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limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$.
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.
| If you are familiar with the Maclaurin series for the basic trig functions, you can solve this limit and others like it very quickly or even in your head. It also avoids differentiation mistakes from L'Hospital's.
In your case, second order expansion is enough. Taking
$$
\sin(x)\sim x-\frac{x^3}{6}\\
x\cos(x)\sim x(1-\frac{x^2}{2})=x-\frac{x^3}{2}
$$
Your limit becomes
$$
\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}=
\lim_{x\to 0} \frac{(x- \frac{x^3}{6})^2- (x-\frac{x^3}{2}) ^ 2}{x^4}
$$
And now you can notice that lower order terms terms dominate, and fortunately the $x^2$ cancel (or it would diverge), and pluck out the coefficients on $x^4$, which is $2/3$. If this is uncomfortable, you can always expand and check.
| {
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Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - 1) = 2(\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}})$$
And as I solved it, the equation came down to:
$2\operatorname{cosec}^{2}{B}+2\cot^{2}{B}.$
This can be written as $2(\operatorname{cosec}^{2}{B} + \cot^{2}{B}),$ which can further be written as $2\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}}.$
But I can't seem to get my mind around the middle part of the question, that is, $2(\operatorname{cosec}^{2}{B} - 1)$
Is it possible to write it like this, or is this an error in the question paper itself?
| Get rid of cosecants and cotangent:
$$
\frac{\csc B-\cot B}{\csc B+\cot B}=
\frac{\dfrac{1}{\sin B}-\dfrac{\cos B}{\sin B}}
{\dfrac{1}{\sin B}+\dfrac{\cos B}{\sin B}}=
\frac{1-\cos B}{1+\cos B}
$$
Thus your left-hand side is
$$
\frac{1-\cos B}{1+\cos B}+\frac{1+\cos B}{1-\cos B}=
\frac{(1-\cos B)^2+(1+\cos B)^2}{(1+\cos B)(1-\cos B)}=
\frac{2(1+\cos^2B)}{1-\cos^2B}
$$
and hence it equals the right-hand side.
On the other hand, the middle term is
$$
2(\csc^2B-1)=2\left(\frac{1}{\sin^2B}-1\right)=
2\frac{1-\sin^2B}{\sin^2B}=2\frac{\cos^2B}{1-\cos^2B}
$$
which is definitely not equal to the right-hand side.
Just apply it with $B=\pi/2$: the left-hand side and the right-hand side are both $2$, whereas the middle term is $0$.
The identity would be true if the middle term is
$$
2\csc^2B-1
$$
because
$$
2\csc^2B-1=\frac{2}{\sin^2B}-1=\frac{2-\sin^2B}{\sin^2B}
=\frac{2-1+\cos^2B}{1-\cos^2B}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polar to cartesian form of r=sin(3θ) $$\sin(2θ) = \sinθ\cosθ + \cosθ\sinθ = 2\sinθ\cosθ $$
$$\cos(2θ) = \cosθ\cosθ - \sinθ\sinθ = \cos²θ - \sin²θ $$
$$\sin(3θ) = \sin(2θ+θ) $$
$$= \sin(2θ)\cosθ + \cos(2θ)\sinθ $$
$$= 2\sinθ\cos²θ + \cos²θ\sinθ - \sin³θ
$$ $$= 2\sinθ(1-\sin²θ) + (1-\sin²θ)\sinθ - \sin³θ
$$ $$= 2\sinθ - 2\sin³θ + \sinθ - \sin³θ - \sin³θ
$$ $$r=\sin(3θ)= 3\sinθ - 4\sin³θ$$
How to deal with the differing exponents? I could add an $r$ to both sides or $r^3$ but not both.
$$r^4= 3r^3\sinθ - 4r^3\sin^{3}θ $$
$$r^4= 3r^2y - 4y^3$$
$$(x^2+y^2)^{4/2}= 3(x^2+y^2)^{2/2}y - 4y^3$$
$$(x^2+y^2)^{2}= 3(x^2+y^2)y - 4y^3$$
$$(x^2+y^2)^{2}= 3yx^2-y^3$$
| Hint
$$\sin3\theta=3\sin(\theta)\cos^2(\theta)-\sin^3(\theta)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$ What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$?
My effort:
We have $x+y-1\lt \lfloor x+y\rfloor \leq x+y$. What can I do after this? Please help
| As by definition, $\,\lfloor x\rfloor\le x <\lfloor x\rfloor+1$, we deduce
$$\lfloor x\rfloor+\lfloor y\rfloor\le x+y<\lfloor x\rfloor+\lfloor y\rfloor+2, $$
whence
$$\lfloor x\rfloor+\lfloor y\rfloor\le \lfloor x+y\rfloor\le\lfloor x\rfloor+\lfloor y\rfloor+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now we apply this to cosines in limit
$$\cos 2x = 1 - 2x^2 + O(x^4)$$
$$\cos 3x = 1 - \frac{9x^2}{2} + O(x^4)$$
Then use equality
$$ (1 + x)^n = 1 + xn + o(x)$$
It yields us
$$\lim_{x\to 0} \frac{1 - (1 - \frac{x^2}{2})(1 - x^2)(1 - \frac{3x^2}{2})}{x^2}$$
After simplification 1s cancel and we get $-(-1/2 - 1 - 3/2) = 3$.
| More generally, for any integer $N\geq 1$,
$$\lim_{x\to 0}\frac{1-\prod_{n=1}^N \sqrt[n]{\cos(nx)}}{x^2}=\frac{N(N+1)}{4},$$
because
$$\prod_{n=1}^N \sqrt[n]{\cos(nx)}=\prod_{n=1}^N \sqrt[n]{1+\frac{(nx)^2}{2}+o(x^2)}=\prod_{n=1}^N \left(1+\frac{(nx)^2}{n2}+o(x^2)\right)\\
=\prod_{n=1}^N \left(1+\frac{nx^2}{2}+o(x^2)\right)=
1+\frac{x^2}{2}\sum_{n=1}^N n+o(x^2)=1+\frac{N(N+1)}{4}\cdot x^2+o(x^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square of both sides and I tried to factor and failed.
| \begin{align}
& \sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=.25 \\
& \sqrt{3.5^2+x^2}=.25+\sqrt{3.0^2+x^2} \\
\end{align}
Next, square both sides. You get:
\begin{align}
& 3.5^2+x^2=(.25+\sqrt{3.0^2+x^2})^2 \\
3.5^2+x^2=.0625+.50(\sqrt{3.0^2+x^2})+3.0^2+x^2
\end{align}
Simplifying further gets you: $3.1875=.50(\sqrt{3.0^2+x^2})$, $6.375=\sqrt{3.0^2+x^2}$.
$x≈±5.62$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Question about Diophantine equations? Solve the Diophantine equation.
$6x+ 9y = 15$
We can see that $x = 1$ and $y=1$ solves the equation, so all solutions are given by $x = 1 + 3n$ and $y = 1 - 2n, n \in \mathbb{Z}$. By using:
$$x=x_0 + n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 - n \times \frac{a}{gcd(a,b)}$$
However, solving it as my textbook says I get:
I get that $gcd(9,6)=3$
$9 = 1 \times 6 + 3$
$6 = 2 \times 3 + 0 $
Then,
$3 = 9 - (1\times 6)$
So,
$3=6(-1) + 9(1) \implies 15 = 6(-5) + 9(5)$
All solutions are given by (as it is stated in my textbook) $$x=x_0 - n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 + n \times \frac{a}{gcd(a,b)}$$
Thus,
$x = - 5 - 3n$ and $y = 5 + 2n, n \in \mathbb{Z}$
My textbook give the answer as $ x=1+3n, y= 1-2n$, how did they arrive at that by using $$x=x_0 - n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 + n \times \frac{a}{gcd(a,b)}$$
and $$3=6(-1) + 9(1)?$$
All I get is $x=-1 -3n, y= 1 + 2n,n \in \mathbb{Z}$ which gives $3$ and not $15$, or $x = - 5 - 3n, y = 5 + 2n, n \in \mathbb{Z}$ which gives $15$ but looks ugly. What am I missing?
| Given $6x+9y=15\implies 2x+3y=5$
\begin{equation}
x = \frac{(5 - 3 y)}{2}\quad\land\quad y =\frac{(5 - 2 x)}{3}
\end{equation}
A pattern in a spreadsheet suggests $\quad x = (3 n + 1), n\in\mathbb{Z}\quad$
and, by substitution we get
\begin{equation}
y =\frac{(5 - 2 (3n-1)}{3}=\frac{(5 - 6n+1)}{3}=\frac{(6 - 6n)}{3}=(1-2n)
\end{equation}
$$\therefore\quad x = (3 n + 1),\quad y = (1 - 2 n), \quad n \in\mathbb{Z}$$
$$(x,y)\in\{\cdots, (-5,5), (-2,3),(1,1),(4,-1),(7,-3),\cdots\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| Hint:
$$\sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
$$\cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Conjecture $\sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}=\frac{1}{12\pi^2}$ Wolfram Alpha numerical calculation shows that the quantity
$$
\sum_{n=1}^\infty\frac{12\pi^2n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}
$$
is 1 with high accuracy. Can anybody prove the resulting conjecture:
$$
\sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}\overset{?}{=}\frac{1}{12\pi^2}
$$
| Take the function
$$
f(z) = \frac{z^2}{\sin (\pi z) \left(\cosh \left(\sqrt{3} \pi z\right)-\cos (\pi z) \right)}.
$$
Observe that $\sin (\pi z) \left(\cosh \left(\sqrt{3} \pi z\right)-\cos (\pi z) \right)=2\sin (\pi z) \sin (\pi z \omega) \sin (\pi z\omega^2) $, where $\omega=e^{2\pi i/3}$. This means that $f(z)dz=\frac{z^3}{\sin (\pi z) \left(\cosh \left(\sqrt{3} \pi z\right)-\cos (\pi z) \right)}\frac{dz}{z}$ is symmetric under $z\to z\omega$.
The rest is a routine procedure in application of residue theorem. Consider a closed contour $C$ composed of 2 rays, $z=x$, and $z=x\omega$ $(x>0)$ closed by a circle of large radius $R$. The poles inside the contour are $z=ne^{\pi i/3}$. Note that the integral along the circle vanishes in the limit $R\to\infty$, and the integrals along straight lines cancel each other out. Now one carefully calculates the contribution of residues at $z=n$ and $z=n\omega$ and the contribution of the residue at $z=0$:
$$
\lim_{R\to\infty}\int_Cf(z)dz=-\pi i~\sum_{n=1}^\infty\underset{z=n}{\text{res}}f(z)\cdot 2-\frac{2\pi i}{3}\underset{z=0}{\text{res}}f(z)=2\pi i~\sum_{n=1}^\infty\underset{z=ne^{\pi i/3}}{\text{res}}f(z).
$$
As a result
$$
2\sum_{n=1}^\infty\frac{n^2}{\pi(-1)^n \left(\cosh(\pi n\sqrt{3})-(-1)^n\right)}=-\frac{1}{3}\frac{1}{2\pi^3}
$$
Finally
$$
\sum_{n=1}^\infty\frac{n^2}{1-(-1)^n \cosh(\pi n\sqrt{3})}{=}\frac{1}{12\pi^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
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How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ I have prove this inequality
$$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$
because
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\dfrac{2}{\sqrt{k}+\sqrt{k+1}}=2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=2\sqrt{n+1}-2$$
so
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>2\sqrt{n+1}-2$$
Now I conjecture：
if postive arithmetic progression $\{a_{n}\}$ ,and the common difference of successive members is $d>0$.have
$$\dfrac{1}{\sqrt{a_{1}}}+\dfrac{1}{\sqrt{a_{2}}}+\cdots+\dfrac{1}{\sqrt{a_{n}}}\ge\dfrac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$$
| Note that
$$
\begin{align}
\frac1n\sum_{k=1}^n\frac2{\sqrt{a+kd}+\sqrt{a+(k-1)d}}
&=\frac1n\sum_{k=1}^n\frac2d\left(\sqrt{a+kd}-\sqrt{a+(k-1)d}\right)\\
&=\frac2{nd}\left(\sqrt{a+nd}-\sqrt{a}\right)\\[4pt]
&=\frac2{\sqrt{a+nd}+\sqrt{a}}\tag{1}
\end{align}
$$
Let $a_k=\sqrt{a+kd}$.
$$
\begin{align}
\frac1{n+1}\sum_{k=0}^n\frac1{a_k}
&=\frac1{n+1}\left[\frac12\left(\frac1{a_n}+\frac1{a_0}\right)+\sum_{k=1}^n\frac12\left(\frac1{a_k}+\frac1{a_{k-1}}\right)\right]\tag{2}\\
&\ge\frac1{n+1}\left[\frac2{a_n+a_0}+\sum_{k=1}^n\frac2{a_k+a_{k-1}}\right]\tag{3}\\
&=\frac2{a_n+a_0}\tag{4}
\end{align}
$$
Explanation:
$(2)$: algebra
$(3)$: harmonic mean is less than the arithmetic mean
$(4)$: apply $(1)$
Inequality $(4)$ is the inequality in question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\cos^n(x) - \sin^n(x)=1$
Solve the equation $\cos^n(x) - \sin^n(x)=1,n \in \mathbb{N}-\{0\}$
If $n$ is even then $\cos^n(x) = \sin^n(x)+1$ is only possible if $\sin(x)=0$ therefore the solution is $x=k\pi, k \in \mathbb{Z}$.
I'm having problems with $n$ odd case.
UPDATE
For $n=1$ we have $\cos(x) - \sin(x)=1$ and by squaring we get $sin(x)cos(x)=0$ which leads to $x=k\pi$ or $x=\pm \frac \pi 2 + k\pi$. From these solutions, only $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are valid.
Also $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are solutions for all odd $n$
| It suffices to investigate the solutions over $[0,2\pi)$, as all solutions will be some integer multiple of $2\pi$ away from a solution in $[0,2\pi)$. Let $f(x) = \cos^n(x)-\sin^n(x)$, with $n$ odd and $n\ge 3$. Then
$$f'(x) = n(\cos^{n-1}(x)(-\sin x) - \sin^{n-1}(x)\cos(x)) = -n\sin(x)\cos(x)(\cos^{n-2}(x)+\sin^{n-2}(x)) $$
It can easily be seen that $f'$ has zeros at $0$, $\frac{\pi}{2}$, $\frac{3\pi}{4}$, $\pi$, $\frac{3\pi}{2}$, and $\frac{7\pi}{4}$, with $f'<0$ on $(0,\frac{\pi}{2})\cup(\frac{3\pi}{4},\pi)\cup(\frac{3\pi}{2},\frac{7\pi}{4})$ and $f'>0$ on $(\frac{\pi}{2},\frac{3\pi}{4})\cup(\pi,\frac{3\pi}{2})\cup(\frac{7\pi}{4},2\pi)$.
Now
\begin{align} f(0) &= 1 \\
f(\frac{\pi}{2}) &= -1 \\
f(\frac{3\pi}{4}) &= -2\left(\frac{1}{\sqrt{2}}\right)^n \\
f(\pi) &= -1 \\
f(\frac{3\pi}{2}) &= 1 \\
f(\frac{7\pi}{4}) &= 2\left(\frac{1}{\sqrt{2}}\right)^n
\end{align}
From this, we see that $0$ and $\frac{3\pi}{2}$ are solutions. Furthermore, since $f'<0$ on $(0,\frac{\pi}{2})$, we have that $f(x) < f(0) = 1$ for $x\in(0,\frac{\pi}{2})$. Similarly, since $f'>0$ on $(\frac{\pi}{2},\frac{3\pi}{4})$, we have that $f(x)<f(\frac{3\pi}{4}) = -2\left(\frac{1}{\sqrt{2}}\right)^n < 1 $ for $x\in(\frac{\pi}{2},\frac{3\pi}{4})$. Applying similar reasoning to the other four intervals, we find that $f(x)<1$ for all $x\in[0,2\pi)$ except $x=0$ and $x=\frac{3\pi}{2}$.
It follows that the only solutions in $[0,2\pi)$ are $x=0$ and $x=\frac{3\pi}{2}$, so the only solutions in $\mathbb{R}$ are $\boxed{x = 2k\pi}$ and $\boxed{x = \frac{3\pi}{2} + 2k\pi}$ for $k\in\mathbb{Z}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new and probably better function to work with: $$\lim_{x\to \infty}\frac{9x}{\sqrt{x^2 + 4x} + \sqrt{x^2-5x}}$$
| If you know Differential Calculus, you may also use L'hopital's Rule to solve for the limit. Just as a recap, one instance of L'hopital's Rule states:
Given $\frac{f(x)}{g(x)}$ with some arbitrary limit to $c$ imposed on it, if all of the following constraints are met:
$\lim_{x\rightarrow{c}} f(x) = 0 $
$\lim_{x\rightarrow{c}} g(x) = 0$
$\lim_{x\rightarrow{c}}\frac{f'(x)}{g'(x)}=L$
then
$\lim_{x\rightarrow{c}}\frac{f(x)}{g(x)} = L$
Now, we have
$$\lim_{x\rightarrow\infty}\sqrt{x^2+4x}-\sqrt{x^2-5x}$$
which is obviously not in the form of $\frac{f(x)}{g(x)}$ as needed. However, we can put it into one, like so, as long as $x>0$:
$$\lim_{x\rightarrow\infty}\sqrt{x^2+4x}-\sqrt{x^2-5x} = $$
$$\lim_{x\rightarrow\infty}x\Bigg(\sqrt{1+\frac{4}{x}}-\sqrt{1-\frac{5}{x}}\Bigg)=$$
$$\lim_{x\rightarrow\infty}\frac{\sqrt{1+\frac{4}{x}}-\sqrt{1-\frac{5}{x}}}{x^{-1}} = \frac{0}{0}$$
Now we have the form we want and we found the limit, which is in the indeterminite form $\frac{0}{0}$. We also have our two functions, which are
$$f(x)=\sqrt{1+\frac{4}{x}}-\sqrt{1-\frac{5}{x}}$$
$$g(x)=x^{-1}=\frac{1}{x}$$
Now we have to do find the derivative of $f(x)$ and $g(x)$, which are:
$$f'(x)=\frac{d}{dx}\bigg(1+\frac{4}{x}\bigg)^{0.5}\frac{d}{dx}\bigg(1+\frac{4}{x}\bigg)-\frac{d}{dx}\bigg(1-\frac{5}{x}\bigg)^{0.5}\frac{d}{dx}\bigg(1-\frac{5}{x}\bigg)=$$
$$-\frac{4}{2x^{2}\sqrt{1+\frac{4}{x}}}-\frac{5}{2x^{2}\sqrt{1-\frac{5}{x}}}=$$
$$\frac{-4\sqrt{1-\frac{5}{x}}-5\sqrt{1+\frac{4}{x}}}{2x^{2}\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}$$
$$g'(x)=-x^{-2}=-\frac{1}{x^2}$$
And finally
$$\lim_{x\rightarrow\infty}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow\infty}\frac{\Bigg(\frac{-4\sqrt{1-\frac{5}{x}}-5\sqrt{1+\frac{4}{x}}}{2x^{2}\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}\Bigg)}{-\frac{1}{x^2}}=$$
$$\lim_{x\rightarrow\infty}x^{2}\Bigg(\frac{4\sqrt{1-\frac{5}{x}}+5\sqrt{1+\frac{4}{x}}}{2x^{2}\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}\Bigg)=$$
$$\lim_{x\rightarrow\infty}\frac{4\sqrt{1-\frac{5}{x}}+5\sqrt{1+\frac{4}{x}}}{2\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}=$$
$$\frac{4\sqrt{1}+5\sqrt{1}}{2\sqrt{1}\sqrt{1}}=\frac{9}{2}$$
We have our answer based on L'hopital's Rule. I know this was not required, as there is a much easier algebraic manner to solve for the limit, but I thought it was nice trying to find the limit this way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
2 Related Questions About Finding A Closed Form
*
*Consider the sequence defined by
$$
\begin{cases}
s_0=0\\
s_1=3\\
s_n=6s_{n-1}-9s_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $s_n$.
*Consider the sequence defined by
$$
\begin{cases}
t_0=5\\
t_1=9\\
t_n=6t_{n-1}-9t_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $t_n$.
I am having trouble with these questions about closed forms, could someone walk me step by step through each problem? Thanks!
| Hint:
For you first problem you could rewrite it as a matrix equation
\begin{align}
\begin{pmatrix}
s_n\\
s_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
6s_{n-1}-9s_{n-2}\\
s_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
6 &-9\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
s_{n-1}\\
s_{n-2}
\end{pmatrix}
=
\begin{pmatrix}
6 &-9\\
1 & 0
\end{pmatrix}^{n-1}
\begin{pmatrix}
s_{1}\\
s_{0}
\end{pmatrix}.
