Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Determinant of group representation p4m The group p4m consists of all compositions of translations,mirror reflections, and rotations by 90 degrees about any center of rotation in the grid.
$g(m, r, u, v)=\begin{bmatrix}
(-1)^{m}\cos \frac{r \pi }{2} & -(-1)^{m}\sin \frac{r \pi }{2} & u\\
\sin \frac{r \pi }{2} & \cos \frac{r \pi }{2} & v\\
0 & 0 & 1
\end{bmatrix}$
In the paper Group Equivariant Convolutional Networks (Section 4.3) they have mentioned that:
$m = \frac{1}{2}(1-\det(g)) $
That means, $\det(g) = 1 - 2m$
But when I tried to find the determinant manually I got the following:
$\det(g) = \begin{vmatrix}
(-1)^{m}\cos \frac{r \pi }{2} & -(-1)^{m}\sin \frac{r \pi }{2} & u\\
\sin \frac{r \pi }{2} & \cos \frac{r \pi }{2} & v\\
0 & 0 & 1
\end{vmatrix}\\
\Rightarrow \det(g) = 1.\begin{vmatrix}
(-1)^{m}\cos \frac{r \pi }{2} & -(-1)^{m}\sin \frac{r \pi }{2}\\
\sin \frac{r \pi }{2} & \cos \frac{r \pi }{2}\\
\end{vmatrix} \\
\Rightarrow \det(g) = (-1)^{m}\cos^{2} \frac{r \pi }{2} + (-1)^{m}\sin^{2} \frac{r \pi }{2}\\
\Rightarrow \det(g) = (-1)^{m}(\cos^{2} \frac{r \pi }{2} + \sin^{2} \frac{r \pi }{2}) \\
\Rightarrow \det(g) = (-1)^{m}$
Clearly, $1-2m \neq (-1)^{m}$
Am I doing anything wrong? If yes please correct me. If no then how can I find the value of m?
| $1-2m$ is indeed equal to $(-1)^m$. In that paper it is mentioned that $m\in \{0,1\}$. $1-2\cdot 0=1=(-1)^0$, and $1-2\cdot 1=-1=(-1)^1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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the sum of the reciprocal of the squares of roots $$x^3-6x^2+5x-7=0$$
Find $\frac{1}{r^2}$+$\frac{1}{s^2}$+$\frac{1}{t^2}$ where $r,s,t$ are roots of the equation.
*
*First I got the reciprocal of the equation above and I got $-7x^3+5x^2-6x+1=0$
*Using newtons identity/sum the sum of all the roots of this equation is $5/7$. The sum of the squares of the equations can be found by solving $-7s+5(5/7)-12=0$, for $s$ (s is the sum of the squares of the roots). I get $59/-49$
Question: Am I correct or are my steps wrong. I know that if you simplify $\frac{1}{r^2}$+$\frac{1}{s^2}$+$\frac{1}{t^2}$ you get $\frac{s^2t^2+r^2t^2+r^2s^2}{r^2s^2t^2}$ but I don't know how to calculate the numerator of the fraction.
Thanks in advance.
| Hint:
$$\begin{align*} \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}&=\frac{r^2s^2+s^2t^2+t^2r^2}{r^2s^2t^2} \\ \\
(rs+st+tr)^2&=r^2s^2+s^2t^2+t^2r^2+2rst(r+s+t).
\end{align*}$$
Also,
$$\begin{align*}
r+s+t&=6 \\
rs+st+tr&=5 \\
rst&=7
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$ The Equation
How can I analytically show that there are no real solutions for $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$?
My attempt
With $u = -x+2$
$\sqrt[3]{u-1}-\sqrt[3]{u+1}=1$
Raising to the power of $3$
$$(u+1)^{2/3}(u-1)^{1/3} - (u+1)^{1/3}(u-1)^{2/3}=1\\(u+1)^{1/3}(u^2-1)^{1/3} - (u-1)^{1/3}(u^2-1)^{1/3}=1\\(u^2-1)^{1/3}\cdot\boxed{\left[(u+1)^{1/3}-(u-1)^{1/3}\right]}=1$$
Raising to the power of $3$:
$$(u^2-1)\cdot\left[3(u+1)^{1/3}(u-1)^{2/3}-3(u+1)^{2/3}(u-1)^{1/3}+2\right]=1\\(u^2-1)\cdot\left[3(u^2-1)^{1/3}(u-1)^{1/3}-3(u+1)^{1/3}(u^2-1)^{1/3}+2\right]=1$$
Thus: $(u^2-1)\cdot\left[3(u^2-1)^{1/3}\boxed{\left[(u-1)^{1/3}-(u+1)^{1/3}\right]}+2\right]=1$
And with $y = (u-1)^{1/3}-(u+1)^{1/3}$, we can say that:
$y^3=-3y(u^2-1)^{1/3}+2$
I am stuck... Any tips for this radical equation?
| Use the fact that $a^3+b^3+c^3-3abc$ is divisible by $a+b+c$.
Let
$$a=\sqrt[3]{x-3}$$
$$b=\sqrt[3]{1-x}$$
$$c=-1$$
Then you have $a+b+c=0$ so
$$a^3+b^3+c^3=3abc$$ or
$$x-3+1-x -1=-3\sqrt[3]{(x-3)(1-x)}$$
and
$$1=\sqrt[3]{(x-3)(1-x)}$$
So
$$(x-3)(1-x)=1$$
and you get $x=2$. If you substitute you find it does not satisfy the equation.
Indeed, this show that $x-3=2-x=-1$
Now the cube roots of $-1$ are roots of $$x^3+1=(x+1)(x^2-x+1)$$ and if $\alpha$ and $\beta$ are the two non real roots, then $$\alpha+\beta =1.$$
So by taking these non real roots, the equation can be solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many ways can we draw three balls such that at least two are red? We have $7$ balls, three red, two white and two blue. How many ways can we select three of them such that at least two are red?
So, my answer was:
If the balls are identical, then there are $3$ ways: $RRR$, $RRB$ and $RRW$.
If the balls are distinguishable, then there are ${7}\choose{3} $=$35$ total possibilities, but there are ${4}\choose{3}$=$4$ ways that have no red balls, and $2 \cdot$${4}\choose{2}$=$12$ with only one ball so there are 19 with 2 red balls or more.
But the answer given is 13.
Could anyone explain this to me?
| You are correct that there are $\binom{7}{3}$ ways to select three of the seven balls and that there are $\binom{4}{3}$ ways to select none of the red balls. However, there are $\binom{3}{1}\binom{4}{2}$ ways to select exactly one of the three red balls and two of the remaining four balls, which yields
$$\binom{7}{3} - \binom{4}{3} - \binom{3}{1}\binom{4}{2} = 13$$
ways to select at least two red balls.
Alternatively, we can select exactly two red balls in $\binom{3}{2}\binom{4}{1}$ ways since we must select two of the three red balls and one of the other four balls, and we can select all three red balls in $\binom{3}{3}$ ways. Since these cases are mutually exclusive, we obtain
$$\binom{3}{2}\binom{4}{1} + \binom{3}{3} = 13$$
selections with at least two red balls.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve: $x^2\equiv 1 \pmod{20},x^2\equiv 6 \pmod{15},x^2\equiv 9 \pmod{18}.$ I want to solve: $x^2\equiv 1 \pmod{20}, x^2\equiv 6 \pmod {15}, x^2\equiv 9\pmod{18}.$ This is a system of congruence equations, but these are not linear and moduli are not coprime. So,we cannot apply chinese remainder theorem here. However, I think I can solve for $y\equiv 1 \pmod{20}, y\equiv 6 \pmod{15}, y\equiv 9 \pmod{18}$ by using extended chinese remainder theorem,then we have to search for the square numbers out of them.How to do that?
| Let $m,n,k,r,s\in\mathbb Z_{≥0}^{+}$ then we have,
$$\begin{cases}x^2=20m+1\\ x^2=15n+6\\ x^2=18k+9\end{cases}$$
First restriction:
$$\begin{align} &15n+6=18k+9\\
\implies &5n+2=6k+3\\
\implies &5(n-k)-k=1\\
\implies &k+1=5r\\
\implies &k=5r-1\\
\implies &n=6r-1\\
\end{align}$$
Second restriction:
$$\begin{align}&20m+1=90r-9\\
\implies &20m=10(9r-1)\\
\implies &2m=r+8r-1\\
\implies &r-1=2s\\
\implies &r=2s+1 \\
\implies &m=9s+4\end{align}$$
We conclude that,
$$\begin{align}&m=9s+4\\
&n=12s+5\\
&k=10s+4\end{align}$$
This implies,
$$\begin{align}&x^2=180s+81=9(20s+9)\\\
\implies &u^2=20s+9,\thinspace u\in\mathbb Z^{+}.\end{align}$$
Final answer:
Let $u=20m-n; m,n\in\mathbb Z^{+}$ with $0<n<20$ (we can also write: $x=20m±n; m,n\in\mathbb Z^{+}$ with $0<n<10$)
Then we have,
$$\begin{align}(20m-n)^2-9&\equiv n^2-9\\
&\equiv 0 \thinspace \thinspace \thinspace \text{(mod 20)}\end{align}$$
We see that the number $n^2$ must end with $9$. Thus, we can restrict the $n$, such that
$$n\in\left\{3,7,13,17\right\}$$
Therefore, we obtain the all possible solutions as follows:
$$x=60m-3n,\thinspace n\in\left\{3,7,13,17\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274290",
"timestamp": "2023-03-29T00:00:00",
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Solve for $x$, $\ x^2+(4x^3-3x )^2=1$, given that $x=\cos\alpha$ I know that it would be:
$\cos^2\alpha+3\cos^2\alpha=1$ by identities
then it would be:
$4\cos^2\alpha=1$
dividing $4$ to both sides:
$\cos^2\alpha=\frac{1}{4}$
square both sides:
$\cos\alpha=\pm\frac{1}{2}$
After this, what should I do?
| As mentioned by op, using $\ (a-b)^2=a^2-2ab+b^2$ to solve for the values of x, given the equation $\ cos^2\alpha+\cos^23\alpha=1$ is:
$\pm\frac{\sqrt 2}{2}, \pm0.9238..., \pm0.3826...$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Getting the wrong value from $\lim_\limits{x\to-\infty}x-\sqrt{x^2+7x}$ Consider the following limit
$$
\lim_{x\to-\infty}x-\sqrt{x^2+7x}
$$
Going through some algebra leads to
$$\begin{align}
\lim_{x\to-\infty}x-\sqrt{x^2+7x}&=\lim_{x\to-\infty}\frac{(x-\sqrt{x^2+7x})(x+\sqrt{x^2+7x})}{(x+\sqrt{x^2+7x})}\\
&=\lim_{x\to-\infty}\frac{-7x}{x+\sqrt{x^2+7x}}\\
&=\lim_{x\to-\infty}\frac{-7}{1+\sqrt{1+7/x}}=-\frac72
\end{align}$$
Using WolframAlpha's step-by-step solution, however, gives this limit to be $-\infty$. Here's what it's doing
$$\begin{align}
\lim_{x\to-\infty}x-\sqrt{x^2+7x}&=\lim_{x\to-\infty}x-\lim_{x\to-\infty}\sqrt{x^2+7x}
\end{align}$$
where, by the power rule,
$$\begin{align}
\lim_{x\to-\infty}\sqrt{x^2+7x}&=\sqrt{\lim_{x\to-\infty}(x^2+7x)}\\
&=\sqrt{\lim_{x\to-\infty}x^2}\\
&=\sqrt{\left(\lim_{x\to-\infty}x \right)^2}=\infty
\end{align}$$
and so
$$
\lim_{x\to-\infty}x-\lim_{x\to-\infty}\sqrt{x^2+7x}=-\infty-\infty=-\infty
$$
What is wrong in this solution? I wondered if it was the power rule failing for undefined values of the square root, but not sure what to argue.
| $$\lim_{x\to-\infty}\sqrt{ax^2+bx+c}=\lim_{x\to-\infty}\sqrt{a(x^2+\frac{b}{a}x+\frac{c}{a})}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \sqrt{(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid\times\sqrt{1+\frac{\frac{4ac-b^2}{4a}}{(x+\frac{b}{2a})^2}}=$$
$$=\lim_{x\to-\infty}\sqrt{a}\times \mid {x+\frac{b}{2a}}\mid$$
$$=\lim_{x\to-\infty}\sqrt{a}\times (-x-\frac{b}{2a})$$
$$\lim_{x\to-\infty}(x-\sqrt{x^2+7x})=\lim_{x\to-\infty}(x-\sqrt{1}\times \mid {x+\frac{7}{2}}\mid)=$$
$$=\lim_{x\to-\infty}x-(-x-\frac{7}{2})=$$
$$=\lim_{x\to-\infty}(2x+\frac{7}{2})=-\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computation of the succession $\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right)$ For training I have decided to solve this limit of a succession
$$\lim\limits_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right).$$
My first attempt:
\begin{split}
\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\
&=\lim_{n\to \infty } 3^{n+1}\left(1-\frac{3^{n\cdot \sqrt{1-\frac{1}{n^2}}}}{3\cdot 3^{n}}\right)\\
&=\ldots
\end{split}
I have abandoned this approach because I wanted to use a notable limit (if $\{b_n\}\to 0$) that could probably be useful; but seeing the rounded brackets I have thought that I have occurred many times:
$$\lim_{n\to \infty} \frac{a^{b_n}-1}{b_n}$$
My second attempt:
\begin{split}
\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\
&=\lim_{n\to \infty } 3^{\sqrt{n^2-1}}\left(\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}-1\right)\\
&=\ldots
\end{split}
Considering only
$$\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}=3^{(n+1-\sqrt{n^2-1})}$$
Taking only the exponent I have:
$$n+1-\sqrt{n^2-1}=\frac{((n+1)-\sqrt{n^2-1})\cdot ((n+1)+\sqrt{n^2-1})}{(n+1)+\sqrt{n^2-1}}$$
$$=\frac{2n+2}{n+1+\sqrt{n^2+1}} \tag 1$$
If I take the limit of the $(1)$, easily:
$$\lim_{n\to \infty}\frac{2n+2}{n+1+\sqrt{n^2+1}}=1$$
i.e.
$$\lim_{n\to \infty}3^{(n+1-\sqrt{n^2-1})}=3$$
Definitely I think that this limit is $\infty\cdot 2=+\infty$.
I think that is correct. Do you users think there are better alternatives? Thanks in advance.
| For $n>1$ you have
$$3^{n+1}-3^{\sqrt{n^2-1}}>3^{n+1}-3^{\sqrt{n^2}}=3^{n+1}-3^n=2\cdot3^n.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why do the sum-of-angles and half-angle formulas return different answers for $\cos\frac\pi{12}$? Suppose I want to calculate $\cos\frac{\pi}{12}$. Using the sum-of-angles formula and using the half-angle formula results in different answers.
$$
\cos\frac{\pi}{12} = \cos\frac{\pi/6}{2} = \sqrt{\frac{1}{2}+\frac{\cos\frac{\pi}{6}}{2}} = \sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}
$$$$
\cos\frac{\pi}{12} = \cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4} = \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}+\sqrt{6}}{4}
$$
Why does this happen? Shouldn't these be the same?
Context: I'm writing up a common student question nicely as a Q&A here. If anyone has ideas how this could be re-tagged or edited to be more easily find-able by a person with this question (that is, more appropriately be indexed by search engines), please feel free to edit.
| Those may not look the same, but they are equal! Notice the first one is wrapped completely in a square root, so rewrite the second one so that it too is wrapped in a single square root:
$$
\frac{\sqrt{2}+\sqrt{6}}{4} =
\sqrt{\left(\frac{\sqrt{2}+\sqrt{6}}{4}\right)^2} =
\sqrt{\frac{2 + 2\sqrt{2}\sqrt{6} + 6}{16}} =
\sqrt{\frac{8 + 4\sqrt{3}}{16}} =
\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}
$$
| {
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"url": "https://math.stackexchange.com/questions/4289202",
"timestamp": "2023-03-29T00:00:00",
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Showing $\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$ by using multiplication of complex numbers Imagine three right triangles on top of each other. The legs of the first two triangles have the ratio $1/3$ and the third triangle $1/7$. The total angle of all three angles is $\pi/4$.
I can get that with the arctan, of course, but I don't know how to use the complex numbers. Because the task is:
Show that
$$\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$$
by using multiplication of complex numbers.
| This question is equivalent to finding the product
$$(3+i) \cdot (3+i) \cdot (7+i)$$
and expecting it in the form $n(1+i)$.
Do you see why?
This is because $\arg(3+i) = \tan^{-1} \frac{1}{3}$, $\arg(7+i) = \tan^{-1} \frac{1}{7}$ and $\arg \lambda(1+i)=\frac{\pi}{4}$. On complex multiplication, principal arguments add up. So it can be concluded that
$$2\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{7}=n\pi + \frac{\pi}{4}$$
As noted by @MarkSaving, complex multiplication in general helps only upto multiples of $\pi$. We need to take into account $0 < \tan^{-1} \frac{1}{7} < \frac{\pi}{4}$ and $0 < \tan^{-1} \frac{1}{3} < \frac{\pi}{4}$ so that the sum
$$0 < 2\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{7}<\frac{3\pi}{4}$$
from where it can be seen that our sum can be equal to $\pi/4$ only.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that one of three expressions is a perfect square $a$, $b$, $c$ are natural numbers such that
$a^2 + b^2 + c^2 = 1 + 2abc$
Prove that one of $\frac{a+1}{2}, \frac{b+1}{2}, \frac{c+1}{2} $ is a perfect square.
Since at least one of $a,b,c$ has to be odd, WLOG that $a = 2x-1$, where $x$ is a natural number. Then I tried to split the problem into the cases depending on the parity of $b,c$ but I haven't found anything yet.
| If we solve the quadratic in $c$, we obtain
$$
c=ab \pm \sqrt{(a^2-1)(b^2-1)} \tag{1}
$$
We see that $(a^2-1)(b^2-1)$ is a perfect square.
Let $d \gt 0$ be the square-free kernel of $a^2-1$, so that $a^2-1=dA^2$ for some positive integer $A$. Then $b^2-1=dB^2$ for some positive integer $B$.
If $x_0^2-dy_0^2=1$ is the fundamental solution of $x^2-dy^2=1$, then we have two exponents $n$ and $m$ such that
$$
a+A\sqrt{d}=z_0^n, b+B\sqrt{d}=z_0^m \tag{2}
$$
where $z_0=x_0+y_0\sqrt{d}$.
Multiplying or dividing the two relations above, we obtain
$$
ab-dAB +(aB+bA)\sqrt{d} = z_0^{n+m},
ab+dAB +(-aB+bA)\sqrt{d} = z_0^{n-m} \tag{3}
$$
If $n$ is even, say $n=2p$ for some integer $p$, we can write $a+A\sqrt{d}=z_1^2$ where $z_1=z_0^p$. If we write $z_1=x_1+y_1\sqrt{d}$, then $x_1^2-dy_1^2=1$, and $a=x_1^2+dy_1^2=2x_1^2-1$, so $\frac{a+1}{2}=x_1^2$ is a perfect square.
Similarly, if $m$ is even we deduce that $\frac{b+1}{2}$ is a perfect square. Finally, if $n$ and $m$ are both odd, then $n\pm m$ are both even and we deduce from (3) and (1) that $\frac{c+1}{2}$ is a perfect square. This finishes the proof.
| {
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For which values of c does the system have a solution? I was doing my homework and stumbled upon this question where I have a system of 3 equations with 5 unknowns/variables. I got stuck trying to make it into an echelon form of this matrix.
The problem is stated as follows:
Find which values of c does the system have a solution? For each c, find all solutions as linear combinations of certain vectors.
In the image below I tried to solve it but got stuck and I'm not sure how to continue this equation..
| Set $x_5 = c $ and $x_6 = c^2 $, then the system becomes
$\begin{bmatrix} 2&&1&&5&&4&&0&&-1 \\ 1 && -1&& 1 &&-1&&-4&&0 \\ 1 && 0 && 2 && 1 && -1 && 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 3 \end{bmatrix} $
Solving using Gauss-Jordan elimination, the solutions are
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} 5 \\ -8\\ 0 \\ 0 \\ 2 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t_3 \begin{bmatrix} -1 \\ 3 \\ 0 \\ 0 \\ -1 \\ 1 \end{bmatrix} $
From which it follows that
$x_5 = c = 2 - t_3 $ and $ x_6 = c^2 = t_3 $
Hence, we must have $2 - c = c^2 $, i.e. $c^2 + c - 2 = 0$ which can be factored as $ (c + 2)(c - 1) = 0 $. Therefore, the two values of $c$ for which there is a solution are $c = -2 $ and $c = 1$.
For $c = -2$, we get $t_3 = 4 $ and for $c = 1$ we get $t_3 = 1 $
Therefore, there are two corresponding sets of solutions:
Case I: for $c = -2$ we get,
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ 0 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \end{bmatrix} $
Case II: for $c = 1$, we get,
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \\ 0 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \end{bmatrix} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Evaluating $\tan\frac\pi{12}-\cot\frac\pi{12}+\tan\frac{5\pi}6-\cot\frac{5\pi}6$ knowing the roots $3x^4+4\sqrt{3}x^3-18x^2-4\sqrt{3}x+3=0$ I know that the roots of this equation
$$3x^4+4\sqrt{3}x^3-18x^2-4\sqrt{3}x+3=0$$ are $x=\tan{\frac{\pi}{12}}, \tan{\frac{\pi}{3}}, \tan{\frac{7\pi}{12}}, \tan{\frac{5\pi}{6}}$.
I need to use this information to find the value of
$$\tan{\frac{\pi}{12}}-\cot{\frac{\pi}{12}}+\tan{\frac{5\pi}{6}}-\cot{\frac{5\pi}{6}}$$
I know that I can rearrange $\tan{\frac{\pi}{12}}-\cot{\frac{\pi}{12}}$ into $-2\left(\frac{2\tan{\frac{\pi}{12}}}{1-\tan{\frac{\pi}{12}}^2}\right)^{-1}$ and then use double angle to find the value but this doesn't use the part beforehand.
Also from playing on the calculator I know $-\cot{\frac{\pi}{12}}=\tan{\frac{7\pi}{12}}$ and that $-\cot{\frac{5\pi}{6}}=\tan{\frac{\pi}{3}}$ so if I can prove these two results I can just use sum of roots but I don't know how.
| Like @WA Don: indicated, you use the formula
$$\tan(\theta + \frac{\pi}{2}) = \tan(\frac{\pi}{2}- (-\theta)) = \cot(-\theta) = -\cot \theta$$
and you are done.
Explaining the equation at the beginning: if $\tan \theta = x$, it expresses the equality
$$\tan 4 \theta = \tan \frac{\pi}{3} = \sqrt{3}$$
Indeed, we have
$$\tan 2 \theta = \frac{2 \tan \theta}{1- \tan^2 \theta}$$
and so
$$\tan 4 \theta = \frac{2 \tan 2 \theta}{1- \tan^2 2 \theta}= \frac{ 4(x-x^3)}{x^4 - 6 x^2 + 1}$$
Therefore, the equation
$$\tan 4 \theta = \tan \frac{\pi}{3}$$
is equivalent to
$$4(x-x^3)= \sqrt{3}( x^4 - 6 x^2 + 1)$$
or
$$3 x^4 + 4 \sqrt{3}x^3 - 18 x^2 - 4 \sqrt{3}x + 3 =0$$
Note that apriori, if $x$ is a root of the above equation, then $-\frac{1}{x}$ is also a root.
The fact that the numbers indicated are the roots follows from simple trigonometry.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $B^{10}$
Calculate $B^{10}$ when $$B = \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$$
The way I did it was
$$ B = I + A $$
where
$$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
and $A^2=-I$. Since $A$ and $I$ are commutative,
$$\begin{aligned} B^2 &= (I+A)^2 = 2A \\ B^3 &= (I+A)2A = 2A-2I\\ B^4 &= (I+A)(2A-2I) = -4I\\ \vdots \\ B^{10} &= 32A \end{aligned}$$
Is there a simpler method or a smarter approach if you want to do this for, e.g., $B^{100}$?
| $B^{100}$ $=(B^4)^{25}$ $=(-4I)^{25}=(-4)^{25}I$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ Given the function $f: \mathbb{R}\to \mathbb{R}$ satisfies:
\begin{cases}
f(1)=1 \\
f(x+y)=f(x)+f(y)+2xy,\quad&\forall x,y \in \mathbb{R}\\
f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4},&\forall x\neq0
\tag{1}\label{eqn1}
\end{cases}
Then find the limit of:
\begin{align}
L=\lim_{x \to 0} \frac{\frac{1}{e^{2f(x)}}-\sqrt[3]{1+f(x)}}{\ln\bigl(1+f(x)\bigr)}
\end{align}
My attempts
I can easily guess that $f(x)=x^2$, which totally satisfies \eqref{eqn1}. What I did was that I set $g(x)=f(x)-x^2$, then I got:
\begin{cases}
g(x)+g(y)=g(x+y)\\
g\left(\frac{1}{x}\right)x^4=g(x)
\tag{2}\label{eqn2}
\end{cases}
From \eqref{eqn1}, substitute $x=y=0$ we have $f(0)=0$
Now $g(1)=g(0)=0$, then I substitute $y=1-x$ in \eqref{eqn2} and got that
\begin{align}
g\left(\frac{1}{x}\right)x^4+g\left(\frac{1}{1-x}\right)(1-x)^4=0
\end{align}
Because the values of either $\frac1x$ or $\frac1{1-x}$ must lie between $0$ and $1$, I suspect that we can prove $g \equiv 0$.
Even if I have $f$ then, I'm still clueless about how to calculate the limit.
Any idea or solution or suggestion on the tools?
Any help is appreciated!
| First to prove $g(x) = 0$:
We have $g(0) = 0$, $g(1) = f(1) -1^2 = 0$, so $g(x+1) = g(x)$. Also $g(x) + g(-x) = 0$ so $g(-x) = -g(x)$. We have:
$g(x) = g(x+1) = (x+1)^4 g(\frac{1}{x+1}) = -(x+1)^4 g(-\frac{1}{x+1}) = -(x+1)^4 g(1- \frac{1}{x+1}) = -(x+1)^4 g(\frac{x}{x+1}) = -(x+1)^4 (\frac{x}{x+1})^4 g(\frac{x+1}{x}) = -x^4 g(\frac{x+1}{x}) =-x^4 g(1 + \frac{1}{x}) = -x^4 g( \frac{1}{x}) = -x^4 \frac{1}{x^4} g(x) = -g(x)$
Therefore $g(x) = 0$, $f(x) = x^2$. As for the limit, we can substitute $x^2$ with $y$ and apply L'Hôpital's rule .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the measure of the radius of the circle inscribed in the triangle ATB? For reference: Calculate the radius of the circle inscribed
in the triangle ATB
My progress: I put the trigonometric solution...would a geometric solution be possible?
|
First note that $\triangle ATB$ is right triangle with $\angle ATB = 90^\circ$.
Drop a perp from $P$ to $BQ$. Then, $AB = PD = 2 \sqrt2$ (as $PQ = 3, QD = 1$).
Next drop a perp from $B$ to $PQ$. Then, $ \displaystyle \triangle QBH \sim \triangle QPD \implies QH = \frac{2}{3}, BH = \frac{4 \sqrt2}{3}$
$ \displaystyle TH = TQ - HQ = \frac43$
$ \displaystyle BT = \sqrt{TH^2 + BH^2} = \frac{4}{\sqrt3}$
Now for a right triangle, inradius is $ \displaystyle \frac{a + b - c}{2} ~, ~ $ where $c$ is hypotenuse and $a$ and $b$ are perpendicular sides.
