Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $(a,b)=1$, then $ \gcd(a^2+b^2,a^3+b^3)\mid (a-b)$
If $(a,b)=1$, then $\gcd(a^2+b^2,a^3+b^3)\mid (a-b)$
The only way that can help is to find some common factor of $a^2+b^2$ and $a^3+b^3$.
That does not seem obvious enough, so will directly try to divide $a^3+b^3$ by $a^2+b^2$.
This leads to nowhere too.
It seems that need to use the fact that $a,b$ are relatively prime, but am unable to use that.
Let for some suitable integers $x, y$, have $ax +by =1$.
| Note that
$$
\left(a^3+b^3\right)-b\left(a^2+b^2\right)=(a-b)a^2
$$
and
$$
a\left(a^2+b^2\right)-\left(a^3+b^3\right)=(a-b)b^2
$$
and if
$$
ax+by=1
$$
then
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2\left(ax^3+3x^2by\right)+b^2\left(3axy^2+by^3\right)
\end{align}
$$
Thus,
$$
\begin{align}
a-b
&=(a-b)\left(\color{#C00}{a^2}\left(ax^3+3x^2by\right)+\color{#090}{b^2}\left(3axy^2+by^3\right)\right)\\
&=\color{#C00}{\left(\left(a^3+b^3\right)-b\left(a^2+b^2\right)\right)}\left(ax^3+3x^2by\right)\\
&+\color{#090}{\left(a\left(a^2+b^2\right)-\left(a^3+b^3\right)\right)}\left(3axy^2+by^3\right)\\
&=\boldsymbol{\left(a^3+b^3\right)}\left(ax^3+3x^2by-3axy^2-by^3\right)\\
&+\boldsymbol{\left(a^2+b^2\right)}\left(3a^2xy^2+aby^3-abx^3-3x^2b^2y\right)
\end{align}
$$
which means that
$$
\left.\left(a^2+b^2,a^3+b^3\right)\,\middle|\,a-b\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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I would like to find the center of this group
Show that the center of $SL(2, \mathbb{C})$ is $\pm Id.$ Hint: Use $
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}
$.
The center of a group is $C(G) = \{ g \in G : g\ast h = h \ast g \quad \forall h \in G \}.$ And it is true that $
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
-1 & 0 \\
0 & -1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}
$, have determinant $1$, hence, they are in the group, and particularily, the suggested matrices commute with both $\pm Id.$ But this doesn't prove that only two matrices, $Id, -Id$, are in the center. And It doesn't show that, $Id, -Id$ commutes with all the matrices in the group. So why would I pick those matrices in the first place? What am I missing?
| Let $g = \pmatrix{a & b \\ c & d} \in C(G)$. One then has $$\pmatrix{a & b \\ c & d}\pmatrix{1 & 1 \\ 0 & 1} = \pmatrix{1 & 1 \\ 0 & 1}\pmatrix{a & b \\ c & d}$$ $$\pmatrix{a & a+b \\ c & c+d} = \pmatrix{a + c & b+d \\ c & d}$$
One then has $c=0$ and $a=d$.
With matrix $\pmatrix{1 & 0 \\ 1 & 1}$, you will get $b=0$.
So now you have $g = \pmatrix{a & 0 \\ 0 & a}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$A^n = I$ only if $n\equiv 0 \pmod{4}$ Is the following proof correct? In addition could you pleases suggest a more cleaner or shorter proof.
Theorem. Given any $n\in\mathbb{Z^+}$ then
$$\begin{pmatrix}
0&0&-1\\
0&1&0\\
1&0&0\\
\end{pmatrix}^n
=
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
$$
only if $4 \mid n$.
Proof. Assume that $n\in\mathbb{Z^+}$ and let $A$ denote the matrix
$$
\begin{pmatrix}
0&0&-1\\
0&1&0\\
1&0&0\\
\end{pmatrix}
$$
We prove the contrapositive. Given the standard choice of basis for $\mathbb{R}^3$ we can see $A$ represents a $90^{\circ}$ clockwise rotation about the $y$-axis consequently we may write out $A^2$, $A^3$ and $A^4$ as follows:
$$
A^2 =
\begin{pmatrix}
-1&0&0\\
0&1&0\\
0&0&-1\\
\end{pmatrix}
$$
$$
A^3 =
\begin{pmatrix}
0&0&1\\
0&1&0\\
-1&0&0\\
\end{pmatrix}
$$
$$
A^4 =
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
$$
Now assume that $4\not\mid n$ then the division theorem implies that $n$ is of the form $4q+1$, $4q+2$ or $4q+3$. Considering the case where $4q+1$ we can see that
$$A^{4q+1} = (A^{4})^q\cdot A = I^q\cdot A = I\cdot A\neq I.$$
By similar reasoning we can make the same conclusion in the other two cases.
$\blacksquare$
| You are correct and your argument handles all cases with minor change.
If natural $n>1$ and $k \in \{0,1,2,3\}$
$$A^{4n+k} = (A^4)^n \cdot A^k = I \cdot A^k = A^k$$
and so $A^{4n+k}=I$ only when $k=0$. That is $A^n=I$ only when $n$ is a multiple of $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
5 heads or 5 tails in a row on 10 toss : why this answer? An acquaintance of mine came up with this question:
What is the probability of having 5 heads or 5 tails in a row, when tossing a fair coin 10 times.
Using this answer, I came up with this solution
$\dfrac{10 \choose 5}{2^{10}} = \dfrac{252}{1024} = \dfrac{63}{256}$.
Yet when thinking of it, I cannot explain in plain words my solution.
Can anyone help?
Update:
My initial answer was flawed
It should have been this instead
$\dfrac{112}{1024}$
| $\binom{10}{5}=\frac{10!}{5!5!}$ counts all the anagrams of the word $HHHHHTTTTT$ which has $10$ letters, $5$ $H$s and $5$ $T$s. $(1/2)^{10}$ is the probability that one of these anagrams appears after $10$ tosses.
However in your question it is asked to count the arrangements where the 5 heads or 5 tails are in a row such as $TTHHHHHTTT$ or $HHHHTTTTTH$, but also $HTHHHHHHTH$ (which is not an anagram of $HHHHHTTTTT$). So your answer is not correct.
In order to have at least $5$ heads in a row, we consider the following cases:
1) $HHHHHTXXXX$, $THHHHHTXXX$, $XTHHHHHTXX$, $XXTHHHHHTX$, $XXXTHHHHHT$, $XXXXTHHHHH$ $\implies 2^{4}+4\cdot 2^3+2^4=64$;
2) $HHHHHHTXXX$, $THHHHHHTXX$, $XTHHHHHHTX$, $XXTHHHHHHT$, $XXXTHHHHHH$ $\implies 2^{3}+3\cdot 2^2+2^3=28$;
3) $HHHHHHHTXX$, $THHHHHHHTX$, $XTHHHHHHHT$, $XXTHHHHHHH$ $\implies 2^{2}+2\cdot 2^1+2^2=12$;
4) $HHHHHHHHTX$, $THHHHHHHHT$, $XTHHHHHHHH$ $\implies 2^{1}+2^0+2^1=5$;
5) $HHHHHHHHHT$, $THHHHHHHHH$ $\implies 2^0+2^0=2$;
6) $HHHHHHHHHH$ $\implies 2^0=1$;
and the total number is $64+28+12+5+2+1=112$.
Similarly the total number of ways to have at least 5 tails in a row is $112$. For a row of $5$ heads OR $5$ tails the number of ways is
$112+112-2=222$ (we subtract $2$ because otherwise $TTTTTHHHHH$ and $HHHHHTTTTT$ are counted twice).
Finally the probability of having $5$ heads or $5$ tails in a row, when tossing a fair coin $10$ times is
$$\frac{222}{2^{10}}=\frac{111}{512}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
| As for real $\alpha, x^2\le1$
WLOG $x^2=\cos2y\implies0\le2y\le\dfrac\pi2\implies\cos y,\sin y\ge0$
Using $\cos2y=1-2\sin^2y=2\cos^2y-1,$
$$\tan\alpha=\dfrac{\cos y-\sin y}{\cos y+\sin y}=\tan\left(\dfrac\pi4-y\right)$$
$\implies\alpha=m\pi+\dfrac\pi4-y$ where $m$ is any integer
$x^2=\cos2y=\cos\left(2m\pi+\dfrac\pi2-2\alpha\right)=?$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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About splitting fields and primitive elements. Question 1: Find the splitting field $K$ of $x^4 + x^2 + 1$ over $\mathbb{Q}$. What is the degree of $K/\mathbb{Q}$? ($\textbf{Hint:}$ Recall that $\xi = e^{2\pi i/3}$ is such that $\xi^2 + \xi + 1 = 0$ and $cos(\pi/3) = 1/2$ and $sin(\pi/3) = \sqrt{3}/2)$
Question 2: Find a primitive element $\xi$ for $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Prove your claim.
My proof:
Let $x^2 = \xi$, then $\xi^2 + \xi + 1= 0$. Now the roots of this polynomial can be written as
$$
\psi_1 = e^{2\pi i/3},\psi_2 = e^{5\pi i/3}
$$
Hence,
$$
x^2 = e^{2\pi i/3}, \mbox { or } x^2 = e^{5\pi i/3}
$$
and
$$
x = \pm e^{\pi i/3}, x = \pm e^{5\pi i/6}
$$
Therefore, we can write the solutions of the equation as
$$\begin{align*}
x_1 =& \frac{1}{2} + i\frac{\sqrt{3}}{2}\\
x_2 =& - \frac{1}{2} - i\frac{\sqrt{3}}{2} \\
x_3 =& -\frac{\sqrt{3}}{2} + i\frac{1}{2} \\
x_4 =& \frac{\sqrt{3}}{2} - i\frac{1}{2}
\end{align*}$$
Now the desired splitting field is $K = \mathbb{Q}(\sqrt{3},i)$. The degree of this extension $\mathbb{Q}(\sqrt{3},i)/\mathbb{Q}$ can be calculated using this well-known result
$$
[\mathbb{Q}(\sqrt{3},i):\mathbb{Q}] = [\mathbb{Q}(\sqrt{3},i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}] = 2\times2 = 4.
$$
Alternatively, we can see that
$$
\mathbb{Q}(\sqrt{3},i) = \{a1 + b\sqrt{3} + ci + d\sqrt{3}i: a,b,c,d \in \mathbb{Q} \}
$$
Hence, $\{1,\sqrt{3},{i},\sqrt{3}i\}$ is a basis for $\mathbb{Q}(\sqrt{3},i)$ and the degree of this extension is the number of elements in it's basis which is four. $\hspace{5pt}\square$
I just want to know if my solution for question 1 is correct.
About question 2: I know that $\alpha = \sqrt{2} + \sqrt{3}$ must be a primitive element but I did not know how to prove it. Maybe the primitive element theorem can help since this extension is finite? Or I have to find some minimal polynomial to prove that $Q(\alpha) = Q(\sqrt{2},\sqrt{3})$? I was looking in Wikipedia and I found this:
"If the minimal polynomial of $\alpha$ in $Q[x]$ is $a(x) = x^4 − 10x^2 + 1 = (x - \sqrt{3} - \sqrt{2})(x + \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} + \sqrt{3})$.
The minimal polynomial in Q[x] of the sum of the square roots of the first n prime numbers is constructed analogously, and is called a Swinnerton-Dyer polynomial."
This is very interesting for me but in my class they never discuss this theorem.
Any other way to prove this?
| We want to prove that $\mathbb{Q}(\sqrt{2} + \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Obviously, we have $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$, so it's enough to prove the opposite inclusion.
Consider $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \ni (\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2 \sqrt{6}$. Therefore, $\sqrt{6} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$.
Now, $\sqrt{6}(\sqrt{2}+\sqrt{3}) = \sqrt{12} + \sqrt{18} = 2\sqrt{3} + 3 \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Therefore, $(2\sqrt{3} + 3 \sqrt{2}) - 2 \cdot (\sqrt{2} + \sqrt{3}) = \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$, and so also obviously $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show $\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}$ I am trying to show $\displaystyle{\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}}.$
Any help?
(I am having troubles using the half circle infinite contour)
Or more specifically, what is the residue $\text{res} \left(\frac{1}{z^3+1},z_0=e^\frac{\pi i}{3} \right )$
Thanks!
| $$\int_{0}^{+\infty}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{dx}{1+x^3}+\int_{1}^{+\infty}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{1+x}{1+x^3}\,dx$$
clearly equals $\int_{0}^{1}\frac{dx}{1-x+x^2}$, which is an elementary integral. We may also state
$$\int_{0}^{1}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{1+x-x^3-x^4}{1-x^6}\,dx = \sum_{n\geq 0}\left[\frac{1}{6n+1}+\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+5}\right]$$
and by denoting as $\chi_3$ and $\chi_6$ the non-principal Dirichlet's characters $\!\!\pmod{3}$ and $\!\!\pmod{6}$ the RHS can be written as $L(\chi_6,1)+\frac{1}{2}\,L(\chi_3,1)$, which is related to $\sqrt{3}$ and $\pi$ via the class number formula.
Another approach is given by the reflection formula for the $\psi$ (digamma) function, since
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(x)-\psi(1-x)=-\pi\cot(\pi x)$$
ensure
$$ \sum_{n\geq 0}\left[\frac{1}{6n+1}-\frac{1}{6n+5}\right]=\tfrac{\pi}{6}\cot\left(\tfrac{\pi}{6}\right),\qquad \sum_{n\geq 0}\left[\frac{1}{3n+1}-\frac{1}{3n+2}\right]=\tfrac{\pi}{3}\cot\left(\tfrac{\pi}{3}\right). $$
| {
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"url": "https://math.stackexchange.com/questions/2645842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Continued fraction of binomial function $(1+z)^{1/4}$ If I ask WolframAlpha for the continued fraction of $(1+z)^{1/4}$, I get
and I wonder what would be the representation for any $k = n$? Could you give me any hints or a reference?
| This is a particular case of Gauss' continued fraction for the ratio of hypergeometric functions:
$$\frac{{_2 F_1}(A,B+1;C+1;Z)}{{_2 F_1}(A,B;C;Z)}=\cfrac{1}{1-\cfrac{\frac{A(C-B)}{C(C+1)}Z}{1-\cfrac{\frac{(B+1)(C-A+1)}{(C+1)(C+2)}Z}{1-\cfrac{\frac{(A+1)(C-B+1)}{(C+2)(C+3)}Z}{1-\cfrac{\frac{(B+2)(C-A+2)}{(C+3)(C+4)}Z}{1- \dots}}}}}$$
The choice of functions gets a little tricky, so I will just write without proof (see also the article for the hypergeometric function):
$$(1+z)^a=1+az \frac{{_2 F_1}(1,1+a;2;-z)}{{_2 F_1}(1,a;1;-z)}$$
Substituting the parameters, we obtain:
$$(1+z)^a=1+\cfrac{az}{1+\cfrac{\frac{1 \cdot (1-a)}{1 \cdot 2}z}{1+\cfrac{\frac{1 \cdot (1+a)}{2 \cdot 3}z}{1+\cfrac{\frac{2 \cdot (2-a)}{3 \cdot 4}z}{1+\cfrac{\frac{2 \cdot (2+a)}{4 \cdot 5}z}{1+ \dots}}}}}$$
I hope it's clear how to continue.
Now substitute $a=\frac{1}{4}$ to get your continued fraction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$ As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$
I don't know how to calculate the limit. I am pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$
| Hint:
$$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{2}{n^2+1}+...+\frac{n}{n^2+1}=\frac{1}{2}$$ and
$$\lim_{n\to\infty} \frac{1}{n^2+n}+\frac{2}{n^2+n}+...+\frac{n}{n^2+n}=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Series convergence or divergence . $\sum_{n=1}^\infty \left(\frac{n^4}{n^5+7}\right)$ $$\sum_{n=1}^\infty \left(\frac{n^4}{n^5+7}\right)$$
I try this series by the comparison test with $a_n\le b_n,$ $a_n=\frac{n^4}{n^5+7}$ and $b_n=\frac{n^4}{n^5}=\frac{1}{n}$
then $b_n$ diverges, dose the series diverges ?
| A bit more formally:
Let $n \ge 2:$
$a_n=\dfrac{n^4}{n^5+7} \gt \dfrac{n^4}{n^5+n^5} =\dfrac{1}{2n}.$
Since $(1/2) \sum \dfrac{1}{n}$ diverges,
$\sum \dfrac{n^4}{n^5+7}$ diverges (comparison test).
| {
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Prove the inequality $\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}$
Let $a,b,c$ are nonnegative real numbers such that $a^2+b^2+c^2=1$.
Prove the inequality
$$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}$$
I tried the method of Lagrange multipliers and Jensen's inequality but I have not been proved this inequality
| Consider the function $f(x)= \frac{\sqrt x}{1+x}$,
with the substitution that $x=a^2,y=b^2,z=c^2$, $\implies x+y+z=1$
we have a new expression $\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2} \longrightarrow \frac{\sqrt x}{1+x}+ \frac{\sqrt y}{1+y} +\frac{\sqrt z}{1+z } \longrightarrow f(x)+f(y)+f(z)$
Differentiating $f(z)$ twice, we get negative second derivatives for $0 \le x\le 1$, so it is concave
by Jensen's inequality, we get $ f(\frac{x+y+z}{3}) \ge \frac{f(x)}{3} +\frac{f(y)}{3}+\frac{f(z)}{3} \implies \frac{\sqrt \frac{1}{3}}{1+\frac{1}{3}}\ge \frac{f(x)}{3} +\frac{f(y)}{3}+\frac{f(z)}{3}$
$\implies \frac{3\sqrt3}{4} \ge f(x)+f(y)+f(z)= \frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve a system of equation
Solve this system of equations.
$$\begin{cases}
x^{2} + y^{2} = 4 \\
z^{2} + t^{2} = 9 \\
xt + yz = 6
\end{cases}$$
My attempts:
*
*$(x^{2}+y^{2})(z^{2} + t^{2})=36 \xrightarrow{\text{by applying Lagrange formula}} (xt + zy)^{2} + (zx -ty)^{2}=36 \xrightarrow{xt + yz = 6} 36 + (zx -ty)^{2} =36\Rightarrow (zx -ty)^{2} = 0 \Rightarrow zx = ty \Rightarrow z = \frac{ty}{x}$
*$xt + yz = 6 \xrightarrow{z = \frac{ty}{x}} xt + \frac{ty^{2}}{x} = 6 \Rightarrow x^{2}t + y^{2}t = 6x \Rightarrow t(x^{2} + y^{2})=6x \xrightarrow{x^{2} + y^{2} = 4} t=\frac{3}{2}x$
*$z= \frac{ty}{x} \xrightarrow{t=\frac{3}{2}x} z = \frac{3}{2}y$
Can we go further and determine the exact value of variables?
| We have $4$ unknowns with three equations. So, we should not have deterministic solutions
$zx=ty\implies \dfrac xy=\dfrac tz=k$(say)
$$4=y^2(1+k^2)$$ $$9=z^2(k^2+1)$$
$$\text{and }6=yz(k^2+1)$$
$$yz\ne0,\dfrac96=\dfrac{z^2}{yz}=\dfrac zy$$
$\implies \dfrac z3=\dfrac y2=u$(say)
$$\implies\{x,y,z,t\}=\{k(2u),2u, 3u,k(3u)\}$$ where $u(\ne0),k$ are arbitrary numbers such that $$4=4u^2(1+k^2)$$
$$u^2(k^2+1)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the coefficient of $x^{18}$ in $(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $
Find the coefficient of $x^{18}$ in $$(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $$
This is the first time coming across a generating function question and am not quite sure how to solve this and am looking for some help, thanks!
| Edit: So it looks like this could have been done much easier by factoring first. I'll leave this answer here in the event that this may be useful to someone with a related problem
Well we can get $x^{18}$ in our equation whenever we multiply terms together who's exponents will add to $18$. The left part of the equation will add in $1,2,3,4,5$ so we need to find the coefficients on the left side who's terms are $x^{17}, x^{16}, x^{15}, x^{14}, x^{13}$
Now to find each coefficient, we need to find all the sets of $5$ numbers in $[2,k]$ that sum to our exponents listed prior. I'll do $13$, for example.
$$
13 = 2+2+2+2+5
$$
$$
13 = 2+2+2+3+4
$$
$$
13 = 2+2+3+3+3
$$
Those are (hopefully I missed none), all the sums that sum to our exponent. now since these 13's can be got by multiplying any order of these terms, we need to see how many combinations these have by using the formula $\frac{n!}{k_1!k_2!...k_n!}$ so,
$$
2+2+2+2+5 \rightarrow \dfrac{5!}{4!1!} = 5
$$
$$
2+2+2+3+4 \rightarrow \dfrac{5!}{3!1!1!} = 20
$$
$$
2+2+3+3+3 \rightarrow \dfrac{5!}{2!3!} = 10
$$
So summing this up we get that $5+20+10=35$, therefore the coefficient in front of our term becomes $35x^{13}$. When we finally multiply that out by the right side to get $x^5\cdot35x^{13} = 35x^{18}$.
Now in order to finish the problem, you must apply this process to $x^{14}, x^{15}, x^{16}, x^{17}$, and sum their respective coefficients. Good luck!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A problem about $\epsilon-\delta$ definition of a limit where the target is to find $\delta$ The fuction is:
$f(x)=2-\frac{1}{x}$
It requires to find $\delta$ such that $0<\mid x-1 \mid<\delta$ else, $\mid f(x)-1 \mid<0.1$
My process so far, has been trying to compare the stuff that I have, because I found the $\delta$ like that before, but I don't know how to continue with this especific problem.
What I have:
$\mid 2-\frac{1}{x}-1\mid<0.1$
$\mid 1-\frac{1}{x}\mid<0.1$
$\mid\frac{x-1}{x}\mid<0.1$
$\frac{\mid x-1\mid}{\mid x\mid}<0.1$
$\mid x-1 \mid<0.1\mid x \mid$
From here, I don't know how to proceed. The answer in the book says that $\delta=\frac{1}{11}$
| First, choose $\delta<1$ so that we have
\begin{eqnarray}
|x-1|&<&\delta\\
-\delta<x-1&<&\delta\\
1-\delta&<&x<1+\delta\\
0&<&x
\end{eqnarray}
Then we have
\begin{eqnarray}
\left\vert f(x)-1\right\vert&<&\frac{1}{10}\quad\text{ iff}\\
\left\vert 2-\frac{1}{x}-1\right\vert&<&\frac{1}{10}\quad\text{ iff}\\
\left\vert 1-\frac{1}{x}\right\vert&<&\frac{1}{10}\quad\text{ iff}\\
-\frac{1}{10}<1-\frac{1}{x}&<&\frac{1}{10}\quad\text{ iff}\\
-\frac{1}{10}-1<-\frac{1}{x}&<&\frac{1}{10}-1\quad\text{ iff}\\
-\frac{11}{10}<-\frac{1}{x}&<&-\frac{9}{10}\quad\text{ iff}\\
\frac{11}{10}>\frac{1}{x}&>&\frac{9}{10}\quad\text{ iff}\\
\frac{9}{10}<\frac{1}{x}&<&\frac{11}{10}
\end{eqnarray}
which is equivalent to the previous inequality.
So the reciprocals are related in the reverse order
$$ \frac{10}{11}<x<\frac{10}{9} $$
Subtract $1$ to get
$$ -\frac{1}{11}<x-1<\frac{1}{9}$$
So we should choose $|x-1|<\dfrac{1}{11}$
This gives us
$$ -\frac{1}{11}<x-1<\frac{1}{11} $$
And since $\dfrac{1}{11}<\dfrac{1}{9}$ we have
$$-\frac{1}{11}<\frac{1}{x}<\frac{1}{9}$$
which is equivalent to
$$ \left\vert f(x)-1\right\vert<\frac{1}{10} $$
provided it is true that
$$|x-1|<\dfrac{1}{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^2+xy+y^2 \equiv 0\pmod{n^2} $ When I was dealing with the equation
$x^2+xy+y^2 \equiv 0 \pmod{n^2}$
At $n$ = 3, I found that x, y must be $\equiv 0 \pmod n$. Then I wonder and did $n =$ 3 to 10k, using a mini computer program. Interestingly, I found that only when n is in this OEIS sequence, the above statement is true.
Can someone give some hints or tips on how to proof this?
In other words, how should I prove (if this is even true) that:
For all n without prime factor congruent to 1 mod 6, the equation $x^2+xy+y^2 \equiv 0 \pmod{n^2}$ has only solutions $(x, y) = (0, 0) \pmod n$?