\end{align}
Thus, to find the closed form of $s_n$ it suffices to compute the $n-1$ power of your matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative vanishes at $x=0$ and $x=2\pi$.
| A simple way:
$$ I=\int_{0}^{2\pi}\frac{dx}{5-4\cos x}=\int_{0}^{\pi}\left(\frac{1}{5-4\cos x}+\frac{1}{5+4\cos x}\right)\,dx\tag{1}$$
$$ I = 10\int_{0}^{\pi}\frac{dx}{25-16\cos^2 x}=20\int_{0}^{\pi/2}\frac{dx}{25-16\cos^2 x} \tag{2}$$
and now, by setting $x=\arctan t$,
$$ I = 20 \int_{0}^{+\infty}\frac{dt}{25(1+t^2)-16} = 20 \int_{0}^{+\infty}\frac{dt}{9+25 t^2}\tag{3}$$
so by setting $t=\frac{3}{5}u$,
$$ I = 20\cdot\frac{3}{5}\cdot\frac{1}{9}\int_{0}^{+\infty}\frac{du}{1+u^2}=\frac{4}{3}\cdot\frac{\pi}{2}=\color{red}{\frac{2\pi}{3}}.\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Nth power of the following matrix The matrix is
$ \begin{pmatrix}0&1\\-1&0\end{pmatrix}^n $
for $n=2 \implies \left(\begin{matrix}-1 & 0\\ 0 &-1\end{matrix}\right)$ for $n=3 \implies \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for $n=4 \implies \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$
so I assume that for every $n=2k $, where k is a natural number and bigger than $0$ the matrix will be
$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ and for every $n=2k+1$ where k is a natural number and bigger than $0 $the matrix will be $\begin{pmatrix}1&0\\0&1\end{pmatrix}$
How can I prove it? probably with induction and how can I get easily the inverses of the matrices?
| You just have to calculate the first power:
$$A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}\;,\;\;A^2=\begin{pmatrix}\!-1&0\\0&-1\end{pmatrix}=-I$$
and that's all we need, since then
$$A^3=A\cdot A^2=-A\;,\;\;A^4=I\;,\;\;A^5=A;,\ldots\text{etc.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Arrangements of a,a,a,b,b,b,c,c,c in which no three consecutive letters are the same Q: How many arrangements of a,a,a,b,b,b,c,c,c are there such that
$\hspace{5mm}$ (i). no three consecutive letters are the same?
$\hspace{5mm}$ (ii). no two consecutive letters are the same?
A:(i). 1314. ${\hspace{5mm}}$ (ii). 174.
I thought of using the General Principle of Inclusion and Exclusion along with letting $p_i$ denote a property that.. , and doing so, I will evaluate $E(0)$, which gives us the number of arrangements without any of the properties.
How do I go about doing that? I am trying to use the method stated above in solving this question, but I am unable to generalize the properties $p_i$.
| This answer is based upon a generating function of generalized Laguerre polynomials
\begin{align*}
L_k^{(\alpha)}(t)=\sum_{i=0}^k(-1)^k\binom{k+\alpha}{k-i}\frac{t^i}{i!}
\end{align*}
The Laguerre polynomials have some remarkable combinatorial properties and one of them is precisely suited to answer problems of this kind. This is nicely presented in Counting words with Laguerre series by Jair Taylor. We find in section $3$ of this paper:
Theorem: If $m_1,\ldots,m_k,n_1,\ldots,n_k$ are non-negative integers, and $p_{m,n}(t)$ are polynomials defined by
\begin{align*}
\sum_{n=0}^\infty p_{m,n}(t)x^n=\exp\left(\frac{t\left(x-x^m\right)}{1-x^m}\right)
\end{align*}
then the total number of $k$-ary words that use the letter $i$ exactly $n_i$ times and do not contain the subwords $i^{m_i}$ is
\begin{align*}
\int_0^\infty e^{-t}\prod_{j=1}^k p_{m_j,n_j}(t)\,dt
\end{align*}
Here we consider a $3$-ary alphabet $\{a,b,c\}$ and words built from the characters $$a,a,a,b,b,b,c,c,c$$
First case: Bad words $\{aaa,bbb,ccc\}$
We have $m_1=m_2=m_3=3, n_1=n_2=n_3=3$
We obtain with some help of Wolfram Alpha
\begin{align*}
p_{3,3}(t)&=[x^3]\exp\left(\frac{t\left(x-x^3\right)}{1-x^3}\right)\\
&=[x^3]\left(1+tx+\frac{1}{2}t^2x^2+\left(\frac{1}{6}t^3-t\right)x^3+\cdots\right)\\
&=\frac{1}{6}t^3-t
\end{align*}
It follows
\begin{align*}
\int_0^\infty e^{-t}\left(\frac{1}{6}t^3-t\right)^3\,dt=1314
\end{align*}
Second case: Bad words $\{aa,bb,cc\}$
We have $m_1=m_2=m_3=2, n_1=n_2=n_3=3$ and obtain
\begin{align*}
p_{2,3}(t)&=[x^3]\exp\left(\frac{t\left(x-x^2\right)}{1-x^2}\right)\\
&=[x^3]\left(1+tx+\left(\frac{1}{2}t^2-t\right)x^2+\left(\frac{1}{6}t^3-t^2+t\right)x^3+\cdots\right)\\
&=\frac{1}{6}t^3-t^2+t
\end{align*}
It follows
\begin{align*}
\int_0^\infty e^{-t}\left(\frac{1}{6}t^3-t^2+t\right)^3\, dt=174
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
What time are the minute and hour hands of a clock perpendicular?
At noon the minute and hour hands of a clock coincide. Assuming the hands of the clock move continuously:
a) What is the first time $T_1$ when they are perpendicular?
b) What is the next time $T_2$ when they again coincide?
I believe I have solved this problem, but I would like someone to verify my answer.
I got $T_1$ to be 12:16 P.M. and 21.818181... seconds ($\frac{180}{11}$ minutes after noon).
I got $T_2$ to be 1:05 P.M. and 27.272727... seconds ($\frac{720}{11}$ minutes after noon).
If there are any questions relating to this problem, please feel free to ask them, as I will do my best to answer them.
| At noon the hands coincide. In $15 + \delta$ minutes, the angle between them will be $90^\circ$.
The minute hand moves at $\dfrac{360^\circ}{60\ \text{min}} = 6$ degrees per minute. The hour hand moves at $\dfrac{360^\circ}{12\ \text{hr}} = \dfrac 12$ degrees per minute. So we need to solve
\begin{align}
(15+\delta)\left(6 - \frac 12 \right) &= 90 \\
15 + \delta &= \frac{180}{11} \\
\delta &= \frac{15}{11} \text{min}
\end{align}
So $T_1 = 12:15 + 0:01\frac{4}{11} = 12:16\frac{4}{11}$.
The hands of the clock will be $90^\circ$ apart every $1 + \epsilon$ hours after that, and they will return to their original position after $11$ changes.
\begin{align}
11(1 + \epsilon) &= 12 \\
1 + \epsilon &= \frac{12}{11} \\
\epsilon &= \frac{1}{11} \text{hr} \\
\epsilon &= 5\frac{5}{11} \text{min}
\end{align}
So
$T_2 = 12:16\frac{4}{11} + 1:05\frac{5}{11} = 1:05\frac{9}{11}$
$T_3 = 1:05\frac{9}{11} + 1:05\frac{5}{11} = 2:11\frac{3}{11}$
$\vdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving $\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}$ without L'Hospital or Sandwiching Using the L'Hospital I got: $\lim\limits_{x \to 0} \frac{\ln(x+1)}{\ln(1+4x+x^2)} = \lim\limits_{x \to 0}\frac{\frac{1}{1+x}}{\frac{4+2x}{1+4x+x^2}}= \frac{1}{4}$,
I then wondered, if I could sandwich it andobserved that for: $\frac{x-1}{x} \le \ln(x) \le x-1 $, that:$ \frac{1}{4} \xleftarrow{x \to 0} \frac{\frac{x+1-1}{x+1 }}{1+4x+x^2-1} \le \frac{\ln(x+1)}{\ln(1+4x+x^2)} \le \frac{x+1-1}{\frac{1+4x+x^2-1}{1+4x+x^2}} \xrightarrow{x \to 0}\frac{1}{4} $
I then however wondered, if it were directly possible to get the result by cleverly manipulating the variables, so that the term simplifies itself to $\frac{1}{4}$ in the limit. Simply substituting $x+1$ or $1+4x+x^2$ with $e^u$ seems rather cumbersone, if (!) it is constructive at all. I am afraid, that there is just no real nice simplification of above. But maybe someone knows another nice, elegant way to get this limit :)
I am always happy to expand and practice my toolkit.
As always thanks in advance.
| Hint:
You can evaluate it based on this limit, $$\lim_{x\to0}\frac{\ln(x+1)}x=1,$$ by writing,
$$\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}=\lim\limits_{x \to 0} \frac{\ln(1+x)}{x}\cdot\frac{4x+x^2}{\ln(1+4x+x^2)}\cdot\frac{x}{4x+x^2}=\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are real numbers then $(x-y-z)^2 \geqslant 0$. That is $x^2 + y^2 +z^2 -2xy-2xz-2yz \geqslant 0$. But this implies that $\frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$, so because $x^2 + y^2 +z^2 \geqslant \frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$ we get the desired result.
| Your concerns were already addressed in the comments, so there is no need to keep underlying your actual sign mistake. But on a meta-mathematical point of view, something should have struck you, since your method would have derived a symmetric inequality from a non-symmetric one, kind of strange.
A working and fast proof:
$$\color{red}{0\leq} (x-y)^2+(x-z)^2+(y-z)^2 = 2\color{red}{\left(x^2+y^2+z^2-xy-xz-yz\right)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
| If you expand the squares, you'll see that every column is a linear combination of
$$
\begin{bmatrix}a^2 \\ b^2 \\ c^2 \\ d^2\end{bmatrix},
\begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}, \text{ and }
\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}.
$$
There are four columns that are linear combinations of three vectors. So they can't be linearly independent, thus the determinant is $0$.
(Or: The column space of the matrix is spanned by $3$ vectors. So the rank of the matrix is at most $3$, which means the matrix is singular. So the determinant is $0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 2
} |
If $x*y$ is the operation $x*y = xy/(x+y+1)$, then how does $x*y*z = xyz(x+y+1)/(x+y+z+xy+xz+1)$? $$x*y = \frac{xy}{x+y+1}$$
What I got was,
$$x*(y*z) = x* \frac{yz}{y+z+1} =x\frac{\frac{yz}{y+z+1}}{x+\frac{yz}{y+z+1}+1}$$
The book's answer is
$$x*(y*z) = x*\frac{yz}{y+z+1} = \frac{xyz(y+z+1)}{x+y+z+xy+xz+1}$$
| You can go on from where you stopped:
$$
\frac{x\dfrac{yz}{y+z+1}}{x+\dfrac{yz}{y+z+1}+1}=
\frac{\dfrac{xyz}{y+z+1}}{\dfrac{(x+1)(y+z+1)+yz}{y+z+1}}=
\frac{xyz}{(x+1)(y+z+1)+yz}
$$
Thus you get
$$
x*(y*z)=\frac{xyz}{x+y+z+xy+yz+zx+1}
$$
The book's answer is wrong. Set $x=y=1$. Then
$$
y*z=\frac{1}{3}
$$
and so
$$
x*(y*z)=1*\frac{1}{3}=\frac{\dfrac{1}{3}}{1+\dfrac{1}{3}+1}=\frac{1}{7}
$$
The book's formula would give $1/2$, while
$$
\frac{xyz}{x+y+z+xy+yz+zx+1}=\frac{1}{7}
$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Proof $\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$
$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$
I've tried converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual convergence value. Help and hints are both appreciated
| $$
\begin{aligned}
& S_{0} = \sum_{n=0}^{\infty} \frac{n^{0}}{4^{n}} = \sum_{n=0}^{\infty} \frac{1}{4^{n}} = \sum_{n=0}^{\infty} (1/4)^{n} = \frac{1}{1 - (1/4)} \Rightarrow \color{red}{S_{0} = \frac{4}{3}} \\ \\
& S_{1} = \sum_{n=0}^{\infty} \frac{n^{1}}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1 - 1}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1}{4^{n}} - \sum_{n=0}^{\infty} \frac{1}{4^{n}} = 4 \sum_{n=0}^{\infty} \frac{n + 1}{4^{n + 1}} - S_{0} \\
& \qquad = 4 \sum_{n=1}^{\infty} \frac{n}{4^{n}} - S_{0} = 4 \left[ - 0 + \sum_{n=0}^{\infty} \frac{n}{4^{n}} \right] = 4 S_{1} - S_{0} \Rightarrow \color{red}{S_{1} = \frac{1}{3} S_{0} = \frac{4}{9}} \\ \\
& S_{2} = \sum_{n=0}^{\infty} \frac{n^{2}}{4^{n}} = \sum_{n=0}^{\infty} \frac{(n + 1)^{2} - 2 n - 1}{4^{n}} = 4 \sum_{n=0}^{\infty} \frac{(n + 1)^{2}}{4^{n+1}} - 2 \sum_{n=0}^{\infty} \frac{n}{4^{n}} - \sum_{n=0}^{\infty} \frac{1}{4^{n}} \\
& \qquad = 4 S_{2} - 2 S_{1} - S_{0} = 4 S_{2} - \frac{5}{3} S_{0} \Rightarrow \color{red}{S_{2} = \frac{5}{9} S_{0} = \frac{20}{27}} \\ \\
& \sum_{n=0}^{\infty} \frac{n^{2} + 3 n + 2}{4^{n}} = S_{2} + 3 S_{1} + 2 S_{0} = \frac{20}{27} + \frac{12}{9} + \frac{8}{3} = \frac{20 + 36 + 72}{27} = \frac{128}{27} \\ \\
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
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Find all $0^\circ\leq A\leq360^\circ$ with $\tan A + \tan 2A + \tan 3A = 0$ solve:
$\tan A+ \tan 2A+ \tan 3A=0$
My Attempt:
$$\tan A+\tan 2A+\tan 3A=0$$
$$\tan A+\frac {2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
$$\frac {\tan A-\tan^3A+2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
What should I do further. Please suggest.
| $$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(2a) = 2 \cdot \frac{\tan(a)}{ (1 - \tan(a)^2) }$$
$$\tan(3a) =\tan(2a + a)$$
$$= \frac{\tan(a) + \tan(2a }{ 1 - \tan(a) \cdot \tan(2a)}$$
$$= \frac{(\tan(a) + \frac{2 \cdot \tan(a }{1 - \tan(a)^2)} }{ \frac{(1 - tan(a) \cdot 2 \cdot \tan(a) }{ 1 - \tan(a)^2}}$$
$$\frac{\tan(a) - \tan(a)^3 + 2\tan(a) }{(1 - \tan(a)^2 - 2\tan(a)^2}$$
$$ = \frac{3\tan(a) - \tan(a)^3}{ 1 - 3\tan(a)^2} $$
$$=\tan(a) \cdot \frac{3 - \tan(a)^2 }{ 1 - 3\tan(a)^2} $$
Given
$$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(a) + \frac{2\tan(a) }{ (1 - \tan(a)^2)} + \frac{\tan(a) \cdot (3 - \tan(a)^2) }{ (1 - 3\tan(a)^2)) } = 0$$
$$\tan(a) = 0 $$
$$a = \pi \cdot k $$
$$1 +\frac{2}{(1 -\tan(a)^2)}+\frac{(3 -\tan(a)^2) }{(1 - 3\tan(a)^2)} = 0 $$
$$\frac{((1 - \tan(a)^2) \cdot (1 - 3\tan(a)^2) + 2\cdot (1-3\tan(a)^2) + (3 -\tan(a)^2) \cdot (1 -\tan(a)^2)) }{ ((1 -\tan(a)^2)\cdot (1-3\tan(a)^2)) }= 0$$
$$(1 - \tan(a)^2) \cdot (1 - 3\tan(a)^2) + 2 \cdot (1 - 3\tan(a)^2) + (3 - \tan(a)^2) \cdot (1 - \tan(a)^2) = 0 $$
$$1 - 4\tan(a)^2 + 3\tan(a)^4 + 2 - 6\tan(a)^2 + 3 - 4\tan(a)^2 + \tan(a)^4=0 $$
$$6 - 14\tan(a)^2 + 4\tan(a)^4 = 0 $$
$$2\tan(a)^4 - 7\tan(a)^2 + 3 = 0 $$
$$\tan(a)^2 = \frac{(7 \pm \sqrt{(49 - 24}}{ 4 }$$
$$\tan(a)^2 = \frac{(7 \pm 5)}{4} $$
$$\tan(a)^2 = \frac{12}4 , \frac24 $$
$$\tan(a)^2 = 3 , \frac1 2 $$
$$\tan(a) = \pm \sqrt(3) , \pm \sqrt\frac{(2)}{ 2} $$
$$\tan(a) = 0, -\sqrt(3), \sqrt(3), -\sqrt\frac{(2)}{ 2}, \sqrt\frac{(2)}{ 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit to infinity using $\delta$-$\epsilon$ language. I'm trying to prove:
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2 + x - 1}{3x^2 + 2x + 1} = \frac{1}{3}$$
using $\delta$-$\epsilon$ language. Here is my attempt:
Proof: We wish to show that as $x$ gets arbitrarily large that the difference between $f(x)$ and $\frac{1}{3}$ gets arbitrarily small. In other words, we show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that $$x > \delta \Rightarrow \bigg| \frac{x^2 + x - 1}{3x^2 + 2x + 1} - \frac{1}{3} \bigg| < \epsilon$$
Now observe for $x > 2$,
\begin{align*}
\bigg| \frac{x^2 + x - 1}{3x^2 + 2x + 1} - \frac{1}{3} \bigg| &= \bigg| \frac{x-4}{9x^2 + 6x + 3} \bigg| \tag{$\star$} \\
&\leq \bigg| \frac{x-4}{3(x^2 + 2x + 1)} \bigg|\\
&= \bigg| \frac{x-4}{3(x + 1)^2} \bigg| \\
&= \bigg| \frac{1}{3(x+1)} - \frac{5}{3(x+1)^{2}} \bigg|\\
&< \frac{1}{x+1}
\end{align*}
Let $\delta = \max\left\{ 2, \dfrac{1}{\epsilon} - 1 \right\}$. Then, if $x > \delta$ we have that
$$\star < \dfrac{1}{x+1} < \frac{1}{\delta + 1} = \dfrac{1}{\left(\frac{1}{\epsilon} - 1 \right) + 1} = \epsilon$$
I just want to make sure that I did this correctly. Thank you.
| It's correct, but it can be simplified.
When you are at
$$
\left|\frac{x-4}{9x^2+6x+3}\right|
$$
you can observe that it's not restrictive to assume $x>4$; next, $x-4<x$ and $9x^2+6x+3>9x^2$, so
$$
\frac{x-4}{9x^2+6x+3}<\frac{x}{9x^2}<\frac{1}{9x}
$$
Therefore we can take
$$
\delta=\max\left\{4,\frac{1}{9\varepsilon}\right\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
If the value of $xy$ , If $ x^2 + y^2 = 34$ and $x + y =8$ If $ x^2 + y^2 = 34$ and $x + y =8$ find the value of xy
So what should I be considering here the sum of two squares or..
I'm struggling here & any help to figure this out will be kindly appreciated
| Use:
$$x+y=8\Longleftrightarrow x=8-y\Longleftrightarrow x^2=\left(8-y\right)^2$$
So:
$$x^2+y^2=34\Longleftrightarrow\left(8-y\right)^2+y^2=34$$
So, for $y$ we find $y=3$ or $y=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Finding features of parabolic formula I have a question that requires me to find the intercepts and the turning point of a parabola from it's formula, but I'm not really sure how to find all this information out from the equation below. I tried using completing the square but I ended up with a weird answer that I don't believe is right. The equation is:
$$y=-3x^2-5x+2$$
Thanks!!
| There are two intercepts. The $y$-intercept, and the $x$-intercepts.
$y$-intercept:
To find the $y$-intercept, you simply plug in $x=0$ into the equation to get the $y$ value. So thus, we have$$y=-3\cdot 0^2-5\cdot 0+2=2$$ so the $y$-intercept is $(0,2)$.
$x$-intercepts:
The $x$ intercepts are points that give $y=0$ when you plug in $y$. So plugging in $y=0$, we get $$-3x^2-5x+2=0\Leftrightarrow -(3x-1)(x+2)=0\implies x=\frac 13,x=-2$$
Therefore, the $x$-intercepts are $\left(\frac 13,0\right)$ and $(-2,0)$.
To find the Vertex, you can substitute $x$ with $\frac {-b}{2a}$ where given the quadratic of the form $ax^2+bx+c=0$.
In your example, the quadratic is $y=-3x^2-5x+2$ so we see that $a=-3,b=-5,c=2$. So we have $$-\frac b{2a}=-\frac 56\tag1$$
Substituting $x$ with $(1)$ gives the $y$ value:$$f\left(\frac 56\right)=-3\left(-\frac 56\right)^2+5\left(\frac 56\right)+2=\frac {49}{12}$$
And so the vertex lies on the point $\left(-\frac 56,\frac {49}{12}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is this inequality true? And can it be shown? If $x$ and $y$ are both positive, odd integers, greater than $1$, is it true that,
$(y-1) \left( \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) - \ln(4) \right) + \displaystyle\sum_{i=\frac{x+1}{2}}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) > 0$?
It seems to be true, but I am having difficulty showing it for all relevant $x, y >1$.
| By multiplying and assembling the sums
$(y-1) \left( \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) - \ln(4) \right) + \displaystyle\sum_{i=\frac{x+1}{2}}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) = $
$= \displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) - \sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right)- (y-1)\ln(4) >0 \implies $
$\implies \displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) > \sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right)+ (y-1)\ln(4).$
Since $\sum_{i=1}^{\infty} \left( \frac{1}{i(2i-1)} \right) = \ln(4),$ we have
$\displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) > \sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) + \sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y-1}{i(2i-1)} \right).$
Hence,
$\displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{1}{i(2i-1)} \right) > \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem regarding inequality There is a question in an algebra textbook that asks to prove $|x^2-1|\lt\frac{21}{100}$, given that $|x-1|\lt\frac{1}{10}$.
Is the question correct(solvable)? If it is, how?
| Suppose $|x - 1| < \frac{1}{10}$. Then
$$|x^2 - 1| = |(x-1)(x+1)| = |x-1|\cdot|x+1| < \frac{1}{10} |x+1|$$
We now need to show $|x+1| < \frac{21}{10}$ in order to show $|x^2 - 1| < \frac{21}{100}$.
