθ=e^{iθ}$, so I would appreciate if this question can be solved in the simplest way possible. But other methods are welcome.
| \begin{align}
\frac1{1-z\cos\theta}&=\frac1{1-(\cos\theta+i\sin\theta)\cos\theta}\\
&=\frac1{1-\cos^2\theta-i\sin\theta\cos\theta}\\
&=\frac1{\sin^2\theta-i\sin\theta\cos\theta}\color{blue}{\cdot\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^2\theta+i\sin\theta\cos\theta}}\\
&=\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^4\theta+\sin^2\theta\cos^2\theta}\\
&=\frac{\sin^2\theta+i\sin\theta\cos\theta}{\sin^2\theta(\sin^2\theta+\cos^2\theta)}\\
&=\frac{\sin^2\theta}{\sin^2\theta}+\frac{i\sin\theta\cos\theta}{\sin^2\theta}\\
&=1+i\cot\theta
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Same roots as coefficients in quadratic equation.
Coefficients of quadratic equation $ax^2+bx+c=0$ are different
integers. Roots of the equation are numbers $a$ and $b$. Find all
equations that follow that condition.
After substitution $x_0=a,b$ we have two equations:
$$ a^3+ab+c=0 $$
$$ab^2+b^2+c=0$$
$$a^3+ab+c=ab^2+b^2+c$$
$$a^3+ab-ab^2-b^2=(a-b)(a^2+ab+b)=0$$
Solution $a=b$ doesn't count because of the assumption different integers.
$$a^2+ab+b=0 $$
$$a^2=-b(a+1) $$
The only conclusion I can think of is that one of $b$ or $a+1$ must be negative to prevent from contradiction $a^2<0$. Substition $a$ and $b$ into formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ leads to the same pair of equations.I don't know how to work this out.
Solution: $-2x^2+4x+16$
| You already got
$$a^2+ab+b=0$$
from which we have
$$b=\frac{-a^2}{a+1}=\frac{(a+1)(-a+1)-1}{a+1}=-a+1-\frac{1}{a+1}$$
Since $\frac{1}{a+1}$ has to be an integer, we have to have $$a+1=\pm 1\implies a=0,-2$$
When $a=0$, we get $b=0=a$, so $a\not=0$.
So, the answer is $(a,b)=(-2, 4)$ and $c=16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
probability density function: Finding c as upper limit domain of f(x)
So nowhere in my textbook does it show how to determine the value of c for any probability density function f(x).
However, I just want to check my work to see if I got it right.
We know that for any pdf f(x) the Integral from infinity to negative infinity = 1, that is:
${\int_{-\infty}^{\infty} f(x)dx\ =} 1 $
However, since we are given two boundaries, 1 and c,
${\int_{1}^{c} f(x)dx\ =} 1 $
since, for all x between 1 and c has the probability density function x.
The definite integral becomes:
$ {1/2c^2 - 1/2 = 1} $
Thus,
$ 1/2c^2 = 1 + 1/2 $
$ c^2 = 3 $
$ c = \sqrt{3} $
Am i correct?
| You're correct that
$$\int_{-\infty}^{\infty} f(x)dx = 1 \tag{1}\label{eq1A}$$
Since $f(x) = 0$ for $x \le 0$ and $x \gt c$, you need to split the integral into $2$ parts where it's non-zero, i.e.,
$$\begin{equation}\begin{aligned}
1 & = \int_{0}^{1} x^2dx + \int_{1}^{c} xdx \\
& = \left. \frac{x^3}{3}\right\rvert_{0}^{1} + \left. \frac{x^2}{2}\right\rvert_{1}^{c} \\
& = \frac{1}{3} + \frac{c^2}{2} - \frac{1}{2} \\
& = \frac{c^2}{2} - \frac{1}{6}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note you can't ignore the part where $0 \lt x \lt 1$ as you did in your question. This leads to
$$\begin{equation}\begin{aligned}
1 & = \frac{c^2}{2} - \frac{1}{6} \\
\frac{7}{6} & = \frac{c^2}{2} \\
c^2 & = \frac{7}{3} \\
c & = \sqrt{\frac{7}{3}} \\
c & \approx 1.53
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
I've shown the result accurate to $2$ decimal places, as requested.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is it possible to find where two curves have a common value? Suppose I have;
$$f(x) = x^2 - 42x + 364$$
$$g(y) = y^2 - 35y + 364$$
Computing out values I find they have a common value
$$f(6) = g(8) = g(27) = f(36) = 148$$
Is there a way of finding these values?
Working through based on received help...
(1) $$f(x) = g(y)$$
(2) $$f(x) - g(y) = x^2 - 42x + 364 - y^2 + 35y - 364 = 0$$
(3) $$x^2 - 42x - y^2 + 35y = 0$$
Complete the squares...
(4) $$(x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4}$$
(5) $$4 ((x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4} )$$
(6) $$(4x^2 - 168x + 42^2) - (4y^2 - 140y + 35^2) = 42^2 - 35^2$$
(7) $$(2x - 42)^2 - (2y - 35)^2 = 539$$
Considering (7) as Fermat's factorization method;
(8) $$a^2 - b^2 = c$$
(9) $$(a + b)(a - b) = 539$$
(10) $$(2x - 42 + 2y - 35)(2x - 42 - 2y + 35) = 539$$
(11) $$(2x + 2y - 77)(2x - 2y - 7) = 539$$
Then, considering (11) as a simple product of two terms gives;
(12) $$d * e = 539$$
(13) $$(2x - 2y - 7) = d$$
(14) $$(2x + 2y - 77) = 539/d$$
| I presume you're looking for integers $x$ and $y$ such that $f(x) = g(y)$. By completing the square, the equation $f(x) = g(y)$ can be written as
$$ (2x-42)^2 - (2y-35)^2 = (2x + 2y - 77)(2x - 2y -7) = 539$$
If $d$ is a (positive or negative) integer that divides $539$,
you can solve
$$ \eqalign{2x + 2y - 77 &= d\cr
2x - 2y - 7 &= 539/d\cr}$$
to get
$$ \eqalign{x &= \frac{d^2 + 84 d + 539}{4d}\cr
y &= \frac{d^2 + 70 d - 539}{4d}\cr}$$
and this will give all the solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I evaluate this limit without using L'hospital, derivatives or series expansion? I have troubles solving this particular limit without using series expansion, L'Hospital rule or derivatives.
$$\lim_{x \to 0}{\frac{x\sin{x}+2(\cos{x}-1)}{x^3}}$$
I have tried multiplying by $(\cos{x}+1)$ and got rid of one $x$ in the denominator, but got arguably something even messier. I got this:
$$\lim_{x \to 0}{\frac{x(\cos{x}+1)-2\sin{x}}{x^2(\cos{x}+1)}}$$
Any help would be appreciated.
| As J.G. already shown it holds $$\begin{align}\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{2x\sin\frac{x}{2}\cos\frac{x}{2}-4\sin^2\frac{x}{2}}{x^3}\\&=\frac{4\sin^2\frac{x}{2}}{x^2}\frac{\frac{x}{2}\cot\frac{x}{2}-1}{x} \\ &= \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \frac{f(x) - f(0)}{x - 0} \end{align}$$
for $f(x) = x\cot(x)$
So we get by the mean value theorem: $$\lim_{x\to 0} \frac{x\sin x+2(\cos x-1)}{x^3} = 0\cdot\lim_{x\to 0} f'(x) = 0\cdot1 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Calculating harmonic mean of $f(i)$, where $f(i)$ represent the number of times the number $i$ is selected. $i \in \{1,2,3\}$.
Let $X =\{ 1,2,3 \}$. A number is selected from the set $X$ with replacement and this process is repeated $1100$ times. For each $i,i \in \{1,2,3\}$, let $f(i)$ represent the number of times the number $i$ is selected. Also, let $S$ denote the total sum of $1100$ numbers selected. If $S^3 = 162 f(1)f(2)f(3)$. What will be the harmonic mean of $f(1),f(2),f(3)$ ?
If There are $f(1)$ red balls, $f(2)$ blue balls, $f(3)$ white balls and ‘$x$’ green balls (balls of the same color are identical). The balls are arranged in a row such that no two balls of the same color are consecutive. Let $x_1$ and $x_2$ be the minimum and maximum values of $x$ respectively for which the above
arrangement is possible. Find the value of $x_1+x_2$.
My attempt: Let $f(3)=n$ and $f(2)=m$ where $m$ and $n$ are positive integers, then $f(1)=1100-n-m$ and $S=3n+2m+1100-n-m$ $= 1100+2n+m$. Given $(1100+2n+m)^3 = 162nm(1100-m-n)$, We need to calculate $\frac{3}{\frac{1}{n}+\frac{1}{m}+\frac{1}{1100-n-m}}$ $= \frac{3nm(1100-n-m)}{1100(n+m)-(n^2+m^2+nm)}$. There must be a trick involved somehwere, i don't think we actually need to find $m$ and $n$. I am also stuck with the second part. Why is $162$ special ? it could have been $S^3 = 100 f(1)f(2)f(3)$ or some other number.
| We were given three equations:
*
*$f(1)+f(2)+f(3)=1100$
*$f(1)+2f(2)+3f(3)=S$
*$\left(f(1)+2f(2)+3f(3)\right)^{3}=162f(1)f(2)f(3)$
Let us define $x=\sqrt[3]{f(1)}$, $y=\sqrt[3]{2f(2)}$, and $z=\sqrt[3]{3f(3)}$ then substitute it back to the third equation
$$
\begin{aligned}
\left(x^{3}+y^{3}+z^{3}\right)^{3}&=27x^{3}y^{3}z^{3}\\
x^{3}+y^{3}+z^{3}&=3xyz\\
\\
x^{3}+y^{3}+z^{3}-3xyz&=0\\
(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-xz)&=0
\end{aligned}
$$
Since $x+y+z\neq0$ then $x^{2}+y^{2}+z^{2}-xy-yz-xz=0$. But for real numbers $x,y,z$ $x^{2}+y^{2}+z^{2}\geq xy+yz+sz$ with equality if $x=y=z$.
Then $f(1)=2f(2)=3f(3)=6x\rightarrow f(1)=6x, f(2) = 3x, f(3)=2x$. Substitute to equation $1$ to obtain $6x+3x+2x=1100 \rightarrow x=100$
There You go, $f(1)=600, f(2)=300, f(2)=200$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help to understand how to deduce the value of $\cos^2\displaystyle\big(\frac{\pi}{5}\big)+\cos^2\big(\frac{2\pi}{5}\big)$ I have some trouble understanding the steps provided in the book on solving the last part of the question.
For context purposes (and to provide the full picture), the whole question (with solution) is as follows:
Prove de Moivre's theorem for positive integer exponents.
(I have no problem with this)
Using de Moivre's theorem, show that $\sin5\theta=a \sin^5\theta+b \cos^2\theta\sin^3\theta+c\cos^4\theta\sin\theta$ , where $a,b$ and $c$ are integers to be determined. (No issue with this either)
Express $\displaystyle\frac{\sin5\theta}{\sin\theta}$ in terms of $\cos\theta$, where $\theta$ is not a multiple of $\pi$. (Still no problem)
Hence, find the roots of the equation $16x^4-12x^2+1=0$ in trigonometric form. (I understand this part)
Deduce the value of $\displaystyle\cos^2\Big(\frac{\pi}{5}\Big)+\cos^2\Big(\frac{2\pi}{5}\Big)$. (I don't understand the working given by the book)
My question is, since it is found that $\displaystyle\cos^2\theta=\frac{3\pm\sqrt5}{8}$, how to tell that $\displaystyle\cos^2\frac{\pi}{5}=\frac{3+\sqrt5}{8}$ and how is it known that $\displaystyle\cos^2\frac{2\pi}{5}=\frac{3-\sqrt5}{8}$?
| $$t=\pi/10 \implies 5t=\pi/2 \implies 2t=\pi/2-3t \implies \sin 2t= \sin(\pi/2-3t)$ \implies \sin 2t= \cos 3t$$
$$\implies 2\sin t \cos t=4 \cos^2 t-3 \cos t \implies 4\sin^2t+2 \sin t+1=0 \implies \sin t =\frac{\pm\sqrt{5}-1}{4}$$
Since $t=\pi/10$ is acute angle so its $\sin$ value should be positive. hence.
$$\sin \pi/10=\frac{\sqrt{5}-1}{4}~~~~(1)$$
using $$\cos 2t=1-\sin^2 t \implies \cos \pi/5=\frac{\sqrt{5}+1}{4}~~~(2)$$
Let $$F=\cos^2(\pi/5)+\cos^2(2\pi/5)=\cos^2(\pi/5)+[\cos(\pi/2-\pi/10)]^2=\cos^2(\pi/5)+\sin^2(\pi/10)$$
Using (1) and (2), we get
$$F=\left(\frac{\sqrt{5}+1}{4}\right)^2+\left(\frac{\sqrt{5}-1}{4}\right)^2=2(5/16+1/16)=3/4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $\triangle ABC$ with circumradius $R$, orthocenter $H$, and nine-point center $O_9$, show $O_9A^2+O_9B^2+O_9C^2+O_9H^2=3R^2$.
Prove that, if $O_9$ is the nine-point center of triangle $ABC$ with $H$ as its orthocenter, then
$$O_9A^2+O_9B^2+O_9C^2+O_9H^2=3R^2\,.$$
What I tried to do draw $O$ as $O_9$ is the midpoint of $OH$ and tried stuff as $AO=BO=CO=R$.
| Write $N$ for the nine-point center of the triangle $ABC$, and $O$ for the circumcenter.
Let $M_a$, $M_b$, and $M_c$ be the midpoints of $BC$, $CA$, and $AB$, respectively. Use Stewert's Theorem or Law of Cosines to establish that
$$NB^2\cdot M_aC-NM_a^2\cdot BC+NC^2\cdot BM_a-M_aC\cdot BC\cdot BM_a=0\,.$$
Since $M_aC=BM_a=\dfrac12BC$, we have
$$NM_a^2=\frac{NB^2+NC^2}{2}-\frac14BC^2\,.$$
However, $N$ is the circumcenter of the triangle $M_aM_bM_c$, which is similar to the triangle $ABC$ via a homothethy centered at the centroid $G$ of the triangle $ABC$ with the scaling factor $-\dfrac12$. Therefore, $NM_a$ is the circumradius of the triangle $M_aM_bM_c$, which is half the circumradius of the triangle $ABC$, namely, $NM_a=\dfrac12 R$. This gives
$$\frac{R^2}{4}=\frac{NB^2+NC^2}{2}-\frac14BC^2\,.$$
We also have two similar equations:
$$\frac{R^2}{4}=\frac{NC^2+NA^2}{2}-\frac14CA^2$$
and
$$\frac{R^2}{4}=\frac{NA^2+NB^2}{2}-\frac14AB^2\,.$$
Combining all three of them yields
$$NA^2+NB^2+NC^2-\frac{1}{4}(BC^2+CA^2+AB^2)=\frac{3R^2}{4}\,.$$
From this past question,
$$BC^2+CA^2+AB^2=9R^2-OH^2\,.$$
Therefore,
$$NA^2+NB^2+NC^2-\frac14(9R^2-OH^2)=\frac{3R^2}{4}\,,$$
which is equivalent to
$$NA^2+NB^2+NC^2-\frac{OH^2}{4}=3R^2\,.$$
By Euler's Theorem, $N$ bisects $OH$; therefore, $\dfrac{OH}{2}=NH$, making
$$NA^2+NB^2+NC^2-NH^2=3R^2\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597306",
"timestamp": "2023-03-29T00:00:00",
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Minimum and maximum of $y=\cos\theta(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})$ The minimum and maximum values of expression $y=\cos\theta(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})$ is:
My attempt is as follows:-
Attempt $1$:
$$y=\cos\theta\sin\theta+\sqrt{\sin^2\theta\cos^2\theta+\sin^2\alpha\cos^2\theta}$$
$$y=\dfrac{\sin2\theta}{2}+\sqrt{\dfrac{\sin^22\theta}{4}+\sin^2\theta\cos^2\theta}$$
Here the term $\cos^2\theta$ created the issue and I didn't find any way to proceed further.
Attempt $2$:
Rationalizing
$$y=\dfrac{\cos\theta\cdot\sin^2\alpha}{\sin\theta-\sqrt{\sin^2\theta+\sin^2\alpha}}$$
$$y=\dfrac{\sin^2\alpha}{\tan\theta-\sqrt{\tan^2\theta+\sin^2\alpha\sec^2\theta}}$$
Here I was not able to comprehend how to get minimum and maximum value.
Attempt 3:
$$\dfrac{dy}{dx}=-\sin\theta(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})+\cos\theta\left(\cos\theta+\dfrac{1}{2\sqrt{\sin^2\theta+\sin^2\alpha}}\cdot2\sin\theta\cos\theta\right)$$
$$\dfrac{dy}{dx}=-\sin\theta(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})+\cos^2\theta\cdot\dfrac{\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha}}{\sqrt{\sin^2\theta+\sin^2\alpha}}$$
$$\dfrac{dy}{dx}=(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})\left(-\sin\theta+\dfrac{\cos^2\theta}{\sqrt{\sin^2\theta+\sin^2\alpha}}\right)$$
Now from here it will be very difficult to decide about the maximum and minimum value.
