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Matrix transformation shearing the triangle to be right triangle If we have a triangle at $(1,1), (5,3), (7,1)$, how to find the sheared matrix to transform the triangle to be right triangle at $(1,1)$.
Is it that we need to find $i$ in
$\begin{pmatrix}
1 && i \\
0 && 1 \\
\end{pmatrix}$
But how to ensure the right triangle.
| Definition
Coordinate Transformation
Given a 2D Point $(x,y)$, to transform the point to another 2D coordinate space is defined through the following set of equations
$$x'=a\cdot x + b\cdot y$$
$$y'=c\cdot c + d\cdot y$$
where $a,b,c,d$ are real value constant with the constraint $a\cdot d - b\cdot c \ne 0$
And can be written in the matrix form as
$$\begin{pmatrix}
x' \\
y' \\
\end{pmatrix}
=
\begin{pmatrix}
a && b \\
c && d \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$
or in short
$$T( \vec x ) = \mathbf{A} \vec x $$
where $\vec x$ is a column vector.
For multiple vectors, the same can be written as
$$T( \vec X ) = \mathbf{A} \vec X \tag1$$
where
$$T(\vec X) = \left [\vec x_1\, \vec x_2\, ....\, \vec x_n \right ]$$
In your case
$$A=\begin{pmatrix}
1 && i \\
0 && 1 \\
\end{pmatrix}$$
$$X=\begin{pmatrix}
1 && 5 && 7 \\
1 && 3 && 1\\
\end{pmatrix}$$
Representing as $(1)$ we get
$$T(\vec X)=\begin{pmatrix}
1 && i \\
0 && 1 \\
\end{pmatrix}
\begin{pmatrix}
1 && 5 && 7 \\
1 && 3 && 1\\
\end{pmatrix}$$
$$\Rightarrow T(\vec X)=\begin{pmatrix}i + 1 & 3 i + 5 & i + 3\\1 & 3 & 1\end{pmatrix}\tag2$$
$$\Rightarrow T(\vec X)=\left [ \vec x_1\,\vec x_2\,\vec x_3\right ]$$
Once we determine the transposed vector, we need to determine the dot product of the respective vectors to determine, which of them equates to 0 satisfy the condition where in one of the points is $(1,1)$
$$\vec x_1 \cdot \vec x_2 = \left(i + 1\right) \left(3 i + 5\right) + 3$$
Solving which gives
$$\begin{bmatrix}- \frac{4}{3} - \frac{2}{3} \sqrt{2} \mathbf{\imath}, & - \frac{4}{3} + \frac{2}{3} \sqrt{2} \mathbf{\imath}\end{bmatrix}$$
$$\vec x_2 \cdot \vec x_3 = \left(i + 3\right) \left(3 i + 5\right) + 3$$
Solving which gives
\begin{bmatrix}- \frac{7}{3} - \frac{1}{3} \sqrt{5} \mathbf{\imath}, & - \frac{7}{3} + \frac{1}{3} \sqrt{5} \mathbf{\imath}\end{bmatrix}
$$\vec x_3 \cdot \vec x_1 = \left(i + 1\right) \left(i + 3\right) + 1$$
Solving which gives
\begin{bmatrix}-2\end{bmatrix}
Replacing each of these values in $(2)$, its evident that only the last solution satisfies the conditions
\begin{pmatrix}-1 & -1 & 1\\1 & 3 & 1\end{pmatrix}
So $i=-2$ is the solution to the problem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/335832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help me to prove this integration Where the method used should be using complex analysis.
$$\int_{c}\frac{d\theta}{(p+\cos\theta)^2}=\frac{2\pi p}{(p^2-1)\sqrt{p^2-1}};c:\left|z\right|=1$$
thanks in advance
| i do my self like this
on $|z| = 1, z = e^{i\theta}, d\theta=\frac{dz}{iz}$
using substitution $\cos\theta=\frac{z+z^{-1}}{2}$
$\frac{1}{i}\int\frac{\frac{dz}{z}}{(p+\frac{z+{z}^{-1}}{2})(p+\frac{z+{z}^{-1}}{2})} or \frac{1}{i}\int\frac{1}{(2pz+z^2+1)(p+\frac{z+{z}^{-1}}{2})} $
and then multiple by $\frac{z}{z}$
$\frac{1}{i}\int\frac{z}{(2pz+z^2+1)(2pz+z^2+1)}$
the roots of $(2pz+z^2+1)$, $z_{1}=-p+\sqrt{p^2-1};z_{2}=-p-\sqrt{p^2-1}$
let $z_{1}=a$ and $z_{2}=b$,
so $a-b=2\sqrt{p^2-1}$ and $ a+b=-2p$
and then i use residue theorm, but the result on the right $\frac{\pi p}{2(p^2-1)\sqrt{p^2-1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/335930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.
Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
Here's my idea:
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
And I'm stuck here.
I need to prove that:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$
In the first case using Cauchy-Schwarz Inequality I prove that:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$3(a^2 + b^2 + c^2) \ge (a+b+c)^2$
Now I need to prove that:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$
I need I don't know how to continue.
In the second case I tried proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and
$a^2 + b^2 + c^2 \ge a+b+c$
Using Cauchy-Schwarz Inequality I proved:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$
$a^2 + b^2 + c^2 \ge a+b+c$
But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$
So please help me with this problem.
P.S
My initial idea, which is proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
maybe isn't the right way to prove this inequality.
| From the given inequality $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ observe that $$2(ab+bc+ac)=(a+b+c)^2-a^2-b^2-c^2$$ We can rewrite the original inequality as
$$a^2+2\sqrt{a}+ b^2+2\sqrt{b}+ c^2+2\sqrt{c}\ge9$$ since $(a+b+c)=3$. Using AM-GM
set the LHS up as follows:
$$a^2+\sqrt{a}+\sqrt{a}\ge3\sqrt[3]{a^2 \sqrt{a}\sqrt{a}}=3a$$
$$b^2+\sqrt{b}+\sqrt{b}\ge3\sqrt[3]{b^2 \sqrt{b}\sqrt{b}}=3b$$
$$c^2+\sqrt{c}+\sqrt{c}\ge3\sqrt[3]{c^2 \sqrt{c}\sqrt{c}}=3c$$
Adding the three inequalities yields
$$a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 3(a+b+c) =9 $$ with equality if an only if $a$=$b$=$c$=$1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
A problem with limit How to attack this one?
Does the following limit exists:
$$\lim_{x\to +\infty}\dfrac {\cos^5x\sin^5x} {x^8\sin^2x-2x^7\sin x\cos^2x+x^6\cos^4x+x^2\cos^8x}$$
| The denominator can be rewritten as the sum of the two nonnegative terms
$$
(x \sin x - \cos^2 x)^2 x^6 + x^2 \cos^8 x.
$$
For all integral $n\ge 1$, when $x$ is in the interval $[2\pi n-\frac{\pi}{4}, 2\pi n+\frac{\pi}{4}]$ or $[2\pi n+\frac{3\pi}{4}, 2\pi n+\frac{5\pi}{4}]$, $|\cos x|\ge 2^{-1/2}$, so the second term will be at least $x^2/16$. On the other hand, if $x$ is in the interval $[2\pi n+\frac{\pi}{4}, 2\pi n+\frac{3\pi}{4}]$ or $[2\pi n+\frac{5\pi}{4}, 2\pi n+\frac{7\pi}{4}]$, $|x \sin x|\ge |\sin x|\ge 2^{-1/2}$ and $|\cos^2 x|\le\frac 12$, so the first term will be at least $\frac 12 x^6 (1-2^{-1/2})^2$. Assume that $x\ge 2\pi$. Then $\frac 12 x^6 (1-2^{-1/2})^2\ge x^2/16$, so, regardless of the interval $x$ belongs to, the denominator is bounded below by $x^2/16$. Therefore, the limit exists and is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Convergence of the sequence $x_{k+1}=\frac{1}{2}(x_k+\frac{a}{x_k})$
Consider the sequence $x_{k+1}=\frac{1}{2}(x_k+\frac{a}{x_k}), a\gt 0, x\in\mathbb{R}$. Assume the sequence converges, what does it converge to?
I'm having trouble seeing how to start,
Any help would be appreciated
Thanks
| This sequence has a closed form.
First, if $x_0=\sqrt{a}$, $x_n=\sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0\ne\sqrt{a}$ in the following.
Now, given $x_n\ne\sqrt{a}$ for some $n$, we have
$$x_{n+1}-\sqrt{a}=\frac{1}{2}\left(x_n+ \frac{a}{x_n} \right) - \sqrt{a}=\frac{1}{2}\left( \sqrt{x_n} - \frac{\sqrt{a}}{\sqrt{x_n}}\right)^2 > 0$$
Hence $x_n>\sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > \sqrt{a}$ too.
We can thus write $x_n = \sqrt{a} \ \coth{t_n}$. Then
$$x_{n+1} = \frac{1}{2}\left( \sqrt{a} \ \coth{t_n} + \frac{a}{\sqrt{a} \ \coth{t_n}} \right) = \frac{1}{2}\sqrt{a} \left( \coth{t_n} + \mathrm{th}\ t_n\right) = \sqrt{a} \ \coth \ 2 t_n$$
We have thus $t_{n+1}=2\ t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
$$x_n=\sqrt{a} \coth \left( 2^n \arg \coth \frac{x_0}{\sqrt{a}}\right)$$
And the value inside parentheses tends to infinity, so $\lim(x_n)=\sqrt{a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338098",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is $\lim_{n\to +\infty} \frac{\sqrt{n+1}}{2n^2+1}$ convergent? I don't know how to simplify this expression to apply the ratio test to determine if the sequence converges or diverges
$$\lim_{n\to +\infty} \frac{\sqrt{n+1}}{2n^2+1}$$
| Estimate sequence terms above and below
$$\frac{1}{2(n+1)^{\frac{3}{2}}}=\frac{\sqrt{n+1}}{2(n+1)^2} \leqslant \frac{\sqrt{n+1}}{2n^2+1} \leqslant \frac{2\sqrt{n}}{2n^2}=\frac{\sqrt{n}}{n^2}=\frac{1}{n^{\frac{3}{2}}}.$$
Hence $$\frac{\sqrt{n+1}}{2(n+1)^2}=O({n^{-\frac{3}{2}}}), \;\;\;n\to\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$ The question I've been given is this:
Using both sides of this equation:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
Find an expression for $$\sum_{n=0}^{\infty} n^2x^n$$
Then use that to find an expression for
$$\sum_{n=0}^{\infty}\frac{n^2}{2^n}$$
This is as close as I've gotten:
\begin{align*}
\frac{1}{1-x} & = \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \sum_{n=2}^{\infty} n(n-1)x^{n-2} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n(n-1)\frac{x^n}{x}(x+1) \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} (n^2x + n^2 - nx - n)\frac{x^n}{x} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n^2x^n + n^2\frac{x^n}{x} - nx^n - n\frac{x^n}{x} \\
\end{align*}
Any help is appreciated, thanks :)
| We can write any monomial, or polynomial in $n$, as a falling factorial polynomial using Stirling numbers of the second kind or this simple algorithm.
After we can use the following identities of finite calculus
$$\Delta_n n^\underline m=(n+1)^\underline m- n^\underline m=m n^\underline{m-1},\quad \sum n^\underline m\delta n=\frac{n^\underline{m+1}}{m+1}+C\tag{1}$$
$$\Delta_n x^n=(x-1)x^n,\quad\sum x^n\delta n=\frac{x^n}{x-1}+C\tag{2}$$
$$\sum f(n)\Delta_n g(n)\delta n=f(n)g(n)-\sum \Delta_n f(n)\mathrm E_n g(n)\delta n\tag{3}$$
$$\sum_{n=0}^h f(n)=\sum\nolimits_0^{h+1} f(n)\delta n\tag{4}$$
where $\mathrm E_n g(n):=g(n+1)$ is the shift operator and $C$ is any $1$-periodic function. This case is particularly easy, we have that $n^2=n^\underline 2+n^\underline 1$, hence
$$\sum n^2 x^n\delta n=\sum n^\underline 2 x^n\delta n+\sum n^\underline 1 x^n\delta n=\frac{n^\underline 2x^n}{x-1}-\frac2{x-1}\sum n^\underline 1x^{n+1}\delta n+\sum n^\underline 1 x^n\delta n+C=\\=\frac{n^\underline 2x^n}{x-1}+\left(1-\frac{2x}{x-1}\right)\left(\frac{n^\underline1 x^n}{x-1}-\frac1{x-1}\sum x^{n+1}\delta n\right)+C=\\=\frac{n^\underline 2x^n}{x-1}-\frac{x+1}{(x-1)^3}\left((n+1)x^{n+1}-nx^n\right)+C$$
Then for $|x|<1$ from above we have that
$$\sum_{n=0}^\infty n^2x^n=\sum\nolimits_0^\infty n^2x^n\delta n=\frac{(x+1)x}{(x-1)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Solving linear recurrence relation Solve the following linear recurrence relation:
$$h_n=4h_{n-1}-4h_{n-2}+n^2 2^n$$
for $n\geq2$ and $h_0=h_1=1$
| $$h_n - 2h_{n-1} = 2(h_{n-1} - 2h_{n-2}) + n^2 \cdot 2^n$$
Let $h_n - 2h_{n-1} = f_n$. We then have
$$f_n = 2f_{n-1} + n^2 \cdot 2^n = 2(2f_{n-2} + (n-1)^2 \cdot 2^{n-1}) + n^2 \cdot 2^n = 4 f_{n-2} + 2^n \cdot (n^2 + (n-1)^2)$$
Hence, by induction, we have
$$f_n = 2^n \cdot (n^2 + (n-1)^2 + \cdots + 2^2 + 1^2) + 2^n \cdot f_0 = 2^n \cdot \left( f_0 + \dfrac{n(n+1)(2n+1)}6\right)$$
Hence,
\begin{align}
h_n & = 2h_{n-1} + 2^n \cdot \left( c + \dfrac{n(n+1)(2n+1)}6\right)\\
& = 4h_{n-2} + 2^n \cdot \left(2c + \dfrac{n(n+1)(2n+1) + (n-1)n(2n-1)}6\right)\\
& = 2^n \cdot h_{0} + 2^n \cdot \left(nc + \dfrac{\displaystyle \sum_{k=1}^n k(k+1)(2k+1)}6\right)
\end{align}
Hence,
$$h_n = 2^n \left(an^4+bn^3+cn^2+dn+e\right)$$
where the $a,b,c,d$ and $e$ can be obtained using the recurrence in terms of the initial values $h_0$ and $h_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339975",
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"source": "stackexchange",
"question_score": "5",
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} |
$\lim _{n \rightarrow \infty }a_{n}=\sqrt{a_{n-2}a_{n-1}}$ , $a_1=1, a_2=2$ I was asked to find the limit of:
$a_{n+2}=\sqrt{a_na_{n+1}}$
$a_1=1, a_2=2$
It seems as if the sequence is constant from n=4 and it's value is $a_n=\sqrt{2\sqrt{2}} -\forall{n>3}$
I'd just like to double-check I did thing right.
Thanks!
| Take logs of both sides and define $b_n = \log{a_n}$. Then
$$2 b_n - b_{n-1}-b_{n-2} = 0$$
$$b_0=0$$
$$b_1=\log{2}$$
This is a constant coefficient difference equation with solution $b_n=A r^n$,where $r$ satisfies
$$2 r^2-r-1=0$$
with solutions $r_+ = 1$ and $r_-=-1/2$. Thus
$$b_n = A + B \left (-\frac{1}{2} \right )^n$$
with
$$A+B=0$$
$$A-\frac{1}{2} B = \log{2}$$
The $$b_n=\frac{2}{3} \log{2} \left [ 1 + \left (-\frac{1}{2} \right )^{n+1} \right]$$
Then going back to the original sequence:
$$a_n = 2^{\frac{2}{3}\left [ 1 + \left (-\frac{1}{2} \right )^{n+1} \right]}$$
$$\lim_{n \rightarrow \infty} a_n = 2^{2/3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/340408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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References to integrals of the form $\int_{0}^{1} \left( \frac{1}{\log x}+\frac{1}{1-x} \right)^{m} \, dx$ While extending my calculation techniques, with aid of Mathematica, I found that
\begin{align*}
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{3} \, dx
&= -6 \zeta '(-1) -\frac{19}{24}, \\
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{4} \, dx
&= -10 \zeta '(-2)-2 \zeta '(-1)-\frac{37}{72}, \\
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{5} \, dx
&= -\frac{35}{3} \zeta '(-3)-\frac{15}{2} \zeta '(-2)-\frac{5}{3} \zeta '(-1)-\frac{3167}{8640}, \\
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{6} \, dx
&=-\frac{21}{2} \zeta '(-4)-14 \zeta '(-3)-\frac{31}{4} \zeta '(-2)-\frac{3}{2} \zeta '(-1)-\frac{1001}{3600}.
\end{align*}
I conjectures that these relations extends also to higher degrees:
My Guess. For $m \geq 3$, we can write
$$ \int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{m} \, dx = - \Bigg( q + \sum_{k=1}^{m-2} q_k \zeta'(-k) \Bigg) $$
for some positive rational numbers $q$ and $q_k$.
Is there any reference regarding this problem?
Addendum. Following i707107's advice, I obtained the following formula
\begin{align*}
& \int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{m} \, dx \\
&= -\frac{H_{m-1}}{(m-1)!} + \frac{1}{(m-1)!} \sum_{k=1}^{m-1} \left[{{m-1}\atop{k}}\right] \zeta(1-k) \\
&\quad - \frac{1}{(m-2)!}\sum_{j=1}^{m-1}\sum_{l=m-j}^{m-1} \binom{m}{j} \binom{m-2}{j-1} \left[{{j-1}\atop{l+j-m}}\right] \{ \zeta'(1-l) + H_{m-j-1} \zeta(1-l) \}
\end{align*}
valid for $m \geq 2$, where $\left[{{n}\atop{k}}\right]$ denotes the unsigned Stirling's number of the first kind.
| You can try
$$
\int_0^1\left(\frac{1}{\log x}+\frac{1}{1-x}\right)^m (-\log x)^{s-1}\,dx
$$
for $s$ with sufficiently large real part.
This will give you an expression involving $\Gamma$ and $\zeta$ functions.
Then use the analytic continuation to $\sigma>0$ and plug in $s=1$.
I checked this method for $m=2$ and got the right answer. I am sure that this will work in $m\geq 3$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "24",
"answer_count": 1,
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Does a function exist with the property $f(-n^2+3n+1)=(f(n))^2+1$? Let $f:\mathbb{R}\to \mathbb{R}$ be a function which fulfills for every $n \in \mathbb{N}$
$$f(-n^2+3n+1)=(f(n))^2+1$$
Is it possible that such a function exists?
| The main idea here is writing at first some of the equations you get and look if they have common terms. Indeed here the terms for $n=3$ and $n=1$ are very interessting, as in both only occur $f(1)$ and $f(3)$.
\begin{align*}
f(3)&=1+f(1)^2 \tag{$i$}\\
f(1)&=1+f(3)^2 \tag{$ii$}\\
\end{align*}
As we don't know that much lets try to get an equation only having $f(1)$.
At first we have this equation:
$$f(1)=1+f(3)^2$$
No we use $(i)$ to express $f(3)$ in terms of $f(1)$
$$f(1)= 1+(1+f(1)^2)^2=1+1^2+2f(1)^2+f(1)^4$$
is that possible?
Note that $f(1)$ is a solution of
$$0= 2-x+2x^2+x^4$$
but this one has no real solution, hence your function can't exist,
as for $x \in [0,1]$
$$2-x+2x^2+x^4\geq 2-x> 0 $$
and for $x\in [1,\infty)$ we know that $x<x^2$ and hence
$$2-x+2x^2+x^4 \geq 2+x^2+x^4>0$$
and for $x\in (-\infty,0]$
$$2-x+2x^2+x^4 \geq 2+2x^2 +x^4 >0$$
| {
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"url": "https://math.stackexchange.com/questions/341479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor $(a^2+2a)^2-2(a^2+2a)-3$ completely I have this question that asks to factor this expression completely:
$$(a^2+2a)^2-2(a^2+2a)-3$$
My working out:
$$a^4+4a^3+4a^2-2a^2-4a-3$$
$$=a^4+4a^3+2a^2-4a-3$$
$$=a^2(a^2+4a-2)-4a-3$$
I am stuck here. I don't how to proceed correctly.
| If you assign $$ a^2 + 2a = x $$ you'll get: $$ x^2 - 2x - 3 $$
Considering that $$ x^2 - 2x - 3 = (x - 3)(x+1) $$ you'll get: $$ (a^2 + 2a - 3)(a^2 + 2a+1) = (a + 3)(a - 1)(a + 1)^2 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ $$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
| Here is an interesting different way: Let
$$x = \cos(A+B)\cos(A-B) \\
y = \sin(A+B)\sin(A-B)$$
Then,
$$x+y = \cos(A+B-A+B) = \cos(2B) \\
x-y = \cos(A+B+A-B) = \cos(2A)$$
You can add these to get $2x$ or subtract them to get $2y$. Then expand using the double-angle formulas. This gives you two trig product formulas at the same time.
| {
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"url": "https://math.stackexchange.com/questions/345703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Conditional probabilities, urns I found this question interesting and apparently it has to do with conditional probabilities:
An urn contains six black balls and some white ones. Two balls are drawn simutaneously. They have the same color with probability 0.5. How many with balls are in
the urn?
