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Showing that the inflection points of $x\sin x$ lie on a certain curve
Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$
I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?
| $f(x)$ has an inflection point only if $f''(x)=0.$
$$f'(x) = \sin x + x \cos x$$
$$f''(x) = \cos x + \cos x -x \sin x=2\cos x - f(x)$$
$$f''(x) = 2\sqrt{1-\sin^2 (x)} - f(x)$$
$$f''(x) = 2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$0=2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$y=2\sqrt{1-y^2/x^2}$$
$$y^2 = 4-4y^2/x^2$$
$$...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Inequality $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$
prove that for $a,b,c$ being positives and $a+b+c=1$:$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$$
This is a very interesting inequality which i came upon accidentally.We also see that the condition $a+b+c=1$ is not needed.I modified the inequality slightly using Nesbitt's inequality and hence it boils down to prove an even stronger inequality $$\sum_{cyc}\left(\frac{b}{a}-\frac{2c}{a+b}\right)\ge 0$$ Since this was a stronger version i checked WA which shows that it is valid.I have tried to get an SOS but failed.The problem is that even after complete expanding and cross-multiplying we get a cyclic inequality and hence Muirheads theorem fails.
| First inequality:
$$ LHS = \sum_{cyc} \dfrac{bc}{a(a+c)} = \sum_{cyc} \dfrac{(bc)^2}{a^2bc+abc^2} \ge \dfrac{(ab+bc+ca)^2}{ 2abc(a+b+c)} \ge \dfrac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$ I have tried to wrap my head around this for some time now, and quite frankly I am stuck.
Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$
Then what is the value of $a/b$?
I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?
| Here is a solution using the observation in Cauchy's proof of Basel problem.
Lemma. We have
$$ \prod_{k=1}^{n} \left( t - \tan^2\left(\frac{k\pi}{2n+1}\right)\right) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k. \tag{*} $$
Using this, we immediately know that
$$ a = \sum_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{2} = n(2n+1) $$
and
$$ b = \prod_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{1} = 2n+1. $$
So the ratio $a/b$ is exactly $n$. In OP's case, $n = 8$ and hence the answer is $8$.
Proof of Lemma. Write $m = 2n+1$. Then by the de Moivre's formula,
\begin{align*}
\frac{\cos(mx) + i\sin(mx)}{\cos^m x} = \left( \frac{\cos x + i\sin x}{\cos x} \right)^m = (1 + i\tan x)^m = \sum_{k=0}^{m} \binom{m}{k} (i\tan x)^k.
\end{align*}
Taking the imaginary parts only and plugging $m = 2n+1$ back,
\begin{align*}
\frac{\sin((2n+1)x)}{\cos^{2n+1} x} = \sum_{k=0}^{n} (-1)^k \binom{m}{2k+1} \tan^{2k+1} x,
\end{align*}
and so,
\begin{align*}
(-1)^n \frac{\sin((2n+1)x)}{\sin x \cos^{2n} x} = \sum_{k=0}^{n} (-1)^{n-k} \binom{m}{2k+1} \tan^{2k} x.
\end{align*}
However, the left-hand side becomes zero if $x_k = \frac{k\pi}{2n+1}$ for $k = 1, \dots, n$, and also note that $\tan^2(x_k)$ for $k = 1, \dots, n$ are all different. So, denoting the right-hand side of $\text{(*)}$ by $P(t)$, i.e., writing
$$ P(t) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k, $$
then this implies that $P(\tan^2(x_k)) = 0$ for $k = 1, \dots, n$, hence $\tan^2(x_k)$ for $k = 1, \dots, n$ are $n$ distinct zeros of $P(t)$. Since the leading coefficient of $P(t)$ is $1$, this proves $\text{(*)}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can you find the limit of this: $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$? $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$
I would start with expanding it with $* \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}} \;$ but I don't know how to progress from there.
I can't use L'Hospital's rule. I also have 3 other exercises that are like this but if I can see one solved, I think I will be able to do the other ones as well.
Thanks
| I'm going to write it out.
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}} * \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}$
$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{1+x\sin(x) \; - \; \cos(x)}$
$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{x^2*\frac{\sin(x)}{x} \; x^2*\frac{1- \cos(x)}{x^2}}$
$\lim\limits_{x \to \; 0} \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\frac{\sin(x)}{x} \; +\frac{1- \cos(x)}{x^2}}$
$\frac{\sqrt{1+0}+1}{1+\frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of a product with growing number of factors I'm trying to solve the following limit:
$$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$
Since originally, the limit yields an $1^\infty$ indeterminate expression, my first idea was taking logarithms:
$$\log L = \lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^{n}\log\left(\frac{n^2-k}{n^2+2k-1}\right)=\\=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac{1-\frac{k}{n^2}}{1+\frac{2k-1}{n^2}}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)-\log\left(1+\frac{2k-1}{n^2}\right)\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)^{n^2}-\log\left(1+\frac{2k-1}{n^2}\right)^{n^2}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)$$
Now if I could do the limit of the terms inside the sum I'd have:
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log e^{-k}-\log e^{2k-1}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-k-2k+1\right)=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-3k+1\right)$$
and then the limit would be trivial. The question is: is that allowed? Why? If not, how could I proceed?
Thanks!
| $\require{cancel}$
$$L=\lim_{n\to\infty}\frac{\cancel{n^{2n}}(1-\frac1{n^2})(1-\frac2{n^2})\cdots(1-\frac{n}{n^2})}{\cancel{n^{2n}}(1+\frac1{n^2})(1+\frac3{n^2})\cdots(1+\frac2n-\frac1{n^2})}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}}e^{\frac{-2}{n^2}}...e^{\frac{-n}{n^2}}}{e^{\frac{1}{n^2}}e^{\frac{3}{n^2}}...e^{\frac{2n-1}{n^2}}}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}{(\sum_{r=1}^n r)}}}{e^{\frac{1}{n^2}{((2\sum_{r=1}^n r)-n)}}} \\=\lim_{n\to\infty}e^{\frac{-1}{n^2}((3\sum_{r=1}^n r)-n)}=\lim_{n\to\infty}e^{((-3\frac{1}{n}\sum_{r=1}^n \frac{r}{n})+\frac1n)}=\lim_{n\to\infty}e^{((-3\int_0^1xdx+\frac1n)}\to e^{\frac{-3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $n!>2^n$ holds for all $n>3$? I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$
| The number $n!$ is the product of the $n-1$ numbers $2,3,\ldots,n$, each of which is greater than or equal to $2$ (and all of them but $2$ is actually greater than $2$) and, since $n>3$, at least one of them is greater than or equal to $4$. So, $n!>2^{n-2}\times4=2^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$ $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.
The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:
$ \lim_{x \to +\infty} a_{n+1} = \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$
$g = \frac{\sqrt{1+g^2+1}}{g} \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$
| $$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}=\\ \frac{\sqrt{1+a_n^2}-1}{a_n}.\frac{\sqrt{1+a_n^2}+1}{\sqrt{1+a_n^2}+1}=\\\frac{1+a_n^2-1}{a_n(\sqrt{1+a_n^2}+1)}=\\
\frac{a_n}{(\sqrt{1+a_n^2}+1)}\leq \frac{a_n}{(\sqrt{1}+1)}=\frac{a_n}{2} $$so $$0<a_{n+1}\leq \frac 12a_n\leq\frac14 a_{n-1}\leq\frac18a_{n-2}\cdots<1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimizing $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ over positive reals with $a+b=1$. Why is the minimum not $18$?
If $a,b \in R^+$ such that $a+b=1$, then find the minimum value of $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$
We can write $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}=\left(a+\frac{a+b}{a}\right)^{2}+\left(b+\frac{a+b}{b}\right)^{2}=\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}$$
Using $Q.M\geq A.M\geq G.M$ we have
$$\sqrt{\frac{\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}}{2}} \geq \frac{2+\frac{a}{b}+2+\frac{b}{a}}{2}=2+\frac{\frac{a}{b}+\frac{b}{a}}{2} \geqslant 2+1=3$$
So
$$\begin{array}{l}
\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2} \geqslant 18 \\
\Rightarrow \quad\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant 18
\end{array}$$
But in an alternate approach i got the correct minimum as $12.5$
| The issue is in the first line of simplification.
$$a+\frac{a+b}{a}=a+1+\frac{b}{a}\neq 2+\frac{b}{a}.$$
If you make the correct simplification, you can continue
$$\sqrt{\frac{\left(a+1+\frac ba\right)^2+\left(b+1+\frac ab\right)^2}{2}}\geq \frac{a+b+2+\frac ab+\frac ba}{2}=\frac{3+\frac ab+\frac ba}{2}\geq \frac 52,$$
and get the correct result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why is this method right to solve some equations of degree 2? Given this equation
$-x^2+x+6=4$
We can write it as
$(3-x)(x+2) = 4$
So weridly if I take
$(3-x) = 4$ and $(x+2) = 4$, I can get $x=-1$ and $x=2$ that are indeed the correct solutions.
Why does this happen? And in which cases can this occur?
| Consider $(a-x)(x-b) = c$
With roots $r_1$ and $r_2$.
We need four equalities to be true
$$a-r_1 = c, \quad r_1-b = 1, \quad a-r_2 = 1, \quad r_2-b = c$$
We get
$$\text{$r_1=a-c=b+1$ and $r_2=a-1 = b+c$}$$
Which implies $c=a-b-1$
So for any $a$ and $b$
$$(a-x)(x-b) = a-b-1$$
will behave as you described.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$ I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$
Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$
Since $e^{x^4-2x^3+x^2-1}>0$ for all $x\in \Bbb R$ and $2x-1<0\iff x<\frac{1}{2}$. $f$ is decreasing on $(-\infty,\frac{1}{2})$.
Furthermore, $f$ is increasing on $(\frac{1}{2},\infty)$, $f$ is differentiable at $x=\frac{1}{2}$, and $f'(\frac{1}{2})=0$, $f$ attains its minimum value at $x=\frac{1}{2}$.
Am I right?
| $f(x)=\int_{0}^{x^2-x} e^{t^2-1} dt \implies f'(x)= (2x-1) e^{(x^2-x)^2-1} >0 ~if~ x>1/2$. Hence $f(x)$ in increasing for $x>1/2$ and decreasing for $x<1/2$. Yes you are right. there is a min at $x=1/2$. This one point does not matter, you may also say that $f(x)$ is increasing in $[1/2,\infty)]$ and decreasing on 4(-\infty, 1/2]$.
Note: whether a function increasing or decreasing is decided by two points (not one). For instance, $x_1>x_2 \leftrightarrows f(x_1) > f(x_2).$ If $f(x)$ is decreasing.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find maxima and minima of $f(x,y) = x^3 + y^3 -3x -3y$ in $x + 2y=3$ I need to find max and min of $f(x,y)=x^3 + y^3 -3x -3y$ with the following restriction: $x + 2y = 3$.
I used the multiplier's Lagrange theorem and found $(1,1)$ is the minima of $f$. Apparently, the maxima is $(-13/7, 17/7)$ but I could not find it via Lagrange's theorem.
Here's what I did:
I put up the linear system:
$\nabla f(x,y) = \lambda \, \nabla g(x,y)$
$g(x,y) = 0$
then,
$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$
$x + 2y -3 = 0$
Solving for $\lambda$, I got $\lambda = 0$, which gave me $x = 1$ and $y = 1$.
How can I find the maxima if lambda only gives one value which is $0$?
| This is from your working -
$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$
$3x^2 - 3 = \lambda, 3y^2-3 = 2\lambda$
Equating $\lambda$ from both equations,
$6x^2-6 = 3y^2-3 \implies 2x^2 - y^2 = 1$
Substitute $x$ from $x+2y = 3$
$2(3-2y)^2 - y^2 = 1$
$\implies 7y^2 - 24y + 17 = 0 \, $ or $(7y-17)(y-1) = 0$
Can you take it from here and find possible points for extrema?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Determine the linear function $f: R \rightarrow R$ There's a linear function $f$ whose correspondence rule is $f(x)=|ax^2-3ax+a-2|+ax^2-ax+3$. State the values of the parameter $a$ that fully define the function $f$.
According to the problem condition the function $ f (x) $ is linear by
so the coefficient that accompanies $ x ^ 2 $ must be equal to 0, this is
meets when:
\begin{align*}
ax^2-3ax+a-2 & < 0\\
a(x^2-3x) &<2-a \\
\
a\underbrace{\left(x^2-3x+\frac{9}{4}\right)}_{
\left(x-\frac{3}{2}\right)^2β₯ 0} &<2-a +\frac{9}{4}a\\
a \left(x-\frac{3}{2}\right)^2 &<2+\frac{5}{4}a\\
0 & <2+\frac{5}{4}a \\
-\frac{8}{5} & <a
\end{align*}
What else is missing to analyze?
| If there's some $x_0$ such that $ax_0^2-3ax_0+a-2>0$, then there's some neighborhood around $x_0$ such that $\forall x \in (x_0-\delta, x_0+ \delta)$, $ax^2-3ax+a-2>0$, $f(x)$ is quadratic unless $a=0$. On the other hand if $a=0$ then $f(x)$ is indeed linear.
If $a\ne 0$, it follows that $ax^2-3ax+a-2 \le 0, \forall x \in \mathbb R$. Therefore $a<0$ and the discriminant is non-negative, i.e.
$$(-3a)^2 \le 4 a (a-2) \iff a(5a+8) \le 0 \iff -\frac{8}{5} \le a <0.$$
Therefore $-\frac{8}{5} \le a \le 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
| According to the Vieta's formulas and the proposed relation, one has that
\begin{align*}
\begin{cases}
x_{1} + x_{2} + x_{3} = 3\\\\
2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = k\\\\
2x_{1}x_{2}x_{3} = 1\\\\
x^{2}_{1} + x^{2}_{2} + x^{2}_{3} = 6
\end{cases}
\end{align*}
Consequently, we have that
\begin{align*}
3^{2} = (x_{1} + x_{2} + x_{3})^{2} = x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = 6 + k \Rightarrow \boxed{k = 3}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.
What I Tried: Here is a picture :-
I made an attempt solving this and found rather a weird conclusion.
We have that $A,P,C,R$ are collinear, and also $AP = CR$.
Let $DL = LS = BK = KQ = x$ . We have $AP = x\sqrt{2}$ . Since $PC$ = $(4 - 2\sqrt{2})$ , $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ .
Now, $PF = FC$ . So by Pythagoras Theorem on $\Delta PFC$ , $PF = FC = \sqrt{12 - 8 \sqrt{2}}$.
$AL = LS = \sqrt{12 - 8 \sqrt{2}} + 2x$ and $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ . So now by Pythagoras Theorem :-
$$2\bigg(\sqrt{12 - 8 \sqrt{2}} + 2x\bigg)^2 = (4 - 2\sqrt{2} + 2x\sqrt{2})$$
$$\rightarrow 2\bigg(12 - 8\sqrt{2} + 4\sqrt{12 - 8\sqrt{2}}x + 4x^2\bigg) = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow 24 - 16\sqrt{2} + 8\sqrt{12 - 8\sqrt{2}}x + 8x^2 = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}}x = 2\sqrt{2}x - 16x$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}} = 2\sqrt{2} - 16$$
But this is obviously false.
So I literally ended up in a completely false equation, might be making a mistake in this complicated calculations, and I cannot find it. So can anyone tell me where I made the mistake, or I am solving this in the correct way or not?
| Using the fact that the side of the outer square and diagonal of the inner square are equal to the diameter of the circle,
$AD = \frac{AL + PE}{2} = \frac{2r + r\sqrt2}{2} = \frac{(2 + \sqrt2)(2 - \sqrt2)}{2} = 1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A question of Roots of unity
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
\cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.
I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.
| Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
The roots of $$\dfrac{z^9-1}{z-1}=0$$ are $e^{2\pi ir/9}; r=1,2,3,4,5,6,7,8$
Dividing both sides $z^4,$
$$z^4+\dfrac1{z^4}+z^3+\dfrac1{z^3}+z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$
$z+\dfrac1z=2\cos\dfrac{2\pi r}9=2y$(say)
Use
$\left(z+\dfrac1z\right)^2=z^2+\dfrac1{z^2}+2$
$\left(z+\dfrac1z\right)^3=z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$
$z^4+\dfrac1{z^4}=\left(z^2+\dfrac1{z^2}\right)^2-2=\left(\left(z+\dfrac1z\right)^2-2\right)^2-2$
to form a bi-quadratic equation in $y$
Now use Vieta's formula
See also: factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
| {
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"url": "https://math.stackexchange.com/questions/3937022",
"timestamp": "2023-03-29T00:00:00",
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Finding recursive formula for repeated term on sequence I have a sequence
$$3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,...$$
The pattern is so difficult. The best i can do is knowing that after jumping to the "next number" it has length $2+2n$. I mean i have $3$ as initial value. And first repeated term has lenght $2+2(1)=4$. ($4$ repeated term) The second number is $4$ and its length is $2+2(2)=6$. ($6$ repeated term). The third number is $5$ and its length is $2+2(3)=8$. ($8$ repeated term)
How to find the recursive formula? I'm really stuck. Please help me.
| $$a_0=3;\;a_n=a_{n-1}+\left\lfloor -\text{frac}\left(\frac{1}{2} \left(\sqrt{4 n+9}-3\right)\right)\right\rfloor +1$$
where $\text{frac}(x)$ is the fractional part of $x$.
From $0$ to $20$ we have
$$\{3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6\}$$
This works because $\frac{1}{2} \left(\sqrt{4 n+9}-3\right)$ is integer when $n=k^2+3k$ which give exactly the sequence $4,10,18,\ldots$ when the number of the given sequence must increase by one.
I mean $3,3,3,3,\mathbf{4},4,4,4,4,4,\mathbf{5},5,5,5,\ldots$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$
$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$
Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$
Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$
However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.
So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!
| You can correct the error that Ryan points out. Instead, note that
$$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 8 - 6ab.$$
So,
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
becomes
$$a^5 + b^5 + 5ab(8 - 6ab) + 20a^2b^2 = 32.$$
Using $a^5 + b^5 = 32$, then expanding and collecting like terms,
$$40a^2b^2 - 10ab = 0 \implies ab = 0 \text{ or } ab = 4.$$
So, now we know that
$$(r - a)(r - b) = r^2 - (a + b)r + ab$$
equals either $r^2 - 2r + 4$, or $r^2 - 2r$, depending on the value of $ab$. Note that the former has no real roots (though it does have complex solutions, but that's whole other kettle of fish), but the latter does. The two numbers would have to be the roots of the former polynomial: $0$ and $2$. We must therefore have, putting the numbers in ascending order,
$$\sqrt[5]{16 - \sqrt{x}} = 0 \quad \sqrt[5]{16 + \sqrt{x}} = 2.$$
It's not hard to see that $\sqrt{x} = 16 \implies x = 256$.
Note: this is the same answer as your method. The reason is, because as it turned out $b = 0$, we actually had $(a + b)^3 = a^3 + b^3$ for that particular $a$ and $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Evaluating $\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} dx$ This question is not a homework question, but I just searched it on WolframAlpha and just want to know how it's done. WolframAlpha lists the following:
$$\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} \mathrm dx = \frac{\sqrt{\pi}\Gamma (\frac{4}{3})}{\Gamma (\frac{5}{6})}$$
How can this be evaluated? I'm not too sure where to start.
| Substituting $v=-x$ and $u=v^3$ gives
$$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}=\int_0^1\frac{dv}{\sqrt{1-v^3}}=\frac{1}{3}\int_0^1 u^{-\frac{2}{3}}(1-u)^{-\frac{1}{2}}du=\frac{1}{3}\beta\left(\frac{1}{3},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{2}\right)}{3\Gamma\left(\frac{5}{6}\right)} $$
Notice that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma\left(\frac{4}{3}\right)=\frac{1}{3}\Gamma\left(\frac{1}{3}\right)$, we then get
$$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}dx=\frac{\sqrt{\pi}\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{5}{6}\right)} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving $x^d \equiv a \pmod{29}$ using primitive roots Can anyone please detail the general approach to questions of the form $x^d \equiv a \pmod{29}$? For example, Wolfram Alpha states that $x^5 \equiv 8 \pmod{29}$ has one solution, $x^4 \equiv 4 \pmod{29}$ has no solution, and $x^7 \equiv 12 \pmod{29}$ has many solutions, but I have no idea how I'd go about proving that no solutions exist, or that any solutions I find are comprehensive.
