Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Principal value of complete elliptic integral of third kind Q: How to evaluate the following integral?
$$
PV\int_k^1\frac{dt}{(x^2-t^2)\sqrt{(t^2-k^2)(1-t^2)}}
$$
where $k<x<1$. For the sake of definiteness we may assume $x=0.8, k=0.5$.
Attempt: The substitution $t^2=k^2\cos^2\theta+\sin^2\theta$ can be used to transform the integral into
$$
{1\over x^2-1}\int_0^{\pi\over2}\frac{d\theta}{\left(1-\frac{1-k^2}{1-x^2}\sin^2\theta\right)\sqrt{1-(1-k^2)\sin^2\theta}}
$$
which can be expressed as the complete elliptic integral of third kind
$$
{1\over x^2-1}\Pi\left(\frac{1-k^2}{1-x^2},\sqrt{1-k^2}\right)
$$
if $$\frac{1-k^2}{1-x^2}<1\iff x^2<k^2$$
in which case there would not be any singularity in the denominator. But in our case this is not true and the integrand is singular at
$$\theta=\arcsin\left(\sqrt{\frac{1-x^2}{1-k^2}}\right)$$
Can someone show me the right contour for this integral and the nuances that come with it?
| Using the parameter notation,
$$PV \int_k^1
\frac {dt} {(x^2 - t^2) \sqrt {(t^2 - k^2) (1 - t^2)}} =\\
\frac 1 {1 - x^2} \left( \Pi(1 - x^2, 1 - k^2) - K(1 - k^2) \right), \\
0 < k < x < 1.$$
This is independent of how $K$ and $\Pi$ are extended to arguments outside $(0, 1)$. Can be derived in a similar way to this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute $\sum\limits_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$ I have troubles finding the limit of the following series: $\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$
So far I figured it'd easier to split the sum into:
$\sum_{n=1}^\infty \frac{5}{3^{2n+1}} \sum_{n=1}^\infty \frac{4n-1}{3^{2n+1}}$
= $\sum_{n=1}^\infty 5 \cdot\frac{1}{3^{2n+1}} +\sum_{n=1}^\infty 4n-1 \cdot \frac{1}{3^{2n+1}}$
And with $\sum_{n=1}^\infty \frac{1}{w^n} = \frac{1}{w-1}$ you get the following terms:
$5\cdot \frac{1}{3^{2n+1}-1} + 4n-1\cdot \frac{1}{3^{2n+1}-1}$
which is bascially a sequence, but im stuck right here.. help is very appreciated!
| Let $F_m=\dfrac{Am+B}{3^{2m+1}}$ and
$\dfrac{4+4n}{3^{2n+1}}=F_{n-1}-F_n$
$\implies\dfrac{4+4n}{3^{2n+1}}=\dfrac{A(n-1)+B}{3^{2n-1}}-\dfrac{An+B}{3^{2n+1}}=\dfrac{9\{A(n-1)+B\}-(An+B)}{3^{2n+1}}$
$\implies4+4n=8B-9A+8An$
$\implies8A=4\iff A=\dfrac12,4=8B-9A\iff B=?$
Clearly, $$\sum_{n=1}^\infty\dfrac{4+4n}{3^{2n+1}}=\sum_{n=1}^\infty\left(F_{n-1}-F_n\right)=F_0-\lim_{n\to\infty}F_n$$
Now can you establish $$\lim_{n\to\infty}F_n=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| a) Since $\sin^2(\theta) + \cos^2(\theta) =1 $ we can write: $1=1^2= (\sin^2(\theta) + \cos^2(\theta))^2= \cos^4(\theta)+\sin^4(\theta)+2 \sin^2(\theta) \cos^2(\theta)$. Hence $\cos^4(\theta)+\sin^4(\theta) = 1-2\sin^2(\theta)\cos^2(\theta) = 1 - \frac{1}{2} \sin^2(2\theta) = 1 - \frac{1}{2} \frac{1-\cos(4\theta)}{2} $, the claim follows. b) is similar, even there you just have to remember basic goniometric ids (in particular bisection and dupliaction).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Prove $\sqrt{3} + \sqrt{5}$ is irrational If it is assumed that $\sqrt{3}$ is known to be irrational (not the case for $\sqrt{5}$), then prove that $\sqrt{3}+\sqrt{5}$ is irrational.
My approach:
Assume that $\sqrt{3}+\sqrt{5}$ is rational. Then there exist coprime integers $p$ and $q$ so that $\frac{p}{q}$ is rational and $\frac{p}{q}=\sqrt{3}+\sqrt{5}$. Thus $(\sqrt{3}+\sqrt{5})^2=2(4+\sqrt{15})=\frac{p^2}{q^2}$, which implies that $p^2$ is even, so $p$ is also even. Let $p:=2m$ for some integer $m$, then $p^2=4m^2$. Thus $q^2(4+\sqrt{15})=2m^2$, which implies that $q^2$ is even, and so $q$ is even. But this contradicts that $p$ and $q$ are coprime, and we arrive at a contradiction.
My way of proving this does not use the fact that $\sqrt{3}$ is
irrational. Please let me know if my proof is correct, and how to use
the above mentioned fact.
| If $\sqrt3+\sqrt5=x$ is rational, then $5=(x-\sqrt3)^2=x^2-2\sqrt3x+3$, and
$$\sqrt3=\frac{x^2-2}{2x}$$
But then $\sqrt3$ is also rational. However, we know that $\sqrt3$ is irrational, so $x$ is also irrational.
And we know $\sqrt3$ is irrational by the usual argument: is $\sqrt3=p/q$, with coprime integers $p$ and $q$, then $p^2=3q^2$. Hence $3$ divides $p$, but then $9$ divides $p^2$ and $3$ divides $q$, so $p$ and $q$ are not coprime, contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove that $\sum\limits_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$ using the Binomial Theorem? I have this proposition:
$$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$
How can I prove that? How to use the Binomial Theorem to solve that?
| Show that $\sum\limits_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$
Solution:
$\sum\limits_{k=0}^{n} \binom{n}{k} = \sum\limits_{k=0}^{n}\binom{n}{k+1} + \binom{n}{0}$
$\sum\limits_{k=0}^{n} \binom{n}{k} = \sum\limits_{k=0}^{n}\binom{n}{k+1} +1$
therefrore,
$\sum\limits_{k=0}^{n} \binom{n}{k+1} =\sum\limits_{k=0}^{n} \binom{n}{k} -1 $
$\sum\limits_{k=0}^{n+1} \binom{n+1}{k+1} =\sum\limits_{k=0}^{n+1} \binom{n+1}{k} -1 $ .....(1)
we know,
$\sum\limits_{k=0}^{n} \binom{n}{k} = 2^{n} $
so,
$\sum\limits_{k=0}^{n+1} \binom{n+1}{k} = 2^{n+1} $
finally (1) can be written as...
$\sum\limits_{k=0}^{n+1} \binom{n+1}{k+1} =2^{n+1} -1 $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Number of perfect squares in $ 4^{6} \cdot 6^{7} \cdot 21^{8}$ Why can't we use this approach:
rewriting, $2^{19} \cdot 7^{8} \cdot 3^{15}$, and then
using $(3\cdot 2)\cdot(4^9)\cdot(49^4)\cdot(9^7)$.
Hence, number of perfect squares is $10\cdot 5\cdot 8$.
The answer seems to differ.
| As you've noted:
$4^{6}\cdot6^{7}\cdot21^{8}=2^{\color\red{19}}\cdot3^{\color\green{15}}\cdot7^{\color\orange{8}}$
Any perfect square diving this number must be of the following type:
$(2^{0\text{ or }2\text{ or }\dots\text{ or }\lfloor\color\red{19}/2\rfloor\cdot2})\cdot(3^{0\text{ or }2\text{ or }\dots\text{ or }\lfloor\color\green{15}/2\rfloor\cdot2})\cdot(7^{0\text{ or }2\text{ or }\dots\text{ or }\lfloor\color\orange{8}/2\rfloor\cdot2})$
Hence the amount of perfect squares dividing this number is:
$(\lfloor\color\red{19}/2\rfloor+1)\cdot(\lfloor\color\green{15}/2\rfloor+1)\cdot(\lfloor\color\orange{8}/2\rfloor+1)=400$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$
Prove that the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is $\frac{x^2}{a^2-b^2}-\frac{z^2}{b^2}=1, y=0$ or $\frac{y^2}{a^2-b^2}+\frac{z^2}{a^2}=-1, z=0$.
EDIT:
Here $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is the base of the cone. Let the vertex of the cone be $(x_1,y_1,z_1)$. Let the generator be $$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}.$$
Then $z=0$ implies any point on ellipse be $(x_1-lz_1/n, y_1-mz_1/n,0)$. It lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then the point satisfies the ellipse we get $$\frac{(nx_1-lz_1)^2}{a^2}+\frac{(ny_1-mz_1)^2}{a^2}=n^2$$
Eliminating $l,m,n$ we get the equation of the cone as $$\frac{1}{a^2}(zx_1-xz_1)^2+\frac{1}{b^2}(zy_1-yz_1)^2=(z-z_1)^2.$$ How to get the locus of vertex $(x_1,y_1,z_1)$?
Edit 2
How to get the locus of the vertices in the given two forms using purely mathematical way?
I not able to solve the problem. Please help.
| You have a good start of getting an enveloping elliptic cone:
$$\frac{(Z x-X z)^2}{a^2}+\frac{(Y z-Z y)^2}{b^2}=(z-Z)^2$$
Rearrange to
$$
\begin{pmatrix}
x & y & z
\end{pmatrix}
\begin{pmatrix}
\frac{Z^2}{a^2} & 0 & -\frac{XZ}{a^2} \\
0 & \frac{Z^2}{b^2} & -\frac{YZ}{b^2} \\
-\frac{ZX}{a^2} & -\frac{ZY}{b^2} & \frac{X^2}{a^2}+\frac{Y^2}{b^2}-1
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}+
2Zz-Z^2=0$$
Due to circular symmetry of right circular cone, the quadratic terms can be diagonalized into the form of $$\lambda x'^2+\lambda y'^2+\mu z'^2$$
which means the matrix has two equal eigenvalues.
That is
$$\det
\begin{pmatrix}
\frac{Z^2}{a^2}-\lambda & 0 & -\frac{XZ}{a^2} \\
0 & \frac{Z^2}{b^2}-\lambda & -\frac{YZ}{b^2} \\
-\frac{ZX}{a^2} & -\frac{ZY}{b^2} & \frac{X^2}{a^2}+\frac{Y^2}{b^2}-1-\lambda
\end{pmatrix}=0$$
has a double root.
Also the vertex should lie on the mirror plane of the ellipse (i.e. either $X=0$, $Y=0$ or $Z=0$; that's only the case for right circular cone).
Case I: $X=0$
$$
\left( \frac{Z^2}{a^2}-\lambda \right)
\left[
\lambda^2+
\left( 1-\frac{Y^2+Z^2}{b^2} \right)\lambda-
\frac{Z^2}{b^2}
\right]=0$$
We have two positive roots and one negative root, both factors share the same root $\lambda=\dfrac{Z^2}{a^2}$.
Now
$$\frac{Z^4}{a^4}+
\left( 1-\frac{Y^2+Z^2}{b^2} \right) \frac{Z^2}{a^2}-
\frac{Z^2}{b^2}=0 $$
$$\frac{Y^2}{a^2-b^2}+\frac{Z^2}{a^2} = -1$$
which is an imaginary ellipse provided $a^2 > b^2$.
Case II: $Y=0$
$$
\left( \frac{Z^2}{b^2}-\lambda \right)
\left[
\lambda^2+
\left( 1-\frac{X^2+Z^2}{a^2} \right)\lambda-
\frac{Z^2}{a^2}
\right]=0$$
We have two positive roots and one negative root, both factors share the same root $\lambda=\dfrac{Z^2}{b^2}$.
Now
$$\frac{Z^4}{b^4}+
\left( 1-\frac{X^2+Z^2}{a^2} \right) \frac{Z^2}{b^2}-
\frac{Z^2}{a^2}
=0 $$
$$\frac{X^2}{a^2-b^2}-\frac{Z^2}{b^2} = 1$$
which is a hyperbola provided $a^2 > b^2$.
Case III: $Z=0$
$$\lambda^2
\left( \frac{X^2}{a^2}+\frac{Y^2}{b^2}-1-\lambda \right) = 0$$
The double root is $\lambda=0$, the cone degenerates to a flat plane $z=0$.
The vertex is any point on
$$Z=0$$
Further points to be noticed:
The conics $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \cap z=0$ and $\frac{x^2}{a^2-b^2}-\frac{z^2}{b^2}=1 \cap y=0$ are known as the focal ellipse and focal hyperbola of confocal quadrics
$$\frac{x^2}{a^2+t}+\frac{y^2}{b^2+t}+\frac{z^2}{t}=1$$
where $a^2>b^2$.
See also another answer with picture of mine
.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there are infinitely many natural numbers $a$ with a specific property
Prove that there are infinitely many natural numbers $a$ with the
following property: the number $n^4 + a$ is not prime for any
natural number $n$.
I only need a hint, how to start with.
| $(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$
We would like this to be of the form $x^4+w$, so we need $a=-c$.
So the product is equal to: $x^4+(b+d-a^2)x^2+(ad-ab)x+bd$, so we need $b=d$ or $a=0$. But if $a=0$ we'll need $b=-d$ later on, and thatwould just yield $x^4-d^2$.
So the product is equal to: $x^4+(2b-a^2)x^2+bd$.
So the polynomial $(x^2+ax+b)(x^2-ax+b)=x^4+(2b-a^2)x^2+b^2$ works whenever $2b=a^2$.
Therefore the polynomial we want is:
$(x^2+2ax+2a^2)(x^2-2ax+2a^2)=x^4+4a^4$
We just have to find the values of $a$ so that each of the factors never takes on the value $1$ or $-1$ when $x$ is an integer.
So we find the discriminants of $x^2+2ax+2a^2\pm1$ and $x^2-2ax+2a^2\pm1$.
They take on values $4a^2-4(2a^2+1)=4(-1-a^2)$ and $4a^2-4(2a^2-1)=4(1-a^2)$.
Clearly both of these discriminants are negative if $a>1$.
Therefore $x^4+4a^4$ is not an integer for any $x\in\mathbb Z$ and $a\geq 2\in \mathbb Z$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $a^2\equiv b^2\pmod m$ then $a\equiv b \pmod m$ or $a\equiv-b \pmod m$ Prove that if $a^2\equiv b^2\pmod m$ then $a\equiv b \pmod m$ or $a\equiv-b \pmod m%$
since $a^2\equiv b^2\pmod m$
then $m\mid a^2-b^2$ and $a^2-b^2=mc$ for some integer $c$.
I am not really sure where to go from here and was thinking maybe to use contradiction?
| $a^2 \equiv b^2 \pmod{m}$ means $m|(a+b)(a+b)$.
It does not mean that $m|a+b$ or $m|a-b$. (This happens only when either $\gcd(m,a+b)=1$ or $\gcd(m,a-b)=1$).
And hence generally we cannot say $a\equiv b\pmod m$ or $a \equiv -b\pmod m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$ Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$
I came across this result
while trying to solve this:
inductive proof for $\binom{2n}{n}$
My proof is cumbersome,
so I hope that
someone can come up
with a more elegant proof.
Note:
I know that
$\sum_{k=0}^{n}\binom{n}{k}^2
=\binom{2n}{n}
$.
| Vandermonde's Identity
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}^{\large2}
&=\sum_{k=0}^n\binom{n}{k}\binom{n}{n-k}\\
&=\binom{2n}{n}
\end{align}
$$
and
$$
\begin{align}
\sum_{k=1}^n\binom{n}{k}\binom{n}{k-1}
&=\sum_{k=1}^n\binom{n}{k}\binom{n}{n-k+1}\\
&=\binom{2n}{n+1}\\
&=\frac{n}{n+1}\binom{2n}{n}
\end{align}
$$
which proves the result.
Another Approach Copied From This Answer
Lemma:
$$
\sum_{k=1}^n\binom{n}{k}\binom{n}{k-1}=\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2\tag{1}
$$
Proof:
Since $\binom{n}{k-1}=\frac{k}{n-k+1}\binom{n}{k}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{n-k+1}\binom{n}{k}$. Therefore,
$$
\frac{n-k+1}{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]\binom{n}{k-1}=\binom{n}{k}\binom{n}{k-1}\tag{2}
$$
Since $\binom{n}{k}=\frac{n-k+1}{k}\binom{n}{k-1}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{k}\binom{n}{k-1}$. Therefore,
$$
\frac{k}{n+1}\left[\binom{n}{k-1}+\binom{n}{k}\right]\binom{n}{k}=\binom{n}{k-1}\binom{n}{k}\tag{3}
$$
Adding $(2)$ and $(3)$ and cancelling yields
$$
\frac{n-k+1}{n+1}\binom{n}{k-1}^2+\frac{k}{n+1}\binom{n}{k}^2=\binom{n}{k-1}\binom{n}{k}\tag{4}
$$
Summing $(4)$ over $k$, and substituting $k\mapsto k+1$ in the leftmost sum, gives
$$
\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2=\sum_{k=1}^n\binom{n}{k-1}\binom{n}{k}\tag{5}
$$
QED
| {
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Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .
| We show the right side of the inequality, that is,
\begin{align*}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2,
\end{align*}
which is equivalent to
\begin{align*}
\frac{b+c-a}{b+c}+\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b} > 1.
\end{align*}
Let $x=b+c-a$, $y=c+a-b$, and $z=a+b-c$. Then $x>0$, $y>0$, and $z>0$. Moreover,
\begin{align*}
\frac{b+c-a}{b+c}+\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b} &= \frac{x}{x+\frac{y+z}{2}}+\frac{y}{y+\frac{x+z}{2}}+\frac{z}{z+\frac{x+y}{2}}\\
&=\frac{x^2}{x^2+x\frac{y+z}{2}}+\frac{y^2}{y^2+y\frac{x+z}{2}}+\frac{z^2}{z^2+z\frac{x+y}{2}}\\
&\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+xy+xz+yz} > 1.
\end{align*}
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find area of the triangle ABC, given the coordinates of vertices in plane
$A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.
I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.
I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.
I do not understand how this answer is achieved.
| Follow-through
As you have said before, the side lengths of $\triangle ABC$ is $AB=AC=5\sqrt{5}$, $BC=\sqrt{10}$, using Heron's formula, we can compute the answer.
Heron's formula states that given side lengths $a,b,c$ of $\triangle ABC$, the area is given $$\sqrt{s(s-a)(s-b)(s-c)}\tag{1}$$
Where $s$ Is the semi perimeter. ($s=\frac {a+b+c}{2}$).
So in your case, we have $$a=5\sqrt{5},b=5\sqrt{5},c=\sqrt{10}\tag{2}$$
The semi perimeter is $$\frac {10\sqrt{5}+\sqrt{10}}{2}\tag{3}$$ and plugging in the values, we have $$\sqrt{\frac {10\sqrt{5}+\sqrt{10}}{2}\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-\sqrt{10}\right)}=\boxed{17.5}\tag{4}$$
| {
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"answer_id": 0
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Changing form of equation of ellipse $4x^2 + 9y^2 - 32x - 36y + 64 = 0$ into standard form My homework question is:
Find the vertices and foci of the ellipse whose equation is given by:
$$4x^2 + 9y^2 - 32x - 36y + 64 = 0.$$
I'm trying to convert it into the standard form so I can get $c^2$.
currently I have: $$4(x-4)^2 + 9(y-2)^2 = -164$$
(youtube told me to add $64$ and $36$ to make perfect square)
after this point I am confused because $\frac{9(y-2)^2}{-164}$ seems out of the scope of my precalc course and I have a feeling I made a mistake.
any help is appreciated :D
| $4x^2 +9y^2 -32x -36y +64 = 4(x^2 - 8x +16-16) +9(y^2 - 4y +4 - 4) +64 = 4(x-4)^2 + 9(y-2)^2 - 64 -36 + 64 = 0$.
Finally, $4(x-4)^2 + 9(y-2)^2 = 36$, where the foci are deduced by dividing by 36:
\begin{equation}
\frac{1}{9}(x-4)^2 + \frac{1}{4}(y-2)^2 = 1
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $abc$ where $a^2+b^2+c^2=144$ and $ab+bc+ca=144$ Total surface area of a cuboid is 288 sq.cm. and length of a diagonal of it is 12 cm. Find its volume.
This is the question. We know that diagonal =$a^2+b^2+c^2$ and surface area =$2(ab+bc+ca)$ and volume =$abc$ . So finally we have to find the value of $abc$ where $$a^2+b^2+c^2=144$$ and $$ ab+bc+ca =144 .$$ Somebody please help me.
| Note that
$$(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)=2(144-144)=0$$
which implies that $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Given that $n$ is even, find a closed-form expression for $\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k}$ Given that $n$ is even, find a closed-form expression for
$$\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k},$$
or, in other words, for the sum
$$2^2\cdot 2\cdot\binom n2 + 2^4\cdot 4\cdot\binom n4 + 2^6\cdot 6\cdot\binom n6 + \cdots + 2^n\cdot n\cdot\binom nn.$$
You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful.