As far as I am concerned I would say it is two white balls...
| probability of 2 black balls: $\frac{6}{8}*\frac{5}{7}=\frac{30}{56}$
of 2 white balls: $\frac{2}{8}*\frac{1}{7}=\frac{2}{56}$
$\frac{30}{56}+\frac{2}{56}=\frac{32}{56}\neq\frac{1}{2}$
Assume the number of white balls is $n$:
2 black balls = $\frac{6}{6+n}*\frac{5}{5+n}=\frac{30}{n^2+11n+30}$
2 white balls = $\frac{n}{6+n}*\frac{n-1}{5+n}=\frac{n^2-n}{n^2+11n+30}$
total probability $=\frac{1}{2}$ (as per problem setup) and also $=\frac{30}{n^2+11n+30}+\frac{n^2-n}{n^2+11n+30}=\frac{n^2-n+30}{n^2+11n+30}$
Moving sides, $2(n^2-n+30)=n^2+11n+30$
$2n^2-2n+60=n^2+11n+30$
$n^2-13n+30=0$
$(n-10)(n-3)=0$
$n=\{10,3\}$
Thus, this problem is solved when there are either 3 or 10 white balls.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$?
| the equation:
$X^2+Y^2=Z^2-1$
Solutions can be written using the solutions of Pell's equation: $p^2-2k(k-1)s^2=1$
$k$ - given by us.
Solutions hav e form:
$X=2kps-2(k-1)s^2$
$Y=2(k-1)ps+2ks^2$
$Z=p^2+2(k^2-k+1)s^2$
And more:
$X=2p^2-2(3k-2)ps+2(2k-1)(k-1)s^2$
$Y=2p^2-2(3k-1)ps+2k(2k-1)s^2$
$Z=3p^2-4(2k-1)ps+2(3k^2-3k+1)s^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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To find Area of rectangular with given 3 parameters
$a,b,c$ are given parameters . I would like to find Area of (ABCD) rectangular.
I can find $d$ from $a,b,c$.
$$(x-m)^2+(y-n)^2=a^2$$
$$(x-m)^2+n^2=b^2$$
$$m^2+(y-n)^2=c^2$$
$$m^2+n^2=d^2$$
$$m^2+n^2+(x-m)^2+(y-n)^2=a^2+d^2=b^2+c^2$$
$$d=\sqrt {b^2+c^2-a^2}$$
Let's define
$\angle AEB =\alpha$,
$\angle DEC =\beta$ ,$\angle AED =\gamma$ , $\angle BEC =\phi$
$$x^2=a^2+c^2-2ac \cos (\alpha)$$
$$x^2=b^2+d^2-2bd \cos (\beta)$$
$$y^2=c^2+d^2-2cd \cos (\gamma)$$
$$y^2=a^2+b^2-2ab \cos (\phi)$$
$$b^2+d^2-2bd \cos (\beta)=a^2+c^2-2ac \cos (\alpha)$$
$$a^2+b^2-2ab \cos (\phi)=c^2+d^2-2cd \cos (\gamma)$$
And also we know that
$$\alpha + \beta + \phi + \gamma = 2 \pi $$
$$\cos (\alpha + \beta + \phi + \gamma)= \cos (2 \pi)=1$$
Area of $ABCD =\frac{1}{2} [ac \sin (\alpha) + bd \sin (\beta) + cd \sin (\gamma))+ ab \sin (\phi)]=xy$
I am stuck to solve the equations and find the area by given $a,b,c$, Is it possible to find area of ABCD rectangular via given 3 parameters $a,b,c$ ?
Thanks for hints and answers.
UPDATE: Nov, 14th 2014:
I proved that Area of ABCD does not depend on only $a,b,c$
$$y=a.\sin P +b \sin Q$$
$$|EF|=a \cos P=b \cos Q$$
$$x=a.\cos P +\sqrt{c^2-a^2 \sin^2 P}$$
$$y=a.\sin P +b \sin Q=a.\sin P +b \sqrt{1-\frac{a^2 \cos^2 P}{b^2}}$$
$$y=a.\sin P +b \sin Q=a.\sin P + \sqrt{b^2-a^2 \cos^2 P}$$
Area of $ABCD=x.y=(a.\cos P +\sqrt{c^2-a^2 \sin^2 P})(a.\sin P + \sqrt{b^2-a^2 \cos^2 P})$
The formula shows that The Area also depends on an angle not only $a,b,c$
|
look at the above picture, you can see orange and yellow rectangles are all satisfy a,b,c if m,n is not fixed. and there are many such rectangles so you can't find the area. but you can find max area by a,b,c which might be another exercise you can do which is very hard to find out the final result.
for m,n is fixed, $d=\sqrt{m^2+n^2}$ .
BTW, if one of m and n is known,then the area is fixed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating the integral $\int_{0}^{\pi /6}\sqrt{1-\left(\frac{R_s\sin \theta }{C_L}\right)^2} d\theta$ I want to integrate $I=\int\limits_{0}^{\pi /6}{\sqrt{1-{{\left( \frac{{{R}_{s}}\sin \theta }{{{C}_{L}}} \right)}^{2}}}d \theta}$.
I get incomplete elliptic integral $E(z\mid m)$ in the calculation by mathematica. I need some simple calulation for including the function in further calulations along with other functions.
Any way to proceed directly, without the help of Elliptic Integral?
| For the binomial series of $\sqrt{1-x}$ , $\sqrt{1-x}=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^n}{4^n(n!)^2(1-2n)}$
$\therefore\int_0^{\frac{\pi}{6}}\sqrt{1-\left(\dfrac{R_s\sin\theta}{C_L}\right)^2}~d\theta=\int_0^{\frac{\pi}{6}}\sum\limits_{n=0}^\infty\dfrac{(2n)!R_s^{2n}\sin^{2n}\theta}{4^n(n!)^2(1-2n)C_L^{2n}}d\theta=\int_0^{\frac{\pi}{6}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(2n)!R_s^{2n}\sin^{2n}\theta}{4^n(n!)^2(1-2n)C_L^{2n}}\right)d\theta$
Now for $\int\sin^{2n}\theta~d\theta$ , where $n$ is any natural number,
$\int\sin^{2n}\theta~d\theta=\dfrac{(2n)!\theta}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int_0^{\frac{\pi}{6}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(2n)!R_s^{2n}\sin^{2n}\theta}{4^n(n!)^2(1-2n)C_L^{2n}}\right)d\theta$
$=\left[\theta+\sum\limits_{n=1}^\infty\dfrac{((2n)!)^2R_s^{2n}\theta}{4^{2n}(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sin^{2k-1}\theta\cos\theta}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}\right]_0^{\frac{\pi}{6}}$
$=\left[\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2R_s^{2n}\theta}{4^{2n}(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sin^{2k-1}\theta\cos\theta}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}\right]_0^{\frac{\pi}{6}}$
$=\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2R_s^{2n}\pi}{4^{2n}6(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sin^{2k-1}\dfrac{\pi}{6}\cos\dfrac{\pi}{6}}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}$
$=\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2R_s^{2n}\pi}{2^{2n+1}3(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sqrt3}{4^{2n+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}$
| {
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How to integrate $f( \theta ) = \frac{1}{a + \sin( \theta ) }$? Let $a > 1$. I am wondering how evaluate the integral: $$ \int_{0}^{2 \pi } \frac{1}{a + \sin( \theta) } d \theta $$ by means of methods of complex analysis. In the homework assignment, the following hint is given: write $\sin( \theta ) = (e^{i \theta } - e^{- i \theta} ) / 2i $ and interpret the integral (after some algebraic manipulations) as a complex line integral of a rational function over the positively oriented unit circle.
I write $\theta := t$, so I'll have to type less.
This is how I approached the question: We know, from the definition of the complex line integral, that $$ \int_{ \alpha } f(t) dt = \int_{a}^{b} f( \alpha (t) ) \alpha ' (t) .$$
In our case, we have $\alpha(t) = e^{i t}$. So, if we want to rewrite our "ordinary" integral as a complex line integral, we have the equation $$ f( \alpha (t) ) \cdot e^{i t} = \frac{1}{a + \frac{ e^{i t} - 1/e^{i t} }{2i} } $$. If we divide both sides by $e^{i t}$, and rewrite the denominator of the resulting fraction a bit, we obtain: $$f( \alpha (t)) = \frac{2i}{e^{2 i t} + 2 i a e^{i t} -1 } . $$ Since we already noted, that $ \alpha(t) = e^{i t} $, I thought that, based on this, we can deduce that $$f(t) = \frac{2i}{t^2 + 2 i a t - 1} $$.
From here, I'm not entirely sure how to proceed. One possibility is to find the roots of the polynomial in the denominator of the fraction in the integral, by means of the (abc)-rule. We obtain the roots $t_1 =i a - \sqrt{1 -a} $ and $t_2 = i a + \sqrt{ 1 - a }$. We know, that $a >1 $, so we can rewrite these roots: $t_1 = i a - i \sqrt{a-1} = i(a - \sqrt{a-1} $ , and $t_2 = i a + i \sqrt{a-1} = i (a + \sqrt{a-1} ) $, so we can rewrite our integral as follows: $$ \int_{ \alpha } \frac{1}{ (t - t_1) (t-t_2) } $$ . But how do we proceed from here? We don't know the value of $a$, so we don't know how "big" the roots are. Could we use the Cauchy Integral Formula? Or something else? How do we use that we integrate over a positively orientated unit circle?
| As you did, let $z=e^{i\theta}$. Then $dz=izd\theta$ and
$$ \sin\theta=\frac{1}{2i}\left(z-\frac{1}{z}\right) $$
and hence
\begin{eqnarray*}
\int_0^{2\pi}\frac{1}{a+\sin\theta}d\theta&=&\int_{|z|=1}\frac{1}{a+\frac{1}{2i}\left(z-\frac{1}{z}\right)}\frac{dz}{iz}\\
&=&2\int_{|z|=1}\frac{1}{z^2+2aiz-1}dz\\
&=&2\int_{|z|=1}\frac{1}{(z+ai)^2+a^2-1}dz\\
&=&2\int_{|z|=1}\frac{1}{(z+ai+\sqrt{a^2-1}i)(z+ai-\sqrt{a^2-1}i)}dz\\
&=&2\cdot 2\pi i\text{Res}\left(\frac{1}{(z+ai+\sqrt{a^2-1}i)},z=-ai+\sqrt{a^2-1}i)\right)\\
&=&4\pi i\frac{1}{2\sqrt{a^2-1}i}\\
&=&\frac{2\pi}{\sqrt{a^2-1}}.
\end{eqnarray*}
Another simple way is to use trigonometric transforms instead of complex analysis. Let $t=\tan\frac{\theta}{2}$ and then $\sin\theta=\frac{2t}{t^2+1}$ and $d\theta=\frac{2}{t^2+1}dt$. Thus
\begin{eqnarray*}
\int_0^{2\pi}\frac{1}{a+\sin\theta}d\theta&=&\int_{-\infty}^\infty\frac{1}{a+\frac{2t}{t^2+1}}\frac{2}{t^2+1}dt\\
&=&2\int_{-\infty}^\infty\frac{1}{a(t^2+1)+2t}dt\\
&=&\frac{2}{a}\int_{-\infty}^\infty\frac{1}{(t+\frac{1}{a})^2+1-\frac{1}{a^2}}dt\\
&=&\frac{2}{a}\frac{1}{\sqrt{1-\frac{1}{a^2}}}\left.\arctan\frac{t+\frac{1}{a}}{\sqrt{1-\frac{1}{a^2}}}\right|_{-\infty}^\infty\\
&=&\frac{2\pi}{\sqrt{a^2-1}}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the maximum of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$ If $a,b,c,d$ are distinct real numbers such that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}=4$ and $ac=bd$.
Then how would we calculate the maximum value of $$\dfrac{a}{c}+\dfrac{b}{d}+\dfrac{c}{a}+\dfrac{d}{b}.$$
I was unable to proceed due to the 'distinct'.
| Let $w=\frac{a}{b}$, $x=\frac{b}{c}$, $y=\frac{c}{d}$, $z=\frac{d}{a}$. Then
$$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=wx+xy+yz+zw=(x+z)(w+y)\leqslant\Bigl(\frac{w+x+y+z}{2}\Bigr)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to generalize Ramanujan's lower bound for factorials that he used in his proof of Bertrand's Postulate? I am starting to feel more confident in my understanding of Ramanujan's proof of Bertrand's postulate. I hope that I am not getting overconfident.
In particular, Ramanujan's does the following comparison in step (8):
$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x}{2} + \frac{1}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$
It occurs to me that this can be generalized to:
$$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) - \ln\Gamma(\frac{x}{b_3} + \frac{1}{2})\le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
when:
$$\frac{x}{b_1} = \frac{x}{b_2} + \frac{x}{b_3}$$
I would really appreciate it if my argument could be reviewed and someone could call out any mistakes either in the answer or in comments. :-)
Here's the argument for this generalization:
Let:
$$\{\frac{x}{b_i}\} = \frac{x}{b_i} - \lfloor\frac{x}{b_i}\rfloor$$
where:
$$0 \le \{\frac{x}{b_i}\} < 1$$
Since:
$$\{\frac{x}{b_1}\} + \lfloor\frac{x}{b_1}\rfloor = \{\frac{x}{b_2}\} + \lfloor\frac{x}{b_2}\rfloor + \{\frac{x}{b_3}\} + \lfloor\frac{x}{b_3}\rfloor$$
We have:
$$\{\frac{x}{b_1}\} \le \{\frac{x}{b_2}\} + \{\frac{x}{b_3}\}$$
So that:
$$-\{\frac{x}{b_1}\} \ge -\{\frac{x}{b_2}\} + -\{\frac{x}{b_3}\}$$
$$2-\{\frac{x}{b_1}\} \ge 1-\{\frac{x}{b_2}\} + 1-\{\frac{x}{b_3}\}$$
$$1-\{\frac{x}{b_1}\} \ge 1-\{\frac{x}{b_2}\} - \frac{1}{2} + 1-\{\frac{x}{b_3} \} - \frac{1}{2}$$
$$\lfloor\frac{x}{b_1}\rfloor + 1 - \frac{x}{b_1} \ge (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - \frac{1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})$$
$$\lfloor\frac{x}{b_1}\rfloor + 1 \ge (\lfloor\frac{x}{b_2}\rfloor + 1) + (\lfloor\frac{x}{b_3}\rfloor + 1)$$
If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,
Using the logic in the answer here:
$$\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}\frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}$$
Let:
$x_1 = \frac{x}{b_1}$, $\Delta{t_1} = 1 - \{\frac{x}{b_1}\}$,
$x_2 = \frac{x}{b_2}+\frac{1}{2}$, $\Delta{t_2} = \frac{1}{2} - \{\frac{x}{b_2}\}$
$x_3 = \frac{x}{b_3}+\frac{1}{2}$, $\Delta{t_3} = \frac{1}{2} - \{\frac{x}{b_3}\}$
where $\frac{x}{b_2} \ge \frac{x}{b_3}$ (Otherwise, switch the two values).
Then:
$$\frac{\Gamma(\lfloor\frac{x}{b_1}\rfloor+1)}{\Gamma(\frac{x}{b_1})} \ge \frac{\Gamma(\lfloor\frac{x}{b_2}\rfloor+1)}{\Gamma(\frac{x}{b_2} + \frac{1}{2})}\frac{\Gamma(\lfloor\frac{x}{b_3}\rfloor+1)}{\Gamma(\frac{x}{b_3}+\frac{1}{2})}$$
So then it follows:
$$\ln\Gamma(\lfloor\frac{x}{b_1}\rfloor + 1) - \ln\Gamma(\frac{x}{b_1}) \ge \ln\Gamma(\lfloor\frac{x}{b_2}\rfloor + 1) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) + \ln\Gamma(\lfloor\frac{x}{b_3}\rfloor + 1) - \ln\Gamma(\frac{x}{b_3} + \frac{1}{2})$$
And we have shown:
$$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2}+\frac{1}{2}) - \ln\Gamma(\frac{x }{b_3}+\frac{1}{2}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
Please let me know if you see any mistakes.
Thanks,
-Larry
Edit: Based on reviewing Zander's answer, I believe that this argument can be saved. The revision requires two separate arguments:
*
*one for: $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$
*another for: $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$
The link for the first argument is here. The link for the second argument is here.
| After this step:
$$
\lfloor\frac{x}{b_1}\rfloor + 1 - \frac{x}{b_1} \ge (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - \frac{1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})
$$
we have
$$
\lfloor\frac{x}{b_1}\rfloor + 1 \ge (\lfloor\frac{x}{b_2}\rfloor + 1) + (\lfloor\frac{x}{b_3}\rfloor + 1)-1
$$
but what you have is incorrect.
For example, let $x=77.1,b_1=6,b_2=7,b_3=42$, then $\lfloor x/b_1\rfloor=12, \lfloor x/b_2\rfloor=11, \lfloor x/b_3 \rfloor=1$ and $13\not\ge 14$.
And in this case your inequality is also violated:
$$
\ln \Gamma\left(\frac{77.1}{6}\right)-\ln\Gamma\left(\frac{77.1}{7}+\frac{1}{2}\right)-\ln \Gamma\left(\frac{77.1}{42}+\frac{1}{2}\right) = 3.10698\cdots
\\
\ge \ln(12!)-\ln(11!)-\ln(1!) = 2.4849\cdots
$$
| {
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Basis of complex matrix vector space over $\Bbb{R}$ I understand that the basis of the vector space $$Mat_2(\Bbb{R}) = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}$$ over $\Bbb{R}$ is $$e = \left\{ \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix} \right\}$$
However, I can't figure out the basis or the vector space $Mat_2(\Bbb{C}) = \begin{pmatrix}a_{11}+b_{11}i & a_{12}+b_{12}i \\ a_{21}+b_{21}i & a_{22}+b_{22}i\end{pmatrix}$ over $\Bbb{R}$.
Thank You.
| That would be $e = \left\{ \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix} \right\}$together with $e' = \left\{ \begin{pmatrix}i & 0\\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & i\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ i & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & i\end{pmatrix} \right\}$.
Think of it this way. You want elements such that you can write matrix of the form
$Mat_2(\Bbb{C}) = \begin{pmatrix}a_{11}+b_{11}i & a_{12}+b_{12}i \\ a_{21}+b_{21}i & a_{22}+b_{22}i\end{pmatrix}$. But you can only multiply the elements with real numbers. Therefore you will need two elements for every position in the matrix. One element takes care of the complex part, one element for the real part.
Could you find base elements for $\mathbb{C}$ over $\mathbb{R}$? That would be the familiar $1$ and $i$. A square matrix of dimension $n\times m$ has as dimension the dimension of the underlying field to the power $n\times m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles:
$\tan75^\circ = 2 + \sqrt{3}$
He gave this proof (via a text message!)
$1. \tan75^\circ$
$2. = \tan(60^\circ + (30/2)^\circ)$
$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $
$4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$
Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?
