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Sum of an infinite series $\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}$ I want to compute this series, $$\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}.$$ I don't know how to contiune after factoring, which yields $\frac{1}{k(k-1)^2(k+1)^2(k-2)^2(k+2)^2}.$
I know that $\frac{1}{k(k-1)}$ telescopes. However, I can't apply this here, as the product of the sum is not equal to the sum of the product. I only want hints for now.
| Perform a partial fraction decomposition; specifically, find constants $A_1, A_2, \ldots, A_9$ such that $$\begin{align}\frac{1}{k(k^4-5k^2+4)^2} &= \frac{A_1}{k} + \frac{A_2}{k-1} + \frac{A_3}{k+1} + \frac{A_4}{k-2} + \frac{A_5}{k+2} \\&\quad+ \frac{A_6}{(k-1)^2} + \frac{A_7}{(k+1)^2} + \frac{A_8}{(k-2)^2} + \frac{A_9}{(k+2)^2}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4103684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| First, let us substitute $x := u + v, y := u - v$ in the constraint to get
$$(u+v)^2 + (u+v)(u-v) + (u-v)^2 = 3u^2 + v^2 = 1.$$
(Note that this substitution is a standard one to try once you notice the fact that the constraint equation is symmetric with respect to swapping $x$ and $y$. It is closely related to the operation of rotating the plane clockwise by $45^\circ$ to convert a curve symmetric about the line $y = x$ to a curve symmetric about the line $v = 0$.)
Thus, the constraint curve can be parameterized by $u = \frac{1}{\sqrt{3}} \cos \theta, v = \sin \theta$ for $0 \le \theta < 2\pi$, and so the parameterization in terms of $x, y$ is
$$(x, y) = \left(\frac{1}{\sqrt{3}} \cos\theta + \sin\theta, \frac{1}{\sqrt{3}} \cos\theta - \sin\theta \right).$$
Now, substituting into the objective function, we have
$$x^2 - 3xy - 2y^2 = 2 \sin^2\theta + 2\sqrt{3} \sin\theta \cos\theta - \frac{4}{3} \cos^2\theta = \\ \frac{1}{3} - \frac{5}{3} (\cos^2\theta - \sin^2\theta) + \sqrt{3} (2\sin\theta \cos\theta) = \frac{1}{3} - \frac{5}{3} \cos(2\theta) + \sqrt{3} \sin(2\theta).$$
From here, using the standard fact that the function $A \cos t + B \sin t$ has maximum value $\sqrt{A^2 + B^2}$, we see that $x^2 - 3xy - 2y^2$ has maximum value $\frac{1}{3} + \sqrt{(-5/3)^2 + (\sqrt{3})^2} = \frac{1}{3} + \frac{1}{3} \sqrt{52}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sqrt{2} + \sqrt{3}$ is not a rational number. I am trying to prove that $\sqrt{2} + \sqrt{3}$ is not a rational number.
I have used the following Theorems to prove that $\sqrt{2} + \sqrt{3}$ is not a rational number.
Theorem 3.2) Let $x$ be an integer. $x^2$ is odd if and only if $x$ is odd.
Theorem 5.6) If $x$ is a rational number and $y$ is an irrational number, then $x+y$ is irrational.
Theorem 5.9) Let $n$ be an integer. If $n^2$ is divisible by $3$, then $n$ is divisible by $3$.
The following is my proof.
The number $\sqrt{2} + \sqrt{3}$ is not a rational number.
Proof. Suppose that $\sqrt{2} + \sqrt{3}$ is a rational number. It follows that there exist an integer $p$ and a nonzero integer $q$ such that
\begin{align}
\sqrt{2} + \sqrt{3} = \frac{p}{q} \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
Because the number $\sqrt{2} + \sqrt{3}$ is a nonzero number, the number $\frac{p}{q}$ is nonzero.
Multiplying both sides of the above equation by $(\sqrt{2} - \sqrt{3})$, one obtains
\begin{align}
(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = \frac{p}{q}(\sqrt{2} - \sqrt{3}) \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
By multiplying $-1$ on both sides of the above equation and making $\frac{p}{q}\sqrt{3} - \frac{p}{q}\sqrt{2}$ as the subject one obtains
\begin{align}
\frac{p}{q}\sqrt{3} - \frac{p}{q}\sqrt{2} = 1 \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
Squaring both sides of the above equation, one obtains
\begin{align}
5\frac{p^2}{q^2} - 2\sqrt{2}\sqrt{3}\frac{p^2}{q^2} = 1 \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
By multiplying both sides of the above equation by $\frac{q^2}{2p^2}$, one obtains
\begin{align}
\frac{5}{2} - \sqrt{2}\sqrt{3} = \frac{q^2}{2p^2} \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
To show that $\sqrt{2}\sqrt{3}$ is irrational, suppose that $\sqrt{2}\sqrt{3}$ is on the contrary rational.
It follows that there exist an integer $r$ and a nonzero integer $s$ such that
\begin{align}
\sqrt{2}\sqrt{3} = \frac{r}{s} \text{ for some $r,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Assuming that $r$ and $s$ has no common factor. Squaring both sides of the above equation one obtains
\begin{align}
2 \cdot 3 = \frac{r^2}{s^2} \text{ for some $r,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Making $\frac{r^2}{3}$ as the subject, one obtains
\begin{align}
\frac{r^2}{3} = 2s^2 \text{ for some $r,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Because the number on the $RHS$ of the above equation is an integer, $r^2$ is divisible by $3$. It follows from Theorem 5.9) that $r$ is divisible by $3$.
Thus,
\begin{align}
r = 3i \text{ for some $i \in \mathbb{Z}$}
\end{align}
Substituting the above equation into the equation $\frac{r^2}{3} = 2s^2$ one obtains
\begin{align}
3i^2 = 2s^2 \text{ for some $i,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Because the number on the $RHS$ of the above equation is even, $3i^2$ must be even. Since $3$ is odd, the factor $2$ must be from the number $i^2$. Therefore, $i^2$ is even. It follows from Theorem 3.2) that $i$ is even.
Thus,
\begin{align}
i = 2u \text{ for some $i,u \in \mathbb{Z}$}
\end{align}
Substituting the above equation into the equation $r = 3i$, one obtains
\begin{align}
r = 2 \cdot 3u \text{ for some $u \in \mathbb{Z}$}
\end{align}
Substituting the above equation into the equation $2 \cdot 3 = \frac{r^2}{s^2}$ and making $s^2$ as the subject, one obtains
\begin{align}
s^2 = 2 \cdot 3 u^2 \text{ for some $s,u \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Because $s^2$ is even and divisible by $3$, it follows from Theorem 3.2) and Theorem 5.9) that $s$ is even and divisible by $3$.
Thus,
\begin{align}
s = 2 \cdot 3 v \text{ for some $v \in \mathbb{Z}$ and $v \ne 0$}
\end{align}
Substituting the equation $r = 2 \cdot 3u$ and the equation $s = 2 \cdot 3 v$ into the equation $\sqrt{2}\sqrt{3} = \frac{r}{s}$, one obtains
\begin{align}
\sqrt{2}\sqrt{3} = \frac{2 \cdot 3u}{2 \cdot 3 v} \text{ for some $u,v \in \mathbb{Z}$ and $v \ne 0$}
\end{align}
Because $r$ and $s$ have common factors, one obtains a contradiction with the assumption that $r$ and $s$ have no common factor.
Hence, the number $\sqrt{2}\sqrt{3}$ is irrational.
Since on the equation $\frac{5}{2} - \sqrt{2}\sqrt{3} = \frac{q^2}{2p^2}$ the number $\frac{5}{2}$ is rational and $\sqrt{2}\sqrt{3}$ is irrational, by Theorem 5.6), the number $\frac{5}{2} - \sqrt{2}\sqrt{3}$ is irrational. However, because $\frac{p}{q}$ is a nonzero rational number, the number $\frac{q^2}{2p^2}$ on the $RHS$ of the equation $\frac{5}{2} - \sqrt{2}\sqrt{3} = \frac{q^2}{2p^2}$ is a nonzero rational number.
Thus, one obtains a contradiction.
Therefore, the number $\sqrt{2} + \sqrt{3}$ is not a rational number.
Is the proof correct?
Reference
Daepp, U. and Gorkin, P., 2011. Reading, Writing, and Proving. 2nd ed. pp.29, 54, 55.
| The proof in the original question is more work than is necessary to prove something so simple.
All you need to do is show that $(\sqrt{2} + \sqrt{3})^2$ is irrational${}^\dagger$, which amounts to showing that $\sqrt{6}$ is irrational, which you can prove in the same way you prove $\sqrt{2}$ is irrational.
$\dagger$ Note that $x^2$ irrational $\implies x\ $ is irrational. This is easy to see by considering the contrapositive.
| {
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"url": "https://math.stackexchange.com/questions/4111510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the limit of the sum $\sum_{k=n}^{2n-1}\frac{1}{2k+1}$ Find
$$\lim_{n\to+\infty}\sum_{k=n}^{2n-1}\frac{1}{2k+1}$$
i tried to write as
$$\sum_{k=0}^{n-1}\frac{1}{2k+2n+1}$$
$$=\frac 1n\sum_{k=0}^{n-1}\frac{1}{\frac{2k+1}{n}+2}$$
like a Riemann sum, but i can't find the corresponding function.
any idea will be appreciated.
| As a general approach, sums like this often can be evaluated as a double limit
$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 + \frac{1}{n}} = \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 + \frac{1}{m}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 }= \int_0^2 \frac{dx}{2x+2}$$
The first step, $\lim_{n \to \infty} S_{nn} = \lim_{n\to \infty} \lim_{m \to \infty} S_{mn}$ is valid for a double sequence $(S_{mn})$ when we have $S_{mn} \to T_n$ as $m \to \infty$ uniformly for all $n$.
In this case, convergence is uniform as $m \to \infty$ for all $n$, since
$$\left| \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 + \frac{1}{m}}- \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 } \right| = \left| \frac{1}{n}\sum_{k=0}^{n-1}\frac{-\frac{1}{m}}{\left(\frac{2k}{n} + 2 + \frac{1}{m}\right)\left(\frac{2k}{n} + 2 \right)}\right| \\ \leqslant \frac{1}{nm}\sum_{k=0}^{n-1}\frac{1}{\left|\frac{2k}{n} + 2 + \frac{1}{m}\right|\left|\frac{2k}{n} + 2 \right|}\leqslant \frac{1}{nm}\sum_{k=0}^{n-1}\frac{1}{4} = \frac{1}{4m}\underset{m\to \infty}\longrightarrow 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Angle between vectors by using cross product Given $A=(0,10,2)$ and $B=(0,-4,0.5)$, find the angle between $A$ and $B$ using the cross product.
My attempt by using dot product:
$$\cos (\theta)=\frac{A\cdot B}{|A||B|}=\frac{(0)(0)+10(-4)+2(0.5)}{\sqrt{0^2+10^2+2^2}\sqrt{0^2+(-4)^2+(0.5)^2}}=\frac{-39}{41.1}$$
Then $\theta=\arccos(-0.948)= 161.44$
Now using the cross product:
$$\sin(\theta)=\frac{|A\times B|}{|A||B|}=\frac{ \begin{vmatrix}
i & j & k \\
0 & 10 & 2 \\
0 & -4 & 0.5 \\
\end{vmatrix}}{\sqrt{0^2+10^2+2^2} \sqrt{0^2+(-4)^2+(0.5)^2}}=
\frac{|10(0.5)i+0(-4)k+0(2)j-0(10)k-2(-4)i-(0.5)(0)j|}{\sqrt{1690}}=\frac{|13i|}{41.1}=\frac{13}{41.1}=0.316$$
but then I get $\theta=\arcsin(0.316)=18.42$
What am I doing wrong?
| The cross product isn’t a conclusive means to determine the angle $\theta$ between two vectors, since within $\theta$’s range $[0,180^\circ],$ $$\sin\theta=x\implies\\\theta=\arcsin x\text{ or }180^\circ-\arcsin x.$$
(On the other hand, for $\theta\in[0,180^\circ],$ $$\cos\theta=x\implies\theta=\arccos x.)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit cycle within $\frac{1}{4}Show that the system
\begin{align}
x'&=-y+x(1-2x^2-3y^2)\nonumber\\
y'&=x+y(1-2x^2-3y^2)\nonumber
\end{align}
has a limit cycle in $\frac{1}{4}<r<1$.
Here's what I've done so far:
\begin{align}
xx'&=-xy+x^2(1-2x^2-3y^2)\nonumber\\
yy'&=xy+y^2(1-2x^2-3y^2)\nonumber
\end{align}
\begin{align}
xx'+yy'&=-xy+x^2(1-2x^2-3y^2)+xy+y^2(1-2x^2-3y^2)\nonumber\\
\frac{1}{2}(x^2+y^2)'&=x^2(1-2x^2-3y^2)+y^2(1-2x^2-3y^2) \nonumber\\
\frac{1}{2}(x^2+y^2)'&=(x^2+y^2)(1-2x^2-3y^2) \nonumber
\end{align}
Knowing that $r^2=x^2+y^2$:
\begin{align}
\frac{1}{2}(x^2+y^2)'&=(x^2+y^2)(1-2x^2-3y^2) \nonumber\\
\frac{1}{2}(r^2)'&=(r^2)(1-2x^2-3y^2) \nonumber
\end{align}
I'm now stuck here and I'm not really sure what to do next because of this part $(1-2x^2-3y^2)$.
| Noting that on $r=1$,
$$ \frac{1}{2}(r^2)'=(r^2)(1-2x^2-3y^2)=r^2(1-2r^2-y^2)<0 $$
and on $r=\frac14$,
$$ \frac{1}{2}(r^2)'=(r^2)(1-2x^2-3y^2)=r^2(1-3r^2+y^2)>0 $$
by the Poincare-Bendixson Theorem, there is a closed trajectory lying in $\frac14<r<1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum ... Given a positive integer $n$, find the minimum value of $$\frac{x_1^3+x_2^3+...+x_n^3}{x_1+x_2+...+x_n}$$
subject to the condition that $x_1,x_2,...,x_n$ be distinct positive integers
I tried and this is what happened
Order the numbers $x_1 < x_2 < ··· < x_n$ and call the expression from the statement
$E(x_1, x_2,...,x_n)$. Note that $E(x1, x_2,...,x_n) > x^2_n/n$ , which shows that as the variables
tend to infinity, so does the expression. This means that the minimum exists. Assume
that the minimum is attained at the point $(y_1, y_2,...,y_n)$. If $y_n − y_1 > n$ then there
exist indices $i$ and $j$ , $i<j$ , such that $y_1,...,y_{i}+1,...,y_j − 1,...,y_n$ are still distinct integers.
| I think you have got the important first step down in your post. Say the minimum is attained at $x_1<x_2<\dots <x_n$ and assume that $x_n-x_1> n.$ Then like you said, there exists $1\leq i<j\leq n$ such that $x_1<x_2<\dots x_i+1<x_{i+1}<\dots x_{j-1}<x_j-1<\dots x_n.$ Call this new tuple $(y_1,y_2,\dots y_n)$ and observe that:
$$E(y_1,y_2,\dots y_n) - E(x_1,x_2,\dots x_n) = \dfrac{(x_i+1)^3+(x_j-1)^3 - x_i^3-x_j^3}{\sum x_i} = \dfrac{3(x_i^2-x_j^2)+3(x_i+x_j)}{\sum x_i} = $$
$$ = \dfrac{3(x_i+x_j)(x_i-x_j+1)}{\sum x_i} < 0$$
since $x_j-1>x_i+1\implies x_i-x_j+1<-1.$
This severely restricts the possibilities into either $x_n = x_1+n$ or $x_n = x_1+n-1$, both of which should be easily bashable.
EDIT: First consider $x_n = x_1+n-1.$ This means, $x_k = a+(k-1), \,1\leq k\leq n.$ The denominator is easily $x_1+x_2+\dots+x_n = na+\frac{n(n-1)}{2}$ and the numerator is:
$$\sum_{k=0}^{n-1}(a+k)^3 = na^3+\dfrac{3n(n-1)}{2}a^2+\dfrac{(n-1)n(2n-1)}{2}a+\dfrac{n^2(n-1)^2}{4}$$
and the whole expression miraculously turns out to be:
$$E = \dfrac{na^3+\dfrac{3n(n-1)}{2}a^2+\dfrac{(n-1)n(2n-1)}{2}a+\dfrac{n^2(n-1)^2}{4}}{na+\frac{n(n-1)}{2}} = a^2+(n-1)a + \dfrac{n(n-1)}{2}$$
and so clearly, the minimum is attained at $a = 1,$ which would be $E = \dfrac{n(n+1)}{2}.$
Now, if $x_n = x_1 + n,$ then we can assume our tuple is the whole $a,a+1,\dots a+n$ except for one $a+j.$ The denominator is then $(n+1)a + \frac{n(n+1)}{2} - a - j = na+\dfrac{n(n+1)}{2}-j$ and the numerator is:
$$\sum_{k=0}^n(a+k)^3 - (a+j)^3=na^3+3\left(\dfrac{n(n+1)}{2}-j\right)a^2+3\left(\dfrac{(n+1)n(2n+1)}{6}-j^2\right)a+\dfrac{n^2(n+1)^2}{4}-j^3$$
and the rest should follow similarly with a bit more algebra.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.
| If $t$ is a solution, then so is $-t$, hence it is worth tring to write things in terms of $u:=t^2$. Note that
$$\frac1{t-2}+\frac1{t+2}=\frac{(t+2)+(t-2)}{(t-2)(t+2)}=\frac{2t}{u-4} ,$$
$$\frac1{t-1}+\frac1{t+1}=\frac{(t+1)+(t-1)}{(t-1)(t+1)}=\frac{2t}{u-1} ,$$
and of course
$$\frac1t=\frac{2t}{2u} $$
so that we want to solve
$$\frac1{u-4}+\frac1{u-1}+\frac1{2u}=0 $$
This gives you only a quadratic numerator.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$ Theorem. $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$.
I'm tasked with proving the given theorem by induction. Here's where I've gotten so far...
Proof. Clearly, the theorem is true for $n=1$, establishing the base case. Moreover, it can easily be shown that the theorem works for $n=2,3,4,5,\ldots, 18$.
Suppose, now, that $k$ is an integer for which the given theorem holds. Consider, then, $k+18$.
$$2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}}$$
By Fermat's Little Theorem, if $n \equiv m \pmod{p-1}$, then $a^m \equiv a^n \pmod{p}$. Clearly, $k+18 \equiv k \pmod{18}$. Then,
$$2^{2^{k+18}} \equiv 2^{2^k} \pmod{19}$$
$$3^{2^{k+18}} \equiv 3^{2^k} \pmod{19}$$
$$5^{2^{k+18}} \equiv 5^{2^k} \pmod{19}$$
Hence, $2^{2^{k+18}}$ may be expressed as $\;2^{2^k} + 19n$ for some integer $n$, $\;3^{2^{k+18}}$ as $\;3^{2^k}+ 19m$ for some integer $m$, and $\;5^{2^{k+18}}$ as $\;5^{2^k} + 19q$ for some integer $q$.
Thus, $2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}} = (2^{2^{k}} + 3^{2^{k}} + 5^{2^{k}}) + 19(m+n+q)$, a sum of integers divisible by $19$, and thus clearly divisible by $19$.
By the principle of induction, we've thus shown that $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$, concluding the proof.
$\blacksquare$
Is this approach valid?
| I don't think it's valid. $k \equiv k + 18 \pmod {18}$ so $2^k \equiv 2^{k+18} \pmod {19}$. And that means $a^{2^k} \equiv a^{2^{k+18}} \pmod m$ for an $m$ where $\phi(m) = 19$ and $\gcd(a,m) =1$ but... that doesn't work for $a^{2^{k+18}}\equiv a^{2^k} \pmod {19}$
But you could be onto something.
$2^{k}\pmod{18}$ cycles. $\phi 9 = 6$ so $2^{k+ 6} \equiv 2^k \pmod 9$ and $2^m \equiv 0 \pmod 2$ so for $k \ge 1$ we know $2^{k+ 6}\equiv 2^k$ and so $a^{2^{k+6}}\equiv a^{2^{k}} \pmod {19}$.
So we can do $6$ base cases. Easier than $18$.
For $k = 1, ....6$ we have $2^k \equiv 2, 4,8,16, 14, 10 \pmod {18}$ and
But we can do better.
$2^2\equiv 4; 2^4 \equiv 16\equiv -3; 2^8 \equiv 9 \equiv 3^2\pmod {19}$
$3^2 \equiv 9; 3^4 \equiv 81 \equiv 5; 2^8 \equiv 5^2\pmod {19}$.
And $5^2 \equiv 6; 5^4 \equiv 36 \equiv -2$ and $5^{8}\equiv 2^2\pmod {19}$.
SO for $k \ge 1$ we have
$2^{2^{k+2}} + 3^{2^{k+2}} + 5^{2^{5+2}} \equiv 3^{2^k} + 5^{2^k} + 2^{2^k}\equiv 2^{2^k} + 3^{2^k} + 5^{2^k}\pmod {19}$
So we just need 2 base cases:
$2^2 + 3^2 + 5^2 \equiv 4+9 + 6 \equiv 0 \pmod {19}$
$2^{2^2} + 3^{2^2} + 5^{2^2} \equiv 16 + 81 + 36 \equiv -3+5 -2\equiv 0 \pmod{19}$
That's it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $(x+1)^p \equiv x^p +1 \pmod{p^2}$ given conditions on $x$ and $p$. The problem is as follows.
Let $p$ be a prime of the form $p = 3k + 1$ for some $k \in \mathbb{Z}$. Let $x$ be an integer such that $p \nmid x$ and $[x]_p$ has multiplicative order 3 in the ring $(\mathbb{Z}/p\mathbb{Z})^{\times}$. Show that
$$
(x+1)^p \equiv x^p+1 \pmod{p^2}
$$
My attempt:
Rewrite $(x+1)^p$ in its binomial expansion. We get $(x+1)^p = x^p+ {p\choose 1}x^{p-1} + {p\choose 2}x^{p-2} + \dots + {p\choose p-1}x + 1$. Now it is sufficient to prove that ${p\choose 1}x^{p-1} + {p\choose 2}x^{p-2} + \dots + {p\choose p-1}x \equiv 0 \pmod {p^2}$. However I do not know how to prove this last equivalence.
Any hints on how to procede would be greatly appreciated!
My knowledge consists of basic number theory and $p$-adic numbers and related theorems.
| I don't offhand see any way to use the binomial expansion to solve your question. Instead, here's an alternate approach. Because $[x]_p$ has multiplicative order $3$, then $x \not\equiv 1 \pmod{p}$ and
$$x^3 \equiv 1 \pmod{p} \implies x^3 - 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
Note the lifting-the-exponent lemma states that, for odd primes $p$ where $p \not\mid x$ and $p \not\mid y$, then
*
*If $p \mid x - y$, $\; \nu _{p}(x^{n}-y^{n}) = \nu _{p}(x - y) + \nu _{p}(n)$.
*If $n$ is odd and $p \mid x + y$, $\; \nu _{p}(x^{n} + y^{n}) = \nu _{p}(x + y) + \nu _{p}(n)$.
Thus, using \eqref{eq1A} and the first option above gives
$$\nu_p((x^{3})^{p} - 1) = \nu_p(x^3 - 1) + \nu_p(p) \ge 2 \tag{2}\label{eq2A}$$
We therefore have
$$x^{3p} - 1 \equiv 0 \pmod{p^2} \tag{3}\label{eq3A}$$
Using Fermat's little theorem gives $x^{p} - 1 \equiv x - 1 \not\equiv 0 \pmod{p}$. Since $x^{3p} - 1 = (x^p - 1)(x^{2p} + x^{p} + 1)$, we thus get from \eqref{eq3A} that
$$x^{2p} + x^{p} + 1 \equiv 0 \pmod{p^2} \implies x^{p} + 1 \equiv -x^{2p} \pmod{p^2} \tag{4}\label{eq4A}$$
Since $x^3 = (x - 1)(x^2 + x + 1)$ and $x \not\equiv 1 \pmod{p}$, we thus get from \eqref{eq1A} that
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{5}\label{eq5A}$$
Next, using the second option of the lifting-the-exponent lemma statement mentioned earlier gives
$$\nu_p((x^{2})^{p} + (x + 1)^p) = \nu_p(x^2 + x + 1) + \nu_p(p) \ge 2 \tag{6}\label{eq6A}$$
From this and using \eqref{eq4A}, we get
$$\begin{equation}\begin{aligned}
x^{2p} + (x + 1)^p & \equiv 0 \pmod{p^2} \\
(x + 1)^p & \equiv -x^{2p} \pmod{p^2} \\
(x + 1)^p & \equiv x^p + 1 \pmod{p^2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find prime numbers with given property I've been working in the following number theory problem: find all prime numbers $p,q,r$ such that
$$p^2+q^2+r^2-1$$
is a perfect square. Does anyone has some hint for the problem? Thanks in advance.
| The triple $p=q=2, r = 3$ is the only possible solution (up to reordering). To show this first reduce the expression modulo $4$. Since a square must always be congruent to $0$ or $1$ mod $4$ we get
$$
p^2+q^2+r^2 \equiv 1 \pmod{4} \quad \text{or} \quad p^2+q^2+r^2 \equiv 2 \pmod{4}.
$$
Since any odd number squares to $1 \pmod{4}$ we may assume $p = 2$. Hence we need to find all primes $q,r$ such that $r^2+q^2+3$ is a square. Now reduce the equation mod $3$ to find
$$
r^2 + q^2 \equiv 0 \pmod{3} \quad \text{or} \quad r^2 + q^2 \equiv 1 \pmod{3},
$$
since $0$ and $1$ are the only quadratic residues modulo $3$. Therefore at least one of $q$ or $r$ must be divisible by $3$. Since we assume them to be prime this forces without loss of generality $r = 3$. Finally we look for primes $q$ such that $q^2 + 12$ is a square. The quadratic residues modulo $8$ are $0,1,4$. Reducing mod $8$ therefore gives
$$
q^2 \equiv 4 \pmod{8} \quad \text{or} \quad q^2 \equiv 5 \pmod{8} \quad \text{or} \quad q^2 \equiv 0 \pmod{8}.
$$
Since $5$ is a non-residue we must have $q^2 \equiv 0 \pmod{8}$ or $q^2 \equiv 4 \pmod{8}$. In either case $q$ must be even. Since we assumed it to be prime it follows $q=2$.
| {
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"url": "https://math.stackexchange.com/questions/4131645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Almost there?? Help with inequality.(updated) I was recently doing a combinatorics problem from MMO given in book Arthur Engel. I might have solved it (though unlikely) as I arrived at following inequality:-
Both $x$ and $y$ are positive. the first term is not in the pattern, while rest are.