So inradius of $ \displaystyle \triangle BTH = \frac{2 (\sqrt2 + 1 - \sqrt3)}{3} $
As $\triangle BTH \sim \triangle BAT, ~ $ inradius of $\triangle BAT$,
$ \displaystyle r = \sqrt{\frac32} \cdot \frac{2 (\sqrt2 + 1 - \sqrt3)}{3} = \frac{2 + \sqrt2 - \sqrt6}{\sqrt3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why do I get incorrect solutions $ x = 0 $ and $ x = 2 $ for $ x + 1 + \sqrt{4x + 1} = 0 $? Here is my incorrect attempt at solving
$$ x + 1 + \sqrt{4x + 1} = 0. $$
Subtracting $ \sqrt{4x + 1} $ from both sides,
$$ x + 1 = -\sqrt{4x + 1} $$
Squaring both sides,
$$ x^2 + 2x + 1 = 4x + 1 $$
Subtracting $ 4x + 1 $ from both sides,
$$ x^2 - 2x = 0 $$
We have obtained
$$ x(x - 2) = 0 $$
It has two solutions: $ x = 0 $ and $ x = 2 $.
But if we substitute $ x = 0 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 1 + \sqrt{1} = 2 $$
If we substitute $ x = 2 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 3 + \sqrt{9} = 6 $$
Apparently I have made a mistake in some step that has led to this contradiction. Which step is incorrect in my solution above? It must be the squaring step that changes the minus sign to positive sign.
But squaring both sides is an often used step in many equations. What rules I need to keep in mind while solving such equations so that I do not get incorrect solution after squaring both sides?
| Here is one correct way to solve this problem I learnt from comments.
I should write
$$ x + 1 = -\sqrt{4x + 1} \iff \begin{cases} (x + 1)^2 = 4x + 1 \\ (x + 1) \le 0 \end{cases} \iff \begin{cases} x(x - 2) = 0 \\ x \le -1 \end{cases} $$
Then I should solve $ x(x - 2) = 0 $ to get $ x = 0 $ and $ x = 2 $. Then I should reject both solutions because both solutions do not satisfy $ x \le 1 $.
Therefore the equation has no solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the maximum integer value of the segment (IG) joining the incenter to the centroid in the triangle below? For reference:
Given the obtuse triangle $ABC$, obtuse at $B$,
where $IG \parallel BC$, $I$ being the incenter and $G$ the centroid of this triangle and the perimeter of the triangle is $144$.
Calculate the maximum integer value of $IG$.
My progress:
My draw and the relationships I found
$IG \parallel BC \implies a = \dfrac{b+c}{2}\\
\dfrac{AI}{IJ} = \dfrac{b+c}{2p}=\dfrac{b+c}{144}\\
c^2+a^2=2BM^2+\dfrac{b^2}{2}$
From angle bissector theorem ($\triangle ABC-AN::$),
$\dfrac{BJ}{CJ}=\dfrac{c}{b}\\
AJ^2 =bc-CJ\cdot BJ \\
\triangle ADK \sim \triangle ACB \implies:\\
\dfrac{AM}{CM}=\dfrac{AK}{BK}=\dfrac{b}{c}$
but I don't see where to fit $IG$..
| $ \displaystyle \frac{AI}{IJ} = \frac{b+c}{a} = 2 ~ $ (given $IG \parallel BC$)
$b+c = 2a \implies a = 48 ~ $ (using $a+b+c = 144$)
$ \displaystyle IG = \frac23 \cdot JN = \frac23 \left(BN - BJ \right)$
$ \displaystyle = \frac23 \left(\frac a2 - \frac{ac}{b+c} \right) = \frac{a (b-c)}{3 (b+c)} = \frac{b-c}{6}$
(given $a = 48, b + c = 96$)
Using Triangle inequality, we must have $b - c \lt 48$ so the next biggest value of $(b-c)$ that gives an integer value for $IG$ is $42$ and hence max integer value of $IG$ is $7$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ Let $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ find the value of $\left[ {I + \sqrt 3 } \right] $ {where [.] represent greatest integral function}
Let $T(x)={{x^3} - 3{x^2} + 5}$, $T'\left( x \right) = {x^3} - 3{x^2} + 5 = 3{x^2} - 6x = 3x\left( {x - 2} \right) > 0,x \in \left( {2,3} \right)$
$T(x)$ is increasing for $x\in(2,3)$
Not able to proceed further
| Well, I have an idea to compute that integral. Let,
$$I(t)=\int_{2}^{3}\sqrt{x^3-3x^2+5t} dx$$
Notice that $I'(1)$ gives the required integral but a constant term in front.
Now we are only required to evaluate $I(t)$.
Note that for all $t\in\mathbb{N}$ and $x\in [2,3]$, $5t>x^{3}-3x^{2} \implies \frac{x^3-3x^2}{5t} <1$.
$$I(t)=\sqrt{5t}\int_{2}^{3}\sqrt{1+\frac{x^3-3x^2}{5t}} dx$$
Since, $\frac{x^3-3x^2}{5t} <1$ we can use fractional binomial theorem
Now at last you will be left with a summation, whose bounds are easily predictable by adjusting the sum to another well known convergent sums. This easily enables you to calculate your required floor value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ .
My approach is as follow
$y = \sqrt {{x^2} + 4x + 4 + 9} + \sqrt {{x^2} - 8x + 16 + 25} \Rightarrow y = \sqrt {{{\left( {x + 2} \right)}^2} + 9} + \sqrt {{{\left( {x - 4} \right)}^2} + 25} $
Within the square roots the minimum values are 3 and 5 hence the minimum value should be greeter than 8 but not able to find the actual minimum value.
| I'll approach as geometry.
Let three point as $A(-2,-3), P(x,0), B(4,5)$. Then $y=\overline{AP}+\overline{PB}$.
By triangle inequality, $y\ge\overline{AB}=10$.
So the minimum value is $10$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What's the measure of the $\angle BAC$ in the triangle below? For reference: In the right triangle $ABC$, right at $B$, the corner $AF$ is drawn such that $AB = FC$ and $\angle ACB = 2 \angle BAF$. Calculate $\angle BAC$.
My progress:
$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: Law ~of~ cosines:\\
AF^2 = x^2+AC^2-2.x.AC.cosC\\
\triangle ABC:\\
cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\
x^2+(x+BF)^2 = AC^2\\
Th.Stewart \triangle ABC:\\
AC^2.BF+x^3=AF^2BC+BC.x.BF$
...??
| Here is a construction that makes things simple. Extend $CB$ such that $BE = BF = y$
Now $ \displaystyle \angle BAE = \frac{\angle C}{2} \implies \angle CAE = 90^0 - \frac{\angle C}{2}$
And we notice that $\triangle ACE$ is isosceles so $AC = x + 2y$
Applying Pythagoras in $\triangle ABC$,
$(x+2y)^2 = x^2 + (x+y)^2$
$4y^2 = x^2 + y^2 - 2xy = (x-y)^2$
That leads to $x = 3y$ and sides of $\triangle ABC$ are in the ratio $3:4:5$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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If $A^3-2A^2-A+2I=O$, then $A$ is equal to If $A^3-2A^2-A+2I=O$, then $A$ is equal to
$(A) I$ $(B) 2I$ $(C)\begin{bmatrix} 2 & -1 & 2 \\-1 & 0 & 0 \\\ 0 & 1 & 0
\end{bmatrix}$ $(D)\begin{bmatrix} 2 & 1 & -2 \\1 & 0 & 0 \\\ 0 & 1 & 0
\end{bmatrix}$
My Attempt:
$A^3-2A^2-A+2I=O$
$A^2(A-2I)-I(A-2I)=O$
$(A^2-I)(A-2I)=O$
Either $(A^2-I)=O$ or $(A-2I)=O$
I got $A=I$ or $A=-I$ or $A=2I$ but the correct answers are (A), (B), (C), (D).
Please help.
| The eignevalues of $C,D$ are $\pm1,2$ with the characteristic equation is:
$x^3-2x^2-x+2=0$. The eigenvalues of $A,B$ are $1$'s which also obey this equation, so they are all possible answers.
| {
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Conditional Probability - truth/lie The probability that two poker players speak truth is 3/5 and 2/3. A card is drawn from the pack and they both agree that it is a queen. What is the probability that it is really a queen ?
My approach:-
4 cases possible for A and B based on one saying Truths and Lies (TT,TF,FT,FF)
Case 1:- Both speaking truth and it is really a queen = 3/5 * 2/3 * 4/52
Case 2 and 3:- One person saying truth and other lie will not be possible as it would lead to an inconsistency because it is given that both agree it to be queen
Case 4:-Both speaking a lie, and it is not a queen = 2/5 * 1/3 * 48/52 = 8/65
Answer= 12/65
Answer given in book is 1/5
Can this be solved using probability tree diagram ?
| If, as lulu suggested, A and B choose their lie with a uniform distribution, we wind up with a vastly different answer. The problem can be broken down as follows:
Define $Q, A_Q, B_Q$ to be the events that the drawn card is a queen, $A$ says the card is a queen, and $B$ says the card is a queen respectively.
Then, you have the following truth table:
$$\begin{array}{c|c|c|c}Q & A_Q & B_Q & \text{Prob} \\ \hline T & T & T & \tfrac{1}{13}\cdot \tfrac{3}{5}\cdot \tfrac{2}{3} = \tfrac{2}{65} \\ T & T & F & \tfrac{1}{13} \cdot \tfrac{3}{5} \cdot \tfrac{1}{3} = \tfrac{1}{65} \\ T & F & T & \tfrac{1}{13} \cdot \tfrac{2}{5}\cdot \tfrac{2}{3} = \tfrac{4}{195} \\ T & F & F & \tfrac{1}{13}\cdot \tfrac{2}{5}\cdot \tfrac{1}{3} = \tfrac{2}{195} \\ F & T & T & \tfrac{12}{13}\cdot \left(\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\cdot \left(\tfrac{1}{3}\cdot \tfrac{1}{12}\right) = \tfrac{1}{1170} \\ F & T & F & \tfrac{12}{13}\left(\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(1-\tfrac{1}{3}\cdot \tfrac{1}{12}\right)=\tfrac{7}{234} \\ F & F & T & \tfrac{12}{13}\left(1-\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(\tfrac{1}{3}\cdot \tfrac{1}{12}\right)=\tfrac{29}{1170} \\ F & F & F & \tfrac{12}{13}\left(1-\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(1-\tfrac{1}{3}\cdot \tfrac{1}{12}\right) = \tfrac{203}{234}\end{array}$$
Note: I am assuming a uniform distribution of lies So, if the card is not a queen, and they say it is a queen, not only is it the probability of them lying, it is also the probability that their lie happened to be a queen. So, $P(Q^C\cap A_Q^C)$ is the probability that it is not a queen and $A$ says a different card. $A$ will tell the truth with probability $\tfrac{3}{5}$. $A$ will say it is a queen with probability $\tfrac{2}{5}\cdot \tfrac{1}{12}$. $A$ will lie and say it is another card (that is not a queen) with probability $\tfrac{2}{5}\cdot \tfrac{11}{12}$. For example, if the actual card is a king, he will say it is not a king with probability $\tfrac{2}{5}$, but the chosen lie (ace through queen) is uniformly distributed over that event. So, each individual card that is not a king has its own $\tfrac{2}{5}\cdot \tfrac{1}{12} = \tfrac{1}{30}$ chance of being told.
Next, we add up the probabilities to ensure we have a total probability space:
$$\dfrac{2}{65}+\dfrac{1}{65}+\dfrac{4}{195}+\dfrac{2}{195}+\dfrac{1}{1170}+\dfrac{7}{234}+\dfrac{29}{1170}+\dfrac{203}{234} = 1$$
And indeed we do. So, unless I made a calculation error, we have:
$$P(Q|A_Q\cap B_Q) = \dfrac{P(Q\cap A_Q\cap B_Q)}{P(A_Q \cap B_Q)} = \dfrac{\left(\dfrac{2}{65}\right)}{\left(\dfrac{2}{65}\right)+\left(\dfrac{1}{1170}\right)} = \dfrac{36}{37}$$
Another note: This also assumes all events are independent (players choose to lie independently of reality and each other). This seems like another valid answer to the original problem since it was not clearly worded.
| {
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Find $\textstyle{\frac{1\cdot 2}{3!} +\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\frac{4\cdot2^4}{6!}+\cdots}$ up to n terms? $$
\frac{(1)2}{3!} + \frac{(2)2^2}{4!} + \frac{(3)2^3}{5!} + \frac{(4)2^4}{6!} + \cdots =\sum\limits_{k=1}^{n}\frac{k\cdot 2^k}{(k+2)!}
$$
My attempt:
$$
\begin{align}
e^x&=\sum_{n=1}^{\infty}\frac{x^n}{n!} \\
(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!} \\
x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n}}{n!} \\
x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{n(n+1)(n+2)x^{n}}{(n+2)!}
\end{align}
$$
After this attempt I realized exponential series is for infinite terms whereas the question concerns finite terms so approach may not work.
Can you please give any hints on the right approach to be tried?
| This is not an answer: Just mistakenly thought that we need to find the sum to infinity.
Given, $\displaystyle \sum_{m=1}^\infty \frac{m\cdot 2^m}{(m+2)!}$. We put $m=n-2 $ and thus the series becomes $\displaystyle\sum_{n=3}^\infty \frac{(n-2)\cdot 2^{n-2}}{n!}=\frac{1}{4}\sum_{n=3}^\infty \frac{(n-2)\cdot 2^n}{n!}=\frac{1}{4}\sum_{n=3}^\infty \frac{n\cdot 2^n}{n!}-\frac{1}{4}\sum_{n=3}^\infty \frac{2\cdot 2^n}{n!}=A+B(say)$. Where $A,B$ denotes the first and second sum respectievly.
Now, $$A=\frac{1}{4}\sum_{n=3}^\infty\frac{n\cdot 2^n}{n!}=\frac{1}{2}\sum_{n=3}^\infty\frac{2^{n-1}}{(n-1)!}=\frac{1}{2}(e^2-1-2).$$ Recall that $\displaystyle e^x=\sum_{n=1}^\infty\frac{x^n}{n!}$ now put $x=2$.
Similarly, $$\displaystyle B=\frac{1}{2}\sum_{n=3}^\infty \frac{2^n}{n!}=\frac{1}{2}(e-1-2-\frac{2^2}{2!})=\frac{1}{2}(e^2-5).$$
So $\displaystyle A+B=\frac{1}{2}(e^2-3-e^2+5)=1.\square$
Hope this works.
| {
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How to calculate summation of $\frac{n^2 + n + 1 }{n^2 + n}$ over $1$ to $25$? Can we calculate the summation of numerator, and denominator and then divide them both? But that does not seems to be a plausible way. So, I reduced it to $S = 1+ \frac{1}{n^2+n}$. But now I do not how to compute this sum.
| Using telescoping sum and the simplification you found,
$$\\\sum_1^{25} \frac{n^2+n+1}{n^2+n}\\
= \sum_1^{25}\left(1+\frac{1}{n(n+1)}\right)\\
= \sum_1^{25}1 + \sum_1^{25}\frac{1}{n(n+1)}\\
= 25 + \sum_1^{25}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
= 25 + \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{25}-\frac{1}{26}\\
=25\frac{25}{26}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{aligned}
$$
I find the solution is
$$\frac{\ln{\left(x + \sqrt{1 - x^{2}} \right)}}{2} + \frac{\sin^{-1}{\left(x \right)}}{2}+C$$
How can I get this without trigonometric substitution?
| Let $y= x+\sqrt{1-x^2}$. Then, $dx=\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}-x}dy$ and
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}}
& =\int \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}-x} \frac{dy}y\\
&=\ \frac12\int \bigg( \frac1y + \frac{1}{\sqrt{1-x^2}-x}\bigg)dy\\
&=\ \frac12\int \frac{dy}y+\frac12 \int \frac{dx}{\sqrt{1-x^2}}\\
&= \frac12 \ln y +\frac12\sin^{-1}x+C
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Evaluate $\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \sqrt{\frac{x}{y z}} \,d z \,d y \,d x$ $$
\text { Evaluate } \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \sqrt{\frac{x}{y z}} \,d z\, d y\, d x
$$
How to solve this ?
I tried this
$\begin{aligned} & \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \sqrt{\frac{x}{y z}} d z d y d x \\
=& \int_{0}^{1} \int_{0}^{1-x} \frac{\sqrt{x}}{\sqrt{y}}[2 \sqrt{z}]{0}^{1-x-y} d y d x \\
=& \int_{0}^{1} \int_{0}^{1-x} \frac{\sqrt{x}}{\sqrt{y}} 2 \sqrt{1-x-y} d y d x \\
=& \int_{0}^{1} \int_{0}^{1-x} 2 \sqrt{x} \frac{\sqrt{1-x-y}}{\sqrt{y}} d y d x \\=& \int_{0}^{1} \int_{0}^{1-x} 2 \sqrt{x} \sqrt{\frac{1-x-y}{y}} d y d x \end{aligned}$
| First make a substitution $1-y\to y$ to get that this is equivalent to
$$\int_0^1\int_x^1 2\sqrt{x}\sqrt{\frac{y-x}{1-y}}\, dy\, dx$$
Next, switch the order of integetration,
$$\int_0^1\int_0^y \frac{2}{\sqrt{1-y}}\sqrt{xy-x^2}\, dx\, dy$$
$$=\int_0^1\int_0^y \frac{1}{\sqrt{1-y}}\sqrt{4xy-4x^2-y^2+y^2}\, dx\, dy$$
$$=\int_0^1\int_0^y \frac{1}{\sqrt{1-y}}\sqrt{y^2-(2x-y)^2}\, dx\, dy$$
$$=\int_0^1\frac{1}{\sqrt{1-y}}\int_0^y \sqrt{y^2-(2x-y)^2}\, dx\, dy$$
You can make a quick sketch of the graph of $\sqrt{y^2-(2x-y)^2}$ where y is a parameter ranging from $0$ to $1$ to see that geometrically
$$\int_0^y \sqrt{y^2-(2x-y)^2}\, dx=\frac{\pi y^2}{4}$$
Hence, our integral is equivalent to
$$\frac{\pi}{4}\int_0^1 \frac{y^2}{\sqrt{1-y}}\, dy$$
Make a substitution of $u=\sqrt{1-y}$ to get
$$=\frac{\pi}{2}\int_0^1(1-u^2)^2\, du$$
$$=\frac{\pi}{2}\int_0^1 u^4-2u^2+1\, du$$
$$=\frac{\pi}{2}\left(\frac{1}{5}-\frac{2}{3}+1\right)$$
$$=\frac{4\pi}{15}$$
I'm not sure how much of this is necessary, but I believe that switching the order of integration is an essential step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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if $Ax^2+Bxy+Cy^2$ is positive definite, then $A>0$ and $B^2-4AC<0$ if $Ax^2+Bxy+Cy^2$ is positive definite, then $A>0$ and $B^2-4AC<0$.
$Ax^2+Bxy+Cy^2=A(x+\frac{B}{2A}y)^2-\frac{B^2-4AC}{4A}y^2$, hence under the condition $A>0$ and $B^2-4AC<0$, the quadratic form is positive definition. But I am wondering why it is necessary for $A>0$ and $B^2-4AC<0$ if the quadratic form is positive definite.
| $f(x, y) = Ax^2 + Bxy + Cy^2$ must be positive for $(x, y) \ne (0, 0)$, in particular:
*
*$f(1, 0) = A > 0$,
*$f(B, -2A) = A B^2 -2AB^2 +4A^2C = A ( 4AC - B^2) > 0$.
Note that in the second line $(x, y)$ is chosen such that the quadratic term $(x+\frac{B}{2A}y)^2$ vanishes.
Alternatively you can argue that the matrix
$$
\begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}
$$
is positive definite if and only if all leading principal minors are positive, that is Sylvester's criterion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Is it possible to improve on these bounds for $\frac{\varphi(n)}{n}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the Euler-totient function of the positive integer $x$ by $\varphi(x)$, and the classical sum of divisors of $x$ by $\sigma(x)=\sigma_1(x)$.
MOTIVATION
From the following source:
Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}.$$
However, we also have
$$\frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}.$$
Notice that
$$\frac{4}{5} \leq \frac{\varphi(q^k)}{q^k} = \frac{q^k \bigg(1 - \frac{1}{q}\bigg)}{q^k} = \frac{q - 1}{q} < 1.$$
Therefore, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{\varphi(n)}{n},$$
and
$$\frac{4}{5}\cdot\frac{\varphi(n)}{n} \leq \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{1}{2},$$
which implies that
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}.$$
WolframAlpha gives the rational approximation
$$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$
Harking back from this answer to a closely related question, we have the following proposition:
THEOREM A: Let $q^k n^2$ be an odd perfect number with special prime $q$. Then we have the following implication:
$$\frac{\varphi(n)}{n} > \frac{1}{2} \implies q = 5.$$
PROOF: Let $q^k n^2$ be an odd perfect number with special prime $q$, and suppose that
$$\frac{\varphi(n)}{n} > \frac{1}{2}.$$
From the equation and lower bound for $\varphi(N)/N$
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}$$
and the equation
$$\frac{\varphi(q^k)}{q^k} = \frac{q - 1}{q},$$
we get the lower bound
$$2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1} < \frac{2\varphi(n)}{n} = x.$$
This implies that we have the upper bound
$$q < \frac{x}{x-1} < \frac{2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}}{\bigg(2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}\bigg) - 1}$$
which can be solved using WolframAlpha, yielding the upper bound
$$q < \frac{217\zeta(3)}{217\zeta(3) - 240} \approx 12.5128,$$
from which it follows that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$. QED
Note that the estimate
$$\frac{\varphi(n)}{n} > \frac{1}{2}$$
was used to justify
$$0 < q < \frac{x}{x-1}.$$
We now endeavor/attempt to prove the estimate
$$\frac{\varphi(n)}{n} > \frac{1}{2}.$$
To do so, consider the equation
$$\frac{2q}{q+1}=\frac{n}{\varphi(n)}.$$
(This is not random guesswork - it emanates from considering the inequalities
$$\frac{\sigma(n^2)}{n^2} < \frac{n^2}{\varphi(n^2)} = \frac{n}{\varphi(n)}$$
$$\frac{\sigma(n^2)}{n^2} = \frac{2q^k}{\sigma(q^k)} \leq \frac{2q}{q+1},$$
and determining whether the right-hand sides are equal.)
Since $q \equiv 1 \pmod 4$, then $(q+1)/2$ is an integer, so that we may rewrite
$$\frac{2q}{q+1}=\frac{n}{\varphi(n)}$$
as
$$q\varphi(n)=\bigg(\frac{q+1}{2}\bigg)\cdot{n}.$$
Since $\gcd(q,n)=1$, then $n \mid \varphi(n)$. This contradicts
$$\frac{1}{2}<\frac{\varphi(n)}{n}=\frac{q+1}{2q}=\frac{1}{2}+\frac{1}{2q} \leq \frac{3}{5}.$$
An alternative proof goes by observing that
$$\gcd(q,n)=\gcd\Bigg(q,\frac{q+1}{2}\Bigg)=\gcd\Bigg(q,\frac{q+1}{2}\cdot{n}\Bigg)=1$$
which contradicts
$$q \mid \Bigg(\frac{q+1}{2}\cdot{n}\Bigg).$$
Thus, either
$$\frac{1}{2} < \frac{q+1}{2q} < \frac{\varphi(n)}{n} < \frac{5}{8}$$
or
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{q+1}{2q} < \frac{3}{5}$$
holds.
Here are our:
QUESTIONS
(1) Do you see a quick way to improve on these bounds, using the ideas we have given in this post, and possibly even more?
(2) If the bounds can be improved, can you show how?
(3) If the bounds cannot be improved, can you explain why?
| Note: This is not really an answer, just some thoughts that recently occurred to me, that would be too long to include in the Comments section.
Let $q^k n^2$ be an odd perfect number with special prime $q$.
It is known that
$$\varphi(x) < \sigma(x)$$
for all $x > 1$. Therefore, we are sure that
$$\frac{\varphi(n)}{n} < \frac{\sigma(n)}{n} < \frac{\sigma(n^2)}{n^2} \leq \frac{2q}{q+1}.$$
Now, in the original post, we have proved that
$$\frac{\varphi(n)}{n} > \frac{1}{2} \implies q = 5.$$
Hence, we have
$$\frac{\varphi(n)}{n} > \frac{q+1}{2q} \implies \frac{\varphi(n)}{n} > \frac{1}{2} \implies q = 5 \implies \frac{\varphi(n)}{n} > \frac{q+1}{2q} = \frac{5+1}{10} = \frac{3}{5}.$$
Therefore, the biconditional
$$q = 5 \iff \frac{\varphi(n)}{n} > \frac{3}{5}$$
holds, assuming the inequality
$$\frac{\varphi(n)}{n} > \frac{q+1}{2q}$$
is true.
But we are sure that the inequality
$$\frac{\varphi(n)}{n} < \frac{2q}{q+1}$$
holds.
Assuming that
$$\frac{\varphi(n)}{n} > \frac{q+1}{2q}$$
is true, we obtain
$$\frac{q+1}{2q} < \frac{2q}{q+1}$$
$$(q + 1)^2 < 4q^2$$
which implies that
$$3q^2 - 2q - 1 > 0$$
$$(3q + 1)(q - 1) > 0$$
from which we finally conclude that
$$q > 1,$$
which is trivial.
Again, we know that the inequality
$$\frac{\varphi(n)}{n} < \frac{2q}{q+1}$$
holds.
Assuming that
$$\frac{\varphi(n)}{n} < \frac{q+1}{2q}$$
then we obtain
$$\Bigg(\frac{\varphi(n)}{n}\Bigg)^2 < \Bigg(\frac{2q}{q+1}\Bigg)\cdot\Bigg(\frac{q+1}{2q}\Bigg) = 1$$
from which we get
$$\frac{\varphi(n)}{n} < 1$$
which, again, is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$ Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$
A) $\frac 23$
B) $\frac 43$
C) $\frac 89$
D) $\frac {16}{9}$
E) $\frac{32}{27}$
My attempts:
I think that the all possible answers are wrong. Because, by Am-Gm inequality we have
$$\frac{\sin x}{\cos y}+\frac{\cos y}{\sin x}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}≥2+2=4.$$
But, Wolfram Alpha gives us a different result : The Global Minimum doesn't exist. However, the local minimum must be $6.$
But, still the problem is not solved. Because Wolfram's graph shows that the minimum can be less than $6$.
I also tried
Let $$\sin x=a,\cos y=b,\cos x=c,\sin y=d$$ with
$$ab+cd=\frac 23≥2\sqrt{abcd}\implies abcd≤\frac 19\\ a^2+c^2=b^2+d^2=1 $$
then I need
$$\min \left(\frac ab+\frac ba+\frac cd+\frac dc\right)$$
But, I can't do anything from here. Finally, I attach the graph drawn by WA.
| Simplifying gives: $$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}=\frac{\sin^2 x+\cos^2y}{\sin x\cos y}+\frac {\cos^2 x+\sin^2y}{\sin y\cos x}=\frac{\sin(x+y)\sin(x-y)+1}{\sin x\cos y}+\frac {1-\sin(x+y)\sin(x-y)}{\sin y\cos x}=\frac {8}{3\sin2x\sin 2y}+\frac{2\sin (x-y)}{3}\left(\frac{1}{\sin x\cos y}-\frac{1}{\sin y\cos x}\right)=\frac {8}{3\sin2x\sin 2y}-\frac 83\frac{\sin^2(x-y)}{\sin2x\sin 2y}=\frac{8\cos^2(x-y)}{3\sin 2x\sin2y}=\frac{16\cos^2(x-y)}{3(\cos 2(x-y)-\cos 2(x+y))}=\frac{16}{3(2-10/9\sec^2(x-y))}$$
The denominator is maximum when $\sec^2(x-y)$ is minimum, which happens when $x=y$, in which case the quantity has value $6$. So the required global minimum should be $6$.
Edit: The edit shows that the given expression (let's call it $f(x,y)$ for brevity) can't take any value smaller than $6$. To see that, let's first note (Refer Note) that $\color{blue}{ 2-\frac {10}9\sec^2(x-y)\gt 0 \text{ for all } (x,y)\in (0,\frac \pi 2)\times (0,\frac \pi 2)}$.
Suppose on the contrary that there exist some $a$ and $b$ in $(0,\frac \pi 2)$ such that $f(a,b)<6$. It follows that $$\frac{16}{3(2-10/9\sec^2(x-y))}\lt 6\implies 8<18-10\sec^2(x-y)\implies \sec^2(x-y)\lt 1 $$ and this is a contradiction. So the assumption that $f$ takes any value less than $6$ is wrong. Hence global minimum value of $f$ under the given conditions is $6$.