Thanks a lot.
| Claim:
If $n$ is a positive integer such that, for some integers $x,y$, not both congruent to zero, mod $n$, we have
$$x^2 + xy + y^2\equiv 0\;(\text{mod}\;n^2)$$
then $n$ has a prime factor congruent to $1$, mod $6$.
Proof:
For $n=1$, the claim holds vacuously, since the hypothesis fails (for any integers $x,y$, it's automatic that $x,y$ are both congruent to zero, mod $1$).
Thus, assume $n > 1$.
Let $n$ be the least positive integer for which a counterexample exists.
Thus, $n$ is the least positive integer such that
*
*There are integers $x,y$, not both congruent to zero, mod $n$, for which $x^2 + xy + y^2\equiv 0\;(\text{mod}\;n^2)$.$\\[4pt]$
*$n$ has no prime factor congruent to $1$, mod $6$.
Our goal is to derive a contradiction.
Let $d=\gcd(x,y,n)$.
Since $x,y$ are not both zero, $d$ is a positive integer.
Then we can write
*
$x=dx_1$
$y=dy_1$
$n=dn_1$
where $x_1,y_1,n_1$ are integers.
Since $x,y$ are not both congruent to zero, mod $n$, we have $d < n$, hence $n_1 > 1$.
If $d>1$, then $n_1 < n$, but then
$$x^2 + xy + y^2\equiv 0\;(\text{mod}\;n^2)$$
can be reduced to
$$x_1^2 + x_1y_1 + y_1^2\equiv 0\;(\text{mod}\;n_1^2)$$
where
*
*$x_1,y_1$ are integers, not both congruent to zero, mod $n_1$.$\\[4pt]$
*$n_1$ has no prime factor congruent to $1$, mod $6$.
so we have a counterexample with $n_1 < n$, contradiction.
Hence, we must have $d=1$, so $\gcd(x,y,n)=1$.
If $n$ is even, then
\begin{align*}
&x^2 + xy + y^2\equiv 0\;(\text{mod}\;n^2)\\[4pt]
\implies\;&x^2 + xy + y^2\equiv 0\;(\text{mod}\;2)\\[4pt]
\end{align*}
but then $x,y$ must both be even, contrary to $\gcd(x,y,n)=1$.
Next, let $e=\gcd(y,n)$.
Suppose $e > 1$
Since $\gcd(x,y,n)=1$, it follows that $\gcd(x,e)=1$.
\begin{align*}
\text{Then}\;\;&x^2 + xy + y^2\equiv 0\;(\text{mod}\;n^2)\\[4pt]
\implies\;&x^2 + xy + y^2\equiv 0\;(\text{mod}\;e)\\[4pt]
\implies\;&x^2\equiv 0\;(\text{mod}\;e)\\[4pt]
\end{align*}
contrary to $\gcd(x,e)=1$.
Hence, we must have $e=1$, so $\gcd(y,n)=1$.
Analogously, we get $\gcd(x,n)=1$.
Let $p$ be a prime factor of $n$.
From $\gcd(y,n)=1$, we get $\gcd(y,p)=1$.
\begin{align*}
\text{Then}\;\;&x^2 + xy + y^2\equiv 0\;(\text{mod}\;n^2)\\[4pt]
\implies\;&x^2 + xy + y^2\equiv 0\;(\text{mod}\;p)\\[4pt]
\implies\;&(x-y)(x^2 + xy + y^2)\equiv 0\;(\text{mod}\;p)\\[4pt]
\implies\;&x^3-y^3\equiv 0\;(\text{mod}\;p)\\[4pt]
\implies\;&x^3\equiv y^3\;(\text{mod}\;p)\\[4pt]
\implies\;&\bigl({\small{\frac{x}{y}}}\bigr)^3\equiv 1\;(\text{mod}\;p)\\[4pt]
\end{align*}
Thus, the order of ${\large{\frac{x}{y}}}$, mod $p$, is either $1$ or $3$.
Consider two cases . . .
Case $(1)$:$\;$The order of ${\large{\frac{x}{y}}}$, mod $p$, is $3$.
But the order of ${\large{\frac{x}{y}}}$, mod $p$, must divide $(p-1)$, hence $3{\,\mid\,}(p-1)$, so $p\equiv 1\;(\text{mod} 3)$.
Since $n$ is odd, so is $p$, hence from $p\equiv 1\;(\text{mod}\;3)$, we get $p\equiv 1\;(\text{mod}\;6)$, contradiction.
Case $(2)$:$\;$The order of ${\large{\frac{x}{y}}}$, mod $p$, is $1$.
Then from ${\large{\frac{x}{y}}}\equiv 1\;(\text{mod}\;p)$, we get $x\equiv y\;(\text{mod}\;p)$, hence
\begin{align*}
&x^2 + xy + y^2\equiv 0\;(\text{mod}\;p)\\[4pt]
\implies\;&y^2 + (y)y + y^2\equiv 0\;(\text{mod}\;p)\\[4pt]
\implies\;&3y^2\equiv 0\;(\text{mod}\;p)\\[4pt]
\implies\;&p{\,\mid\,}3y^2\\[4pt]
\implies\;&p{\,\mid\,}3\qquad\text{[since $\gcd(y,p)=1$]}\\[4pt]
\implies\;&p=3\\[4pt]
\implies\;&3{\,\mid\,}n\\[4pt]
\implies\;&9{\,\mid\,}n^2\\[4pt]
\implies\;&x^2 + xy + y^2\equiv 0\;(\text{mod}\;9)\\[4pt]
\end{align*}
Since $p=3$, then from
*
*$\gcd(x,p)=1$, and $\gcd(x,p)=1$.$\\[4pt]$
*$x\equiv y\;(\text{mod}\;p$.
we get $x\equiv 1\;(\text{mod}\;3)$, and $x\equiv 1\;(\text{mod}\;3)$.
Note that we can't have $x\equiv y\;(\text{mod}\;9)$, else
\begin{align*}
&x^2 + xy + y^2\equiv 0\;(\text{mod}\;9)\\[4pt]
\implies\;&y^2 + (y)y + y^2\equiv 0\;(\text{mod}\;9)\\[4pt]
\implies\;&3y^2\equiv 0\;(\text{mod}\;9)\\[4pt]
\implies\;&y^2\equiv 0\;(\text{mod}\;3)\\[4pt]
\end{align*}
contradiction, since $p=3$, and $\gcd(y,p)=1$.
Testing the congruence $x^2 + xy + y^2\equiv 0\;(\text{mod}\;9)$, subject to the constraints
*
*$x\equiv 1\;(\text{mod}\;3$, and $x\equiv 1\;(\text{mod}\;3$.$\\[4pt]$
*$x\not\equiv y\;(\text{mod}\;9)$.$\\[4pt]$
*$1 \le x,y \le 8$.
we find that there are no solutions, so case $(2)$ yields a contradiction.
Thus, both cases yield a contradiction, which completes the proof of the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder when $3^{29}$ is divided by $12$.
Find the remainder when $3^{29}$ is divided by $12$.
a) $2$ ; b) $3$ ; c) $7$ ; d) $9$ ; e) $12$
Since $\dfrac{3^{29}}{12} = \dfrac{3^{28}}{4} = \dfrac{9^{14}}{4}$, and $9 \equiv 1 \mod 4$, I thought I could do $9^{14} \equiv 1^{14} \mod 4$, so that the answer is $1$. However, that is not one of the answer choices... Where did I go wrong?
| $$3^2\equiv1\pmod4,3^{2n}=(3^2)^n\equiv1^n$$
$$3^{2n+1}\equiv3\pmod{4\cdot3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $A+B+C=0$, then prove that the value of the determinant is $0$. I'll state the question from my textbook below:
If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.
This is how I tried solving the problem:
$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$
$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$
$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$
I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used.
Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?
Any help would be appreciated.
| Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2A+\cos^2B+\cos^2C =1+\cos^2A-\sin^2B+\cos^2C$$
$$=1+\cos(A+B)\cos(A-B)+\cos^2C$$
$$=1-\cos C\cos(A-B)+\cos^2C\text{ as }\cos(A+B)=\cos(\pi-C)=?$$
$$=1-\cos C[\cos(A-B)-\cos C]$$
$\cos(A-B)-\cos C=\cos(A-B)+\cos(A+B)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Split $\prod_{i=0}^n (z+i)^{-1}$ into $\sum_{i=0}^n a_i (z+i)^{-1}$ I'm trying to find a general formula to split
$$\prod_{i=0}^n (z+i)^{-1} = \sum_{i=0}^n a_i (z+i)^{-1}$$
If $n=1$, we have
$$
\frac{1}{z(z+1)} = \frac{1}{z}+\frac{-1}{z+1}
$$
If $n=2$,
$$
\frac{1}{z(z+1)(z+2)} = \frac{1/2}{z}+\frac{-1}{z+1}+\frac{1/2}{z+2}
$$
If $n=3$,
$$
\frac{1}{z(z+1)(z+2)(z+3)} = \frac{1/6}{z}+\frac{-1/2}{z+1}+\frac{1/2}{z+2} + \frac{-1/6}{z+3}
$$
Motivation. I was calculating residues at $z=-m$ of $f(z)=\sum_{n=0}^\infty\prod_{i=0}^n (z+i)^{-1} $
| Hint: Induction using
\begin{align}
& \frac{1}{(z+a)(z+a+1)\cdots(z+a+n)} \\
= & \frac{1/n}{(z+a)\cdots(z+a+n-1)}-\frac{1/n}{(z+a+1)\cdots(z+a+n)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that for a given point on an ellipse, the sum of the distances from each focal point is constant
An ellipse has equation $\frac{x^2}{25}+\frac{y^2}{9} = 1$ and $P(p,q)$ is a point on the ellipse. Points $F_1$ and $F_2$ have coordinates $(-4,0)$ and $(4,0)$. Show that the sum of the distances $|PF_1|$ +$|PF_2|$ does not depend on the value of p.
First, to find distance $|F_1P|$
$|F_1P| = \sqrt{(p+4)^2 + q^2}$
as $\frac{p^2}{25}+\frac{q^2}{9} = 1$
$\frac{9p^2}{25}+q^2 = 9$
$q^2 = 9-\frac{9p^2}{25}$
$|F_1P| = \sqrt{p^2+8p+16 + 9-\frac{9p^2}{25}}$
$= \sqrt{\frac{16}{25}p^2+8p-25}$
$= \sqrt{\frac{1}{25}(16p^2+200p-625)}$
$= \frac{1}{5}\sqrt{(4p+25)^2}$
$= \frac{1}{5}(4p+25)$
Finding the distance $|F_2P|$
$|F_2P| = \sqrt{(4-p)^2 + q^2}$
$= \sqrt{p^2-8p+16 + 9-\frac{9p^2}{25}}$
$= \sqrt{\frac{16}{25}p^2-8p-25}$
$= \sqrt{\frac{1}{25}(16p^2-200p-625)}$
$= \frac{1}{5}\sqrt{(4p-25)^2}$
$= \frac{1}{5}(4p-25)$
Therefore
$|F_1P|+|F_2P|= \frac{1}{5}(4p-25)+\frac{1}{5}(4p+25) = \frac{8}{5}p$
This is exactly what we're trying to disprove. I must have made a mistake somewhere. Please can someone explain where I went wrong. My answer appears to be the negative of what it should be. I expect the correct answer will be 10 units but for some reason it does not seem to work.
| Note that $y=0$ gives $x=\pm5$, so $-5 \le p \le 5$ for all points $P(p,q)$ on the ellipse. Therefore:
$$
|F_1P| = \frac{1}{5}\sqrt{(4p+25)^2} = \frac{1}{5}|4p+25| = \frac{1}{5}(25+4p) \\
|F_2P| = \frac{1}{5}\sqrt{(4p-25)^2} = \frac{1}{5}|4p-25| = \color{red}{\frac{1}{5}(25-4p)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Prove that $1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$ I am in search of a correct and shortcut techniques to prove this.
Otherwise I have calculated each remainder according to the power of $20$ to prove this :-
$1 \equiv 1 \bmod 23$
$\Rightarrow 20 \equiv -3\bmod 23$
$\Rightarrow 20^2 \equiv (-3)^2 \equiv 9\bmod 23$
$20^{2n}\equiv (-3)^{2n}\bmod 23$
$20^{2n-1}\equiv (-3)^{2n-1}\bmod 23$.
Therefore
$1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$
For $\sum_{n=1}^{11} 20^{2n-1} \equiv r \bmod 23$
we have $$-3-4+10-2+5-1-9+11+7-6-8 \equiv 0 \bmod 23$$
For $\sum_{n=0}^{10} 20^{2n} \equiv s \bmod 23$
we have $$1+9+12+16+6+8+3+4+13+2-5 \equiv 0\bmod 23$$
Therefore $$\sum_{n=1}^{11} 20^{2n-1}+\sum_{n=0}^{10} 20^{2n} \equiv r+s \bmod 23$$
$$\Rightarrow 1+20+20^2+20^3+....+20^{21} \equiv 0+0 \equiv 0\bmod 23$$
If possible just show any short-cut correct way to prove this. Any help is appreciated.
| $\sum_{i=0}^{21}{20}^i=\frac{1-{20}^{22}}{1-20}$... But ${20}^{22}\cong 1\pmod{23}$ by Fermat's little theorem...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$ Let $x,y,z$ rationals Show that if $(x+y+z)^3=9(x^2y+y^2z+z^2x)$ then $x=y=z$
I tried this :
Let $x$ be the smallest variable
Write $y=a+x$ and $z=b+x$ Prove $a=b=0$
by factoring the equation as the sum of three squares.
any suggestions?
| CONCLUSION:The roots of $\eta^3 - 3 \eta - 1 = 0$ are
$$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$
Note that we do have
$$ A+B+C=0 \; . \; $$
We get identity
$$ \color{magenta}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} $$
which confirms that the surface is three planes sharing the line $x=y=z\; . \;$
ORIGINAL:
Take $$ x = r-s-t \; , \; \; \; y = r+s-t \; , \; \; \; z = r + 2 t \; , $$ so
$$ 3r = x+y+z \; \; , \; \; 2s = y - x \; , \; \; \; 6t = 2 z - x - y \; \; . $$
Note $(x,y,z)$ is a rational triple if and only if $(r,s,t)$ is a rational triple.
Then
$$ x^2 y + y^2 z + z^2 x = 3 r^3 + \left( s^3 + 3 s^2 t - 9 s t^2 - 3 t^3 \right) $$
and
$$ 9 \left( x^2 y + y^2 z + z^2 x\right) - (3r)^3 = 9 \left( s^3 + 3 s^2 t - 9 s t^2 - 3 t^3 \right) $$
If, for example, $t \neq 0,$ divide by $t^3$ and we must have a root of $p^3 + 3 p^2 - 9p-3$ which is irreducible. If $s \neq 0$ use the reciprocal. Insisting on rational values, we find that both $s,t$ are zero, so
$$ y-x = 0 \; , \; \; \; 2z - x - y = 0 \; , \; $$
and $$ x=y=z $$
Meanwhile, the method answers a simple question, what kind of surface are we describing in $\mathbb R^3 \; ?$ If we have an irrational root $p$ of $p^3 + 3 p^2 - 9p-3=0$ we have some other irrational real $q$ such that
$$ 2z-x-y = q(y-x) \; , $$
$$ (q-1)x + (-q-1) y + 2 z = 0 \; \; , $$
which is evidently a plane containing the line $x=y=z.$ I mostly think the surface is three planes, arranged around the line $x=y=z$ at equal angles, like radii of a circle. Indeed, compared with axes given by vectors $v_1 = (-1,1,0)/ \sqrt(2)$ and $v_2 = (-1,-1,2)/ \sqrt(6),$ it seems the three planes are rotated from $v_1$ in the direction of $v_2$ at exactly the three angles $40^\circ, 100^\circ, 160^\circ,$ repetitions at $220^\circ, 280^\circ, 340^\circ,$ so we see every $60^\circ \; .$ As $360/9 = 40$ this has a bit of plausibility.
Next day: confirming the nature of the surface: first, it is defined by the "curve" gotten by intersecting the surface with the plane $x+y+z = 0,$ as it is a "cylinder" over that curve, with axis of translation the expected line $x=y=z.$ If $$ x = X + t \; , \; \; y = Y + t \; , \; \; z = Z + t \; \; , $$
we find
$$ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) = (X+Y+Z)^3 - 9 \left( X^2 Y + Y^2 Z + Z^2 X \right) $$
Wednesday, finally got it. also Tottenham just scored on Juventus in the Champions League. The roots of $\eta^3 - 3 \eta - 1 = 0$ are
$$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$
We get identity
$$ \color{red}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} $$
which confirms that the surface is three planes sharing the line $x=y=z,$ as $A+B+C = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Compute the derivative
Compute the derivatives of: $$ y = {\sqrt{x + \sqrt{x}}} $$
My attempt:
$$=\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+\sqrt{x}\right)$$
$$\frac{d}{du}\left(\sqrt{u}\right)$$
$$=\frac{d}{du}\left(u^{\frac{1}{2}}\right)$$
$$=\frac{1}{2}u^{\frac{1}{2}-1}$$
$$ \frac{1}{2}u^{-\frac{1}{2}} $$
$$ =u^{-\frac{1}{2}} $$
$$ =\frac{1}{2}u^{-\frac{1}{2}}$$
$$=\frac{1}{2}\cdot \frac{1}{\sqrt{u}}$$
$$=\frac{1\cdot \:1}{2\sqrt{u}}$$
$$=\frac{1}{2\sqrt{x}}$$
$$=1+\frac{1}{2\sqrt{x}}$$
$$=\frac{1}{2\sqrt{u}}\left(1+\frac{1}{2\sqrt{x}}\right)$$
$$=\frac{1}{2\sqrt{x+\sqrt{x}}}\left(1+\frac{1}{2\sqrt{x}}\right)$$
$$=\frac{1\cdot \left(1+\frac{1}{2\sqrt{x}}\right)}{2\sqrt{x+\sqrt{x}}}$$
$$=\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}$$
$$1+\frac{1}{2\sqrt{x}}$$
$$=\frac{1}{1}+\frac{1}{2\sqrt{x}}$$
$$\frac{1}{1}+\frac{1}{2\sqrt{x}}$$
$$=\frac{1\cdot \:2\sqrt{x}}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}$$
$$=\frac{1\cdot \:2\sqrt{x}+1}{2\sqrt{x}}$$
$$=2\sqrt{x}+1$$
$$=\frac{2\sqrt{x}+1}{2\sqrt{x}}$$
$$=\frac{\frac{2\sqrt{x}+1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}$$
$$=\frac{2\sqrt{x}+1}{2\cdot \:2\sqrt{x}\sqrt{x+\sqrt{x}}}$$
$$=\frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}$$
$$\frac{d}{dx}\left(y\right)=\frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}$$
I did not get full marks on this question, and I would like to know where I have gone wrong.
(Please note that I cannot get feedback from my professor as not every student has submitted their assignment). Thank you.
| You need to use the chain rule, and properly identify the components.
You have $f(x)=g\Big(h(x)\Big)$, where $g(x)=\sqrt{x}$ and $h(x)=x+\sqrt{x}$.
Then $f'(x)=g'\Big(h(x)\Big)h'(x)$.
The end result is thus $f'(x)=\frac{1 + \frac{1}{2 \sqrt{x}}}{2 \sqrt{\sqrt{x} + x}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Are there solutions to the following simultaneous equations? I have been looking into solutions to simultaneous equations of the form
$$\sum_{r=1}^{n}a_{r} = \sum_{r=1}^{n}b_{r}$$
$$\sum_{r=1}^{n}a_{r}^{2} = \sum_{r=1}^{n}b_{r}^{2}$$
where $\forall i$, $a_{i}$ and $b_{i}$ are positive integers and no $a_{i}=b_{j}$, $\forall i, j$. This has infinitely many solutions for example
$$n^{2}+(n+k+4)^{2}+(n+2k+5)^{2} \equiv (n+1)^{2}+(n+k+2)^{2}+(n+2k+6)^{2}$$
$$n+(n+k+4)+(n+2k+5) \equiv (n+1)+(n+k+2)+(n+2k+6)$$
I was wondering if there are solutions to the simultaneous equations
$$\sum_{r=1}^{n}a_{r} = \sum_{r=1}^{n}b_{r}$$
$$\sum_{r=1}^{n}a_{r}^{2} = \sum_{r=1}^{n}b_{r}^{2}$$
$$\sum_{r=1}^{n}a_{r}^{3} = \sum_{r=1}^{n}b_{r}^{3}$$
or possibly for even higher powers.
Edit: Could you give example solutions?
| Yes, you can do it for arbitrarily high powers. We can move the $b$s to the left and just ask for solutions to $\sum (\pm 1)a_r^k=0$. We can note that $(n+3)-(n+2)-(n+1)+n=0$ gives a solution to the first equation, then that $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$, so $-(n+7)^2+(n+6)^2+(n+5)^2-(n+4)^2+(n+3)^2-(n+2)^2-(n+1)^2+n^2=0$ gives a solution to the first two. Extending to $16$ terms, changing the signs the same way gives $$(n+15)^3-(n+14)^3-(n+13)^3+(n+12)^3-(n+11)^3+(n+10)^3+(n+9)^2-(n+8)^3-(n+7)^3+(n+6)^3+(n+5)^3-(n+4)^3+(n+3)^3-(n+2)^3-(n+1)^3+n^3=0$$ will allow us to satisfy the first three and so on.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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2016 ARML Individual Problem #8. Tricky Substitutions to find coefficients of a polynomial Let $P(x)$ be the polynomial $x^3+Ax^2+Bx+C$ for some constants $A$, $B$, and $C$. There exists constants $D$ and $E$ such that for all $x$, $P(x+1)=x^3+Dx^2+54x+37$ and $P(x+2)=x^3+26x^2+Ex+115$. Compute the ordered triple $(A,B,C)$.
Try to substitute values of $x$ into $P(x+1)$ and $P(x+2)$ to make your life easier.
| In $P(x+2)$ we notice that we have that nasty $Ex$. We want to get rid of that term with a substitution for $x$. The way we do this is to substitute $0$. This gives us that $P(2)
= 0 + 0 + 0 + 115 = 115$. By substituting $1$ into $P(x+1)$, you get $P(2)=92 + D$. We can solve the equation $92+D=115$. This gives us $D=23$. Now, we can substitute $(x+1)$ for all $x$ in $P(x+1)$. This gives us another polynomial (which I don't want to write out cuz LaTeX). This gives us the final result of $x^3+20x^3+11x+5$ for $P(x)$. Our ordered pair is $(A,B,C) = (20,11,5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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transformation changing basis Let $T:R^3->R^3 $ be defined by $T(a_1,a_2,a_3)=(3a_1+a_2,a_1+a_3,a_1-a_3)
$
so here first i do transformation to the bases $T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}3\\1\\1\end{pmatrix}$, $T \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$ , $T \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix}0\\1\\-1\end{pmatrix}$
then express it as linear combination of standard basis and find coordinate respect to it such as $T \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}=a_1\begin{pmatrix}1\\0\\0\end{pmatrix}+a_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}+a_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and do it for three of them then got the coefficient for $a_1,a_2,a_2$
this matrix with respect to standard basis is $A=\begin{bmatrix}3&1&0\\1&0&1\\1&0&-1\end{bmatrix}$
is this the right idea behind with respect to standard basis?
suppose we choose as a basis $V=R^3$ set ${\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$
the matrix representation of T with respect to this basis is? so if I'm doing the same thing as above such as transform, $T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}3\\1\\1\end{pmatrix}$, $T \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}$, $T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}=\begin{pmatrix}4\\2\\0\end{pmatrix}$ and it turns out this is the matrix
A=\begin{bmatrix}3&4&4\\1&1&2\\1&1&0\end{bmatrix}
but why can t i represent the transformation as linear combination of basis b?
$\begin{pmatrix}3\\1\\1\end{pmatrix}=a_1{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+a_2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}+a_3 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$ for matrix column? i got different answer
but i know also the detour way $V^{-1}AV$
| For $A$ in the standard basis you are correct.
The matrix $\bar A$ in the new basis is given by $V^{-1}AV$ where $V$ is the matrix which has for columns the vectors of the new basis. Can you see why and how it works?
Simply note that matrix $V$ changes the coordinates from the new basis to the standard and $V^{-1}$ changes the coordinates from standard to the new basis.