$($$\frac{(x^3+y^3)}{(x+y)^3}$$)$ + $($$\frac{(x^4+y^4+x^3y+y^3x+x^2y^2)}{(x+y)^4}$ + $\frac{(x^5+y^5+x^4y+y^4x+x^3y^2+x^2y^3)}{(x+y)^5}$ ....$)$ $<1$ ?
I don't know if it is true (not good with inequalities). Please verify it.
Any help is appreciated :)
I can post the question if you want to give it a try but please don't tell the solution here.
My method:
$(1)$ All triangles are similar and each triangle has unique a unique area. Two triangles are congruent iff their areas are equal. Then, I constructed an ''area-based representation''
$(2)$ If the bigger RAT(Right angled triangle) into area $x$ and $y$, Then when we divide the RAT with area $x$ we end up with triangles of areas: $x^2/(x+y)$ and $xy/(x+y)$. So if we just keep dividing the RAT (not caring about congruency at this point) We end up with the nth row of pascals triangle/$(x+y)^{(n-2)}$.
1(x+y)
1(x) 1(y)
1(x^2/(x+y)) 2(xy/(x+y)) 1(y^2/(x+y))
1(x^3/(x+y)^2) 3(x^2y/(x+y)^2) 3(xy^2/(x+y)^2) 1(y^3/(x+y)^2)
$(3)$ If we keep the first triangle unchanged, then divide the second triangle into $x$ and $y$, then divide the third into $x^2/(x+y)$,$y^2/(x+y)$,$xy/(x+y)$,$x^2y/(x+y)^2$,$xy^2/(x+y)^2$, then the fourth triangle must be
$($$\frac{(x^3+y^3)}{(x+y)^2}$$)$ + $($$\frac{(x^4+y^4+x^3y+y^3x+x^2y^2)}{(x+y)^3}$ + $\frac{(x^5+y^5+x^4y+y^4x+x^3y^2+x^2y^3)}{(x+y)^4}$ ....$)$
which I wanted to show is less than $x+y$
$Attempt(promising?)$
I noticed that in case of maximum area we can't get $(x^4,y^4,x^5,y^5...)$ , then I mordified the inequality but was only able to give the required proof for $0.276 < x < 0.723 $
Reason: Notice that when we are dividing one single triangle we can get many terms but from the each of the sets $S_x$ and $S_y$ we can only get one term from each ( so in total $2$ distinct terms and think about this yourself )
$S_x = (1,x,x^2,x^3,x^4...) $
$S_y = (1,y,y^2,y^3,y^4...) $
And since there are four triangles we can get (in max area case) ${1,x,y,x^2,y^2,x^3,y^3}$
| (This is more of a comment regarding OP's approach to the original question. It's not a complete solution.)
With regards to your method, you're on the right track.
*
*Note that $ x + y = 1$, so you could get rid of that.
*Note that the triangles are split up into areas of size $ x^a y^b$.
*For $ (x, y) \neq (1/2, 1/2)$, 2 triangles are congruent iff $ a_1 = a_2, b_1 = b_2$.
*
*Hence, if we have distinct triangles, then their areas will be a subset of $ 1, x, y, x^2, xy, y^2, x^3, x^2y , xy^2, y^3, ... $
*Notice that this is different from your value.
*For $ (x, y) = (1/2, 1/2)$, w triangles are congruent iff $a_1 + b_1 = a_2 + b_2$.
*
*(For now, I will ignore this case and leave it to you).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that ${n \choose 1}-\dots+(-1)^{n-1}{n \choose n} \left(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}\right)=\frac{1}{n}$ Show that
${n \choose 1}$ - $ {n \choose 2}$ (1 + $ \frac{1}{2} $ ) ..............+ $(-1)^{n-1} {n \choose n}$ (1 + $ \frac{1}{2} $ + $ \frac{1}{3} $ +$ \frac{1}{4} $ ............+$ \frac{1}{n} $) = $ \frac{1}{n} $
MY ATTEMPT : I tried by taking the general term
$T_r$ = $(-1)^{r-1}$ ${n \choose r}$ (1 + $ \frac{1}{2} $ + $ \frac{1}{3} $ +$ \frac{1}{4} $ ............+$ \frac{1}{r} $)
It looked some what like ${n \choose r}$ $\int x^{r-1} $ = ${n \choose r}$ $\frac {x^r}{r} $ so I took an expansion say
$T_1(r)$ = $(-1)^{r-1}$ ${n \choose r}$ $(1+ x+x^2+...........+x^{r-1})$ this simplifies to
$T_1(r)$ = $(-1)^{r-1}$ ${n \choose r}$ $(\frac{1-x^r}{1-x})$
I am struck after this I tried to rearrange this and done some simplification but I failed
Please help me
| \begin{align}
\sum_{r=1}^n (-1)^{r-1} \binom{n}{r} \sum_{k=1}^r \frac{1}{k}
&= \sum_{k=1}^n \frac{1}{k} \sum_{r=k}^n (-1)^{r-1} \binom{n}{r} \\
&= \sum_{k=1}^n \frac{1}{k} (-1)^{k-1} \binom{n-1}{k-1} \\
&= \frac{1}{n}\sum_{k=1}^n (-1)^{k-1} \frac{n}{k} \binom{n-1}{k-1} \\
&= \frac{1}{n}\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \\
&= \frac{1}{n}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.
| When expanding the product $(x^0 + x^1 + x^2)^6$, each term looks like $x^{k_1} x^{k_2} \cdots x^{k_6} = x^{k_1+k_2+\dots+k_6}$ where all $k_i\in \{0,1,2\}$. Hence, the coefficient of $x^6$ in the expansion is the number of six-tuples $(k_1,k_2,\dots,k_6)$ such that $k_1+k_2+\dots+k_6=6$ and all $k_i\in\{0,1,2\}$.
The number of $2$s in such a partition can be $0$, $1$, $2$ or $3$ and for each there is exactly one possible partition up to permuation:
\begin{align}
6 &= 1+1+1+1+1+1 \qquad\text{1 permutation}, \\
&= 1+1+1+1+2+0 \qquad\text{$\frac{6!}{4!\,1!\,1!}=30$ permutations}, \\
&= 1+1+2+2+0+0 \qquad\text{$\frac{6!}{2!\,2!\,2!}=90$ permutations}, \\
&= 2+2+2+0+0+0 \qquad\text{$\frac{6!}{3!\,3!}=20$ permutations.}
\end{align}
Hence, in total there are $141$ summands equal to $x^6$ in the expansion, so the term you are looking for is $141 x^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 7
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What is it to solve an equation forward? I'm reading a book in Monetary Economics and I don't understand a step.
I have this expression:
$$ \dfrac{\lambda_{t}}{P_{t}} = \beta \left( \dfrac{\lambda_{t+1} + \mu_{t+1}}{P_{t+1}} \right) $$
And then it says "solving this equation forward implies that":
$$ \dfrac{\lambda_{t}}{P_{t}} = \sum_{i=1}^{\infty} \beta^{i} \left( \dfrac{\mu_{t+i}}{P_{t+i}} \right) $$
I dont't know what they're doing. What is it to solve an equation forward?
| We can begin to solve this equation in two ways, an idea that generalizes to every time we have a relationship between the $t^{\text{th}}$ and $(t+1)^{\text{th}}$ term of a sequence.
One way is to start with the equation for $\frac{\lambda_{t}}{P_{t}}$, and use substitution to eliminate $\frac{\lambda_{t+1}}{P_{t+1}}$, then $\frac{\lambda_{t+2}}{P_{t+2}}$, and so on:
\begin{align}
\frac{\lambda_t}{P_t} &= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta \frac{\lambda_{t+1}}{P_{t+1}} \\
&= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta \left(\beta \frac{\mu_{t+2}}{P_{t+2}} + \beta \frac{\lambda_{t+2}}{P_{t+2}}\right) \\
&= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta^2 \frac{\mu_{t+2}}{P_{t+2}} + \beta^2 \frac{\lambda_{t+2}}{P_{t+2}} \\
&= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta^2 \frac{\mu_{t+2}}{P_{t+2}} + \beta^2 \left(\beta \frac{\mu_{t+3}}{P_{t+3}} + \beta \frac{\lambda_{t+3}}{P_{t+3}}\right) \\
&= \beta \frac{\mu_{t+1}}{P_{t+1}} + \beta^2 \frac{\mu_{t+2}}{P_{t+2}} + \beta^3 \frac{\mu_{t+3}}{P_{t+3}} + \beta^3 \frac{\lambda_{t+3}}{P_{t+3}} \\
&= \dots
\end{align}
The other way is to rewrite this equation as $\frac{\lambda_{t+1}}{P_{t+1}} = \beta^{-1} \frac{\lambda_t}{P_t} - \frac{\mu_{t+1}}{P_{t+1}}$ or better yet $\frac{\lambda_t}{P_t} = \beta^{-1} \frac{\lambda_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t}$, and then use substitution to eliminate $\frac{\lambda_{t-1}}{P_{t-1}}$, then $\frac{\lambda_{t-2}}{P_{t-2}}$, and so on:
\begin{align}
\frac{\lambda_t}{P_t} &= \beta^{-1} \frac{\lambda_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
&= \beta^{-1} \left(\beta^{-1} \frac{\lambda_{t-2}}{P_{t-2}} - \frac{\mu_{t-1}}{P_{t-1}}\right)- \frac{\mu_t}{P_t} \\
&= \beta^{-2} \frac{\lambda_{t-2}}{P_{t-2}} - \beta^{-1}\frac{\mu_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
&= \beta^{-2} \left(\beta^{-1} \frac{\lambda_{t-3}}{P_{t-3}} - \frac{\mu_{t-2}}{P_{t-2}}\right)- \beta^{-1}\frac{\mu_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
&= \beta^{-3} \frac{\lambda_{t-3}}{P_{t-3}} - \beta^{-2}\frac{\mu_{t-2}}{P_{t-2}}- \beta^{-1}\frac{\mu_{t-1}}{P_{t-1}} - \frac{\mu_t}{P_t} \\
&= \dots
\end{align}
I feel like it makes sense to say that in the first method, we're solving the equation "forward", since we're getting higher and higher indices on $\lambda$. In the second method, we're solving the equation "backward", getting lower and lower indices.
In the first method, assuming that $\beta^i \frac{\lambda_{t+i}}{P_{t+i}} \to 0$ as $i \to \infty$, we're eventually left with just an infinite sum with no $\lambda$'s in it. In the second method, we're left with a finite sum with no $\lambda$'s in it, assuming that we have an initial condition for $\frac{\lambda_0}{P_0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of a (rather general) recursive sequence So I am struggling with the following problem:
Let $F$ be a vector (sub)space of recursive sequences satisfying $x_{i+2} = bx_{i+1} + ax_{i}$ in field $\mathbb{K}$, with a vector space endomorphism $L: F \to F, (x_{i})_{i} \to (x_{i+1})_i $. Show that if $a>0$ and $b \neq 0$, $\lim_{i \to \infty} \frac{x_{i+1}}{x_{i}}$ exists and is equal to one of the eigenvalues of the endomorphism $L$.
I have shown that $F$ is indeed a vector subspace of $\mathbb{K}^\mathbb{N}$ and that the function $L$ is and an endomorphism of this vector subspace. I have proven that the $\dim F = 2$ and that we have a "natural" basis of two sequences $f_0 = (1, 0, ...)$ and $f_1 = (0, 1, ...)$. The endomorphism $L$ can then be described by a matrix:
\begin{equation*} A=
\begin{pmatrix}
0 & 1 \\
a & b
\end{pmatrix}
\end{equation*}
in this natural basis. I have then found two eigenvalues, $e_1 = \frac{b + \sqrt{b + 4a}}{2}$ and $e_2 = \frac{b - \sqrt{b + 4a}}{2}$. The related eigenvectors are then
\begin{equation*} v_{1,2} =
\begin{pmatrix}
2 \\
2e_{1,2}
\end{pmatrix}
\end{equation*}
We also see that:
\begin{equation*}
\begin{pmatrix}
x_i \\
x_{i+1}
\end{pmatrix} = A^i
\begin{pmatrix}
x_0 \\
x_1
\end{pmatrix}
\end{equation*}
Since $A$ is diagonalizable, we can write:
\begin{equation*}
\begin{pmatrix}
x_i \\
x_{i+1}
\end{pmatrix} = S D^i S^{-1}
\begin{pmatrix}
x_0 \\
x_1
\end{pmatrix}
\end{equation*}
with
\begin{equation*}
S =
\begin{pmatrix}
2 & 2\\
2e_1 & 2e_2
\end{pmatrix}, \; D =
\begin{pmatrix}
e_1 & 0\\
0 & e_2
\end{pmatrix}, \; S^{-1} = \frac{-1}{4\sqrt{b^2 + 4a}}\begin{pmatrix}
2e_2 & 2e_1\\
2 & 2
\end{pmatrix}
\end{equation*}
We can then find the explicit formula for $x_i$, we get:
\begin{equation}
x_i = \frac{-1}{4\sqrt{b^2 + 4a}}((4e_1^i e_2 + 4e_2^i)x_0 + (4e_1^{i+1} + 4e_2^{i})x_1)
\end{equation}
And so
\begin{equation*}
\lim_{i \to \infty} \frac{x_{i+1}}{x_i} = \lim_{i \to \infty} \frac{(4e_1^{i+1} e_2 + 4e_2^{i+1})x_0 + (4e_1^{i+2} + 4e_2^{i+1})x_1)}{(4e_1^i e_2 + 4e_2^i)x_0 + (4e_1^{i+1} + 4e_2^{i})x_1)}
\end{equation*}
And that is where I am stuck, if I could prove that this limit exists, I could simply use the fact that \begin{equation*}
\theta_{i+1} = \frac{x_{i+1}}{x_{i}} = \frac{bx_i + ax_{i-1}}{x_i} = b + a \frac{x_{i-1}}{x_i} = b + \frac{a}{\theta_{i-1}}
\end{equation*}
and then $\theta = \lim_{i \to \infty}\frac{x_{i+1}}{x_{i}}$ satisfies
\begin{equation*}
\theta = b + \frac{a}{\theta}
\end{equation*}
and so
\begin{equation*}
\theta = \frac{b \pm \sqrt{b^2 + 4a}}{2}
\end{equation*}
which is equal to the eigenvalues of $L$. But how do I prove that this limit exists in the first place? I tried to evaluate the explicit formula that I got, but I didn't find a way to do it. Could you please help?
| The limit in question can be calculated by simply dividing both numerator and denominator by $e_{0}^i$, where $e_{0}$ is the eigenvalue with the largest absolute value. $e_{0}$ is the limit in that case, hence, the point is proven.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $x^8-4x^6+14x^4+44x^2+25$ has three turning points Find a polynomial $f(x)$ with real coefficients that satisfies the following conditions:
*
*$f(x)$ has three turning points
*zeros are $i$ and $2 + i$
*the coefficient of the highest power of $x$ is $1$.
$f(x)$ has zeros $i,2+i$, hence $-i$ and $2-i$ are also zeros of that polynomial. But
$$\left(x-2-i\right)\left(x+2+i\right)\left(x^2+1\right)=\left(x^4-2x^2-3\right)+i\left(-4-4x^2\right)$$
contain complex coefficient. I try to remove the complex coefficient by
$$\left(\left(x^4-2x^2-3\right)+i\left(-4-4x^2\right)\right)\left(\left(x^4-2x^2-3\right)-i\left(-4-4x^2\right)\right)$$
$$
=x^8-4x^6+14x^4+44x^2+25$$
Which have roots $-i,i,2+i,2-i,-2+i,-2-i$ with the coefficient of the highest power of $x$ is $1$. But how to prove it has three turning points? I can show the existence of two turning points as $i$ and $-i$ has even multiplity $\implies$ $2\times 1$ bounce.
| Your original approach works best:
$$p(x)=\left(x-(2-i)\right)\left(x-(2+i)\right)\left(x^2+1\right) = (x^2-4x+5)(x^2+1) =x^4-4x^3+6x^2-4x+5$$
and multiplying this by a conjugate $q(x) = x^4 - 4x^3 + 6x^2 - 4x - 5$, we get:
$$f(x) = x^8 - 8 x^7 + 28 x^6 - 56 x^5 + 68 x^4 - 48 x^3 + 16 x^2 - 25$$
Multiplying by the conjugate gives three turning points! This is true as $f'(x) = p'(x) q(x) + p(x) q'(x)$, and since $p'(x) = q'(x)$ as only the constant term differs, we have $p'(x)(p(x) + q(x))$. $p'(x)$ must have at least one root as it is a cubic ($p'(x) = 4(x-1)^3$), $x$ is a factor of $p(x) + q(x)$ as the constant term is missing, and we are left with another cubic, whose root is not $0$ or $1$, so $f'(x)$ has at least three roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$
I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$
| It is very easy to use double integrals to handle this kinds of infinite sums. Let
$$ f(x)=\sum_{n=1}^\infty\frac1{n(2n+1)}x^{2n+1} $$
and then
$$ f'(x)=\sum_{n=1}^\infty\frac1{n}x^{2n}, f''(x)=\sum_{n=1}^\infty2x^{2n-1}=\frac{2x}{1-x^2}. $$
Then
\begin{eqnarray}
f(1)&=&\int_0^1f'(x)dx=\int_0^1\int_0^xf''(t)dtdx\\
&=&\int_0^1\int_t^1f''(t)dxdt=2\int_0^1\int_t^1\frac{t}{1-t^2}dxdt\\
&=&2\int_0^1\frac{t}{1-t^2}(1-t)dt=2\int_0^1\frac{t}{1+t}dt\\
&=&2(1-\ln2).
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Can any odd positive integer greater than $3$ be expressed as the sum of $2$ perfect square numbers (excluding $0$) plus $1$ prime number? I have been reading about the "odd Goldbach conjecture" which states that:
Every odd integer greater than $7$ can be written as the sum of three odd primes
I have also been reading about the fact that for large $$ there are considerably more primes than perfect squares numbers.
At first I have tried to run an experiment and see if any odd number can be written as the sum of one perfect square and one prime number, and although most attempts (in the beginning) were true, some others where false.
So since the "odd Goldbach conjecture" involves three different numbers, I have allowed myself to check if any odd number can be written as the sum of $2$ perfect squares and $1$ prime number.
I have manually checked it for the first few hundreds' results, and so far I have found no counter example for any odd positive integers greater than $3$
A prime number (greater than $3$) can be written as $6n+1$ or $6n-1$ and a perfect square number can be written as $4n$ or $4n + 1$.
Also, even though perfect squares are more rare than prime numbers, unlike prime numbers, perfect squares have $1$ at their disposal ($1^1 = 1$), and unfortunately for primes $1$ is not a prime number.
So my question is:
If there are any counter examples or any proofs? and if not what is the likelihood for this to be true?
Edit per comment: I have not considered $0$ as a perfect square.
Here are some examples of the early results:
$5 = 1^2 + 1^2 + 3$
$7 = 2^2 + 1^2 + 2$
$9 = 1^2 + 1^2 + 7$
$11 = 2^2 + 2^2 + 3$
$13 = 2^2 + 2^2 + 5$
$15 = 2^2 + 2^2 + 7$
$17 = 1^2 + 3^2 + 7$
$19 = 2^2 + 2^2 + 11$
$21 = 2^2 + 2^2 + 13$
$23 = 3^2 + 3^2 + 5$
$29 = 3^2 + 3^2 + 11$
$31 = 5^2 + 2^2 + 2$
$33 = 3^2 + 1^2 + 23$
$35 = 3^2 + 3^2 + 17$
$37 = 3^2 + 3^2 + 19$
$39 = 5^2 + 1^2 + 13$
$41 = 1^2 + 3^2 + 31$
$43 = 4^2 + 2^2 + 23$
$45 = 6^2 + 2^2 + 5$
$47 = 6^2 + 2^2 + 7$
$49 = 1^2 + 5^2 + 23$
$51 = 6^2 + 2^2 + 11$
$53 = 6^2 + 2^2 + 13$
$55 = 3^2 + 3^2 + 37$
,,,
| In 1923, Hardy and Littlewood [Some problems of “partitio numerorum”, III,
Acta Math., 44 (1923), pp. 1-70] conjectured that all sufficiently large integers $n$ can be written in the form $n=p+a^2+b^2$
where $p$ is a prime, and $a,b$ are integers. In 1959–1960, Linnik [Hardy–Littlewood problem on the representation as the sum of a prime and two squares,
Dokl. Akad. Nauk SSSR, 124 (1959), pp. 29-30 and
An asymptotic formula in an additive problem of Hardy and Littlewood,
Dokl. Akad. Nauk SSSR, 24 (1960), pp. 629-706] confirmed this conjecture.
See also Hooley, On the representation of a number as the sum of two squares and a prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Determine $\cos(\alpha+\theta)$ given $\sin \theta=3/5$ and $\cos \alpha=12/13$ I'm having some trouble with this question. I don't really understand what it is asking.
$\theta$ and $\alpha$ are acute angles in standard position. $\sin\theta=\frac{3}{5}$ and $\cos\alpha=\frac{12}{13}$
What is the exact value of $\cos(\alpha+\theta)$?
I've tried to find a trig identity that will allow us to solve this and haven't had any success.
All Help is appreciated! :)
I'm somewhat familiar with the cosine addition formula.
So far, I have
$\cos(\alpha+\theta) = \cos(\frac{12}{13})\cdot\cos(y)-\sin(\frac{3}{5})\cdot\sin(y)$
I now have
$\cos^2\theta+\sin^2\theta=1$
$\cos^2\theta=1-\frac{9}{25}$
$\sqrt{\cos^2\theta}=\sqrt{\frac{16}{25}}$
$\cos\theta=\frac{4}{5}$
Edit:
Thanks to everyone who helped out!
Here is the answer I came up with:
(cont. from above)
$\sin^2\theta=1-\cos^2\theta\\\sin^2\theta=1-\frac{144}{169}\\\sqrt{sin^2\theta}=\sqrt{\frac{25}{169}}\\\sin\theta=\frac{5}{13}\\\cos(\alpha+\theta)=(\frac{12}{13})(\frac{4}{5})-(\frac{5}{13})(\frac{3}{5})\\\therefore\cos(\alpha+\theta)=\frac{33}{65}$
| Thanks everyone!