EDIT
Using the answers given I'm almost there:
*
*$\tan75^\circ$
*$\tan(45^\circ + 30^\circ)$
*$\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$
*$(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$
*$\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$
*$\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$
*multiply throughout by $(\sqrt{3}) + 1)$
Another alternative approach:
*
*$\tan75^\circ$
*$\tan(45^\circ + 30^\circ)$
*$\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$
*$\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$
*at point 6 in above alternative
| A proof without words (but it uses some geometry). Is that OK?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Linear algebra on Fibonacci number Consider the sequence $\{a_n\}_{n\ge 0}$ given by the recurrence relation
$$a_0=1,\ a_1=-1,\ a_{n+1}=3a_n+10a_{n-1}\ \ \text{for } n\ge2$$
And I am asked to work out the closed form expression for an in the same fashion as the proof for Fibonacci numbers by using linear algebra way
Thanks all for the help!
| We can use the generating function $f(x)$ of $\{a_n\}$ to solve. Let $f(x)=\sum_{n=0}^\infty a_nx^n$. Then
\begin{eqnarray*}
f(x)&=&\sum_{n=0}^\infty a_nx^n=1-x+\sum_{n=2}^\infty a_nx^n\\
&=&1-x+\sum_{n=1}^\infty a_{n+1}x^{n+1}\\
&=&1-x+\sum_{n=1}^\infty (3a_{n}+10a_{n-1})x^{n+1}\\
&=&1-x+3\sum_{n=1}^\infty a_{n}x^{n+1}+10\sum_{n=1}^\infty a_{n-1}x^{n+1}\\
&=&1-x+3x\sum_{n=1}^\infty a_{n}x^{n}+10x^2\sum_{n=1}^\infty a_{n-1}x^{n-1}\\
&=&1-x+3x(f(x)-1)+10x^2f(x)
\end{eqnarray*}
from which we obtain the following function equation
$$ f(x)=1-x+3x(f(x)-1)+10x^2f(x). $$
Hence
\begin{eqnarray*}
f(x)&=&\frac{4x-1}{10x^2+3x-1}\\
&=&\frac{6}{7}\frac{1}{1+2x}+\frac{1}{7}\frac{1}{1-5x}\\
&=&\frac{6}{7}\sum_{n=0}^\infty (-2)^nx^n+\frac{1}{7}\sum_{n=0}^\infty 5^nx^n
\end{eqnarray*}
and so
$$ a_n=\frac{6}{7}(-2)^n+\frac{1}{7}5^n. $$
| {
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"url": "https://math.stackexchange.com/questions/362148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
$c$ is a complex number that satisyfing $(c+\frac{1}{c}+1)(c+\frac{1}{c}) = 1$ Let $c$ is complex-number satisfying :
$(c+\frac{1}{c}+1)(c+\frac{1}{c}) = 1$
So, how could i get
$(3c^{100}+\frac{2}{c^{100}}+1)(c^{100}+\frac{2}{c^{100}}+3)$ ?
| HINT:
On simplification, $c^4+c^3+c^2+c+1=0$
Clearly, $c\ne1$
Multiply either sides the $(c-1),$ we get $c^5-1=(c-1)\cdot0=0$
$\implies c^5=1\implies c^{100}=(c^5)^{20}=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Summing $ \sum _{k=1}^{n} k\cos(k\theta) $ and $ \sum _{k=1}^{n} k\sin(k\theta) $ I'm trying to find
$$\sum _{k=1}^{n} k\cos(k\theta)\qquad\text{and}\qquad\sum _{k=1}^{n} k\sin(k\theta)$$
I tried working with complex numbers, defining $z=\cos(\theta)+ i \sin(\theta)$ and using De Movire's, but so far nothing has come up.
| All identities are coming from Wikipedia page for trigonometric identities. The main identity that we start with is:
$$
\sum_{k=1}^n \cos(kx)= \frac{ \sin((n+\frac{1}{2}) x)}{2\sin(x/2)} -1
$$
Get a derivative with respect to $x$, and we have
\begin{eqnarray*}
\sum_{k=1}^n k \sin(kx)&=& \frac{ \cos(x/2) \sin((n+\frac{1}{2})x)- (2n+1) \cos((n+\frac{1}{2})x) \sin(x/2)}{4\sin^2(x/2)}\\
&=& \frac{-2n \cos((n+\frac{1}{2})x) \sin(x/2) + \left( \cos(x/2) \sin((n+\frac{1}{2})x)- \cos((n+\frac{1}{2})x) \sin(x/2)\right)}{4\sin^2(x/2)}\\
&=& \frac{-2n \cos((n+\frac{1}{2})x) \sin(x/2) + \sin(nx)}{4\sin^2(x/2)}\\
&=& \frac{ -n\sin( (n+1)x) + n\sin(nx) + \sin(nx)}{4\sin^2(x/2)}\\
&=& \frac{ -n\sin( (n+1)x) + (n + 1) \sin(nx)}{4\sin^2(x/2)}.
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
A counting problem involving ternary sequences
A ternary sequence is a sequence all of whose elements are the digits 0, 1 or 2. Find the number of ternary sequences of length 8 in which the digits 0 and 1 each occur an even number of times.
The first case that should be considered is when we have all 2s, in which case there are certainly an even number of 1s and 0s. For each 'dilution' of 2s thereafter there are two options: two 1s are placed or two 0s are placed. There are four such dilutions that can take place so we have $4\times 2 + 1 = 9$ possibilities so far. However, this method makes counting more difficult than it should be. Is there a more efficient way of continuing the count?
| Do as follows: Define $a_n$ as the number of $n$-sequences with an even number of 0 and an even number of 1, $b_n$ for even/odd, $c_n$ for odd/even and $d_n$ for odd/odd. Then $a_0 = 1$, and $b_0 = c_0 = d_0 = 0$. You can set up recurrences for all for them (think how you can get e.g. odd/even by adding a symbol):
$$
\begin{align*}
a_{n + 1} &= a_n + b_n + c_n \\
b_{n + 1} &= a_n + b_n + d_n \\
c_{n + 1} &= a_n + c_n + d_n \\
d_{n + 1} &= b_n + c_n + d_n
\end{align*}
$$
Define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and so on, by the properties of generating functions:
$$
\begin{align*}
\frac{A(z) - a_0}{z} &= A(z) + B(z) + C(z) \\
\frac{B(z) - b_0}{z} &= A(z) + B(z) + D(z) \\
\frac{C(z) - c_0}{z} &= A(z) + C(z) + D(z) \\
\frac{D(z) - d_0}{z} &= B(z) + C(z) + D(z)
\end{align*}
$$
The solution of this nice linear system is:
$$
\begin{align*}
A(z) &= \frac{1 - 2 z - z^2}{(1 - z) (1 + z) (1 - 3 z)} \\
B(z) &= \frac{z}{(1 + z) (1 - 3 z)} \\
C(z) &= \frac{z}{(1 + z) (1 - 3 z)} \\
D(z) &= \frac{2 z}{1 - z) (1 + z) (1 - 3 z)}
\end{align*}
$$
We are really only interested in the $a_n$. Expanding $A(z)$ as partial fractions:
$$
A(z) = \frac{1}{4} \cdot \frac{1}{1 - 3 z}
+ \frac{1}{4} \cdot \frac{1}{1 + z}
+ \frac{1}{2} \cdot \frac{1}{1 - z}
$$
Reading the geometric series:
$$
a_n = \frac{1}{4} \cdot 3^n + \frac{1}{4} \cdot (-1)^n + \frac{1}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as:
If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.
The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left(-\frac{\pi}{2} - x\right)\\
&= \cos\left(\frac{\pi}{2} + x\right)\\
&= -\sin(x)
\end{align*}
Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$?
Similarly, if $\cos(x) = 0.6$, find $\sin\left(\frac{3\pi}{2} + x\right)$, the solution says $\sin\left(\frac{3\pi}{2} + x\right)$ is equal to $\cos(\pi+x)$. How did they get $\cos(\pi+x)$?
| Remember that cosine is $2\pi$ periodic this. This means that
$$\cos(x + 2n\pi) = \cos(x)$$
for any $x\in\mathbb{R}$ and for any integer $n\in\mathbb{Z}$. Basically, this says that if you can shift the graph of cosine left or right by $2\pi$ without changing it. Shifting it once gives you your desired equality since
$$\cos\left(-\frac{\pi}{2} -x\right)=\cos\left(2\pi - \frac{\pi}{2} - x\right)=\cos\left(\frac{3\pi}{2}-x\right)$$
Your second question is similar. Sine and cosine satisfy the shift equality
$$\sin\left(\frac{\pi}{2} + x\right)=\cos(x)$$
Graphically, this means that shifting the graph of sine by $\frac{\pi}{2}$ to the left gives the graph of cosine. Applying this property will give you your second equality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a nice way to interpret this matrix equation that comes up in the context of least squares So I am working on this problem with fitting a second degree polynomial of the form $y=a_1x^2+a_2x+a_3$ to four points using least squares. One of the parts of the problem is to write out the matrix equation that describes the least squares problem. Basically we have the equation
$$\begin{bmatrix}
x_1^2 & x_1 & 1 \\
x_2^2 & x_2 & 1 \\
x_3^2 & x_3 & 1 \\
x_4^2 & x_4 & 1
\end{bmatrix}
\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}
=
\begin{bmatrix}
y_1 \\
y_2 \\
y_3 \\
y_4
\end{bmatrix}$$
Let's call the matrix $A$ so that the problem is $A\mathbf{a}=\mathbf{y}$. Then the least squares equation is $A^TA\mathbf{a}=A^T\mathbf{y}$. If we write that out explicitly we get
$$\begin{bmatrix}
\sum_{i=1}^4x_i^4 & \sum_{i=1}^4x_i^3 & \sum_{i=1}^4x_i^2 \\
\sum_{i=1}^4x_i^3 & \sum_{i=1}^4x_i^2 & \sum_{i=1}^4x_i \\
\sum_{i=1}^4x_i^2 & \sum_{i=1}^4x_i & 4
\end{bmatrix}
\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}
=
\begin{bmatrix}
\sum_{i=1}^4x_i^2y_i \\
\sum_{i=1}^4x_iy_i \\
\sum_{i=1}^4y_i
\end{bmatrix}
$$
This new matrix makes no intuitive sense to me whatsoever but is striking. Why do we care about, for instance, the sum of the 4th powers of $x_i$? Is there a way to interpret why this is the way it is, maybe in terms of calculus or something else?
| To augment the insightful post of @Christopher A. Wong, we offer two other perspectives.
First, why do we form the normal equations? To form a consistent linear system. Typically a linear algebra course begins with problems like
$$
\mathbf{A} a - y = 0.
$$
We learn Gaussian elimination, $\mathbf{L}\mathbf{U}$ decomposition, etc. Then the course presents problem where the above equation has no solution. We generalize the concept of a solution and ask that $r(x) = \mathbf{A} a - y$ be made a small as possible. To discuss the size of a vector, we need a norm. The $2-$norm is the overwhelming favorite and this leads to the method of least squares to minimize $\lVert r\rVert_{2}^{2}.$
Rewrite the problem as
$$
\mathbf{A} a = y.
$$
There is no solution; the vector $y$ is not in the range space of $\mathbf{A}$. The first solution attempt is to craft a consistent problem with the same solution. Multiply both sides of the equality by $\mathbf{A}^{*}$ to create the normal equations:
$$
\mathbf{A}^{*} \left( \mathbf{A}a \right) = \mathbf{A}^{*}\left( y \right).
$$
The parentheses emphasize the fact that we have a consistent linear system because the data vector $\mathbf{A}^{*}\left( y \right)$ in certainly in the column space of $\mathbf{A}^{*}$.
Second, look at the problem in terms of column vectors.
$$
\begin{align}
\mathbf{A} a &= y \\
\left[
\begin{array}{cccc}
\mathbf{1} & x & x^{2} & \dots & x^{d}
\end{array}
\right]
\left[
\begin{array}{cccc}
a_{0} \\
a_{1} \\
\vdots \\
a_{d}
\end{array}
\right]
&=
\left[
\begin{array}{c}
y
\end{array}
\right]
\end{align}
$$
The normal equations take the form
$$
\begin{align}
\mathbf{A}^{*}\mathbf{A} a &= \mathbf{A}^{*}y \\
\left[
\begin{array}{cccc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x & \mathbf{1} \cdot x^{2} & \dots & \mathbf{1} \cdot x^{d} \\
x \cdot \mathbf{1} & x \cdot x & x \cdot x^{2} & \dots & x \cdot x^{d} \\
\vdots & \vdots & \vdots && \vdots \\
x^{d} \cdot \mathbf{1} & x^{d} \cdot x & x^{d} \cdot x^{2} & \dots & x^{d} \cdot x^{d} \\
\end{array}
\right]
\left[
\begin{array}{c}
a_{0} \\
a_{1} \\
\vdots \\
a_{d}
\end{array}
\right]
&=
\left[
\begin{array}{r}
\mathbf{1} \cdot y \\
x \cdot y \\
x^{2} \cdot y \\
\vdots\quad \\
x^{d} \cdot y
\end{array}
\right]
\end{align}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate this integral
Let $$f\left(x\right)= \mbox{ the antiderivative } \frac{x^2}{1-x^5}, $$
$f\left(1\right)=0$
Find $f\left(4\right)$
I know that:
$F\left(x\right) = I\left(x\right) + C $ where $I\left(x\right)$ is the antiderivative (integral) of $f\left(x\right)$
Thus at $x = 1\; F\left(1\right) = 0 = I\left(1\right) + C$, thus $C = -I\left(x\right)$
Thus $F\left(4\right) = f\left(4\right) - I\left(1\right)$
My problem, is how to find the integral of $$ \int \frac{x^2}{1-x^5}\, \mathrm{d}x$$
| Equation $z^5-1=0$ has one real root $z_0=1$ and two pairs of complex-conjugated roots $z_k=e^{\tfrac{2k\pi i}{5}},\;\; k=1,\,\ldots\,,4.$ Then
$$
z^5-1=(z-z_0)(z-z_1)(z-\bar{z}_1)(z-z_2)(z-\bar{z}_2)=\\
=(z-1)(z^2-2\Re{(z_1)} z+1)(z^2-2\Re{(z_2)} z+1)=\\
=(z-1)\left(z^2-2\cos{\left(\dfrac{2\pi}{5}\right)} z+1\right)\left(z^2-2\cos{\left(\dfrac{4\pi}{5}\right)} z+1\right).$$
It is a factorization suggested by Fixed Point.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is going on with this constrained optimization? I'd like to figure out what is going on when trying to maximize a function (below $a_i$ are real numbers)
$F = a_1a_2 + a_2a_3 + \cdots + a_{n - 1}a_n + a_na_1;$
When we have active constraints
$h_1 = a_1 + a_2 + \cdots + a_n = 0;$
$h_2 = a_1^2 + a_2^2 + \cdots +a_n^2 = 1;$
So my gradients
$\nabla F = (a_2+a_n,a_3+a_1, \ldots , a_{n-2}+a_n,a_{n-1}+a_1);$
$\nabla h_1 = (1, 1, \ldots , 1);$
$\nabla h_2 = (2a_1,2a_2,\ldots , 2a_n);$
Kuhn-Tucker should provide necessary conditions in this case which is I guess pretty much the same as the method of Lagrange multipliers:
$$\begin{cases}
a_2+a_n = \lambda_1 + 2\lambda_2a_1; \\
a_3+a_1 = \lambda_1 + 2\lambda_2a_2; \\
a_4+a_2 = \lambda_1 + 2\lambda_2a_3; \\
\ldots \\
a_{n-1} + a_1 = \lambda_1 + 2\lambda_2 a_n;
\end{cases}$$
In a matrix form this is
$$\begin{pmatrix}
2\lambda_2 & -1 & 0 & 0 & \ldots & 0 & -1 \\
-1 & 2\lambda_2 & -1 & 0 & \ldots & 0 & 0 \\
0 & -1 & 2\lambda_2 & -1 & \ldots &0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 &0 &0 & 0 & \ldots & 2\lambda_2 & -1 \\
-1 &0 &0 & 0 & \ldots & -1 & 2\lambda_2
\end{pmatrix}
\begin{pmatrix}
a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_{n-1} \\ a_n
\end{pmatrix} =
- \lambda_1\begin{pmatrix}
1 \\ 1 \\ 1 \\ \vdots \\ 1 \\ 1
\end{pmatrix}
$$
How should I proceed?
| Consider what happens for $n=2$ and $n=3$.
$n=2$: Only two points satisfy the constraints, $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$, and $F$ takes the same value on these two points.
$n=3$: The constraints describe a circle of radius 1 contained in the plane perpendicular to the vector $(1,1,1)$. You have
$2F(a_1,a_2,a_3) = (a_1+a_2+a_3)^2-(a_1^2+a_2^2+a_3^2)=-1$,
for $(a_1,a_2,a_3)$ a vector on this circle. The function $F$ is again constant on the set described by the constraints.
EDIT: Ok, the general case. Note that $F$ is a quadratic form,
$F(a_1,\ldots,a_n) = \left\langle\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_n \end{array}\right),\left(
\begin{array}{ccccc}
0 & \frac{1}{2} & 0 & \cdots & \frac{1}{2}\\
\frac{1}{2} & 0 & \frac{1}{2} & \cdots & 0 \\
0 & \frac{1}{2} & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots \\
\frac{1}{2} & 0 & 0 & \cdots \end{array}\right)
\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_n \end{array}\right)\right\rangle$
For $n\ge 3$, the biggest eigenvalue of the matrix of this quadratic form is $\lambda_1=1$ (it is a stochastic matrix), with eigenvector $v_1=(1,1,\cdots,1)$. Your constraints mean that you want to maximise this quadratic form on the intersection of the unit sphere with the hyperplane perpendicular to $v_1$. I.e., you are looking for its second biggest eigenvalue.
For $n=3$ that is $-\frac{1}{2}$, as we already knew.
For $n=4$, the eigenvalues are $\lambda_1=1$, $\lambda_2=\lambda_3=0$, $\lambda_4=-1$. So the maximum you are looking for is $0$, it is attained on a circle. The minimum is $-1$, it is attained in two antipodal points.
For bigger $n$ you now have to do a little bit more linear algebra. I think the eigenvalues of these matrices must be known.
EDIT: Let $v_2,\ldots,v_n$ be an orthonormal basis of the subspace perpendicular to $v_1=(1,\cdots,1)$ consisting of eigenvectors of the matrix of the quadratic form $F$. Then we have
$F(a_1,\ldots,a_n)=\sum_{i=2}^n \lambda_i x_i^2$,
if $(a_1,\ldots,a_n)=\sum_{i=2}^n x_i v_i\in \{v_1\}^\perp$. If $(a_1,\ldots,a_n)$ is furthermore a unit vector, then we have $\sum_{i=2}^n x_i^2=1$, and $F(a_1,\cdots,a_n)$ is a weighted average of the eigenvalues $\lambda_2,\ldots,\lambda_n$.
| {
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What is the polar form of $ z = 1- \sin (\alpha) + i \cos (\alpha) $? How do I change $ z = 1- \sin (\alpha) + i \cos (\alpha) $ to polar? I got $r = (2(1-\sin(\alpha))^{\frac{1}{2}} $. I have problems with the exponential part. What should I do now?
| $1-\sin\alpha\ge 0$
If $1-\sin\alpha=0,\cos\alpha=0, z=0+i\cdot0$
If $1-\sin\alpha> 0$
$1-\sin\alpha=(\cos\frac\alpha2-\sin \frac\alpha2)^2$ and
$\cos\alpha=\cos^2\frac\alpha2-\sin^2\frac\alpha2$
So, $z=(1-\sin\alpha)+i(\cos\alpha)=(\cos\frac\alpha2-\sin \frac\alpha2)^2+i(\cos^2\frac\alpha2-\sin^2\frac\alpha2)$
If $\cos\frac\alpha2-\sin \frac\alpha2>0,$
$z=(\cos\frac\alpha2-\sin \frac\alpha2)\left(\cos\frac\alpha2-\sin \frac\alpha2+i(\cos\frac\alpha2+\sin \frac\alpha2)\right)$
$=\sqrt2\cos(\frac\alpha2+\frac\pi4)\left(\sqrt2\cos(\frac\alpha2+\frac\pi4)+i\sqrt2\sin(\frac\alpha2+\frac\pi4)\right)$
$=2\cos(\frac\alpha2+\frac\pi4)\left(\cos(\frac\alpha2+\frac\pi4)+i\sin(\frac\alpha2+\frac\pi4)\right)$
If $\cos\frac\alpha2-\sin \frac\alpha2<0,$
$z=(\sin \frac\alpha2-\cos\frac\alpha2)\left(\sin \frac\alpha2-\cos\frac\alpha2-i(\cos\frac\alpha2+\sin \frac\alpha2)\right)$
$=-2\cos(\frac\alpha2+\frac\pi4)\left(-\cos(\frac\alpha2+\frac\pi4)-i\sin(\frac\alpha2+\frac\pi4)\right)$
$=2\cos(\frac\alpha2+\frac{5\pi}4)\left(\cos(\frac\alpha2+\frac{5\pi}4)+i\sin(\frac\alpha2+\frac{5\pi}4)\right)$ as $\cos(\pi+y)=-\cos y,\sin(\pi+y)=-\sin y$
Now, $\cos\frac\alpha2-\sin \frac\alpha2=\sqrt2\cos(\frac\pi4+\frac\alpha2)$ which will be $<0$
if $2n\pi+\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac{3\pi}2\iff 4n\pi+\frac{\pi}2<\alpha<4n\pi+\frac{5\pi}2$
Similarly, $\cos\frac\alpha2-\sin \frac\alpha2$ which will be $>0$
if $2n\pi-\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac\pi2\iff 4n\pi-\frac{3\pi}2<\alpha<4n\pi+\frac\pi2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
| A similar triple completing-the-square is helpful. Using your notation, $(a-1)^2+(b-1)^2+(c-1)^2+2(a+b+c)-3=a+b+c$. This rearranges to $(a-1)^2+(b-1)^2+(c-1)^2=3-(a+b+c)$. Hence $a+b+c\le 3$ since otherwise the sum of three squares would be negative. Consequently $a^2+b^2+c^2\le 3$.