Then we have
*
*$w=Av$
*$v=V\bar v \qquad w=V\bar w$
and then
$$w=V\bar w=Av=AV\bar v\iff \bar w=V^{-1}AV\bar v$$
that is
$$\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}\begin{bmatrix}3&1&0\\1&0&1\\1&0&-1\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}2&3&2\\0&0&2\\1&1&0\end{bmatrix}$$
To obtain the result with a different method note that if we indicate with
$$u={\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, v=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, w=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$$
the vectors of the new basis, we need first to find the expression of the vectors of the standard basis in term of the new basis vectors that is
*
*$e_1=u$
*$e_2=-u+v$
*$e_3=-v+w$
Then we need to find the expression for $T(e_1)$, $T(e_2)$, $T(e_3)$ with respect to the new basis that is
*
*$a_1u+a_2v+a_3w=T(e_1)=(3,1,1)\implies (a_1,a_2,a_3)=(2,0,1)$
*$b_1u+b_2v+b_3w=T(e_2)=(1,0,0)\implies (b_1,b_2,b_3)=(1,0,0)$
*$c_1u+c_2v+c_3w=T(e_3)=(0,1,-1)\implies (c_1,c_2,c_3)=(-1,2,-1)$
Then for the transformation in the new basis we know that
*
*$T(e_1)=T(u)=2u+w$
*$T(e_2)=T(-u+v)=-T(u)+T(v)=u$
*$T(e_3)=T(-v+w)=-T(v)+T(w)=-u+2v-w$
from wich we obtain
*
*$T(u)=2u+w$
*$T(v)=3u+w$
*$T(w)=2u+2v$
which leads to the same result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I find the dimension of an eigenspace? I have the following square matrix
$$ A = \begin{bmatrix} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 &-1 \end{bmatrix} $$
I found the eigenvalues:
*
*$2$ with algebraic and geometric multiplicity $1$ and eigenvector $(1,2,7/3)$.
*$-1$ with algebraic multiplicity $2$ and geometric multiplicity $1$; one eigenvector is $(0,0,1)$.
Thus, matrix $A$ is not diagonizable. My questions are:
*
*How can I find the Jordan normal form?
*How I can find the dimension of the eigenspace of eigenvalue $-1$?
*In Sagemath, how can I find the dimension of the eigenspace of eigenvalue $-1$?
| Most Jordan Normal Form questions, in integers, intended to be done by hand, can be settled with the minimal polynomial. The characteristic polynomial is $\lambda^3 - 3 \lambda - 2 = (\lambda -2)(\lambda + 1)^2.$ the minimal polynomial is the same, which you can confirm by checking that $A^2 - A - 2 I \neq 0.$ Each linear factor of the characteristic polynomial must appear in the minimal polynomial, which exponent at least one, so the quadratic shown is the only possible alternative as minimal.
Next,
$$
A+I =
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
6 & 0 & 0 \\
1 & 3 & 0
\end{array}
\right)
$$
with genuine eigenvector $t(0,0,1)^T$ with convenient multiplier $t$ if desired.
$$
(A+I)^2 =
\left(
\begin{array}{rrr}
9 & 0 & 0 \\
18 & 0 & 0 \\
21 & 0 & 0
\end{array}
\right)
$$
The description I like is that we now take $w$ with $(A+I)w \neq 0$ and $(A+I)^2 w = 0.$ I choose
$$
w =
\left(
\begin{array}{r}
0 \\
1 \\
0
\end{array}
\right)
$$
This $w$ will be the right hand column of $P$ in $P^{-1}A P = J.$ The middle column is $$ v = (A+I)w, $$
so that $v \neq 0$ but $(A+I)v = (A+I)^2 w = 0$ and $v$ is a genuine eigenvector. You already had the $2$ eigenvector, I take a multiple to give integers. i like integers.
$$
P =
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
6 & 0 & 1 \\
7 & 3 & 0
\end{array}
\right)
$$
with
$$
P^{-1} =
\frac{1}{9}
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
-7 & 0 & 3 \\
-18 & 9 & 0
\end{array}
\right)
$$
leading to
$$
\frac{1}{9}
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
-7 & 0 & 3 \\
-18 & 9 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
6 & -1 & 0 \\
1 & 3 & -1
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
6 & 0 & 1 \\
7 & 3 & 0
\end{array}
\right) =
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & -1 & 1 \\
0 & 0 & -1
\end{array}
\right)
$$
It is the reverse direction $PJP^{-1} = A$ that allows us to evaluate functions of $A$ such as $e^{At},$
$$
\frac{1}{9}
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
6 & 0 & 1 \\
7 & 3 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
0 & -1 & 1 \\
0 & 0 & -1
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
-7 & 0 & 3 \\
-18 & 9 & 0
\end{array}
\right) =
\left(
\begin{array}{rrr}
2 & 0 & 0 \\
6 & -1 & 0 \\
1 & 3 & -1
\end{array}
\right)
$$
| {
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How to prove that $e=3-\frac{1}{2!1\cdot 2}-\cdots-\frac{1}{n!(n-1)n}-\cdots$? How to prove that $e=3-\frac{1}{2!1\cdot 2}-\cdots-\frac{1}{n!(n-1)n}-\cdots$? I’ve proved that
$1+1+\frac{1}{2!}+\cdots+\frac{1}{n!}+\frac{1}{n!n}= 3-\frac{1}{2!1\cdot 2}-\cdots-\frac{1}{n!(n-1)n}$.
| Use $\frac{1}{n!(n-1)n}=\frac{1}{n!(n-1)}-\frac{1}{n!n}$
Thus,
$$\sum_{n=2}^{\infty}\frac{1}{n!(n-1)n}=\sum \frac{1}{n!(n-1)}-\frac{1}{n!n}$$
$$=(\frac{1}{2!\cdot 1}-\frac{1}{2!\cdot 2})+(\frac{1}{3!\cdot 2}-\frac{1}{3!\cdot 3})+...$$
$$=\frac{1}{2!\cdot 1}+(-\frac{1}{2!\cdot 2}+\frac{1}{3!\cdot 2})+(-\frac{1}{3!\cdot 3}+\frac{1}{4!\cdot 3})+....$$
$$=\frac{1}{2}+\sum_{n=2} \frac{-1}{n!n}+\frac{1}{(n+1)!n}$$
$$=\frac{1}{2}-\sum \frac{1}{(n+1)!}$$
Thus,
$$3-\sum_{n=2}^{\infty}\frac{1}{n!(n-1)n} = 1+1 +1-\frac{1}{2}+\sum_{n=2}^{\infty}\frac{1}{(n+1)!}=e$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Are my proofs correct for basic addition properties for natural numbers? Are my proofs correct?
Additive Identity: $a + 0 = a$
Definition of Addition: $a + S(b) = S(a + b)$
where $S(a)$ is the successor of $a$.
Claim: $0 + a = a$.
Base Case: When $a=0$, we have $0 + 0 = 0$ which is true by the additive identity definition.
Inductive Step: Suppose $0 + a = a$. We wish to show that $0 + S(a) = S(a)$. By definition of addition, $0 + S(a) = S(0 + a)$. Then $S(0 + a) = S(a)$ by the inductive hypothesis, and we are done.
Claim: $S(a) + b = S(a+b)$.
Base Case: $S(a) + 0 = S(a + 0) = S(a)$ by additive identity and definition of addition.
Inductive Step: Suppose $S(a) + b = S(a+b)$. We want to show that $S(a) + S(b) = S(a + S(b))$. By definition of addition, $S(a) + S(b) = S(S(a) + b)$ which is also equal to $S(S(a+b))$ by inductive hypothesis. Then $S(S(a+b)) = S(a + S(b))$ by definition of addition, and we are done.
Claim: $a + b = b + a$
Base Case: $a + 0 = 0 + a = a$ which is true by our additive identities.
Inductive Step: Suppose $a + b = b + a$. We need to show that $a + S(b) = S(b) + a$. By definition of addition, $a + S(b) = S(a+b)$. Our new addition definition also gives us $S(b) + a = S(b+a)$, which by inductive hypothesis is equal to $S(a+b)$. Therefore both sides are equal and we are done.
Claim: $a + c = b + c$ implies $a = b$.
Base Case: If $c=0$ then we have $a + 0 = b + 0$ which reduces to $a = b$ by our additive identities.
Inductive Step: Suppose that $a + c = b + c$ implies $a=b$. We need to show that $a + S(c) = b + S(c)$ implies $a=b$. By definition of addition, $a + S(c) = S(a+c)$, and $b + S(c) = S(b+c)$. By Peano's axiom stating that S is injective, we have $a+c=b+c$. By inductive hypothesis, $a=b$, so we are done.
Claim: $(a+b)+c = a+(b+c)$.
Base Case: Setting $c=0$, by additive identities the equivalence $(a+b)+0 = a+(b+0)$ reduces to $a+b = a+b$ which is true.
Inductive Step: Suppose $(a+b)+c = a+(b+c)$. We need to show that $(a+b)+S(c) = a+(b+S(c))$. By definition of addition, $(a+b)+S(c) = S((a+b)+c)$, and $a+(b+S(c)) = a + S(b+c) = S(a+(b+c))$. By inductive hypothesis, $S(a+(b+c)) = S((a+b)+c)$. Now both sides are equal, so we are done. What this also suggests is that parentheses can be omitted since they have no impact on the end result, so $(a+b)+c = a+(b+c) = a+b+c$ can be considered equivalent.
| They all look OK to me except this one:
If $a + c = b + c$, then $a = b$.
Note that this is the only theorem that invokes the phrasing "if ... then."
Given the statement $P\implies Q,$ one cannot immediately infer $P.$
The correct inductive hypothesis would be that for some $c,$
$a + c = b + c \implies a = b.$
Instead, you assumed $a + c = b + c,$ which is a different statement.
You end up showing that $a + c = b + c \implies a + S(c) = b + S(c),$
but even if you combine this with the correct inductive hypothesis,
it is not correct to infer from $P\implies Q$ and $P\implies R$
that $R\implies Q.$
You need your inductive step to conclude that
$a + S(c) = b + S(c) \implies a = b.$
| {
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Alternative for calculating the nth of quadratic sequence Given the quadratic sequence
$$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$
$$f(2)=4a+2b+c$$
$$f(3)=9a+3b+c$$
Now solve the simultaneous equations
$$a+b+c=1\tag1$$
$$4a+2b+c=7\tag2$$
$$9a+3b+c=19\tag3$$
$(2)-(1)$ and $(3)-(2)$
$$3a+b=6\tag4$$
$$5a+b=12\tag5$$
$(5)-(4)$
$$a=3$$
$$b=-3$$
$$c=1$$
$$f(n)=3n^2-3n+1$$
This method is very long. Is there another easy of calculating the $f(n)$?
| A better explained version of @trancelocation's answer:
$$\begin{array}{}
1 & & 7 & & 19 & & 37 \\
& 6 & & 12& &18 & \\
& & 6 & & 6 & & \\
\end{array}$$
The second row has equation $6n$. Therefore, $\big(a(n+1)^2+b(n+1)+c \big) - \big(an^2+bn+c \big)$ $ = 6n$, and so:
$$\big(a(n^2+2n+1)+b(n+1)+c \big) - \big(an^2+bn+c \big) = 6n$$
$$(2n+1)a + b = 6n$$
$a=3$ gives $6n+3$, and so $b=-3$. 'Plugging' $n=1$ into $3n^2-3n$ gives $0$, $c=1$, which gives $3n^2-3n+1$.
| {
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Limit: $\lim_{x\to 0}\frac{x^3 - 3x^2+1}{x^3}$ I want find the limit of $\frac{x^3 - 3x^2+1}{x^3}$ as $x\to 0$.
So I write
$$\frac{x^3 - 3x^2+1}{x^3} = 1 - \frac{3x^2 +1}{x^3}$$ and
$$\lim_{x \to 0^- }(\frac{x^3 - 3x^2+1}{x^3})= \lim_{x \to 0^-}(1 - \frac{3x^2 +1}{x^3})$$ Here, the limit must be $\pm \infty$ and I think when $x \to 0^-$ as $x^2$ is something positive, $x^3$ is something negative and since we have $- \frac{3x^2 +1}{x^3}$ we should have $+\infty$. However, it must be $-\infty$. Similarly I think $\displaystyle\lim_{x \to 0^-}(1 - \frac{3x^2 +1}{x^3}) =-\infty$ but it's $+ \infty$.
What do I miss, how should I think?
| You made a sign mistake, $$\frac{x^3 - 3x^2+1}{x^3} = 1 - \frac{3x^2 \color{red}{-}1}{x^3}$$
Hence your conclusion is opposite.
Intuitively, as $x \to 0^-$, $x^2$ is small in magnitude and hence $(3x^2-1)$ approaches $-1$, $x^3$ is something negative and it is small, so we have $- \frac{3x^2 +1}{x^3}$ goes to $-\infty$.
The left hand side limit goes to $-\infty$ while the right hand side limit goes to $+\infty$, hence the limit doesn't exist.
| {
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"url": "https://math.stackexchange.com/questions/2714007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Consider a family of lines (4a + 3)x -(a+1)y -(2a+1) = 0
Consider a family of lines $(4a + 3)x -(a+1)y -(2a+1) = 0$ where $a \in R$. The locus of the foot of the perpendicular from the origin of each member of this family is?
My Attempt:
Let the foot of the perpendicular from the origin on these lines be $(\alpha,\beta)$
Applying the formula of the foot of the perpendicular, I get
$$\alpha = \frac{8a^2 + 10a +3}{17a^2+26a+10}$$
$$\beta = -\biggl(\frac{2a^2+3a+1}{17a^2+26a+10}\biggr)$$
Now, I know that I need to eliminate '$a$' in order to get the locus. But, how am I supposed to do that? It's like quadratic on the top and bottom. Using the quadratic formula would be messy.
Any help would be appreciated.
| One approach: write the two equations as polynomials in $a$ and compute their resultant. To wit, rewrite the equations $$x = {8a^2+10a+3 \over 17a^2+26a+10} \\ y = -{2a^2+3a+1 \over 17a^2+26a+10}$$ as $$(17a^2+26a+10)x-(8a^2+10a+3) = (17x-8)a^2+(26x-10)a+(10x-3) = 0 \\
(17a^2+26a+10)y +(2a^2+3a+1) = (17y+2)a^2+(26y+3)a+(10y+1) = 0.$$ Now form the Sylvester matrix of these two polynomials and compute its determinant (a symbolic algebra program is helpful here): $$\det\begin{bmatrix}17x-8&26x-10&10x-3&0 \\ 0&17x-8&26x-10&10x-3 \\ 17y+2&26y+3&10y+1&0 \\ 0&17y+2&26y+3&10y+1 \end{bmatrix} = 5(x^2-x+y^2-2y).$$ The resulting implicit Cartesian equation is therefore $$x^2+y^2-x-2y=0,$$ which is the equation of a circle of radius $\frac{\sqrt5}2$ centered at $\left(\frac12,1\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding natural domain of real functions involving square root and absolute value I am trying to find the natural domain of the following 2 functions:
*
*$ \sqrt{\frac{x+1}{x-2}+3} $
*$ \sqrt{|x-5| - |x+1|} $
For question 1 I got $[-4,\infty) \cap (-\infty,1]$, but I'm not sure that it is correct. However, I cannot get a solution for the second question.
Steps for question 1:
$\sqrt{\frac{x+1}{x-2}+3} $.
The formula will be well defined if:
$\frac{x+1}{x-2}+3 \ge 0 $
$\frac{x+1}{x-2} \ge -3 $
Hence,
$(x+1 \ge -3)\land(x-2>-3)$ or $(x+1\le-3)\land(x-2<-3)$
Their intersection means the domain is $[-4,\infty) \cap (-\infty,1]$ but I assume that that is wrong.
| For the first question, note that $x \ne 2$,
$$\frac{x+1}{x-2} \ge -3 $$
multiply both sides by $(x-2)^2$,
$$(x+1)(x-2) \ge -3 (x-2)^2$$
$$(x-2)((x+1)+3(x-2)) \ge 0$$
$$(x-2)(4x-5) \ge 0$$
Try to complete it?
For the second question,
We need $$|x-5|-|x+1| \ge 0$$
$$|x-5| \ge |x+1|$$
squaring booth sides
$$x^2-10x+25 \ge x^2+2x +1$$
Can you continue from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the inequality $\sqrt{\frac{4x-1}{x-a}}>a$
Solve the inequality
$$\sqrt{\frac{4x-1}{x-a}}>a$$
My work so far:
If $a<0$ then $$\frac{4x-1}{x-a}\ge0$$
$$x\in(-\infty;a)\cup[\frac14;+\infty)$$
Let $a\ge0$ then
1) $$\frac{4x-1}{x-a}\ge0$$
and
2) $$\frac{4x-1}{x-a}>a^2$$
I need help here.
| If $a \ge 0$, then we can square both sides.
$$\frac{4x-1}{x-a} > a^2$$
We need $x \ne a$.
Multiply both sides by $(x-a)^2$,
$$(4x-1)(x-a) > a^2(x-a)^2$$
$$(x-a)(4x-a^2x-1+a^3) >0$$
$$(x-a)((4-a^2)x-(1-a^3)) >0$$
To solve if further, we need to figure out when does $$a> \frac{1-a^3}{4-a^2}$$
Multiply $(4-a^2)^2$ on both sides, we have
$$a(4-a^2)^2 > (1-a^3)(4-a^2)$$
which is equivalent to
$$(4-a^2)(4a-a^3-1+a^3)>0$$
$$(2-a)(2+a)(4a-1)>0$$
$$(a-2)(2+a)(4a-1)<0$$
$$(a-2)(4a-1)<0$$
$$\frac14 < a< 2$$
We consider $4$ cases:
Case $1$： if $a \le \frac14$, then $x<a$ or $x > \frac{1-a^3}{4-a^2}$.
Case $2$: if $\frac14<a<2$, then $x < \frac{1-a^3}{4-a^2}$ or $x > a$.
Case $3$: if $a =2$, we have $x > 2$.
Case $4$: if $a > 2$, we have $$(x-a)((a^2-4)x-(a^3-1))<0$$
Hence $$a<x<\frac{1-a^3}{4-a^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
how to write the polar form of $x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2$? $$x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2\\
r=(x^2+y^2)^{1/2}\\
x^2/r^4-y^2/r^4$$
Convert $x^2-y^2$ to polar form
$$x=r\cos(\theta)$$
$$y=r\sin(\theta)$$
$$(r\cos(\theta))^2-(r\sin(\theta))^2=r^2 \cos(2 θ)$$
$$r^2 \cos(2 θ)/r^4= \cos(2\theta)/r^2$$
Is that correct?
| Hint
It should look like this in the end:
$$E=x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2=\frac {x^2- y^2}{(x^2 + y^2)^2}=\frac {\cos^2(\theta)- \sin^2(\theta)}{r^2}$$
$$E=\frac {2\cos^2(\theta)- 1}{r^2}=\frac {\cos(2\theta)}{r^2}$$
So it's correct...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof By Induction: $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ Here's what I have so far: $$\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$$ where $n\ge2$.
Test for base case
$$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} > \sqrt 2$$
Simplify
$$ 1 + .707... > 1.414... $$
$$ 1.707... > 1.414... $$
This statement is true.
Thus we can assume $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ is true for $\forall n$
Now prove $n + 1$
Expand
$$\sum_{i=1}^n \frac{1}{\sqrt i} = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} \gt \sqrt n $$
Add the next n in the series, and add it to both sides of the inequality.
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} \gt \sqrt n + \frac{1}{\sqrt{n + 1}} $$
Rewrite
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} = \sum_{i=1}^{n + 1} \frac{1}{\sqrt i}$$
LHS is complete now we have to try to make RHS = $\sqrt {n + 1}$
$$\sqrt n + \frac{1}{\sqrt{n + 1}} = \frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}}$$
Given that n > 0
$$\frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}} = \frac{\sqrt{n(n+1)} + 1} {\sqrt{n+1}}$$
Now since this is my first induction involving inequalities or summation, I was hoping you guys could show me what the next step would be, since I'm really not sure. Thanks in advance!
| Well. Observe that
\begin{align}
\sqrt{n}+\frac{1}{\sqrt{n+1}} = \frac{\sqrt{n^2+n}+1}{\sqrt{n+1}}\geq \frac{n+1}{\sqrt{n+1}} =\sqrt{n+1}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Residue Complex Function I've applied the formula $$\lim_{z \to i} \frac{d}{dz}{(z-z_0)^2.f(z)}$$ and I keep getting a value of 0. Could someone prove my sanity by starting this off?
| Hint: Since $f(z)$ has a double pole at $z=i$, then we see that
\begin{align}
\operatorname{Res}_{z=i}f(z)=\lim_{z\rightarrow i}\frac{1}{2}\frac{d}{dz}\left((z-i)^2\frac{z^2-1}{(z^2+1)^2}\right).
\end{align}
Edit: To put you at ease, observe the power series expansion at $z=i$ of the function is given by
\begin{align}
\frac{z^2-1}{(z^2+1)^2} =&\ \frac{1}{(z-i)^2}\frac{z^2-1}{(z+i)^2} =\frac{-1}{4(z-i)^2}\frac{(z-i)^2+2i(z-i)-2}{\left(1+\frac{z-i}{2i}\right)^2}\\
=&\ \frac{-1}{4(z-i)^2}\left((z-i)^2+2i(z-i)-2\right)\sum_{k=0}^\infty(-1)^k(k+1)\left(\frac{z-i}{2i} \right)^k\\
=&\ \left(\frac{-1}{4} -\frac{i}{2(z-i)}+\frac{1}{2(z-i)^2}\right)\sum_{k=0}^\infty(-1)^k(k+1)\left(\frac{z-i}{2i} \right)^k\\
=&\ \frac{1}{2(z-i)^2}-\frac{1}{8}-\frac{1}{8}i(z-i)+\text{higher order terms}.
\end{align}
Hence it is clear that the residue is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $2x+3[x]-4\{-x\}=4$? I am beginner to greatest integer and fractional part functions, and I just came across this problem:
$$ 2x+3[x]-4\{-x\}=4,$$
where $[x]$ and $\{x\}$ are the greatest integer and the fractional part functions, respectively.
What I have tried is taken $[x] = x $ as $ x-1≤[x]≤x$ but no ideas for fractional part of $x$. However, I know that $ \{x\} = x-[x]$.
How do I proceed further?
| $$2x +3 \lfloor x \rfloor - 4 \{-x\} = 4$$
Let $x = n+\alpha$ where $n$ is an integer and $0 \le \alpha < 1$.
Then ${-x} =
\begin{cases}
0 & \text{If $\alpha = 0$.} \\
1-\alpha & \text{If $\alpha \ne 0$.}
\end{cases}$
So, when $\alpha = 0$
\begin{align}
2x +3 \lfloor x \rfloor - 4 \{-x\} &= 4 \\
2n +3n - 4(0) &= 4 \\
5n &= 4
\end{align}
Which has no integer solution.
When $0 < \alpha < 1$,
\begin{align}
2x +3 \lfloor x \rfloor - 4 \{-x\} &= 4 \\
2(n + \alpha) +3n - 4(1-\alpha) &= 4 \\
5n+6\alpha &=8
\end{align}
The only possible solution is $n=1$ and $\alpha = \dfrac 12$
So $x=1 \frac 12$
Check:
\begin{align}
\left \lfloor 1 \frac 12 \right \rfloor &= 1 \\
\left\{ -1 \frac 12 \right\} &= \frac 12 \\
\end{align}
So
\begin{align}
2x +3 \lfloor x \rfloor - 4 \{-x\}
&= 2\left( 1 \frac 12 \right)
+ 3 \left \lfloor 1 \frac 12 \right \rfloor
- 4 \left\{ -1 \frac 12 \right\} \\
&= 3 + 3(1) - 4\left( \frac 12 \right) \\
&= 3 + 3 - 2 \\
&= 4
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the sequence converges to using the $\varepsilon$-$N$ definition of a limit Question: Prove that the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_n=\frac{2n}{3n+1}$ converges to $\frac{2}{3}$ using the $\varepsilon$-$N$ definition of a limit.
I'm still a beginner when it comes to these types of questions. I've had a go at it but I'm pretty sure this is nonsense. Would appreciate some feedback and suggestions. Thanks!
Solution: Fix $\varepsilon>0$. We need to find $N\in\mathbb{N}$ such that $n>N \implies|a_n-\frac{2}{3}|<\varepsilon.$
We have $|a_n-\frac{2}{3}| < \varepsilon.$
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
Now $2n\ge2$ for all $n\ge1$ and $3n+1\ge3n$ for all $n$, so we have
$\iff |\frac{2}{3n}-\frac{2}{3}| < \varepsilon$
$\iff |\frac{2}{3}(\frac{1}{n}-1)| < \varepsilon$
The sequence is positive for all $n$ so the absolute values are redundant
$\iff \frac{2}{3}(\frac{1}{n}-1) < \varepsilon$
$\iff \frac{1}{n}-1 < \frac{3\varepsilon}{2}$
$\iff \frac{1}{n} < \frac{3\varepsilon}{2}+1$
$\iff \frac{1}{n} < \frac{3\varepsilon+2}{2}$
$\iff n> \frac{2}{3\varepsilon+2}$
So choose any $N>\frac{2}{3\varepsilon+2}$ and the definition is satisfied. Q.E.D.
| HINT: it is $$\frac{2n}{3n+1}-\frac{2}{3}=\frac{-2}{3(3n+1)}$$
and $$\left|\frac{-2}{3(3n+1)}\right|=\frac{2}{3(3n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiation using product rule I'm having trouble simplifying these questions, particularly when they involve square roots of $x$.