Any inputs?
| Continue with the derivative approach,
$$\frac{dy}{dx}=\frac{(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})
(\cos^2\theta-\sin\theta\sqrt{\sin^2\theta+\sin^2\alpha})}{\sqrt{\sin^2\theta+\sin^2\alpha}}=0$$
which leads to
$$\cos^2\theta-\sin\theta\sqrt{\sin^2\theta+\sin^2\alpha}=0$$
Write $\cos^2\theta = 1-\sin^2\theta$ and the locations of the extrema are found at
$$\sin\theta = \frac1{\sqrt{2+\sin^2\alpha}},\>\>\>\>
\cos\theta = \pm \sqrt{\frac{1+\sin^2\alpha}{2+\sin^2\alpha}}$$
Then, it is straightforward to determine that the extrma of the expression
$$y(\theta)=\cos\theta(\sin\theta+\sqrt{\sin^2\theta+\sin^2\alpha})$$
are
$$y_{max} = \sqrt{1+\sin^2\alpha},\>\>\>\>\>y_{min} =- \sqrt{1+\sin^2\alpha}$$
| {
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What is area of circle given areas of $4$ squares
I can't find a way to solve this, I know that by using power of the point the other square area is 64cm^2 and that its side length is 8, but for the area of the circle I'm not too sure where to start, can someone help?
|
The area of the triangle $ABC$ is
$$Area_{ABC} = \frac 12 BX\cdot AC = \frac12 \cdot 5\cdot (4+10) = 35$$
Then, the circumradius of $ABC$ is,
$$R = \frac{AB\cdot BC\cdot CA}{4Area_{ABC}}
=\frac{\sqrt{4^2+5^2}\cdot \sqrt{10^2+5^2}\cdot 14}{4\cdot35}
=\frac{\sqrt{205}}{2}$$
Thus, the area of the circle is
$$Area_{circle} = \pi R^2= \frac{205\pi}4$$
| {
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"url": "https://math.stackexchange.com/questions/3601104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the equation $a(f(x))^2 - af(kx)f(x) + xf’(x) = 0$ What function(s) f satisfies the equation
$$a(f(x))^2 - af(kx)f(x) + xf’(x) = 0$$
where $a$ and $k$ are constants, $k>0$ and $x$ is a real number?
I’m aware that that the function $f(x)=c$ satisfies the equation i.e. $f(x)$ is a constant function. But can it be proven that there is a non constant function f that can satisfy the equation?
| If $f(x) = ux+v$
with $u \ne 0$
then
$\begin{array}\\
0
&=a(f(x))^2 - af(kx)f(x) + xf’(x)\\
&=a((ux+v)^2 - (kux+v)(ux+v)) + xu\\
&=a(u^2x^2+2uvx+v^2 -(ku^2x^2+uv(k+1)x+v^2))+ xu\\
&=a(u^2(1-k)x^2+uv(2-(k+1))x+v^2)+ xu\\
&=au^2(1-k)x^2+(auv(1-k)+u)x+av^2\\
&=au^2(1-k)x^2+u(av(1-k)+1)x+av^2\\
\end{array}
$
so $v = 0$
and the coefficient of $x$ is $1$
which doesn't work.
If $f(x)
=cx^n + O(x^{n-1})$
then
$\begin{array}\\
0
&=a(f(x))^2 - af(kx)f(x) + xf’(x)\\
&=a(cx^n+O(x^{n-1}))^2 - a(cx^n+O(x^{n-1}))(cx^nO(x^{n-1})) + x(nx^{n-1}+O(x^{n-2}))\\
&=a(c^2x^{2n}+O(x^{2n-1})) - a(c(kx)^n+O(x)^{n-1}))(cx^n+O(x^{n-1})) + x(nx^{n-1}+O(x^{n-2}))\\
&=ac^2x^{2n}+O(x^{2n-1}) - ac^2k^nx^{2n}+O(x^{2n-1})\\
&=ac^2(1-k^n)x^{2n}+O(x^{2n-1}) \\
\end{array}
$
so $k^n = 1$.
I'll stop here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem with polynomial divisilibity proof.
Show that for any natural $k$, $m$, $n$ polynomial $x^{3k}+x^{3m+1}+x^{3m+2}$ is divisible by $x^2+x+1$.
My observation:
$$x^a=(x^{a-2}-x^{a-3})(x^2+x+1)+x^{a-3}$$
so,
$$x^{3k}+x^{3m+1}+x^{3m+2}=(x^{3k-2}-x^{3k-3})(x^2+x+1)+x^{3k-3}+(x^{3m-1}-x^{3m-2})(x^2+x+1)+x^{3m-2}+(x^{3n}-x^{3n-1})(x^2+x+1)+x^{3n-1}$$
We can factor $x^2+x+1$, but i don't see any conclusion after that.
| Note that one way is to show that the roots of $x^2+x+1$ satisfy $x^{3k}+x^{3m+1}+x^{3n+2}=0$, as suggested in comments. Here is an easy approach without using complex numbers:
$$x^{3k}+x^{3m+1}+x^{3n+2}=x^{3k}-1+x(x^{3m}-1)+x^2(x^{3n}-1)+x^2+x+1$$
Now note that $$x^2+x+1\bigg|x^3-1\bigg|x^{3a}-1\ \ \forall a\in\Bbb N$$
Therefore, RHS of the first equation is divisible by $x^2+x+1$ and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorize : $P(x)=x^6+x^2+1$ Factorize in $\mathbb{R}[x]$ or $\mathbb{Q}[x]$ :
$$P(x)=x^6+x^2+1$$
I know all roots of $P$ are complex and I don't know if there a factorization
But My try as following :
$$P(x)=x^6+x^2+1=(x^3)^2+1+2x^3-2x^3+x^2$$
$$=(x^3+1)^2+x^2-2x^3$$
I don't know how I complete?
| Note that we can try if we have a purely imaginary root, and indeed we have this particular factorisation: $$(x^2+a)(x^4-ax^2+a^2+1)=x^6+x^2+a^3+a$$
Which fits the orignal polynomial for $\ a^3+a=1$, and this cubic has at least one real root that we can calculate by Cardan's method (I skip to the result though...)
Le set $\alpha=\sqrt[3]{108+12\sqrt{93}}\ $ then $\ a=\dfrac{\alpha}6-\dfrac 2{\alpha}$
Now we get to factorize the remaining quartic, there are formulas for this that should lead you to a factorization in $\mathbb R[x]$ in product of quadratics, but considering there are no odd powers we can try to factorize under the form:
$$(x^2+bx+c)(x^2-bx+c)=x^4+(2c-b^2)x^2+c^2$$
This gives us by identification of the coefficients (note that we are interested in one possible factorization, so I just take the positive square root each time) $\begin{cases}c=\sqrt{a^2+1}\\b=\sqrt{2c+a}\end{cases}$
And there you have it:
$$x^6+x^2+1=(x^2+a)(x^2+bx+c)(x^2-bx+c)$$
| {
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"url": "https://math.stackexchange.com/questions/3607827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
| Your way gives a right inequality!
Indeed, by AM-GM
$$\sum_{cyc}\frac{1+a^2}{1-a^2}\geq3\sqrt[3]{\prod_{cyc}\frac{1+a^2}{1-a^2}}.$$
Thus, it's enough to prove that
$$64\prod_{cyc}(1+a^2)\geq125\prod_{cyc}(1-a^2)$$ or
$$\sum_{cyc}\left(\ln(1+a^2)-\ln(1-a^2)+2\ln2-\ln5\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(1+a^2)-\ln(1-a^2)+2\ln2-\ln5-\frac{27}{20}\left(a-\frac{1}{3}\right)\right).$$
Now, let $$f(a)=\ln(1+a^2)-\ln(1-a^2)+2\ln2-\ln5-\frac{27}{20}\left(a-\frac{1}{3}\right).$$
Thus, $$f'(a)=\frac{2a}{1+a^2}-\frac{-2a}{1-a^2}-\frac{27}{20}=\frac{(3a-1)(9a^3+3a^2+a+27)}{20(1-a^4)},$$
which gives $$a_{min}=\frac{1}{3}$$ and since $$f\left(\frac{1}{3}\right)=0,$$ we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $a, b, c>0$, prove $\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b)$
Given $a,b,c>0$, prove that $$\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b).$$
My attempt:
I have that $$\frac{a^4}{a+b}+\frac{c^2(a+b)}{4}\geq a^{2}c$$
$$\frac{b^4}{b+c}+\frac{a^2(b+c)}{4}\geq b^{2}a$$
$$\frac{c^4}{c+a}+\frac{b^2(c+a)}{4}\geq c^{2}b$$
Adding them up yields $$\sum \frac{a^4}{a+b}\geq\frac{3}{4}\sum a^2c-\frac{1}{4}\sum a^2b.$$
So it boils down to proving that $a^2b+b^2c+c^2a\leq ab^2+bc^2+ca^2$, which isn't correct.
Could you help me with this problem?
| $$\dfrac{a^3}{1+\dfrac{b}{a}}+\dfrac{b^3}{1+\dfrac{c}{b}}+\dfrac{c^3}{1+\dfrac{c}{b}} \ge (a^3+b^3+c^3)\dfrac{1}{1+\dfrac{a^2b+b^2c+c^2a}{a^3+b^3+c^3}}\ge \dfrac{1}{2}(a^2c+b^2a+c^2b)$$
$$\Leftrightarrow 2(a^6+b^5+c^6)+3(a^3b^3+b^3c^3+c^3a^3) \ge \sum_{cyc}(a^5c)+\sum_{cyc}(a^4b^2)+\sum_{cyc}(a^3c^2b)+\sum_{cyc}(a^4bc)+3a^2b^2c^2$$
$$\dfrac{2}{3}(a^6+b^6+c^6)+\dfrac{1}{3}(a^3c^3+b^3a^3+c^3b^3)\ge\sum_{cyc}(a^5c)$$
$$\dfrac{1}{3}(a^6+b^6+c^6)+\dfrac{2}{3}(a^3b^3+b^3c^3+c^3a^3)\ge\sum_{cyc}(a^4b^2)$$
$$\dfrac{1}{3}\sum_{cyc}(a^6+a^3c^3+c^3b^3)\ge\sum_{cyc}(a^3c^2b)$$
$$\dfrac{1}{3}\sum_{cyc}(a^6+a^3c^3+a^3b^3)\ge\sum_{cyc}(a^4bc)$$
$$ \dfrac{1}{3}(a^6+b^6+c^6+2(a^3b^3+b^3c^3+c^3a^3)) \ge 3a^2b^2c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of 4th powers modulo 125 I have been trying to evaluate $\sum_{i=1}^{125} i^4\pmod {125}$. My attempt has been somewhat like this so far:
We know that 125 has a primitive root. Let's call it $r$. Now we know that $$r,r^2,\ldots, r^{\phi(125)}$$
is actually congruent to the set of positive integers that are less than 125 and relatively prime to it, i.e. all the numbers not divisible by 5. Also note that $\phi(125)=100$. Thus I write the sum as follows:
$$\sum_{i=1}^{125} i^4=(5^4+10^4+15^4+\ldots + 125 ^4)+(1+2^4+3^4+\ldots+124^4)$$
$$=5^4(1+2^4+3^4+\ldots+25^4)+(1+r^4+r^8+\ldots+r^{396})$$
$$\equiv \frac{r^{400}-1}{r^4-1}\pmod {125}$$
$$\equiv 0\pmod {125}$$
But when I calculate the expression using wolphram alpha, I get the answer is 100. Where am I going wrong? Please point out. Thanks in advance.
| As $2^2\equiv-1\pmod5,2$ is a primitive root of $5$
Now from Order of numbers modulo $p^2$
ord$_{5^2}2=4$ or $4\cdot5 $
Now as $2^4\not\equiv1\pmod{25},$ord$_{5^2}2=4\cdot5=\phi(25)$
So, $2$ is a primitive root of $5^2$
Using If $g$ is a primitive root of $p^2$ where $p$ is an odd prime, why is $g$ a primitive root of $p^k$ for any $k \geq 1$?,
$2$ is a primitive root of $5^n, n\ge1$
Now $$\sum_{n=0}^{99}2^{4n}=\dfrac{16^{100}-1}{16-1}$$
Now $16^{100}\equiv?\pmod{125(16-1)}$
Now as $125\cdot15=5^4\cdot3, \displaystyle16^{100}=(1+15)^{100}\equiv1+\binom{100}115+\binom{100}215^2\pmod{5^4\cdot3}$
As $\displaystyle(15^2,125\cdot15)=75,\binom{100}2\equiv0\pmod{25}$
$\displaystyle\implies16^{100}-1\equiv1500\pmod{125\cdot15}$
$\displaystyle\implies\dfrac{16^{100}-1}{16-1}\equiv\dfrac{1500}{16-1}\pmod{125}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If the normal to the point P on the ellipse intersects the major and minor axis and $G$ and $g$, then find the relation ...
If the normal to the point P on the ellipse intersects the major and minor axis and $G$ and $g$, then find the relation between $CG$, $Cg$, $a$ and $b$, where C is centre
The normal at $P(x_1,y_1)$ is
$$\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$$
The intercepts will be
$$CG=\frac{x_1(a^2-b^2)}{a^2}$$
$$Cg=\frac{y_1(a^2-b^2)}{b^2}$$
Unfortunately I am not able to eliminate the terms $x_1$ and $y_1$. How should I do it?
| We know that $(x_1,y_1)$ must lie on the ellipse, so, we have:
$$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1 \leftrightarrow y_1^2=b^2\left(1-\frac{x_1^2}{a^2}\right)$$
Substituing, we obtain:
$$Cg^2=\frac{b^2\left(1-\frac{x_1^2}{a^2}\right)(a^2-b^2)^2}{b^4}$$
While $CG^2$ is:
$$CG^2=\frac{x_1^2(a^2-b^2)^2}{a^4}\leftrightarrow x_1^2=\frac{CG^2\cdot a^4}{(a^2-b^2)^2} $$
Substituing again, you arrive at:
$$Cg^2=\frac{b^2\left(1-\frac{CG^2\cdot a^4}{a^2(a^2-b^2)^2}\right)(a^2-b^2)^2}{b^4} \rightarrow Cg^2=\frac{(a^2-b^2)^2-CG^2\cdot a^2}{b^2}$$
From here, it's very simple. In fact:
$$Cg=\sqrt{Cg^2}=\sqrt{\frac{(a^2-b^2)^2-CG^2\cdot a^2}{b^2}}=\frac{\sqrt{(a^2-b^2)^2-CG^2\cdot a^2}}{b}$$
Note that, as @Blue suggested, you can rewrite the relation for $Cg^2$ as:
$$\frac{Cg^2}{a^2} + \frac{CG^2}{b^2} =\frac{ (a^2-b^2)^2}{a^2b^2}$$
This follows from here:
$$Cg^2=\frac{(a^2-b^2)^2-CG^2\cdot a^2}{b^2} \leftrightarrow Cg^2b^2-CGa^2=(a^2-b^2)^2 \leftrightarrow \frac{Cg^2}{a^2} + \frac{CG^2}{b^2} =\frac{ (a^2-b^2)^2}{a^2b^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $DE\perp EF$.
Question: Point $D$ lies inside $\Delta ABC$ such that $\angle DAC=\angle DCA=30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE\perp EF$.
My approach: Let $\angle CDF=\alpha$ and $\angle EDC=\beta$.
Now in $\Delta FDA$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FA}{\sin(120^\circ-\alpha)}\\\implies FD=\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}.$$
Again in $\Delta FDC$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FC}{\sin \alpha}\\\implies FD=\frac{1}{2}.\frac{FC}{\sin \alpha}.$$
Thus, $$\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}=\frac{1}{2}.\frac{FC}{\sin \alpha}\\\implies \frac{\sin \alpha}{\sin(120^\circ-\alpha)}=\frac{FC}{FA}=\frac{1}{2}\\\implies \tan \alpha=\frac{1}{\sqrt 3}\implies \alpha=30^\circ.$$
Thus $\angle ADF=90^\circ$. Now let $CD$ extended meet $AB$ at $J$. Thus $\angle ADJ=60^\circ.$ Now observe that if we can prove that points $A,D$ and $E$ are collinear, then we can conclude that $\angle EDC=\beta=60^\circ$. Hence we will be done.
I tried to use Menalaus Theorem to prove the same, but it was of no use.
Also I tried to use coordinate bash. Consider $\Delta CDA$. Observe that $\Delta CDA$ is isosceles with $CD=BA$. Let $DO$ be the angular bisector of $\angle ADC$. Thus $DO$ is also the perpendicular bisector of $AC$. Now let $O$ be the origin and let $AC$ be the x-axis. Thus clearly $DO$ represents the y-axis. Now let $DA=s$. Thus clearly $A=\left(-\frac{\sqrt 3}{2}s,0\right),C=\left(\frac{\sqrt 3}{2}s,0\right)$ and $D=\left(0,\frac{s}{2}\right)$. Now let $B=(a,b)$, thus $$E=\left(\frac{a}{2}+\frac{\sqrt{3}}{4}s, \frac{b}{2}\right).$$
Now slope of $AB=m_1=\frac{2b}{2a+\sqrt 3s}$ and slope of $DB=m_2=\frac{2b-s}{2a}$. Now since the angle between $AB$ and $DB=60^\circ,$ thus we have $$\sqrt 3=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$ After this I haven't found anything significant.
So, how to proceed after this?
| Arbitrarily set $A=(0,0)$, $C=(1,0)$, $D=\left(\frac12,\frac1{2\sqrt3}\right)$ and $F=\left(\frac23,0\right)$. Now define new points $O=\left(\frac16,\frac1{2\sqrt3}\right)$, $G=\left(\frac13,0\right)$ and $H=\left(0,\frac1{\sqrt3}\right)$, then let $\Gamma$ be the circle with centre $O$ passing through $A$.