Note: If $\sec^2(x-y)\ge \frac{18}{10}$ then $\cos^2(x-y)\le \frac {10}{18}\implies 0\lt \cos (x-y)\le \sqrt {\frac{10}{18}}$ (because $x-y \in (-\frac \pi 2, \frac \pi 2)$ so $\cos $ is +ve) so let's consider two cases: 1) $0\le x-y)\lt \frac \pi 2 \text{ and } 2) -\frac \pi 2\lt x-y \lt 0$.
Case 1: it follows that $x-y\ge \arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is decreasing on $[0,\pi/2))$. Given that $x+y=\arcsin \frac 23$, it follows that $y\le 0$ which is not possible.
Case 2: it follows that $x-y\le -\arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is increasing on $(-\frac \pi 2, 0))$. Then proceeding as in case 1) results in a contradiction.
This establishes the blue colored part.
| {
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"url": "https://math.stackexchange.com/questions/4320041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I find $S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2 p}}{(2 p) !}+\frac{x^{3 p}}{(3 p) !}+\cdots$ by complex numbers, where $p \in N$? In my post on Quora, I had expressed the three series in terms of sine and cosine:
$$1+\frac{x^{3}}{3 !}+\frac{x^{6}}{6 !}+\frac{x^{9}}{9!}+\cdots \cdot=\frac{1}{3}\left[e^{x}+2 e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3}}{2} x\right)\right]\\
\frac{x^{2}}{2 !}+\frac{x^{5}}{5 !}+\frac{x^{8}}{8 !}+\cdots=\frac{1}{3}\left[e^{x}-e^{-\frac{x}{2}}\left(\sqrt{3} \sin \left(\frac{\sqrt{3}}{2} x\right)+\cos \left(\frac{\sqrt{3}}{2} x\right)\right)\right]\\
x+\frac{x^{4}}{4 !}+\frac{x^{7}}{7 !}+\cdots=
\frac{1}{3}\left[e^{x}+e^{-\frac{x}{2}}\left(\sqrt{3} \sin \left(\frac{\sqrt{3}}{2} x\right)-\cos \left(\frac{\sqrt{3}}{2} x\right)\right)\right].$$
They all have the same “length” $3$. I then think that the proof can be extended for a general series $S(p)$ with an arbitrary length $p$ and try to prove similarly that for any natunal number $p$,
$$
S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2 p}}{(2 p) !}+\frac{x^{3 p}}{(3 p) !}+\cdots=\frac{1}{p}\left(e^{x}+e^{\omega x}+e^{\omega^{2} x}+\cdots+e^{\omega^{n-1} x}\right) \tag*{(*)}
$$
where $\omega$ is the complex $p$-th root of unity satisfying $$1+\omega+\omega^{2}+\cdots+\omega^{p-1}=0.$$
My question is how to prove (*) and express , in terms of sine and cosine, the series
$$e^{x}+e^{\omega x}+e^{\omega^{2} x}+\cdots+e^{\omega^{p-1} x}$$
| We are going to evaluate the series
$\displaystyle S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2p}}{(2p) !}+\frac{x^{3p}}{(3p )!}+\cdots $
Let’s start with the series
$ \displaystyle e^{z}=\sum_{x=0}^{\infty} \frac{z^{k}}{k !} \tag*{(*)} $
Now consider the complex $p^{th} $ root of unity $\omega=e^{\frac{ 2\pi}{p}i} $ satisfying
$$ 1+\omega+\omega^{2}+\cdots+\omega^{p-1} =0 \text{ and }\omega^{p}=1.$$
Putting $ z=x$ yields
$ \displaystyle e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} \tag*{(1)}$
Putting $ z=\omega x $ yields
$ \displaystyle e^{\omega x}=\sum_{k=0}^{\infty} \frac{\omega^{k} x^{k}}{k !}\tag*{(2)}$
Putting $\displaystyle z=\omega^{2} x $ yields
$ \displaystyle e^{\omega^{2} x}=\sum_{k=0}^{\infty} \frac{\omega^{2 k} x^{k}}{k !} \tag*{(3)}$
$$\vdots$$
Putting $\displaystyle z=\omega^{p-1} x $ yields
$ \displaystyle e^{\omega^{p-1} x}=\sum_{k=0}^{\infty} \frac{\omega^{(p-1)k} x^{k}}{k !} \tag*{(p-1)}$
$(1)+(2)+(3)+\cdots +(p-1)$ gives
$ \displaystyle \sum_{k=0}^{\infty} \frac{1+\omega^{k}+\omega^{2 k}+\cdots+\omega^{(p-1)k}}{k !}x^{k}=e^{x}+e^{\omega x}+e^{\omega^{2} x}+ \cdots +e^{\omega^{p-1} x}\tag*{}$
$ \begin{aligned} \displaystyle \because 1+\omega^{k}+\omega^{2 k}+\cdots+ \omega^{(p-1) k} &=\left\{\begin{array}{cl}\frac{1-\left(\omega^{p}\right)^{k}}{1-\omega^{p}} & \text { if } p\not | k \\ p & \text { if } p|k\end{array}\right. =\left\{\begin{array}{ll}0 & \text { if } p\not|k \\p & \text { if } p | k\end{array}\right.\\\displaystyle \therefore \sum_{k=0}^{\infty} \frac{p}{(p k) !} x^{p k} &=e^{x}+e^{\omega x}+e^{\omega^{2} x} +\cdots + e^{\omega^{p-1} x} \end{aligned} \tag*{} $
$$ S(p)= \sum_{k=0}^{\infty} \frac{x^{p k}}{(p k) !} =\frac{1}{p} \left(e^{x}+e^{\omega x}+e^{\omega^{2} x} +\cdots + e^{\omega^{p-1} x} \right)\tag*{(**)} $$
When $p$ is odd,
$$
\begin{aligned}
S(p) &=\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}}\left(e^{\omega^{k} x}+e^{\omega^{p-k} x}\right)\right]\\
&=\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}}\left(e^{x e^{\frac{2 k \pi}{p}i} }+e^{x e^{\frac{2(p-k) \pi}{p} i}}\right)\right]\\& =\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}} e^{x \cos \frac{2 k x}{p}}\left(e^{i x \sin \frac{2k\pi}{p}}+e^{-i x \sin \frac{2k \pi)}{p}}\right)\right]\\ &=\frac{1}{p}\left[e^{x}+2 \sum_{k=1}^{\frac{p-1}{2}} e^{x \cos \frac{2 k x}{p}}\left(\cos \left(x \sin \frac{2 k \pi}{p}\right)\right)\right]
\end{aligned}
$$
When $p$ is even,
$$\begin{aligned}
S(p) &=\frac{1}{p}\left[e^{x}+e^{-x}+\sum_{k=1}^{\frac{p}{2} -1}\left(e^{\omega^{k} x}+e^{\omega^{p-k} x}\right)\right]\\ &=\frac{1}{p}\left[e^{x} +e^{-x} +2 \sum_{k=1}^{\frac{p}{2}-1} e^{x \cos \frac{2 k x}{p}}\left(\cos \left(x \sin \frac{2 k \pi}{p}\right)\right)\right]
\end{aligned}$$
By the way, by successive differentiation on $S(p)$, we can find, for any natural number $n<p$, the series,
$$
\frac{x^{n}}{n!}+\frac{x^{n+p}}{(n+p) !}+\frac{x^{n+2 p}}{(n+2 p) !}+\cdots
$$ whose length is $p$ and starting terms other than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What type of curve is described by $4(\cos{x}+\cos{y})-6(\cos{2x}+\cos{2y})+8\cos{x}\cos{y}=7$? Does the curve by the function $$4(\cos{x}+\cos{y})-6(\cos{2x}+\cos{2y})+8\cos{x}\cos{y}=7\\x=[-2\pi/3,2\pi/3],\; y=[-2\pi/3,2\pi/3]$$ belong to any known curve family? A collection of curves is found in Wikipedia (Link1,Link2). A large collection of $\approx 1000$ curves can be found here.
| Negative. I do not see the reason to have a name for the curve defined implicitly by the given condition
$$
4(\cos x + \cos y)
-
6(\cos 2x + \cos 2y )
+
8\cos x\cos y
=7\ ,
$$
where $x,y$ run only in the interval $I$ between the two points $\pm2\pi/3$.
However, we have a name for the curve isolated after using the substitutions
$$
\begin{aligned}
s &=\cos x\ ,\\
t &=\cos y\ ,
\end{aligned}
$$
which is the conic
$$
4(s+t) - 6(2s^2-1+2t^2-1) + 8st-7=0\ .
$$
Alternatively:
$$
12s^2 + 12t^2 - 8st - 4s - 4t -5 = 0\ .
$$
It is an ellipse, since the part in degree two $12s^2+12t^2-8st$ is always $\ge0$, and it is not degenerated
since its determinant is negative
$$
\begin{vmatrix}
12 & -4 &-2 \\
-4 & 12 &-2 \\
-2 & -2 & -5
\end{vmatrix}
=
2^2
\begin{vmatrix}
6 & -2 &-1 \\
-2 & 6 &-1 \\
-2 & -2 & -5
\end{vmatrix}
=
2^4
\begin{vmatrix}
3 & -1 &-1 \\
-1 & 3 &-1 \\
-1 & -1 & -5
\end{vmatrix}
=
2^4
\begin{vmatrix}
4 & -4 & 0 \\
-1 & 3 &-1 \\
-1 & -1 & -5
\end{vmatrix}
=
2^6
\begin{vmatrix}
1 & -1 & 0 \\
-1 & 3 &-1 \\
-1 & -1 & -5
\end{vmatrix}
=
2^6
\begin{vmatrix}
1 & 0 & 0 \\
-1 & 2 &-1 \\
-1 & -2 & -5
\end{vmatrix}
=
2^7(-5-1)=-2^8\cdot 3<0\ .
$$
Since the switch $s\leftrightarrow t$ invariates the equation of the ellipse, one of its axes is $s=t$. This axis intersects the ellipse in the points with $s=t$ and $16s^2-8s-5=0$,
so $s=t=\frac 14(1\pm\sqrt6)$. The mid point between the two intersections is the center of the ellipse, $s=t=1/4$, and the other axis is the perpendicular in this point on the line $s=t$, so its equation is $s+t-\frac 12=0$.
The desmos picture looks like:
As seen from the picture, the lines $s=1$ and $t=1$ are tangent to the ellipse, indeed, setting $t=1$ in the equation leads to $0=12s^2-12s+30=3(2s+1)^2$, having a double solution. So the points $(s,t)=(1/2,1)$ and $(s,t)=(1,1/2)$ are the two tangent points.
This is relevant for us, since $\cos$ takes this maximal value $1$.
The corresponding $(x,y)$-points are $(x,y)=(\pm\pi/3, 0)$ and $(x,y)=(0,\pm\pi/3)$, which are the four intersections of the graph with the coordinate axes. (Further points of the given graph can be traced back in this way to points of the ellipse.)
So the given curve has no name, but if a "name" should be given, it is the preimage of the above ellipse by the map $(s,t)\to(\cos s,\cos t)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Summation of a binomial series Sum of the series $\left( {\begin{array}{*{20}{c}}
{20}\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{20}\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
6
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{20}\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
7
\end{array}} \right) - .... - \left( {\begin{array}{*{20}{c}}
{20}\\
{15}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
{20}
\end{array}} \right)$ is____
My approach is as follow
$T = \left( {\begin{array}{*{20}{c}}
{20}\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{20}\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
6
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{20}\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
7
\end{array}} \right) - .... - \left( {\begin{array}{*{20}{c}}
{20}\\
{15}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
{20}
\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}
n\\
p
\end{array}} \right) = {}^n{C_p}$
$T = \sum\limits_{r = 0}^{15} {{{\left( { - 1} \right)}^r}.{}^{20}{C_r}.{}^{20}{C_{r + 5}}} \Rightarrow T = \sum\limits_{r = 1}^{15} {{{\left( { - 1} \right)}^r}.\frac{{20!}}{{r!\left( {20 - r} \right)!}}.\frac{{20!}}{{\left( {r + 5} \right)!\left( {15 - r} \right)!}}} \Rightarrow T = \frac{{20!}}{{15!}} \times \frac{{20!}}{{25!}}\sum\limits_{r = 0}^{15} {{{\left( { - 1} \right)}^r}.\frac{{15!}}{{r!\left( {15 - r} \right)!}}.\frac{{25!}}{{\left( {r + 5} \right)!\left( {20 - r} \right)!}}} $
How do we approach from here
| Hint :
Consider the product $(x-1)^{20}(1+x)^{20}$. Verify by expansion and multiplication that coefficient of $x^{25}$ is
$$\binom{20}{0}\binom{20}{5} - \binom{20}{1}\binom{20}{6}+\ldots - \binom{20}{15}\binom{20}{20}$$
Now find the coefficient of $x^{25}$ in $(x-1)^{20}(1+x)^{20}$ using a different way. The two must be equal.
| {
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"url": "https://math.stackexchange.com/questions/4332615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is my work transforming $9(x+2)^2$ to $(3x+6)^2$ correct? What is this method called? $$9(x+2)^2 = 3^2(x+2)^2=(3(x+2))^2=(3x+6)^2$$
I want to know if the use of brackets in this problem has been done correctly. What is this method called?
| Yes, all steps are correct.
You can also verify that the result is correct by expanding both expressions.
The first expression expands out to
$$9(x+2)^2=9(x^2+2\cdot 2\cdot x+2^2)=9(x^2+4x+4)=9x^2+36x+36$$
while the final expression expands out to
$$(3x+6)^2=(3x)^2+2\cdot(3x)\cdot 6 + 6^2 = 3^2x^2+2\cdot3\cdot6\cdot x + 36=9x^2+36x+36$$
The two expansions match, confirming the original expressions are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evalute $\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}$ Evaluate the given expression $$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$ The given answer is $\dfrac{1}{4}$. My attempt:
$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$ This is as far I as I am able to reach. Thank you!
PS I don't see how one can get $\dfrac14$. For that we have to get something like $\sqrt[n]{A^n}$.
| Since $2^{2n+4}+2^{2n+2}=2^{2n}(2^4+2^2)=20\times4^n$,$$\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\frac{20}{20\times4^n}}=\frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x$: $\frac{x^2-10x+15}{x^2-6x+15}=\frac{3x}{x^2-8x+15}$ Solve the equation: $\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{3x}{x^2-8x+15}$.
When $x\ne3$ and $x\ne5$ we get $$(x^2-10x+15)(x^2-8x+15)=3x(x^2-6x+15)\\(x^2-9x+15-x)(x^2-9x+15+x)-3x(x^2-6x+15)=0\\(x^2-9x+15)^2-x^2-3x(x^2-6x+15)=0.$$ I am stuck here. The Rational Root Theorem won't be useful as the equation does not have such roots.
I got $-9x$ by averaging $-10x$ and $-8x$.
I don't know if it makes sense.
| Multiplying to a common denominator and factoring we obtain
$$
0=x^4 - 21x^3 + 128x^2 - 315x + 225=(x^2 - 7x + 15)(x^2 - 14x + 15).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\sum_{k=0}^{n-1} {n-1-k\choose k}\left(\frac{1}{2}\right)^{n-1-k}+\sum_{k=0}^{n-2} {n-2-k\choose k}\left(\frac{1}{2}\right)^{n-2-k} $ How can you find
$$
\sum_{k=0}^{n-1} {n-1-k\choose k}\left(\frac{1}{2}\right)^{n-1-k}+\sum_{k=0}^{n-2} {n-2-k\choose k}\left(\frac{1}{2}\right)^{n-2-k}
$$
?
I found the value via interpritting the above formula combinatorially. (If I am correct, it is $\frac{2}{3}+\frac{1}{3}\left(-\frac{1}{2}\right)^n$)
But I want to know how to solve it by way of complex integrals or formal power series or any algebraic manipulations.
| We will consider the generating function
$$F_n(x)=\sum_{k=0}^\infty \binom{n-k}{k} x^k$$
Note that since $\binom{n-k}{k}=\binom{n-k-1}{k-1}+\binom{n-k-1}{k}$, we can multiply both sides by $x^k$ and sum from $k=0$ to $\infty$ to get
$$F_n(x)=xF_{n-2}(x)+F_{n-1}(x)$$
$$F_n(x)-F_{n-1}(x)-xF_{n-2}(x)=0$$
We can solve this linear recursion to get that
$$F_n(x)=\left(\frac{1+\sqrt{1+4x}}{2}\right)^nA(x)+\left(\frac{1-\sqrt{1+4x}}{2}\right)^nB(x)$$
For some functions $A(x)$ and $B(x)$. We can verify that $F_0(x)=1$ and $F_1(x)=1$, so we get the following system
$$\begin{cases}1=A(x)+B(x)\\
1=\left(\frac{1+\sqrt{1+4x}}{2}\right)A(x)+\left(\frac{1-\sqrt{1+4x}}{2}\right)B(x)\end{cases}$$
$$\begin{cases}1=A(x)+B(x)\\
2=(1+\sqrt{1+4x})A(x)+(1-\sqrt{1+4x})B(x)\end{cases}$$
$$\begin{cases}1=A(x)+B(x)\\
1=\sqrt{1+4x}A(x)-\sqrt{1+4x}B(x)\end{cases}$$
$$\begin{cases}1=A(x)+B(x)\\
\frac{1}{\sqrt{1+4x}}=A(x)-B(x)\end{cases}$$
$$\begin{cases}\frac{1+\sqrt{1+4x}}{2\sqrt{1+4x}}=A(x)\\
\frac{-1+\sqrt{1+4x}}{2\sqrt{1+4x}}=B(x)\end{cases}$$
Our desired sum is the value of
$$\frac{F_{n-1}(2)}{2^{n-1}}+\frac{F_{n-2}(2)}{2^{n-2}}$$
From our recursive definition of $F_n(x)$, we get that this is equivalent to
$$\frac{F_n(2)}{2^{n-1}}$$
We can use our explicit formula for $F_n(x)$, to get that this is
$$\frac{\left(\frac{1+\sqrt{1+8}}{2}\right)^nA(2)+\left(\frac{1-\sqrt{1+8}}{2}\right)^nB(2)}{2^{n-1}}$$
$$=\frac{(2)^nA(2)+(-1)^nB(2)}{2^{n-1}}$$
$$=\frac{\frac{2^{n+1}}{3}+\frac{(-1)^n}{3}}{2^{n-1}}$$
$$=\frac{4}{3}-\frac{1}{3}\left(-\frac{1}{2}\right)^{n-1}$$
This result (verified computationally) is double the result you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For $a, b, c$ with $a+b+c=0$ prove $\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$ and $\frac17\sum a^7=\frac15\sum a^5\cdot\frac12\sum a^2$ Consider the following problem:
Problem. Suppose that real numbers $a$, $b$ and $c$ satisfy the condition $a+b+c=0$. Prove the following identities:
$$
\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2},
\\
\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}.
$$
Perhaps, the shortest solution I can think of is as follows: plug $c=-a-b$ into the equation but instead of expanding everything use the following identities
$$
(a+b)^3-a^3-b^3=3ab(a+b),
\\
(a+b)^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2),
\\
(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2.
$$
However, these identies are coming out of nowhere and moreover, in order to prove them one still needs to do some computations.
Question. Is it possible to solve this problem in a "smart" way (i.e. avoiding computations and preferrably elementary since it is almost a high school problem)? Any other solutions are also welcome.
Comment. It should be also noted that it is unclear (at least for me) how those identies were invented. It seems there is no nice similar identities for $\frac{a^p+b^p+c^p}{p}$ for $p$ other than 2, 3, 5, 7.
| Here's another way:
Let $a,b,c$ be the roots of the cubic equation $$x^3+px+q=0$$
(The $x^2$ coefficient is zero since $\sum a=0$)
Then $$\sum a^2=(\sum a)^2-2\sum ab=-2p$$
$$\sum a^3=-p\sum a-q\sum 1=-3q$$
$$\sum a^4=-p\sum a^2-q\sum a=2p^2$$
$$\sum a^5=-p\sum a^3-q\sum a^2=5pq$$
Therefore $$pq=\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$$
Also,
$$\sum a^7=-p\sum a^5-q\sum a^4=-7p^2q$$
$$\implies-p^2q=\frac17\sum a^7=(pq)(-p)=\frac15\sum a^5\cdot\frac12\sum a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4348852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Why is this solution incorrect?
Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then
$\frac{1}{PK^2}
+
\frac{1}{QK^2}$
is the same for all positions of the chord.
If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for $y^2=4ax$.
let the point K be (c,0)
Equation of line PQ using parametric coordinates:
$x=c+rcos\theta$ (Equation 1), $y=rsin\theta$ (Equation 2)
From equation 1 and 2: $(x-c)^2+y^2=r^2$ (Equation 3)
Using Equation 3 and $y^2=4ax$, we get this quadratic in r: $r^2- 4ax -(x-k)^2=0$ (Equation 4)
Roots of this quadratic $(r_1 \ and \ r_2)$ would be the lengths of $PK$ and $QK$
From Equation 4, $r_1 + r_2=0$ and $r_1r_2=-(4ax+(x-k)^2)$
We know, $\frac{1}{PK^2}+\frac{1}{QK^2}=\frac{1}{r_1^2}+\frac{1}{r_2^2}=\frac{(r_1 + r_2)^2-2r_1r_2}{(r_1r_2)^2}=\frac{-2}{r_1r_2}$
As value of $r_1r_2$ is not constant thus $\frac{1}{PK^2}+\frac{1}{QK^2}$ does not turn out to be constant. Hence, this solution is incorrect.
I've seen the correct solution but I wanted to know $\mathbf {why \ this \ solution \ is \ incorrect?}$
|
Roots of this quadratic ($r_1$ and $r_2$) would be the lengths of PK and QK.
This part is incorrect. Roots of this equation correspond to the same value of $x$. P and Q can have different $x$.
Update:
$x=c+r \cos \theta$, $y=r \sin \theta$
$$y^2=4ax \Rightarrow r^2 \sin^2\theta - 4ar \cos\theta - 4ac=0$$
Absolute values of roots of this equation would be the PK and QK.
$r_1+r_2=\frac{4a\cos\theta}{\sin^2\theta}$, $r_1r_2=-\frac{4ac}{\sin^2\theta}$
$$\frac{1}{PK^2}+\frac{1}{QK^2}=\frac{(r_1+r_2)^2-2r_1r_2}{(r_1 r_2)^2}=\frac{16a^2\cos^2\theta+8ac\sin^2\theta}{16a^2c^2}$$
This value does not depend on $\theta$ at $c=2a$. This is the answer.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Are there infinitely many $n$ such that $\omega(n^2+2n)=2$? $\omega:\Bbb N\to\Bbb N$ is prime omega function such that $\omega(n)=\sum_{p\mid n}1$. Is there are infinitely many $n$ such that $\omega(n^2+2n)=2$?
If twin prime conjecture is true, then there exists infinitely many $p\in\Bbb P$ such that $p+2\in\Bbb P$, so $\omega(p^2+2p)=2$.
And also Mersenne prime conjecture is true, then there exists infinitely many $p\in\Bbb P$ such that $2^p-1\in\Bbb P$. So when I let $n=2(2^p-1)$ for Mersenne prime $2^p-1$, then $n+2=2^{p+1}$. So $n^2+2n=2^{p+2}(2^p-1)$, that means $\omega(n^2+2n)=2$.
Since twin prime conjecture and Mersenne prime conjectures are both likely true, so I suspect there are infinitely many $n$ such that $\omega(n^2+2n)=2$, but I cannot go further.
I'll appriciate with your help.
+) There are another way to make $\omega(n^2+2n)=2$ as $n=p^r, n+2=q^s$ as $n=7, 25\cdots$. I'm try to prove this part, but totally stuck.
| If gcd$(m,n)=1$, $\omega(mn)=\omega(m)+\omega(n). $
If $n$ is odd:
then gcd$(n,n+2)=1$, so $\omega(n^2+2n)=\omega(n)+\omega(n+2). $
$\omega(n)=0 \iff n=1$. $\omega(3) <> 2 \implies n>1, \omega(n)>0$ .
So if $n$ odd, then $n$ and $n+2$ must both be prime or powers of primes, the former case$\ \omega(n^2+2n)=2 $ if $ n$ is lesser of two twin primes.
If n is even:
$2=$gcd$(n,n+2)$. $n=2a$. $n=2(b-1) \implies a+1=b$
$\omega(n^2+2n)=\omega(4a(a+1))$
So only one odd prime can divide $a(a+1)$. This can only happen if $a$ or $a+1$ is a power of $2$.
If $a+1=2^p$ then $a=2^p-1$ and $n^2+2n=2^{p+2}(2^p-1)$
, the Mersenne Prime scenario you mention. Interestingly, this is also $8$ times a Perfect Number. This should also work if $2^p-1$ is a power of a prime, i.e. $2^q-p^z=1 \iff 2^q \equiv 1 \pmod{p^z}$. So $q \equiv 0 \pmod{\phi(p^z)} $ by Euler's Totient rule. $\phi(p^z)=p^z-p^{z-1}=p^{z-1}(p-1)$ So $q=kp^{z-1}(p-1)$.
A third possible scenario:
Suppose $a=2^q$. Then $n^2+2n=2^{q+2}(2^q+1)$
$0=4 \cdot 2^{2q}+4\cdot2^q-n(n+2)$
$2^q=\frac{-4+4\sqrt{1+n(n+2)}}{8}=n/2$
so $n=2^{q+1}$
We need $2^q+1$ be a power of a prime. This gives us 24, 80, 288, 1088,...
We need solutions of the form $p^z-2^q=1$
$-1\equiv2^q \pmod{p^z}, z>=1 $
Does such a $q$ only exist if $2$ is a primitive root for prime $p$?
Looks like there are 4 scenarios, twin primes and twin power of primes for odd $n$, Mersenne Primes/Perfect numbers, and powers of primes 1 greater than a power of 2 for even $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356430",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 1,
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Find the solution of the system $x'=x-2y\text{ and }y'=4x+5y$ I have to find the solution of the system $x'=x-2y\:\:\&\:\:y'=4x+5y$ using eigenvalues and eigenvectors of the matrix.
If I rewrite this system I get $$\begin{bmatrix}
x'\\
y'
\end{bmatrix}=\begin{bmatrix}
1 & -2 \\
4 & 5
\end{bmatrix} \begin{bmatrix}
x\\
y
\end{bmatrix}$$
So I got eigenvalue $\lambda_1=3+2i$ for eigenvector $v_1=\begin{bmatrix}
1\\
-1-i
\end{bmatrix}$ and eigenvalue $\lambda_2=3-2i$ for eigenvector $v_1=\begin{bmatrix}
1\\
-1+i
\end{bmatrix}$
Then I got solution for this system $$\begin{bmatrix}
x(t)\\
y(t)
\end{bmatrix}=C_1e^{(3+2i)t}\begin{bmatrix}
1\\
-1-i
\end{bmatrix}+C_2e^{(3-2i)t}\begin{bmatrix}
1\\
-1+i
\end{bmatrix}=\begin{bmatrix}
e^{3t}\cos(2t) & e^{3t}\sin(2t)\\
e^{3t}(-\cos(2t)+\sin(2t)) & e^{3t}(-\cos(2t)-\sin(2t))
\end{bmatrix} \begin{bmatrix}
C_1\\
C_2
\end{bmatrix}$$
Is this correct answer?
| The solution for such systems (according to Perko2001 Differential Equations And Dynamical Systems) is as follows: Let
$$A= \begin{bmatrix}
1 & -2 \\
4 & 5
\end{bmatrix}$$
We have $\lambda_{1} = 3 + 2i$, $\lambda_2 = \overline {\lambda_1}= 3 - 2i$ and
$$v_1 = \begin{bmatrix}
-1+i \\
2
\end{bmatrix} = \begin{bmatrix}
-1 \\
2
\end{bmatrix} + i\begin{bmatrix}
1 \\
0
\end{bmatrix}$$
and
$$v_2 = \begin{bmatrix}
-1-i \\
2
\end{bmatrix} = \begin{bmatrix}
-1 \\
2
\end{bmatrix} + i\begin{bmatrix}
-1 \\
0
\end{bmatrix}$$
Now we construct the matrix $P$ from the imaginary and real parts of the eigenvectors;
$$P=
\begin{bmatrix}
1 & -1 \\
0 & 2
\end{bmatrix}
\Rightarrow P^{-1} =\frac 1 {13} \begin{bmatrix}
3 & 2 \\
-2 & 3
\end{bmatrix}$$
and
$$P^{-1} A P =\begin{bmatrix}
3 & -2 \\
2 & 3
\end{bmatrix} $$
And the final solution
$$\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix} = P e^{3t} \begin{bmatrix}
\cos 2t & -\sin 2t \\
\sin 2t & \cos 2t
\end{bmatrix} P^{-1} \begin{bmatrix}
x_0 \\
y_0
\end{bmatrix}.$$
and
$$\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix} = \frac {e^{3t}} {13} \begin{bmatrix}
1 & -1 \\
0 & 2
\end{bmatrix} \begin{bmatrix}
\cos 2t & -\sin 2t \\
\sin 2t & \cos 2t
\end{bmatrix} \begin{bmatrix}
3 & 2 \\
-2 & 3
\end{bmatrix} \begin{bmatrix}
x_0 \\
y_0
\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating $P(X \le 0)$ from the MGF of $X+Y$, where $X$ and $Y$ are independent and identically distributed random variables Let $X$ and $Y$ be identically distributed independent random variables such that the moment generating function of $X+Y$ is
$M(t) = 0.09 e^{−2t} + 0.24 e^{−t} + 0.34 + 0.24 e^t + 0.09 e^{2t}$.