I am stuck on this problem! I tried using the identity but it only confuses me more. All solutions are greatly appreciated! Thanks!
| $$
S = \sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k} = 2 n \sum_{k=1}^{n/2} 2^{2k-1}\cdot \binom{n-1}{2k-1}
$$
We have by the Binomial theorem:
$$
B = 2n \sum_{k=0}^{n-1} 2^{k}\cdot \binom{n-1}{k} = 2n \cdot 3^{n-1}
$$
and also (n even)
$$
R = 2n \sum_{k=0}^{n-1} (-1)^{n-1-k}2^{k}\cdot \binom{n-1}{k} \\= 2n [ - 2^0 \cdot \binom{n-1} {0} + 2 \cdot \binom{n-1} {1} - 2^2 \cdot \binom{n-1} {2} + \cdots]= 2n
$$
Adding the two gives $B+R = 2S$, hence
$S = n (3^{n-1} +1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of solutions of $2^u=3^l+1$ I want to prove that the unique solution for this equation with $u,l\ge1$ are $u=2$ and $l=1$. I only can prove there doesn't exist a solution if $3\mid u$ or $4\mid u$.
| We have one solution $2^2 - 3^1 =1 $ which is also $2^2 -3^1 =3^2-2^3 $
Looking for more solutions, with $u,v \ge 1$
$$2^{2+u} -3^{1+v} =3^2-2^3 $$
Looking at this modulo $8$
$$2^{2+u} -3^{1+v} \overset{?}\equiv 3^2-2^3 \pmod 8 \\
0 -3^{1+v} \overset{?}\equiv 3^2- 0 \pmod 8 \\
-3^{1+v} \overset{?}\equiv 1 \pmod 8 $$
For odd and even $v$ separated shows, there is no solution:
$$ \begin{array} {rlll}
-9^t &= -1 &\not \equiv 1 \pmod 8 &\qquad \text{for odd $v=1+2t$} \\
-3\cdot 9^t &= -3& \not \equiv 1 \pmod 8 &\qquad \text{for even $v=2t$} \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$
So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{b^2}\int\frac{d \theta}{\cos^2\theta \left(\frac{a^2}{b^2} \tan^2\theta+1 \right)} =\left[u=\frac{a}{b}\tan\theta, du=\frac{a}{b\cos^2\theta} d\theta \right ]\\=\frac{1}{b^2}\int\frac{b}{a\left(u^2+1 \right)}du=\frac{1}{ab}\int\frac{du}{u^2+1}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )+C$$
Then:
$$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan (2\pi) \right )-\frac{1}{ab} \arctan \left(\frac{a}{b}\tan 0 \right )=0$$
Which is incorrect (the answer should be $2\pi/ab$ for $a>0,b>0$).
On the one hand, the substitution is correct, as well as the indefinite integral itself (according to Wolfram it is indeed $\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )$ ), but on the other hand I can see that had I put the limits during the substitution I'd get $\int\limits_{0}^{0} \dots = 0$ because for $\theta = 0 \to u=0$ and for $\theta = 2\pi \to u=0$.
Why is there a problem and how can I get the correct answer?
Edit: Here is Wolfram's answer:
Wolfram is correct because $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$ is the area of an ellipse (defined by $x=a\cos t , y=b\sin t$), that is $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\pi ab$$
| You have everything right up to
$$
\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))-\frac{1}{ab}\arctan(\frac{a}{b}\tan(0))
$$
Now $\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))$ is $2\pi$ because the $\arctan$ and the $\tan$ are inverse functions.
So we get
$$
\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))-\frac{1}{ab}\arctan(\frac{a}{b}\tan(0))=\frac{1}{ab}2\pi-0
$$
or
$$
\frac{2\pi}{ab}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find 7-tuples of pairwise distinct positive integers such that the sum of squares of first 4 equals sum of squares of last 3 As already stated in title, find 7-tuples ($a_1,a_2,a_3,a_4,b_1,b_2,b_3$) of pairwise distinct positive integers such that
$$a_1^2+a_2^2+a_3^2+a_4^2=b_1^2+b_2^2+b_3^2$$
This came in RMO 2016 Delhi paper where one was asked to prove that infinite such tuples exist. I have no idea how to do so.
| For Diophantine equation.
$$a^2+b^2+c^2+d^2=x^2+y^2+z^2$$
You can record a parameterization.
$$a=2(p+s+r-t-q)k$$
$$b=k^2+t^2+q^2-p^2-s^2-r^2$$
$$c=p^2+s^2+r^2+t^2-k^2-q^2-2(p+s+r-q)t$$
$$d=p^2+s^2+r^2+q^2-k^2-t^2-2(p+s+r-t)q$$
$$x=p^2+k^2+t^2+q^2-s^2-r^2+2(s+r-t-q)p$$
$$y=s^2+k^2+t^2+q^2-p^2-r^2+2(p+r-t-q)s$$
$$z=r^2+k^2+t^2+q^2-p^2-s^2+2(p+s-t-q)r$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Power series representation of $f(x) = \frac{x-1}{x+2}$ What I did so far:
\begin{align*}
\frac{x-1}{x+2}&=(\frac{x-1}{2})\frac{1}{1-(-\frac{x}{2})} \\
&=(\frac{x-1}{2})\sum_{n=1}^\infty (-\frac{x}{2})^n\\
&= (x-1)\sum_{n=1}^\infty \frac{-x^n}{2^{n+1}}
\end{align*}
But i'm stuck there.
| But then you're already done...That can't be put in any form that will be nicer imo.
You could also try
$$\frac{x-1}{x+2}=1-\frac3{x+2}=1-\frac32\frac1{1+\frac x2}=1-\frac32\sum_{n=0}^\infty\frac{(-1)^nx^n}{2^n}=\sum_{n=0}^\infty a_nx^n$$
with
$$a_0=1-\frac32=-\frac12\;,\;\;a_n=\frac{3(-1)^{n+1}}{2^{n+1}}\;,\;\;n\ge1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$ Given that $x^n - \frac1{x^n}$ is expressible as a polynomial in $x - \frac1x$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$
My brain is not working with me today! I don't know how to start this problem. Solutions are greatly appreciated.
| $$\left(x-\frac{1}{x}\right)^5 = x^5-5x^3+10x-\frac{10}{x}+\frac{5}{x^3}-\frac{1}{x^5} $$
$$\left(x-\frac{1}{x}\right)^3 = x^3-3x+\frac{3}{x}-\frac{1}{x^3}$$
hence
$$ \left(x-\frac{1}{x}\right)^5 + 5\left(x-\frac{1}{x}\right)^3 +5\left(x-\frac{1}{x}\right)=x^5-\frac{1}{x^5}$$
and
$$ P(z) = \color{red}{z^5+5z^3+5z}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$
I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this:
Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.
For $n = 1$,
$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$
and
$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$
Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.
| Another method (maybe overkill :D): The sum $S(n)=\sum_{k=1}^nP(k)$ in which $P(k)$ is a polynomial of $l^{th}$ degree can be expressed by a polynomial in $n$ of degree $l+1$. For your problem $P(k)=(k+2)(k+4)$ and has degree 2. Hence the sum can be expressed as a polynomial $G(n)=a_3n^3+a_2n^2+a_1n+a_0$. Now calculate four terms in your sum and solve the system $G(i)=a_3i^3+a_2i^2+a_1i+a_0=S(i)$. This method is more powerful, as it allows you to derive arbitrary sums over polynomial expresssions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Understanding Burnside's theorem and using it to solve an elementary problem. I would like to know how to solve this problem using Burnside's theorem:
A bracelet is to be made by threading four identical red beads and four identical yellow beads onto a hoop. How many different bracelets can be made?
Furthermore how do I know when to apply this theorem? How exactly does it work?
| Before we start let us observe that OP asks for a bracelet rather than
a necklace which means we have dihedral symmetry rather than just
rotational symmetry. We can use either the Burnside lemma or the Polya
Enumeration Theorem on this, both require the cycle index $Z(D_8)$ of
the dihedral group $D_8$.
Now to compute the cycle index we start with the rotations. Let the
slots be numbered zero to seven. There is the identity for a
contribution of $a_1^8.$ The rotation that takes $0$ to $1$ makes a
contribution of $a_8.$ The one that takes $0$ to $2$ makes a
contribution of $a_4^2.$ The one that takes $0$ to $3$ contributes
$a_8.$ The one that takes $0$ to $4$ contributes $a_2^4$. From $0$ to
$5$ we get $a_8.$ From $0$ to $6$ we get $a_4^2.$ Finally $0$ to $7$
contributes $a_8.$
Next do the reflections. We have four reflections about an axis
passing through the centers of two opposite edges, each giving
$a_2^4.$ We also have four reflections passing through the centers of
opposite vertices for a contribution of $a_1^2 a_2^3.$
Collect the reflections and the rotations to obtain the cycle
index
$$Z(D_8) = \frac{1}{16}
(a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8 +
4 a_2^4 + 4 a_1^2 a_2^3)
\\ = \frac{1}{16}
(a_1^8 + 5 a_2^4 + 2 a_4^2 + 4 a_8 +
4 a_1^2 a_2^3).$$
Now to apply Burnside we must compute the number of colorings
containing four red beads and four yellow beads fixed by each type of
permutation. For $a_1^8$ we must choose four red singletons, giving
${8\choose 4}.$ For $a_2^4$ we must choose two of the four cycles to
be red, for a contribution of $5\times {4\choose 2}.$ For $a_4^2$ we
have two possiblities giving $2\times 2.$ No contribution from $a_8.$
Finally for $a_1^2 a_2^3$ the singletons are either both red or both
yellow, giving $2\times 4\times {3\choose 2}.$ We get for the answer
$$\frac{1}{16}
\left({8\choose 4} + 5\times {4\choose 2}
+ 2\times 2 + 2\times 4\times {3\choose 2}\right)
= 8.$$
Here we used the fact that to be fixed by a given permutation an assignment must be constant on the cycles.
The same result can be obtained from the Polya Enumeration Theorem
(PET). We have
$$Z(D_8)(R+Y) =
1/16\, \left( R+Y \right) ^{8}+{\frac {5\, \left( {R}^{2}+{Y}^{
2} \right) ^{4}}{16}}\\+1/8\, \left( {R}^{4}+{Y}^{4} \right) ^{2}
+1/4\,{R}^{8}+1/4\,{Y}^{8}+1/4\, \left( R+Y \right) ^{2}
\left( {R}^{2}+{Y}^{2} \right) ^{3}
\\= {R}^{8}+{R}^{7}Y+4\,{R}^{6}{Y}^{2}+5\,{R}^{5}{Y}^{3}+8\,{R}^{4}
{Y}^{4}\\+5\,{R}^{3}{Y}^{5}+4\,{R}^{2}{Y}^{6}+R{Y}^{7}+{Y}^{8}.$$
We see that $$[R^4 Y^4] Z(D_8)(R+Y) = 8.$$
Here we have used the standard substitution $a_d = R^d + Y^d.$
Remark. When manual computation becomes too cumbersome there is
the formula
$$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}$$
for the cycle index of the cyclic group $Z(C_n)$ of order $n$ which
leaves the reflections, which are easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Let $n \in \mathbb{Z}^+$, prove the identity $ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$ Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$
First of all $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$
$$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$
$$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k} \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$
We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$
For the second sum
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$
Integrating both sides from $0$ to $x$, we see that
$$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$
Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$
Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
| Here's a direct proof (no integration, no induction, just the binomial theorem and algebra):
\begin{align}\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}&=\sum_{k=1}^{n-1} \binom {n} {k} \big(1-\frac1{k+1}\big)n^{n-k}
\\&=\sum_{k=1}^{n-1} \binom {n} {k} n^{n-k}-\sum_{k=1}^{n-1} \binom {n} {k} \frac1{k+1}n^{n-k}
\\&=\Big(\sum_{k=0}^{n} \binom {n} {k} n^{n-k}\Big)-\binom{n}{0}n^{n-0}-\binom{n}{n}n^{n-n}-\sum_{k=1}^{n-1} \frac{n!}{k!\, (n-k)!} \frac1{k+1}n^{n-k}
\\&=(1+n)^n-n^n-1-\sum_{k=1}^{n-1} \frac{n!}{(k+1)!\, (n-k)!} n^{n-k}
\\&=(n+1)^n-n^n-1-\sum_{k=1}^{n-1} \frac{n!}{(k+1)!\, (n-k)!} n^{n-k}
\\&=(n+1)^n-n^n-1-\sum_{k=1}^{n-1} \frac{(n+1)!/(n+1)}{(k+1)!\, (n-k)!} n^{n-k}
\\&=(n+1)^n-n^n-1-\frac1{n+1}\sum_{k=1}^{n-1} \frac{(n+1)!}{(k+1)!\, ((n+1)-(k+1))!} n^{n-k}
\\&=(n+1)^n-n^n-1-\frac1{n+1}\sum_{k=1}^{n-1} \binom{n+1}{k+1} n^{n-k}
\\&=(n+1)^n-n^n-1-\frac1{n+1}\sum_{k=2}^{n} \binom{n+1}{k} n^{n-k+1}
\\&=(n+1)^n-n^n-1-\frac1{n+1}\sum_{j=1}^{n-1} \binom{n+1}{n+1-j} n^j\scriptsize{\quad\text{(where }j=n-k+1)}
\\&=(n+1)^n-n^n-1-\frac1{n+1}\sum_{j=1}^{n-1} \binom{n+1}{j} n^j
\\&=(n+1)^n-n^n-1-\frac1{n+1}\left(-\binom{n+1}{0}n^0-\binom{n+1}{n}n^n-\binom{n+1}{n+1}n^{n+1}+\sum_{j=0}^{n+1} \binom{n+1}{j} n^j\right)
\\&=(n+1)^n-n^n-1-\frac1{n+1}\Big(-1-(n+1)n^n-n^{n+1}+(n+1)^{n+1}\Big)
\\&=(n+1)^n-n^n-1-\frac1{n+1}\Big(-1-n^{n+1}\Big)-\Big(-n^n+(n+1)^{n}\Big)
\\&=\require{cancel}\cancel{(n+1)^n}-\bcancel{n^n}-1+\frac1{n+1}\Big(1+n^{n+1}\Big)+\bcancel{n^n}-\cancel{(n+1)^{n}}
\\&=\frac{n^{n+1}+1}{n+1}-1
\\&=\frac{n(n^{n}-1)}{n+1},
\end{align}
as desired.
$$$$
| {
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"url": "https://math.stackexchange.com/questions/1968896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Find values of $a$ and $b$ that make matrix orthogonal
Given the matrix
$$A=\begin{bmatrix}1/2&a\\b&1/2\\ \end{bmatrix}$$
find the values of $a$ and $b$ that make it orthogonal.
So far I have tried using dot product $$(1/2)a+(1/2)b=0$$ and we can conclude that $a=-b$ and $b=-a$. I also tried the following theorem
$$A^T=A^{-1}$$
so
$$\begin{bmatrix}1/2&b\\a&1/2\\ \end{bmatrix}=
\begin{bmatrix}
(2+\frac{4ab}{1-4ab})&\frac{-4a}{1-4ab}\\
\frac{-4b}{1-4ab}&\frac{2}{1-4ab}\\ \end{bmatrix}$$
Can someone tell if I'am on the right track and point me in the right direction? Thanks!
| A square matrix $A$ is orthogonal iff $A^TA=AA^T=I$. Computing $A^TA$, we have
$$
A^TA = \begin{bmatrix} \frac12& b\\ a &\frac12
\end{bmatrix}\begin{bmatrix} \frac12& a\\ b &\frac12
\end{bmatrix} = \begin{bmatrix} \frac14 + b^2& \frac12(a+b)\\ \frac12(a+b) &\frac14 + a^2.
\end{bmatrix}
$$
From $A^TA$ we obtain the system of equations
\begin{align}
\frac14 + b^2 &= 1\\
\frac14 + a^2 &= 1\\
\frac12 (a+b) &= 0,
\end{align}
which has solution set $$(a,b) = \pm\left(\frac{\sqrt3}2, -\frac{\sqrt3}2 \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $ \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x$ I am trying to evaluate the exact value of the following definite integral:
\begin{align} \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x \end{align}
Since $ \int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x$ has symmetry, I did the following:
\begin{align}
\frac{91\pi}{6} \cdot \frac{2}{\pi} = \frac{91}{3}
\end{align}
\begin{align}
\int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x = [\sin(x)]^{\frac{\pi}{2}}_0 = 1
\end{align}
\begin{align}
\therefore \text{Bounded Area } = 1 \cdot \frac{91}{3} = \frac{91}{3}
\end{align}
Apparently, the correct answer is $ \frac{61}{2} $ which is very close to my answer. I cannot understand why my answer is wrong. Could someone please advise me?
| You can't do so, but you can $\frac{90\pi}{6}⋅\frac{2}{π}+\int_0^\frac{\pi}{6}|cosx|dx=30+1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving Inequalities similar to Nesbitt's Is it possible to use inequalities like Cauchy-Schwarz or QM-AM-GM-HM to find the minimum value of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}$ for $a,b,c\gt0$?
From just trying different values, the minimum seems to be $11\over6$,
but how would one prove this? I tried setting
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}=S$, giving $S+6=(a+b+c)(\frac{1}{a+b}+\frac{1}{a+c}+\frac{2}{a+b})$, but I'm not sure how to proceed from here, or if this is even the right first step.
| Enforcing the substitution $A=b+c, B=a+c, C=a+b$ the problem boils down to finding the minumum of
$$\begin{eqnarray*}&& \frac{B+C-A}{2A}+\frac{A+C-B}{2B}+\frac{A+B-C}{C}\\&=&-2+\left(\frac{B}{2A}+\frac{A}{2B}\right)+\left(\frac{C}{2A}+\frac{A}{C}\right)+\left(\frac{B}{C}+\frac{C}{2B}\right)\end{eqnarray*}$$
By setting $\frac{A}{B}=x$ and $\frac{B}{C}=y$, that boils down to studying
$$ f(x,y)= -2+\frac{1}{2x}+\frac{x}{2}+\frac{1}{2xy}+xy+y+\frac{1}{2y} $$
over $\mathbb{R}^+\times\mathbb{R}^+$. Such function has a unique stationary point at $(x,y)=\left(1,\frac{1}{\sqrt{2}}\right)$, hence the minimum of our expression is achieved at $(A,B,C)=(1,1,\sqrt{2})$ and it equals $\color{red}{2\sqrt{2}-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of infinite series $\cos \theta - \frac{\cos 5\theta}{5} + \frac{\cos 7\theta}{7} - \frac{\cos 11\theta}{11} ..., \theta \in (-\pi/3, \pi/3)$ I'm stumped at the following exercise on series summation:
$$\cos \theta - \frac{\cos 5\theta}{5} + \frac{\cos 7\theta}{7} - \frac{\cos 11\theta}{11} ..., \theta \in (-\pi/3, \pi/3)$$
The range gave me a hint that I should (probably?) split this into two summations:
$$\cos\theta+\frac{\cos7\theta}{7}+\frac{\cos13\theta}{13}+...$$
and
$$\frac{\cos5\theta}{5}+\frac{\cos11\theta}{11}+\frac{\cos17\theta}{17}...$$
However I'm stumped at this point. The only hint I was given was that
$$r\cos\theta-r^2\frac{\cos2\theta}{2}+r^3\frac{\cos3\theta}{3}-+...=\frac{1}{2}ln(1+2r\cos\theta+r^2)$$
and that
$$r\sin\theta-r^2\frac{\sin2\theta}{2}+r^3\frac{\sin3\theta}{3}-+...=\arctan\frac{r\sin\theta}{1+r\cos\theta}$$
(This can be obtained by the Taylor expansion of $\ln(1-z)$ and setting $z=re^{i\theta}$, then solving for the real and imaginary parts separately.)
Can anyone give me a hint here please? Thanks!
| For any $x$ such that $|x|<1$ we have
$$ S(x)=\sum_{n\geq 0}\left(\frac{x^{6n+1}}{6n+1}-\frac{x^{6n+5}}{6n+5}\right) = \int_{0}^{x}\sum_{n\geq 0}\left(t^{6n}-t^{6n+4}\right)\,dx = \int_{0}^{x}\frac{1-t^4}{1-t^6}\,dt $$
hence
$$\begin{eqnarray*} S(x) = \int_{0}^{x}\frac{1+t^2}{1+t^2+t^4}\,dt &=& \frac{1}{\sqrt{3}}\left(\arctan\frac{2x-1}{\sqrt{3}}+\arctan\frac{2x+1}{\sqrt{3}}\right)\\&=&\frac{1}{\sqrt{3}}\,\arctan\left(\frac{x\sqrt{3}}{1-x^2}\right)\end{eqnarray*}$$
and $\lim_{x\to 1^-}S(x)=\frac{\pi}{2\sqrt{3}}$. If $|\theta|<\frac{\pi}{3}$, the function $f(t)=\frac{1+t^2}{1+t^2+t^4}$ is holomorphic in a neighbourhood of the circle sector delimited by the angles $0$ and $\theta$ in the unit disk, hence
$$ S(e^{i\theta}) = \frac{\pi}{2\sqrt{3}}+\int_{0}^{\theta}\frac{1-e^{4it}}{1-e^{6it}} ie^{it}\,dt = \frac{\pi}{2\sqrt{3}}+2i\int_{0}^{\theta}\frac{\cos t}{1+2\cos(2t)}\,dt$$
where the last integral is a purely imaginary number. It follows that by considering the real parts of both sides of the last identity,
$$ \forall\theta\in\left(-\frac{\pi}{3},\frac{\pi}{3}\right),\qquad \sum_{n\geq 0}\left(\frac{\cos((6n+1)\theta)}{6n+1}-\frac{\cos((6n+5)\theta)}{6n+5}\right)=\color{red}{\frac{\pi}{2\sqrt{3}}}$$
holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How continue with this proof? I'm trying to proof the following statement:
$7+6 \cdot (7+7^1+7^2+\dots+7^n) = 7^{n+1}$
I try the next:
$n=k$
$7+6 \cdot (7+7^1+7^2+\dots+7^k) = 7^{k+1}$
$7+6 \cdot (7+7^1+7^2+\dots+7^k)-7 = 7^{k+1}-7$
$\frac{6 \cdot (7+7^1+7^2+\dots+7^k)}{6} = \frac{7^{k+1}-7}{6}$
$7+7^1+7^2+\dots+7^k = \frac{7^{k+1}-7}{6}$
$n=k+1$
$7+7^1+7^2+\dots+7^k+7^{k+1} = \frac{7^{k+2}-7}{6}$
Proof:
$\frac{7^{k+1}-7}{6}+7^{k+1} = \frac{7^{k+2}-7}{6}$
$\frac{7^{k+1}-7+6\cdot7^{k+1}}{6}$
Any idea?
| It seems you want to demonstrate by induction that
$$
7 + 6 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right) = 7^{\,n + 1\,}
$$
Now, that's true for $n=1$
$$
\begin{gathered}
n = 1 \hfill \\
7 + 6 \cdot \left( 7 \right) = 7 \cdot \left( {6 + 1} \right) = 7^{\,2\,} \hfill \\
\end{gathered}
$$
and we want to demonstrate that if it is valid for $n$ it is also valid for $n+1$.