Since $\angle DOA=120^\circ$, $B$ lies on $\Gamma$. We also have the equalities
$$GF=FC,BE=EC,HD=DC$$
so we can show $\triangle HBG$ is similar to $\triangle DEF$. Now $GH$ happens to be a diameter of $\Gamma$, so $\angle HBG=90^\circ$ and hence $\angle DEF=90^\circ$ too, or $DE\perp EF$.
| {
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"url": "https://math.stackexchange.com/questions/3615048",
"timestamp": "2023-03-29T00:00:00",
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Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$. My attempt:
Let $x, y$ be roots of
$$t^2-5t+p=0$$
I got
$$p=\frac{5 \pm \sqrt{-431}}{2}$$
using Vieta's relations.
Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$.
But I got two very hideous complex values of x and y. So I think my answer may be wrong.
The values I got are
$$x= \frac{5+\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}$$ or
$$x= \frac{5+\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}$$
Please post your own solutions too if you have any different solutions.
| Hint: Substituting $y=5-x$ the equation $x^5+y^5=275$ is equivalent to
$$
(x^2 - 5x + 19)(x - 2)(x - 3)=0
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
My Attempt:
Number of zeros at end of $n!$ is
$$\left\lfloor \frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^2}\right\rfloor+\dots$$
But is there a method to get converse
| As in my comment, let $f(n)$ denote the number of ending zeros of $n!$, for each positive integer $n$. We define also $r(n)$ to be the largest positive integer $k$ such that $5^k\leq n$. Then,
$$f(n)=\sum_{k=1}^{r(n)}\,\left\lfloor\frac{n}{5^k}\right\rfloor\,.$$
We see that
$$\frac{n+1}{5^k}-1\leq\left\lfloor\frac{n}{5^k}\right\rfloor \leq \frac{n}{5^k}$$
for every $k=1,2,\ldots,r(n)$. Thus,
$$f(n)\leq \sum_{k=1}^{r(n)}\,\frac{n}{5^k}<\sum_{k=1}^\infty\,\frac{n}{5^k}=\frac{n}{4}\,.$$
This shows that $n>4\,f(n)$, or $n\geq 4\,f(n)+1$.
On the other hand,
$$f(n)\geq \sum_{k=1}^{r(n)}\,\left(\frac{n+1}{5^k}-1\right)=\frac{n+1}{4}\,\left(1-\frac{1}{5^{r(n)}}\right)-r(n)\,.$$
Because $5^{r(n)}\leq n<5^{r(n)+1}$, we see that
$$\frac{1}{5^{r(n)}}<\frac{5}{n}\,.$$
Thus,
$$f(n)>\frac{n+1}{4}\,\left(1-\frac{5}{n}\right)-r(n)=\frac{n-4}{4}-\frac{5}{4n}-r(n)>\frac{n-5}{4}-\frac{5}{16\,f(n)}-r(n)$$
(recalling that $n>4\,f(n)$). That is, $n<4\,f(n)+4\,r(n)+5+\dfrac{5}{4\,f(n)}$.
If $n\geq 10$ or $f(n)\geq 2$, we obtain $\dfrac{5}{4\,f(n)}<1$; therefore,
$$n\leq 4\,f(n)+4\,r(n)+5\,.$$
Because $n\leq 5\,f(n)$ clearly, we get that $$r(n)=\big\lfloor\log_5(n)\big\rfloor\leq 1+\Big\lfloor\log_5\big(f(n)\big)\Big\rfloor\,.$$
Ergo,
$$4\,f(n)+1\leq n \leq 4\,f(n)+4\,\Big\lfloor\log_5\big(f(n)\big)\Big\rfloor +9\,.$$
Note that, for a given positive integer $m$, there are at most $5$ positive integer $n$ such that $$f(n)=m\,,$$ and such integers $n$ are of the form $5t$, $5t+1$, $5t+2$, $5t+3$, and $5t+4$ for some positive integer $t$. Therefore, if we are given the value of $m\geq 2$, then we can simply search for $t$ from
$$\left\lceil\frac{4m+1}{5}\right\rceil \leq t \leq \left\lfloor\frac{4m+4\,\big\lfloor\log_5(m)\big\rfloor+9}{5}\right\rfloor\,.$$
(The inequality above is good for $m=1$ as well.)
When $m=1998$, we have $5^4\leq m<5^5$, so that $4\leq \log_5(m)<5$. Ergo,
$$1599\leq t \leq 1603\,.$$
Therefore, we just need to check which $t\in\{1599,1600,1601,1602,1603\}$ works (there will be at most one possible value). It turns out that $t=1600$ leads to a solution. Therefore, all values of $n$ are
$$8000\,,\,\,8001\,,\,\,8002\,,\,\,8003\,,\text{ and }8004\,.$$
It is definitely interesting to find a relationship between $m$ and $t$. I expect that, for the values $m$ such that there exists a positive integer $t$ such that $m=f(5t)$, we have
$$t=\frac{4m}{5}+\lambda\,\log_5(m)+o\big(\log_5(m))$$
for some positive constant $\lambda\leq \dfrac45$. What is the value of $\lambda$? (Gareth Ma kindly helped with the numerical simulation. It seems we can only say $t=\dfrac{4m}{5}+\mathcal{O}\big(\log_5(m)\big)$.)
| {
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A die is thrown $n$ times. What is the probability that $6$ appears even number of times? A die is thrown $n$ times. What is the probability that $6$ appears even number of times (for the purpose of task $0$ is even number)?
The solution from my textbook is:
We have two hypotheses $H_1$ that the array starts with 6 and $H_2$ array does not start with 6; $p_n$ is the probability. $p_n=P(H_1)(1-p_{n-1})+P(H_2)p_{n-1}=\frac{1}{6}+\frac{2}{3}p_{n-1}$. $p_1=\frac{5}{6}$
After solving this recursive relation we get $p_n=\frac{1}{2}(1+(\frac{2}{3})^n)$.
I know how to solve recursion I just don't understand how they got it (I know probabilities of hypothesis). Also, is there any other way to approach this task?
| Hint: Make use of Binomial Expansion.
Assuming n to be even
$(a + b)^n = {n \choose 0}a^n + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^{2} + {n \choose 3}a^{n-3}b^{3} ... + {n \choose n}b^{n}$
and,
$(a - b)^n = {n \choose 0}a^n - {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^{2} - {n \choose 3}a^{n-3}b^{3} ... + {n \choose n}b^{n}$
Adding these two, we get:
$(a + b)^n + (a - b)^n = 2({n \choose 0}a^n + {n \choose 2}a^{n-2}b^{2}...)$,
thus the quantity on the right of the above equation represents the probability of getting an even number of events $a$. Bring the $2$ from RHS to LHS,
$\frac{1}{2}((a + b)^n + (a - b)^n) = ({n \choose 0}a^n + {n \choose 2}a^{n-2}b^{2}...)$
We know from a fair, unbiased die, we have P(Die Value = 6) = 1/6 and P(Die Value != 6) = 5/6. Then, substitute $a$ for 1/6 and $b$ for 5/6 and solve.
| {
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How prove this determinant is $0?$ find the value
$$A_{n}=\begin{vmatrix}
1-\dfrac{1}{(n+1)^2}&\dfrac{1}{2}&\dfrac{1}{3}&\cdots&\dfrac{1}{n+1}\\
\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}&\cdots&\dfrac{1}{n+2}\\
\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}&\cdots&\dfrac{1}{n+3}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\dfrac{1}{n+1}&\dfrac{1}{n+2}&\dfrac{1}{n+3}&\cdots&\dfrac{1}{2n+1}
\end{vmatrix}$$
show that $\det(A_{n})=0$
I have prove
$$\det(A_{1})=\begin{vmatrix}
\dfrac{3}{4}&\dfrac{1}{2}\\
\dfrac{1}{2}&\dfrac{1}{3}
\end{vmatrix}=0$$
and
$$\det(A_{2})=\begin{vmatrix}
\dfrac{8}{9}&\dfrac{1}{2}&\dfrac{1}{3}\\
\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}\\
\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}
\end{vmatrix}=0$$
$$\det(A_{3})=\begin{vmatrix}
\dfrac{15}{16}&\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}\\
\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}\\
\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}&\dfrac{1}{6}\\
\dfrac{1}{4}&\dfrac{1}{5}&\dfrac{1}{6}&\dfrac{!}{7}
\end{vmatrix}=0$$
| Shifting the index, we have
$$A_n = H_n -(1/n^2) e_1 e_1^T$$
where $H_n$ is the Hilbert matrix and $e_1$ is the column vector $[1,0,0,\dots,0]^T$.
By the matrix determinant lemma,
$$
\det(A_n)=(1-(1/n^2)e_1^T H_n^{-1} e_1) \det(H_n)
=(1-(1/n^2) (H_n^{-1})_{11}) \det(H_n) = 0
$$
because the $1,1$ entry of $H_n^{-1}$ is $n^2$.
| {
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Show there exist integers $a$ and $b$ such that $a^2+b^2\equiv -1\mod p$ I'm asked to show there exist integers $a$ and $b$ such that $a^2+b^2\equiv -1\mod p$ for any odd prime $p$.
The solution starts by saying that the subsets $\{a^2\}$ and $\{-1-b^2\}$ of $\mathbb Z/p\mathbb Z$ both contain $\frac{p-1}{2}$ elements. Why is this? What even is the subset $\{a^2\}$? Is this related to how there are $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ quadratic non-residues $\mod p?$
| This is not nessesarily true. First, Fermat's two square theorem states that if
$a^2 + b^2 = p$ where $p$ is prime, then $p = 1\pmod 4$.
In particular if $p$ is not prime, then it is a product of prime powers congruent to ${0,1,2} \pmod 4$. Consider the solutions of
$a^2 + b^2 \mod 8$.
It is trivial that ${0, 1, 4, 5} \mod 8$ are solutions. ${2,6} \mod 8$ are possible solutions. Any integer $p=1\pmod 4$ gives $2p=2\pmod 8$, and
$p=3\pmod 4$ gives $2p=6\pmod 8$. The latter, however, cannot be a solution since there are factors $p=3\pmod 4$ dividing $a^2+b^2$, a contradiciton.
The solutions are therefore $a^2 + b^2 = {0, 1, 2, 4, 5} \mod 8$ and
$a^2 + b^2 + 1 = {1, 2, 3, 5, 6} \mod 8$.
Which does not represent all numbers.
| {
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If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$
But I can't progress after this.
| Let $a=x$, $b:=2y$, and $c:=4z$. Then,
$$\begin{align}a^2+b^2+c^2&=x^2+4y^2+16z^2=48=2\cdot 24\\&=2\cdot(xy+4yz+2zx)=bc+ca+ab\,.\end{align}$$
It is well known that, for real numbers $a$, $b$, and $c$, $a^2+b^2+c^2=bc+ca+ab$ if and only if $a=b=c$. That is, $$x=2y=4z\,.$$
Since $x^2+4y^2+16z^2=48$, we get $$(x,y,z)=\pm(4,2,1)\,.$$
That is,
$$x^2+y^2+z^2=16+4+1=21\,.$$
| {
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Multiplying inequalities of 2 different variable with ranges Given $-\frac{7}{2}\le y <\frac{9}{2}$ and $-1 < x \le 3$, find a and b in $a < xy < b$!
My approach was to find the lowest possible solution for $xy$ which is $3\times-\frac{7}{2}=-\frac{21}{2}$ and to find the highest possible solution for $xy$ which is $3\times\frac{9}{2}$ which is $\frac{27}{2}$. I end up with $-\frac{21}{2} < xy < \frac{27}{2}$.
Is this approach correct?
I also tried multiplying
$0 < x+1 \le 4$ and $0 \le y + \frac{7}{2} < 8$ but this way I get different results. Can I multiply them both if I've made sure that both inequalities are positive, so the signs don't change?
Also, I would love to know if there's another approach I can use in general! Thanks!
| I think, your approach is not enough.
Here is one of the possible proofs.
The right estimation:
Since $|x|\leq3$ and $|y|<\frac{9}{2},$ we obtain:
$$xy\leq|xy|<\frac{9}{2}\cdot3=\frac{27}{2}.$$
Also, $$xy\geq-\frac{21}{2}.$$
For $x\geq0$ we obtain:
$$xy=x\left(y+\frac{7}{2}\right)-\frac{7}{2}x\geq-\frac{7}{2}\cdot3=-\frac{21}{2}.$$
For $x\leq0$ we obtain:
$$xy=x\left(y-\frac{9}{2}\right)+x\cdot\frac{9}{2}\geq-1\cdot\frac{9}{2}>-\frac{21}{2}.$$
Id est, the best estimation it's: $$-\frac{21}{2}\leq xy<\frac{27}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.
My proof is the following:
\begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}
Could anyone find a simpler non-calculator proof without using big numbers?
| Since $17^3 = 4913 < 492 × 10$, then$$
17^6 < 492^2 × 10^2 = 242064 × 10^2 < 243000 × 10^2 = 3^5 × 10^5.
$$
Now it suffices to prove that $(3^5 × 10^5)^{23} < (10^{85})^2$, or $3^{23} < 10^{11}$. Note that $3^9 = 27^3 = 19683 < 2 × 10^4$ and $3^5 = 243 < 25 × 10$, thus$$
3^{23} = (3^9)^2 × 3^5 < (2 × 10^4)^2 × (25 × 10) = 10^{11}.
$$
| {
"language": "en",
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Is $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ correctly proved by induction? I have this statement:
Prove by induction that $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ for $n \in \mathbb{N}$
My attempt was:
Base case: $\frac{1}{1} \leq 1$
Assume that $\displaystyle \sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$ is true for some $m \in \mathbb{N}$
To prove: $\displaystyle \sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$
Since $$\sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$$
$$\sum_{k = 1}^m\frac{1}{k^2} + \frac{1}{(m+1)^2}\leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$
$$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$
Now, I'll prove that (1) $$\displaystyle 2 -\frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m + 1}$$
Developing this inequation, gives $$m^2 + 2m + 1 \geq m^2 + 2m$$ getting that (1) is true.
And by transitivity $$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$$ was proved.
But I don't know if this is a valid induction proof and if not, what is the type of this proof?
Thanks in advance.
| If we know that it is valid for $n$, we only need to show that $\frac{1}{(n+1)^{2}}<\frac{1}{n}-\frac{1}{n+1}$
Therefore
$$
\sum_{k=1}^{n}{\frac{1}{k^{2}}}+\frac{1}{(n+1)^{2}}<2-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}
$$
| {
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Solve $\lfloor{\sin x}\rfloor+\lfloor{\cos x}\rfloor=2^{1-|\sin x|}$
Solve $$\lfloor{\sin x}\rfloor+\lfloor{\cos x}\rfloor=2^{1-|\sin x|}$$
Now, what I did notice is that $\lfloor{\sin x}\rfloor$ can only have three different values, and the same goes for $\lfloor{\cos x}\rfloor$. Moreover, if either is $1$ then the other is $0$, and $|\sin x|$ can only have one value. I am still not sure what are the solutions. Any help will be appreciated!
| You have $1 = 2^0 \le 2^{1-|\sin x|} \le 2^1 = 2$
$$\begin{array}{c|c|c} & \lfloor \sin x \rfloor & \lfloor \cos x \rfloor \\ \hline x = 0 & 0 & 1 \\ 0 < x < \dfrac{\pi}{2} & 0 & 0 \\ x = \dfrac{\pi}{2} & 1 & 0 \\ \dfrac{\pi}{2} < x \le \pi & 0 & -1 \\ \pi < x < \dfrac{3\pi}{2} & -1 & -1 \\ \dfrac{3\pi}{2} \le x < 2\pi & -1 & 0 \end{array}$$
So, the LHS intersects the possible range of the RHS at $x=0$ or $x=\dfrac{\pi}{2}$. So, we need only try those two values:
$$2^{1-|\sin 0|} = 2^1 = 2 \neq 1 \\ 2^{1-\left|\sin \tfrac{\pi}{2}\right|} = 2^0 = 1 = \left\lfloor \sin \dfrac{\pi}{2} \right\rfloor + \left\lfloor \cos \dfrac{\pi}{2}\right\rfloor$$
Thus, the solution is $x = \dfrac{\pi}{2}+2\pi n, n\in \mathbb{Z}$.
| {
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How would one go about solving $\int_{0}^{\pi/2} \sqrt{\cos(x)} dx$? Why does the gamma function show up in the answer? Wolfram Alpha says the answer is $\sqrt{\frac{2}{\pi}} \Gamma(\frac{3}{4})^2$, but that just seems so strange. How does an integral with finite bounds relate to a function where one of the bounds is infinity? How would you go about solving this? Any and all help is greatly appreciated!
| Substituting $u=\sqrt{\cos(x)}$ and $v=u^4$ gives
$$ \int_0^{\pi/2}\sqrt{\cos(x)}dx=\int_0^1\frac{2u^2}{\sqrt{1-u^4}}du=\frac{1}{2}\int_0^1\frac{v^{-1/4}}{\sqrt{1-v}}dv=\frac{1}{2}\beta\left(\frac{3}{4},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{5}{4}\right)} $$
But $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{4}\right)=\frac{\Gamma\left(\frac{3}{2}\right)}{\sqrt{2}}=\frac{1}{2}\sqrt{\frac{\pi}{2}}$ because of $\Gamma(s)\Gamma\left(s+\frac{1}{2}\right)=2^{1-2s}\Gamma(2s)$ so that
$$ \int_0^{\pi/2}\sqrt{\cos(x)}dx=\sqrt{\frac{2}{\pi}}\Gamma^2\left(\frac{3}{4}\right)$$
| {
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Maximise area of ellipse inscribed inside an isosceles triangle
Find the maximum area(in terms of $A$) of the ellipse inscribed inside an isosceles triangle of area $A$, having one axis lying along the perpendicular from the vertex of the triangle to its base.