Calculate $P(X \le 0)$.
My attempt
Let $u=e^t$
$M_{X+Y}(t)=\frac{1}{(10u)^2}(9+24u+34u^2+24u^3+9u^4)$
Now let $v=u+\frac{1}{u}$ and $p(u)=9u^4+24u^3+34u^2+24u+9$
$p(u)=u^2[9(u^2+\frac{1}{u^2})+24(u+\frac{1}{u})+34]=u^2[9(v^2-2)+24v+34]=u^2(9v^2-18+24v+34)=u^2(9v^2+24v+16)=u^2(3v+4)^2=u^2(3u+\frac{3}{u}+4)^2=[u(3u+\frac{3}{u}+4)]^2=(3u^2+4u+3)^2$
Thus $M_{X+Y}(t)=\frac{1}{(10u)^2}(3u^2+4u+3)^2=[\frac{(3u^2+4u+3)}{10u}]^2=(0.3u+0.4+0.3u^{-1})^2=(0.3e^t+0.4+0.3e^{-t})^2$
Because $X$ and $Y$ are independent and identically distributed, the MGF of $X+Y$ equals $(0.3e^t+0.4+0.3e^{-t})^2$, where $0.3e^t+0.4+0.3e^{-t}$ is the MGF common to $X$ and $Y$.
Thus $M_X(t)=0.3e^t+0.4+0.3e^{-t}$, and since $M_X(t)$ uniquely determines the distribution of $X$, it follows that $X$ is a discrete random variable whereby $X=-1, X=0$, and $X=1$ with respective probabilities $0.3,0.4$, and $0.3$.
$\therefore P(X \le 0)=0.3+0.4=0.7$.
When is it appropriate to use the change of variable technique (i.e. $v=u+\frac{1}{u}$) to factor polynomials and in what situations would it not work?
| The method works for Quasi-palindromic quartic functions as $$f(x)=a_0\cdot x^4+a_1\cdot x^3+a_2\cdot x^2+a_1\cdot m\cdot x+a_0 \cdot m^2$$
In you case the parameters are $m=1, a_0=9, a_1=24$ and $a_2=34$. So you have a (real) palindromic function. For reference see here.
If you have not a palindromic quartic function, then you can multiply out $g(x)=(ax^2+bx+c)^2$ and then you compare the coefficients.
| {
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"url": "https://math.stackexchange.com/questions/4363214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Derive an upper bound for $\log(x)$ A paper has given that for $x_0 > 0$,
$$
\log x \leq \frac{x}{x_0} + \log x_0 - 1.
$$
The way I tried to prove this is:
For $|t| < 1$,
\begin{align*}
\log(1 - t) = -\sum_{k=1}^n \frac{t^k}{k}
\end{align*}
We can replace $t = 1 - \frac{x}{x_0}$,
\begin{align*}
\log x - \log x_0 & = - [(1 - \frac{x}{x_0}) + \frac{1}{2}(1 - \frac{x}{x_0})^2 + ...]\\
& = -1 + \frac{x}{x_0} - [\frac{1}{2}(1 - \frac{x}{x_0})^2 + \frac{1}{3}(1 - \frac{x}{x_0})^3 + ...]\\
& = -1 + \frac{x}{x_0} - c
\end{align*}
How can we show that $c$ is positive and what's the range of $x$ (is it $0 < x < 2x_0$)?
| For any real $w$, by the Taylor formula with Lagrange remainder, we have
$$
e^{w - 1} = 1 + (w - 1) + \frac{{(w - 1)^2 }}{2}e^\xi \ge 1 + (w - 1) = w
$$
with some $\xi$ between $0$ and $w-1$. Thus, for any $w>0$, $w-1 \ge \log w$. With $w=x/x_0$, where $x$, $x_0>0$, we get
$$
\log \left( {\frac{x}{{x_0 }}} \right) \le \frac{x}{{x_0 }} - 1 \Longleftrightarrow \log x \le \frac{x}{{x_0 }} + \log x_0 - 1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$
I've been asked to solve the limit.
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Here's my approach:
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Using the identity, $\cos(x) =\sin(90^{\circ} - x)$
\begin{aligned}\implies \lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
& = \lim_{x\to0}\frac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
\\& = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)\cdot\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}{\sin(\sin(x^2))}
\\ & = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}\cdot \underbrace{\lim_{x\to0}\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}_{1}
\\ & = \dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(\sin(x^2))}{\sin(x^2)}}_1\cdot\sin(x^2)}
\\ & =\dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(x^2)}{x^2}}_1\cdot x^2}
\\ & = \color{blue}{\boxed{\lim\limits_{x\to0}\dfrac{\pi}{2x^2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}}
\end{aligned}
Now, I'm unable to think of anything to do with this boxed part. Can anyone check my above method and tell me what to do further with this question? Any other shorter method is also most welcomed!
| I found another way to compute the final boxed limit.
$$\begin{aligned}&\lim_{x\to0} \frac{\pi}{2x^2}\cdot\frac{\cos(x)-1}{\cos(x)}\\=&\lim_{x\to0}\frac{-(1-\cos(x))}{x^2}\cdot\frac{\pi}{2\cos(x)}\end{aligned}$$
The limit $\boxed{\lim\limits_{x\to0}\dfrac{1-\cos(x)}{x^2} = \dfrac12}$
$$\begin{aligned}\implies& \lim_{x\to0}\frac{-(1-\cos(x))}{x^2}\cdot\frac{\pi}{2\cos(x)} = \lim_{x\to0}\frac{-1}{2}\cdot\frac{\pi}{2\cos(x)}\\=&\frac{-\pi}{4\cos(0)} = \frac{-\pi}{4}\end{aligned}$$
So the limit is $-\pi/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Using factor theorem on a multivariable function Q: Use the factor theorem to show that $a + b - c$ is a factor of $f(a, b, c) = (a+b+c)^3 - 6(a + b + c)(a^2 + b^2 + c^2) + 8(a^3 + b^3 + c^3)$ and hence factorise completely.
I know that the factor theorem states for $f(x)$ if $f(a) = 0$ then $x-a$ is a factor but when there are multiple variables how do you use factor theorem?
| We are given that
$$ f(a,b,c) := (a+b+c)^3 - 6(a+b+c)(a^2+b^2+c^2) + 8(a^3+b^3+c^3).$$
Define three power sum polynomials
$$ p_1(a,b,c) := a+b+c, \quad p_2(a,b,c) := a^2+b^2+c^2, \quad p_3(a,b,c) := a^3+b^3+c^3. $$
Notice that
$$ p_1(a,b,a+b) = 2(a+b), \qquad p_3(a,b,a+b) = (a+b)(2a^2+ab+2b^2). $$
Define a one variable polynomial $\,P(x) := f(a,b,x)\,$
and rewrite $\,P(a+b) = f(a,b,a+b)\,$ in terms of power sums to get
$$ P(a+b) = 8(a+b)^3 - 12(a+b)(a^2+b^2+(a+b)^2) + 8(a+b)(2a^2+ab+2b^2) $$
which factors further as
$$ P(a+b) = (a+b)\left( 8(a+b)^2 - 12(-2ab+2(a+b)^2) +
8(-3ab+2(a+b)^2)\right) $$
and simplify to get
$$ P(a+b) = (a+b)\left( (8-24+16)(a+b)^2 + (24-24)ab \right) = 0. $$
Use the factor theorem to get that $\, P(x) \,$ is divisible
by $\,x-(a+b).\,$ In particular, this implies that $\,P(c) = f(a,b,c)\,$
is divisible by $\,c-(a+b) = -(a+b-c).\,$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sec x=\frac{a}{b}$, find the exact value of $\cot x-\frac1{\sin x}$, for positive $a$ and $b$, and for $x\in(\frac{3\pi}{2},2\pi)$
If $\sec(x) = \frac{a}{b}$, find the exact value of
$$\cot(x) - \frac{1}{\sin(x)}$$ and $a$, $b$ are positive real values and $x\in(\frac{3\pi}{2},2\pi)$
*
*I know that $\sec(x) = \frac{1}{\cos(x)}, $ so: $\cos(x) = \frac{b}{a}$
*I used the right angled triangle to find the third side, which is $\sqrt{a^2-b^2}$
*I used reciprocal functions of $\tan$ and $\sin$ to obtain the values:
$cosec(x) = \frac{a}{(\sqrt{a^2-b^2})}$ and $\cot(x) = \frac{b}{\sqrt{a^2-b^2}}$
and finally getting $\frac{(b-a)}{\sqrt{a^2-b^2}}$ as my answer. But that is not correct.
The correct answer is: $\frac{\sqrt{a^2-b^2}}{a+b}$. Now I am wondering how.
| *
*Remember, when using a right-angled triangle to determine
trigonometric values, we are actually just determining their
absolute values (i.e., dropping any negative sign).
This is evidenced by the fact that the process evaluates ratios of triangle lengths, each of which is expressed as a nonnegative value in the form $\sqrt{u^2\pm v^2}.$
In the given example, since $x$ is located in the fourth quadrant, $\operatorname{cosec}x$ must be negative and so equals $\displaystyle\mathbf-\left(\frac a{\sqrt{a^2-b^2}}\right)$ rather than $\displaystyle\frac a{\sqrt{a^2-b^2}}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate the shaded area of this 1/4 circle?
First I calculate the area of 1/4 circle, that is $\frac{1}{4}\pi.r^2=\frac{1}{4}\pi.4^2=4\pi.$
Then I know the area of each 1/2 circles inside the 1/4 circle is $\frac{1}{2}\pi.2^2=2\pi.$ But I know it's not as simple as 1/4 circle $-$ 1/2 circles inside. I'm having a hard time determining the shaded area that is the intersection of two 1/2 circles inside. Any idea of the steps for this solution? Thanks in advance.
|
Let origin $O(0,0)$ be the center of quarter circle (as shown in the figure).
Now we have following equations of quarter and semi-circles
$$x^2+y^2=16$$
$$x^2+(y-2)^2=4$$ $$(x-2)^2+y^2=4$$
It's worth noticing that the straight line $y=x$ divides the required area into two equal parts. Therefore the total required area is twice the area of one part under $y=x$ which is twice the sum of areas $A_1, A_2$ & $A_3$ (as shown in above figure) given as follows
$$2(A_1+A_2+A_3)=2\left(\int_0^2\left(x-\left(2-\sqrt{4-x^2}\right)\right)dx+\int_2^{2\sqrt2}\left(x-\sqrt{4-(x-2)^2}\right)dx+\int_{2\sqrt2}^4\left(\sqrt{16-x^2}-\sqrt{4-(x-2)^2}\right)dx\right)$$ (Rearranging the terms) $$=2\left(\int_0^{2\sqrt2}xdx+\int_0^2(\sqrt{4-x^2}-2)dx+\int_{2\sqrt2}^4\sqrt{16-x^2}\ dx-\int_{2}^4\sqrt{4-(x-2)^2}dx\right)$$
$$=2\left(\left[\frac{x^2}{2}\right]_0^{2\sqrt2}+\left[\frac x2\sqrt{4-x^2}+2\sin^{-1}\left(\frac x2\right)-2x\right]_0^2+\left[\frac x2\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_{2\sqrt2}^4-\left[\frac{(x-2)}{2}\sqrt{4-(x-2)^2}+2\sin^{-1}\left(\frac{x-2}{2}\right)\right]_{2}^4\right)$$ $$=2\left(4+\pi-4+4\pi-4-2\pi-\pi\right)$$
$$=4\pi-8$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to show $\frac{1}{\omega-3\omega^2}=\frac{4\omega^2+3}{13}$. I was solving a cubic equation with two different methods and one of the root I got with first method is $\frac{1}{\omega-3\omega^2}$ and with second method I got $\frac{4\omega^2+3}{13}$.
$\omega$ is the cube root of unity
If I set one equals another then I got an equality $13=13$ which indicates that they are equal.
$$\frac{1}{\omega-3\omega^2}=\frac{\omega^2}{1-3\omega}$$
After this I think we have to do rationalization kind of thing to the make denominator real but I could not figure out How.
Thank you for your help!
| The conjugate of $\omega$ is $\omega^2$, so multiply top and bottom by $\omega^2-3\omega$. This gives
$$\frac{1}{\omega-3\omega^2}$$
$$=\frac{1}{\omega-3\omega^2}\cdot\frac{\omega^2-3\omega}{\omega^2-3\omega}$$
$$=\frac{\omega^2-3\omega}{1-3\omega^2-3\omega^4+9}$$
Use $\omega^3=1$ and $1+\omega+\omega^2=0$ to give
$$=\frac{\omega^2-3\omega}{1-3(\omega^2+\omega)+9}$$
$$=\frac{\omega^2-3\omega}{13}$$
$-3\omega=3+3\omega^2$ then gives your result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there $B \in M_4({\mathbb{R}})$ such that $B^2$ is similar to $A$ ? If it is possible, find it. $A =\left ( \begin{matrix}
0 & 0 & 1 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{matrix} \right )$
Is there $B \in M_4({\mathbb{R}})$ such that $B^2$ is similar to $A$ ? If it is possible, find it.
The Jordan normal form of $A$ is $J=\begin{pmatrix}
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$, $P=\begin{pmatrix}
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
1 & 0 & 0 & 1\\
\end{pmatrix}$ such that $P^{-1}AP=J$
Then, $J=B^2$ but how can I find B? How do I know if it is possible to find $B$?
| Your initial step of finding the Jordan form of $A$ is a helpful. With that, we can see that $A^3 = 0$, but $A^2 \neq 0$. This is enough for us to figure out that there is no $B$ such that $B^2$ is similar to $A$.
Indeed, if $B^2$ were similar to $A$, then it would follow that $B^6 = (B^2)^3$ similar to $A^3 = 0$, which means that $B$ is nilpotent, which means that the characteristic polynomial of $B$ must be $p(x) = x^4$. From this, it follows that $p(B) = B^4 = 0$. On the other hand, $B^4 = (B^2)^2$ must be similar to $A^2 \neq 0$. Thus, $B^4$ is similar to both the zero matrix and $A^2 \neq 0$, which means that $A^2$ is a non-zero matrix that is similar to the zero matrix, which is impossible.
The question of finding a general rule that determines whether $A$ has a square root is a bit trickier. As it turns out, this is true iff the blocks of the Jordan form of $A$ with length greater than $1$ can be paired off such that the sizes of the blocks in each pair are at most one apart.
| {
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"url": "https://math.stackexchange.com/questions/4375571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hilbert's Identity for solving Waring's Problem Showing that $g(4)$ exist is quite easy using this identity
\begin{eqnarray}
6(a^2 + b^2 + c^2 + d^2)^2 &=& (a + b)^4 + (a - b)^4 + (c + d)^4
\nonumber
\\ & &+(c-d)^4+(a+c)^4+(a-c)^4
\nonumber
\\ & &+(b+d)^4+(b-d)^4+(a+d)^4
\nonumber
\\& &+(a-d)^4+(b+c)^4+(b-c)^4
\nonumber
\end{eqnarray}
For $g(6)$:
$$60(a^2_1+a^2_2+a_3^2+a_4^2)^3=\sum_{i>j>k}(a_i\pm a_j\pm a_k)^6+2\sum_{i>j}(a_i\pm a_j)+36\sum_i a_i^6$$
And for $g(8)$
$$
5040(a_1^2+a_2^2+a_3^2+a_4^2)^4=6\sum(a_1\pm a_2\pm a_3\pm a_4)^8+\sum_{i>j>k}(2a_i\pm a_j\pm a_k)^8+60\sum_{i>j}(a_i\pm a_j)^8+6\sum_i(2a_i)^8
$$
Does anyone know if there`s a general formula for $N(a^2+b^2+c^2+d^2)^k$, for some $N$ depending on $k$, such that it is the sum of perfect powers of $2k$?
| You should find the bounds of $a,b,c$ and show that $(a,b,c)=(1,2,3)$ and its permutations.
If you consider that $a\geq b\geq c$, then $(abc)^2 = a^3+b^3+c^3 \leq 3a^3\Leftrightarrow (bc)^2 \leq 3a$
Also, we see that $$a^2 \mid a^3+b^3+c^3\Rightarrow b^3+c^3 \geq a^2\Leftrightarrow bc^4 \leq 18$$
Therefore, $c=1$.
We now have to solve the equation $$a^3+b^3+1=a^2b^2$$
Of course $a\neq b$. If $b=2$, then $a=3$.
For $b\geq 2$, then $a>b+1$, so since $a^2|b^3+1$, we take that $a^2\leq b^2-b+1< b^2\Leftrightarrow a< b$, contradiction.
Therefore $$(a,b,c)=(1,2,3)$$ and its permutations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all complex number z which satistfy the equation $|z + 1| = |z - i|$ I am trying to solve the aforementioned problem but have previously only worked with complex number inequalities where the resulting set is (the inside/outside of) a circle.
I am asked to determine all complex numbers z which satisfy the equation
$$|z + 1| = |z - i|$$
I have tried writing $z$ as $x + iy$ and grouping the real and imaginary parts together:
$|x + iy + 1| = |x + iy - i|$
but this just leads to
$|(x + 1) + i(y - 0)| = |(x + 0) + i(y - 1)|$
$\sqrt{(x + 1)^{2} + (y - 0)^{2}} = \sqrt{(x + 0)^{2} + (y - 1)^{2}}$
$(x + 1)^{2} + (y - 0)^{2} = (x + 0)^{2} + (y - 1)^{2}$
$x^2 + 2x + 1 + y^2 = x^2 + y^2 - 2y + 1$
$x + y = 0$
| Let $z=x+iy$ with $x, y\,\in\,\bar{\mathbb{R}}$. We have:
$$|(x+1)+iy|=|x+i(y-1)|$$
So, using the definition of modulus:
$$\sqrt{(x+1)^2+y^2}=\sqrt{x^2+(y-1)^2}\implies x^2+2x+1+y^2=x^2+y^2-2y+1\implies 2x+1=-2y+1\implies x+y=0$$
Simply, we have a line that is the bisector of II and IV quadrant.
Also, there is a very nice geometrical interpretation, shown in this graph:
| {
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Solve $I=\int_{0}^{1}\int_{0}^{1}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}\arctan{\frac{y^2}{x^2}}dxdy$ I tried to solve this integral by using polar coordinate system, but it becomes harder:
$$I=\int_{0}^{1}\int_{0}^{1}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}\arctan{\frac{y^2}{x^2}}dxdy$$
After transformation
$$I=2\int_{0}^{\frac{\pi}{4}}\arctan{(\tan^2\theta)}\sin\theta \cos\theta d\theta\int_{0}^{\frac{1}{cos\theta}}r^2\arctan{(r)}dr$$
And after IBP, $$\int r^2\arctan{(r)}dr=\frac{1}{6}\left(2r^3\arctan(r)-r^2+\ln(r^2+1)\right)$$ and replace upper and lower limits, it becomes a mess.
Did I misdirect? Need some help from everyone, thank you very much.
| The integrand function is not symmetric with respect to the line $y=x$, so $I$ is not two times the integral over the triangle $\{(x,y):0\leq x\leq y\leq 1\}$.
On the other hand, note that
$$
\begin{align*}
I&=\int_{0}^{1}\int_{0}^{x}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}\arctan\left(\frac{y^2}{x^2}\right)\,dydx\\
&\qquad+\int_{0}^{1}\int_{0}^{y}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}\left(\frac{y^2}{x^2}\right)\,dxdy\\
&=\int_{0}^{1}\int_{0}^{x}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}\left(\arctan\left(\frac{y^2}{x^2}\right)+\arctan\left(\frac{y^2}{x^2}\right)\right)dydx.
\end{align*}$$
Now recall that $\arctan(t)+\arctan(1/t)=\pi/2$ for $t>0$, therefore
$$\begin{align*}
I&=\frac{\pi}{2}\int_{0}^{1}\int_{0}^{x}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}dydx\\
&=\frac{\pi}{2}\int_{0}^{\pi/4}\cos(\theta)\sin(\theta)\left(\int_{0}^{1/\cos(\theta)}r^2\arctan(r)\,dr\right)d\theta\\
&=\frac{\pi}{2}\int_{0}^{\pi/4}\cos(\theta)\sin(\theta)\left[\frac{1}{6}\left(2r^3\arctan(r)-r^2+\ln(r^2+1)\right)\right]_{0}^{1/\cos(\theta)}\,d\theta\\
&=\frac{\pi}{12}\int_{1}^{\sqrt{2}}\left(\frac{\ln(1+t^2)}{t^3}+2\arctan(t)-\frac{1}{t}\right)\,dt\\
&=\frac{\pi}{48}\left(8\ln(2)-7\ln(3)+8\sqrt{2}\arctan(\sqrt{2})-2\pi\right)
\end{align*}$$
where $t=1/\cos(\theta)$.
For the second part, we have also another way which seems to be easier
$$\begin{align*}
I&=\frac{\pi}{2}\int_{0}^{1}\int_{0}^{x}\frac{xy}{\sqrt{x^2+y^2}}\arctan{\sqrt{x^2+y^2}}dydx\\
&=\frac{\pi}{2}\int_{0}^{1}r^2\arctan(r)\left(\int_{0}^{\pi/4}\cos(\theta)\sin(\theta)d\theta\right)\,dr\\
&\qquad+\frac{\pi}{2}\int_{1}^{\sqrt{2}}r^2\arctan(r)\left(\int_{\arccos(1/r)}^{\pi/4}\cos(\theta)\sin(\theta)d\theta\right)\,dr\\
&=\frac{\pi}{2}\int_{0}^{1}r^2\arctan(r)\left[-\frac{\cos^2(\theta)}{2}\right]_{0}^{\pi/4}\,dr+\frac{\pi}{2}\int_{1}^{\sqrt{2}}r^2\arctan(r)\left[-\frac{\cos^2(\theta)}{2}\right]_{\arccos(1/r)}^{\pi/4}\,dr\\
&=\frac{\pi}{8}\int_{0}^{1}r^2\arctan(r)\,dr+\frac{\pi}{8}\int_{1}^{\sqrt{2}}\arctan(r)\left(2-r^2\right)\,dr\\
&=\frac{\pi}{48}\left(8\ln(2)-7\ln(3)+8\sqrt{2}\arctan(\sqrt{2})-2\pi\right).
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378529",
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"question_score": "6",
"answer_count": 1,
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} |
Find a recursive formula for $\int_0^2x^n(4-x^2)^{1/2}dx$
Find a recursive formula for $I_n=\int_0^2x^n(4-x^2)^{1/2}dx$ for $n\ge2$.
Source : Cambridge A-level Further Mathematics 9231_s16_qp_13 Question 6
My work
$\frac{d}{dx}(x^n(4-x^2)^{3/2})=-3x^{n+1}(4-x^2)^{1/2}+nx^{n-1}(4-x^2)^{3/2}$
Integrating each term gives,
$0=-3I_{n+1}+n\int _0^2 x^{n-1}(4-x^2)^{3/2} dx$
I got stuck on this integral : $ n\int _0^2 x^{n-1}(4-x^2)^{3/2} dx$
Integration by parts will give $3I_{n+1}$ which is not helpful. I also tried the substitution $x=2\sin \theta$ but got stuck again with $n\int (2\sin \theta)^{n-1}(18\cos^4 \theta)d\theta$. The mark scheme was not helpful as well.
The final answer should be $(n+3)I_{n+1}=4nI_{n-1}$. I feel like I'm missing something obvious because the question is not worth many marks.
| Method 1
Rationalization followed by integration by parts yields
$$
\begin{aligned}
I_{n} &=\int_{0}^{2} x^{n}\left(4-x^{2}\right)^{\frac{1}{2}} d x \\
&=\int_{0}^{2} \frac{x^{n}\left(4-x^{2}\right)}{\left(4-x^{2}\right)^{\frac{1}{2}}} d x \\
&=-\int_{0}^{2} x^{n-1}\left(4-x^{2}\right) d \left(4-x^{2}\right)^{\frac{1}{2}} \\
&=-\left[x^{n-1}\left(4-x^{2}\right)^{\frac{3}{2}}\right]_{0}^{2}+\quad \int_{0}^{2}\left[4(n-1) x^{n-2}-(n+1) x^{n}\right](4-x)^{\frac{1}{2}}\\&=4(n-1) I_{n-2}-(n+1) I_{n}
\end{aligned}
$$
Rearranging yields
$$
\boxed{I_{n}=\frac{4(n-1)}{n+2} I_{n-2}}
$$
In particular, $$
\begin{array}{l}
I_{0}=\pi, \quad I_{1}=\dfrac{8}{3}\quad \Rightarrow \quad I_{2}=\dfrac{4}{4} I_{0}=\pi \textrm{ and } I_{3}=\dfrac{4(2)}{5} \cdot \dfrac{8}{3}=\dfrac{64}{15}.
\end{array}\\\\
$$
Method 2
Integration by parts yields $$
\begin{aligned}
{I}_{n} &=\int_{0}^{2} x^{n}\left(4-x^{2}\right)^{\frac{1}{2}} d x \\
&=-\frac{1}{3} \int_{0}^{2} x^{n-1} d\left(4-x^{2}\right)^{\frac{3}{2}} \\
&=-\left[\frac{x^{2}\left(4-x^{2}\right)^{\frac{3}{2}}}{3}\right]_{0}^{2}+\frac{n-1}{3} \int_{0}^{2} x^{n-2}\left(4-x^{2}\right)^{\frac{3}{2}} d x\\&= \frac{n-1}{3} \int_{0}^{2} x^{n-2}\left(4-x^{2}\right)\left(4-x^{2}\right)^{\frac{1}{2}} d x\\&=\frac{n-1}{3}\left(4 I_{n-2}-I_{n}\right)
\end{aligned}
$$
We can now conclude that $$
\boxed{I_{n}=\frac{4(n-1)}{n+2} I_{n-2}}
$$
| {
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"url": "https://math.stackexchange.com/questions/4378717",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying algebraic expression with fraction of polynomials I cannot seem to find a way to simplify the following expression:
$$\frac{9{a}^{4}-{a}^{2}{ b }^{ 4 } +16 { b }^{ 8 } }{ 3 { a }^{ 2 } -5a { b }^{ 2 } +4 { b }^{ 4 } }$$
I have tried factoring by $a$ and rewriting the division as multiplication with inverse.
| A straightforward solution, simply use long division.
\begin{array}{rrrrrrr}
& & 3a^2 &+5a b^2&+4b^4&&\\
\hline
3a^2-5ab^2+4b^4 & | & 9a^4 & & -a^2b^4 &&+ 16 b^8\\
&&-9a^4&+15a^3b^2&-12a^2b^4\\
\hline
&&&15a^3b^2&-13a^2b^4&&+16b^8\\
&&&-15a^3b^2&+25a^2b^4&-20ab^6\\
\hline
&&&&12a^2b^4&-20a b^6&+16b^8\\
&&&&-12a^2b^4&+20ab^6&-16b^8\\
\hline
&&&&&&0
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac{1}{1+\sin x}\: dx$ using substitution $u=1+\sin x$ I'm trying to work out what went wrong with the following method and how I can precisely correct it.