Differentiate the following with respect to $x$ and simplify:
$y=(x+2)x^\frac{3}{2}$
My attempt:
Using product rule: $u=x^\frac{3}{2}, v=(x+2)$ therefore $\frac{du}{dx}=\frac{3}{2}x^\frac{1}{2}, \frac{dv}{dx}=1\\$
$\frac{dy}{dx}= \frac{3\sqrt{x}}{2}(x+2)+(\sqrt{x})^3$
Factorise:$\sqrt{x}[\frac{3}{2}(x+2)+(\sqrt{x})^2]\\\sqrt{x}[\frac{3}{2}x+3+x]$
The given answer is $\frac{\sqrt{x}}{2}(5x+6)$, which I can't achieve and I can't understand why the denominator of 2 is a common factor.
| A general rule: first simplify, then differentiate. The function in question is a product, namely $$y=(x+2)x^\frac{3}{2}.$$
First simplify it to a sum
$$y=x^{\frac{5}{2}}+2x^\frac{3}{2}$$
and then differentiate
$$y’=\frac{5}{2}x^{\frac{3}{2}}+3x^{\frac12}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What method can be used to solve a numeric finding puzzle involving getting the right sum? I've been trying to solve this puzzle at no avail. Can somebody help me?
The problem is as follows:
In each circle from the figure below write down an integer number
between $1$ to $9$. Each one of them must be different from each
other. The resulting sum from those numbers joined by three circles in
a straight line and those joined by arrows in the direction indicated be equal to $18$. Which
number must be written in the circle painted by an orange shade?
$\textrm{The alternatives given were:}$
*
*7
*4
*5
*3
*6
The only thing I could come up with was to make a sequence of dividing $18$ into different choices for summing that number between $1$ to $9$.
Therefore:
$1+8+9=18$
$2+7+9=18$
$3+6+9=18$
$4+5+9=18$
and the rest would be repeating those numbers in different order like
$5+4+9=18$
$6+3+9=18$
$7+2+9=18$
$8+1+9=18$
So the preceding can't be:
But this other combination can also reach to $18$:
$8+7+3=18$
$8+6+4=18$
However at this point I got tangled up with many choices therefore end up with clear clue to take to get $18$.
How can I solve this problem without just guessing?. Is there an algorithm, formula or method that can be followed orderly to avoid wasting much time?.
I would like someone could re drawn the question and explain me with much details possible.
Please do not just fill out the drawing right away. I know maybe you have the ability to do this, but this is not what I'm looking for but rather a more step-by-step solution which has proven that it works in many similar scenarios.
Edit: I currently replaced the previous picture to this version which is the right one.
| Fellow the image Below:
$A + B + C + D +E +F + G+ H+ X = 1 + 2+3+4+5+6+7+8+9 = 45$
so
$X+ C +F = (A+B+C+D+E+F+G+H+X) - (B+E+H) - (A+D+G) = 45 - 18 -18 = 9$.
so
$2X + A+E +C +F = (X+C+F)+(A+E+X)= 9 + 18 = 27$
so
$A+B+C+D+E+F + 3X = (2X+A+E+C+F) + (B+E+D) = 27 + 18 = 45$
so
$A+B+C + 3X = (A+B+C+D+E+F) - (D+E+F) = 45-18 = 27$.
so
$3X = (A+B+C + 3X) - (A+B+C) = 27-18 = 9$.
So $X = \frac {3X}3 = \frac 93 = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one establish continuity at a point for a multivariable scalar field? I am trying to understand continuity for multivariable scalar fields, $f: \Bbb R^n \rightarrow \Bbb R$.
In single variable calculus, I understand the $\delta-\epsilon$ formulation
of continuity.
For a function $f: \Bbb R \rightarrow \Bbb R$, $f$ is continuous if
for any arbitrary $\epsilon >0$, $\exists \delta$ such that
$$|x - x_0 | < \delta \Rightarrow |f(x) - f(x_0)| <\epsilon$$
For example, suppose $f(x) = 3x - 6$.
\begin{align*}
|f(x)-f(a)|&<\epsilon\\
|(3x-6)-(3a-6)|&<\epsilon\\
|3x-3a|&<\epsilon\\
3|x-a|&<\epsilon\\
|x-a|&<\frac{\epsilon}{3}\\
\end{align*}
So for $a \in \Bbb R$, then $\delta = \frac{\epsilon}{3}$ implies $|f(x) - f(a)|<\epsilon$
For scalar fields, Tom Apostol says
A function $f: \Bbb R^n \rightarrow \Bbb R$ is continuous at $\mathbf{a}$ if
*
*$f$ is defined at $\mathbf{a}$
*$\displaystyle \lim_{\mathbf{x}\rightarrow \mathbf{a}} f(\mathbf{x})=f(\mathbf{a})$
But I don't understand how to apply the $\delta-\epsilon$ formulation here.
Suppose I want to show $f(x,y) = x^4 + y^4 - 4 x^2 y^2$ is continuous at some point, say $\mathbf{a} = (3,4)$.
\begin{align*}
|f(x,y)-f(3,4)|&<\epsilon \\
|x^4 + y^4 - 4 x^2 y^2 - 3^4 - 4^4 + 4\times 3^2 4^2|&<\epsilon\\
\end{align*}
But this expression is so much uglier than the single variable case. Should I be approaching this just like the single variable case? In other words, should the final simplification look like
\begin{align*}
||\mathbf{x}-\mathbf{a}|| &<\delta\\
\sqrt{(x-x_0)^2 + (y-y_0)^2}&<\delta\\
\end{align*}
Or is this the wrong approach? I have googled a lot, but I can't find examples that show continuity for multivariable functions, just ones that disprove it.
| Not that bad. For
\begin{align*}
&|f(x,y)-f(3,4)|\\
&=|x^{4}+y^{4}-4x^{2}y^{2}-3^{4}-4^{4}+4\cdot 3^{2}4^{2}|\\
&=|(x^{2}+3^{2})(x+3)(x-3)+(y^{2}+4^{2})(y+4)(y-4)-4(xy+3\cdot 4)(xy-3\cdot 4)|\\
&\leq|x^{2}+3^{2}||x+3||x-3|+|y^{2}+4^{2}||y+4||y-4|+4|xy+3\cdot 4||(x-3)y+3(y-4)|\\
&\leq(x^{2}+9)(|x|+3)|x-3|+(y^{2}+16)(|y|+4)|y-4|+4(|xy|+12)(|x-3||y|+3|y-4|).
\end{align*}
For $(x-3)^{2}+(y-4)^{2}<\delta^{2}=\min\{1,\epsilon\}$, then $|x-3|<1$ and $|y-4|<1$ and we have $|x|+3\leq|x-3|+6<7$, $x^{2}+9<25$, $|y|+4\leq|y-4|+8<9$, $y^{2}+16<41$, $|xy|+12<32$, so
\begin{align*}
|f(x,y)-f(3,4)|\leq M(|x-3|+|y-4|)\leq 2M\sqrt{|x-3|^{2}+|y-4|^{2}}<2M\epsilon,
\end{align*}
where $M>0$ is a universal constant, essentially, it is the sum of all those bounds, one may take $M=25\cdot 7+41\cdot 9+4\cdot 32(5+3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$
Find $\frac{dy}{dx}$ if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1\leq x\leq 1$
My Attempt
$$
x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\
2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\
\frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\
\frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy}
$$
How do I proceed further and find the derivative ?
| $$
x\sqrt{1+y}+y\sqrt{1+x}=0\implies x\sqrt{1+y}=-y\sqrt{1+x}\\
\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+y^2x\\
\implies \boxed{(1+x)\color{red}{y^2}-x^2\color{red}{y}-x^2=0}
$$
$B^2-4AC=x^4+4x^2(1+x)=x^4+4x^2+4x^3=x^2(x^2+4x+4)=x^2(x+2)^2$
So,
$$
\begin{align}
y&=\frac{x^2\pm\sqrt{x^2(x+2)^2}}{2(x+1)}=\frac{x^2\pm|{x(x+2)}|}{2(x+1)}\\
&=\frac{x^2+x(x+2)}{2(x+1)}\text{ or }\frac{x^2-x(x+2)}{2(x+1)}\\
&=\frac{x^2+x^2+2x}{2(x+1)}\text{ or }\frac{x^2-x^2-2x}{2(x+1)}\\
&=\frac{2x(x+1)}{2(x+1)}\text{ or }\frac{-2x}{2(x+1)}\\
y&=x \text{ or } \frac{-x}{x+1}
\end{align}
$$
As $y\neq x$ we have the derivative,
$$\color{blue}{
y'=\frac{-1}{(1+x)^2} }
$$
Note: $x+\sqrt{1+y}=−y\sqrt{1+x}\implies(x−y)(x+y+xy)=0\implies x−y=0$ or $(x+y+xy)=0$. Note that $x−y=0\implies x=y$ is equivalent to the original function only when $x=0$, and $x\sqrt{1+y}=−y\sqrt{1+x}\Leftrightarrow(x+y+xy)=0$ for all $x∈D_\text{original function}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that the sequence $3+\frac{1}{n^2}$ is Cauchy Question: Prove that the sequence defined by ${x_n}=3+\frac{1}{n^2}$ is Cauchy.
My attempt: Need to show that $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $m,n>N \implies|x_n-x_m|<\varepsilon.$
Fix $\epsilon>0$. (Edit: and let $m>n$.)
We have $|x_n-x_m|=|3+\frac{1}{n^2}-(3+\frac{1}{m^2})|$
$=|\frac{1}{n^2}-\frac{1}{m^2}|$
$=|\frac{m^2}{m^2n^2}-\frac{n^2}{m^2n^2}|$
$=|\frac{m^2-n^2}{m^2n^2}|$
$\le|\frac{m^2}{m^2n^2}|$
$=|\frac{1}{n^2}|=\frac{1}{n^2}.$
Then $\frac{1}{n^2}<\epsilon \implies n^2>\frac{1}{\epsilon}\implies n>\sqrt\frac{1}{\epsilon}$.
So choose $N=\lfloor\sqrt\frac{1}{\epsilon}\rfloor+1$ and the definition is satisfied. Q.E.D.
Is this correct? Thanks.
| Just want to add my thought to this as you did most of it. Often times, you replace the minus $-$ by the plus $+$. So this means:
$\left|\dfrac{m^2-n^2}{m^2n^2}\right| \le \dfrac{m^2+n^2}{m^2n^2}= \dfrac{1}{m^2}+\dfrac{1}{n^2}\le \dfrac{1}{m}+\dfrac{1}{n} < \dfrac{1}{N}+\dfrac{1}{N}=\dfrac{2}{N}< \epsilon$, if $N > \dfrac{2}{\epsilon}$. Thus you can say: choose $N > \dfrac{2}{\epsilon}$ and for $m, n > N$ then you have $|x_m-x_n| < \epsilon$, proving it a Cauchy Sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle Given that roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle, find condition on $b$.
If $a$ is a root then $a(\frac{1}{2}+i\frac{\sqrt 3 }{2})$ i also a root. Sum of root is $-1 = a(\frac{3}{2}+i\frac{\sqrt3}{2})$ and product is $a^2(\frac{1}{2}+i\frac{\sqrt 3 }{2}) = \frac{b}{3}$
I am asking is this correct and also does veeta's formula work in complex numbers? What is your method and how to proceed using this, find $a$ and put in second equation to get $b$ is complicated.
Thanks a lot!!
| Obviously the roots shall be complex, so that $12b>9$.
$$
z_{1,2}=\frac{-3\pm i\sqrt{12b-9}}{6}\Rightarrow |z_1|=|z_2|=\sqrt{\frac{b}{3}},\quad |z_2-z_1|=\frac{\sqrt{12b-9}}{3}\\
\Rightarrow 3b=12b-9\Rightarrow b=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequality proof by induction of $n!>2^n+3^n$ Give a proof by induction:
$\forall n\ge n_0 \in \mathbb{N}$ we have $n!>2^n+3^n$.
Attempt:
Base case: $n_0=7$
Induction step:
$k!>2^k+3^k$
Prove $(k+1)!>3^{k+1}+2^{k+1}$
We know $(k+1)!>(k+1)(2^k+3^k)$
as $(k+1)>6 \because k\ge 7$:
$(k+1)!>6(2^k+3^k)$
$(k+1)!>3*2^{k+1}+2*3^{k+1}$
as $3*2^{k+1}+2*3^{k+1}>3*2^{k+1}+2*3^{k+1}-2*2^{k+1}-3^{k+1}$
$(k+1)!>2^{k+1}+3^{k+1}$
still getting my head around this method of induction with inequalities.
Any help is greatly appreciated, thank you!
| From here for $k> 2$ we have
$$(k+1)!\stackrel{I.H.}>(k+1)(2^k+3^k)=(k+1)2^k+(k+1)3^k>2\cdot 2^k+3\cdot 3^k=2^{k+1}+3^{k+1}$$
and this prove the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$ a+b+c=0,\ a^2+b^2+c^2=1$ implies $ a^4+b^4+c^4=\frac{1}{2}$ This is a high school problem. It is solved by algebraic calculation
(not difficult) - $\ast$.
But I have a problem in interpretation the result.
Here intersection of two surfaces $a+b+c=0,\ a^2+b^2+c^2=1$ is a
circle $C$. That is, by $\ast$, $a^4+b^4+c^4$ is constant on it. What does make
it so ?
Proof : Note that $S:=\{
(a,b,c)|a^4+b^4+c^4=\frac{1}{2}\}$ is a convex surface. In further,
there is $x,\ y\in S$ s.t. $|x|<1<|y|$.
And I guess that $C$ is not a geodesic : Consider a norm on $\mathbb{R}^3$. For instance $\| (a,b,c) \|_1 := |a|+|b|+|c|$ In this case unit ball wrt this norm is octahedron. In further, unit balls of standard Euclidean norm, infinite norm are unit sphere $S_2$, a cube $S_3$, respectively. Clearly, intersection of $S_2$ and plane $\{ (a,b,c)| a+b+c=0\}$ is a geodesic (So is $S_3$). Here considering $S$ and $S_3$, we can assume that $S$ is like $S_3$. That is, $S$ has largest Gaussian curvature at eight points. Hence geodesic is far from these points, but $C$ is not.
I enumerate some these exercises, but how can we interpret the answer ?
[add] Consider $ \Delta':=\{ (x,y,z)\in S_2 | x,\ y,\ z\geq 0\} $, which
is an equilateral geodesic triangle of side length $\frac{\pi}{2}$.
In further $\Delta \subset \Delta'$ is also an equilateral geodesic
triangle of side length $\frac{\pi}{3}$, which touches mid points of
sides in $\Delta'$. Clearly it is not in a plane.
Then we have a claim that if $ f (x,y,z)=(x^4,y^4 ,z^4)$, then
$f(\partial \Delta )$ is in a plane.
[my answer] Consider $SO(3)$-action on $S^2(1)$. Here what is $smallest$ invariant set ? It is $S^2$. Consider a finite subgroup of $SO(3)$. For instance, a group $H$ which acts on cube whose center is origin. So smallest invariant set of $H$ is not $S^2$. It may be union of finite number of great circles, $T$. And note that $F(x,y,z)=x^4+y^4+z^4$ is invariant under $H$. Hence we can guess that $F$ is constant on $T$.
| HINT:
From the identity
$$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
the equality $a^4+b^4+c^4=1/2$ is a consequence of $a+b+c=0$ and $a^2+b^2+c^2=1$. One can also look at gradients and see a geometric reason from the equality.
Note that conversely $a^4+b^4+c^4=1/2$ and $a+b+c=0$ implies $(a^2+b^2+c^2)^2=1$ and so (reals!) $a^2+b^2+c^2=1$. So indeed the plane cuts the surface along that circle.
One can cook up different surfaces of equation
$$P(a,b,c)(a^2+b^2+c^2-1)+Q(a,b,c)(a+b+c)=0$$ that will contain the same circle.
If one looks at the surface $a^4+b^4+c^4=1/2$, it has the shape of a rounded cube. How is it that the plane $a+b+c=0$ cuts it along a circle?
Recall that a cube $\max(|a|,|b|,|c|)=1$ is cut by this plane along a regular hexagon with vertices the permutations of $(-1,0,1)$. Now the permutations of $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$ lie on both of the surfaces $a^2+b^2+c^2=1$ and $a^4+b^4+c^4=\frac{1}{2}$. That should make it more intuitive.
Note that each of the $4$ planes $a\pm b\pm c = 0$ cuts our surface along a circle, as the identity tells us.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Expressing the integral as the limit of sums and evaluating Hello I'm having some issues with a homework problem.
I need to express the integral as the limit of sums and evaluate
$$\int_0^\pi \sin(7x)\,dx$$
I can find
$$ \Delta x = \frac{\pi-0}{n} = \frac{\pi}{n} \text{ and }x_i = 0 + \frac{\pi}{n}i$$
so I get
$$\int_0^\pi \sin(7x)\,dx = \lim_{n\to \infty}\sum_{i=1}^n\sin\left(\frac{7\pi}{n}i\right)\frac{\pi}{n} $$
I know that $$\lim_{n\to \infty}\sum_{i=1}^n\sin\left(\frac{7\pi}{n}i\right)\frac{\pi}{n} = \frac{2}{7}$$
but I need to understand how to get there only using limits and sums.
this is where I'm stuck, I don't know how to proceed while expressing the limit and sum.
Any help would be greatly appreciated.
| $$\begin{align}
& \text{using}\ \text{lagrange trigonometric identity}\ \text{which reads:} \\
& \sum\nolimits_{i=1}^{n}{\operatorname{Sin}\left( i\theta \right)}=\frac{1}{2}Cot\left( \frac{\theta }{2} \right)-\frac{\operatorname{Cos}\left( \left( n+\frac{1}{2} \right)\theta \right)}{2\operatorname{Sin}\left( \frac{\theta }{2} \right)}, \\
& Now \\
& \sum\nolimits_{i=1}^{n}{\operatorname{Sin}\left( i\frac{7\pi }{n} \right)}=\frac{1}{2}Cot\left( \frac{7\pi }{2n} \right)-\frac{\operatorname{Cos}\left( \left( n+\frac{1}{2} \right)\frac{7\pi }{n} \right)}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)} \\
& \quad \quad \quad \quad \quad \quad =\frac{1}{2}Cot\left( \frac{7\pi }{2n} \right)-\frac{\operatorname{Cos}\left( 7\pi +\frac{7\pi }{2n} \right)}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)} \\
& \quad \quad \quad \quad \quad \ \ =\frac{\operatorname{Cos}\left( \frac{7\pi }{2n} \right)}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)}-\frac{\operatorname{Cos}\left( 7\pi +\frac{7\pi }{2n} \right)}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)} \\
& \quad \quad \quad \quad \quad \ \ =\frac{1}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)}\left( \operatorname{Cos}\left( \frac{7\pi }{2n} \right)-\operatorname{Cos}\left( 7\pi +\frac{7\pi }{2n} \right) \right) \\
& So \\
& \underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=1}^{n}{\operatorname{Sin}\left( i\frac{7\pi }{n} \right)}\frac{\pi }{n} \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{\pi }{n}\sum\nolimits_{i=1}^{n}{\operatorname{Sin}\left( i\frac{7\pi }{n} \right)}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\pi }{n}\left( \frac{1}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)}\left( \operatorname{Cos}\left( \frac{7\pi }{2n} \right)-\operatorname{Cos}\left( 7\pi +\frac{7\pi }{2n} \right) \right) \right) \\
& \quad \quad \quad \quad \quad \ \ \quad \quad \quad =\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{\pi }{n}}{2\operatorname{Sin}\left( \frac{7\pi }{2n} \right)}\ \underset{n\to \infty }{\mathop{\lim }}\,\left( \operatorname{Cos}\left( \frac{7\pi }{2n} \right)-\operatorname{Cos}\left( 7\pi +\frac{7\pi }{2n} \right) \right) \\
& \quad \quad \quad \quad \quad \ \ \quad \quad \quad =\frac{\pi }{2\times \frac{7\pi }{2}}\times \left( \operatorname{Cos}\left( 0 \right)-\operatorname{Cos}\left( 7\pi \right) \right)=\frac{1}{7}\times 2=\frac{2}{7} \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the indefinite integral $\frac{x\sin x}{1+\cos^2 x}$
Evaluate $$\int\frac{x\sin x}{1+\cos^2 x}dx$$
My attempt:
$$I=\int\frac{x\sin x}{1+\cos^2 x}dx=\int x\frac{\sin x}{1+\cos^2 x}dx=\\x\int \frac{\sin x}{1+\cos^2 x}dx-\int \left[\frac{d}{dx}x\int\frac{\sin x}{1+\cos^2 x}dx\right]dx$$
$I'=\displaystyle \int \frac{\sin x}{1+\cos^2 x}dx$
Let $u=\cos x$
$\therefore \dfrac{du}{dx}=-\sin x$
$\implies du=(-\sin x)dx$
$\therefore \displaystyle I'=-\int \frac{du}{1+u^2} $
$\implies I'=-\dfrac{1}{1}\tan^{-1}\left(\dfrac{u}{1}\right)+C \implies I'=-\tan^{-1}(u)+C$
$\implies I'=-\tan^{-1}(\cos x)+C$
$\therefore \displaystyle \int \frac{\sin x}{1+\cos^2 x}dx=-\tan^{-1} (\cos x)+C$
$\therefore \displaystyle I=x\cdot[-\tan^{-1} (\cos x)]-\int [-\tan^{-1} (\cos x)] dx$
$\implies \displaystyle I=-x \tan^{-1} (\cos x)+\int \tan^{-1} (\cos x)dx$
I cannot understand how to proceed further. Please help.
| I wrote this in response to the comment by the OP that he/she actually was trying to solve the definite integral over $[0, \pi]$.
Using the fact that
$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(b + a - x) dx$$
we find
\begin{align}
I &= \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^{2}x} dx \\
&= \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^{2}(\pi - x)} dx \\
&= \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^{2}x} dx \\
&= \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} dx - I \\
\implies I &= \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2}x} dx \qquad \text{(use substitution $u = \cos(x)$ to evaluate)} \\
&= \frac{\pi}{2} \cdot \frac{\pi}{2} \\
&= \frac{\pi^{2}}{4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What is the expansion of $\sqrt{1+ux+vx^2}$ in powers of $x$? What is the expansion of $\sqrt{1+ux+vx^2}$ in powers of $x$?
This came up
in my answer here:
How to solve this recurrence$f(n)=A\cdot f(n-1)+B\sum{f(i)f(n-i)},\;1\leq i\leq n-1,$ and $f(1)=K$?
I just did a
straightforward expansion
and wondered if
there is a better way to do it.
Here is my work:
Since
$\sqrt{1+y}
=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}y^k
$,
Then
$\begin{array}\\
\sqrt{1+ux+vx^2}
&=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}(ux+vx^2)^k\\
&=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}x^k(u+vx)^k\\
&=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}x^k\sum_{j=0}^k \binom{k}{j}v^jx^ju^{k-j}\\
&=\sum_{k=0}^{\infty} \sum_{j=0}^k\dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}x^{k+j} \binom{k}{j}v^ju^{k-j}\\
&=\sum_{n=0}^{\infty} x^n\sum_{j=0}^n\dfrac{(-1)^{n-j+1}}{4^{n-j}(2(n-j)-1)}\binom{2(n-j)}{n-j} \binom{n-j}{j}v^ju^{n-2j}
\qquad(k = n-j)\\
\end{array}
$
We can play with
$\binom{2(n-j)}{n-j} \binom{n-j}{j}
=\dfrac{(2n-2j)!(n-j)!}{(n-j)!^2j!(n-2j)!}
=\dfrac{(2n-2j)!}{(n-j)!j!(n-2j)!}
$
but I don't see
anything beyond this.