WLOG, Let the vertices of the isosceles triangle be $A(p,0),B(0,q),C(0,-q)$
$A=1/2*p*2q=pq$.
Let the centre of the ellipse be $(a,0)$, Let the ellipse have axis of lengths $2a,2b$.
The equation of the inscribed ellipse is $\frac{(x-a)^2}{a^2}+\frac{y^2}{b^2} = 1$.
The ellipse has tangent equations $\frac{x}{p} \pm \frac{y}{q}=1$.
General tangent equation for ellipse(with shifted centre, centre at(a,0)) is
$y=m(x-a) \pm \sqrt{a^2m^2+b^2}$.
So relating any of the lines with general equation,
Gives $q(p-a)^2=a^2q^2+b^2p^2$,
Or $pq^2-2aq=bp^2$.
Or $A^2-2aA=bp^3$
So area of ellipse=$\pi ab=
\pi a \frac{A^2-2aA}{p^3}$.
However I'm unable to eliminate p in this.
How to proceed?
| SteinerInellipse:
"The Steiner inellipse of a triangle has the largest area of any
inellipse of that triangle; its area is $\tfrac\pi{3\sqrt3}$ times the
area of the triangle".
The Steiner inellipse
is tangent to the sides of the triangle at their midpoints,
so for the isosceles triangle one of its axes
will be localed along the perpendicular from the vertex of the triangle to its base,
hence it is indeed a solution.
For the isosceles $\triangle ABC$ with side lengths
$|AC|=|BC|=a$ and the base $|AB|=c$
the lengths of semi-axes
of the Steiner inellipse are
\begin{align}
s_a&=
\max\Big(\tfrac{\sqrt3}6\,c,\, \tfrac16\,\sqrt{4a^2-c^2}\Big)
,\\
s_b&=
\min\Big(\tfrac{\sqrt3}6\,c,\, \tfrac16\,\sqrt{4a^2-c^2}\Big)
.
\end{align}
Update:
Let WLOG $A=(-1,0)$, $B=(1.0)$,
$C=(0,q)$,
$s_a$ and $s_b$-
the major and minor semi-axes of the ellipse, respectively.
Then
the center of the ellipse $O=(0,s_a)$
and
the equation of the ellipse is
\begin{align}
\frac{x^2}{b^2}+\frac{(y-a)}{a^2}&=1
\tag{1}\label{1}
,
\end{align}
and the equation for the upper arc is
\begin{align}
y(x)&=\frac{s_a}{s_b}\cdot\Big(s_b+\sqrt{s_b^2-x^2}\Big)
\tag{2}\label{2}
,\\
y'(x)&=
-\frac{s_a}{s_b}\cdot \frac{x}{\sqrt{s_b^2-x^2}}
\tag{3}\label{3}
.
\end{align}
Equation for the tangent line $AC$ is
\begin{align}
y_t(x)&=q\,x+q
\tag{4}\label{4}
.
\end{align}
Let the tangent point be $B_m=(x_0,y_0)$.
Then
\begin{align}
y'(x_0)&=q
,\\
x_0&=
-\frac{q\,s_b^2}{\sqrt{q^2\,s_b^2+s_a^2}}
.
\end{align}
Also
\begin{align}
y_0 &= q\,x_0+q
,\\
y_0&=
\frac{s_a}{s_b}\cdot\Big(s_b+\sqrt{s_b^2-x_0^2}\Big)
.
\end{align}
This is enough to express $s_a,\,s_b$ in terms of $x_0$:
\begin{align}
s_a&=q\cdot\frac{1+x_0}{2+x_0}
,\\
s_b&=\sqrt{-\frac{x_0}{2+x_0}}
.
\end{align}
The area of the inellipse is then
\begin{align}
S_e(x_0)&=\pi\cdot s_a\cdot s_b
\\
&=
q\cdot \pi\,(1+x_0)\cdot\sqrt{-\frac{x_0}{(2+x_0)^3}}
,\\
S_e'(x_0)&=
-\frac{q\cdot\pi\cdot(1+2\,x_0)}{\sqrt{-x_0\,(2+x_0)^5}}
,
\end{align}
and the maximal area is reached at $x_0=-\tfrac12$,
as expected.
| {
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Show that $e^{\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)}}= 2 + \sqrt{3}$ How from $e^{\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)}}$ do I get $2+\sqrt{3}$?
This is a simple problem that I am unable to solve. I tried using properties of logarithms, but I cannot find $2+\sqrt{3}$, as Wolfram says... how can I do it?
Thanks.
| Note that, as Matti P.'s question comment indicates, you have
$$(2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 \implies 2-\sqrt{3} = (2+\sqrt{3})^{-1} \tag{1}\label{eq1A}$$
To make the algebra simpler, use
$$x = 2+\sqrt{3} \implies x^{-1} = 2-\sqrt{3} \tag{2}\label{eq2A}$$
The exponent of $e$ in your expression now becomes
$$\begin{equation}\begin{aligned}
\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)} & = \frac{3\ln(x)\ln(x^{-1})}{2\ln(x^{-1}) - \ln(x)} \\
& = \frac{3\ln(x)(-1)\ln(x)}{-2\ln(x) - \ln(x)} \\
& = \frac{-3\ln^2(x)}{-3\ln(x)} \\
& = \ln(x)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This now results in
$$\begin{equation}\begin{aligned}
e^{\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)}} & = e^{\ln(x)} \\
& = x \\
& = 2 + \sqrt{3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
| {
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Calculus Spivak Chapter 14 problem 7; f such that $\int_{1}^{x} f=f^2(x)+C$ i´m having trouble understanding why is $f'(x)=0$ in the solution of this problem of Calculus by Spivak.
find a continuous fuction f such that:
$\int_{0}^{x} f=f^2(x)+C$
Here´s the solution that i´m talking about:
Clearly $f^2$ is differentiable everywhere. So $f$ is differentiable at x whenever $f(x)\neq 0$ and:
$f(x)=2f(x)f'(x)$.
So $f'(x)=0$ at such points.Thus $f(x)$ is constant on any interval where it is non-zero. So $f(x)=0$
Can someone help understand this?
Note: I found this solution at the Answer book for calculus Third edition, I guess it is a mistake or something.
| The question is pretty old but I'll just leave the actual answer to the question if anyone looks for it, since it's pretty cool and it's a shame the answer book made the mistake.
First notice that, by the Fundamental Theorem of Calculus, the condition is equivalent to
$$(f^2)' = f,$$
so that we'll just deal with that and find all continuous functions which satisfy this.
First, as in the answer, if $f(x)\neq 0 $ then, by continuity,
\begin{align}
f'(x) &= \lim_{h\to0} \frac{f(x+h)-f(x)}{h}\\
&= \lim_{h\to0} \frac{f^2(x+h)-f^2(x)}{h} \cdot \frac{1}{f(x+h)+f(x)}\\
&= (f^2)'(x)\cdot \frac{1}{2f(x)} = \frac{1}{2}.
\end{align}
Therefore, if $f(x) \neq 0$ for all $x$ in an interval $I$, then there is a number $c$ such that $f(x) = \frac{1}{2} x + c$ for all $x$ in $I$. We'll use this fact multiple times.
First, a trivial solution is $f = 0$, so we'll look for others.
Suppose $f(x_0) > 0$ for some $x_0$. The first step is to show that $f(x) > 0$ for all $x$ in $[x_0, \infty)$: If we had $f(x) = 0$ for some $x > x_0$, we could let $x_1$ be the smallest number $> x_0$ which satisfies $f(x_1)=0$, so that $f(x)>0$ for $x$ in $[x_0, x_1)$. (Note that we are using continuity here.)
This implies that there is a number $c$ such that $f(x) = \frac{1}{2} x + c$ for all $x$ in $[x_0, x_1)$. But then
$$f(x_1) = \lim_{x \to x_1^+} f(x) = \frac{1}{2} x_1 + c > 0,$$
a contradiction. Therefore $f(x) > 0$ for all $x$ in $(x_0, \infty)$. This means that we can find a number $c$ such that $f(x) = \frac{1}{2} x + c$ for all $x > x_1$.
Next step is to show that $f(x) = \frac{1}{2}x + c$ for all $x$ in $[-2c, \infty)$. First, note that we must have $f(x) = 0$ for some $x$ in $[-2c, x_0]$: If we had $f(x) > 0$ for all $x$ in $[-2c, x_0]$, there'd be a number $\bar{c}$ such that $f(x) = \frac{1}{2} x + \bar{c}$ for all $x$ in $[-2c, x_0]$. But $f(x_0) = \frac{1}{2} x_0 + c$, so we must have $c = \bar{c}$. However, this means $f(-2c) = \frac{1}{2} (-2c) + c = 0$, a contradiction. We conclude that there is some $x$ in $[-2c, x_0]$ such that $f(x) = 0$; let $x1$ be the largest one. Then $f(x) > 0$ for all $x$ in $(x_1, x_0]$, so, as before, $f(x) = \frac{1}{2} x + c$ for all $x$ in $(x1, x0]$. But then
$$0 = f(x_1) = \lim_{x \to x_1^+} f(x) = \frac{1}{2} x_1 + c,$$
so $x_1 = -2c$.
In conclusion: If $f(x) > 0$ for some $x$, then there is a number $c$ such that $f(x) = \frac{1}{2} x + c$ for all $x$ in $[-2c, \infty)$.
A similar analysis shows that if $f(x) < 0$ for some $x$ then there is a number $c$ such that $f(x) = \frac{1}{2} x + c$ for all $x$ in $(-\infty, -2c]$.
Note: If there are numbers $c_1, c_2$ with $f(x) = \frac{1}{2} x + c_1$ for $x \in (-\infty, -2c_1]$ and $f(x) = \frac{1}{2} x + c_2$ for $x \in [-2c_2, \infty)$, we must have $-2c_1 \leq -2c_2$. Otherwise, we'd have $-2c_1 \in (-2c_2, \infty)$, so $0 = f(-2c_1) = \frac{1}{2} (-2c_1) + c_2 = c_2 - c_1 \neq 0$, which is false.
Finally, we can list all solutions.
*
*$f = 0$.
*$f(x) =
\begin{cases}
\frac{1}{2} x + c, & x \geq -2c\\
0, & \text{otherwise}
\end{cases}$, where $c$ is any number.
*$f(x) =
\begin{cases}
\frac{1}{2} x + c, & x \leq -2c\\
0, & \text{otherwise}
\end{cases}$, where $c$ is any number.
*$f(x) =
\begin{cases}
\frac{1}{2} x + c_1, & x \leq -2c_1\\
0, & -2c_1 < x < -2c_2\\
\frac{1}{2} x + c_2, & x \geq -2c_2\\
\end{cases}$, where $c_1$ and $c_2$ are any numbers with $-2c_1 \leq -2c_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find natural number solutions to $x^2-2y^2=2$ One solution I found is $x=2, y=1$. But my question is:
Is this the only natural solution?
| With an equation of the form $x^2-Dy^2=N$, whenever $(x, y)$ is a solution, $(ax+by, cx+dy)$ is a solution for some $a, b, c, d$ which depend on the coefficients of the original equation but not on $x$ or $y$. Here's one way to find those coefficients $a, b, c, d$ in the OP's case:
\begin{align*}
(ax+by)^2-2(cx+dy)^2 &= 2\\
\implies (a^2-2c^2)x^2+2(ab-2cd)xy+(b^2-2d^2)y^2 &= 2\\
\implies a^2-2c^2 &= 1;\\
ab &= 2cd;\\
b^2-2d^2 &= -2.
\end{align*}
So if you can find a solution to $a^2-2c^2 = 1$, you may take $d=a$ and $b=2c$. $a=3, b=2$ works, so we get the recurrence which Robert Israel pointed out.
More generally, if the original equation is $x^2-Dy^2=N$,
\begin{align*}
(ax+by)^2-D(cx+dy)^2 &= N\\
\implies (a^2-Dc^2)x^2+2(ab-Dcd)xy+(b^2-Dd^2)y^2 &= N\\
\implies a^2-Dc^2 &= 1;\\
ab &= Dcd;\\
b^2-Dd^2 &= -D\\
\implies d^2-\frac{b^2}{D} &= 1\\
\implies d^2-D\Big(\frac{b}{D}\Big)^2&=1
\end{align*}
So again, if you can find a solution to $a^2-2c^2 = 1$, you may take $d=a$ and $b=Dc$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doubt in solution of problem in secrets in inequalities by pham kim hung Question -
Suppose that $a, b, c$ are three side-lengths of a triangle with perimeter
3. Prove that
$$
\frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \geq \frac{9}{a b+b c+c a}
$$
Solution. Let $x=\sqrt{b+c-a}, y=\sqrt{c+a-b}, z=\sqrt{a+b-c} .$ We get $x^{2}+y^{2}+z^{2}=$
3. The inequality becomes
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{36}{9+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}
$$
Let $m=x y, n=y z, p=z x .$ The inequality above is equivalent to
$$
(m+n+p)\left(m^{2}+n^{2}+p^{2}+9\right) \geq 36 \sqrt{m n p}
$$
which is obvious by AM-GM because
$$
m+n+p \geq 3 \sqrt[3]{m n p}, m^{2}+n^{2}+p^{2}+9 \geq 12 \sqrt[6]{m n p}
$$
now i did not understand how they got this last inequality $m^{2}+n^{2}+p^{2}+9 \geq 12 \sqrt[6]{m n p}
$
by am-gm ...
thankyou
| Hint: Write $9=1+1+\cdots +1$ and then apply AM-GM
$$m^2+n^2+p^2+1+\cdots +1\geq 12\sqrt[12]{m^2n^2p^2}=12\sqrt[6]{mnp}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find maximum of $|f(z)| = |\frac{(z-1)^2(z+1)}{z^2}|$ on $|z|=\sqrt2, z\in\mathbb{C}$
Find maximum of $|f(z)| = |\frac{(z-1)^2(z+1)}{z^2}|$ on $|z|=\sqrt2, z\in\mathbb{C}$
I tried to solve it by parametrization $z = \sqrt{2}(cos(t)+isin(t))$ but it seems too complicated to figure out by differentiation in real line(with variable $t$). Is there another way of figuring out?
note: I know so far that two saddle points of $f(z)$ lie on $|z|=\sqrt2$ but that's all I know.
| Note that $f(z) \ne 0$ on $|z|=\sqrt 2$, so representing $f(re^{i\theta})=Re^{i\Phi}$ the local extrema of $|f|=R$ for $r$ constant are given by $\frac{\partial R}{\partial \theta}=0$.
Note that locally where $f \ne 0$ we have $\log f =\log R +i\Phi$, so using $\frac{\partial \log f}{\partial \theta}=i\frac{zf'}{f}$ we have $\frac{zf'}{f}=\frac{\partial \Phi}{\partial \theta}-i\frac{1}{R}\frac{\partial R}{\partial \theta}$ so we need $\Im \frac{zf'}{f}=0$.
(the two critical points $\frac{-1 \pm i\sqrt 7}{2}$ which are on the given circle automatically satisfy that)
Since $f(z)=\frac{(z^2-1)(z-1)}{z^2}=z-1-\frac{1}{z}+\frac{1}{z^2}$ we have $zf'=z+\frac{1}{z}-\frac{2}{z^2}=\frac{z^3+z-2}{z^2}=\frac{(z-1)(z^2+z+2)}{z^2}$, so
$\frac{zf'}{f}=\frac{z^2+z+2}{z^2-1}=1+\frac{z+3}{z^2-1}$, so $\Im \frac{zf'}{f}=0$ implies $\Im ((z+3)(\bar z^2-1))=0$ or (remembering that $|z|^2=2$), $\Im (2\bar z-z+3\bar z^2)=0$ or $(2\bar z-z+3\bar z^2)-(2z-\bar z+3z^2)=0$ which means $(\bar z-z)(\bar z+z+1)=0$ so either we get the two critical points which are precisely the roots of $\bar z+z+1$ or $z=\bar z$ so $z=\pm \sqrt 2$
By inspection $f(\pm \sqrt 2)=\frac{\pm \sqrt 2 -1}{2}$ so the absolute values are $\frac{\sqrt 2 +1}{2}, \frac{\sqrt 2 -1}{2}$, while at the critical points we get $\frac{5}{2}$ hence that is the global maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac1{x^4} < \frac1{x^3} - \frac1{(x+1)^3}$
Prove that for $x \ge 2$, $$\frac1{x^4} < \frac1{x^3} - \frac1{(x+1)^3}.$$
What I have so far is:
$${1\over x^3} - {1\over (x+1)^3} = {(x+1)^3-x^3\over x^3(x+1)^3} = {3x^2+3x+1\over x^3(x+1)^3} > {(x+1)^2\over x^3(x+1)^3} = {1\over x^3(x+1)}.$$
As seen, this won't lead to the correct expression, so could anyone give me any hints on how should I approach the question algebraically?
I've thought of using the graph of ${1\over x^4}$ and using the area-under the graph but I was thinking if there is a trick to solving it via algebraic means.
| \begin{align*}{1\over x^3} - {1\over (x+1)^3}-\frac{1}{x^4}=\frac{2x^3-2x-1}{x^4(x+1)^3}>0 \qquad(x\geq 2)
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/3649633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\langle x_n\rangle $ be a sequence such that $x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2} $for all $n \in \mathbb{N}$, prove that $x_n$ is convergent If $\langle x_n\rangle $ is a sequence of positive real numbers such that $$x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2}$$ for all $n \in \mathbb{N},\ $ let $x_1 <x_2$
then subsequence of odd terms is increasing and subsequence of even terms is decreasing .But how to prove it mathematically? We have
$$x_{(n+2)}-x_n=\frac{(x_{n+1}- x_{n})}{2}= \frac{(x_{n}- x_{n-2})}{4}$$ How to proceed from here? Any hint please.
| (I'm pretty sure that
I and many others
have done this before
but I'll work it out again.)