Unfortunately, I don't have a cute trick for the opposite inequality. Let $x=a^2, y=b^2$. I want to minimize $f(x,y)=x+y+\frac{1}{xy}$, assuming $x,y>0$. Setting the partials equal to zero, I get $1-\frac{1}{xy^2}=0=1-\frac{1}{yx^2}$, which has unique solution $x=y=1$. (as $x,y$ approach either 0 or $\infty$, $f(x,y)$ grows without bound, so this is a minimum). Consequently $f(x,y)\ge f(1,1)=3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing the unit normal vector - Simplifying help I have a surface
$$X(u,v) = \left(3uv^2 - u^3 - \frac{u}{3}, 3u^2v - v^3 - \frac{v}{3}, 2uv \right), $$
and the cross product
$$(X_u \times X_v) = \left(3(u^2 + v^2) \frac{1}{3} \right) \cdot \left(2v, 2u, \frac{1}{3} - 3(u^2 + v^2) \right).$$
I have to show that the unit normal vector is
$$N(u,v) = \frac{1}{3(u^2 + v^2) + \frac{1}{3}} \cdot \left( 2u, 2v, \frac{1}{3} - 3(u^2 + v^2) \right).$$
So, to find the unit, I just need to divide by $| X_u \times X_v|$ which should end up being $\left( 3(u^2 + v^2) + \frac{1}{3} \right)^2$. So I do this:
$$\sqrt{4u^2 + 4v^2 + \left(\frac{1}{3} - 3(u^2 + v^2) \right)^2}$$
$$ = \sqrt{4u^2 + 4v^2 + \frac{1}{9} -2(u^2 + v^2) + 9(u^2 + v^2)^2}$$
$$ = \sqrt{2u^2 + 2v^2 + \frac{1}{9} + 9(u^2 + v^2)^2}$$
but I'm stuck on how to simplify this. Expanding the $9(u^2 + v^2)^2$ doesn't really help me much. What should I do?
| Expanding $(3(u^{2} + v^{2}) + \frac{1}{3})^{2}$, we see:
$(3(u^{2} + v^{2}) + \frac{1}{3})*(3(u^{2} + v^{2}) + \frac{1}{3}) = 9(u^{2}+v^{2})^{2} + 2*3*\frac{1}{3}(u^{2} + v^{2}) + \frac{1}{9}$;
Does this help?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving quadratic equations by completing the square.
Graphing $y=ax^2+ bx + c$ by completing the square
*
*Add and subtract the square of half the coefficent of $x$.
*Group the perfect square trinomial.
*Write the trinomial as a square of a binomial.
Rewrite $y = x^2 + 6x + 8$ into $y = a(x-h)^2 + k$.
I've tried solving this but I get a bit confused at the step where I have to "write the trinomial as a square of a binomial". Not exactly sure how to do that.
| To complete the square for the equation:
$y = x² + 6x + 8$
Start by:
1) $x^2+6x+8$
2) $x^2+\dfrac 62x +8$ *divide the bx by 2
3) $(x^2+\dfrac 62x + (\dfrac 62)^2)+8$ *square the bx term to form a perfect square
4) $(x^2+ 3x + (36/4))+8$
5) $(x^2+3x + 9)+8-9$ *add the inverse of the c term
6) $(x^2+3x+9)-1$ *factor the perfect square
7) $(x+3)^2-1$
8) $y=(x+3)^2-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $S_n = 1+ 2 +3 + \cdots + n$, then prove that the last digit of $S_n$ is not 2,4 7,9. If $S_n = 1 + 2 + 3 + \cdots + n,$ then prove that the last digit of $S_n$ cannot be 2, 4, 7, or 9 for any whole number n.
What I have done:
*I have determined that it is supposed to be done with mathematical induction.
*The formula for an finite sum is $\frac{1}{2}n(n+1)$.
*This means that since we know that $n(n+1)$ has a factor of two, it must always end in 4 and 8.
*Knowing this, we can assume that $n(n+1)\bmod 10 \neq 4$ or $n(n+1)\bmod 10 \neq 8$.
| Note that $1+2+3+4+5$ is divisible by $5$, and for the same reason $5k+1+5k+2+5k+3+5k+4+5k+5$ is divisible by $5$ for any integer $k$.
Let $5m$ be the largest multiple of $5$ which is $\le n$. Then $n=5m$ or $n=5m+1$ or $n=5m+2$ or $n=5m+3$ or $n=5m+4$.
The sum of the integers from $1$ to $5m$ is divisible by $5$. So when we sum from $1$ to $n$, the remainder on division by $5$ is $0$, or $1$, or $1+2$, or the remainder when $1+2+3$ is divided by $5$, or the remainder when $1+2+3+4$ is divided by $5$. So the possibilities are $0$, $1$, $3$, $1$, and $0$. In particular, the remainders cannot be $2$ or $4$. So the last digit cannot be $2$, $7$, $4$, or $9$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Expectation and children What is the expected number of children if the probability of having a child(boy or girl) = 1/2 and if you want a boy and a girl?
E(No of children) = ?
| Whatever sex the first-born kid is, our waiting time after that until a child of the opposite sex is born is $\frac{1}{1/2}=2$. So the expected number of children is $3$.
For more detail, we can assume without loss of generality that the first-born is a boy. Let $y$ be the expected additional number of children until a girl is born.
The second child is a girl with probability $\frac{1}{2}$. In that case the additional number of children was $1$. And with probability $\frac{1}{2}$, the second child is a boy, in which case the expected number of additional children is $1+y$. Thus
$$y=\frac{1}{1}+\frac{1}{2}(1+y).$$
Solve for $y$. We get $y=2$, so the expected total number of children is $1+2$.
Or else we can do the problem the hard way. Let $X$ be the total number of children. Then $X=2$ with probability $\frac{1}{2}$, $3$ with probability $\frac{1}{2^2}$, $4$ with probability $\frac{1}{2^3}$, and so on. Thus
$$E(X)=2\cdot\frac{1}{2}+3\cdot\frac{1}{2^2}+4\cdot\frac{1}{2^3}+\cdots.$$
We leave it at that. The infinite sum turns out to be $3$, but a summation of series argument for that is quite a bit longer than the first way we solved the problem.
Solve for $y$. We get $y=2$.
| {
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Is $1/z$ differentiable on $\mathbb{C}\setminus\{0\}$? The way I usually solve these kind of questions is by using the Cauchy Riemann Equations...
$h(z)=\dfrac{1}{z}$
$=\dfrac{1}{x+iy}$
$=\dfrac{1}{x+iy} \dfrac{x-iy}{x-iy}$
$=\dfrac{x-iy}{(x+iy)(x-iy)}$
$=\dfrac{x-iy}{x^2+y^2}$
$=\dfrac{x}{x^2+y^2}-\dfrac{iy}{x^2+y^2}$
Does this make $u(x,y)=\dfrac{x}{x^2+y^2}$ and $v(x,y)=-\dfrac{y}{x^2+y^2}$
Im a little confused with the example.. or is there a better way for me to show whether it is differentiable or not?
| Hint:
$$\frac{1}{x+iy}\not=\frac{1}{x}+\frac{1}{iy}.$$
However,
$$
\frac{1}{x+iy}=\frac{1}{x+iy}\frac{x-iy}{x-iy}
$$
Does this help?
| {
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Approximating a log-power function I can't figure out how the following approximation has been done, I would appreciate any guidance:
$$y=-60+10\log_{10}\left[\frac{\left(\frac{99}{100}\right)^m}{\frac{1}{11}\left(\frac{1}{3}\right)^m+\frac{10}{11}\left(\frac{1}{6}\right)^m}\right]$$
is approximated to: $y=-49.6+4.73m$ when $m>5$ and $y=-60+7.73m$ when $m<3$. Thanks in advance.
| Let's put $y$ in logarithm and exponential form (factorizing $\left(\frac 16\right)^m=\left(\frac 13\right)^m\left(\frac 12\right)^m$) :
\begin{align}
y&=-60+\frac {10}{\ln(10)}\;\ln\left[\frac{e^{m\ln\left(\frac{99}{100}\right)}}{\frac{1}{11}e^{m\ln\left(\frac{1}{3}\right)}\left(1+\frac {10}{2^m}\right)}\right]\\
&=-60+\frac {10}{\ln(10)}\;\left[m\,\ln\left(\frac{99}{100}\right)-m\,\ln\left(\frac{1}{3}\right)-\ln\left(1+\frac {10}{2^m}\right)+\ln(11)\right]\\
\end{align}
At this point observe that :
\begin{align}
a&:=\frac {10}{\ln(10)}\left(\ln\left(\frac{99}{100}\right)+\ln(3)\right)\approx 4.73\\
b&:=-60+\frac {10}{\ln(10)}\;\ln\left(11\right)\approx -49.6\\
\end{align}
If $\;m\gg 1\;$ then $\ \ln\left(1+\frac {10}{2^m}\right)\sim \frac {10}{2^m}\ $ may be neglected and you get $\ b+a\,m\ $ as wished.
For small $m$ rewrite $y$ as :
\begin{align}
y&=-60+\frac {10}{\ln(10)}\;\ln\left[\frac{e^{m\ln\left(\frac{99}{100}\right)}}{\frac{10}{11}e^{m\ln\left(\frac{1}{6}\right)}\left(1+\frac {2^m}{10}\right)}\right]\\
\end{align}
so that using :
\begin{align}
c&:=\frac {10}{\ln(10)}\left(\ln\left(\frac{99}{100}\right)+\ln(6)\right)\approx 7.74\\
d&:=-60-\frac {10}{\ln(10)}\;\ln\left(\frac{10}{11}\right)\approx -59.6\\
\end{align}
you will get the approximation $\ d+c\,m\ $ (this one is not very good for 'moderate' values since valid only for $m\ll 1$.
| {
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Find no. of points where $f$ and $g$ meet.
If $f(x)=x^2$ and $g(x)=x \sin x+ \cos x$ then
(A) $f$ and $g$ agree at no points
(B) $f$ and $g$ agree at exactly one point
(C) $f$ and $g$ agree at exactly two points
(A) $f$ and $g$ agree at more than two points.
Trial: I think I need to find the values of $x$ for which $f(x)=g(x)$. So, I have $x^2-x \sin x- \cos x=0$. How can I solve this ? Please help
| Note that $x^2$ and $x\sin x+\cos x$ are both even functions, so the graphs of $y=x^2$ and $y=x\sin x+\cos x$ are symmetric about the $y$ axis. So if we kow the story for $x\ge 0$, we know the story everywhere.
At $x=0$, the curve $y=x^2$ is below the curve $y=x\sin x+\cos x$.
But $x^2$, in the long run, is far bigger than $x\sin x+\cos x$. This is clear, but if we want formal proof, observe that for positive $x$, we have $|x\sin x+\cos x|\le x+1$. Thus if $x\gt 2$, then $x^2\gt x+1$.
So the two curves meet at some positive $x$. If you want to be formal, use the Intermediate Value Theorem.
Do they meet at more than one positive $x$? The derivative of $x^2$ is $2x$, while the derivative of $x\sin x+\cos x$ is $x\cos x$. Since $x\cos x\lt x$, at any positive $x$ the function $x^2$ grows faster than the function $x\sin x+\cos x$. So once $x^2$ gets above $x\sin x+\cos x$, the function $x\sin x+\cos x$ can never again catch up to $x^2$. A formal proof here would involve the Mean Value Theorem.
Thus the two functions are equal at exactly one positive $x$, and, by symmetry, at exactly one negative $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What have I done wrong in solving this problem with indices rules? The question asks to simplify:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$
So I used $(a^m)^n=a^{mn}$ to get
$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$
However, this isn't the answer, and I can't see what I've done wrong.
This is what the mark scheme says:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$
To me, my answer looks more simple than theirs, and I can't see what I've done wrong.
| $$\Big(\frac{25x^4}{4}\Big)^{-\frac{1}{2}}=\Big(\frac{4}{25x^4}\Big)^{\frac{1}{2}}=\frac{4^{\frac{1}{2}}}{25^{\frac{1}{2}}(x^4)^{\frac{1}{2}}}=\frac{2}{5x^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!
| I am a bit surprised that the answers given so far that invoke cyclotomic polynomials don't go all the way to exhibit a factorisation of $1+x^2+\cdots+x^{2n}$. Using $x^m-1=\prod_{d\mid m}\Phi_d(x)$ one gets
$$
1+x^2+\cdots+x^{2n}=\frac{x^{2n+2}-1}{x^2-1}=\prod_{\substack{d\mid 2(n+1) \\ d\notin\{1,2\}}}\Phi_d(x),
$$
and since cyclotomic polynomials are irreducible over the rational numbers, this is precisely the factorisation of your polynomial in $\Bbb Q[x]$. The product is empty for $n=0$, has a single factor $\Phi_4(x)=x^2+1$ for $n=1$; in all other cases it has at least two factors, namely the polynomials $\Phi_{n+1}(x)$ and $\Phi_{2(n+1)}(x)$. So in particular it is always reducible for $n\geq2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help
| You can solve the square of a sum by writing out the square as a product of two sums and writing out the multiplication for each pair of terms.
$$(1+\sqrt{x-1})^2 =\\
(1+\sqrt{x-1})(1+\sqrt{x-1})=\\
1\times1 + 1\times\sqrt{x-1}+\sqrt{x-1}\times1+\sqrt{x-1}\times\sqrt{x-1}=\\
1+\sqrt{x-1}+\sqrt{x-1}+x-1=\\
2\sqrt{x-1}+x$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$ I need to find a closed-form for the following integral. Please give me some ideas how to approach it:
$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$
| My calculation shows that
\begin{align*}
\int_{0}^{\infty} \frac{x^3}{(x^4 + 1)(e^x - 1)} \, dx
&= \frac{\gamma}{2} - \log\sqrt{2\pi} + \frac{\pi}{4} \frac{\sin\frac{1}{\sqrt{2}}}{\cosh\frac{1}{\sqrt{2}} - \cos\frac{1}{\sqrt{2}}} \\
&\quad - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n\left(1 + (2\pi n)^4\right)} \\
&\approx 0.389075976914101580976629 \cdots.
\end{align*}
My solution is divided into several steps:
Step 1. Let us introduce the function
$$ I(s) = \int_{0}^{\infty} \frac{x^{s}}{(x^4+1)(e^{x}-1)} \, dx $$
It is easy to see that
$$ I(s) + I(s+4) = \Gamma(s+1)\zeta(s+1). \tag{1} $$
This allows us to extend $I(s)$ as a meromorphic function on $\Bbb{C}$.
Step 2. Consider a contour $C$ starting from $\infty + \epsilon i$, making a counter-clockwise turn around $z = 0$ and going back to $\infty - \epsilon i$ as follows:
If we use a logarithm function with the branch cut $[0, \infty)$, we see that
$$ (e^{2\pi i s} - 1)I(s) = \int_{C} \frac{z^{s}}{(z^4+1)(e^z - 1)} \, dz. \tag{2} $$
Also, for $ 1 < s < 2$ we can confirm that $(2)$ is rewritten as
$$ (e^{2\pi i s} - 1)I(s) = \lim_{n\to\infty} \int_{C_{n}} \frac{z^{s}}{(z^4+1)(e^z - 1)} \, dz, $$
where $C_n$ is the contour given by
Here, the condition $1 < s < 2$ is introduced in order to create an appropriate decay speed for the integral along the contour $C - C_{n}$. By applying the Cauchy integration formula, we have
$$ (e^{2\pi i s} - 1)I(s) = -2\pi i \sum_{\omega^4 = -1} \operatorname{Res}_{z=\omega} \frac{z^{s}}{(z^4+1)(e^z - 1)} - 2\pi i \sum_{n\neq 0} \operatorname{Res}_{z=2\pi i n} \frac{z^{s}}{(z^4+1)(e^z - 1)}. $$
Simplifying,
$$ I(s) = \frac{\pi}{\sin \pi s} \frac{e^{-i\pi s}}{4} \sum_{\omega^4 = -1} \frac{\omega^{s+1}}{e^{\omega} - 1} - \frac{2^{s}\pi^{s+1}}{\sin \frac{\pi s}{2}} \sum_{n=1}^{\infty} \frac{n^s}{1+(2\pi n)^4} \tag{3} $$
Since both sides define a meromorphic function for $\Re s < 3$, they coincide on this range.
Step 3. Combining $(1)$ and $(3)$, we have
\begin{align*}
I(s+3)
&= \Gamma(s)\zeta(s) - I(s-1) \\
&= \Gamma(s)\zeta(s) - \frac{\pi}{\sin \pi s} \frac{e^{-i\pi s}}{4} \sum_{\omega^4 = -1} \frac{\omega^{s}}{e^{\omega} - 1} - \frac{(2\pi)^{s}}{2\cos \frac{\pi s}{2}} \sum_{n=1}^{\infty} \frac{n^s}{n \left( 1+(2\pi n)^4 \right)}
\end{align*}
Taking $s \to 0$, we obtain the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 3,
"answer_id": 1
} |
How to solve this integral for a hyperbolic bowl? $$\iint_{s} z dS $$ where S is the surface given by $$z^2=1+x^2+y^2$$ and $1 \leq(z)\leq\sqrt5$ (hyperbolic bowl)
| A related problem. Note that,
$$ z=\sqrt{ 1+x^2+y^2 } \implies z_x=\frac{x}{\sqrt{ 1+x^2+y^2 }},\quad z_y=\frac{y}{\sqrt{ 1+x^2+y^2 }} $$
$$ \iint_{s} z dS = \iint_{D} z \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA $$
$$ = \iint_{D} \sqrt{1+(x^2+y^2)}\sqrt{{\frac {1+2\,{x}^{2}+2\,{y}^{2}}{1+{x}^{2}+{y}^{2}}}}\,dxdy $$
$$ =\iint_{D} \sqrt{{{1+2\,{x}^{2}+2\,{y}^{2}}{}}}\,dx dy $$
Now, $D\equiv \left\{ x^2+y^2 \leq 4 \right\}$. To see this notice that
$$ 1 \leq z\leq\sqrt5 \implies 1 \leq \sqrt{1+x^2+y^2} \leq\sqrt5 \implies x^2+y^2\leq 4. $$
So, we can use polar coordinates as
$$ = \int_{0}^{2\pi}\int_{0}^{2} \sqrt{1+2 r^2}\,r\, dr d\theta = \frac{26\pi}{3} . $$
Added: if you want to parametrize the surface, you go this way,
$$ x=r\cos(\phi),\quad y = r\sin(\phi), \quad z^2 = 1+x^2+y^2= 1+r^2 \implies z=\sqrt{1+r^2}. $$
You can write it in a vector form as
$$ \textbf{T}(r,\phi)= r\cos(\phi)\textbf{i}+ r\sin(\phi)\text{j}+ \sqrt{1+r^2}\, \text{k} $$
$$ \implies T_r = \cos(\phi)\textbf{i}+ \sin(\phi)\text{j} + \frac{r}{\sqrt{1+r^2}} \text{k}, $$
$$ T_\phi = -r\sin(\phi)\textbf{i} + r\cos(\phi)\text{j}+ 0 \,\text{k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What is the function given by $\sum_{n=0}^\infty \binom{b+2n}{b+n} x^n$, where $b\ge 0$, $|x| <1$ For a nonnegative integer $b$, and $|x|<1$, what is the function given by the power series
$$
\sum_{n=0}^\infty \binom{b+2n}{b+n} x^n.
$$
For $b=0$, this post shows
$$
\sum_{n=0}^\infty \binom{2n}{n}x^n = (1-4x)^{-1/2}.
$$
How do we proceed for an integer $b>0$?
I tried to start from the power series of
$$
(1-x)^{-(b+2n)},
$$
but it doesn't work.
| Note that the quoted identity is not difficult to verify using a
variant of Lagrange Inversion. Introduce
$$T(z) = w = \sqrt{1-4z}$$ so that
$$z = \frac{1}{4} (1-w^2)$$ and
$$dz = -\frac{1}{2} w \; dw$$
Then we seek to compute
$$[z^n] \frac{2^b}{T(z)} (1+T(z))^{-b}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{2^b}{T(z)} (1+T(z))^{-b} \; dz.$$
Using the substitution this becomes
$$- \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}}
\frac{2^b}{w} (1+w)^{-b} \frac{1}{2} w \; dw
\\ = - \frac{1}{2\pi i}
\int_{|w-1| = \epsilon}
\frac{4^{n+1}}{(1-w)^{n+1} \times (1+w)^{n+1+b}}
2^{b-1} \; dw
\\ = - \frac{1}{2\pi i}
\int_{|w-1| = \epsilon}
\frac{(-1)^{n+1} 4^{n+1}}{(w-1)^{n+1} \times (1+w)^{n+1+b}}
2^{b-1} \; dw.$$
It follows that the value of the integral is given by
$$(-1)^n 4^{n+1} 2^{b-1}
[(w-1)^n] \frac{1}{(1+w)^{n+1+b}}
\\ = (-1)^n 4^{n+1} 2^{b-1}
[(w-1)^n] \frac{1}{(2+(w-1))^{n+1+b}}
\\ = (-1)^n 4^{n+1} 2^{b-1} \frac{1}{2^{n+1+b}}
[(w-1)^n] \frac{1}{(1+(w-1)/2)^{n+1+b}}
\\ = (-1)^n 2^{2n+b+1} \frac{1}{2^{n+1+b}}
\frac{(-1)^n}{2^n} {n+n+b\choose n+b}
\\= {2n+b\choose n+b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expressing $\sqrt{n +m\sqrt{k}}$ Following this answer, is there a simple rule for determining when:
$$\sqrt{n +m\sqrt{k}}$$
Where $n,m,k \in \mathbb{N}$, can be expressed as:
$$a + b\sqrt{k}$$
For some natural $a,b$?