Here is the answer I got:
$\cos^2\theta+\sin^2\theta=1\\\cos^2\theta=1-\sin^2\theta\\\cos^2\theta=1-(\frac{3}{5})^{2}\\\cos^2\theta=1-\frac{9}{25}\\\sqrt{\cos^{2}\theta}=\sqrt{\frac{16}{25}}\\\cos\theta=\frac{4}{5}$
$\sin^2\theta=1-\cos^2\theta\\\sin^2\theta=1-\frac{144}{169}\\\sqrt{sin^2\theta}=\sqrt{\frac{25}{169}}\\\sin\theta=\frac{5}{13}\\\cos(\alpha+\theta)=(\frac{12}{13})(\frac{4}{5})-(\frac{5}{13})(\frac{3}{5})\\\therefore\cos(\alpha+\theta)=\frac{33}{65}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Intermediate problem using Chain Rule
If $y=\frac{d}{dx} [\sin \sqrt{1+\cos (x)}]$ than, differentiate $x$.
$$\frac{d}{dx} [\sin \sqrt{1+\cos (x)}]$$
$$=\frac{d}{dx} [\sin (1+\cos (x))^{\frac{1}{2}}]$$
$$=\frac{1}{2} \cos (1+\cos (x))^{-\frac{1}{2}} (-\sin x)$$
$$=-\frac{\sin (x)}{2\cos (1+\cos (x))^\frac{1}{2}}$$
I found that the answer is wrong. I found the answer which was solved using
$$\frac{dx}{dy}=\frac{dx}{du} \frac{du}{dy} $$
But, I want to figure it out using
I was trying to solve above question as I did for this
$$\frac{d}{dx}[\tan(x^2+1)]=\sec^2 (x^2+1) \cdot 2x$$
| $$\frac{d}{dx}\sin (\sqrt{1+ \cos x })$$
Note : differentiate the outer function first which is $~\sin x $ and then inner function which is $\sqrt{1+ \cos x} $
Step 1 differentiating the outer function :
$$~\cos(\sqrt{1+cos x }) ~\frac{d}{dx}(\sqrt{1+ \cos x }) $$
Step 2 Differentiating the inner function :
$$= \frac{d}{dx}(\sqrt{1+ \cos x }) = \frac{1}{2}(1+\cos x)^{\frac {-1}{2}} { \frac{d}{dx}(1+ \cos x }) $$
step 3 :
$$ = { \frac{d}{dx}(1+ \cos x })= 0 -\sin x $$
$$ = \frac{d}{dx}(\sqrt{1+ \cos x }) = \frac{1}{2}(1+\cos x)^{\frac {-1}{2}} (-\sin x) $$
$$ = \frac{-\sin x}{2 {\sqrt{1 +\cos x }}} $$
Final step :
$$ \frac{d}{dx}\sin (\sqrt{1+ \cos x }) = ~- \frac {\cos ({\sqrt{1+ \cos x})}~\sin x}{2 {\sqrt{1 +\cos x }} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$ I am trying to compute the following double sum
$$\boxed{\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)}$$
I proceeded as following
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{\left(m+n \right)\left(m-n \right)}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{2m}\left[\frac{1}{\left(m+n \right)}+\frac{1}{\left(m-n \right)}\right]$$
$$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m+n \right)}}_{2\zeta(3)}+ \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n} \sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m-n \right)}$$
The first sum is given here and I have also evaluated here, therefore we get that
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \sum_{m=1}^{\infty}\frac{1}{m}-\frac{1}{\left(m-n \right)}$$
$$=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \int_{0}^{1}\frac{1-x^{-n}}{1-x}dx$$
$$=\zeta(3)-\frac{1}{2} \int_{0}^{1}\frac{1}{1-x}\left[ \sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{x^{-n}}{n^2}\right]dx$$
$$=\zeta(3)-\frac{1}{2} \int_{0}^{1}\frac{\zeta(2)-Li_{2}\left( \frac{1}{x} \right)}{1-x}dx$$
From this post we can use the relation
$$Li_{2}\left( \frac{1}{x} \right)-Li_{2}\left( x \right)=\zeta(2)-\frac{1}{2}\log^2(-x)$$
To get
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{\zeta(2)}{2} \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2} \int_{0}^{1}\frac{Li_{2}\left( x \right)+\zeta(2)-\frac{1}{2}\log^2(-x)}{1-x}dx$$
$$=\zeta(3)+\frac{\zeta(2)}{2} \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2} \int_{0}^{1}\frac{Li_{2}\left( x \right)}{1-x}dx-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$=\zeta(3)+\frac{\zeta(2)}{2} \log(1-x)\Big|_{0}^{1}+\frac{1}{2}\left\{ -Li_{2}(x)\log(1-x)\Big|_{0}^{1}-\int_{0}^{1}\frac{\log^2(1-x)}{x}dx\right\} -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$=\zeta(3)-\frac{1}{2}2\zeta(3) -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
My questions are the following:
(1)Is the last equality correct?
(2)If so, what is the meaning of $\log^2(-x)=\log^2(e^{i\pi}x)$?
(3) How do we solve the last integral?
| This doesn't really answer your questions, however consider another approach perhaps as a way of side-lining somewhat difficult integrals
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{n'=1}^{\infty }{\sum\limits_{m'=1}^{\infty }{\frac{1}{{{n}^{2}}\left( m-n \right)}}}-\frac{1}{2}\sum\limits_{n=1}^{\infty }{\sum\limits_{m=1}^{\infty }{\frac{1}{{{n}^{2}}\left( m+n \right)}}}+\frac{1}{4}\sum\limits_{m=1}^{\infty }{\frac{1}{{{m}^{3}}}}$$
Primes on sums denote excluding $n=m$. Note
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}},n\ge 0$$
So we have by the theorem that the sum over all residues is zero
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ -\underset{z=0}{\mathop{res}}\,\frac{\psi \left( -z \right)+\gamma }{{{z}^{2}}\left( m-z \right)}+\underset{z=-m,0}{\mathop{res}}\,\frac{\psi \left( -z \right)+\gamma }{{{z}^{2}}\left( m+z \right)}+\frac{1}{2{{m}^{3}}} \right\}}$$
The first residue calculation specifically excludes the residue at $z=m$ to avoid the singularities at $n=m$. Using the expansion above to determine the residues we find
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}\\=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ \frac{{{m}^{2}}{{\pi }^{2}}-6}{6{{m}^{3}}}+\frac{-1+\gamma +\psi \left( m \right)}{{{m}^{2}}}+\frac{6+6m-{{m}^{2}}{{\pi }^{2}}}{6{{m}^{3}}}+\frac{1}{2{{m}^{3}}} \right\}}$$
Cleaning this up
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ \frac{\gamma +\psi \left( m \right)}{{{m}^{2}}}+\frac{1}{2{{m}^{3}}} \right\}}=\frac{1}{2}\left\{ \zeta \left( 3 \right)+\frac{1}{2}\zeta \left( 3 \right) \right\}=\frac{3}{4}\zeta \left( 3 \right)$$
Where we’ve used the reasonably well known representation
$$\zeta \left( 3 \right)=\sum\limits_{m=1}^{\infty }{\frac{\gamma +\psi \left( m \right)}{{{m}^{2}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}$ How to evaluate this definite integral from MIT Integration Bee 2006?
$$\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}.$$
So far, I have shown that the indefinite integral is
$$\frac{2x^3 + 3x - 2(1+x^2)^{3/2}}{3}.$$
At $x = 0$, the expression above equals $-\dfrac{2}{3}.$
Using WolframAlpha, I also know that the definite integral equals $\dfrac{2}{3}$.
So the only thing left to show is $$\lim_{x\rightarrow \infty} \frac{2x^3 + 3x - 2(1+x^2)^{3/2}}{3}=0.$$
I'm not sure how to calculate this limit.
| Let’s tackle the limit by L’Hospital Rule and Rationalization.
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[2 x^{3}+3 x-2\left(1+x^{2}\right)^{\frac{3}{2}}\right] \\
=& \lim _{x \rightarrow \infty} \frac{2+\frac{3}{x^{2}}-2\left(\frac{1}{x^{2}}+1\right)^{\frac{3}{2}}}{\frac{1}{x^{3}}} \quad \left(\frac{0}{0}\right) \\
=& \lim _{x \rightarrow \infty} \frac{-\frac{6}{x^{3}}-3 \sqrt{\frac{1}{x^{2}}+1}\left(-\frac{2}{x^{3}}\right)}{-\frac{3}{x^{4}}} \\
=& 2 \lim _{x \rightarrow \infty} \frac{\left(x-\sqrt{1+x^{2}}\right)\left(x+\sqrt{1+x^{2}}\right)}{x+\sqrt{1+x^{2}}} \\
=&-2 \lim _{x \rightarrow \infty} \frac{1}{x+\sqrt{1+x^{2}}} \\
=& 0
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Directly Computing Posterior Distribution for Gaussian Likelihood and Prior my first question so apologies in advance.
I'm currently studying posterior distributions and for practice sake I'm trying to directly compute the posterior distribution for the mean of a normal random variable (unit variance) with standard normal prior. I'm trying to validate if this is correct (note that this is for observing ONE new data point).
The posterior distribution is defined as
$$p(\theta\mid x)=\frac{p(x\mid\theta)p(\theta)}{\int_\theta p(x\mid\theta) p(\theta) \, d\theta}$$
and I am getting
$$p(\theta\mid x)=\frac{1}{\sqrt{\pi}}e^{-\frac{1}{4}x^2+x\theta-\theta^2}$$
I did a couple sanity checks,
$$\int_{-\infty}^\infty p(\theta\mid X=0)\,d\theta=\int_{-\infty}^\infty \frac{1}{\sqrt{\pi}} e^{-\theta^2} \, d\theta=1$$
$$\int_{-\infty}^\infty p(\theta\mid X=2) \, d\theta=\int_{-\infty}^\infty \frac{1}{\sqrt{\pi}} e^{-1+2\theta-\theta^2} \, d\theta=1$$
Is this the correct analytical solution?
My derivation is as follows:
$$\text{Let } X \sim N(\theta,1)$$
$$\text{Let } \theta \sim N(0,1)$$
The likelihood (dependent on theta):
$$p(x\mid\theta)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-\theta)^2}$$
The prior:
$$p(\theta)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\theta^2}$$
The joint density:
$$p(x\mid\theta)p(\theta)=\frac{1}{2\pi}e^{-\frac{1}{2}(x-\theta)^2}e^{-\frac{1}{2}\theta^2}$$
$$=\frac{1}{2\pi}e^{-\frac{1}{2}x^2+x\theta-\frac{1}{2}\theta^2}e^{-\frac{1}{2}\theta^2}$$
$$=\frac{1}{2\pi}e^{-\frac{1}{2}x^2+x\theta-\theta^2}$$
sanity check:
$$\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{1}{2\pi}e^{-\frac{1}{2} x^2 + x\theta-\theta^2} \, dx \, d\theta=1$$
The marginal:
$$p(x)=\int_\theta p(x\mid\theta)p(\theta) \, d\theta$$
$$=\frac{1}{2\pi}e^{-\frac{1}{2}x^2} \int_{-\infty}^\infty e^{-\theta^2+\theta x} \, d\theta$$
$$=\frac{1}{2\pi}e^{-\frac{1}{2}x^2}\Biggl[\sqrt{\pi}e^{\frac{1}{4}x^2}\Biggl] \text{by Gaussian integral}$$
$$=\frac{1}{2\sqrt{\pi}}e^{-\frac{1}{4}x^2}$$
sanity check:
$$\int_{-\infty}^\infty \frac{1}{2\sqrt{\pi}}e^{-\frac{1}{4}x^2} \, dx=1$$
The posterior:
$$p(\theta\mid x)=\frac{p(x\mid\theta)p(\theta)}{\int_\theta p(x\mid \theta) p(\theta) \, d\theta} = \frac{\frac{1}{2\pi}e^{-\frac{1}{2}x^2+x\theta-\theta^2}}{\frac{1}{2\sqrt{\pi}}e^{-\frac{1}{4}x^2}}$$
$$=\frac{\sqrt{\pi}}{\pi}e^{-\frac{1}{2}x^2+x\theta-\theta^2+\frac{1}{4}x^2}$$
$$=\frac{1}{\sqrt{\pi}}e^{-\frac{1}{4}x^2+x\theta-\theta^2}$$
| \begin{align}
& (x-\theta)^2 + \theta^2 = \big(x^2-2x\theta+\theta^2\big) + \theta^2 \\[8pt]
= {} & 2\theta^2 - 2x\theta + x^2 \\[6pt]
= {} & 2(\theta^2 - x\theta) + x^2 \\[6pt]
= {} & 2\left( \theta^2 - x\theta + \tfrac 1 4 x^2 \right) + x^2 -\tfrac 1 2 x^2 \quad\text{(completing the square)} \\[6pt]
= {} & 2\left( \theta - \tfrac 1 2 x \right)^2 + \tfrac 1 2 x^2 \\[8pt]
\text{So } & e^{-\frac 1 2(x-\theta)^2} e^{-\frac 1 2 \theta^2} = \exp\left(-\frac 1 {2\sigma^2} \left( \theta-\frac 1 2 x \right)^2 \right) \times \text{constant} \\[6pt]
\text{where } & \sigma^2 = 1/2 \text{ and “constant” means not depending on } \theta.
\end{align}
Thus the posterior mean is $x/2$ and the posterior variance is $1/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I wrote the series of $e^{a \sin^{-1}x}$ but it contained terms like $(\sin^{-1}x)^2$, $(\sin^{-1}x)^3$ and so on so I could not find the series. My idea was to find the series of the anti derivative of the function and then to derivate the obtained series. Any other way to do it?
| Let
$$
f(x) = e^{a\arcsin x} = \sum\limits_{n = 0}^\infty {f_n x^n } .
$$
As you already observed, $f(x)$ satisfies the non-linear ODE $(1 - x^2 )(f'(x))^2 = a^2f^2 (x)$. Differentiating this equation and dividing through by $2f'(x)$ yields
$$
(1 - x^2 )f''(x) - xf'(x) - a^2 f(x) = 0.
$$
Substituting the power series into this equation gives $f_0 = 1$, $f_1 = a$ (you can see from the definition that $f(0) = 1$, $f'(0) = a$) and
$$
f_{n + 2} = \frac{{a^2+n^2}}{{(n + 1)(n + 2)}}f_n
$$
for $n\geq 0$. The power series expansion you are asking for then follows since
$$
\frac{{e^{a\arcsin x} }}{{\sqrt {1 - x^2 } }} = \frac{1}{a}f'(x) = \sum\limits_{n = 0}^\infty {\frac{{(n + 1)f_{n + 1} }}{a}x^n } .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
} |
Root of unity , find $Re(z^{2557})+Im(z^{2014})$
Let $x = e^{\frac{2i\pi}{2n+1}}$ and let $z = \frac{1}{2} + x + x^2 + \cdots + x^n$. Find $Re(z^{2557}) + Im(z^{2014})$.
My Work
$1 + x + x^2 + \cdots + x^{2n} = 0$
$(1 + x + x^2 + \cdots + x^n) + x^n(1 + x + x^2 + \cdots + x^n) = x^n$
$z +\frac{1}{2} = \frac{x^n}{1+x^n}$
$z = \frac{x^n - 1}{2(1+x^n)}$.
Now, how can I proceed?
Thanks in advance!
| Let $\frac {2n\pi}{2n+1}=a$
Then we have, from Euler's formula:
$$z=\frac {x^n-1}{2(x^n+1)}=\frac {(\cos a-1)+i\sin a}{2((\cos a+1) +i\sin a)}=\frac {((\cos a-1)+i\sin a)((\cos a+1)-i\sin a)}{2((\cos a +1)^2+ \sin^2 a)}=\frac {2i\sin a}{4(1+\cos a)}=\frac {\tan {\frac {a}{2}}}{2} i$$
So, $z$ is purely imaginary, hence real part will be zero and you will get a closed form for the imaginary part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?
| $x=t^2+2t$
$y=3t^4+4t^3$
$\displaystyle\frac{dx}{dt}=2t+2$ and $\frac{dy}{dt}=12t^3+12t^2$
$\displaystyle\frac{dy}{dx}=\frac{12t^3+12t^2}{2t+2}=\frac{6t^3+6t^2}{t+1}=6t^2$
$\displaystyle\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}6t^2=12t\frac{dt}{dx}=\frac{12t}{\frac{dx}{dt}}=\frac{12t}{2t+2}=\frac{6t}{t+1}$
Now when $x=8$, $t=-4,2$ and therefore corresponding values of $y=512,80$, Since we need to calculate $\frac{d^2y}{dx^2}$ at the point $(8,80)$ which corresponds to $t=2$ therefore ,
$\displaystyle\frac{d^2y}{dx^2}=\frac{6t}{t+1}=\frac{6*2}{2+1}=\fbox4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Solve $y' + y \cot x = 2 \cos x$ on $(0,\pi)$, and prove that there's a solution on $(-\infty,+\infty)$ Problem
This question concerns exercise #8 from section 8.5 of Calculus, Vol. I (Apostol):
Find all solutions of $y' + y \cot x = 2 \cos x$ on the interval $(0,\pi)$. Prove that exactly one of these is also a solution on $(-\infty,+\infty)$.
My specific doubt is about the portion "Prove that exactly one of these is also a solution on $(-\infty,+\infty)$", but I will give the full solution attempt for completeness.
Solution attempt
This equation has the form $y' + P(x)y + Q(x)$, where $P(x)$ and $Q(x)$ are continuous on the interval $(0,\pi)$. Assuming the initial condition that $y = b$ when $x = a$, the solution $y = f(x)$ is given by
$$f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$$
where $A(x) = \int_a^x P(t) dt$.
To apply this formula, we choose $a = \pi/2$, so that all solutions will be expressed in terms of $b$. First, we compute:
$$A(x) = \int_{\pi/2}^x \cot tdt = \log (\sin x) - \log(\sin (\pi/2)) = \log(\sin x)$$
So, we get:
$$\begin{aligned}
f(x) &= be^{-A(x)} + e^{-A(x)} \int_{\pi/2}^x 2(\cos t) e^{A(t)} dt \\
&= be^{-\log(\sin x)} + 2e^{-\log(\sin x)} \int_{\pi/2}^x (\cos t) e^{\log(\sin t)} dt \\
&= \dfrac{b}{\sin x} + \dfrac{2}{\sin x} \int_{\pi/2}^x \cos t \sin t dt \\
&= \dfrac{b}{\sin x} + \dfrac{2}{\sin x} \int_1^{\sin x} u du & (u = \sin t,\ du = \cos t dt) \\
&= \dfrac{b}{\sin x} + \dfrac{2}{\sin x} \dfrac{\sin^2 x - 1}{2} \\
&= \dfrac{b}{\sin x} + \dfrac{\sin^2 x - 1}{\sin x} \\
&= \dfrac{b - 1}{\sin x} + \sin x \\
&= \sin x + \dfrac{C}{\sin x} & (C = b - 1)
\end{aligned}$$
The doubt arises in the part of proving that exactly one of these is also a solution on $(-\infty,+\infty)$.
If we let $C = 0$, we get the solution $f(x) = \sin x$. This function is defined for all $x$, so, at first, it appears that this is the answer. However, the original equation is not defined in $x=0$ to begin with, because $\cot x$ is not defined at $x = 0$. So, it seems that there shouldn't be a solution valid on the whole interval $(-\infty,+\infty)$.
What am I missing?
| This is
$y'\sin x + y \cos x = 2 \cos x \sin x
$
or
$(y\sin x)' =\sin(2x)
$
so
$y\sin(x)
=-\frac12\cos(2x)+c
$
so
$y
=-\dfrac{\cos(2x)-c}{2\sin(x)}
=-\dfrac{1-2\sin^2(x)-c}{2\sin(x)}
=\sin(x)+\dfrac{c_1}{\sin(x)}
$.
For this to be defined
for all $x$,
we must have
$c_1 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
probability of rolling 3 dice numbers which product is a multiple of 6 As the question states, I need to find the probability of rolling 3 dice numbers which product is a multiple of 6.
what I have tried:
$\omega={(6,k,k), (3,2,k), (3,4,k))}$
$P(A) = \frac{3\times(1^1\times6^2) + 6\times(1^1\times1^1\times6^1) + 6\times(1^1\times1^1\times6^1)}{6^3}=\frac{180}{216}=\frac{5}{6}$
I know writing $1^1$ seems silly, but just trying to show what I'm doing
Thanks for helping and sorry if it is a bad question
| Let $a_2,b_2,c_2$ ($a_3,b_3,c_3$) denote the exponents of $2$ ($3$) in the prime factorization of the number shown on the three dice. Then the probability you want is
$$P(a_2+b_2+c_2>0, a_3+b_3+c_3>0) = 1-P(a_2+b_2+c_2=0\ or \ a_3+b_3+c_3=0)$$
$$ = 1-P(a_2+b_2+c_2=0)-P(a_3+b_3+c_3=0)+P(a_2+b_2+c_2=0, a_3+b_3+c_3=0)$$
$$ (by\ independence)= 1-P(a_2=0)^3-P(a_3=0)^3+P(a_2=0,a_3=0)^3$$
$$ = 1-P(1,3,\ or \ 5)^3-P(1,2,4, \ or \ 5)^3+P(1 \ or \ 5)^3$$
$$ = 1 - (\frac{3}{6})^3 - (\frac{4}{6})^3+(\frac{2}{6})^3 = \frac{133}{216}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
If $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ and $\cos3\alpha=\dfrac34$ then find $\cos^8\alpha+\cos^8\beta+\cos^8\gamma$
If $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ and $\cos3\alpha=\dfrac34$ then $\cos^8\alpha+\cos^8\beta+\cos^8\gamma=$
*
*A) $\dfrac3{128}$
*B) $\dfrac{27}{32}$
*C) $\dfrac3{16}$
*D) $\dfrac5{128}$
ATTEMPT $1$: $e^{i\alpha}=\cos\alpha+i\sin\alpha$, $e^{i\beta}=\cos\beta+i\sin\beta$, $e^{i\gamma}=\cos\gamma+i\sin\gamma$
Adding, we get $e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0$
Now, if $a+b+c=0\implies a^3+b^3+c^3=3abc$
Therefore, $e^{i3\alpha}+e^{i3\beta}+e^{i3\gamma}=3e^{i(\alpha+\beta+\gamma)}$
$\implies \cos3\alpha+i\sin3\alpha+\cos3\beta+i\sin3\beta+\cos3\gamma+i\sin3\gamma=3\cos(\alpha+\beta+\gamma)+3i\sin(\alpha+\beta+\gamma)$
Sure, $\cos3\alpha$ is given but what about the rest? Also, we need $\cos^8\alpha$. How to get that? On RHS, can we use the following formula?
$\cos(A+B+C)=\cos A\cos B\cos C-\cos A\sin B \sin C-\cos B\sin C\sin A-\cos C\sin A\sin B$
ATTEMPT $2$: $$\cos3\alpha=\frac34\\4\cos^3\alpha-3\cos\alpha=\frac34\\\cos\alpha(4\cos^2\alpha-3)=\frac34\\4\cos^2\alpha-3=\frac34\sec\alpha$$
Also, $0\le4\cos^2\alpha\le4\implies-3\le4\cos^2\alpha-3\le1\implies-3\le\frac34\sec\alpha\le1\implies-4\le\sec\alpha\le\frac43$
But $\sec\alpha\ge1\implies1\le\sec\alpha\le\frac43$
Does that help?
ATTEMPT $3$: Hit and try. If $\beta=\frac{\pi}2\implies\alpha+\gamma=\pi$
But it doesn't fit sine equation. Have tried with $0,\pi$ etc. Not working.
| $e^{i\alpha}$, $e^{i\beta}$ and $e^{i\gamma}$ represent the three vertices of an equilateral triangle inscribed in the unit circle. WLOG, let $\beta=\alpha+\dfrac{2\pi}3$ and $\gamma=\alpha+\dfrac{4\pi}3$.
\begin{align*}
\cos^8\alpha+\cos^8\beta+\cos^8\gamma&=
\dfrac{(e^{i\alpha}+e^{-i\alpha})^8}{256}+\dfrac{(e^{i\beta}+e^{-i\beta})^8}{256}+\dfrac{(e^{i\gamma}+e^{-i\gamma})^8}{256}
\end{align*}
Note that $e^{ik\alpha}+e^{ik\beta}+e^{ik\gamma}=0$ if $k$ is not a multiple of $3$.
Using the binomial theorem,
\begin{align*}
&\;\cos^8\alpha+\cos^8\beta+\cos^8\gamma\\
=&\;\dfrac{(e^{i\alpha}+e^{-i\alpha})^8}{256}+\dfrac{(e^{i\beta}+e^{-i\beta})^8}{256}+\dfrac{(e^{i\gamma}+e^{-i\gamma})^8}{256}\\
=&\;\frac{8e^{6i\alpha}+8e^{6i\beta}+8e^{6i\gamma}+8e^{-6i\alpha}+3(70)+8e^{-6i\beta}+8e^{-6i\gamma}}{256}\\
=&\;\frac{24e^{6i\alpha}+24e^{-6i\alpha}+210}{256}\\
=&\;\frac{48\cos6\alpha+210}{256}\\
=&\;\frac{48(2\cos^23\alpha-1)+210}{256}\\
=&\;\frac{27}{32}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
*
*A) $x^2-2x+2=0$
*B) $x^2-5x+5=0$
*C) $x^2-7x+7=0$
*D) $x^2-9x+9=0$
Method $1$:$$\sec^2\theta+\csc^2\theta=\frac1{\cos^2\theta}+\frac1{\sin^2\theta}\\=\frac1{\sin^2\theta\cos^2\theta}\\=\frac4{\sin^22\theta}\ge4$$
Also, $\sec^2\theta\csc^2\theta=\dfrac1{\sin^2\theta\cos^2\theta}$
So, options $B),C),D)$ are correct.
Method $2$: Let the quadratic equation be $x^2-px+q=0$
So, $\sec^2\theta+\csc^2\theta=p, \sec^2\theta\csc^2\theta=q\implies \csc^2\theta=\dfrac{q}{\sec^2\theta}$
Putting that in the sum of roots, we get $$\sec^2\theta+\frac{q}{\sec^2\theta}=p\\\implies\sec^4\theta-p\sec^2\theta+q=0\\\implies\sec^2\theta=\frac{p\pm\sqrt{p^2-4q}}2\ge1\\\implies p\pm\sqrt{p^2-4q}\ge2\\\implies\pm\sqrt{p^2-4q}\ge2-p\\\implies p^2-4q\ge4+p^2-4p\\\implies p-q\ge1$$
What's wrong in this method?
| See my comment under the question and go on from there, as follows:
$$
\frac {-p\pm\sqrt{p^2+4p}} 2 = \begin{cases} -1\pm\sqrt3 & \text{if } p=2, \\
\quad\cdots & \text{if } p=5, \\ \quad\cdots & \text{if } p=7, \\ \quad\cdots & \text{if } p=9. \end{cases}
$$
When $p=2,$ one of the solutions is negative and so cannot be $\sec^2\theta$ or $\csc^2\theta$ if $\theta$ is real. The other one is negative and so cannot be $\sec^2\theta$ or $\csc^2\theta$ if $\theta$ is real.