If
$x(n+1) = ax(n)+(1-a)x(n-1)
$
where
$0 < a < 2$
then
$\begin{array}\\
x(n+1)-x(n)
&= ax(n)+(1-a)x(n-1)-x(n)\\
&= (a-1)x(n)+(1-a)x(n-1)\\
&= (a-1)(x(n)-x(n-1))\\
\text{so}\\
x(n+k)-x(n+k-1)
&= (a-1)^k(x(n)-x(n-1))\\
\text{and}\\
x(n+k)-x(n)
&=\sum_{j=1}^{k}(x(n+j)-x(n+j-1)\\
&= \sum_{j=1}^{k}(a-1)^j(x(n)-x(n-1))\\
&= (x(n)-x(n-1))\sum_{j=1}^{k}(a-1)^j\\
&= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{1-(a-1)}\\
&= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\
\text{so}
&\text{putting } n = 1\\
x(k+1)-x(1)
&= (x(1)-x(0))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\
&= (x(1)-x(0))(\dfrac{a-1}{2-a}-\dfrac{(a-1)^{k+1}}{2-a})\\
\end{array}
$
Since
$0 < a < 2$,
we have
$-1 < a-1 < 1$
so
$(a-1)^{k+1} \to 0$
and
$\begin{array}\\
x(k+1)
&\to
x(1)+(x(1)-x(0))\dfrac{a-1}{2-a}\\
&=(1+\dfrac{a-1}{2-a})x(1)-\dfrac{a-1}{2-a}x(0)\\
&=\dfrac{2-a+a-1}{2-a}x(1)-\dfrac{a-1}{2-a}x(0)\\
&=\dfrac{1}{2-a}x(1)+\dfrac{1-a}{2-a}x(0)\\
\end{array}
$
This is the case
$a = \frac12$
so
$\begin{array}\\
x(k)
&\to
\dfrac{1}{2-1/2}x(1)+\dfrac{1-1/2}{2-1/2}x(0)\\
&=\dfrac{1}{3/2}x(1)+\dfrac{1/2}{3/2}x(0)\\
&=\dfrac23 x(1)+\dfrac13 x(0)\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/3652879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of Equations: Distances from the Torricelli Point to the Vertices of a Triangle I am calculating the length between the Torricelli point and the triangle apex. With the law of cosines I get this system:
$$\left\{\begin{array}{cc}
x^2+y^2+xy&=a^2\,,\\x^2+z^2+xz&=b^2\,,\\z^2+y^2+zy&=c^2\,.\end{array}\right.$$
Please help me solve it.
| In @Batominovski's symmetric notation, we have
$$\begin{align}
a^2&=q^2+r^2+q r\\
b^2&=r^2+p^2+r p\\
c^2&=p^2+q^2+p q
\end{align} \tag{1}$$
Using Mathematica's Resultant function (or, equivalently, Groebner bases), we can eliminate, say, $q$ and $r$ to get
$$\begin{align}
0 &= 9 p^4 (a^4 - a^2 b^2 + b^4 - a^2 c^2 - b^2 c^2 + c^4) \\[4pt]
&+3 p^2 (a^6 - 4 a^4 b^2 + 5 a^2 b^4 - 2 b^6 - 4 a^4 c^2 + 3 a^2 b^2 c^2 -
b^4 c^2 + 5 a^2 c^4 - b^2 c^4 - 2 c^6) \\[4pt]
&+ \phantom{9p^4}(a^2 - b^2 - b c - c^2)^2 (a^2 - b^2 + b c - c^2)^2 \\
&= 9H (p^2)^2 + 3J (p^2) + K
\end{align} \tag{2}$$
This is ugly, but at least it's a quadratic in $p^2$ that can be readily solved. Conveniently, the discriminant factors nicely.
$$\begin{align}
(3J)^2-4(9H)K &= 27 (a^2 - b^2)^2 (a^2 - c^2)^2 (-a + b + c) (a + b - c) (a - b + c) (a + b + c) \\[4pt]
&= 27 (a^2-b^2)^2(a^2-c^2)^2\cdot 16|\triangle ABC|^2 \tag{3}
\end{align}$$
so that we have
$$p^2 = \frac{-J \pm 4\,|\triangle ABC|\,|a^2-b^2|\,|a^2-c^2|\,\sqrt{3}}{6H} \tag{4}$$
I suspect there's an clean way to write $(4)$, and perhaps even a clever way to derive it, but that'll take a bit more thinking.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Exponential Diophantine Equations Find all positive integers $a, b, c > 1$, all relatively prime with respect to each other such that $b\mid 2^{a} +1, c\mid 2^{b} +1, a\mid 2^{c} +1$. $$$$ It is easy to see that all of $a, b, c$ are odd. Suppose $3\mid a$ then we have $3\mid 2^c+1$ and as $(a, c)=1$ so we have $c=6k+1$ or $c=6k+5$ for some $k$. But $9\mid a$ implies $9\mid 2^c+1$ which is not possible. This is true for all the three $a, b, c$. Hence either all of them are equal to $3$ or have some prime factor other than $3$. Suppose the later one is true. Let $p$ be a smallest prime factor of $b$ other than $3$ then we have $$2^{2a} \equiv 1 \mod p$$ and also we have $$2^{p-1} \equiv 1 \mod p$$
So we have $$2^d \equiv 1 \mod p$$ where $d=\gcd (2a, p-1)$. Now $d \neq 2$ as $p \neq 3$. Also $d \neq 3$ because that would have implied $p=7$ and $3|a$ and looking modulo $7$ we arrive at a contradiction. So let $q$ be a prime factor other than $3$ dividing both $a$ and $p-1$ then we have $q<p$ and also as $a$ divides $2^c+1$ we have $$2^{2c} \equiv 1 \mod q$$ and also $$2^{q-1} \equiv 1 \mod q$$. Again by the same reasoning as above we can show that there is a prime factor $r$ other than $3$ dividing both $c$ and $q-1$ and hence we have $r<q$ and as $c|2^b+1$ we have $$2^{2b} \equiv 1 \mod r$$ and also we have $$2^{r-1} \equiv 1 \mod r$$. Again by the same reasoning we can show that there exists a prime $s$ other than $3$ dividing both $b$ and $r-1$ and hence $s<r<q<p$ and $s$ divides $b$ bot as $p$ was the smallest prime factor other than $3$. So a contradiction and hence there do not exist such positive integers.
$$$$Is My Proof Correct?????
| Something's wrong, because e.g. for $a=3, b=9$, I get solutions with $c = 3, 9, 19, 27, 57, 171$, and $513$. Why do you think $(a,c)=1$?
| {
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Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\frac{1}{\sqrt{2}}\left(\sqrt{46-6\sqrt{5}}-2\sqrt{6-2\sqrt{5}}\right)=$$
$$=\frac{1}{\sqrt{2}}\left(\sqrt{(3\sqrt{5}-1)^2}-2\sqrt{(\sqrt5-1)^2}\right)=\frac{1}{\sqrt{2}}\left(3\sqrt{5}-1-2(\sqrt{5}-1)\right)=$$
$$=\frac{1}{\sqrt{2}}(\sqrt{5}+1)=\frac{1}{\sqrt{2}}\sqrt{6+2\sqrt{5}}=\sqrt{3+\sqrt{5}}.$$
But this way does not work for the starting identity.
How to prove the starting identity without squaring?
Thank you!
| Thumb rule to merge any two irrationals: $\boxed{\color{red}{\sqrt{a}\pm\sqrt{b}=\sqrt{(\sqrt{a}\pm\sqrt{b})^2}} \quad \forall \ \color{blue}{a>b}}$
$$\therefore \sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}$$
$$=\sqrt{\left(\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}\right)^2}$$
$$=\sqrt{23-\sqrt{17}+4(7-\sqrt{17})-4\sqrt{(23-\sqrt{17})(7-\sqrt{17})}}$$
$$=\sqrt{23-\sqrt{17}+4(7-\sqrt{17})-4\sqrt{(23-\sqrt{17})(7-\sqrt{17})}}$$
$$=\sqrt{51-5\sqrt{17}-4\sqrt{178-30\sqrt{17}}}$$
$$=\sqrt{51-5\sqrt{17}-4\sqrt{178-2\cdot 5\cdot 3\sqrt{17}}}$$
$$=\sqrt{51-5\sqrt{17}-4\sqrt{(3\sqrt{17}-5)^2}}$$
$$=\sqrt{51-5\sqrt{17}-4(3\sqrt{17}-5)}$$
$$=\sqrt{51-5\sqrt{17}-12\sqrt{17}+20}$$
$$=\sqrt{71-17\sqrt{17}}$$
Proved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Please solve this equation for r. I have the following equation:
$$A^2t^2\space dt^2=-c^2dt^2+\frac{dr^2}{1-Kr^2}+r^2(d\theta^2+\sin^2 \theta\space d\phi^2)$$
The angular separation isn't used for the measurements we're taking, so $d\theta$ and $d\phi$ are zero. So the first simplification is easy:
$$A^2t^2\space dt^2=-c^2dt^2+\frac{dr^2}{1-Kr^2}$$
Can someone walk me through the solution to this equation for r? I'm pretty sure it will have the general form of
$$r=\int_{t_1}^{t_0}(c+A t)\space dt$$
But I'm not exactly sure how the curvature, (${1-Kr^2}$), will factor into the solution.
| $$(A^2t^2+c^2) dt^2=\frac{dr^2}{1-kr^2}$$
The ODE is separable :
$$\pm\int \sqrt{A^2t^2+c^2}dt=\pm\int\frac{dr}{\sqrt{1-kr^2}}+\text{constant}$$
$$\frac12 t \sqrt{A^2t^2+c^2}+\frac{c^2}{2A}\ln\left(A^2t+A\sqrt{A^2t^2+c^2}\right)=\pm\frac{1}{\sqrt{k}}\sin^{-1}(r\sqrt{k})+C$$
$$\frac12 t \sqrt{A^2t^2+c^2}+\frac{c^2}{2A}\ln\left(At+\sqrt{A^2t^2+c^2}\right)=\pm\frac{1}{\sqrt{k}}\sin^{-1}(r\sqrt{k})+C'$$
$$r=\pm\sin\left(\frac{\sqrt{k}}{2} t \sqrt{At^2+c^2}+\frac{c^2\sqrt{k}}{2A}\ln\left(At+\sqrt{A^2t^2+c^2}\right)-C'\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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show this sum $\sum_{i=0}^{n}(-1)^{i-j}\binom{n+i}{i}\binom{n+k-i}{k-i}\binom{2n}{n+j-i}=\binom{2n}{n}$
Let $n,k$ be positive integers. Show that
$$\sum_{i=0}^{n}(-1)^{i-j}\binom{n+i}{i}\binom{n+k-i}{k-i}\binom{2n}{n+j-i}=\binom{2n}{n}$$
for all $j=0,1,2,\cdots,k$.
I don't know if this is an existing conclusion. After thinking for a long time, I found it difficult to deal with it.
Batominovski's edit:
However, when $k=0$ and $j=0$, the required sum is
$$S=\sum_{i=0}^n(-1)^i\binom{n+i}{i}\binom{n-i}{-i}\binom{2n}{n-i}=\binom{n}{0}\binom{n}{0}\binom{2n}{n}=\binom{2n}{n}.$$
When $k=1$ and $j=0$, the required sum is
$$S=\binom{n}{0}\binom{n+1}{1}\binom{2n}{n}-\binom{n+1}{1}\binom{n}{0}\binom{2n}{n-1}=(n+1)\binom{2n}{n}-(n+1)\binom{2n}{n-1}.$$
Because $\binom{2n}{n-1}=\binom{2n}{n+1}$, we have
$$(n+1)\binom{2n}{n-1}=\binom{n+1}{1}\binom{2n}{n+1}=\binom{n}{1}\binom{2n}{n}=n\binom{2n}{n}.$$
Hence, $S=(n+1)\binom{2n}{n}-n\binom{2n}{n}=\binom{2n}{n}$. When $k=1$ and $j=1$, the required sum is
$$S=-\binom{n}{0}\binom{n+1}{1}\binom{2n}{n+1}+\binom{n+1}{1}\binom{n}{0}\binom{2n}{n}=-(n+1)\binom{2n}{n+1}+(n+1)\binom{2n}{n}.$$
So, similar to the previous case, $S=\binom{2n}{n}$. How to prove this for generalized pairs $(k,j)$? Is there a combinatorial proof?
| Rewrite the binomial coefficients in terms of factorials
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \frac{(n+i)!}{\color{red}{n!}i!} \frac{(n+k-i)!}{\color{red}{n!}(k-i)!} \frac{\color{red}{(2n)!}}{(n+j-i)!(n+i-j)!} =\color{red}{\frac{ (2n)!}{(n!)^2}}.
\end{eqnarray*}
$\color{red}{\text{Cancel}}$ and $\color{blue}{\text{create}}$
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \frac{(n+i)!}{(n+i-j)!\color{blue}{j!}} \frac{(n+k-i)!}{(n+j-i)!\color{blue}{(k-j)!}} \frac{\color{blue}{k!}}{i!(k-i)!} = \color{blue}{\frac{k!}{j!(k-j)!}}.
\end{eqnarray*}
Thus, it suffices to show
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} \binom{n+i}{j} \binom{n+k-i}{k-j} =\binom{k}{j}.
\end{eqnarray*}
We shall use $2$ coefficient extractors
\begin{eqnarray*}
\binom{n+i}{j} &=& [x^j]: (1+x)^{n+i} \\
\binom{n+k-i}{k-j} &=& [y^{k-j}]:(1+y)^{n+k-i}.
\end{eqnarray*}
So
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} \binom{n+i}{j} \binom{n+k-i}{k-j}
&=& \sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} [x^j]: (1+x)^{n+i} [y^{k-j}]:(1+y)^{n+k-i}\\
&=&[x^j][y^{k-j}]: \sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} (1+x)^{n+i} (1+y)^{n+k-i}\\
&=&[x^j][y^{k-j}]: (-1)^j \left( 1-\frac{1+x}{1+y} \right)^k (1+x)^{n} (1+y)^{n+k}\\
&=&[x^j][y^{k-j}]:(-1)^j \left( y-x \right)^k (1+x)^{n} (1+y)^{n}\\
&=& \binom{k}{j}. \\
\end{eqnarray*}
| {
"language": "en",
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"answer_id": 1
} |
Why use elimination method vs. substitution method in linear algebra? What is the motivation to use the elimination method as opposed to the substitution method? I've always found the latter much easier. Are there any examples where the substitution method fails or when it is inferior to the elimination method?
For an example of the substitution method see:
Substitution method review (systems of equations)
| Note that the elimination method is equivalent to the substitution method, where the choice of the variable for which we solve is constrained.
Example: Consider the $3 \times 3$ system
$$
\begin{cases}
x + y + z = 2\\
x + 2y + 3z = 5\\
x + 4y + 9z = 15
\end{cases}
$$
With elimination, we would proceed as follows:
$$
\pmatrix{1&1&1&2\\
1&2&3&5\\
1&4&9&15} \leadsto
\pmatrix{1&1&1&2\\
0&1&2&3\\
0&3&8&13} \leadsto
\pmatrix{1&1&1&2\\
0&1&2&3\\
0&0&2&4} \leadsto
\pmatrix{1&1&1&2\\
0&1&2&3\\
0&0&1&2}.
$$
At this point, we see that the original system of equations can be rewritten as
$$
\begin{cases}
x + y + z = 2\\
\quad \ \ y + 2z = 3\\
\quad \ \ \quad \ \ z = 2.
\end{cases}
$$
If we solve the first equation for $x$, substitute, solve the first of the resulting two equations for $y$, then the result is exactly the same. In particular,
$$
x+y+z = 2 \implies x = 2-y-z.
$$
Substituting this into the second and third equations yields
$$
\begin{cases}
(2-y-z) + 2y + 3z = 5\\
(2-y-z) + 4y + 9z = 15
\end{cases} \implies
\begin{cases}
y + 2z = 3\\
3y + 8z = 13.
\end{cases}
$$
This is exactly what we have after the first step of elimination. Solving the second equation gives us
$$
y + 2z = 3 \implies y = 3 - 2z,\\
3(3-2z) + 8z = 13 \implies 2z = 4.
$$
This is exactly what we have after the second step of elimination.
| {
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"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{x\to a} \frac{1}{(a^2-x^2)^2}(\frac{a^2+x^2}{ax}-2\sin( \frac{a\pi}{2})\sin(\frac{x\pi}{2}))$
Without using the L'Hopital Rule, evaluate the following limit.
$$\lim_{x\to a} \frac{1}{(a^2-x^2)^2}\left(\dfrac{a^2+x^2}{ax}-2\sin\left( \frac{a\pi}{2}\right)\sin\left(\frac{x\pi}{2}\right) \right)$$
Where $a$ is an odd integer.