This boils down to asking for what $n,m,k \in \mathbb{N}$ there exist $a,b\in \mathbb{N}$ such that:
$$2ab= m,\ \ \text{and}\ \ a^2+b^2k = n$$
| Hint: Please see this link. To translate it to your problem, note that $ b = m^2k $. From here, one of the square roots on the right hand side of their expression must reduce.
In the case that $ a + \sqrt{a^2 - b} = 2p^2, p \in \mathbb{N} $, $ p^2 = \frac{a + \sqrt{a^2 - b}}{2} $. For the second case $ q^2 = \frac{a - \sqrt{a^2 - b}}{2} $. Hence, if $ \frac{a \pm \sqrt{a^2 - b}}{2} $ is a perfect square then the nested square root is reducible.
For example, let $ a = 30, b = 896 $. In this case, $ \frac{a + \sqrt{a^2 - b}}{2} = 16 $ so $ \sqrt{30 + \sqrt{896}} $ is reducible, being equal to $ 4 + \sqrt{14} $. Also, note that if $ a = 30, b = 756 $, it is reducible because $ \frac{a + \sqrt{a^2 - b}}{2} = 9 $, being equal to $ 3 + \sqrt{21} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality $xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$ Let $x,y,z$ is real numbers,and such that $x+y+z=1$,and in $\Delta ABC$,prove that
$$xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$$
I think this inequality maybe use $x^2+y^2+z^2\ge 2yz\cos{A}+2xz\cos{B}+2xy\cos{C},x,y,z\in R$
Thank you everyone.
| We need to prove that $$(x+y+z)^2\geq4xy\sin^2\gamma+4xz\sin^2\beta+4yz\sin^2\alpha$$ or
$$x^2+y^2+z^2+2xy\cos2\gamma+2xz\cos2\beta+2yz\cos2\alpha\geq0$$ or
$$z^2+2(x\cos2\beta+y\cos2\alpha)z+x^2+y^2+2xy\cos2\gamma\geq0,$$
for which it's enough to prove that
$$(x\cos2\beta+y\cos2\alpha)^2-(x^2+y^2+2xy\cos2\gamma)\leq0$$ or
$$x^2\sin^22\beta+2(\cos2\gamma-\cos2\beta\cos2\alpha)xy+y^2\sin^22\alpha\geq0$$ or
$$(x\sin2\beta-y\sin2\alpha)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
| \begin{align*} \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{1}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx \\
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{0}^{1} \frac{\log (x^{-1} + 1)}{x^2 + 1} \, dx \quad (x \mapsto x^{-1}) \\
&= 2 \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx - \int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx
\end{align*}
For the first integral, we plug
$$ u = \frac{1-x}{1+x}, \quad dx = - \frac{2}{(u+1)^2} \, du. $$
Then it is easy to find that
$$ \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx = \int_{0}^{1} \frac{\log 2 - \log (u + 1)}{u^2 + 1} \, du = \frac{\pi}{4}\log 2 - \int_{0}^{1} \frac{\log (u + 1)}{u^2 + 1} \, du $$
and hence
$$ \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx = \frac{\pi}{8}\log 2. $$
For the second integral, we plug $x = e^{-t}$ and we have
\begin{align*}
\int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx
&= - \int_{0}^{\infty} \frac{t e^{-t}}{1 + e^{-2t}} \, dt
= - \sum_{n=0}^{\infty} (-1)^{n} \int_{0}^{\infty} t \, e^{-(2n+1)t} \, dt \\
&= - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = - G,
\end{align*}
where $G$ is the Catalan constant.
Therefore we have
$$ \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx = \frac{\pi}{4} \log 2 + G. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 5
} |
Chinese remainder theorem issue Let's say I have the following equations:
$$x \equiv 2 \mod 3$$
$$x \equiv 7 \mod 10$$
$$x \equiv 10 \mod 11$$
$$x \equiv 1 \mod 7$$
And I need to find the smallest x for which all these equations are correct. So:
$N = 3\times10\times11\times7 = 2310$
$N_1 = \frac{2310}{3} = 770$
$N_2 = \frac{2310}{10} = 231$
$N_3 = \frac{2310}{11} = 210$
$N_4 = \frac{2310}{7} = 330$
Then:
$\gcd (3, 770) = 1 = 257\times3 - 1\times770; \qquad x_{1} = -1 \equiv 2 \mod 3$
$\gcd(10, 231) = 1 = -23\times10 + 1\times231; \quad x_{1} = 1$
Etc., and finally:
$x = 2\times2\times N_1 + 7\times1\times N_2 + ...$
I understand the whole method except for one thing - when I solved $\gcd(3, 770)$ I got $x_{1} = -1$...why is there $-1 \equiv 2 \mod 3$, from which the final result is 2?
| More simply: $ $ mod $3,11\!:\ x\equiv -1,\ $ so $ $ mod $33\!:\ x\equiv -1\ $ so $\ x = -1 + 33j.$
mod $7\!:\ 1 \equiv x \equiv -1 + 33j\ \Rightarrow\ j\equiv \frac{2}{33} \equiv \frac{2}{{-}2} \equiv -1\ $ so $\ j = -1 + 7k.$
Substituting: $\ x = -1 + 33j = -1 + 33(-1 + 7k) = -34 + 231k.$
mod $10\!:\ 7 \equiv x \equiv -34 + 231k \equiv -4 + k\ $ so $\ k\equiv 7+4\equiv 1,\ $ i.e. $\ k = 1+10n.$
Substituting: $\ x = -34 + 231k = -34 + 231(1+10n) = 197 + 2310n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A simple 2 grade equations system If we have:
$$x^2 + xy + y^2 = 25 $$
$$x^2 + xz + z^2 = 49 $$
$$y^2 + yz + z^2 = 64 $$
How do we calculate $$x + y + z$$
| $$x^2 + xy + y^2 = 25 \dots (1)$$
$$x^2 + xz + z^2 = 49 \dots(2)$$
$$y^2 + yz + z^2 = 64 \dots(3)$$
$(2)-(1)$
$x(z-y)+(z+y)(z-y)=24$
$\Rightarrow (x+y+z)(z-y)=24$
Similarly we get ,
$(x+y+z)(y-x)=15$ by $(3)-(2)$
$(x+y+z)(z-x)=39$ by $(3)-(1)$
Clearly Let $1/\lambda=(x+y+z)\ne 0$
Then we have,
$(z-y)=24\lambda$
$(y-x)=15\lambda$
$\Rightarrow x+z-2y=9\lambda$
$\Rightarrow 3y=1/\lambda+9\lambda\Rightarrow y=1/3\lambda+3\lambda$
Now solve for x and put in the first equation to find the value of $\lambda$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/399488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to use "results from partial fractions"? Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies
$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$
The denominator can be factored into $(1-2x)(1+x)^2$. Using results from partial fractions, it can be shown that there exists constants $C_1,C_2,C_3$ such that
$$A(x)=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$$
Determine these constants and find $a_5$ using this new expression.
What exactly are "results from partial fractions" and how should I use them?
| We describe what we do once we have the coefficients $C_1,C_2,C_3$. It all comes from the expansion
$$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.\tag{$1$}$$
Putting $t=2x$, we find that $\frac{C_1}{1-2x}$ has expansion
$$C_1+2C_1 x+4C_1x^2+8C_1x^3+\cdots.$$
Putting $t=-x$, we can in a similar way get the power series expansion of $\frac{C_2}{1+x}$.
For $\frac{C_3}{(1+x)^2}$ we need an additional idea. Look at Equation $(1)$, and differentiate both sides. We get
$$\frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+\cdots.$$
Put $t=-x$, and multiply by $C_3$.
Now that we have the three power series expansion, we can read off the coefficient of $x^n$ in their sum.
Details: Almost all the work in finding the $C_i$ has been done by DonAntonio. Bringing $\frac{C_1}{1-2x}=\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$ to the common denominator $(1-2x)(1+x)^2$, we find that the numerator is $C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)$. Thus, identically, we must have
$$C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)=6-x+5x^2.$$
There are various ways to find the constants. Put $x=-1$. We get $C_3(3)=12$, so $C_3=4$. Put $x=1/2$. We get $C_1(9/4)=27/4$, so $C_1=3$. Finally, the constant term on the left is $C_1+C_2+C_3$, while on the right it is $6$. Since $C_1+C_3=7$, we get $C_2=-1$. So our original function is equal to
$$\frac{3}{1-2x}-\frac{1}{1+x}+\frac{4}{(1+x)^2}.\tag{$2$}$$
Finally, we compute the coefficient $a_5$ of $x^5$. By the discussion in the main answer, the coefficient of $x^5$ in the expansion of $\frac{1}{1-2x}$ is $2^5$.
The coefficient of $x^5$ in the expansion of $\frac{1}{1+x}$ is $(-1)^5$.
The expansion of $\frac{1}{(1-t)^2}$ was obtained by differentiating $1+t+t^2+\cdots$. so the $t^5$ term is $6t^5$. Putting $t=-x$ we get that the coefficient of $x$ is $6(-1)^5$. Putting things together, and remembering our $C_i$, we get
$$a_5=(3)(2^5)+(-1)(-1)^5+(4)(6)(-1)^5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/399633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Being inside or outside of an ellipse Let $A$ be a point $A$ not belonging to an ellipse $E$. We say that $A$ lies inside
$E$ if every line passing trough $A$ intersects $E$. We say that $A$ lies otside $E$
if some line passing trough $A$ does not intersect $E$. Let $E$ be the ellipse
with semi-axes $a$ and $b$. Show that
(a) a point $A=(x,y)$ is inside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}< 1$
(b) a point $A=(x,y)$ is outside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}> 1$.
I wrote down the equation of an arbitrary line $L$ passing through $A$. If $A$ is inside $E$ then clearly there is another point, say $B$, that belongs to $A\cap L$. I ended up with a bunch of big equations which didn't give any result.
| Case $a:$
Let the point be $P(h,k)$ and any arbitrary line passing through $P$ be $y=mx+c$
$\implies k=m\cdot h+c\iff c=k-m\cdot h$
Let us find the intersection of the line with the given ellipse
$${So,}\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\implies \dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}$$
$$\text{On simplification,}x^2(b^2+m^2)+2a^2cm x+a^2(c^2-b^2)=0$$
For intersection, there will be $2$ distinct values of $m$
i.e., the discriminant $(2a^2cm)^2-4(b^2+m^2)a^2(c^2-b^2)=4a^2(a^2m^2+b^2-c^2)>0$
i.e., $a^2m^2+b^2-c^2>0$
Putting the value of $c, a^2m^2+b^2-(k-m\cdot h)^2>0$
Putting $m=0, b^2-k^2>0$
Put $m=\frac1n, \frac{a^2}{n^2}+b^2-(k- \frac hn)^2>0 \iff a^2+b^2n^2-(k\cdot n-h)^2>0$
Putting $n=0, a^2-h^2>0$
$$\implies (b^2-k^2)(a^2-h^2)>0 \implies a^2b^2> a^2k^2+ b^2h^2\implies 1> \frac{k^2}{b^2}+\frac{h^2}{a^2}$$
Case $b:$ there will be no intersection i.e., the discriminant $<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this system of equation. $x^2-yz=a^2$
$y^2-zx=b^2$
$z^2-xy=c^2$
How to solve this equation for $x,y,z$. Use elementary methods to solve (elimination, substitution etc.).
Given answer is:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$
| Observe that $$(a^2)^2-b^2\cdot c^2=(x^2-yz)^2-(y^2-zx)(z^2-xy)=x(x^3+y^3+z^3-3xyz)$$
Similarly, $$b^4-c^2a^2=y(x^3+y^3+z^3-3xyz)\text{ and }c^4-a^2b^2=z(x^3+y^3+z^3-3xyz)$$
So, $$\frac x{a^4-b^2c^2}=\frac y{b^4-c^2a^2}=\frac z{c^4-a^2b^2}=\frac1{x^3+y^3+z^3-3xyz}=k\text{(say)}$$
Now, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)\frac{\{(x-y)^2+(y-z)^2+(z-x)^2\}}2$
$x+y+z=k(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)=\frac k2\{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2\}$
and $x-y=k\{a^4-b^2c^2-(b^4-c^2a^2)\}=k(a^2+b^2+c^2)(a^2-b^2)$
$\implies (x-y)^2+(y-z)^2+(z-x)^2=k^2(a^2+b^2+c^2)^2\{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2\}=k^2(a^2+b^2+c^2)^2\cdot 2(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)$
$\implies x^3+y^3+z^3-3xyz=k^3\{(a^2+b^2+c^2)(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)\}^2=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2 $
$$\implies\frac1k=x^3+y^3+z^3-3xyz=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2$$
$$\implies k^2=\frac1{a^6+b^6+c^6-3a^2b^2c^2}$$ as $a^6+b^6+c^6-3a^2b^2c^2=(a^2+b^2+c^2)\frac{\{(bc-a^2)^2+(ca-b^2)^2+(ab-c^2)^2\}}2\ge 0$ for real $a,b,c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Calculate the limit of two interrelated sequences? I'm given two sequences:
$$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$
as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.
Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:
$$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$
Seeing no pattern, I did the same once more:
$$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$
While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.
How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!
Edit:
While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.
| It's obvious that $a_n$ and $b_n$ are in the same situation, so their limits highly depend on the initial values. Following are some points we can obtain from $a_1=1$ and $b_1=2$:
*
*$a_n>0$ and $b_n>0\ ;a_{n+1}-a_{n}=\frac{1+a_n}{b_n}>0$ and similarly $b_{n+1}-b_n>0$. Therefore, $\{a_n\}$ and $\{b_n\}$ are strictly increasing sequences;
*$b_{n+1}-a_{n+1}=\frac{(b_n-a_n)+(b_n^2-a_n^2)+a_n\cdot b_n(b_n-a_n)}{a_n\cdot b_n}$, and thus by induction $b_n>a_n$ for every $n$;
*$b_{n+1}-b_{n}=\frac{1+b_n}{a_n}>\frac{b_n}{a_n}>1$, which implies $b_n$ increases to $+\infty$;
*$\frac{a_{n+1}}{a_n}=1+\frac{1}{b_n}+\frac{1}{a_n\cdot b_n}$ converges to $1$ as $n\to \infty$.
Now we prove $\lim a_n$ exists and find its closed form. To show the existence, it suffices to show $\{a_n\}$ is bounded. First assume $a_n$ increases to infinity, and we will derive a contradiction with the last point listed above.
From the fact
$$b_n(a_{n+1}+1)=(1+a_n)(1+b_n)=a_n(b_{n+1}+1)$$
Denote $c_n:=a_n+1$ and $d_n:=b_n+1$, then we obtain
$$
\begin{cases}
\frac{c_n-1}{c_n}=\frac{d_n}{d_{n+1}}\\
\frac{d_n-1}{d_n}=\frac{c_n}{c_{n+1}}
\end{cases}
\Rightarrow
\begin{cases}
\frac{d_{n+1}}{d_n}=\frac{c_n}{c_n-1}\\
\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}
\end{cases}
$$
For $n\ge 2$,
$d_n=d_1\cdot \frac{d_2}{d_1}\cdots \frac{d_n}{d_{n-1}}=d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}$, which implies
$$
d_1(1-\frac{c_n}{c_{n+1}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}})
$$
In fact,$$\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}=\frac{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}}{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}-1} \Rightarrow \frac{c_{n}}{c_{n+1}}=1-\frac{1}{d_1}((1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}}))$$
Together with $d_1(1-\frac{c_{n+1}}{c_{n+2}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n}})$, we get $(1-\frac{1}{c_n})(1-\frac{c_n}{c_{n+1}})=1-\frac{c_{n+1}}{c_{n+2}}$. That is
$$
\frac{c_{n+1}}{c_{n+2}}-\frac{c_{n}}{c_{n+1}}=\frac{1}{c_{n}}-\frac{1}{c_{n+1}}
$$
Hence, for $n\ge 2$
$$
\frac{c_{n}}{c_{n+1}}-\frac{c_2}{c_3}=\frac{1}{c_2}-\frac{1}{c_{n}} $$
which is equivalent to
$$\frac{c_{n}}{c_{n+1}}+\frac{1}{c_{n}}=c(constant):=\frac{c_2}{c_3}+\frac{1}{c_2}=\frac{7}{6} \\ (c_1=2,d_1=3;c_2=3,d_2=6;c_3=\frac{18}{5})
$$
Now it is clear that $$\frac{c_{n+1}}{c_{n}}=\frac{c_n}{\frac{7}{6}c_n-1}=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} $$
Well, the problem has been reduced to solve $c_n(=a_n+1)$, and it's you can use the same method to solve $b_n$, and I would like to leave this open to you, but note as I mentioned before if $a_n\to +\infty$, then $c_n\to +\infty$ and $$
\frac{c_{n+1}}{c_{n}}(=\frac{\frac{a_{n+1}}{a_n}+\frac{1}{a_n}}{1+\frac{1}{a_n}})=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} \text{converges to } \frac{6}{7} \text{instead of } 1
$$ this contradicts the last point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Prove ${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $
Let $a,b,c \ge0$, prove the on equality: $${\frac
{1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac
{1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $$
I tried: $$LHS = \sum\frac 1{1+ab^2}+\sum \frac {a^4}{a+a^2b^2} \ge\frac 9{3+\sum ab^2} + \frac {(a^2+b^2+c^2)^2}{\sum a+ \sum (ab)^2}\ge...$$
but it seem like useless.
Look like better inequality is $ \left( 1+{a}^{3} \right) \left( 1+{b}^{3} \right) \left( 1+{c}^{3}
\right) \ge \left( 1+a{b}^{2} \right) \left( 1+b{c}^{2} \right)
\left( 1+c{a}^{2} \right)$ (but still can't prove it)
| By AM-GM and Holder we obtain:
$$\sum_{cyc}\frac{1+a^3}{1+ab^2}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(1+a^3)}{\prod\limits_{cyc}(1+ab^2)}}=3\sqrt[3]{\frac{\sqrt[3]{\prod\limits_{cyc}(1+a^3)^3}}{\prod\limits_{cyc}(1+ab^2)}}=$$
$$=3\sqrt[3]{\frac{\sqrt[3]{\prod\limits_{cyc}(1+a^3)(1+b^3)^2}}{\prod\limits_{cyc}(1+ab^2)}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(1+ab^2)}{\prod\limits_{cyc}(1+ab^2)}}=3.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)
Is this the right way?
| To add to your step, observe the following:
$$\sqrt{n}+1 = \sqrt{(\sqrt{n}+1)^2} = \sqrt{n+1+2\sqrt{n}} > \sqrt{n+1+1} = \sqrt{n+2}.$$
The "$>$" part comes from your assumption $n \ge 5$, so $2\sqrt{n} \ge 2\sqrt{5} > 1$. Now, we have:
$$n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 = \sqrt{n+1} + (\sqrt{n}+1) > \sqrt{n+1} + \sqrt{n+2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 1
} |
Show that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively. As the title says, I need help showing that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively, where $x$ is a natural number.
In order to do this, I have to use the fact that if $x$ is any natural number >1, any natural number $n\geq x$ can be represented uniquely in the form
$n=c_0+c_1x+c_2x^2+...+c_mx^m$,
where $0\leq c_i \leq x-1$ for $i=0,1,2,...,m-1$ and
$0< c_m \leq x-1$.
| Hint: IF either of these numbers were squares, then you could express them as $N^2$, where $N$ is some number. By what you know, you can write $N=c_0+c_1x+c_2x^2+\ldots+c_mx^m$ uniquely. Notice that $N$ uses at most three powers of $x$: $N=c_0+c_1x+c_2x^2$, since when you square it you need an $x^4$ to appear as the highest power. Now square $N$ out and show you can't have nonnegative $c_i$ which give you the coefficients of your numbers. For example, for $x^4+x^3+x^2+x+1$, you immediately get $c^2_0\equiv 1\pmod{x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$ $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.
| $$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots \sqrt{2^{2^{n-1}}}}}}\\&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots }}}\\&=\sqrt{1+\sqrt{2}\cdot\sqrt{1+\sqrt{1+\cdots }}}\\&=\sqrt{1+\sqrt{2}\phi}\\&<2\end{aligned}$$
We can get a tighter bound for the limit by breaking the pattern a little further down the line:
$$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2^2+\cdots}}}}\\&=\sqrt{1+\sqrt{2+\sqrt{3+2\sqrt{1+\cdots }}}}\\&=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{5}}}}\approx 1.7665398\end{aligned}$$
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots }}}\approx 1.7579327$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| $1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$
So, $1*0=0$ witch means that we Are Left with
$1+1+1+1+1+1+1+1+1+1+1+1(+0)$ witch is $12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
} |
Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$
Let $x,y,z\in \mathbb R^+$ prove that:
$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz \le RHS$. How to prove above inequality ? It is too hard for me.
| $$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{AM-GM}{\ge} 3xyz$$
DOES help, since you can also prove:
$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
Note that this is the same as
$$2^6(x^2y+ y^2z+z^2x)^3(y^2x+z^2y+x^2z)^3 \geq 3^6x^2y^2z^2(x^2+yz)^2(y^2+xz)^2(z^2+xy)^2 \,.$$
which after long computations and many tears can be reduced to this all junk being positive. If you look to the few negative terms , you can get easely rid of them by invoking the AM-GM inequality.