So we take our identity, multiply both sides by $7$, do some massage
$$
\begin{gathered}
n \to n + 1 \hfill \\
7 + 6 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right) = 7^{\,n + 1\,} \hfill \\
7 \cdot \left( {7 + 6 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right)} \right) = 7 \cdot 7^{\,n + 1\,} = 7^{\,n + 2\,} \hfill \\
\left( {7 \cdot 7 + 6 \cdot 7 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\
\left( {7 \cdot \left( {6 + 1} \right) + 6 \cdot 7 \cdot \left( {7 + 7^{\,2\,} + \cdots + 7^{\,n\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\
\left( {7 + 6 \cdot 7 + 6 \cdot \left( {7^{\,2\,} + 7^{\,3\,} + \cdots + 7^{\,n + 1\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\
\left( {7 + 6 \cdot \left( {7 + 7^{\,2\,} + 7^{\,3\,} + \cdots + 7^{\,n + 1\,} } \right)} \right) = 7^{\,n + 2\,} \hfill \\
\end{gathered}
$$
and get the demonstration
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Which number is greater, $2^\sqrt2$ or $e$?
Claim: $\color{red}{2^\sqrt2<e}$
Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus
$$ f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln x \implies e^{x-1} > x^{\sqrt{x}} $$
Putting $x=2$, we obtain
$$e>2^\sqrt2$$
What do you think about this approach? Is there an easier way?
| The inequality $e>2^\sqrt2$ is equivalent to
$$e^{\sqrt{2}}>(2^\sqrt2)^{\sqrt{2}}=2^2=4.$$
Now
$$e^{\sqrt{2}}>\sum_{k=0}^5\frac{(\sqrt{2})^k}{k!}=\frac{13}{6}+\frac{41\sqrt{2}}{30}>\frac{13}{6}+\frac{4}{3}\cdot\frac{7}{5}=\frac{121}{30}>4$$
because $\sqrt{2}>7/5$ (equivalent to $2\cdot 5^2>7^2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
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} |
Simplify $26! \cdot x \equiv 730^{171} \pmod{232}$ I want to simplify the expression
\begin{align}
26! \cdot x \equiv 730^{171} \pmod{232}
\end{align}
My first step is to solve for $26! \cdot a \equiv 1 \pmod{232}$, that is, find the inverse of $26!$ modulo $232$.
To simplify, I should first calculate $26! \bmod{232}$. Using Wilson's theorem seems to be the best approach. But since $232 = 2^3 \cdot 29$ is not a prime number, I have to split up the expression using the Chinese Remainder Theorem:
\begin{align*}
\begin{cases}
26! \equiv a \pmod{29}\\
26! \equiv b \pmod{8}
\end{cases}
\end{align*}
I can use Wilsons theorem to solve the first equation as follows:
\begin{align*}
28! = 26!\cdot 27 \cdot 28 \equiv 26!(-2)(-1) \equiv 26!\cdot 2 \equiv -1 \pmod{29}
\end{align*}
Using that $2\cdot 15 \equiv 1 \pmod{29}$, we find
\begin{align*}
26! \equiv -15 \equiv 14 \pmod{29}
\end{align*}
However, I cannot use Wilsons theorem to solve the second equation. Any suggestions on how to solve that? Or might there be a better way to calculate $26! \bmod{232}$ altoghether?
| Hint: Divide by $2^3$ on both sides (and the modulus) to obtain the equivalent congruence: $$(26)(25)(3)(23!) x \equiv (365)^3(730)^{168} \pmod{29}$$
Note that the left side contains $(26)(25)(3)(23!)$, which is relatively prime to $29$ and thus invertible modulo $29$ (like in the work you have done, Wilson's Theorem can be useful here). Then note that $730$ and $29$ are relatively prime; this permits you to use Fermat's Little Theorem (or Euler's Theorem) to simplify $730^{168}$.
It should be more straightforward from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding the minimum value of $(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2$ We assume $\alpha,\beta$ are real numbers, I would like to find the minimum value of
$$(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2.$$
After expanding $(\alpha+5)^2+9\cos^2 \beta-2(\alpha+5)|3\cos \beta|+\alpha^2+4\sin^2 \beta-4\alpha|2\sin \beta|$, I would have some help with this.
| Using Geometry
Let $(\alpha +5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2 = d^2$
and Let $x_{1}=\alpha+5\;\;,x_{2}=3|\cos \alpha|$ and $y_{1}=\alpha$ and $y_{2} = 2|\sin \beta|$
$\Rightarrow (x_{1},y_{1})$ lie on the line $x-y-5=0$ and $(x_{2},y_{2})$ lie in the first quadrant,s
well on the ellipse $\displaystyle \frac{x^2}{9}+\frac{y^2}{4} = 1$
and $d$ is the distance between $(x_{1},y_{2})$ and $(x_{2},y_{2})$
Now minimum $$d = \bf{Distance\; b/w\; (3,0)\; and \; line = \frac{|3-0-5|}{\sqrt{1^2+1^2}}} = \sqrt{2}$$
So $$\min (\alpha +5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2 = \min(d^2) = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equation of a circle tangent which makes a triangle of area $a^2$ with the coordinate axes What is the equation of the tangent to the circle $x^2+y^2-2ax-2ay+a^2=0$ which makes with the coordinate axes a triangle with area $a^2$?
Attempt: The equation of the circle can be re-written as $(x-a)^2+(y-a)^2=a^2$. Let's say the tangent equation is $\frac{x}{c}+\frac{y}{d}=1$. From the condition that it is a tangent I get
$|\frac{\frac{a}{c}+\frac{a}{d}-1}{\sqrt{\frac{1}{c^2}+\frac{1}{d^2}}}|=a$. Simplifying I get $a(c+d-\sqrt{c^2+d^2})=cd$ and the area condition gives $\frac{1}{2}cd=a^2$. But I am not able to solve this two equation and get values of $c$ and $d$.
| Without loss of generality fix $a=1$. The trick here is to realise that $c$ and $d$ must have different signs, or else the formed triangle is either too big or too small (as pointed out by Jean Marie). With this restriction, the condition that the triangle has area 1 must be modified to
$$-\frac12cd=1\text{ or }cd=-2$$
Substituting this into the other equation we get
$$c+d-\sqrt{c^2+d^2}=-2$$
$$(c+d)+2=\sqrt{(c+d)^2-2cd}=\sqrt{(c+d)^2+4}$$
$$(c+d)^2+4(c+d)+4=(c+d)^2+4$$
$$4(c+d)=0$$
$$c=-d$$
It is easy to see from this and $cd=-2$ that $c=\sqrt2$ and $d=-\sqrt2$, or the other way around. Therefore we have the two lines $\frac x{\sqrt2}-\frac y{\sqrt2}=\pm1$. The solution lines for general $a$ merely multiply the constant on the RHS by $a$:
$$\frac x{\sqrt2}-\frac y{\sqrt2}=\pm a$$
$$x-y=\pm a\sqrt2$$
Here is a plot of the solution lines and the triangles they define.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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On a polynomial with roots in $[1,3]$ that are also of the form $2+\csc\theta$
If the roots of the equation $x^2+bx+c=0$ are $\alpha,\beta$ such that $1\leq \alpha,\beta\leq 3$ and also
given that $(\csc \theta+2)^2+b(\csc \theta+2)+c=0$ satisfy for two different values of $\csc \theta$
Then value of $|b+c|$ and $\min (x^2+bx+c)$
$\bf{My\; Try::}$ We can write it as $x^2+bx+c = (x-\alpha)(x-\beta)=x^2-(\alpha+\beta )x+\alpha\beta$
So $\alpha +\beta = -b$ and $\alpha \beta = c$
and $(\csc \theta+2)^2+b(\csc \theta +2)+c = (\csc \theta-\alpha')(\csc \theta-\beta')=\csc^2 \theta -(\alpha'+\beta' )\csc \theta+\alpha'\beta'$
So $$\csc^2 \theta+(4+b)\csc \theta+(4+2b+c) = \csc^2 \theta -(\alpha'+\beta' )\csc \theta+\alpha'\beta'$$
So $\alpha '+\beta'=-(4+b)$ and $\alpha'\beta'=(4+2b+c)$
Now how can i solve it after that, Help required, Thanks
| Note that $|\csc\theta|\ge1$, and hence $2+\csc\theta$ must lie outside $(1,3)$. Yet two distinct $2+\csc\theta$ values are the roots of the given quadratic, which is given to have roots within $[1,3]$. Hence the roots are 1 and 3 and the quadratic is
$$(x-1)(x-3)=x^2-4x+3=(x-2)^2-1$$
Thus $|b+c|=|-4+3|=1$ and $\min(x^2+bx+c)=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $x+y+z$ divides $x^2+y^2+z^2$ where x,y,z are consecutive terms in a geometric series The natural numbers $x,y,z$ are consecutive terms in a geometric series. Proves that $x+y+z$ divides $x^2+y^2+z^2$.
$x = ar\\
y = ar^2 = xr\\
z = ar^3 = xr^2$
So..
$\begin{align}
x + y + z & = x + xr + xr^2 \\
& = x(1+r+r^2)\end{align}$
$\begin{align}
x^2 + y^2 + z^2 & = x^2 + x^2r^2 + x^2r^4\\
& = x^2(1+r^2+r^4)\end{align}$
I can see that x is a factor for both but I've no idea where to go from here
| Remark that $1+r^2+r^4=(1+r+r^2)(1-r+r^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986614",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Is there a formula describing the length of line subtended from a chord in a circle? (Picture Included) Is there a formula that describes the length of line x in a circle, assuming you know the chord length (c), perpendicular distance of the chord from the circumference (d), circle radius (r), and the angle that line x subtends with the chord (a).
|
*
*Observe that $MO = r -d$ and since the triangle $AMO$ is right angled, $$AM =\frac{c}{2}= \sqrt{OA^2 - OM^2} = \sqrt{r^2 - (r-d)^2}$$
*Triangle $MNO$ is right angled and $\angle \, OMN = a$ so
$$MN = MO \,\cos(a) = (r-d) \,\cos(a), \,\,\,\, ON = MO \,\sin(a) = (r-d)\, \sin(a)$$
*Triangle $OND$ is right angled so by Pythagoras' theorem
$$ND = \sqrt{OD^2 - ON^2} = \sqrt{r^2 -(r-d)^2 \sin^2(a)}$$
*By the power of point relation
\begin{align}
\frac{c^2}{4} &= AM^2 = AM \cdot MB = CM \cdot MD = CM \cdot \Big(MN + ND\Big)\\
&= x \, \Big( (r-d) \,\cos(a) + \sqrt{r^2 -(r-d)^2 \sin^2(a)}\Big)
\end{align} thus
$$x = \frac{c^2}{4 \, \Big( (r-d) \,\cos(a) + \sqrt{r^2 -(r-d)^2 \sin^2(a)}\Big)}$$ or equivalently, if you prefer
$$x = \frac{r^2 - (r-d)^2}{(r-d) \,\cos(a) + \sqrt{r^2 -(r-d)^2 \sin^2(a)}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the following integration "trick" valid to reduce my integrand to a constant? $$\int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(9-x)}} +1}dx = 1$$
I noticed when $x$ went from $2$ to $4$, $3+x$ went from $5$ to $7$, and $9-x$ went from $7$ to $5$.
I noticed that if we reverse the interval of $9-x$ we obtain $-(9-x)$ and whose interval goes from $5$ to $7$. In short, I concluded that reversing the interval of $9-x$ yielded the expression $3+x$.
Therefore I allowed the above integral to be $$\int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(9-x)}} +1} dx = \int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(3+x)}} +1} dx.$$
Effectively, the logs cancelled and I was left with $\int_2^4 {1\over 1+1}dx =1$.
| Counterexample:
$$\int_2^4\frac{3+x}{9-x}dx=\int_2^4\left(1-\frac{6}{9-x}\right)dx=\left.\left(x+6\log(9-x)\right)\right|_2^4=2+6\log\frac37\ne\int_2^4dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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FInd the missing digit in $2^{29}$ given all nine digits differ The number $2^{29}$ has (in base $10$) $9$ digits, all different. Which digit is missing?
I think about using fermats theorem dosen't know how to begin
| $2^{29}\equiv2^{-1}\equiv\frac{1}{2}\equiv\frac{10}{2}\equiv5\equiv \text{s} \pmod 9$ where $\text{s}$ denotes the sum of its digits.
The sum of the ten digits is $\text{S}=0+1+2+3+4+5+6+7+8+9=\frac{1}{2}\times10\times9=45\equiv0 \pmod 9$
Therefore, if $2^{29}$ contains $9$ of these and the remaing one is $n$, $2^{29}\equiv S-n\equiv-n \pmod{9}$
so $s\equiv-n\equiv5 \pmod{9}$ and the missing digit is $n=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989618",
"timestamp": "2023-03-29T00:00:00",
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} |
General proof for combinatorics Prove that ${n \choose k} + {n \choose k+1} = {n+1 \choose k+1}$.
I can prove it using numeric examples, but i need a broader proof. I think it might have to do with Pascal's Triangle.
|
Argumentative Method:
We know that the number of ways to choose $k$ objects from $n$ objects is $\binom{n}{k}$. Let us add 1 more object into the group. Now we have to choose $k+1$ objects from $n+1$ objects.
Lets say we include the newest item in the group. Then the number of ways to choose from the remaining $n$ objects is $\binom{n+1-1}{k+1-1} = \binom{n}{k}$.
Lets say we do not include the newest item in the group. Then the number of ways to choose from the remaining $n$ objects is $\binom{n+1-1}{k+1} = \binom{n}{k+1}$.
Since there are no overlaps between the 2 cases, $\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$.
Mathematical Method:
$$
\begin{align*}
\binom{n}{k} + \binom{n}{k+1} & = \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\
& = \frac{n!(k+1)}{(k+1)!(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-k)!}\\
& = \frac{n!(n-k+k+1)}{(k+1)!(n-k)!}\\
& = \frac{n!(n+1)}{(k+1)!(n-k)!}\\
& = \frac{(n+1)!}{(k+1)!(n-k)!}\\
& = \binom{n+1}{k+1}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989813",
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If $\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c}$, then find $a,b,c$.
If
$$\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c} $$
then find $a, b, c$.
Now, using partial fractions I calculated
$$a = 62-10\ln (2) \qquad\qquad b = -15 \qquad\qquad c = 20$$
but it took me more than 45 minutes to do all the work. The question was asked in an MCQ exam where only 4-5 minutes are available. I am probably missing something which can help to solve it. Thanks!
Note it was asked in an exam for students of grade 12 so I basically don't know very complex integrations thus I am searching for integration via elementary functions only.
| $$\frac {x^3+3}{x^6(x^2+1)}=\frac{3}{x^6}-\frac{3}{x^4}+\frac{1}{x^3}+\frac{3}{
x^2}-\frac{1}{x}+\frac{x}{x^2+1}-\frac{3}{x^2+1}$$ So $$\int\frac {x^3+3}{x^6(x^2+1)}\,dx=-\frac{3}{5 x^5}+\frac{1}{x^3}-\frac{1}{2 x^2}+\frac{1}{2} \log
\left(x^2+1\right)-\frac{3}{x}-\log (x)-3 \tan ^{-1}(x)$$ that is to say $$-\frac{3}{5 x^5}+\frac{1}{x^3}-\frac{1}{2 x^2}-\frac{3}{x}+\frac{1}{2} \log
\left(\frac{x^2+1}{x^2}\right)-3 \tan ^{-1}(x)$$ Now, using the bounds, it seems to be quite fast.
| {
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Prob. 11, Chap. 3, in Baby Rudin: If $a_n > 0$ and $\sum a_n$ diverges, then how do we show that $\sum \frac{a_n}{1+a_n}$ too diverges? Here's Prob. 11, Chap. 3, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $a_n > 0$, $s_n = a_1 + \cdots + a_n$, and $\sum a_n$ diverges.
(a) Prove that $\sum \frac{a_n}{1+ a_n}$ diverges. [ I have no clue of how to prove this!]
(b) Prove that $$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq 1- \frac{s_N}{s_{N+k}}$$ [I can show this.] and deduce that $\sum \frac{a_n}{s_n}$ diverges. [How to?]
(c) Prove that $$ \frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}} - \frac{1}{s_n}$$ and deduce that $\sum \frac{a_n}{s_n^2}$ converges. [This I can show, I think.]
(d) What can be said about (the convergence or divergence of) $$\sum \frac{a_n}{1+ n a_n} \ \ \ \mbox{ and } \ \ \ \sum \frac{a_n}{1+ n^2 a_n}?$$ [ How to answer this?]
I would prefer those answers that use only the machinary developed by Rudin himself upto this point in the book.
Here's what I can show:
Since $a_n > 0$, we have $0 < s_{N+1} < \cdots < s_{N+k}$ and so
$$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq \frac{a_{N+1} + \cdots + a_{N+k}}{s_{N+k}} = 1- \frac{s_N}{s_{N+k}}.$$
As $a_n > 0$, so, for all $n = 2, 3, 4, \ldots$, we have $0 < s_{n-1} < s_n$ and therefore $$ \frac{a_n}{s_n^2} = \frac{s_n - s_{n-1}}{s_n^2} \leq \frac{s_n - s_{n-1}}{s_n s_{n-1}} = \frac{1}{s_{n-1}} - \frac{1}{s_n},$$ and hence $$ 0 \leq \sum_{k=1}^n \frac{a_k}{s_k^2} = \frac{1}{a_1} + \sum_{k=2}^n \frac{a_k}{s_k^2} \leq \frac{1}{a_1} + \sum_{k=2}^n \left( \frac{1}{s_{k-1}} - \frac{1}{s_k} \right) = \frac{1}{a_1} + \frac{1}{s_1} - \frac{1}{s_n} \to \frac{2}{a_1} + 0 $$ as $n \to \infty$ because $a_n > 0$ and $\sum a_n$ diverges and hence $s_n = a_1 + \cdots + a_n \to \infty$ as $n \to \infty$. So if $\lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} $ exists [ but how to show this?}, then we must have $$ 0 \leq \lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} \leq \frac{2}{a_1}.$$
Am I right?
| Part (b).
For any $N$ choose $k$ such that $S_N / S_{N+k} < 1/2$. This is possible because $S_n \to \infty$.
Hence, in violation of Cauchy criterion,
$$\sum_{k=N+1}^{N+k} \frac{a_k}{S_k} \geqslant \frac{S_{N+k} - S_N}{S_{N+k}} = 1 - \frac{S_N}{S_{N+k}} > \frac{1}{2},$$
and $\sum (a_n/S_n)$ must diverge.
| {
"language": "en",
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From $32\cdot (\frac{1-.5^n}{1-.5})$ to $\frac{64(2^n-1)}{2^n}$ I have to prove that $32\cdot (\frac{1-.5^n}{1-.5})$ is equal to $\frac{64(2^n-1)}{2^n}$
I know that they are equal because when I put them in my calculator the graphs are the same. However, I'd like to do this manually.
So far I have done:
$$32\cdot (\frac{1-.5^n}{1-.5}) = 32 \cdot (\frac{.5}{.5} + \frac{.5^n}{.5}) = 32 \cdot (\frac{.5\cdot.5^n}{.5}) = \frac{32\cdot.5\cdot.5^n}{.5}= \frac{8^n}{.5} $$
How do I solve this?
| Easier to start from the right.
$$\begin{align}\frac {64(2^n-1)}{2^n}
&=64\left(\frac {2^n}{2^n}-\frac 1{2^n}\right)\\\\
&=\frac {32}{0.5}\left(1-\left(\frac 12\right)^n\right)\\\\
&=\frac{32(1-0.5^n)}{1-0.5}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for infinitely many $ n$ the greatest prime factor of $ a^n-1$ is greater than $ n\log_a n$ This problem from Andreescu Problems from the book Page 331 problem 29 :
Let $ a$ be an integer greater than 1. Prove that for
infinitely many $ n$ the greatest prime factor of $ a^n-1$ is
greater than $ n\log_a n$
| By Zsigmondy's theorem, for every pair $(a,n)$ with $a\ge2$ and $n\ge7$, there is a prime $P$ such that $P$ divides $a^n-1$, but $P$ does not divide $a^k-1$ for any $1\le k<n$; in other words, the multiplicative order of $a$ modulo $P$ is precisely $n$. Due to Fermat's theorem, the multiplicative order divides $P-1$, so $P\equiv1\pmod{n}$.
Hence, for $n\ge7$ there is a prime divisor $P$ of $a^n-1$ with $P\in\{n+1,2n+1,3n+1,\ldots\}$. Our goal is to make the small elements in this set composite.
Take three large consecutive primes, $q<q_1<q_2$ and let $Q=\prod_{p<q}p$ be the product of primes up to $q-1$. Chose a positive integer $n$ such that
\begin{align*}
qn+1 &\equiv 0 \pmod{Q} \\
(q-1)n+1 &\equiv 0 \pmod{q_1} \\
(q+1)n+1 &\equiv 0 \pmod{q_2} \\
\end{align*}
By the Chinese Remainder Theorem, this system has a solution modulo $Qq_1q_2$. Instead of the smallest positive solution, let $n$ be the second one with
$Qq_1q_2 < n < 2 Qq_1q_2$.