I know that $2$ is a primitive root $\bmod 29$, so $2^{28} \equiv 1 \pmod{29}$. For the first one I can manipulate to get $(2^{23})^5 \equiv 8 \pmod{29}$ to give $x \equiv 2^{23} \equiv 10 \pmod{29}$ as a solution, but this doesn't prove that no other solutions exist, does it? Also, I don't know how this approach would work with the third problem because $12$ cannot be written $2^n$.
Furthermore, I'm unsure of the general technique to show that the second is not soluble?
| $(x^5)^{17}\equiv x^{85}\equiv (x^{28})^3x\equiv x\bmod 29$,
so if $x^5\equiv8\bmod29$, then $x\equiv8^{17}=2^{51}=2^{28}2^{23}\equiv2^{23}\equiv10\bmod29$.
If $x^4\equiv4=2^2\bmod29$ had solutions, then $1\equiv x^{28}\equiv (x^4)^7\equiv(2^2)^7=2^{14}\bmod29$, so $2$ would not be a primitive root.
$2^7\equiv12\bmod29$, so if $x^7\equiv12\bmod29$, then $x^7\equiv2^7\bmod29$,
so $x\equiv2a\bmod29$, where $a^7\equiv1\bmod29$.
$a^7\equiv1\bmod29$ when $a\equiv2^0, 2^4, 2^8, 2^{12}, 2^{16}, 2^{20}, $ or $2^{24}$; i.e., $a\equiv1, 16, 24, 7, 25, 23, $ or $20$;
so $x\equiv2, 3, 19, 14, 21, 17, $ or $11\bmod29$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$ Let $z=e^{i\frac{2\pi}{7}}$
Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$
How do I solve from here?
| Since $$100\equiv _7 2$$ and $$300\equiv _7 -1$$ and $$500\equiv _7 3$$ We have \begin{align} &=z^{2}+z^{-2} + z^{-1}+z^{1} + z^{3}+z^{-3} \\
&= {z^5+z+z^2+z^4+z^6+1\over z^3}\\
& ={z^6+z^5+z^4+\color{red}{z^3}+z^2+z+1-\color{red}{z^3}\over z^3}\\ &= {{z^7-1\over z-1} -z^3\over z^3} \\
&= {0-z^3\over z^3} =-1\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum. I need to find supremum of:
$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$
I found out that:
$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} = \frac{m^2(n^3 + 3n^2) + 2m(n^3+3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
$$\frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
And from inequality of means (AM - GM) I know that:
$$\frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} \leq \frac{1}{2}$$
Therefore I found a limitation of my set A. Now I need a sequence of elements belonging in $A$ that would be convergent to $\frac{1}{2}$. That is the part I can't do myself.
From graph I know that if I take $m = n$ then for $n \in [1 ; \infty)$ I get decreasing function and for that function $\frac{1}{2}$ is reached by $n < 1$. A maximal value of that function for $n \in [1 ; \infty)$ is equal to $\frac{12}{25}$ and is reached by $n = 1$. But that is not a solution.
There is no ANY fact that would legitimise value $\frac{12}{25}$ as the supremum of that set since there always might by something bigger if $n \neq m$.
What I need is a prove that in the supremum of set $A$: $m = n$ (I don't know why that would be the case) or that there is a bigger value reached for $n$ and $m$ that are not equal.
It might be important to add that I can not use integrals nor derivatives.
| This is not an answer, but something you could try. You could still use differentiation. Replace the function with a continuous one,
$$f(x,y)=\dfrac{x^2y^3+3x^2y^2+2xy^3+6xy^2}{(x^2+2x)^2+(y^3+3y^2)^2},$$
Calculate $\max_{(x,y)\in\mathbb{R}^2}f(x,y)$. Then calculate the values near it which correspond to integers and pick amongst them.
PS:- Please try to make the questions a little less "wordy". At least please try to keep that in mind for all future posts.
| {
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3D transform cylinder to make center line z axis I have an array of points that represent the surface of a cylinder, I was able to calculate the center line of this cylinder but it is translated and rotated away from origin. I want to transform this cylinder to make the center line become the Z axis, I remember it was done by multiplying with a transform matrix but I don't remember how.
here is an example of the cylinder, I have the center line 2 points (start and end), bottom point is supposed to become the new (0,0,0) origin, I added some numbers for simplification.
What I want:
any help is much appreciated
| With the rotation matrix
$$R=\left(
\begin{array}{ccc}
\frac{1}{6} \left(\sqrt{3}+3\right) & \frac{1}{6} \left(\sqrt{3}-3\right) & -\frac{1}{\sqrt{3}} \\
\frac{1}{6} \left(\sqrt{3}-3\right) & \frac{1}{6} \left(\sqrt{3}+3\right) & -\frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\end{array}
\right)$$
the points $(2,2,2)$ and $(3,3,3)$ become $\left(0,0,2 \sqrt{3}\right)$ and $\left(0,0,3 \sqrt{3}\right)$.
Thus, given a point $P=(x,y,z)$ the transformation
$$P'=RP+(0,0,-2\sqrt 3)$$
should do the job.
Edit
The rotation matrix for a generic vector $(d x,d y,d z)=(x_2-x_1,y_2-y_1,z_2-z1)$ is the following. To simplify the notation I substituted $d=\sqrt{dx^2+dy^2+dz^2}$
$$
R=\left(
\begin{array}{ccc}
\frac{\frac{dx^2 dz}{d}+dy^2}{dx^2+dy^2} & \frac{dx dy \left(\frac{dz}{d}-1\right)}{dx^2+dy^2} & -\frac{dx}{d} \\
\frac{dx dy \left(\frac{dz}{d}-1\right)}{dx^2+dy^2} & \frac{\frac{dy^2 dz}{d}+dx^2}{dx^2+dy^2} & -\frac{dy}{d} \\
\frac{dx}{d} & \frac{dy}{d} & \frac{dz}{d} \\
\end{array}
\right)
$$
| {
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Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$ I found the following exercise in a problem book (with no solutions):
Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$
I tried AM-GM for the fraction on the LHS but got stuck from there.
| By Schur's inequality and AM-GM $$\begin{align*}x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(x+z)&\\ \ge 2( {(xy)}^{3/2}+{(yz)}^{3/2}+{(zx)}^{3/2})\end{align*}\tag S$$
Now as $$\frac{1}{1+x}\le \frac{1}{2\sqrt{x}}$$ we have to prove $$a+b+c+3\sqrt[3]{abc}\ge 4\left(\frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{b}}+\frac{1}{2\sqrt{c}}\right)$$ or it suffices to prove $$a+b+c+3\sqrt[3]{abc}\ge 2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}$$ which is just a direct consequence of inequality (S) and so we are done!
Note here $\sqrt[3]{abc}=\sqrt{abc}=1$ is used to normalize inequality
| {
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"timestamp": "2023-03-29T00:00:00",
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Application of Induction in the analyze of the convergence a sequence defined recursive.
Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.
My approach: Note that $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly I would like to find $a_{n}$. Now, I was trying to find a pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.
How can find $a_{n}$?
Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.
| Update: Thanks Brian M. Scott for your insight.
I'll add the case where some $a_k=\frac 43$. Per Brian, we need to solve for the sequence $b_k$ such that $b_1=\frac 43$, $b_{k+1}=\frac{4b_k-1}{3b_k}$. This can be solved in a similar fashion, but easier because $b_1$ is given.
Note that
$$
b_{k+1} - 1 = \frac{b_k-1}{3b_k}$$$$
b_{k+1} - \frac 13 = \frac{b_k-\frac{1}{3}}{b_k}\tag 1
$$
From $(1)$ we conclude $b_k>\frac 13, \forall k$ via induction.
Then $\frac{b_{k+1}-1}{b_{k+1}-\frac 13} = \frac{1}{3} \frac{b_k-1}{b_k-\frac 13} \implies \frac{b_k-1}{b_k-\frac 13} = \frac{1}{3^{k-1}} \left( \frac{b_1 - 1}{b_1 - \frac 13}\right) = \frac{1}{3^k}$
Therefore $b_k = \frac{1 - \frac{1}{3^{k+1}}}{1-\frac{1}{3^l}} = \frac{3^{k+1} -1}{3^{k+1}-3}$ which is the same as Brian's results.
Original answer:
Since $1$ and $\frac 13$ are roots of the characteristic equation $x=\frac{1}{4-3x}$, we have
$$a_{n+1}-1 = \frac{3(a_n-1)}{4-3a_n}$$
$$a_{n+1}-\frac 13 = \frac{a_n-\frac 13}{4-3a_n}$$
So if no $a_n = \frac 13$ you have
$$\frac{a_{n+1}-1}{a_{n+1}-\frac 13} = 3 \frac{a_n-1}{a_n-\frac 13} = 3^n \frac{a_1-1}{a_1-\frac 13}$$
Of course you need to take care of the case where $a_1=\frac 13$.
| {
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Surface integral on $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$ Let $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$, $F(x,y,z)=(x,0,-x)$ and $n(x,y,z)$ be the unit normal vector of $S$ such that $n(0,0,-1)=(0,0,-1)$.
I want to evaluate $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS$.
My Attempt
Let $f(x,y,z)=x^2+y^2+z^2-1$. Then $n$ can be calculated by $n=\frac{\nabla f}{|\nabla f|}=(x,y,z)$. This satisfies the condition stated in the problem.
Therefore we have $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS=\iint_{S}(x^2-zx)dS$. Now we need to calculate this surface integral, but I'm encountering issues.
According to this website, I have two options. One option is to find an orthogonal projection of $S$. The other option is to find a parameterization of $S$. However, I couldn't do either of them. Is there a simple expression for them? Any help is appreciated.
| I will present three ways of tackling this problem.
$\textbf{Option 1}$: Directly
Parametrizing spherical coordinates as usual we can find the bounds by examining the plane equation
$$x+y+z = 1 \implies \sin\phi\cos\theta+\sin\phi\sin\theta+\cos\phi = 1$$
which after a little manipulation becomes
$$\cos\theta+\sin\theta = \frac{1-\cos\phi}{\sin\phi} = \tan\left(\frac{\phi}{2}\right)$$
It's obvious from looking at the sphere with the planar cap cutoff that doing the $\theta$ integral first would require two integrals (one each for the upper and lower parts), but $\phi$ first would only require one. Writing the integral gets us
$$\int_0^{2\pi}\int_{2\cot^{-1}\left(\frac{1}{\sin\theta+\cos\theta}\right)}^\pi \sin^3\phi\cos^2\theta-\sin^2\phi\cos\phi\cos\theta\:d\phi d\theta$$
$\textbf{Option 2}$: Rotate then directly
Once we have the scalar surface integral $\iint_S x^2-xz\:dS$ we can consider a change of variables to rotate the plane so it is completely horizontal from the top. The closest distance between the plane and the origin is $\frac{1}{\sqrt{3}}$ so hopefully that motivates the following rotational coordinate change:
$$\begin{cases}u = \frac{x-y}{\sqrt{2}} \\ v = \frac{x+y-2z}{\sqrt{6}} \\ w = \frac{x+y+z}{\sqrt{3}}\\ \end{cases}$$
This change of variables is a pure rotation so it has Jacobian $1$. This was obtained by choosing a direction for the new '$z$' and '$x$' ($w$ and $u$, respectively) then taking their cross product to find the third orthogonal vector (then including the factor to make them unit vectors), which means
$$u^2+v^2+w^2 = x^2+y^2+z^2$$
Since this is a rotation matrix, inverting the system of equations is as easy as taking the transpose:
$$\begin{cases}x = \frac{u}{\sqrt{2}} + \frac{v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ y = -\frac{u}{\sqrt{2}} + \frac{v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ z = -\frac{2v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ \end{cases}$$
which gives us a new surface integral on the same sphere, only rotated
$$\iint_S \frac{u^2+v^2}{2}+\frac{2uv}{\sqrt{3}}+\frac{uw}{\sqrt{6}}+\frac{vw}{\sqrt{2}}\:dS$$
The nice thing about this is that with this rotated sphere cutoff at $w=\frac{1}{\sqrt{3}}$, we can now exploit symmetry. $u$ and $v$ are both odd functions, so any term with just an odd power of either will vanish, leaving us with
$$\iint_S \frac{u^2+v^2}{2}\:dS = \int_0^{2\pi}\int_{\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)}^\pi \frac{1}{2}\sin^3\phi\:d\phi d\theta$$
which easy enough to evaluate since the integrand will end up being only cosines.
$\textbf{Option 3}:$ Divergence theorem
We close off the surface with a plane and consider the equation
$$\iint_{\text{sphere}}F\cdot dS + \iint_{\text{plane}}F\cdot dS = \iiint_V \nabla \cdot F dV$$ $$ = \int_0^{2\pi}\int_{\frac{1}{\sqrt{3}}}^1 \int_{\cos^{-1}\left(\frac{1}{\sqrt{3}\rho}\right)}^\pi\rho^2\sin\phi\:d\phi d\rho d\theta + \frac{4}{3}\pi\left(\frac{1}{\sqrt{3}}\right)^3$$
Since $\nabla\cdot F = 1$, we only want the volume, thus we have implicitly taken advantage of the rotation from option 2 to set up our bounds. The remaining surface integral can be calculated as follows
$$\iint_{\text{plane}}F\cdot dS = \iint_{\text{plane}}(x,0,-x)\cdot\frac{(1,1,1)}{\sqrt{3}}dS = 0$$
therefore the answer will only be the volume integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the coefficient of ${x^9}$ in the expansion of $ (1 + x)( 1 + x^2)( 1 + x^3)..(1 + x^{100})$ Find the coefficient of ${x^9}$ in the expansion of $\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^3}} \right)..\left( {1 + {x^{100}}} \right)$.
The official answer is 8.
How do I find the general term,
Dividing the above equation by $(1-x)$ is not generating the required result.
| Hint:
$$\begin{align}9=9+0\\=8+1\\=7+2\\=6+3\\=6+2+1\\=5+4\\=5+3+1\\=4+3+2\end{align}$$
We don't have to worry about 4 summands, since $1+2+3+4>9$. There is no known closed form for the general term.
| {
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Nice geometry question to prove tangency Let $\triangle ABC$ be a scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at a point $P$. Let $AP$ meet $BC$ at a point $K$ and let $M$ be the midpoint of $BC$. Let $X$ be a point on $AC$ such that $AB$ is tangent to the circumcircle of $\triangle BXC$. Let $BX$ meet $AM$ at a point $T$. Let the line through $C$ and parallel to $AM$ meet $TK$ at a point $V$.
Prove that $AV$ is tangent to $\omega$.
I tried the question with $\sqrt{bc}$ inversion with reflection over angle bisector. This approach does not seem to help much. Please provide hint?
| Here you have an approach employing barycentric coordinates. For more information, have a look at this awesome handout by Evan Chen. Following his advice, I will use , when referring to normalized coordinates, and : for homogeneous coordinates.
Let $\triangle ABC$ be the reference triangle, i.e. $A=(1,0,0); B=(0,1,0);C=(0,0,1)$ with side lengths $a=BC, b=CA, c=AB$. Observe that $M=(0:1:1).$ Furthermore, notice that $(\triangle BXC)$ being tangent to $AB$ is equivalent to $$\angle XBA=\angle ACB\implies \triangle ABX\sim \triangle ACB\implies \frac{AX}{AC}=\frac{c^2}{b^2} $$ Therefore $$\overrightarrow{AX}=\frac{c^2}{b^2}\cdot \overrightarrow{AC}=\frac{c^2}{b^2}\cdot (-1, 0, 1)=\left(-\frac{c^2}{b^2}, 0, \frac{c^2}{b^2}\right)\implies X=\left(1-\frac{c^2}{b^2},0, \frac{c^2}{b^2}\right)\equiv \left(b^2-c^2:0:c^2\right)$$
Thus, we infer that $AM: z-y=0$ and $BX: c^2\cdot x-\left(b^2-c^2\right)\cdot z=0$. It follows that $T=\left(b^2-c^2:c^2:c^2\right)$.
On the other hand, looking at the definition of $K$, it is well-known that $AK$ is but the $A$-symmedian. Hence, $$AK:b^2z-c^2y=0, \; BC:x=0\implies AK\cap BC=K=\left(0:b^2:c^2\right)$$ It follows that \begin{align*}TK&: (b^2-c^2)b^2\cdot z+c^4\cdot x-c^2b^2\cdot x-(b^2-c^2)c^2\cdot y=0\\\iff TK&: (b^2-c^2)\cdot (b^2z-c^2y)+c^2x\cdot (c^2-b^2)=0\\\iff TK&: b^2z-c^2y-c^2x=0\end{align*} In order to determine the equation of the line $l$ through $C$ parallel to $AM$, I will use the method described here (see page $76$). This yields $$l: x+2y=0$$ Thus $l\cap TK=V=(-2b^2:b^2:-c^2)$. Besides, the equation of the tangent $t$ to $\omega$ at $A$ is given by $b^2z+c^2y=0$ (cf. Chen's handout, corollary $15$). We are almost there, since $$V\in t\iff b^2\cdot(-c^2)+c^2\cdot b^2=0 $$ Which is clearly true.
| {
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Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."
If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?
| As another approach, factor the two quadratics as
$$x^2 + x + 1 = (x-\phi)(x+\psi)$$ $$x^2 - x + 1 = (x + \phi)(x - \psi)$$
Where $\phi$ and $\psi$ are solutions to $\phi\psi=-1$, $\psi - \phi=1$
So problem reduces to solving for $A$ and $B$
$$(x-\phi)^n(x+\psi)^n = P(x)(x+\phi)(x-\psi) + Ax + B$$
Evaluate this for $x=-\phi$ and $x=\psi$ to get
$$(-2\phi)^n(\psi - \phi)^n = -A\phi + B$$
$$(\psi - \phi)^n(2\psi)^n = A\psi + B$$
Remember that $\psi - \phi = 1$ and the rest should be straightforward
| {
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Dividing a polynomial with integer coefficients by a quadratic In the following question, one of the coefficients of the divisor quadratic polynomial is unknown; so, it cant be factored for performing synthetic division. I tried to perform long division hoping to get some clues but ran into a conflict.
Appreciate your help. Thanks!!
Here is the question:
$x^2 -x +a$ evenly divides $x^8 + 5x^6 + 13x^4 + 20x^2 +36$. Determine positive integer $a$.
A few things I noted are:
*
*$a$ must be a factor of 36 so, I am thinking $a$ would be from $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$
*All the coefficients of the dividend polynomial are integers and degrees of all the terms are even. Not sure how to use this info.
*$P(x) = Q(x)(x^2 -x +a) + mx + n$. Since the remainder is zero, $m$ and $n$ are zero. When I tried long division, I got two equations; but the value of $a$ would come as imaginary which conflicts with the first observation and the question asks for real integer $a$.
| Some quick observations : if $x$ is a root of $P(x)$ then so is $-x$. Thus if $(x^2-x+a)$ is a factor of $P(x)$, so is $(-x)^2-(-x)+a=x^2+x+a.$
Then $(x^2-x+a)(x^2+x+a)=x^4+(2a-1)x^2+a^2$ is a factor of $P(x)$. $\, a^2|36 \,$ $\Rightarrow a \in \{1,2,3,6\}$. (Here @Anwesha1729 observe in their answer that as $P(x)$ takes odd values also for integer $x$, $a$ must be odd.)
Let $P(x)=(x^4+(2a-1)x^2+a^2)(x^4+bx^2+c)$.
Comparing coefficients :
$$b+(2a-1)=5$$
$$a^2+c+(2a-1)b=13$$
$$a^2b+(2a-1)c=20$$
$$a^2c=36$$
It can be readily checked $a=3$ $\, (b=0,c=4)$
| {
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Help solving a recurrence relation I'm quite stuck on how to solve recurrence/difference equations.
For example I have the following linear inhomogeneous equation:
$a_n = 2a_{n-1} + 2^n n$, and $a_0 = 1/2$
I know that $2a_{n-1} = 1/2(2^n)$, and then we have $a_n = 1/2(2^n) + n(2^n)$. This is where I'm stuck though and the inhomogeneous part is throwing me off.
Am I in the right direction thinking that: $a_n = \frac12 \prod 2^n n$ ?