Also,
any corrections appreciated.
|
We obtain a slightly simplified representation by using the binomial series expansion
\begin{align*}
\sqrt{1+ux+vx^2}&=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k}(ux+vx^2)^k\\
&=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k}(ux)^k\left(1+\frac{v}{u}x\right)^k
\end{align*}
and the coefficient $[x^n]$ of the series expansion is
\begin{align*}
\color{blue}{[x^n]\sqrt{1+ux+vx^2}}&=\sum_{k=0}^n\binom{\frac{1}{2}}{k}u^k[x^{n-k}]\sum_{j=0}^k\binom{k}{j}\left(\frac{v}{u}x\right)^j\\
&=\sum_{k=0}^n\binom{\frac{1}{2}}{k}u^k\binom{k}{n-k}\left(\frac{v}{u}\right)^{n-k}\\
&\,\,\color{blue}{=\sum_{k=0}^n\binom{\frac{1}{2}}{k}\binom{k}{n-k}u^{2k-n}v^{n-k}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $a,b,c > 0 \Rightarrow \frac{b+c}{\sqrt{a^2 + bc}} + \frac{a+c}{\sqrt{b^2 + ac}} + \frac{a+b}{\sqrt{c^2 + ab}} \geq 4$ Prove:
$a,b,c > 0 \Rightarrow \frac{b+c}{\sqrt{a^2 + bc}} + \frac{a+c}{\sqrt{b^2 + ac}} + \frac{a+b}{\sqrt{c^2 + ab}} \geq 4$
I was able to obtain a loose bound. My method:
\begin{align}
\frac{b+c}{\sqrt{a^2 + bc}} + \frac{a+c}{\sqrt{b^2 + ac}} + \frac{a+b}{\sqrt{c^2 + ab}} &\geq 3\left(\frac{(b+c)(a+c)(a+b)}{\sqrt{(a^2 + bc)(b^2 + ac)(c^2 + ab)}}\right)^{\frac{1}{3}} \\
&> 3\cdot2^{\frac{1}{3}} \sim 3.78
\end{align}
The first step follows from the AM-GM inequality. The second step follows from the inequality: $4(a^2+bc)(b^2+ac)(c^2+ab) \leq [(b+c)(a+c)(b+a)]^2 $ which is an inequality which has been asked as a question previously. (I am unable to find it's link somehow). Also the second step has a '$>$' instead of a '$\geq$' because both inequalities do not hold simultaneously.
How to prove the tight bound?
[I had tried Jensen's inequality to $\frac{1}{\sqrt{x}}$ and C-S, but it didn't help.]
| By Holder and Schur
$$\sum_{cyc}\frac{b+c}{\sqrt{a^2+bc}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{b+c}{\sqrt{a^2+bc}}\right)^2\sum\limits_{cyc}(b+c)(a^2+bc)}{\sum\limits_{cyc}(b+c)(a^2+bc)}}\geq$$
$$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(b+c)\right)^3}{\sum\limits_{cyc}(b+c)(a^2+bc)}}=\sqrt{\frac{8(a+b+c)^3}{2\sum\limits_{cyc}(a^2b+a^2c)}}=$$
$$=2\sqrt{\frac{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}{\sum\limits_{cyc}(a^2b+a^2c)}}\geq2\sqrt{\frac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{\sum\limits_{cyc}(a^2b+a^2c)}}\geq4.$$
I used the following Schur.
For positives $a$, $b$ and $c$ prove that
$$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0.$$
In our case
$$\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=$$
$$=\sum_{cyc}(a^3-a^2b-a^2c+abc+4a^2b+4a^2c+abc)\geq$$
$$\geq\sum_{cyc}(4a^2b+4a^2c+abc).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What's the Jordan form of $J^2$? $$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$
What's the Jordan form of $J^2$?
I know that it has two blocks and $2$ independent eigenvectors.
So it could be
$$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}
\text{ or }
\begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$
But why can't the Jordan chain have the length of $4$ and $1$?
| Nice question! Here's another perspective:
You have an ordered basis $\{v_1,v_2,v_3,v_4,v_5\}$ for $\mathbb{R}^{5}$ such that $Jv_{5} = v_{4}$, $Jv_{4} = v_{3}$, $Jv_3 = v_{2}$, $Jv_2 = v_{1}, Jv_{1} = 0$. This is called a cycle of generalized eigenvectors.
If you consider $J^{2}$, you get two cycles:
$$J^{2}v_{5} = v_{3}; \quad J^{2}v_{3} = v_{1}; \quad J^{2}v_{1} = 0$$
$$J^{2}v_{4} = v_{2}; \quad J^{2}v_{2} = 0$$
One cycle has length 3 and the other has length 2, meaning the Jordan form has a block of size 3 and a block of size 2. Also, $\{v_1,v_3,v_5,v_2,v_4\}$ is a Jordan basis for $J^{2}$.
| {
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"answer_id": 0
} |
Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$ How should I approach this type of a system equation?
$$
\begin{cases}
\dfrac xy-\dfrac yx=\dfrac{15}4\\
2x-5y=9
\end{cases}
$$
I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I got this system:
$$
\begin{cases}
4x^2 - 4y^2 = 15xy\\
x - 2.5y = 4.5 \Longrightarrow x = 4.5 + 2.5y
\end{cases}
$$
Then I put $x$ from the second equation in the first one:
$$4(4.5 + 2.5y)^2 - 4y^2 = 15y(4.5 + 2.5y)$$
When I solved it I got these results:
*
*$$x_1 = \frac{189}{22},\ y_1 = \frac{18}{11}.$$
*$$x_2 = -3,\ y_2 = -3.$$
But these results are incorrect.
These are the answers from my book:
*
*$$x_1 = 12,\ y_1 = 3.$$
*$$x_2 = \frac{9}{22},\ y_2 = -\frac{18}{11}.$$
| The substitution is correct. Continuing:
\begin{align*}
4(4.5 + 2.5y)^2 - 4y^2 &= 15y(4.5 + 2.5y) \\
(25 y^2 + 90 y + 81) - 4y^2 &= 37.5 y^2 + 67.5 y \\
-16.5 y^2 + 22.5 y + 81 &= 0 \\
-33 y^2 + 45 y + 162 &= 0 \\
11 y^2 - 15 y - 54 &= 0 \\
(11 y + 18) (y - 3) &= 0
\end{align*}
For $y = -18/11$, we obtain $x = 9/22$. For $y = 3$, we obtain $x = 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the Taylor series about $z=0$ and the Laurent series about $z=-3$
*
*Let $f(z)=\frac{10z}{z^2+z-6}$, find the coefficient of $z$ in the Taylor series of $f$ expanded about $z=0$ and state the open set in $\mathbb C$ where the series converges.
*Find the Laurent series of $f$ about $z=-3$ and state the open set where this series converges.
*
*For the first part:
$$\begin{align}f(z)&=\frac{6}{z+3}+\frac{4}{z-1}\\
&=\frac{6}{3}\frac{1}{1+\frac{z}{3}}-4\frac{1}{2-z}\\
&=2\frac{1}{1-(-\frac{z}{3})}-2\frac{1}{1-\frac{z}{2}}\end{align}$$
So Taylor series would be
$$2\sum_{n=0}^{\infty}\left(-\frac{z}3\right)^n-2\sum_{n=0}^{\infty}\left(\frac{z}2\right)^n$$
Where $|-\frac{z}{3}|<1$ and $|\frac{z}{2}|<1$, so $|z|<3$ and $|z|<2$, is this correct so far and how do I deduce the open set from this? Would it be $B_3(0)\setminus B_2(0)$?
*
*For the second part:
Let $z=-3+w$ and input into $f$,
$$\begin{align}f(z)&=\frac{6}{(-3+w)+3}+\frac{4}{(-3+w)-2} \\
&=\frac{6}{w}+\frac{4}{-5+w}\\
&=\frac{6}{1-1+w}+\frac{-4}{5-w}\end{align}$$
....and then I get a little lost from there.
Any corrections, tips or solutions would be great!!
| HINT: let $g(z):=\frac1{z-1}$. Then there are two distinct Laurent expansions of $g$ around $z=-3$ depending of the considered domain.
Observe that
$$g(z)=\frac1{z-1}=\frac1{(z+3)-4}=\begin{cases}\frac1{z+3}\cdot\frac1{1-\frac4{z+3}}=\frac1{z+3}\sum_{k=0}^\infty\left(\frac4{z+3}\right)^k,& |z+3|>4\\
-\frac14\cdot\frac1{1-\frac{z+3}4}=-\frac14\sum_{k=0}^\infty\left(\frac{z+3}4\right)^k,&|z+3|<4\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get the value of $a^{-3} - b^{-3}$ I have a statement that says:
If $a - b = 3$, and $a$ is the reciprocal of $b$, that is equal to $a = \frac{1}{b}$, get the value of $a^{-3} - b^{-3}$
So i have,
$a = 3 + b$, $b = a - 3$, $a = \frac{1}{b}$
$3 + b = \frac{1}{b}$, then ($b^2 + 3b - 1 = 0$ )
or
$a = \frac{1}{a-3}$, then ($a^2 -3a -1 = 0$ )
So, i need the value of $a^{-3} - b^{-3}$
Now i use the Quadratic formula for $a$:
$a = \frac{3 \pm\sqrt{13}}{2}$, then $a^-3 = \frac{8}{(3\pm\sqrt{13})^3}$
Now i use the Quadratic formula for $b$:
$b = \frac{-3 \pm\sqrt{13}}{2}$, then $b^-3 = \frac{8}{(-3\pm\sqrt{13})^3}$
Finally,
= $\frac{8}{(3\pm\sqrt{13})^3} - \frac{8}{(-3\pm\sqrt{13})^3}$, I will use $+$ instead of the $\pm$, so
= $\frac{8}{(3+\sqrt{13})^3} - \frac{8}{(-3+\sqrt{13})^3}$
= $\frac{8}{40\sqrt{13}+144} - \frac{8}{40\sqrt{13}-144}$
= $-36$
In fact, the result is correct but the development, has been very long, for a question that I am supposed to answer in less than 2 minutes, So, what other way do you see to solve this exercise? I've been looking for it on my own, but I do not clarify with another form.
| Since $a=1/b$ and $b=1/a$ we get
$$a^{-3}-b^{-3} = b^3-a^3 = (b-a)(b^2+ab+a^2)=-3((a-b)^2+3ab) = -3(9+3)=-36$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate without L'Hopital: $\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$ Evaluate the following limit without using L'Hospital method
$$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$
My turn is
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{(x^{14}-1)-(x^9-1)-(x^5-1)}\right\}$$
then divide the numerator and the denominator by $$x-1$$
But I got again $$\frac{0}{0}$$
| Noting
$$ x^9-1=(x-1)\sum_{n=0}^8x^n, x^5-1=(x-1)\sum_{n=0}^4x^n $$
one has
\begin{eqnarray}
&&\lim_{x\to1}\frac{9}{x^9-1}-\frac{5}{x^5-1}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4x^n-5\sum_{n=0}^8x^n}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4(x^n-1)-5\sum_{n=0}^8(x^n-1)}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4\sum_{k=0}^n(x-1)x^k-5\sum_{n=0}^8\sum_{k=0}^n(x-1)x^k}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4\sum_{k=0}^nx^k-5\sum_{n=0}^8\sum_{k=0}^nx^k}{\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\frac{9\sum_{n=0}^4(n+1)-5\sum_{n=0}^8(n+1)}{5\times9}\\
&=&-2.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
solution to recursive equation $f(n)=2^{n-1}-f(n-1)$ how do I solve this recursive equation:
$f(n)=2^{n-1}-f(n-1)$ when $f(0)=1$
I tried the iteration method but got to a series with changing +/- signs which I had hard time solve.
| One approach via generating functions:
$$\begin{align}
G(x) &= \sum_{n=0}^{\infty} f(n) x^n \\
&= 1x^0 + \sum_{n=1}^{\infty} f(n) x^n \\
&= 1 + \sum_{n=1}^{\infty} 2^{n-1}x^n-\sum_{n=1}^{\infty}f(n-1) x^n \\
&= 1 + x\sum_{n=0}^{\infty} 2^{n}x^n-x\sum_{n=0}^{\infty}f(n) x^{n} \\
&= 1 + \frac{x}{1-2x}-xG(x) \\
&= \frac{x - 1}{2 x^2 + x - 1} \\
&= \frac{2}{3} \cdot \frac{1}{1 + x} + \frac{1}{3} \cdot \frac{1}{1 - 2 x}\end{align}$$
Implying:
$$f(n) = \frac{2}{3} \cdot (-1)^n + \frac{1}{3} \cdot 2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2749225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Simplify this simple expression? So the expression is $-(1/3)(1/2)^{3/2} + 1/3 - (1/3)(1/\sqrt{2})^3$.
The answer in my book is $1/3(1 - 1/\sqrt{2})$. But that simplifies to $1/3 - 1/3\sqrt{2}$. How come it isn't $1/3 + 2(-1/\sqrt{2})$, which is what I got.
| $ −\frac 13(\frac 12)^{\frac 32}+\frac 13 −\frac 13(\frac 1{\sqrt 2})^3=$
$\frac 13[-(\frac 12)^{\frac 32}+1 - (\frac 1{\sqrt 2})^3]=$
$\frac 13[-(\frac 12)^{\frac 32}+1 - (\frac 1{2})^{\frac 32}]=$
$\frac 13[1 - 2*(\frac 1{2})^{\frac 32}]=$
$\frac 13[1 - 2*(\frac 1{2})^{1+\frac 12}]=$
$\frac 13[1 - 2*(\frac 1{2})^{1}(\frac 12)^{\frac 12}]=$
$\frac 13[1- \frac 12^{\frac 12}]=$
$\frac 13 - \frac 1{3\sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $xyz$
$x$, $y$, $z$ are real numbers such that
$①$ $ x+y+z=\sqrt{3}$
$②$ $xy+yz+zx=1$
In this case, find the value of $xyz$.
$$$$
Here's my attempt.
$$$$
$3=(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = x^2 + y^2 + z^2 + 2$
$\therefore \ \ x^2 + y^2 + z^2 = 1 $
$\\$
$ x^3 +y^3 + z^3 - 3xyz=(x+y+z)(x^2 + y^2 + z^2 - xy-yz-zx) = 0$
$\therefore \ \ x^3 + y^3 + z^3 = 3xyz $
$\\$
What can I do... after these?
$$$$
I guessed the answer like this.
If $ x=y=z = \frac{\sqrt{3}}{3}$ , then these satisfy $①, ②$.
So, $\ $ $xyz=\frac{\sqrt{3}}{9}$....
Thanks in advance.
| If $x,y,z$ are the roots of the equation $x^3 + ax^2 + bx + c = 0$ then $a = -\sqrt 3$ and $b = 1$.
So, we want $x^3 - \sqrt 3 x^2 + x + c = 0 \equiv C$ to have all real roots.
This is nicely described by the cubic determinant $$
D_C = a^2b^2 - 4b^3 - 4a^3c - 27c^2 + 18abc
$$
being greater than or equal to $0$.
In our case, $D_C = 3 - 4 + 12\sqrt 3c - 27c^2 - 18\sqrt 3 c = -27c^2 - 6\sqrt 3 c- 1$.
It turns out that $D_C = -27\left(c + \frac 1{3\sqrt 3}\right)^2$, giving $c = \frac {-1}{3 \sqrt 3}$.
Hence the product of the roots is $-c = \frac{1}{3 \sqrt 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Range of a function, with contradictory restriction I have this function:
$$f(x)=\frac{x}{x^2+1}$$
it's clear that the domain is all real numbers, because the denominator never become zero. I've done the following procedure to found the range of $f(x)$:
$\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \operatorname{Dom} f,y=f(x)} \right \} \\
\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},y=\dfrac{x}{x^2+1}} \right \} \\
\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},x=\dfrac{1\pm \sqrt{1-4y}}{2y}} \right \} \\
\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\left (1-4y^2 \right )\geq 0\, \wedge\, y\neq 0 } \right \}\\
\operatorname{Rec} f=[-\tfrac{1}{2},\tfrac{1}{2}]-\left \{ 0 \right \}$
But, the function is zero when $x=0$ $(f(0)=0)$, and the range of the function is $[-\tfrac{1}{2}, \tfrac{1}{2}]$. So I don't know where is my error in the procedure.
| $$f(x)=\frac{x}{x^2+1}$$
Since the function is an odd function $(f(x)=-f(-x))$, then the minimum value is equal to the opposite of the maximum value. We see that $f(x) \ge 0$ when $x \ge 0$, so we need to find the maximum value of $f(x)$ when $x \ge 0$.
Note that
$\dfrac 12 - f(x) = \dfrac 12 - \dfrac{x}{x^2+1}
= \dfrac{x^2+1-2x}{2(x^2+1)}
= \dfrac{(x-1)^2}{2(x^2+1)}$
So, when $x \ge 0$, we see that $\dfrac 12 - f(x) \ge 0$ and, when $x=1$, we see that $\dfrac 12 - f(x) = 0$.
Since $\dfrac 12 - f(x) \ge 0 \iff f(x) \le \dfrac 12$ and $f(1)= \dfrac 12$, we see that $\displaystyle \max_{x \ge 0} f(x) = \dfrac 12$. Hence, because $f(x)$ is an odd function, $\operatorname{Range} f(x) = \left[ -\dfrac 12, \dfrac 12 \right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Finding invert for complicated function I'm looking for an invert for this function (when x > 0): wolframalpha
Just wondering if there is any chance it exists..
Maybe some numerical methods could help finding it?
| Your function is
$f(x)
=log((x*x+9)/(x*x))*sqrt((3*x-x*x*atan(3/x))/atan(3/x))/(-2*x*atan(3/x)+6)\\
=\log\left(\dfrac{x^2+9}{x^2}\right)\dfrac{\sqrt{\dfrac{3x-x^2\arctan(3/x)}{\arctan(3/x)}}}{-2x\arctan(3/x)+6}\\
=\log\left(1+\dfrac{9}{x^2}\right)\dfrac{\sqrt{\dfrac{3x}{\arctan(3/x)}-x^2}}{-2x\arctan(3/x)+6}\\
=\log\left(1+(3/x)^2\right)\dfrac{\sqrt{\dfrac{3}{x\arctan(3/x)}-1}}{-2\arctan(3/x)+6/x}\\
=\frac12\log\left(1+(3/x)^2\right)\dfrac{\sqrt{\dfrac{3}{x\arctan(3/x)}-1}}{-\arctan(3/x)+3/x}\\
$
Letting
$3/x=y$, so $x = 3/y$,
$f(3/y)
=\frac12\log\left(1+y^2\right)\dfrac{\sqrt{\dfrac{y}{\arctan(y)}-1}}{-\arctan(y)+y}\\
=\frac12\log\left(1+y^2\right)\dfrac{\sqrt{\dfrac{y-\arctan(y)}{\arctan(y)}}}{y-\arctan(y)}\\
=\frac12\log\left(1+y^2\right)\sqrt{\dfrac{1}{\arctan(y)(y-\arctan(y))}}\\
=\dfrac{\log\left(1+y^2\right)}{2\sqrt{\arctan(y)(y-\arctan(y))}}\\
$
Letting $y = \tan(z)$,
$f(3/\tan(z))
=\dfrac{\log\left(1+\tan^2(z)\right)}{2\sqrt{z(\tan(z)-z)}}\\
$
Considering that
$\tan(z)-z$
can't be inverted
(at least I don't think it can),
there is,
in my opinion,
no hope of
inverting this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $\int_0^{\pi/2}\frac{1}{\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)}\,\mathrm{d}x$ I got stuck at this definite integral, any idea?
$$\int_0^{\pi/2}\frac{1}{\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)}\,\mathrm{d}x$$
Thank you.
| \begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin ^{2} x-\sin x \cos x+\cos ^{2} x} d x =& \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\tan ^{2} x-\tan x+1} d x \\
=& \int_{0}^{\infty} \frac{d t}{t^{2}-t+1}, \text { where } t=\tan x \\
=& \int_{0}^{\infty} \frac{d t}{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}} \\
=& \frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]_{0}^{\infty} \\
=& \frac{2 \pi}{\sqrt{3}}
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz? The question states:
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$
Prove that
$x^2yz + xy^2z + xyz^2 ≤ 1/3$
I have a proof of this relying on the fact that:
$x^2/3 +y^2/3 + z^2/3 \geq (x+y+z)^3/9 $ (A corollary of C-S I believe)
Is there an elementary proof without this fact (or C-S in general)?
| Yes, we can can get a sum of squares here.
We need to prove that
$$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or
$$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or
$$\sum_{cyc}(2x^4-2x^2y^2+6x^2y^2-6x^2yz)\geq0$$ or
$$\sum_{cyc}(x^4-2x^2y^2+y^4+3(x^2z^2-2z^2xy+y^2z^2))\geq0$$ or
$$\sum_{cyc}((x^2-y^2)^2+3z^2(x-y)^2)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
if $|az^2+bz+c|\le 1$, find the maximum of $|a|+|b|$ Let $a,b,c$ be complex numbers, and $f(z)=az^2+bz+c$, such that if the complex number $|z|\le 1$, then we have $|f(z)|\le 1$. Find the maximum of $|a|+|b|$.
if we $a,b,c$ be real and $z$ be real, I can find the maximum is $2$,because $f(z)=2z^2-1$ such it. and
$$a=\dfrac{1}{2}[f(1)+f(-1)]-f(0),b=\dfrac{1}{2}[f(1)-f(-1)]$$
so
$$|a|+|b|=\dfrac{1}{2}|f(1)+f(-1)-2f(0)|+\dfrac{1}{2}|f(1)-f(-1)|\le 2$$But for complex, maybe this answer is also $2?$,I'm not sure.
| By replacing $f(z)$ with
$$
\tilde f(z) = e^{i\phi} f(e^{i\psi} z) = e^{i(\phi+2\psi)}a z^2 + e^{i(\phi+\psi)}b z + c = \tilde a z^2 + \tilde b z + c
$$
with suitably chosen $\phi, \psi \in \Bbb R$ we can assume that both
$a$ and $b$ are real and $\ge 0$.
Now let $\omega = e^{i\pi/3} = \frac 12 + \frac i2 \sqrt 3$ be the primitive $6^{\text{th}}$ root of unity. Then $\omega^2 = \omega - 1$ and $ \omega^{10} = \omega^5 - 1$.
Therefore
$$
f(\omega) = (a+b)\omega + c - a \\
f(\omega^5) = (a+b)\omega^5 + c - a
$$
which implies
$$
a + b = \frac{f(\omega)- f(\omega^5)}{\omega - \omega^5}
= \frac{f(\omega)- f(\omega^5)}{i \sqrt 3} \, .
$$
Now use that $|f(z)| \le 1$ on the unit circle, this gives
the estimate
$$
|a| + |b| = a + b \le \frac{2}{\sqrt 3} \, .
$$
And this is the actual maximum. Credit for the following example
goes to achille hui: Let $p(z) = 2z^2+4z - 1$, then
$$
\begin{align}|p(e^{it})|^2 &= |2e^{it} - e^{-it} + 4|^2 = |\cos(t) + 4 + 3i\sin(t)|^2\\ &=(\cos(t)+4)^2 + 9\sin(t)^2 = 25 +8\cos(t)(1-\cos(t))\\ &\le 25 + \frac{8}{4} = 27 \, ,\end{align}
$$
so that
$$
f(z) = \frac{2z^2 + 4z - 1}{3\sqrt{3}}
$$
satisfies $|f(z)| \le 1$ on the unit circle and therefore – due to
the maximum principle – for all $z$ in the unit disk.
Remark: This is how I came up with above approach: In order to
compute $a+b$ from two equations
$$
f(z_1) = a(z_1^2-z_1) + (a+b)z_1 + c \\
f(z_2) = a(z_2^2-z_2) + (a+b)z_2 + c
$$
we need different $z_1, z_2$ with $z_1^2-z_1 = z_2^2-z_2$,
or $z_1 + z_2 - 1 = 0$. Also $z_1, z_2$ should be of absolute
value $\le 1$, and their difference as large as possible.
This eventually led to the choice $z_1 = \frac 12 + \frac i2 \sqrt 3$ and $z_2 = \frac 12 - \frac i2 \sqrt 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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If the range of $f(x)= \frac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\frac 56, \frac 32\right]$ then what is $c$?
If the range of the function $f(x)= \dfrac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\dfrac 56, \dfrac 32\right]$ then $c$ is equal to?
Attempt:
$y= \dfrac{x^2+x+c}{x^2+2x+c}$
For real values of $x$, $\Delta \ge 0$
$\implies 4y^2(1-4c) +1-4y(1-2c) - 4c \ge 0$
What do I do next? I am really unable to understand the concept to be followed after this. Could someone explain that?