MY ATTEMPT:
\begin{align*}
\text{Let } u&=1+\sin x\\
\therefore du&=\cos x\:dx\\
&= \pm\sqrt{1-\sin^2x}\:dx\\
&= \pm\sqrt{1-(u-1)^2}\:dx\\
&=\pm \sqrt{2u-u^2}\:dx\\
\therefore dx&= \dfrac{du}{\pm \sqrt{2u-u^2}}
\end{align*}
\begin{align*}
\therefore \int \dfrac{1}{1+\sin x}\: dx &= \pm \int \dfrac{1}{u} \times \dfrac{1}{\sqrt{2u-u^2}} \: du\\
&= \pm \int \dfrac{1}{u^2\sqrt{2/u -1}} \: du\\
&= \pm \int \dfrac{1}{u^2} \times \left( \dfrac{2}{u}-1 \right)^{-1/2}\: du\\
&=\pm \dfrac{1}{2} \int \dfrac{2}{u^2} \times \left( \dfrac{2}{u}-1 \right)^{-1/2}\: du\\
&= \mp \dfrac{1}{2} \times 2\times \sqrt{\left(\dfrac{2}{u}-1\right)} + C\\
&= \mp \sqrt{\dfrac{2}{1+\sin x}-1}+C
\end{align*}
My final answer seems to be the absolute value of one of the primitives of $\dfrac{1}{1+\sin x}$. I know that having the $\pm$ in the $u$-sub is not ideal, but I left it there because I wasn't sure how to address it correctly. From my experience, these problems usually occur because the substitution was of the form $x=f(u)$ instead of $u=g(x)$. But in my method, the latter is used...
Question: How can I properly address this issue?
P.S. Yes, I know there are better ways (Weierstrass sub and clever manipulations) to approach this integral. I just want to know how to properly use this approach.
| Note that $
du=\cos x\:dx= \text{sgn}(\cos x)\sqrt{2u-u^2}dx
$
and your solution is
\begin{align*}
\int \dfrac{1}{1+\sin x}\: dx &= \text{sgn}(\cos x) \int \dfrac{1}{u} \times \dfrac{1}{\sqrt{2u-u^2}} \: du\\
&= - \text{sgn}(\cos x)\sqrt{\dfrac{2}{1+\sin x}-1}+C\\
&= - \text{sgn}(\cos x)\sqrt{\dfrac{1-\sin x}{1+\sin x}}+C\\
&= - \text{sgn}(\cos x){\dfrac{|\cos x|}{1+\sin x}}+C\\
&= - {\dfrac{\cos x}{1+\sin x}}+C
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4380682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\left(\frac{a-b}{a+b}\right)^{11}+\left(\frac{b-c}{b+c}\right)^{11}+\left(\frac{c-a}{c+a}\right)^{11}\leq 1$ where $a,b,c >0$ I have to prove:
$ \displaystyle \tag*{} \left(\frac{a-b}{a+b}\right)^{11}+\left(\frac{b-c}{b+c}\right)^{11}+\left(\frac{c-a}{c+a}\right)^{11}\leq 1$ where $a,b,c >0$
My approach:
$ \displaystyle \tag*{}x \mapsto a-b \\\\ y\mapsto b-c \\\\ z \mapsto c-a$
We get, $x+y+z=0$ and then the inequality becomes:
$\displaystyle \tag*{} \left(\frac{x}{x+2b}\right)^{11}+\left(\frac{y}{y+2c}\right)^{11}+\left(\frac{z}{z+2a}\right)^{11}\leq 1$
I don't know how to proceed from this; Can we use AM-GM inequality to prove? Any hints would be appreciated, thanks.
| Your inequality is
$$\left(\frac{a-b}{a+b}\right)^{11} + \left(\frac{b-c}{b+c}\right)^{11} + \left(\frac{c-a}{c+a}\right)^{11} \leq 1 \tag{1}\label{eq1A}$$
If any $2$ of the variables, e.g., $a$ and $b$, are equal, the LHS of \eqref{eq1A} becomes $0$. Thus, consider they are all different. Next, for simpler algebra & discussions, let
$$d = \frac{a-b}{a+b}, \; \; e = \frac{b-c}{b+c}, \; \; f = \frac{c-a}{c+a} \tag{2}\label{eq2A}$$
Note that $d, e, f \lt 1$, so $d^{11}, e^{11}, f^{11} \lt 1$. There are $3! = 6$ orderings of $a$, $b$ and $c$ into decreasing sizes. With $d^{11}$, $e^{11}$ and $f^{11}$, note $3$ of the size orderings have $1$ value being positive & $2$ values are negative, while $3$ of the orderings have $1$ value being negative & $2$ values being positive. For the first case, \eqref{eq1A} is obviously true.
For the second case, a first size ordering gives
$$a \gt b \gt c \; \; \to \; \; d \gt 0, \, e \gt 0, \, f \lt 0 \tag{3}\label{eq3A}$$
Since $x \ge y \ge 0 \; \to \; x^{11} \ge y^{11}$, then if $-f \ge d$, we get $(-f)^{11} \ge d^{11} \; \to \; d^{11} + f^{11} \le 0$, so \eqref{eq1A} is true, and similarly if $-f \ge e$. Thus, \eqref{eq1A} can only possibly not be true if both $-f \lt d$ and $-f \lt e$. With $-f \lt d$, we have
$$\begin{equation}\begin{aligned}
\frac{a - c}{c + a} & \lt \frac{a - b}{a + b} \\
(a - c)(a + b) & \lt (a - b)(a + c) \\
a^2 + ab - ac - bc & \lt a^2 + ac - ab - bc \\
2ab & \lt 2ac \\
b & \lt c
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, this contradicts \eqref{eq3A}, so this ordering always satisfies \eqref{eq1A}. A second size ordering is with
$$b \gt c \gt a \; \; \to \; \; d \lt 0, \, e \gt 0, \, f \gt 0 \tag{5}\label{eq5A}$$
Here, $-d \lt e$ gives
$$\begin{equation}\begin{aligned}
\frac{b - a}{a + b} & \lt \frac{b - c}{b + c} \\
(b - a)(b + c) & \lt (b - c)(a + b) \\
b^2 + bc - ab - ac & \lt ab + b^2 - ac - bc \\
2bc & \lt 2ab \\
c & \lt a
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
but this contradicts \eqref{eq5A}, so this size ordering also works. Finally, there's
$$c \gt a \gt b \; \; \to \; \; d \gt 0, \, e \lt 0, \, f \gt 0 \tag{7}\label{eq7A}$$
With $-e \lt d \; \to \; e \gt -d \; \to -d \lt e$, we get the same result as in \eqref{eq6A}, i.e., $c \lt a$. Since this contradicts \eqref{eq7A}, this means \eqref{eq1A} also holds with this third size ordering.
This concludes showing \eqref{eq1A} is always true. Note there is nothing particularly special about the power of $11$ as any odd positive integer exponent would work as well, and also the $\le$ in \eqref{eq1A} can be replaced with just $\lt$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$(x^2+1)(y^2-1)dx+xydy=0$
Solve $(x^2+1)(y^2-1) \, dx+xy \, dy=0$
My solution :
$$\begin{align}
\frac{x^2+1}{-x} \, dx &= \frac{-y}{y^2-1} \, dy \\
\left(-x-\frac{1}{x}\right) \, dx &= -\frac{1}{2}\frac{2y}{y^2-1} \, dy \\
\int \left(-x-\frac{1}{x}\right) \, dx &= -\frac{1}{2} \int \frac{2y}{y^2-1} \, dy \\
\frac{x^2}{2}+\ln|x| &= \frac{1}{2}\ln|y^2-1| \\
y^2 &= e^{x^2+2\ln|x|}+1 \\
y &= \pm\sqrt{e^{x^2+2\ln|x|}+1}
\end{align}$$
I got these solutions using ode calculator, and I don't get why it is correct and where am I wrong?
$$y = \pm \frac{\sqrt{e^{-x^2+c_1}+x^2}}x$$
Help please.
Thanks !
| Note that $$e^{-x^2+2\ln |x|+c}+x^2=e^{-x^2+c}e^{2\ln|x|}+x^2=e^{-x^2+c}e^{\ln x^2}+x^2=x^2(e^{-x^2+c}+1)$$
So your answer is the same, just in a different form.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the permutation representation of $S_3$. I don't know how to write a solution. Is my solution ok? (Herstein "Topics in Algebra 2nd Edition") I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
The following problem is Problem 20 on p.81 in this book:
Let $G$ be the group $S_3$. Find the permutation representation of $S_3$. (Note: This gives an isomorphism of $S_3$ into $S_6$.)
I don't know how to write a solution for this problem.
Is the following solution ok?
$$S_3=\{e,\phi,\psi,\psi^2,\phi\psi,\psi\phi\},\text{ }\phi^2=e,\psi^3=e,\phi\psi=\psi^{-1}\phi.$$
$$\tau_e=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\end{pmatrix}.$$
$$\tau_\phi=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\phi&e&\psi\phi&\phi\psi&\psi^2&\psi\end{pmatrix}.$$
$$\tau_\psi=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\psi&\phi\psi&\psi^2&e&\psi\phi&\phi\end{pmatrix}.$$
$$\tau_{\psi^2}=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\psi^2&\psi\phi&e&\psi&\phi&\phi\psi\end{pmatrix}.$$
$$\tau_{\phi\psi}=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\phi\psi&\psi&\phi&\psi\phi&e&\psi^2\end{pmatrix}.$$
$$\tau_{\psi\phi}=\begin{pmatrix}e&\phi&\psi&\psi^2&\phi\psi&\psi\phi\\\psi\phi&\psi^2&\phi\psi&\phi&\psi&e\end{pmatrix}.$$
The permutation representation of $S_3$ is $\{\tau_e,\tau_\phi,\tau_\psi,\tau_{\psi^2},\tau_{\phi\psi},\tau_{\psi\phi}\}$.
If my solution is ok, then I have a next question:
If we get the Cayley Table of a group $G$, then I think we can easily (almost automatically) get the permutation representation of $G$.
Many similar problems are on pp.81-82 in this book.
What is the intent of these problems?
| The way you wrote $S_3$ it is in the form of Dihedral group $D_3$.
For $n=3$, $D_3$ is isomorphic to $S_3$.
Symmetric group $S_n$ can be defined as set of all permutations of a set having n elements or we can say that it is a collection of all one-one and onto functions from a set(n elements) to itself.
The permutation representation of $S_3$ can be given as
$$S_3=\bigg\{
\begin{pmatrix}
1 & 2 & 3 \\
1 & 2 & 3
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 \\
1 & 3 & 2
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 \\
3 & 2 & 1
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 \\
2 & 1 & 3
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 \\
2 & 3 & 1
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 \\
3 & 1 & 2
\end{pmatrix}\bigg\}$$
This is a permutation representation of $S_3$
Now, There exist a subgroup of $S_6$ which is isomorphic to $S_3$. Let us denote that subgroup of $S_6$ by H then
$$H=\bigg\{
\begin{pmatrix}
1 & 2 & 3 & 4 &5 &6 \\
1 & 2 & 3 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 & 4 &5 &6 \\
1 & 3 & 2 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 & 4 &5 &6 \\
3 & 2 & 1 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 & 4 &5 &6\\
2 & 1 & 3 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 & 4 &5 &6 \\
2 & 3 & 1 & 4 &5 &6
\end{pmatrix},\begin{pmatrix}
1 & 2 & 3 & 4 &5 &6\\
3 & 1 & 2 & 4 &5 &6
\end{pmatrix}\bigg\}$$
Here $S_3$ is isomorphic to H.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the length of the latera recta of the ellipse $3x^2+3y^2+2xy-12x+12y+4=0$? Context:
"Find the equation of the ellipse one of whose focus is $(1,-1)$ and the equation of whose directrix is $x-y+2=0$ and eccentricity is $\frac{1}{\sqrt{2}}$. Find also the length of its latera recta."
I was able to find the equation of the ellipse correctly, but I'm unable to find the length of the latera recta.
$$3x^2+3y^2+2xy-12x+12y+4=0$$
The given ellipse can't be expressed in the form $\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1$. If that was possible, I could've found out the length easily using $\frac{2a^2}{b}$ or $\frac{2b^2}{a}$.
My book only taught me how to find the length of the latera recta when the ellipse can be expressed in the form $\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1$, so I have no idea why it's asking me to find the length of the latera recta in this case.
What do I do now?
| The equation can be re-written as
$\begin{bmatrix}x&y\end{bmatrix}\color{blue}{\begin{bmatrix}3&1\\1&3\end{bmatrix}}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}-12&12\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+4=0\tag 1$
Now, by diagonalization of blue colored matrix:
$\overbrace{\begin{bmatrix}3&1\\1&3\end{bmatrix}}^{M}=\overbrace{\begin{bmatrix}\frac 1{\sqrt 2}& \frac 1{\sqrt 2}\\\frac 1{\sqrt 2}&-\frac 1{\sqrt 2}\end{bmatrix}}^{P}\overbrace{\begin{bmatrix}4&0\\0&2\end{bmatrix}}^{D}\begin{bmatrix}\frac 1{\sqrt 2}& \frac 1{\sqrt 2}\\\frac 1{\sqrt 2}&-\frac 1{\sqrt 2}\end{bmatrix}$
So $M=PDP^{-1}$. Substitute for M in $(1)$ and put $\begin{bmatrix}X\\Y\end{bmatrix}:=\begin{bmatrix}\frac 1{\sqrt 2}& \frac 1{\sqrt 2}\\\frac 1{\sqrt 2}&-\frac 1{\sqrt 2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$ to transform $(1)$ to the following familiar form of the ellipse:
$\frac{X^2}8+\frac{(Y-3\sqrt 2)^2}{16}=1$
| {
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"url": "https://math.stackexchange.com/questions/4387334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does $a_{n+1} = a_n + \frac{n^2}{a_n}$ grow? Let $\{a_n\}$ be a positive sequence such that $a_{n+1} = a_n + \frac{n^2}{a_n}$. What is the rate of growth of $a_n$?
I have seen this similar question which deals with $a_{n+1} = a_n + \frac{n^{1/2}}{a_n}$, but I am not sure how to apply that theorem of Cesaro-Stolz here.
| The answer is now completely reviewed thanks to Gary's useful comments.
*
*First note that
$$a_{n+1} = f_n(a_n)$$ where
$$f_n(x) = x + \frac{n^2}x\geq 2n,$$
for $x>0$, so that $$a_{n}\geq 2(n-1)\geq n,\tag{1}\label{1}$$
for $n\geq 2$, and the sequence is divergent.
*Now observe that $$a_{n+1}^2 = a_n^2\left(1+\frac{n^2}{a_n^2}\right)^2,$$ so that letting $b_n = a_n^2$ yields
$$b_{n+1} = g_n(b_n),$$
where
$$g_n(x) = x\left(1+\frac{n^2}x\right)^2.$$
The above function is monotonically increasing for $x\geq n^2$, and has slant asymptote $$h_n(x) = x + 2n^2.$$ The distance $g_n(x)-h_n(x)=\frac{n^4}x$ is positive and monotonically decreasing.
*Iterating on $n$ the inequality $$b_{n+1}\geq b_n + 2n^2,$$ valid, by \eqref{1}, for $n\geq 2$ we obtain that
$$b_{n+1} \geq b_2 -2 + 2\frac{n(n+1)(2n+1)}6\geq \frac{2n^3}3$$
*Using 2. we have, for large enough $n$, \begin{eqnarray}0\leq b_{n+1}-b_n-2n^2&=&g_n(b_{n})-h_n(b_n)\leq g_n\left(\frac{2(n-1)^3}3\right)-h_n\left(\frac{2(n-1)^3}3\right)\leq \\
&\leq&\frac{3n^4}{2(n-1)^3}\leq 3n.\end{eqnarray}
*From 4., thus, we get $$b_n + 2n^2 \leq b_{n+1}\leq b_n + 2n^2 + 3n.$$ Iterating on $n$ yields $$b_2 -5 + \frac{n(n-1)(2n-1)}3 \leq b_n \leq b_2 -5 + \frac{n(n-1)(2n-1)}3 + \frac{3n(n-1)}2,$$
and by the Squeeze Theorem $$b_n \sim \frac23 n^3.$$
Hence $$a_n \sim \sqrt{\frac23} n^{3/2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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What's the area of the shaded regions in the triangle below? For reference: In triangle ABC, $S_1$ and $S_2$ are areas of the shaded regions. If $S_1 \cdot{S}_2=16 cm^4$, calculate $MN$.
My progress:
$\frac{AM.DM}{2}.\frac{CN.FN}{2}=16 \implies AM.DM.CN.FN=64\\
\frac{S1}{S2} = \frac{AM.MD}{CN.FN}\\
\frac{S1}{\frac{MI.DM}{2}}=\frac{AM}{MI}\implies S1 = \frac{AM.DM}{2}\\
\frac{S2}{\frac{NI.FN}{2}}=\frac{CN}{NI}\implies S2 = \frac{CN.FN}{2}$
.....????
| $$\angle A=\angle IDM = \angle DIB = \angle IBF = \angle FIN = \angle CFN$$
$$\frac{DM}{AM}=\frac{IM}{DM}=\frac{FI}{BF}=\frac{FN}{IN}=\frac{CN}{FN}=x$$
$$DM=x AM, IM=x DM = x^2 AM,$$
$$BF=ID=\sqrt{IM^2+DM^2}=x AM \sqrt{1+x^2}, FI=x BF=x^2 AM \sqrt{1+x^2}$$
$$FI^2=IN^2+FN^2=IN^2+(x IN)^2=(1+x^2) IN^2 \Rightarrow IN=\frac{FI}{\sqrt{1+x^2}}=x^2 AM$$
$$FN=x IN=x^3 AM, CN=x FN=x^4 AM$$
$$S_1=AM\cdot DM / 2 = x AM^2 / 2, S_2=FN\cdot CN / 2= x^7 AM^2 /2$$
$$S_1 S_2 = x^8 AM^4 / 4 \Rightarrow x^2 AM = \sqrt[4]{4S_1 S_2}$$
$$MN=IM+IN=2x^2 AM=\sqrt[4]{64 S_1 S_2}=4\sqrt{2} \rm{\ cm}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Simplification of $\frac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$
Simplify
$$\dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$$
Final solution should have rational denominators.
Suppose the solution is $X$, I have tried to make up an equation for $X^2$
$$X^2 = \frac{200}{-2\sqrt{15}+2\sqrt{35}-\sqrt{21}+10}$$
My idea is that solving for $X^2$, if can be simply done, can easily give us $X$.
Please help!
| The solution below is a lot of work to do by hand but at least it does not need any bright ideas for algebraic manipulation.
The problem reduces to finding $\dfrac{1}{\sqrt7+2\sqrt5-\sqrt3}$.
Now, $\theta=\sqrt7+2\sqrt5-\sqrt3$ is a root of $x^8 - 120 x^6 + 3632 x^4 - 28800 x^2 + 256$. The extended Euclidean algorithm gives
$$
1 = \frac{1}{256}(x^8 - 120 x^6 + 3632 x^4 - 28800 x^2 + 256)+ \frac{1}{256}(-x^7 + 120 x^5 - 3632 x^3 + 28800 x)x
$$
Therefore,
$$
\dfrac{1}{\sqrt7+2\sqrt5-\sqrt3} = \frac{1}{\theta} =\frac{1}{256}(-\theta^7 + 120 \theta^5 - 3632 \theta^3 + 28800 \theta)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum and maximum value of $F=|a-2b|+|b-2c|+|c-2a|$
Let $a, b, c \in \Bbb R$ satisfy $a^2+b^2+c^2=21$. Find the minimum and maximum value of $$F=|a-2b|+|b-2c|+|c-2a|$$
I found $7\le F \le \sqrt{399}$ but couldn't prove it. I was thinking of the following inequality:
$$|x_1+x_2+\cdots+x_n|\le|x_1|+|x_2|+\cdots+|x_n|\le \sqrt{n \left(x_1^2+x_2^2+\cdots+x_n^2\right)}$$
but they are not really efficient. Does anyone know how to solve this problem or know where it first appeared?
| Maple:
"extrema(F,g,{a,b,c},'s')"
where
$F:=|a-2b|+|b-2c|+|c-2a|$,
$g:=a^{2}+b^{2}+c^{2}=21$
‘s’ are the solutions.
$min=3\sqrt{7}$,
$max=\sqrt{399}$.
The solutions are:
$a=\sqrt{7}, b=\sqrt{7}, c=\sqrt{7}$,
$a=-\sqrt{7}, b=-\sqrt{7}, c=-\sqrt{7}$;
$a=-\frac{3\sqrt{399}}{19}, b=+\frac{\sqrt{399}}{19}, c=+\frac{3\sqrt{399}}
{19}$,
$a=-\frac{3\sqrt{399}}{19}, b=+\frac{3\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$,
$a=-\frac{\sqrt{399}}{19}, b=-\frac{3\sqrt{399}}{19}, c=+\frac{3\sqrt{399}}{19}$,
$a=+\frac{\sqrt{399}}{19}, b=+\frac{3\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$,
$a=+\frac{3\sqrt{399}}{19}, b=-\frac{3\sqrt{399}}{19}, c=+\frac{\sqrt{399}}{19}$,
$a=+\frac{3\sqrt{399}}{19}, b=-\frac{\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Probability of an error going undetected with a single parity bit Problem:
A byte (which is $8$ bits) is being transmitted over an unreliable computer network. After the last bit is transmitted a parity bit is sent. The probability that a given bit is sent correctly is $0.9$. What is the probability that in transmitting a byte (including the parity bit) an error will occur and it will go undetected?
Answer:
If exactly one bit is transmitted incorrectly then the network will generate a parity error. That is, the system will detect the problem. If exactly three bits are wrong, the system will generate a parity error. However, if there are exactly two bits wrong or exactly four
bits wrong or exactly six bits wrong no parity error will be detected exactly eight bits wrong no parity error will be detected. What we have here is a binomial distribution. Let $p_e$ be the probability we seek.
\begin{align*}
n &= 9 \\
p &= \dfrac{ 9 } { 10 } \\
q &= \dfrac{ 1 } { 10 } \\
p_e &= { 9 \choose 2 }p^7 q^2 + { 9 \choose 4 }p^5 q^4
+ { 9 \choose 6 }p^3 q^6 + { 9 \choose 8 } p q^8 \\
{ 9 \choose 2 } &= \dfrac{ 9(8) } {2} = 36 \\
{ 9 \choose 4 } &= \dfrac{ 9 } { 4 } = \dfrac{ 9(8)(7)(6) } {4(3)(2)} \\
{ 9 \choose 4 } &= \dfrac{ 9(7)(6) } { 3 } = 3(7)(6) = 126 \\
{ 9 \choose 6 } &= \dfrac{ 9(8)(7) } { 3(2) } = 3(4)(7) = 84 \\
{ 9 \choose 8 } &= \dfrac{ 9 } { 1 } = 9 \\
%
p_e &= (45){\left( \dfrac{ 9 } { 10 }\right) }^7 {\left( \dfrac{ 1 } { 10 }\right) }^2 \\
&+ (126){\left( \dfrac{ 9 } { 10 }\right) }^5 {\left( \dfrac{ 1 } { 10 }\right) }^4
+ (84){\left( \dfrac{ 9 } { 10 }\right) }^3 {\left( \dfrac{ 1 } { 10 }\right) }^6 +
9\left( \dfrac{ 9 } { 10 }\right) \left( {\dfrac{ 1 } { 10 }} \right) ^8\\
%
\left( 10^9 \right) p_e &= 36(9^7) + 126(9^5) + 84(9^3) + 9(9) \\
\end{align*}
\begin{align*}
\left( 10^9 \right) p_e &= 36(4782969) + 126( 59049) + 84(729) + 81 \\
\left( 10^9 \right) p_e &= 179688375 \\
p_e &= \dfrac{ 179688375 }{ 10^9 } \\
p_e &= 0.179688375
\end{align*}
Is my solution correct?
| As requested in comments:
Your calculation of the probability of a positive even number of bit errors looks correct, assuming these are independent:
$$\sum\limits_{m=2,4,6,8} {9 \choose m}\left(\frac{1}{10}\right)^m\left(\frac{9}{10}\right)^{9-m}= 0.179688375$$
and is the probability that there is an error and it goes undetected.
The conditional probability that errors go undetected given that there is one or more errors would need to divide that by $1-\left(\frac{9}{10}\right)^{9}$ to get about $0.293331$
| {
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"url": "https://math.stackexchange.com/questions/4398331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1$, $$A = \frac{1}{4}$$
However, $$\frac{1}{4(1 + x)} + \frac{1}{2 (1 - x)^2}$$ doesn't seem to equal $$\frac{1}{(1 + x)(1 - x)^2}$$
How do we decompose this into partial fractions?
Thanks
| try this one $$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{Bx+C}{(1 - x)^2}
$$ and find $A,B,C$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $(1+\frac{1}{n^3})^{n^2}$ I have been trying to solve the limit of $y_n = (1+\frac{1}{n^3})^{n^2}$. Through graphical analysis, I have found that $$\lim_{n \to \infty} y_n = 1$$ Which can also be intuitively be understood as $n^3 \geq n^2$. Using Bernoulli's inequality, you can easily find that $$y_n \geq (1+\frac{n^2}{n^3}) \geq 1$$ I have also found that $$y_n - y_{n+1} \geq \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} - \left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^2} = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left(1 - \left( 1+\frac{1}{(n+1)^3} \right) ^{2n+1} \right) = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left( \frac{1}{(n+1)^3} \right) \left( 1 + \left( 1 + \frac{1}{(n+1)^3} \right) + \cdots + \left( 1+\frac{1}{(n+1)^3}\right)^{2n} \right) \geq 1*0*2n\geq 0 $$$$\implies yn \geq y_{n+1}$$ Thus, by using the monotone convergence theorem, we know $y_n$ converges and has a lower bound of $1$. I am however stuck at showing that $\inf{\{y_n | n \geq 1\}} = 1$, which would show that $\lim_{n \to \infty} y_n = 1$. Could I get a hint or a nudge in the right direction ?
PS: I cannot use exponential and logarithmic properties, nor l'hopital's rule, as we have not defined all these things in class
| You can use the (easily to verify) fact that $\left(1+\frac 1m \right)^m < 3$ for all $m \in \mathbb{N}$ as follows:
\begin{eqnarray*} \left(1+\frac 1{n^3} \right)^{n^2} -1
& = & \frac 1{n^3}\sum_{k=0}^{n^2-1}\left(1+\frac 1{n^3}\right)^k\\
& \leq & \frac 1{n^3}n^2\left(1+\frac 1{n^3}\right)^{n^2-1} \\
& \leq & \frac 1{n}\left(1+\frac 1{n^3}\right)^{n^3} \\
& \leq & \frac 3{n} \stackrel{n\to\infty}{\longrightarrow} 0\\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Quadratic equation question i have a multiple choice question
Let $x_1$ and $x_2$ be the real root of The quadratic equation $x^2-(c+3)x+9=0$ if $x_1<-2$ and $x_2<-2$, then ......