Go on from there is a similar way.
| {
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"url": "https://math.stackexchange.com/questions/4173180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing $\frac{\sin((2n+1)\theta)}{\sin\theta}=(2n+1)\prod_{k=1}^n\left(1-\frac{\sin^2\theta}{\sin^2\frac{k\pi}{2n+1}}\right)$ A problem I solved previously I get
$$\frac{\sin(n\theta)}{\sin(\theta)} = 2^{n-1}\prod_{k=1}^{n-1}\bigg(\cos(\theta)-\cos{\bigg(\frac{k\pi}{n}}\bigg)\bigg)$$
So considering $z^{4n+2} = 1$ leads to
$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = 2^{2n}\prod_{k=1}^{2n}\bigg(\cos(\theta)-\cos\bigg(\frac{k\pi}{2n+1}\bigg)\bigg)......................(1)$$
Now
$$\cos\bigg(\frac{2n\pi}{2n+1}\bigg) = \cos\bigg(\pi-\frac{\pi}{2n+1}\bigg)
=-\cos\bigg(\frac{\pi}{2n+1}\bigg)$$
$$\cos\bigg(\frac{(2n-1)\pi}{2n+1}\bigg) = \cos\bigg(\pi-\frac{2\pi}{2n+1}\bigg) =-\cos\bigg(\frac{2\pi}{2n+1}\bigg)$$
$$...........................................................$$
$$\cos\bigg(\frac{(n+1)\pi}{2n+1}\bigg) = \cos\bigg(\pi-\frac{n\pi}{2n+1}\bigg) =-\cos\bigg(\frac{n\pi}{2n+1}\bigg)$$
multiplying terms $k = 1$ with $k = 2n, k =2$ with $k = 2n-1\dots k = n$ with $k = n+1$ terms, from$......(1)$ I get
$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = 2^{2n}\cdot\bigg(\cos(\theta)+\cos\bigg(\frac{\pi}{2n+1}\bigg)\bigg)\cdot \bigg(\cos(\theta)-\cos\bigg(\frac{\pi}{2n+1}\bigg)\bigg)\hspace{82pt}\cdot\bigg(\cos(\theta)+\cos\bigg(\frac{2\pi}{2n+1}\bigg)\bigg)\cdot\bigg(\cos(\theta)-\cos\bigg(\frac{2\pi}{2n+1}\bigg)\bigg)........\hspace{56pt}\bigg(\cos(\theta)+\cos\bigg(\frac{n\pi}{2n+1}\bigg)\bigg)\cdot\bigg(\cos(\theta)-\cos\bigg(\frac{n\pi}{2n+1}\bigg)\bigg)$$
$$ \hspace{35pt}= 2^{2n}\prod_{k = 1}^{n} \bigg(\cos^{2}(\theta)-\cos^{2}\bigg(\frac{k\pi}{2n+1}\bigg)\bigg).............(2)$$
How can I prove
$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = (2n+1)\prod_{k = 1}^{n}\bigg(1-\frac{\sin^2(\theta)}{\sin^2\big(\frac{k\pi}{2n+1}\big)}\bigg)$$
from my calculation? Is my way of approaching this problem is ok or not? If it isn't give me one or a few hints.
| I figured how to do it
from $.....(2)$ we get-
$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = 2^{2n}\prod_{k=1}^{n}\bigg(1-sin^{2}(\theta)-1+sin^2 \bigg(\frac{k\pi}{2n+1}\bigg)\bigg)$$
$$\hspace{60pt} = 2^{2n}\prod_{k=1}^{n}\sin^{2}\bigg(\frac{k\pi}{2n+1}\bigg)\bigg(1-\frac{\sin^2(\theta)}{\sin^{2}\big(\frac{k\pi}{2n+1})}\bigg)$$
$$\hspace{79pt}=2^{2n}\prod_{k=1}^{n}\sin^{2}\bigg(\frac{k\pi}{2n+1}\bigg)\prod_{k=1}^{n}\bigg(1-\frac{\sin^2(\theta)}{\sin^{2}\big(\frac{k\pi}{2n+1})}\bigg)$$
$$\hspace{320pt}...........(3)$$
Now from here we know $$\prod_{k=1}^{n-1}\sin \bigg(\frac{k\pi}{n}\bigg) = \frac{n}{2^{n-1}}$$
putting $2n+1$ instead of $n$ we get
$$\hspace{5pt}\prod_{k=1}^{2n}\sin \bigg(\frac{k\pi}{2n+1}\bigg) = \frac{2n+1}{2^{2n}}$$
$$\Rightarrow2^{2n}\prod_{k=1}^{2n}\sin \bigg(\frac{k\pi}{2n+1}\bigg) = 2n+1$$
$$\hspace{260pt}..............(4)$$
$$\sin\bigg(\frac{2n\pi}{2n+1}\bigg) = \sin\bigg(\frac{(2n+1)\pi-\pi}{2n+1}\bigg) = \sin\bigg(\pi-\frac{\pi}{2n+1}\bigg) = \sin\bigg(\frac{\pi}{2n+1}\bigg)$$
$$\sin\bigg(\frac{(2n-1)\pi}{2n+1}\bigg) = \sin\bigg(\frac{(2n+1)\pi-2\pi}{2n+1}\bigg) = \sin\bigg(\pi-\frac{2\pi}{2n+1}\bigg) = \sin\bigg(\frac{2\pi}{2n+1}\bigg)$$
$$.............................................................................$$
$$\sin\bigg(\frac{(n+1)\pi}{2n+1}\bigg) = \sin\bigg(\frac{(2n+1)\pi-n\pi}{2n+1}\bigg) = \sin\bigg(\pi-\frac{n\pi}{2n+1}\bigg) = \sin\bigg(\frac{n\pi}{2n+1}\bigg)$$
So from...(4) we get
$$2^{2n}\prod_{k=1}^{n}\sin^{2}\bigg(\frac{k\pi}{2n+1}\bigg) = 2n+1$$
and
$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = (2n+1)\prod_{k=1}^{n}\bigg(1-\frac{\sin^2(\theta)}{\sin^{2}\big(\frac{k\pi}{2n+1})}\bigg)$$
$$[proved]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Approach to calculating limit via polynomial division Given the following exercise:
$ \lim_{x \to \frac{1}{3}} (\frac{27x^2-6x-1}{6x^2+x-1}) $
The rule of L'Hospital with the first derivative yields:
$\lim_{x \to \frac{1}{3}} \frac{27 x^2 - 6 x - 1}{6 x^2 + x - 1} = \frac{12}{5}$
Another approach would be factoring out the x which also yields:
$\frac{12}{5}$
However, I'm asking myself whether polynomial division, too, could help solving the limit.
Polynomial division yields:
$27 x^2 - 6 x - 1 = \frac{9}{2} \cdot (6 x^2 + x - 1) + \frac{7}{2} - \frac{21x}{2}$
polynomial division on Wolfram Alpha
Thanks.
| Hint: To use polynomial division, I would first write (for $x \neq \frac13$):
$$\frac{27x^2-6x-1}{6x^2+x-1} = \frac92 - \frac7{2(2x+1)}$$
Now taking the limit is trivial...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find a general term of a trinomial? I want to find the coefficient of $x^0$ in the expansion of $(x + 1 + 1/x)^4$.
Without an expansion, I keep getting "nested binomial" terms if I group two terms together:
$$\binom{4}{k}\binom{k}{m} (x)^{k-m}(1/x)^{m}$$
For which $k-2m = 0$. I cannot solve further without guessing and checking a value. What is a better way?
| One approach is to multiply by $x^4$ and then find the coefficient of $x^4$ in the result.
Notation: $[x^n]f(x)$ denotes the coefficient of $x^n$ in $f(x)$. Then
$$\begin{align}
[x^0](x+1+1/x)^4 &= [x^4]x^4 (x+1+1/x)^4 \\
&=[x^4](x^2+x+1)^4 \\
&=[x^4] \left( \frac{1-x^3}{1-x} \right)^4 \\
&=[x^4](1-x^3)^4 \;(1-x)^{-4} \\
&=[x^4](1 -4x^3 +O(x^6)) \sum_{i=0}^{\infty} \binom{4+i-1}{i} x^i \tag{*} \\
&= \binom{4+4-1}{4} - 4 \binom{4+1-1}{1} \\
&= 19
\end{align}$$
where at $(*)$ we have used the Binomial Theorem for negative exponents.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving a recursive inequality I have the following inequality for $\{x_t\}_{t\geq 0}$ and trying to find an upper bound for $x_t$. Let $c$ be a constant.
$$x_{t+1}\leq c^2\left[x_t+\frac{2t-3}{4}-\frac{tc^t}{2}+\frac{3c^{2t}}{4}\right]+\frac{1}{4}\tag{1}$$
My attempt: If we had an equality instead of inequality in $(1)$, I would have used generating function of $x_t$ to solve for $x_t$ but for an inequality I don't think I can use generating functions (?). So I tried to think of $x(t)$ as a continuous function of $t$ and $(1)$ being a finite difference approximation. Then:
\begin{align*}
x_{t+1}-x_t &\leq (c^2-1)x_t+c^2\left[\frac{2t-3}{4}-\frac{tc^t}{2}+\frac{3c^{2t}}{4}\right]+\frac{1}{4} \\ \\
x_{t+1}-x_t - (c^2-1)x_t &\leq c^2\left[\frac{2t-3}{4}-\frac{tc^t}{2}+\frac{3c^{2t}}{4}\right]+\frac{1}{4} \\ \\
\frac{dx}{dt}-(c^2-1)x &\leq c^2\left[\frac{2t-3}{4}-\frac{tc^t}{2}+\frac{3c^{2t}}{4}\right]+\frac{1}{4} \\ \\
\frac{d}{dt}\left( e^{-(c^2-1)t}x\right) &\leq e^{-(c^2-1)t}c^2\left[\frac{2t-3}{4}-\frac{tc^t}{2}+\frac{3c^{2t}}{4}\right]+\frac{1}{4} \\ \\
\end{align*}
and then integrate to find an upper bound on $x=x(t)$ but I'd think this is an approximation for an actual upper bound for $x_t$ and am not sure how to solve $(1)$ with or without using what I did above.
| Following the hint in the comments, write $y_t = \frac{x_t}{c^{2t}}$. Then if we divide both sides of your inequality by $c^{2t+2}$ we find
$$
y_{t+1} \leq y_t + \frac{2t - 3}{4}c^{-2t} - \frac{t}{2} c^{-t} + \frac{3}{4} + \frac{1}{4}c^{-2t-2}.
$$
We can clean this up a bit more if we set $u_t = 4 y_t$, and multiply everything in sight by $4$. If we also write $\Delta u$ for the forward difference, we get
$$
(\Delta u)_t = u_{t+1} - u_t \leq (2t-3)c^{-2t} - 2tc^{-t} + 3 + c^{-2t-2}.
$$
Now we use the fundamental theorem of finite calculus. Everything on the right hand side of the inequality is summable in $t$, but if we sum the left hand side we get something that telescopes. Precisely, if we sum $t$ from $0$ to $n-1$ we get (using sage):
$$
\frac{4 x_n}{c^{2n}} =
u_n \leq
\frac{3 \, c^{4} - {\left(3 \, c^{4} - 5 \, c^{2}\right)} c^{2 n} - 5 \, c^{2} - 2 \, {\left(c^{4} - c^{2}\right)} n}{{\left(c^{4} - 2 \, c^{2} + 1\right)} c^{2 n}}
-2 \left (
- \frac{{{\left(c^{2} - c\right)} n + c - c^{n + 1}}}{{\left(c^{2} - 2 \, c + 1\right)} c^{n}}
\right )
+ 3n
+ \frac{c^{2 n} - 1}{{\left(c^{2} - 1\right)} c^{2 n}}.
$$
In the interest of not making silly mistakes I haven't done any simplification here -- these $4$ terms are what you get when you ask sage to sum each of the $4$ terms from the previous expression. You can tell this can be condensed by a fair bit, and it's also worth noting that if $|c| < 1$ a lot of these terms become small, or can be bounded by limits. I'll leave these simplifications and considerations to you, though.
I hope this helps ^_^
| {
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"url": "https://math.stackexchange.com/questions/4178659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Elements of the sequence have a prime factors an element of the sequence I am reading the following problem:
For the sequence $T=3, 7, 11, 15, 19, 23, 27 ...$ prove that every
number in $T$ has a prime factor that is also in $T$
My approach:
The sequence is of the form $4\cdot n + 3$
Each of the numbers in the sequence has a unique decomposition of prime factors $2^a\cdot 3^b \cdot 5^b \cdot 7^d ...$
We can express the prime factors as follows:
$(4\cdot n + 2)^a \cdot (4\cdot n + 1)^b \cdot (4\cdot n + 3)^c$ for $n \ge 0$
If we assume that we do not have any prime factor that is also part of $T$ then we would have prime factors of the form:
$(4\cdot n + 2)^a \cdot (4\cdot n + 1)^b$
For the simple case that $a = 1 \space b = 1$ we have:
$(4\cdot n + 2) \cdot (4\cdot n + 1) = 16 \cdot n^2 + 4 \cdot n + 8 \cdot n + 2 = 4n(4n + 3) + 2 = 4k + 2$ (where $k = 4\cdot n+ 3)$
Which means that we can not get a number in $T$ without having a prime factor of the form $(4\cdot n + 3)^c$
Update based on the comments of @Peter and @Asher2211:
The prime factors can be only of the form $(4\cdot n + 1)$ or $(4\cdot n + 3)$
If we assume that we can have a number in $T$ with prime factors not in $T$ we would have:
$x = (4\cdot n + 1)^a = 1 + a\cdot (4n) + \frac{a(a-1)}{2!}\cdot (4n)^2 +... = 1 + 4nk\space$ where $k = a + \frac{a(a-1)}{2!}\cdot (4n)....$
hence we can not get a number that is in $T$
| An experimental approach:
We rewrite expression as:
$4n+3=3(n+1)+n$
$[n, (n+1)]=1$
So if 3 divides n we can have numbers which are divisible by 3 in the sequence. These numbers make following progression:
$15, 27, 39=3\times 13, 51=3\times 17, \cdot\cdot\cdot 12k+3$
That is the terms of this progression have at least one prime other than 3 as a factor.
Now suppose 3 does not divide n, so we may consider following cases:
1): $\begin {cases}n=3m+1\\n=3m-2\end {cases}or: n\equiv 1\bmod 3\equiv -2\bmod 3$
which gives this progression:
$19, 31, 43, 55=5\times 11, \cdot\cdot\cdot 12k_1-5$
The terms of these progression are a multiple of 5 or have at least a prime other than 3 and 5.
2): $\begin {cases}n=3m-1\\n=3m+2\end {cases}or: n\equiv -1\bmod 3\equiv 2\bmod 3$
which gives this progression:
$23, 35=5\times 7, 47, 59, 71, 83, 95, \cdot\cdot\cdot 12k_2-1$
similar to case 1 the terms of this progression are a multiple of 5 or have at least a prime other than 3 and 5.
We can see that the factors of $4n+3$ cover all primes and we can conclude that any term of resulting sequence from $4n+3$ contains a prime already exist in the factors of previous terms.
| {
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"question_score": "1",
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Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$ Let $a,b,c>0$ satisfy $a^2+b^2+c^2=3$ . Prove that: $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$$
My idea is to use a well-known inequality (We can prove by Schur) $$(a^2+2)(b^2+2)(c^2+2)\ge 9(ab+bc+ca)$$ and the problem is prove $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge(a^2+2)(b^2+2)(c^2+2)$$
Anyone have a better idea ? please help me
| Since $ab+ac+bc\leq a^2+b^2+c^2=3,$ it's enough to prove that:
$$\prod_{cyc}(a^5-2a+4)\geq27$$ or
$$\sum_{cyc}\left(\ln(a^5-2a+4)-\ln3\right)\geq0$$ or
$$\sum_{cyc}\left(\ln(a^5-2a+4)-\ln3-\frac{1}{2}(a^2-1)\right)\geq0.$$
Now, prove that $$\ln(a^5-2a+4)-\ln3-\frac{1}{2}(a^2-1)\geq0$$ for any $0<a<\sqrt3$.
| {
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Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$ I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.
Prove the following inequality without using calculator:
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Can you check my proof? It would be really grateful.
First, we can define function $f(x)$ as following:
$$f(x) = \sqrt[3]{x+1} - \sqrt[3]{x}\space(x > 0)$$
$$f(3) = \sqrt[3]{3+1} - \sqrt[3]{3} = \sqrt[3]{4} - \sqrt[3]{3}$$
$$f(2) = \sqrt[3]{2+1} - \sqrt[3]{2} = \sqrt[3]{3} - \sqrt[3]{2}$$
Then, we will differentiate $f(x)$ to check whether $f(x)$ is a decreasing function or not. $f '(x)$ must be a falling function if $f '(x)$ < 0.
$$f'(x) = \frac{1}{3\sqrt[3]{(x+1)^2}} - \frac{1}{3\sqrt[3]{x^2}}$$
Since the minuend is smaller than the subtrahend (minuend has a bigger denominator than the denominator of subtrahend), we can say $f '(x)$ is less than 0 which makes $f(x)$ a decreasing function. Falling function means that $f(a) > f(a+1)$. Substitute $a=2$ and we get:
$$f(2) > f(3)$$
$$\sqrt[3]{3} - \sqrt[3]{2} > \sqrt[3]{4} - \sqrt[3]{3}$$
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Thank you for reading this text, but it would be more grateful if you check my solution, and comment my solution.
I wish you a beautiful day, and stay safe.
| Consider also a simpler function $g(x) = \sqrt[3]x$. For $x > 0$, $g'(x) = \dfrac1{3x^{2/3}}$ is decreasing.
By mean value theorem, there exists some $c_1 \in (2,3)$ and some $c_2 \in (3,4)$ that satisfy
$$g'(c_1) = \frac{\sqrt[3]3-\sqrt[3]2}{1};\quad g'(c_2) = \frac{\sqrt[3]4-\sqrt[3]3}{1}$$
And since $c_2 > c_1$ and so $g'(c_2) < g'(c_1)$,
$$\sqrt[3]4-\sqrt[3]3 < \sqrt[3]3-\sqrt[3]2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Recurrence relation $a_{n} = a_{n-1}+a_{n-2}+n$ How to solve this recurrence relation?
$a_{n} = a_{n-1}+a_{n-2}+n,a_{1}=a_{0}=1$
What I have tried:
$r^2 = r + 1 \rightarrow r = \frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}$
non-homogeneous part:
$$
\begin{split}
a_{n} &= cn+b \implies c(n)+b = c(n-1) + b + c(n-2) + b + n\\
0 &= -3c+b+n(c+1) \implies c+1 = 0 \iff c=-1 \implies b = -4 \\
a_{n} &= c_{0} \left(\frac{1+\sqrt{5}}{2}\right)^n
+ c_{1} \left(\frac{1-\sqrt{5}}{2}\right)^n -n-4
\end{split}
$$
putting $a_{0}$ and $a_{1}$ in recurrence relation gives $c_{0} = (\frac{1}{2})(5+\frac{7}{\sqrt{5}})$ and $c_{1} = (\frac{1}{2})(5-\frac{7}{\sqrt{5}})$.
I am sure my answer is wrong because for n greater than 1 it doesn't work. help me correct my mistake, please.
| the last equation
$$
0 = n(c+1) + (b-3c),
$$
which implies $c+1 = 0 \iff c=-1$ and $b = 3c = -3 \ne -4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x)$ has a zero at $x=\frac{1}{2}$. But can we conclude that this is unique？
| Note that the derivative is
$$f’(x) = \frac{\ln(1 - x)}{x} - \frac{\ln x}{1 - x}$$
for $0 < x < 1.$ We want to show that $f(x)$ is maximized when $x = 1/2.$
Since we already know that $f’(1/2) = 0,$ it remains to show that $f’’(1/2) < 0$ and that $f’(x) \neq 0$ for all other $x$ in the interval.
We have
$$\begin{align}
f’’(x) &= \frac{1}{x}\cdot \frac{-1}{1 - x} - \frac{\ln(1 - x)}{x^2} - \frac{1}{x}\cdot\frac{1}{1 - x} - \frac{\ln x}{(1 - x)^2} \\
&= -\frac{2}{x (1 - x)} - \frac{\ln(1 - x)}{x^2} - \frac{\ln x}{(1 - x)^2},
\end{align}
$$
from which we can compute that $f’’(1/2) = 8 \ln 2 - 8.$ Since $\ln 2 < 1,$ it follows that $f’’(1/2) < 0.$
If we can show that the function $g$ given by $$g(x) = \frac{\ln(1 - x)}{x}$$ for $0 < x < 1$ is injective, then that will be sufficient to prove that $f’(x)$ is only $0$ for $x = 1/2.$ To do that, we will apply the inequality $$\ln x > 1 - \frac{1}{x}$$ for $x$ in this interval. If you are unfamiliar with this inequality, you can see this question.
From
$$\begin{align}
g’(x) &= -\frac{1}{x(1 - x)} - \frac{\ln(1 - x)}{x^2} \\
&< -\frac{1}{x(1 - x)} - \frac{1}{x^2}\left(1 - \frac{1}{1 - x}\right) \\
&= -\frac{1}{x(1 - x)} + \frac{1}{x(1 - x)} \\
&= 0,
\end{align}
$$
it follows that $g$ is strictly decreasing. Thus, the only zero of $f’(x)$ occurs when $x = 1/2$.
By the second derivative test, we have $$\ln x \ln(1 - x) \leq \ln^2(1/2) = \ln^2 2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If , $\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$ . For , $a_i\in\mathbb Z^+$ & $a_i
$Q.$ If , $$\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$$ For , $a_i\in\mathbb Z^+$ & $a_i<i$ . Find $a_2+a_3+......+a_7$ ?
MY APPROACH :
We have , $$\frac{5}{7}=\frac{3!-1}{3!+1}=\frac{3!+1-2}{3!+1}=1-\frac{2}{3!+1} \Rightarrow 1-\frac{2}{7}$$
Now I've : $$1-\frac{2}{7}=\underbrace{\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}}_{\Lambda}$$ $$\Lambda=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\left(\frac{1}{4!}-\frac{1}{5!}\right)+\left(\frac{1}{5!}-\frac{1}{6!}\right)+\left(\frac{1}{6!}-\frac{2}{7}\right)$$ OR ,$$\frac{a_2}{2!}+\frac{a_3}{3!}+ \frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac{1}{2!}+\frac{2}{3!}+ \frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!}-\frac{1433}{7!}$$
DOUBT : In my solution , all the terms are following the conditions from the Question except $a_7$ not satisfy $a_7\in\mathbb Z^+$ and $a_7\in\{1,2,3,4,5,6\}$ only by this condition $a_i<i$
$\Rightarrow$ Where am I wrong ?
| Another way, starting from the other end.
*
*Multiply by $\,7!\,$ then $\,3600 = 7! \cdot \frac{5}{7} = 7 \cdot (\dots)+a_7\, \implies a_7 \equiv 3600 \equiv 2 \pmod{7}\,$, so $\bbox[border:1px solid black]{\,a_7=2\,}$.
*Move the known term to the left-hand side $\,\frac{5}{7}-\frac{2}{7!} = \frac{257}{360}\,$ and multiply by $\,6!\,$, then $\,514 = 6!\cdot \frac{257}{360} = 6 \cdot (\dots) + a_6 \implies a_6 \equiv 514 \equiv 4 \pmod{6}\,$, so $\bbox[border:1px solid black]{\,a_6 = 4\,}$.
*Move the known term to the left-hand side $\,\frac{257}{360}-\frac{4}{6!}=\frac{17}{24}\,$ and multiply by $\,5!\,$, then $\,85 = 5!\cdot \frac{17}{24} = 5 \cdot (\dots) + a_5 \implies a_5 \equiv 85 \equiv 0 \pmod{5}\,$, so $\bbox[border:1px solid black]{\,a_5 = 0\,}$.
*Nothing to move to the left-hand side now, multiply by $\,4!\,$, then $\,17 = 4!\cdot \frac{17}{24} = 4 \cdot (\dots) + a_4 \implies a_4 \equiv 17 \equiv 1 \pmod{4}\,$, so $\bbox[border:1px solid black]{\,a_4 = 1\,}$.
*Move the known term to the left-hand side $\,\frac{17}{24}-\frac{1}{4!}=\frac{2}{3}\,$ and multiply by $\,3!\,$, then $\,4 = 3!\cdot \frac{2}{3} = 3 \cdot (\dots) + a_3 \implies a_3 \equiv 4 \equiv 1 \pmod{3}\,$, so $\bbox[border:1px solid black]{\,a_3 = 1\,}$.