Answer: $\dfrac{\pi^2a^2+4}{16a^4}$
My attempt:
I considered the substitution $x=\lim_\limits{h\to 0}a\cos h$ and got the limit to be:
$$\lim_\limits{h\to 0}\frac{1}{a^4h^4}\left(\frac{1+\cos^2 h}{\cos h}-2\cos^2(ah^2\pi)\right)$$
Aaaand I'm stuck. Please nudge me in the right direction. Profuse thanks.
| I shall focus on the key term
$$y=\frac{1+\cos^2 (h)}{\cos (h)}-2\cos^2(ah^2\pi)$$ and only use
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ Squaring
$$\cos^2(x)=1-x^2+\frac{x^4}{3}-\frac{2 x^6}{45}+O\left(x^8\right)$$
For the first piece
$$\frac{1+\cos^2 (h)}{\cos (h)}=\frac{2-h^2+\frac{h^4}{3}-\frac{2 h^6}{45}+O\left(h^8\right) } {1-\frac{h^2}{2}+\frac{h^4}{24}-\frac{h^6}{720}+O\left(h^8\right) }$$ Use the long division to get
$$\frac{1+\cos^2 (h)}{\cos (h)}=2+\frac{h^4}{4}+\frac{h^6}{12}+O\left(h^8\right)$$ Now, work the second term replacing $x$ by $ah^2\pi$ and ... finish !
| {
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} |
Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$ For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$
Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left(
a+b-2\,c \right) ^{2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left(
119\,a+119\,b+30\,c \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2}
\right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right) \left( a+b \right) \left( 16\,{a}^{2}+a
b+16\,{b}^{2} \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right) \left( a+b \right) ^{
2} \Big] \geqq 0$
which is clearly true for $c=\max\{a,b,c\}$
I wish to see another proof without $uvw$! Thanks for a real lot!
You can see also here.
| BW helps!
Let $a\leq b\leq c$, $b=a+u$ and $c=a+u+v.$
Thus, we need to prove that:
$$9(u^2+uv+v^2)a^4+3(14u^3+21u^2v+3uv^2-2v^3)a^2+$$
$$+(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)a^2+$$
$$+u(56u^4+140u^3v+114u^2v^2+31uv^3+v^4)a+$$
$$+4u^2(2u^2+3uv+v^2)^2\geq0,$$
for which it's enough to prove that:
$$4(u^2+uv+v^2)(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)\geq(14u^3+21u^2v+3uv^2-2v^3)^2$$ or
$$292u^6+876u^5v-85u^4v^2+610u^3v^3+391u^2v^4+8uv^5+20v^6\geq0,$$ which is obvious.
| {
"language": "en",
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"answer_id": 3
} |
Is it possible to evaluate $\iiint \frac{2x^2+z^2}{x^2+z^2} dxdydz$ using cylindrical coordinates instead of spherical? I know that this integral is way easier with spherical coordinates, but I would like to understand my mistakes; evaluate
$$\iiint_D \frac{2x^2+z^2}{x^2+z^2} dxdydz$$
Where $D=\{(x,y,z)\in\mathbb{R}^3 \ \text{s.t.} \ 1 \leq x^2+y^2+z^2 \leq 4, \ x^2-y^2+z^2 \leq 0\}$.
Letting $x=\rho \cos \theta$, $y=y$ and $z=\rho \sin \theta$ it follows that
$$\iiint_E (2\cos^2 \theta+\sin^2 \theta)\rho d\rho dyd\theta=\iiint_E (1+\cos^2 \theta)\rho d\rho dyd\theta$$
Where $E=\{(\rho,y,\theta)\in\mathbb{R}^3 \ \text{s.t.} \ 1 \leq \rho^2+y^2 \leq 4, \rho^2 \leq y^2\, \rho \geq 0, 0 \leq \theta < 2\pi\}$.
The point is that now I have a lot of conditions on $y$, because $\sqrt{1-y^2} \leq \rho \leq \sqrt{4-y^2}$, $-y\leq\rho\leq y$ and $\rho \geq 0$.
From the existence conditions of the roots we get $-1 \leq y \leq 1$ and $-2 \leq y \leq 2$, so it follows that $-1 \leq y \leq 1$.
So it remains to discuss the cases of $\max\left\{\sqrt{1-y^2},-y\right\} \leq \rho$ and $\rho \leq \min\left\{y,\sqrt{4-y^2}\right\}$; it is $y \leq \sqrt{4-y^2}$ for $-1 \leq y \leq 1$ and it is always $\sqrt{1-y^2} \leq \sqrt{4-y^2}$, we have that
$$\max\left\{\sqrt{1-y^2},-y\right\}=\begin{cases} -y, \ \text{if} -1 \leq y \leq -\frac{1}{\sqrt2} \\ \sqrt{4-y^2}, \ \text{if} \ -\frac{1}{\sqrt2} \leq y \leq 1 \end{cases}$$
So I end up with
$$\iiint_E (1+\cos^2 \theta)\rho d\rho dyd\theta=\int_0^{2\pi} \left(\int_{-1}^{-\frac{1}{\sqrt2}} \left(\int_{-y}^{\sqrt{4-y^2}} (1+\cos^2 \theta)\rho d\rho\right)dy \right)d\theta+$$
$$+\int_0^{2\pi} \left(\int_{-\frac{1}{\sqrt2}}^{1} \left(\int_{\sqrt{1-y^2}}^{\sqrt{4-y^2}} (1+\cos^2 \theta)\rho d\rho\right)dy \right)d\theta$$
But I get the wrong answer, am I missing some more conditions (maybe the discussion of $\rho \geq 0$ too) or am I making other mistakes? Thanks.
| In the $y-\rho$ plane the condition $1\le \rho^2+y^2\le 4$ defines a ring centered at $(0,0)$
with inner radius 1 and outer radius 2. The conditions $\rho^2<y^2$ and $\rho\ge 0$ subselect two
45$^\circ$ sections delimited by $0\le \rho \le y$.
$$
I
=\int\int\int\frac{2x^2+z^2}{x^2+z^2} dxdydz
$$
$$
=
\int_0^{2\pi} d\theta
\int_0^{\sqrt 2} \rho d\rho
\int_{1\le \rho^2+y^2\le 4} dy
\frac{2\rho^2\cos^2\theta+\rho^2\sin^2\theta}{\rho^2}.
$$
Because there is no dependence on $y$ in the integrand and the limits are symmetric functions
of $y$, we may look only at $y\ge 0$
and introduce a factor 2 to collect the part of $y<0$:
$$
I
=
2\int_0^{2\pi} d\theta
\int_0^{\sqrt 2} \rho d\rho
\int_{1\le \rho^2+y^2\le 4,y\ge 0} dy
(2\cos^2\theta+\sin^2\theta)
$$
$$
=
2\int_0^{2\pi} d\theta
\int_0^{\sqrt 2} \rho d\rho
\int_{1\le \rho^2+y^2\le 4,y\ge 0} dy
(1+\cos^2\theta)
$$
$$
=
6\pi
\int_0^{\sqrt 2} \rho d\rho
\int_{1\le \rho^2+y^2\le 4,y\ge 0} dy
$$
The ring geometry suggests to chop this into 2 sections of $\rho$ where the $y-$limits
differ as they are delimited by the circles and the diagonal:
$$
=
6\pi
[
\int_0^{1/\sqrt 2} \rho d\rho
\int_{\sqrt{1-\rho^2}}^{\sqrt{4-\rho^2}} dy
+\int_{1/\sqrt 2}^{\sqrt 2} \rho d\rho
\int_\rho^{\sqrt{4-\rho^2}} dy
]
$$
$$
=
6\pi
[
\int_0^{1/\sqrt 2} \rho d\rho
(\sqrt{4-\rho^2}
-\sqrt{1-\rho^2})
+\int_{1/\sqrt 2}^{\sqrt 2} \rho d\rho
(\sqrt{4-\rho^2}-\rho)
]
$$
$$
=
6\pi
[
\int_0^{\sqrt 2} \rho d\rho
\sqrt{4-\rho^2}
-\int_0^{1/\sqrt 2} \rho d\rho
\sqrt{1-\rho^2}
-\int_{1/\sqrt 2}^{\sqrt 2} \rho^2 d\rho
]
$$
$$
=
6\pi
[
\frac12 \int_0^ 2 du \sqrt{4-u} -\frac12 \int_0^{1/2} du \sqrt{1-u} -\frac{7\surd 2}{12}
]
$$
$$
=
6\pi
[
\frac12 (-\frac{4\surd 2}{3}+\frac{16}{3})
-\frac12 (-\frac{\surd 2}{6}+\frac23)
-\frac{7\surd 2}{12}
]
= 6\pi [ \frac{7}{3}-\frac{7\surd 2}{6} ] =7\pi(2-\sqrt 2)
$$
| {
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} |
Proving that $\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)}$ converges How can I prove that the following summation converges?
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)}$$
I tried to prove that by proving that the following summation in in absolute value converges so the original one converges too, but that's incorrect.
Any other ideas?
| If we write it this way
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{(k+1)}\times \frac{(-1)^{n-k}}{(n-k+1)}$$
Then this is the Cauchy product (discrete convolution) of $\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)}$ with itself. We know that $\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k}$, by changing the index $k \rightarrow k+1$ we get $\ln(1+x)=\sum_{k=0}^\infty \frac{(-1)^{k}x^{k+1}}{k+1}$ and we have $\ln(2)=\ln(1+1)=\sum_{k=0}^\infty \frac{(-1)^{k}}{k+1}$ and we can conclude that
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)} = \left(\ln(2)\right)^2$$
For the convergence part, this series $\color{red}{\text{does not}}$ converge absolutely. To prove this, we consider the tail's sum
$$ \sum_{r=m}^n \sum_{k=0}^r \Bigl|\frac{(-1)^r}{(k+1) (r-k+1)} \Bigr| = \sum_{r=m}^n\sum_{k=0}^r \frac{1}{(k+1) (r-k+1)} $$
Now consider the inner sum
$$\sum_{k=0}^r \frac{1}{(k+1) (r-k+1)} = \frac{1}{r+2}\sum_{k=0}^r \frac{(r-k+1) + (k+1)}{(k+1) (r-k+1)} \\ = \frac{1}{r+2}\sum_{k=0}^r\Bigl( \frac{1}{k+1}+\frac{1}{r-k+1} \Bigr) = \frac{2}{r+2}\sum_{k=0}^r \frac{1}{k+1} > \frac{2\ln(r+2)}{r+2}$$
using the integral test, because $ \sum_{k=0}^r \frac{1}{k+1} = \sum_{k=1}^{r+1} \frac{1}{k} > \int_1^{r+2} \frac{1}{x} dx = \ln(r+2)$
and overall we have
$$\sum_{r=m}^n\sum_{k=0}^r \frac{1}{(k+1) (r-k+1)} > 2\sum_{r=m}^n \frac{\ln(r+2)}{r+2}$$
which we know diverges when $n\rightarrow \infty$ regardless to $m$ (Rudin, Principles of Mathematical Analysis, Theorem 3.29. for $p=-1$). But this series is alternating and it converges non-absolutely (again, Rudin, Principles of Mathematical Analysis, Theorem 3.43.). To show this, the only thing we need to show is $\sum_{k=0}^r \frac{1}{(k+1) (r-k+1)} \ge \sum_{k=0}^{r+1} \frac{1}{(k+1) (r-k+2)}$ because the alternating series is of the form of $\sum_{n=0}^\infty (-1)^n \Bigl( \sum_{k=0}^n \frac{1}{(k+1)\times (n-k+1)}\Bigr)$. Based on the above calculation we have to show
$$\frac{2}{r+2}\sum_{k=0}^r \frac{1}{k+1} \ge \frac{2}{r+3}\sum_{k=0}^{r+1} \frac{1}{k+1} \\ \Rightarrow \frac{1}{r+2}\sum_{k=0}^r \frac{1}{k+1} \ge \frac{1}{r+3}\sum_{k=0}^{r} \frac{1}{k+1} + \frac{1}{(r+2)(r+3)} \\ \Rightarrow \Bigl( \sum_{k=0}^{r} \frac{1}{k+1} \Bigr)\Bigl(\frac{1}{r+2} - \frac{1}{r+3}\Bigr)\ge \frac{1}{(r+2)(r+3)} \\ \Rightarrow \Bigl( \sum_{k=0}^{r} \frac{1}{k+1} \Bigr)\Bigl(\frac{1}{(r+2)(r+3)}\Bigr)\ge \frac{1}{(r+2)(r+3)} $$
But that's obviously true $\square$.
Therefore the value of the series can be anything according to the Riemann's rearrangement theorem, but in this formation, the value of the series converges to $\left(\ln(2)\right)^2$
(P.S. if you want to get an intuition, think of it as $\int\int \frac{1}{x^2}$ that gives you $\ln x$ which we know is unbounded.)
| {
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Long division of polynomial I am reading a math history book which says:
"Gregory could have learned in Italy that the area under the curve $y=1 / (1+x^2)$, from
$x=0$ to $x=x$, is arctan x, and a simple long division converts $
1 / (1+x^2)$ to $1 - x^2 + x^4 - x^6+...$"
Can someone please explain how is
$1/(1+x^2) = 1 - x^2 + x^4 - x^6+...$
And
How do you perform long divison of this kind?
Thanks.
| $$(1+x^2)(1-x^2+x^4+\dots)$$
$$=(1-x^2+x^4+\dots)$$
$$+x^2(1-x^2+x^4+\dots)$$
$$=(1-x^2+x^4+\dots)$$
$$+(x^2-x^4+x^6+\dots)$$
$$=1$$
| {
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Let $a+b=2 \text{ and } ab = -1.$ Determine $a^{10} + b^{10}$
Let $a,b \in \mathbb{R}$ for which $$a+b=2 \text{ and } ab = -1.$$ Determine $a^{10}+b^{10}.$
I tried to approach this the following way:
From $ab=-1$ we get that $a = -\frac{1}{b}$ and substituting this to $a+b=2$ yields in $b^2-2b-1=0$. This quadratic has the solutions $b= 1 \pm \sqrt{2}$
Subsituting now $b$ we have that $a= - \frac{1}{1\pm \sqrt{2}}$
I have the feeling that I'm not going in the right direction here. Is there something I'm not seeing since the quadratic formula seems to yield a bit too much work?
| You have $4 = (a+b)^2 = a^2+2ab+b^2=a^2+b^2-2$,
so $a^2+b^2 = 6$. Square and cube this to get
$$a^4+2a^2b^2+b^4 = 36$$
$$a^4+b^4= 36-2(-1)^2 = 34$$
and
$$a^6+3a^4b^2+3a^2b^4+b^6 $$ $$ = a^b+b^6+3a^2(-1)^2+3b^2(-1)^2 = 6^3=216$$
$$a^6+b^6 = 216 - 3(a^2+b^2) = 216-3(6) = 198.$$
Multiply these two results together:
$$(a^4+b^4)(a^6+b^6) = 34\cdot 198$$
$$a^{10} +b^{10}+a^6b^4+a^4b^6 = 6732$$
$$a^{10}+b^{10} = 6732 - a^2(-1)^4-b^2(-1)^4$$ $$ =6732 -(a^2+b^2) = 6732-6 = 6726.$$
| {
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"answer_id": 1
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Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}),$$
i.e. we need next to show that $(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge (a + b + c)$, but i don't know how to do it.
Any help appreciated
| By Minkowski (triangle inequality) $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sum_{cyc}\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2}\geq$$
$$\geq\sqrt{\left(\sum_{cyc}\left(a+\frac{b}{2}\right)\right)^2+\frac{3}{4}\left(\sum_{cyc}b\right)^2}=\sqrt3(a+b+c).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the volume of a specific region of a solid of revolution about the y-axis I have a given polynomial, f(x)=c1*x^7+c2*x^6+c3*x^5+c4*x^4+c5*x^3+c6*x^2+c7*x+c8 rotated about the y axis, which results in a nice surface:
What I would like to be able to do is evaluate the volume of a portion of this solid for a given square area, say a square defined by the coordinates (200,200) (200,400) (400,200) (400, 400) for the coordinates in the image. Basically the portion of a rectangle of arbitrary height as long as the height is greater than the maximum value reached by the curve:
I have a sneaking suspicions I'm wildly overthinking a somewhat straightforward calculus problem but throwing in coordinate systems and the rotated curve is throwing me off.
| The rotated function you're trying to integrate is
$$f(r,\theta) = \sum_{k=0}^n c_k r^k$$
For your case, $n=7$, so the fully written out form would be
$$f(r,\theta) = c_0 + c_1r + c_2r^2+c_3r^3+c_4r^4+c_5r^5+c_6r^6+c_7r^7$$
Note that these $c_i$ are not the same as yours, you can compare the above expression to your own to see which coefficients are which.
To convert to rectangular coordinates we set $r = (x^2+y^2)^\frac{1}{2}$, so
$$f_{\text{rect}}(x,y)= \sum_{k=0}^n c_k (x^2+y^2)^\frac{k}{2}$$
Plugging in $n=7$ and evaluating the integral term-by-term you should be able to get an exact solution,
$$ \sum\limits_{k=0}^7 I_k $$
I'll give an example of an even and an odd term because they have different ways of evaluating.
First, the easier one is the even coefficient, since the term
$$(x^2+y^2)^\frac{k}{2} = (x^2+y^2)^i$$
is a just a polynomial when $k$ is even. Take $k=6$ for instance. The term we need to integrate is then
$$(x^2+y^2)^3 = x^6 + 3x^4y^2 + 3x^2y^4 + y^6$$
This expansion can be done by hand, I won't go into the details of shortcuts but you can search for "binomial expansion shortcut" for some nice tricks. Otherwise, you can write out
$$(x^2+y^2)(x^2+y^2)(x^2+y^2)$$
and use the distributive property.
To integrate, we make use of the fact that integrals are linear (so we can distribute among sums and factor out constants) and a nice property of double integrals
$$\iint f(x)g(y)\text dx \text dy = \int f(x) \text dx \int g(y) \text dy$$
which lets us integrate any polynomial in any number of variables. Very nice.