(a) $c<\frac{-19}{2}$ or $c>9$
(b) $\frac{-19}{2}<c\le-9$
(c) $\frac{-19}{2}<c\le-7$
(d) $-9<c<3$
(e) $c>3$
and the correct answer is (b) , here is the solution, we know that based from the quadratic equations,
$x^2-(c+3)x+9=0$ $\dots(L_1)$
$x_1+x_2=c+3$ and $x_1 x_2=9$
then we know that $x_1<-2$ and $x_2<-2$ this implies
$x_1+2 < 0$ and $x_2+2<0$
then when we add it we obtain ,
$x_1+x_2+4<0$
$c+3+4<0$ , then we get
$c<-7$ $\dots (H_1)$
when we multiply it we obtain ,
$(x_1+2)(x_2+2)>0$
$x_1x_2+2(x_1+x_2)+4>0$
$9+2(c+3)+4>0$ then we will obtain
$c>\frac{-19}{2}$ $\dots (H_2)$
Now, since we are given the roots must be $x_1<-2$ and $x_2<-2$ of $(L_1)$ this means it can be either have a twin roots or two distinct root so this means the discriminant must be larger or equal to zero
$D\geq0$
$(-(c+3))^2-4.1.9 \geq 0$
$c^2+6c+9-36 \geq 0$
$c^2+6c-27 \geq 0$
$(c-3)(c+9) \geq 0$
so to satisfy the equation , $c\leq-9$ or $c\geq3$ $\dots(H_3)$
So the value of $c$ must be the intersection of $(H_!)$,$(H_2)$,$(H_3)$ which is $\frac{-19}{2}<c\le-9$
so the answer is (b)
However, i am still not clear about the question and the logic order of this solution , this question seems like asking what is the value of c so that the value of $x_1$ and $x_2$ will be always less than-2.However the question says "if $x_1<-2$ and $x_2<-2$" that means , that we are given $x_1<-2$ and $x_2<-2$ then it will give a lot of implication of $c$ . This question seems like we do the converse proof , we finding the value of c then it imply the roots are less than 2. i am really confuse about it . the second one is , we are given this condition $x_1+2<0$ and $x_2+2<0$ . then we try to solving the value of c by adding it and multiply it , my question is , how does this step ("by adding it and multiply it") can guarantee the value of c so that the value of $x_1$ and $x_2$ will be always less than -2 ? i dont get it . When i analyze the solution and graph seems like they asking what condition of c so that the roots are always less than -2 .
|
$x^2 - (c+3)x + 9 = 0.$
The roots are $x_1, x_2,$ which may or may not be distinct.
$x_1 < -2, x_2 < -2.$
As I interpret the question, it is asking for all values of $c$ that will satisfy the above constraints.
I would have worded the solution differently.
First of all, you know that
$x_1 + x_2 = (c + 3) \implies (c + 3) < -4 \implies c < -1.$
Second of all, you know that the two roots are real, which implies that
$(c + 3)^2 \geq (4 \times 9) = 6^2 \implies c \leq -9,$
since it has been established that $c$ must be negative.
So, the problem is reduced to determining for which values of $c$ that are $\leq -9,$ will the two roots, $x_1$ and $x_2$ both be $< -2.$
To determine this, the quadratic formula must be applied.
Then,
$$x_1, x_2 = \frac{1}{2} \left[(c + 3) \pm \sqrt{(c+3)^2 - 36}\right]$$
$$= \frac{1}{2} \left[(c + 3) \pm \sqrt{c^2 + 6c - 27}\right]. \tag1 $$
At this point, since $c \leq -9,$ it is obvious that the root represented by
$$\frac{1}{2} \left[(c + 3) - \sqrt{c^2 + 6c - 27}\right]$$
will definitely be $< -2.$
Consequently, the entire problem has been reduced to determining for which values of $c \leq -9$, will the following root also be $< -2$:
$$\frac{1}{2} \left[(c + 3) + \sqrt{c^2 + 6c - 27}\right]. \tag2 $$
To attack (2) above, I first clear the $~\displaystyle \frac{1}{2}$ fraction, and isolate the expression under the radical.
Thus,
$$(c + 3) + \sqrt{c^2 + 6c - 27} < -4 \implies $$
$$\sqrt{c^2 + 6c - 27} < -4 - (c + 3) = -(7+c). \tag3 $$
Examining (3) above, I note that under the constraint that $c \leq -9,$ the LHS and RHS are both non-negative numbers. Further, I know that (in general), if $0 \leq r,s$ then $r < s \iff r^2 < s^2$. So, I can attack (3) above by squaring both sides.
This implies that I am looking for the values of $c \leq -9$ such that
$$c^2 + 6c - 27 < c^2 + 14c + 49 \iff -76 < 8c.$$
This simplies to a lower bound for $c$ of
$$\frac{-19}{2} = \frac{-76}{8} < c.$$
| {
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"url": "https://math.stackexchange.com/questions/4405656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to tackle the integral $\int_{0}^{1} \sqrt{-1+\sqrt{\frac{4}{x}-3}} d x$? $ \text {Let } y=\sqrt{-1+\sqrt{\frac{4}{x}-3}}\textrm{ then ,}$
$ \displaystyle \begin{aligned}I&=16 \int_{0}^{\infty} \frac{y^{2}\left(y^{2}+1\right) d y}{\left(y^{4}+2 y^{2}+4\right)^{2}}\\&=4\left[3 \underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}+2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}} d y}_{J}+\underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}-2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}}}_{K} d y\right] \end{aligned}\tag*{} $
Now let’s play a little trick on the integral $ J$.
$\displaystyle \begin{aligned}J &=\int_{0}^{\infty} \frac{1+\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{\infty} \frac{d\left(y-\frac{2}{y}\right)}{\left[\left(y-\frac{2}{y}\right)^{2}+6\right]^{2}} \\&=\int_{-\infty}^{\infty} \frac{d u}{\left(u^{2}+6\right)^{2}}\\ &\stackrel{u=\sqrt6 \tan \theta}{=}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sqrt{6} \sec ^{2} \theta d \theta}{\left(6 \sec ^{2} \theta\right)^{2}}\\&=\frac{\pi}{12 \sqrt{6}} \end{aligned} \tag*{} $
For the integral $ K$ , we first split the interval into two.
$ \displaystyle \begin{aligned}K &=\int_{0}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{1-\frac{1}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y+\int_{1}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}}+\int_{3}^{\infty} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}} d y \\&=\int_{\infty}^{3} \frac{d u}{\left(u^{2}-2\right)^{2}}+\int_{3}^{\infty} \frac{d v}{\left(v^{2}-2\right)^{2}} \\&=0 \end{aligned} \tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=4\left(3 \cdot \frac{\pi}{12 \sqrt{6}}\right)=\frac{\pi}{\sqrt{6}}}\tag*{} $
My Question
Is there any other substitution or method to tackle the integral?
| We can rewrite the integrand as a rational function with just a few substitutions:
$$\begin{align*}
I &= \int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}} \, dx \\[1ex]
&= 8 \int_0^1 \sqrt x \sqrt{1-x} \, \frac{dx}{(3x^2+1)^2} \tag{1} \\[1ex]
&= 16 \int_0^\infty \frac{x^2 (x^2+1)}{\left(x^4+2x^2+4\right)^2} \, dx \tag{2}
\end{align*}$$
*
*$(1)$ : substitute $x\mapsto\frac{4x^2}{3x^2+1}$
*$(2)$ : substitute $x\mapsto\sqrt{\frac{1-x}x}$
and this could probably be made even simpler. But we can evaluate the current form quite readily with complex analysis.
Let $\mathcal I=2I$ be the same integral but taken over the entire real line. Replace $x$ with complex $z$ and integrate along a semicircular contour with radius $R>\sqrt2$. It's easy to see the integral over the circular arc will vanish as $R\to\infty$, and all we need to do is add up the residues at the order-$2$ poles $z=\frac{\pm1+i\sqrt3}{\sqrt2}$.
| {
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"source": "stackexchange",
"question_score": "6",
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How many distinct straight lines can be formed that are given by equation $ax + by = 0$, where $a$, $b$ are numbers from the set {$0,1,2,3,4,5,6,7$}? My answer is 37.
*
*If $a = 0$, then only possible equation is $ y= 0$
*If $b = 0$, then only possible equation is $ x= 0$
For {$1,2,3,4,5,6,7$}, possible pairs are $7*7 = 49$. Therefore, $49+2 = 51$ equations in total but it includes several repeats:
*
*$(1,1), ..., (7,7)$ are all $x + y = 0$. Therefore, $51 - 6 = 45$
*$(1,2), (2,4), (3,6)$ are all $x + 2y = 0$. Therefore, $45 - 2 = 43$
*$(1,3), (2,6)$ are all $x + 3y = 0$. Therefore, $43 - 1 = 42$
*$(2,1), (4,2), (6,3)$ are all $2x + y = 0$. Therefore, $42 - 2 = 40$
*$(2,3),(4,6)$ are all $2x + 3y = 0$. Therefore, $40 - 1 = 39$
*$(3,1),(6,2)$ are all $3x + 1y = 0$. Therefore, $39 - 1 = 38$
*$(3,2),(6,4)$ are all $3x + 2y = 0$. Therefore, $38 - 1 = 37$
Thus, we can make $37$ unique equations.
Is my reasoning correct? Is there any generalized/better method available?
| $37$ is the right answer. Here is how I approached it -
The given equation is $ax + by = 0$; $a, b \in (0, 7)$
$(a)$ Let's start with $a = 1$. All eight values of $b$ give unique lines. Similarly for $b = 1$, we get eight unique lines but $a = b = 1$ is counted twice. That leads to $15$ unique lines.
$(b)$ Then for $~2 \leq a, b \leq 7$, we choose $a$ and $b$ such that they are relatively prime, otherwise they would give lines already counted in $(a)$.
So,
For $a = 5, 7$, we have five value of $b$.
For $a = 3$, we have four values of $b$.
For $a = 2, 4$, we have three values of $b$.
For $a = 6$, we have two value of $b$.
That leads to number of unique lines as $~15 + 22 = 37$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Minimize $\frac{x^2}{x-9} $ given that $x > 9$ Its given that $ x > 9 $ and I have to find minima of
$$ y = \frac{x^2}{(x-9) } $$
I did this using three methods.
Method 1). Let
$$ f(x) = \frac{x^2}{(x-9) } $$
Using, calculus, we get
$$ f'(x) = \frac{2x}{(x-9)} - \frac{x^2}{(x-9)^2} $$
Using this, we can find that the critical point is $ x = 18$ and $ f'(x) < 0 $ for $ 9 < x < 18 $ and $ f'(x) > 0 $ for $ x > 18 $. So, using first derivative test, $ f(x) $ has absolute minimum at $x = 18$. So, we can find that $f(18) = 36 $ is the absolute minimum.
Method 2). Since we have
$$ y = \frac{x^2}{(x-9) } $$
$$ \Longrightarrow x^2 - xy +9y = 0 $$
This is a quadratic equation. Now $x > 9$, so it follows that $ y > 0 $. Now the discriminant of this quadratic equation is
$$ D = y^2 - 36 y $$
Since roots of this quadratic equation are values of $x$ and since $x > 9$, we know that the roots are real and so we have discriminant $ D \geqslant 0 $, which means that
$$ y ( y-36) \geqslant 0 $$
Since $ y > 0 $, it follows that $ y \geqslant 36 $. So, 36 is the minimum value of $y$.
Method 3.
Now, we have
$$ \frac{(x-9) }{x^2} = \frac{1}{x} - \frac{9}{x^2} $$
$$ \frac{(x-9) }{x^2} = \Big( \frac{1}{9} \Big) \Big( \frac{9}{x} \Big) \Big( 1- \frac{9}{x} \Big) $$
We should note that since $ x > 9$, all quantities above are positive. So, we can now use AM-GM inequality
$$ \frac{9}{x} + \Big( 1- \frac{9}{x} \Big) \geqslant 2 \bigg[ \frac{9}{x}\Big( 1- \frac{9}{x} \Big) \bigg]^{1/2} $$
$$ 1 \geqslant 6 \bigg [ \frac{1}{x} \Big( 1- \frac{9}{x} \Big) \bigg]^{1/2} $$
$$ \therefore \; \bigg[ \frac{1}{x} - \frac{9}{x^2} \bigg ] \leqslant \frac{1}{36} $$
$$ \therefore \; \frac{(x-9) }{x^2} \leqslant \frac{1}{36} $$
$$ \therefore \; \frac{x^2}{x-9} \geqslant 36 $$
So, the minimum value is $36$
Are these good approaches ?
| All the methods stated by you are quite good.(I personaly would have preferred the discrimant one as it is a comparitively shorter method).
Also,I would like to state another method which is similar(but simpler)to your last method:
$\frac{x^2}{(x-9) }=\frac{x^2-81+81}{(x-9) }=18+(x-9) +\frac{81}{(x-9) }$
Following which, AM-GM can be applied as $x>9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding the symbolic expression for a particular matrix raised to the $n^{\text{th}}$ power Consider the following matrix, with $a$ being a real number between $0$ and $1$.
\begin{equation}
\text{M}=\left[
\begin{array}{cc}
1-\frac{a^2}{3} & \frac{1}{3} \left(2 a-a^2\right) \\
\frac{2 a^2}{3} & 1-\frac{2}{3} \left(2 a-a^2\right) \\
\end{array}
\right].
\end{equation}
For a general integer $n$, I am trying to compute the closed form expression for the entries of:
\begin{equation}
\text{FinalVector}= \text{M}^{n} \left(
\begin{array}{c}
1\\
1 \\
\end{array}
\right).
\end{equation}
It is very easy to compute what $\text{FinalVector}$ is like for a few specific values of $n$, but is there a way to find a formula for a general $n$?
Here are the values for $\text{FinalVector}$ for the first few $n$ (from Mathematica.)
$n = 1$
$$\left(
\begin{array}{c}
-\frac{2 a^2}{3}+\frac{2 a}{3}+1 \\
\frac{4 a^2}{3}-\frac{4 a}{3}+1 \\
\end{array}
\right)$$
$n = 2$
$$\left(
\begin{array}{c}
-\frac{2 a^4}{9}+\frac{10 a^3}{9}-\frac{20 a^2}{9}+\frac{4 a}{3}+1 \\
\frac{4 a^4}{9}-\frac{20 a^3}{9}+\frac{40 a^2}{9}-\frac{8 a}{3}+1 \\
\end{array}
\right)$$
$n = 3$
$$\left(
\begin{array}{c}
-\frac{2 a^6}{27}+\frac{2 a^5}{3}-\frac{22 a^4}{9}+\frac{122 a^3}{27}-\frac{14 a^2}{3}+2 a+1 \\
\frac{4 a^6}{27}-\frac{4 a^5}{3}+\frac{44 a^4}{9}-\frac{244 a^3}{27}+\frac{28 a^2}{3}-4 a+1 \\
\end{array}
\right)$$
$n = 4$
$$\left(
\begin{array}{c}
-\frac{2 a^8}{81}+\frac{26 a^7}{81}-\frac{16 a^6}{9}+\frac{440 a^5}{81}-\frac{812 a^4}{81}+\frac{308 a^3}{27}-8 a^2+\frac{8 a}{3}+1 \\
\frac{4 a^8}{81}-\frac{52 a^7}{81}+\frac{32 a^6}{9}-\frac{880 a^5}{81}+\frac{1624 a^4}{81}-\frac{616 a^3}{27}+16 a^2-\frac{16 a}{3}+1 \\
\end{array}
\right)$$
| There is no need to do any eigendecomposition. As observed by the others, $M$ is a rank-one update of the identity matrix:
$$
M=\pmatrix{
1-\frac{a^2}{3}&\frac{1}{3}(2a-a^2)\\
\frac{2a^2}{3}&1-\frac{2}{3}(2a-a^2)}
=I+\pmatrix{1\\ -2}
\pmatrix{-\frac{a^2}{3}&\frac{1}{3}(2a-a^2)}
=I+uv^T.
$$
Therefore, when $n\ge1$,
\begin{aligned}
M^n
&=(I+uv^T)^n\\
&=\sum_{k=0}^n\binom{n}{k}(uv^T)^k\\
&=I+\sum_{k=1}^n\binom{n}{k}(uv^T)^k\\
&=I+\sum_{k=1}^n\binom{n}{k}(v^Tu)^{k-1}uv^T\\
&=\begin{cases}
I+\frac{(1+v^Tu)^n-1}{v^Tu}uv^T&\text{if }v^Tu\ne0,\\
I+nuv^T&\text{if }v^Tu=0.\\
\end{cases}
\end{aligned}
It follows that
$$
M^n\mathbf1
=\begin{cases}
\mathbf1+\left[\frac{(1+v^Tu)^n-1}{v^Tu}(v^T\mathbf1)\right]u&\text{if }v^Tu\ne0,\\
\mathbf1+n(v^T\mathbf1)u&\text{if }v^Tu=0.\\
\end{cases}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$
Let, $x,y,z>0$ such that $xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$$
My progress:
Using the Cauchy-Schwars inequality I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(xy+yz+xz)(x+y+z)≥2(x+y+z)^2 \implies xy+yz+xz≥x+y+z$$
But, this is not always true.
I also tried
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥(xy+yz+xz)\left(\frac{(x+y+z)^2}{3}+x+y+z\right)≥2(x+y+z)^2 \implies (xy+yz+xz)(x+y+z+3)≥6(x+y+z)$$
But, again I failed.
| pqr method:
Let $p = x + y + z, q = xy + yz + zx, r = xyz = 1$.
Using AM-GM, we have $p \ge 3\sqrt[3]{r} = 3$ and $q \ge 3\sqrt[3]{r^2} = 3$.
The inequality is written as
$$q(p^2 - 2q + p) \ge 2p^2$$
or
$$(q - 2)p^2 + q(-2q + p) \ge 0.$$
Using $q\ge 3$ and $p^2 \ge 3q$, it suffices to prove that
$$(q - 2)\cdot 3q + q(-2q + p) \ge 0$$
or
$$q(q + p - 6) \ge 0$$
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Geometric sequence $a,b,c,d$ and arithmetic sequence $a, \frac{b}{2},\frac{c}{4}, d-70$ The first four terms, given in order, of a geometric sequence $a,b,c,d$ and arithmetic sequence $a, \frac{b}{2},\frac{c}{4}, d-70$, find the common ratio $r$ and the values of each $a,b,c,d$.
What I have tried:
$b=ar, c=ar^2,d=ar^3$, we have find the value of $b$ by calculating the mean from the values $1$ and $3$.$$ar=a+\frac{ar^2}{4} \implies 0 = a(r^2-4r+4a) \\0 = a(r-2)(r+2)$$
To calculate for $c$ we take the second and fourth term of the sequence.
$$\frac{ar^2}{2}=\frac{ar}{2}+ar^3-70 \implies 70 = a(r^3-\frac{r^2}{2}+\frac{r}{2})$$
However, this seems a little messy because it involves complex numbers - am I still on the right track?
| There is a concealed "joke" in this problem which is obscured by choosing to solve directly for $ \ r \ \ . $ If we instead seek the common difference $ \ \Delta \ $ between the terms of the arithmetic sequence, we might start by writing those terms as $ \ a \ \ , \ \ a + \Delta \ = \ \frac{b}{2} \ \ , \ \ a \ + \ 2 \Delta \ = \ \frac{c}{4} \ \ , \ \ a \ + \ 3 \Delta \ = \ d - 70 \ \ . $ In the geometric sequence then, the common ratio between terms is found from
$$ r \ \ = \ \ \frac{2a \ + \ 2 \Delta}{a} \ \ = \ \ \frac{4a \ + \ 8 \Delta}{2a \ + \ 2 \Delta} \ \ = \ \ \frac{a \ + \ 3 \Delta \ + \ 70}{4a \ + \ 8 \Delta} \ \ . $$
Cross-multiplication of the first two ratios (assuming $ \ a \neq 0 \ $ and $ \ a \ \neq \ -\Delta \ , \ $ which in fact cannot work for both sequences simulatneously) produces
$$ a·(4a + 8 \Delta) \ \ = \ \ (2a + 2 \Delta)^2 \ \ \Rightarrow \ \ 4a^2 \ + \ 8a \Delta \ \ = \ \ 4a^2 \ + \ 8a \Delta \ + \ 4 \Delta^2 \ \ \Rightarrow \ \ 4 \Delta^2 \ = \ 0 \ \ (!) \ \ . $$
[Yeah, I didn't believe this the first time I ran through a solution either...]
Since $ \ a \ = \ 0 \ $ will not "work" in the arithmetic sequence, we must indeed have $ \ \Delta \ = \ 0 \ \ , $ which of course means that all of the arithmetic sequence terms are identical. The equation for the geometric series ratio "collapses" to
$$ r \ \ = \ \ \frac{2a }{a} \ \ = \ \ \frac{4a }{2a } \ \ = \ \ 2 \ \ = \ \ \frac{a \ + \ 70}{4a } \ \ , $$
with the equality of the last two segments of the equation yielding $ \ 8a \ = \ a + 70 \ \Rightarrow \ a \ = \ 10 \ \ . $ Hence, the geometric sequence is $ \ 10 \ , \ 20 \ , \ 40 \ , \ 80 \ \ , $ making the arithmetic sequence
$ \ 10 \ \ , \ \ \frac{20}{2} \ \ , \ \ \frac{40}{4} \ \ , \ \ 80 - 70 \ \ . $
(Very amusing, Problem-Poser Person...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that the solutions to $(z+1)^7 = z^7$ have same real parts Prove that all the solutions to the equation $(z+1)^7=z^7$ have equal real parts, and find what this real part is equal to.
I found that $z=\frac{1}{e^{2\pi i/7}-1}, \frac{1}{e^{4\pi i/7}-1}, \frac{1}{e^{6\pi i/7}-1}, \frac{1}{e^{8\pi i/7}-1}, \frac{1}{e^{10\pi i/7}-1}, \frac{1}{e^{12\pi i/7}-1},$ but I'm not sure what to do next.
| Let $~\displaystyle w = \left(z + \frac{1}{2}\right).$
Then, you have that
$$\left(w + \frac{1}{2}\right)^7 = \left(w - \frac{1}{2}\right)^7.$$
This implies that
$$\left|w + \frac{1}{2}\right|^7 = \left|w - \frac{1}{2}\right|^7.$$
This implies that
$$\left|w + \frac{1}{2}\right| = \left|w - \frac{1}{2}\right|.$$
This implies that any satisfying value for $w$ must be equidistant between the two complex values $[(1/2) + i(0)]$ and $[-(1/2) + i(0)].$
This implies that any satisfying value for $w$ must be on the perpendicular bisector of the line segment running from $[(1/2) + i(0)]$ to $[-(1/2) + i(0)].$
This implies that any satisfying value for $w$ must have it's real portion equal to $0$.
Therefore, since $~\displaystyle w = \left(z + \frac{1}{2}\right),$ you must have that any (corresponding) satisfying value for $z$ must have its real portion equal to $(-1/2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Need pure geometric solution for proof on 10-20-40-50 angle problem $D$ is a point in $\triangle{ABC}$ so that $\angle{ABD}=10^{\circ}$, $\angle{DBC}=20^{\circ}$, $\angle{BCD}=40^{\circ}$, $\angle{DAC}=50^{\circ}$.
Find $\angle{BAD}$.
This problem is easily done with trigonometric Ceva theorem as:
$$
\begin{aligned}
&\dfrac{\sin x}{\sin50^{\circ}}
\cdot
\dfrac{\sin(60^\circ-x)}{\sin40^{\circ}}
\cdot
\dfrac{\sin20^\circ}{\sin10^\circ}=1
\\
\implies
&
\sin x \cdot \sin(60^\circ-x)
=
\dfrac
{\sin50^\circ\cdot \sin40^\circ\cdot \sin10^\circ}
{\sin20^\circ}
\\
&\qquad
=
\dfrac
{ \cos40^\circ\cdot \sin40^\circ \cdot \sin10^\circ}
{2\sin10^\circ\cdot \cos10^\circ}
=
\dfrac{\sin80^\circ}{4\cos10^\circ}
=\dfrac{1}{4}
\\
\implies
& -\dfrac{\cos60^\circ - \cos(2x-60^\circ)}2 = \dfrac{1}{4}
\\
\implies &
\cos(2x-60^\circ)=1
\\
\implies
&2x-60^\circ=0
\\
\implies
&x=\boxed{30^\circ}\ .
\end{aligned}
$$
Any idea on how to solve this problem in pure geometric approach? Thanks.
|
I just got this solution (no phantom point and few verbal words):
Let $E$ be the circumcenter of $\triangle{ACD}$, so $AE=CE=DE, \angle{DEC}=2\angle{DAC}=100^{\circ}$
$\implies \angle{EDC}=\angle{ECD}=40^{\circ}=\angle{BCD} \implies DE \parallel BC$
Extend $ED$ and intersects with $AB$ at $F$, so $\angle{AFE}=\angle{ABC}=30^{\circ}$. Make $G$ on $BC$ so that $\angle{EFG}=100^{\circ}=\angle{FEC} \implies CEFG$ is an isosceles trapezoid $\implies FG=CE$
Let $H$ be the circumcenter of $\triangle{BFG}$, so $BH=GH=FH, \angle{FHG}=2\angle{FBG}=60^{\circ} \\ \implies \triangle{FGH} \text{ is an equilateral triangle}\\ \implies FH=GH=BH=FG=CE=DE=AE, \angle{HFG}=60^{\circ} \\ \implies \angle{HBF}=\angle{HFB}=\angle{HFG}-\angle{BFG}=\angle{HFG}-(180^{\circ}-\angle{FBG}-\angle{FGB}) \\ =\angle{HFG}-(180^{\circ}-\angle{FBG}-\angle{GFE}) =60^{\circ}-(180^{\circ}-100^{\circ}-30^{\circ})=10^{\circ} \\ \implies FH \parallel BD, \angle{HBD}=20^{\circ}=\angle{DBC}=\angle{FDB} \\ \implies BDFH \text{ is an isosceles trapezoid} \\ \implies BH=DF \implies DF=DE=AE$
Check $\triangle{AEF}, \angle{AFE}=30^{\circ}, AE=DE=DF \implies EF=2AE$.
Let $O$ be the circumcenter of $\triangle{AEF}$, so $\angle{AOE}=2 \angle{AFE}=60^{\circ}, AO=FO=EO$
$\implies \triangle{AOE} \text{ is equilateral triangle}\\ \implies AO=OE=AE=DE=DF=FO=EO \\ \implies AD=AO=FD \\ \implies \angle{BAD}=\angle{AFE}=\boxed{30^{\circ}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
find all possible triples of real numbers $(a, b, c)$.
Given the real numbers $a, b, c$ so that the equation
$$x^2 + ax + b = 0$$
has exactly two different solutions, namely $p$ and $q$, and the
equation
$$x^3 + ax^2 + bx +c= 0$$
has exactly two distinct solutions, namely $p_0$ and $q_0$. If it
turns out that $p_0- p = q_0 -q$, find all possible triples of real
numbers $(a, b, c)$.
Suppose
$f(x)=x^3+ax^2+bx+c$ and $g(x)=x^2+ax+b$
Obviusly, $f(p)=f(q)=c$
$WLOG$, $p_0$ is double root of $f(x)$
Using Vieta
$$p+q=-a$$
$$pq=b$$
$$2p_0+q_0=-a$$
$$p_0^2+2p_0q_0=b$$
$$p_0^2q_0=-c$$
Observe that another form of $f(x)$ and $g(x)$
$f(x)=(x-p_0)^2(x-q_0)$ and $g(x)=(x-p)(x-q)$
$$f(p)-f(q)=(p-p_0)^2(p-q_0)-(q-p_0)^2(q-q_0)$$
$$0=(q-q_0)[(p-p_0)(p-q_0)-(q-p_0)^2]$$
$q-q_0=0$ or $(p-p_0)(p-q_0)-(q-p_0)^2=0$
For $q-q_0=0$
So we get that $p=p_0$, $q=q_0$ and $c=0$. Cause $c=0$, $f(x)=x(x^2+ax+b)$. So the roots of $f(x)$ is $p, q, 0$, since $p-q\neq 0$ so $p=0$ and $q=-a$, caused $b=0$. $(a, b, c)=(a, 0, 0)$ for $q\neq 0$
I'm confused how to use $(p-p_0)(p-q_0)-(q-p_0)^2=0$
| WLOG, $p_0$ is double root of $x^3+ax^2+bx+c$, then $2p_0+q_0=-a=p+q \Rightarrow p_0+(p_0-p)+(q_0-q)=0$. $p_0-p=q_0-q \Rightarrow p_0+2(p_0-p)=0 \Rightarrow p=\frac32 p_0 \Rightarrow q=q_0+\frac{p_0}2$.
Then $b=pq=\frac32 p_0q_0+\frac34 p_0^2$. Also $b=2p_0q_0+p_0^2$. Then $\frac32 p_0q_0+\frac34 p_0^2=2p_0q_0+p_0^2 \Rightarrow$ $\frac12 p_0q_0+ \frac14 p_0^2 \Rightarrow$ $2p_0q_0+p_0^2=0 \Rightarrow b=0$.
Then one of $p$ and $q$ is 0.