Now consider the numbers
$$
n+1, \quad 2n+1, \quad \ldots, \quad
(q-1)n+1, \quad qn+1, \quad (q+1)n+1, \quad
\ldots, \quad (2q-1)n+1. \tag{1}
$$
By the definition, the middle three terms, $(q-1)n+1$, $qn+1$, and $(q+1)n+1$
are divisible by $q_1$, $Q$ and $q_2$, respectively, so they are composite. The remaining elements are of the form $qn+1\pm kn$ with some $2\le k<q$; every prime divisor $p$ of $k$ divides $qn+1\pm kn$ as well. So the numbers in (1) are all composite. Therefore, $P \ge 2qn+1$.
From the Prime Number Theorem we can get
\begin{gather*}
n < 2Qq_1q_2 < e^{(1+\varepsilon)q} \\
P > 2qn > \frac{2\ln n}{1+\varepsilon} n > n\cdot \log_a n.
\end{gather*}
| {
"language": "en",
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"source": "stackexchange",
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Solutions for $\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0$ with a certain value range. To proof
How prove this equation has only one solution $\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$
I need first an analytically (not: numerically) proof for the following problem:
Be $\enspace\displaystyle 0<\beta<\frac{2\pi}{3}<\alpha<2\pi$ .
Then it exists exactly $\,$ one $\,$ solution $\,(\alpha;\beta)\,$ for $\enspace \cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0 $ ?
(The answers below show: No.)
Known: $\enspace\displaystyle (\alpha_0;\beta_0):=\left(\pi;\arccos\left(\frac{1}{3}\right)\right)\enspace$ is a solution.
| It is not true that $(\alpha_0,\beta_0)=(\pi,\arccos(\frac{1}{3}))$ is the only solution. Consider for example $\alpha_1=\frac{5}{6}\pi$ and $g(x)=\cos(\alpha_1)+\cos(x)-2\cos(\alpha_1+x)$. We have
$$
g(\frac{\pi}{2})=
\cos(\frac{5}{6}\pi)+\cos(\frac{\pi}{2})-2\cos(\frac{4\pi}{3})=
-\frac{\sqrt{3}}{2}+0-2(\frac{-1}{2})=\frac{2-\sqrt{3}}{2} >0. \tag{1}
$$
Also, $$g(\frac{5\pi}{8})=
\cos(\frac{5}{6}\pi)+\cos(\frac{5\pi}{8})-2\cos(\frac{35\pi}{24})
\leq -\frac{\sqrt{3}}{2}-\frac{1}{3}+1=\frac{4-\sqrt{27}}{6}<0. \tag{2} $$
To justify (2), we need (a) $\cos(\frac{5\pi}{8}) \lt -\frac{1}{3}$ and (b)
$\cos(\frac{35\pi}{24}) \gt -\frac{1}{2}$. The easiest one is (b) : it follows from
$\frac{35\pi}{24}\in (\frac{4\pi}{3},2\pi)$ and the fact that $\cos$ is decreasing on that interval. Regarding (a), note first that $\cos(\frac{5\pi}{8})=-\sin(\frac{\pi}{8})$.
Next, using Taylor's formula with Lagrange remainder, we have
$$
f(x)-f(0)-f'(0)x-f''(0)\frac{x^2}{2}-f'''(0)\frac{x^3}{6}=\frac{f^{(4)}(\xi)}{24}x^{4}
$$
for some $\xi\in[0,x]$. Applying this to $f(x)=\sin(x)$, we deduce $\sin(x)\geq x-\frac{x^3}{6}$ for $x\in [0,\frac{\pi}{2}]$. In particular,
$\sin(\frac{\pi}{8}) \geq h(\frac{\pi}{8})$. Since $h'(x)=1-2x^2$, we see that $h$
is increasing on $[0,\frac{1}{2}]$, so $h(\frac{\pi}{8})\geq h(\frac{3}{8})=\frac{375}{1024} \gt \frac{1}{3}$. This finishes the proof.
| {
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Describe the multiplication in the ring $\Bbb Q[x]/(x^2+x+1)$. Describe the multiplication in the ring $\Bbb Q[x]/(x^2+x+1)$. What is the multiplicative inverse of $[x]$?
To figure this out, I found that $x = -\frac {1}{2} + i \sqrt {\frac{3}{4}}$. So there clearly aren't any real solutions to this. But I am trying to figure out what elements of this ring look like in order to find an inverse and to see what multiplication looks like. I am new to field extensions and am a little confused as to how to work with them. Any help is appreciated.
| There are a few ways to think about it.
The first is quite literal. The elements of $\frac{\mathbb{Q}[x]}{(x^2+x+1)}$ are residues of polynomials, $\overline{f(x)}$. In this ring $\overline{f(x)} = \overline{g(x)}$ if and only if $f(x)-g(x)$ is divisible by $x^2+x+1$.
To multiply $\overline{f(x)}$ and $\overline{g(x)}$, multiply them as polynomials in $\mathbb{Q}[x]$, then divide by $x^2+x+1$ and take the remainder.
Another way: $$\frac{\mathbb{Q}[x]}{(x^2+x+1)} \cong \{a+b\alpha |\ a,b \in \mathbb{Q}, \text{ and } \alpha^2 + \alpha + 1 = 0 \}$$
$\alpha$ is some "thing" that satisfies the polynomial $x^2 + x +1$. Then multiplication becomes:
\begin{align*} (a + b\alpha)(c + d\alpha) &= ac + (ad + bc)\alpha + bd\alpha^2\\
&= ac + (ad + bc)\alpha + bd(-\alpha - 1)\\
&= (ac-bd) + (ad + bc - bd)\alpha.
\end{align*}
Per user1952009's suggestion:
How could we express $\alpha^3$ and $\alpha^4$ as something of the form $a+b\alpha$?
\begin{align*}
\alpha^3 &= \alpha \alpha^2\\
&= \alpha(-\alpha-1)\\
&= -\alpha^2 - \alpha\\
&= 1.
\end{align*}
In this cool turn of events, we see that $\alpha^3 = 1$. Another way to see that this would happen is that $x^3-1 = (x-1)(x^2+x+1)$ and since $\alpha$ satisfies $x^2+x+1$, it will then satisfy $x^3-1$.
From this we see that $\alpha^4 = \alpha$ and in fact for $n \in \mathbb{N}$:
$$
\alpha^n = \begin{cases}
1 \ &\text{ if } n \equiv 0 \pmod 3 \\
\alpha \ &\text{ if } n \equiv 1 \pmod 3 \\
-1 - \alpha \ &\text{ if } n \equiv 2 \pmod 3
\end{cases}
$$
| {
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Is this a correct demonstration that 13 divides $2^{70} + 3^{70}$? Is it true that: $13|(2^{70} + 3^{70})$ iff 13 mod(n) divides $2^{70} + 3^{70}$ mod(n)?
I assumed that it is the case. Being that correct, then 13 = 1 (mod(2)) and $2^{70} = 0$ (mod(2)); $3^{70} = 3x3x...x3 = 1x1x...x1 $ (mod(2)). Therefore $2^{70} + 3^{70}$ = 0 + 1 = 1 = 13 (mod(2)), and it follows that $13|2^{70} + 3^{70}$.
I missed a lot of classes on modular arithmetic and didn't compensate it at home, so I probably got something wrong. But what I used to conceive this demonstration is that, as far as I know, the set of congruence classes forms a subring of the integers, therefore it satisfies all ring operation properties.
| A trickier approach. Since $4$ and $9$ are quadratic residues $\!\!\pmod{13}$, for $p=13$ we have
$$ 4^{\frac{p-1}{2}}\equiv 9^{\frac{p-1}{2}}\equiv 1\pmod{13} $$
hence
$$ 4^{36}\equiv 9^{36}\equiv 1\pmod{13} $$
and
$$ 2^{70}+3^{70} = 4^{35}+9^{35} \equiv 4^{-1}+9^{-1} \equiv 0\pmod{13} $$
since $\frac{1}{4}+\frac{1}{9}=\frac{\color{red}{13}}{36}$. Even simpler: $35$ is odd, hence $13=2^2+3^2$ is a divisor of $2^{2\cdot 35}+3^{2\cdot 35}$.
| {
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$\int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}$ I want to evaluate following integral
\begin{align}
\int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}
\end{align}
Simple try on this integral is using branch cut and apply residue theorem.
Usual procedure gives for $0 < \alpha < 1$, with $Q(x)$ deg $n$ and $P(x)$ deg $m$, for $x>0$, $Q(x) \neq 0$
\begin{align}
\int_0^{\infty} \frac{x^\alpha P(x)}{Q(x)} dx = \frac{2\pi i}{1- e^{i\alpha 2 \pi}} \sum_j Res[\frac{z^\alpha P(z)}{Q(z)} , z_j]
\end{align}
where $z_j$ are poles which does not make $\frac{P}{Q}$ be zero.
This formula comes from Mathews and Howell's complex analysis textbook.
And this is nothing but applying branch cut to make $x^{\frac{1}{4}}$ singled valued function. I think this formula works for above improper integral but results seems different.
Apply $\alpha=\frac{1}{4}$ and take poles $z_0=-1$, $z_1 = e^{\frac{i \pi}{3}}$
, $z_2 = e^{\frac{i5 \pi}{3}}$, i got different things.
Am i doing right?
\begin{align}
\frac{2\pi i}{1-i}\frac{1}{3} \left( e^{\frac{1}{4} \pi i} + e^{-\frac{7}{12}\pi i} + e^{-\frac{35}{12} \pi i}\right)
\end{align}
| You may simply remove the branch cut by setting $x=z^4$:
$$ I = 4 \int_{0}^{+\infty}\frac{z^4\,dz}{1+z^{12}} = 2\int_{-\infty}^{+\infty}\frac{z^4}{1+z^{12}} \tag{1}$$
and by evaluating the residues at the roots of $1+z^{12}$ in the upper half-plane,
$$ I = \frac{2\pi}{3\sqrt{2+\sqrt{3}}}=\frac{\pi}{3}\left(\sqrt{6}-\sqrt{2}\right)\tag{2} $$
follows. By setting $\frac{1}{1+z^{12}}=u$, the integral $(1)$ can also be evaluated through Euler's Beta function and the reflection formula for the $\Gamma$ function, since:
$$ I = \frac{1}{3}\int_{0}^{1}u^{-5/12}(1-u)^{-7/12}\,du = \frac{1}{3}\,\Gamma\left(\frac{5}{12}\right)\,\Gamma\left(\frac{7}{12}\right)=\frac{\pi}{3\sin\frac{5\pi}{12}}.\tag{3}$$
| {
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Prove that $13^n -4^n$ is divisible by $9$ for every integer $\geq $
Prove that $13^n - 4^n$ is divisible by $9$ for every integer $n$ is greater than or equal to $1$.
I have written down my answer but I don't know if it is correct.
$ 13(13^k) - 4^{(k+1)}$
$= 9(13^k) + 4(13^k) - 4{(4^k)}$
$= 9(13^k) + 4(13^k - 4^k)$
$= 9(13^k) + 4(9a)$ where $9a = 13^k -4^k$
$= 9(13^k + 4a)$
$= 9b$ where $b=13^k + 4a$
| You can prove this by induction.
First, show that this is true for $n=1$:
$13^{1}-4^{1}=9$
Second, assume that this is true for $n$:
$13^{n}-4^{n}=9k$
Third, prove that this is true for $n+1$:
$13^{n+1}-4^{n+1}=$
$13\cdot13^{n}-4\cdot4^{n}=$
$13\cdot13^{n}-(13-9)\cdot4^{n}=$
$13\cdot13^{n}-13\cdot4^{n}+9\cdot4^{n}=$
$13\cdot(\color\red{13^{n}-4^{n}})+9\cdot4^{n}=$
$13\cdot\color\red{9k}+9\cdot4^{n}=$
$9\cdot13k+9\cdot4^{n}=$
$9\cdot(13k+4^{n})$
Please note that the assumption is used only in the part marked red.
| {
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Euler product formula Why is it that
$$\prod_{p\,\le\, y}\sum_{k=0}^\infty p^{-ks} = \sum_{n\,\in\,\{n \,:\, p \,\mid\, n \, \Rightarrow \, p\,\le\,y\}} n^{-s} \text{ ?}$$
I understand that we have a product of finitely many absolute convergent series and therefore rearrangement is allowed, but I do not see how exactly the terms are being rearranged.
| Your product $\displaystyle\prod_{p\,:\,p\,\le\,y}$ specifies $p\le y$, so let's try $y=5$, so that $p\in\{2,3,5\}.$ Your sum $\displaystyle \sum_{n \, : \, p \, \mid \, n,\ p\,\le\,y}$ requires $n$ to be divisible only by those primes, so where looking at integers divisible only by $2$, $3$, or $5$ and by no other primes.
\begin{align}
& \left( 1 + \frac 1 {2^s} + \frac 1 {4^s} + \frac 1 {8^s} + \cdots \right) \\
{}\times {} & \left( 1 + \frac 1 {3^s} + \frac 1 {9^s} + \frac 1 {27^s} + \cdots \right) \\
{}\times{} & \left( 1 + \frac 1 {5^s} + \frac 1 {25^s} + \frac 1 {125^s} + \cdots \right) \\[10pt]
= {} & \text{a sum in which, for example, } \frac 1 {7200^s} \text{ appears because} \\[5pt]
& 7200 = 2^5 \times3^3\times5 \text{ has no prime} \\[5pt]
& \text{factors besides 2, 3, and 5;} \\[5pt]
& \text{thus } \frac 1 {7200^s} = \frac 1 {2^{5s}} \times \frac 1 {3^{2s}} \times \frac 1 {5^s}. \tag a
\end{align}
Within the expression $\displaystyle \left( 1 + \frac 1 {2^s} + \frac 1 {4^s} + \frac 1 {8^s} + \cdots \right)$ one finds the term $1/2^{5s}$; specifically, it is the sixth term in that sum. Likewise, within the sum $\displaystyle \left( 1 + \frac 1 {3^s} + \frac 1 {9^s} + \frac 1 {27^s} + \cdots \right)$ one finds $1/3^{2s}$; specifically, it is $1/9^s$. And within $\displaystyle \left( 1 + \frac 1 {5^s} + \frac 1 {25^s} + \frac 1 {125^s} + \cdots \right)$ one finds $1/5^s$. Hence the term in the line labeled $(\text{a})$ above appears. And so do all other numbers $1/n^s$ where $n$ is a number divisible only by primes that are $\le y$.
It is necessary to rely on the fact that each number has only one prime factorization: If there were two prime factorizations of $7200$, then we would see $\dfrac 2 {7200^s}$ instead of $\dfrac 1 {7200^s}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of integer solutions to $Ax + By + Cz = D$ How many integer solutions exits for the equations, $Ax + By + Cz = D$, where
$B = (A + 1), C = (A + 2)$, and $x, y, z$ are non-negative i.e $x, y, z, >= 0$
I require a general solution which can be implemented using code as well. We would be given the values of $A, D$.
| First note that $\gcd(A, A + 1, A + 2) = 1 $. Since this gcd while divide $D$ this Diophantine equation will always have solutions. So first solve $Au + (A+1)v = \gcd(A, A+1) = 1$. Then solve $(1)w + (A+2)s = gcd(1,A+2) = 1$. After finding the integer solutions(values of $u,v,w$ and $s$) using the Extended Euclidean Algorithm we then have
\begin{align}
w + (A +2)s &= 1\\
(Au + (A+1)v)w + (A+2)s &= 1\\
A(uw) + (A+1)(vw) + (A+2)s &= 1\\
A(Duw) + (A+1)(Dvw) + (A +2)(Ds) &= D
\end{align}
Thus we have our solutions $x = Duw$, $y = Dvw$ and $z = Ds$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a binomial term/general formula for recurrence relation We know that Pascal's triangle obeys the recurrence relation $\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} $
And we can simply $\binom{n}{k}$ by $\frac{n!}{k!\,(n - k)!}$
I have a recurrence relation where
$$ f(n, k) = f(n - 1, k) + f(n - 2, k - 1) $$
How can I get a generel formula for that?
Thanks in advance!
Edited:
Base Case:
$f(n, 1) = n$ and $f(n, k) = 0$ when $n < k$
| Let us compute $f(n, 2)$ for any integer $n\ge2$.
First see that $$f(2, 2) = f(1, 2) + f(0, 1) = 0$$
$$f(3, 2) = f(2, 2) + f(1, 1) = 1$$
$$f(4, 2) = f(3, 2) + f(2, 1) = 3$$
$$f(5, 2) = f(4, 2) + f(3, 1) = 6$$
We can conjecture that $f(n, 2) = {(n-2)(n-1)\over2} = \sum_{k=0}^{n-2}k$
Indeed $f(2,2)$ is our base step, and if $f(n, 2) = \sum_{k=0}^{n-2}k$, then $$f(n+1) = f(n, 2) + f(n-1, 1) = \sum_{k=0}^{n-2}k + n-1 = \sum_{k=0}^{n-1}k$$
Thus $f(n, 2) = \sum_{k=0}^{n-2}k$ for all $n$.
Now compute $f(n, 3)$ for any integer $n\ge 3$ :
$$f(3, 3) = f(2, 3) + f(1, 2) = 0$$
$$f(4, 3) = f(3, 3) + f(2, 2) = 0$$
$$f(5, 3) = f(4, 3) + f(3, 2) = 1$$
$$f(6, 3) = f(5, 3) + f(4, 2) = 4$$
$$f(7, 3) = f(6, 3) + f(5, 2) = 10$$
We have $$f(n, 3) = \sum_{i=1}^{n-2} f(i, 2) = \sum_{i=1}^{n-2}\sum_{k=1}^{i-2}k = {(n-2)(n-3)(n-4)\over6}$$
What follows is that $$f(n, k) = \sum_{i=1}^{n-2} f(i, k-1) =\sum_{i_1=1}^{n-2} \sum_{i_2=1}^{i_1-2} {\cdot \cdot \cdot} \sum_{i_{k-1}=1}^{i_{k-2}-2} i_{k-1}$$
Computing this for different values of $k$, we can conjecture that for $k\ge 3$:
$$f(n,k) = (c_k + n(n-2k-1))(n-2){(n-k-3)!\over(n-2k)!k!}$$
Where $c_k$ is a sequence of integers. Now since :
$$f(n+2,k+1) = (c_{k+1} + (n+2)(n-2k-1))n{(n-k-4)!\over(n-2k)!(k+1)!}$$
and $$f(n+1,k+1) + f(n,k) = (c_{k+1} + (n+1)(n-2k-2))(n-1){(n-k-3)!\over(n-2k-1)!(k+1)!} + (c_k + n(n-2k-1))(n-2){(n-k-3)!\over(n-2k)!k!}$$
We get : $$(n-k-3)[(c_{k+1}+(n+1)(n-2k-2))(n-1)(n-2k) + (c_k + n(n-2k-1))(n-2)(k+1)] = (c_{k+1} + (n+2)(n-2k-1))n$$
Therefore $$-(k+3)[2k(c_{k+1}-2k-2)-2(k+1)c_k] = 0$$ and thus $$c_{k+1} = {k+1\over k}c_k + 2(k+1)$$
And $c_1 = 0$
We figured : $$f(n,k) = (c_k + n(n-2k-1))(n-2){(n-k-3)!\over(n-2k)!k!}$$ where $c_1 = 0$ and $c_{k+1} = {k+1\over k}c_k + 2(k+1)$ for any natural $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding coefficients in binomials raised to a power Recently was taught the Binomial Theorem and have begun doing practice questions. I have come across a problem which I think is more to do with the gaps in my knowledge, rather than a misunderstanding of the Binomial Theorem.
Here is the question: The coefficient of $x^3$ in the expansion of $(2+x)(3-ax)^4$ is $30$. Find the values of the constant $a$
My working: $$(3-ax)^4 = \displaystyle\sum_{i=0}^4 \displaystyle \binom{4}{i} 3^{4-i}(ax)^i $$
$$ =\displaystyle \binom{4}{0}(3)^4+\displaystyle \binom{4}{1}3^3(-ax)^1+\displaystyle \binom{4}{2}3^2(-ax)^2+\displaystyle \binom{4}{3}3(-ax)^3+\displaystyle \binom{4}{4}(-ax)^4$$
$$ = 81 -108ax + 54a^2x^2 - 12a^3x^3 + a^4x^4 $$
$$ (2-x)(81 -108ax + 54a^2x^2 - 12a^3x^3 + a^4x^4)$$ From here I noticed the only way of making an $x^3$ term is to multiply $2$ by $-12a^3x^3$ and to multiply $x$ by $54a^2x^2$ From here, I worked out that $$ -24a^3 + 54a^2 = 30 $$
$$ a^2(-24a+54)=30$$
From here, I worked out that that: $$ a = \pm\sqrt30 $$$$a = 1 $$
From the solutions to this question, I got only 1 correct solution. Which is a = 1. However, my other two solutions are incorrect. What have I done wrong here??
| $$
\begin{align}
30
&=\left[x^3\right](2+x)(3-ax)^4\\[6pt]
&=2\left[x^3\right](3-ax)^4+\left[x^2\right](3-ax)^4\\
&=2\cdot3^1\cdot(-a)^3\binom{4}{3}+3^2\cdot(-a)^2\binom{4}{2}\\
&=-24a^3+54a^2
\end{align}
$$
Therefore,
$$
\begin{align}
0
&=4a^3-9a^2+5\\
&=(a-1)\left(4a^2-5a-5\right)
\end{align}
$$
Solve for $a$.
| {
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"url": "https://math.stackexchange.com/questions/2002592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help evaluating $ \int_{0}^{π} \frac{1}{({3+cosθ})^2}dt $ $\int_{0}^{π} \frac{1}{({3+cosθ})^2}dt $
so I approached
$\frac{1}{2}\int_{0}^{2π} \frac{1}{({3+cosθ})^2}dt $
$z=e^{iθ}$,
$\\dθ=\frac{1}{iz}dz\\\cosθ = \frac{e^{iθ}+e^{-iθ}}{2} =\frac{z+z^{-1}}{2}$
$\frac{2}{i}\oint\frac{z}{(z^2+6z+1)^2}dz$
then I cannot solve the problem....
| The poles of the function are $z_1 = 2\sqrt{2}-3$ and $z_2 = -3-2\sqrt{2}$. Only $z_1$ lies inside the unit circle.