Here's the full calculation for the $k=6$ term:
$$
\begin{equation}\begin{split}
I_6 & = \iint c_6(x^2+y^2)^\frac{6}{2}\text dx \text dy \\
& = \iint c_6(x^2+y^2)^3\text dx \text dy \\
& = c_6 \iint (x^2+y^2)^3\text dx \text dy \\
& = c_6 \iint x^6 + 3x^4y^2 + 3x^2y^4 + y^6\text dx \text dy \\
& = c_6 \left(\iint x^6\text dx \text dy + 3 \iint x^4y^2\text dx \text dy + 3 \iint x^2y^4\text dx \text dy + \iint y^6\text dx \text dy\right) \\
& = c_6 \left(\int x^6\text dx \int \text dy + 3 \int x^4 \text dx \int y^2\text dy + 3 \int x^2 \text dx \int y^4 \text dy + \int \text dx \int y^6 \text dy\right) \\
& = c_6 \left(\frac{x^7 y}{7} + \frac{x^5 y^3}{5} + \frac{x^3 y^5}{5} + \frac{xy^7}{7}\right) \\
\end{split}\end{equation}
$$
You can evaluate the last step with limits instead to get the numerical values you're looking for.
Okay, now the odd coefficients. We'll use 1 for the example. The basic idea is to take an integral that looks like
$$\iint (x^2+y^2)^\frac{1}{2} \text dx \text dy$$
and make a substitution $x = y\tan \theta$, so $\text dx = y\sec^2 \theta \text d\theta$. Since $\tan^2 \theta + 1 = \sec^2 \theta$, this integral becomes
$$\iint y^{2}\sec^{3} \theta\text d\theta \text dy$$
using the neat property we used before, this is just
$$\int y^{2}\text dy \int \sec^{3} \theta \text d\theta$$
The first part evaluates easily to $\frac{y^3}{3}$, the second part takes some work. We'll make use of integration by parts with $u = \sec \theta$ and $v = \sec^2 \theta \text d\theta$. This gives $\text du = \sec \theta \tan \theta \text d\theta$ and $v = \tan \theta$. Plugging into the integration by parts formula and simplifying we get
$$\int \sec^{3} \theta \text d\theta = \sec \theta \tan \theta - \int \sec^{3} \theta \text d\theta + \int \sec \theta \text d\theta$$
The integral we're looking for appears on both sides so we can solve for it, and using the known integral $\int \sec \theta \text d\theta$ (can be evaluated by substitution), we have
$$\int \sec^{3} \theta \text d\theta = \frac{1}{2}(\sec \theta \tan \theta + \log|\sec \theta + \tan \theta|)$$
and so finally
$$ I_1 = \frac{1}{6}y^3(\sec \theta \tan \theta + \log|\sec \theta + \tan \theta|) $$
You can reverse the substitution then plug in limits, or you can convert the $(x,y)$ limits to $\theta$ and evaluate that way.
Note that for higher odd powers, you will have to repeat the process of integration by parts and solving for your integral multiple times to reduce to the known integral of $\sec \theta$. You can find some formulae that may help by searching for "integral of odd powers of secant"
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_{-1}^{1} [ \frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}] dx $ Evaluate $$\int_{-1}^{1} \left[ \frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}\right] dx $$
My attempt :$$ \frac{2}{3} \left[\frac{x^4}{4}\right]_{x=-1}^{x=1} + \frac{2}{3} \left[\frac{(2-x^2)^\frac{-1}{2}}{-1/2}\right]_{x=-1}^{x=1}=0$$
Is its True ?
| Using property of even and odd functions,
$$\int_{-1}^{1} \left(\frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}\right) dx $$
$$=\frac{4}{3}\int_{0}^{1}(2-x^2)^{3/2} dx $$
Let $x=\sqrt2\sin\theta\implies dx=\sqrt2\cos\theta d\theta$
$$=\frac{4}{3}\int_{0}^{1}(2-2\sin^2\theta)^{3/2} \sqrt2\cos\theta d\theta $$
$$=\frac{4}{3}\int_{0}^{\pi/4}4\cos^4\theta d\theta $$
$$=\frac{16}{3}\int_{0}^{\pi/4}\left(\frac{\cos4\theta+4\cos2\theta+3}{8}\right)d\theta $$
$$=\frac{2}{3}\int_{0}^{\pi/4}(\cos4\theta+4\cos2\theta+3)d\theta $$
$$=\frac{2}{3}\left(\frac{\sin4\theta}{4}+2\sin2\theta+3\theta\right)_0^{\pi/4} $$
$$=\frac{4}{3}+\frac{\pi}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality question. Let $a,b,c>0$ with $a+b+c=1$. Show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{\left(a^3 + b^3 + c^3\right)}{abc}$$
Ohhhkk. So first off,
\begin{align} a^3 + b^3+ c^3 & =a^3 + b^3+ c^3- 3abc +3abc\\
& =\ (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))+3abc\\
& = \ (1-3(ab+bc+ca)) + 3abc \\
\end{align}
Using this the inequality becomes,
$$7 \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 9+ \frac{2}{abc}$$
How do i proceed from here? Was this the right approach? Is there a better one?
| Hint: Just expand and you're done.
(As per Michael's first step) Expanding out the denominator and homogenizing, we WTS
$$ (a+b+c)(ab+bc+ca) \leq 3 abc + 2(a^3+b^3+c^3)$$
Expanding terms and cancelling $abc$, we WTS
$$ a^2b + b^2 a + c^2 b + b^2 a + c^2 b + a^2 c \leq 2 (a^3 + b^3 + c^3).$$
This should be obvious / well known e.g.
$a^2b + b^2 a + c^2 b \leq a^3 + b^3 + c^3$ and $b^2 a + c^2 b + a^2 c \leq a^3 + b^3 + c^3$.
Unsurprisingly, since you applied an identity (as opposed to an inequality), we can follow the same steps of homogenizing using your approach.
We WTS
$$ 7 (ab+bc+ca)(a+b+c) \leq 9 abc + 2(a+b+c)^3$$
Expanding, and cancelling common terms, this becomes:
$$ a^2b + b^2 a + c^2 b + b^2 a + c^2 b + a^2 c \leq 2 (a^3 + b^3 + c^3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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All integer solutions of $x^3-y^3=2020$.
Find all integer pairs $(x,y)$ satisfying $$x^3-y^3=2020\,.$$
First,
$x^3-y^3=(x-y)(x^2+xy+y^2)=2020$
and $2020=2^2\cdot 5 \cdot 101$.
But what next? Can it be worked out by using modulo? Or how?
Any idea? Thanks in advance.
| Alternatively, if $p$ is a prime natural number that divides $x^2+xy+y^2$, then $$(2x+y)^2+3y^2=4(x^2+xy+y^2)\equiv 0\pmod{p}\,.$$ Thus, either $p$ divides both $x$ and $y$, or $\left(\dfrac{-3}{p}\right)=1$. Now, by quadratic reciprocity,
$$1=\left(\dfrac{-3}{p}\right)=\left(\dfrac{p}{-3}\right)=\left(\dfrac{p}{3}\right)\,,$$
whence $p\equiv 1\pmod{3}$. Because $$(x-y)(x^2+xy+y^2)=x^3-y^3=2020=2^2\cdot 5\cdot 101$$ with $5\not\equiv 1\pmod{3}$ and $101\not\equiv 1\pmod{3}$, we conclude that $5$ and $101$ cannot divide $x^2+xy+y^2$. Thus, the only possible prime divisor of $x^2+xy+y^2$ is $2$, and if $2\not\equiv 1\pmod{3}$ is a factor of $x^2+xy+y^2$, we must have $2\mid x$ and $2\mid y$. Since $x^2+xy+y^2\geq 0$, this implies
$$x^2+xy+y^2=1\text{ or }x^2+xy+y^2=4\,.$$
The only solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $x^2+xy+y^2=1$ are $$(x,y)=\pm (1,0),\pm(0,1),\pm(1,-1)\,.$$
The only solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $x^2+xy+y^2=4$ are $$(x,y)=\pm (2,0),\pm(0,2),\pm(2,-2)\,.$$
None of these solutions satisfies $x^3-y^3=2020$.
| {
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$\sin x = \cos y, \sin y = \cos z, \sin z = \cos x$ For real numbers $x,y,z$ solve the system of equations:
$$\begin{align} \sin x = \cos y,\\
\sin y = \cos z,\\
\sin z = \cos x\end{align}$$
Source: high school olympiads, from a collection of problems for systems of equations, no unusual tricks involved.
So far I found that if we square two equations and use the $\sin^2 x + \cos^2 x=1$ we get $\sin^2 y + \cos^2 z=1$ which yields $\sin^2 y = \sin^2 z$. Is this correct or am I missing something? I don't know how to continue
| As you have written we can form the following equations-
$$\sin^2(x)+\sin^2(z)=1$$
$$\sin^2(x)+\sin^2(y)=1$$
$$\sin^2(y)+\sin^2(z)=1$$
By eliminating the $\sin^2(x)$term from the second bracket, we now get these three equations
$$\sin^2(x)+\sin^2(z)=1$$
$$\sin^2(y)-\sin^2(z)=0$$
$$\sin^2(y)+\sin^2(z)=1$$
Now by adding the 2nd and 3rd equation we get the following:
$$\sin^2(x)+\sin^2(z)=1$$
$$\sin^2(y)-\sin^2(z)=0$$
$$2\sin^2(y)=1$$
Now we get that $y=\sin^{-1}(\pm\frac{\sqrt2}{2})$
Similarly through back substitution we get that both $x$ and $z$ are also equal to $\sin^{-1}(\pm\frac{\sqrt2}{2})$
Now you can find ALL the values of $x, y$ and $z$.
*NOTE: $x, y$ and $z$ need not be equal. For example-
$$x=\frac{3\pi}{4}$$
$$y=\frac{\pi}{4}$$
$$z=\frac{7\pi}{4}$$
are also solutions
| {
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"url": "https://math.stackexchange.com/questions/3705362",
"timestamp": "2023-03-29T00:00:00",
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How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in this form: $\sqrt{16x^2 - 3^2}$ so I know to use $x = 3\sec u$:
$$
\begin{align}
& \int \frac{\sqrt{16x^2 - 9}}{x} \, dx \\
= {} & \int \frac{\sqrt{16x^2 - 3^2}}{x} \, dx \\
= {} & \int \frac{\sqrt{16(3\sec u)^2 - 3^2}}{3\sec u} 3\sec u\tan u \, du \\
= {} & \int \frac{(\sqrt{16(3\sec u)^2 - 3^2)}(3\sec u\tan u)}{3\sec u} \, du \\
= {} & \int \sqrt{(16(3\sec u)^2 - 3^2)}(\tan u) \, du
\end{align}
$$
This doesn't seem to make it easy. However, using a calculator online, it suggests I instead use $x = \dfrac{3}{4}\sec{u}$ which simplifies the integral to a crisp $\int 3\tan^2 u \, du$.
My question is, how did the calculator get $a = \dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
| Note: $\sqrt{16x^2-3^2}$ is a difference of squares. Draw a picture of a right triangle suggested by this: $4x$ the hypotenuse, $3$ one of the legs (say the side opposite angle $\theta$), and $\sqrt{16x^2-3^2}$ the side adjacent to angle $\theta$.
Do it, don't just rely on my description.
So then:
$$
\sin\theta = \frac{3}{4x},
\\
\cos\theta = \frac{\sqrt{16x^2-3^2}}{4x},
\\
\tan\theta = \frac{3}{\sqrt{16x^2-3^2}}.
$$
Use the simplest one to suggest the substitution:
$$
x = \frac{3}{4}\csc \theta,
\\
dx = -\frac{3}{4}\csc\theta\cot\theta\;d\theta
$$
Then substitute back into your integral, looking at your picture to find how to move between $x$ and $\theta$. Here
$$
\frac{\sqrt{16x^2 - 9}}{x} = 4\cos \theta
$$
so we get
\begin{align}
\int\frac{\sqrt{16x^2 - 9}}{x}\;dx &= -\int 4\cos \theta
\frac{3}{4}\csc\theta\cot\theta\;d\theta
\\ &= -3\int\frac{\cos^2\theta}{\sin^2\theta}\;d\theta =
3\big(\cot \theta + \theta\big)+C
\end{align}
and then look at the picture to get
$$
3\big(\cot \theta + \theta\big)+C=
3 \left[\frac{\sqrt{16x^2-3^2}}{3} + \arcsin\frac{3}{4x}\right]+C
$$
this method also works for "sum of squares". Draw the right triangle
suggested by that particular sum of squares.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Sum of complex roots' fractions According to this:
If $\omega^7 =1$ and $\omega \neq 1$ then find value of
$\displaystyle\frac{1}{(\omega+1)^2} +
\frac{1}{(\omega^2+1)^2} +
\frac{1}{(\omega^3+1)^2} +
... + \frac{1}{(\omega^6+1)^2}=?$
First I try like
$\displaystyle\frac{1}{\omega+1} +
\frac{1}{\omega^2+1} +
\frac{1}{\omega^3+1} +
... + \frac{1}{\omega^6+1} = 3
$
I have done distribution them and finally got the solution $\dfrac{5}{3}$
However, this is, without a doubt, a time-consuming way.
Can someone please suggest easier way to solve this one.
| Note that $\omega, \ldots, \omega^6$ are precisely the roots of the sixth degree polynomial:
$$p(x) = x^6 + \cdots + 1 = \dfrac{x^7-1}{x-1}.$$
Thus, we can write
$$p(x) = (x-\omega)\cdots(x-\omega^6).$$
Taking (natural) $\log$ on both sides and differentiating gives us
$$\dfrac{p'(x)}{p(x)} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$
Note that
\begin{align}
\log p(x) &= \log(x^7 - 1) - \log(x-1)\\
\implies \dfrac{p'(x)}{p(x)} &= \dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1}.
\end{align}
This gives us that
$$\dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$
Differentating both sides again gives us
$$\dfrac{(x^7-1)(42x^5) - (7x^6)(7x^6)}{(x^7-1)^2} + \left(\dfrac{1}{x-1}\right)^2 = -\left(\dfrac{1}{x-\omega}\right)^2-\cdots-\left(\dfrac{1}{x-\omega^6}\right)^2.$$
Now, we simply substitute $x = -1$ both sides. It is clear that the RHS transforms to the negative of what we want, whereas the LHS becomes
\begin{align}
\dfrac{(-2)(-42) - (7)(7)}{(-2)^2} + \left(\dfrac{1}{-2}\right)^2 &= \dfrac{84-49}{4} + \dfrac{1}{4}\\
&= \dfrac{36}{4} = 9
\end{align}
This gives us the answer as $-9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $ For $a,b,c \in (0,\infty).$ Prove$:$
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $$
My proof by SOS$:$
$$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) \left( a+b \right) \left( b+c \right) \left( c+a \right)\, \cdot \,(\text{LHS}-\text{RHS})$$
$$=\frac{5}{4} abc \sum\limits_{cyc} c^2 (a+b-2c)^2 (a-b)^2 +\frac{1}{4} \sum\limits_{cyc} {c}^{3} \left( 4\,{a}^{2}+3\,ab+4\,{b}^{2} \right) \left( a-b
\right) ^{4}$$
However$,$ it's hard to find this SOS's form without computer.
So I am looking for alternative solution without $uvw.$ Thanks very much!
| We can prove this inequality as an inequality in the Schur's form typing.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a(b+c)}{a^2+bc}\geq1+\frac{16abc}{\prod\limits_{cyc}(a+b)}$$ or
$$2-\frac{16abc}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\left(1-\frac{a(b+c)}{a^2+bc}\right)$$ or
$$\frac{2\sum\limits_{cyc}(a^2b+a^2c-2abc)}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}$$ or
$$\frac{2\sum\limits_{cyc}(b+c)(a-b)(a-c)}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}$$ or
$$\sum_{cyc}(a-b)(a-c)\left(\frac{2}{(a+b)(a+c)}-\frac{1}{a^2+bc}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(a-c)^2}{(a+b)(a+c)(a^2+bc)}\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Number theory and perfect squares Let $p,q,w$ be positive integers such that: $$ 1+p+q\sqrt{3} = (2+\sqrt{3})^{2w-1}$$ $\texttt{ }$ Prove that $p$ is a perfect square.
I have observed that the $RHS$ is raised to an odd power, and that the number of terms of $RHS$ equal to $2w\texttt{ }$
So we get that, $$1+p+q\sqrt{3} = 4^w + 3^w +(2w-2\texttt{ } terms)$$
And I can't proceed further.
| First off, you can prove that $1+p-q\sqrt{3}=(2-\sqrt{3})^{2w-1}$. By using binomial to expand the right-handed side and compare the coefficients, we obtain $$p=2^{2w-1}+\binom{2w-1}{2}2^{2w-3}\cdot 3+\binom{2w-1}{4}2^{2w-5}\cdot 3^2+...+\binom{2w-1}{2w-2}2\cdot 3^{w-1}-1$$
Thus, $p$ is odd. Note also that $$(1+p)^2-3q^2=(1+p+q\sqrt{3})(1+p-q\sqrt{3})=(2+\sqrt{3})^{2w-1}(2-\sqrt{3})^{2w-1}=1$$
Let $p=l-1$ the equality above becomes, since $p$ is odd we have $l$ is even,$(l-1)(l+1)=3q^2$ and $\gcd(l-1,l+1)=1$, since $l$ is even. So we have $2$ cases, as follows.