Now, all this suggests that the inequality
$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
is true, all you have to do is find a neater solution :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to solve integrals of type $ \int\frac{1}{(a+b\sin x)^4}dx$ and $\int\frac{1}{(a+b\cos x)^4}dx$ $$\displaystyle \int\frac{1}{(a+b\sin x)^4}dx,~~~~\text{and}~~~~\displaystyle \int\frac{1}{(a+b\cos x)^4}dx,$$
although i have tried using Trg. substution. but nothing get
| Use $\sin(x)= \frac{2\tan\frac{x}{2}}{1 +\tan^2 \frac {x}{2}}$ and $\cos(x)= \frac{1-\tan ^{2}\frac{x}{2}}{1 +\tan^{2}\frac {x}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Limits problem to find the values of constants - a and b If $\lim_{x \to \infty}(1+\frac{a}{x}+\frac{b}{x^2})^{2x}=e^2$ Find the value of $a$ and $b$. Problem :
If $\lim\limits_{x \to \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}=e^2$ Find the value of $a$ and $b$.
Please suggest how to proceed this problem :
If we know that $\lim\limits_{x \to \infty}\left( 1+\frac{1}{x}\right)^x= e$
Will this work somehow... please suggest
| Take logs of both sides to get
$$2 = \lim_{x \to \infty} 2 x \log{\left ( 1+ \frac{a}{x} + \frac{b}{x^2}\right)}$$
Use $\log{(1+y)} \sim y - (y^2)/2$ as $y \to 0$ to get
$$2 = \lim_{x \to \infty} \left ( 2 a + \frac{2 b-a^2}{x}\right)$$
So we require $a=1$. The value of $b$ is unimportant.
ADDENDUM
Details of the expansion of the log term to $O(1/x^2)$:
$$\begin{align}\log{\left ( 1+ \frac{a}{x} + \frac{b}{x^2}\right)} &= \frac{a}{x} + \frac{b}{x^2} - \frac12 \left ( \frac{a}{x} + \frac{b}{x^2} \right )^2 + O\left(\frac{1}{x^3}\right)\\ &= \frac{a}{x} + \frac{b}{x^2} - \frac{a^2}{2 x^2} - \frac{a b}{x^3} - \frac{b^2}{2 x^4}+ O\left(\frac{1}{x^3}\right)\\ &= \frac{a}{x} + \frac{b}{x^2} - \frac{a^2}{2 x^2} + O\left(\frac{1}{x^3}\right) \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove that for every $n\in \mathbb{N}^{+}$, there exist a unique $x_{n}\in[\frac{2}{3},1]$ such that $f_{n}(x_{n})=0$ Let $f_{n}(x)=-1+x+\dfrac{x^2}{2^2}+\dfrac{x^3}{3^2}+\cdots+\dfrac{x^n}{n^2}$,
(1) Prove that for every $n\in \mathbb{N}^{+}$, then there exist unique $x_{n}\in[\frac{2}{3},1]$ such that
$f_{n}(x_{n})=0$
(2) Show that the sequence $(x_{n})$ of (1) is such that
$$0<x_{n}-x_{n+p}<\dfrac{1}{n}$$
for all $p\in \mathbb{N}$.
(3):$x_{n}=A+\dfrac{B}{n}+\dfrac{C}{n^2}+O(\dfrac{1}{n^2})$, find $A,B,C$?
My attempts :
For $(1)$, I have prove it :
$$f_{n}(1)=-1+1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}>0,$$
\begin{align}
f_{n}(\frac{2}{3})&=-1+\dfrac{2}{3}+\cdots+\dfrac{(\dfrac{2}{3})^n}{n^2}\\
&\le-\dfrac{1}{3}+\dfrac{1}{4}\sum_{k=2}^{n}\left(\dfrac{2}{3}\right)^k\\
&=-\dfrac{1}{3}\cdot\left(\dfrac{2}{3}\right)^{n-1}<0
\end{align}
and
$$f'_{n}(x)=1+\dfrac{x}{2}+\cdots+\dfrac{x^{n-1}}{n}>0$$
But for $(2)$, I have methods:
since
$$x>0, f_{n+1}(x)=f_{n}(x)+\dfrac{x^{n+1}}{(n+1)^2}>f_{n}(x)$$
so $$f_{n+1}(x_{n})>f_{n}(x_{n})=f_{n+1}(x_{n+1})=0$$
since $f_{n+1}(x)$ is increasing, so $x_{n+1}<x_{n}$
since
$$f_{n}(x_{n})=-1+x_{n}+\dfrac{x^2_{n}}{2^2}+\cdots+\dfrac{x^n_{n}}{n^2}=0$$
$$f_{n+p}(x_{n})=-1+x_{n+p}+\dfrac{x^2_{n+p}}{2^2}+\cdots+\dfrac{x^{n+p}_{n+p}}{(n+p)^2}=0$$
and use $0<x_{n+p}<x_{n}\le 1$ we have
\begin{align}
x_{n}-x_{n+p}&=\sum_{k=2}^{n}\dfrac{x^k_{n+p}-x^k_{n}}{k^2}+\sum_{k=n+1}^{n+p}\dfrac{x^k_{n+p}}{k^2}\le\sum_{k=n+1}^{n+p}\dfrac{x^k_{n+p}}{k^2}\\
&\le\sum_{k=n+1}^{n+p}\dfrac{1}{k^2}<\sum_{k=n+1}^{n+p}\dfrac{1}{k(k-1)}=\dfrac{1}{n}-\dfrac{1}{n+p}<\dfrac{1}{n}
\end{align}
But for part $(3)$ I can't prove it,Thank you everyone can help,and for part $(2)$ have other nice methods?
| The $f_n$ converge to $f(x) = f_\infty(x) = -1 + \displaystyle \sum_{n=1}^{\infty} \frac{x^k}{k^2} $ = Li_2$(x)-1$ (the dilogarithm, minus $1$) with remainder terms $R_n(x) = \displaystyle \sum_{k=n}^{\infty} \frac{x^k}{k^2}$. Then:
*
*$x_n$ is a decreasing sequence that converges to $A \in (0,1)$ with $f(A)=0$, the inverse dilogarithm of $1$.
*The inverse of $f$ has a power series expansion near $0$ with real coefficients and a positive radius of convergence (due to analyticity of $f$ for $|z|<1$ and nonzero derivative of $f$ on $(0,1)$), $$ f^{-1}(z)= A + \beta z + \gamma z^2 + \dots$$ where $\beta$ and $\gamma$ can be calculated from $A$, $f'(A) = -\log(1-A)$, and $f''(A)=\frac{1}{1-A}$ by formal inversion of series.
*$f(x_{n}) = R_{n+1}(x)$ so that $x_n = f^{-1}(R_{n+1}(x_n))$ once $n$ is large enough that $R_{n+1}$ is in the circle of convergence.
*$R_p(x)$ is between $\frac{x^p}{p^2}$ and $\frac{x^p}{p^2} (1-x)^{-1}$ by comparing terms and is within a factor of $1 + O(\frac{1}{p})$ of the upper bound for constant positive $x < 1$
*From the formula in (3), $\quad x_{n-1} $ differs from $A$ by $O(x_n^n / {n^2}$) which is much smaller than a polynomial in $\frac{1}{n}$. This difference is dominated, for large $n$ and any positive $u$, by $(A+u)^n / {n^2}$ which implies that the asymptotic expansion of $x_n$ in powers of $\frac{1}{n}$ with constant coefficients is $A + \frac{0}{n} + \frac{0}{n^2} + \frac{0}{n^3} + \cdots$. Better than that, to any finite number of terms in the power series of $f^{-1}(z)$ there is no asymptotic difference between use of $x_n$ and $A$ in the formulas, and the first two terms of the asymptotics are $$x_{n} = f^{-1}(R_{n+1}(A)) = A + \beta \frac{A^{n+1}}{(1-A)(n+1)^2} + O(R^2) = A + (\frac{\beta A}{1-A}) \frac{A^n}{n^2}$$ plus terms of higher order in $A^n$ or $(1/n))$.
Conclusions:
*
*The answer to the question as written is $B=C=0$ and $A = ($ inverse_ dilogarithm of $1$) $= 0.761542944532045588068051872410222873413830030301381136443779....$, but the true magnitude of $(x_n - A)$ is about $\frac{A^n}{n^2}$. Exactly as foreseen in the earlier answer from Hagen von Eitzen.
*$\beta = \frac{1}{\log(\frac{1}{1-A})}$
*going to order $\gamma z^2$ looks challenging.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Evaluating $\lim_{n \to \infty} (1 + 1/n)^{n}$ I was recently thinking about how I could evaluate the famous limit of 'e' as I haven't ever seen a proof. I can't really find anything online so I've tried to evaluate the limit myself. And I was also thinking it would be nonsensical to use L'Hopital's rule, am I right?
So I did the following:
$$\lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = \exp \left(\lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n} \right) \right)$$
$$=\lim_{n \to \infty} \exp \left( n \cdot \left( \frac{1}{n} - \frac{(1/n)^{2}}{2} + \cdots \right) \right)$$
$$= \lim_{n \to \infty} \exp \left( 1 - \frac{(1/n)}{2} + \cdots \right)$$
$$= e$$
I am not sure, is my logic correct or does it create circularity by taking logarithms and assuming $$f(x) = \exp \left( \ln f(x) \right)$$ ?
| Here is the "standard" proof:
Let $a_{n}= \left(1 + \frac{1}{n} \right)^{n}= \frac{(n+1)^n}{n^n}$.
Then
$$\frac{a_{n+1}}{a_n}=\frac{(n+2)^{n+1}}{(n+1)^{n+1}}\frac{n^n}{(n+1)^n}=\frac{(n+2)^{n+1}n^{n+1}}{(n+1)^{2n+2}}\frac{n+1}{n}$$
$$=\left( \frac{n^2+2n}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}=\left( 1-\frac{1}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}$$
Now, by Bernoulli
$$\left( 1-\frac{1}{(n+1)^{2}}\right)^{n+1}\geq 1-\frac{n+1}{(n+1)^{2}}=\frac{n}{n+1}$$
Thus
$$\frac{a_{n+1}}{a_n} \geq \frac{n}{n+1}\frac{n+1}{n}=1 \,.$$
This shows that $a_n$ is increasing.
Now, let
Let $b_{n}= \left(1 + \frac{1}{n} \right)^{n+1}= \frac{(n+1)^{n+1}}{n^{n+1}}$.
Then
$$\frac{b_n}{b_{n+1}}=\frac{(n+1)^{n+1}}{n^{n+1}}\frac{(n+1)^{n+2}}{(n+2)^{n+2}}=\left( \frac{(n+1)^2}{n(n+2)} \right)^{n+2}\frac{n}{n+1}$$
$$\geq \left(1+ \frac{n+2}{n(n+2)} \right)\frac{n}{n+1}=1$$
Thus $b_n$ is decreasing.
As $a_n \leq b_n$ it follows that $a_n$ is increasing and bounded from above by $b_1$..
Now, if we denote the limit of this sequence by $e$ (which is the historic definition), it can be proven from
$$\left(1 + \frac{1}{n} \right)^{n}=\left(1 + \frac{1}{n} \right)^{n+1} =e$$
that $(e^x)'=e^x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let $f(\frac ab)=ab$, where $\frac ab$ is irreducible, in $\mathbb Q^+$. What is $\sum_{x\in\mathbb Q^+}\frac 1{f(x)^2}$?
Let $f(\frac ab)=ab$, where $\frac ab$ is irreducible, in $\mathbb Q^+$. What is $\sum_{x\in\mathbb Q^+}\frac 1{f(x)^2}$?
Club challenge problem. I don't think it's possible to do with only high school calculus. Help, please?
| Note first that $f$ is multiplicative over the rationals in some sense. More precisely, if $\frac{a}{b}$ and $\frac{c}{d}$ are in reduced form and $(a,d) = (b,c) = 1$, we have that $f(\frac{ac}{bd}) = f(\frac{a}{b}) \cdot f(\frac{c}{d})$.
You have to justify this next step, but (if the sum behaves nicely) we can thus split it up into an infinite product using unique factorization of the rationals:
$$
\begin{align*}
\sum_{x \in \mathbb{Q}^+} \frac{1}{f(x)^2} &=
\prod_{p \text{ prime}} \left(\cdots + \frac{1}{f(p^{-2})^2} + \frac{1}{f(p^{-1})^2} + \frac{1}{f(1)^2} + \frac{1}{f(p^1)^2} + \frac{1}{f(p^2)^2} + \cdots\right)
\\
&= \prod_{p \text{ prime}} \left( 1 + \frac{2}{p^2} + \frac{2}{p^4} + \frac{2}{p^6} + \cdots \right)
\\
&= \prod_{p \text{ prime}} \left( \frac{2}{1 - \frac{1}{p^2}} - 1 \right)
= \prod_{p \text{ prime}} \left( \frac{p^2 + 1}{p^2 - 1} \right)
\end{align*}
$$
I admit I am unsure of how to evaluate this product, but this seems helpful.
Update: This product evaluates to $\frac52$! See the comments for an explanation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find maximum value of $f(x)=2\cos 2x + 4 \sin x$ where $0 < x <\pi$ Find the maximum value of $f(x)$ where
\begin{equation}
f(x)=2\cos 2x + 4 \sin x \ \
\text{for} \ \ 0<x<\pi
\end{equation}
| $f(x)=2\cos 2x+4\sin x$
$=2(1-2\sin^2x)+4\sin x$
$=2-4\sin^2x+4\sin x$
$=2-(4\sin^2 x-2(2\sin x)+1)+1$
$=2+1-(2\sin x-1)^2\le (2+1)=3$
Equality holds iff $2\sin x-1=0\Rightarrow \sin x=1/2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proving there are no integer solutions for $3x^2=9+y^3$
Prove there are no $x,y\in\mathbb{Z}$ such that $3x^2=9+y^3$.
Initial proof
Let us assume there are $x,y\in\mathbb{Z}$ that satisfy the equation, which can be rewritten as $$3(x^2-3)=y^3.$$ So, $$3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3.$$ As $3 \mid -3$, it follows that $$3 \mid x^2 \Rightarrow 3^2 \mid x^2.$$ Say $x^2=3^2 \cdot a^2$ for some $a \in \mathbb{Z}$. Then $$\begin{align*}x^2-3 &= 3^2 \cdot a^2 - 3 \\&= 3 \cdot (3 \cdot a^2 - 1)\end{align*}$$ As $3^2 \mid x^2-3$, it follows that $3 \mid 3 \cdot a^2 - 1$. It is obvious that $3 \mid 3 \cdot a^2$, so it follows that $3 \mid -1$, which is false. Therefore, the assumption that there are $x,y\in\mathbb{Z}$ was wrong.
Alternative proof
$$
\begin{align*}
3x^2&=9+y^3\\
3(x^2-3)&=y^3
\end{align*}
$$
So, $3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3 \Rightarrow 3^2 \mid x^2 \Rightarrow 3^2 \not\mid x^2 - 3$ and we are done.
Both proofs are essentially the same, except for the fact that the second proof is much shorter. I think the first proof is much more clear, but it takes a bit longer to write down.
*
*Would the second proof suffice in a math competition, or is it too short?
*This problem was part of the Dutch finals of 1978. I figured it out in under a minute, which is quite unusual for me, so now I wonder: is my proof correct?
| Assume that the expression has integer solutions, or atleast one for that matter:
Work in modulus $3$, if:
$$3(x^2-3)=y^3$$
$y^3$ must be a multiple of $3$. Then we can set $y=3k$ where $k$ is an integer. Now we have:
$$27k^3=3(x^2-3)$$
Dividing both sides by $3$ gives us $9k^3=x^2-3$
Taking $-3$ to the other side shows us that $x^2$ is a mutiple of $3$. Thus $x$ is a multiple of $3$. Then we can set $x=3t$ where $t$ is an integer. This gives us:
$$9k^3+3=9t^2$$
Dividing both sides by $3$ again leaves us with this:
$$3k^3+1=3t^2$$
Which is impossible because $3t^2$ is a multiple of $3$ and $3k^3+1$ is not.
This is a contradiction.
Thus the proof is complete.
| {
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Decimal pattern in division of two digit numbers by 9 Can some one explain how this is possible ?
1) 13 / 9 = 1.(1 + 3) = 1.444 ...
2) 23 / 9 = 2.(2 + 3) = 2.555 ...
3) 35 / 9 = 3.(3 + 5) = 3.888 ...
4) 47 / 9 = 4.(4 + 7) = 4.(11) → 4.(11 - 9) = 5.222 ...
5) 63 / 9 = 6.(6 + 3) = 6.(9) → 6.(9 - 9) = 7.0 = 7
| Let $a$ be the tens digit and $b$ be the ones digit so that our two digit number is $10a+b$. In general, the pattern seems to be:
$$
\dfrac{10a+b}{9} = (a+j).kkk...
$$
where:
$$(j,k)=
\begin{cases}
(0,a+b) & \text{if }a+b \in \{0,1,2,...,8\} \\
(1,a+b-9) & \text{if }a+b \in \{9,10,11,...,17\} \\
(2,a+b-18) & \text{if }a+b =18 \\
\end{cases}$$
In other words, $j$ and $k$ are the quotient and remainder (respectively) upon dividing $a+b$ by $9$. To see why this makes sense, recall that $\boxed{0.kkk... = k/9}$. This can be proven a number of ways (for example, it is a convergent geometric series). Hence, observe that:
$$
\dfrac{10a+b}{9} =\dfrac{9a+(a+b)}{9}=a+\dfrac{a+b}{9}=a+j+\dfrac{k}{9} = (a+j).kkk...
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find domain of $\log_4(\log_5(\log_3(18x-x^2-77)))$
Problem:$\log_4(\log_5(\log_3(18x-x^2-77)))$
Solution:
$\log_3(18x-x^2-77)$ is defined for $(18x-x^2-77) \ge 3$
$(x^2-18x+77) \le -3$
$(x^2-18x+80) < 0$ {As it can't be 0}
$(x-8)(x-10)<0$
$8<x<10$
So domain is $(8,10)$
Am I doing right ??
| HINT:
For for real $m>0$ $\log_ma,$ is defined if $a>0$ as if $\log_ma=b\iff a=m^b>0$
$\log_4(\log_5(\log_3(18x-x^2-77)))$ is defined if $\log_5(\log_3(18x-x^2-77))>0$
$\implies \log_3(18x-x^2-77)>5^0=1$
$\implies (18x-x^2-77)>3^1=3$
$\implies x^2-18x+80<0$
Now, the roots of $x^2-18x+80=0$ are $\frac{18\pm\sqrt{18^2-4\cdot1\cdot 80}}2=8,10$
Now, if $(x-a)(x-b)<0$ where $a,b(<a)$ are real
Either $x-a<0$ and $x-b>0\implies b<x<a$
or $x-a>0\implies x>a$ and $x-b<0\implies x<b$ which needs $a<x<b$ which is impossible as $a>b$
So, the domain here will be $(8,10)$
| {
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A question on geometry? I wanted to know, given quadrilateral ABCD such that $AB^2+CD^2=BC^2+AD^2$ , prove that $AC⊥BD$ .
Help.
Thanks.
|
Here is Geometry method:
$AB^2=BE^2+AE^2, DC^2=DF^2+CF^2, BC^2=BE^2+CE^2,AD^2=AF^2+DF^2$
$AB^2+CD^2=BC^2+AD^2 \implies BE^2+AE^2+DF^2+CF^2=BE^2+CE^2+AF^2+DF^2$
$\implies AE^2+CF^2=CE^2+AF^2 \implies AE^2- AF^2=CE^2-CF^2$ .....<1>
$AE \ge AF, CE \le CF $ so for <1>:
LHS $\ge 0$, RHS $ \le 0$, the only condition is both sides $=0 \ \implies E=F \ \implies AC\perp BD$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| As several have shown above, a, b and c are roots of the cubic $abc=t^3-.5t$. In order for this to have three real roots, abc must lie between the maximum and minimum of $t^3-.5t$. At one of these extrema, two of the roots are equal to the location of the extrema, and the third is found from the requirement that the sum of the roots be zero. One set is $a=b=\frac{1}{\sqrt{6}}$, $c=\frac{-2}{\sqrt{6}}$. Notice that it is not necessary to calculate the discriminant, which can sometimes be messy if the sum of the roots is nonzero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Sum of greatest common divisors As usually, let $\gcd(a,b)$ be the greatest common divisor of integer numbers $a$ and $b$.