You must find the residue of this pole. It's easy to see that its a pole of order $2$, so, the residue has a value of $\displaystyle \frac{-3i}{16\sqrt{2}}$
Hence, the value of your integral is $\displaystyle \pi i \left(\frac{-3i}{16\sqrt{2}}\right) = \frac{3\pi}{16\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arithmetic-geometric progression;Sn? $S_n=1+ 3x^2+ 5x^4 +7x^6 +...+(2n-1)x^{(2n-2)}$
a)Write down the series for $x^2S_n$.
Hence show that $S_n=\frac{1}{(1-x^2)} + (2x^2)\frac{(1-x^{(2n-2)})}{(1-x^2)}- (2n-1)\frac{(x^{2n})}{(1-x^2)}$, where $x$ cannot be $\pm1$
The series for $x^2S_n = x^2 +3x^4 + 5x^6 +7x^8+...
+(2n-3)(x^{(2n)})$
So how to show?
| So here is the way to go about it:
$$S_n = 1 + 3x^2 + 5x^4 + \cdots + (2n-1)x^{(2n-2)} \rightarrow$$
$$x^2S_n = x^2 + 3x^4 + 5x^6 + \cdots + (2n-1)x^{2n}$$
Now if we subtract $x^2S_n$ from $S_n$ term by term we get:
$$S_n - x^2S_n = 1 + (3x^2 - x^2) + (5x^4 - 3x^4) + \cdots + [(2n-1)x^{2n-2} - (2n-3)x^{2n-2}] - (2n-1)x^{2n} =$$
$$= 1 + 2x^2 + 2x^4 + \cdots + 2x^{2n-2} - (2n-1)x^{2n}$$
Now, on the left hand side we factor $S_n$ out to get $(1-x^2)S_n$. On the right hand side we group the terms in 3 parts:
$$(1-x^2)S_n = [1] + [2x^2 + 2x^4 + \cdots + 2x^{2n-2}] - [(2n-1)x^{2n}]$$
Now the middle group is a geometric series:
$$G_n = 2x^2 + 2x^4 + \cdots + 2x^{2n-2} = 2(x^2 + x^4 + \cdots + x^{2n-2}) = 2\times \frac{x^2 - x^{2n}}{(1 - x^2)} = 2x^2\times \frac{1 - x^{2n-2}}{(1 - x^2)}$$
Plugging in we have $$(1-x^2)S_n = [1] + [2x^2\times \frac{(1 - x^{2n-2})}{(1 - x^2)}] - [(2n-1)x^{2n}]$$
Now dividing both sides by $(1 - x^2)$ you get:
$$S_n = \frac{1}{(1 - x^2)} + \frac{2x^2(1-x^{2n-2})}{(1-x^2)^2} - \frac{(2n-1)x^{2n}}{(1-x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of Equations with Two variables I am stuck on the following question:
Consider the system of equations
$$2x + by = 1$$
$$3x + y = c$$
For which values of b and c does the system have a unique solution?
For which values of b and c is the system inconsistent?
The internet tells me that the determinant cannot equal $0$ but I'm not sure how to apply this with the two constants
| Make the system of equation is a matrix-vector equation
$$ \begin{vmatrix} 2 & b \\ 3 & 1 \end{vmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ c \end{bmatrix} $$
Now you can see under what conditions you have two identical equations and when you have two parallel lines (each equation being a line in the xy plane).
Edit 1
If the determinant is zero $2-3b=0$, $b=\frac{2}{3}$ the two equations have equivalent left hand size
$$\left. \begin{aligned} 2 x + \frac{2}{3} y & =1 \\ 3 x + y & = c \end{aligned} \right\} \left. \begin{aligned} \frac{3}{2} \left( 2 x + \frac{2}{3} y \right) & =\frac{3}{2} (1) \\ 3 x + y & = c \end{aligned} \right\} \begin{aligned} 3 x + y & = \frac{3}{2} \\ 3 x + y & = c \end{aligned} $$
So your two cases are $c=\frac{3}{2}$ and $c \neq \frac{3}{2}$.
The first makes both equations the same with infinite solutions. For example $$\begin{aligned} x & = \frac{t}{3} \\ y & = \frac{1}{2}-t \end{aligned}$$
The second makes the equations inconsistent.
When $b \neq \frac{2}{3}$ all values of $c$ produce a result.
| {
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"url": "https://math.stackexchange.com/questions/2005505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is this trigonometric identity true? Suppose $$\frac{{{{\sin }^4}(\alpha )}}{a} + \frac{{{{\cos }^4}(\alpha )}}{b} = \frac{1}{{a + b}}$$ for some $a,b\ne 0$.
Why does $$\frac{{{{\sin }^8}(\alpha )}}{{{a^3}}} + \frac{{{{\cos }^8}(\alpha )}}{{{b^3}}} = \frac{1}{{{{(a + b)}^3}}}$$
| Solving the system
$$\begin{cases}X^2+Y^2=1\\\dfrac{X^4}{a}+\dfrac{Y^4}{b}=\dfrac{1}{a+b}\end{cases}$$ where obviously $X$ and $Y$ are the sinus and cosinus respectively, one has $$X=-\sqrt{\dfrac{a}{a+b}}\\Y=\pm\sqrt{\dfrac{b}{a+b}}$$
This gives directly the equalities $\dfrac{X^8}{a^3}=\dfrac{a^4}{a^3(a+b)^4}$ and
$\dfrac{Y^8}{b^3}=\dfrac{b^4}{b^3(a+b)^4}$ hence the result
$$\dfrac{X^8}{a^3}+\dfrac{Y^8}{b^3}=\dfrac{1}{(a+b)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Find $f(1)+\cdots+f(60)$
Let $f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$. Find $f(1)+\cdots+f(60)$.
I considered rationalizing the denominator, but that seems to make the fraction more complicated. We get $$\dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} = \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right).$$ is there an easier way?
| $$f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$$ $$= \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right)$$
Now say $a=\sqrt{2x+1}$ and $b=\sqrt{2x-1}$
Then $f(x)=\dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right)=\frac{1}{2}\left(a^2+b^2+ab\right)\left(a-b\right)=\frac{1}{2}(a^3-b^3)$
Therefore, $$\boxed{f(x)=\frac{1}{2}\left[(2x+1)^\frac{3}{2}-(2x-1)^\frac{3}{2}\right]}$$
So, $f(1)+\cdots+f(60)=\frac{1}{2}\left[(2\cdot 60+1)^\frac{3}{2}-(2\cdot 1-1)^\frac{3}{2}\right]=665$
Hope this helps you.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot 18^n - 54 \cdot 2^n = 0 \Leftrightarrow ???$$
What do I do next? Am I doing it right?
| $18^{n+1}=2^{n+1}\times 27$
$3^{2n+2}\times 2^{n+1}=2^{n+1}\times3^{3}$
$3^{2n+2}=3^3$
$2n+2=3$
$n=1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding Residue I am having difficulty with with calculating the residue for $$\text{res}[\frac{\exp(\frac{1}{z})}{z^{2}-16,},z=0]$$
I was able to calculate the residues when $z=4$ and $z=-4$. However Im not sure how to approach this part of the question.
| $e^z$ is an entire function, so $z = 0$ is an essential singularity for $e^{1/z}$.
$$e^{1/z} = \sum_{k = 0}^{+\infty} \frac{1}{k!}(z^{-k})$$
$$\frac{1}{z^2-16} = \frac{1}{16\left(\frac{z^2}{16} - 1\right)} = \frac{-1}{16} \sum_{n = 0}^{+\infty} \frac{z^{2n}}{16^n}$$
You can see the first terms of the expansion ad $z = 0$:
$$-\frac{1}{16}\left(1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3!z^3} + \frac{1}{4!z^4} + \frac{1}{5!z^5} + \frac{1}{6!z^6} + \frac{1}{7!z^7} + \frac{1}{8!z^8} + \frac{1}{9!z^9} + \ldots \right) \times \\\\ \times \left(1 + \frac{z^2}{16} + \frac{z^4}{16^2} + \frac{z^6}{16^3} + \frac{z^8}{16^4} + \ldots\right)$$
Now you know that the residue is the coefficient of the term $z^{-1}$ and you can easily obtain it by finding all the term depending only on $z^{-1}$: for example the second terms on the left times the first term on the right bracket, plus the the fourth term on the left bracket times the second on the right bracket and so on, and you will get
$$-\frac{1}{16}\left(\frac{1}{z} + \frac{1}{16\times 3! z} + \frac{1}{16^2\times 5! z} + \frac{1}{16^3\times 7! z} + \ldots\right)$$
It's easy to guess the series for that expansion:
$$\frac{-1}{16z} \sum_{k = 0}^{+\infty} \frac{1}{16^k\cdot (2k + 1)!}$$
The series converges to
$$4\sinh \left(\frac{1}{4}\right)$$
Hence the residue is
$$\frac{-1}{4} \sinh \left(\frac{1}{4}\right)$$
That is indeed the value of your residue
$$-0.06315307920204207...$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the coefficient of a term in an expansion. I was looking over some problems from the American Invitational Mathematics Examination (AIME) and this problem from the 2004 AIME caught my eye:
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Evidently, the coefficient of $x$ is $-8$, since summing the coefficients of $x$ within the binomials gives $-1+2-3+\cdots+14-15 = -8$, but how would I go about finding the coeffecient of $x^2$?
| FIRST METHOD
$$\begin{align}
&\;\;\;\;\underbrace{\color{magenta}{(1-x)(1+2x)}\color{blue}{(1-3x)(1+4x)}\cdots \color{green}{(1-13x)(1+14x)}}_{14\text{ terms}}(1-15x)\\
&=\underbrace{\color{magenta}{(1+x-2x^2)}\color{blue}{(1+x-12x^2)}\cdots \color{green}{(1+x-182x^2)}}_{7\text{ terms}} (1-15x)\\
&=(1+ax+bx^2+\cdots+\bullet\; x^{14})(1-15x)\\
&=1+(a-15)x+(b-15a)x^2+\cdots
\end{align}$$
Equating coefficients gives
$$a=7\\
b=-(\underbrace{1\cdot 2}_2+\underbrace{3\cdot 4}_{12}+\cdots +\underbrace{13\cdot 14}_{182})+\binom 72=-483$$
The coefficient of $x^2$ is
$$C=b-15a=-588\\
|C|=\color{red}{588}$$
SECOND METHOD
Consider combination of two factors at a time.
Coefficient if $x^2$ is
$$\begin{align}
\sum_{i=2}^{15}(-1)^i\ i\color{green}{\sum_{j=1}^{i-1}(-1)^j\ j}
&=\sum_{i=2}^{15}(-1)^i\ i\cdot \color{green}{(-1)^{i-1}\bigg\lfloor \frac i2\bigg\rfloor}\\
&=-\sum_{i=2}^{15}i\bigg\lfloor \frac i2\bigg\rfloor\\
&\color{lightgrey}{=-\left[(2+3)\cdot 1+(4+5)\cdot 2+(6+7)\cdot 3+\cdots+(14+15)\cdot 7\right]}\\
&\color{lightgrey}{=\sum_{r=1}^7 (2r+(2r+1))r}\\
&=\color{red}{-588}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle
If $a$, $b$, and $c$ are the three sides of a triangle, then
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
lies in what interval?
| As expression
$$\tag{1}f(a,b,c):=\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
is invariant in any change of scale, we introduce the following (value) constraint:
$$\tag{2}g(a,b,c):=a+b+c=1$$
This means that the subsequent work is in this simplex, or more exactly in the subset of this simplex with supplementary (domain) constraints:
$$\tag{3} a \leq b+c, \ \ b \leq c+a, \ \ c \leq a+b$$
(in case of equality, we have flat triangles).
Under all these constraints, $(1)$ becomes:
$$\tag{4}f(a,b,c)=\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$$
maximizing/minimizing $(4)$ under constraint $(2)$ is done by writing Lagrange equations:
$$grad (f) \ = \ \lambda \ grad(g) \ \ \ \iff \ \ \ \cases{\dfrac{1}{(1-a)^2}=\lambda 1\\ \dfrac{1}{(1-b)^2}=\lambda 1\\ \dfrac{1}{(1-c)^2}=\lambda 1},$$
from which $a=b=c=\frac13$ (equilateral triangle), giving $f(\frac13,\frac13,\frac13)=\frac32.$
This extremum is a minimum because the Hessian matrix,
diag $\left(\dfrac{2}{(1-a)^3},\dfrac{2}{(1-b)^3},\dfrac{2}{(1-c)^3}\right)$ is positive definite.
Now, the maxima(s) should be found on the borders. As all the borders play a symmetrical role, consider only boundary $c=a+b$, which implies (using $(2)$)
$$1=a+b+c=2(a+b)\ \ \ \Longrightarrow \ \ \ a+b=\frac12$$
$$\tag{5} \Longrightarrow \ \ \ 0 \leq a\leq 1/2, b=\dfrac12-a, c=\frac12.$$
Thus:
$$f(a,b,c)=f(a,\frac12-a,\frac12)=\dfrac{a}{1-a}+2-2a.$$
A rapid study of this function for $a \in [0,\frac12]$ shows that the value of its maximum is $2$, occurring for $a=0$ or $a=\frac12$. Both associated with a "flat" triangle with sides $(\frac12,\frac12,0) $.
Conclusion: the values taken by $f$ are situated in the interval $[\frac32,2]$. As $f$ is continuous on a compact domain, it is the whole interval $[\frac32,2].$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Solve a given trigonometric equation Solve the following equation:
$$\sin x + \sin \left( x + \frac{7\pi}{24} \right) = \frac{\sqrt{2 - \sqrt 2}}{2} + \frac{\sqrt 6 + \sqrt 2}{4}$$
So far, I found out that $\frac{\sqrt{2 - \sqrt 2}}{2} = \sin \frac{\pi}{8}$ and $\frac{\sqrt 6 + \sqrt 2}{4} = \sin \frac{7\pi}{12}$.
Thank you!
| Let $f(x)=\sin x +\sin (x+7\pi /24).$
We have $$f(\pi /8)=\sin (\pi /8)+\sin (\pi /8+7\pi /24)=\sin (\pi /8)+\sin (10\pi /24)=$$ $$=\sin (\pi /8)+\sin (14\pi /24)=\sin (\pi /8)+\sin (7\pi /12)$$ which is the RHS of your equation.
Now $f'(x)=\cos x +\cos (x+7\pi /24).$
So $f(x)$ is strictly increasing for $-\pi /2-7\pi /48\leq x\leq \pi /2-7\pi /48.$
And $f(x)$ is strictly decreasing for $\pi/2 -7\pi /48\leq x< 3\pi/2-7\pi /48.$
So there are at most 2 values of $x$ in $[-\pi/2-7\pi /48,\;\; 3\pi /2-7\pi /48)$ for which $f(x)=f(\pi /8).$ And there do exist 2 values: $x=\pi /8$ and $x=\pi-\pi /8=7\pi /8.$
So $x=\pi /8 +2\pi n$ or $x=7\pi /8+2\pi n$ for some (any) $n\in \mathbb Z.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Periodicity of Sine from infinite Product Formula: $z\prod_{n=1}^{\infty}\left( 1 - \frac{z^2}{n^2\pi^2}\right)$ Let $\sin z$ be defined by the following infinite product:
$$
\sin (z) = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2\pi^2} \right)
$$
How can one derive that $\sin(z + 2\pi) = \sin(z)$?
| One may write
$$
\begin{align}
&(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{(z+2\pi)^2}{n^2\pi^2}\right)
\\\\=\:&(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{z+2\pi}{n\pi}\right)\prod_{n=1}^{N} \left(1+\frac{z+2\pi}{n\pi}\right)
\\\\=\:&(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{z}{n\pi}-\frac{2}{n}\right)\prod_{n=1}^{N} \left(1+\frac{z}{n\pi}+\frac{2}{n}\right)
\\\\=\:&(z+2\pi)\prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right) \prod_{n=1}^{N} \left(1-\frac{2}{n\left(1-\frac{z}{n\pi} \right)}\right)\left(1+\frac{2}{n\left(1+\frac{z}{n\pi} \right)}\right)
\\\\=\:&(z+2\pi)\prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right) \cdot\prod_{n=1}^{N} \frac{((n-2) \pi -z) ((n+2) \pi +z)}{(n \pi -z) (n \pi +z)}
\end{align}
$$ then factors telescope in the latter product giving
$$
\begin{align}
&(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{(z+2\pi)^2}{n^2\pi^2}\right)
\\\\=\:&(z+2\pi)\cdot \prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right)\cdot \frac{z(\pi+N\pi +z) ((2+N)\pi+z)}{(z+2\pi)((N-1)\pi-z)(N\pi-z)}
\\\\=\:&z\cdot \prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right)\cdot \frac{(\pi+N\pi+z)((2+N)\pi+z)}{((N-1)\pi-z) (N\pi-z)}
\end{align}
$$ then letting $N \to \infty$ gives the desired identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving limits of two sequences I've trouble proving that:
a) $\lim(\frac{1}{\sqrt[n]{2}-1} - \frac{2}{\sqrt[n]{4}-1})=\frac{1}{2}$
b) $\lim \frac{2n-1}{2n+1}\cdot \frac{2n-2}{2n+2}\cdot ... \cdot \frac{n}{3n}=0$.
In a) I've tried using formula $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ which gives me $\lim(\frac{1}{\sqrt[n]{2}-1} - \frac{2}{\sqrt[n]{4}-1})=\lim \sum_{k=1}^{n}\frac{2}{\sqrt[n]{2^k}}(1- \frac{4}{3\sqrt[n]{2^k}})$ and I can't go any further.
In b) I've tried using Squeeze Theorem $\lim \frac{2n-1}{2n+1}\cdot \frac{2n-2}{2n+2}\cdot ... \cdot \frac{n}{3n}>(\frac{1}{3})^n$ but i cant find second inequality.
| *
*One may recall that, by the Taylor series expansion, as $n \to
\infty$, $$ \sqrt[n]{2}=e^{\large\frac{\ln 2}{n}}=1+\frac{\ln
2}{n}+\frac{\ln^2 2}{2n^2}+\mathcal{O}\left(\frac1{n^3}\right) $$
giving $$ \frac{1}{\sqrt[n]{2}-1}=\frac{n}{\ln
2}-\frac12+\mathcal{O}\left(\frac1{n}\right) $$ similarly $$
\frac{2}{\sqrt[n]{4}-1}=\frac{n}{\ln
2}-1+\mathcal{O}\left(\frac1{n}\right) $$ then, as $n \to \infty$,
$$
\frac{1}{\sqrt[n]{2}-1}-\frac{2}{\sqrt[n]{4}-1}=\frac12+\mathcal{O}\left(\frac1{n}\right)
$$ and one gets the announced limit.
*One may observe that, as $n \to \infty$, $$ \frac{2n-1}{2n+1}\cdot
\frac{2n-2}{2n+2} \cdots \frac{n}{3n}=\frac12\cdot\frac{(2n)!}{n!} \cdot
\frac{(2n)!}{(3n)!} \sim \left(\frac{16}{27} \right)^n\cdot \frac{2
n}{\sqrt{3}} \to 0$$ by using Stirling's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve $a^2 + b^2 = t, 2ab = 1/t$ for $a,b$ I want to solve
$$
a^2 + b^2 = t \\
2ab = 1/t
$$
for $a$ and $b$ in terms of $t$. Wolfram solves it but the solution is very dirty.
| $$
\left \{ \begin{array}{rclr}
a^2+b^2 &=& t & \qquad \cdots \cdots (1) \\
2ab &=& \dfrac{1}{t} & \qquad \cdots \cdots (2) \\
\end{array}
\right.$$
From $(1)$, $$t \ge 0$$
By $AM \ge GM$, $$t\ge \frac{1}{t} \implies t\ge 1$$
$(1)+(2)$, $$(a+b)^2=t+\frac{1}{t} \implies |a+b|=\sqrt{t+\frac{1}{t}}$$
$(1)-(2)$, $$(a-b)^2=t-\frac{1}{t} \implies |a-b|=\sqrt{t-\frac{1}{t}}$$
Since $2ab=\dfrac{1}{t} \in (0,1]$,
$$
\begin{pmatrix} a \\ b \end{pmatrix}=
\begin{pmatrix}
\frac{1}{2} \left( \sqrt{t+\frac{1}{t}} \pm \sqrt{t-\frac{1}{t}} \right) \\
\frac{1}{2} \left( \sqrt{t+\frac{1}{t}} \mp \sqrt{t-\frac{1}{t}} \right)
\end{pmatrix}
\quad \text{or} \quad
\begin{pmatrix}
\frac{1}{2} \left( -\sqrt{t+\frac{1}{t}} \pm \sqrt{t-\frac{1}{t}} \right) \\
\frac{1}{2} \left( -\sqrt{t+\frac{1}{t}} \mp \sqrt{t-\frac{1}{t}} \right)
\end{pmatrix}$$
P.S.
The solution by Wolfram Alpha isn't "dirty", it takes care the sign of $2ab$ after simplification.