Since $|x-1| < \frac{1}{10}$, this means
$$ -\frac{1}{10} < x-1 < \frac{1}{10}.$$
Adding 2 to each side gives us:
$$ -\frac{1}{10} + 2 < x-1+2 < \frac{1}{10} + 2.$$
We can do this without any issues since 2 is positive. Now, add everything together to get:
$$ \frac{19}{10} < x + 1 < \frac{21}{10}.$$
Since $-\frac{21}{10} < \frac{19}{10}$ and $\frac{19}{10} < x+1$, we have by the transitive property that ....
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hard problem I can not prove properly. I managed to concoct a weak proof for the below problem but I would love to see if there are any different ways to solve it.
Find the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3 .... +2^3-1^3$
the options listed are
a)$1^2$
b)$2^2$
c)$1007^2$
d)$2014^2$
I think the answer is c) but i just got that by calculating the last digit of the sum to check the divisibility. But I doubt that the proof is concrete. Help would be appreciated.
| Consider:
\begin{align}
A_{2014} &= \sum_{n=1}^{1007} [(2n)^3 - (2n-1)^3] = \sum_{n=1}^{1007} ((2n) -(2n-1))((2n)^2 + (2n)(2n-1) + (2n-1)^2) \\
&= \sum_{n=1}^{1007} (12n^2 - 6n + 1) \\
&= 4087631519 = 19^{2} \cdot 29^{1} \cdot 53^{2} \cdot 139 = 29 \cdot 139 \cdot (1007)^2.
\end{align}
In comparison: $2014^{2} = 2^2 \cdot 19^2 \cdot 53^2$, $1007^{2} = 19^2 \cdot 53^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Probability that number is divisible by 5 Two numbers $x$ and $y$ are chosen at random without replacement from the set $${1,2,\cdots,5n}$$.then the probability that $x^4-y^4$ is divisible by $5$ is----
I could write down $x^4-y^4$ as $(x-y)(x+y)(x^2+y^2)$ and then i could easily figure out the possoble cases for first and second brackets but i couldnot figure out for the third bracket.
Please help me in this regard.
Thanks.
| Note that if $x$ is not a multiple of $5$ then $x^4-1$ is divisible by $5$ (why?).
This implies that if $x$ and $y$ are not a multiple of $5$ then $x^4-y^4=(x^4-1)-(y^4-1)$ is divisible by $5$.
Can you take it from here?
P.S. Finally you will find that $x^4-y^4$ is NOT divisible by 5 iff the set $\{x,y\}$ ($5n(5n-1)/2$ choices) contains a multiple of $5$ ($n$ choices) and a not-multiple of $5$ ($4n$ choices). Hence the probability is
$$1-\frac{(n)\cdot (4n)}{\frac{5n(5n-1)}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
| Considering the problem geometrically, then
$$
f(x) = x^{\,2} - ax + 1
$$
is a vertical parabola, convex (upward concave), with axis $x=a/2$ and vertex in $V=(a/2,1-a^2/4)$.
Thus the parameter $a$ is then just a translation parameter.
So, from the above and with the help of the sketch it is clear that, to satisfy the requirements we shall have
$$
\eqalign{
& \exists \,x\;:\;\left\{ \matrix{
\left| {f(x)} \right| \le 1/4 \hfill \cr
\left| {f(x + 2)} \right| \le 1/4 \hfill \cr} \right.\quad \Leftrightarrow \cr
& \Leftrightarrow \quad \left\{ \matrix{
2 \le \overline {A_{\,1} A_{\,2} } \hfill \cr
- 1/4 \le V_{\,y} \le 1/4 \hfill \cr} \right.\quad \cup \quad \left\{ \matrix{
2 \le \overline {A_{\,1} A_{\,2} } \hfill \cr
V_{\,y} \le - 1/4 \hfill \cr
\overline {B_{\,1} B_{\,2} } \le 2 \hfill \cr} \right. \cr}
$$
which means
$$
\left\{ \matrix{
2 \le \overline {A_{\,1} A_{\,2} } \hfill \cr
\overline {B_{\,1} B_{\,2} } = \emptyset \quad \vee \quad \overline {B_{\,1} B_{\,2} } \le 2 \hfill \cr} \right.
$$
i.e.
$$
\left\{ \matrix{
2 \le \sqrt {a^{\,2} - 3} \hfill \cr
a^{\,2} - 5 \le 0\quad \vee \quad \sqrt {a^{\,2} - 5} \le 2 \hfill \cr} \right.
$$
and finally
$$ \bbox[lightyellow] {
\left\{ \matrix{
7 \le a^{\,2} \hfill \cr
a^{\,2} \le 9 \hfill \cr} \right.\quad \Rightarrow \quad \sqrt 7 \le \left| a \right| \le 3
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 3
} |
A Cauchy–Schwarz-like inequality involving dot products Let $\mathbf{a}_1,\ldots,\mathbf{a}_N$ and $\mathbf{b}_1,\ldots,\mathbf{b}_N$ be vectors in $\mathbb{R}^n$. Prove that:
\begin{equation}
\bigg(\sum_{i=1}^{N} \mathbf{a}_i^\top\mathbf{a}_i\bigg) \bigg(\sum_{i=1}^{N} \mathbf{b}_i^\top\mathbf{b}_i\bigg)
\geq
N \sum_{i=1}^N \bigg( (\mathbf{a}_i^\top\mathbf{a}_i) (\mathbf{b}_i^\top\mathbf{b}_i) - (\mathbf{a}_i^\top\mathbf{b}_i)^2 \bigg).
\end{equation}
| The inequality doesn't seem to be true. Consider the following example in $\mathbb{R}^2$
\begin{align}
\mathbf{a}_1 = 2\mathbf{e}_1,\ \ \mathbf{a}_2 = \mathbf{e}_2 \ \ \text{ and } \ \ \mathbf{b}_1 = 2\mathbf{e}_2,\ \ \mathbf{b}_2 = \mathbf{e}_1.
\end{align}
Observe
\begin{align}
(\mathbf{a}_1^T\mathbf{a}_1+\mathbf{a}_2^T\mathbf{a}_2)(\mathbf{b}_1^T\mathbf{b}_1+\mathbf{b}_2^T\mathbf{b}_2) = (4+1)(4+1)=25
\end{align}
and
\begin{align}
2(\mathbf{a}_1^T\mathbf{a}_1\mathbf{b}_1^T\mathbf{b}_1+\mathbf{a}_2^T\mathbf{a}_2\mathbf{b}_2^T\mathbf{b}_2)=2(16+1) = 34
\end{align}
which means the above inequality doesn't hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Internal angle bisectors and right-angled triangle? Imagine a right angled triangle. Two of its internal bisectors are of length $7cm$ and $4cm$, respectively. The first internal angle bisector belongs to the angle of the side $a$ and the second one belongs to the angle of the side $b$ Calculate the length of the hypotenuse of the right angled triangle.
| Here is a sketch of the triangle and its two internal bisectors $\overline{AP}=7$ , $\overline{BQ}=4$ :
We shall need a cosine law:
$$
\cos(A) = 2\cos^2(\frac{1}{2}A)-1 \quad ; \quad
\cos(B) = 2\cos^2(\frac{1}{2}B)-1
$$
Three highly non-linear equations with three unknowns can be derived then:
$$
2\cos^2(\frac{1}{2}A)-1 = 2\left(\frac{b}{7}\right)^2-1 = \frac{b}{c} = \cos(A) \\
2\cos^2(\frac{1}{2}B)-1 = 2\left(\frac{a}{4}\right)^2-1 = \frac{a}{c} = \cos(B) \\
c^2 = a^2 + b^2
$$
Since finding a solution by hand seems to be hopeless, I decided to feed this
into some computer algebra system (MAPLE) and solve anything numerically:
> eqns := {2*(b/7)^2-1=b/c,2*(a/4)^2-1=a/c,c^2=a^2+b^2};
> sols := {a,b,c}; fsolve(eqns,sols,0..infinity);
{ a = 3.402041954, b = 6.813171845, c = 7.615326654 }
Where $c$ is the length of the hypotenuse that is asked for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Find a line which is tangent to the curve $y=x^4-4x^3$ at 2 points How can I solve this? Should I set the two points as $a^4-4a^3$ and $b^4-4b^3$?
| Let $(t,t^4-4t^3)$ be a point of contact, then
$$\frac{dy}{dx} = \frac{\dot{x}}{\dot{y}} = 4t^3-12t^2$$
The equation of tangent is
$$(y-t^4+4t^3)=4t^2(t-3)(x-t)$$
Let $(s^4-4s^3,s)$ be the self-intersection of its tangent.
\begin{align*}
(s^4-4s^3)-(t^4-4t^3) &= 4t^2(t-3)(s-t) \\
(s^4-t^4)-4(s^3-t^3) &= 4t^2(t-3)(s-t) \\
(s^2-t^2)(s^2+t^2)-4(s-t)(s^2+st+t^2) &= 4t^2(t-3)(s-t) \\
(s-t)[(s+t)(s^2+t^2)-4(s^2+st+t^2)] &= 4t^2(t-3)(s-t) \\
(s-t)[(s+t)(s^2+t^2)-4(s^2+st+t^2)-4t^2(t-3)] &= 0 \\
(s-t)[s^3+(t-4)s^2+t(t-4)s+t^2(8-3t)] &= 0 \\
(s-t)^2[s^2+2(t-2)s+t(3t-8)] &= 0 \\
\end{align*}
The double root $s=t$ is due to tangency, thus another intersections are given by $$s=2-t\pm \sqrt{2(2+2t-t^2)}$$
provided $1-\sqrt{3} \le t \le 1+\sqrt{3}$.
The line will be a common tangent if $\Delta=0$, that is
$$(s,t)=(1\pm \sqrt{3},1\mp \sqrt{3})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find a number $\alpha > 1$ such that the following holds
We want to find a real number $\alpha > 1$ such that
$$ \int\limits_{0}^{\infty} \frac{ dx }{(1+x^{\alpha})^{\alpha}} = 1
$$
What I got so far
First, I used Integration by parts to obtain
$$ \int\limits_{0}^{\infty} \frac{ dx }{(1+x^{\alpha})^{\alpha}} = \underbrace{\frac{x}{(1+x^{\alpha})^{\alpha}} \bigg|_0^{\infty}}_{=0} + \int\limits_0^{\infty} \frac{ \alpha^2 x^{\alpha} }{(1+x^{\alpha})^{\alpha+1}} dx $$
Now, notice
$$ \int\limits_0^{\infty} \frac{ \alpha^2 (x^{\alpha} +1 -1)}{(1+x^{\alpha})^{\alpha+1}} dx = \int\limits_0^{\infty} \frac{ \alpha^2 }{(1+x^{\alpha})^{\alpha}} dx - \int\limits_0^{\infty} \frac{ \alpha^2 }{(1+x^{\alpha})^{\alpha+1}} dx $$
Therefore, for our original integral to be $1$, we must have
$$ \int\limits_0^{\infty} \frac{ \alpha^2 }{(1+x^{\alpha})^{\alpha}} dx =1+ \int\limits_0^{\infty} \frac{ \alpha^2 }{(1+x^{\alpha})^{\alpha+1}} dx $$
But, notice
$$ 1 = - \int\limits_0^{\infty} \frac{ \alpha x^{\alpha - 1} dx }{(1+ x^{\alpha} )^{\alpha}} $$
Thus, the above expression is equivalent to
$$ \int\limits_0^{\infty} \frac{\alpha^2+\alpha x^{\alpha-1}} {(1+x^{\alpha})^{\alpha}} = \int\limits_0^{\infty} \frac{ \alpha^2 }{(1+x^{\alpha})^{\alpha+1}} dx $$
Therefore,
$$ (1+x^{\alpha})^{\alpha}(\alpha^2 + \alpha x^{\alpha-1} ) = \alpha^2 $$
holds for all $x>0$. In particular, if $x=1$, we have
$$ 2(\alpha^2 + \alpha) = \alpha^2 $$
but this equation has no solution $\alpha>1$. What am I doing wrong?
| If $\alpha = \phi\textrm{( the golden ratio )}, \textrm{ then }\alpha^2-\alpha -1=0,$
$$
\begin{aligned}
\int_0^{\infty} \frac{1}{\left(1+x^\alpha\right)^\alpha} d x & \stackrel{x\mapsto\frac{1}{x}}{=} \int_{\infty}^0 \frac{1}{\left(1+\frac{1}{x^\alpha}\right)^\alpha} \frac{d x}{-x^2} \\
& =\int_0^{\infty} \frac{x^{\alpha^2-2}}{\left(x^\alpha+1\right)^\alpha} d x \\
& =\int_0^{\infty} \frac{x^{\alpha-1}}{\left(x^\alpha+1\right)^\alpha} d \alpha \\
& =\frac{1}{\alpha} \int_0^{\infty} \frac{d\left(x^\alpha+1\right)}{\left(x^\alpha+1\right)^\alpha} \\
& =\frac{1}{\alpha}\left[\frac{\left(x^\alpha+1\right)^{\alpha+1}}{-\alpha+1}\right]_0^{\infty} \\
& =\frac{1}{\alpha(1-\alpha)} {(-1)} \\
& =1
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute the determinant of circulant matrix with entries $\cos j\theta$ compute the determinant
$
\begin{vmatrix}
\cos \theta & \cos 2\theta & \cos 3\theta &\cdots &\cos n\theta \\
\cos n\theta & \cos \theta & \cos 2\theta & \cdots & \cos (n-1)\theta \\
\cdots \\
\cos 2\theta & \cos 3\theta & \cos 4\theta & \cdots & \cos \theta
\end{vmatrix}
$
I have tried to apply the formula for circulant matrix $\text{det} =\prod_{i=1}^n f(\omega _i)$ with $\omega _i$ the nth root of unity.This gives an ugly result which I failed to simplify. Should I try another way? Any help will be greatly appreciated on how to obtain a neater result.
By the way, wolfram alpha gives when $n=3$ it's $\cos ^3 \theta + \cos ^3 2\theta + \cos ^3 3\theta -3 \cos \theta \cos 2\theta \cos 3\theta$, and for $n=4$ it also can be factored to a nice form, so I think simplification is possible.
| let $ A = \pmatrix{\cos t & \cos 2t &\ldots & \cos nt\\
\cos nt & \cos t & \ldots &\cos(n-1)t\\\ldots & \ldots &\ldots &\ldots\\
\cos 2t & \cos 3t &\ldots & \cos t}
$ and $\omega$ the $n$ the root of unity. that is $\omega^n = 1.$ you can verify that
$\pmatrix{1 & \omega & \ldots & \omega^{n-1}}^\top$ is an eigenvector corresponding to the eigenvalue $\lambda_\omega = \cos t + \omega \cos 2t + \ldots + \omega^n \cos nt.$ therefore the determinant of $A$ is the product of of the $n$ eignevalues $\{\lambda_\omega : \omega^n = 1\}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving complex series is convergent Question 1: Test the following series for convergence
$$\sum_{n=0}^\infty \left(\frac{2+i}{3-4i}\right)^{2n}$$
I got that it was convergent by doing a ratio test and getting
$$p = \frac{\sqrt{3}}{5} < 1 $$
which checks out for convergence. Wondering if that was the correct way to approach solving this series.
Question 2: Same as Question 1
$$\sum_{n=0}^\infty \left(\frac{1+i}{2-i}\right)^{n}$$
I am unsure how to approach solving this question and any advice would be much appreciated.
| The first series is a geometric series of the form $\sum_{n=0}^{\infty} q^n$ with $q = \left( \frac{2 + i}{3 - 4i} \right)^2$. Since
$$ \left| \frac{2 + i}{3 - 4i} \right| = \left| \frac{2 + i}{3 - 4i} \frac{3 + 4i}{3 + 4i}\right| = \left| \frac{2 + 11i}{25} \right| = \frac{\sqrt{4 + 121}}{25} = \frac{1}{\sqrt{5}} $$
we have $|q| = \frac{1}{5} < 1$ and so the series converges.
Similarly, for the second series we have $q = \frac{1 + i}{2 - i}$ and
$$ |q| = \left| \frac{1 + i}{2 - i} \right| = \left| \frac{1 + i}{2 - i} \frac{2+i}{2+i} \right| = \left| \frac{1 + 3i}{5}\right| = \frac{\sqrt{1 + 9}}{5} = \sqrt{\frac{2}{5}} $$
and agan, the series converges.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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If $n > 1$ is an integer not of the form $6k + 3$, prove that $n^{2} + 2^{n}$ is composite. I try to divide in to $5$ case $6k, 6k+1, 6k+2, 6k+4, 6k+5$
and put them in $n^{2}$ and consider the possible last digit 0f $2^n$
and sum with those form
I can only prove $6k, 6k+2, 6k+4$ case .the last digit is even number and can divide by $2$.
But I can't prove $6k+1, 6k+5$ case because might be odd and not guarantee Ican divide by some numbers
Anyone can help me by give a hint ?
| The hint is $n > 1$. When $n = 1$, $n^2 + 2^n = 3$ which is prime. This hints that $n^2 + 2^n$ may be divisible by $3$.
We know that $6k+1$ can be written as $3(2k)+1$ and $6k+5$ can be written as $3(2k+1)+2$. Thus if $n$ is of the form $6k+1$ or $6k+5$, then $n^2 \equiv 1\pmod3$.
The next part is to prove that $2^n \equiv 2 \pmod 3$ if $n$ is of such form. Notice that $2^0 \equiv 1 \pmod 3$, $2^1 \equiv 2 \pmod 3$ and $2^2 \equiv 1 \pmod 3$. Thus for all $k \ge 0$, $2^{2k+1} \equiv 2 \pmod 3$. Since $6k+1$ can be written as $2(3k)+1$ and $6k+5$ can be written as $2(3k+2)+1$, $2^n \equiv 2 \pmod 3$.
Thus $n^2 + 2^n \equiv 1+2 \pmod 3 \equiv 0 \pmod 3$. This implies that $n^2 + 2^n$ is divisible by 3 if $n = 6k+1$ or $n = 6k+5$.
| {
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prove that if x is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor+\lfloor x + \frac{1}{2}\rfloor$ prove that if $x$ is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor + \lfloor x + \frac{1}{2}\rfloor$
I've seen the proof in my discrete math book, but it was very confusing. I don't understand why they considered those cases and why they chose $\frac{1}{2}$ specifically. It would be great if someone can explain better or provide a much clearer proof.
Proof from the book:
Let x = n + $\epsilon$ where n is an integer and 0 $\le\epsilon \lt 1$
case 1: Consider 0 $\le\epsilon \lt \frac{1}{2}$
2x = 2n + 2$\epsilon$ and $\lfloor$2x$\rfloor$ = 2n because 0 $\le 2\epsilon \lt 1$
Similarly, x + $\frac{1}{2}$= n + ( $\frac{1}{2}$+ $\epsilon$), so $\lfloor$x + $\frac{1}{2}$$\rfloor$ = n,
because 0 < $\frac{1}{2}$ + $\epsilon$ < 1.
Consequently,$\lfloor$2x$\rfloor$ = 2n and $\lfloor$x$\rfloor$ + $\lfloor$x + $\frac{1}{2}$$\rfloor$ = n + n = 2n.
case 2: $\frac{1}{2}$ ≤ $\epsilon$< 1.
2x = 2n + 2$\epsilon$ =(2n + 1) + (2$\epsilon$ − 1).
Because 0 ≤ 2$\epsilon$ − 1 < 1, it follows that $\lfloor$2x$\rfloor$ = 2n + 1. Because$\lfloor$x + $\frac{1}{2}$$\rfloor$ = $\lfloor$n + ( $\frac{1}{2}$+ $\epsilon$)$\rfloor$ = $\lfloor$n + 1 + ($\epsilon$ − $\frac{1}{2}$ )$\rfloor$ and 0 ≤ $\epsilon$ − $\frac{1}{2}$ < 1,
it follows that $\lfloor$x + $\frac{1}{2}$$\rfloor$ =n + 1.
Consequently, $\lfloor$2x$\rfloor$ = 2n + 1 and $\lfloor$x$\rfloor$ + $\lfloor$x + $\frac{1}{2}$$\rfloor$= n + (n + 1) = 2n + 1.
This concludes the proof.
| One has to distinguish these cases since in this example "something special" happens with the floor function when the variable $x$ passes an integer multiple of ${1\over2}$. Nevertheless it can be said that the proof in your book is much too complicated.
Instead begin with the remark that both sides of the stated equation increase by $2$ when $x$ increases by $1$. It is therefore sufficient to prove it for $0\leq x<1$.
If $0\leq x<{1\over2}$ then $x+{1\over2}<1$ and therefore $\lfloor2x\rfloor=0=\lfloor x\rfloor+\bigl\lfloor x+{1\over2}\bigr\rfloor$.
Similarly: If ${1\over2}\leq x<1$ then $\lfloor2x\rfloor=1=0+1=\lfloor x\rfloor+\bigl\lfloor x+{1\over2}\bigr\rfloor$.
| {
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Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get.
$$\begin{bmatrix} & & & \\ & & & \\ & & &\\& & & & \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$
| Preliminary: I assume it is clear that equation $AB=C'$ (where $C'$ is a column-shuffled version of $C$) is equivalent to 7 relationships between the 7 columns of $B$ and the corresponding 7 columns of $C'$:
$$A.\begin{pmatrix}2 \\-1\\ -4\\ -1\end{pmatrix}=\begin{pmatrix} . \\ .\\ . \\ . \end{pmatrix}, \ \ A.\begin{pmatrix}0 \\1\\ 0\\ 0\end{pmatrix}=\begin{pmatrix} . \\ .\\ . \\ . \end{pmatrix}, \ \ etc...$$
Here is a solution:
$$A=\pmatrix{-1& 0& 0& -1\\-1& 0& -1& 2\\1& 0& 1& -1\\-2&
1& -2& 3},$$
which is such that
$$\tag{1}\pmatrix{-1& 0& 0& -1\\-1& 0& -1& 2\\1& 0& 1& -1\\-2&
1& -2& 3}\cdot\begin{pmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{pmatrix}=\begin{pmatrix}-1 & 0 &1 & 1&2&3\\0 & 0 & 0 & 1& 0 & 1\\ -1& 0 & 0 & 0&2&0\\ 0&1&0&0&0&0\end{pmatrix}$$
One sees that the last matrix is matrix $C$ with its columns shuffled.