No answers please but some nudges in the right direction would be greatly appreciated. Thanks!
| You didn't use the generating-functions tag, so here is more than a nudge for that approach. Let $A(z)=\sum_{n=0}^\infty a_n z^n$ be the ordinary generating function for $(a_n)$. Then the recurrence and initial condition imply that
\begin{align}
A(z)
&= \frac{1}{2} z^0 + \sum_{n=1}^\infty \left(2 a_{n-1} + 2^n n\right)z^n \\
&= \frac{1}{2} + 2z \sum_{n=1}^\infty a_{n-1} z^{n-1} + 2z\sum_{n=1}^\infty n (2z)^{n-1} \\
&= \frac{1}{2} + 2z A(z) + 2z \frac{1}{(1-2z)^2},
\end{align}
so
\begin{align}
A(z) &= \frac{1/2 + 2z /(1-2z)^2}{1-2z} \\
&= \frac{1+4z^2}{2(1-2z)^3} \\
&= \frac{1}{(1 - 2 z)^3} - \frac{1}{(1 - 2 z)^2} + \frac{1/2}{1 - 2 z} \\
&= \sum_{n=0}^\infty \binom{n+2}{2} (2 z)^n - \sum_{n=0}^\infty \binom{n+1}{1} (2 z)^n + \frac{1}{2}\sum_{n=0}^\infty \binom{n+0}{0} (2 z)^n \\
&= \sum_{n=0}^\infty \frac{(n+2)(n+1)-2(n+1)+1}{2} (2 z)^n \\
&= \sum_{n=0}^\infty (n^2+n+1) 2^{n-1} z^n,
\end{align}
which immediately implies that $a_n = (n^2+n+1) 2^{n-1}$.
| {
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$24ml+1=k^2$ has no solution for all $l=1 \dots m$ Investigating solutions of $$24ml+1=k^2$$ for $l=1\dots m$
The question is to find the $m$-s for which the above equation has no solution for all $l=1..m$-s.
The first few $m$-s are:
$$3, 9, 24, 27, 81, 192, 243, 432, 729$$
Actually have found that the $m$ should be of form $2^a3^b$. Seems hang on the simple task, of finding general formula for this.
One interesting addition. Just tested the cases when the $$12ml+1=k^2$$ has no solution for all $$l=1..m$$ Seems it has no solution iff $m=3^a$ But has no proof yet.
| The OP also conjectures that $12ml+1=k^2$ has no solutions for $1\le l\le m$ if and only if $ m=3^a$. This is true and can be proved as follows.
Only if
For $12ml=(k-1)(k+1)$ we must have $k=6x\pm 1$. Then $ml=x(3x\pm 1).$
Let $m=3^aM$, where $M>1$ is coprime to $3$. Let $x=3^aX$ and then $$Ml=X(3^{a+1}X\pm 1).$$
We can choose $X$, $0< X <M$, such that $3^{a+1}X\equiv 1$ modulo $M$. Replacing $X$ by $M-X$ if necessary we can therefore choose $X$, $0< X \le \frac{M}{2},$ such that $(3^{a+1}X\pm 1)=MY$ for some integer $Y$.
Then $l=XY$.
If $m<l$ then $$3^aM<\frac{M}{2}\frac{(3^{a+1}M\pm 2)}{2M} \text{ i.e. } 3^a<\frac{3^{a+1}\pm \frac{2}{M}}{4}. $$
Then $a=0$ and $M=2$. But then $l=1$ gives the solution $12ml+1=5^2$.
If
Let $m=3^a$ and $x=3^aX$, then we must solve $l=X(3^{a+1}X\pm 1)>3^a=m,$ a contradiction.
| {
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Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $ The limit can be rewritten as
$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$
Recall the inequality,
$$ \cos x < \frac{\sin x}{x} < 1$$
holds for $ x\in (-\pi/2, \pi/2) $.
This provides the new inequalities
$$ 0 < \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] < \frac{1-\cos x}{x}, \text{ with } x>0 $$
$$ \frac{1-\cos x}{x} < \frac{1}{x} \left[1- \frac{\sin x}{x} \right]< 0, \text{ with } x<0 $$
Applying the Squeeze Theorem, we get
$$ \lim_{x \rightarrow 0^{\pm}} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] = 0$$
Note:
$ \lim_{ x \rightarrow 0} \frac{1-\cos x}{x} = 0 $
| Your proof is right. But I would have used Taylor developments :
$$\frac{x-\sin x}{x^2} \underset{x\rightarrow 0}{=} \frac{x-(x-x^3/6+o(x^3))}{x^2}\underset{x\rightarrow 0}{=}\frac{x}{6}+o(x)\underset{x\rightarrow 0}{\longrightarrow} 0$$
| {
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How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?
| By drawing triangles it should be easy to see that \begin{align*}
\cos \arcsin \left(\frac{3}{5}\right)&=\frac{4}{5}\\
\cos \arctan \left(\frac{7}{24}\right)&=\frac{24}{25}\\
\sin \arcsin\left(\frac{3}{5}\right)&=\frac{3}{5}\\
\sin \arctan\left(\frac{7}{24}\right)&=\frac{7}{25}
\end{align*}
(note the Pythagorean triples (3,4,5), and (24,7,25).) Putting this all together: $$\cos \arcsin \left(\frac{3}{5}\right)\cos \arctan \left(\frac{7}{24}\right)-\sin \arcsin\left(\frac{3}{5}\right)\sin \arctan\left(\frac{7}{24}\right)=\frac{4}{5}\cdot \frac{24}{25}-\frac{3}{5}\cdot \frac{7}{25}=\frac{3}{5}$$
| {
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Approach ideas for the integral $\int\frac{dx}{(x^4-16)^2}$ Well, the title sums it up pretty well. I'm in search for some smart approach ideas for solving this indefinite integral:
$$\int\frac{dx}{(x^4-16)^2}$$
I know one that would work for sure, namely partial fraction decomposition, but it gets really heavy when regrouping the coefficients for the powers of $x$ and then solving an $8\times8$ system of linear equations. It will eventually work, but I suspect there is something more ingenuine behind this problem.
I also tried all sorts of trigonometric substitutions and formulations, but that added square power really is a bummer to it all.
I'm generally open to any exchange on the topic and would be glad to hear some advice in such situations. Many thanks in advance!
| As $x^4-2^4=(x^2-2^2)(x^2+2^2)$ and $(x^2+2^2)-(x^2-2^2)=8$
$$\dfrac{8^2}{(x^4-16)^2}=\dfrac{(x^2+4-(x^2-4))^2}{(x^2-4)^2(x^2+4)^2}=\dfrac1{(x^2-4)^2}+\dfrac1{(x^2+4)^2}-\dfrac2{(x^2-4)(x^2+4)}$$
For the first, write the numerator as $\dfrac{(x+2-(x-2))^2}{16}$ and expand
For the second either $x=2\tan t$
or integrate by parts $$\int\dfrac1x\cdot\dfrac x{(x^2+4)^2}\ dx$$
Again for the last, $\dfrac{(x^2+4)-(x^2-4)}{(x^2-4)(x^2+4)}$
| {
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How to solve the recurrence $()=5(/2)+^3()$ using iteration method How do we solve the recurrence $()=5(/2)+^3()$ using the Iteration method?
I solved the recurrence using a master method - master method
Now using the iteration method
$()=5(/2)+3() = 5(5T(/4)+(n/2)^3)*log(n/2))+n^3*logn$ $=$ ... $= 5^i(n/2^i)+n^3*log(n) * β(n^3*(5^k) *log(n/2^k) ) / 8^k) $
How is it equal to $Ξ(n^3log(n))$ ?
| Do the substitution:
$$\begin{array} {rcl}
T(n) &=& n^3\log(n) \\
&+& 5(n/2)^3\log(n/2)\\
&+& 5^2(n/2^2)^3\log(n/2^2)\\
&+& 5^3(n/2^3)^3\log(n/2^3)\\
&+& \dots \\
\\
&=& n^3\log(n) \\
&+& 5/2^3 n^3 (\log n - \log 2)\\
&+& 5^2/2^6 n^3 (\log n - 2\log 2)\\
&+& 5^3/2^9 n^3 (\log n - 3 \log 2)\\
&+& \dots \\
\\
&=& n^3\log(n) (1 + 5/2^3 + 5^2/2^6 + 5^3/2^9 + \dots) \\
&-& n^3\log(2) (0 + 1\cdot 5/2^3 + 2\cdot 5^2/2^6 + 3 \cdot 5^3/2^9 + \dots )\\
\end{array}$$
That's a geometrics series and an arithmetico-geometric series. But you don't actually need to calculate them to get the asymptotic behavior, since it is just going to be some positive constants.
| {
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Is the solution that Wolfram Alpha gives really the only solution to this problem? My formula
$$(z-y)(z^2+zy+y^2)=(y-x)(y^2+yx+x^2)$$
$$z>=y+2$$
$$y>=x+2$$
$$x>=2$$
I am trying to understand a geometric problem better by plugging my formula into Wolfram Alpha. Normally I get a result which helps me understand the issue more graphically, but this time it is only spitting out a single solution.
This formula on Wolfram Alpha
The solution from Wolfram Alpha written out below:
$$x>=2$$
$$y>=\frac{\sqrt[3]{x^3+\sqrt{x^6+96x^3+256}+48}}{\sqrt[3]{2}}+\frac{8\sqrt[3]{2}}{\sqrt[3]{x^3+\sqrt{x^6+96x^3+256}+48}}+2$$
$$z=(-1)^{2/3}\sqrt[3]{2y^3-x^3}$$
My questions
*
*Is this the only solution?
*How would I know or check this?
*Since z has a complex component (-1)^{2/3} in this solution it can only ever be a complex number, correct?
Anything that improves my understanding is greatly appreciated!
| Your equation says $x^3 + z^3 = 2 y^3$. Note that you want $x,y,z$ to be all $\ge 2$.
With $y = ((x^3 + z^3)/2)^{1/3}$, your inequalities are equivalent to
$$ \eqalign{\frac{x^3}{2} + \frac{z^3}{2} &\le (z-2)^3\cr
(x+2)^3 &\le \frac{x^3}{2} + \frac{z^3}{2}\cr
x \ge 2\cr}$$
Now the resultant of the polynomials $(x^3 + z^3)/2 - (z-2)^3$ and $(x+2)^3 - (x^3+z^3)/2$ with respect to $x$ is $-108 (3 z^2 - 6 z + 4) (z-2)^3$ whose only real root is $2$. Thus in the region $z
> 2$, the boundaries of the regions represented by the first and second inequalities don't intersect. We can then check that the second inequality is implied by the first in this region. So the answer can be written as
$$ \eqalign{y &= ((x^3 + z^3)/2)^{1/3}\cr
2 \le x &\le (2(z-2)^3 - z^3)^{1/3} \cr}$$
Note finally that $2 \le (2(z-2)^3 - z^3)^{1/3}$ if and only if $z$ is greater than the real root of $4 + z^2/2 = (z-2)^3$, which is
$$ (28 + 4 \sqrt{17})^{1/3} + \frac{8}{(28+4 \sqrt{17})^{1/3}} + 4$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $x^3 + \frac{1}{x^3}$ Question
$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 + \frac{1}{x^3}$ and choose the correct answer from the following options:
*
*198
*216
*200
*186
What I have did yet
I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$
I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!
| Let have a look at this: finding $x^6+y^6$ given $x+y$ and $xy$
$$U_n=x^n+\frac 1{x^n}$$
$s=x+\frac 1x=U_1$ and $p=x\times\frac 1x=1$
also we have $\begin{cases}U_2=34\\U_0=x^0+\frac 1{x^0}=1+1=2\end{cases}$
We have the relation $$U_{n+1}=U_1U_n-U_{n-1}$$
$U_2=U_1^2-U_0\implies U_1^2=34+2=36\implies U_1=\pm 6$
And $U_3=U_1U_2-U_1=\pm 6(34-1)=\pm 198\quad$ (among proposed answers the positive one fits).
Note: $x$ cannot be integer, since $x+\frac 1x=\pm 6\iff x=\pm(3\pm 2\sqrt{2})$
This method can be generalized to most $x^n+y^n$ types of questions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $2\le U_n \le 3$ for all $n$ Given $U_n = (1 + 1/n)^n$ , $n = 1,2,3,\ldots\;.$
Show that $2 \le U_n \le 3$ for all n
This is what I've done. Can anyone help?
$$\begin{align*}
a_n=\left(1+\frac1n\right)^n&=\sum_{r=0}^n{^nC_r}(1)^r\left(\frac1n\right)^{n-r}\\
&=\sum_{r=0}^n{^nC_r}\left(\frac1n\right)^{n-r}\\
&=1+\frac{n}n+\frac1{2!}\frac{n(n-1)}{n^2}+\frac1{3!}\frac{n(n-1)(n-2)}{n^3}\\
&\quad\,+\ldots+\frac1{n!}\frac{n(n-1)\ldots(n-n+1)}{n^n}
\end{align*}$$
Since $\forall k\in\{2,3,\ldots,n\}$: $\frac1{k!}<\frac1{2^k}$, and $\frac{n(n-1)\ldots\big(n-(k-1)\big)}{n^k}<1$,
$$\begin{align*}
a_n&<1+\left(1+\frac12+\frac1{2^2}+\ldots+\frac1{2^{n-1}}\right)\\
&<1+\left(\frac{1-\left(\frac12\right)^n}{1-\frac12}\right)<3-\frac1{2^{n-1}}<3
\end{align*}$$
| Here is another way to look at it that might be helpful, if the question is why for all $n$ it holds. Consider $f(x)=(1+\frac{1}{x})^x$. It is an increasing function of $x$:
$$ f'(x)=\frac{d}{dx}e^{x \ln(1+\frac{1}{x})}
= \left[\ln\left(1+\frac{1}{x}\right) - \frac{1}{1+x}\right]\left(1+\frac{1}{x}\right)^x >0, \forall x>0$$
Now:
$f(1)=2$, and $\lim_\limits{n\rightarrow+\infty}f(n)=e<3$, so $2\leq f(n) < e < 3, \forall n$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
proving limits exist Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.
I basically need to determine how large $n$ has to be to imply
$$\frac{3n+24}{n^3-6}<\epsilon$$
the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 β€ 27n$, it suffices for us to get $\frac{27n}{n3-6} < Ξ΅$.
it is inferred, thereafter, that all we need is $\frac{n^3}2 β₯6$ or $n^3 β₯12$ or $n>2$.
this is where I have a problem. I don't understand how we got to the point where all we need is $\frac{n^3}2β₯6$.
|
I basically need to determine how large $n$ has to be to imply
$$\frac{3n+24}{n^3-6}<\epsilon$$
the idea is to upper bound the numerator and lower bound the denominator.
For $n > 9$, $0 < (n^3 - 64n),$ which is a lower bound for the denominator.
For $n > 9$, $(n-8) > 1 \implies$ that the numerator is
less than $3(n+8)(n-8) = 3(n^2 - 64).$
Therefore, the fraction is less than
$$\frac{3(n^2 - 64)}{n^3 - 64n} = \frac{3}{n}.$$
Therefore, choose $n$ such that $n > 9$ and
$\frac{3}{n} < \epsilon \implies \frac{3}{\epsilon} < n.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding unique number and value of p. Function = f(x) = $\frac{x^2+px+1}{x^2+x+1}.$
Q 1For any value of p , a unique number will be there in the range . Find it.
Q 2 If nine integers are in the range of f(x) , then solve for p.
Q3If range of function is non negative , then find maximum no of integers in the range can be .
For the 1st one , i thought unique number as neither a prime number nor a composite number. It has only one factor, i.e, the number itself. Is it correct ?
For the 2nd , I am not getting
For 3rd , no of integers will change as value of p will change . Then how to find the amswer
| Question 1
let $x=0$. No matter what the value of $p$ is, this value will be in common with all possible functions.
Question 2
Question 2 is asking the following:
If we have the function $f(x)=\frac{x^2+px+1}{x^2+x+1}$ and we look at range of $f(x)$, for some value of $p$ the range of $f(x)$ includes exactly $9$ integers. Find this value of $p$.
For question 2, note that
$$\frac{x^2+px+1}{x^2+x+1}=\frac{x^2+x+1-x+px}{x^2+x+1}=1+\frac{x(p-1)}{x^2+x+1}$$
I will assume that you want an integer value for $p$ (unless you want an inequality; if you do please specify), and that you mean exactly $9$ integers are in the range of $f(x)$.
If you actually look at this graph for different values of $p$ it is clear that the function always has a clear maximum and minimum point. We can find the coordinates of these stationary points by differentiating with respect to $x$:
$$\frac{d}{dx}\left(1+\frac{x(p-1)}{x^2+x+1}\right)=\frac{(p-1)(1-x^2)}{(x^2+x+1)^2}$$
after some rearranging and using the quotient rule. Now let this derivative be equal to $0$ for the stationary points.
$$\frac{(p-1)(1-x^2)}{(x^2+x+1)^2}=0\implies 1-x^2=0\implies x+\pm1$$
Hence the stationary points have coordinates $(1,f(1))$ and $(-1,f(-1))$ which comes to the coordinates $(-1,1-(p-1))$ and $(1,1+\frac{p-1}{3})$ which simplifies to
$$(-1,~2-p)~~\text{and}~~(1,\frac{p+2}{3})$$
If the range only includes exactly $9$ integers, then let the difference between the maximum point and minimum point of the function be $8$, as this is the smallest possible amount that could include $9$ integers. Hence
$$\frac{p+2}{3}-(2-p)=8\implies p+2-6+3p=24\implies 4p=28$$
so $p=7$. Check this indeed includes $9$ integers by finding the $y$-values of the coordinates of the stationary points for this value of $p$, and we are done.
Question 3
Let $f(x)>0$:
$$\frac{x^2+px+1}{x^2+x+1}>0$$
Note that $x^2+x+1>0$ for all real $x$, so we can multiply through by $x^2+x+1$:
$$x^2+px+1>0$$
But $x^2+px+1=(x+\frac{p}{2})^2+1-\frac{p^2}{4}$. So if $f(x)>0$ for all values of $x$ then $\frac{p^2}{4}<1$
$$\implies (p+2)(p-2)<0$$
Hence $-2<p<2$.
Now plug in $p=2$ and $p=-2$ respectively into your general coordinates for the stationary points and see how many integers it includes.
| {
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Given the minimal polynomial of a matrix $A^2$, what could the minimal polynomial of $A$ be? It is given that the minimal polynomial of $A^2$ is $\phi_{A^2}(x) = (x-1)^2$, where $A$ is a complex $4\times4$ matrix. The question is, what are the possible minimal polynomials for $A$?
From the given that $\phi_{A^2}(x) = (x-1)^2$, I can derive several things. First off, I know that the characteristic polynomial $f_{A^2}$ of $A^2$ must have the same irreducible factors, which implies that the only eigenvalue of $A^2$ is $1$. Secondly, it is easily seen that $(A^2 -I)^2 = O$ and $A^2 - I \neq O$. Therefore, the matrix $A^2 - I$ is nilpotent with index 2. This implies that the invariant system is either $\{2,1,1\}$ or $\{2,2\}$. Putting this together, there are two possible Jordan forms of $A^2$:
$$ J_{A^2,1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \quad\text{or}\quad J_{A^2,2} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix} $$
Great, all these things I know about $A^2$... but how does any of it lead to information about the minimal polynomial of $A$? Or just any information about $A$ at all? I'm kind of stuck on this.
| HINT
You know a polynomial satisfied by $A$, namely $(A^2-1)^2$. So the minimal polynomial has to be a factor of $(x-1)^2(x+1)^2$.
Which of the possible factors could give the correct minimal polynomial for $A^2$?
If neither power of $(x-1)^i(x+1)^j$ is $2$ then we would have one of $A-I=0,A+I=0,(A+I)(A-I)=0$. In all these cases $A^2=I$, a contradiction. This cuts your work considerably.
Finally, to check the remaining possibilities easily, use simple 2x2 Jordan blocks e.g.
$A=\begin{pmatrix}1&1&0&0 \\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}$ has minimal polynomial $(x-1)^2$. What is the minimal polynomial for $A^2$?
| {
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"timestamp": "2023-03-29T00:00:00",
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For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$
I tried the most rudimentary method ie. differentiating wrt x and hoped to get at least 2 solutions for $fβ(x)$, unfortunately I wasnβt able to find a solution for implying function is monotonous . Now whether itβs sis increasing or decreasing still remains a mystery, given that we donβt know what $a$ is.
What method can be used to solve this?
| I think that the derivative gives a good indication.
$$f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$$
$$f'(x)=4 \sqrt{\frac{a}{x (a x+8)^3}}-\frac{1}{2 (x+1)^{3/2}}$$ Without any constraint, the derivative cancels at four points
$$x_{1,2}=\pm \frac{2 \sqrt{2}}{\sqrt{a}}\qquad x_{3,4}=\frac{2 \left(16-6a\pm \sqrt{2(a-8)^2 (2-a)}\right)}{a^2}$$ $x_{3,4}$ do not exist in the real domain if $a >2$ but, in any case, remains $x_{1}= \frac{2 \sqrt{2}}{\sqrt{a}}$.