The answer is:
$c= 4$
| Hint. Note that
$$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2}\quad\mbox{and}\quad \lim_{x\to \pm \infty}f(x)=1.$$
Moreover $f$ is bounded over the real line if and only if $x^2+2x+c=(x+1)^2+c-1>0$ that is $c>1$. In the bounded case, in order to determine the range of the continuous function $f$, consider the values of $f$ at $\pm \sqrt{c}$ (where the derivative $f'$ is zero):
$$f(\sqrt{c})=1-\frac{1}{2(\sqrt{c}+1)}<1<f(-\sqrt{c})=1+\frac{1}{2(\sqrt{c}-1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ using the Taylor series of $e^{x+2 }$.
Answer:
$$
e^{x+2}=1+(x+2)+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\ldots
$$
Integrating, we get
$$
\int e^{x+2} dx=x+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\ldots
$$
Again integrating, we get
$$
\int e^{x+2} dx=\frac{x^2}{2}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\frac{(x+2)^5}{5!}+\ldots\\
\Rightarrow \int e^{x+2} dx=\frac{x^2}{2}+(x+2)^3 \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}\\
\Rightarrow \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}=\frac{\int e^{x+2} dx-\frac{x^2}{2}}{(x+2)^3}
$$
Is this the sum of the series ?
| Your approach is sound, but you should be clear about one thing: when you write $\int e^{x+2}\,\mathrm dx$, what you mean is $\int_0^xe^{t+2}\,\mathrm dt$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I find an explicit formula, using backtracking, for $C_n =3C_{n-1} -2C_{n-2}$ with initial values $C_1=5$ and $C_2=3$? I have the recurrence $C_n =3C_{n-1} -2C_{n-2}$ with initial values $C_1=5$ and $C_2=3$, and want to find an explicit formula for $C_n$.
I made a table with the terms up to the $6$th term and noticed:
\begin{align}
C_3 &= C_2-4\\
C_4 &= C_3-8\\
C_5 &= C_4-16\\
C_6 &= C_5-32
\end{align}
But I still can't find the correct formula.
| Those four equations you wrote mean: $$C_n = C_{n-1} -2^{n-1}.$$
Then you can use the recurrence directly: \begin{align} C_n &= C_{n-1} -2^{n-1}\\ &=C_{n-2} -2^{n-2} -2^{n-1}\\ &=C_{n-3} -2^{n-3} -2^{n-2} -2^{n-1}\\ &\vdots\\ &= C_1 -\sum_{k=1}^{n-1} 2^k\\ &= C_1 \bbox[yellow]{+1} -\sum_{k=\bbox[yellow]{0}}^{n-1} 2^k\\ &= C_1 +1 -(2^n -1)\\ &= C_1 +2 -2^n \qquad\qquad(C_1=5)\\ &=7 -2^n.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why is $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$ defined at -1, 0 and not at 1 Why does the function:
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$
Is defined at $x = -1, x = 0$ (we will have $1/0$ in the fraction then) and not at $x = 1$ according to the following graph?
Live example
| $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$
$=\frac{\frac{2x}{(x-1)(x+1)}}{\frac{3x+1}{x(x+1)}}$
$=\frac{2x^2(x+1)}{(x-1)(x+1)(3x+1)}$
$=\frac{2x^2}{(x-1)(3x+1)}$
so $f(x)$ can be defined except at $x=1$ and $x=-\frac{1}{3}$. The apparent singularities at $x=0$ and $x=-1$ are removable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Express the sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ in terms of ${x_0, x_1, n}$ Given a sequence $\{x_n\}$ where $x_n = 2x_{n-1} + 3x_{n-2}$ how can one express it in terms of ${x_0, x_1, n}$. Can this be generalized for ${x_n = \alpha x_{n-1} + \beta x_{n-2}}$
I've tried to use the following approach:
$$\eqalign{
& 1x_0 = x_01 \\
& zx_1 = x_1z \\
& z^2x_2 = (2x_1 + 3x_0)z^2 \\
& z^3x_3 = (2x_2 + 3x_1)z^3 \\
& z^4x_4 = (2x_3 + 3x_2)z^4 \\
& ... \\
& z^nx_n = (2x_{n-1} + 3x_{n-2})z^n
}$$
Then sum LHS with RHS which will produce:
$${
x_0 + \sum\limits_{k = 1 }^na_nz^n = x_0 +zx_1 + 2\sum\limits_{k = 2}^nx_{n-1}z^n + 3 \sum\limits_{k = 2}^n x_{n-2}z^n
}$$
Let
$${
G(z) = x_0 + \sum\limits_{k = 1 }^na_nz^n
}$$
Then RHS may be expressed in terms of ${G(z)}$. For example
$${
2\sum\limits_{k = 2}^na_{n-1}z^n = 2z\sum\limits_{k = 2}^na_{n-1}z^{n-1} = 2z(\sum\limits_{k = 1}^na_{n}z^{n} + x_0 - x_0) = 2z(G(z) - x_0)
}$$
Applying those transformations I eventually got ${G(z)}$ expressed in terms of z and ${x_1, x_0}$. But at this point I got stuck.
I got a sum of fractions:
$${
\frac{3x_0 - x_1}{4(1+z)}+\frac{x_0+x_1}{4(1-3z)}
}$$
I guess i could expand the fractions into series and find their sum, but i am not supposed to know about such expansions at the point of the book i took the problem from.
All of the above feels like a wrong approach. So the question is whether this can be done in a more elegant way.
| Consider the matrix:
$$ X_n =
\begin{bmatrix}
x_{n+2} & x_{n+1} \\
x_{n+1} & x_{n}
\end{bmatrix}
$$
Then there is a recurrence relation:
$$X_{n+1} =
X_n
\begin{bmatrix}
2 & 1 \\
3 & 0
\end{bmatrix}
$$
So we may say that
$$X_{n} = X_0
\begin{bmatrix}
2 & 1 \\
3 & 0
\end{bmatrix}^{n} = X_0A^n$$
Through whatever method you'd like to use, realize the eigenvectors of $A$ are $(-1, 3)$ and $(1, 1)$, with values $-1$ and $3$. Thus through a change of basis,
$$A =
\begin{bmatrix}
-1 & 1 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 \\
0 & 3
\end{bmatrix}
\begin{bmatrix}
-1 & 1 \\
3 & 1
\end{bmatrix}^{-1}
=
\begin{bmatrix}
-1 & 1 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 \\
0 & 3
\end{bmatrix}
\begin{bmatrix}
-1/4 & 1/4 \\
3/4 & 1/4
\end{bmatrix}
$$
and
$$
A^n =
\begin{bmatrix}
-1 & 1 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 \\
0 & 3
\end{bmatrix}^n
\begin{bmatrix}
-1/4 & 1/4 \\
3/4 & 1/4
\end{bmatrix}
=
\begin{bmatrix}
-1 & 1 \\
3 & 1
\end{bmatrix}
\begin{bmatrix}
(-1)^n & 0 \\
0 & 3^n
\end{bmatrix}
\begin{bmatrix}
-1/4 & 1/4 \\
3/4 & 1/4
\end{bmatrix}
$$
Finally, let $x_0 = 0$ (or whatever, it don't matter) so that
$$
X_0=
\begin{bmatrix}
x_2 & x_1 \\
x_1 & x_0
\end{bmatrix}$$
Then multiplying, (the top-right and bottom-left elements are actually equal)
$$
X_n=
\begin{bmatrix}
\dfrac{3^{n+1}(x_2 + x_1) - (-1)^{n}(3x_1 - x_2)}{4} & \dfrac{3^{n}(x_2 + x_1) + (-1)^{n}(3x_1 - x_2)}{4} \\
\dfrac{3^{n+1}(x_1 + x_0) - (-1)^{n}(3x_0 - x_1)}{4} & \dfrac{3^{n}(x_1 + x_0) + (-1)^{n}(3x_0 - x_1)}{4}
\end{bmatrix}
$$
The bottom right element is $x_n$. Thus,
$$x_n = \frac{3^{n}(x_1 + x_0) + (-1)^{n}(3x_0 - x_1)}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
change boundary to theta $y=1-x^2$,
find length of curve in first quadrant
$\int ds= \int \sqrt{ (dx^2+dy^2)}=\int \sqrt{ (1+dy^2)}dx$
$\int_{0} ^1\sqrt{1+(-2x)^2}=\sqrt{1+4x^2}=\sqrt{4({\frac{1}{4}+x^2})}=2\sqrt{({\frac{1}{4}+x^2})}$
$x=\frac{1}{2}tan \theta$
$\frac{dx}{d\theta}=\frac{1}{2}cos \theta$
$\int_{0} ^{tan^-2}2\sqrt{({\frac{1}{4}+\frac{1}{4}tan^2\theta})}\frac{1}{2}sec^2 \theta d\theta$= $\frac{1}{2}sec ^3\theta d\theta$
is this right?
and also when i changed to $\theta$, is the boundary right?
i got very nice integral but the boundary is bad
| Sorry Vik, I was being too careless. The problem is first that your are choosing $x = (1/2)\sin(\theta)$. But this way, $x$ has a max of only $1/2$, when we need it to reach all the way to $1$. Plus, your integration is wrong. You are trying to say that $1+\sin^2(x) = \cos^2(x)$, when the correct expression is $1-\sin^2(x) = \cos^2(x)$. I would recommend that you use the Pythagorean identity with $\sec$ and $\tan$ for the trig substitution. $$1+\tan^2(\theta) = \sec^2(\theta)$$.
| {
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"answer_count": 1,
"answer_id": 0
} |
Why is $\frac{0.5}{ \cos^2(30)} = \frac{\tan(30)}{\cos(30)} $, but $\frac{0.5}{ \cos^2(13) } \neq \frac{\tan(13)}{\cos(13)} $? Is just it a coincidence that
$$\frac{0.5}{ \cos^2(30)} = \frac{\tan(30)}{\cos(30)} $$
However
$$\frac{0.5}{ \cos^2(13) } \neq \frac{\tan(13)}{\cos(13)} $$
is not equal ? And if not does anyone know a reason why they just so happen to be equatable?
| If we use $x$ as a variable, we get the following:
$$\begin{align}
& \frac {0.5}{\cos^2x} = \frac {\tan x}{\cos x} \\
\implies& \frac {1}{2} \cos x = \cos^2 x \tan x\\
\implies& \cos x \tan x\ = \frac {1}{2} \\
\implies& \cos x \frac {\sin x}{\cos x} = \frac {1}{2}\\
\implies& \sin x\ = \frac {1}{2}\
\end{align}$$
Thus the only solutions to the above equation are $x = 30^\circ \pm 360^\circ k$ or $x = 150^\circ \pm 360^\circ k$ where $k$ is any natural number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving inverse function problems The following is my (likely completely incorrect) attempt.
$$f(x)= 3x + 5, \qquad \qquad g(x)= 1 - 2x$$
*
*Show that $f{^{-1}}(2) < g(-3)$
$$\frac{1}{3\cdot 2+5} < (1-2\cdot(-3))\implies\frac{1}{11} < 7$$
*Find the value of $a$ such that $f(2a) = g{^{-1}}(a)-2$
$$6a+5 = \frac{1}{-1-2a}$$
$$6a = \frac{1}{4-2a}$$
$$8a = \frac{1}{4}\implies a = \frac{1}{32}$$
*Solve $\frac{f(x)}{2} + \frac{g(x)}{3} = -23$
$$\frac{3x+5}{2} + \frac{1-2x}{3} = -23$$
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23$$
$$2x+17 = \frac{-23}{6}\implies x = -\frac{125}{12}$$
| The first one is already stated in other answers.
For the second, the incorrect step is:
$$g(x)=1-2x\Rightarrow x=\dfrac{1-g(x)}{2}\Rightarrow g^{-1}(x)=\dfrac{1-x}{2}$$
So we need to solve $6a+5=\dfrac{1-a}{2}-2\Leftrightarrow 12a+14=1-a\Leftrightarrow a=-1.$
For the third, the incorrect step is:
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23 (2)$$
$$2x+17 = \frac{-23}{6}$$
This is because from $(2)$
$\Leftrightarrow 9x+15+2-4x=-23\times6$
$\Leftrightarrow 5x+17=-138$
$\Leftrightarrow x=-31$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Example 14, Chapter 1, Higher Algebra by Henry Sinclair. I stumbled across the question 14 in Higher Algebra by Henry Sinclair and S.R Knight:
If $$(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2$$
Show $$x:a=y:b=c:z$$
I started by expanding the trinomials on both sides:
$$a^2x^2+b^2y^2+c^2z^2 + (a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = a^2x^2 + b^2y^2 + c^2z^2 + (axby + axcz) + (byax + bycz) + (czax +czby)$$
That simplifies to $$(a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = (axby + axcz) + (byax + bycz) + (czax +czby)$$
Adding like terms, I got
$$(a^2y^2 + a^2z^2) + (b^2x^2 + b^2z^2) + (c^2x^2 + c^2y^2) = 2(axby + 2axcz + 2bycz)$$
Rearranging, I finally obtained:
$$(ay-bx)^2 + (az-cx)^2 + (bz-cy)^2 = 0$$
Now, I realized that if $\frac{x}{a} = \frac{y}{b}$ then $ay = bx$. It then follows that $ay - bx = 0$. This also applies for the other fractions.
However, in the context of this problem, we do not know that $x:a = y:b = z:c$. So, I'm not sure how I can proceed to prove it with just the equation above.
Any help would be appreciated.
| This is the Cauchy-Schwarz inequality,$$(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges How to prove that $$u_n=\sum_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$$
is a Cauchy sequence?
The exercise I am reading gives as a hint that we should use the inequality: $$0\leq\cos\left(\frac kn\right)^{2n^2/k}\leq e^{-k},$$
for all $k\leq n$.
I tried to estimate $|u_m-u_n|$, but I don't know how to deal with the $n$ inside the sum.
| First of all,
$\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}\leq e^{-k}
$
is equivalent to
$\cos\left(\frac{k}{n}\right)
\leq e^{-k^2/2n^2}
$
which follows from
$e^{-x}
\ge 1-x+x^2/2$
for $\frac12 \ge x \ge 0$
(so
$e^{-k^2/2n^2}
\ge 1-k^2/(4n^2)+k^4/(8n^4)
$)
and
$\cos(x)
\le 1-x^2/2+x^4/24$
(so
$\cos(k/n)
\le 1-k^2/(2n^2)+k^4/(24n^4)
$).
Note that we have to go
to the $x^4$ term.
Then
$u_n
=\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}
\le\sum_{k=1}^n e^{-k}
\lt \frac{1}{e-1}
$
so
$u_n$ is a bounded sequence.
However,
we have not yet shown
that
$u_n$ is increasing.
Instead
I will show that
$u_n
\to \frac{1}{e-1}
$
with too much computation.
We have
$\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}
\ge \left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k}}
= \left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k
$.
For $0 < x \le \frac12$,
$\begin{array}\\
-\ln(1-x)
&=\sum_{k=1}^{\infty} \dfrac{x^k}{k}\\
&=x+\sum_{k=2}^{\infty} \dfrac{x^k}{k}\\
&\lt x+\sum_{k=2}^{\infty} \dfrac{x^k}{2}\\
&=x+\dfrac{x^2}{2(1-x)}\\
&\le x+x^2/4\\
\end{array}
$
or
$\ln(1-x)
\ge -x-x^2/4
$.
Therefore,
for $0 < x \le \frac12$,
$(1-x)^{1/x}
=\exp((1/x)\ln(1-x))
\ge\exp(-(1/x)(x+x^2/4))
=\exp(-1-x/4)
$
so
$\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}
\ge \exp(-1-k^2/(8n^2))
$
so
$\left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k
\ge \exp(-k-k^3/(8n^2))
$
so
$u_n
=\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}
\ge\sum_{k=1}^n \exp(-k)\exp(-k^3/(8n^2))
$.
I will now split the sum
into two parts:
$[1, n^c]$
and
$(n^c, n]$
where
$0 < c < 1$.
If
$k > n^{c}$,
all the terms are positive,
so that sum is positive.
For the rest of $u_n$,
since
$e^{-x}
\gt 1-x
$,
$\begin{array}\\
\sum_{k=1}^{n^{c}} \exp(-k)\exp(-k^3/(8n^2))
&\gt \sum_{k=1}^{n^{c}} \exp(-k)(1-k^3/(8n^2))\\
&= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)k^3/(8n^2)\\
&= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)n^{3c-2}/8\\
&\gt \sum_{k=1}^{n^{c}} \exp(-k)-\frac{n^{3c-2}}{8}\sum_{k=1}^{n^{c}} \exp(-k)\\
\end{array}
$
and the first term
approaches
$\sum_{k=1}^{\infty} \exp(-k)
=\frac{1/e}{1-1/e}
=\frac{1}{e-1}
$
and the second term
is less than
$\frac{n^{3c-2}}{8(e-1)}
$
which goes to zero
if
$c < \frac23$.
Therefore
the sum approaches
$\frac{1}{e-1}$.
Whew.
That was harder than I
thought it would be.
Hope that it's correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
$p^{q+1}+ q^{p+1}$ is a perfect square. Find all the prime numbers $p$ and $q$ such that $p^{q+1} + q^{p+1}$ is a perfect square. I tried this but I can't imagine where should I start fromCan some one give me a hint please.
| Lemma: $a^2\equiv 0\text{ or }1 \mod 4$ for all $a\in\Bbb{Z}$.
In fact,
*
*if $a$ is even, then $a^2=(2k)^2=4k^2\equiv 0 \mod 4$
*if $a$ is odd, then $a^2=(2k+1)^2=4k^2+4k+1\equiv 1 \mod 4$.
Case 1: $p,q$ both odds.
Then $p+1$ and $q+1$ are even and $p^{q+1}\equiv q^{p+1}\equiv 1\mod 4$. Their sum can not be a square because is $2 \mod 4$.
Case 2: $p=q=2$.
Then $2^3+2^3=16=4^2$ is a square. Thus, $(p,q)=(2,2)$ is solution.
Case 3: wlog $p=2$ and $q$ is odd.
We have $2^{q+1}+q^3=t^2$. Because $q+1$ is even,
$q^3=t^2-2^{q+1}=\underset{A}{(t-2^{(q+1)/2})}\underset{B}{(t+2^{(q+1)/2})}$
Now, we have 2 subcases:
a) $q^3=B$ and $1=A$.
b) $q^2=B$ and $q=A$.
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction.
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$.
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus:
$3^{2(k+1)+1}+2^{(k+1)+2} = 3^{2k+3}+2^{k+3} = (3)(3)(3^{2k+1})+(2)(2^{k+2}) = 9(3^{2k+1})+2(2^{k+2})$
I feel like I'm almost there, if I could've factor 9 and 2 somehow I could say that $3^{2k+1}+2^{k+2}=7m$ for some integer $m$, but I can't find a way to do it. What am I missing? Or did I do a blunder somewhere along the road?
Thanks in advance.
| the trick to induction is put the $k+1$ expression in terms of the $k$ expression.
If $ 3^{2k+1} + 2^{k+2}=7m$ then
$3^{2(k+1)+1}+2^{(k+1)+2} = 9(3^{2k+1})+2(2^{k+2}) = 2[ 3^{2k+1} + 2^{k+2}] + 7*3^{2k+1}$
$= 2(7m) + 7*3^{2k+1} = 7(2m + 3^{2k+1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How do I show that a complex number $z$ is a root of $z^6-1$? This problem would be less confusing to me if one was not subtracted from $z^6$.
Here is the problem:
If $z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$, show that $z$ is a root of $z^6-1$.
My approach to this problem involved using de Moivre's theorem to convert $z$ to it's complex polar form. I think converting it to complex polar form first makes it easy to identify the root:
$$z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$$
$$a=cos(\frac{\pi}{3}), b=sin(\frac{\pi}{3})$$
$$r=\sqrt{cos(\frac{\pi}{3})^2+sin(\frac{\pi}{3})^2}$$
$$r=1$$
$$\theta=Tan^1(\frac{b}{a})$$
$$\theta=Tan^-1(\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})})$$
$$\theta=\frac{\pi}{6}$$
$$\therefore z=1\angle{\frac{\pi}{6}}$$
This part is the confusing bit. Because $1$ is subtracted from $z^6$, I decided to expand the whole thing and then show that $z$ is a root of $z^6-1$.
$$(cos(\frac{\pi}{3})+jsin(\frac{\pi}{3}))^6-1$$
$$(\frac{1}{2}+\frac{\sqrt{3}}{2}j)^6-1$$
Noting that $j^6=-1$...
$$(\frac{1}{2}-\frac{\sqrt{3}}{2})^6-1$$
The equation becomes...
$$(-\frac{13}{32})-1$$
Which equates to...
$$-\frac{45}{32}$$
Convert to complex polar form...
$$z^6=-\frac{45}{32}\angle{0rad}$$
$$z^6=(-\frac{45}{32}\angle{0rad})^6$$
$$z^6=(-\frac{45}{32})^6\angle{(0)}^6$$
$$z^6=(-\frac{45}{32})^6\angle{6(0+2\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(12\pi-11\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(\pi n)}$$
Noting that $z$ is raised to the first power, make $n=1$
$$z^6=(-\frac{45}{32})^6\angle{\pi (1)}$$
I get...
$$z^6=(-\frac{45}{32})^6\angle{\pi}$$
Clearly the complex polar form I got is not equal to that of $z$. Can someone lead me to the right path?
| Not really sure what our OP is trying to do here; what the intended method might be. But I'm pretty sure that
$\theta = \text{Tan}^{-1} \left ( \dfrac{\sin(\pi/3)}{\cos (\pi/3)} \right ) = \text{Tan}^{-1} ( \tan (\pi/3)) = \dfrac{\pi}{3} \ne \dfrac{\pi}{6}, \tag 1$
so it seems like the first wrong turn was made at this point. And I can't see the logical connection 'twixt
$\left ( \dfrac{1}{2} + j \dfrac{\sqrt 3}{2} \right)^6 \tag 2$
and
$\left (\dfrac{1}{2} - \dfrac{\sqrt3}{2} \right )^6; \tag 3$
so the OP's remarks from this point on are pretty much difficult for me to decipher.
Nevertheless, we do have de Moivre's formula,
$(\cos \theta + j\sin \theta)^n = \cos (n\theta) + j \sin(n\theta); \tag 4$
and if we set
$z = \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3}, \tag 5$
then
$z^6 = \left ( \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3} \right)^6 = \cos \left (6 \dfrac{\pi} 3 \right ) + j \sin \left (6 \dfrac{\pi}{3} \right) = \cos 2 \pi + j \sin 2\pi = \cos 2 \pi = 1, \tag 6$
whence
$z^6 - 1 = 0. \tag 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes. Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$
By induction: $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$. Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note: This argument was taken from Wikipedia.
(2) $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$
If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$
(3) For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$
From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$
(4) Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:
${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.
| Your thinking seems correct to me. Here's a similar approach: The cases $n=4,5$ can be checked by hand. For $n\geq6$ we have $n!\leq\left(\frac{n}{2}\right)^n$, which is easily proved by induction, and hence
$$\binom{3n}{n}=\frac{(3n)!}{(2n)!n!}=\prod_{i=k}^n\frac{2n+k}{k}=\prod_{k=1}^n\left(1+\frac{2n}{k}\right)>2^n\frac{n^n}{n!}\geq4^n,$$
where the first inequality comes from the fact that $1+\frac{2n}{k}>\frac{2n}{k}$ for all $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Set of equalities If $x, y$ and $z$ are natural numbers for which
$x^{2}-y = k^{2},$
$y^{2}-z = m^{2},$
$z^{2}-x = n^{2},$
(where $k, m, n$ are integers),
how can we prove that the only solution is $x=y=z=1$?
I have tried the basics, $x^{2}-k^{2}=y$ and $y^{2}=m^{2}+z$ etc but I can't get anywhere.
Any ideas are most welcome!
| I assume when you say $x,y,z \in \Bbb N$, you define $\Bbb N=\Bbb Z^+=\{ 1,2,3,\dots \} $.
WLOG, we can assume that $k,m,n≥0$. Then
$$ z^2-x=n^2 \implies x=(z-n)(z+n)≥z+n≥z$$
This is because $x>0 \implies \text{sgn}(z-n)=\text{sgn}(z+n)$. But $z+n>0$, so $z-n>0$. And since $z-n \in \Bbb Z$ it follows that $z-n≥1$ which is why the inequality holds.