The way I have obtained matrix $A$ is in 3 steps:
*
*a) by observing first that, in matrix $B$ as well as in matrix $C$, there are 2 column vectors (say $U$ and $-2U$ in $B$, and $V$ and $-2V$ in $C$) that are proportional with the same proportionality coefficient, namely columns 1 and 5 in $B$, and columns $4$ and $6$ in $C$. They have to be mapped the ones onto the others.
*b) by reducing, in a second step, matrices $B$ and $C$ to $4 \times 4$ matrices denoted $B_1$, resp. $C_1$ by removing, in each one, the 2 columns that have been found in a). I was looking for linear combinations of columns of $B_1$ and $C_1$ giving $0$ (because neither $B_1$ nor $C_1$ is full rank). As I didn't find something significant, using a CAS, I obtained the kernels (nullspaces) of $B_1$ and $C_1$ : their bases are strikingly similar: $(0,-2,-1,1)$ for $B_1$ and
$(-2,-1,1,0)$ for $C_1$; this provides the coefficients of null linear combinations of columns I hadn't found by hand computation. This was an "invitation" to place in correspondence columns numbered 2,3,4 of $B_1$ with columns 1,2,3 of $C_1$.
*c) finally, by creating 2 new $4 \times 4$ matrices $B_2$ and $C_2$ through the following operations : i) removal in $B_1$ and in $C_1$ of one of the columns that has been used by the linear combination $-2,-1,1$, for example column 2 for $B_1$ and column 1 for $C_1$ (these columns being "charged" of the non-invertibility of $B_1$ and $C_1$ !), then ii) "augmenting" $B_2$ and $C_2$ by "concatenation" of $U$ (as defined in part a)) as fourth column for $B_2$ and $V$ for $C_2$. As we must have $A.B_2=C_2$, matrix $A$ is obtained by computing:
$$A=C_2.B_2^{-1}$$
($B_2$ is checked invertible).
As we haven't worked by necessary and sufficient conditions, we have to check (this has been done in (1)) that product $A.B$ gives a shuffled version of $C$.
| {
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Which number is greater, $2^\sqrt2$ or $e$?
Claim: $\color{red}{2^\sqrt2<e}$
Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus
$$ f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln x \implies e^{x-1} > x^{\sqrt{x}} $$
Putting $x=2$, we obtain
$$e>2^\sqrt2$$
What do you think about this approach? Is there an easier way?
| Another way:
since $e^x$ is increasing and $\sqrt2\approx1.414213$ we have
$$(2^{\sqrt2})^{\sqrt2}=4\lt e^{1.4}\approx4.055199\lt e^{\sqrt2}$$It follows
$$\left((2^{\sqrt2})^{\sqrt2}\right)^{\frac{1}{\sqrt2}}\lt(e^{\sqrt2})^{\frac{1}{\sqrt2}}\iff2^{\sqrt2}\lt e
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic equation with absolute values Solve $|x^{2}+x-2|+|x^{2}-x-2|=2$
My attempt:
$|x^{2}+x-2|=
x^{2}+x-2, x\in (-\infty,-2]\cup[1,\infty)$
$|x^2+x-2|=-x^{2}-x+2, x\in(-2,1)\\$
$|x^{2}-x-2|=x^{2}-x-2, x\in (-\infty,-1]\cup[2,\infty)$
$|x^{2}-x-2|=-x^{2}+x+2, x\in (-1,2)\\$
$1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$
$x^{2}+x-2+x^{2}-x-2=2$
$\Rightarrow 2x^{2}-6=0$
$\Rightarrow x^{2}=3$
$\Rightarrow x_1=\sqrt{3}, x_2=-\sqrt{3}$
$2)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-1,2) \Rightarrow x\in[1,2)$
$x^{2}+x-2-x^{2}+x+2=2$
$\Rightarrow 2x=2$
$\Rightarrow x=1$
$3)$ $x\in(-2,1)$ and $x\in(-\infty,-1]\cup[2,\infty)$ $\Rightarrow x\in(-2,-1]$
$-x^{2}-x+2+x^{2}-x-2=2$
$\Rightarrow x=-1$
$4)$ $x\in(-2,1)$ and $x\in(-1,2)$ $\Rightarrow x\in(-1,1)$
$-x^{2}-x+2-x^{2}+x+2=2$
$\Rightarrow -2x^{2}+2=0$
$\Rightarrow -2x^{2}=-2$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm{1}$ but that's not in the interval $(-1,1)$ so I can throw away that solution.
My question is how do I find the intersection in $1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$ since $-\sqrt{3}$ and $\sqrt{3}$ should not be the solutions.
| Your work is essentially correct but, as you noted, the case 1) is not well solved. In this case the range is
$x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$, that is :
$$
\{(-\infty,-2]\cup[1,\infty)\}\cap\{(-\infty,-1]\cup[2,\infty)\}=(-\infty,-2]\cup [2,\infty)=D_1
$$
that is: $x\le -2 \mbox{ or } x\ge 2$
since $-2<-\sqrt{3}<2$ and $-2<\sqrt{3}<2$ , these two values are not in $D_1$ and are not roots of the equation.
| {
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How to Solve Summation by Hand I am having trouble solving the following summation by hand.
$$ \sum_{i=0}^{n-1} 2^i (n-i) $$
Can someone guide me in the right direction, especially for the $i \cdot 2^i$ part?
| Here's a hint:
$$ \sum_{i=0}^k i 2^i = 2 + 2*2^2+ 3*2^3 + \dots + k* 2^k \\
= ( 2 + 2^2 + 2^3 + \dots + 2^k ) \\
+ (2^2+2^3+\dots + 2^k) \\
+ \dots + (2^{k-1}+2^k) + (2^k) \\
= 2(1+ \dots + 2^{k-1}) + \dots + 2^{k-1}(1+2) + 2^k \\
= 2(2^k-1)+2^2(2^{k-1}-1) + \dots + 2^{k-1}(2^2-1) \\
= (k-1) 2^{k+1} + 2^k - 2(1+2+\dots+2^{k-2}) \\
= (k-1) 2^{k+1} + 2.$$
Can you get it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I solve this equation of functions? I need to find two functions continuous on $(-a, a)$, such that:
*
*$f(0) = 0$
*$f'(0) = 0$
*$g(0) = 0$
*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$
*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).
*The $\sin$ and $\cos$ functions are not used
How can I do this?
| Let
$$ f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=0}^{\infty} b_{n} \, x^{n} $$
then by the given conditions, $f(0) = 0$, $f'(0) = 0$, $g(0) = 0$, it is found that $b_{0} = 0$, $a_{0} = a_{1} = 0$ and lead to
$$ f(x) = \sum_{n=1}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=2}^{\infty} b_{n} \, x^{n}. $$
There is a lot of freedom to choose the coefficients $a_{n}$ and $b_{n}$.
What remains is:
*
*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$
*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).
In this view let $b_{n} = \frac{(-1)^{n+1}}{n}$ such that $g(x) = \ln(1+x)$. Now,
\begin{align}
\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} &= \lim_{x \to 0} \frac{(a_{2} \, x^{2} + \cdots + a_{n} \, x^{n} + \cdots)}{(1+x) \, \ln^{2}(1+x) } \, \to \frac{0}{0} \\
&= \lim_{x \to 0} \frac{(2 \, a_{2} \, x + \cdots + n \, a_{n} \, x^{n-1} + \cdots)}{2 \, \frac{\ln(1+x)}{1+x} } \, \to \frac{0}{0} \\
&= \lim_{x \to 0} \frac{(2 \, a_{2} + \cdots + n(n-1) \, a_{n} \, x^{n-2} + \cdots)}{2 \, \frac{1}{1+x} } \, \to a_{2} \neq 0 \\
\end{align}
This allows for
$$ f(x) = \sum_{n=2}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \ln(1+x). $$
| {
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Show that the 6 lines can be represented by the following equation: $(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$
Question: Figure $3$ shows six lines passing through the origin. The lines are separated by equal angles. Some
exact values of $\tan(t)$ are given in Table $1$.
$(i)$ Show that the lines can be represented by the following equation:
$$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$
$(ii)$ Find an equation for a hyperbola that does not cross any of the six lines in Figure $3$, giving
reasons for your answer.
I'm just really stuck , how do I even start this question! My approach has been this:
Let $y=mx+c$ for an equation of any line since all lines pass through the origin $(0,0)$ then $y=mx$ and because $m=\tan(t)$ we have the equation of any of these lines is $y=\tan(t)x$
And since there are $6$ lines passing through the origin there are $12$ sub divisions which means the graph is separated into $12$ parts and the angle between each of the parts will be $\frac{2\pi}{12}=\frac{\pi}{6}$
But I am confused , how should I continue? Am I even on the right track?
| First you can see that there are six lines on your graph and the given equation is the product of three terms in $x^2$ and $y^2$. So each of those three terms gives two of those lines :
$$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$
Is equivalent to :
$$x^2-y^2=0$$
$$x^2-(7-4\sqrt{3})y^2=0$$
$$x^2-(7+4\sqrt{3})y^2=0$$
Then separating $x$ and $y$ :
$$x^2=y^2\rightarrow y=\pm x$$
$$x^2=(7-4\sqrt{3})y^2\rightarrow y=\pm x\frac{1}{\sqrt{7-4\sqrt{3}}}$$
$$x^2=(7+4\sqrt{3})y^2\rightarrow y=\pm x\frac{1}{\sqrt{7+4\sqrt{3}}}$$
It appears that what confuses you is this, the plane is actually equally divided by those lines:
$$\frac{1}{\sqrt{7+4\sqrt{3}}} = 2-\sqrt{3}$$
It comes from :
$$7+4\sqrt{3}=4+4\sqrt{3}+3=(2+\sqrt{3})^2$$
| {
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Using the Chinese Remainder Theorem to solve the following linear congruence: $17x \equiv 9 \pmod{276}$ The book I am following (Elementary Number Theory by David Burton) uses the Chinese Remainder Theorem to solve $17x \equiv 9 \pmod{276}$ by breaking it up into a system of three linear congruences,
$$17x \equiv 9 \pmod{3}$$
$$17x \equiv 9 \pmod{4}$$
$$17x \equiv 9 \pmod{23}$$
I realize that the latter system is guaranteed to have a unique solution modulo $3 \times4\times23 = 276$. What I fail to understand is how and why the solution of this system of congruences the same as the solution of the initial congruence, $17x \equiv 9 \pmod{276}$. How can I solve this linear congruence using the Chinese Remainder Theorem?
| First Bit - why the solution of the system is a valid answer
If we have a number $y$ such that $y\equiv9\pmod{3}$ and $y\equiv9\pmod{4}$ and $y\equiv9\pmod{23}$ then we know that $y-9\equiv0\pmod{3}$ and $y-9\equiv0\pmod{4}$ and $y-9\equiv0\pmod{23}$. In other words $y-9$ divides by $3$, $4$ and $23$. As these are all coprime then $y-9$ must divide by their product which is $276$.
Second Bit - actually doing the math
$17x\equiv9\pmod{3}$
$17x\equiv0\pmod{3}$
So $x=3y,y\in\mathbb{Z}$
So the second equation becomes
$17x\equiv9\pmod{4}$
$51y\equiv1\pmod{4}$
$3y\equiv1\pmod{4}$
$y\equiv3\pmod{4}$
So $y=4z+3,z\in\mathbb{Z}$
Putting this into the third equation:
$17x\equiv9\pmod{23}$
$51y\equiv9\pmod{23}$
$5y\equiv9\pmod{23}$
$5(4z+3)\equiv9\pmod{23}$
$20z+15\equiv9\pmod{23}$
$20z+6\equiv0\pmod{23}$
This leads easily to $z=2$ which then gives $y=11$ and hence $x=33$.
Checking: $17\times33=561=2\times276+9$
| {
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Show that $\{a_n\}$ is an increasing sequence and bounded from above Question
Let $a_1$ = 1 and $a_{n+1} = \sqrt{1+2a_n}$. Show that $\{a_n\}$ is increasing and bounded from above.
Attempted solution
showing that the sequence is increasing:
\begin{align*}
a_n &\geq 1 \quad \forall n = {1,2,3,...} \\
\Rightarrow 1+2a_n &\geq 3 \\
\Rightarrow \sqrt{1+2a_n} &\geq \sqrt{3} \\
\Rightarrow a_{n+1} &\geq \sqrt{3} \quad \forall n = {1,2,3,...} \\
\Rightarrow a_{n+1} &\geq a_n
\end{align*} Is this a valid approach?
Assuming that I have shown that the sequence is increasing, I show that it is bounded by doing the following
\begin{align*}
a_{n+1} &\geq a_n \\
\Rightarrow \sqrt{1+2a_n} &\geq a_n \\
\Rightarrow 1+2a_n &\geq a^2_n \\
\Rightarrow (a_n+1)^2 - a^2_n &\geq a^2_n \\
\Rightarrow \frac{a_n + 1}{a_n} &\geq \sqrt{2} \\
\Rightarrow a_n &\leq 1+\sqrt{2} \quad \forall n = 1,2,3,...
\end{align*}
Is there another more straightforward way to show that this sequence is bounded?
| We prove by induction that the generalized sequence
$$a(1)=1,a(n+1)=\sqrt{2 c \;a(n)+1},\;c>0$$
is
(i) increasing.
The quotient of two successive terms of the sequence is
$$q(n+1)=\frac{a(n+1)}{a(n)}$$
Obviously
$$q(2)=\sqrt{2 c+1}>1$$
Now
$$q(n+1)=\sqrt{\frac{2 c a(n-1) q(n)+1}{2 c a(n-1)+1}}>\sqrt{\frac{2 c a(n-1)+1}{2 c a(n-1)+1}}=1$$
QED.
(ii) the sequence is bounded.
In fact we have just proven that
$$a(n+1)=\sqrt{2 c a(n)+1}>a(n)$$
Hence
$$2 c a(n)+1>a(n)^2$$
$$1>a(n)^2-2 c a(n)=(a(n)-c)^2-c^2$$
or
$$c^2+1>(a(n)-c)^2$$
Which means that $\left| a(n)\right|$ cannot grow unbounded.
QED.
(iii) the sequence converges to one of the fixed points, i.e. solutions of
$$a^2=2 a c+1$$
giving
$$\text{a1}=\sqrt{c^2+1}+c,\text{a2}=c-\sqrt{c^2+1}$$
Since $a(1)>0$ and the sequence increases it canot become negative. Hence the limit is $\text{a1}$.
In the case of the OP we have $c=1$ and hence $a1 = 1+\sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,x,y>0$ and $a^{11}+b^{11}\leq 1$ and $x^{11}+y^{11}\leq 1\;,$ Then prove that $a^{5}x^6+b^5y^6\leq 1$
If $a,b,x,y>0$ and $a^{11}+b^{11}\leq 1$ and $x^{11}+y^{11}\leq 1\;,$ Then prove that $a^{5}x^6+b^5y^6\leq 1$
$\bf{My\; Try::} $Using $\bf{A.M\geq G.M}$
$$\frac{a^{11}+a^{11}+a^{11}+a^{11}+a^{11}+x^{11}+x^{11}+x^{11}+x^{11}+x^{11}+x^{11}}{11}\geq \sqrt[11]{a^{55}x^{66}}$$
and $$\frac{b^{11}+b^{11}+b^{11}+b^{11}+b^{11}+y^{11}+y^{11}+y^{11}+y^{11}+y^{11}+y^{11}}{11}\geq \sqrt[11]{b^{55}y^{66}}$$
Now Adding These two inequalities, We get
$$a^{5}x^6+b^5y^6\leq \frac{5(a^{11}+b^{11})+6(x^{11}+y^{11})}{11}\leq \frac{5+6}{11}=1$$
My question is can we solve in some short way, If yes then plz explain me, Thanks
| Using Hölder's inequality with $p=\frac{11}{5}$ and $q=\frac{11}{6}$:
\begin{align}a^5x^6+b^5y^6 & \leq ((a^5)^p+(b^5)^p)^{\frac{1}{p}}((x^6)^q+(y^6)^q)^{\frac{1}{q}}\\ & \leq (a^{11}+b^{11})^{\frac{5}{11}}(x^{11}+y^{11})^{\frac{6}{11}}=1.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$ Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$
My attempt:
$a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)-6abc\geq 0$
$\implies a^{2}+a^{2}b^{2}+b^{2}+b^{2}c^{2}+c^{2}+c^{2}a^{2}-2abc-2abc-2abc\geq 0$
$\implies (a-bc)^{2}+(b-ac)^{2}+(c-ab)^{2}\geq 0$
Each of these terms must be non-negative, thus the sum is also non-negative.
I'm new to writing proofs, so I don't know whether this proof is fine.
| Alternatively by $AM \ge GM$,
\begin{align*}
a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) &=
a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2 \\
& \ge 6\sqrt[6]{a^6b^6c^6} \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is $x*y = \frac{1}{\frac{1}{x} + \frac{1}{y}}$ closed? Injective? Surjective? Let $x*y = \frac{1}{\frac{1}{x} + \frac{1}{y}}$.
*
*Is $x*y$ closed in $(0, \infty)$?
*For a given $x > 0$, is the function $f(y) = x*y, (0, \infty) \to (0,\infty)$ injective?
*For a given $x > 0$, is the function $f(y) = x*y, (0, \infty) \to (0,\infty)$ surjective?
| What I tried:
*
*Yes.
If $x,y > 0$, then $x*y > 0$.
QED
*Yes
Let $y_1,y_2 > 0$ s.t. $f(y_1) = f(y_2)$.
Then
$$\frac{1}{\frac{1}{x} + \frac{1}{y_1}} = \frac{1}{\frac{1}{x} + \frac{1}{y_2}}$$
$$\to \frac{1}{x} + \frac{1}{y_1} = \frac{1}{x} + \frac{1}{y_2}$$
$$\to \frac{1}{y_1} = \frac{1}{y_2}$$
$$\to y_1 = y_2$$
QED
*No
Let $a \in \text{codomain}(f) = (0,\infty)$. Suppose $\exists y \in \text{domain}(f) = (0, \infty)$ s.t.
$$\frac{1}{\frac{1}{x} + \frac{1}{y}} = a$$
$$\to \frac{1}{x} + \frac{1}{y} = \frac{1}{a}$$
$$\to \frac{1}{y} = \frac{1}{a} - \frac{1}{x}$$
$$\to \frac{1}{y} = \frac{x}{xa} - \frac{a}{xa}$$
$$\to \frac{1}{y} = \frac{x-a}{xa}$$
$$\to y = \frac{xa}{x-a}$$
Thus, there is no such $y$ if $a=x$.
QED
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$. Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$.
$x=\frac{23}{3}-\frac{2}{3} y$ so when I plug it in I get
$\sqrt{\frac{23}{3}-\frac{2}{3}y+y} +\sqrt{2\times\frac{23}{3}-\frac{2}{3}y+y+2}=7$
$\sqrt{\frac{23}{3}+\frac{y}{3}} + \sqrt{\frac{52}{3}-\frac{y}{3}}=7$
$2\sqrt{\frac{1196}{9}+\frac{29}{9}y-\frac{y^{2}}{9}} = 24$
$\frac{1196}{9}+\frac{29}{9}y-\frac{y^{2}}{9}=144$
$y^{2}-29y+100=0$
$y_1=25, x_1=-9$ or $y_2=4, x_2=5$.
The solution is correct, but is everything written properly? Should I put $\iff$ at the beginning of each row?
| Put $a=x+y,\; b=2x+y+2$.
Equations:
$$\begin{cases}\begin{align}
&\sqrt{a}+\sqrt{b}=7&\Rightarrow a+b+2\sqrt{ab}&=49\\
&a+b-2=23&\Rightarrow a+b&=25\end{align}\bigg\rbrace\Rightarrow 25+2\sqrt{ab}=49\Rightarrow ab=144\\
\end{cases}\\$$
$$\begin{align}
a+\frac {144}a&=25\\
a^2-25a+144&=0\\
(a-16)(a-9)&=0\\
a&=16, 9\\
b&=9, 16\\
x=b-a-2&=-9,5\\
y=a-x&=25,4\\
\Rightarrow (x,y)&=(-9,25), (5,4)\;\blacksquare\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why do these two summations not equate? Why is the below equation false?
(1) $\sum_{i=0}^n i^2 = \ (1/4)\sum_{i=0}^{2n} {i}^2 $
When If we let $\ j=2i $ , and substitute $\ i=(j/2)$ into the leftmost equation above, then:
(2) $\sum_{j=0}^{2n} {(j/2)}^2 $
Which gives the same sum as:
$\ (1/4)\sum_{i=0}^{2n} {i}^2$
Note: I know the equation on line (1) is wrong, by actually summing the adjacent summations, but I can't see why this would be so.
| $$\begin{align}
\sum_{i=0}^n i^2&=0^2+1^2+2^2+3^2+4^2+\cdots +n^2\qquad\qquad\qquad\qquad\qquad (1)\\
\sum_{j=0}^{2n} (j/2)^2&=\left(\frac 02\right)^2+\left(\frac 12\right)^2+\left(\frac 22\right)^2+\left(\frac 32\right)^2+\left(\frac 42\right)^2\cdots+\cdots +\left(\frac n2\right)^2+\cdots+\left(\frac {2n}2\right)^2\\
&=\frac 14 \left(0^2+1^2+2^2+3^2+4^2+\cdots+(2n)^2\right)\\
&=0^2+\left(\frac 12\right)^2+1+\left(\frac 32\right)^2+2^2+\left(\frac 52\right)^2+\cdots+\left(\frac {n-1}2\right)^2+n^2\qquad \qquad (2)\end{align}$$
You can see clearly that $(1)\neq (2)\\$ as well as identify the additional terms in $(2)$.