On the other side $f''(x_1)$ is positive if $0 <a <2$ and negative if $a>2$.
Add all of that to @mathcounterexamples's answer.
| {
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Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $ Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $
I did these steps:
*
*Basis step
$$P(10): \ \ 1024 \geq 1000 \ (True)$$
*Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.
does this demonstration work?
I know there's a similar question but the solution is different, I want to know if my demonstration is valid as well.
|
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.
does this demonstration work?
No, your analysis (in and of itself) hasn't shown that
$(n+1)^3 \leq 2^{n+1}$, assuming that I am not overlooking anything. This is what you are trying to prove.
The easy way, is:
after noting that $2^{10} > 10^3$, and assuming that $2^n > n^3$, simply note that for $n \geq 10$,
$$\left(\frac{n+1}{n}\right)^3 < 2. \tag1$$
Note that $2^{1/3} > 1.25$, and for $n \geq 10, \frac{n+1}{n} < 1.25$.
This justifies equation (1) above.
Thus:
$$2^{n+1} = 2 \times 2^n > 2 \times n^3 > \left(\frac{n+1}{n}\right)^3 \times n^3 = (n+1)^3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Differential equation with partial fraction How do you separate this differential equation into a partial fraction?
Solve the following differential equation:
$$\frac{dy}{dx}=\frac{2y^2-xy+x^2}{xy-x^2}$$
| Let $\frac{y}{x}=v$
$$\frac{{\rm d}y}{{\rm d}x} = v + x\frac{{\rm d}v}{{\rm d}x} =\frac{2v^2-v+1}{v-1} \\
x\frac{{\rm d}v}{{\rm d}x} = \frac{v^2+1}{v-1} \\
\frac{v-1}{v^2+1}dv = \frac {{\rm d}x}{x} \\
\text{Integrating both sides} \\
\int \frac{v-1}{v^2+1}dv =\int \frac {{\rm d}x}{x} \\
\int \frac{v}{v^2+1}dv - \int \frac{1}{v^2+1}dv= \frac {{\rm d}x}{x} \\
\frac{1}{2} \ln(v^2+1) - \tan^{-1}v = lnx + C\\
\frac{1}{2} \ln \left(\frac{y^2+x^2}{x^4}\right) = \tan^{-1}\left( \frac{y}{x}\right) + C $$
Where $C$ is the constant of integration.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$ Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?
| You have the following problem: $$\begin{bmatrix}x^2+y^4=20\\ x^4+y^2=20\end{bmatrix}$$
In the first equation, you can isolate $x$ to get $x=\sqrt{20-y^4}$ and $\:x=-\sqrt{20-y^4}$. You then substitute the $x$ values into the second equation to get:
$$\left(\sqrt{20-y^4} \right)^4 +y^2=20, \left(-\sqrt{20-y^4} \right)^4 +y^2=20$$
Simplifying the equations will result in $y=2, y=-2$. You can then substitute the solution into the first equation and you will get $x=2,-2$ depending on which value of $y$ you substitute. Therefore, the final solutions are:
$$\begin{pmatrix}x=2,\:&y=2\\ x=-2,\:&y=2\\ x=2,\:&y=-2\\ x=-2,\:&y=-2\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
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Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$ Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.
Here is what I have done so far
$$ \cos x + \cos y = -\cos z $$
$$ (\cos x + \cos y)^{2} = (-\cos z)^{2} $$
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y = \cos^{2}z $$
likewise,
$$ \sin x + \sin y = -\sin z $$
$$ (\sin x + \sin y)^{2} = (-\sin z)^{2} $$
$$ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y = \sin^{2}z $$
from this, you get
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y + \sin^{2} x + \sin^{2} y + 2\sin x \sin y = \sin^{2}z + cos^{2}z $$
$$ 1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1 $$
$$ \cos x \cos y + \sin x \sin y = -\frac{1}{2} $$
$$ \cos(x-y) = -\frac{1}{2} $$
$$ x-y = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
likewise,
$$ \cos (x-z) = -\frac{1}{2} $$
$$ x-z = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
where do I go from here?
| Continuing from your work, we have $y-z\equiv z-x\equiv x-y\equiv\frac23 r\pi\;\,(\!\!\!\!\mod2\pi)$, with $r\in\{1,2\}$ ($r$ is constant, since the three differences sum to zero, modulo $2\pi$). So $$\cos2x+\cos2y+\cos2z=\cos2x+\cos(2x-\tfrac43r\pi)+\cos(2x+\tfrac43r\pi)\qquad\qquad$$
$$=\cos2x+2\cos2x\cos\tfrac43r\pi$$
$$=0.\qquad\qquad\qquad\qquad\quad$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.
\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}
so $x > 100$
| Working a little differently, you get an exact result:
$$
\begin{align}
x &= \left(\frac 21 \times \frac 22\right) \times \left(\frac 43 \times \frac 44\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{10000}\right) \\
&= \frac{1}{10000\, !} [2\times 4 \times \cdots \times 10000]^2\\
&= \frac{1}{10000\, !} [2^{5000}\times 1\times 2 \times \cdots \times 5000]^2\\
&= \frac{1}{10000\, !} 4^{5000}[5000 \, !]^2\\
&= \frac{4^{5000}}{\binom{10000}{5000}}
\end{align}
$$
Now proceed as in @Thomas Andrews's solution above.
| {
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Integration by substitution: $\displaystyle\int\frac{dx}{(1+x^2)^2}$ I came across this solution of a tricky integral using implicit substitution (i.e., manipulating differentials without actually declaring $u=\cdots$).
\begin{align*}
\int \frac{dx}{(1+x^2)^2} &= \int \frac{1 + x^2 - x^2}{(1+x^2)^2}\,dx\\
&= \int \frac{dx}{1+x^2} - \int \frac{x^2}{(1+x^2)^2}\,dx\\
&=\tan^{-1}x - \frac{1}{2}\int \frac{x}{(1+x^2)^2}\,d(x^2+1)\\
&= \tan^{-1}x+\frac 12\int x\, d\Big(\frac 1{1+x^2}\Big)\\
&=\tan^{-1}x + \frac12\cdot \frac{x}{1+x^2}- \frac12\int \frac{dx}{1+x^2}\\
&= \frac12\Big(\tan^{-1}x + \frac{x}{1+x^2}\Big)+c.
\end{align*}
Can someone help me understand how to go from the third line to the fourth? It doesn't seem like an obvious step. I'm guessing that it's because
$$d\Big(\frac 1{u(x)}\Big) = -\frac{d(u(x))}{u(x)^2},$$
but it doesn't seem like a natural next step, it's just the way I justified it to myself after already having seen the solution. Why should it be the next step?
| Here's something you might like.
Writing the indefinite integral as a definite one and applying Feynman's method
\begin{align}I(a)&=\int_c^x\dfrac{1}{a^2+x^2}\,\mathrm dx=\dfrac1a\arctan\left(\dfrac xa\right)\bigg|_c^x=\dfrac1a\arctan\left(\dfrac xa\right)+C\\I'(a)&=\int_c^x-\dfrac{2a}{(a^2+x^2)^2}\,\mathrm dx=-\dfrac1{a^2}\arctan\left(\dfrac xa\right)-\dfrac x{a^2+x^2}+K\\-\dfrac1{2a}I'(a)&=\int_c^x\dfrac1{(a^2+x^2)^2}\,\mathrm dx=\dfrac1{2a^3}\arctan\left(\dfrac xa\right)+\dfrac1{2a(a^2+x^2)}+K\\-\dfrac12I'(1)&=\int_c^x\dfrac1{(1+x^2)^2}\,\mathrm dx=\dfrac12\arctan x+\dfrac1{2(1+x^2)}+K\end{align}
$$\boxed{\boxed{\int\dfrac1{(1+x^2)^2}\,\mathrm dx=\dfrac12\left[\arctan x+\dfrac1{1+x^2}+K\right]}}$$
where $C=\dfrac1a\arctan\left(\dfrac ca\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Expressing a projective conic in standard form I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:
$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$
into the form:
$$X^2 + Y^2 + Z^2 = 0$$
Then my attempt to do so was to first find the matrix of the symmetric bilinear form which is given by:
$$ \left[ \begin {array}{ccc} 2&1&-3/2\\ 1&0&2
\\ -3/2&2&1\end {array} \right]
$$
Then I wanted to orthogonally diagonalise this matrix to obtain an expression that doesn't have any cross terms in it. However the eigenvectors of this matrix are not "nice" (I've checked with Maple and they involve a lot of square roots).
In the solution given to me, the matrix that expresses this conic in standard form was given by the following matrix:
$$ \left[ \begin {array}{ccc} 1/2\,\sqrt {2}&1/10\,\sqrt {10}&-1/3\,
\sqrt {3}\\ 0&1/10\,\sqrt {10}&-i/6\sqrt {3}
\\ 0&1/5\,\sqrt {10}&i/3\sqrt {3}\end {array}
\right]
$$
I can see that this matrix is orthogonal, but how does it simplify the expression of the conic above? And how do I find it from scratch? And how do this in general?
| I found some notes online, the task is to begin with symmetric $H$ and solve $P^T HP = D$ diagonal, which can be done over the reals. Then forcing the identity over the complexes requires multiplying by an additional diagonal complex matrix on both sides.
Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 1 }{ 2 } & 1 & 0 \\
- 2 & \frac{ 11 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
4 & 2 & - 3 \\
2 & 0 & 4 \\
- 3 & 4 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 1 }{ 2 } & - 2 \\
0 & 1 & \frac{ 11 }{ 2 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
4 & 0 & 0 \\
0 & - 1 & 0 \\
0 & 0 & 30 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 \\
- \frac{ 3 }{ 4 } & - \frac{ 11 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
4 & 0 & 0 \\
0 & - 1 & 0 \\
0 & 0 & 30 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 1 }{ 2 } & - \frac{ 3 }{ 4 } \\
0 & 1 & - \frac{ 11 }{ 2 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
4 & 2 & - 3 \\
2 & 0 & 4 \\
- 3 & 4 & 2 \\
\end{array}
\right)
$$
Alright, given this $P$ and the resulting $D,$ we multiply on the right and left by
$$
P_1 = \left(
\begin{array}{rrr}
\frac{1}{2} & 0 & 0 \\
0 & i & 0 \\
0 & 0 & \frac{1}{\sqrt {30}} \\
\end{array}
\right)
$$
to reach $I.$ I began with double your symmetric matrix, so you may need to multiply throughout by $\sqrt 2$ to take your actual matrix to $I.$ Note how completely the answer changes based on choices made; mine looks nothing like your given answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral question with cube roots Evaluate the following integral:
$$\int\sqrt[3]{x^3+\frac 1 {x^3}} dx$$
For $x^3=u$ and $1+\frac 1 {u^2}=v$, getting $$-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv$$, what is the next step?
| I would chose a different substitution:
$$I =\int \big(x^3 +\frac{1}{x^3} \big)^{\frac{1}{3}}dx$$
$$=\int \frac{(x^6 +1)^{\frac{1}{3}}}{x} dx $$
Let: $ u=(x^6+1)^{\frac{1}{3}} $, then:
$x^6=u^3-1 \text{ , and } dx= \frac{(x^6 +1)^{\frac{2}{3}}}{2x^5} du$ .
$$I=\frac{1}{2} \int \frac{u^3}{u^3-1}du$$
$$=\frac{1}{2} \int \frac{1}{u^3-1}du \text{ }+ \text{ } \frac{1}{2}u$$
The first integral can easily be solved now with partial fraction decomposition; and using the derivative of the arctan function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| $b-a=4^{2\cdot2^k}-4^{2^k}+2^{2\cdot2^k}-2^{2^k}$
$=4^{2^k}(4^{2^k}-1)+2^{2^k}(2^{2^k}-1)$
$=4^{2^k}(2^{2^k}-1)(2^{2^k}+1)+2^{2^k}(2^{2^k}-1)$
$=(2^{2^k}-1)(8^{2^k}+4^{2^k}+2^{2^k})$
$=(2^{2^k}-1)2^{2^k}\color{blue}{(4^{2^k}+2^{2^k}+1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
Asymptotic Expansion of Digamma Function While reading the wikipedia page of the Digamma function (https://en.wikipedia.org/wiki/Digamma_function#Asymptotic_expansion)
I noticed that it said the asymptotic expansion for the digamma function ($\psi(z)$) can be obtained from using
\begin{equation} \psi(z+1) = -\gamma + \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+z}) \end{equation} (where $\gamma$ is the EulerβMascheroni constant) combined with EulerβMaclaurin formula to conclude
\begin{equation} \psi(z) \approx \log(z) - \frac{1}{2z} \end{equation} My main confusion about this is that when I tried to use the EulerβMaclaurin formula to approximate the sum as an integral, I could not figure out how to obtain a $\gamma$ to cancel out the $\gamma$ term in the series expansion. Any help on how to obtain the above relationship using the formula or just where the $\gamma$ comes from when I apply the Euler-Maclaurian formula would be greatly appreciated!
| I shall assume that $|\arg (z+1)|<\pi$. By the EulerβMaclaurin formula,
\begin{align*}
\psi (z + 1) = & - \gamma + \int_1^{ + \infty } {\left( {\frac{1}{t} - \frac{1}{{t + z}}} \right)dt} + \frac{1}{2}\left( {1 - \frac{1}{{1 + z}}} \right) \\ & + \int_1^{ + \infty } {\left( {\left\{ {1 - t} \right\} - \frac{1}{2}} \right)\left( {\frac{1}{{t^2 }} - \frac{1}{{(t + z)^2 }}} \right)dt}
\\ = & \; \log (z + 1) - \gamma + \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} - \frac{1}{{2(z + 1)}} - \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt}.
\end{align*}
Thus, it remains to show that
$$
\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} = \gamma
$$
and
$$
\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt} = \mathcal{O}\!\left( {\frac{1}{{(z + 1)^2 }}} \right).
$$
The first one can be shown as follows:
\begin{align*}
\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} & =\sum\limits_{k = 1}^\infty {\int_k^{k + 1} {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} } = \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{1 - t}}{{(k + t)^2 }}dt} } \\ & = \sum\limits_{k = 1}^\infty {\left[ {\frac{1}{k} - \log \left( {1 + \frac{1}{k}} \right)} \right]} = \mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\left[ {\frac{1}{k} - \log \left( {1 + \frac{1}{k}} \right)} \right]}
\\ &
= \mathop {\lim }\limits_{n \to + \infty } \left(\sum\limits_{k = 1}^n {\frac{1}{k}} - \log (n + 1) \right)= \gamma .
\end{align*}
For the second claim, we use integration by parts and some estimations to obtain
\begin{align*}
& \left| {\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt} } \right| = \left| {\int_1^{ + \infty } {\frac{{\left\{ t \right\} - \left\{ t \right\}^2 }}{{(t + z)^3 }}dt} } \right| \\ & \le \frac{1}{4}\int_1^{ + \infty } {\frac{1}{{\left| {t + z} \right|^3 }}dt} = \frac{1}{4}\int_0^{ + \infty } {\frac{1}{{\left| {t + (z + 1)} \right|^3 }}dt}
\\ &
\le \frac{1}{4}\int_0^{ + \infty } {\frac{1}{{(t + \left| {z + 1} \right|)^3 }}dt} \sec ^3 \left( {\frac{{\arg (z + 1)}}{2}} \right) = \frac{1}{{8\left| {z + 1} \right|^2 }}\sec ^3 \left( {\frac{{\arg (z + 1)}}{2}} \right).
\end{align*}
In summary, replacing $z+1$ by $z$,
$$
\psi (z) = \log z - \frac{1}{{2z}} + R(z)
$$
where
$$
\left| {R(z)} \right| \le \frac{1}{{8\left| z \right|^2 }}\sec ^3 \left( {\frac{{\arg z}}{2}} \right)
$$
and $|\arg z|<\pi$. With more careful considerations, the constant $\frac{1}{8}$ can be replaced by $\frac{1}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Set of points that satisfy the sum of the distances between a point P and points A(-2,0) and B(2,0) = 7 So, I have to find the set of points whose distance to point $A(-2, 0)$ plus the distance to point $B(2, 0)$ is equal to $7$.
I am solving it this way:
$ d(P, A) + d(P, B) = 7 \Leftrightarrow $
$ \sqrt{(x+2)^2+y^2}+\sqrt{(x-2)^2+y^2}=7 \Leftrightarrow $
$ (x+2)^2+y^2+(x-2)^2+y^2=49 \Leftrightarrow $
$ x^2+4x+4+y^2+x^2-4x+4+y^2=49 \Leftrightarrow $
$ 2x^2+2y^2=49-8 \Leftrightarrow $
$x^2+y^2=\frac{41}{2} $
So, we have a circle centered in the Origin and radius $\sqrt{\frac{41}{2}}$
Now, when I drew this circle, and measured the distance from any point belonging to the circle to A adding to the distance of that same point to B, I get the result of around $9$
And effectively, $2\sqrt{\frac{41}{2}} \approx 9.05$
Where am I going wrong?
| When you expand $(a+b)^2$, you get $\color{blue}{\text{cross terms}}$ also as
$$(a+b)^2=a^2+b^2+\color{blue}{2ab}$$
You have missed
$$(x+2)^2+y^2+(x-2)^2+y^2+\color{blue}{2\cdot \sqrt{(x+2)^2+y^2}\cdot\sqrt{(x-2)^2+y^2}}=49$$
One more squaring will clear the square roots and finally give correct equation (of an ellipse).
If this is tedious, one can recognize the curve with given property of constant sum of distances from two points as an ellipse. And try to find its parameters.
The ellipse will be centered at midpoint of two foci, $(0,0)$. Major axis $2a=7 \Rightarrow a^2=49/4$. Distance of center from foci, $ae=2$. Minor axis is $2b$ where $b^2=a^2-a^2e^2=33/4$. Therefore required ellipse is
$$\frac{4x^2}{49}+\frac{4y^2}{33}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Moment of inertia of hollow paraboloid A hollow paraboloid of length $l$ and radius $r$ with mass $m$ and uniform density $\rho$ has the origin at the base of its circular cross section, and at the center of its cross section. I am trying to find the moment of inertia for the axis that cuts through the paraboloid's center of gravity (its x axis) and the remaining axis y and z which should yield the same values of $I$ since it is axisymmetrical about the y and z axes. I have worked out the solution for what I believe is a filled paraboloid, but I am unsure how to calculate it for a hollow paraboloid, whether it be an infinitesimally small thickness or a uniform thickness $t$. I consider $x$ to be the distance from the origin of the paraboloid along the x axis towards the tip of the paraboloid.
$dm=\rho \pi r^2 dx$
$dI_x=\frac{1}{2}r^2dm$
$I_x=\int_0^l \frac{1}{2}r^2 \rho \pi r^2 dx$
$ = \frac{1}{2}r^4\rho \pi x|_0^l$
$I_y=I_z=\int_0^l (\frac{1}{4}r^2+x^2)\rho \pi r^2 dx$
$=(\frac{1}{4}\rho \pi r^4 x+\frac{1}{3}x^3\rho \pi r^2)|_0^l$
Revised solution (thanks to Math Lover)
$a$ is the constant that determines the shape of the nose cone of the paraboloid
$dI_x=\frac{1}{2}axdm$
$I_x=\int_0^l \frac{1}{2}a^2x^2 \rho \pi dx$
$I_x=\frac{1}{6}a^2x^3 \rho \pi$
$I_y=I_z=\int_0^l (\frac{1}{4}ax+x^2)\rho \pi ax dx$
$I_y=I_z=\int_0^l (\frac{1}{4}a^2x^2\rho \pi+ax^3\rho \pi ) dx$
$I_y=I_z=(\frac{1}{12}a^2x^3\rho\pi+\frac{1}{4}ax^4\rho \pi )$
Second revised solution (once again thanks to Math Lover)
Being a hollow paraboloid, we need to take the surface integral. For a hollow paraboloid we consider the surface area of everything but the base. This results in the equation:
$A=\frac{\pi b}{6h^2} [(b^2+4h^2)^{\frac{3}{2}}-b^3]$
Substituting base radius $b$ and height $h$ with our equivalent variables:
$A=\frac{\pi \sqrt{ax}}{6x^2} [(ax+4x^2)^{\frac{3}{2}}-(ax)^{\frac{3}{2}}]$
| We have a hollow paraboloid of length $L$ and radius $R$ along $x-$axis. If the equation of the paraboloid is $y^2 + z^2 = r^2 = ax$,
Then $R^2 = aL \implies a = \frac{R^2}{L}$. We will continue our working with $a$ and substitute this value of $a$ in the end.