Eliminating $b$ gives biquadratic in $a$:
$$4t^2a^4-4t^3a^2+1=0 \implies a^2=\frac{t^2 \pm \sqrt{t^4-1}}{2t}$$
$$
\begin{pmatrix} a \\ b \end{pmatrix}=
\begin{pmatrix}
\sqrt{\dfrac{t^2 \pm \sqrt{t^4-1}}{2t}} \\
\sqrt{\dfrac{t^2 \mp \sqrt{t^4-1}}{2t}}
\end{pmatrix}
\quad \text{or} \quad
\begin{pmatrix}
-\sqrt{\dfrac{t^2 \pm \sqrt{t^4-1}}{2t}} \\
-\sqrt{\dfrac{t^2 \mp \sqrt{t^4-1}}{2t}}
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Probability of point inside of rectangle $(0,0),(2,0),(2,1),(0,1)$ closer to $(0,0)$ than $(3,1)$. A point $P$ is randomly selected from the rectangular region with vertices $(0,0)$, $(2,0)$, $(2,1)$, $(0,1)$. What is the probability that $P$ is closer to the origin than it is to the point $(3,1)$?
| Distance from $(0,0) \le$Distance from $(3,1)$
i.e. let the point be $(x,y)$ then
$$\sqrt{x^2+y^2}\le \sqrt{(x-3)^2+(y-1)^2}$$
Squaring both sides canceling out returns following:-
$$y\le5-3x \tag{1}$$
The point should be on the left side of line $5-3x$
So we calculate the area bounded by inequality $(1)$ and rectangle and divide it by total area.
Line $5-3x$ cuts rectangle at $(\frac{4}{3},1)$ and $(\frac{5}{3},0)$. So $$\frac{\text{Required Area}}{\text{Total Area}}=\frac{1/2\times1/3\times 1+4/3+1/3}{2}=\frac{11}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\sqrt{6}\cdot x^4 - (\sqrt{3}+\frac{3}{2}\sqrt{2})x^2 +\frac{3}{2} = 0 $ Of course I could put this is mathematica/wolframalpha or use a formula, but I think in here is a trick how to solve it very simply, but I can't figure it out. Help/Hints very appreciated
$\sqrt{6}\cdot x^4 - (\sqrt{3}+\frac{3}{2}\sqrt{2})x^2 +\frac{3}{2} = 0 \Leftrightarrow $
$x^2(\sqrt{6}x^2-\sqrt{3}+ \frac{3}{2}\sqrt{2})= -\frac{3}{2}\Leftrightarrow $
| As there is no $x^3$ or $x$ term you can treat it as a quadratic in $x^2$.
Let $t=x^2$ as dvix suggested.
$$\sqrt{6}\cdot t^2-\sqrt{3}t-\frac{3}{2}\sqrt{2}t+\frac{3}{2}=0$$
Then factorize as suggested by G.Sassetelli.
$$\sqrt{3}t(\sqrt{2}t-1)-\frac{3}{2}(\sqrt{2}t-1)=0$$
$$\left(\sqrt{3}t-\frac{3}{2}\right)\left(\sqrt{2}t-1\right)=0$$
$$t=\frac{\sqrt{3}}{2}\text{ or }t=\frac{1}{\sqrt{2}}$$
Note: This makes me think the original question was trig related so it might have been useful to include your steps which let up to the equation you were trying to solve.
$$x^2=\frac{\sqrt{3}}{2}\text{ or }x^2=\frac{1}{\sqrt{2}}$$
$$x=\frac{\sqrt[4]{3}}{\sqrt{2}}\text{ or }x=\frac{1}{\sqrt[4]{2}}$$
Or with rational denominators:
$$x=\frac{\sqrt[4]{12}}{2}\text{ or }x=\frac{\sqrt[4]{8}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
There are no rational $a, b, c $ with $a+b\sqrt{u}+c\sqrt{v} =\sqrt{uv} $ If $u$ and $v$
are rationals such that
$\sqrt{u}, \sqrt{v}, $ and $\sqrt{uv}$
are all irrational,
show that there are no
rational $a, b, $ and $c$
such that
$a+b\sqrt{u}+c\sqrt{v}
=\sqrt{uv}
$.
This is a generalization
of my answer
to this question:
Proof by contradiction: finding integers that satisfy $a+b\sqrt{2}+c\sqrt{3}=\sqrt{6}$.
I will post a solution
in two days
if there are
no posted solutions.
| Lemma: $\sqrt{u}$ and $\sqrt{v}$ must be linearly independent over the rationals.
Else $a \sqrt{u} + b \sqrt{v} \in \mathbb{Q}$ with $a,b \in \mathbb{Q} \setminus \{0\}$ $\implies$ $(a \sqrt{u} + b \sqrt{v})^2 = a^2 u + b^2 v + 2ab \sqrt{uv} \in \mathbb{Q}$ $\implies$ $\sqrt{uv} \in \mathbb{Q}$, contrary to the hypothesis.
Proof:
$$
\sqrt{uv}=a+b\sqrt{u}+c\sqrt{v} \quad\quad \Big| \cdot \sqrt{u}
$$
$$
\begin{align}
u \sqrt{v} & =a\sqrt{u}+bu+c\sqrt{uv} \\
& = a\sqrt{u}+bu+c(a+b\sqrt{u}+c\sqrt{v}) \\
& = (a+bc)\sqrt{u} + c^2 \sqrt{v} + ac + bu
\end{align}
$$
By the linear independence proved in the lemma, the coefficients of $\sqrt{v}$ must match i.e. $\,u = c^2\,$ but that contradicts the irrationality of $\sqrt{u}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Tiling of different sized chess boards. Combo Problem Show that a 444-by-77 chessboard can be completely tiled by 1-by-6 pieces
but a 44-by-777 chessboard cannot.
I tried to show this by taking a multiplication table of both boards and trying to find the area that a tile covers.
This was not the right way to approach this problem. Anyone know how to do it?
| HINT: Since $444$ is a multiple of $6$, it’s easy to tile the $444\times 77$ board. Color the $44\times 777$ board diagonally with $6$ colors, like this:
$$\begin{array}{|c|c|c|} \hline
1&2&3&4&5&6&1&2&3&\ldots\\ \hline
2&3&4&5&6&1&2&3&4&\ldots\\ \hline
3&4&5&6&1&2&3&4&5&\ldots\\ \hline
4&5&6&1&2&3&4&5&6&\ldots\\ \hline
5&6&1&2&3&4&5&6&1&\ldots\\ \hline
6&1&2&3&4&5&6&1&2&\ldots\\ \hline
1&2&3&4&5&6&1&2&3&\ldots\\ \hline
2&3&4&5&6&1&2&3&4&\ldots\\ \hline
3&4&5&6&1&2&3&4&5&\ldots\\ \hline
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ \hline
\end{array}$$
Each tile must cover exactly one square of each color. Show that this is impossible, because there are not the same numbers of squares of each color.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Nice try of yours. You can further continue...:
$$
x^{2000} + \frac{1}{x^{2000}} = \left(x^{1000} + \frac{1}{x^{1000}}\right)^2 - 2 \\
x^{1000} + \frac{1}{x^{1000}} = \left(x^{500} + \frac{1}{x^{500}}\right)^2 - 2 \\
x^{500} + \frac{1}{x^{500}} = \left(x^{250} + \frac{1}{x^{250}}\right)^2 - 2 \\
\vdots \\
x^{n} + \frac{1}{x^n} = \left(x^{n/2} + \frac{1}{x^{n/2}}\right)^2 - 2 \\
$$
In the case of $2000 = 2^4 \cdot 5^3$. That is, when you are done with all the $2$s, you will be left with $5$:
$$
x^{125} + \frac{1}{x^{125}} =
\left[\left(x^{25}\right)^5 + \left(\frac{1}{x^{25}}\right)^5\right] =
$$
We do basically the same you did:
$$
(a+b)^5 = {{5}\choose{0}}a^5 + {{5}\choose{1}}a^4 b + \cdots \\
(a+a^{-1})^5 = {{5}\choose{0}}a^5 + {{5}\choose{5}}b^{-5} + \cdots \\
$$
Notice the intermediate terms:
$$
{{5}\choose{1}}a^{4} a^{-1} +
{{5}\choose{2}}a^{3} a^{-2} +
{{5}\choose{3}}a^{2} a^{-3} +
{{5}\choose{4}}a^{1} a^{-4}
=
{{5}\choose{1}}a^{3} +
{{5}\choose{2}}a +
{{5}\choose{3}}a^{-1} +
{{5}\choose{4}}a^{-3}
$$
That is:
$$
\left[x^{5} + \frac{1}{x^{5}}\right]^5 =
\left[x^{5} + \frac{1}{x^{5}}\right] +
10\left[\left(x^{5}\right)^3 + \left(\frac{1}{x^{5}}\right)^3\right] +
5\left[\left(x^{5}\right)^5 + \left(\frac{1}{x^{5}}\right)^5\right]
$$
You can keep doing so and simplifying to smaller and smaller terms. You can do this because:
$$
{{n}\choose{k}} = {{n}\choose{n-k}} = \frac{n!}{k!(n-k)!}
$$
That is, $(a + a^{-1})^n$ will have the terms $a^{-k} a^{n-k}$ and $a^k a^{-(n-k)}$ with the same coeficients. And this term forms:
$$
a^k a^{-n+k} = a^{k-n+k} = a^{2k-n} \\
a^{-k} a^{n-k} = a^{-k+n-k} = a^{-2k+n} = a^{-(2k-n)}
$$
So, the coeficients opposite terms will be equal, then you will always be able to build terms like $x^k + 1/x^k$. Thus you can keep doing it until you are left only with the $x + 1/x$. It might take a while..... =D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 5
} |
Elliptic Integral $$\int_{1}^{x}\frac{dt}{\sqrt{t^3-1}}$$ does this have a closed form involving jacobi elliptic functions of parameter $k$?
N.B I tried with the change of variables $t=1+k\frac{1-u}{1+u}$. But this leads no where. http://mathworld.wolfram.com/JacobiEllipticFunctions.html
update the above integral is equivalent to $$\int\limits_{0}^{\sec^{-1}x^{\frac{3}{2}}}\sec^{\frac{2}{3}}tdt$$.
| Here is the proof of Ng Chung Tak's solution.
The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and
the lower limit of integration equal to the real root. So we have
\begin{equation}
\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(x-a)}{m+(x-a)},k \right)
\end{equation}
where
\begin{equation}
m^{2} = (r - a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} - \frac{a-r}{2m}
\end{equation}
$\mathrm{cn}^{-1}z$ is the inverse of one of the Jacobi elliptic functions and $k$ is the modulus.
To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}
We now have
\begin{equation}
a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} - \frac{3}{4\sqrt{3}}
\end{equation}
and thus
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \, \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 - x}{\sqrt{3} \, - 1 + x},\sqrt{\frac{1}{2} - \frac{3}{4\sqrt{3}}} \right)
\end{equation}
Note that the value of $k$ here is equal to that of Ng Chung Tak's solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the $\lim\limits_{x \to 0+} (\frac{1}{x} - \frac{1}{\arctan x})$ Right so, so far I have gotten the denominators the same and have the achieved the indeterminate 0/0 as $\lim\limits_{x \to 0+} \frac{\arctan x - x}{x\arctan x}$ once I apply L' Hospital's Rule though, it gets really messy. So I am wondering if there is a more elegant or efficient way to find the answer without having to deal with the nesting fractions.
| \begin{align}
& \lim_{x\to0} \frac{\arctan x - x}{x\arctan x} = \lim_{x\to0} \frac{\left(\dfrac 1 {1+x^2} \right) - 1}{x\left( \dfrac 1 {1+x^2} \right) + 1\cdot\arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac{1 - (1+x^2)}{ x + (1+x^2) \arctan x } = \lim_{x\to0} \frac{-x^2}{x + (1+x^2)\arctan x} \\[10pt]
= {} & \left( \lim_{x\to0} \frac {\left(\dfrac x {\arctan x}\right)} {\left( \dfrac x {\arctan x} \right) + (1+x^2)} \right) \cdot \left( \lim_{x\to0} (-x) \right) \\[10pt]
\end{align}
Now use the fact that $\displaystyle\lim_{x\to0} \frac x {\arctan x} = 1.$
This is to be expected because $x \arctan x$ crosses the axis with a slope of $1$ and is an odd function, so no $x^2$ term can appear in its power series, so in $\arctan x - x$ the $x$ term cancels and there's no $x^2$ term, so it's like $x^3 + \text{higher-degree terms}$, whereas $x\arctan x$ is like $x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2^{\sqrt{n}}>n^2$ inequality proof I am trying to do some trick with lim but get stuck on $2^{\sqrt{n}}>n^2$. I want to prove it for n large enough but don't know how to do induction step. Maybe this fact does not require a proof at all...
| Taking for example $n=400$ as a base case
$$2^\sqrt{400}=2^{20}>10^6>400^2=16\cdot 10^4,x=\sqrt{n}\\2^x>x^4\\2^{x+1}>(x+1)^4\\2\cdot 2^x>x^4+4x^3+6x^2+4x+1\\x^4+15x^3>x^4+4x^3+6x^2+4x+1\\2^x+15x^3>x^4+15x^3\\\frac{3}{4}x^4\geq15 x^3\\2\cdot2^x>2^x+\frac{3}{4}2^x>x^4+4x^3+6x^2+4x+1$$
The $\frac{3}{4}x^4\geq15 x^3$ comes since $x\geq 20$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lagrange Multiplier Three Dimension (2). Consider the surface produced by the equation $xy^2z^3 = 2$ . Find the points in this surface closest to the origin?
So far here is what I have,
Let
$$ f(x,y,z) = x^2 + y^2 + z^2 $$
Where f is the square of the distance from the point $(x,y,z)$ to the origin. We want to minimize $f$, so our constraint is
$$ g(x,y,z) = xy^2z^3 - 2 $$
We must solve
$$\nabla f = \lambda \nabla g \quad \text{ and } \quad g = 0$$
By the former we must have
$$
\Bigg( \begin{matrix} 2x \\ 2y \\ 2z \\ \end{matrix} \Bigg) =
\lambda \Bigg( \begin{matrix}
y^2z^3 \\
2xyz^3 \\
3xy^2z^2 \\
\end{matrix} \Bigg) $$
Solving this system we get equations
$$ 2x = \lambda \text{ } y^2z^3 $$ $$ $$
$$ 2y = 2\lambda \text{ } xyz^3 $$ $$ $$
$$ 2z = 3\lambda \text{ } xy^2z^2$$ $$ $$
This is where I start to get stuck. From what I gathered from websites and youtube videos is that I would try to solve it using either cases, or by having the left side equal the same thing. I tried the latter and believe the answers were
$$ y^2 = 2x^2 \to y = \pm x\sqrt(2) \text{ and}$$
$$ 2z^2 = 3y^2 \to z = \pm y \sqrt{\frac{3}{2}} $$
I got these solutions by multiplying the first, second, and third equation by yz, xz, xy, respectively. Then having the first equation equal the second and then having the second equation = third equation. So it should look like this:
$$ 2xyz_1 = \lambda \text{ } y^3z^4 $$ $$ $$
$$ 2xyz_2 = 2\lambda \text{ } x^2yz^4 $$ $$ $$
$$ 2xyz_3 = 3\lambda \text{ } x^2y^3z^2$$ $$ $$
But from here I am stuck and I have no idea what to do or even if I am on the right path. Any help would be greatly appreciated.
| Hint
Note that $xyz\ne 0$ because if one is zero the other two must be zero. Thus we have that
$$\dfrac{2x}{y^2z^3}=\frac{2y}{2xyz^3}=\frac{2z}{3xy^2z^2}.$$ Simplifying the fractions we arrive at
$$\dfrac{2x}{y^2z^3}=\frac{1}{xz^3}=\frac{2}{3xy^2z}.$$ Now we cancel a $z$ in each denominator getting
$$\dfrac{2x}{y^2z^2}=\frac{1}{xz^2}=\frac{2}{3xy^2}.$$ From the first and the third we have $$6x^2y^2=2y^2z^2\implies z^2=3x^2.$$ From the second and the third $$3xy^2=2xz^2\implies 3y^2=2z^2\implies y^2=2x^2.$$ Thus
$$y=\pm\sqrt2 x,\quad z=\pm \sqrt 3 x.$$ Now use the constraint to get
$$\pm x\cdot 2x^2 \cdot 3\sqrt 3 x^2=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the exact value of the series $\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot.....}}}$ Kindly tell me
what is the value of:
$\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot.....}}}$
According to ramanujan ,it is equal to 3
I want to know how...
| This formula makes no sense. When u write "$\ldots$" it's like saying "and so on" where you expect the reader to understand from context what you are talking about.
For example if commonly when u see $1+2+3+4+\ldots$ you would think about a limit of sequence of partial sums therefore you would said $1+2+3+4+\ldots=\infty$ since that sequence does not converge.
However, speaking of ramanujan, he used different context where he stated that $1+2+3+4+\ldots=-{1 \over 12}$ simply because he meant something different by "$\ldots$"
Returning to your question. Your "$\ldots$" does not point to any scheme.
ramanujan wrote something like it's 3 because
$3=\sqrt{9}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\cdot5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\ldots=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}}$
Unfortunately such argument would work for any positive value.
For example $10=\sqrt{100}=\sqrt{1+2\cdot{99 \over 2}}=\sqrt{1+2\sqrt{1+3\cdot{ 9797 \over 12}}}=\ldots$
the weird fraction will become irrelevant
| {
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Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$ Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$
i know that $ \sum _{k=1} 1/k^2 = \pi^2/6$ how to do prove this
| You should note that $$\frac{\pi^2}{6}=\sum_{k=1}^\infty\frac{1}{k^2}=\sum_{k=1}^\infty\frac{1}{(2k)^2}+\sum_{k=1}^\infty\frac{1}{(2k-1)^2}$$
$$=\frac{1}{4}\sum_{k=1}^\infty\frac{1}{k^2}+\sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{24}+\sum_{k=1}^\infty\frac{1}{(2k-1)^2}$$
This shows that
$$\sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$$
Now recognize that
$$\sum_{k=1}^\infty\frac{1}{\left(k-\frac{1}{2}\right)^2}=\sum_{k=1}^\infty\frac{4}{(2k-1)^2}$$
THis should help lead you to your goal.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove induction when n is part of the sum How to Prove
$$\sum_{k=1}^{2n} (-1)^{k-1}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{k+n}$$
by induction
we did a few of induction but the n was never a part of the sum so im kinda lost on where to start because if I would try to use n+1 i have no idea how the second sum changes
| When you are evaluating a complex sum, it is often helpful to try a few cases.
We can have the case $n=1$, and we get
$$\sum_{k=1}^{2}(-1)^{k-1}\frac{1}{k}=\sum_{k=1}^{1}\frac{1}{k+n}$$
$$\frac{1}{1}-\frac{1}{2}=\frac{1}{2}\text{.}$$
Indeed, the equation holds.
Let's try a few more values of $n$ to see if there is a clear pattern:
$$n=2$$
$$\sum_{k=1}^{4}(-1)^{k-1}\frac{1}{k}=\sum_{k=1}^{2}\frac{1}{k+n}$$
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{1}{3}+\frac{1}{4}$$
$$n=3$$
$$\sum_{k=1}^{6}(-1)^{k-1}\frac{1}{k}=\sum_{k=1}^{3}\frac{1}{k+n}$$
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$$
Looking at this last equation, we see some clear patterns in change. The left side is pretty easy to see, but the right side is a little more subtle. In particular,
$$\left(\sum_{k=1}^{n}\frac{1}{k+n}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{k=1}^{n+1}\frac{1}{k+n+1}\text{.}$$
On the left hand side, we have that
$$\left(\sum_{k=1}^{2n}(-1)^{k-1}\frac{1}{k}\right)+\frac{1}{2n+1}-\frac{1}{2n+2}=\sum_{k=1}^{2(n+1)}(-1)^{k-1}\frac{1}{k}\text{.}$$
Formulate the inductive step, use a base case, and you are done!
| {
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Integer solutions of $n^2+n = 2x^2+2x$ I know that the integers solutions of the equation:
$$ n^2+n = 2x^2+2x $$
are
$$n = \frac{1}{4} (-(3 - 2 \sqrt{2})^m - \sqrt{2} (3 - 2 \sqrt{2})^m - (3 + 2 \sqrt{2})^m + \sqrt{2} (3 + 2 \sqrt{2})^m + 2),$$
$$x = \frac{1}{8} (2 (3 - 2 \sqrt{2})^m + \sqrt{2} (3 - 2 \sqrt{2})^m + 2 (3 + 2 \sqrt{2})^m - \sqrt{2} (3 + 2 \sqrt{2})^m + 4),$$
$m \in \mathbb{Z}, m\ge0$
but I don't understand how.
Someone can point me in the right direction to solve this problem.
| One way is to set $m=2n+1$ and $y=2x+1$ and get the equivalent Pell equation $m^2=2y^2-1$.
The fundamental solution is $m=1, y=1$ and the general solution comes from the odd powers of $1+\sqrt2$. You go from one solution to another by multiplying by $(1+\sqrt2)^2=3+2\sqrt2$. This gives
$$
\pmatrix{ m_{k+1} \\ y_{k+1}}
=
\pmatrix{ 3 & 4 \\ 2 & 3}
\pmatrix{ m_{k} \\ y_{k}},
\qquad
\pmatrix{ m_{0} \\ y_{0}}
=\pmatrix{ 1 \\ 1}
$$
whose eigenvalues are $3 \pm 2 \sqrt2$. The general solution is a linear combination of powers of these eigenvalues.
You can avoid eigenvalues etc by considering the powers of $1-\sqrt2$, which helps extract both parts.
| {
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"timestamp": "2023-03-29T00:00:00",
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Do $z^{3/4}=-1 $ solutions exist? I want to get some perspective on complex exponentiation and the best possible approach to solve them.