*Move the known term to the left-hand side $\,\frac{2}{3}-\frac{1}{3!}=\frac{1}{2}\,$, so $\bbox[border:1px solid black]{\,a_2 = 1\,}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Split $\{1,...,3n\}$ into triples with $x+y=5z$ - no solutions? Following on from Split $\{1,2,...,3n\}$ into triples with $x+y=4z$ and For which $n\in\Bbb N$ can we divide $\{1,2,3,...,3n\}$ into $n$ subsets each with $3$ elements such that in each subset $\{x,y,z\}$ we have $x+y=3z$?
I have a rickety proof there is no way to split the numbers $\{1,...,3n\}$ into triples $(x,y,z)$ with $x+y=5z$. Perhaps someone could fix it, find their own proof or disprove it.
Thanks to Rob Pratt, $(2,3,1)$ is a solution. Are there any others?
For the moment, ignore the need to keep all numbers between $1$ and $3N$. Suppose we have a set of $N$ disjoint triples. Let $M$ be the highest $z$.
I thought that the smallest possible sum of $z$ would come when none of the $z$ was $M/5$ or less. Suppose a triple is $(x_0,y_0,z_0)$ with $z_0\le M/5$. Then both $x_0$ and $y_0$ are less than $M$. Replace the two triples whose $z$ values are $z_0$ and $M$, with others whose $z$ values are $x_0$ and $y_0$. (This is the bit that might not be possible.)
A lower sum of $z$ would result because $x_0+y_0=5z_0\le M\lt M+z_0$.
There must be at least $N$ values for the $z$ between $M/5+1$ and $M$ so $M\ge5N/4$. The sum of the $z$ is at least the sum of the numbers from $(N/4)+1$ to $5N/4$, which is $(3N^2+2N)/4$.
But the sum of all $3N$ numbers is $3N(3N+1)/2$, which must be exactly $6$ times the sum of the $z$. The lowest possible sum of $z$ is too high, so there is no solution.
□
| Modulo $6$ you need $\frac{3n(3n+1)}2$ divisible by $6.$ It is obviously divisible by $3,$ but it is divisible by $2$ only when $n\equiv 0,1\pmod 4.$
(This is because if $x+y=5z,$ then $x+y+z\equiv 0\pmod 6.$)
Any $m\geq \frac{6n}5$ must be used as an $x$ or $y,$ because no $x+y=5m\geq 6n.$
Assume we have a solution.
Let $S_{xy}$ be all the values used as $x$ or $y$ values, and $S_z$ are all the values used as $z,$ then $$\sum_{m\in S_{xy}} m=5\sum_{m\in S_z} m$$
This means if $T\subseteq S_{xy}$ you must have $$\sum_{m\in T} m\leq 5\sum_{m\notin T} m$$
Or, adding $\sum_{m\notin T}m$ to both sides:
$$\sum_{m=1}^{3n} m\leq 6 \sum_{m\notin T} m$$
But by the last note above, $T=\left\{m\mid m\geq \frac{6n}{5}\right\}$ is such a set.
So you need:
$$\sum_{k=1}^{3n} m\leq 6\sum_{k=1}^{\lceil 6n/5\rceil -1} m$$
Or:
$$\begin{align}9n^2+3n&=3n(3n+1)\\&\leq 6\lceil 6n/5\rceil (\lceil 6n/5\rceil -1)\\&\leq 6\cdot (6n/5)(6n/5+1)\quad(*)\\&=\frac{216}{25}n^2+\frac{36}{5}n \end{align}$$
(When we know $n$ specifically, we can do better than the (*) step, just by calculating $\lceil 6n/5\rceil.$ We’ll need to do to eliminate a few cases.)
Subtracting, you want $$\frac9{25}n^2\leq \frac{21}{5}n$$ or $$n\leq \frac{35}{3}.$$
So you’ve cut down to the cases $n\leq11.$
We’ve already eliminated $n=2,3,6,7,10,11$ by the modulo $4$ criterio.
We know $n=1$ is a counterexample.
We are left with $n=4,5,8,9.$ We can do better by skipping the (*) step above, and using exact value of $\lceil 6n/5\rceil$ for these $n.$ When $n=4,5,8,9$ we get $5,6,10,11,$ respectively.
Then $$\begin{align}9\cdot 4^2+3\cdot 4=156&>120=6 \cdot 5\cdot 4\\
9\cdot 5^2+3\cdot5=240&>180=6\cdot6\cdot 5\\
9\cdot 8^2+3\cdot 8=600&>540=6\cdot 10\cdot 9\\
9\cdot 9^2+3\cdot 9=756&>660=6\cdot 11\cdot 10\end{align}$$
So there is no solution other than $n=1.$
| {
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Does the sum of the series $\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt n}{n}$ have an analytic expression? Just out of curiosity, I'd like to know whether or not the sum of the series
$$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}$$
has a known analytic expression.
I stumbled across this series while trying to evaluate
$$\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx$$
The convergence of this integral can be seen by making use of the inequality $\lfloor x\rfloor > x-1$ and the fact that $\coth^{-1}(t)\to 0$ as $t\to\infty$:
\begin{align*}
\int_1^t\frac{1}{\lfloor x^2\rfloor}dx &= \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{\lfloor x^2\rfloor}dx\\
&< \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{x^2-1}dx\\
&= \int_\sqrt{1}^\sqrt{2}\frac{1}{1}dx-\int_\sqrt{2}^t\frac{1}{1-x^2}dx\\
&= \sqrt{2}-1-\left[\coth^{-1}(t)-\coth^{-1}\left(\sqrt 2\right)\right]\\
&= \sqrt{2}-1-\coth^{-1}(t)+\coth^{-1}\left(\sqrt 2\right)\\
&\to \sqrt{2}-1+\coth^{-1}\left(\sqrt 2\right)\text{ as }t\to\infty\\
\end{align*}
Since this implies that $\int_1^t 1/\lfloor x^2\rfloor dx$ is strictly increasing ($1/\lfloor x^2\rfloor >0$ for every $x\geq 1$) and bounded above, the integral necessarily converges. By breaking up the integral
$$\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx$$
into integrals indexed by the intervals $\left[\sqrt{i},\sqrt{i+1}\right]$ for $i=1,2,3,...,k$ and simplifying the resulting sum, I was able to show that
$$\int_1^{\sqrt{k+1}}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^{k} \frac{\sqrt{n+1}-\sqrt n}{n}$$
is true for every $k\geq 0$, which yields
$$\int_1^\infty \frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}$$
after letting $k\to\infty$. This equality is the main reason why I'm interested in the sum of the aforementioned series.
After some (unsurprisingly) futile attempts to evaluate the integral, I expect there to be no closed-form expression for the sum, which is why I'm open to an analytic expression (gamma function, Bessel functions, Riemann zeta function, etc.). Any help is appreciated.
Edit: after seeing the bounds provided by Markus Scheuer and Jorge, I thought I'd share some of my own.
From the fact that $x-1<\lfloor x\rfloor<x$ is true for every non-integer $x\geq 1$, we can infer that for every integer $k\geq 1$,
$$\int_\sqrt{k+1}^\infty \frac{1}{x^2}dx<\int_\sqrt{k+1}^\infty \frac{1}{\lfloor x^2\rfloor}dx<\int_\sqrt{k+1}^\infty \frac{1}{x^2-1}dx$$
Using
$$\int_{x}^{\infty}\frac{1}{t^2-1}dt=\coth^{-1}(x)$$
and
$$\int_\sqrt{k+1}^\infty\frac{1}{\lfloor x^2\rfloor}dx=\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx-\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}-\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
we deduce that
$$\frac{1}{\sqrt{k+1}}+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}<\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}<\coth^{-1}\left(\sqrt{k+1}\right)+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
| I do not know but to me it is so nice a candidate for Euler-Maclaurin. I promise no closed form just kind of close evaluation. When I write $\infty$ I mean $\lim\limits_{n \to \infty}$
$$\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=\int\limits_{1}^{\infty}\frac{\sqrt{x+1}-\sqrt{x}}{x}\, dx + \sqrt{2}-1 + D = 1 - \sqrt{2} + 2 \log(1+\sqrt{2}) + D$$
$$f(x)=\frac{\sqrt{x+1}-\sqrt{x}}{x}$$
$$D=-\sum_{k=1}^{\infty}\frac{B_{2k}}{(2k)!}f^{(2k-1)}(1)$$
Now it is the matter of finding the odd derivatives of $f(x)$ at $1$.
Take $u(x)=\sqrt{x+1}-\sqrt{x}$ and $v(x)=\frac{1}{x}$ and apply
$$(uv)^{(n)}=\sum_{k=0}^n {n \choose k} u^{(n-k)} v^{(k)}$$
since derivatives of $u$ and $v$ are straightforward and you have it
$$u^{(n)}(x)=(-1)^{n+1}\frac{(2n-3)!!}{2^n}(\frac1{(x+1)^{\frac{2n-1}{2}}}-\frac1{x^{\frac{2n-1}{2}}})$$
$$v^{(n)}(x)=(-1)^{n}n!\frac{1}{x^{n+1}}$$
Or
$$u^{(n)}(1)=(-1)^{n+1}\frac{(2n-3)!!}{2^n}(\frac1{2^{\frac{2n-1}{2}}}-1)$$
$$v^{(n)}(1)=(-1)^{n}n!$$
Finally
$$(uv)^{(n)}(1)=(-1)^{n+1}\sum_{k=0}^n {(n)_k}(2n-2k-3)!!(\frac{\sqrt{2}}{2^{2n-2k}}-\frac1{2^{n-k}})$$
$$\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=1 - \sqrt{2} + 2 \log(1+\sqrt{2})+$$ $$\sum_{p=1}^{\infty}(-1)^{2p+1}\frac{B_{2p}}{(2p)!}\sum_{k=0}^{2p-1} {(2p-1)_k}(4p-2k-5)!!(\frac{\sqrt{2}}{2^{4p-2k-2}}-\frac1{2^{2p-k-1}})$$
Not closed, but still calculable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$
I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}-\frac{(7x)^6}{6!}+\dots\}}{3x^2}\\ =\lim\limits_{x\to 0}\frac{x^2(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1}{3}(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)\\ =\frac {49}{6}$
Can this be solved without using the Taylor series?
| It is already known that:
$$\lim\limits_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$
Now Manipulating,
$$=\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.\frac{49}{49}$$
$$=\lim\limits_{x\to 0}\frac{1-\cos 7x}{(7x)^2}.\frac{49}{3}$$
$$=\frac{49}{3\cdot2}=\frac{49}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
$
| Lower bound: use AM-QM inequality:
$$\frac{h^2}{2}=\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2=\frac{(30-h)^2}{4}$$
Solving the quadratic inequality for positive $h$ will give to you $h\ge 30(\sqrt{2}-1)= 12.42\ldots$ (lowest $h$ is achieved when the triangle is isosceles)
Upper bound: use $a+b>h$
$30=a+b+h>h+h=2h$
Therefore $h<15$
So the integral values of $h$ are $13$ and $14$
You can verify by solving $a+b=30-h$ and $a^2+b^2=h^2$
For $h=13$ you will get the well-known Pythagorean triplet $5,12,13$
For $h=14$ you will get $8-\sqrt{34},8+\sqrt{34},14$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $\displaystyle\prod_{i=1}^{n}\frac{n+i}{2i-3}=2^n(1-2n)$. Prove by induction that $\displaystyle\prod_{i=1}^{n}\frac{n+i}{2i-3}=2^n(1-2n)$.
What I did:
It's easy to see that the base case is true. Then I have to prove that if the statement is true for $n$ then it must be true for $n+1$. That is to say, if true for $n$, prove that $\displaystyle\prod_{i=1}^{n+1}\frac{n+1+i}{2i-3}=2^{n+1}(1-2(n+1))=\left(\prod_{i=1}^{n}\frac{n+1+i}{2i-3}\right)\frac{2n+2}{2n-1}=-2\cdot2^n(2n+1)$. Of course the $+1$ inside the product is blocking my progress, because I can't replace it by the inductive hypothesis. I've figured that $-2\cdot2^n(2n+1)=-2\cdot2^n(1-2n)-n2^{n+3}$, but I don't know if that helps.
Help is much appreciated!
| The inductive hypothesis is:
$$\prod_{i=1}^{k} \frac {k+i}{2i-3}=2^k(1-2k)$$
LHS can be written as:
$$\frac {k+1}{-1} \cdot \frac {k+2}{1} \cdot \frac {k+3}{3}...\cdot \frac {k+k}{2k-3}=\left(\frac {k+2}{-1} \cdot \frac {k+3}{1} \cdot \frac {k+4}{3}...\cdot \frac {k+k}{2k-5}\right)\cdot \left(\frac {k+1}{2k-3}\right)=\left(\left(\prod_{i=1}^{k+1} \frac {k+1+i}{2i-3}\right)\left(\frac {2k-3}{2k+1} \cdot \frac {2k-1}{2k+2}\right)\right) \cdot \left(\frac {k+1}{2k-3}\right)=2^k(1-2k)$$
Now simplify the terms other than the $\prod$.
You are left with:
$$\prod_{i=0}^{k+1} \frac {k+1+i}{2i-3}=-2^{k+1}(1+2k)$$
This completes the inductive step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions of $\sqrt{x^2+5x-14} + |x^2+4x-12|=0$ My attempt:
Given,
$$\sqrt{x^2+5x-14} + |x^2+4x-12|=0 \tag{1}$$
Since $|a|=\sqrt{a^2}$, $$\sqrt{x^2+5x-14}=-\sqrt{(x^2+4x-12)^2}$$
Squaring both sides, $$x^2+5x-14=(x^2+4x-12)^2$$
When I simplify the above, I get two real solutions: $x=2$ and $x=2.138 \text{ (approximately)}$. However, there is only one solution to equation $(1)$ according to Wolfram Alpha, $x=2$.
Is my solution incorrect? If so, where did I go wrong?
| When you square, you may get additional solutions. You need to test those solutions in the original equation.
Alternatively,
$\sqrt{x^2+5x-14} + |x^2+4x-12|=0$
$\implies x^2+5x-14 = 0 \ \text { and } x^2+4x-12 = 0$
$\implies (x-2) (x+7) = 0 \ \text { and } (x-2)(x+6) = 0$
So only solution that works is $x = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the limit $\lim_{x \to 0}\frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\cosh x}{x(\sqrt{1+2 x}-\sqrt[3]{1+3 x})}$ Evaluate the limit $$L=\lim _{x \to 0}\frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\cos h x}{x(\sqrt{1+2 x}-\sqrt[3]{1+3 x})}$$
By generalized binomial expansion we have
$$\sqrt{1+2 x}-\sqrt[3]{1+3 x}=\left[1+\frac{1}{2}(2 x)+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}(2 x)^{2}+\cdots\right]-\left[1+\frac{1}{3}(3 x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2}(3 x)^{2}+\cdots\right]$$
$\implies$
$$\sqrt{1+2x}-\sqrt[3]{1+3x}=\frac{x^2}{2}+O(x^3)$$
$\implies$
$$L=\lim _{x \rightarrow 0} \frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\cos h x}{x^{3}\left(\frac{\frac{x^{2}}{2}+O\left(x^{3}\right)}{x^{2}}\right)}$$
$\implies$
$$L=2 \lim _{x \rightarrow 0} \frac{2+\tan \left(e^{x}-\cos x\right)-e^{x}-\left(\frac{e^{x}+e^{-x}}{2}\right)}{x^{3}} \to (1)$$
We have $$e^{x}=1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4}+\ldots$$ and
$$\cos x=1-\frac{1}{2} x^{2}+\frac{1}{24} x^{4}-\frac{1}{720} x^{6}+.$$
Thus we have
$$e^x-\cos x=x+x^2+\frac{x^3}{6}+O(x^4)$$
Now we have the Maclaurin's series expansion of $\tan x$ as:
$$\tan x=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+...$$
So we get
$$\tan(e^x-\cos x)=x+x^2+\frac{x^3}{6}+\frac{1}{3}x^3+O(x^4) \to (2)$$
Also we have
$$e^{x}+\cosh x=e^{x}+\frac{e^{x}+e^{-x}}{2}$$
$$e^x+\cos hx=\begin{aligned}
\frac{3}{2}(1+x+&\left.\frac{x^{2}}{2!}+\frac{x^{3}}{3 !}+..\right)
+\frac{1}{2}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2!}-\frac{x^{3}}{3 !}+\cdots^{}\right)
\end{aligned}$$
$\implies$
$$e^x+\cos hx=2+x+x^2+\frac{x^3}{6}+O(x^4) \to (3)$$
Using $(2),(3)$ in $(1)$ we get
$$L=\frac{2}{3}$$
Is there any alternate way?
| There is one right way to do this limit:
$$ \lim_{x \to 0} \frac{
\frac{1- \cosh x}{x}+ \left[ \frac{\tan(e^x - \cos x)}{x} -\frac{1-e^x}{x} \right]}{
\sqrt{1 +2x}- (1+3x)^{\frac13}}=\lim_{x \to 0} \frac{\left( - \frac{x}{2!} - \frac{x^3}{4!}+O(x^5)\right)+ \left[ \frac{\left(2e^x -1- \cos x \right)+\frac{(e^x - \cos x)^3}{3}}{x}.. \right]}{\left[ 2-1\right] \frac{x^2}{2!} +O(x^3)}$$
Note that:
$$ (2e^x - 1 - \cos x)= \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$$:
Hence,
$$ \lim_{x \to 0}\frac{ \left( - \frac{x}{2!} - \frac{x^3}{4!}+O(x^5) \right)+ \left[ \frac{\frac{x^2}{2} + \frac{x^3}{3} + O(x^4) }{x}.\right]}{ \frac{x^2}{2!} +O(x^3)} $$
$$ \lim_{x \to 0}\frac{\frac{x^2}{3} + O(x^3)}{\frac{x^2}{2!}+O(x^3)}= \frac23$$
Idea: I first rearranged the expression to make them all look like derivatives, for the denominator I tried to find the first non zero term of in it's expansion and then expanded numerator in that.
It is still a mess, but it reduces solving the question in an algorithim.
*Arrange the expressions to look like known limits on numerator and denominator
*Find deg denominator
*Expand numerator till deg denominator
*Careful algebra
*Take the limit
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that the limit of $f(x,y)=\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ is $0$ as $(x,y)$ approaches $(0,0)$? I need to show that $\displaystyle\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ exists. I know that it is equal to zero. Until now, all I know how to do is to prove using the $\epsilon-\delta$ definition. I tried to prove it by the following way:
Let $\delta>0$ be a real number such that $0<\sqrt{x^2+y^2}<\delta$. We have $0< x^2+y^2<\delta^2$.
We know that $\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq\left\vert x^2\sin(y)\right\vert+\left\vert y^2\sin(x)\right\vert=x^2\left\vert\sin(y)\right\vert+y^2\left\vert\sin(x)\right\vert\leq x^2+y^2$.
Thus
$$0<\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq x^2+y^2<\delta^2$$
and dividing the inequality by $x^2+y^2>0$ we have
$$ 0<\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}<1<\dfrac{\delta^2}{x^2+y^2}. $$
I thought that this could be useful because $\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}=\left\vert\dfrac{x^2\sin(y)+y^2\sin(x)}{x^2+y^2}\right\vert$, but I don't know how to proceed.
| If the limit exists, it will exist for any metric (so long as it is a valid metric).
While it may seem natural to use the Euclidean metric, it is not always the easiest to work with. I frequently prefer to work with the maximum metric.
That is $d((x_1,y_1),(x_2,y_2)) = \max (|x_2-x_1|,|y_2-y_1|)$
In this example, $\delta = \max (|x|,|y|)$
and $\left|\frac {x^2\sin y + y^2\sin x}{x^2 + y^2}\right| < \frac {2\delta^3}{\delta^2} = 2\delta$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding modular inverse of $x^4$ in $GF(2^5)\mod (x^5+x^2+1)$ I cannot spot where I am going wrong in this. I am using Extended Euclidean's algorithm here.
$(x^5+x^2+1) = (x^4)(x) + (x^2+1)$
$(x^4) = (x^2+1)(x^2+1) + 1$
let $P(x)=x^4$ and $Q(x)=x^5+x^2+1$
$(x^5+x^2+1) + (x^4)(x) = (x^2+1)$
$(x^4) + (x^2+1)(x^2+1) = 1$
$P(x)+[Q(x)+P(x)(x)][Q(x)+P(x)(x)]=1$
$P(x)(1+xQ(x)+P(x)x^2+xQ(x)) +Q(x)(...)=1$
$P(x)(1+P(x)x^2)+Q(x)(...)=1$
$P(x)(1+x^6) + Q(x)(...)=1$
$(1+x^6) \bmod (x^5+x^2+1)=x^3+x+1$
I am getting the inverse of $x^4 \mod (x^5+x^2+1)$ as $x^3+x+1$ but it seems the right answer is $x^4+x^2+1$. Can anyone tell me where it is wrong?
| Your answer is correct since, taking into account that the characteristic of the field is $2$, so that $x^5=x^2+1$,
\begin{align}
x^4(x^3+x+1)&=x^7+x^5+x^4=x^7+x^4+x^2+1 \\
&=x^2\cdot x^5+x^4+x^2+1=\color{red}{\not x^4}+\color{blue}{\not x^2}+\color{red}{\not x^4}+\color{blue}{\not x^2}+1\\
&=1.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Linearity of Variance and Expected value I know Expected value has the property of linearity $E(X+a)=E(X) + a$, but it also seems to hold for $E(X^3+a)$. But Variance also has the property $V(X+a)=V(X) + 0$, but it does not hold for $V(X^3+a)=V(X^3) + 0$.
I discovered this from a question that asked to find the expected value adn variance for $Y = X^3 + 2$ given a distribution for $X$.
$$\begin{array}{|r|r|r|r|r|}\hline x&-1&0&1&2\\\hline p(x)&0.1&0.4&0.3&0.2\\\hline\end{array}$$
The table shows the distribution for $X$, so I worked out $E(Y) = E(X^3 - 2) = 1.8 - 2 =-0.2$.
For variance I am thinking $V(Y) = V(X^3 - 2) = E(X^{3\times2}) - E(X^3)^2 - 0 =E(X^6)-0.2^2$ from which I get 11.16
| From the definition of variance and the Linearity of Expectation:$$\begin{align}\mathsf V(X^3+a) &= \mathsf E((X^3+a)^2)-(\mathsf E(X^3+a))^2 \\ &= \mathsf E(X^6+2aX^3+a^2)-(\mathsf E(X^3)+a)^2\\ &=\bigl(\mathsf E(X^6)+2a\mathsf E(X^3)+a^2\bigr)-\bigl(\mathsf E(X^3)^2+2a\mathsf E(X^3)+a^2\bigr)\\&=\mathsf E(X^6)-\mathsf E(X^3)^2\\&=\mathsf V (X^3)\end{align}$$
$$\begin{array}{|r|r|r|r|r|}\hline x&-1&0&1&2\\\hline p(x)&0.1&0.4&0.3&0.2\\\hline\end{array}$$
$\mathsf E(X^3) = -1\cdot 0.1+1\cdot 0.3+0+2^3\cdot 0.2 = 1.8\\\mathsf E(X^6) = 1\cdot(0.1+0.3)+2^6\cdot0.2 = 13.2\\\mathsf {V}(X^3)= 13.2-1.8^2 = 9.96$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of the error of the $\arctan(x)$ function I was wondering about the following problem: is it possible to find the best estimation possible of the remainder between the function $\arctan(x)$ and its Taylor polynomial?
I was thinking about expressing the $(n+1)$-th derivative of the $\arctan(x)$ function. I came up with this:
$$\dfrac{\text{d}^n}{\text{d}x^n} \arctan(x) = \dfrac{(-1)^{n-1}(n-1)!}{(1+x^2)^{n/2}}\sin\left(n\arcsin\left(\dfrac{1}{\sqrt{1+x^2}}\right)\right)$$
Then the idea could be another derivative and then useing the integral remainder but this idea seems impracticable.
The Taylor polynomial was easy to compute, but I believe that neither the Lagrange remainder nor the integral remainder are good ideas.
Is there some deeper or more useful ways to evaluate this difference?
| For any $N\geq 0$ and real $t$, we have
$$
\frac{1}{{1 + t^2 }} = \sum\limits_{k = 0}^{N - 1} {( - 1)^k t^{2k} } + ( - 1)^N \frac{{t^{2N} }}{{1 + t^2 }}.
$$
Integrating both sides form $0$ to $x$ ($x$ being real), we find
\begin{align*}
\arctan x & = \sum\limits_{k = 0}^{N - 1} {( - 1)^k \frac{{x^{2k + 1} }}{{2k + 1}}} + ( - 1)^N \int_0^x {\frac{{t^{2N} }}{{1 + t^2 }}dt}
\\ &
= \sum\limits_{k = 0}^{N - 1} {( - 1)^k \frac{{x^{2k + 1} }}{{2k + 1}}} + ( - 1)^N x^{2N + 1} \int_0^1 {\frac{{s^{2N} }}{{1 + x^2 s^2 }}ds} .