What you probably wanted was
$$\sum_{i=0}^n i^2=\frac 14\sum_{i=0}^n (2i)^2$$
You might be interested to know that
$$\sum_{i=0}^{2n} i^2=\sum_{i=1}^{2n} i^2=\sum_{i=1}^n (2i-1)^2+(2i)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Reduction Integration of $I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$ How do I derive this using reduction formula?
$$I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$$
I got $\sqrt{x^2+a^2}(x^n) - \frac{1}{n-1}(x^{n-1})\sqrt{x^2+a^2} + \frac{1}{n-1}(I_n) $ . I'm not too sure if this is correct. I would appreciate any form of help or correction to my answer. Thank you.
| For starters, I would get rid of the $a$ parameter:
$$ I_n = a^n \int \frac{x^n\,dx}{\sqrt{x^2+1}} = a^n J_n \tag{1} $$
then apply integration by parts:
$$ J_n = \int x^{n-1}\frac{x\,dx}{\sqrt{1+x^2}} = x^{n-1}\sqrt{1+x^2}-(n-1)\int x^{n-2}\sqrt{1+x^2}\,dx\tag{2}$$
and notice that $\sqrt{1+x^2}=\frac{1}{\sqrt{1+x^2}}+\frac{x^2}{\sqrt{1+x^2}}$, leading to:
$$ n\,J_n = x^{n-1}\sqrt{1+x^2} - (n-1)\,J_{n-2}.\tag{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Nontrivial rational solution of $x^3-3x^2+3x+1=y^2$ So, I'm looking for a rational solution of $$x^3-3x^2+3x+1=y^2$$ that is different from $(0,\pm 1)$. I tried to substitute $y=xt+1, t\in\mathbb Q$, getting following equation $$
x(x^2-(t^2+3)x+(-2t+3))=0$$ with the discriminant of the quadratic term $t^4+6t^2+8t-3$. The next step would be to find a $t$ for that this discriminant is a square of a rational number. Could you give me a tip how to do it or propose another solution?
| As hinted in the comments, we may consider the equivalent equation
$$y^2 = x^3 + 2$$
This is an elliptic curve. Finding rational points is an active domain in research and many books have been written, e.g. Silverman's Rational Points on Elliptic Curves.
Now, you already have some rational points on your curve: $(-1,-1)$ and $(-1,1)$. We can construct new rational points on the elliptic curve by taking the tangent line through one of these points, and calculating the intersection with the elliptic curve.
The tangent line to $(-1,1)$ is
$$y = \frac{3}{2}(x+1) + 1 = \frac{3}{2}x+\frac{5}{2}$$
Now, taking this $y$ in the original equation, we get:
$$\left(\frac{3}{2}x+\frac{5}{2}\right)^2 = x^3 + 2$$
Solving this cubic for $x$, we obtain two solutions: $x = -1$ and $x = \frac{17}{4}$. You can then calculate $y = \pm \sqrt{\left(\frac{17}{4}\right)^3+2} = \pm \frac{71}{8}$.
So, two more points on your curve are $\left(\frac{17}{4},\frac{71}{8}\right)$ and $\left(\frac{17}{4},-\frac{71}{8}\right)$.
You can then repeat this with the new solution(s) to find more of them, e.g.
$$(x,y) = \left(\frac{66113}{80656},\frac{36583777}{22906304}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Factorize $x^{2n}+1$ to evaluate $\int\tan^{1/n}(x)dx$ I've been attempting to find a formula for $\int\tan(x)^{1/n}dx, n\in\Bbb N$.
I started out by performing the substitution $$u=\tan(x)^{1/n}$$
$$dx=\frac {nu^{n-1}}{1+u^{2n}}du$$ to transform the integral into $$n\int \frac{u^n}{1+u^{2n}}du$$
This is a relatively simple rational function, so it should be possible to solve via a partial fraction expansion. For the simple case where $n=2$, I was able to rewrite the integral as $$\frac{\sqrt2}{2}\left(\int\frac{u}{u^2-\sqrt2u+1}du - \int\frac{u}{u^2+\sqrt2u+1}du\right)$$
Which can then be solved by completing the square to yield $$\frac{\sqrt2}{4}\ln\left(\left(\sqrt{\tan{x}}-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\over\left(\sqrt{\tan{x}}+\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\right)+\frac{\sqrt2}{2}\left(\arctan\left({\sqrt{2\tan{x}}-1}\right)+\arctan\left({\sqrt{2\tan{x}}+1}\right)\right)+C$$
However, I can't figure out how to factor the denominator for the partial fraction expansion of the general case. It will always be a product of $n$ irreducible quadratic factors, clearly, but I don't know how to locate them. I have managed to find the complex linear factors, using Euler's identity to find that they're of the form $$x\pm e^{im\pi\over{2n}}$$ where m is allowed to vary from 1 to $2n$. How can I combine these linear factors into real, irreducible quadratics for use in my partial fraction expansion? Alternatively, is there a simpler method of evaluating this integral? Any help would be vastly appreciated.
| I do not think that this is an answer but it is too long for a comment.
Adressing this problem, you enter in the world of the gaussian hypergeometric function since $$I_n=n\int \frac{u^n}{1+u^{2n}}\,du=\frac{n \,u^{n+1} }{n+1}\,\, _2F_1\left(1,\frac{n+1}{2 n};\frac{3n+1}{2 n};-u^{2 n}\right)$$ which, as you showed, can simplify for $n=2$.
In other words, back to $x$, $$I_n=\int{\tan^{\frac 1n} (x)}\,dx=\frac{n \tan ^{\frac{n+1}{n}}(x)}{n+1}\,\,\, _2F_1\left(1,\frac{n+1}{2 n};\frac{3n+1}{2 n};-\tan
^2(x)\right)$$
Edit
In your comment, you reported that $$u^6+1=(u^2+\sqrt{3}u+1)(u^2-\sqrt{3}u+1)(u^2+1)$$ So, using partial fraction decomposition $$\frac{u^3}{1+u^6}=\frac{u}{3
\left(u^2+1\right)}+\frac{u}{6 \left(u^2-\sqrt{3} u+1\right)}+\frac{u}{6 \left(u^2+\sqrt{3} u+1\right)}$$ which can be integrated leading to $$\int\frac{u^3}{1+u^6}\,du=-\frac{1}{6} \log \left(u^2+1\right)+\frac{1}{12} \log \left(u^2-\sqrt{3}
u+1\right)+\frac{1}{12} \log \left(u \left(u+\sqrt{3}\right)+1\right)-\frac{\tan
^{-1}\left(\sqrt{3}-2 u\right)}{2 \sqrt{3}}-\frac{\tan ^{-1}\left(2
u+\sqrt{3}\right)}{2 \sqrt{3}}$$ Here again, the logarithms can be combined together as well as the arctangents.
Much more tedious is the factorization of $u^8+1$; it leads to
$$u^8+1=\left(u^2-2 u \sin \left(\frac{\pi }{8}\right)+1\right) \left(u^2+2 u \sin
\left(\frac{\pi }{8}\right)+1\right)$$ $$ \left(u^2-2 u \cos \left(\frac{\pi
}{8}\right)+1\right) \left(u^2+2 u \cos \left(\frac{\pi }{8}\right)+1\right)$$ leading to a small monster I shall not reproduce here (too long !).
Similarly, we could write $$u^{10}+1=(1+u^2)(u^8-u^6+u^4-u^2+1)$$ and the last term again can be factorized as the product of four quadratic polynomials.
$$\left(u^2-\sqrt{\frac{1}{2} \left(5-\sqrt{5}\right)} u+1\right)
\left(u^2+\sqrt{\frac{1}{2} \left(5-\sqrt{5}\right)} u+1\right)$$
$$ \left(u^2-\sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)} u+1\right)
\left(u^2+\sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)} u+1\right)$$
Looking for something else, I found this post where Dr.MV's answer would be very interesting for yur work.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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System of equations with inequality
Question Statement:-
If $$\begin{eqnarray}
x=cy+bz\\
y=az+cx\\
z=bx+ay
\end{eqnarray}$$ where $x,y,z$ are not all zero, prove that $a^2+b^2+c^2+2abc=1$
Further if at least one of $a,b,c$ is a proper fraction, prove that:-
(i) $a^2+b^2+c^2\lt 3$
(ii) $abc\gt-1$
Attempt at a solution:-
The first part (i.e the part where we have to prove $a^2+b^2+c^2+2abc=1$) of the question gets completed pretty easily.
As the given system of equations is homogeneous and as $x,y,z$ are not all zero so the solutions are non-trivial. And for a homogeneous system of equations to have non-trivial solutions we have :-
$$\begin{vmatrix}
1 & -c & -b\\
-c & 1 & -a\\
-b & -a & 1\\
\end{vmatrix}=0$$
And on expanding the determinant we get
$(1-a^2)+c(-c-ab)-b(ac+b)=0\implies a^2+b^2+c^2+2abc=1$
I am not able to come up with any way to solve the remaining two portions of the questions.
| Assuming $a$ is a proper fraction, then
$$a^2<1$$
Considering quadratic in $c$,
$$c^2+2abc+a^2+b^2-1=0$$
Admitting real value of $c$,
\begin{align*}
\Delta & \ge 0 \\
4(a^2-1)(b^2-1) &\ge 0 \\
b^2-1 & \le 0
\end{align*}
Hence,
$$b^2 \le 1$$
Similarly,
$$c^2 \le 1$$
Hence,
$$a^2+b^2+c^2<3$$
Also
$$abc=\frac{1-a^2-b^2-c^2}{2}>-1$$
| {
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"url": "https://math.stackexchange.com/questions/1999251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Value of $\sin(\pi/18)$ by nested radicals I recently asked a question about how to find value of $\sin(\pi/18)$ and I understand that there is no expression for $\sin(\pi/18)$ that uses the ordinary arithmetical operations. but I know that we have exact value by nested radicals. I want to know how we get this value ?
http://mathworld.wolfram.com/TrigonometryAnglesPi18.html
| Deriving $\sin\frac{\pi}{18}$ will be simpler when we implement Half angle cosine formula
$$2\sin\frac{\theta}{2} = \sqrt{2-2\cos\theta}$$
$$2\cos\frac{\theta}{2} = \sqrt{2+2\cos\theta}$$
and cosine angle identity
$$2\cos(\pi-\theta) = -2\cos\theta$$
We can derive as follows
$2\sin(\frac{\pi}{18})$ = $\sqrt{2-2\cos(\frac{\pi}{9})}$ = $\sqrt{2-\sqrt{2+2\cos(\frac{2\pi}{9}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{4\pi}{9}}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\frac{8\pi}{9}}}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{9}}}})}$
Now you can observe $2\cos(\frac{\pi}{9})$ repeats in cycle infinitely
$$\therefore 2\sin\frac{\pi}{18} = \sqrt{2-\sqrt{2+\sqrt{2+...}}}$$ $[-++]$ is the repeating pattern in infinite nested radical. Refer here. Welcome to the world of infinite nested square roots of 2 as solution to cosine angles!
| {
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Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
$$41-4\cdot17=1$$
So the inverse of $17$ is $-4$.
That is, the inverse of $17$ is $37$
Am I right?
| Under the Extended Euclidean algorithm, you annotate the process of finding a GCD with how you got there. This can be used for finding multiplicative inverses:
$$
\begin{array}{c|c|c|c|l}
q & r & a & b & \text{implied equation} \\ \hline
& 41 & 1 & 0 & 41 = 1\cdot 41 + 0\cdot 17\\
& 17 & 0 & 1 & 17 = 0\cdot 41 + 1\cdot 17\\
2 & 7 & 1 & -2 & \;\; 7 = 1\cdot 41 + (-2)\cdot 17\\
2 & 3 & -2 & 5 & \;\;3 = (-2)\cdot 41 + 5\cdot 17\\
2 & 1 & 5 & \color{red}{-12} & \;\;1 = 5\cdot 41 + (-12)\cdot 17\\
\end{array}
$$
The first two lines setup the numbers under consideration as the bases. Then at each stage, $q$ gives the integer division result for the preceding two $r$ values, which then determines how to construct the $r$, $a$ and $b$ values for the current row: subtract off $q$ times the row above value from the row above that. (The final column here is not needed, it's just to aid understanding if you haven't seen this before).
This gives us the relevant Bezout's identity, $5\cdot 41 + (-12)\cdot 17 = 1 $, which immediately gives us $-12$ as the inverse of $17 \bmod 41$:
$5\cdot 41 + (-12)\cdot 17 \equiv (-12)\cdot 17 \equiv 29\cdot 17 \equiv 1 \bmod 41$
that is, $17^{-1} \equiv 29 \bmod 41$ .
The second question is then easily solved since $17n\equiv 14 \bmod 41$ $\implies$ $n\equiv -12\cdot 14 \bmod 41$ (i.e. $n\equiv -168 \equiv -4 \equiv 37 \bmod 41$ ).
| {
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Calculating the probability of reaching each absorbing state in Markov Chain I'm starting with a Matrix that looks like this:
[[ 0 , 1/2, 0, 0 , 0 , 1/2],
[4/9, 0 , 0, 3/9, 2/9, 0 ],
[ 0 , 0 , 1, 0 , 0 , 0 ],
[ 0 , 0 , 0, 1 , 0 , 0 ],
[ 0 , 0 , 0, 0 , 1 , 0 ],
[ 0 , 0 , 0, 0 , 0 , 1 ]]
And I'm being asked to calculate the probability of reaching each of the absorbing states C, D, E, and F starting from A. Obviously C is 0, and the other probabilities are given as:
A to D -> 3/14
A to E -> 1/7
A to F -> 9/14
but I don't know what steps or formulas I need to arrive at those values.
I don't really know where to start with this so any help would be greatly appreciated!
| The state transition diagram for the given problem is given by:
Consider transitions from A to D:
It could happen in several ways, as listed in the following. All these paths are mutually exclusive. Invoking the definition of of first passage time, we can obtain the required probability.
\begin{eqnarray*}
A\rightarrow B \rightarrow D &=& p_{AB}\cdot p_{BD}\\
&=&\left(\frac{1}{2}\right)\left(\frac{3}{9}\right)=\frac{3}{18}\\
A\rightarrow B\rightarrow A\rightarrow B \rightarrow D &=& p_{AB}\cdot p_{BA}\cdot p_{AB}\cdot p_{BD}\\
&=&\left(\frac{1}{2}\right)\left(\frac{4}{9}\right)\left(\frac{1}{2}\right)\left(\frac{3}{9}\right)=\left(\frac{1}{2}\right)\left(\frac{2}{9}\right)\left(\frac{3}{9}\right)\\
A\rightarrow B\rightarrow A\rightarrow B\rightarrow A\rightarrow B \rightarrow D &=& p_{AB}\cdot p_{BA}\cdot p_{AB}\cdot p_{BA}\cdot p_{AB}\cdot p_{BD}=p_{AB}\cdot \left(p_{BA}p_{AB}\right)^{2}\cdot p_{BD}\\
&=&\left(\frac{1}{2}\right)\left(\frac{2}{9}\right)^{2}\left(\frac{3}{9}\right)\\
\end{eqnarray*}
and several more such transitions. The probability of ever reaching from state A to D is obtained from the following probability:
\begin{eqnarray*}
&&\frac{3}{18}+\left(\frac{1}{2}\right)\left(\frac{2}{9}\right)\left(\frac{3}{9}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{9}\right)^{2}\left(\frac{3}{9}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{9}\right)^{3}\left(\frac{3}{9}\right)+\cdots\\
&=&\frac{3}{18}+\left(\frac{1}{2}\right)\left(\frac{2}{9}\right)\left(\frac{3}{9}\right)\left\{1+\frac{2}{9}+\left(\frac{2}{9}\right)^{2}+\cdots\right\}\\
&=&\frac{3}{18}+\frac{1}{3\times 9}\left\{\dfrac{1}{1-2/9}\right\}=\frac{27}{126}=\frac{3}{14}
\end{eqnarray*}
In a similar manner, we can find other probabilities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $x$ is a real and $0 < x < 4$, then $\frac{4}{x(4-x)} \geq 1$ I got this exercise from the textbook Book of Proof, CH4 E12. I've tackle this problem in the following manner:
Suppose x is a real and $0 < x < 4$, it follows that,
\begin{align*}
&\Rightarrow 0 - 2 < x - 2 < 4 - 2 \\
&\Rightarrow 4 < (x - 2)^2 < 4\\
&\Rightarrow 0 \leq (x - 2)^2 < 4
\end{align*}
Since, $x(4 - x) = 4x - x^2 = 4 - (x - 2)^2$, then
$$\dfrac{4}{x(4 - x)} = \dfrac{4}{4 - (x - 2)^2}.$$
This expression is greater or equal to $1$ for
$0 \leq (x - 2)^2 < 4$. Thus,
$$\dfrac{4}{x(4 - x)} \geq 1.$$
I'm quite new to proof technique and I'm using this book to self-learn logic and proofing writing. My question is: is the solution stated above logically sound? Would my arguments be considered sufficient to prove that $P \Rightarrow Q$?
| Hint See that: $0\le x\le 4 $ implies $0\le 4-x\le 4$ and then,
$$0\le x(4-x) =-(x-2)^2+4\le4$$
Thus directly implies your inequality.$$\dfrac{4}{x(4 - x)} \geq 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the linear factors of $x^3 - 2$? The polynomial is defined as $$x^3 - 2 \in \mathbb{C}[x]$$
How do you figure out the linear factors of the term? If I evaluate the function in polar coordinates I get some weird terms that can't possibly be correct. Wolfram says that I have to multiply the real solution by the third roots of $i$ (as that's the unity vector), but if I write those solutions as linear factors and multiply the term out I get something different from my polynomial.
Can someone please help me?
Thank you!
Edit: My attempt at solving with polar coordinates: $$x^3 = 2 = 2+0i = 2(\cos(0) + i\sin(0)) | \sqrt[3]{}$$ $$x = \sqrt[3]{2}(\cos(L*\frac{2\pi}{3}) + i\sin(L*\frac{2\pi}{3}) | L \in \{0, 1, 2\}$$ $$x_1 = \sqrt[3]{2}$$ $$x_2 = \sqrt[3]{2}(-\frac{1}{2} + i\frac{\sqrt{3}}{2})$$ $$x_3 = \sqrt[3]{2}(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$$
| See that any root that you could have of $x^3-2$ is of the style $\sqrt[3]{2}\ e^{2it\pi/3}$ with $t=0,1,2$, so it factors into
$$(x-\sqrt[3]{2})(x-\sqrt[3]{2}\ e^{2i\pi/3})(x-\sqrt[3]{2}\ e^{4i\pi/3})$$
If you want to know what is $e^{2ti\pi/3}$ use the polar coordinates:
$$e^{2it\pi/3}=\cos\left(\frac{2\pi t}{3}\right)+i\sin\left(\frac{2\pi t}{3}\right)$$
And all the roots are:
$$\sqrt[3]{2}\ e^{2i\pi/3}=\sqrt[3]{2}\left( \cos\left(\frac{2\pi }{3}\right)+i\sin\left(\frac{2\pi }{3}\right)\right)=\sqrt[3]{2}\left(-\frac{1}{2}+-\frac{i\sqrt{3}}{2}\right)$$
$$\sqrt[3]{2}\ e^{4i\pi/3}=\sqrt[3]{2} \left(\cos\left(\frac{4\pi }{3}\right)+i\sin\left(\frac{4\pi }{3}\right)\right)=\sqrt[3]{2}\left(-\frac{1}{2}--\frac{i\sqrt{3}}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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image of circle after inversion Given a circle $C$ of radius $r$ and center $(x_0,y_0)$, what is the center and radius of $C$ after inversion in the unit circle?
A simpler question has been asked:
Image of a circle under conformal map $1/z$
| You have $(x-x_0)^2+(y-y_0)^2=r^2$ as the original circle. Performing the inversion means replacing $x$ by $\frac x{x^2+y^2}$ and $y$ by $\frac y{x^2+y^2}$. Abbreviating $d:=x^2+y^2$ and $f:=x_0^2+y_0^2-r^2$ one can write the circle after inversion as
\begin{align*}
\left(\frac{x}{d}-x_0\right)^2+\left(\frac{y}{d}-y_0\right)^2 &= r^2 \\
(x-x_0d)^2 + (y-y_0d)^2 &= (rd)^2 \\
x^2-2xx_0d+x_0^2d^2+y^2-2yy_0d+y_0^2d^2 &= r^2d^2 \\
d-2xx_0d+x_0^2d^2-2yy_0d+y_0^2d^2 &= r^2d^2 \\
1-2xx_0+x_0^2d-2yy_0+y_0^2d &= r^2d \\
(x^2+y^2)(x_0^2+y_0^2-r^2)-2xx_0-2yy_0+1 &= 0 \\
(x^2+y^2)f-2xx_0-2yy_0 &= -1 \\
(x^2+y^2)-2x\frac{x_0}f-2y\frac{y_0}f &= -\frac1f \\
\left(x^2-2x\frac{x_0}f\right) +
\left(y^2-2y\frac{y_0}f\right) &= -\frac1f \\
\left(x^2-2x\frac{x_0}f+\left(\frac{x_0}f\right)^2\right) +
\left(y^2-2y\frac{y_0}f+\left(\frac{y_0}f\right)^2\right) &=
\left(\frac{x_0}f\right)^2 + \left(\frac{y_0}f\right)^2 -\frac1f \\
\left(x-\frac{x_0}f\right)^2 + \left(y-\frac{y_0}f\right)^2 &=
\frac{x_0^2+y_0^2-f}{f^2} \\
\left(x-\frac{x_0}{x_0^2+y_0^2-r^2}\right)^2 + \left(y-\frac{y_0}{x_0^2+y_0^2-r^2}\right)^2 &=
\left(\frac{r}{x_0^2+y_0^2-r^2}\right)^2 \\
\end{align*}
So the new center would be at
$$\left(\frac{x_0}{x_0^2+y_0^2-r^2}, \frac{y_0}{x_0^2+y_0^2-r^2}\right)$$
and the new radius would be
$$\frac{r}{\left\lvert x_0^2+y_0^2-r^2\right\rvert}$$
Some sanity checks. If the denominator is zero, that means your circle passes through the origin, so its image will pass through infinity. Makes sense. If the original radius is zero, then so is the new one, as points map to points. Sane as well. And the direction of the new center matches that of the original center, as expected.
| {
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"url": "https://math.stackexchange.com/questions/2009953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove that $21|a^2 +b^2 \implies 441|a^2+b^2$
Prove that $$21|a^2+b^2 \implies 441|a^2+b^2$$
I can show that if $3$ divides $a^2+b^2$ than so does $9$, so the part that's left to do is to prove that $7$ and $49$ divide $a^2+b^2$ and I am stuck at this point.
| Note that
\begin{align}
1^2&\equiv 1\pmod{7},\\
2^2&\equiv 4\pmod{7},\\
3^2&\equiv 2\pmod{7},\\
4^2&\equiv 2\pmod{7},\\
5^2&\equiv 4\pmod{7},\\
6^2&\equiv 1\pmod{7}.