As we know, moment of inertia of mass $dm$ around an axis = $dm \cdot r^2$
For a hollow paraboloid, $dm = \rho \ dS$ (where $\rho$ is the uniform density represented as mass per unit surface area)
$dS = \sqrt{1 + (\frac{\partial x}{\partial y})^2 + (\frac{\partial x}{\partial z})^2} \ dA = \sqrt{1 + (\frac{2y}{a})^2 + (\frac{2z}{a})^2} \ dA$
where $dA$ is the area of the projection of $dS$ in $YZ$ plane.
$dS = \frac{1}{a} \sqrt{a^2 + 4y^2 + 4z^2} \ dy \ dz = \frac{1}{a} \sqrt{a^2 + 4r^2} \ r \ dr \ d\theta$
So, $\displaystyle I_x = \frac{2 \pi \rho}{a} \int_0^R r^3 \ \sqrt{a^2 + 4r^2} \ dr$
For moment of inertia around $y$ or $z$ axis, we note that the radius is $\sqrt{r^2 + x^2}$.
So $\displaystyle I_y = I_z = \frac{2 \pi \rho}{a} \int_0^R r \big(r^2 + \frac{r^4}{a^2}\big) \ \sqrt{a^2 + 4r^2} \ dr$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Problem by proving a statement by induction I want to prove the following statement by induction for $n\leq 1$:
\begin{align*}
\sum_{k=n}^{2n-1}\frac{1}{k}= \sum_{k=1}^{2n-1}\frac{(-1)^{k+1}}{k}
\end{align*}
For $n=1$ its easy to see:
\begin{align*}
\sum_{k=1}^{2-1}\frac{1}{k}= \sum_{k=1}^{1}\frac{1}{k}= 1 = \sum_{k=1}^{1}\frac{(-1)^{2}}{1}
\end{align*}
For $n\mapsto n+1$
\begin{align*}
\sum_{k=n+1}^{2(n+1)-1}\frac{1}{k}= \sum_{k=n+1}^{2n+1}\frac{1}{k} = \sum_{k=n-1}^{2n-1}\frac{1} {k}+\frac{1}{n+1}+\frac{1}{n}\dots \ ?
\end{align*}
I am not really sure how to deal with the $k=n+1$ in the sum, to use the case that the statement holds for n. Could anybody help me? Thank you very much!
| Call the left sum $L(n)$ and the right sum $R(n)$. Then we have
$$L(n + 1) - L(n) = \frac{1}{2n}+\frac{1}{2n+1} - \frac{1}{n},$$
and
\begin{align}
R(n + 1) - R(n) &= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+2}}{2n+1} \\
&= - \frac{1}{2n} + \frac{1}{2n+1}.
\end{align}
So by comparison, we need to check that
$$\frac{1}{2n} - \frac{1}{n} = - \frac{1}{2n}, $$
which is easy enough.
Therefore $L(n+1) - L(n) = R(n + 1) - R(n)$. So if $L(n) = R(n)$ then $L(n + 1) = R(n + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$ For $a,b,c>0;abc=1.$ Prove that $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$$
I will post my solution in the answer. Now I'm looking forward to another solution.
| My solution. Assume $(b-1)(c-1)\ge 0 \rightarrow 1=abc\ge (a+b-1)c\rightarrow a+b \le \dfrac{1}{c}+1.$
\begin{align*} \text{LHS}&=\sqrt{\left(a^2+b^2\right)-(a+b)+2+2\sqrt{\left(a^2-a+1\right)\left(b^2-b+1\right)}}+\sqrt{c^2-c+1}\\&\ge \sqrt{t^2-2t+4}+\sqrt{c^2-c+1}=\text{P}\,(\text{where}\,t=a+b)\end{align*}
Let $f(t)=P-c-t$ then prove $f(t)\ge 0.$
Since $t-1<\sqrt{(t-1)^2+3}=\sqrt{t^2-2t+4},$
we have $$f'(t)=\dfrac{t-1}{\sqrt{t^2-2t+4}}-1<1-1=0\rightarrow f(t)\ge f\left(\dfrac{1}{c}+1\right)\ge 0.$$
By some simple calculations, we need to prove $3c^2(c-1)^2\ge 0,$ which is true!
Done. Equality holds when $a=b=c=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Any neat way to solve the integral $\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$? Straight to the point: given the integral
$$\iint_Q \frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$$
where $Q=[-a,a]\times[-b,b]$, can you think of any neat way to solve it?
At a first glance it looked quite innocent. No need to say that Iβve changed my mind.
EDIT:
Integrating first w.r.t. y, using the integral
$$\int \frac{1}{\left(\xi^2+\alpha^2\right)^{3/2}}\,d\xi = \frac{\xi}{\alpha^2\sqrt{\xi^2+\alpha^2}} + \text{constant}, $$
one gets to
$$ 2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx $$
and here I get pretty stuck to be honest, so if someone could give me any hints that would be appreciated. Still, I think there should be some nicer way to approach this from the start.
Since I was asked for the source of this problem: I simply asked myself a basic physics question, and trying to find an answer led to this integral.
| Substitute $x = \sqrt{b^2+z^2}\sinh t$ after integrating over $y$
\begin{align}
&\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy \\
= &2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx \\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\sinh^2 t +z^2}dt\\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\cosh^2 t -b^2}dt\\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac{d(\tanh t)}{
z^2+b^2\tanh^2 t }\\
=&\frac 4z \tan^{-1} \left(\frac bz \tanh t\right)\bigg|_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}} }\\
=&\frac 4z \tan^{-1} \left(\frac{ab}{z\sqrt{a^2+b^2+z^2}}\right)
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Partitioning 18 into odd number of even components I need to find a total number of ways $18$ can be partitioned into odd number of even components.
Does the function below satisfy the "odd number of components" requirement?
$$
P_{1,3,5,7,9}(X) = \frac{1}{1-x} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^5} \cdot \frac{1}{1-x^7} \cdot \frac{1}{1-x^9}
$$
Given that $n$ is even, then every partition of $n$ has an even number of odd parts.
Can it be said that if $n$ is even, then every partition of $n$ has an odd number of even parts?
How can I add to that the requirement that components are supposed to be even?
I have spent 6 hours reading about self-conjugation, odd and distinct parts and generating functions and I'm clueless.
| It may be quickest to just write the numbers down: you should be able to find $16$ cases.
$$18 =2+2+14=2+4+12=2+6+10=2+8+8=4+4+10 \\=4+6+8=6+6+6 =2+2+2+2+10=2+2+2+4+8\\=2+2+2+6+6=2+2+4+4+6
=2+4+4+4+4 \\= 2+2+2+2+2+2+6=2+2+2+2+2+4+4 \\
=2+2+2+2+2+2+2+2+2$$
One approach could be to say that, by dividing each part by $2$, the number of ways $18$ can be partitioned into an odd number of even parts is equal to the number of ways $9$ can be partitioned into an odd number of parts of any parity. That just pushes the issue further down to a question which has been considered slightly more. OEIS A027193 enumerates these but does not in my view give a simple expression for finding the number of such partitions.
The easiest one to use that I know is that the number of ways $n$ can be partitioned into an odd number of parts is $p(n-1^2)-p(n-2^2)+p(n-3^2)-\cdots$ where $p(n)$ is the total number of partitions of $n$. So here in the case of $n=9$ you get $p(8)-p(5)+p(0) = 22-7+1=16$.
| {
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confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.
$$\frac{n(n+1)}{(n+3)^3}$$
Using the ratio test, then simplifying in stages:
$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+1)}$$
$$\frac{(n+1)(n+2)(n+3)^3}{(n+4)^3n(n+1)}$$
$$\frac{(n+2)(n+3)^3}{(n+4)^2(n+4)n}$$
$$\frac{(n+3)^3}{(n+4)^22n}$$
The concluding remark:
$$\frac{1}{2}\lim_{n \to \infty}\frac{(n+3)^3}{(n+4)^2n}$$
Hence the series is divergent, unless I went wrong somewhere?
| Ratio test is not effective because
$$\lim_{n\to \infty }\frac{(n+2) (n+3)^3}{n (n+4)^3}=1$$
You can use comparison test as
$$\frac{n (n+1)}{(n+3)^3}\sim \frac1n ;\text{ as }n\to\infty$$
and $\sum\frac1n $ diverges
| {
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Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):
What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:
*
*$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$
*$$ x_{3/2} = x_{5}^2 $$
*the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$
*the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$
| Let $\alpha$ denote the real root of $X^3+X^2-1$. Then $\alpha^3 = 1-\alpha^2$. If we plug in $\alpha^2$ into $X^3-X^2+2X-1$, we obtain
\begin{align*}
(\alpha^2)^3-(\alpha^2)^2+2(\alpha^2)-1 &= (1-\alpha^2)^2-\alpha^4+2\alpha^2-1\\
&=(1-\alpha^2)^2-(1-\alpha^2)^2=0,
\end{align*}
so $\alpha^2$ is a root of $X^3-X^2+2X-1$.
| {
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Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.
The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
x &= 1\\
x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\
x &= (-1)^{2/9} \\
x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\
x &= (-1)^{4/9} \\
x &= -(-1)^{5/9} \\
x &= (-1)^{2/3} \\
x &= -(-1)^{7/9} \\
x &= (-1)^{8/9}
\end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?
| Your elements are basically $(e^{ik\pi/9})_{k=0}^8$. They form the cyclic group of $9$ elements under ordinary multiplication. The most obvious generator is $r=e^{i\pi/9}$ (with $k=1$).
However, any such element with $k$ relatively prime to $9$ (e.g. $4,5,8$) would generate the group as well. For example, starting with $r^4$, you get a cycle
$$
\left< r^4,r^8,r^{12} = r^3,r^7,r^{11}=r^2,r^6,r^{10} = r,r^5, r^9 = 1 \right>
$$
| {
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Proof by induction: Inductive step struggles Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$
where $ n $ is a nonnegative integer.
Preforming the basis step where $ n $ is equal to 0
$$
1 = \frac{2^{1}+(-1)^{0}}{3\times2^{0}} = \frac{3}{3} = 1
$$
Now the basis step is confirmed.
Then I started the inductive step where $ n = k $ is assumed true and I needed to prove $ n = k+1 $
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left(- \frac{1}{2}\right)^{k} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times2^{k+1}}
$$
Using the inductive hypothesis
$$
\frac{2^{k+1}+(-1)^{k}}{3\times2^{k}} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times 2^{k+1}}
$$
After this I am struggling here trying to get around to the end. I would appreciate any guidance.
| \begin{align}
\frac{2^{k+1}+(-1)^k}{3 \times 2^k}+\left(-\frac{1}{2}\right)^{k+1}
&=\frac{2 \times 2^{k+1}+2 \times(-1)^k}{3 \times 2^{k+1}}+\frac{3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}+2 \times (-1)^k+3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}-2 \times (-1)^{k+1}+3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}+(-1)^{k+1}}{3 \times 2^{k+1}}
\end{align}
| {
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If $\tan x\tan y=\frac{b}{a},\,a,\,b\ne 0$, prove that $\frac{\sec^2x}{a\tan^2x+b}+\frac{\sec^2y}{a\tan^2y+b}=\frac{a+b}{ab}$ If $\tan x\tan y=\frac{b}{a},\,a,\,b\ne 0$, prove that $\frac{\sec^2x}{a\tan^2x+b}+\frac{\sec^2y}{a\tan^2y+b}=\frac{a+b}{ab}$
I solved it in the following way:
$\frac{1+\tan^2x}{a\tan^2x+b}+\frac{1+\tan^2y}{a\tan^2y+b}$
$=\frac{a\tan^2y+b+a\tan^2x\tan^2y+b\tan^2x+a\tan^2x+a\tan^2x\tan^2y+b+b\tan^2y}{2b^2+ab(\tan^2x+\tan^2y)}$
$=\frac{a(\tan^2y+\tan^2x)+b(\tan^2x+\tan^2y)+tb+2a\tan^2x\tan^2y}{2b^2+ab(\tan^2x+\tan^2y)}$
$=\frac{(\tan^2x+\tan^2y)(a+b)+2a\tan^2x\tan^2y+2b}{2b^2+ab(\tan^2x+\tan^2y)}$
$2b^2=2ab\tan x\tan y$
$\therefore \frac{1+\tan^2x}{a\tan^2x+b}+\frac{1+\tan^2y}{a\tan^2y+b}=\frac{(\tan^2x+\tan^2y)(a+b)+2a\tan^2x+\tan^2y+2b}{ab(\tan^2x+\tan^2y+2\tan x\tan y)}$
It remains to prove that:
$2a\tan^2x\tan^2y+2b=2(a+b)\tan x\tan y$
$2a\tan^2x\tan^2y+2b=ta\tan x\tan y+2b\tan x\tan y$
$2a\tan^2x\tan^2y+2b=2b+2b\tan x\tan y$
$2a\tan^2x\tan^2y=2b\tan x\tan y$
$\tan^2x\tan^2y=\frac{b}{a}\tan x\tan y$
Which is true.
$\therefore \frac{1+\tan^2x}{a\tan^2x+b}+\frac{1+\tan^2y}{a\tan^2y+b}=\frac{(a+b)(\tan x+\tan y)^2}{ab(\tan x+\tan y)^2}=\frac{a+b}{ab}$
Which completes the proof.
This proof is overly complex and took me over an hour to think of and write. Could you please explain to me a simpler, less messy and more intuitive proof?
| Let $t:=\tan x$ so $\tan y=\frac{b}{at}$ and we want$$\frac{1+t^2}{at^2+b}+\frac{1+b^2/(a^2t^2)}{b^2/(at^2)+b}=\frac{b(1+t^2)+at^2+b^2/a}{b(at^2+b)}=\frac{a+b}{ab},$$taking out a factor of $t^2+b/a$.
| {
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Question regarding a proof of a inequality Given $a,b,c, \in (0,\infty)$, then the following inequality holds
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 2 \sqrt6 (a+b+c)$$
What I've tried:
First, I noticed that we can factor
$$5a^2+12ab+7b^2 = (a+b)(5a+7b)$$
1.) Therefore we can use AM-GM with $a+b$as the first term and $(5a+7b)$ as the second term
$$\sqrt{(a+b)(5a+7b)} \leq \frac{(a+b)+(5a+7b)}{2}$$
So
$$\sqrt{5a^2+12ab+7b^2} \leq 3a+4b$$
Analogous we get
$$\sqrt{5b^2+12bc+7c^2} \leq 3b+4c$$
$$\sqrt{5c^2+12ca+7a^2} \leq 3c+4a$$
Summing all three inequalities,
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 7(a+b+c)$$
However, since $7>2\sqrt6$, this approach doesn't yield the wanted result.
2.) Another way was to see what constraints are on $a,b$ are if we have that
$$\sqrt{{5a^2+12ab+7b^2}} \leq \sqrt6 (a+b)$$
and analogue for $b,c$ and $c,a$.
Squaring both sides, and knowing that $a,b$ are positive real numbers, we can only consider the positive branch of the square-root,
$(a+b)(5a+7b) \leq 6(a+b)^2$ then $5a+7b \leq 6a+6b$, thus $b \leq a$.
Analogous,
$c \leq b \leq a$, but also $a \leq c$ which would lose generality because by the transitive property, $c \leq a$ and $a \leq c$ which is true if and only if $a=c$. Therefore this approach is also not correct.
3.) I don't think we can apply Minkowski's Inequality, I can't find a way to write $7b^2$ in a way to write our expression as a sum of two squares of binomials.
How can this inequality be proven? Can we use any of these ideas?
| Your inequality become the equality when $a=b=c.$ But for $a=b,$ then
$$\sqrt{(a+b)(5a+7b)} \ne \frac{(a+b)+(5a+7b)}{2}.$$
By the AM-GM inequality, we have
$$\sqrt{(a+b)(5a+7b)} = \frac{1}{\sqrt6} \cdot \sqrt{6(a+b) \cdot (5a+7b)} \leqslant \frac{6(a+b)+(5a+7b)}{2\sqrt6} = \frac{11a+13b}{2\sqrt6}.$$
Equality hold for $a=b.$
Therefore
$$\sum \sqrt{(a+b)(5a+7b)} \leqslant \sum \frac{11a+13b}{2\sqrt6} = 2\sqrt 6 (a+b+c).$$
| {
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Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$ given two positive numbers $a, b$ so that $a+ b= 1$
Sui Zhen Lin ; @szl6208 gave a very beautiful proof for the following inequality
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}+ \sqrt{\frac{4b}{a+ 4}}\leq 1$$
Source: AoPS/@szl6208_ on.AoPS
Proof. We have
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}\leq\frac{\sqrt{a\left ( 9b^{2}- 8b+ 4 \right )\cdot a\left ( a+ 1 \right )^{2}}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}\leq\frac{a\left ( 9b^{2}- 8b+ 4 \right )+ a\left ( a+ 1 \right )^{2}}{2\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}$$
$$\sqrt{\frac{4b}{a+ 4}}\leq\frac{\sqrt{4b\left ( a+ 4 \right )\cdot\left ( 2b+ 2 \right )^{2}}}{\left ( a+ 4 \right )\left ( 2b+ 2 \right )}\leq\frac{4b\left ( a+ 4 \right )+ \left ( 2b+ 2 \right )^{2}}{2\left ( a+ 4 \right )\left ( 2b+ 2 \right )}$$
Hence, we need to prove that
$$\frac{a\left ( 9b^{2}- 8b+ 4 \right )+ a\left ( a+ 1 \right )^{2}}{2\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}+ \frac{4b\left ( a+ 4 \right )+ \left ( 2b+ 2 \right )^{2}}{2\left ( a+ 4 \right )\left ( 2b+ 2 \right )}\leq 1$$
$$\Leftrightarrow RHS- LHS= \frac{a\left ( ab+ 11a+ 1 \right )\left ( a- b \right )^{2}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )\left ( a+ 4 \right )\left ( b+ 1 \right )}\geq 0$$
wait a minute, actually
$$RHS- LHS- \frac{a\left ( ab+ 11a+ 1 \right )\left ( a- b \right )^{2}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )\left ( a+ 4 \right )\left ( b+ 1 \right )}= \left ( a+ b- 1 \right )f\left ( a, b \right )= 0$$
but how.. why is that so perfect ? I think there maybe are solutions constructed by substitutions like $g\left ( a \right )- g\left ( 1- b \right )\!, g\left ( 2a \right )- g\left ( a+ 1- b \right )\!= \left ( a+ b- 1 \right )f\left ( x \right ).$ What's the best perfect substitution here ? Someone teaching me, huh ?
| Using algebra.
From the constraint $b=1-a$. So, we face
$$f(a)=\sqrt{\frac{a^2}{9 a^2-10 a+5}}+2 \sqrt{\frac{1-a}{a+4}}$$
$$f'(a)=\frac{5-5 a}{(9 a^2-10 a+5)^{3/2}}-\frac{5}{\sqrt{1-a} (a+4)^{3/2}}$$
Square and factor; the numerator is just
$$(2 a-1) (5 a-1) \left(a^2-a+1\right) \left(73 a^2-118 a+61\right)$$ So, the derivative only cancels for $a=\frac 15$ and $a=\frac 12$. The second derivative test shows the the first is a minimum and the second a maximum. So, we have
$$f\left(\frac{1}{5}\right)=\frac 32\sqrt{\frac{3}{7}} \qquad \text{and} \qquad f\left(\frac{1}{2}\right)=1$$
Around $a=\frac 12$, we have
$$f(a)=1-\frac{20}{27} \left(a-\frac{1}{2}\right)^2-\frac{1480}{729}
\left(a-\frac{1}{2}\right)^3+O\left(\left(a-\frac{1}{2}\right)^4\right)$$
| {
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Calculation in Routh's theorem The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:
Option 1
The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.
That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.
Option 2
This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.
The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.
The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.
| A variant on my option 2: as @RiverLi notes, the numerator is of at most degree $2$ in $z$, so is of the form $(c-1)(Aa+Bb+Cc+D)$. That $A=B=0,\,C=D=1$ follows from the case $x=y=z$.
| {
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What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.
Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.
Any help is appreciated!
| You can distinguish between the two possibilities you found by starting modulo $2000$:
because the Carmichael function of $2000$ is $100$, $7^{2000}\equiv(7^{100})^{20}\equiv1\bmod2000$,
so $7^{2005}\equiv7^5\equiv807\bmod 2000$.
Therefore, $7^{2005}-1\equiv806\bmod2000$.