\end{align*}
Thus, for example,
\begin{align*}
\left| {( - 1)^N x^{2N + 1} \int_0^1 {\frac{{s^{2N} }}{{1 + x^2 s^2 }}ds} } \right| & = \left| x \right|^{2N + 1} \int_0^1 {\frac{{s^{2N} }}{{1 + x^2 s^2 }}ds} \\ & \le \left| x \right|^{2N + 1} \int_0^1 {s^{2N} ds} = \frac{{\left| x \right|^{2N + 1} }}{{2N + 1}}
\end{align*}
and
\begin{align*}
\left| {( - 1)^N x^{2N + 1} \int_0^1 {\frac{{s^{2N} }}{{1 + x^2 s^2 }}ds} } \right| & = \left| x \right|^{2N + 1} \int_0^1 {\frac{{s^{2N} }}{{1 + x^2 s^2 }}ds} \\ & \ge \frac{\left| x \right|^{2N + 1}}{1+x^2} \int_0^1 {s^{2N} ds} = \frac{{\left| x \right|^{2N + 1} }}{{(1+x^2)(2N + 1)}},
\end{align*}
for any $N\ge 0$ and real $x$.
The remainder can also be represented in the form
$$
( - 1)^N \frac{{x^{2N + 1} }}{{2N + 1}}C_N (x),
$$
where
$$
C_N (x) = (2N + 1)\int_0^1 {\frac{{s^{2N} }}{{1 + x^2 s^2 }}ds} = (2N + 1)\int_0^{ + \infty } {\frac{{e^{ - (2N + 1)t} }}{{1 + x^2 e^{ - 2t} }}dt} .
$$
Watson's lemma then leads to the asymptotic expansion
\begin{align*}
C_N (x) & \sim \sum\limits_{k = 0}^\infty {\frac{{(2x^2 )^k A_k ( - 1/x^2 )}}{{(1 + x^2 )^{k + 1} }}\frac{1}{{(2N + 1)^k }}}
\\ & =
\frac{1}{{1 + x^2 }} - \frac{{x^2 }}{{1 + x^2 }}\sum\limits_{k = 1}^\infty {( - 1)^k \frac{{2^k A_k ( - x^2 )}}{{(1 + x^2 )^k }}\frac{1}{{(2N + 1)^k }}}
\end{align*}
as $N\to +\infty$ uniformly on compact subsets of $\mathbb R$. Here $A_k$ denotes the $k$th Eulerian polynomial. At leading order
$$
C_N (x) \sim \frac{1}{{1 + x^2 }},
$$
showing that the lower bound above is sharp for sufficiently large $N$.
| {
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Partial fraction expansion question I have to integrate following expression (but integration is not the problem):
$$\frac{x^2+3x-2}{(x-1)(x^2+x+1)^2}$$
It is pretty obvious that:
$$\frac{x^2+3x-2}{(x-1)(x^2+x+1)^2}=\frac{A}{x-1} + \frac{Mx+N}{x^2+x+1} + \frac{Px+Q}{(x^2+x+1)^2}$$
The first and the easiest step is to find an $A$:
$$A=\frac{x^2+3x-2}{(x^2+x+1)^2}, x=1$$
$$A=\frac{2}{9}$$
And then there comes a problem - I don't know how to do the rest. I tried to multiply the whole thing by $(x^2+x+1)^2$ and differentiate, but it didn't seem to be useful at all, especially because $(x^2+x+1)^2$ doesn't have real roots.
As popping900 suggested. I can take just four different x values and solve system of for equations, but i would like to see a more elegant or shorter solution, if one exists
| I have been thinking about it for around 15 minutes and couldn't see any elegant or very short solutions. So decided to use pretty common and obvious method, that i already used before. I multiplied both sides of equation by $(x-1)(x^2+x+1)^2$ and here is what i got:
$$x^2+3x-2=\frac{2}{9}(x^2+x+1)^2+(Mx+N)(x-1)(x^2+x+1)+(Px+Q)(x-1)$$
Then i got system of four equations(according to coefficients of the first polynomial) and solved them. Got:
$M=-\frac{2}{9}, N=-\frac{4}{9}, P=\frac{1}{3}, Q=\frac{8}{3}$
But I didn't really wanted to use it, because i was told by multiple people, that it is not elegant and very straightforward
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the value of the $CH$ segment in the figure below? For reference: In the figure, $ABCDE$ is a regular pentagon with $BD = BK, AB = BT ~and ~TK = 2\sqrt5$. Calculate $CH$ (If possible by geometry instead of trigonometry)
My progress:
$Draw KD \rightarrow \triangle DBK(isosceles)\\
Draw TAB \rightarrow \triangle BTA(isosceles)\\
a_i = \frac{180(5-2)}{5} = 108^\circ\\
\angle A EH= 360 -2(108)-2(90)=54^\circ$
but I can't finish...
i I made the figure of peterwhy
| The following is based on the property of regular pentagons, that the ratio of their diagonals to their sides are in golden ratio $\varphi = \frac{1 + \sqrt 5}{2}$. For example,
$$\frac{BK}{BT} = \frac{BD}{BA} = \varphi$$
Rotate $\triangle KBT$ about $B$ clockwise by $90^\circ$ to $\triangle DBT'$. Then $ABT'$ is a straight line. Because $\triangle ABD$ is isosceles, drop altitude $DM$ onto $AB$, and $M$ would be the midpoint of $AB$. (Diagram by OP peta arantes)
Let $s$ be the side length of the regular pentagon. Then consider $\triangle DBT'$ and the foot of altitude $M$,
$$\begin{align*}
DT'^2 &= BT'^2 + BD^2 + 2 BT'\cdot BM\\
(2\sqrt 5)^2 &= s^2 + (\varphi s)^2 + 2s \cdot \frac s2\\
20 &= s^2 + \frac{3 + \sqrt 5}{2} s^2 + s^2\\
&= \frac{7 + \sqrt 5}{2} s^2\\
s^2 &= \frac{40}{7 + \sqrt 5}
\end{align*}$$
Next, consider $\triangle CDH$. Drop altitude $HN$ onto side $CD$, where $N$ is the foot of the altitude. Since $\triangle CDA$ is isosceles and $DH = \frac 12 DA$, so $ND = \frac 14 CD = \frac 14 s$.
The required length $CH$ can be found by
$$\begin{align*}
CH^2 &= CD ^2 + DH^2 - 2 CD \cdot ND\\
&= s^2 + \left(\frac{\varphi s}{2}\right)^2 - 2s\cdot \frac{s}{4}\\
&= \left[1 + \frac 14\cdot \frac{3 + \sqrt 5}{2} - \frac 12 \right]s^2\\
&= \frac {7 + \sqrt 5}8\cdot \frac{40}{7 + \sqrt 5}\\
&= 5\\
CH &= \sqrt 5
\end{align*}$$
| {
"language": "en",
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Show that $\frac{P_a}{a^2}+\frac{P_b}{b^2}+\frac{P_c}{c^2}\ge\frac{3}{4R}$. When is the equality reached? Show that $\dfrac{P_a}{a^2}+\dfrac{P_b}{b^2}+\dfrac{P_c}{c^2}\ge\dfrac{3}{4R}$. When is the equality reached?
We're dealing with an acute triangle $ABC$ where $h_a,h_b$ and $h_c$ are altitudes. $P_a,P_b$ and $P_c$ are the distances from the vertices of the triangle to the segments joining the feet of the altitudes. $R$ is the radius of the circumscribed circle of $ABC$.
I discovered a few things in the given configuration. $$\dfrac{P_a}{h_a}=\cos\measuredangle BAC$$ For the proof, we can look at the similar triangles $ABC$ and $AB_1C_1$. In them $\dfrac{AB_1}{AB}=\dfrac{h_{B_1C_1}}{h_{BC}}=\dfrac{P_a}{h_a}=\cos\measuredangle BAC$ (using the definition of cosine of an acute angle in the right triangle $AB_1B$). To show that the mentioned triangles are similar we can use $\triangle AB_1B\sim\triangle AC_1C$ and then SAS.
Similarly, $$\dfrac{P_b}{h_b}=\cos\measuredangle ABC\\ \dfrac{P_c}{h_c}=\cos\measuredangle ACB.$$ Another thing I was able to prove is $$P_a+P_b+P_c=\dfrac{a^2+b^2+c^2}{4R}.$$ We can derive that by simply substituting $$P_a=h_a\cos\alpha,P_b=h_b\cos\beta,P_c=h_c\cos\gamma,h_a=\dfrac{2S}{a},h_b=\dfrac{2S}{b},h_c=\dfrac{2S}{c},$$ where $S$ is the area of triangle $ABC$ and $\alpha,\beta$ and $\gamma$ are the angles of the triangle, and the law of cosines for simplification. Thank you in advance!
| As you mentioned, $\triangle AB_1C_1 \sim \triangle ABC \ $.
So, $AC_1 = \cfrac{P_a}{\sin C} = \cfrac{2R \cdot P_a}{c}$
Similarly show that $\triangle A_1BC_1 \sim \triangle ABC \ $, $C_1B = \cfrac{2R \cdot P_b}{c}$
$c = AC_1 + C_1 B = \cfrac{2R (p_a + P_b)}{c} \implies P_a + P_b = \cfrac{c^2}{2R}$
Also, $P_b + P_c = \cfrac{a^2}{2R}, P_c + P_a = \cfrac{b^2}{2R}$
So we get, $P_a = \cfrac{b^2+c^2-a^2}{4R}$
$\sum_{cyc} \cfrac{P_a}{a^2} = \cfrac{1}{4R} \sum_{cyc} \cfrac{b^2+c^2-a^2}{a^2}$
$ = \cfrac{1}{4R} \sum_{cyc} \left (\cfrac{a^2}{b^2} + \cfrac{b^2}{a^2} - 1 \right)$
$\geq \cfrac{3}{4R} \ $ (by AM-GM)
| {
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Need help to evaluate $\int\limits_{\frac{1}{\sqrt{2}}}^{1} \int\limits_{\sqrt{1-x^{2}}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$ I have this double integral $\int\limits_{\frac{1}{\sqrt{2}}}^{1} \int\limits_{\sqrt{1-x^{2}}}^{x} \frac{1}{\sqrt{x^2+y^2}}dydx$
I tried to transform into polar coordinates using $x = r\cos \theta$ , $y=r\sin\theta$ with $\left | J \right |= r$. Getting something like
$\int \int \frac{1}{r}rdrd\theta$ , but unable to define the upper and lower bounds of the integral.
Any help with that?
| Bounds come from inequalities
$$\left\{\begin{array}{l}
\frac{1}{\sqrt{2}} \leqslant x \leqslant 1 \\
\sqrt{1-x^{2}} \leqslant y \leqslant x
\end{array}\right\}$$
putting polar coordinates we obtain
$$\left\{\begin{array}{l}
\frac{1}{\sqrt{2}} \leqslant r \cos \theta \leqslant 1 \\
\sqrt{1-r^{2}\cos^2 \theta} \leqslant r \sin \theta \leqslant r \cos \theta
\end{array}\right\}$$
from first line we have two inequalities $\frac{1}{\sqrt{2} \cos \theta} \leqslant r $ and $ r \leqslant \frac{1}{\cos \theta}$. Second line gives $1 \leqslant r$ and $\sin \theta \leqslant \cos \theta$. Putting together we have
$$\int\limits_{0}^{\frac{\pi}{4}}\int\limits_{1}^{\frac{1}{\cos \theta}}$$
And, of course, it can be solved without polar coordinates using
$$\int \frac{1}{\sqrt{x^2+y^2}}dy=\ln\left(\frac{|y+\sqrt{x^2+y^2}|}{|x|} \right)+C$$
but, I prefer polar.
| {
"language": "en",
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Inequality for a decreasing, concave function on [0,1] Suppose I have a function $f:[0,1]\rightarrow \mathbb{R}_{\geq0}$ that is decreasing and concave. Let $a,b,c,d\in(0,1)$, with $a>b$ and $a+b>1$. I am trying to prove the following inequality:
$$
f\Bigg(\frac{(1-a)\cdot c}{(1-a)\cdot c + b\cdot d }\Bigg)\cdot \big( (1-a)\cdot c + b\cdot d \big)
+
f\Bigg(\frac{a\cdot c}{a\cdot c + (1-b)\cdot d }\Bigg)\cdot \big( a\cdot c + (1-b)\cdot d \big) \leq \\
f\Bigg(\frac{(1-b)\cdot c}{(1-b)\cdot c + a\cdot d }\Bigg)\cdot \big( (1-b)\cdot c + a\cdot d \big)
+
f\Bigg(\frac{b\cdot c}{b\cdot c + (1-a)\cdot d }\Bigg)\cdot \big( b\cdot c + (1-a)\cdot d \big)
$$
My approach was the following. Since $f$ is also subadditive (it's concave and $f(0)\geq0$), you can show that
$$
f\Bigg(\frac{(1-a)\cdot c}{(1-a)\cdot c + b\cdot d }\Bigg)\cdot \big( (1-a)\cdot c + b\cdot d \big) \leq f\big( (1-a)\cdot c \big),
$$ and
$$
f\Bigg(\frac{a\cdot c}{a\cdot c + (1-b)\cdot d }\Bigg)\cdot \big( a\cdot c + (1-b)\cdot d \big) \leq f\big(a\cdot c \big),
$$
after which I got stuck.. Does anyone know how to proceed, or perhaps know a better way to approach this?
EDIT: Similarly, (using the subadditivity), you can also show that
$$
f\Bigg(\frac{(1-b)\cdot c}{(1-b)\cdot c + a\cdot d }\Bigg)\cdot \big( (1-b)\cdot c + a\cdot d \big)\geq f(1) \cdot (1-b)\cdot c,
$$
and
$$
f\Bigg(\frac{b\cdot c}{b\cdot c + (1-a)\cdot d }\Bigg)\cdot \big( b\cdot c + (1-a)\cdot d \big) \geq f(1) \cdot b\cdot c.
$$
If you combine these inequalities you will get the sufficient condition:
$$f\big( (1-a)\cdot c \big) + f\big(a\cdot c \big) \leq f(1)\cdot c,$$
which is not satisfied because of the decreasing property of $f$. So this condition is too "loose", probably because the inequalities are too loose..
| A counterexample:
Let $f(x) = 2 - x^2$.
Let $a = 2/3, ~ b = 1/2, ~ c = 1/3, ~ d = 3/4$.
We have
$$\mathrm{LHS} = \frac{35}{72}f(8/35) + \frac{43}{72}f(16/43) = \frac{18589}{9030}$$
and
$$\mathrm{RHS} = \frac{2}{3}f(1/4) + \frac{5}{12}f(2/5) = \frac{247}{120}.$$
We have $\mathrm{LHS} - \mathrm{RHS} = \frac{3}{12040}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Geometric proof of $\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$
Prove:
$$\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$$
Geometrically.
I found the algebraic solutions here but I want to figure out how to calculate the above using geometrical arguments of complex numbers/ vectors.
Some hints of geometry I figured out already:
*
*$\frac{A+B}{2}$ is equidistant in angle mangitude from $A$ and $B$, that is $|\frac{A+B}{2} -A| = | \frac{A+B}{2} - B|$, similar results hold for the other angle averages.
*We may think of the above as summing the projections of three unit vectors onto the x-axis: $ \left( \tau(A) + \tau(B) + \tau(C) + \tau( A+B+C)\right) \cdot \hat{i}$, where $\tau(\phi)$ is the unit vector making angle $\phi$ with the x-axis
Related , related
|
Consider the eight points on the unit circle at angles $±α ± β ± γ$ from $(1, 0)$, shown in blue. We can construct their centroid in two ways.
*
*Pair them vertically, yielding the four red midpoints
$$(\cos (-α + β + γ), 0), \quad (\cos (α - β + γ), 0), \\
(\cos (α + β - γ), 0), \quad (\cos (α + β + γ), 0),$$
and then average these.
*Pair them hierarchically, as shown in green.
*
*First, pair the points separated by angles of $2α$. The resulting level-1 midpoints lie on a circle of radius $\cos α$ at angles $±β ± γ$ from $(\cos α, 0)$.
*Then, pair the level-1 midpoints separated by angles of $2β$. The resulting level-2 midpoints lie on a circle of radius $\cos α \cos β$ at angles $±γ$ from $(\cos α \cos β, 0)$.
*Finally, pair the level-2 midpoints separated by an angle of $2γ$. The resulting level-3 midpoint is $(\cos α \cos β \cos γ, 0)$.
Since the two constructions must be equivalent,
$$\begin{multline*}
\frac{\cos (-α + β + γ) + \cos (α - β + γ) + \cos (α + β - γ) + \cos (α + β + γ)}{4} \\
= \cos α \cos β \cos γ,
\end{multline*}$$
which yields the desired identity when $α = \frac{A + B}{2}, β = \frac{B + C}{2}, γ = \frac{A + C}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Integration by Substitution with $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ The angle was halved I've been encountering the problem of the below integration.
$$ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } \tag{1} $$
The official description states that the above integration formula can be calculated using substitution of integration.
$$ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \therefore ~~ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } = \frac{ 2\pi }{ \sqrt{ R ^{2} -r ^{2} } } $$
Currently I can't derive the above RHS.
I think firstly find out the form of result of calculations of indefinite integral of eqn1 is wiser way.
What I tried so far are as below.
$$ t= \tan^{}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ \theta_{} }{ 2 }= \tan^{-1} \left( t \right) ~~ \leftarrow~~ \text{Thought that this approach won't work} $$
$$ \frac{ dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) \cdot \frac{1}{2} $$
$$ \frac{ 2dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ d\theta }{ 2 dt } =\sec^{-2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ d\theta }{ 2 dt } = \left( \sec^{}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2} $$
$$ \frac{ d\theta }{ 2 dt } = \left( \cos^{-1}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2} $$
$$ \frac{ d\theta }{ 2 dt } = \cos^{2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ d\theta = 2 dt \cdot \cos^{2}\left(\frac{ \theta_{} }{ 2 } \right) $$
First things to first, the equation1 has $~ \cos^{}\left(\theta_{} \right) ~$ however how can I handle $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ ??
| Note that:
$$\cos(\theta)=\cos(2\frac{\theta}2)=\cos^2(\frac{\theta}2)-\sin^2(\frac{\theta}2)=\frac{\cos^2(\frac{\theta}2)-\sin^2(\frac{\theta}2)}{\cos^2(\frac{\theta}2)+\sin^2(\frac{\theta}2)}$$
Simplify by $\displaystyle{\cos^2(\frac{\theta}2)}$:
$$\cos(\theta)=\displaystyle{\frac{1-\tan^2(\frac{\theta}2)}{1+\tan^2(\frac{\theta}2)}}$$.
With the above, the integrand becomes:
$$\displaystyle{\frac{1}{R+r\cdot\frac{1-\tan^2(\frac{\theta}2)}{1+\tan^2(\frac{\theta}2)}}=\frac{1+\tan^2(\frac{\theta}2)}{R(1+\tan^2(\frac{\theta}2))+r(1-\tan^2(\frac{\theta}2))}}$$
Substitute $t=\tan(\frac{\theta}2)$. It follows that $dt=\frac12(1+\tan^2(\frac{\theta}2))d\theta$ and the integral becomes:
$$2\cdot2\int_0^\infty\frac{1}{R(1+t^2)+r(1-t^2)}dt$$ which you should be able to easily solve.
| {
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$(1+x+x^2)^n=P_0+P_{1}x + P_{2}x^2+ \cdots +P_{2n}x^{2n}$ Prove that,$ P_0+P_{3}+P_{6}+ \cdots =3^{(n-1)}$ Let's say $$ S_n = (1+x+x^2)^n $$
n=1 $$S_1=1+x+x^2$$
n=2 $$S_2=1+2x+3x^2+2x^3+x^4$$
n=3 $$S_3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6$$
n=4 $$S_4=1+4x+10x^2+16x^3+19x^4+16x^5+10x^6+4x^7+x^8$$
By taking coefficients of the $S_n$ we can form this type of triangle similar to Pascal's Trinagle
$$\begin{matrix}
&&&&&&&&&1\\
&&&&&&&1&&1&&1\\
&&&&&1&&2&&3&&2&&1\\
&&&&1&&3&&6&&7&&6&&3&&1\\
&&1&&4&&10&&16&&19&&16&10&&4&&1
\end{matrix}$$
| Let us denote $\omega$ as the third root of unity .
Therefore , ${\omega}^3 = 1$.
And ${\omega}^2 + \omega + 1=0$.
Now , $$(1+x+x^2)^n= P_0 + P_1x + P_2{x^2} + \ldots + P_{2n}x^{2n}....(1)$$
Putting $x=1$ in equation (1) ,we get
$$3^n= P_0 + P_1x + P_2{x^2} + \ldots + P_{2n}x^{2n}....(2)$$
Putting $x= \omega$ in (1) we get
$$(1+\omega + {\omega}^2)^n= 0 = P_0 + P_1\omega + P_2{\omega^2} + \ldots + P_{2n}\omega^{2n}....(3)$$
Putting $x=\omega^2$ in equation (1) we get ,
$$(1+\omega^2 + \omega^4)^n = 0 =P_0 + P_1\omega^2 + P_2{\omega^4} + \ldots + P_{2n}\omega^{4n} ....(4)$$
Adding equations (2) ,(3) and (4) we get ,
$$3^n = 3( P_0 + P_3 + P_6 + \ldots )$$
Therefore , $$(P_0 + P_3 + P_6 + \ldots ) = \frac{3^n}{3}= 3^{n-1}$$
| {
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Let $a_0,a_1,a_2$ be rationals, $\det\begin{pmatrix}a_0&a_1&a_2\\a_2&a_0+a_1&a_1+a_2\\a_1&a_2&a_0+a_1\end{pmatrix}=0$, show $a_0=a_1=a_2=0$
Let $a_0,a_1,a_2$ be rationals, $\det\begin{pmatrix}a_0&a_1&a_2\\a_2&a_0+a_1&a_1+a_2\\a_1&a_2&a_0+a_1\end{pmatrix}=0$, show $a_0=a_1=a_2=0$.
This is a problem in PHD candidate to PKU (2019).
The matrix is not invariant under permutations of $1,2,3$. So I could not just let the last column being the linear combination of the first two. Soundly, I have no idea. Any hint is grateful.
| HINT:
I suggest first to permute $a_1$ and $a_2$, getting the matrix
$$A=\begin{pmatrix}a_0&a_2&a_1\\a_1&a_0+a_2&a_1+a_2\\a_2&a_1&a_0+a_2\end{pmatrix}$$
Now, consider the ring
$$\mathbb{Q}[x]/(x^3-x-1)$$
Since the polynomial $x^3 - x -1$ is irreducible, we get in fact a field.
The multiplication by an element $a_0 + a_1 x + a_2 x^2$ has the above matrix in the basis $1$, $x$, $x^2$.
$\bf{Added:}$ Another solution in the style of linear algebra.
Consider the matrix
$$X=
\begin{pmatrix}0 & 0 & 1\\
1 & 0& 1\\
0 & 1 & 0\end{pmatrix}$$
( the companion matrix of the polynomial $x^3 - x -1$)
Then our matrix equals
$$A = a_0 + a_1 X + a_2 X^2$$
Let's diagonalize $X$. Consider $\alpha$, $\beta$, $\gamma$ the roots of the polynomial $x^3 - x -1$. Then one can check that
$$ \begin{pmatrix}1 & \alpha & \alpha^2\\
1 & \beta & \beta^2\\
1 & \gamma & \gamma^2 \end{pmatrix} \cdot X = \operatorname{Diag}(\alpha, \beta, \gamma) \cdot \begin{pmatrix}1 & \alpha & \alpha^2\\
1 & \beta & \beta^2\\
1 & \gamma & \gamma^2 \end{pmatrix}$$
Therefore, $X$ is diagonalizable with eigenvalues $\alpha$, $\beta$, $\gamma$, and $A$ is diagonalizable with eigenvalues $a_0 + a_1 \alpha + a_2 \alpha^2, \ldots$. We conclude that
$$\det A= (a_0 + a_1 \alpha + a_2 \alpha^2)(a_0 + a_1 \beta + a_2 \beta^2) ( a_0 + a_1 \gamma + a_2 \gamma^2)$$
Now we use that $x^3 - x -1$ is irreducible over $\mathbb{Q}$, so neither of $\alpha$, $\beta$, $\gamma$ is root of a rational polynomial of degree $\le 2$.
Let's note that the determinant equals the norm of the element $a_0 + a_1 x + a_2 x^2$ in the field $K = \mathbb{Q}[x]/(x^3 - x -1)$
$$N^K_{\mathbb{Q}}(a_0 + a_1 x + a_2 x^2) = \det A $$
$\bf{Added:}$ Another solution:
Note that we may assume $a_0$, $a_1$, $a_2$ integers (with $\gcd = 1$). We used the polynomial $x^3 - x -1$ which is irreducible over $\mathbb{Q}$. But the polynomial is also irreducible $\mod 2$. We could have worked over the basic field $\mathbb{Z}/2$ instead of $\mathbb{Q}$ and still get the same result. This suggests that we can show the result $\mod 2$. Therefore, we can show: if $a_0$, $a_1$, $a_2$ are integers, not all even then the above determinant is odd. Indeed the determinant equals
$$\det A = a_0^3 + 2 a_0 ^3 a_3 - a_0 a_1^2 - 3 a_0 a_1 a_2 + a_0 a_2^2 + a_1^3 - a_1 a_2^2 + a_3^3 $$
(see this WA calculation). Now, if the $a_i$ are in $\mathbb{Z}/2$ then the expression equals
$$a_0 + a_1 + a_2 + a_0 a_1 + a_0 a_2 + a_1 a_2 + a_0 a_1 a_2 = (1+a_0)(1+ a_1)(1+a_2) + 1$$
and this equals $1 \pmod 2$ if at least one of the $a_i$ is $1 \pmod 2$. We are done.
| {
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Expressing the $n$–th derivative of $y=\frac{1}{(1+x^2)^2}$ In an attempt to find a way to express the n-th derivative of $y=\frac{1}{(1+x^2)^2}$ using the binomial theorem I got stuck in the last computation.