\end{align}
It is impossible to get 2 numbers with replacement from $\{1,2,4\}$ to get a sum that is divisible by $7$. Thus, to have $a^2+b^2\equiv 0\pmod{7}$, it must be that $a,b\equiv 0\pmod{7}$ and so $a^2,b^2\equiv 0\pmod{49}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
the diophantine forms of the equation $a^2 +b^3 = c^5$ I just want to know the forms of the numbers a,b and c in the Diophantine equation:
$a^2 +b^3 = c^5$
| See OEIS sequence A178130
$$\eqalign{104^2 + 28^3 &= 8^5\cr
654^2 + 127^3 &= 19^5\cr
2816^2 + 32^3 &= 24^5\cr
3912^2 + 124^3 &= 28^5\cr
4096^2 + 256^3 &= 32^5\cr
6048^2 + 288^3 &= 36^5\cr
48500^2 + 275^3 &= 75^5\cr
19683^2 + 1458^3 &= 81^5\cr
65216^2 + 1008^3 &= 88^5\cr
77824^2 + 1280^3 &= 96^5\cr
\ldots}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving System of Linear Equations with LU Decomposition of $4 \times 3$ matrix The following is all confirmed to be true:
Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$
U =
$
\begin{bmatrix}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
$
L =
$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-2 & 0 & 1 & 0\\
-1 & -1 & -1 & 0\\
\end{bmatrix}
$
Okay so using that I need to solve the following system:
$
x_2 - 2x_3 = 0 \\
-x_1 + 2x_2 - x_3 = -2 \\
2x_1 -4x_2 + 3x_3 = 5 \\
x_1 - 3x_2 + 2x_3 = 1
$
So step one is solving $Ly = b$, where $y = Ux$
So that is:
$
y_1 = 0\\
y_2 = -2\\
-2y_1 + y_3 = 5 \\
-y_1 - y_2 -y_3 = 1 \\
$
How can we find $y_3$ in the last two equations? Because,
$
-2(0) + y_3 = 5 \\
-(0) - (-2) - y_3 = 1 \\
$
So in the second to last equation $y_3 = 5$, but in the last equation $y_3 = 1$. Very confused.
| $$
\begin{align}
\mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\
% P
\left[
\begin{array}{cccc}
0 & \boxed{1} & 0 & 0 \\
\boxed{1} & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
%
&=
% L
\left[
\begin{array}{rrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-2 & 0 & 1 & 0 \\
-1 & -1 & -1 & 1 \\
\end{array}
\right]
% U
\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}
\right]
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the Jordan Canonical Form of a Classical Adjoint of a Jordan Block Let $A$ be a size $n$ Jordan matrix with $0$ on its diagonal, that is
$$A = J_n(0) = [a_{ij}] = \begin{cases}
1, &j=i+1\\
0, &\text{elsewhere}
\end{cases}
$$
What is the Jordan Canonical Form of the classical adjoint of A, $\text{adj} A$?
Can we start with the fact that $A$ is singular and $A (\text{adj} A) = 0_n?$
| If you just start computing the classical adjoint for $n=2,3,4...$ you should notice a pattern as to what they look like.
$$adj\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & - 1\\ 0 & 0\end{pmatrix}$$
$$adj\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
$$ adj \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$$
Once you prove that this pattern holds, the Jordan Form is straightforward to compute.
| {
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"url": "https://math.stackexchange.com/questions/2013297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Fibonacci numbers solution to this recurrence relation Prove that the Fibonacci number's are the solutions the following recurrence relation, $$S_n=5S_{n-4}+3S_{n-5}$$
For all n greater than or equal to 5, where we have
$$S_0=0$$ $$S_1=1$$ $$S_2=1$$ $$S_3=2$$ $$S_4=3$$
Then use the formula to show that the Fibonacci number's satisfy the condition that $f_n$ is divisible by 5 if and only if n is divisible by 5.
| hint:characteristic equation$$x^5=5x+3\\x^5-5x-3=0$$ then $$x^2-x-1|x^5-5x-3\\ \to $$divide $x^5-5x-3$ by $x^2-x-1$
$$x^5-5x-3=(x^2-x-1)(x^3+x^2+2x+3)$$
and $x^2-x-1$ is the characteristic equation of $f_n=f_{n-1}+f_{n-2}\\----\\$
Another solving :
$$f_{n+5}=f_{n+4}+f_{n+3}\\
=(f_{n+3}+f_{n+2})+f_{n+3}=\\2f_{n+3}+f_{n+2}$$ now put $f_{n+3}$
$$f_{n+5}=2f_{n+3}+f_{n+2}=\\2(f_{n+2}+f_{n+1})+f_{n+2}=\\3f_{n+2}+2f_{n+1}\\$$and now
$$f_{n+5}=3f_{n+2}+2f_{n+1}=\\3(f_{n+1}+f_{n})+2f_{n+1}=\\5f_{n+1}+3f_n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $a$, $b$, $c$ the sides of a triangle, prove $\left| \frac{a}b-\frac{b}a+\frac{b}c-\frac{c}b+\frac{c}a-\frac{a}c\right|< 1$ The inequality I have to prove that if $a$, $b$, and $c$ are the sides of a triangle, then
$$\left| \frac{a}b-\frac{b}a+\frac{b}c-\frac{c}b+\frac{c}a-\frac{a}c\right|< 1$$
I initially thought of changing it in terms of sine using the law of sines, which gives me
$$\left| \frac{\sin A}{\sin B}-\frac{\sin B}{\sin A}+\frac{\sin B}{\sin C}-\frac{\sin C}{\sin B}+\frac{\sin C}{\sin A}-\frac{\sin A}{\sin C}\right| < 1$$
However, from there I'm not sure how to proceed. Any tips?
| Hint:
Using the substitution $a = x+y, b=y+z, c = z+x$ (if $a, b, c$ are sides of a triangle then $x, y, z$ always exist and are positive) we get
$$\left| \sum_{cyc} \left( \frac{a}b-\frac{b}a \right)\right| = \left|\frac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)} \right| < 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y \sqrt{x z},\\
x z +y z \geq 2 z \sqrt{xz}.
\end{gather*}
Add and get
$$
xy+xz+yz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}.
$$
By condition
$$
xyz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}
$$
Dividing by $\sqrt{xyz}$ we obtain the inequality.
Question. Are there another ways to prove it?
| From your assumption, we obtain that
\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le 1.
\end{align*}
Since
\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} &=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z} +\frac{1}{y} + \frac{1}{z} \right)\\
&\ge\frac{1}{2}\left(\frac{2}{\sqrt{xy}} + \frac{2}{\sqrt{xz}} +\frac{2}{\sqrt{yz}} \right),
\end{align*}
that is,
\begin{align*}
\frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{xz}} +\frac{1}{\sqrt{yz}}\le 1,
\end{align*}
the required inequality follows immediately by multiplying $\sqrt{xyz}$ to both sides.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show the coefficient of $x^3$ in the expansion of $(1+x+x^2+x^3)(1+x^2)(1+x^3)$ is the same as the number of partitions of $3$ (let a partition of $n \in \Bbb N, n>0 $ be a sum of positive integers)
How would you show that the number of different partitions of $3$ is the coefficient of
$x^3$
in the
expansion of:
$(1+x+x^2+x^3)(1+x^2)(1+x^3)?$
My attempt:
since there are $3$ brackets to choose factors of $x^3$ from, we can say the following:
$ x^3 = x^a x^b x^c, a \in \{0,1,2,3\}, b \in \{0,2\}, c \in \{0,3\}$
The coefficient of $x^3$ in the expansion will equal the number of solutions satisfying $ x^3 = x^a x^b x^c$.
$a=0 \Rightarrow b=0 $ and $ c=3$,
$a=1 \Rightarrow b=2 $ and $ c=0$,
$a=2 \Rightarrow$ no solution.
$a=3 \Rightarrow b=0 $ and $ c=0$,
$\Rightarrow$ the coefficient of $x^3$ is $3$ which equals the number of partitions of $3$.
Do I understand what the question is asking? Because I'm wondering if just multiplying the brackets and obtaining the coefficient is an equivalent solution to the question.
Thanks.
| You can see the "situation" as I try and depict in this sketch
For what concerns the partitions of $3$ (and of $2$ and $1$) , all the necessary "pieces" are provided.
Explanation of the sketch
When you expand the first polynomial, this will provide a factor $x^k$ to multiply the rest, and $k$ is either $0$ or $1$ or $2$ or $3$.
The second polynomial will provide a factor $x^j$ with $j=0,2$, and the third a factor $x^n$ with $n=0,3$.
When then you collect the terms $x^{(j+k+n)}$, grouping together all the terms with the exponent $j+k+n=q$, then each of the various terms will correspond to each one of the partitions of $q$ composed of $0..3$ parts $=1$ +$0..1$ parts $=2$ + $0..1$ parts $=3$, and the resulting coefficient of $x^q$ will be the number of those partitions.
| {
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"url": "https://math.stackexchange.com/questions/2019205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find all positive three - digit integers $\overline{abc}$ such that $\overline{abc}=5abc$ Number is divisible by $5$ if the last digit is $0$ or $5$. Since $c\neq 0$, $c=5$.
Using brute force (checking each combination with three digits), I get that the only solution is $175$ : $a=1,b=7,c=5$.
How to solve a problem efficiently?
| You already know that $c=5$.
Hence $100a+10b+5=25ab$.
Hence $20a+2b+1=5ab$.
$20a+2b+1$ is odd.
Hence $5ab$ is odd.
Hence $5ab$ ends with $5$.
Hence $20a+2b+1$ ends with $5$.
Hence $2b+1$ ends with $5$.
Hence $2b$ ends with $4$.
Hence $b=2$ or $b=7$.
Since $5ab$ is odd, $b$ must be odd.
Hence $b=7$.
Hence $20a+2\cdot7+1=5a\cdot7$.
Hence $a=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020594",
"timestamp": "2023-03-29T00:00:00",
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Solve for $x$ in $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$ $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$
Note that I didn't forget any parenthesis. I copied it as it is in my book.
I tried:
$$3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0 \Leftrightarrow 3\cdot 5^{x^2} - 5 \cdot 3^{2x} = 0 \Leftrightarrow 5 \cdot 3^{x^2} = 3 \cdot 5^{x^2} \Leftrightarrow ???$$
What do I do next?
Also, this $3^{x^2+1}$ confuses me. Does it expand to $3^{x^2}\cdot 3$ or $3^{x^2}\cdot 3^x$?
| $3^{x^2+1} = 3^{x^2} \cdot 3$ is correct.
Furthermore,$$ 5^{x^2} \cdot 3 = 3^{x^2} \cdot 5 \iff 5^{x^2-1} = 3^{x^2-1} \iff \Big(\frac{5}{3} \Big)^{x^2-1} = 1$$.
This happens exactly when $x^2-1=0$,or when $x = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Simplify $\frac {2\sqrt{x^2+x+1}-2-x} {x^2} $ In order to find the limit of the function at $0$ I need to simplify the fraction $$\frac {2\sqrt{x^2+x+1}-(2+x)} {x^2} $$
so that at $0$ it didn't look like $\frac 0 0$. I'm almost certain that it can be solved with a clever substitution of the variable but I didn't manage to find one.
I've also tried multiplying numerator and denominator by $2\sqrt{x^2+x+1}+2+x$, but without any success.
| Multiplying numerator and denominator by $$2\sqrt{x^2+x+1}+2+x$$ we get
after simplification $$\frac{3x^2}{x^2(2\sqrt{x^2+x+1}+2+x)}$$ and after cancelling $$x^2$$ you will get the searched limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is that:
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} \tag{$\ast$} $$
is always an integer? On the same page I found the formula:
$$ \left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\
1 & b & b^2 & b^4 \\
1 & c & c^2 & c^4 \\
1 & d & d^2 & d^4 \\ \end{array} \right| = P \times (a+b+c+d)$$
My guess is that $P$ is short-hand for the product of all the different pairs of numbers:
$$ P = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) $$
This is not the right formula I'm afraid. However, I do see:
$$ \left|\begin{array}{cccc} 1 & 2 & 16 \\
1 & 3 & 81 \\
1 & 5 & 625 \end{array} \right| = P \times \stackrel{2}{C}(2,3,5)$$
which could mean anything. The book is in German - which I do not understand, and the man has invented this wonderful symbol $\stackrel{2}{C}$.
Mainly I am wondering if other integer identities can be found this way. Or is it only the case for $(a,b,c) = (2,3,5)$.
| Because
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} = a^2 + a b + a c + b^2 + b c + c^2 \text{,} $$
the result is certainly an integer for distinct integers $a$,$b$,$c$. (It's undefined if any pair of them are equal.)
Your guess about $P$ being the product of differences of alphabetical pairs is correct.
Up to a minus sign, $\overset{2}{C}(2,3,5) = -69 $, is the negative of the first example you gave.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The number of paths of length $n$ that end at a specific vertex in a grid-like graph Suppose you have a graph consisting of nine vertices laid out in a $3\times3$ grid where each vertex is connected to each of its horizontal, vertical, and diagonal neighbors. So the corner vertices have degree three, the edge vertices have degree five, and the center vertex has degree eight.
I want to count the number of paths of length $n$ that end at a given node $v$. You can make the observation that this number is simply the sum of the number of paths of length $n-1$ that end at each neighbor of $v$. With this in mind we can calculate this number iteratively for each vertex and for each $n$.
$$
\begin{matrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
& {n=1} & \\
\end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
3 & 5 & 3 \\
5 & 8 & 5 \\
3 & 5 & 3 \\
& {n=2} & \\
\end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
18 & 24 & 18 \\
24 & 32 & 24 \\
18 & 24 & 18 \\
& {n=3} & \\
\end{matrix}
\quad\longrightarrow\quad\dotsb
$$
Is there a (nice) formula for the number of paths of length $n$ that end at a corner, edge, and center vertex?
| Instead of using normal $3\times 3$ matrix, write it out as a $9\times 1$ vector : $$\left[\begin{array}{c}1&2&3\\4&5&6\\7&8&9\end{array}\right] \to [1,2,3,4,5,6,7,8,9]^T$$
and use adjacency matrix for multiplication:
$$
\left[
\begin{array}{c}
0&1&0&1&1&0&0&0&0\\
1&0&1&1&1&1&0&0&0\\
0&1&0&0&1&1&0&0&0\\
1&1&0&0&1&0&1&1&0\\
1&1&1&1&0&1&1&1&1\\
0&1&1&0&1&0&0&1&1\\
0&0&0&1&1&0&0&1&0\\
0&0&0&1&1&1&1&0&1\\
0&0&0&0&1&1&0&1&0
\end{array}
\right]$$
The matrix is symmetric, and according to Wolfram Alpha it has rather nice diagonal form.
All combined, that gives you a closed-form solution for any position in your matrix.
To make it a bit easier, it's enough to have $3\times 1$ vector (corner, side, center), as all the other values are symmetric, and the simplified matrix also has a nice diagonal form, no surprises here. For example:
$$
\left[\begin{array}{c}0&2&1\\ 2&2&1 \\ 4&4&0\end{array}\right]^2 \left[\begin{array}{c}1\\1\\1\end{array}\right] = \left[\begin{array}{c}18\\24\\32\end{array}\right]
$$
I hope this helps $\ddot\smile$
| {
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"url": "https://math.stackexchange.com/questions/2025395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula? To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ , but I do not see how we can get the antiderivate of $\ \frac{1-x^2}{(x^2+1)^2}\ $ either without any guess (If we know this antiderivate , we can express $\frac{1}{(x^2+1)^2}$ as a linear combination of $\ f(x):=\frac{1}{x^2+1}\ $ and $\ g(x):=\frac{1-x^2}{(x^2+1)^2}\ $ , namely $\ \frac{1}{(x^2+1)^2}=\frac{f(x)+g(x)}{2}\ $)
How can I calculate $\int \frac{dx}{(x^2+1)^2}$ only by using intgration by parts and the substitution rule as well as other basic integration rules ? I am looking for a solution not containing a guess or the iteration formula.
| For substitution method, see How do I solve this trigonometric substitution integral?.
For intgration by parts,
$$\int\dfrac{dx}{(1+x^2)^2}=\int\dfrac{2x}{(1+x^2)^2}\cdot\dfrac1{2x}\ dx$$
$$=\dfrac1{2x}\int\dfrac{2x}{(1+x^2)^2}dx-\int\left(\int\dfrac{2x}{(1+x^2)^2}dx\cdot \dfrac{d(1/2x)}{dx}\right)dx$$
$$=-\dfrac1{2x(1+x^2)}-\dfrac12\int\dfrac{dx}{(1+x^2)x^2}$$
Now $$\int\dfrac{dx}{(1+x^2)x^2}=\int\dfrac{(1+x^2)-x^2}{(1+x^2)x^2}dx=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $x>0$, then find the greatest value of the expression $ \frac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$
If $x>0$, then find the greatest value of the expression $ \dfrac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$
This expression simplifies to $ \frac{(x^{100})(x-1)}{x^{200}-1}$ using sum of n terms of GP. Now one can find the maxima by equating the derivative to zero. But is there any other way not involving calculus to get the maximum value?
| AM-GM inquelity $=>$
$$
\\1+x+\cdots+x^{2n}\geq (2n+1)(x^{0+1+\cdots+2n})^{1/{2n}}=(2n+1){x^{\frac{2n+1}{2}}}
\\f(x)=\frac{x^n}{1+x+\cdots+x^{2n}}\leq\frac{x^n}{(2n+1){x^{\frac{2n+1}{2}}}}=\frac{1}{(2n+1)\sqrt{x}}
\\
\\\max_{x>0}(f(x))=\frac{1}{(2n+1)\sqrt{x}}
$$
then $f(x)=\frac{1}{(2n+1)\sqrt{x}}\>$ if and only if $x=1$
$$\\$$
ANSWER $\>\frac{1}{2n+1}=\frac{1}{201}\>$.
(Sorry english my second language)
| {
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"url": "https://math.stackexchange.com/questions/2028324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Calculate the exact value of $\sin\frac{11\pi}{8}$ Calculate the exact value of $\sin\frac{11\pi}{8}$.
The formula $\sin^2x=\frac12(1–\cos2x)$ may be helpful.
I was thinking of using the Angle-Sum and -Difference Identity:
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
For instance: $\alpha=\frac{20\pi}{8}$ and $\beta=\frac{9\pi}{8}$
Am I on the right track?
| Another idea: Let be $x = 3\pi/8$
$$0 = \sin(3\pi) = \sin(8x) = 2\sin(4x)\cos(4x) = \cdots $$
$$= (-128\sin^7 x + 192\sin^5 x - 80\sin^3 x + 8\sin x)\cos x.$$
Then, $s = \sin x$ is a root of
$$0 = 128s^7 - 192s^5 + 80s^3 - 8s = 8(16s^6 - 24s^4 + 10s^2 - 1)s = 8(8s^4 - 8s^2 + 1)(2s^2 - 1)s.$$
The roots of the biquadratic are
$$\pm\sqrt{\frac12\pm\frac{\sqrt2}4}.$$
Finally, you must discard the wrong solutions using that $s>1/2$ (why this is true?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
$$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
| Let $z=a-bi$, then $z^2-z=(a^2-b^2-a)+(-2ab+b)i$
\begin{align*}
z^2-z &= \frac{i}{2} \\
z &= \frac{1\pm \sqrt{1+2i}{}}{2} \\
a-bi &=
\frac{1}{2}
\left(
1 \pm \sqrt{\frac{\sqrt{5}+1}{2}} \pm i\sqrt{\frac{\sqrt{5}-1}{2}} \,
\right) \\
a &= \frac{1}{2} \pm \sqrt{\frac{\sqrt{5}+1}{8}} \\
b &= \mp \sqrt{\frac{\sqrt{5}-1}{8}} \\
\end{align*}
N.B.:
$$x+yi=
\left(
\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}+
i\frac{y}{|y|} \sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} \,
\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
How to find the general formula of the recurrence sequence? Let two sequences $\{a_n\},\{b_n\}$ such that
$a_{2n-1}=a_{2n}=a_{2n-2}+\frac{1}{b_{2n-2}}$,
$b_{2n}=b_{2n+1}=b_{2n-1}+\frac{1}{a_{2n-1}}$,
$a_1=b_1=1$.