Therefore, $\dfrac{7^{2005}-1}2\equiv403\bmod1000$.
Therefore, $\dfrac{7^{2005}-1}6\equiv403\times3^{-1}\bmod{1000}.$
In general, if $n=3k-1$ then $3^{-1}\equiv k\pmod n$,
and if $n=3k+1$ then $3^{-1}\equiv -k\pmod n$,
so $3^{-1}\equiv-333\equiv667\bmod{1000}$.
Therefore, $\dfrac{7^{2005}-1}6\equiv403\times667\equiv801\bmod1000$.
| {
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Determining the limit: What does it mean when one obtains zero in the calculation? Consider the sequence $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\\$. To determine the limit I did the following:
\begin{aligned}
a_{n} &=\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right) \frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}} \\[10pt]
&=\frac{2 \sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}=\frac{2 \sqrt{n}}{\sqrt{n} \sqrt{1+\frac{1}{\sqrt{n}}}+\sqrt{n} \sqrt{1-\frac{1}{\sqrt{n}}}} \\[10pt]
&=\frac{2}{\sqrt{1+\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}+\sqrt{1-\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}} \\[10pt] &= \dfrac{2}{2} = 1.
\end{aligned}
However, my first thought was $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}} = \sqrt{n}\left(\sqrt{1 + \dfrac{1}{\sqrt{n}}}\space - \sqrt{1 - \dfrac{1}{\sqrt{n}}}\right) = \sqrt{n}\space (1-1) = 0$
with the same argument as above. Is it correct that one can't make a statement about the convergence in the latter calculation because we have $\infty \cdot 0$ ? If yes, why exactly is this the case?
Edit: To see that $\lim\limits_{n\to \infty} \dfrac{1}{\sqrt{n}} = 0$ pick some arbitrary $\epsilon > 0$. We want to find a $N$ s.t. $\forall n\geq N\colon |\dfrac{1}{\sqrt{n}} - 0|< \epsilon \Longleftrightarrow \dfrac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow n >\dfrac{1}{\epsilon^2}$ meaning we can choose $N = \dfrac{1}{\epsilon^2} + 1$ for example.
| Your second approach is wrong.
To save writing let $x=\sqrt{n}$, and expand square roots in power series in $\frac{1}{x}$, so expression becomes $x(\sqrt{1+\frac{1}{x}}-\sqrt{1-\frac{1}{x}})$
$=x(1+\frac{1}{2x}+O(\frac{1}{x^2})-1+\frac{1}{2x}+O(\frac{1}{x^2}))$
$=1+O(\frac{1}{x})\to 1$ as $x\to \infty$
| {
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} |
Simplify the sum of the square of a polynomial having a fourth degree. Today I learn about polynomial. Because I want to improve my knowledge. Thank you for your support and time for sharing information and experience.
From question :
If $a, b, c$ and $d$ are the roots of polynomial $Ax^4+Bx^3+Cx^2+Dx+E$ then find the value of
$a^2+b^2+c^2+d^2$
What I know :
*
*$a+b+c+d=\frac{-B}{A}$
*$ab+ac+ad+bc+bd+cd=\frac{C}{A}$
*$abc+abd+acd+bcd=\frac{-D}{A}$
*$abcd=\frac{E}{A}$
What I try :
$$(a+b+c+d)^2=(a+b+c+d)(a+b+c+d)$$
$$\left(\frac{-B}{A}\right)^2=(a^2+ab+ac+ad)+(ab+b^2+bc+bd)+(ac+bc+c^2+cd)+(ad+bd+cd+d^2)$$
$$\left(\frac{-B}{A}\right)^2=2(ab+ac+ad)+2(bc+bd)+2(cd)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2=2(ab+ac+ad+bc+bd+cd)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2=2\left(\frac{C}{A}\right)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2-2\left(\frac{C}{A}\right)=a^2+b^2+c^2+d^2$$
$$\frac{B^2}{A^2}-\frac{2AC}{A^2}=a^2+b^2+c^2+d^2$$
$$a^2+b^2+c^2+d^2=\frac{B^2-2AC}{A^2}$$
My Question:
*
*Is my work correct ?
*Is it possible trying from $abcd=\frac{E}{A}$ ?
Thank you for your help and your time. God bless you.
| Indeed, we know that
$$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd$$ and you find these sums from Vieta's formulas.
Solving from $abcd$ alone is obviously impossible, because there are infinitely many ways to obtain the same value from different $a,b,c,d$, and these result in different sums $a^2+b^2+c^2+d^2$. To make $a,b,c,d$ uniquely determined, you need more information, i.e. more polynomial coefficients.
(The complete discussion of why all other possible combinations of the polynomial coefficients do not work is out of my reach/time.) Note that you can't retrieve the four roots from less than four coefficients, but we are not looking for the individual roots but for the sum of squares.
| {
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How do I evaluate $\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx$? I need to calculate the following definite integral:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx.$$
The only thing that I've found is:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx = \int_{1/3}^3 \frac{\arctan \frac{1}{x}}{x^2 - x + 1} \; dx,$$
but it doesn't seem useful.
| $$\begin{eqnarray*}\int_{1/3}^{3}\frac{\arctan x}{x^2-x+1}\,dx&=&\int_{1/3}^{1}\frac{\arctan x}{x^2-x+1}\,dx+\int_{1}^{3}\frac{\arctan x}{x^2-x+1}\,dx\\&=&\int_{1}^{3}\frac{\arctan x+\arctan\frac{1}{x}}{x^2-x+1}\,dx\\&=&\frac{\pi}{2}\int_{1}^{3}\frac{dx}{x^2-x+1}\\&=&2\pi\int_{1}^{3}\frac{dx}{(2x-1)^2+3}\\&=&\pi\int_{1}^{5}\frac{dt}{t^2+3}\\&=&\color{red}{\frac{\pi}{\sqrt{3}}\,\arctan\frac{5}{\sqrt{3}}-\frac{\pi^2}{6\sqrt{3}}}.\end{eqnarray*}$$
| {
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but IΒ΄m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I canΒ΄t conclude our first inequality.
Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I donΒ΄t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.
| Answer
$$
\begin{align}
0
&\le\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\\
&=(x-1)^2+(y-1)^2-(x-1)(y-1)\\
&=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)-(xy-x-y+1)\\
&=\left(x^2+y^2+1\right)-(xy+x+y)
\end{align}
$$
Motivation
To prove $x^2+y^2+1\ge xy+x+y$, look at the difference
$$
\begin{align}
\left(x^2+y^2+1\right)-(xy+x+y)
&=(x-y)^2+(x-1)(y-1)\tag1\\
&=(u-v)^2+uv\tag2\\[2pt]
&=(u-v)^2+\tfrac14\left((u+v)^2-(u-v)^2\right)\tag3\\
&=\tfrac34(u-v)^2+\tfrac14(u+v)^2\tag4\\
&=\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\tag5\\
&\ge0\tag6
\end{align}
$$
Explanation:
$(1)$: subtract $2xy+1$ from both sides of the minus sign
$(2)$: $u=x-1$ and $v=y-1$
$(3)$: $uv=\frac14\left((u+v)^2-(u-v)^2\right)$
$(4)$: combine terms
$(5)$: $u=x-1$ and $v=y-1$
$(6)$: sum of two non-negative numbers is non-negative
| {
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"question_score": "3",
"answer_count": 9,
"answer_id": 7
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Quick questions of infinite series Are these true:
$$1.\sum_{n=1}^{\infty}\frac{1}{n^2+88n}=\frac{1}{88}\sum_{n=1}^{88}\frac{1}{n}$$
$$2.\sum_{n=1}^{\infty}\frac{1}{n^2+4n}=\frac{25}{48}$$
For first one, it's factored. ΒΏIt's not factored correctly? Because the bound is from $\infty$ to $88$ and if you multiply $\frac{1}{8}$ with the other fraction, it's not the same.
For second one, it's basic finding if it's converge. I find pattern $\frac{1}{5}+\frac{1}{12}+\frac{1}{21}$, etc.
| $$\frac{1}{n^2 + 88n} = \frac{1}{n(n+88)} = \frac{1}{88}\left( \frac{1}{n} - \frac{1}{n+88} \right).$$ Therefore, $$\sum_{n=1}^\infty \frac{1}{n^2 + 88n} = \frac{1}{88} \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+88} = \frac{1}{88} \sum_{n=1}^{88} \frac{1}{n}.$$
The other sum is handled in a similar fashion.
| {
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What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$? What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)
I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.
Is there any way by which I can determine where these functions intersect?
| First of all $x > 0$.
If $x \le 0$ then $x^3 < 0$ so $[x] = x^3 -3 \le -3$ so $[x] \le -3$ and $x \le -2$. (Obviously I'm being overly cautious. $[x]\ne x$ unless $x$ is on integer so $x \le -3$ and obviously $x$ can't be $0$ so $[x]\le -4$ and ... so on but... this just stick with $x \le -2$.)
This means that $|x^3| > |x|$ and that $x^3 - x < 0$. So $x^3 - [x] \ge x^3 -x > x^3 -[x] -1$ so $x^3 -[x] < 1< 3$.
So $x > 0$.
....
Now $[x] \le x$ so $[x]^3 - [x] \le x^3-[x] = 3$.
We must have $[x]^3 -[x] = [x]([x]^2 - 1) = [x]([x]-1)([x]+1) \le 3$.
If $[x] \ge 2$ then $[x]([x]-1)([x]+1)\ge 2\cdot 1\cdot 3 > 3$ so we must have
$[x]=1$
Okay. So $1\le x < 2$. Let $r = x-1$ and $x = 1+ r$.
Then $x^3 -[x] = (1+r)^3 =1 = 3r + 3r^2 + r^3 = 3$.
Now $r$ is monotonically increasing on $r\in [0,1)$ so there is at most one solution.
And $3\cdot 0 + 3\cdot 0^2 + 0^3 < 3$ and $\lim_{r\to 1} 3r + 3r^2 + r^3 = 7 > 3$.
So by continuity there is one solution.
We might be able to solve for $r$ but we don't have to. $r$ is the unique and only solution to $3r + 3r^2 + r^3 = 3$ and $x = 1+r$.
And... oh, googly moogly......
If $[x]=1$ then $x^3 - 1 =3$ and..... $x =\sqrt[3] 4$ and $r = \sqrt[3] 4 -1$ is the unique solution to $3r+3r^2 +r^3 = 3$ and ....
Ah... geeze.... that was trivial to solvi
$3r + 3r^2 + r^3 = 3$
$1+3r + 3r^2 + r^3 = 3+1 = 4$.
$(1+r)^3 = 4$.
.... forest for the trees, forest for the trees.
| {
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Epsilon-delta proofs $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I am getting back to $\delta-\epsilon$ arguments after a break, so I'd like ask if my $\delta$-response to the below $\epsilon$-challenges is technically correct and rigorous.
Problem 4.2.2. For each stated limit, find the largest possible $\delta-$neighbourhood that is a proper response to the given $\epsilon$-challenge.
(a) $\lim_{x \to 3}(5x - 6) = 9$, where $\epsilon = 1$
(b) $\lim_{x \to 4}\sqrt{x} = 2$, where $\epsilon = 1$
(c) $\lim_{x \to \pi} [[x]] = 3$, where $\epsilon = 1$. (The function $[[x]]$ returns the greatest integer less than or equal to $x$)
(d) $\lim_{x \to \pi} [[x]] = 3$, where $\epsilon = 0.01$
Proof.
(a) We would like to make the distance $\absval{(5x - 6) - 9}<1$, so $\absval{5x - 15} < 1$, thus $\absval{x - 3} < 1/5$. Thus, $\delta = 1/5$.
(b) The expression $\absval{\sqrt{x} - 2}$ can be written as,
\begin{align*}
\sqrt{x} - 2 &= (\sqrt{x} - 2) \times \frac{(\sqrt{x} + 2)}{(\sqrt{x} + 2)}\\
&= \frac{\absval{(x - 4)}}{\absval{\sqrt{x} + 2}}\\
&\le \frac{\absval{(x - 4)}}{2}\\
&< \frac{\delta}{2}
\end{align*}
So, if we pick $\delta = 2$, the distance $\absval{\sqrt{x} - 2}$ would be smaller than $\epsilon = 1$.
(c) Consider the expression $\absval{[[x]] - 3}$.
We are interested to make the distance $\absval{[[x]] - 3}$ smaller than $\epsilon = 1$. So, $2 < [[x]] < 4$. Therefore, $[[x]]=3$. This is true, if and only if, $3 \le x < 4$. Consequently, $3 - \pi \le x - \pi < 4 - \pi$. This inequality is satisfied, if the distance $\absval{x - \pi} < \pi - 3$. So, $\delta = \pi - 3$.
(d) Again, $\delta = \pi - 3$ is a proper response to the given $\epsilon-$ challenge.
| all good except part (b), which is wrong because you are requested to find the largest $\delta$, which must be $(\delta = 3)$, since
$|\sqrt{4-3} - 2| = 1.$
| {
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Rank inequality
Let $ A, B \in M_n (\mathbb{C})$ such that $(A-B)^2 = A -B$. Then $\mathrm{rank}(A^2 - B^2) \geq \mathrm{rank}( AB -BA)$.
I tried to apply the basic inequalities without results. How to start? Thank you.
| $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
$\newcommand{\M}{\begin{pmatrix} I_{(r)} & 0 \\ 0 & 0 \end{pmatrix}}$
The key to this problem is to use the property of idempotent matrix $A - B$. As @Berci suggested, denote $A - B = P$, then $P^2 = P$, which implies there exists an order $n$ invertible matrix $T$, such that $P = T\diag(I_{(r)}, 0)T^{-1}$, where $r
= \rank(P)$ (this is a standard result for idempotent matrix, if you need a proof for that, I can add it as appendix).
Now $A = B + P$, substituting it to $A^2 - B^2$ and $AB - BA$ respectively, we have $A^2 - B^2 = P + BP + PB, AB - BA = PB - BP$. Equivalently, we have
\begin{align*}
& T^{-1}(A^2 - B^2)T = \M + T^{-1}BT\M + \M T^{-1}BT, \tag{1} \\
& T^{-1}(AB - BA)T = \M T^{-1}BT - T^{-1}BT\M. \tag{2}
\end{align*}
Partition $T^{-1}BT$ as $\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}$, where $A_{11}$ is of order $r$, then $(1)$ and $(2)$ become
\begin{align*}
& T^{-1}(A^2 - B^2)T = \begin{pmatrix} I_{(r)} + 2A_{11} & A_{12} \\ A_{21} & 0
\end{pmatrix}, \\
& T^{-1}(AB - BA)T = \begin{pmatrix} 0 & A_{12} \\ -A_{21} & 0 \end{pmatrix}.
\end{align*}
In view of this, to show that $\rank(A^2 - B^2) \geq \rank(AB - BA)$, it suffices to show
\begin{align*}
\rank\begin{pmatrix} B_1 & B_2 \\ B_3 & 0\end{pmatrix}
\geq \rank\begin{pmatrix} 0 & B_2 \\ -B_3 & 0\end{pmatrix}
\end{align*}
for any order $r$ matrix $B_1$, $r \times (n - r)$ matrix $B_2$, and $(n - r) \times r$ matrix $B_3$. To this end, let $\begin{pmatrix} 0 \\ e_1 \end{pmatrix},
\ldots, \begin{pmatrix} 0 \\ e_i \end{pmatrix}, \begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix}, \ldots, \begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix}$ be the maximal linear independent group of $\begin{pmatrix} 0 & B_2 \\ -B_3 & 0\end{pmatrix}$, where $e_1, \ldots, e_i$ are columns of $B_3$, and $e_{i + 1}, \ldots, e_{i + j}$ are columns of $B_2$, $0 \leq i \leq r$, $0 \leq j \leq n - r$.
It can then be shown that columns $\begin{pmatrix} f_1 \\ e_1 \end{pmatrix},
\ldots, \begin{pmatrix} f_i \\ e_i \end{pmatrix}, \begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix}, \ldots, \begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix}$ are linearly independent, where $f_1, \ldots, f_i$ are corresponding columns of matrix $B_1$. To wit, suppose
\begin{align*}
a_1\begin{pmatrix} f_1 \\ e_1 \end{pmatrix} +
\cdots + a_i\begin{pmatrix} f_i \\ e_i \end{pmatrix} + a_{i + 1}\begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix} + \cdots + a_{i + j}\begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix} = 0,
\end{align*}
then
\begin{align*}
& a_1f_1 + \cdots + a_if_i + a_{i + 1}e_{i + 1} + a_{i + j}e_{i + j} = 0 \tag{3} \\
& a_1e_1 + \cdots + a_ie_i = 0. \tag{4}
\end{align*}
By linear independence of $e_1, \ldots, e_i$, from $(4)$ we obtain $a_1 = \cdots = a_i = 0$. Substituting these back to $(3)$, we have $a_{i + 1}e_{i + 1} + \cdots +
a_{i + j}e_{i + j} = 0$, which implies $a_{i + 1} = \cdots = a_{i + j} = 0$ by linear independence of $e_{i + 1}, \ldots, e_{i + j}$. This shows $\begin{pmatrix} f_1 \\ e_1 \end{pmatrix},
\ldots, \begin{pmatrix} f_i \\ e_i \end{pmatrix}, \begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix}, \ldots, \begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix}$ are linearly independent, whence
\begin{align*}
\rank\begin{pmatrix} B_1 & B_2 \\ B_3 & 0\end{pmatrix} \geq i + j =
\rank\begin{pmatrix} 0 & B_2 \\ -B_3 & 0\end{pmatrix}.
\end{align*}
This completes the proof.
| {
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How to deal with subtraction of sigma? How to solve this?
$$\displaystyle\sum_{n=1}^\infty
\frac{n}{3^{n-1}}-\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}$$
My work
$$S=\displaystyle\sum_{n=1}^\infty \left(\frac{n}{3^{n-1}}-\frac{n}{3^n}\right)$$
$$S=\displaystyle\sum_{n=1}^\infty\frac{3n-n}{3^n}=\displaystyle\sum_{n=1}^\infty \frac{2n}{3^n}$$
which seems to yield nothing.
| I suspect that you're supposed to shift the index of one of the series. What you've done is 100% valid, but as you said, it's not yielding something "useful". Instead, take the series
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}},$$
and shift the index. Let $m = n - 1$. Then when $n = 1$, $m = 0$, so
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}} = \sum_{m=0}^\infty \frac{m+1}{3^{m}} = \sum_{n=0}^\infty \frac{n + 1}{3^n} = 1 + \sum_{n=1}^\infty \frac{n + 1}{3^n}.$$
Thus,
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}}-\sum_{n=1}^\infty \frac{n}{3^n} = 1 + \sum_{n=1}^\infty \frac{n + 1}{3^n} - \sum_{n=1}^\infty \frac{n}{3^n} = 1 + \sum_{n=1}^\infty \frac{1}{3^n},$$
which is a geometric series that you can calculate easily.
| {
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Reduction formula for $\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $ I'm struggling to go any further. Anyone have any hints?
$$ I _ n = \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $$
I used integration by parts where I differentiated $ x ^ n $, but it resulted in an $ \arcsin x $ term which didn't get me anywhere.
$$ u = x ^ n $$
$$ v = \arcsin x $$
$$ \mathrm d v = \frac { \mathrm d x } { \sqrt { 1 - x ^ 2 } } $$
$$ I _ n = \int u \ \mathrm d v = u v - \int v \ \mathrm d u = x ^ n \arcsin x - ( n - 1 ) \int x ^ { n - 1 } \arcsin x \ \mathrm d x $$
| Integrate by parts as follows
\begin{align}
\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } d x
&= -\int \frac { x ^{n-1} } { (1 - x ^ 2 )^{\frac{n-1}2}} d \left(\frac{(1-x^2)^{\frac n2}}n\right)\\
&=-\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n\int \frac { x ^ {n-2} } { \sqrt { 1 - x ^ 2 } } d x \\
\end{align}
hence the reduction formula
$$I_n = -\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n I_{n-2}$$
| {
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Intuition for polynomial long division So I am trying to learn some basic algebra and go over some precalc material and I am terribly confused on why or how polynomial long division works.
I have looked at many sources online and they seem to all suggest that division of numbers is identical to division of polynomials since a number in its decimal expansion can be viewed as a sort of polynomial e.g $a_n10^n+\cdots+a_2100+10a_1+a_0$
What i don't understand is why we only use the term with the highest degree in the denominator as the divisor for the whole polynomial as shown here Polynomial Long Division whereas with normal division we divide the whole divisor into the dividend for example when calculating $88Γ·32$ we look at how many times $32$ goes into $88$ and not how many times $30$ does whilst with polynomial division only the highest degree term is considered which corresponds to 30 in this example ignoring 2.