I'll explain what I did:
$$f(x)=\frac{1}{(1+x^2)^2}=\frac{1}{(x+i)^2(x-i)^2}=\\
\frac{i}{4}\left( \frac{1}{x+i} \right)-\frac{1}{4}\left( \frac{1}{(x+i)^2} \right)-\frac{i}{4}\left( \frac{1}{x-i} \right)-\frac{1}{4}\left( \frac{1}{(x-i)^2} \right)$$
Then I used $y=\frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^nn!a^n}{(ax+b)^{n+1}}$ and $y=\frac{1}{(ax+b)^2} \Rightarrow y_{n}=\frac{(-1)^n(n+1)!a^n}{(ax+b)^{n+2}}$ to get:
$$\frac{(-1)^n}{4}\left( in!\left( \frac{(x-i)^{n+1}-(x+i)^{n+1}}{(x^2+1)^{n+1}} \right)-(n+1)!\left( \frac{(x-i)^{n+2}+(x+i)^{n+2}}{(x^2+1)^{n+2}} \right) \right)$$
Then I stopped for a second and I analized $(x-i)^{n+1}-(x+i)^{n+1}$ and $(x-i)^{n+2}+(x+i)^{n+2}$ using the binomial theorem:
-
$(x-i)^{n+1}-(x+i)^{n+1}$ = $$\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}\cdot i^k((-1)^k-1)$$
$i^k((-1)^k-1) =0$ for $k=2n$ ,whereas for $k=2n+1$ $\longrightarrow$$-2i(-1)^{\frac{k-1}{2}}$
-
$(x-i)^{n+2}+(x+i)^{n+2}$ = $$\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}\cdot i^k((-1)^k+1)$$
$i^k((-1)^k+1) =0$ for $k=2n+1$ , whereas for $k=2n$ $\longrightarrow$$2(-1)^{\frac{k}{2}}$.
And now I got stuck sorting out the indeces $n,k$ of summations; what I obtain is:
$$\frac{d^n}{dx^n}(f(x))=\\
\frac{(-1)^nn!}{2(x^2+1)^{n+1}}\cdot \sum_{k=1}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} - \frac{(-1)^n(n+1)!}{2(x^2+1)^{n+2}}\cdot \sum_{k=2}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}$$
But testing it I can see it’s wrong.
Can anyone help me please sorting the $n,k$ indices and perhaps have a look at my computations also?
Thank you
| I will check later about your indexes and the calculations.
I know you asked explicitly for a Binomial Series solution, yet I just wanted to point out that the $n$-th derivative of the function above can be easily found with
$$\dfrac{\text{d}^n}{\text{d}x^n} \dfrac{1}{(1+x^2)^2} = \dfrac{n!}{4}\left[-\dfrac{2 + n + ix}{(-x + i)^{n+2}} - \dfrac{2+n - ix}{(-x-i)^{n+2}}\right]$$
| {
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How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But it didn't work.
Please Help me.
| $\tan y \approx y + \frac{y^3}{3}$ from the Taylor series. We can discard the $y^3/3$ term to get $\arctan x \approx x$, as we only need the $x^2$ term and below.
Then:
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x} = \lim_{x \to 0} \frac{x^2}{1-\cos x} \frac{1 + \cos x}{1 + \cos x} = 2.$$
| {
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Calculate $ \int \int_{R^2} \frac{dxdy}{(1+4x^2+9y^2)^2}$. Calculate: $ \int \int_{R^2} \frac{dxdy}{(1+4x^2+9y^2)^2}$.
Solution from book:
We form a set $D_n=\{ (x,y)|4x^2+9y^2\leq n^2 \}$.
Then $4x^2+9y^2=r^2$; $x=\frac{1}{2}rcost$ and $y=\frac{1}{3}rsint$ where $0\leq r \leq n$ and $0\leq t \leq 2\pi$. After that I know how to solve the integral. I just don't understand why is it done this way, why do we form this set $D_n$. The theory behind this is bugging me..
And another example is: $ \int \int_{D} \frac{dxdy}{(1-x^2-y^2)^2}$. Where D is a unit circle.
Here $D_n=\{ (x,y)|x^2+y^2\leq (1-\frac{1}{n})^2 \}$. Why?
Also $ \int \int_{R^2} \frac{dxdy}{1+(x^2+y^2)^2}$ what would $D_n$ be for this one?
| An alternative approach.
For any $A>0$ we have $\int_{0}^{+\infty}\frac{dx}{A+x^2}=\frac{\pi}{2\sqrt{A}}$, hence by differentiating with respect to $A$ we have
$$ \int_{0}^{+\infty}\frac{dx}{(A+x^2)^2}=\frac{\pi}{4 A^{3/2}},\qquad \int_{-\infty}^{+\infty}\frac{dx}{(1+y^2+x^2)^2}=\frac{\pi}{2(1+y^2)^{3/2}} $$
and by Fubini's theorem
$$ \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2+y^2)^2}=\frac{\pi}{2}\int_{-\infty}^{+\infty}\frac{dy}{(1+y^2)^{3/2}}=\frac{\pi}{2}\left[\frac{y}{\sqrt{1+y^2}}\right]_{-\infty}^{+\infty}=\pi, $$
so by letting $x=2u$ and $y=3v$ we have
$$ \iint_{\mathbb{R}^2}\frac{du\,dv}{(1+4u^2+9v^2)^2}=\color{red}{\frac{\pi}{6}}.$$
| {
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Limit at (0,0) of $\frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$ $$\text{Let}\ f(x,y) = \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$$
WolframAlpha tells me that $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist. To prove the non-existence of the limit, I tried three different paths ($y=kx, y=kx^2, y=kx^3)$ and they all equal zero. From the graph of the function it looks like the limit is indeed zero. Also I thought I was able to prove the limit does indeed exist at $(0,0)$ with the epsilon-delta definition:
$$y^2 \leq x^2+y^2 \leq x^2 + y^4 \iff \frac{y^2}{x^2+y^4} \leq 1 \overset{|x|<1}{\iff} \frac{|x|y^2}{x^2+y^4} \leq 1 \iff \frac{|x|y^2\sqrt{x^2+y^2}}{x^2+y^4} \leq \sqrt{x^2+y^2} \iff \left|\frac{xy^2\sqrt{x^2+y^2}}{x^2+y^4} - 0\right|\leq \left|\sqrt{(x-0)^2+(y-0)^2}\right| \lt δ := ε $$
So for $δ = ε$ we have that $\|x -(0,0)\| < δ \implies |f(x,y) - 0| < ε$
I'm confused
| Hint: $\dfrac{xy^2}{x^2+y^4} \le \dfrac{1}{2}$. This is true by the AM-GM inequality. Note that the inequality you claimed is wrong because the line $y^2 \le y^4$ is not true !
| {
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Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers. Several formulae that solve the diophantine equation
$$w^4 + x^4 = y^4+z^4 \tag{1}$$
are presented in this collection. The simplest one bases on
$$f_1 = a^7 + a^5 - 2 a^3 + 3 a^2 + a \tag{2}$$
and sets$\def\hf{{}^h\!f}$
$$ (w, x, y, z) = (\hf_1,\ \hf_1(b,-a),\ \hf_1(a,-b),\ \hf_1(b,a)) $$
where $\hf(a,b) = f(a/b)\cdot b^{\deg f}$ denotes the homogenized version of $f$.
Question: How does one find such formulae?
The collection refers to Hardy & Wright, and MathWorld mentions that parametric solutions to (1) were already known to Euler in 1802, but without mentioning which one.
Note: $f_1$ appears to be — in some sense — the simplest of such functions: It produces (158, 59, 134, 133) from (1,2), the smallest solution of (1). All solutions that I checked using Sage were non-trivial except for the case $a=b=1$, and with $\gcd(a,b)=1$ all solutions satisfied $\gcd(w,x,y,z)\in\{1,2\}$. The last two properties also showed up for the next 2 more complicted functions from the collection that I tried like
$$f_2 = -a^{13} + a^{12} + a^{11} + 5 a^{10} + 6 a^{9} - 12 a^{8} - 4 a^{7} + 7 a^{6} - 3 a^{5} - 3 a^{4} + 4 a^{3} + 2 a^{2} - a + 1
$$
| $$w^4 + x^4 = y^4 + z^4 \tag{1}$$
Let $w=mt+a, x=t-b, y=t+a, z=mt+b,$ then we get
$(-4a-4b-4bm^3+4am^3)t^3+(-6b^2m^2-6a^2+6b^2+6a^2m^2)t^2+(-4b^3m+4a^3m-4a^3-4b^3)t=0$
Hence let $$m = \frac{a^3+b^3}{-b^3+a^3}$$ then we get $$t = \frac{3(-b^3+a^3)b^2a^2}{a^6-2b^2a^4-2b^4a^2+b^6}$$
Thus we get a parametric solution as follows.
$w = a(a^6+b^2a^4-2b^4a^2+3b^5a+b^6)$
$x = b(a^6-3ba^5-2b^2a^4+b^4a^2+b^6)$
$y = a(a^6+a^4b^2-2a^2b^4-3ab^5+b^6)$
$z = b(a^6+3ba^5-2b^2a^4+b^4a^2+b^6)$
Let $b=1$, we get
$w = a^7+a^5-2a^3+3a^2+a$
$x = a^6-3a^5-2a^4+a^2+1$
$y = a^7+a^5-2a^3-3a^2+a$
$z = a^6+3a^5-2a^4+a^2+1$
This solution is not a complete solution.
Several solutions are found on A^4+B^4=C^4+D^4
| {
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"source": "stackexchange",
"question_score": "4",
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Question about exact ODE. Why in $h'(y)$ contain $x$? Find the solution ODE
$$\left(x^3e^xy+4x^2e^xy+2xe^xy\right)dx+(x^3e^x+x^2e^x)dy=0.$$
Let $M(x,y)=x^3e^xy+4x^2e^xy+2y$ and $N(x,y)=x^3e^x+x^2e^x$.
\begin{align*}
\dfrac{\partial M}{\partial y}&=\dfrac{\partial}{\partial y}\left(x^3e^xy+4x^2e^xy+2xe^xy\right) \\
&=x^3e^x+4x^2e^x+2xe^x.\\
\dfrac{\partial N}{\partial x}&=\dfrac{\partial}{\partial x}\left(x^3e^x+x^2e^x\right) \\
&=3x^2e^x+x^3e^x+2xe^x+x^2e^x\\
&=x^3e^x+4x^2e^x+2xe^x.
\end{align*}
This is exact ODE since $\dfrac{\partial M}{\partial y}= \dfrac{\partial N}{\partial x}$.
Now,
\begin{alignat}{2}
&&\dfrac{\partial F(x,y)}{\partial x}&=M(x,y)\nonumber\\
\Longleftrightarrow\quad
&&\dfrac{\partial F(x,y)}{\partial x}&=x^3e^xy+4x^2e^xy+2y\nonumber\\
\Longleftrightarrow\quad
&&\int\partial F(x,y)&=\int\left(x^3e^xy+4x^2e^xy+2y\right) \partial x\nonumber\\
\Longleftrightarrow\quad
&&\int\partial F(x,y)&=\int x^3e^xy\partial x+\int4x^2e^xy \partial x+\int2y \partial x.\label{ijoet}
\end{alignat}
Consider that
\begin{align}
\int x^3e^xy\partial x&=x^3e^xy-\int e^xy 3x^2 \partial x\nonumber\\
&=x^3e^xy-\int 3x^2 e^x y \partial x\nonumber\\
&=x^3e^xy-\left(3x^2e^x y-\int e^x y 6x \partial x\right)\nonumber\\
&=x^3e^xy-3x^2e^x y+\int 6x e^x y\partial x\nonumber\\
&=x^3e^xy-3x^2e^x y+6x e^x y-\int e^x y 6\partial x\nonumber\\
&=x^3e^xy-3x^2e^x y+6x e^x y- 6 e^x y+h_1(y).\label{meong}
\end{align}
\begin{align}
\int 4x^2e^xy\partial x&=4x^2e^xy-\int e^xy 8x \partial x\nonumber\\
&=4x^2e^xy-\int 8x e^xy \partial x\nonumber\\
&=4x^2e^xy-\left(8x e^xy-\int e^xy 8 \partial x\right)\nonumber\\
&=4x^2e^xy-8x e^xy+\int 8e^xy \partial x\nonumber\\
&=4x^2e^xy-8x e^xy+ 8e^xy +h_2(y).\label{meong1}
\end{align}
\begin{align}
\int2y \partial x&= 2xy+h_3(y).\label{meong2}
\end{align}
So, we have
\begin{alignat}{2}
&&\int\partial F(x,y)&=x^3e^xy-3x^2e^x y+6x e^x y- 6 e^x y+h_1(y)\nonumber\\
&&&\quad +4x^2e^xy-8x e^xy+ 8e^xy +h_2(y)+2xy+h_3(y)\nonumber\\
\Longleftrightarrow\quad
&&F(x,y)&=x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+2xy +h(y)\nonumber
\end{alignat}
which $h(y)=h_1(y)+h_2(y)+h_3(y)$.
Next, consider that
\begin{alignat*}{2}
&&\dfrac{\partial F(x,y)}{\partial y}&=N(x,y)\\
\Longleftrightarrow\quad
&&\dfrac{\partial}{\partial y}\left(x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+2xy +h(y)\right)&=x^3e^x+x^2e^x\\
\Longleftrightarrow\quad
&&x^3e^x+x^2e^x -2 x e^x +2e^x +2x +h'(y)&=x^3e^x+x^2e^x\\
\Longleftrightarrow\quad
&&h'(y)&=2 x e^x -2e^x -2x \\
\end{alignat*}
When I want to find $h(y)$, I have found $h'(y)=2 x e^x -2e^x -2x$.
Why in $h'(y)$ contain $x$? What my mistake?
| It is much easier than what you make it to be: If
$$
M(x,y)\,\mathrm{d}x+N(x)\,\mathrm{d}y=0
$$
is exact and $M$ contains no terms independent of $y$, then LHS must be $\mathrm{d}(N(x)y)$, since that is what the product rule gives you for the $\mathrm{d}y$. If there are terms independent of $y$ in $M(x,y)$, then you can (theoretically) integrate them to get $\mathrm{d}(N(x)y+f(x))$.
| {
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Proving $\frac{\cos A - \cos B}{\sin A + \sin B} = \frac{\sin B - \sin A}{\cos A + \cos B}$
Prove $$\dfrac{\cos A - \cos B}{\sin A + \sin B} = \dfrac{\sin B - \sin A}{\cos A + \cos B}$$
I tried as shown below and am not sure how to do it. Your help is appreciated. Thanks.
Proving from left hand side:
$$\dfrac{\cos A}{\sin A + \sin B} - \dfrac{\cos B}{\sin A + \sin B}$$
$$=\dfrac{\dfrac{\cos A}{\sin A}} {\dfrac{\sin A+\sin B}{\sin A}} - \dfrac{\dfrac{\cos B}{\sin B}}{\dfrac{\sin A+\sin B}{\sin B}}$$
$$= \dfrac{\dfrac{1}{\tan A}}{1+\dfrac{\sin B}{\sin A}}- \dfrac{\dfrac{1}{\tan B}}{1+\dfrac{\sin A}{\sin B}}$$
| We have that (for $\sin A + \sin B \neq 0$ and $\cos A + \cos B \neq 0$)
$$\dfrac{\cos A - \cos B}{\sin A + \sin B} = \dfrac{\sin B - \sin A}{\cos A + \cos B} $$
$$\iff (\cos A - \cos B)(\cos A + \cos B)=(\sin B - \sin A)(\sin B + \sin A)$$
$$\iff \cos^2 A - \cos^2 B=\sin^2 B - \sin^2 A$$
$$\iff \cos^2 A + \sin^2 A= \cos^2 B + \sin^2 B$$
Edit
As an alternative, by sum to product identities we have
*
*LHS
$$\dfrac{\cos A - \cos B}{\sin A + \sin B} =\frac{-2\sin\left(\frac{A+B}2\right)\sin\left(\frac{A-B}2\right)}{2\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}=-\tan\left(\frac{A-B}2\right)$$
*
*RHS
$$\dfrac{\sin B - \sin A}{\cos A + \cos B} =\frac{2\sin\left(\frac{B-A}2\right)\cos\left(\frac{A+B}2\right)}{2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}=-\tan\left(\frac{A-B}2\right)$$
| {
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How to Prove $\lim_{(x, y) \rightarrow (0, 0)}\frac{x^3-y^3}{x^2+y^2}=0$ I wanted to prove how
$$
\lim_{(x, y) \to (0, 0)} \frac{x^3-y^3}{x^2+y^2} = 0.
$$
Specifically, I want to use the Squeeze Theorem for multivariable calculus. Then I know that I should pick an arbitrary function $g(x, y)$ where $|f(x, y)-L| \leq g(x, y)$ and the limit of $g(x, y)$ is $0$. Since the limit of $f(x, y) =0$, I just have to make
$$
\Biggl\lvert \frac{x^3-y^3}{x^2+y^2} \Biggr\rvert \leq g(x, y),
$$
where the limit of $g(x,y)=0$. I was given a hint that $|x^3-y^3| \leq |x|^3+|y|^3$. So I divided both sides by $|x^2+y^2|$,
$$
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}
\leq \frac{\lvert x\rvert^3+ \lvert y\rvert^3}{\bigl\lvert x^2+y^2 \bigr\rvert},
$$
meaning that
$$
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}
\leq \lvert x\rvert + \lvert y\rvert.
$$
What do I do next?
| Now,
$$
0\leqq\left|\frac{x^3-y^3 }{x^2+y^2} \right|
\leq \lvert x\rvert + \lvert y\rvert$$
Letting $(x,y)\to(0,0)$, you get
$$
\lim_{(x,y)\to(0,0)} \ \ \ 0\leq\lim_{(x,y)\to(0,0)}\ \ \left|\frac{x^3-y^3 }{x^2+y^2} \right|
\leq \lim_{(x,y)\to(0,0)}\lvert x\rvert + \lvert y\rvert$$
The LHS and RHS are $0$ so due to the squeeze theorem, you get $$\lim_{(x,y)\to(0,0)}\ \ \left|\frac{x^3-y^3 }{x^2+y^2} \right| =0.$$
This means $$\lim_{(x,y)\to(0,0)}\ \ \ \frac{x^3-y^3 }{x^2+y^2} =0.$$
But I think that you should show why $\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}
\leq \frac{\lvert x\rvert^3+ \lvert y\rvert^3}{\bigl\lvert x^2+y^2 \bigr\rvert}
$ means $
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}
\leq \lvert x\rvert + \lvert y\rvert.
$
| {
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A committee of $4$ has to be formed from $6$ men and $4$ women must contain at least one woman
From a group of $6$ men and $4$ women a committee of $4$ persons is to be formed in such a way that committee has at least one woman.
I tried to divide this work into 2 tasks-
*
*Selecting $1$ woman out of $4$ to fulfill the requirement
*Then selecting $3$ members from rest of the people
So,the answer which I get from this is,
$$9C3 × 4C1 = \frac{9 \cdot 8 \cdot 7}{6} \cdot 4 = 336$$
But this is not the correct answer. What am I doing wrong here??
| Your method counts each committee with exactly $k$ women $k$ times, once for each way you could designate one of those $k$ women as the woman on the committee.
There are
$$\binom{4}{k}\binom{6}{4 - k}$$
ways to select a committee of four people with exactly $k$ women and $4 - k$ men, so the number of four-person committees with exactly $k$ women is
$$\sum_{k = 1}^{4} \binom{4}{k}\binom{6}{4 - k} = \binom{4}{1}\binom{6}{3} + \binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} + \binom{4}{4}\binom{6}{0} = 195$$
which, as indicated in the comments of Math Lover and Thomas Andrews, can be found more simply by subtracting the number of committees with only men from the number of four-person committees.
$$\binom{10}{4} - \binom{6}{4} = 195$$
Your method counts each committee with two women twice. For instance, if the women Angela and Barbara and the men Charles and Douglas are chosen to serve on the committee, your method counts this committee twice:
$$
\begin{array}{c | c}
\text{designated woman} & \text{additional people}\\ \hline
\text{Angela} & \text{Barbara, Charles, Douglas}\\
\text{Barbara} & \text{Angela, Charles, Douglas}
\end{array}
$$
Similarly, your method counts each committee with three women three times and the committee with all four women four times. Notice that
$$\binom{4}{1}\binom{6}{3} + \color{red}{\binom{2}{1}}\binom{4}{2}\binom{6}{2} + \color{red}{\binom{3}{1}}\binom{4}{3}\binom{6}{1} + \color{red}{\binom{4}{1}}\binom{4}{4}\binom{6}{0} = 336$$
where the factors in red denote the number of ways you could designate one of the $k$ women on the committee as the woman on the committee.
| {
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Find all possible values of $f(2018)$ Find all possible values of $f(2018)$ where $f$ is a function from $\mathbb{Q}$ to $\mathbb{R_+}$ satisfying the
following three conditions:
$f(2) = \frac{1}{2}$;
for any rational $x$ if $f(x) ⩽ 1$, then $f(x + 1) ⩽ 1$;
$f(xy) = f(x)f(y)$ for all rationals x and y.
I'm not quite sure how to tackle this sort of problem. Since $f(2018) = \frac{f(1009)}{2} \leq \frac{1}{2}$, and $1009$ is prime, does this mean all values below half are possible? How do I solve this type of problem?
| $f$ can only be the 2-adic valuation on $\mathbb{Q}$. That is, $f(0) = 0$ (a simple deduction from $f(0 \times 2) = f(0) f (2)$), and if $q = \pm 2^k \frac{a}{b}$ with $a$ and $b$ both odd, then $f(q) = 2^{-k}$. This means that $f(2018) = \frac{1}{2}$.
A few simple deductions to start:
*
*$\frac{1}{2} = f(2) = f(1) f(2) = \frac{f(1)}{2}$, so $f(1) = 1$.
*$1 = f(1) = f((-1)^2) = f(-1)^2$, and $f$ has nonnegative range, so $f(-1) = 1$.
*$f(-x) = f(-1) f(x) = f(x)$ for all $x$.
*$f(n) \leq 1$ for all integers $n$ (proof: we have it for $n = 0, 1, 2$, and the case $|n| \geq 3$ follows from $f(x) \leq 1 \implies f(x+1) \leq 1$ and the evenness of $f$).
*If $0 < x < y$, $f(y) \leq 1$, and $y - x$ is an integer, then $f(x) \leq 1$. (Proof: $f(-y) = f(y) \leq 1$, so $f(-y + (y-x)) = f(-x) \leq 1$, so $f(x) \leq 1$.)
Now let $p \neq 2$ be an odd prime. For any sufficiently large integer $n$, we have $f \left( \frac{2^n}{p} \right) = \frac{1}{2^n f(p)} < 1.$ Furthermore, $2^n \equiv 1 \bmod p$ if $n$ is a multiple of $p-1$ by Fermat's little theorem. So there's some integer $n$ such that $f\left(\frac{2^n}{p}\right) \leq 1$ and $\frac{2^n}{p} - \frac{1}{p}$ is an integer, proving $f\left( \frac{1}{p} \right) \leq 1$. As $f(p) \leq 1$ and $f\left( \frac{1}{p} \right) f(p) = 1$, so $f(p) = 1$. The result that $f$ must be the 2-adic valuation follows immediately from the multiplicativity of $f$.
| {
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Prove that ${f(1);f(2);f(3)...;f(3^{k})}$ is a complete set of residues mod $3^{k}$
Let a function $ f : \Bbb N^* \to \Bbb N^* $ ($\mathbb N^*$ is the set of positive integers)
$f(1) = 1 $;
$f(n+1) = f(n) +2^{f(n)}$ for all positive integers $n$.
Prove that ${f(1),f(2),f(3),\dots,f(3^{k})}$ is a complete set of residues mod $3^{k}$ (where $k$ is a positive interger).