Similarly, we have
$$y^2-z=m^2 \implies z=(y-m)(y+m)≥y+m≥y$$
Finally,
$$x^2-y=k^2 \implies y=(x-k)(x+k)≥x+k≥x$$
Combining the three results, we get
$$x≥z≥y≥x$$
which implies that $x=y=z$. Changing all variables to $x$:
\begin{align}
& x^2-x=k^2 \qquad x^2-x=m^2 \qquad x^2-x=n^2 \\
\implies & x^2-k^2=x^2-m^2=x^2-n^2=x \\
\implies & x=(x+k)(x-k)=(x+m)(x-m)=(x+n)(x-n) \\
\implies & x≥x+k \qquad x≥x+m \qquad x≥x+n \\
\implies & k≤0 \qquad m≤0 \qquad n≤0 \\
\implies & k=m=n=0 \\
\implies & x=y=z=1
\end{align}
Personally, I define the natural numbers to be $\Bbb N =\{ 0,1,2, \dots \} $.
But if one of $x,y$ or $z$ is $0$ (say $x=0$), then
$$y=x^2-k^2=0-k^2=-k^2 \implies y=0$$
$$z=y^2-m^2=0-m^2=-m^2 \implies z=0$$
So indeed the only extra solution we get is $x=y=z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Is this the best route to go in this implicit derivation problem? I have this equation:
$$e^{\frac{x}{y}} = x - y$$
I seem to be going down the wrong path. Is this right so far?
$$\frac{dy}{dx} = e^{\frac{x}{y}} \cdot \frac{dy}{dx} (x \cdot y^{-1}) = 1 - y'$$
$$e^{\frac{x}{y}} ( x \cdot -y^{-2} \cdot y' + y^{-1}) = 1 - y'$$
$$ -e^{\frac{x}{y}} x \cdot y^{-2} \cdot y' + e^{\frac{x}{y}} y^{-1}) = 1 - y'$$
WolframAlpha has this as
$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$
| Why is nobody suggesting logarithmic differentiation ?
First take log base $e$ of both sides to get
$$\frac{x}{y(x)}=\ln (x-y(x))\Rightarrow x=y(x)\cdot \ln (x-y(x)).$$
Now differentiate implicitly:
$$\begin{align}
1&=y\cdot \frac{1}{x-y}\cdot \left(1-\frac{dy}{dx}\right)+\ln(x-y)\cdot \frac{dy}{dx}
\\ \Rightarrow 1&=\frac{y}{x-y}-\frac{y}{x-y}\cdot \frac{dy}{dx}+\ln(x-y)\cdot \frac{dy}{dx}
\\ \Rightarrow \frac{dy}{dx}\left(\ln(x-y)-\frac{y}{x-y}\right)&=\frac{x-2y}{x-y}
\\ \Rightarrow \frac{dy}{dx}&=\frac{\frac{x-2y}{x-y}}{\ln(x-y)-\frac{y}{x-y}}
\\ \Rightarrow \frac{dy}{dx}&=\frac{x-2y}{(x-y)\ln(x-y)-y}.
\end{align}$$
OK, maybe now I know why...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
If $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$ , then $\triangle ABC$ is a right triangle
Show that the triangle whose angles satisfy the equality
$$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$$
is right-angled.
I've tried many times, but was unsuccessful.
| \begin{align}
\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}
&=2
.
\end{align}
There are known identities
for any $\triangle ABC$
with sides $a,b,c$, angles $A,B,C$
radius of inscribed circle $r$,
radius of circumscribed circle $R$
and semiperimeter $\rho=\tfrac12(a+b+c)$:
\begin{align}
a^2+b^2+c^2
&=
2\rho^2-8r\,R-2r^2
,\\
\cos A \cos B \cos C
&=\frac{\rho^2-(r+2R)^2}{4R^2}
.
\end{align}
\begin{align}
{\sin^2 A + \sin^2 B + \sin^2 C}
&=2(\cos^2 A + \cos^2 B + \cos^2 C)
,\\
{\sin^2 A + \sin^2 B + \sin^2 C}
&=2(1-\sin^2 A + 1-\sin^2 B + 1-\sin^2 C)
,\\
3(\sin^2 A + \sin^2 B + \sin^2 C)
&=6
,\\
\sin^2 A + \sin^2 B + \sin^2 C
&=2
,\\
4R^2\sin^2 A + 4R^2\sin^2 B + 4R^2\sin^2 C
&=2\cdot4R^2
,\\
a^2+b^2+c^2&=8R^2
,\\
2\rho^2-8r\,R-2r^2&=8R^2
,\\
\rho^2-(r+2R)^2&=0
,\\
\frac{\rho^2-(r+2R)^2}{4R^2}
=0
.
\end{align}
And since
\begin{align}
\cos A \cos B \cos C
&=\frac{\rho^2-(r+2R)^2}{4R^2}
,
\end{align}
one of
$\cos A,\cos B,\cos C$
must be 0,
hence, one of the angles must be $\tfrac\pi2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Elliptic Integral from Gradshteyn and Ryzhik In my course of work I came across the following elliptic integral, and found its solution in Gradshteyn and Ryzhik:
$$\int_{b}^{u} \frac{\mathrm{d} x}{\sqrt{(a-x)(x-b)(x-c)(x-d)}} = \frac{2}{\sqrt{(a-c)(b-d)}}\,F(\lambda, r),$$
where
$$\lambda = \arcsin \left(\sqrt{\frac{(a-c)(u-b)}{(a-b)(u-c)}}\right), \quad r = \sqrt{\frac{(a-b)(c-d)}{(a-c)(b-d)}},$$
for $a \geq u > b > c > d$.
Now, I trust Gradshteyn and Ryzhik that this is correct, but I am curious how to derive it. I know that the elliptic integral of the first kind can be defined as
$$F(\lambda, r) = \int_{0}^{\sin\lambda} \frac{\mathrm{d} t}{\sqrt{(1-r^{2} t^{2})(1-t^{2})}},$$
but I struggle to see how to reconcile this with the formula above, where one has:
$$\sqrt{(a-x)(x-b)(x-c)(x-d)} = \sqrt{(-ac+(a+c)x-x^{2})(bd-(b+d)x+x^{2})}.$$
If someone can give me a hint how to proceed with the derivation, I'd be very grateful.
| By transforming the variable from $x$ to $t$ using the relation,
\begin{equation}
\begin{split}
t=\sqrt{\frac{(a-c)(x-b)}{(a-b)(x-c)}}
\end{split}
\end{equation}
you will obtain
\begin{equation}
\begin{split}
&x=\frac{-(a-c)b+(a-b)c\,t^2}{-(a-c)+(a-b)\,t^2} \\
&dx=\frac{2(a-b)(a-c)(b-c)\,t}{[-(a-c)+(a-b)\,t^2]^2}\,dt
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)
\end{split}
\end{equation}
Inserting the above relation to the integral
\begin{equation}
\begin{split}
&\int^u_b\frac{dx}{\sqrt{(a-x)(x-b)(x-c)(x-d)}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{split}
\end{equation}
gives
\begin{equation}
\begin{split}
&\int^u_b\frac{dx}{\sqrt{(a-x)(x-b)(x-c)(x-d)}} \\
&=\frac{2}{\sqrt{(a-c)(b-d)}}
\int^{a_1}_{a_2}\frac{dt}{\sqrt{(1-t^2)\left[1-\frac{(a-b)(c-d)}{(a-c)(b-d)}t^2\right]}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)
\end{split}
\end{equation}
Transforming from $t$ to $\theta$ using the relation $t=\sin\theta$, the integral in (3) is expressed as
\begin{equation}
\begin{split}
&\frac{2}{\sqrt{(a-c)(b-d)}}
\int^{\alpha}_{\beta}\frac{d\theta}{\sqrt{1-k^2\,\sin^2\theta}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)
\end{split}
\end{equation}
Here
\begin{equation}
\begin{split}
&a_1=\sqrt{\frac{(a-c)(u-b)}{(a-b)(u-c)}} \\
&a_2=0 \\
&\alpha=\sin^{-1}\left[\sqrt{\frac{(a-c)(u-b)}{(a-b)(u-c)}}\right] \\
&\beta=0 \\
&k^2=\frac{(a-b)(c-d)}{(a-c)(b-d)} \\
\end{split}
\end{equation}
Therefore the integral becomes
\begin{equation}
\begin{split}
&\frac{2}{\sqrt{(a-c)(b-d)}}
\int^{\alpha}_{\beta}\frac{d\theta}{\sqrt{1-k^2\,\sin^2\theta}}=\frac{2}{\sqrt{(a-c)(b-d)}}\,F[\alpha,k]
\end{split}
\end{equation}
| {
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"url": "https://math.stackexchange.com/questions/2776577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use prime factorization.
The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}
$$
is a rational $\sqrt{2}$ with a smaller denominator.
The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.
I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the rational root theorem (which uses unique factorization
in its proof).
| If $\sqrt[3]2$ we rational, it would be $a/b$ for $a$, $b\in\Bbb N$. Then there is $n\in\Bbb N$ with $n\sqrt[3]2$ and $n\sqrt[3]4$ integers, say $n=b^2$. Let $n$ be the least positive integer with $n\sqrt[3]2$, $n\sqrt[3]4\in\Bbb N$. But $(\sqrt[3]2-1)n$ would be an even smaller one...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of real roots of a polynomial Find the number of real roots of the polynomial $f(x)=x^5+x^3-2x+1$
( I think similar questions have already been asked. However, I am not sure whether the exact question has already been raised by some user. I urge the members of this mathematical community to mark my question as duplicate if it has already been asked. I tried an attempt below:-)
What I attempted:- I know a little bit of theory of equations but not all. While I was in the examination, I did not ponder much more on any theorems relating to real roots. Rather I tried to do it with some intuition.
We have $f(1)=1$,$f(0)=1$,$f(2)=1$, $f(\frac{1}{2})=\frac{5}{32}$,$f(-\frac{1}{2})=\frac{59}{32}$,$f(-2)=-35$
It just helped me to trace a root between $-1$ and $-2$. Then I tried to study the nature of the graph.
Here $f'(x)=5x^4+3x^4-2$
Let $f'(x)=5x^4+3x^2-2=0$
$\Rightarrow 5y^2+3y-2=0$
(where $x^2=y$)
$\Rightarrow (5y-2)(y+1)=0$
Therefore, $y=\frac{2}{5}$, or
$y=-1$ (It is not of our interest, as we are dealing with real roots)
Therefore $x=\sqrt{\frac{2}{5}},-\sqrt{\frac{2}{5}}$
These two values divides the real line into three disjoint subsets $(-\infty,-\sqrt{\frac{2}{5}})$, $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$, $(\sqrt{\frac{2}{5}},\infty)$
It s quite easy to notice that $f$ is decreasing in $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$ and increasing in $(-\infty,-\sqrt{\frac{2}{5}})$, $(\sqrt{\frac{2}{5}},\infty)$
Again, $f(-\sqrt{\frac{2}{5}})=1.911$ and $f(\sqrt{\frac{2}{5}})=0.089$
Looking at all these details it is clear that $f(x)$ comes close to zero in $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$ but is not equal to zero. There is no possibility for $f(x)$ to be zero in $(\sqrt{\frac{2}{5}},\infty)$ as it is increasing here. Again, we have already discovered one root between $-1$ and $-2$. Apart from that there is of course no root as $f$ is again increasing in $(-\infty,-\sqrt{\frac{2}{5}})$.
So, te graph should look something like this
Therefore there is only one real root of the polynomial.
Is my approach correct? Is there any way to do the same thing using some more beautiful as well as advanced method ? (I have just made a rough plot of the graph in my answer sheet). I am not sure if the examiner would expect much better method than this one.
| Sturm's theorem is a very powerful tool used to count real roots, but may be overkill here since your method also works and is arguably simpler. We take $f(x)$ and $f^\prime(x)$ and then compute successive (negated) remainders $r_1(x),r_2(x),\dots$ of polynomial long division (similar to the Euclidean algorithm) to get the sequence of polynomials $f(x), f^\prime(x), r_1(x), \dots$.
So if $f(x) = x^5 + x^3 - 2x +1$ and $f^\prime(x) = 5x^4 + 3x^2 - 2$, then
$$f(x) = \frac{x}{5} \cdot f^\prime(x) - (-\frac{2}{5}x^3+\frac{8}{5}x - 1)$$
$$f^\prime(x) = (-\frac{25}{2}x)(-\frac{2}{5}x^3 + \frac{8}{5}x - 1) - (-23x^2 + \frac{25}{2}x + 2)$$
$$-\frac{2}{5}x^3 + \frac{8}{5}x - 1 = (\frac{2}{115}x + \frac{5}{529})(-23x^2 + \frac{25}{2}x + 2) - (-\frac{1531}{1058}x + \frac{539}{529})$$
$$-23x^2 + \frac{25}{2}x + 2 = (\frac{24334}{1531}x+\frac{5984577}{2343961})(-\frac{1531}{1058}x + \frac{539}{529}) - (+\frac{1409785}{2343961})$$
Now, for really large numbers, this sequence of polynomials gives the signs $+,+,-,-,-,+$, which has $2$ sign changes. For really large negative numbers, the sequence has the signs $-,+,+,-,+,+$, which has $3$ sign changes. The single additional sign change means that $f(x)$ has a single real root.
| {
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"url": "https://math.stackexchange.com/questions/2780189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find value of $\tan\big(\frac{\pi}{25}\big)\cdot \tan\big(\frac{2\pi}{25}\big)\cdots\tan\big(\frac{12\pi}{25}\big)$ Find value of:
$$\displaystyle \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg)$$
The solution I tried:
Assume
$$P = \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg),$$
with the help of $\tan(\pi-\theta)=-\tan \theta$, then
$$P=\tan\bigg(\frac{13\pi}{25}\bigg)\cdot \tan\bigg(\frac{14\pi}{25}\bigg)\cdot \tan\bigg(\frac{15\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{24\pi}{25}\bigg)$$
which gives
$$P^2=\prod^{24}_{r=1}\tan\bigg(\frac{r\pi}{25}\bigg).$$
How do I proceed from here?
| Consider the polynomial
$$f(t) = \frac{1}{2i}\left((1+it)^{25} - (1-it)^{25}\right) = t^{25} + \cdots + 25 t$$
When $t = \tan\theta$, we have
$$f(\tan\theta) = \frac{1}{2i}\left[\left(\frac{e^{i\theta}}{\cos\theta}\right)^{25} - \left(\frac{e^{-i\theta}}{\cos\theta}\right)^{25}\right]
= \frac{1}{\cos^{25}\theta}\sin(25\theta)$$
which vanishes at $\theta = 0, \pm \frac{\pi}{25},\ldots,\pm\frac{12\pi}{25}$.
This implies the $25$ roots of $f(t)$ are $0, \pm \tan\frac{\pi}{25},\ldots,\tan\frac{12\pi}{25}$.
Apply Vieta's formula to the coefficient of $t$ in $f(t)$ and notice $\tan\frac{\pi}{25}, \ldots, \tan\frac{12\pi}{25}$ are positive, we obtain:
$$\prod_{k=1}^{12}\left(-\tan^2\frac{k\pi}{25}\right) = (-1)^{24}25
\quad\implies\quad \prod_{k=1}^{12} \tan\frac{k\pi}{25} = \sqrt{25} = 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational.
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?
| Let $x^2 - 2 x = \frac{m}{n}$ for $m\in \mathbb{Z}, n\in \mathbb{N}$. We can solve this quadratic equation and obtain
$$
x = 1 \pm \sqrt{1 + \frac{m}{n}}
$$
Than we can substitute this $x$ this to $x^3 - 5x$
$$
x^3 - 5x = 1 \pm 3\sqrt{1 + \frac{m}{n}} + 3 \bigg(1 + \frac{m}{n}\bigg) \pm \sqrt{1 + \frac{m}{n}} \bigg(1 + \frac{m}{n}\bigg) -
$$
$$
- 5\bigg(1 \pm \sqrt{1 + \frac{m}{n}}\bigg) =
$$
$$
= -1 + \frac{3m}{n} + \sqrt{1 + \frac{m}{n}} \bigg(\pm 3 \pm 1 \pm \frac{m}{n} \mp 5\bigg)
$$
Since this number should be rational we must have either
$$
3+1+\frac{m}{n} - 5 = -1 + \frac{m}{n} = 0
$$
or
$$
-3-1-\frac{m}{n} + 5 = 1- \frac{m}{n} = 0.
$$
Both cases give us $\frac{m}{n} = 1$ and
$$
x^3 - 5 x = -1 + 3 \frac{m}{n} = 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Calculate the determinant of this $5 \times 5$ matrix
Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha
& \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0
& 0 & 1 & a^2
& b^2 \\ 0 & 0 & 0 & a
& b \\ 0 & 0 & 0 & -b
& a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$
I have trouble solving the determinant.. But what is immediately visible are those zeroes in the matrix, just one more zero is needed such that this matrix is a triangular matrix (the element $a_{54}$ must be zero for this but it is $-b$ instead). If it was zero we could just multiply the diagonal and the product would be our determinant.
I have tried various ways to form this matrix such that $a_{54}$ is zero but the way I formed harmed the matrix and I got a wrong determinant as solution : /
As example, I have multiplied row $4$ with $b$, multiply row $5$ with $a$ and then add row $4$ to row $5$. Because I multiplied row $5$ with $a$, I need to divide the determinant by $a$ at the end.
So then I have the matrix
$$\begin{pmatrix}
\sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab\\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2\\
0 & 0 & 1 & a^2 & b^2\\
0 & 0 & 0 & a & b\\
0 & 0 & 0 & 0 & a^2+b^2
\end{pmatrix}$$
$$\text{Thus }\det = \frac{\sin\alpha \cdot \sin\alpha \cdot 1 \cdot a \cdot (a^2+b^2)}{a}=\sin^2\alpha \cdot (a^2+b^2)$$
But this is wrong and I don't see how to get the correct determinant...?
| Using determinant of block matrix
$$\det\begin{pmatrix}B&C\\ 0& D\end{pmatrix}=\det(B)\det(D)$$
we get that the desired determinant is
$$\det\begin{pmatrix}\sin\alpha&\cos\alpha\\ -\cos\alpha& \sin\alpha\end{pmatrix}\det(1)\det\begin{pmatrix}a&b\\ -b& a\end{pmatrix}=a^2+b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Equation of a parabola, given the vertex and the axis I get totally stuck in this exercise:
Consider the set of parabolas with vertex in $V=(1,1)$ and, as axis, the line $y=2x-1$. Write the equation of the generic parabola of that set.
Thank you for your help!
| Consider a parabola $y-1=A(x-1)^2$ rotated clockwise by $\alpha$ about the point $(1,1)$.
The equation of the rotated parabola is
$$v=Au^2$$
where
$$\begin{align}
\left(\begin{array} .
u\\v\end{array}\right)
&=
\left(\begin{array}.\;\ \cos(-\alpha)&\sin(-\alpha)\\-\sin(-\alpha)&\cos(-\alpha)\end{array}\right)
\left(x-1\atop y-1\right)\\
&=\left(\begin{array}. \cos\alpha&-\sin\alpha\\\sin\alpha&\;\;\ \cos\alpha\end{array}\right)
\left(x-1\atop y-1\right)\\
&=\left(\begin{array}. \sin\theta&-\cos\theta\\\cos\theta&\;\;\ \sin\theta\end{array}\right)
\left(x-1\atop y-1\right)
&&\scriptsize(\alpha+\theta=\tfrac\pi 2)\\
\end{align}$$
and $\tan\theta=2$ (slope of $y=2x-1$).
Hence equation of rotated parabola is
$$\begin{align}
(x-1)\cos\theta+(y-1)\sin\theta
&=A\big[(x-1)\sin\theta-(y-1)\cos\theta\big]^2\\
\tfrac 1{\sqrt{5}}(x-1)+\tfrac 2{\sqrt{5}}(y-1)
&=A\big[\tfrac 2{\sqrt{5}}(x-1)-\tfrac 1{\sqrt{5}}(y-1)\big]^2\\
\tfrac 1{\sqrt{5}}\big[(x-1)+2(y-1)\big]&=\tfrac 15 A\big[2(x-1)-(y-1)\big]^2\\
x+2y-3
&=\tfrac 1{\sqrt{5}}A\big[2x-y-1\big]^2\\
\color{red}{x+2y-3}
&\color{red}{=A'\big[2x-y-1\big]^2}\\
\end{align}$$
where $A, A'$ are constants.
See Desmos implementation here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| You made a little mistake
$$\int \frac{1}{(x^2+1)^2} \, dx = \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\$$
It should be $\sin(2u)$ and not $\sin(u)$
$$\displaystyle \int \cos(2u)du=\frac {\sin(2u)}2+K$$
And
$$\sin(2u)=\frac {2\tan(u)}{1+\tan^2(u)}=\frac {2x}{1+x^2}$$
$$u=\arctan(x) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
How to evaluate this integral $\int_0^1 \frac{x\log ^2(\sqrt{x^2+1}+1)}{\sqrt{1-x^2}} \, dx$ How to evaluate $$A=\int_0^1 \frac{x \log ^2\left(\sqrt{x^2+1}+1\right)}{\sqrt{1-x^2}} \, dx$$See the details here from a similar question: Evaluating $\int_0^1 \frac{z \log ^2\left(\sqrt{z^2+1}-1\right)}{\sqrt{1-z^2}} \, dz$.
By applying the same process, I find:
$$A=\int_0^1\ln^2\left(\sqrt{2-y^2}+1\right)dy=\int_1^{\sqrt2}\frac {x\log^2 (x+1)}{\sqrt{2-x^2}}dx=\sqrt2 \int_{0}^{\pi/4}\cos\varphi\ln^2\left(\sqrt2 \cos\varphi+1\right)d\varphi.$$
$$A=\ln^22+32(1+\sqrt2)\int_{0}^{\sqrt2-1}\frac{t^2}{\left(1+t^2\right)^2(t+1+\sqrt2)(-t+1+\sqrt2)}\ln\left(\sqrt2\, \frac{1-t^2}{1+t^2}+1\right)dt.$$
Now,put $$A=\int_0^1 \frac{x \log ^2\left(\sqrt{x^2+1}+1\right)}{\sqrt{1-x^2}} \, dx, B=\int_0^1 \frac{x \log ^2\left(\sqrt{x^2+1}-1\right)}{\sqrt{1-x^2}} \, dx.$$
We have:
$$A-B=4\int_0^{1}\frac {x\log x\log(\sqrt{1+x^2}+1)}{\sqrt{1-x^2}}dx-4\int_0^{1}\frac {x\log^2 x}{\sqrt{1-x^2}}dx$$
$$A+B=-2\int_0^{1}\frac {x\log (\sqrt{1+x^2}-1)\log(\sqrt{1+x^2}+1)}{\sqrt{1-x^2}}dx+4\int_0^{1}\frac {x\log^2 x}{\sqrt{1-x^2}}dx$$
$$\int_0^{1}\frac {x\log^2 x}{\sqrt{1-x^2}}dx=\ln^22-\frac{\pi^2}{12}-2\ln2+2.$$
But how to deduce these integrals.(using Mathematica?)
| I can at least get you going on the last one. Let
$$I(p)=\int_0^1 \frac{x^p}{\sqrt{1-x^2}}dx$$
Then we have
$$\begin{align}
I(p)
&=\int_0^1 \frac{x^p}{\sqrt{1-x^2}}dx\\
&=\frac{1}{2}\int_0^1 \frac{x^{\frac{p-1}{2}}}{\sqrt{1-x}}dx\\
&=\frac{1}{2}\mathrm{B}\bigg(\frac{p+1}{2},\frac{1}{2}\bigg)\\
&=\frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{p+1}{2})}{\Gamma(\frac{p+2}{2})}\\
\end{align}$$
where $\mathrm{B}$ is the Beta function and $\Gamma$ is the Gamma function. This gives us
$$I''(1)=\int_0^1 \frac{x\ln^2(x)}{\sqrt{1-x^2}}dx=\frac{\sqrt{\pi}}{2} \frac{d^2}{dp^2}\frac{\Gamma(\frac{p+1}{2})}{\Gamma(\frac{p+2}{2})}$$
You can easily finish off this double derivative by brute force (or Mathematica, as you proposed) using a special value of the Digamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
A specific question about a factor theorem proof I know there are tons of answers to questions about the factor theorem, however, I couldn't find what I was looking for so apologies if someone has already answered this before.
Since
$\frac{x^k - y^k}{x-y} = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$
${x^k - y^k} = (x-y)(x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1})$
hence ${x^k - y^k} = (x-y)q_k(x)$
As usual, we write our polynomial $ p$ as
$p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 $
$p(x)-p(y) = $
$a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 $
$ -(a_ny^n + a_{n-1}y^{n-1} + ... + a_1y + a_0)$
$= a_n(x^n-y^n) + a_{n-1}(x^{n-1}-y{n-1}^) + ... + a_1(x-y)$
The terms have the form $a_k(x^k-y^k)$. But $x^k-y^k = (x-y)q_k(x)$, and substitute this in:
$p(x)-p(y)=$
$a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)$
$(x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1)$
$= (x-y)q(x)$ where $q(x) = a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1$
If $p(a) = 0$ and $y = a$ then
$p(x)-p(a) = (x-a)q(x) $ Q.E.D
My question is: why we don't write $p(x) - p(y)$ as
$a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)q_1(x)$
$(x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1q_1(x))$.