So how are they the same? Some intuition behind why polynomial division works would be greatly appreciated.
Thanks in advance.
| If you really want to understand the mathematics, do not do the kind of long division that you were taught. Instead, you must learn to write down true equalities. For example:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
β $\lfrac{4x^3-7x^2-6}{2x^2+5}$
β $ = \lfrac{\color{blue}{2x}(2x^2+5)-10x-7x^2-6}{2x^2+5}$
β $ = 2x + \lfrac{-7x^2-10x-6}{2x^2+5}$
β $ = 2x + \lfrac{\color{blue}{-3.5}(2x^2+5)+17.5-10x-6}{2x^2+5}$
β $ = 2x-3.5 + \lfrac{17.5-10x-6}{2x^2+5}$
β $ = 2x-3.5 + \lfrac{-10x+11.5}{2x^2+5}$.
Where did the blue terms come from? First you need to recognize that what I wrote is true. Next the reason for choosing those blue terms was to eventually get an equal expression where the fractional part is 'simpler'. You should try other terms in place of the blue terms and see why they do not work, and then you will understand why those blue terms work.
Also, there is absolutely nothing forcing you to 'simplify' the given polynomial fraction in that way. The following is also true:
β $\lfrac{4x^3-7x^2-6}{2x^2+5}$
β $= \lfrac{-6-7x^2+4x^3}{5+2x^2}$
β $= \lfrac{\color{blue}{-1.2}(5+2x^2)+2.4x^2-7x^2+4x^3}{5+2x^2}$
β $= -1.2 + \lfrac{2.4x^2-7x^2+4x^3}{5+2x^2}$
β $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$.
Why might you want to do this? Well, if you want to know the behaviour of $\lfrac{4x^3-7x^2-6}{2x^2+5}$ as $xβ0$, then you want to extract out the 'big' terms, and leave the 'residual' fraction with a 'small' numerator, so that it is clear what the asymptotic behaviour is. For instance as $xβ0$ we have:
β $\lfrac{4x^3-7x^2-6}{2x^2+5}$
β $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$
β $= -1.2 + 0.2Β·\lfrac{-4.6x^2+4x^3}{1+0.4x^2}$
β $β -1.2 + 0.2Β·(-4.6x^2+4x^3)(1-0.4x^2+O(x^4))$
β $β -1.2 + 0.2Β·(-4.6x^2+4x^3+O(x^4))$
β $= -1.2-0.92x^2+0.8x^3+O(x^4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
$b=a^p+1$ is a perfect square. Show that $p|(b-9)$ $p$ is a prime, and $a$ is a positive integer, $b=a^p+1$ is a perfect square. Show that $p|(b-9)$
It seem very interesting problem.if let $a^p+1=x^2$,it is clear $p\neq 2$
I have prove : $x$ is odd
proof:if $x$ is even number,then $(x+1,x-1)=1$,and note $a^p=(x+1)(x-1)$,then exist $r>s\ge 1\in N^{+}$ such $x+1=r^p,x-1=s^p$
so
$$2=(x+1)-(x-1)=r^p-s^p=(r-s)(r^{p-1}+r^{p-2}s+\cdots+s^{p-1})
\ge p\ge 3$$which is contradiction
which is a pretty interesting and nice result. I wonder in which ways we may approach it.
| From the comments we have the elementary proofs that $x^2 = a^p + 1$ has no solutions for prime $p > 3$, so it suffices to prove that when $p = 3$, $x$ is divisible by $3$.
We have $\gcd(a+1, a^2-a+1) \mid 3$, so we can assume that they are coprime and hope for a contradiction. If they are coprime, then both are squares, say
$$\begin{align*}a+1 &= n^2 \\a^2-a+1 &= m^2 \,. \end{align*}$$
But $a^2 - a + 1 = m^2$ is impossible when $a > 1$, because
$$(a-1)^2 = a^2 - 2a + 1 < a^2 - a - 1 < a^2 \,.$$
This is our contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$? If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?
I tried to solve it as follows:
I state that $p=ab$
$p^2=(2b+15)(2a+15)$
$p^2=4ab+30(a+b)+225$
$p^2=4p+30(a+b)+225$
and this is where I got stuck. I don't know how to get over this hurdle. could you please explain to me how to solve the question?
| This is not a straightforward way to answer the question, but illustrates a matter that the problem touches on.
We can treat the given equations as "solutions" to the problem: if $ \ y^2 \ = \ 2x + 15 \ $ describes a transformation such that $ \ T(a) \ = \ b \ $ and $ \ T(b) \ = \ a \ \ , $ and so $ \ T( \ T(a) \ ) \ = \ a \ $ and $ \ T( \ T(b) \ ) \ = \ b \ \ , $ with $ \ a \neq b \ \ , $ find the product $ \ ab \ \ . $
Of course the original equation $ \ y^2 \ = \ 2x + 15 \ $ is not a function of $ \ x \ \ . $ Further, using either of its associated square-root functions alone do not suffice: $ \ f(x) \ = \ \sqrt{2x + 15} \ $ has the fixed point $ \ x \ = \ 5 \ $ and $ \ g(x) \ = \ -\sqrt{2x + 15} \ \ , $ the fixed point $ \ x \ = \ -3 \ \ , $ as is easily computed and is seen in the graph below.
But the specification in the equation can be satisfied if we "link" the two functions so as to determine $ \ f(a) \ = \ b \ , \ g(b) \ = \ a \ $ and thus $ \ g( \ f(a) \ ) \ = \ a \ $ and $ \ f( \ g(b) \ ) \ = \ b \ \ . $ This produces a pair of transformations
$$ f( \ g(x) \ ) \ = \ \sqrt{ \ 2Β·(-\sqrt{2x + 15}) \ + \ 15} \ \ \ \text{and} \ \ \ g( \ f(x) \ ) \ = \ -\sqrt{ \ 2Β·(\sqrt{2x + 15}) \ + \ 15} \ \ . $$
So we now seek to solve $ \ g( \ f(a) \ ) \ = \ -\sqrt{ \ 2Β·(\sqrt{ \ 2a + 15 \ }) \ + \ 15} \ = \ a \ \ $ [solving $ \ f( \ g(b) \ ) \ = \ b \ \ $ will take us to the same place]. This looks like it will be spectacularly unhelpful; let us proceed:
$$ - \ \sqrt{ \ 2Β·(\sqrt{ \ 2a + 15 \ }) \ + \ 15} \ \ = \ \ a \ \ \Rightarrow \ \ 2Β· \sqrt{ \ 2a + 15 \ } \ \ = \ \ a^2 \ - \ 15 $$
$$ \Rightarrow \ \ 4Β·( 2a + 15 ) \ \ = \ \ (a^2 \ - \ 15)^2 \ \ \Rightarrow \ \ a^4 \ - \ 30a^2 \ - \ 8a \ + \ 165 \ \ = \ \ 0 \ \ . $$
Certainly we didn't want to have to deal with a quartic equation. But it turns out (not coincidentally) that the fixed points of $ \ f(x) \ $ and $ \ g(x) \ $ that we found earlier are roots of this equation. We thus find that the polynomial can be factored as
$$ a^4 \ - \ 30a^2 \ - \ 8a \ + \ 165 \ \ = \ \ (a + 3)Β·(a - 5)Β·(a^2 + 2a - 11) \ \ = \ \ 0 \ \ , $$
for which the zeroes of the quadratic factor are the sought-after $ \ a \ = \ -1 + 2\sqrt3 \ $ and $ \ b \ = \ -1 - 2\sqrt3 \ \ , $ the product of these being the constant term of that factor, $ \ \mathbf{ab \ = \ -11} \ \ . $ [These zeroes are also seen in the graph.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$
Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$
For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$
Am I correct, especially for the right direction ?
| The case where $p \leqslant 0$ is quite simple to establish, so I will look only at the case where $p > 0$. One way you can establish the convergence/divergence result is by using upper and lower bounds on the summation. For all values $n \geqslant 1$ and $p>0$ we can easily establish the bounds:
$$\int \limits_{n}^{n+1} \frac{1}{r^p} \ dr \leqslant \frac{1}{n^p} \leqslant \int \limits_{n-1}^n \frac{1}{r^p} \ dr.$$
Applying these bounds to each term in the sum (except for the first term for the upper bound) gives:
$$\int \limits_1^\infty \frac{1}{r^p} \ dr \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1 + \int \limits_1^\infty \frac{1}{r^p} \ dr.$$
Solving the definite integral we get:
$$R(p) \equiv \int \limits_1^\infty \frac{1}{r^p} \ dr
= \begin{cases}
\infty & & \text{if } 0 < p \leqslant 1, \\[8pt]
\frac{1}{p-1} & & \text{if } p > 1. \\[6pt]
\end{cases}$$
Substituting this form for the integral we therefore obtain:
$$R(p) \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1+R(p).$$
We can now look at the conditions on $p$. If we have $0 < p \leqslant 1$ then we get:
$$\infty \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \infty.$$
If we have $p>1$ then we get:
$$\frac{1}{p-1} \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \frac{p}{p-1}.$$
This establishes that the sum is divergent when $0 < p \leqslant 1$ and convergent when $p>1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$ I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.
Do you have any ideas? :)
EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to
$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$
By expanding with 1:
$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$
$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$
$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$
And using the substitution: $t=\tan\left(x\right)$
$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$
$$t^2=\tan^2\left(x\right)$$
$$t^2=\frac{\sin^2x}{\cos^2x}$$
$$t^2=\frac{1-\cos^2x}{\cos^2x}$$
$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$
$$t^2+1=\frac{1}{\cos^2x}$$
Using it:
$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$
I don't think I got to the expected result but I can't seem to be able to find whyβ¦
| Double-check my arithmetic, but here's the strategy.
@Bernard's suggested substitution $t=\tan x$ gives
\begin{align}
& \int\left(1-\frac{t^4}{(t^2\sqrt{2}+1)^2-2(\sqrt{2}-1)t^2}\right) dt \\
= {} & t-\int\frac{t^4}{(t^2\sqrt{2}+ct+1)(t^2\sqrt{2}-ct+1)} \, dt\end{align}
with $c:=\sqrt{2\sqrt{2}-1}$. You can do the rest with partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When is that last statement true?
| You can use induction to prove
$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$
is an integer when $n = 3^k$. The base case of $k = 0$ gives $3^k = 1$, with \eqref{eq1A} becoming $\frac{10-1}{9} = 1$. Next, assume \eqref{eq1A} is an integer for $k = m$ for some integer $m \ge 0$. Then, for $k = m + 1$,
$$\begin{equation}\begin{aligned}
\frac{10^{3^{m+1}} - 1}{9(3^{m+1})} & = \frac{\left(10^{3^{m}}\right)^3 - 1}{\left(9(3^{m})\right)3} \\
& = \frac{(10^{3^{m}} - 1)(10^{3^{2m}} + 10^{3^{m}} + 1)}{\left(9(3^{m})\right)3} \\
& = \left(\frac{10^{3^{m}} - 1}{9(3^{m})}\right)\left(\frac{10^{3^{2m}} + 10^{3^{m}} + 1}{3}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The first factor above is an integer based on the induction hypothesis, i.e., \eqref{eq1A} being an integer for $n = 3^{m}$. For the second factor, note
$$10 \equiv 1 \pmod{3} \implies 10^{3^{2m}} \equiv 10^{3^{m}} \equiv 1 \pmod{3} \tag{3}\label{eq3A}$$
This means
$$10^{3^{2m}} + 10^{3^{m}} + 1 \equiv 3 \equiv 0 \pmod{3} \implies 3 \mid 10^{3^{2m}} + 10^{3^{m}} + 1 \tag{4}\label{eq4A}$$
i.e., the second factor is also an integer. With the product of $2$ integers being an integer, \eqref{eq2A} shows \eqref{eq1A} is an integer for $n = 3^{m+1}$. This proves by induction that \eqref{eq1A} is an integer for $n = 3^k$ for all integers $k \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here? Am I going correct?
| Without AM-GM inequality:
Let, $x=a+b$ and $y=a-b$, then you get
$$6(a+b)^2+2(a+b)(a-b)+6(a-b)^2=9$$
$$14a^2+10b^2=9$$
$$x^2+y^2=(a+b)^2+(a-b)^2=2a^2+2b^2$$
$$x^2+y^2=2a^2+2 \times \frac{9-14a^2}{10}$$
$$x^2+y^2=\frac{9-4a^2}{5}β€\frac95$$
$$\color {gold}{\boxed {\color{black} {\text{max}[x^2+y^2]=\frac 95.}}}$$
If you want to find the minimum value of the expression, you can write in the same way:
$$x^2+y^2=2a^2+2b^2$$
$$x^2+y^2=2b^2+2 \times \frac{9-10b^2}{14}$$
$$x^2+y^2=\frac{4b^2+9}{7}β₯\frac 97$$
Then, you get
$$\color {gold}{\boxed {\color{black} {\text{min}[x^2+y^2]=\frac 97.}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$ How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$
What I tried
Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamma \left(n+\frac{3}{4}\right)=\Gamma\left(\frac{3}{4}\right)\prod_{k=0}^{n-1}\left(k+\frac{3}{4}\right).$$
Then
$$\lim_{n\to\infty}\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
Since
$$\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\prod_{k=1}^n \frac{4k(4k-2)}{(4k-1)^2}$$
for all $n\in\mathbb{N}$, it follows that
$$\prod_{k=1}^\infty \frac{4k(4k-2)}{(4k-1)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
But note that this actually gives an interesting Wallis-like product:
$$\frac{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdots}{3\cdot 3\cdot 7\cdot 7\cdot 11\cdot 11\cdots}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
I'm stuck at the Wallis-like product, though.
| $$a_n=2^{4 n}\frac{ \Gamma^4 \left(n+\frac{3}{4}\right)}{\Gamma^2 (2 n+1)
}$$
$$\log(a_n)=4 n \log (2)+4 \log \left(\Gamma \left(n+\frac{3}{4}\right)\right)-2 \log (\Gamma (2 n+1))$$
Applying Stirling approximation twice and continuing with Taylor series
$$\log(a_n)=\log (\pi )-\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=\pi \left(1-\frac{1}{8 n}+\frac{5}{128 n^2} \right)+O\left(\frac{1}{n^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Series $A= \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$ Consider convergence of series:
$A=\displaystyle \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$
I have a idea
\begin{align*}
a_n&=n^\frac{1}{n^2+1}-1
\\&=e^{\frac{\ln n}{n^2+1}}-1 \sim b_n= \dfrac{\ln\,n}{n^2+1} \text{ when } n \to \infty
\end{align*}
I wanna consider convergence of series $B=\displaystyle \sum_{n=1}^{\infty}b_n$. I have trouble here.
| OP -- Consider convergence of series:
$A=\displaystyle \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$
$$ n^\frac{1}{n^2+1}-1\ =
\ \frac{n-1}{\sum_{k=0}^{n^2}\,n^{\frac k{n^2+1}}\ }\ < $$
$$ \frac{n-1}{\left(n^2-\lfloor\frac{n^2+1}2\rfloor\right)
\cdot n^{\frac 12}}\ <
\ \frac{n^{\frac 12}}{n^2-\lfloor\frac{n^2+1}2\rfloor}\ \le $$
$$ \frac 2{n^\frac 32} $$
hence the series is summable (is convergent).
Great
| {
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"url": "https://math.stackexchange.com/questions/4086336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ I am trying to prove a statement about the decomposition field of a polynomial that has both $\sqrt{2-\sqrt{2}}$ and $\sqrt{2+\sqrt{2}}$ as roots. I cannot find a way to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. I have tried writing it in the basis $1,\sqrt{2+\sqrt{2}}(\sqrt{2+\sqrt{2}})^2,(\sqrt{2+\sqrt{2}})^3$ but nothing works and without this I cannot prove that the decomposition field of $t^4-4t^2+2$ is $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ over $\mathbb{Q}$
| Alternatively (without using conjugates),
Observe that,
*
*If $\sqrt{2-\sqrt{2}}=x$, then
$$(2-x^2)^2=2$$
*
*If $\sqrt{2+\sqrt{2}}=x$, then
$$(x^2-2)^2=2$$
Finally, we can observe that both polynomial equations are equal to each other:
$$(2-x^2)^2=2 \iff (x^2-2)^2=2$$
This implies, if $\sqrt{2-\sqrt{2}}$ is the root of your minimal polynomial over $\mathbb Q$, then $\sqrt{2+\sqrt{2}}$ also must be root of the polynomial.
So, your minimal polynomial equals to: ($x_{1,2}=\sqrt{2Β±\sqrt{2}}$)
$$(x^2-2)^2=2$$
which gives
$$x^4-4x^2+2=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all odd functions of the form $f(x) = \frac{ax + b}{x + c}$ I am working through a pure maths book and am stuck on odd and even functions.
Let $f(x) = \frac{ax + b}{x + c}$ where x, a, b, c are real and $x \ne \pm c$. Show that if $f$ is an even function then $ac = b$. Deduce that if $f$ is an even function then $f(x)$ must reduce to the form $f(x) = k$, where k is constant. Find all odd functions of the form $\frac{ax + b}{x + c}$
I have solved the first part. I am fairly sure of the second part. But I cannot solve the third part.
My calculations are as follows:
If $f(x)$ is even $f(x) = f(-x)$
$\implies \frac{ax + b}{x + c} = \frac{-ax + b}{-x + c}$
$\implies (ax + b)(c-x)= (x+c)(b-ax)$
$ac = b$
$\implies \frac{ax + b}{x + c} = \frac{ax + ac}{x + c} = \frac{a(x + c)}{x + c} = a$
So I am assuming that the a is the k to which the question refers.
Now, if $\frac{ax + b}{x + c}$ is odd, $f(-x) =-f(x)$. So
$\frac{-ax + b}{-x + c} = \frac{-(ax + b)}{x + c}$
$\implies (b-ax)(c+x)= (c-x)(-ax-b)$
$\implies 2bc=2ax^2 \implies bc=ax^2 \implies x = \sqrt\frac{bc}{a}$
but this does not lead to the answer in the book, which is $f(x)=\frac{k}{x}$
| A polynomial is odd iff all its exponents are odd and even iff all its exponents are even (including the exponent $0$).
A fraction is odd iff the numerator is odd an the denominator is even or vice versa.
The denominator in question can't be even; it's odd iff $c=0$. Now the numerator must be even, that is $a=0$. Hence all solutions are $b/x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $P=\sqrt{a^2-2ab+b^2}+\left(\frac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\frac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$ is rational Show that the number $$P=\sqrt{a^2-2ab+b^2}+\left(\dfrac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\dfrac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$$ is a rational number ($P\in\mathbb{Q})$ if $a\in\mathbb{Q},b\in\mathbb{Q},a>0,b>0$ and $a\ne b$. Find the value of $P$ if $a=1.1$ and $b=1.22$.
My try: $$P=\left|a-b\right|+\dfrac{a-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}:\dfrac{b\sqrt{a}+\sqrt{b}\left(a-\sqrt{ab}\right)}{a-\sqrt{ab}}=\\=|a-b|+\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a-\sqrt{ab}}{b\sqrt{a}+a\sqrt{b}-\sqrt{b^2a}}.$$
| Just keep chewing on it until it falls apart
$|a-b|+\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a-\sqrt{ab}}{b\sqrt{a}+a\sqrt{b}-\sqrt{b^2a}}=$
$|a-b| +\frac {\sqrt {ab}}{\sqrt a - \sqrt b}\cdot \frac {\sqrt a(\sqrt a -\sqrt b)}{b\sqrt a + a\sqrt b - b\sqrt a}=$
$|a-b| + \frac {\sqrt{ab}}{\sqrt a-\sqrt b}\cdot \frac {\sqrt a(\sqrt a-\sqrt b)}{a\sqrt b}=$
$|a-b| + \frac {\sqrt {ab}\sqrt a}{a\sqrt b}=$
$|a-b| + \frac {\sqrt{a^2b}}{\sqrt{a^2b}}=$
$|a-b| + 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find value of $x^3+y^3+z^3$ if $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$ If $x,y$ and $z$ are positive real numbers such that $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$, then what is the value of $x^3+y^3+z^3$?
I first thought of making them all equal because in that case the product is maximum, but obviously that was wrong and that is only valid for integers. By trial-and-error, I found that $(x,y,z)=(3,4,5)$ does satisfy the conditions, but I can't get a solid proof and method for this, other than just trial-and-error.