$ z^{3/4}=-1 ;z \in \mathbb{C}$
Let $z=x+iy=\rho e^{i \theta} $ where $\rho =|z|$ and tan$\theta=\frac xy$
$w=-1=(1).e^{i(\pi+2k\pi)} $
According to me, there is a simple way to prove there is a solution to this equation. I may very well be wrong, but here I go.
$$ \frac{3\theta}{4}=\pi+2k\pi $$
$$ \theta =\frac{4\pi}{3}+\frac{8k\pi}{3}$$
Which gives me 3 unique solutions for $\theta$: $\frac{4\pi}{3},\frac{2\pi}{3},0$
I'd then like to check that they verify $|z|=1$, I can disqualify $\theta=0$ in a jiffy. I then check:
$$tan\frac{4\pi}{3}=\frac{y}{x}$$
$$x tan\frac{4\pi}{3}=y$$
So from $|z|=1$:
$$ \sqrt{(x tan\frac{4\pi}{3})^2 +x^2}=1$$
$$x=\frac{+}{-}\frac{1}{ \sqrt{ (tan^2\frac{4\pi}{3} +1)} } $$
So nothing seems out of place to me so far... Wolfram Alpha insists there are no solutions and my calculator doesn't make the cut for complex numbers.
| Let $z=r(\cos(\theta)+i\sin(\theta))$, by De-Moivre's theorem, we have: $z^3=r^3(\cos(3\theta)+i\sin(3\theta))$ with $0\leq\theta\leq2\pi$. Now we have to compute the value of: $$\left(z^3\right)^{1/4}=\left(r^3(\cos(3\theta)+i\sin(3\theta)\right)^{1/4}$$
with $0\leq\phi_k\leq2\pi$. Using again De-Moivre' theorem, we have:
$$w^4=z^3\leftrightarrow R^4(cos(4\phi)+i\sin(4\phi))=r^3(\cos(3\theta)+i\sin(3\theta))$$
From here, we deduce that: $$\phi_k=\frac{1}{4}(3\theta+2k\pi), k=0,1,2,3$$ and $R^4=r^3$. Now we can impose $w=-1$ and obtain:
$$(r^3)^{1/4}\left(\cos\left(\frac{3\theta+2k\pi}{4}\right)+i\sin\left(\frac{3\theta+2k\pi}{4}\right)\right)=1(\cos(\pi)+i\sin(\pi))$$
This leads to:
$$3\theta=4\pi+8k_1\pi-2k\pi,k_1=0,1,2$$
The solutions are: $$\theta=\pi/3 \: \vee \theta=5\pi/3 \: \vee \theta=\pi$$
In conclusion, I think that the solutions to this equations are the all $z$ such that $w=z^3=1$ because in that case the equation $w^{1/4}=-1$ has four solutions that in Gauss plane are the vertex of a square.
| {
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How to prove there are no solutions to the equation $a^2-4ab+b^2=0$ if $a$ and $b$ are real numbers and $b \neq 0$? I am trying to prove $a^2-4ab+b^2=0$ has no solutions for all real numbers $a$ and $b$ and $b \neq 0$
My attempt:
We know that $a^2 \geq 0$ and $b^2 > 0$ since $b \neq 0$. So then $a^2 + b^2 >0$. Now I'm stuck as I'm not sure how to show that $a^2 + b^2 = 4ab$ has no solutions given the above conditions. A little assistance would be greatly appreciated.
| This isn't true. Consider $a = 1,$ and $b = 2 + \sqrt{3}$.
Then $a^{2} -4ab + b^{2} = 1 -4(2 + \sqrt{3}) + (2 + \sqrt{3})^{2} = 1 - 8 -4\sqrt{3} + 4 +4\sqrt{3} +3 = 0$.
Perhaps the question meant to ask over the integers instead of the reals?
| {
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Prove $\sum\limits_\text{cyc}\frac{a}{a+(n-1)b}\geq 1$ For $a_i> 0$, $n \in \mathbb{N}$ prove or disprove $$\frac{a_1}{a_1+(n-1) a_2} + \frac{a_2}{a_2+(n-1)a_3}+\dots+\frac{a_n}{a_n+(n-1)a_1}\geq 1.$$
Written in a cyclic notation
$$
\sum_\text{cyc} \frac{a}{a+(n-1)b} \geq 1.
$$
I have conjectured this one after running into couple similar ones. Perhaps it is known, but I could not find it solved anywhere. For small values of $n$ it apparently holds, for $n=1$ we have:
$$\frac{a}{a} = 1 \geq 1
$$
For $n=2$ it is:
$$
\frac{a}{a+b} + \frac{b}{b+a} = \frac{a+b}{a+b} = 1 \geq 1
$$
For $n=3$ it starts to be interesting:
$$
\frac{a}{a+2b} + \frac{b}{b+2c} + \frac{c}{c+2a} \geq 1
$$
Here the Cauchy-Schwartz inequality seems to do the trick:
\begin{align}
\left(\sum_\text{cyc} \frac{a} {a+2b}\right)\left(\sum_\text{cyc} a(a+2b)\right) &\geq \left(\sum_\text{cyc} a\right)^2\\
\left(\sum_\text{cyc} \frac{a}{a+2b}\right) (a+b+c)^2&\geq (a+b+c)^2 \\
\left(\sum_\text{cyc} \frac{a}{a+2b}\right) &\geq 1
\end{align}
However for higher $n$ I'm stuck. I have tried Cauchy-Schwartz inequality, as well as Holder's and Jensen's, but no luck. Also considered induction but it did not lead to anything nice.
It appears that the equality holds whenever $a_1=a_2=\cdots=a_n$. Also the inequality can be equivalently written in a form
\begin{align}
\sum_\text{cyc} \frac{a}{a+(n-1)b} = \sum_\text{cyc}\frac{a+(n-1)b-(n-1)b}{a+(n-1)b} = n-(n-1)\sum_\text{cyc}\frac{b}{a+(n-1)b} &\geq 1\\
\end{align}
so
\begin{align}
1 &\geq \sum_\text{cyc}\frac{b}{a+(n-1)b}.\\
\end{align}
Anyone knows how to prove/disprove this for generic $n$?
| For $n\in\{1,2\}$ our inequality is obviously true.
Let $n\geq3$ and $\frac{a_2}{a_1}=e^{x_1}$, $\frac{a_3}{a_2}=e^{x_2}$,..., $\frac{a_1}{a_n}=e^{x_n}$.
Hence, $x_1+x_2+...+x_n=0$ and we need to prove that
$\sum\limits_{i=1}^nf(x_i)\geq1$, where $f(x)=\frac{1}{1+(n-1)e^x}$.
But $f''(x)=\frac{(n-1)e^x((n-1)e^x-1)}{(1+(n-1)e^x)^3}>0$ for all $x\geq0$.
Thus, by Vasc's RCF Theorem it's enough to prove our inequality for
$e^{x_1}=e^{x_2}=...=e^{x_{n-1}}=a$ and $e^{x_n}=\frac{1}{a^{n-1}}$, which gives
$$(n-1)a\left(a^{n-1}-(n-1)a+n-2\right)\geq0,$$ which is obvious by AM-GM.
Done!
| {
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Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ and then solving but always I end up with complicated things which are nowhere near the answer.
How do I solve?
Thanks for any help!!
| $$\begin{align*}\lim_{x\to y} \frac{\sin^2(x)-\sin^2(y)}{x^2-y^2}&=\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\frac{\sin x+\sin y}{x+y}\\&=\frac{2\sin y}{2y}\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\\&=\frac{\sin y}{y}(\sin y)'\\&=\frac{\sin y \cos y}{y}\end{align*}$$
$$=\frac{\sin2y}{2y}$$
| {
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Prove the given cubic inequality $a,b,c> 0$ prove $$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
I tried AM-GM but can't reached to solution please help me
| Hint If you want to use AM-GM,
$$\frac{a^3}{b+c}+\frac{a(b+c)}4\ge a^2$$
Add three such inequalities to sum and then use the rearrangement $ab+bc+ca\le a^2+b^2+c^2$.
| {
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Need help with RHS of Induction Problem about $1^{3} + 2^{3} + ... + n^{3}$ = $(1 + 2 + ... + n)^{2}$ Prove $1^{3} + 2^{3} + ... + n^{3}$ = $(1 + 2 + ... + n)^{2}$ for all positive integers n.
I've tried to work with this problem using mathematical induction. However, I really don't understand how to manipulate the right side of the equation. Any help would be greatly appreciated.
Thank you in advance.
| It works for $n=1$.
$$\begin{align} 1^3+\cdots+n^3+(n+1)^3 & =(1+\cdots +n)^2+(n+1)^3 \quad \mathrm{induction} \ \mathrm{hypothesis} \\ &= \frac{n^2(n+1)^2+4(n+1)^3}{4} \\ &=\frac {(n+1)^2(n^2+4(n+1))}{4} \\ &=\frac {(n+1)^2((n+1)+1)^2}{4} \\ &=(1+\cdots +n+(n+1))^2\end{align}$$
| {
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rotation of hyperbola How can we rotate the rectangular hyperbola
xy=c ( c is any constant)
Into a form of standard hyperbola that is
(x/a)$^2$ - (y/b)$^2$ = 1
By rotating the hyperbola .
| $$
\begin{bmatrix}X\\Y\end{bmatrix}=\mathcal{Rot}_{\text{cw}}(45)\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\cos45&-\sin45\\\sin45&\cos45\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\
=\frac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\
\implies X=\frac{x-y}{\sqrt{2}}\quad;\quad Y=\frac{x+y}{\sqrt{2}}
$$
$$
xy=c\implies \frac{x-y}{\sqrt{2}}.\frac{x+y}{\sqrt{2}}=\frac{x^2-y^2}{\sqrt{2}}=c\\\frac{x^2}{\sqrt{2}c}-\frac{y^2}{\sqrt{2}c}=1
$$
$$
\color{blue}{xy=c \iff \frac{x^2}{\sqrt{2}c}-\frac{y^2}{\sqrt{2}c}=1}
$$
| {
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Need help with proof for Dedekind cuts on $\mathbb{Q}^+$ I am working on a proof about Dedekind cuts on the positive rational numbers. I have been stuck for a while on the following point and would appreciate any help.
Given $x\in \mathbb{Q}^+$ such that $x^2<2$, how to find $y\in \mathbb{Q}^+$ such $y>x$ and $y^2<2$?
| The Stern-Brocot tree is one of my favorite tools. It is assumed to be well-known in what follows. Let $m$ and $n$ be positive integers. Initialize (zeroth iteration):
$$
x = \frac{m}{n} < \sqrt{2} \quad ; \quad y = 2/x = \frac{2n}{m} > \frac{2}{\sqrt{2}} \quad \Longrightarrow \quad y > \sqrt{2}
$$
Now walk through the Stern-Brocot tree iteratively until $\,y < \sqrt{2}$ . From the properties of the tree, we know that always will be $\,y > x$ .
First iteration:
$$
y := \frac{m+2n}{n+m} = \frac{x+2}{x+1} < \sqrt{2} \quad \mbox{?}
$$
No, because a contradiction with $\,x<\sqrt{2}\,$ is derived:
$$
(x+2)^2 < 2(x+1)^2 \quad \Longleftrightarrow \quad x^2+4x+4 < 2x^2+4x+2 \quad \Longleftrightarrow \quad x^2 > 2
$$
Second iteration:
$$
y := \frac{m+(m+2n)}{n+(n+m)} = \frac{2m+2n}{m+2n} = \frac{2x+2}{x+2} < \sqrt{2} \quad \mbox{?}
$$
Yes, because:
$$
(2x+2)^2 < 2(x+2)^2 \quad \Longleftrightarrow \quad 4x^2+8x+4 < 2x^2+8x+8 \quad \Longleftrightarrow \quad x^2 < 2
$$
So the outcome is:
$$
y(x) = 2\frac{x+1}{x+2}
$$
Check, check, double check .. If we are allowed to have an embedding of the rationals in the reals, then the derivative is:
$$
y'(x) = \frac{2(x+2)-(2x+2)}{(x+2)^2} = \frac{2}{(x+2)^2} > 0
$$
And some function values are:
$$
y(0) = 1 \quad ; \quad y(\sqrt{2}) = 2\frac{\sqrt{2}+1}{2+\sqrt{2}} = \sqrt{2}
$$
But $y(x)$ is monotonically increasing, so $\sqrt{2}$ is the maximum at the interval $\left[0,\sqrt{2}\right]$ and all other values of $y(x)$ are smaller than this, but greater than $x$, as requested.
BONUS. Let $N$ be any positive integer. Now generalize the question as follows.
Given $x\in \mathbb{Q}^+$ such that $x^2 < N$, how to find $y\in \mathbb{Q}^+$ such that $y>x$ and $y^2 < N$ ?
It's left as an exercise for the reader to prove that this is a solution:
$$
y(x) = N\frac{x+1}{x+N}
$$
Can't resist to display some members of the $y$ - family in a $[0,3]\times[0,3]$ picture:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all $p$ primes such that $-8$ is quadratic residue modulo $p$ I need help with finding all $p$ primes such that $-8$ is quadratic residue modulo $p$.
I know that :
$(\frac{-8}{p})=(\frac{-1}{p})(\frac{8}{p})=(\frac{-1}{p})(\frac{2^2}{p})(\frac{2}{p})=(\frac{-1}{p})(\frac{2}{p})=(-1)^{\frac{p-1}{2}}
(-1)^{\frac{p^2-1}{8}}$
But I'm not sure how to continue, I wanted to use CRT but 4 and 8 are not coprime....
How can I continue from here?
| Hint 1: Show that $(-1/p) = 1$ if and only if $p \equiv 1 \pmod{4}$ (consider using Euler's Criterion).
Hint 2: Show that for $p \equiv 1 \pmod{4}$, $2^{(p - 1)/2} \equiv (-1)^{(p - 1)/4} \pmod{p}$, and thus $(2/p) \equiv (-1)^{(p - 1)/4} \pmod{p}$ by Euler's Criterion. Similarly, for $p \equiv 3 \pmod{4}$, show that $2^{(p - 1)/2} \equiv (-1)^{(p + 1)/4} \pmod{p}$, and thus $(2/p) \equiv (-1)^{(p + 1)/4} \pmod{p}$. One way to do this is start with $2^{(p - 1)/2} \left(\frac{p - 1}{2} \right)!$ on one side of an equivalence modulo $p$, expand the factorial term by term, and show that the other side evaluates to $(-1)^{(p - 1)/4} \left(\frac{p - 1}{2} \right)!$ and $(-1)^{(p + 1)/4} \left(\frac{p - 1}{2} \right)!$ for $p \equiv 1 \pmod{4}$ and $p \equiv 3 \pmod{4}$, respectively. (Gauss's Lemma is also possible for this part).
Hint 3: If you complete both the above steps, you will have classifications of primes $p$ for which $-1$ and $2$ are quadratic residues or nonresidues modulo $p$. This gives you a classification of $p$ where $-8$ is a quadratic residue, since as per your equation, we consider the case where $(-8/p) = (-1/p)(2/p) = 1$, or $(-1/p) = (2/p) = \pm 1$.
Also: if the moduli in a system of congruences are not relatively prime, only consider the cases where the congruences are consistent with each other. For example, if we have the system $x \equiv 1 \pmod{4}$ and $x \equiv 5 \pmod{8}$, the second congruence is a stronger version of the first (since $x = 8k + 5 = 4(2k + 1) + 1 \equiv 1 \pmod{4}$). On the other hand, the system $x \equiv 1 \pmod{4}$ and $x \equiv 3 \pmod{8}$ is inconsistent and not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In a determinant prove $xyz = -1$ If we are given the following determinant
$$\begin{vmatrix}
x^3+1 & x^2 & x \\
y^3+1 & y^2 & y \\
z^3+1 & z^2 & z \\
\end{vmatrix}=0
$$
and $x, y, z$ are all different, then we have to prove that $xyz = -1$.
I tried to expand the determinant, but using that, it is getting too complicated.
| This is my answer (for sure it isn't the best approach). Applying cofactors we have
$$(x^3+1)\begin{vmatrix}y^2 & y\\z^2 & z\end{vmatrix}-x^2\begin{vmatrix}y^3+1 & y\\z^3+1 & z\end{vmatrix}+x\begin{vmatrix}y^3+1 & y^2\\z^3+1 & z^2\end{vmatrix}=0.$$
After some operations we get $$(x^3+1)yz(y-z)-x^2(y-z)(yz(y+z)-1)+x(y-z)(y^2z^2-y-z)=0.$$ Then $$(y-z)(x^3yz+yz-x^2y^2z-x^2yz^2+x^2+xy^2z^2-xy-xz)=0.$$
Since $y\neq z$, we deduce that $x^3yz+yz-x^2y^2z-x^2yz^2+x^2+xy^2z^2-xy-xz=0$. Finally we factorize the LHS and we get $$xyz(x^2-xy-xz+yz)+(x^2+yz-xy-xz)=(x^2-xy-xz+yz)(xyz+1)=$$ $$=(x-z)(y-z)(xyz+1)=0.$$
But $x\neq y$ and $x\neq z$, hence it must be $xyz+1=0$, i.e., $xyz=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove the series $(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}=\pi-4(...)$
Show that for $k\ge1$,
\begin{align}
\\&\quad(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}\\[10pt]&=\pi-4\left(1-{1\over 3}+{1\over 5}-{1\over 7}+\cdots+{1\over 2k-1}\right).
\end{align}
I try:
The RHS sort of the favourite Leibniz $\pi$ series.
We could split into composite of fractions
${A\over 2n+1}+{B\over 2n+3}+{C\over 2n+5}+\cdots$ and then take the sum individually.
We know that
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+1)}=\pi$
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+3)}=3\pi-8$
We don't know the general of
$\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+2k+1)}=F(k)$
I am sure what to do next? Help please!
| This is probably not an answer but it is too long for a comment.
Concerning the last question $$F(k)=\sum_{n=0}^{\infty}{2^{n+1}\over {2n\choose n}(2n+2k+1)}=\frac{2 }{2 k+1}\,\,
_3F_2\left(1,1,k+\frac{1}{2};\frac{1}{2},k+\frac{3}{2};\frac{1}{2}\right)$$ where appears the generalized hypergeometric function. Unfortunately, the expression seems to simplify only for small values of $k$
$$\left(
\begin{array}{cc}
k & F(k) \\
0 & \pi \\
1 & -8+3 \pi \\
2 & -\frac{208}{9}+\frac{23 \pi }{3} \\
3 & -\frac{4232}{75}+\frac{91 \pi }{5}
\end{array}
\right)$$
Concerning the general problem $$S_k=(-1)^k\sum_{n=0}^{\infty}{2^{n+1}(2k-1)!!\over {2n\choose n}(2n+1)(2n+3)\cdots(2n+2k+1)}$$ it reduces to $$S_k=(-1)^k\left(\psi \left(\frac{k}{2}+\frac{3}{4}\right)-\psi
\left(\frac{k}{2}+\frac{1}{4}\right)\right)$$ where appears the digamma function
$$\left(
\begin{array}{cc}
k & S_k \\
0 & \pi \\
1 & \pi-4 \\
2 & \pi-\frac{8}{3} \\
3 & \pi-\frac{52}{15} \\
4 & \pi-\frac{304}{105} \\
5 & \pi-\frac{1052}{315} \\
6 & \pi-\frac{10312}{3465} \\
7 & \pi-\frac{147916}{45045} \\
8 & \pi-\frac{135904}{45045}
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Probability of rolling $3$ identical dice out of $6$. If $6, 6$ sided dice are rolled, what is the probability that any three are the same?
| Probability that all are same $=6\dbinom{6}{6}\left(\dfrac{1}{6}\right)^6$
Probability that exactly $5$ are same $=6\dbinom{6}{5} \left(\dfrac{1}{6}\right)^5 \left(\dfrac{5}{6}\right)$
Probability that exactly $4$ are same $=6\dbinom{6}{4} \left(\dfrac{1}{6}\right)^4 \left[\left(\dfrac{5}{6}\right) \left(\dfrac{4}{6}\right)+ \left(\dfrac{5}{6}\right) \left(\dfrac{1}{6}\right)\right]$
Probability that exactly $3$ are same $=6\dbinom{6}{3} \left(\dfrac{1}{6}\right)^3 \left[\left(\dfrac{5}{6}\right) \left(\dfrac{4}{6}\right)\left(\dfrac{3}{6}\right)+ \dbinom{3}{2}\left(\dfrac{5}{6}\right)\left(\dfrac{4}{6}\right)\left(\dfrac{1}{6}\right)+\left(\dfrac{5}{6}\right)\left(\dfrac{1}{6}\right)\left(\dfrac{1}{6}\right)\right]$
Sum of the these gives required probability
| {
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"url": "https://math.stackexchange.com/questions/2071095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Discrepancy in differentiating: $y=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$ $$y=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$$
$$\sqrt{1+\sin{x}}=\sin{\frac{x}{2}}+\cos{\frac{x}{2}}$$
$$\sqrt{1-\sin{x}}=\sin{\frac{x}{2}}-\cos{\frac{x}{2}}$$
$$y=\tan^{-1}\left(\tan{\frac{x}{2}}\right)$$Differentiating, we get: $$\frac{dy}{dx}=\frac{1}{2}$$
But taking: $$\sqrt{1-\sin{x}}=\cos{\frac{x}{2}}-\sin{\frac{x}{2}}$$
$$y=\tan^{-1}\left(\cot{\frac{x}{2}}\right)$$
Differentiating, we get: $$\frac{dy}{dx}=\frac{-1}{2}$$
I figured that I needed to use absolute value for the simplification of $\sqrt{1-\sin{x}}$, i.e. $\sqrt{1-\sin{x}}=\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|$.