What is the asymptotics of
$$\frac{1}{n^2} \sum_{i=1}^{i=n} \sum_{j=1}^{j=n} \gcd(i,j)$$
as $n \to \infty?$
| We can rewrite the sum as
$$
\begin{align}
S(n)&=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j) \\ &= \frac{1}{n^2}\sum_{g=1}^n\sum_{i\le\lfloor n/g\rfloor}\sum_{\substack{j\le\lfloor n/g\rfloor\\(i,j)=1}} g \\
&= \frac{1}{n^2}\sum_{g=1}^n g\left(-1+2\sum_{i=1}^{\lfloor n/g\rfloor} \varphi(i)\right) \\
&= -\frac{n(n+1)}{2n^2}+\frac{2}{n}\sum_{g=1}^n \frac{g}{n}\sum_{i=1}^{\lfloor n/g\rfloor} \varphi(i)
\end{align}
$$
Write
$$
f(x) = \frac{1}{x}\sum_{i\le x}\varphi(i) = \frac{3x}{\pi^2}+E(x) \\
E(x) = o(\log x)
$$
(see Eric Naslund's exposition) then
$$
\begin{align}
S(n) &= -\frac{1}{2}-\frac{1}{2n}+\frac{2}{n}\sum_{g=1}^{n}f(n/g) \\
&= -\frac{1}{2}-\frac{1}{2n}+\frac{6}{\pi^2}\sum_{g=1}^{n}\frac{1}{g}+\frac{2}{n}\sum_{g=1}^n E(n/g) \\
&= \frac{6}{\pi^2}\log n+\frac{6\gamma}{\pi^2}-\frac{1}{2}+C+o(1) \\
&= \frac{6}{\pi^2}\log n + C' + o(1)
\end{align}
$$
where the constant $C$ arises from
$$
E(x) = o(\log x) \\
\left|\frac{2}{n}\sum_{g=1}^n E(n/g)\right|< \frac{C}{n}\sum_{g=1}^n\log(n/g)=C\left(\log n - \frac{\log n!}{n}\right)=C+o(1)
$$
by Stirling's approximation. Calculations suggest $C=0.39344\cdots, C'=0.24434\cdots$.
| {
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Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
| You could mess about with integration by substitution methods.
For instance, if you change the expression to this:
$$\frac{1}{2}\int_0^\infty \frac{(x^2+1)-(x^2-1)}{x^4+1}dx = \frac{1}{2}\int_0^\infty \frac{(x^2+1)}{x^4 + 1}-\frac{(x^2-1)}{x^4+1}dx$$
which you can then separate out as:
$$\frac{1}{2}\left[\int_0^\infty \frac{(x^2+1)}{x^4+1}dx - \int_0^\infty \frac{(x^2-1)}{x^4+1}dx\right]$$
and then divide the top and bottom in both expressions by $x^2$ to get:
$$\frac{1}{2}\left[\int_0^\infty \frac{1+ \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx - \int_0^\infty \frac{1- \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\right]$$
and then you can change some variables. If you take:
$u= x- \frac{1}{x}$ so $du= 1 + \frac{1}{x^2}$ and $v= x+\frac{1}{x}$ and $dv = 1 - \frac{1}{x^2}$
you can square the functions:
$u^2 = x^2 -2 +\frac{1}{x^2}$ or $u^2 + 2 = x^2+\frac{1}{x^2}$
$v^2 = x^2 +2 +\frac{1}{x^2}$ or $v^2-2 = x^2 +\frac{1}{x^2}$
and when you substitute in you end up with
$$\frac{1}{2}\left[\int_0^\infty \frac{du}{u^2+2} - \int_0^\infty \frac{dv}{v^2-2}\right]$$
and from there it should be easier to do, no? I think this was the way I approached it in a class last semester and I saw someone else in the interwebs had done something similar. :-) so I could be wrong. But I think this should work. Or you could I think rearrange the above substitutions into an integration by parts for each expression. But when I did that it was a long process.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$?
Why $f(x) = x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$ (where $\zeta = \zeta_7$ is a primitive root of the unit) over $\mathbb{Q}$?
Of course it's irreducible by the Eisenstein criterion, however it apparently does not satisfies $1 + 2(\zeta + \zeta^2 + \zeta^4)$ as a root, I tried to calculate several times however I couldn't get $f(1 + 2(\zeta + \zeta^2 + \zeta^4))$ = 0$.
Thanks in advance.
| Compute:
$$\begin{align*}
(1+2(\zeta+\zeta^2+\zeta^4))^2+7&=\bigg[1^2+4(\zeta+\zeta^2+\zeta^4)+4(\zeta+\zeta^2+\zeta^4)^2\bigg]+7\\[0.1in]
&=\bigg[1+4(\zeta+\zeta^2+\zeta^4)+4(\zeta^2+\zeta^4+\zeta^8+2\zeta^3+2\zeta^5+2\zeta^6)\bigg]+7\\[0.1in]
(\mathsf{\text{because }}\zeta^8=\zeta)\quad&=\bigg[1+4(\zeta+\zeta^2+\zeta^4)+4(\zeta^2+\zeta^4+\zeta+2\zeta^3+2\zeta^5+2\zeta^6)\bigg]+7\\[0.1in]
&=\bigg[1+8\zeta+8\zeta^2+8\zeta^3+8\zeta^4+8\zeta^5+8\zeta^6\bigg]+7\\[0.1in]
&=(-7)+8+8\zeta+8\zeta^2+8\zeta^3+8\zeta^4+8\zeta^5+8\zeta^6+7\\[0.2in]
&=(-7)+8\Phi_7(\zeta)+7\\[0.1in]
\left(\begin{array}{c}\mathsf{\text{because }}\Phi_7\mathsf{\text{ is the}}\\ \mathsf{\text{minimal poly of }}\zeta\end{array}\right)\quad&=(-7)+7\\[0.1in]
&=0
\end{align*}$$
| {
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Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
| These kind of questions often benefit from the identity $a^2-b^2=(a-b)(a+b)$ in conjunction with Pythagorean trig identitities. Here,
$$
\begin{align}
\frac{\sin t-\cos t+1}{\sin t+\cos t -1}&=\frac{(\sin t+1)-\cos t}{(\sin t+\cos t) -1}\cdot\frac{(\sin t+1)+\cos t}{(\sin t+\cos t) +1}\\
&=\frac{\sin^2 t+2\sin t+1-\cos^2 t}{\sin^2 t+2\sin t\cos t+\cos^2 t -1}\\
&=\frac{2\sin^2 t+2\sin t}{2\sin t\cos t}\\
&=\frac{\sin t+1}{\cos t}\\
&=\tan t+\sec t\\
&=(\tan t+\sec t)\frac{\tan t-\sec t}{\tan t-\sec t}\\
&=\frac{\tan^2 t-\sec^2 t}{\tan t-\sec t}\\
&=\frac{1}{\sec t-\tan t}\\
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem. How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$
| Let us denote base of right angle triangle as b, perpendicular ( height ) as p, and hypotenuse as h,
$$\cos A = \frac{b}{h} ; \qquad \sin A = \frac{p}{h}\tag{i}$$
Therefore, $$\frac{\cos^2A}{1-\sin A} = \frac{\frac{b^2}{h^2}}{1-\frac{p}{h}}$$
[By putting the values of $\cos A$ and $\sin A$ from $(i)$]
Which after simplification gives you:
$$\frac{b^2}{h(h-p)}\tag{ii}$$
Now as
$$b^2 = h^2-p^2\quad\text{[using pythagoreous theorem]}\tag{iii}$$
By putting the value of $b^2$ from (iii) in (ii) you get :
$$\frac{h^2-p^2}{h(h-p)} = \frac{h+p}{h} = 1+ \frac{p}{h} = 1+ \sin A$$
(hence proved)
| {
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How to find the partial derivative of this complicated function? Find the partial derivative with respect to a (treat other variables as constants):
$\displaystyle f(a,b,c)=\frac {a}{\sqrt{a^2+8bc}}-\frac {a^r}{a^r+b^r+c^r}$
The article I'm reading says it should be:
$\displaystyle \frac {\sqrt{a^2+8bc}-\dfrac{a^2}{\sqrt{a^2+8bc}}}{a^2+8bc}-\frac{ra^{r-1}(a^r+b^r+c^r)-a^r \cdot ra^{r-1}}{(a^r+b^r+c^r)^2}$
but I don't know how to get there.
Thanks in advance!
| Form the partial derivatives quotient rule, we know:
$$\displaystyle \frac{\partial}{\partial a} \left(\frac{g}{h}\right) = \frac{g' h - g h'}{h^2}$$
Given the function:
$$\displaystyle f(a,b,c)=\frac {a}{\sqrt{a^2+8bc}}-\frac {a^r}{a^r+b^r+c^r}$$
Lets do each term separately, so we have:
$$\frac{(1) \sqrt{a^2+8bc}-(a)(\frac{1}{2})(a^2+8bc)^{-1/2}(2a)}{a^2 + 8 bc} = \frac {\sqrt{a^2+8bc}-\dfrac{a^2}{\sqrt{a^2+8bc}}}{a^2+8bc}$$
Of course, we can reduce this further to:
$$\dfrac{8 b c}{(a^2+8 b c)^{3/2}}$$
For the second one we have:
$$\displaystyle \frac{r a^{r-1}(a^r + b^r + c^r) - a^r(r a^{r-1})}{(a^r + b^r + c^r)^2}$$
Of course we can reduce this to:
$$\dfrac{r a^{r-1} (b^r+c^r)}{(a^r+b^r+c^r)^2}$$
The final result would be:
$$\displaystyle \frac{\partial}{\partial a} f(a, b, c,) = \dfrac{8 b c}{(a^2+8 b c)^{3/2}} - \dfrac{r a^{r-1} (b^r+c^r)}{(a^r+b^r+c^r)^2}$$
| {
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A proposed proof by induction of $1+2+\ldots+n=\frac{n(n+1)}{2}$
Prove: $\displaystyle 1+2+\ldots+n=\frac{n(n+1)}{2}$.
Proof
When $n=1,1=\displaystyle \frac{1(1+1)}{2}$,equality holds.
Suppose when $n=k$, we have $1+2+\dots+k=\frac{k(k+1)}{2}$
When $n = k + 1$:
\begin{align}
1+2+\ldots+k+(k+1) &=\frac{k(k+1)}{2}+k+1 =\frac{k(k+1)+2k+2}{2}\\
&=\frac{k^2+3k+2}{2}\\
\text{[step]}&=\displaystyle\frac{(k+1)(k+2)}{2}=\displaystyle\frac{(k+1)((k+1)+1)}{2}
\end{align}
equality holds.
So by induction, the original equality holds $\forall n\in \mathbb{N}$.
Question 1: any problems in writing?
Question 2: Why [step] happen to equal? i.e., why does $k^2+3k+2=(k+1)(k+2)$ hold?
| Nice work! If you want to take a bit more time, you can note that $$\frac{k^2+3k+2}2=\frac{k^2+2k+k+2}2=\frac{k(k+2)+1(k+2)}2=\frac{(k+1)(k+2)}2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I get polynomial $p(x)$? $p(x)$ is divided evenly into $x^{2}+1$, and $p(x)+1$ is divided evenly into $x^{3}+x^{2}+1$. How can I get $p(x)$?
| The given data tells us:
$$\begin{align}r(x)(x^2 + 1) &= p(x)\\
s(x)(x^3 + x^2 + 1) &= p(x) + 1
\end{align}$$
The above equations imply that
$$s(x)(x^3 + x^2 + 1) - r(x)(x^2 + 1) = 1. \tag1$$
Does the structure of the equation ring any bells? Yes! Bezout's identity!
Now use Euclid's division algorithm on $x^2 + 1$ and $x^3 + x^2 + 1$ to obtain:
$$x(x^3 + x^2 + 1) - (x^2 + x - 1)(x^2 + 1) = 1\tag2$$
Compare (1) and (2) and eventually get $p(x)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2$ for $x > 0$
Let $x>0$. Show that
$$(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2.$$
Do you have any nice method?
My idea $F(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$
then we hvae $F'(x)=\cdots$ But it's ugly.
can you have nice methods? Thank you
by this I have see this same problem
let $0<x<1$ we have $$x+\dfrac{1}{x^x}<2$$
this problem have nice methods:
becasue we have
$$\dfrac{1}{x^x}=\left(\dfrac{1}{x}\right)^x\cdot 1^{1-x}<x\cdot\dfrac{1}{x}+(1-x)\cdot 1=2-x$$
| Remark: Bernoulli inequality and AM-GM are enough.
Letting $x = \frac{1}{y}$, it suffices to prove that, for all $y > 0$,
$$\left(1 + \frac{1}{y}\right)^{\frac{y}{1+y}} + y^y > 2.$$
If $y \ge 3$, clearly the inequality is true.
If $0 < y < 3$, we have
\begin{align*}
\left(1 + \frac{1}{y}\right)^{\frac{y}{1+y}} + y^y - 2
&= \frac{(1+y)^{\frac{y}{1+y} + 1}}{(1+y)\cdot (y^y)^{\frac{1}{1+y}}} + y^y - 2\\[5pt]
&\ge \frac{1 + y \left(\frac{y}{1+y} + 1\right)}{(1+y)\cdot \left(1 + (y^y - 1)\cdot \frac{1}{1+y}\right)} + y^y - 2 \tag{1}\\[5pt]
&= \frac{2y^2 + 2y + 1}{(1+y)(y^y + y)} + y^y - 2\\[5pt]
&= \frac{2y^2 + 2y + 1}{(1+y)(y^y + y)} + (y^y + y) - y - 2\\[5pt]
&\ge 2\sqrt{\frac{2y^2 + 2y + 1}{(1+y)(y^y + y)} \cdot (y^y + y)} - y - 2 \tag{2}\\[5pt]
&= 2\sqrt{\frac{2y^2 + 2y + 1}{1 + y}} - y - 2\\[5pt]
&> 0
\end{align*}
where in (1) we have used Bernoulli inequality to obtain
$$(1+y)^{\frac{y}{1+y} + 1} \ge 1 + y \left(\frac{y}{1+y} + 1\right)$$
and
$$(y^y)^{\frac{1}{1+y}} \le 1 + (y^y - 1)\cdot \frac{1}{1+y};$$
and in (2) we have used AM-GM,
and the last inequality follows from
$$\left(2\sqrt{\frac{2y^2 + 2y + 1}{1 + y}}\right)^2 - (y + 2)^2 = \frac{y^2 (3 - y)}{1 + y} > 0.$$
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac {-\sqrt{1 - (t-1)^2}}{t} = \int \frac {-\sqrt{t^2 + 2t}}{t} = \int \frac {-\sqrt t \cdot \sqrt t \cdot \sqrt{t +
2}}{\sqrt t \cdot t} $$
$$= \int \frac {- \sqrt{t + 2}}{\sqrt t } = \int - \sqrt{1 + \frac 2t} = ? $$
What next? I don’t know. Also, I’ve tried another “substitution”, namely $1 + \cos x = 2 \cos^2 \frac x2) $
$$ \int \frac {dx}{1+ \cos x} = \int \frac {dx}{2 \cos^2 \frac x2} = \int \frac 12 \cdot \sec^2 \frac x2 = ? $$
And failed again. Help me, please.
| $$
\begin{array}{lcl}
\int_0^{\pi/2} \frac{1}{1+\cos x}dx &=& \int_0^{\pi/2} \frac{1}{(\cos^2(x/2)+\sin^2(x/2))+(\cos^2 (x/2)-\sin^2(x/2))}dx\\
&=& \int_0^{\pi/2} \frac{1}{2\cos^2(x/2)}dx\\
&=& \int_0^{\pi/4}\frac{1}{\cos^2u}du=\tan\frac{\pi}{4}=1
\end{array}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$
find the value
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}$$
is true?
Yes, I tink this problem can prove
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$
so $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}=1or -3$$
How many nice methods prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ ?
and I see this easy problem
http://zhidao.baidu.com/question/260913315.html
| I think you might be talking about a triangle, with $a$, $b$, $c$ as it's sides of a triangle. If the angles opposite to side a, b, c are A, B,C, then $$\cos A=\frac{b^2+c^2-a^2}{2cb}$$ and similarly, $$\cos B=\frac{a^2+c^2-b^2}{2ac}$$ and $$\cos C=\frac{b^2+a^2-c^2}{2ab}$$ If it is true then $\cos A = \cos B = \cos C =1$ from the first term (since minimum and maximum of cos x with any x is plus minus 1). So the answer is either -3, -2,-1 ,1,2 or 3.
Form here, we are getting a variety of answers, but since here only one of the angle can be negative(obtuse), the answer could only be 1 or 3. (since the sum could only be 1+1-1 or 1+1+1).
Now since all three angles of the triangle cannot be 0, so they must be one angle 180 and the rest 0. So 1 is the only answer.
NOTE: The above is only true when a,b and c are angles of a triangle. And the above is the called the cosines rule of a triangle. Although the triangle discussed above is in fact a straight line, but it can be noticed that the sum of the two sees containing the 180 angle is equal to the third side.
| {
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If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$ If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$.
I have tried to solve this problem in the following way :
\begin{align}&\sin a + \sin b=2 \\
\Rightarrow &2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=2\\
\Rightarrow &\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=1
\end{align}
What will be the next ?
| The maximum value of the $sin$ function is 1, at $\pi / 2 + 2k\pi$, for any integer $k$.
So $a = \pi / 2 + 2k\pi$, and $b = \pi / 2 + 2l\pi$, for any integers $k$, $l$.
Therefore, $\sin(a + b) = \sin(\pi + 2(k + l)\pi) = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $\sum_{k=1}^n \frac{1}{2k+1}$ is not an integer? Let $S=\sum_{k=1}^n \frac{1}{2k+1}$, how can we prove with elementary math reasoning that $S$ is not an integer?
Can somebody help?
| Let $p=2k'+1$ be the largest odd prime number less than or equal to $2n+1$. Now consider the sum:
$$\frac13 + \frac15 + \frac17 + \cdots + \frac{1}{2k'-1} + \frac{1}{p} + \frac{1}{2k'+3} + \cdots + \frac{1}{2n+1}.$$
We can find a common denominator for this fraction by taking the product:
$$3\cdot5\cdots (2k'-1) \cdot p \cdot (2k' +3) \cdots (2n+1) = M.$$
In the numerator we have a sum of several terms:
$$\frac{M}{3} + \frac{M}{5} + \cdots + \frac{M}{p} + \cdots + \frac{M}{(2n+1)}.$$
Each term here is an integer.
I claim that all but $\frac{M}{p}$ is divisible by $p$. This would mean the numerator is not divisible by $p$ and is congruent to $\frac{M}{p}$ modulo $p$. Since $p$ appears in the denominator, this sum cannot be an integer.
Suppose that $\frac{M}{p}$ was indeed divisible by $p$. This means that we must have had another multiple of $p$ appear in one of our denominators. The next valid term would be $3p$. However, Bertrand's postulate tells us that for any natural number $n$ we have a prime between $n$ and $2n$. This tells us that there would have been a new largest prime between $p$ and $3p$. Which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the integral of $\int e^x\,\sin x\,\,dx$? I'm trying to solve the integral of $\left(\int e^x\,\sin x\,\,dx\right)$ (My solution):
$\int e^x\sin\left(x\right)\,\,dx=$
$\int \sin\left(x\right) \,e^x\,\,dx=$
$\left(\sin(x)\,\int e^x\right)-\left(\int\sin^{'}(x)\,\left(\int e^x\right)\right)$
$\left(\sin(x)\,e^x\right)-\left(\int\cos(x)\,e^x\right)$
$\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(\int-\sin\left(x\right)\,e^x\right)\right)$
$\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(-\sin\left(x\right)\,e^x-\int-\cos\left(x\right)\,e^x\right)\right)$
I don't know how to complete because the solution gonna be very complicated.
| Another way via leibniz theorem,
$$ \frac{d^4 (e^x \sin x)}{dx^4} =e^x \left[\binom{4}{0} \sin x+ \binom{4}{1} \cos x - \binom{4}{2} \sin x - \binom{4}{3} \cos x+ \binom{4}{4} \sin x\right]$$
Or,
$$ \frac{1}{ \sum_{k=0}^2 \binom{4}{2k} } \frac{d^4 (e^x \sin x)}{dx^4} = e^x \sin x$$
Integrate both sides and apply leibniz for third derivative again,
$$ \int e^x \sin x dx = \frac{1}{ \sum_{k=0}^2 (-1)^k \binom{4}{2k} } \int \frac{d^4 (e^x \sin x)}{dx^4} dx=e^x \frac{\sum_{j=0}^3 \binom{3}{j} \sin( x + j \frac{\pi}{2})}{ \sum_{k=0}^2 (-1)^k \binom{4}{2k} } $$
Since,
$$ \frac{d^i}{dx^i} \sin x = \sin( x +i \frac{\pi}{2} )$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given that $\cos x =-3/4$ and $90^\circGiven that $\;\cos x =-\frac{3}{4}\,$ and $\,90^\circ<x<180^\circ,\,$ find $\,\tan x\,$ and $\,\csc x.$
This question is quite unusual from the rest of the questions in the chapter, can someone please explain how this question is solved? I tried Pythagorean Theorem, but no luck. Is it possible to teach me how to use the circle diagram?
| The cosine of an angle corresponds to the $x$-coordinate in the unit circle, and the sine of an angle cooresponds to its $y$-coordinate on the unit circle.
Note that $\,\cos = -\frac 34 < 0\,$ if and only if the angle $x$ terminates in either the second or third quadrant, where the angle $x$ is measured with respect to the positive $x$-axis.