Thank you.
| Applying the AM-GM inequality for 3 times $\frac{x}{3}$, 4 times $\frac{y}{4}$ and 5 times $\frac{z}{5}$, we have
\begin{align}
12 &= x+y+z \\ & = 3\times \frac{x}{3} + 4 \times \frac{y}{4} +5 \times \frac{z}{5} \ge (3+4+5)\sqrt[12]{\left(\frac{x}{3}\right)^3\left(\frac{y}{4}\right)^4\left(\frac{z}{5}\right)^5} = 12
\end{align}
The equality occurs if and only if $\frac{x}{3} = \frac{y}{4}= \frac{z}{5}$ and $x+y+z= 12$ or
$$(x,y,z) = (3,4,5)$$
Q.E.D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing $\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)$, given $\sin z=z\prod_{n=1}^{ \infty}\left(1 - \frac{z^2}{n^2\pi^2}\right)$ How to show that
$$\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)$$
using the well-know representation of sine as infinite product that is
$$\sin z=z\prod_{n=1}^{ \infty}\left(1 - \frac{z^2}{n^2\pi^2}\right)$$
I have tried this problem for quite considerable amount of time. Indeed I also found a nice way to prove the former without using the latter. However I'm wondering if we can prove the result using the latter infinite product. Your help would be highly appreciated.
Thanks in advance.
| Since $\sinh z = \frac{1}{i}\sin(iz)$, we have
$$\begin{align*}
\sinh z &= \frac{1}{i}\sin(iz) \\
&= \frac{1}{i}\times (iz)\prod\limits_{n=1}^{\infty}\left(1 - \frac{(iz)^2}{n^2\pi^2}\right) \quad (\text{using the product formula for }\sin(\cdot))\\
&= z\prod\limits_{n=1}^{\infty}\left(1 -\frac{-z^2}{n^2\pi^2}\right)\\
&= z\prod\limits_{n=1}^{\infty}\left(1 +\frac{z^2}{n^2\pi^2}\right), \\
\end{align*}
$$
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Combinatorial Summation I'm trying to solve the following question:
If $s_n$ is the sum of first $n$ natural numbers, then prove that
$$2(s_1s_{2n}+s_2s_{2n-1}+\dots+s_ns_{n+1})=\frac{(2n+4)!}{5!(2n-1)!}$$
This is where I've come so far:
The general term of $(1-x)^{-2n}$ happens to be $t_{r+1}=\frac{(2n+r-1)!}{(2n-1)!r!}x^r$ and therefore the 6th term, i.e, $t_{5+1}=t_6=\frac{(2n+4)!}{(2n-1)!5!}x^5$, which looks like the RHS of the question I'm trying to solve.
Second, I found that $$(1-x)^{-3}=s_1+s_2x+\dots+s_nx^{n-1}+\dots$$
It's clear that the LHS must be something like, $(1-x)^{-a}(1-x)^{-b}$. But I'm not able to find one such combination which would give me $2(s_1s_{2n}+s_2s_{2n-1}+\dots+s_ns_{n+1})$.
Help is greatly appreciated.
| Here is a rather detailed algebraic derivation. We recall that multiplication of a generating function $A(x)$ with $\frac{1}{1-x}$ gives the sum of the coefficients of $A(x)$:
\begin{align*}
A(x)=\sum_{n=0}^\infty \color{blue}{a_n}x^n\qquad\qquad \frac{1}{1-x}A(x)=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^na_k}\right)x^n
\end{align*}
We so derive
\begin{align*}
\frac{1}{1-x}&=\sum_{n=0}^\infty x^n\\
\frac{1}{(1-x)^2}&=\sum_{n=0}^\infty \left(\sum_{k=0}^n 1\right)x^n=\sum_{n=0}^\infty(n+1)x^n\\
\frac{1}{(1-x)^3}&=\sum_{n=0}^\infty \sum_{k=0}^n\left(k+1\right)x^n \\
&=\sum_{n=0}^\infty s_{n+1}x^n=1+3x+6x^2+\cdots\tag{1}
\end{align*}
Squaring $\frac{1}{(1-x)^3}$ we get from (1) thanks to the Cauchy product
\begin{align*}
\frac{1}{(1-x)^6}=\sum_{m=0}^\infty \left(\sum_{k=0}^m s_{k+1}s_{m-k+1}\right)x^m\tag{2}
\end{align*}
Now we are ready to show the validity of
\begin{align*}
2\left(s_1s_{2n}+s_2s_{2n-1}+\ldots+s_ns_{n+1}\right)=\frac{(2n+4)!}{5!(2n-1)!}\qquad\qquad n\geq 1\tag{3}
\end{align*}
We obtain for $n\geq 1$:
\begin{align*}
\color{blue}{2\left(s_1s_{2n}+s_2s_{2n-1}+\cdots+s_ns_{n+1}\right)}
&=2\sum_{k=0}^{n-1}s_{k+1}s_{2n-k}\\
&=\sum_{k=0}^{n-1}s_{k+1}s_{2n-k}+\sum_{k=0}^{n-1}s_{n-k}s_{n+1+k}\tag{4}\\
&=\sum_{k=0}^{n-1}s_{k+1}s_{2n-k}+\sum_{k=n}^{2n-1}s_{2n-k}s_{k+1}\tag{5}\\
&=\sum_{k=0}^{2n-1}s_{k+1}s_{2n-k}\tag{6}\\
&=[x^{2n-1}]\frac{1}{(1-x)^6}\tag{7}\\
&=[x^{2n-1}]\sum_{j=0}^\infty\binom{-6}{j}(-x)^j\tag{8}\\
&=[x^{2n-1}]\sum_{j=0}^\infty\binom{j+5}{j}x^j\tag{9}\\
&=\binom{2n+4}{2n-1}=\binom{2n+4}{5}\tag{10}\\
&\,\,\color{blue}{=\frac{(2n+4)!}{5!(2n-1)!}}
\end{align*}
and the claim (3) follows.
Comment:
*
*In (4) we write the right-hand sum by changing the order of summation: $k\to n-1-k$.
*In (5) we shift the index of the right-hand sum by $n$.
*In (7) we use that (6) is the coefficient of $x^{2n-1}$ in (2).
*In (8) we make a binomial series expansion.
*In (9) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}$.
*In (10) we select the coefficient of $x^{2n-1}$ and use $\binom{p}{q}=\binom{p}{p-q}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
\infty} (a_n-3)6^n$.
First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the form $0 \cdot \infty$.
Moreover, we obtained a similar result here, which is related to the form $\lim\limits_{n \to \infty} (a_n-3)9^n$.
How should I proceed?
| Here's a proof that $b_n:=(3-a_n)6^n$ converges to a finite limit. This is merely a proof of existence of a finite limit (which I don't think is trivial).
It is shown here that $a_n$ is strictly increasing and has limit $3$.
Note that
$$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}$$
hence
$$(3-a_{n+1})6^{n+1} = \frac{6}{3+\sqrt{6+a_n}} (3-a_n)6^n$$
thus
$$\frac{b_{n+1}}{b_n} = \frac{6}{3+\sqrt{6+a_n}}>1$$
Let us show that $\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.
From $$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}\leq \frac 13 (3-a_n)$$
one readily finds that $3-a_n = O(\frac 1{3^n})$, hence $a_n = 3 + O(\frac 1{3^n})$, thus
$$\ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)
= \ln\Big(\frac{6}{3+\sqrt{9+O(\frac 1{3^n})}}\Big)
= \ln(1+O(\frac 1{3^n}))
= O(\frac 1{3^n})$$
Hence$\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.
Thus $\sum_n \log(\frac{b_{n+1}}{b_n})$ converges, hence the sequence $\log b_n$ converges to a finite limit, hence $b_n$ converges as well to some positive real.
Therefore, the sequence $(a_n-3)6^n$ converges to a negative real number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show it is impossible to find a $+ve$ integer such that sum of its sq. and its cube is an integral multiple of the sq. of the next highest integer. Based on the problem statement, we can form the equation as :-
$x^2+x^3 = k(x+1)^2$
where $x$ and $x+1$ are positive integers and $k$ will be the integral multiple.
One doubt that I have here is as per the statement " $k$ will be the integral multiple" , does this mean that $k$ will be a Whole number i.e. $k$ can be a positive integer, negative integer or $0$ ?
Now
$x^2+x^3 = k(x+1)^2$ $\Rightarrow x^2(x+1) = k(x+1)^2$
$\Rightarrow (x+1) [x^2-k(x+1)] = 0$
One solution will be $x =-1$ which is not a valid solution in this case.
Another solution will come from $x^2-k(x+1) =0$ which gives $x$ as :-
$x = \frac{k}{2} \pm \sqrt{\frac{k(k+4)}{4}}$
How can we proceed from here in order to prove the statement? One way I was adopting was to put different integral values of $k$ and checking if we get any $x$ as a positive integer but this does not seem to be a reliable way to prove this. I am stuck at this part. Please help !
Thanks in advance !
| An easier approach might be to say
$$k = \frac{x^2+x^3}{(x+1)^2} = \frac{x^2(1+x)}{(x+1)^2}= \frac{x^2}{x+1}= \frac{x^2-1}{x+1}+\frac{1}{x+1} =x-1+\frac{1}{x+1}$$
and, for positive integer $x$, that is strictly between $x-1$ and $x$, so not an integer
| {
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Evaluate:- $\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
Evaluate:- $\dfrac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
What I Tried:- Let $a = 4 , b = \sqrt{15} , c = 6, d= \sqrt{35}$ . Then I get :-
$$\rightarrow \frac{[a + b]^{3/2} + [a - b]^{3/2}}{[c + d]^{3/2} - [c - d]^{3/2}}$$
Now I can put the formulas $(a^3 + b^3)$ and $(c^3 - d^3)$
$$\rightarrow \frac{[(a + b)^{1/2} + (a - b)^{1/2}][(a + b) - \sqrt{a^2 - b^2} + (a - b)]}{[(c + d)^{1/2} - (c - d)^{1/2}][(c + d) + \sqrt{c^2 - d^2} + (c - d]}$$
$$\rightarrow \frac{7[(a + b)^{1/2} + (a - b)^{1/2}]}{13[(c + d)^{1/2} - (c - d)^{1/2}]}$$
From here, I do not know how to proceed. Can anyone help me?
| Simply square the expression. A lot of cancelation will happen and you'll end up in $490/1690$. Now take square root to get the final answer $7/13$.
| {
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For small $|x|<1$, $\alpha\in \mathbb{R}$, is it true that $e^{\alpha t}-(1+\alpha x+\alpha^2 x^2)^{\frac{t}{x}}=O(x)$? So expanding out the second term gives:
$$(1+\alpha x+\alpha^2 x^2)^{\frac{t}{x}}=e^{\frac{t}{x}\ln(1+\alpha x+\alpha^2 x^2)}$$
$$=e^{\frac{t}{x}((\alpha x+\alpha^2 x^2)-\frac{1}{2}(\alpha^2 x^2+2\alpha^3 x^3+\alpha^4 x^4)+\frac{1}{3}(\alpha^3 x^3+3\alpha^4 x^4+3\alpha^5 x^5+\alpha^6 x^6)+...)}$$
$$=e^{t((\alpha +\alpha^2 x)-\frac{1}{2}(\alpha^2 x+2\alpha^3 x^2+\alpha^4 x^3)+\frac{1}{3}(\alpha^3 x^2+3\alpha^4 x^3+3\alpha^5 x^4+\alpha^6 x^5)+...)}$$
$$=e^{t(\alpha +\frac{1}{2}\alpha^2 x -\frac{2}{3}\alpha^3x^2+\frac{1}{4}\alpha^4 x^3+...)}$$
So $$e^{\alpha t}-e^{t(\alpha +\frac{1}{2}\alpha^2 x -\frac{2}{3}\alpha^3x^2+\frac{1}{4}\alpha^4 x^3+...)}$$
$$=e^{\alpha t}-e^{\alpha t}e^{t(\frac{1}{2}\alpha^2 x -\frac{2}{3}\alpha^3x^2+\frac{1}{4}\alpha^4 x^3+...)}$$
$$=e^{\alpha t}-e^{\alpha t}(1+t(\frac{1}{2}\alpha^2 x -\frac{2}{3}\alpha^3x^2+\frac{1}{4}\alpha^4 x^3+...)+...$$
$$=-e^{\alpha t}t(\frac{1}{2}\alpha^2 x -\frac{2}{3}\alpha^3x^2+\frac{1}{4}\alpha^4 x^3+...)+...$$
And so this appears to be $O(x)$? I feel like I've gone wrong somewhere?
| Your work is perfectly correct but, if I may suggest, it could be done faster.
Considering $$\Delta=e^{\alpha t}-(1+\alpha x+\alpha^2 x^2)^{\frac{t}{x}}$$ let $\alpha t= T$ and $\alpha x= X$ to make
$$\Delta=e^{T}-(1+X+X^2)^{\frac{T}{X}}$$
$$y=(1+X+X^2)^{\frac{T}{X}}\implies \log(y)=\frac T X \log(1+X+X^2)$$ By Taylor
$$\log(1+X+X^2)=X+\frac{X^2}{2}+O\left(X^3\right)$$
$$\log(y)=T+\frac{T X}{2}+O\left(X^2\right)$$
$$y=e^{\log(y)}=e^T \left(1+\frac{1}{2}TX+O\left(X^2\right)\right)$$
$$\Delta=\frac{1}{2}TXe^T+O\left(X^2\right)=O\left(X\right)$$
| {
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Prove $f(x,y) =\sqrt{16-x^2-y^2}$ is continuous using $\epsilon$, $\delta$? I want to prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that
$$(x-x_0)^2 + (y-y_0)^2 < \delta^2 \Rightarrow \left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| < \epsilon.$$
I note that $|x-x_0| \leq \delta$ and $|y-y_0| \leq \delta$.
What I've tried:
$$\left| \sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| = \frac{|x^2-x_0^2 + y^2-y_0^2|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} }$$
$$= \frac{|(x-x_0)(x+x_0) + (y-y_0)(y+y_0)|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} } \leq
\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x^2-y^2}+\sqrt{16-x_0^2-y_0^2} }$$
$$\leq
\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x_0^2-y_0^2} }$$
But then I do not know how to continue.
| First note that if $|x-x_0|+|y-y_0|<1$ then each of $|x+x_0|,\ |y+y_0|$ is bounded by some $M>0\ $ (Proof: triangle inequality). So assume $\delta<1.$ There are two cases:
$1).\ $ If $x_0^2+y_0^2\neq16,\ $ set $N=\sqrt{16-x_0^2-y_0^2}.$ Then
$\displaystyle\left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right|\le \frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{\sqrt{16-x_0^2-y_0^2}}=\frac{|x-x_0| \cdot |x+x_0| + |y-y_0|\cdot |y+y_0|}{N}<\frac{M}{N}\cdot (|x-x_0|+|y-y_0|)$
and we may take $\delta=\min\{1,\epsilon N/M\}.$
$2).\ $ If $x_0^2+y_0^2=16$ then
$\displaystyle|f(x,y)-f(x_0,y_0)|^2=16-x^2-y^2=x_0^2-x^2+y_0^2-y^2<M(|x-x_0|+|y-y_0|)$ so we may take $\delta=\min\{1,\epsilon^2/M\}.$
| {
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Induction on $S(n,k)$ Consider $$S(n,k):= 1^{k}+\ldots+n^{k}.$$
I have to find a formula for $$S(n,3).$$
After trying for some n values I found out that
$$S(1,3)=1^{3}=1^{2}$$
$$S(2,3)=1^{2} +2^{2}=3^{2}$$
$$S(3,3)=1^{2} +2^{2} +3^{3}=6^{2}$$
$$S(4,3)=1^{2} +2^{2} +3^{3}+4^{3}=10^{2}$$
$$S(5,3)=1^{2} +2^{2} +3^{3}+4^{3}+5^{3}=15^{2}$$
Which led to the conjecture that $$S(n,3)=\left(1+\ldots+n\right)^{2}$$
I know for a fact that $$S(n,1)=\frac{n^{2}}{2}+\frac{n}{2}$$
Therefore,
$$S(n,3)=\left(\frac{n^{2}}{2}+\frac{n}{2}\right)^2$$
But when trying to prove the formula using induction I get stuck on the inductive step.
$$S(k+1,3)=1^{3}+\ldots+k^{3}+(k+1)^3=\left(\frac{(k+1)^{2}}{2}+\frac{k+1}{2}\right)^2$$
I don't know how to proceed. I have tried manipulating the RHS, but it's been fruitless.
| On inductive step we can use
$$1^3+2^3+\cdots +k^3+(k+1)^3 = (1+2+\cdots +k)^2 +(k+1)^3 =(1+2+\cdots +k+(k+1))^2$$
Last equality is same with
$$\left( \frac{k(k+1)}{2} \right)^2 +(k+1)^3 = \left( \frac{(2+k)(k+1)}{2} \right)^2$$
and can be easy checked.
| {
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"timestamp": "2023-03-29T00:00:00",
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Right solution to an integration problem One more from me.
Please help me understand the following. How do you get from here:
$\frac{1}{8}\int{1dx}-\frac{1}{8}\int{\cos(2x)dx}-\frac{1}{8}\int{\cos^2(2x)dx}+\frac{1}{8}\int{\cos^3(2x)dx}$
to here:
$=\frac{1}{16}x-\frac{1}{64}\sin(4x)-\frac{1}{48}\sin^3(2x)+C$
Is the above even correct? I found this in a textbook but found another solution online and I am not sure which one to believe anymore. The one i found online is this:
$=\frac{1}{16}x-\frac{1}{64}\sin(2x)-\frac{1}{64}\sin(4x)+\frac{1}{192}\sin(6x)+C$
The starting point was for both the following:
$\int{\sin^4(x)\cos^2(x)dx}$
Thank you.
| We know that $$\cos^2(x) = \frac {1 + \cos(2x)}{2}$$ and $$\sin^2(x) = \frac {1 - \cos(2x)}{2}$$. So let us convert the integrand into something we can integrate:
$$\sin^4(x)\cos^2(x) = (\sin^2(x))^2\cos^2(x)$$ $$=\frac {1-\cos(2x)}{4}^2 \times \frac {1+\cos(2x)}{2}$$ $$=\frac {1 - 2\cos(2x) + \cos^2(2x)}{4} \times \frac {1 + \cos(2x)}{2}$$ $$=\frac {1 - 2\cos(2x) + \cos^2(2x) + \cos(2x) - 2\cos^2(2x) + cos^3(2x)}{8}$$ $$=\frac {1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)}{8}$$ This was all the side work.
Now we can integrate it :
$$\int \sin^4(x)\cos^2(x) = \int \frac {1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)}{8} dx$$
$$=\frac{1}{8} \left[x - \frac {1}{2}\sin(2x) - \left(\frac {x}{2} + \frac{\sin(2x)}{4} + C_1\right) + \left(\frac{1}{2}\sin(2x) - \frac{1}{6}\sin^3(2x) + C_2\right)\right]$$
$$=\frac{1}{8} \left[\frac{x}{2} - \frac{\sin(2x)}{4} - \frac{1}{6}\sin^3(2x)\right] + C\tag{C = C_1 + C_2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{1β
2}+\frac{1}{3β
4}+\frac{1}{5β
6}+....+\frac{1}{199β
200}$ = $\frac{1}{101}+\frac{1}{102}+\frac{1}{103}...+\frac{1}{200}$
Prove that $$\frac{1}{1β
2}+\frac{1}{3β
4}+\frac{1}{5β
6}+....+\frac{1}{199β
200}= \frac{1}{101}+\frac{1}{102}+\frac{1}{103}...+\frac{1}{200}$$
My Approach:
$T_{r}=\frac{1}{\left(2r\right)β
\left(2r-1\right)}$
$\sum_{n=1}^{100}(T_{r}=\frac{1}{\left(2r\right)β
\left(2r-1\right)}=\frac{\left(2r\right)-\left(2r-1\right)}{\left(2r\right)β
\left(2r-1\right)}=\frac{1}{\left(2r-1\right)}-\frac{1}{2r})$
This gives:
$S=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.......+\frac{1}{197}-\frac{1}{198}+\frac{1}{199}-\frac{1}{200}$
which apparently leads to a dead end. Please help me out.
| $S=1-\frac{1}{2}+\frac{1}{3}...-\frac{1}{200}=\sum_{n=1}^{200}\frac{1}{n}-2\sum_{n=1}^{100}\frac{1}{2n}=\sum_{n=101}^{200}\frac{1}{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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