NOTE! That I'm asking whether or not $a_1q_1(x) = a_1$
Can anyone please explain this to me thanks!
| $q_1$ = 1
$q_k(x) = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$
Therefore, letting $k = 1$,
$q_1(x) = x^{1-1}y^0 = x^0y^0 = 1$
So they are right. Rather than write $q_1(x)$, they can just drop it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing the surface integral of the octant of a sphere with polar coordinate substitution Let me first describe where I start:
$$\iint_Sz^2\,dS$$
We want to compute the surface integral of the octant of a sphere $S$.
The radius = 1.
The sphere is centered at the origin.
$$S=x^2+y^2+z^2=1.$$
$$z=f(x,y)=\sqrt{1-x^2-y^2}$$
$R$ is the projection of $S$ on the $xy$-plane.
Now we compute the normalization factor used to project the integral on the $xy$-plane. First we compute the derivative $\frac{\partial f}{\partial x}$:
If
$$m=1-x^2-y^2$$
and
$$n=\sqrt{m}\,,$$
then the derivative of $n$ is
$$n'=\frac12m^{-\frac12},$$
and the derivative of $m$ (with regard to $x$) equal to:
$$m'=-2x.$$
Now we can compute $\frac{\partial f}{\partial x}$ using the chain rule:
$$\frac{\partial f}{\partial x}=n'\cdot m'=\frac12m^{-\frac12}\cdot m'=\frac12(1-x^2-y^2)^{-\frac12}\cdot-2x=-{\frac{x}{\sqrt{(1-x^2-y^2)}}}.$$
Since $\sqrt{(1-x^2-y^2)}=z$:
$$\frac{\partial f}{\partial x}=-{\frac xz}$$
And in the same manner (using the derivative of $m$ with regard to $y$) we can calculate $\frac{\partial f}{\partial y}$:
$$\frac{\partial f}{\partial y}=-{\frac yz}$$
Now we can calculate the normalization factor:
$$\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}=\sqrt{1+\frac{x^2}{z^2}+\frac{y^2}{z^2}}=\frac1z\sqrt{x^2+y^2+z^2}.$$
Since $x^2+y^2+z^2=1$:
$$\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}=\frac1z$$
Now we can calculate the projection $R$ of the sphere $S$ on the $xy$-plane:
$$\iint_Sz^2\,dS=\iint_R z^2\frac1z \,dx\,dy=\iint_R z\;dx\,dy.$$
Substituting for $z$ finishes the conversion of the surface integral (remember $z=f(x,y)=\sqrt{1-x^2-y^2}$):
$$\iint_Sz^2dS=\iint_R \sqrt{1-x^2-y^2}\;dx\,dy.$$
This is where my problem starts:
The book I'm reading says if we convert this to polar coordinates, the integration should be trivial.
So we convert to polar coordinates:
$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$z=f(x,y)=f(r\cos\theta,r\sin\theta)$$
Calculate the Jacobian determinant:
$$\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\\ \end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r\,(\cos^2\theta + \sin^2\theta)=r$$
Substitute:
$$\iint_Sz^2dS=\iint_R \sqrt{1-x^2-y^2}\;dx\,dy=\iint_T\sqrt{1-r^2\cos^2\theta-r^2\sin^2\theta}\cdot r \cdot dr\,d\theta$$
$$\iint_T\sqrt{1-r^2(\cos^2\theta+\sin^2\theta)}\cdot r \cdot dr\,d\theta = \iint_T\sqrt{1-r^2}\cdot r \cdot dr\,d\theta=\iint_T\sqrt{r^2-r^4}\cdot dr\,d\theta$$
Since the radius $r=1$ it is easy to see that:
$$\int^1_0\sqrt{r^2-r^4}\cdot dr=\int^1_0(r^2-r^4)^{\frac12}\cdot dr = \int^1_0 \frac{(r^2-r^4)^{\frac32}}{\frac32}=0$$
According to the book the result of the calculation of the surface of the sphere in the first octant should be $\pi/6$.
That won't happen if $\int^1_0\sqrt{r^2-r^4}\cdot dr=0$.
The domain of $\theta$ is:
$$0\le\theta\le\frac12\pi$$
So where am I going wrong?
| It seems to me that you are using the power rule
$$\int (r^2 - r^4)^{1/2} dr= \frac{1}{3/2} (r^2 - r^4)^{3/2} +C,$$
but this is incorrect. Indeed if you differentiate the right hand side with respect to $r$, you do not get $(r^2 - r^4)^{1/2}$, but instead $(r^2 - r^4)^{1/2} (2r - 4r^3)$.
As noted in the comment, you can use substitution $u = 1-r^2$ to get
$$\int_0^1 \sqrt{1-r^2} rdr= -\frac 12 \int_1^0 \sqrt u du.$$
(we have $\int_1^0$ instead of $\int_0^1$ on the right hand side since you have to plug in the value accordingly: for $r=0$, you have $u= 1-r^2 = 1$, and for $r=1$ you get $u = 1-r^2 = 0$).
So you have
$$\int_0^1 \sqrt{1-r^2} dr = -\frac 12 \int_1^0 \sqrt u du = -\frac 13 u^{3/2}\bigg|_1^0 = \frac 13 $$
and thus $\int z^2 dS = \frac \pi 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
| By parts directly:
$$\begin{cases}u=\arcsin\sqrt x,&u'=\frac1{2\sqrt x\sqrt{1-x}}\\{}\\
v'=1,&v=x\end{cases}\;\;\implies \int\arcsin\sqrt x\,dx=x\arcsin\sqrt x-\frac12\int\sqrt\frac x{1-x}\,dx$$
and now substitute
$$\;u^2=\frac x{1-x}\implies x(-u^2-1)=-u^2\implies x=\frac{u^2}{u^2+1}=1-\frac1{1+u^2}\implies$$
$$ dx=\frac{2u}{(1+u^2)^2}\;$$
and from here your integrals equals
$$x\arcsin\sqrt x-\int\frac{u^2}{(1+u^2)^2}\,du=x\arcsin x-\frac12\left(\arctan u-\frac u{1+u^2}\right)$$
and now go back to $\;x\;$ and etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Trigonometric Inequality $\sin (2x) \gt \sqrt 2 \sin (x)$ I wish to solve this inequality:
$\sin (2x) \gt \sqrt 2 \sin (x)$
My approach:
I tried to isolate the $x$ on the left side by using the sine sum formula:
$2\sin(x)\cos(x) \gt \sqrt2\sin(x)$
then I divided by $\sin(x) \over 2$ both sides:
$\cos(x) \gt {\sqrt2 \over2}$
$x \lt \cos^{-1}({\sqrt2 \over2})$
$x < {\pi \over4}$
From that I can conclude that $x < {7\pi \over4}$, but I know the answer is still incomplete as it should be
$0 \lt x \lt {\pi \over 4}$, $\pi \lt x \lt {7 \over 4}\pi$
As I was able to see on Desmos graph plotter.
Does my approach gives the tools to reach this answer or have I commited a mistake?
| You should not have divided by $\sin x$. That causes you to lose information you need to solve the problem.
\begin{align*}
\sin(2x) & > \sqrt{2}\sin x\\
2\sin x\cos x & > \sqrt{2}\sin x\\
2\sin x\cos x - \sqrt{2}\sin x & > 0\\
\sqrt{2}\sin x(\sqrt{2}\cos x - 1) & > 0
\end{align*}
The inequality is satisfied if $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ or if $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$.
Let's focus on solving the problem in the interval $[0, 2\pi)$ for the moment.
The inequality $\sin x > 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (0, \pi)$. The inequality $$\sqrt{2}\cos x - 1 > 0 \iff \cos x > \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in [0, \frac{\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi)$. Hence, $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ if
$$x \in (0, \pi) \cap \left\{\left[0, \frac{\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right)\right\} = \left(0, \frac{\pi}{4}\right)$$
The inequality $\sin x < 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\pi, 2\pi)$. The inequality $$\sqrt{2}\cos x - 1 < 0 \iff \cos x < \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\frac{\pi}{4}, \frac{7\pi}{4})$. Hence, $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$ if
$$x \in (\pi, 2\pi) \cap \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) = \left(\pi, \frac{7\pi}{4}\right)$$
Thus, $\sin(2x) > \sqrt{2}\sin x$ in the interval $[0, 2\pi)$ if
$$x \in \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{7\pi}{4}\right)$$
Since the sine function has period $2\pi$, the general solution is
$$x \in \bigcup_{k \in \mathbb{Z}} \left\{\left(2k\pi, \frac{\pi}{4} + 2k\pi\right) \cup \left(\pi + 2k\pi, \frac{7\pi}{4} + 2k\pi\right)\right\}$$
| {
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"url": "https://math.stackexchange.com/questions/2792220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A bit of confusion in finding invariant measures I'm currently studying Ergodic theory and I'm a bit confused about one way in which we define invariant measures. A measure $\mu(A)$ on a set $A$ is said to be invariant, if for some measurable function $f$, $\mu(A)= \mu(f^{-1}(A))$. I noticed in the lecture notes that the example using the Gauss map is as follows:
The first line confuses me since we somehow reverted to describing $T^{-1}[a,b]$ in terms of $T[a,b]$ and I can't understand why this is justifiable. Moreover, I'm not sure why this correctly shows that the measure will ultimately be invariant.
| Suppose $0 \leq a \leq b \leq 1$. Observe that if $x \in [0,1]$, then $x \in T^{-1}([a,b])$ if and only if there is a $n \in \mathbb{Z}$ such that $n + a \leq \frac{1}{x} \leq n + b$. Since $x \in [0,1]$, we know that $\frac{1}{x} \geq 1$. Thus, $n \in \{1,2,\dots\}$. In particular, we can conclude by inverting the inequality that $\frac{1}{n + b} \leq x \leq \frac{1}{n + a}$. This proves
\begin{equation*}
T^{-1}([a,b]) \subseteq \bigcup_{n = 1}^{\infty} \left[\frac{1}{n + b}, \frac{1}{n + a}\right]
\end{equation*}
Running the argument backwards, we see that the opposite inclusion also holds so that the two sets are, in fact, equal. Note also that the sets in the union are disjoint, which follows from the fact that the sets $\{[n + a, n + b]\}_{n \in \mathbb{Z}}$ are disjoint.
The disjointness of the sets $\{[(n + b)^{-1},(n + a)^{-1}]\}_{n \in \mathbb{N}}$ justifies the computations in your graphic. Thus, it remains to prove
\begin{equation*}
\frac{1}{\log(2)} \sum_{n = 1}^{\infty} \left[\log \left(1 + \frac{1}{a + n}\right) - \log \left(1 + \frac{1}{b + n}\right) \right] = \frac{1}{\log(2)} \left(\log(1 + b) - \log(1 + a)\right)
\end{equation*}
since the RHS equals $\mu([a,b])$. To do this, we use the properties of the logarithm. In particular,
\begin{align*}
\sum_{n = 1}^{N} \log \left( 1 + \frac{1}{a + n}\right) &= \log \left(\prod_{n = 1}^{N} \left(1 + \frac{1}{a + n} \right) \right) \\
&= \log \left( \prod_{n = 1}^{N} \left(\frac{a + n + 1}{a + n} \right)\right) \\
&= \log \left(\frac{a + N + 1}{a + 1}\right).
\end{align*}
Therefore,
\begin{align*}
\sum_{n = 1}^{N} \left[\log\left(1 + \frac{1}{a + n}\right) - \log\left(1 + \frac{1}{b + n} \right)\right] = \log\left(\frac{a + N + 1}{b + N + 1} \cdot \frac{b + 1}{a + 1} \right).
\end{align*}
Sending $N \to \infty$, we obtain
\begin{equation*}
\sum_{n = 1}^{\infty} \left[\log \left(1 + \frac{1}{a + n}\right) - \log \left(1 + \frac{1}{b + n}\right) \right] = \log(1 + b) - \log(1 + a),
\end{equation*}
which, after dividing by $\log(2)$, is what we were after.
| {
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"url": "https://math.stackexchange.com/questions/2793031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $x^{18} = y^{21} = z^{28}$ If $3, 3\log_yx, 3\log_zy, 7\log_xz$ are in A.P, then prove that thing in the title.
I tried to equate: $3(\log_yx -1) = 3(\log_zy - \log_yx) = 7\log_xz - 3\log_zy$ . Will it be of any help?
| You can rewrite the logs as $\log_y x^3$, $\log_x y^3$, and $\log_x z^7$. The arithmetic progression gives you a number $t$ with the property that \begin{align*} \log_y x^3 &= 3 + t \\ \log_z y^3 & = 3 + 2t \\ \log_x z^7 &= 3 + 3t. \end{align*}
Rewrite in exponential form:
\begin{align*} x^3 &= y^{3+t} \\ y^3 &= z^{3+2t} \\ z^7 &= x^{3+3t} \end{align*}
We can try to get a sense of what $t$ is doing:
$$x^3 = y^{3+t} = (y^3)^{\frac{3+t}{3}} = (z^{3+2t})^{\frac{3+t}{3}} = (z^7)^{ \frac{3+2t}{7} \frac{3+t}{3}} = (x^{3+3t})^{ \frac{3+2t}{7} \frac{3+t}{3}} = (x^3)^{(1+t)\frac{3+2t}{7} \frac{3+t}{3}}. $$
This forces
$$ (1+t)\cdot \frac{3+2t}{7} \cdot \frac{3+t}{3} = 1$$
whose only real solution is $t = \frac 12$.
The first two equations above become $x^3 = y^{7/2}$ and $y^3 = z^4$ which lead immediately to $x^{18} = y^{21}$ and $y^{21} = z^{28}$ as desired.
| {
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"url": "https://math.stackexchange.com/questions/2794526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability that a particular football team wins a two-game series against another team Question
Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2},\frac{1}{6},\frac{1}{3}$ respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games. Find $$P(X>Y)$$
My Approach
I took simple approach that $X>Y$
$T_1~\text{wins and } T_2~\text{loses}=\frac{1}{2} \times\frac{1}{3} $
$T_1~\text{wins and } T_2~\text{draws}\frac{1}{2} \times\frac{1}{6}$
$T_1~\text{draws and } T_2~\text{loose}\frac{1}{6} \times\frac{1}{3}$
Required probability$$\frac{1}{6} +\frac{1}{12}+\frac{1}{18}=\frac{11}{36}$$
I know it is wrong. But I am unable to move on.
Please help!
| Lets think about the full distribution for $X$ and $Y$. What you have done is combined the distributions of $T_1$ and $T_2$. We want the combination of $X$ and $Y$. When $X=6$, $Y=0$. When $X=4$, $Y=1$. When $X=1$, $Y=4$. When $X=3$, $Y=3$. When $X=0$, $Y=6$. If $W$ is win, $D$ is draw and $L$ is loss, we can: $WW, LL, WL, LW, WD, DW, DD$
We want the combinations where $X \gt Y$ strictly. That is $X=6$, $Y=0$ AND $X=4$, $Y=1$. This is represented by $WW$ or $WD$ or $DW$. Sum this:
$2$ Wins: $\frac 12 \times \frac 12$
$1$ Win $1$ Draw: $\frac 12 \times \frac 16$
So
$$P(X>Y)=P(X=6,Y=0)+P(X=4,Y=1)$$
$$=P(WW)+2 \times P(WD)$$
$$=\frac 12 \times \frac 12 \space + \space 2 \times\frac 12 \times \frac 16= \frac 14+ \frac 1{6}= \frac 5{12}$$
Edit ### - Looks like my first guess was right.
So 5/12 I hope is your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Value of $\frac{x^n -a^n}{ x-a}$? Is there a rule that governs the value of $\frac{x^n-a^n}{x-a}$? For instance $\frac{x^{12}-1}{x-1}$ is equal to $x^{11} + x^{10}... +1$. How is this result obtained? There is no limit over here, so I don't think there is calculus used, also I'm still in a beginner stage in calculus.
| Calculus is only required for the limit (if any) as the values of $n$ get infinitely large.
For any finite positive integer value of $n$ then (assuming $x \ne a$ then
$\frac {x^n - a^n}{x-a} = x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}=\sum\limits_{k=0}^{n-1} x^ka^{(n-1) - k}$.
The reason why is that if
$(x-a)K = x^n - x^a$ and if $x\ne a$ then $K = \frac {x^n-x^a}{x-a}$ and:
$(x-a)( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}) = $
$x( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}) - $
$a( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1}) =$
$(x^n + x^{n-1}a + ......+x^2a^{n-2} +xa^{n-1}) -$
$(x^{n-1}a + x^{n-2}a^2 + .......... + xa^{n-1} + a^n)=$
$x^n + (x^{n-1}a - x^{n-1}a) + (x^{n-2}a^2 -x^{n-2}a^2) + ..... + (xa^{n-1} - xa^{n-1}) - a^n = $
$x^n - a^n$.
(This is called "telescoping" as it "folds up like a telescope"; when you multiply one term $x^ia^j$ by $x$ to get $x^{i+1}a^j$ and you multiply the next term $x^{i+1}a^{j-1}$ by $-a$ you get $-x^{i+1}a^j$ and they cancel each other out. And only the very first term, $x^{n-1}$ times $x$ and the very last term $a^{n-1}$ times $-a$ are left.)
This is a handy and very well known "trick". Look at it. Get it. Chuckle at it because it is so clever. And never forget it. It will be very useful later and you will see it a lot[*].
.......
In calculus we figure out what happens when
$n\to \infty$ and $\lim\limits_{n\to \infty} \frac {x^n - a^n}{x-a} = \lim\limits_{n\to \infty}(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ..... + xa^{n-2} + a^{n-1})$
and we can use it too figure out that.
$1 + \frac 12 + (\frac 12)^2 + (\frac 12)^3+ ......... = $
$\lim \frac {1^n - (\frac 12)^n}{1 - \frac 12} = \frac {1 - 0}{\frac 12} = 2$
And $1 + \frac 23 + (\frac 23)^2 + (\frac 23)^3+ ......... = $
$\lim \frac {1^n - (\frac 23)^n}{1 - \frac 23} = \frac {1 - 0}{\frac 23} = 1\frac 13$
And than if $|r| < 1$ then $ 1 + r + r^2 + ..... = \lim \frac{1-r^n}{1-r} = \frac 1{1-r}$.
Which is ... kind of neat.
But that's calculus and you'll learn that later.
======
[*]And don't forget when in a year or two a novice asks you about it, be sure to look down at them with a "well, isn't obvious" look and show the trick and make sure they feel like crap for not coming up with themselves .... and suggest they maybe would be better off dropping mathematics altogether and taking up something a little more "creative" like basket weaving.
This is an important aspect of math learning... Not sure, why but it seems to be.
| {
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"url": "https://math.stackexchange.com/questions/2799086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Maximum Likelihood Estimation for Normal Distribution with parameters $\theta$ and $2\theta$ with $\theta>0$ I'm using log-likelihood and eventually I get to the following quadratic equation: $n{\theta}^{2} + 2n\theta - \sum_{i=1}^{n}X_i^2 = 0$. After solving the equation I get a solution ($\frac{-n + \sqrt{n^2 + n\sum_{i=1}^{n}X_i^2}}{n}$) that seems rather cumbersome and it's hard to check weather the given solution is indeed a maximum of the log-likelihood function. Is the solution correct and there simply isn't an easier way to do it or am I missing something here?
| $$
L(\theta) \propto \frac 1 {\sqrt\theta^n} \exp\left( \frac {-1}{2(2\theta)} \cdot \sum_{i=1}^n(x_i-\theta)^2 \right)
$$
Let $\overline x = \frac 1 n \sum_{i=1}^n x_i$ and $s^2 = \frac 1 n \sum_{i=1}^n
(x_i-\overline x)^2.$ Then algebra tells us that
$$
\sum_{i=1}^n (x_i-\theta)^2 = ns^2 + n(\overline x - \theta)^2.
$$
Hence
$$
L(\theta) \propto \frac 1 {\sqrt\theta^n} \exp\left( \frac{-1}{4\theta} \left( ns^2 + n(\overline x -\theta)^2 \right) \right).
$$
\begin{align}
\ell(\theta) & = \log L(\theta) = \text{constant} - \frac n 2 \log\theta - \frac {ns^2} {4\theta} - \frac{n(\overline x-\theta)^2}{4\theta} \\[10pt]
& = \text{constant} - \frac n 2 \log\theta - \frac{ns^2}{4\theta} - \frac{n\overline x^2}{4\theta} + \text{constant} - \frac{n\theta} 4 \\[10pt]
\ell\,'(\theta) & = - \frac n {2\theta} + \frac{ns^2}{4\theta^2} +\frac{n\overline x^2}{4\theta^2} - \frac n 4 \\[10pt]
& = \frac {-n} {4\theta^2} \left( 2\theta - s^2 - \overline x^2 + \theta^2 \right). \\[10pt]
& = (\text{negative constant}) \cdot \left( (\theta+1)^2 -s^2-\overline x^2 - 1 \right)
\end{align}
Since $\theta$ cannot be negative, the above expression is $0$ only if
$$
\theta = -1 +\sqrt{s^2+\overline x^2 + 1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim\limits \frac{x-\sin\sin\cdots\sin x}{x^3}.$ Problem
Evaluate $$\lim\limits_{x \to 0} \frac{x-\sin\sin\cdots\sin x}{x^3},$$where $\sin\sin\cdots\sin x$ denotes n-fold composite sine function.
Solution
Consider applying Taylor's Formula with 3-order at $x=0$. We may obtain $$f_n(x)=\sin\sin\cdots\sin x=x-\frac{n}{6}x^3+\mathcal{O}(x^3).\tag{*}$$ To prove this, we can apply the mathematical induction. Let $n=1,$ then $$f(x)=\sin x=x-\frac{1}{6}x^3+\mathcal{O}(x^3),$$ It's true and shows that $(*)$ holds for $n=1$. Assume that $(*)$ holds for $n=k$. Then $$\begin{align*}f_{k+1}(x)&=\sin(f_k(x))\\&=x-\frac{k}{6}x^3+\mathcal{O}(x^3)-\frac{1}{6}\left(x-\frac{k}{6}x^3+\mathcal{O}(x^3)\right)^3+\mathcal{O}(x^3)\\&=x-\frac{k+1}{6}x^3+\mathcal{O}(x^3)\end{align*}.$$ This shows that $(*)$ holds for $n=k+1$. As a result, $(*)$ holds for all $n=1,2,\cdots.$
Now,let's go back to deal with the problem. $$\lim\limits_{x \to 0} \frac{x-\sin\sin\cdots\sin x}{x^3}=\lim\limits_{x \to 0} \dfrac{x-\left(x-\dfrac{n}{6}x^3+\mathcal{O}(x^3)\right)}{x^3}=\frac{n}{6}.$$
Please correct me if I'm wrong. Hope to see other solutions. Thanks.
| I would like to share a more tricky method with you.
Use the same notation, and assume $f_0(x) = x$, then we can prove that
$$
\lim_{x \to 0} \frac{x - f(x)}{x^3} = \frac{1}{6}
$$
and
$$
\lim_{x \to 0} \frac{f_k(x)}{x} = 1, ~ \forall k \in \mathbb{N}.
$$
Thus we have
$$
\begin{aligned}
\lim_{x \to 0} \frac{f_0(x) - f_k(x)}{x^3} &= \lim_{x \to 0}\sum_{i = 1}^k \frac{f_{i - 1}(x) - f_{i}(x)}{x^3}\\
&=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f_{i}(x)}{x^3}\\
&=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f(f_{i-1}(x))}{(f_{i-1}(x))^3}\cdot \frac{(f_{i-1}(x))^3}{x^3}\\
&=\sum_{i = 1}^k\lim_{x \to 0} \frac{f_{i - 1}(x) - f(f_{i-1}(x))}{(f_{i-1}(x))^3} \cdot \lim_{x \to 0}\frac{(f_{i-1}(x))^3}{x^3}\\
&= \sum_{i = 1}^k \frac{1}{6}\\
&= \frac{k}{6}.
\end{aligned}
$$
| {
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"url": "https://math.stackexchange.com/questions/2800110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| Hint: Find the value of $x+{1\over x}$ and from that find $x^3+{1\over x^3}$
After that multiply $x^3+{1\over x^3}$ and $x^2+{1\over x^2}$
| {
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"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.