Since we are given that $\,90 \lt x \lt 180,\,$ we know that the angle $x$ terminates in the second quadrant. So $\sin x > 0$.
So, $\,\tan x = \dfrac{\sin x}{\cos x} <0,\;$ and $\,\csc x = \dfrac{1}{\sin x}>0$.
Now, we know that by the Pythagorean Theorem as it relates to trigonometry identities, $${\bf \sin^2 x + \cos^2 x = 1} $$ $$ \begin{align} \iff \sin^2 x & = 1 -\cos^2 x \\ \\ \implies \sin x & = \pm \sqrt{1 -\cos^2 x} \\ \\ & = \pm \sqrt{1 - \left(\dfrac {-3}{4}\right)^2} \\ \\ & = \pm \sqrt{\frac 7{16}} \\ \\ & = \pm \frac{\sqrt 7}{4}\end{align}$$ Since we know that in the second quadrant, $\sin x > 0$, we take the positive root: $$\sin x = \frac{\sqrt 7}{4}$$
and so you have all you need to compute $$\tan x=\dfrac{\sin x}{\cos x} = \dfrac{\sqrt 7/4}{-3/4} = -\left(\dfrac{\sqrt 7}{3}\right)$$ $$\csc x=\dfrac1{\sin x}= \dfrac{1}{\sqrt 7/4} = \dfrac 4{\sqrt 7} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Recursion Question - Trying to understand the concept Just trying to grasp this concept and was hoping someone could help me a bit. I am taking a discrete math class. Can someone please explain this equation to me a bit?
$f(0) = 3$
$f(n+1) = 2f(n) + 3$
$f(1) = 2f(0) + 3 = 2 \cdot 3 + 3 = 9$
$f(2) = 2f(1) + 3 = 2 \cdot 9 + 3 = 21$
$f(3) = 2f(2) + 3 = 2 \cdot 21 + 3 = 45$
$f(4) = 2f(3) + 3 = 2 \cdot 45 + 3 = 93$
I do not see how they get the numbers to the right of the equals sign. Please someone show me how $f(2) = 2f(1) + 3 = 2 \cdot 9 + 3$. I see they get "$2\cdot$" because of $2f$ but how and where does the $9$ come from? I also see why the $+3$ at the end of each equation but how and where does that number in the middle come from?
| Simply use substitution.
We are given the initial value $$\color{blue}{\bf f(0) = 3}\tag{given}$$
Each subsequent value of the function $f$ depends on the preceding value. So the function evaluated at $(n + 1)$ depends (is defined, in part) on the function's value at $n$: That's what's meant by a recursive definition of the function $f$, which here is defined as: $$f(n+1) = 2\cdot f(n) + 3,\quad \color{blue}{f(0) = 3}$$
Knowing $f(0)$ is enough to "get the ball rolling":
$${\bf f(0 + 1) = f(1)} = 2\color{blue}{\bf f(0)} + 3 = 2\cdot \color{blue}{\bf 3} + 3 = 6 + 3 = \bf 9$$
Now, knowing $f(1)$ we can compute $f(2)$
$$f(1 + 1) =f(2) = 2\cdot {\bf f(1)} + 3 = 2\cdot {\bf 9} + 3 = 18 + 3 = {21}$$
Now that we know $f(2) = 21$ we can find $f(2 + 1) = f(3)$:
$$f(3) = 2\cdot f(2) + 3 = 2 \cdot 21 + 3 = 42 + 3 = 45$$
Now that we know $f(3) = 45$, we can compute $f(3+1)=f(4)$:
$$f(4) = 2\cdot f(3) + 3 = 2\cdot 45 + 3$$
And so on...
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $ \lim_{x \to \infty} x \cdot \log(\frac{x+1}{x+10})$ is $-9$ Given this limit:
$$ \lim_{x \to \infty} x \cdot \log\left(\frac{x+1}{x+10}\right) $$
I may use this trick:
$$ \frac{x+1}{x+1} = \frac{x+1}{x} \cdot \frac{x}{x+10} $$
So I will have:
$$ \lim_{x \to \infty} x \cdot \left(\log\left(\frac{x+1}{x}\right) + \log\left(\frac{x}{x+10}\right)\right) = $$
$$ = 1 + \lim_{x \to \infty} x \cdot \log\left(\frac{x}{x+10}\right) $$
But from here I am lost, I still can't make it look like a fondamental limit. How to solve it?
| $$
\begin{align}
\lim_{x\to\infty}x\log\left(\frac{x+1}{x+10}\right)
&=\lim_{x\to\infty}x\log\left(1-\frac{9}{x+10}\right)\\
&=\lim_{x\to\infty}\frac{\log\left(1-\frac{9}{x+10}\right)}{-\frac{9}{x+10}}\left(-\frac{9x}{x+10}\right)\\[9pt]
&=\lim_{u\to0}\frac{\log(1+u)}{u}\lim_{x\to\infty}\left(-\frac{9x}{x+10}\right)\\[9pt]
&=1\cdot(-9)\\[18pt]
&=-9
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding perpendicular bisector of the line segement joining $ (-1,4)\;\text{and}\;(3,-2)$
Find the perpendicular bisector of the line joining the points $(-1,4)\;\text{and}\;(3,-2).\;$
I know this is a very easy question, and the answer is an equation. So any hints would be very nice. thanks
| If the point $P=(x,y)$ lies on the perpendicular bisector of the points $A=(-1,4)$ and
$B= (3, -2)$, then the distances $PA$ and $PB$ must be the same.
Then
$$
PA^2 = PB^2 \\
\Rightarrow (x+1)^2 + (y-4)^2 = (x-3)^2 + (y+2)^2 \\
\Rightarrow x^2 +2x + 1 + y^2 -8y +16 = x^2 -6x +9 +y^2 + 4y + 4 \\
\Rightarrow 2x - 8y +17 = -6x +4y +13 \\
\Rightarrow 8x - 12y + 4 = 0 \\
\Rightarrow 2x -3y + 1 = 0
$$
So, the perpendicular bisector has equation $2x -3y + 1 = 0$.
| {
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Discrete Math Question: arithmetic progression A lumberjack has $4n + 110$ logs in a pile consisting of n layers. Each layer has two more logs than the layer directly above it. If the top layer has six logs, how many layers are there? Write the steps to calculate the equation for the problem and state the number of layers.
It's given that the top layer has 6 logs and each layer below that gains 2 logs per layer.
1st layer: 6
2nd layer: 6 + 1(2) = 6 + 2 = 8
3rd layer: 6 + 2(2) = 6 + 4 = 10
4th layer: 6 + 3(2) = 6 + 6 = 12
5th layer : 6 + 4(2) = 6 + 8 = 14
and so on
the nth layer: $a_n = 6 + 2(n-1)$
1st term 6 and common difference 2 with n terms
1st term: $a_n = 6$
last term: $a_n = 6 + 2n + 4$
$S_n = (n/2)(a_1 + a_n)$
$$4n + 110 = (n/2)(6 + 6 + 2(n-1))$$
Can anyone help break this equation down to solve for n?
| $$6+8+10+\cdots+(6+2n-2)=6n+(0+2+4+\cdots+2n-2)=\\6n+2(0+1+2+\cdots+n-1)=6n+2\frac{(n-1)n}{2}=6n+(n-1)n=n^2+5n$$
You were told that this equals $4n+110$, so $n^2+5n=4n+110$, which is a quadratic equation $n^2+n-110=(n+11)(n-10)=0$. The only positive solution is $n=10$.
| {
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Calculating $\lim_{x\to 0}\left(\frac{1}{\sqrt x}-\frac{1}{\sqrt{\log(x+1)}}\right)$ Find the limit
$$\lim_{x\to 0}\left(\frac{1}{\sqrt x}-\frac{1}{\sqrt{\log(x+1)}}\right)$$
| Hint
$$\frac{\log(1+x)}{x}\to 1$$
$$\frac{\log(1+x)-x}{x^2}\to -\frac{1}{2}$$
Further Hint
$$\displaylines{
\frac{1}{{\sqrt x }} - \frac{1}{{\sqrt {\log (x + 1)} }} = \frac{{\sqrt {\log (x + 1)} - \sqrt x }}{{\sqrt {x\log \left( {1 + x} \right)} }}\left( {\frac{{\sqrt {\log (x + 1)} + \sqrt x }}{{\sqrt {\log (x + 1)} + \sqrt x }}} \right) \cr
= \frac{{\log (x + 1) - x}}{{\sqrt {x\log \left( {1 + x} \right)} }}{\left( {\sqrt {\log (x + 1)} + \sqrt x } \right)^{ - 1}} \cr
= \frac{{\log (x + 1) - x}}{{\sqrt {x\log \left( {1 + x} \right)} }}\frac{1}{{\sqrt x }}{\left( {\sqrt {\frac{{\log (x + 1)}}{x}} + 1} \right)^{ - 1}} \cr
= \frac{1}{{\sqrt x }}\frac{{\log (x + 1) - x}}{x}{\left( {\sqrt {\frac{{\log \left( {1 + x} \right)}}{x}} } \right)^{ - 1}}{\left( {\sqrt {\frac{{\log (x + 1)}}{x}} + 1} \right)^{ - 1}} \cr
= \sqrt x \frac{{\log (x + 1) - x}}{{{x^2}}}{\left( {\sqrt {\frac{{\log \left( {1 + x} \right)}}{x}} } \right)^{ - 1}}{\left( {\sqrt {\frac{{\log (x + 1)}}{x}} + 1} \right)^{ - 1}} \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/444404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are there $a,b>1$ with $a^4\equiv 1 \pmod{b^2}$ and $b^4\equiv1 \pmod{a^2}$? Are there solutions in integers $a,b>1$ to the following simultaneous congruences?
$$
a^4\equiv 1 \pmod{b^2} \quad \mathrm{and} \quad b^4\equiv1 \pmod{a^2}
$$
A brute-force search didn't turn up any small ones, but I also don't see how to rule them out.
| Maybe like this:
First observe that b^2 | (a^4-1) and a^2 | (b^4-1). Hence, we can write a^4-1=kb^2 and b^4-1=ma^2, with k, m being positive integers.
So that gives then that a^4-1 = ((b^4-1)/m)^2-1=(b^8-2b^4+1)/m^2-1 = kb^2
That means that b^8-2b^4+1 = kb^2*m^2+m^2
Hence b^2(b^6-2b^2-km^2)=m^2-1
Now note that b^2 then should be a divider of m^2-1, cause b^6-2b^2-km^2 is still a integer.
That means that m^2 = 1 mod b^2, with m and b being still integers.
Hence, m^2 = 1 +lb^2, with l being integer.
Substitution then gives that b^6-(2+lk)b^2-(k+l)=0
Now solve this equation for b^2 and observe that this gives two complex solutions and one real solution, the only thing to do is to solve that explicitly and try for integer values of l and k to get integer value of b, thereby m^2 also becomes integer and a divider.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/446269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim\limits_{x \to 0} \frac1{1-\cos (x^2)}\sum\limits_{n=4}^{\infty} n^5x^n$ I'm trying to solve this limit but I'm not sure how to do it.
$$\lim_{x \to 0} \frac1{1-\cos(x^2)}\sum_{n=4}^{\infty} n^5x^n$$
I thought of finding the function that represents the sum but I had a hard time finding it.
I'd appreciate the help.
| Of course finding a formula for the sum of the series is not needed to solve this.
$$\begin{align}
\cos(x^2) &= 1 - \frac{x^4}{2} + O(x^8)\qquad\text{as }x \to 0
\\
\frac{1}{1-\cos(x^2)} &= 2 x^{-4} + O(1)
\\
\sum_{n=4}^\infty n^5 x^n &= 4^5 x^4 + O(x^5)
\\
\frac{1}{1-\cos(x^2)}\sum_{n=4}^\infty n^5 x^n &= 2\cdot 4^5 + O(x)
\end{align}$$
and the limit is $2\cdot 4^5 = 2048$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
convert ceil to floor Mathematically, why is this true?
$$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$
Assume $a$ and $b$ are positive integers.
Is this also true if $a$ and $b$ are real numbers?
| This is not true in general, e.g. take $b = 1/2$ and $a = 2,$ so the LHS is $4$ while the RHS is $3.$
Suppose $b\nmid a,$ since that case is trivial. Use the division algorithm to write $a = qb + r,$ where $q \in \mathbb{N}\cup \{0\}$ and $0 < r < b.$ Then the LHS is $q+1$ while the RHS is $\lfloor (q+1) + \frac{r-1}{b} \rfloor.$ Since $r$ is in fact an integer, $b > r \ge 1$ yields $\frac{b-1}{b} > \frac{r-1}{b} \ge 0,$ whence $\lfloor (q+1) + \frac{r-1}{b} \rfloor = q+1,$ as desired.
Edit: I was a bit hasty in giving the general condition. Here it is corrected.
If $b \mid a,$ then the LHS is $q$ and $r = 0,$ so you need $b \ge 1$ for the RHS to also be $q.$ If $b\nmid a,$ we write $a = qb + r,$ where $q \in \mathbb{Z}$ and $0 < r < |b|.$
So we need $0 \le \frac{r-1}{b} < 1.$ If $b > 0,$ this occurs so long as $r \ge 1$ (which in particular means that $b > 1$).
If $b < 0,$ then we need $r > 1 + b.$ Now, we also have $r < |b| = -b,$ which gives $-b > 1 + b,$ or $b < -1/2.$ For $b \le -1,$ $r > 1+b$ holds trivially, and for $-1 < b < -1/2,$ one has to actually check that $r > 1 +b.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The inductive step can be proved as follows.
$2^k < \binom{2k}{k} \implies 2^{k+1} < 2\binom{2k}{k} = \frac{2(2k)!}{k!k!} = \frac{2(2k)!(k + 1)}{k!k!(k + 1)} = \frac{2(2k)!(k+2)}{(k+1)!k!}<\frac{(2k)!(2k+2)(2k+1)}{(k+1)!k!(k+1)} = \binom{2(k+1)}{k+1}$
Second part: $2^{2n} > \binom{2n}{n}$. Again, the base is trivial. We can assume that for some $k$ our statement is satisfied and prove that inductive step as follows:
$2^{2k} > \binom{2k}{k} \implies 2^{2k + 2} > 2^2\binom{2k}{k} = \frac{2\cdot2(2k)!}{k!k!} = \frac{2\cdot2(2k)!(k+1)(k+1)}{k!k!(k+1)(k+1)} = \frac{(2k)!(2k+2)(2k+2)}{(k+1)!(k+1)!} > \frac{(2k)!(2k+1)(2k+2)}{(k+1)!(k+1)!} = \binom{2(k+1)}{k+1}$
Is there a non-inductive derivation for the inequality?
| The central binomial coefficient is largest implies a lower bound of $(2^{2n})/(2n+1)$. This exceeds $2^n$ for $n>1$ and is almost its square for large $n$.
There are more subsets than subsets of a fixed cardinality implies an upper bound of $2^{2n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 6
} |
Is $x^2-y^2=1$ Merely $\frac 1x$ Rotated -$45^\circ$? Comparing their graphs and definitions of hyperbolic angles seems to suggest so aside from the $\sqrt{2}$ factor:
and:
| Write it in polar coordinates:
$$x^2 - y^2 = 1 \ \to \ r^2(\cos^2 \theta - \sin^2 \theta) r^2 = r^2\cos(2\theta) = 1$$
$$xy=1 \ \to \ r^2\sin\theta\cos\theta=\frac{1}{2}r^2\sin(2\theta) =1$$
Or:
$$\frac{1}{2}r^2\sin(2\theta)=\frac{1}{2}r^2\cos(2(\theta + 45^\circ))=\left(\frac{r}{\sqrt{2}}\right)^2\cos(2(\theta + 45^\circ))$$
So you see that $y=1/x$ is indeed a rotated copy, up to a multiplicative factor .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
2 Different integrals of $\int \left ( \tan{x}\right ) ^3 dx $. My friend asked me why this function has 2 different integrals. I'm very confused.
\begin{align}
\int \left ( \tan{x}\right ) ^3 dx &=\int \left ( \tan{x} \right )^2 \tan{x}dx \\
&=\int \left ( \sec^2 {x} -1 \right ) \tan{x} dx \\
&=\int \tan{x} \left ( \tan{x} \right )' -\int \tan{x}dx \\
&=\frac12 \tan^2{x} +\log{|\cos{x}|}+C
\end{align}
Also,
\begin{align}
\int \left ( \tan{x}\right ) ^3 dx &=\int \left ( \tan{x} \right )^2 \tan{x}dx \\
&=\int \left ( \sec^2 {x} -1 \right ) \tan{x} dx \\
&=\int \frac{-1}{t^3}dt-\int \frac{-1}{t}dt \\
&=\frac12 \sec^2{x} +\log{|\cos{x}|}+C
\end{align}
with $x=\cos {t}$.
I don't know why this integral has two answers and how to know that there exist 2 and more answers. Thanks in advance.
| Note that $\frac{1}{2}\sec^2 x$ and $\frac{1}{2}\tan^2 x$ differ by a constant. Since there is always an arbitrary constant of integration, the two answers are the same answer.
In the same way $x^2+17+C$ and $x^2+C$ are both correct integrals of $2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
| \begin{align*}\frac{3}{\frac{\sqrt{5}}{5}} &= 3 \cdot \frac{5}{\sqrt 5}\\
&= 3 \cdot \frac{5}{\sqrt 5} \cdot 1\\
&= 3 \cdot \frac{5}{\sqrt 5} \cdot \frac{\sqrt 5}{\sqrt 5}\\
&= 3 \cdot \frac{5\sqrt 5}{5}\\
&= 3\sqrt 5
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
Interpreting the ; in a series This question is linked to this question.
So, suppose I set $n=5$. Given the following formula:
$$\frac{1}{n}, \dots , \frac{n-1}{n} $$
Am I suppose to get:
$$
\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \hspace{8.2cm}(1)
$$
Or
$$
\frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2},
\frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \hspace{5cm} (2)
$$
?
In other words, what purpose to the ";" in:
$$ \frac{1}{2}; \frac{1}{3}, \frac{2}{3}; \frac{1}{4}, \frac{3}{4}; \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \dots ; \frac{1}{n}, \dots , \frac{n-1}{n}.$$
serve?
Also, the formula does not mention anything about skipping numbers that are not in lowest common terms. Is skipping this assumed given the definition of $f$? For example, in (2), there is no $\frac{2}{4}$ because it is equal to $\frac{1}{2}$ which is already listed earlier.
Thank you in advance for any help provided.
| You are supposed to get (1), the complete enumeration makes it clear. The semicolon is only put for clarity in separating terms with different denumerator. Skipping is implicit because the author is enumerating the values of a set, hence repeating them has no effect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
divisibility test let $n=a_m 10^m+a_{m-1}10^{m-1}+\dots+a_2 10^2+a_1 10+a_0$, where $a_k$'s are integer and $0\le a_k\le 9$, $k=0,1,2,\dots,m$, be the decimal representation of $n$
let $S=a_0+\dots+a_m$, $T=a_0-a_1\dots+(-1)^ma_m$
then could any one tell me how and why on the basis of divisibility of $S$ and $T$ by $2,3,\dots,9$ divisibility of $n$ by $2,3,\dots,9$ depends?
I am not getting the fact why we introduce $S,T$
same question $n=a_m (100)^m+a_{m-1}(1000)^{m-1}+\dots+a_2 1000^2+a_1 1000+a_0$,
$0\le a_k\le 999$
| The presence of many subscripts can make something simple look not so simple. So we deal with a number like $N=a_4\cdot 10^4+a_3\cdot 10^3 +a_2\cdot 10^2 +a_1\cdot 10^1+a_0$, where the $a_i$ are digits. Let $S=a_4+a_3+a_2+a_1+a_0$.
We have
$$N-S=a_4\cdot 9999+a_3\cdot 999+a_2\cdot 99+a_1\cdot 9.\tag{1}$$
The right-hand side of (1) is divisible by $3$, Thus $N-S$ is divisible by $3$. It follows that if $N$ is a multiple of $3$, then so is $S$, and that if $S$ is a multiple of $3$, then so is $N$.
It is easier to find out quickly whether $S$ is a multiple of $3$ than to find out whether $N$ is a multiple of $3$, so finding $S$ is a useful way to determine whether $N$ is a multiple of $3$.
Exactly the same holds for divisibility by $9$. The right-hand side of (1) is divisible by $9$, so $N$ is divisible by $9$ if and only if $S$ is divisible by $9$.
We illustrated the idea with a general $5$-digit number. But the same technique always works, for the same reason: $10^k-1$ is always divisible by $9$,
The numbers $S$ and $T$, by themselves, are not enough to determine divisibility by any of $2$, $4$, $5$, $6$, $7$, or $8$.
The usefulness of the number $T$ is that $N$ is divisible by $11$ if and only if $T$ is. Once you are comfortable with $S$ and $3$ and $9$, please leave a message and I can try to explain what $T$ has to do with divisibility by $11$. The idea is quite similar to the idea we have used, just a bit more messy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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