With the help of OEIS, I find a solution:
$a_{2n-1}=a_{2n}=\frac{(2n-1)!!}{(2n-2)!!}$,
$b_{2n}=b_{2n+1}=\frac{(2n)!!}{(2n-1)!!}$,
where $(2n-1)!!=1\times3\times\cdots\times(2n-1)$ and $(2n)!!=2\times4\times\cdots\times(2n)$.
How can I get it without mathematical induction?
| I only manage to decouple the recurrence relations: We have
\begin{array}{rlcrl}
a_{2n} &= a_{2n-2} + 1/b_{2n-2} \quad (*) & \quad &
b_{2n+1} &= b_{2n-1} + 1/a_{2n-1} \quad (**)\\
a_{2n} &= a_{2n-1} & \quad &
b_{2n+1} &= b_{2n} \\
a_1 &= 1 & \quad &
b_1 &= 1
\end{array}
then solving equation $(**)$ for $a_{2n-1}$ in terms of the $b$ (and similar solving $(*)$ for $b_{2n-2}$ in terms of the $a$), and some index translation, we get
$$
a_{2n} = a_{2n-1} = \frac{1}{b_{2n+1}-b_{2n-1}} \quad\quad
a_{2n-2} = a_{2n-3} = \frac{1}{b_{2n-1}-b_{2n-3}} \\
b_{2n-1} = b_{2n-2} = \frac{1}{a_{2n} - a_{2n-2}} \quad\quad
b_{2n+1} = b_{2n} = \frac{1}{a_{2n+2} - a_{2n}} \\
$$
and inserting back, plus changing $b_{2n-2}\to b_{2n-1}$ and $a_{2n-1} \to a_{2n}$, we get
$$
\frac{1}{b_{2n+1}-b_{2n-1}}
= \frac{1}{b_{2n-1}-b_{2n-3}} + \frac{1}{b_{2n-1}} \\
\frac{1}{a_{2n+2}-a_{2n}}
= \frac{1}{a_{2n}-a_{2n-2}} + \frac{1}{a_{2n}}
$$
A recurrence relation
$$
\frac{1}{c_{k+1} - c_k}
= \frac{1}{c_k-c_{k-1}} + \frac{1}{c_k}
$$
is non-linear. We can try to transform it:
$$
\frac{1}{c_k} = -\frac{\Delta(c_k) - \Delta(c_{k-1})}{\Delta(c_k) \Delta(c_{k-1})} \iff \\
\frac{\Delta(c_k)}{c_k}
= -\frac{\Delta\Delta(c_{k-1})}{\Delta(c_{k-1})}
$$
or
$$
\frac{1}{c_{k+1} - c_k}
= \frac{2c_k-c_{k-1}}{c_k(c_k-c_{k-1})} \iff \\
$$
\begin{align}
c_{k+1}
&= c_k + \frac{c_k(c_k-c_{k-1})}{2c_k-c_{k-1}} \\
&= c_k \left( 1 + \frac{c_k-c_{k-1}}{2c_k-c_{k-1}} \right) \\
&= c_k \frac{3c_k-2c_{k-1}}{2c_k-c_{k-1}} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}$ I'm interested in the integral
$$
I=\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}.\tag{1}
$$
So far I have been able to reduce this integral to an integral of an elementary function in the hope that it will be more tractable
$$
I=\frac{8\pi}{\sqrt{3}}\int_{-\infty}^\infty\frac{e^{ix\sqrt{3}}\ dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^3},\tag{2}
$$
using the approach from this question. In that question it was also proved that
$$
\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3},\tag{3}
$$
which gives some indication that the integral in the right hand side of $(2)$ might be calculable.
Also note that the integrand in $(1)$ can be expressed as
$$
\Gamma(x+1)\left|\Gamma\left(1+e^{\frac{2\pi i}{3}}x\right)\right|^2.
$$
Bending the contour of integration in the integral on the RHS of $(2)$ one obtains an alternative representation
$$
I=8\pi\int_0^\infty\frac{e^{x\sqrt{3}}~dx}{\left(2\cos x+e^{x\sqrt{3}}\right)^3}.\tag{4}
$$
There are some calculable integrals containing the infinite product $\prod\limits_{k=1}^\infty\left(1+\frac{x^3}{k^3}\right)$, e.g.
$$
\int_{0}^\infty\frac{\left(1-e^{\pi\sqrt{3}x}\cos\pi x\right)e^{-\frac{2\pi}{\sqrt{3}}x}\ dx}{x\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}=0.
$$
Q: Is it possible to calculate $(1)$ in closed form?
| define:
$$ Q^n(x) = \prod_{a=1}^n{1-\left(\frac{x}{a}\right)^3}
= \left(1-\left(\frac{x}{a}\right)^3\right)
\cdot\prod_{b\ne a}{1-\left(\frac{x}{b}\right)^3}
= \left(1-\left(\frac{x}{a}\right)^3\right)\cdot Q_a(x)
$$
using:
$$\begin{align*}
(a) && & 1-\left(\frac{x}{a}\right)^3 = 0
\Leftrightarrow x=az_i, \; (z_i)^3=0.\; i=0,1,2 \\
(b) && & \frac{d}{dx}\left(1-\left(\frac{x}{a}\right)^3\right)
= -3\frac{x^2}{a^2}; \; (x=az_i)\; -3\frac{\bar{z}_i}{a} \\
(c) && & (b\ne a) \Rightarrow1-\left(\frac{az_i}{b}\right)^3
= 1-\left(\frac{a}{b}\right)^3 \\
(d) && & \frac{d}{dx}Q^n(x)=\left(1-\left(\frac{x}{a}\right)^3\right)Q'_a(x)
- 3\frac{x^2}{a^3}Q_a(x) \\
(a,b,c,d) \Rightarrow (e) && & Q'(az_i)=0-3\frac{\bar{z}_i}{a}Q_a(az_i)
= -3\frac{\bar{z}_i}{a}
\prod_{b\ne a}{\left(1-\left(\frac{a}{b}\right)^3\right)}
= -3\frac{\bar{z}_i}{a}P(a)
\end{align*}$$
$$\begin{align*}
\frac1{Q^n(x)} && & \stackrel{pfd}{=}
\sum_{a=1}^n{\sum_{i=0}^3{\frac1{(x-az_i)\cdot Q'_a(x)}}} \\
&& & \stackrel{(e)}{=} \sum_{a=1}^n{\sum_{i=0}^3{\frac1{-3\bar{z_i}(x-az_i)}
\cdot \frac{a}{P(a)}}} \\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\sum_{i=0}^3{\frac{z_i}{-3(x-az_i)}}}\\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{a^2}{a^3-x^3}}\\
\end{align*}$$
$$\begin{align*}
\int_0^{\infty}{\frac1{Q_n(-x)}dx} && & = \int_0^{\infty}{
\sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{a^2}{a^3+x^3}}dx} \\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\int_0^{\infty}{\frac{a^2}{a^3+x^3}dx}}\\
&& & = \sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{2\pi}{3\sqrt{3}}} \\
\end{align*}$$
$$\begin{align*}
\lim_{n\to\infty}\int_0^{\infty}{\frac1{Q_n(-x)}dx} && &
= \lim_{n\to\infty}\sum_{a=1}^n{\frac{a}{P(a)}\cdot\frac{2\pi}{3\sqrt{3}}}\\
&& & = \frac{2\pi}{3\sqrt{3}}\lim_{n\to\infty}\sum_{a=1}^n{\frac{a}{P(a)}}\\
\end{align*}$$
Answer:
$$\frac{2\pi}{3\sqrt{3}}\sum_{a=1}^{\infty}{a\prod_{b\ne a}{\left(1-\left(\frac{a}{b}\right)^3\right)^{-1}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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On the integral $\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}$ V. Reshetnikov gave the interesting integral,
$$\int_0^1\frac{\mathrm dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{2-x\,\sqrt3}}=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
After some experimentation, it turns out that more generally, given some integer/rational $N$, we are to find an algebraic number $\gamma$ that solves,
$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$
(Compare to the similar integral in this post.) Equivalently, to find $\gamma$ such that,
$$\begin{aligned}
\frac{1}{N}
&=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)\\[1.8mm]
&= \frac{B\left(\gamma^2;\ \tfrac14,\tfrac14\right)}{B\left(\tfrac14,\tfrac14\right)}
\end{aligned} \tag3$$
with beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$, and $B\left(\tfrac14,\tfrac14\right)=\frac{\sqrt\pi}{\Gamma^2\left(\frac14\right)}$. Reshetnikov's example, after tweaking, was just the case $N=\frac{3}{2}$ and $\gamma=\frac{3^{1/4}}{\sqrt{2}}$.
Solutions for prime $N=2,3,5,7$ are known. Let $v=\gamma$, then,
$$-1 + 2 v^2 = 0\quad\quad N=2\\ - 1 + 2 v + 2 v^2 = 0\quad\quad N=3\\ - 1 + 8 v - 4 v^2 - 8 v^3 + 4 v^4 = 0\quad\quad N=5$$
etc, with $N=7$ using a $12$-deg equation. I found these using Mathematica's FindRoot command but, unlike the other post, I couldn't find a nice common form for $\gamma$. (The pattern of this family is also different. I had expected $N=7$ to also involve a sextic only.)
Q: Is it true one can find algebraic number $\gamma$ for all prime $N$? What is it for $N=11$?
Update, Aug 16, 2019
In this comment, Reshetnikov gave the explicit solution to,
$$I\left(\gamma^2;\ \tfrac14,\tfrac14\right) = \tfrac17$$
as,
$$\small\gamma = \frac16\left(5\cos x-\sqrt3\sin x-1-\sqrt3\sqrt{7+4\sqrt7-(11+2\sqrt7)\cos x+\sqrt3(5+2\sqrt7)\sin x}\right)$$
where $x = \tfrac13\arccos\big(\tfrac{13}{14}\big)$.
P.S. I forgot I also found $\gamma$ in this 2016 post as,
$$\gamma = \tfrac12\left(2\cos\tfrac{2\pi}7-\sqrt{2\cos\tfrac{4\pi}7+\sqrt2\csc\tfrac{9\pi}{28}}\right)$$
| It seems solving $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac1N$ also solves $I(\gamma^2;\tfrac{1}{4},\tfrac{1}{4})=\frac1N$ as they involve reciprocal polynomials. So again following Nemo's lead in this answer, we employ the duplication formula,
$$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{4})=I(1-q^2;\tfrac{1}{2},\tfrac{1}{4})$$
where $p,q$ are related by the $8$-deg,
$$p^2 (-1 + 2 q + q^2)^4 = 16 q (-1 + q^2) (1 + q^2)^2\tag1$$
So given a solution to $\displaystyle I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})= \frac{1}{2^n N}$ for $n=0$, the duplication formula enables us to find infinitely many for $n\geq1$.
For example, the solution to $I(p^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{3}$ is $p = 1-\sqrt3$. One can then solve for $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{6}$ by plugging the known $p$ into $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 1,
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The sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically How could we show that the sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically?
When we take the quotient $\frac{a_n}{a_{n+1}}$ we get $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}$. Howcan we conclude that this quotient is $\geq 1$ ?
| Every term of the sequence is positive. So, the sequence is decreasing if, and only if, the ratio $a_{n+1}/a_{n} < 1$ for every integer $n \geq 2$.
\begin{equation*}
\frac{a_{n+1}}{a_{n}} = \frac{(2n+3)(n-1)^2}{(2n+1)n^{2}} = \frac{(2n+3)(n^{2} - 2n + 1)}{2n^{3} + n^{2}}
= \frac{2n^{3} - n^{2} - 4n + 3}{2n^{3} + n^{2}} .
\end{equation*}
For every integer $n \geq 2$,
\begin{equation*}
2n^{2} + 4n - 3 = \left(n + 2 - \sqrt{\frac{5}{2}}\right)\left(n + 2 + \sqrt{\frac{5}{2}}\right) > 0 .
\end{equation*}
So,
\begin{equation*}
2n^{3} - n^{2} - 4n + 3 < 2n^{3} + n^{2} ,
\end{equation*}
and $a_{n+1}/a_{n} < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that a group is abelian under given conditions Let $(G, \cdot)$ be a group with the property that there exists $a \in G$ such that $ax^3a = x$, $\forall x \in G$. Prove that $(G, \cdot)$ is abelian.
Here is my solution:
By plugging in $x := ax$ in the given condition, we get $a^3x^3a = x$. But $ax^3a$ is also equal to $x$, so $a^3x^3a = ax^3a$ $\Rightarrow$ $a^2 = e$. That means that $a = a^{-1}$.
Now, using $a = a^{-1}$ in the given condition, that is $ax^3a = x$, we get $ax^3 = xa$. By squaring both sides, we get $a^2x^6 = x^2a^2$. But we know that $a^2 = e$, so $x^6 = x^2$, or, equaivalently, $x^4 = e$.
I know that if $x^2 = e$, $\forall x \in G$, then the group is abelian. But does this hold if $x^4 = e$?
Thank you!
| There are a couple of issues with your argument, but it starts with
By plugging in $x := ax$ in the given condition, we get $\color{red}{a^3x^3a = x}$.
The equation in red is incorrect. If you replace $x$ with $ax$, then you would get $a(ax)^3a = ax$.
Replace $x$ with $axa^{-1}$ in the condition to get $a(axa^{-1})^3 a = axa^{-1}$, or $a^2x^3a^{-2} = axa^{-1}$. Cancellation gives $ax^3a^{-1} = x$; in particular, setting $x = a$ yields $a^3 = a$, or $a^2 = e$. Now $$(yx)^3 = a^{-1}(yx)a = (a^{-1}ya)(a^{-1}xa) = y^3x^3,$$ that is, $y(xy)^2x = y^3x^3$. By cancellation, $(xy)^2 = y^2x^2$. Thus,
$$x^2y^2 = ax^6a^{-1} ay^6a^{-1} = ax^6y^6a^{-1} =a(y^2x^2)^3a^{-1} = y^2x^2.$$
Hence $(xy)^2 = x^2y^2$, i.e., $x(yx)y = x^2y^2$. Cancellation results in $yx = xy$. Consequently, $G$ is abelian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum area of a nth-side polygon whose vertices satisfy In the Argand diagram, a square has its vertices satisfying a 4th power equation with coefficients being integers, i.e.
$$z^4+az^3+bz^2+cz+d=0, z\in \mathbb C, a,b,c,d \in \mathbb Z$$
Find the minimum area.
I suspect that the vertices statisfying $$z^4=1$$ will do, but I cannot find a way to prove it.
Apart from square, the question also asks me to find the minimum area of nth-side polygon. Again, I lack some insights to solve this problem.
| \begin{align*}
f(z) &= \left( z+\frac{a}{4} \right)^4+d-\frac{a^4}{256} \\
&= z^4+az^3+\frac{3a^2}{8}z^2+\frac{a^3}{16}z+d \\
\left| z_k+\frac{a}{4} \right| &=
\sqrt[4]{\left| d-\frac{a^4}{256} \right|} \\
\text{area} &= 2\sqrt{\left| d-\frac{a^4}{256} \right|}
\end{align*}
Case I: $a=0$
$$\text{area}=2\sqrt{d}$$ where $d\in \mathbb{Z}[i]\backslash \{ 0 \}$
Case II: $a\ne 0$
$$\dfrac{3a^2}{8},\, \dfrac{a^3}{16},\, d\in \mathbb{Z}[i]$$
$$\implies a=4m+4ni \quad \text{or} \quad a=(4m+2)+(4n+2)i$$
Without loss of generality, considering the cases for $a=4$ and $a=2+2i$ only.
The area is given by
$$2\sqrt{\left| d-1 \right|} \quad \text{and}
\quad 2\sqrt{\left| d+\frac{1}{4} \right|} \quad \text{respectively}$$
Hence, the minimum area is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove $\sin(\cos x) \neq \cos(\sin x)$ Prove that the trigonometric equation
sin(cosx) = cos(sinx)
has no solutions.
I graphed the functions and I can see they are never equal to eachother, but how do I do a formal proof for this?
| $$\cos(\sin x)=\sin(\cos x)=\cos\left(\frac\pi2-\cos x\right)$$
$$\text{So}, \sin x=2n\pi\pm \left(\frac\pi2-\cos x\right)$$
$$\text{Taking the '+' sign,} \sin x=2n\pi+ \left(\frac\pi2-\cos x\right)\implies \sin x+\cos x=\frac{(4n+1)\pi}2$$ which can be $\cdots,-\frac{3\pi}2,\frac{\pi}2,\frac{5\pi}2,\cdots$
$$\text{Taking the '-' sign,} \sin x=2n\pi- \left(\frac\pi2-\cos x\right)\implies \sin x-\cos x=\frac{(4n-1)\pi}2$$ which can be $\cdots,-\frac{5\pi}2,-\frac{\pi}2,\frac{3\pi}2,\cdots$
Now let, $1=r\cos\theta,1=r\sin\theta$ where $r>0$
So, $(r\cos\theta)^2+(r\sin\theta)^2=1+1=2\implies r^2=2\implies r=\sqrt2$
$\sin x\pm\cos x=r\cos\theta\sin x\pm r\sin\theta\cos x=\sqrt2\sin(x\pm \theta)$
So, $-\sqrt2\le \sin x\pm\cos x\le \sqrt 2 $
Now, $\sqrt 2<1.5<\frac\pi2$ as $3<\pi$
$\implies -\sqrt 2>-\frac\pi2$
$\implies -\frac\pi2<-\sqrt2\le \sin x\pm\cos x\le \sqrt 2<\frac\pi2 $
Hence, there is no real soultion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| You could solve it this way:
The second fraction, $\frac {2}{2x+6}$ can be simplified into $\frac 1{x+3}$. Thus,
$$\frac 5{x+3}+\frac 2{2x+6}=4\implies\frac 6{x+3}=4\tag1$$
And cross multiplying, we get $4(x+3)=6\implies x=x=-\frac 32$
As to where you went wrong, nothing has gone wrong yet. Use the quadratic formula and see what you get!
Hint: You will get two solutions. One of them doesn't work (i.e it's extraneous).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving two infinite nested radicals from a condition Kindly please tell me how to prove the following problem:
Solve completely : $$ x^2 =y+a ; y^2=z+a ; z^2 =x+a ...(1) $$ and show that (with or without using the preceding question): $$\sqrt{11-2\sqrt{11+2\sqrt{11-...}}} = 1+4\sin 10^\circ$$ and $$\sqrt{23-2\sqrt{23+2\sqrt{23+2\sqrt{23-...}}}} = 1+4\sqrt{3}\sin 20^\circ$$. The signs under the outward-most radical sign have period of three; they are respectively; $-,+,- ; -,+,+$. I think that equation $(1)$ implies that $x$ satisfies a polynomial equation of degree $8$ but I'm not sure. Any help is appreciated.
| We need re-scaling:
\begin{align*}
bx &= \color{red}{-2}\sqrt{11\color{green}{-2}\sqrt{11 \color{blue}{+2}\sqrt{11 \color{red}{-2}\sqrt{11-\ldots}}}} \\
by &= \color{green}{-2}\sqrt{11 \color{blue}{+2} \sqrt{11 \color{red}{-2} \sqrt{11 \color{green}{-2}\sqrt{11-\ldots}}}} \\
bz &= \color{blue}{+2}\sqrt{11 \color{red}{-2} \sqrt{11 \color{green}{-2} \sqrt{11 \color{blue}{+2}\sqrt{11-\ldots}}}} \\
b^2x^2 &= 4(11+by) \\
b^2 &= 4b \\
b &= 4 \\
a &= \frac{11}{4} \\
\end{align*}
Note that $$\fbox{$y<x<0<z$}$$
Next is plug and play:
\begin{align*}
\xi &= -\frac{1}{2}-2\sin 10^{\circ} \\
\eta &= \xi^2-\frac{11}{4} \\
&= -\frac{1}{2}+2\sin 10^{\circ}-2\cos 20^{\circ} \\
\zeta &= \eta^2-\frac{11}{4} \\
&= -\frac{1}{2}+2\sin 10^{\circ}+2\cos 40^{\circ} \\
\zeta^2-\frac{11}{4}
&= -\frac{1}{2}+16\sin 10^{\circ}-18\cos 20^{\circ}+18\cos 40^{\circ} \\
&= -\frac{1}{2}+16\sin 10^{\circ}-36\sin 10^{\circ}\sin 30^{\circ} \\
&= -\frac{1}{2}-2\sin 10^{\circ} \\
\xi &= \zeta^2-\frac{11}{4}
\end{align*}
Also
\begin{align*}
\xi-\eta
&= 2\cos 20^{\circ}-4\sin 10^{\circ} \\
&= 2(\cos 20^{\circ}-\cos 80^{\circ})-2\sin 10^{\circ} \\
&= 4\sin 30^{\circ} \sin 50^{\circ}-2\sin 10^{\circ} \\
&= 2(\sin 50^{\circ}-\sin 10^{\circ}) \\
&> 0
\end{align*}
Now $$\fbox{$\eta<\xi<0<\zeta$}$$
Therefore $$(x,y,z)=(\xi, \eta, \zeta)$$
That is
$$\fbox{$
\sqrt{11\color{green}{-2}\sqrt{11 \color{blue}{+2}
\sqrt{11 \color{red}{-2}\sqrt{11-\ldots}}}}
= 1+4\sin 10^{\circ}
$}$$
Furthermore,
\begin{align*}
t &=
\left[
\left(
t^2-\frac{11}{4}
\right)^2-\frac{11}{4}
\right]^2-\frac{11}{4} \\
0 &=
\left(
t^2-t-\frac{11}{4}
\right)
\left(
t^3-\frac{1}{2}t^2-\frac{17t}{4}+\frac{25}{8}
\right)
\left(
t^3+\frac{3}{2}t^2-\frac{9t}{4}-\frac{19}{8}
\right)
\end{align*}
*
*The first factor corresponds to $\pm \sqrt{a \pm \sqrt{a \pm \ldots}}$
*The second factor corresponds to the cyclic permutation of $++-++-\ldots$
*The third factor corresponds to the cyclic permutation of $--+--+\ldots$
The second case is left as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Principal part: $ \frac{1}{z(\exp(z)-1)^2} $ (Laurent series) I have to compute from the following, its principal part of the Laurent series around $z=0$.
$$ \frac{1}{z(\exp(z)-1)^2} $$
I have trouble with computing the series of $(\exp(z)-1)^2$. I wanted to take out the $\frac{1}{z}$ and then multiply it by the series of the exp term.