*
*$p=0 \Rightarrow p_0=0\Rightarrow c=0$. Then $q=q_0=-a$ and $a$ is free parameter. $(a,b,c)=(a,0,0)$ with $a\neq 0$.
*$q=0 \Rightarrow p=-a \Rightarrow p_0=-\frac23a \Rightarrow q_0=q-\frac{p_0}2=\frac13a \Rightarrow c=-p_0^2q_0=-\frac4{27}a^3$. $(a,b,c)=(3t,0,-4t^3)$ with $t=\frac{a}3\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I prove that, if $x,y,z>0$ and $xyz=1$, then $2(x^2+y^2+z^2)+9\geq 5(x+y+z)$
How can I prove that, if $x,y,z>0$ and $xyz=1$, then
$$2(x^2+y^2+z^2)+9\geq 5(x+y+z)$$
I used the famous inequality
$$x^2+y^2+z^2+3\geq 2(x+y+z)$$
I got $$2(x^2+y^2+z^2)+9\geq 4(x+y+z)+3\geq 5(x+y+z)$$
But, the last inequality gives $x+y+z\leq3$ which is not correct.
| See the post by @arqady here: https://artofproblemsolving.com/community/c6h1293235p6852442
It's a pretty nice solution, so I'll copy it over here:
We prove the stronger inequality
$$xyz+2(x^2+y^2+z^2)+8\ge 5(x+y+z)$$
WLOG suppose $(x-1)(y-1)\ge 0\implies xy\ge x+y-1\implies xyz\ge xz+yz-z$.
Then we want to show that (by simply plugging in this inequality to the original one)
$$2z^2+(x+y-6)z+2x^2+2y^2-5x-5y+8\geq0$$
Let $x+y=2t$. Using the well-known and easy to prove inequality $2(x^2+y^2)\ge (x+y)^2$, it remains to show that
$$z^2+z(t-3)+2t^2-5t+4\ge 0$$
It suffices to show that
$$(t-3)^2\le 4(2t^2-5t+4)$$
Upon expansion and simplification, this is equivalent to $(t-1)^2\ge 0$, which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving $x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$ for positive values with $xyz=1$
How can you prove that
$$x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$$
given that $x,y,z > 0$ and $xyz = 1$.
We can easily prove that the equality holds when $x = y = z = 1$
I could able to prove the result when one of $x,y,z$ is $1$ considering $x,y,z$ as $x,1/x, 1$ using the inequality $x + 1/x \geq 2$ for any positive number of $x$.
But couldn't able to find a full proof.
| pqr method:
Let $p = x + y + z, \, q = xy + yz + zx, \, r = xyz = 1$.
It suffices to prove that $p^2 - 2q + p \ge 2q$ or
$$p^2 + p - 4q \ge 0.$$
Using $p^3 - 4pq + 9r \ge 0$ (three degree Schur), we have
$$q \le \frac{p^3 + 9}{4p}.$$
It suffices to prove that
$$p^2 + p - 4\cdot \frac{p^3 + 9}{4p} \ge 0$$
or
$$(p - 3)(p + 3)/p \ge 0$$
which is true since $p\ge 3\sqrt[3]{r} = 3$ (AM-GM).
We are done.
Remark: Three degree Schur inequality is
$$x(x - y)(x - z) + y(y - z)(y - x) + z(z - x)(z - y) \ge 0.$$
In pqr language (or substitution), it is $p^3 - 4pq + 9r \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrating the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
Integrate the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
My attempt:
I chose shifted polar coordinates $$\begin{cases}x=2+r\cos\varphi\\ y=r\sin\varphi.\end{cases}$$ Then $$\begin{aligned}x^2+y^2&=4\\\iff (2+r\cos\varphi)^2+(r\sin\varphi)^2&=4\\\iff r(2\cos\varphi+r)&=0\end{aligned}$$ and $$\begin{aligned}(x-1)^2+y^2&=1\\\iff (1+r\cos\varphi)^2+(r\sin\varphi)^2&=1\\\iff r(4\cos\varphi)&=0\end{aligned}$$
So, I'm integrating over the set $$\left\{(\varphi,r),\frac\pi2\le\varphi\le\frac{3\pi}2, -2\cos\varphi\le r\le -4\cos\varphi\right\}.$$
The integral becomes $$\begin{aligned}\int_{\pi/2}^{3\pi/2}\int_{-2\cos\varphi}^{-4\cos\varphi}r(2+r\cos\varphi)drd\varphi&=\int_{\pi/2}^{3\pi/2}\left(r^2+\frac{r^3}3\cos\varphi\Big|_{-2\cos\varphi}^{-4\cos\varphi}\right)d\varphi\\&=\int_{\pi/2}^{3\pi/2}12\cos^2\varphi-\frac{56}3\cos^4\varphi d\varphi\\&=\int_{\pi/2}^{3\pi/2}\frac{56}3\cos^2\varphi(1-\cos^2\varphi)d\varphi-\int_{\pi/2}^{3\pi/2}\frac{20}3\cos^2\varphi d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}(2\sin\varphi\cos\varphi)^2-\frac{10}3\int_{\pi/2}^{3\pi/2}(1+\cos(2\varphi))d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}\frac{1-\cos(4\varphi)}2d\varphi-\frac{10}3\left(\varphi+\frac{\sin(2\varphi)}2\Big|_{\pi/2}^{3\pi/2}\right)\\&=\frac{14}6\left(\varphi-\frac{\sin(4\varphi)}4\Big|_{\pi/2}^{3\pi/2}\right)-\frac{10}3\pi\\&=\frac{14}6\pi-\frac{10}3\pi\\&=\frac{-6}6\pi\\&=-\pi\end{aligned}$$
UPDATE:
I've just found my mistake, instead of dividing by $3,$ I divided by $2$ after the first integration in the $r$ variable.
| In polar coordinates, the inner circle is $r=2\cos \theta$ and the integral is
$$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}r\cos\theta \> rdr d\theta=\frac83 \int_{-\pi/2}^{\pi/2}\cos^4\theta\> d\theta= \pi
$$
Per symmetry, the integration over the outer circle vanishes. Thus, the overall result is $-\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is this question solvable using the law of sines and cosines?
*
*Is it possible to (analytically) calculate the area of the following triangle using the rules of sines and cosines?
*Is it possible to calculate it using only the rule of sines?
The data given is:
$BD$ is a median to $AC$, $\angle ABD = 50^\circ$, $\angle DBC = 18^\circ$ and $BD = 4.2~\text{cm}$.
It seems to me that some data is missing.
Am I right?
| Let $x = AB$ , $ y = BC$, $\beta = \angle BDA $
Then
$\text{Area} = \dfrac{1}{2} (4.2) ( x \sin(50^\circ) + y \sin (18^\circ) ) = \dfrac{1}{2} x y \sin(68^\circ)\hspace{15pt} (1) $
And from the law of sines we have
$\dfrac{ x }{\sin \beta } = \dfrac{AD}{\sin(50^\circ)}$
$\dfrac{y}{\sin \beta } = \dfrac{DC}{\sin(18)^\circ}$
So by dividing these two equations we get
$ \dfrac{x}{y} = \dfrac{ \sin(18^\circ)}{ \sin(50^\circ) } \hspace{15pt} (2)$
Equation $(2)$ implies that
$ y = \left(\dfrac{ \sin(50^\circ)}{ \sin(18^\circ) } \right) x \hspace{15pt} (3)$
Substituting for $y$ from $(3)$ into $(1)$, and dividing by $x$ (because $x$ cannot be zero), results in
$ (4.2) ( 2 \sin(50^\circ) ) = x \dfrac{\sin(68^\circ) \sin(50^\circ) }{\sin(18^\circ)} $
Hence,
$ x = \dfrac{ 8.4 \sin(18^\circ) }{\sin(68^\circ)} $
and, consequently,
$ y = \dfrac{ 8.4 \sin(50^\circ) }{ \sin(68^\circ)}$
And finally, the area is
$\text{Area} = \dfrac{1}{2} x y \sin(68^\circ) = 35.28 \dfrac{ \sin(18^\circ) \sin(50^\circ) }{ \sin(68^\circ)} \approx 9.0074$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444509",
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"source": "stackexchange",
"question_score": "4",
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Solve the integral $\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$ Question
Solve the integral,$$\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$$
Attempt
I converted the equation in terms of $\sin(x)$ and $\cos(x)$ using the definition of $\tan(x)$ and $\cot(x)$, and then applied the substitution $t=\sin(x)$, however, this has proved itself to be quite a difficult integral to resolve in and of itself.
I would be great full for any suggestions of a more compact method. Any hints would be also be greatly appreciated.
| Integrate as follows
\begin{align}
&\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx\\
=& \int_{0}^{\frac{\pi}2} \frac{\sin^3x+ \cos^3x}{\tan^3x+\cot^3x} dx
= \int_{0}^{\frac{\pi}2} \frac{(\sin x+ \cos x)(1-\sin x\cos x)}{(\tan x+ \cot x)(\tan^2x+\cot^2x-1)} dx\\
=& \int_{0}^{\frac{\pi}2} \frac{(\sin x+ \cos x)(1-\sin x\cos x)(\sin x\cos x)^3}{1-3\sin^2x\cos^2x} dx\\
=& \ \frac14\int_{0}^{\frac{\pi}2} \frac{(\sin x+ \cos x)(2-\sin 2x)\sin^32x}{4-3\sin^22x} dx\>\>\>\>\>\>\>t= \sin x-\cos x\\
=& \ \frac14\int_{-1}^1 \frac{(1+t^2)(1-t^2)^3}{4-3(1-t^2)^2} dt
= \int_{-1}^1 \left( \frac{t^4}{12}+\frac1{36}+\frac{3t^2-1}{9(3t^4-6t^2-1)} \right)dt\\
=& \ \frac4{45}+\frac{2^{3/2}}{3^{9/4} }
\left( \tan^{-1}\sqrt{2\sqrt3+3}- \tanh^{-1}\sqrt{2\sqrt3-3}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
About Inequalities of a series I have a question in an old exam that confused me: assuming that we have this equality for all $x\in \mathbb R$
$$\frac{1}{1+e^x}=\frac12-\sum_{n=0}^\infty\frac{2x}{x^2+(2n+1)^2\pi^2}$$
prove that for all $N\in \mathbb N$ and $x>0$
$$\frac1\pi\arctan\left(\frac{(2N-1)\pi}{x}\right)\le \phi_N(x)\le \frac1\pi\arctan\left(\frac{(2N+1)\pi}{x}\right)$$
where
$$\phi_N(x)=\frac{1}{1+e^x}+\sum_{n=0}^{N-1} \frac{2x}{x^2+(2n+1)^2\pi^2}.$$
So, what I tried so far: I know that
$$\phi_N(x)=\frac12-\sum_{n=N}^\infty\frac{2x}{x^2+(2n+1)^2\pi^2}$$
and I know that for $|x|<1$ we have
$$\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$ and I can't find another idea. Any suggestion?
| Hint:
Using MVT for integrals, we have
\begin{align*}
\frac{2x}{x^2+(2n+1)^2\pi^2}
&\ge \int_n^{n+1} \frac{2x}{x^2+(2t + 1)^2\pi^2}\, \mathrm{d} t\\
&= \frac{1}{\pi}\arctan \frac{(2n + 3)\pi}{x} - \frac{1}{\pi}\arctan \frac{(2n + 1)\pi}{x}
\end{align*}
and
\begin{align*}
\frac{2x}{x^2+(2n+1)^2\pi^2}
&\le \int_n^{n+1} \frac{2x}{x^2+(2t - 1)^2\pi^2}\, \mathrm{d} t \\
&= \frac{1}{\pi}\arctan \frac{(2n + 1)\pi}{x} - \frac{1}{\pi}\arctan \frac{(2n - 1)\pi}{x}.
\end{align*}
| {
"language": "en",
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"source": "stackexchange",
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How to find the closed form of $\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…?$ In my post, I had found the exact value of the integral
$$\displaystyle I:=\int_{-\infty}^{\infty} \frac{x^{3} \sin x}{\left(1+x^{2}\right)^{2}}dx= \frac{\pi}{2 e}\tag*{} $$
by differentiating $J(a)$w.r.t. $a$ once.
$\displaystyle J(a)=\int_{-\infty}^{\infty} \frac{x \sin x}{1+a x^{2}} dx =\frac{\pi e^{-\frac{1}{\sqrt{a}}}}{a}, \textrm{ where }a>0\tag*{} $
Then I generalize $I$ to $$
I_{n}=\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…
$$
by differentiating $J(a)$ w.r.t. by $n$ times at $a=1$.
$$
\left. J^{(n)}(1) =\pi \frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}
$$
$$J^{(n)}(1) = \int_{-\infty}^{\infty} \frac{(-1)^{n} n ! x^{2n+1}\sin x}{\left(1+a x^{2}\right)^{n+1}} dx= (-1)^{n} n !I_n \tag*{}$$
Hence we can conclude that
$$ I_{n}=\left.\frac{(-1)^{n}}{n !} J^{(n)}(1) = \frac{(-1)^{n} \pi}{n !}\frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}
\tag*{} $$
Theoretically, we can find the exact value of $I_n$ by $J^{(n)}(1)$.
Urging to the closed form of $I_n$, I use Leibniz’s Rule to find $J^{(n)}(1)$
\begin{aligned}
\left.\frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right) \right|_{a=1} &= \left.\sum_{k=0}^{n}{n\choose k} \left(a^{-1}\right)^{(n-k)}\left(e^{-\frac{1}{\sqrt{a}}}\right)^{(k)}\right|_{a=1} =
\left.\sum_{k=0}^{n} {n\choose k} (-1)^{n-k}(n-k)!( e^{-\frac{1}{\sqrt{a}}} )^{(k)} \right|_{a=1}
\end{aligned}
Hence $$I_{n}=\frac{(-1)^{n}}{n !} J^{(n)}(1)=
\left. \pi\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}( e^{-\frac{1}{\sqrt{a}}} )^{(k)} \right|_{a=1}$$
By Wolframalpha, we have
$$
\frac{\partial^{k} e^{-1 / \sqrt{x}}}{\partial x^{k}}=e^{-1 / \sqrt{x}} x^{-k} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i}\left(-\frac{1}{\sqrt{x}}\right)^{j}\left(\frac{1}{2}(2+i-j-2 k)\right)_{(k)}}{i !(j-i) !}
$$
Putting $x=1$ yields
$$
\left.\frac{\partial^{k} e^{-\frac{1}{\sqrt{x}}}}{\partial x^{k}}\right|_{x=1}=e^{-1} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i}\left(\frac{1}{2}(2+i-j-2 k)\right) _{(k)}}{i !(j-i) !}
$$
Now we can conclude that
$$I_{n} = \pi\sum_{k=0}^{n} \frac{(-1)^{k}}{k!} \left.\frac{\partial^{k} e^{-\frac{1}{\sqrt{x}}}}{\partial x^{k}}\right|_{x=1}=\frac{\pi}{e} \sum_{k=0}^{n} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i+k}\left(\frac{1}{2}(2+i-j-2 k)\right)_{(k)}}{i !(j-i) !k!} $$
The closed form is rather complicated and ugly. Is there a simpler one?
| Only with Mathematica:
$$\int_{-\infty }^{\infty } \frac{x^{2 n+1} \sin (x)}{\left(1+x^2\right)^{n+1}} \, dx=\frac{1}{2} \pi (-1)^{2 n} \, _1F_2\left(n+1;\frac{1}{2},1;\frac{1}{4}\right)+\frac{1}{2} \pi \, _1F_2\left(n+1;\frac{1}{2},1;\frac{1}{4}\right)-\frac{\sqrt{\pi } (-1)^{2 n} \Gamma
\left(n+\frac{3}{2}\right) \, _1F_2\left(n+\frac{3}{2};\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)}{\Gamma (n+1)}-\frac{\sqrt{\pi } \Gamma \left(n+\frac{3}{2}\right) \,
_1F_2\left(n+\frac{3}{2};\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)}{\Gamma (n+1)}$$
MMA code:
Using:
Im[FourierTransform[x^(2 n + 1)/(1 + x^2)^(n + 1), x, -s, FourierParameters -> {1, -1}]] /. s -> 1
We have:
(Pi*HypergeometricPFQ[{1 + n}, {1/2, 1}, 1/4])/2 + ((-1)^(2*n)*Pi*HypergeometricPFQ[{1 + n}, {1/2, 1}, 1/4])/2 - (Sqrt[Pi]*Gamma[3/2 + n]*HypergeometricPFQ[{3/2 + n}, {3/2, 3/2}, 1/4])/ Gamma[1 + n] - ((-1)^(2*n)*Sqrt[Pi]*Gamma[3/2 + n]*HypergeometricPFQ[{3/2 + n}, {3/2, 3/2}, 1/4])/Gamma[1 + n]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .
| $$z^3=-8$$
$$z^3 = r^3\cdot[\cos(3\cdot\theta) + i\sin(3\cdot\theta)]$$
Let's substitute $R = r^3$, $\rho= 3\cdot\theta$ so that:
$$R\cdot\sin(\rho) = 0$$
$$R\cdot\cos(\rho) = -8$$
Since $|z^3| = R = |-8| = 8$ you can get $\rho$ by doing:
$$8\cdot\cos(\rho) = -8$$
$$\cos(\rho) = -1$$
$$\rho = (2k-1)\cdot\pi$$
where $k$ is an arbitrary integer $k \in \mathbb{Z}$.
Let's make the anti-substitution and we obtain:
$$r = \sqrt[3]{R}=2$$
$$\theta = \frac{2k-1}{3}\cdot\pi$$
So,
$$z = 2\cdot\left[\cos\left(\frac{2k-1}{3}\cdot\pi\right) + i\sin\left(\frac{2k-1}{3}\cdot\pi\right)\right]$$
Although it seems that there are infinite solutions for $z(k)$, this is not true, because $\cos()$ and $\sin()$ are periodic. Because of the fundamental theorem of algebra $z^3+8=0$ will have 3 solutions, which also matches with the $\cos()$ and $\sin()$ periodicity for $z$. Thus, possible solutions are defined this way:
For $k=0$, $z = 2\cdot\left[\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right] = 1-\sqrt{3}\cdot i$.
For $k=1$, $z = 2\cdot\left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right] = 1+\sqrt{3}\cdot i$.
For $k=2$, $z = 2\cdot[\cos(\pi) + i\sin(\pi)] = -2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the closed form of summation of binomial coefficients For positive integers $k$, I have gotten that $$\sum\limits_{i=0}^{k}\frac{(-1)^{k+i-1}}{i+1}\binom{k+i}{i}\binom{k}{i}=0.$$
But for positive integers $m$ with $1\leq m<k$, how can I get the closed form of $$\sum\limits_{i=0}^{m}\frac{(-1)^{k+i-1}}{i+1}\binom{k+i}{i}\binom{k}{i}.$$
| Suppose for we seek a closed form of
$$\sum_{q=0}^m \frac{(-1)^{q-1}}{q+1}
{k+q\choose q} {k\choose q}.$$
where $1\le m\lt k.$ This is (Iverson bracket)
$$[z^m] \frac{1}{1-z}
\sum_{q\ge 0} z^q \frac{(-1)^{q-1}}{q+1}
{k+q\choose q} {k\choose q}
\\ = \frac{1}{k+1} [z^m] \frac{1}{1-z}
\sum_{q\ge 0} z^q (-1)^{q-1}
{k+q\choose q} {k+1\choose q+1}
\\ = \frac{1}{k+1} [z^m] \frac{1}{1-z}
[w^k] (1+w)^{k+1}
\sum_{q\ge 0} z^q (-1)^{q-1}
{k+q\choose q} w^q
\\ = \frac{1}{k+1} [z^m] \frac{1}{z-1}
[w^k] (1+w)^{k+1}
\frac{1}{(1+wz)^{k+1}}.$$
The contribution from $w$ is
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{k+1}} (1+w)^{k+1} \frac{1}{(1+wz)^{k+1}}.$$
We put $w/(1+w) = v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$ to
get
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{k+1}} \frac{1}{(1+zv/(1-v))^{k+1}}
\frac{1}{(1-v)^2}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{k+1}} \frac{(1-v)^{k-1}}{(1-v(1-z))^{k+1}}.$$
This is
$$\sum_{q=0}^{k-1} (-1)^q {k-1\choose q}
{2k-q\choose k-q} (1-z)^{k-q}.$$
Applying the coefficient extractor in $z$ we find
$$\frac{(-1)^{m-1}}{k+1} \sum_{q=0}^{k-1} (-1)^q {k-1\choose q}
{2k-q\choose k-q} {k-1-q\choose m}.$$
Observe that
$${k-1\choose q} {k-1-q\choose m}
= \frac{(k-1)!}{q! \times m! \times (k-1-q-m)!}
= {k-1\choose m} {k-1-m\choose q}.$$
This will correctly evaluate to zero when $k-1-m\lt q.$ Continuing we
find
$$\frac{(-1)^{m-1}}{k+1} {k-1\choose m}
\sum_{q=0}^{k-1} (-1)^q
{2k-q\choose k-q} {k-1-m\choose q}.$$
Working with the sum we see that we may lower to $q=k-1-m$ due to the
third binomial coefficient and the condition $1\le m\lt k.$ We thus
obtain
$$\sum_{q=0}^{k-1-m} (-1)^q
{2k-q\choose k-q} {k-1-m\choose q}
\\ = [z^k] (1+z)^{2k}
\sum_{q=0}^{k-1-m} (-1)^q
\frac{z^q}{(1+z)^q} {k-1-m\choose q}
\\ = [z^k] (1+z)^{2k}
\left[1-\frac{z}{1+z}\right]^{k-1-m}
= [z^k] (1+z)^{k+1+m} = {k+1+m\choose k}.$$
Collecting everything we finally have
$$\bbox[5px,border:2px solid #00A000]{
\frac{(-1)^{m+1}}{k+1} {k-1\choose m} {k+1+m\choose k}.}$$
The OP had an extra factor $(-1)^k$ on the sum.
| {
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"url": "https://math.stackexchange.com/questions/4462359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the structure of sequence A094615 obvious? Is the triangular structure described in A094615 proven?
That is, is it proved that the following computation lasts forever?
\begin{eqnarray*}
3 &=& 2\cdot 1 + 1 \\
5 &=& 3\cdot 1 + 2 \\
\ \\
7 &=& 2\cdot 3 + 1 \\
11 &=& 2\cdot 5 + 1 = 3\cdot 3 + 2\\
17 &=& 3\cdot 5 + 2 \\
\ \\
15 &=& 2\cdot 7 + 1 \\
23 &=& 2\cdot 11 + 1 = 3\cdot 7 + 2\\
35 &=& 2\cdot 17 + 1 = 3\cdot 11 + 2\\
53 &=& 3\cdot 17 + 2\\
\vdots
\end{eqnarray*}
Thank you.
| That linked page has an explicit representation of the entries given by Lamine Ngom, namely
$$T(n,k) = 2^{n+1-k}3^{k-1} -1 $$
Should be not too hard to verify by induction that the triangular pattern continues.
| {
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Method to find the unique substitution which interchanges the original polynomials in this problem If $f$ is an $R$ to $R$ function such that $f(x^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17$ find f(x)
One way would be to assume its a polynomial equation and solve for it . Otherwise : Notice that if we replace $x$ by $1-x$ then $f(x^2-3x+5)+2f(x^2+x+3)=6x^2-2x+13$
Now we just need to solve the system of equations:
$\left\{\begin{matrix}
f(x^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17\\
f(x^2-3x+5)+2f(x^2+x+3)=6x^2-2x+13
\end{matrix}\right.$
Solving that system of equations gives $\left\{\begin{matrix}
f(x^2+x+3)=2x^2+2x+3\\
f(x^2-3x+5)=2x^2-6x+7
\end{matrix}\right.$
This gives $f(x)=2x-3$.
*
*My question is how to deduce /calculate and arrive at that there is a way (here its $x \rightarrow 1-x$ ) such that two polynomial inside the function as input can be interchanged into each other via unique substitution ? And what are these pair of polynomials called as ? They seem to be very special in the sense everything about their nature is deducable from its other one .
| You are looking for a substitution $x \mapsto g(x)$ such that $\,g^2(x) + g(x) + 3 = x^2 - 3x + 5\,$ $\,\iff g(x)\left(g(x) + 1\right) = x^2-3x+2=(x-2)(x-1)=(1-x)(2-x)\,$. The factors on both sides differ by $\,1\,$, so the natural substitutions to try are $\,g(x)=x-2\,$ or $\,g(x) = 1-x\,$. They both work, but only $\,g(x) = 1-x\,$ also satisfies $\,g^2(x)- 3 g(x) + 5 = x^2+x+3\,$ which is required for the reverse substitution to work.
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putting the values wont help either but I am not able to factorize the polynomial or manipulate it to help me.
I would be grateful if anybody could help me.
| Note that\begin{align}x^6-2\sqrt3x^5&=x^5\left(x-2\sqrt3\right)\\&=x^5\left(2-\sqrt3\right)\\&=x^4x\left(2-\sqrt3\right)\\&=x^4\end{align}and that therefore$$x^6-2\sqrt3x^5-x^4+x^3-4x^2+2x-\sqrt{3}=x^3-4x^2+2x-\sqrt3.$$Now, note that\begin{align}0&=\left(x-\left(2+\sqrt3\right)\right)\left(x-\left(2-\sqrt3\right)\right)\\&=x^2-4x+1.\end{align}So,\begin{align}x^3-4x^2+2x-\sqrt3&=x\overbrace{(x^2-4x+1)}^{\phantom0=0}+x-\sqrt3\\&=x-\sqrt3\\&=2.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solving $x^2-3-\operatorname{frac}(x)=0$
Problem. Solve the equation
$$ x^2 - 3 - \operatorname{frac}(x) = 0 $$
I tried to solve this question by moving $3$ to the other side:
$$x^2 - \operatorname{frac}(x) = 3$$
Since the fractional part of R.H.S. is $0$,
$$
\operatorname{frac}\left(x^2\right) = \operatorname{frac}(x)
\qquad\text{or}\qquad
\operatorname{frac}\left(x^2\right) = 1 - \operatorname{frac}(x).$$
Since there is only 1 case for $\operatorname{frac}(x^2)=\operatorname{frac}(x)$ which is $\operatorname{frac}\left(x^2\right)=\operatorname{frac}(x)=0$ and no value of $x$ satisfies the equation, we can rule this condition out.
We are left with $\operatorname{frac}\left(x^2\right)=1-\operatorname{frac}(x)$ and $x$ is negative, but I am stuck here.
Can anyone please give me an idea to proceed with this question?
| Since $x^2 = 3 + \operatorname{frac}(x)$, we have $3 \leq x^2 < 4$. Solving this inequality, we have either
$$ \sqrt{3} \leq x < 2 \qquad\text{or} \qquad -2 < x \leq -\sqrt{3}. $$
*
*In the former case, we have $\lfloor x \rfloor = 1$ and hence we must have
$$ x^2 = 3 + (x - 1) = x + 2. $$
However, solving this equation gives $x = -1$ or $x = 2$, none of which satisfying the restriction $\sqrt{3} \leq x < 2$. So, there is no solution of the equation in this case.
*In the latter case, we have $\lfloor x \rfloor = -2$, and so, we have
$$ x^2 = 3 + (x + 2) = x + 5. $$
Solving this equation gives two values $x = \frac{1}{2}(1 \pm \sqrt{21})$, and indeed, the choice
$$ x = \bbox[color:navy;padding:5px;border:1px navy dotted;]{\frac{1}{2}(1-\sqrt{21})} \approx -1.79129 $$
satisfies the equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A couple in a dark room pulls 2 socks each from a shared drawer. What's the probablity of them getting exactly one & two pairs without trading? A couple in a dark room pulls 2 socks each from a shared drawer. The drawer contains 4 blue, 7 red, and 3 yellow socks. The couple cannot trade socks. What is the probability of exactly one pair (of the same colour) being drawn? And what is the probability of both of them each getting a pair?
Similarly to this question I know that the probability of drawing a pair when only one person is drawing 2 socks from this drawer is
$$\frac{\binom{4}{2}}{\binom{14}{2}} + \frac{\binom{7}{2}}{\binom{14}{2}} + \frac{\binom{3}{2}}{\binom{14}{2}} = \frac{6}{91} + \frac{21}{91} + \frac{3}{91} = \frac{30}{91}$$
but I'm completely lost when it comes to drawing two pairs at once.