Case 1 : $l-1 = 3x^2,l+1=y^2$ for some positive integer $x,y$. In this case we have both $x,y$ are odd since $l$ is even and thus $x^2,y^2\equiv 1\pmod 8$ and we have $y^2-3x^2=2$ which is impossible since, this will leads us to $$2=y^2-3x^2\equiv -2\pmod 8$$
Case 2 : $l-1 = x^2,l+1=3y^2$ for some positive integer $x,y$. In this case the result follows $p=l-1=x^2$ as desired $\Box$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $ S=\frac{\cos 2x}{1\cdot 3}+\frac{\cos 4x}{3\cdot 5}+\frac{\cos 6x}{5\cdot 7}+\dots=\sum_{n=1}^\infty\frac{\cos (2nx)}{(2n-1)(2n+1)} $
Find a sum of the series:
$$
S=\frac{\cos 2x}{1\cdot 3}+\frac{\cos 4x}{3\cdot 5}+\frac{\cos 6x}{5\cdot 7}+\dots=\sum_{n=1}^\infty\frac{\cos (2nx)}{(2n-1)(2n+1)}
$$
My attempt:
$$
\begin{aligned}
&z=\cos x+i\sin x\\
&S=\frac{1}{2}\text{Re}\sum_{n=1}^\infty\frac{z^{2n}}{2n-1}-\frac{1}{2}\text{Re}\sum_{n=1}^\infty\frac{z^{2n}}{2n+1}
\end{aligned}
$$
But calculating these sums seems a bit difficult to me. Perhaps there is a better approach to this problem?
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\underline{\underline{\mbox{Note that}}}\quad
\bbox[10px,#ffd]{\left.\sum_{n = 1}^{\infty}{\cos\pars{2nx} \over
\pars{2n - 1}\pars{2n + 1}}
\,\right\vert_{\ \color{red}{\large x\ \in\ \mathbb{C}}}}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}{\expo{2n x\ic} \over
\pars{2n - 1}\pars{2n + 1}} + \pars{x \to - x}\label{1}\tag{1}
\end{align}
\begin{align}
&\bbox[10px,#ffd]{{1 \over 2}\sum_{n = 1}^{\infty}{\expo{2n x\ic} \over
\pars{2n - 1}\pars{2n + 1}}} =
{1 \over 4}\sum_{n = 1}^{\infty}{\expo{2nx\ic} \over 2n - 1} -
{1 \over 4}\sum_{n = 1}^{\infty}{\expo{2nx\ic} \over 2n + 1}
\\[5mm] = &\
{1 \over 4}\sum_{n = 1}^{\infty}{\expo{2nx\ic} \over 2n - 1} -
{1 \over 4}\sum_{n = 2}^{\infty}{\expo{2nx\ic}\expo{-2xi} \over 2n - 1} =
{1 \over 4} + {1 \over 4}\pars{1 - \expo{-2xi}}
\sum_{n = 1}^{\infty}{\expo{2nx\ic} \over 2n - 1}
\\[5mm] = &\
{1 \over 4} + {1 \over 2}\,\ic\sin\pars{x}
\bbox[#eef,1px]{\expo{-xi}\sum_{n = 1}^{\infty}{\expo{2nx\ic} \over 2n - 1}}
\label{2}\tag{2}
\end{align}
Lets evaluate the $\ds{\bbox[5px,#eef]{\mbox{blue}}}$ expression in (\ref{1}):
\begin{align}
&\bbox[#eef,1px]{\expo{-xi}\sum_{n = 1}^{\infty}{\expo{2nx\ic} \over 2n - 1}} =
\expo{-xi}\sum_{n = 2}^{\infty}{\expo{nx\ic} \over n - 1}\,{1 + \pars{-1}^{n} \over 2} =
{1 \over 2}\sum_{n = 1}^{\infty}{\pars{\expo{x\ic}}^{n} \over n}
\bracks{1 - \pars{-1}^{n}}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}{\pars{\expo{x\ic}}^{n} \over n} -
{1 \over 2}\sum_{n = 1}^{\infty}{\pars{-\expo{x\ic}}^{n} \over n} =
-\,{1 \over 2}\ln\pars{1 - \expo{xi}} +
{1 \over 2}\ln\pars{1 + \expo{xi}}\label{3}\tag{3}
\end{align}
Replacing (\ref{3}) in (\ref{2}):
\begin{align}
&\bbox[10px,#ffd]{{1 \over 2}\sum_{n = 1}^{\infty}{\expo{2n x\ic} \over
\pars{2n - 1}\pars{2n + 1}}} =
{1 \over 4} + {1 \over 4}\,\ic\sin\pars{x}
\bracks{\ln\pars{1 + \expo{x\ic}} - \ln\pars{1 - \expo{x\ic}}}
\end{align}
The final result is found by replacing this result in (\ref{1}}:
\begin{align}
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\bbox[10px,#ffd]{\left.\sum_{n = 1}^{\infty}{\cos\pars{2nx} \over
\pars{2n - 1}\pars{2n + 1}}
\,\right\vert_{\ \color{red}{\large x\ \in\ \mathbb{C}}}}
\\[5mm] = &\
{1 \over 2}
\\[2mm] + & {1 \over 4}\,\ic\sin\pars{x}
\bracks{\ln\pars{1 + \expo{x\ic}} - \ln\pars{1 - \expo{x\ic}} -
\ln\pars{1 + \expo{-x\ic}} + \ln\pars{1 - \expo{-x\ic}}}
\label{4}\tag{4}
\end{align}
When $\ds{x \in \mathbb{R}}$, (\ref{4}) is reduced to:
\begin{align}
\mrm{f}\pars{x} & =
\left\{\begin{array}{lcl}
\ds{\mrm{f}\pars{-x}} & \mbox{if} & \ds{x < 0}
\\[2mm]
\ds{{1 \over 2} - {1 \over 4}\,\pi\sin\pars{x}} & \mbox{if} & \ds{0 \leq x \leq \pi}
\\[2mm]
\ds{\mrm{f}\pars{x - \pi}} && \mbox{otherwise}
\end{array}\right.
\end{align}
$\ds{\mrm{f}\pars{x}}$ Graph" />
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A sequence of quadratic equations Let $p\le q$ be roots of the (real) quadratic equation $x^2+ax+b=0$, $|p|+|q|\ne 0.$ Form the new equation $x^2+px+q=0$, find its real roots (if exist), etc. For example, if $a=3, b=2$, then $p=-2,q=-1$, $x^2-2x-1=0$ has two real roots $p_1= 1 - \sqrt{2}$ and $q_1= 1 + \sqrt{2}$ (note that $p_1\le q_1$) but the equation $x^2+p_1x+q_1$ does not have real roots, the process ends.
Question: What is the longest possible sequence of quadratic equations we can get?
| Let's work backwards: given a final state, how many steps would it take to reach the beginning? Let the final state be represented by $r_1, r_2$ with $r_1 \le r_2$.
We can see that the quadratic before the final one must have been $$(x-r_1)(x-r_2) = x^2+(-r_1-r_2)x + r_1r_2$$
In order to have a sequence longer than $1$, it must be true that $$-r_1-r_2 \le r_1r_2 \tag 1$$
To get that quadratic, the previous quadratic must have been $$(x+r_1+r_2)(x-r_1r_2) = x^2 + (r_1+r_2-r_1r_2)x - r_1r_2(r_1+r_2)$$
In order to get a sequence longer than $2$, it must be true that $$r_1+r_2-r_1r_2 \le - r_1r_2(r_1+r_2) \tag 2$$
Taking it one more step, the following condition must also hold in order to get a sequence longer than $3$: $$(r_1r_2-r_1-r_2+r_1r_2(r_1+r_2)) \le (r_1r_2-r_1-r_2)(r_1r_2(r_1+r_2)) \tag 3$$
Finally, this last condition must also hold in order to get a sequence longer than $4$: $$-\left(r_{1}r_{2}-r_{1}-r_{2}+r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)-\left(r_{1}r_{2}-r_{1}-r_{2}\right)\left(r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)\le\left(r_{1}r_{2}-r_{1}-r_{2}+r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)\left(r_{1}r_{2}-r_{1}-r_{2}\right)\left(r_{1}r_{2}\left(r_{1}+r_{2}\right)\right) \tag 4$$
In order to satisfy $(2)$, it must be true that $r_1 < 0$ or that $r_2 < 0$. Letting the $x$-axis be $r_1$ and the $y$-axid be $r_2$, this eliminates the first quadrant. In order to satisfy $(3)$, it must be true that $r_1 > 0$ or that $r_2 > 0$, eliminating the third quadrant. However, in order to satisfy both $(1)$ and $(4)$, $(r_1, r_2)$ can only be in the second and fourth quadrants. Therefore, there are no real $r_1, r_2$ that satisfy $(1), (2), (3), (4)$.
This then means that the maximum length of a sequence of the quadratic equations is $4$, obtained by setting $$a = r_1r_2-r_1-r_2+r_1r_2(r_1+r_2), b = (r_1r_2-r_1-r_2)(r_1r_2(r_1+r_2)$$
for any $r_1, r_2$ that satisfy $(1), (2), (3)$, and $r_1 \le r_2$.
Edit: The conditions $(1), (2), (3), r_1 \le r_2$ can be rewritten as $$-\frac{r_{2}}{r_{2}+1}\le r_{1}\le\frac{-\left(r_{2}^{2}+1-r_{2}\right)+\sqrt{\left(r_{2}^{2}+1-r_{2}\right)^{2}-4r_{2}^{2}}}{2r_{2}}$$ with $r_1 \le 0 \le r_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
| I think we can try with $$\log\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\dots\right)$$ and put $x=1/5$. The calculations are simple and one can estimate the error easily. Estimation of error gives $$\log\frac{1+x}{1-x}<2x+\frac{2x^3}{3}+\frac{2x^5}{5(1-x^2)}$$ With a little calculation of the easy variety (division by 2,3,5 etc) you can conclude that the right hand side of the above inequality for $x=1/5$ is less than $13/32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
} |
The Maclaurin series for $e^{\sin(x)}$ up to the fourth term I am trying to get the Maclaurin series for $e^{\sin(x)}$ up to the fourth term, which should be:
$$e^{\sin(x)}=1+x+\frac{1}{2}x^2-\frac{1}{8}x^4+o(x^4)\tag{1}$$
This is how I go about it:
$$e^{\sin(x)}=1+\sin(x)+\frac{1}{2}(\sin(x))^2+\frac{1}{6}(\sin(x))^3+o(x^4)\tag{2}$$
$$\sin(x)=x-\frac{x^3}{6}+o(x^4)\tag{3}$$
$$(\sin(x))^2=x^2-\frac{x^4}{6}+o(x^5)-\frac{x^4}{6}=x^2-\frac{x^4}{3}+o(x^5)\tag{4}$$
$$(\sin(x))^3=x^3+o(x^4)\tag{5}$$
Then I plug (3),(4) and (5) into (2), which results in:
$$e^{\sin(x)}=1+\left(x-\frac{x^3}{6}\right)+\frac{1}{2}\left(x^2-\frac{x^4}{3}\right)+\frac{1}{6}\left(x^3\right)+o(x^4)=1+x+\frac{1}{2}x^2-\frac{1}{6}x^4+o(x^4)\tag{6}$$
Which is wrong. Please tell me where I go wrong.
| In my humble opinion, you could do it faster
$$A=e^{\sin(x)}\implies \log(A)=\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$
$$A=e^{\log(A)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^5\right
)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. Can someone help me with this sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. I can't find $S_n$
| $$\begin{eqnarray*}S&=&\sum_{n\geq 1}\frac{3^n}{5^n-4^n}=\sum_{n\geq 1}\left(\frac{3}{5}\right)^n\frac{1}{1-\left(\frac{4}{5}\right)^n}=\sum_{n\geq 1}\sum_{k\geq 0}\left(\frac{3}{5}\right)^n\left(\frac{4}{5}\right)^{kn}=\sum_{k\geq 0}\frac{3\cdot 4^k}{5^{k+1}-3\cdot 4^{k}}\\&=&\sum_{m\geq 1}\frac{\frac{3}{4}4^m}{5^m-\frac{3}{4}4^m}\end{eqnarray*} $$
is actually a series deceleration, but there are ways to improve the convergence speed of the LHS.
For instance
$$ S = 3+\sum_{n\geq 2}\frac{3^{n}}{5^n-4^n}=3+\sum_{n\geq 1}\frac{3^{n+1}}{5^{n+1}-4^{n+1}} $$
where the last series is pretty close to $\frac{3}{5}S$, leading to
$$ \frac{2}{5}S = 3 + \sum_{n\geq 1}\left(\frac{3^{n+1}}{5^{n+1}-4^{n+1}}-\frac{3}{5}\cdot\frac{3^{n}}{5^{n}-4^{n}}\right) $$
and
$$ S = \frac{15}{2}-\frac{3}{2}\sum_{n\geq 1}\frac{3^{n}4^n}{(5^{n+1}-4^{n+1})(5^{n}-4^n)}.$$
The same trick can be applied to the last series, whose main term roughly behaves like $\left(\frac{12}{25}\right)^n$.
$$ S = \frac{95}{26}+\frac{162}{13}\sum_{n\geq 1}\frac{3^n 4^{2n}}{(5^{n+2}-4^{n+2})(5^{n+1}-4^{n+1})(5^n-4^n)} $$
After a few steps we get $S\approx 4.92476079$, but I highly doubt there is a nice closed form fo $S$. After all this is pretty much the acceleration method employed by Shingo Takeuchi (and reminiscent of Gosper's works in the seventies) for proving the trascendence of some series like $\sum_{k\geq 1}\frac{1}{2^k-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show $\frac{d}{d x}\left(|x|^{1/2}\right)=\frac{x}{2|x|^{3/2}}$? $$\frac{d}{d x}\left(|x|^{\frac{1}{2}}\right)=\frac{x}{2|x|^{\frac{3}{2}}}$$
and also the second derivative $$\frac{d^{2}}{d x^{2}}\left(\sqrt{|x|}\right)=\frac{\delta(x)}{\sqrt{|x|}}-\frac{x^{2}}{4|x|^{\frac{7}{2}}}$$
I know this may seem basic. But I don't get how the Dirac delta (thanks for the comments for correcting me) comes in. Is it just a way someone has decided to write the result for $x >0$, $x<0$ and $x=0$ cleverly in one line. Is it arbitrary as long as the Kronecker term beats the other term?
If anyone has a clean derivation for these I'd be grateful.
EDIT: thanks for a lot of your answers, I've learnt a lot.
Still not entirely sure with the second derivative, the formula is here: https://www.wolframalpha.com/input/?i=second+derivative+of+%7Cx%7C%5E1%2F2, is this wrong? I still think it is probably correct. From taking something like. d/dx($\operatorname{sign}(x) \frac{1}{2|x|^{\frac{1}{2}}} )=1/2 ($$\frac{\delta(x)}{\sqrt{|x|}} -\frac{sign(x)*x}{2|x|^{5 / 2}})$ Don't know why first summand is missin 1/2. I know its distributions, but can't you still differentiate sign(x), you just have to be careful and when evaluating use test functions and integrals
| This is not a complete answer, but for the case of the absolute function defined on the real numbers, this might help.
Since $|x|$ is non-differentiable only at $0$, all the derivatives we calculate will have domain $D \subseteq\mathbb{R} \setminus \{0\}$.
If we write $|x|=\sqrt{x^2}$ we see that
$$
\frac{d}{dx} \left(|x| ^\frac{1}{2} \right) = \frac{d}{dx} \left(x^2\right)^\frac{1}{4} = \frac{1}{4}\left(x^2\right)^{-\frac{3}{4}} \left[\frac{d}{dx}x^2 \right]= \frac{1}{4}\left(x^2\right)^{-\frac{3}{4}}2x = \frac{x}{2\left(x^2\right)^\frac{3}{4}} = \frac{x}{2\left(\sqrt{x^2}\right)^\frac{3}{2}} = \frac{x}{2|x|^\frac{3}{2}}
$$
As for the second part
$$
\frac{d^2}{dx^2} \left(|x| ^\frac{1}{2} \right) = \frac{d}{dx} \left(\frac{x}{2\left(x^2\right)^\frac{3}{4}}\right) = \frac{2\left(x^2\right)^\frac{3}{4} - 2x\left(\frac{3}{4}(x^2)^{-\frac{1}{4}} (2x)\right)}{4\left(x^2\right)^\frac{3}{2}} = \frac{2\left(x^2\right)^\frac{3}{4}}{4\left[\left(x^2\right)^\frac{3}{4}\right]^2} - \left(\frac{3}{4}\right)\frac{4x^2}{4\left(x^2\right)^\frac{3}{2}(x^2)^{\frac{1}{4}}} = \frac{1}{2\left(x^2\right)^\frac{3}{4}}\left(\frac{2x^2}{2x^2}\right) - \frac{3x^2}{4\left(x^2\right)^\frac{7}{4}} = \frac{2x^2 -3x^2} {4\left(x^2\right)^\frac{7}{4}} = -\frac{x^2}{4 |x|^\frac{7}{2}}
$$
Now, since the Dirac delta turns it's input into $0$ if $x \neq 0$, and the domain on which we're working on doesn't have $0$, our last answer will be equal to
$$
-\frac{x^2}{4 |x|^\frac{7}{2}} = 0 -\frac{x^2}{4 |x|^\frac{7}{2}} = \frac{\delta(x)}{\sqrt{|x|}} -\frac{x^2}{4 |x|^\frac{7}{2}}
$$
since $\frac{\delta(x)}{\sqrt{|x|}}=0$ for all $x \in D$.
I suspect there might be a more general definition of the absolute value you can work with to get the Dirac delta out directly, but I'm not familiar with it. Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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