Thanks already.
| It is convenient to represent $\exp(z)-1$ using the big-O notation.
\begin{align*}
\exp(z)-1&=\sum_{n=1}^\infty\frac{z^n}{n!}\\
&=z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\tag{1}
\end{align*}
We obtain
\begin{align*}
\frac{1}{z\left(\exp(z)-1\right)^2}&=\frac{1}{z\left(z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\right)^2}\tag{2}\\
&=\frac{1}{z\left(z^2+z^3+\frac{7}{12}z^4+\frac{1}{4}z^5+O(z^6)\right)}\tag{3}\\
&=\frac{1}{z^3}\cdot\frac{1}{1+z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)}\tag{4}\\
&=\frac{1}{z^3}\sum_{n=0}^\infty(-1)^n\left(z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)\right)^n\tag{5}\\
&=\frac{1}{z^3}\left(1-\left(z+\frac{7}{12}z^2+\frac{1}{4}z^3+O(z^4)\right)\right.\\
&\qquad\qquad\left.+\left(z^2+\frac{7}{6}z^3+O(z^4)\right)-\left(z^3+O(z^4)\right)+O(z^4)\right)\tag{6}\\
&=\frac{1}{z^3}\left(1-z+\frac{5}{12}z^2-\frac{1}{12}z^3+O(z^4)\right)\tag{7}\\
&=\color{blue}{\frac{1}{z^3}-\frac{1}{z^2}+\frac{5}{12z}}-\frac{1}{12}+O(z)\tag{8}
\end{align*}
with the blue marked part in (8)
being the principal part of $$\frac{1}{z\left(\exp(z)-1\right)^2}$$ at $z=0$.
Comment:
*
*In (2) we use the representation of $\exp(z)-1$ from (1).
*In (3) we multiply out and collect all terms with power greater or equal $6$ in $O(z^6)$.
*In (4) we factor out $\frac{1}{z^3}$.
*In (5) we apply the geometric series expansion.
*In (6) we multiply out, write the terms for $n=0,1,2$ and $n=3$ and collect all other summands in $O(z^4)$.
*In (7) we collect the terms accordingly.
| {
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"url": "https://math.stackexchange.com/questions/2049995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I prove this using mathematical induction? \begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}
Using the mathematical induction step I arrive at this :
\begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}
And I don't see any other way to continue except to divide
\begin{align}
n+1(n+3)
\end{align}
into
\begin{align}
n(n+2)+something
\end{align}
and substitute it with the beginning of the fraction.
But that doesn't get me anywhere.
| All you need to do is realize that
$$\frac{(n+1)(n+2)(2n+9)}6=\frac{(n+1)(n+1+1)(2(n+1)+7)}6$$
which is what you are trying to prove, but with $(n+1)$ everywhere $n$ originally was.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$
I tried Holder, AM-GM and more, but without success.
| Hello Michael Rozenberg . I start my proof with a classical substitution .
Put:
$A=a$;
$AB=b$;
$AC=c$;
$AD=d$
We get :
$$\frac{1}{\sqrt{1+3B}}+\frac{B}{\sqrt{B+3C}}+\frac{C}{\sqrt{C+3D}}+\frac{D}{\sqrt{D+3}}\geq \sqrt{1+B+C+D}$$
Make a new substitution like this :
$x=B$;
$C=xh$;
$D=x\beta$
We get :
$$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\frac{x\beta}{\sqrt{x\beta+3}}\geq \sqrt{1+x(1+h+\beta)}$$
We can rewriting the inequality like this :
$$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}\geq 0$$
Now we introduce the following function :
$$f(x)=\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}$$
The idea is to integrate the function $f$ we get :
$$F(x)=\frac{2}{3}\sqrt{1+3x}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}+\frac{2}{3\beta}(x\beta+3)^{\frac{3}{2}}-\frac{6}{\beta}\sqrt{x\beta+3}-\frac{2}{3(1+h+\beta)}(1+x(1+h+\beta))^{\frac{3}{2}}$$
Now the idea is to note that the function $f$ is positiv if the function $F$ is increasing .
Namely when we have :
$F(X)\leq F(Y)$ with $X\leq Y$
Here we take $X=x-\epsilon$ and $Y=y+\epsilon$
We cut the function $F$ in others to a question of space
We have :
$g(x-\epsilon)-g(x+\epsilon)=\frac{2}{3}\sqrt{1+3(x-\epsilon)}-\frac{2}{3}\sqrt{1+3(x+\epsilon)}$
$u(x-\epsilon)-u(x+\epsilon)=\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}-(\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}})$
$v(x-\epsilon)-v(x+\epsilon)=\frac{2}{3\beta}((x-\epsilon)\beta+3)^{\frac{3}{2}}-\frac{2}{3\beta}((x+\epsilon)\beta+3)^{\frac{3}{2}}$
$q(x+\epsilon)-q(x-\epsilon)=\frac{6}{\beta}\sqrt{(x+\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x+\epsilon)(1+h+\beta))^{\frac{3}{2}}
-(\frac{6}{\beta}\sqrt{(x-\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x-\epsilon)(1+h+\beta))^{\frac{3}{2}}
)$
Now we have the following elementary inequalities with $n\geq x^6$:
$g(x-\epsilon)-g(x+\epsilon)\leq -\frac{2}{3}\frac{\epsilon}{n} x^6$
$u(x-\epsilon)-u(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6(\frac{1}{\sqrt{1+3h}}+\frac{h}{\sqrt{h+3\beta}})$
$v(x-\epsilon)-v(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6$
$q(x+\epsilon)-q(x-\epsilon)\leq \frac{\epsilon}{n}(\frac{2}{3})x^6$
So you have just to add the elementary inequalities to get the result (and derivate)
If you have questions tell me .
| {
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"timestamp": "2023-03-29T00:00:00",
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Check induction proof that $\sum\limits_{i=0}^ni\left(i+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$ claim: $\sum _{i\:=\:0}^n\:i\left(i+1\right)\:=\:\frac{n\left(n+1\right)\left(n+2\right)}{3}$
base case n = 0
$\frac{0\left(0+1\right)\left(0+2\right)}{2}$ ; $0\:=\:0$
inductive hyp (n = k):
$\sum _{i=0}^k\:i\left(i+1\right)\:=\:\frac{k\left(k+1\right)\left(k+2\right)}{3}$
to show: n = k+1
$\sum
_{i=0}^k\:i\left(i+1\right)\:=\:\frac{\left(k+1\right)\left(k+1+1\right)\left(k+1+2\right)}{3}$
$\sum _{i=0}^k\:i\left(i+1\right)\:=\:\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}$
not sure what to do next
| For the case $n=k+1$:
$$\sum_{i=0}^{k+1}\:i\left(i+1\right)\:=\sum_{i=0}^{k}\:i\left(i+1\right)+(k+1)(k+2)$$
Using inductive hyp we get:
$$\sum_{i=0}^{k+1}\:i\left(i+1\right)\:=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=$$
$$(k+1)(k+2)\left(\frac{k}{3}+1\right)=\frac{(k+1)(k+2)(k+3)}{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I calculate surface of ellipse rotating around x axel? How to calculate surface of ellipse rotating around x axel when ellipse equation is :
$y^{2}+4x^{2} = 36$
I started with this formula for calculating surface of rotating object around x axel
$ \int_{a}^{b}f(x) \sqrt{1 + f'(x)^{2}} dx$
but i always end up with complicated solution, I believe there is easier way of solving this but I am missing it. Any hint and help is welcome.
| The surface area is given by
\begin{equation}
A=2\pi\int_{a}^{b}y\sqrt{1+(y^\prime)^2}\,dx
\end{equation}
with $y=2\sqrt{9-x^2}$, $a=-3,\,b=3$. Because of symmetry we can double the integral from $0$ to $3$.
Since $\sqrt{1+(y^\prime)^2}=\sqrt{\dfrac{9-3x^3}{9-x^2}}$ we have
\begin{eqnarray}
A&=&4\pi\int_{0}^{3}2\sqrt{9-x^2}\sqrt{\dfrac{9+3x^2}{9-x^2}}\,dx\\
&=&8\sqrt{3}\pi\int_{0}^{3}\sqrt{3+x^2}\,dx
\end{eqnarray}
which can be done by trigonometric substitution or by using the formula
\begin{equation}
\int\sqrt{a^2+x^2}\,dx=\dfrac{1}{2}\left[x\sqrt{a^2+x^2}+a^2\ln\vert x+\sqrt{a^2+x^2}\vert\right]+c
\end{equation}
If I substituted correctly, this gives $A=\left[72+12\sqrt{3}\ln\left(1+\frac{2\sqrt{3}}{3}\right)\right]\pi$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the length of a parabolic arc if two end points & angle at both ends are specified? First I confess I my mathematics knowledge is not very good. So first can an expert please first confirm if this question is actually solvable? Because elsewhere on net I have read that you actually need to specify 3 points to find the length of a parabolic arc.
Imagine this is a plane which takes off at an angle of 30 degrees & gradually makes an angle of 60 degrees by the time it is at a distance of 3 km from the point where it left the ground.
| Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=\tan 30°=\frac{\sqrt{3}}{3}$ and this gives $b=\frac{\sqrt{3}}{3}$.
Now the problem is to find a point $P=(p,y(p))$ such that the tangent at this point has a slope of $y'(p)=\sqrt{3}=\tan 60°$ and a distance from the origin $PO=3$.
This gives the system
$$
\begin{cases}
\sqrt{p^2+\left(ap^2+\frac{\sqrt{3}}{3}p\right)^2}=3\\
2ap+\frac{\sqrt{3}}{3}=\sqrt{3}
\end{cases}
$$
Find $a=\frac{\sqrt{3}}{3p}$ from the second equation and substitute in the first equation. With a bit of algebra you can find $p$ and solve the problem.
Substituting $a=\frac{\sqrt{3}}{3p}$, the first equation gives $p=\frac{3\sqrt{3}}{\sqrt{7}} $, so we have $a=\frac{\sqrt{7}}{9}$ and the equation of the parabola is:
$$
y=\frac{\sqrt{7}}{9}x^2 + \frac{\sqrt{3}}{3}x
$$
with derivative:
$$
y'=\frac{2\sqrt{7}}{9}x + \frac{\sqrt{3}}{3}x
$$
The ''final'' point have coordinates:
$$
P=(p,y(p))= \left(\frac{3\sqrt{3}}{\sqrt{7}},\frac{6}{\sqrt{7}} \right)
$$
So the arc length from $O$ to $P$ is given by the integral:
$$
\int_0^p \sqrt{1+[y'(x)]^2}dx= \int_0^p\sqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
\int \sqrt{1+(2ax+b)^2}dx =\frac{1}{2a}\int\sqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=\tan v$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| Let
$$f(x)=\sum_{n=0}^{+\infty}\frac{x^n}{2^n}=\frac{2}{2-x}$$
with a radius of convergence $R=2$
$$f'(x)=\sum_{n=0}^{+\infty}n\frac{x^{n-1}}{2^n}=\frac{2}{(2-x)^2}$$
thus, your sum is $f'(1)=2$.
| {
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"url": "https://math.stackexchange.com/questions/2059172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Vector Parametric Form Flat Please see the image with the question and my answer. The thing I'm having an issue with is understanding what this vector description actually is. Is it a 2-dimensional flat in $\mathbb{R}^6$?
In fact, according to the table, we have
$n-2$ Cartesian equations and $2$ free parameters in this example. So since $n−2=4$ then $n=6$? Is this true?
Thank you for taking the time to look at this post. Really appreciate your help.
| The table constructs a formula between the number of free parameters $F$, number of equations $E$, and the dimension of the space $n$ as follows
$$n-F=E\tag{1}$$
however, it should be emphasized that $E$ is the number of linearly independent equations and not just the number of equations! Now, let us see how many linearly independent equations we have in your example. For doing so we may perform the usual elementary row operations on the system of equations. This can easily be done with the augmented matrix
\begin{align}
\left[
\begin{array}{ccccc|c}
1 & 0 & 3 & 2 & 0 & 8\\
0 & 0 & 1 & 0 & 2 & 4\\
0 & 0 & 0 & 1 & 1 & 2\\
3 & 0 & 8 & 5 & -3 & 18
\end{array}
\right]
&\to
\left[
\begin{array}{ccccc|c}
1 & 0 & 3 & 2 & 0 & 8\\
3 & 0 & 8 & 5 & -3 & 18 \\
0 & 0 & 1 & 0 & 2 & 4\\
0 & 0 & 0 & 1 & 1 & 2\\
\end{array}
\right]
\\
&\to
\left[
\begin{array}{ccccc|c}
1 & 0 & 3 & 2 & 0 & 8\\
0 & 0 & \frac{1}{3} & \frac{1}{3} & 1 & 2 \\
0 & 0 & 1 & 0 & 2 & 4\\
0 & 0 & 0 & 1 & 1 & 2\\
\end{array}
\right]
\\
&\to
\left[
\begin{array}{ccccc|c}
1 & 0 & 3 & 2 & 0 & 8\\
0 & 0 & \frac{1}{3} & \frac{1}{3} & 1 & 2 \\
0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3}\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{array}
\right]
\\
\end{align}
So as we can see there are just $3$ linearly independent equations in your example as the last row turned out to be zero. So in your example we will get
\begin{align}
F&=2 \\
E&=3 \\
n&=F+E=5
\end{align}
However, when you want the dimension of the space you are working in, you can simply look at the number of variables in the equations.
The formula $(1)$ is a simple manifestation of the rank-nullity theorem which you may study in a course dealing with theoretical aspects of linear algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is my result to $z^2 - (3+4i)z - 1 + 7i = 0$ right? I solved it much like I would a second degree equation, with a=1, b=-(3+4i) and c=(-1+7i). Is that the correct approach?
My final result was $z = \frac{3+4i \pm i\sqrt{3+4i}}{2}$
Is that correct? There is no way for me to check that.
| Here is the standard way t o compute the square roots of a complex number: $$(x+iy)^2=3+4i\iff \begin{cases}x^2-y^2=3,\\2xy=4.\end{cases}$$
Further the square of the module of $x+iy$ is equal to the module of $3+4i$:
$$x^2+y^2=5.$$
So we first solve the linear system in $x^2$ and $y^2$:
$$\begin{cases}x^2-y^2=3,\\x^2+y^2=5\end{cases}\iff\begin{cases}x^2=4,\\y^2=1\end{cases}\iff\begin{cases}x=\pm2,\\y^2=\pm1.\end{cases}$$
Further,$xy=2>0$ implies $x$ and $y$ have the same sign, and finally we obtain
$$x+iy=\pm(2+i).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$
Vasc's LCF Theorem does not help here. Also I tried MV method, but without success.
| WLOG, assume that $a\ge b \ge c \ge d$.
Let
$$f(u) := \frac{u^2}{u^3 + 2}.$$
Fact 1: $f(u)$ is strictly increasing on $(0, \sqrt[3]{4}]$
and strictly decreasing on $[\sqrt[3]{4}, \infty)$.
We split into four cases.
Case 1: $c > \sqrt[3]{13 - 3\sqrt{17}}$
and $a < \sqrt[3]{13 + 3\sqrt{17}}$
Let
$$g(u) := \frac{\mathrm{e}^{2u}}{\mathrm{e}^{3u} + 2}.$$
We have
$$g''(u) = \frac{\mathrm{e}^{2u}(\mathrm{e}^{6u} - 26\mathrm{e}^{3u} + 16)}{(\mathrm{e}^{3u} + 2)^3}.$$
We have
$g''(u) \le 0$ on $[\ln \sqrt[3]{13 - 3\sqrt{17}}, ~ \ln \sqrt[3]{13 + 3\sqrt{17}}]$.
Thus, using Jensen's inequality, we have
\begin{align*}
f(a) + f(b) + f(c) &= g(\ln a) + g(\ln b) + g(\ln c)\\
&\le 3 \, g\left(\frac{\ln a + \ln b + \ln c}{3}\right)\\
&= 3 \cdot \frac{(abc)^{2/3}}{abc + 2}.
\end{align*}
It suffices to prove that
\begin{align*}
3 \cdot \frac{(abc)^{2/3}}{abc + 2} + \frac{d^2}{d^3 + 2} \le \frac43
\end{align*}
or
$$3 \cdot \frac{(1/d)^{2/3}}{1/d + 2} + \frac{d^2}{d^3 + 2} \le \frac43 $$
which is true for all $d > 0$. (Note: Let $d = x^3$.)
Case 2: $d < 3/5$ and $c \le \sqrt[3]{13 - 3\sqrt {17}}$
Using Fact 1, we have
$$f(a) + f(b) + f(c) + f(d)
\le f(\sqrt[3]{4}) + f(\sqrt[3]{4})
+ f(\sqrt[3]{13 - 3\sqrt{17}}) + f(3/5) < \frac43.$$
Case 3: $d < 3/5$ and $a \ge \sqrt[3]{13 + 3\sqrt{17}}$
Using Fact 1, we have
$$f(a) + f(b) + f(c) + f(d)
\le f(\sqrt[3]{13 + 3\sqrt{17}}) + f(\sqrt[3]{4})
+ f(\sqrt[3]{4}) + f(3/5) < \frac43.$$
Case 4: $d \ge \frac35$
Fact 2:
$\frac{1}{3} - \frac{u^2}{u^3 + 2} + \frac13 \ln u \ge 0$ for all $u \ge 3/5$.
(Note: Take derivative.)
Using Fact 2, we have
$$\sum_{\mathrm{cyc}} \left(\frac{1}{3} - \frac{a^2}{a^3 + 2} + \frac13 \ln a\right) \ge 0.$$
The desired result follows.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove at $(0,0)$, $f_{xy}$ and $f_{yx}$ aren't equal If $f(x, y) =xy(\frac{x^2-y^2}{x^2+y^2})$ when $x$ and $y$ are not simultaneously zero when $x=0,y=0$. Show that at $(0,0),\ f_{xy} \neq f_{yx} $
| First find $f_x$ and $f_y$.
For $(x,y) \neq (0,0)$ you have
$$f_x(x,y) = y \frac{x^2-y^2}{x^2+y^2} + xy \frac{2x(x^2+y^2) - 2x (x^2 - y^2)}{(x^2+y^2)^2} = y \frac{x^2-y^2}{x^2+y^2} + xy \frac{4xy^2}{(x^2+y^2)^2} = y \frac{x^2-y^2}{x^2+y^2} + \frac{4x^2y^3}{(x^2+y^2)^2}. \qquad (1)$$
Similarly,
$$f_y (x,y) = x \frac{x^2-y^2}{x^2+y^2} + xy \frac{-2y(x^2+y^2) - 2y (x^2 - y^2)}{(x^2+y^2)^2} = x \frac{x^2-y^2}{x^2+y^2} + xy \frac{-4yx^2}{(x^2+y^2)^2} = x \frac{x^2-y^2}{x^2+y^2} + \frac{-4x^3y^2}{(x^2+y^2)^2}. \qquad (2)$$
Now find partial derivatives at (0,0), using the definition
$$f_x(0,0) = \lim\limits_{\Delta x \to 0} \frac{f(\Delta x, 0) - f(0,0)}{\Delta x} = \lim\limits_{\Delta x \to 0} \frac{0 - 0}{\Delta x} = 0. \qquad (3)$$
$$f_y(0,0) = \lim\limits_{\Delta y \to 0} \frac{f(0, \Delta y) - f(0,0)}{\Delta y} = \lim\limits_{\Delta y\to 0} \frac{0 - 0}{\Delta y} = 0. \qquad (4) $$
So now you can find second derivatives. Again you have to use the definition.
$$f_{xy} (0,0) = \lim\limits_{\Delta y \to 0} \frac{f_x (0, \Delta y) - f_x(0,0)}{\Delta y} \overset{(1),(3)}{=} \lim\limits_{\Delta y \to 0} \frac{ -\Delta y - 0}{\Delta y} = -1.$$
$$f_{yx} (0,0) = \lim\limits_{\Delta x \to 0} \frac{f_y (\Delta x, 0) - f_y(0,0)}{\Delta x} \overset{(2),(4)}{=} \lim\limits_{\Delta x \to 0} \frac{ \Delta x - 0}{\Delta x} = 1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove an inequality with positive numbers Let $n\in\mathbb N$ and $a_{1},...,a_{n} >0$, such that $a_{1} + \ldots +a_{n} =1$. Show that
$$
\left(\frac{1}{a_{1}^{2}} + \ldots +\frac{1}{a_{n}^{2}} \right)\left(2(a_{1}^{2}+\ldots+a_{n}^{2}) - \frac{1}{n}\right)\ge n^2.
$$
I tried induction, but it does not work.
Any help would be great.
| Using Cauchy-Schwarz, we obtain
$$
a_1^2+\cdots+a_n^2\ge \frac{1}{n}(a_1+\cdots+a_n)=\frac{1}{n},
$$
and hence
$$
2(a_1^2+\cdots+a_n^2)-\frac{1}{n}\ge a_1^2+\cdots+a_n^2.
$$
Next, Cauchy-Schwarz again
$$
(a_1^2+\cdots+a_n^2)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)\ge (1+\cdots+1)^2=n^2,
$$
and hence
$$
\big(2(a_1^2+\cdots+a_n^2)-\frac{1}{n}\big)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)
\ge (a_1^2+\cdots+a_n^2)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)\
\ge n^2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
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