Additionally I know that the total amount of ways to draw these socks is $\binom{14}{2} \times \binom{12}{2} = 91\times66=6006$ since 2 socks are first drawn from 14 and then 2 socks are drawn from the remaining 12. This also means that the first person getting a pair and the second person getting any other 2 socks happen in $30\times\binom{12}{2}=1980$ out of the $6006$ ways (which is the same probability as $\frac{30}{91}$). But I'm not sure if this is useful for solving the problem.
I really need some insight for how to approach this problem as it's not the same thing as getting two pairs when drawing 4 or more socks/cards etc.
| Since true blue anil has answered the second question, I will answer the first question. What is the probability that they get exactly one pair?
The probability is
$$\frac{N}{\binom{14}{2} \times \binom{12}{2}} = \frac{N}{6006}. \tag1 $$
Assume that the people are A and B, and that A pulls two socks out first.
In computing the denominator in (1) above, for convenience, order is construed to be relevant. That is, A pulling a pair of blue socks, then B pulling a pair of red socks is construed to be distinct from A pulling a pair of red socks, then B pulling a pair of blue socks.
Therefore, the numerator, $(N)$ must be computed in a consistent manner.
Further, by symmetry, the number of combinations where A pulls a pair and B does not must equal the number of combinations where A does not pull a pair and B does pull a pair.
Therefore, I will compute twice the number of combinations where A pulls a pair and B does not.
In all of the possibilities below, I will use the syntax:
b-r to denote the number of ways that B can pull a blue and red.
b-y to denote the number of ways that B can pull a blue and yellow.
r-y to denote the number of ways that B can pull a red and yellow.
*
*A pulls a blue pair.
$\binom{4}{2} = 6~$ ways.
Then, there are 2 blue, 7 red, 3 yellow remaining.
Therefore,
b-r + b-y + r-y equals
$(2 \times 7) + (2 \times 3) + (7 \times 3) = 41.$
Overall partial sum is
$6 \times 41 = 246.$
*A pulls a red pair.
$\binom{7}{2} = 21~$ ways.
Then, there are 4 blue, 5 red, 3 yellow remaining.
Therefore,
b-r + b-y + r-y equals
$(4 \times 5) + (4 \times 3) + (5 \times 3) = 47.$
Overall partial sum is
$21 \times 47 = 987.$
*A pulls a yellow pair.
$\binom{3}{2} = 3~$ ways.
Then, there are 4 blue, 7 red, 1 yellow remaining.
Therefore,
b-r + b-y + r-y equals
$(4 \times 7) + (4 \times 1) + (7 \times 1) = 39.$
Overall partial sum is
$3 \times 39 = 117.$
Therefore,
$$N = 2 \times \left(246 + 987 + 117\right) = 2700.$$
Therefore, the probability of this happening is
$$\frac{2700}{6006} = \frac{450}{1001}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve for m in this equation without using the quadratic formula
Tangents drawn from (b,a) to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ makes angles $\theta_1$
$\theta_2$ with the x-axis. If $\tan\theta_1$$\tan\theta_2$
=2 then $b^2-a^2=....$
The equation of the tangent will be $y=mx+(a−mb)$ and since the condition of tangency is $(a-mb)^2=a^2m^2-b^2$ we get after simplify
$$m^2(b^2-a^2)-2abm+a^2+b^2=0$$
How can I find $(b^2-a^2)$ with a calculator or any easier way than the quadratic formula.
Knowing that $m_1m_2=2$
| Converting a comment to an answer, as requested ...
If all you know is that $m_1m_2=2$ (where $m_1$ and $m_2$ are the roots of the quadratic), then you can't find a specific value for $b^2−a^2$.
By Vieta's Formulas, we know that
$$\frac{a^2+b^2}{b^2−a^2}=m_1m_2=2$$
From this, we can deduce $b^2=3a^2$, so that $b^2-a^2=2a^2$, but that's as far as we can go.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\int_{0}^{\pi} \ln (b \cos x+c)$ without using Feynman’s integration technique? I shall find the integral by Feynman’s Technique Integration on a particular integral
$\displaystyle I(a)=\int_{0}^{\pi} \ln (a \cos x+1) d x,\tag*{} $
where $-1\leq a \leq 1.$
$\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\pi} \frac{\cos x}{a \cos x+1} d x, \\&=\frac{1}{a} \int_{0}^{\pi} \frac{(a \cos x+1)-1}{a \cos x+1} d x \\&=\frac{\pi}{a}-\frac{1}{a} \int_{0}^{\pi} \frac{d x}{a \cos x+1} \\&\stackrel{t=\tan \frac{x}{2}}{=} \frac{\pi}{a}-\frac{1}{a} \int_{0}^{\infty} \frac{1}{1+\frac{a\left(1-t^{2}\right)}{1+t^{2}}} \cdot \frac{2 d t}{1+t^{2}} \\&=\frac{\pi}{a}-\frac{2}{a} \int_{0}^{\infty} \frac{d t}{(1-a) t^{2}+(1+a)} \\&=\frac{\pi}{a}-\frac{2}{a \sqrt{1-a^{2}}} \tan^{-1}\left[\frac{\sqrt{1-a} t}{\sqrt{1+a}}\right]_{0}^{\infty} \\&=\frac{\pi}{a}-\frac{\pi}{a \sqrt{1-a^{2}}}\end{aligned}\tag*{} $
Integrating both sides w.r.t. $a$ yields
\begin{aligned}\int I^{\prime}(a) d a &=\pi\int\left(\frac{1}{a}-\frac{1}{a \sqrt{1-a^{2}}}\right) da \\& \stackrel{a=\sin \theta}{=} \pi\int\left(\frac{1}{\sin \theta}-\frac{1}{\sin \theta \cos \theta}\right) \cos \theta d \theta \\&=\pi\int \frac{\cos \theta-1}{\sin \theta} d \theta\\&I(a) =\pi \int \frac{-\sin ^{2} \theta}{\sin \theta(\cos \theta+1)} d \theta\\&=\pi \ln (1+\cos \theta) +C\end{aligned}
Putting $a=0$ gives $C=-\pi\ln 2$ and hence
$$
\boxed{\int_{0}^{\pi} \ln (a \cos x+1) d x =\pi \ln \left[1+\cos \left(\sin ^{-1} a\right)\right]= \pi \ln \left(\frac{1+\sqrt{1-a^{2}}}{2} \right)}
$$
I now want to generalize it to
$$I(b,c)=\displaystyle \int_{0}^{\pi} \ln (b \cos x+c),\tag*{} $$
where $c\neq 0$ and $-1\leq \frac{b}{c} \leq 1.$
$$
\begin{aligned}
I(b,c)&=\int_{0}^{\pi} \ln (b \cos x+c) \\
&=\int_{0}^{\pi} \ln \left[c\left(\frac{b \cos x}{c}+1\right)\right] \\
&=\pi \ln c+\int_{0}^{\pi} \ln \left(\frac{b}{c} \cos x+1\right) d x \\
&=\pi \ln c+I\left(\frac{b}{c}\right)
\end{aligned}
$$
Putting $a=\frac{b}{c}$ yields
$$\boxed{\int_{0}^{\pi} \ln (b \cos x+c) = \pi\left[\ln c+\ln \left(1+\sqrt{1-\frac{b^{2}}{c^{2}}}\right)\right] = \pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right)}
$$
For example,
$$
\int_{0}^{\pi} \ln (\cos x+1)=\pi \ln \left(\frac{1}{2}\right)=-\pi \ln 2;
$$
$$
\int_{0}^{\pi} \ln (\sqrt{3} \cos x+2) d x=\pi\ln \frac{3}{2}
$$
Is there any method other than Feynman’s integration technique?
| You can use Chebyshev polynomials of the second kind. The identity
$$ U_n(x) = 2^n \prod_{k=1}^{n}\left(x-\cos\frac{\pi k}{n+1}\right) $$
leads to
$$\log U_n(A) = n\log 2+\sum_{k=1}^{n}\log\left(A-\cos\frac{\pi k}{n+1}\right) $$
then to
$$ \int_{0}^{\pi}\log(A-\cos\theta)\,d\theta = -\pi\log 2+\lim_{n\to +\infty}\frac{\pi \log U_n(A)}{n+1} $$
for any $A>1$.
| {
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"source": "stackexchange",
"question_score": "7",
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Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| You have the correct answer, but here's a shorter way.
From Vieta's formulas, $a + b = 7$ and $ab = 2$
$$a^2 + b^2 = (a + b)^2 - 2ab = 7^2 - 2(2) = 49-4 = 45$$
$$a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 =45^2 - 2(2^2) = 2017$$
$$a^6 + b^6 = (a^4 + b^4)(a^2 + b^2) - (ab)^2(a^2 + b^2) = 2017 \times 45 - 2^2(45) = 90585$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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What purely real analytic techniques are there to evaluate $\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\,\mathrm{d}x$? $\newcommand{\d}{\,\mathrm{d}}$Last night, I evaluated the following integral: $$\begin{align}I:&=\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\d x\\&=\int_{-1}^1\frac{1}{(1+x^4)\sqrt{1-x^2}}\d x\\&=\frac{\pi}{2^{3/4}}
(\sin(\pi/8)+\cos(\pi/8))\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Using a "double keyhole" (as I phrase it) contour method involving a management of branch cuts and residues at infinity, here. Although I was happy to have succeeded in this, I wondered afterwards if I would have had any hope of evaluating $I$ with real analytic technique only.
The challenge:
Evaluate $I$ without use of complex analysis or even of complex arithmetic (e.g. for partial fraction decompositions involving $i$)
I posed this to some friends and they came up with the following method which I wanted to share with MSE:
$$\begin{align}I&\overset{x\mapsto\tan x}{=}\int_{-\infty}^\infty\frac{1+x^2}{(1+x^2)^2+x^4}\d x\\&\overset{x\mapsto1/x}{=}2\int_0^\infty\frac{1+x^2}{(1+x^2)^2+1}\d x\\&=2\int_0^\infty\int_0^\infty e^{-t(1+x^2)}\cos(t)\d t\d x\quad\text{Repr. with IBP}\\&=\sqrt{\pi}\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\end{align}$$
$$\begin{align}J:&=\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\\J^2&=\int_0^\infty\int_0^\infty\frac{e^{-(t+x)}\cos(t)\cos(x)}{\sqrt{tx}}\d t\d x\\&\overset{x\mapsto tx}{=}\int_0^\infty\int_0^\infty\frac{e^{-t(1+x)}\cos(t)\cos(tx)}{\sqrt{x}}\d x\d t\\&=\frac{1}{2}\int_0^\infty\frac{1}{\sqrt{x}}\cdot\frac{1+x+x^2}{(1+x)(1+x^2)}\d x\\&\overset{x\mapsto x^2}{=}\frac{1}{2}\int_0^\infty\left(\frac{1+x^2}{1+x^4}+\frac{1}{1+x^2}\right)\d x\\&=\frac{1}{2}\left[\frac{\pi}{4}\csc\left(\frac{\pi}{4}\right)+\frac{\pi}{4}\csc\left(\frac{3\pi}{4}\right)+\frac{\pi}{2}\right]\\&=\frac{\pi}{4}(1+\sqrt{2})\end{align}$$
Referencing this answer by Sangchul.
We conclude:
$$\begin{align}I&=\sqrt{\pi}\cdot\sqrt{J^2}\\&=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{4}(1+\sqrt{2})}\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Among those who helped me, who use MSE, I credit @TheSimpliFire and @KStarGamer who are much better at real integration than I am!
My question is less of a question and more of a request for a list - a list of other, purely real, methods to attack this integral. I hope the outcome of this will be an interesting selection of advanced integration techniques that I and others can learn from.
Note 1: I am aware of this posting by Quanto but it uses complex numbers.
Note 2: You must expand the cosine product as a sum of cosines and use the same integral representation (which is classically gotten from complex arithmetic but can be done with integration by parts): $$\int_0^\infty e^{-tx}\cos(t)\d t=\frac{x}{x^2+1},\,x\gt0$$
| Using symmetry of sine and letting $t=\tan x$ yields
$$
\begin{array}{l}
\displaystyle I=:\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1+\sin ^{4} x} d x=2 \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\sec ^{4} x+\tan ^{4} x} d x
=2 \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}+t^{4}} d t
\end{array}
$$
Dividing both numerator and denominator by $t^2$ yields
$$
\begin{aligned}I&=\int_{0}^{\infty } \frac{\frac{1+\sqrt{2}}{\sqrt{2}}\left(\sqrt{2}+\frac{1}{t^{2}}\right)+\frac{1-\sqrt{2}}{\sqrt{2}}\left(\sqrt{2}-\frac{1}{t^{2}}\right)}{2 t^{2}+2+\frac{1}{t^{2}}} d t\\& =\frac{1+ \sqrt{2}}{\sqrt{2}} \int_{0}^{\infty} \frac{d\left(\sqrt{2} t-\frac{1}{t}\right)}{\left(\sqrt{2} t-\frac{1}{t}\right)^{2}+2(\sqrt{2}+1)}+\frac{1-\sqrt{2}}{\sqrt{2}} \int_{0}^{\infty} \frac{d\left(\sqrt{2} t+\frac{1}{t}\right)}{\left(\sqrt{2} t+\frac{1}{t}\right)^{2}-2(\sqrt{2}-1)}\\& =\frac{1+\sqrt{2}}{2 \sqrt{\sqrt{2}+1}}\left[\tan ^{-1} \frac{\left(\sqrt{2} t-\frac{1}{t}\right)}{\sqrt{2(\sqrt{2}+1)}}\right]_{0}^{\infty}+\frac{1-\sqrt{2}}{4 \sqrt{\sqrt{2}-1}} \ln \left|\frac{\left.\sqrt{2} t+\frac{1}{t}-\sqrt{2(\sqrt{2}-1}\right) }{\sqrt{2} t+\frac{1}{t}+\sqrt{2}(\sqrt{2}-1}\right|_0^{\infty}\\&= \frac{\pi}{2} \sqrt{1+\sqrt{2}}\end{aligned} $$
| {
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"url": "https://math.stackexchange.com/questions/4474056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluate the indefinite integral $\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}$ without trigonometric substitution. In order to find
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}
$$
we set $t=\arctan x$. Then $x=\tan t$ and $dt=\frac{dx}{x^2+1}$, so
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int\frac{dt}{\sqrt{\frac{1}{\cos^2t}}}=\int\cos tdt\\
=\sin t+C=\sin(\arctan x)+C
$$
Now, since
$$
\sin(\arctan x)=\sqrt{\frac{\tan^2(\arctan x)}{\tan^2(\arctan x)+1}}
$$
the answer is $\frac{x}{\sqrt{x^2+1}}+C$.
My Question: Is there another way to find this integral without using trigonometry?
| Another way is to use power series:
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n} x^n \tag{1} $$
$$\text{by differentiation and reindexing}\quad \frac{1}{(1-x)\sqrt{1-x}} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(2n+1) x^n \tag{2}$$
$$\text{by }x\mapsto -x^2\quad \frac{1}{(1+x^2)\sqrt{1+x^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(2n+1)(-1)^{n} x^{2n} \tag{3}$$
$$\text{by termwise integration}\quad \int_{0}^{x}\frac{dt}{(1+t^2)\sqrt{1+t^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(-1)^n x^{2n+1}\tag{4} $$
and by comparing $(4)$ and $(1)$ we have that the RHS of $(4)$ is $\frac{x}{\sqrt{1+x^2}}$.
| {
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Prove $\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} dx = \frac{\pi}{2} \sec (\frac{\pi}{2n})$ without complex techniques Prove
$$\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} \,dx = \frac{\pi}{2} \sec \left(\frac{\pi}{2n}\right)$$
for all natural numbers $n \ge 2$.
There are several answers (A1 A2) to this integral but they all involve the gamma function or the beta function or contour integration etc. Can one solve this using only 'real' 'elementary' techniques? For $n = 2$ and $n = 3$ it can be solved using only elementary substitutions and partial fractions.
| Here is to integrate with partial fractions. Utilize the factorization
$ 1+y^{2n} = \prod^n_{k=1} (y^2-2y\cos a_k +1 )$, with $a_k=\frac{(2k-1)\pi}{2n}$
\begin{align}
&\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} \,dx \\
=&\ \frac12 \int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x}+\sqrt[n]{\cot x} \ {dx }
\overset{y^n=\tan x} = \frac 12\int_0^\infty\frac{n(y^n+y^{n-2})}{1+y^{2n}}dy\\
= &\ \frac12 \int_0^\infty \sum_{k=1}^n \frac{(-1)^{k+1}\sin 2a_k}{y^2-2y\cos a_k +1 }dy
=- \sum_{k=1}^n (-1)^{n-k}a_k\cos a_k\\
=&-\frac{d}{dt}\bigg(\sum_{k=1}^n (-1)^{n-k}\sin a_k t\bigg)_{t=1}= - \frac{d}{dt}\bigg(\frac{\sin \pi t}{2\cos\frac{\pi t}{2n}}\bigg)_{t=1}
= \frac\pi2 \sec\frac\pi{2n}
\end{align}
| {
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Integral $\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt$ $$I=\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt=-\frac{4}{3}G $$ G is Catalan's constant.
$$I=\int_0^1 \frac{(1+t^2)^2}{(1+t^2)(1-t^2+t^4)}\ln(t)dt=\int_0^1 \frac{(1+t^2)^2}{(1+t^6)}\ln(t)dt$$
Do series expansion:
$$I=\sum_{n=0}^\infty (-1)^n \int_0^1(1+2t^2+t^4)t^{6n}\ln(t)dt$$
Integrate term by term:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2} \right)$$
Re-organize them:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}-\frac{1}{(6n+3)^2}+\frac{1}{(6n+5)^2}\right)-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2} $$
The first part is Catalan's constant:
$$I=-G-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2}=-G-\frac{3}{9}G=-\frac{4}{3}G$$
Done.
| Addendum: This is in response to the original question asked by the OP, which was essentially why the below is true:
$$\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{1}{(6n+5)^2}\right)=\frac{10}{9}G$$
So the obvious way to split this up would be into the sums
$$\newcommand{\AA}{\mathcal{A}}
\newcommand{\BB}{\mathcal{B}}
\begin{align*}
\AA &:= \sum_{n=0}^{\infty} \frac{(-1)^n}{(6n+1)^2} = 1 - \frac 1{7^2} + \frac{1}{13^2} - \frac{1}{19^2} \cdots \\
\BB &:= \sum_{n=0}^\infty \frac{(-1)^n}{(6n+5)^2} = \frac{1}{5^2} -\frac{1}{11^2} + \frac{1}{17^2} \cdots
\end{align*}$$
Recall, by definition,
$$G := \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} \cdots$$
Notice that
$$\AA + \BB = G - \frac{1}{3^2} + \frac{1}{9^2} - \frac{1}{15^2} + \frac{1}{21^2} \cdots$$
Multiply the bottom of each term in $G$'s summation by $-3^2$. Then
$$-\frac 1 9 G = -\frac{1}{3^2} + \frac{1}{3^2} \frac{1}{3^2} - \frac{1}{3^2} \frac{1}{5^2} + \frac{1}{3^2} \frac{1}{7^2} \cdots$$
This is, once you simplify, precisely the missing terms. You can also see this as
$$\AA + \BB = G + \sum_{n=0}^\infty \frac{(-1)^n}{(6n+3)^2}$$
and
$$\sum_{n=0}^\infty \frac{(-1)^n}{(6n+3)^2}
= \sum_{n=0}^\infty \frac{(-1)^n}{3^2 (2n+1)^2}
= \frac 1 9 \sum_{n=0}^\infty \frac{(-1)^n}{ (2n+1)^2}
= \frac 1 9 G$$
Hence,
$$\AA + \BB = G + \frac 1 9 G = \frac{10}{9} G$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Rate of convergence of ${}_2F_1(an,-r,n+1,\frac{n+1}{an} )$ to $0$ Consider the following function:
\begin{align}
{}_2F_1(an,-r,n+1,\frac{n+1}{an} )
\end{align}
where $r \ge 2$ is a positive integer and $a \in (0,1)$ and ${}_2F_1$ is the hypergeometric function.
The function ${}_2F_1(an,-r,n+1,\frac{n+1}{an} )$ goes to zero as $n \to \infty$. I am interested in finding the rate at which this is happening.
By using Wolram-Alpha I found that
\begin{align}
\lim_{n \to \infty} n \cdot {}_2F_1(an,-2,n+1,\frac{n+1}{an} )&=\frac{1}{a}-1,\\
\lim_{n \to \infty} n^2 \cdot {}_2F_1(an,-3,n+1,\frac{n+1}{an} )&=-4+\frac{-2+6a}{a^2},\\
\lim_{n \to \infty} n^2 \cdot {}_2F_1(an,-4,n+1,\frac{n+1}{an} )&=\frac{3(1-a)^2}{a^2}\\
\end{align}
Basically, from play with Wolram-Alpha the behavior seems to be as follows:
\begin{align}
\begin{array}{cc}
\lim_{n \to \infty} n^{ \frac{r}{2}} \cdot {}_2F_1(an,-r,n+1,\frac{n+1}{an} )=c(r,a), & \text{ if } r \text{ even }\\
\lim_{n \to \infty} n^{\frac{r+1}{2}} \cdot {}_2F_1(an,-r,n+1,\frac{n+1}{an} )=c(r,a), & \text{ if } r \text{ odd }\\
\end{array}
\end{align}
Note that I am not so much interested in the final limit as the rate.
| The relation between contiguous hypergeometric functions
\begin{equation}
(c-a)F\left(a-1,b;c;z\right)+\left(2a-c+(b-a)z\right)F\left(a,b;c;z\right)+a(z
-1)F\left(a+1,b;c;z\right)=0
\end{equation}
can be used to prove the proposed property for $n\to\infty$ by induction
\begin{equation}
{}_2F_1(an,-r,n+1,\frac{n+1}{an} )=
\begin{cases}
\frac{c(r,a)}{n^{r/2}}+o(n^{-r/2}) & \text{ if } r \text{ even }\\ \frac{c(r,a)}{n^{(r+1)/2}}+o(n^{-(r+1)/2}) & \text{ if } r \text{ odd }
\end{cases}
\end{equation}
Denoting
\begin{equation}
f_q={}_2F_1(-q,an;n+1;\frac{n+1}{an} )
\end{equation}
we suppose that for a given positive integer $p$
\begin{align}
f_{2p}&=\frac{c(2p,a)}{n^p}+o(n^{-p})\\
f_{2p+1}&=\frac{c(2p+1,a)}{n^{p+1}}+o(n^{-p-1})
\end{align}
(verified for $p=1$).
Then, with $a=-2p-1,b=an,c=n+1,z=(n+1)/(an)$, the contiguity relation reads
\begin{equation}
(n + 2p + 2)f_{2p+2}-\frac{(2p+1)(2an-n-1)}{an}f_{2p+1}+\frac{(2p+1)(an-n-1)}{an}f_{2p}=0
\end{equation}
and thus
\begin{align}
\frac{1}{2p+1}f_{2p+2}=\left( \frac{2a-1}{an}+o(1/n) \right)\left( \frac{c(2p+1,a)}{n^{p+1}}+o(n^{-p-1}) \right)-\left(\frac{a-1}{an} +o(1/n) \right)\left( \frac{c(2p,a)}{n^p}+o(n^{-p}) \right)
\end{align}
so
\begin{equation}
f_{2p+2}=(2p+1)\left( \frac1a-1 \right)\frac{c(2p,a)}{n^{p+1}}+o(n^{-p-1})
\end{equation}
which is of the expected form with
\begin{equation}
c(2p+2,a)=(2p+1)\left( \frac1a-1 \right)c(2p,a)
\end{equation}
The same procedure for $a=-2p-2$ gives
\begin{equation}
f_{2p+3}=(2p+2)\frac{(1-a)c(2p+1,a)+(2a-1)c(p+2,a)}{an^{p+2}}+o(n^{-p-2})
\end{equation}
which corresponds to the expected form with:
\begin{equation}
c(2p+3,a)=(2p+2)\frac{(1-a)c(2p+1,a)+(2a-1)c(2p+2,a)}{a}
\end{equation}
By induction, the proposed asymptotic behavior is thus verified.
Remark: a simple expression exists for the even parameters:
\begin{equation}
c(2p,a)=(2p-1)!!\left( \frac{1-a}{a} \right)^p
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Are there any other dyadic points on the unit circle? There are four points on the unit circle with integer coordinates
$$
(\pm 1,0)\;, \; (0,\pm 1)
$$
Besides these four, are there any other points on the unit circle with coordinates from $ \mathbb{Z}[1/2] $?
If we extend the coordinate ring to $ \mathbb{Z}[1/\sqrt{2}] $ then we get the eight points on the unit circle
$$
(\pm 1,0)\;, \; (0,\pm 1) \;, \; (\pm 1/\sqrt{2},\pm 1/\sqrt{2})
$$
Besides these eight, Are there any other points on the unit circle with coordinates from $ \mathbb{Z}[1/\sqrt{2}] $?
| Using suggestion from wasn't me we can prove that these are the only four dyadic points on the unit circle. Let $ (x,y) $ be dyadic point on the unit circle then we can write
$$
x=\frac{a}{2^k} \; y=\frac{b}{2^\ell}
$$
for $ a,b \in \mathbb{Z} $ and $ k, \ell $ nonnegative integers. So we must have
$$
\frac{a^2}{2^{2k}}+\frac{b^2}{2^{2\ell}}=1
$$
Without a loss of generality assume $ k \geq \ell $. Then we must have
$$
a^2+(2^{k-\ell}b)^2=2^{2k}
$$
The only solution to this is $ a= \pm 2^k, 2^{k-\ell}b=0 $ or $ a= 0, 2^{k-\ell}b=\pm 2^k $. For a proof of this fact see
Prove there are 4 integer roots on $x^2+y^2=2^n$
So we can conclude that either $ x=\frac{a}{2^k}=\pm 1, y=\frac{b}{2^\ell}=0 $ or $ x=\frac{a}{2^k}=0, y=\frac{b}{2^\ell}=\pm 1 $.
For the second question let
$$
(x,y)=(a+\frac{b}{\sqrt{2}},c+\frac{d}{\sqrt{2}})
$$
be a $ \mathbb{Z}[1/\sqrt{2}] $ point on the unit circle. Then
\begin{align*}
x^2+y^2&=1\\
(a+\frac{b}{\sqrt{2}})^2+(c+\frac{d}{\sqrt{2}})^2&=1\\
a^2+\frac{b^2}{2}+\sqrt{2}ab+c^2+\frac{d^2}{2}+\sqrt{2}cd &=1\\
\sqrt{2}(ab+cd) &=1-(a^2+\frac{b^2}{2}+c^2+\frac{d^2}{2})\\
\end{align*}
The right hand side is an integer. In order for the left hand side to be an integer we must have $ ab+cd=0 $. Which implies
$$
1=a^2+\frac{b^2}{2}+c^2+\frac{d^2}{2}
$$
This is only possible if $ a=\pm 1 $ and everything else is $ 0 $ (2 cases) or $ c=\pm 1 $ and everything else is $ 0 $ (2 cases) or $ a=b=0 $ and $ a=\pm1, b=\pm 1 $ (4 cases). That exactly restricts us to the eight $ \mathbb{Z}[1/\sqrt{2}] $ points on the unit circle listed in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of the reciprocals of the odds (alternating) I've been trying to work out what the following infinite series converges to:
$$\frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11}...$$
But evaluating something like this is still new to me, and I haven't found any answers online (perhaps I'm not googling well enough?).So my question is, what value does it converge to and how is it derived?
| As mentioned by @KentaS, you can proceed as follows (for $|x| < 1$):
\begin{align*}
\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} & \Rightarrow \frac{1}{1 + x^{2}} = 1 - x^{2} + x^{4} - x^{6} + \ldots\\\\
& \Rightarrow \arctan(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \ldots\\\\
& \Rightarrow \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots
\end{align*}
It is worth mentioning that, even though the geometric series does not converge for $x = 1$, the expression deduced for $\arctan$ also holds true when $x = 1$. To conclude so, it suffices to apply the Leibniz test as suggested by @Hussain-Alqatari in the comments.
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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