Subsequently, I put the below functions into Wolfram Alpha's input box to differentiate: $$y_1=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$$
And
$$y_2=\tan^{-1}\left(\frac{\left|\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right|+\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|}{\left|\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right|-\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|}\right)$$
The answers should ideally match, but they dont. $\frac{dy_1}{dx}=\frac{-1}{2}\neq\frac{dy_2}{dx}$
Why don't they?
| $$\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\dfrac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2}{1+\sin x-(1-\sin x)}=\dfrac{1+|\cos x|}{\sin x}$$
Case$\#1:$ If $\cos x\ge0, |\cos x|=+\cos x$
Using $\cos2A=2\cos^2A-1,\sin2A=2\sin A\cos A$
$\tan^{-1}\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\tan^{-1}\dfrac{1+\cos x}{\sin x}=\tan^{-1}\cot\dfrac x2=\tan^{-1}\tan\left(\dfrac{\pi-x}2\right)$
$y=m\pi+\dfrac{\pi-x}2$ where $m$ is an arbitrary constant such that $-\dfrac\pi2\le y\le\dfrac\pi2$
$$\dfrac{dy}{dx}=?$$
Case$\#2:$ If $\cos x<0, |\cos x|=-\cos x$
Please carry on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $\sqrt[3]x-\sqrt[3]{x-14}=2$, find $x - 1/x$
$$ \sqrt[3]x-\sqrt[3]{x-14}=2$$
Then find $x - \frac{1}{x} = ?$
I've tried to take third power of both sides but I couldn't find anything.
| Let $\sqrt[3]x=a,\sqrt[3]{x-14}=b$
$a-b=2, a^3-b^3=14$
Method$\#1:$
As $(a-b)^3=a^3-b^3-3ab(a-b)$
$2^3=14-3ab\cdot2\iff ab=1$
Method$\#2:$
$$a^3-(a-2)^3=14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{r'^3}d\theta$ Let $P$ be a fixed point inside a circle with radius $r$ and $Q$ a non-fixed point on its perimeter as shown below, and let $\alpha=\frac xr$
Is there any closed-form solution for the following integral?
$$\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{{r'}^{3}}d\theta$$
I tried Wolfram-Alpha without any luck. It doesn't look similar to any form of the integrals I have ever seen before. I appreciate if anybody gives it a try.
| Hint. One may recall that
$$
r'=\sqrt{r^2+x^2-2rx\cos \theta}
$$ we are then looking for
$$
\begin{align}
&\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{{r'}^{3}}d\theta
\\&=\frac1{r^3}\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{\left(1+\alpha^2-2\alpha\cos \theta\right)^{3/2}}d\theta
\\&=\frac1{r^3}\frac1{(1+\alpha^2)^{3/2}}\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{\left(1-\frac{2\alpha}{1+\alpha^2}\cos \theta\right)^{3/2}}d\theta
\\&=\frac1{r^3}\frac1{(1+\alpha^2)^{3/2}}\sum_{n=0}^\infty \frac{2\:\Gamma\big(n+\frac32\big)}{\Gamma\big(n+1\big)\Gamma\big(\frac12\big)}\cdot\int_0^{2\pi}\left(\cos^n{\theta}-\alpha\cos^{n+1}{\theta}\right)d\theta\cdot\left(\frac{2\alpha}{1+\alpha^2} \right)^n
\end{align}
$$ setting $a=\dfrac{2\alpha}{1+\alpha^2}$, $0<a<1$, one may then use Wallis' integral
$$
\int_0^{2\pi}\cos^n{\theta}\:d\theta= 2\cdot\frac{\Gamma\big(\frac12\big)\Gamma(\tfrac{n+1}{2})}{\Gamma\big(\tfrac{n}{2}+1\big)}
$$ to obtain
$$
\begin{align}
&\sum_{n=0}^\infty \frac{2\:\Gamma\big(n+\frac32\big)}{\Gamma\big(n+1\big)\Gamma\big(\frac12\big)}\cdot\int_0^{2\pi}\left(\cos^n{\theta}-\alpha\cos^{n+1}{\theta}\right)d\theta\cdot a^n
\\\\&=\sum_{n=0}^\infty \frac{4\:\Gamma\big(n+\frac32\big)\Gamma(\tfrac{n+1}{2})}{\Gamma\big(n+1\big)\Gamma\big(\tfrac{n}{2}+1\big)} \cdot a^n-\sum_{n=0}^\infty \frac{4\:\Gamma\big(n+\frac52\big)\Gamma(\tfrac{n}{2}+1)}{\Gamma\big(n+2\big)\Gamma\big(\tfrac{n}{2}+\tfrac32\big)} \cdot a^n
\\\\&=\frac{4\mathrm{E}\left(\frac{2 a}{1+a}\right)}{(1-a)\sqrt{1+a}}-\frac{4\alpha \mathrm{K}\left(\frac{2 a}{1+a}\right)}{a\sqrt{1+a}}+\frac{4\alpha\mathrm{E}\left(\frac{2 a}{1+a}\right)}{a(1-a)\sqrt{1+a}}
\end{align}
$$ where we have used elliptic integrals $\mathrm{E}(r):=\mathrm{E}(\tfrac\pi2,r)$ and $\mathrm{K}(r):=\mathrm{K}(\tfrac\pi2,r)$.
Finally,
$$
r^3(1+\alpha^2)^{3/2}\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{{r'}^{3}}d\theta=\frac{4\mathrm{E}\left(\frac{2 a}{1+a}\right)}{(1-a)\sqrt{1+a}}-\frac{4\alpha \mathrm{K}\left(\frac{2 a}{1+a}\right)}{a\sqrt{1+a}}+\frac{4\alpha\mathrm{E}\left(\frac{2 a}{1+a}\right)}{a(1-a)\sqrt{1+a}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$. I have found a new problem which asks:
Find the remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$.
I am thinking to find the remainder using Fermet's theorem, but I think I am unable to do it.
Please help.
| Observe that $\gcd(10,7)=1$.
Therefore by Euler's theorem: $10^{\phi(7)}\equiv1\pmod7$.
Since $7$ is prime: $\phi(7)=7-1=6$.
Therefore: $10^6\equiv1\pmod7$.
Therefore:
*
*$10^{10^{1}}\equiv10^{6\cdot1+4}\equiv(\color\red{10^{6}})^{1}\cdot10^{4}\equiv\color\red1^{1}\cdot10^{4}\equiv10000\equiv4\pmod7$
*$10^{10^{2}}\equiv10^{6\cdot16+4}\equiv(\color\red{10^{6}})^{16}\cdot10^{4}\equiv\color\red1^{16}\cdot10^{4}\equiv10000\equiv4\pmod7$
*$10^{10^{3}}\equiv10^{6\cdot166+4}\equiv(\color\red{10^{6}})^{166}\cdot10^{4}\equiv\color\red1^{166}\cdot10^{4}\equiv10000\equiv4\pmod7$
*$\dots$
We can easily prove by induction that $\forall{n\in\mathbb{N}}:10^{10^{n}}\equiv4\pmod7$.
Therefore: $\sum\limits_{n=1}^{10}10^{10^{n}}\equiv\sum\limits_{n=1}^{10}4\equiv40\equiv5\pmod7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Prove that $f(x) = ax^2$ For some function $f \in C(R)$ the following equality holds: $f(x) + f(y) = f(\sqrt{x^2 + y^2})$ for any $x, \ y \in R$.
Prove that $f(x) = ax^2 \ \forall x\in \mathbb{R}$, where $a = f(1)$.
| This assumes that
$f$ is differentiable.
$f(x) + f(y)
= f(\sqrt{x^2 + y^2})
$.
If $y=0$,
then
$f(x)+f(0) = f(x)$,
so
$f(0) = 0$.
Since
$f(x)+f(-y)
=f(\sqrt{x^2 + y^2})
=f(x)+f(y)
$,
$f(y) = f(-y)$.
Differentiating wrt $x$,
$\begin{array}\\
f'(x)
&=(\sqrt{x^2+y^2})'f'(\sqrt{x^2+y^2})\\
&=\dfrac{x}{\sqrt{x^2+y^2}}
f'(\sqrt{x^2+y^2})\\
\text{so}\\
\dfrac{f'(x)}{x}
&=\dfrac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}\\
\end{array}
$
Therefore
$\dfrac{f'(x)}{x}
$
is constant.
Setting
$\dfrac{f'(x)}{x}
=c$,
$f'(x) = cx$
so
$f(x) = cx^2/2+d$.
Since $f(0) = 0$,
$f(x) = cx^2/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove$\sum\frac{a}{a^2+a+1}\leq\sum\frac{1}{a^2+a+1}$ If I asked the obvious because of my lack of knowledge, then I am sorry!
If $abcd=1,a,b,c,d>0$;prove:$$\sum_\limits{cyc}\frac{a}{a^2+a+1}\leq\sum_\limits{cyc}\frac{1}{a^2+a+1}$$
I use $f(x)=\frac{x-1}{x^2+x+1}$,it without success;I want to use$a=\frac{yz}{x^2},b=\frac{zt}{y^2},c=\frac{tx}{z^2},d=\frac{xy}{t^2},$and the Order inequality,but I'm not sure if it's right.Could anyone help me? Thanks a lot!
| $\sum\limits_{cyc}\frac{1-a}{1+a+a^2}=\sum\limits_{cyc}\left(\frac{1-a}{1+a+a^2}+\frac{1}{3}\ln{a}\right)$.
Let $f(x)=\frac{1-x}{x^2+x+1}+\frac{1}{3}\ln{x}$.
Hence, $f'(x)=\frac{(x-1)(x^3+6x^2+3x-1)}{3x(x^2+x+1)^2}$.
Which says that there is unique $0<x_1<1$, for which $f(x)\geq0$ for all $x\geq x_1$.
Easy to see that $x_1=0.0779...$,
which says that for $\min\{a,b,c,d\}\geq0.08$ our inequality is true.
Let $a<0.08$ and $g(x)=\frac{1-x}{1+x+x^2}$.
Hence, $g'(x)=\frac{x^2-2x-2}{(1+x+x^2)^2}$, which says that $\min\limits_{x>0}g=g(1+\sqrt3)$
and $g$ is a decreasing function on $(0,0.08]$.
Id est, $\sum\limits_{cyc}\frac{1-a}{1+a+a^2}\geq g(0.08)+3g(1+\sqrt3)=0.38...>0$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $f'(x)$ where $f(x) = x^\frac{3}{2}$ using $h \to 0$ method $$f(x) = x^{\frac{3}{2}}$$
$$f'(x) = ?$$
How would you solve this using the $h\to 0$ method?
Here's the initial setup.
$$f'(x) = \lim_{h \to 0}\frac{(x+h)^\frac{3}{2}-x^{3/2}}{h}$$
I tried difference of cubes but it didn't get me anywhere.
| ${(\sqrt {x + h} )^3} - {\left( {\sqrt x } \right)^3} = \left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {{{\left( {x + h} \right)}^2}} + \sqrt {\left( {x + h} \right)x} + \sqrt {{x^2}} } \right)$ and $h = x + h - x = \left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {x + h} + \sqrt x } \right)$ so
$$\mathop {\lim }\limits_{h \to 0} \frac{{{{(\sqrt {x + h} )}^3} - {{\left( {\sqrt x } \right)}^3}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {{{\left( {x + h} \right)}^2}} + \sqrt {\left( {x + h} \right)x} + \sqrt {{x^2}} } \right)}}{{\left( {\sqrt {x + h} + \sqrt x } \right)}} = \frac{3}{2}\sqrt x $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $a_{-1}$ in the Laurent series of $f(z)=z^{3}\cdot\cos(\frac{1}{z})\cdot e^{\frac{1}{z^{2}}}$ I need to find the coefficient $a_{-1}$ in the Laurent series of $f(z)=z^{3}\cdot\cos(\frac{1}{z})\cdot e^{\frac{1}{z^{2}}}$.
I tried two methods:
*
*$a_{-1}=\frac{1}{2\pi i}\oint_{C_{r}}f(z)dz$
I chose $r=1$ and wrote $a_{-1}=\frac{1}{2\pi i}\oint_{C_{r}}f(z)dz=\frac{1}{2\pi i}\int_0 ^{2\pi}ie^{it}f(e^{it})dt$, which didn't lead me anywhere.
*Trying to compute directly, I know these Laurent series about $z=0$:
$\cos(\frac{1}{z})=1-\frac{z^{-2}}{2!}+\frac{z^{-4}}{4!}+...$
$e^{\frac{1}{z^{2}}}=1+z^{-2}+\frac{z^{-4}}{2!}+\frac{z^{-6}}{3!}+...$
So $f(z)=(1-\frac{z^{-2}}{2!}+\frac{z^{-4}}{4!}+...)(z^3+z+\frac{z^{-1}}{2!}+\frac{z^{-3}}{3!}+...)$
and $a_{-1}$ is the coefficient of $(\frac 1 z)$, so $a_{-1}=\frac {1}{2!}-\frac {1}{2!}+\frac{1}{4!}=\frac 1 {24}$.
Is my work okay? Did I miss any theoretical point? When is it advised to use each method? And what if I needed to calculate the whole series, not just one coefficient?
Any sharing of your experience will be greatly appreciated.
| Looks fine to me. You could also consider
\begin{align}
e^{w^2}\cos w
&=\frac{e^{w^2+iw}+e^{w^2-iw}}{2} \\
&=\frac{1}{2}\left(1+(w^2+iw)+\frac{(w^2+iw)^2}{2}+\frac{(w^2+iw)^3}{6}+\frac{(w^2+iw)^4}{24}\right)+\\
&\phantom{{}={}}
\frac{1}{2}\left(1+(w^2-iw)+\frac{(w^2-iw)^2}{2}+\frac{(w^2-iw)^3}{6}+\frac{(w^2+iw)^4}{24}\right)+o(w^4)\\
&=1+w^2+w^4-w^2-w^4+\frac{1}{24}w^4+o(w^4)\\
&=1+\frac{1}{24}w^4+o(w^4)
\end{align}
Thus
$$
z^3e^{1/z^2}\cos\frac{1}{z}=z^3+\frac{1}{24}z^{-1}+o(z^{-1})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How did people get the inspiration for the sums of cubes formula? I stumbled upon this neat formula for sums of cubes with arbitrary $x,y\in\mathbb{Z}$$$(x^2+9xy-y^2)^3+(12x^2-4xy+2y^2)^3=(9x^2-7xy-y^2)^3+(10x^2+2y^2)^3\tag1$$
With $1729=1^3+12^3=9^3+10^3$ as its first instance. And I believe that this formula was used by Ramanujan to find a formula for$$a^3+b^3=c^3\pm1$$
So my question?
Questions:
*
*What would be someone's thinking process when finding other formulas such as $(1)$?
*Are there any other formulas similar to $(1)$?
I'm thinking along the lines of starting with $$(x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3=(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3$$
But even Mathematica can't solve the ensuing system that follows. So for the moment, I'm stuck.
| Actually, Mathematica can solve,
$$\small (x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3+(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3 =0\tag2$$
One may be guided by the principle of fait accompli (accomplished fact). Ramanujan and others already found solutions therefore $(2)$, approached the right way, must be solvable.
What you do is expand $(2)$ and collect powers of $x,y$. The Mathematica command is Collect[P(x,y),{x,y}] to get,
$$P_1x^6+P_2x^5y+P_3x^4y^2+P_4x^3y^3+P_5x^2y^4+P_6xy^5+P_7y^6 = 0$$
where the $P_i$ are polynomials in the other variables. The hard part is then solving the system,
$$P_1 = P_2 = \dots =P_7 = 0$$
After much algebraic manipulation (which I don't have the strength to type all down), one ends up with the simple identity (which I gave to Mathworld back in 2005),
$$(ax^2-v_1xy+bwy^2)^3 + (bx^2+v_1xy+awy^2)^3 + (cx^2+v_2xy+dwy^2)^3 + (dx^2-v_2xy+cwy^2)^3 = \color{blue}{(a^3+b^3+c^3+d^3)}(x^2+wy^2)^3\tag3$$
where,
$$v_1= c^2-d^2\\ v_2= a^2-b^2\\ w= (a+b)(c+d)$$
Thus, if the $RHS$ is zero, or you find a single instance of $\color{blue}{a^3+b^3+c^3+d^3 = 0}$, then the $LHS$ yields a quadratic parameterization that guarantees an infinite more. So to answer your question, $(3)$ can be used to generate infinitely many Ramanujan-type formulas like $(1)$.
Example: The two smallest taxicab numbers are,
$$1^3+12^3=9^3+10^3\\
\color{blue}{2^3+16^3=9^3+15^3}$$
Using the second one and formula $(3)$, and after scaling the variable $y' \to y/12$ to reduce coefficient size, one gets,
$$(\color{blue}2 x^2 + 12 x y - 48 y^2)^3 + (\color{blue}{16} x^2 - 12 x y - 6 y^2)^3 + (\color{blue}{-9} x^2 - 21 x y + 45 y^2)^3 + (\color{blue}{-15} x^2 + 21 x y + 27 y^2)^3 = 0$$
and you can see its "parents" in blue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
Algebra problem regarding system of logarithmic equations. The equations are:
$\log_{4}(x)+\log_{4}(y)=5$
$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6$
I attempted to solve this problem by solving the pair of equations for $x$.
For the first equation:
$\Longrightarrow \log_{4}(xy)=5 \Longrightarrow xy=4^{5} \Longrightarrow xy=1024 \Rightarrow x=\dfrac{1024}{y}$
For the second equation:
$\Longrightarrow \log{4}(x)=\dfrac{6}{\log_{4}(y)} \Longrightarrow x=4^{\frac{6}{\log_{4}(y)}}$
Then,
$\Longrightarrow \dfrac{1024}{y}=4^{\frac{6}{\log_{4}(y)}}$
How should I move on from here?
| We have:
$\log_{4}(x)+\log_{4}(y)=5$
$\Rightarrow \log_{4}(x)=5-\log_{4}(y) \hspace{5 mm}$ (i)
$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6 \hspace{4.25 mm}$ (ii)
Substituting (i) into (ii):
$\Rightarrow \big(5-\log_{4}(y)\big)\big(\log_{4}(y)\big)=6$
$\Rightarrow 5\log_{4}(y)-\big(\log_{4}(y)\big)^{2}=6$
$\Rightarrow \big(\log_{4}(y)\big)^{2}-5\log_{4}(y)+6=0$
$\Rightarrow \log_{4}(y)=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(6)}}{2(1)}$
$\Rightarrow \log_{4}(y)=\dfrac{5\pm{1}}{2}$
$\Rightarrow \log_{4}(y)=2,3$
$\Rightarrow y=16,64$
Using (i):
$\Rightarrow \log_{4}(x)=5-\log_{4}(16)$
$\hspace{19.5 mm}=5-2$
$\hspace{19.5 mm}=3$
$\Rightarrow x=64$
or
$\Rightarrow \log_{4}(x)=5-\log_{4}(64)$
$\hspace{19.5 mm}=5-3$
$\hspace{19.5 mm}=2$
$\Rightarrow x=16$
Therefore, the solutions to the system of equations is $x=16$ and $y=64$ or $x=64$ and $y=16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum\limits_{k=1}^{n-1} (n-k) x^k$ is non-decreasing for $x \in ]-1,1[$. Question
I would like to show for arbitrary $n \in \mathbb{N}$, that the polynomial:
$$
p(x) := \sum_{k=1}^{n-1} (n-k) x^k
$$
is non-decreasing.
Start Of Solution
We can write its derivative by:
$$
p'(x) = \frac{(-(x + 1) (x^n-1) + n (x - 1) (x^n + 1))}{(x - 1)^3},
$$
as $(-1 + x)^3 < 0$ for $x \in ]-1,1[$ it suffices to show for arbitrary $n$ that:
$$
n(x-1) (x^n + 1) \leq (x+1)(x^n - 1),
$$
for $n = 1$ this is trivial (as both sides become equal). Thus we assume the inequality to hold for $n$ and show it for $n+1$, here I use $x^{n+1} + 1 = x(x^n+1) + (1-x)$ to use the induction hypothesis but I don't achieve the inequality through this method.
No positive roots
By Descartes' Sign rule we see that $p'(x) = \sum\limits_{k=1}^{n-1}(n-k)k x^{k-1}$ has no positive roots, thus it suffices to show
| I found it I think.
Let's replace $n$ by $n+1$ to obtain:
$$
p(x) = \sum_{k=1}^n(n+1-k)x^k,
$$
now we let the sum go from $0$ to $n-1$:
$$
p(x) = \sum_{k=0}^{n-1} (n-k)x^{k+1},
$$
taking the derivative yields:
$$
p'(x) = \sum_{k=0}^{n-1}(n-k)(k+1)x^k
$$
it now suffices to show that $p'(x) \geq 0$ on $]-1,1[$.
One can check (for example with mathematica) that we have:
$$
p'(x) = \frac{n(x^{n+1} + 1) - 2 (x^n + \dots + x))}{(x-1)^2},
$$
thus we just need to show that
$$\frac{(x^{n+1} + 1)}{2} \leq \frac{x^n+\dots+x}{n}.$$
For $x \in ]-1,0]$ we have $\frac{(x^{n+1} + 1)}{2} \geq 0$ while $ \frac{x^n+\dots+x}{n} = x(1-x^n)/(1-x) \leq 0$, so we may assume $x \in [0,1[$. We do this part by induction skipping the case $n = 1$. We have:
$$
\frac{x^{n+1} + \dots + x}{n+1} = \frac{nx \frac{x^n + \dots + x}{n} + x}{n+1} \leq \frac{nx \frac{x^{n+1} + 1}{2}+x}{n+1} \overset{?}{\leq} \frac{x^{n+2} + 1}{2}.
$$
It remains to show that the last inequality holds. Subtracting the left hand side from the right hand side we obtain:
$$
\frac{(x^{n+2} - ((n+2)x - n-1)}{2(m+1)},
$$
which is positive as $x^{n+2}$ is convex and the tangent line at $1$ is given by $y = ((n+2)x - n -1)$.
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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