$f(n+1) = f(n) +2^{f(n)}$
$\Rightarrow f(n+1) =f(n-1) + 2^{f(n-1)}+2^{f(n)} = ... = 1 + 2^{f(1)} + 2^{f(2)} +...+ 2^{f(n)}$
Suppose there exist $2$ positive integers $a, b$ satisfying $ 1 \le a < b \le 3^{k} $ :
$ f(a) \equiv f(b) \pmod{ 3^{k}} $
$\Rightarrow 1 + 2^{f(1)} + 2^{f(2)} +...+ 2^{f(a-1)} \equiv 1 + 2^{f(1)} + 2^{f(2)} +...+ 2^{f(b-1)} \pmod {3^{k}}$
$\Rightarrow 2^{f(a)} + 2^{f(a+1)} + ... + 2^{f(b-1)} \equiv \pmod {3^{k}}$
It is easy to see that $2$ is the primitive root mod $3^{k}$ and $f(n) \equiv 1 \pmod 2 $
$\Rightarrow 2^1 ; 2^2 ;...; 2^{ 2.3^{(k-1)} } $ reduced set of residues
mod $3^{k} $
So if $ f(a) \equiv f(b)$ (mod $3^{k}) \Rightarrow f(a) \equiv f(b)$ (mod $3^{k-1})$
$\Rightarrow f(a) \equiv f(b)$ (mod $2.3^{k-1})$
$\Rightarrow 2^{f(a)} \equiv 2^{f(b)}$ (mod $3^{k})$
$\Rightarrow f(a) + 2^{f(a)} \equiv f(b)+2^{f(b)}$ (mod $3^{k})$
$\Rightarrow f(a+1) \equiv f(b+1)$ (mod $3^{k})$
$\Rightarrow$ if $f(a) \equiv f(b)$ (mod $3^{k})$ then $f(a+s) \equiv f(b+s) $ (mod $3^{k}) (b+s \le 3^{k} )$
I want to point out the absurdity in the above argument but have no idea, I hope to get help from everyone. Thanks very much ! (This is a problem from a long time ago, I guarantee.)
| can induct on k.
induction hypothesis: $f(n),f(n+1),...,f(n+3^k-1)$ is a complete set of residues modulo $3^k$ for all $n\geq1$
for the base case,
{f(1),f(2),f(3)}={1,3,11}={0,1,2} is a complete residue set mod 3.
$m_1\equiv m_2\pmod{3}\implies f(m_2)-f(m_1)\equiv 2^{f(m_1)} +2^{f(m_1+1)}...+ 2^{f(m_2)}\equiv 2+...+2\equiv 2(m_2-m_1)\equiv 0\pmod{3}$
f(n),f(n+1),f(n+2) is a complete residue set mod 3.
for the induction step,
$\{f(n),f(n+1),...,f(n+3^k-1)\}$ is a complete residue set, so it is $\{1,2,...,3^k\}\pmod{3^k}$
the $3^k$ numbers are all odd.
by CRT, $\{f(n),f(n+1),...,f(n+3^k-1)\}\equiv\{1,3,5,...,2*3^k-1\}\pmod{2*3^k}$
by FLT, $\{2^{f(n)},2^{f(n+1)},...,2^{f(n+3^k-1)}\}\equiv\{2^1,2^2,...,2^{2*3^k-1}\}\pmod{3^{k+1}}$
$f(n+3^k) - f(n) = 2^{f(n)} + 2^{f(n+1)} + \dots + 2^{f(n+3^k-1)}
\\ \equiv 2^1 + 2^3 + \dots + 2^{2 \cdot 3^k-1}
\\ \equiv 2\frac{4^{3^k}-1}{4-1}
\\ \equiv 2*3^k \pmod{3^{k+1}}$
$f(n+3^{k+1}) - f(n)\equiv (f(n+3*3^k) - f(n+2*3^{k+1}))+...\equiv (2*3^k)*3\equiv0\pmod{3^{k+1}}$
$m_1\equiv m_2\pmod{3^{k+1}}\implies f(m_1)\equiv f(m_2)\pmod{3^{k+1}}$
now we show $f:Z/3^{k+1}Z\to Z/3^{k+1}Z$ is injective.
$f(m_1)\equiv f(m_2)\pmod{3^{k+1}}\implies f(m_1)\equiv f(m_2)\pmod{3^k} \implies m_1\equiv m_2\pmod{3^k}\implies m_1\equiv m_2\pmod{3^{k+1}}$
${f(n),…,f(n+3^{k+1}-1)}$ is a complete set of residues mod $3^{k+1}$, completing the induction step.
--
some calculations
$(f(1),...,f(9))\equiv(1,3,2,7,0,8,4,6,5)\pmod{9}$
| {
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Finding all pairs of integer solutions to ${x^y = (x+y)^2}$ How do I find all solutions for ${x^y = (x+y)^2}$, where $x$ and $y$ are positive integers and at least one of them is prime.
For example, ${64 = 2^6 = (2+6)^2}$
A friend of mine pointed out that $x|y^2$ then $x^2|y^2$, so, when $x$ is prime and $y$ is not prime, we can write $y=xk$. Thus $x^{xk} = x^2 (1+k^2)$ and $x^{xk-2} = (1+k^2)$.
| This is not an answer. This is a clarification to a comment I did to the OP.
$y$ must be even if $x$ is not a perfect square because:
Since $y$ is an integer there are two possibilities, either $y=2k$ or $y=2k+1$. If $y=2k+1$(odd) then
$x^y=x^{2k+1}=(x+2k+1)^2$
$x^\frac{2k+1}{2}=x+2k+1$
$x^k\sqrt{x}=x+2k+1$
Notice that here the $RHS$ is an integer and the $LHS$ would also be an integer if $x$ is a perfect square otherwise the $LHS$ would not be an integer which is a contradiction.
This means that if $x$ is not a perfect square then $y$ is even.
| {
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Check if the following series converge $$\sum_{n=1}^{\infty}\frac{1 \cdot 4 \cdot 7 \cdot ... \cdot (3n+1)}{n^2 \cdot 3^n \cdot n!}$$
This is my solution:
Let's rewrite the question:
$$\sum_{n=1}^{\infty}\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}$$
If we use D'Alembert's criterion:
$$\lim_{n \to \infty} \frac{\frac{(3n+4)(3n+1)!!!}{(n+1)^2\cdot3^n \cdot3 \cdot(n+1) \cdot n!}}{\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}} = \lim_{n \to \infty} \frac{3n^3 + 4n^2}{3n^3 + 9n^2 + 9n+3} = 1$$
Which doesn't give an answer. However, if we use Raabe's criterion:
$$\lim_{n \to \infty} n\cdot(1-\frac{3n^3 + 4n^2}{3n^3 + 9n^2 + 9n+3}) = \frac{5}{3}$$
I calculated the limit correctly, I get the same result from WolframAlpha. Since the result is greater than 1, the series converge.
Is this correct?
| Your proof is fine, as an alternative by limit comparison test we have
$$\frac{\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}}{\frac1{n^\frac 43}}=\frac{(3n+1)!!!}{n^\frac23 \cdot 3^n \cdot n!} \to 0$$
indeed by Stirling's approximation we have
$$\frac{(3n+1)!!!}{n^\frac23 \cdot 3^n \cdot n!}=\frac{(n+\frac13)!}{n^\frac23 \cdot n!} \sim \frac{\sqrt {2\pi (n+\frac13)}\cdot \left(\frac {(n+\frac13)} e\right)^{(n+\frac13)}}{n^\frac23 \cdot \sqrt {2\pi n}\cdot \left(\frac n e\right)^n}=$$
$$=\sqrt{1+\frac1{3n}}\cdot \frac{(n+\frac13)^\frac13}{ e^\frac13\cdot n^\frac23}\cdot \left(1+\frac1{3n}\right)^n \to 0$$
therefore the given series converges by limit comparison test with $\sum \frac1{n^\frac43}$.
| {
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Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$ The Equation
How can I analytically show that there are no real solutions for $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$?
My attempt
With $u = -x+2$
$\sqrt[3]{u-1}-\sqrt[3]{u+1}=1$
Raising to the power of $3$
$$(u+1)^{2/3}(u-1)^{1/3} - (u+1)^{1/3}(u-1)^{2/3}=1\\(u+1)^{1/3}(u^2-1)^{1/3} - (u-1)^{1/3}(u^2-1)^{1/3}=1\\(u^2-1)^{1/3}\cdot\boxed{\left[(u+1)^{1/3}-(u-1)^{1/3}\right]}=1$$
Raising to the power of $3$:
$$(u^2-1)\cdot\left[3(u+1)^{1/3}(u-1)^{2/3}-3(u+1)^{2/3}(u-1)^{1/3}+2\right]=1\\(u^2-1)\cdot\left[3(u^2-1)^{1/3}(u-1)^{1/3}-3(u+1)^{1/3}(u^2-1)^{1/3}+2\right]=1$$
Thus: $(u^2-1)\cdot\left[3(u^2-1)^{1/3}\boxed{\left[(u-1)^{1/3}-(u+1)^{1/3}\right]}+2\right]=1$
And with $y = (u-1)^{1/3}-(u+1)^{1/3}$, we can say that:
$y^3=-3y(u^2-1)^{1/3}+2$
I am stuck... Any tips for this radical equation?
| We have that by $A^3-B^3=(A-B)(A^2+AB+B^2)$
$$\sqrt[3]{x-3}+\sqrt[3]{1-x}=\sqrt[3]{x-3}-\sqrt[3]{x-1}=$$
$$=\frac{-2}{\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2}}<0$$
indeed the latter is true for $x= 1,2,3$ and for $x\neq 1,2,3$ by AM-GM
$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-1)^2} > 2\sqrt[3]{|x-3||x-1|}>0$$
and therefore
$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2} >\sqrt[3]{|x-3||x-1|}>0$$
| {
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"source": "stackexchange",
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Finding all $ f: \mathbb{R} \to \mathbb{R}$ such that $ f(x+f(y))=f(x)+2xy^2+y^2f(y) $ $$
f: \mathbb{R} \to \mathbb{R}, f(x+f(y))=f(x) + 2xy^2 + y^2f(y)
$$
How can we solve this problem? This is my try, but can't go more.
$$
P(x, y): f(x+f(y))=f(x) + 2xy^2 + y^2f(y) \\
P(0, 0): f(f(0))=f(0) \\
let \ f(0) = a \Rightarrow f(a)=a. \\
P(a, a): f(2a)=a+2a^3+a^3 \\
P(a, 0): f(2a)=a \\
\Rightarrow 3a^3=0 \Rightarrow a = 0, f(0) = 0. \\
\ \\
Assume) \ \exists \ t \ s.t. \ t \ \neq 0, \ f(t)=0. \\
P(t, t): 0 = 2t^3 \Rightarrow t = 0, \text{Contradiction.} \\
\therefore f(t)=0 \Leftrightarrow t = 0.
\ \\
P(0, y): f(f(y))=y^2f(y). \\
P(f(x), x): f(2f(x))=f(f(x))+2f(x) \cdot x^2 + x^2f(x) = 4f(f(x))=4x^2\cdot f(x) \\
P(2f(x), x): f(3f(x))=f(2f(x))+4f(x)\cdot x^2 + x^2f(x) = 9f(f(x)) = 9x^2 \cdot f(x) \\
\cdot \\
\cdot \\
\cdot \\
f(nf(x))=n^2f(f(x))=n^2x^2f(x) \ for \ \forall n \in \mathbb{N}. \\
\ \\
P(-f(x), x): f(0)=f(-f(x))-2f(x) \cdot x^2 + x^2f(x) \Rightarrow f(-f(x)) = x^2f(x) = f(f(x)) \\
P(-2f(x), x): f(-f(x))=f(-2f(x))-4f(x)\cdot x^2 + x^2f(x) \Rightarrow f(-2f(x))=4x^2f(x)=4f(f(x)) \\
\cdot \\
\cdot \\
\cdot \\
f(mf(x))=m^2f(f(x))=m^2x^2f(x) \ for \ \forall m \in \mathbb{Z}. \\
$$
| $$f(x+f(y))=f(x)+2xy^2+y^2f(y)$$
*
*$(x,y)\equiv (0,0)$
$$f(f(0))=f(0)\implies f(c)=c\;,\;\;\; c=f(0).$$
*$(x,y)\equiv (0,c)$
$$f(f(c))=c+c^2f(c)\implies c^3=0\implies f(0)=c=0. $$
*
*$(x,y)\equiv (0,x)$
$$f(f(x)=x^2f(x)\tag{1} $$
*$(x,y)\equiv (f(x),1)$
$$f(f(x)+f(1))=f(f(x))+2f(x)+f(1)\tag{2}$$
*$(x,y)\equiv (f(1),x)$
$$f(f(1)+f(x))=f(f(1))+2f(1)x^2+x^2f(x)\tag{3}$$
*$(x,y)\equiv (0,1)$
$$ f(f(1))=f(1)\tag{4}$$
Adding $(1)$ and $(2)$, then subtracting $(3)$ and $(4)$ from the sum leads to,
$$f(x)=f(1)x^2$$
Substituting this result in the original F.E, we have,
$$f(1)^2=1\implies f(1)=\pm 1$$
$$\therefore\; \boxed{f(x)=\pm \;x^2} \;\;\forall x\in \mathbb{R} $$
| {
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Prove that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$ if $a^2+b^2+c^2+abc=4$. Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4$. How do you prove that $\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$?
My Approach: I tried basic algebraic techniques, but they didn't work. So, I moved on to a trigonometric approach. Note that $a^2+b^2+c^2+abc=4$ implies that $a<2$, and similarly for $b$ and $c$. So, I can let
$\qquad a=2\cos\alpha \qquad b=2\cos\beta \qquad c=2\cos\gamma \qquad 0<\alpha,\beta,\gamma<90^{\circ}$
where $a,b,c$ are side lengths of an acute triangle. These are the lengths of acute triangle because if any one of the angle was to be greater than or equal to $90^{\circ}$, one of the side would be zero or negative due to our assumption. (By the way, if we put supposed values of $a,b,c$ in L.H.S of the given equation, the result is equal to $4$.)
Now, we need to prove that
$\dfrac{\cos\alpha+\cos\beta}{\cos\gamma}+\dfrac{\cos\beta+\cos\gamma}{\cos\alpha}+\dfrac{\cos\gamma+\cos\alpha}{\cos\beta}\ge 2(\cos\alpha+\cos\beta+\cos\gamma)+\dfrac{1}{2}\left(\dfrac{1}{\cos\alpha}+\dfrac{1}{\cos\beta}+\dfrac{1}{\cos\gamma}\right)$
which can be further simplified into
$\left(\cos\alpha+\cos\beta+\cos\gamma-\dfrac{1}{2}\right)\left(\dfrac{1}{\cos\alpha}+\dfrac{1}{\cos\beta}+\dfrac{1}{\cos\gamma}-2\right)\ge4$.
We need to prove this inequality given that $\alpha+\beta+\gamma=180^{\circ}$ and $0<\alpha,\beta,\gamma<90^{\circ}$.
I tried to look for triangle inequalities which would help me out but was too overwhelmed and frustrated after going through Wikipedia.
After struggling a lot with this question, I'm not even sure if trigonometric approach is the best option. Please help me out!
| By AM-GM $$4=a^2+b^2+c^2+abc=a^2+b^2+c^2+\sqrt{a^2b^2c^2}\leq a^2+b^2+c^2+\sqrt{\left(\frac{a^2+b^2+c^2}{3}\right)^3}.$$
Now, let $a^2+b^2+c^2=3x^2,$ where $x>0$.
Thus, $$4\leq3x^2+x^3$$ or
$$x^3+3x^2-4\geq0$$ or
$$x^3-x^2+4x^2-4x+4x-4\geq0$$ or $$(x-1)(x^2+4x+4)\geq0$$ or $$x\geq1$$ or $$\frac{a^2+b^2+c^2}{3}\geq1,$$ which gives $$a^2+b^2+c^2\geq3.$$
In another hand, it's obvious (for exampleб by the Carnot's theorem) that $$a+b+c\leq3.$$
Id est, $$1\leq\sqrt{\frac{a^2+b^2+c^2}{3}},$$
$$a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which says that it's enough to prove that:
$$\sum_{cyc}\frac{a+b}{c}\geq3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which is homogeneous already.
Can you end it now?
I got that the last inequality is true even for any positives $a$, $b$ and $c$.
Indeed, we need to prove that:
$$\sum_{cyc}(a^2b+a^2c-abc)\geq(ab+ac+bc)\sqrt{\frac{a^2+b^2+c^2}{3}}.$$
Now, let $a+b+c=3u$ and $ab+ac+bc=3v^2$, where $v>0$.
Thus, by AM-GM $abc\leq v^3$ and since $$\sum_{cyc}(a^2b+a^2c)=(a+b+c)(ab+ac+bc)-3abc=9uv^2-3abc,$$ it's enough to prove that:
$$9uv^2-6v^3\geq3v^2\sqrt{3u^2-2v^2}$$ or
$$3u-2v\geq\sqrt{3u^2-2v^2}$$ or
$$\sqrt{3u^2-2v^2+6(u-v)^2}\geq\sqrt{3u^2-2v^2}$$ and we are done!
| {
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Doubt in testing the solution of a quadratic equation So i just stumbled upon something while solving this equation
$\sqrt{6x-2} + 5 - 3x = 0$
$\sqrt{6x-2} = 3x - 5$
Squaring on both sides
$6x-2 = 9x^2 + 25 - 30x$
$9x^2 - 36x + 27 = 0$
$x^2 - 4x + 3 = 0$
$x^2 - 3x - x + 3 = 0$
$x(x - 3) -1( x - 3) = 0$
$(x - 1)( x - 3) = 0$
$x = 1,3$
But when I plug in the values to verify the solution,
$\sqrt{6x-2} + 5 - 3x = 0$
$\sqrt{6(1)-2} + 5 - 3(1) = 0$
$\sqrt{4} + 5 - 3 = 0$
The value of $\sqrt{4}$ is 2 and -2
When I use -2 the equation is satisfied , but when I use +2 the its not
Similarly when I plug in 3
$\sqrt{6(3)-2} + 5 - 3(3) = 0$
$\sqrt{16} + 5 - 9 = 0$
Now if I consider +4 , the equality is satisfied but when I consider -4 it is not.
So what's the correct solution to the problem?
| The original equation we are to solve is $ \ \sqrt{6x-2} \ + \ 5 - 3x \ = \ 0 \ \ , $ which can be written (as you did) as the curve intersection equation
$ \ \sqrt{6x-2} \ = \ 3x - 5 \ \ . $ The left side represents a square-root curve, which looks like the "upper half" ( $ \ y \ \ge \ 0 \ $ ) of a "horizontal" parabola with its vertex at $ \ x \ = \ \frac13 \ $ (since the domain of this function is $ \ x \ \ge \ \frac13 \ \ . $ The right side represents a "steeply-rising" line with its $ \ y-$ intercept at $ \ (0 \ , \ -5) \ . $ So we would expect that these curves intersect only once.
We can look at what happens with the solutions to this equation in a couple of ways. When both sides of the equation are "squared" to produce $ \ 6x-2 \ = \ (3x - 5)^2 \ \ , $ the character of the geometric situation is changed: the square-root curve becomes a straight line and the former line appears as an "upward-opening" parabola. This new line also has a steep positive slope; since its $ \ y-$intercept, $ \ (0 \ , \ -2) \ $ is "to the left" and only slightly "below" the vertex of the parabola at $ \ \left( \frac53 \ , \ 0 \right) \ \ , $ these two curves will have two intersections which you found from $ \ 9x^2 - 36x + 27 \ = \ 9·(x - 1)·(x - 3) \ = \ 0 \ \ . $
However, the "half" of the parabola "to the left" of the parabola's symmetry axis at $ \ x \ = \ \frac53 \ $ is falsely "generated" by the act of squaring the function $ \ y \ = \ 3x - 5 \ \ . $ Therefore, the solution $ \ x \ = \ 1 \ < \ \frac53 \ \ $ is "spurious" and does not constitute a solution to the original equation (as you found upon inserting $ \ \sqrt{6·1 \ - \ 2} \ + \ 5 \ - \ 3·1 \ = \ \sqrt4 + 5 - 3 \ \neq \ 0 \ \ ) \ . $
On a separate question you raised, the single solution $ \ x \ = \ 3 \ $ inserted into this equation yields $$ \sqrt{6·3 \ - \ 2} \ + \ 5 \ - \ 3·3 \ \ = \ \ \sqrt{16} \ + \ 5 \ - \ 9 \ \ = \ \ (+4) \ + \ 5 \ - \ 9 \ \ \overbrace{=}^{!} \ \ 0 \ \ . $$
The square-root operation only produces a non-negative value, so the appearance of $ \ \sqrt{16} \ $ does not also call for $ \ (-4) \ $ to be used in the equation.
We can also describe this in terms of the intersection(s) of the square-root curve and the line $ \ y \ = \ 3x - 5 \ \ . $ As was discussed earlier, the curve $ \ y \ = \ \sqrt{6x - 2} \ \ $ will only have $ \ y \ \ge \ 0 \ \ $ [the red curve in the graph below], so there can only be the single intersection with the line at $ \ (3 \ , \ 3·3 - 5 = 4) \ \ . $ However, when the original equation is squared, it allows the creation of a second equation which is also consistent with that squared equation, namely, $ \ -\sqrt{6x-2} \ = \ 3x - 5 \ \ . $ This new [orange] curve has $ \ y \ \le \ 0 \ \ $ and has a single intersection with the line $ \ y \ = \ 3x - 5 \ $ at $ \ (1 \ , \ 3·1 - 5 = -2) \ \ . $ Again, this "negative square-root" curve is not implied in the original equation, so we simply disregard this result. The only solution for our equation is $ \ \mathbf{ x \ = \ 3 \ } \ \ . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$, $\ x^2+(4x^3-3x )^2=1$, given that $x=\cos\alpha$ I know that it would be:
$\cos^2\alpha+3\cos^2\alpha=1$ by identities
then it would be:
$4\cos^2\alpha=1$
dividing $4$ to both sides:
$\cos^2\alpha=\frac{1}{4}$
square both sides:
$\cos\alpha=\pm\frac{1}{2}$
After this, what should I do?
| Plugging $x = \cos \alpha$ we get
$$\cos^2\alpha + \cos^2 3\alpha = 1$$
(since $\cos 3\alpha = 4\cos^3\alpha - 3\cos \alpha$) which implies
$$\sin^2\alpha = \cos^23\alpha \iff (\sin\alpha - \cos3\alpha)(\sin\alpha + \cos3\alpha) = 0$$
Case 1. Let $$\sin\alpha - \cos3\alpha = 0$$
or $$\cos\left(\frac{\pi}{2} - \alpha\right) - \cos3\alpha = 0$$ or $$\sin\left(\frac{\pi}{4} + \alpha\right)\sin\left(\frac{\pi}{4} - 2\alpha\right) = 0.$$
This gives either
$$\sin\left(\frac{\pi}{4} + \alpha\right) = 0$$
or
$$\sin\left(\frac{\pi}{4} - 2\alpha\right) = 0$$
Solce them individually.
Case 2. Let $$\sin\alpha + \cos3\alpha = 0$$
This case is similar to the Case 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Quickly simplify $(\sqrt{3}+1)^6+(\sqrt{3}-1)^6$ Here's my question:
Simplify $(\sqrt{3}+1)^6+(\sqrt{3}-1)^6$.
I'm aware that I can just use binomial theorem to expand each of the terms individually and then just cancel/add/subtract the like terms however I'm wondering whether there is a quicker way to solve this question.
| One of the quick methods without using the binomial theorem can be constructed as follows:
Let $\sqrt 3
+1=m,\thinspace \sqrt 3-1=n$, then we have
$$\begin{cases}m^2+n^2=8\\mn=2\end{cases}$$
Then using the formula,
$$\begin{align}m^6+n^6=\left(m^2+n^2\right)^3-3\left(mn\right)^2\left(m^2+n^2\right)\tag 1\end{align}$$
we get
$$m^6+n^6=8^3-12\times 8=416.$$
Explanation: $(1)$
I used the following well-known formula:
$$\begin{align}m^3+n^3=(m+n)^3-3mn(m+n)\end{align}$$
Then, we can derive the required equality:
$$\begin{align}m^6+n^6=\left(m^2\right)^3+\left(n^2\right)^3=\left(m^2+n^2\right)^3-3\left(mn\right)^2\left(m^2+n^2\right).\end{align}$$
Small Supplement:
Based on the formula $(1)$, we can also use the following identity:
$$m^6+n^6=\left(m^2+n^2\right)\left(\left(m^2+n^2\right)^2-3(mn)^2\right)$$
where, $m^2+n^2=8$ and $mn=2$.
Thus, we have
$$\begin{align}\left(\sqrt 3+1\right)^6+\left(\sqrt 3-1\right)^6&=8(64-12)\\
&=416.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4287737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate the integral $I = \int_o^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}dx$? $$I = \int_{0}^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}\,dx$$.
I tried to solve by substituting $t = 2\pi\sqrt x \implies t^2 = 4{\pi}^2x \implies tdt = 2\pi^2dx$
$$I = \frac1{8\pi^4}\int_0^{\infty}\frac{t^3}{\sqrt{e^t-1}}dt.$$
But now I'm unable to solve from here.
| $$I = \int_0^{\infty}\frac{x}{\sqrt{e^{2\pi\sqrt x}-1}}dx$$
Let
$$x = \frac{t^2}{4\pi^2} \implies dx = \frac{tdt}{2\pi^2}$$
Thus,
$$I =\frac1{8\pi^4} \int_0^{\infty}\frac{t^3}{\sqrt{e^t-1}}dt$$
Let's consider a more general integral.
$$f(s) = \int_0^{\infty} \frac{e^{-st}}{\sqrt{e^t-1}}dt$$
$$\frac {df(s)}{ds^3}=f_3(s) = -\int_0^{\infty}\frac{t^3e^{-st}}{\sqrt{e^t-1}}dt$$
$$\color{blue}{I = -\frac 1{8\pi^4}\left(\frac {df(s)}{ds^3}\right)_{s=0}}\text{...................#1}$$
Now,
$$u =e^{-t} \implies \ln u = -t \implies -\frac{du}{u} = dt$$
+---------+--------+
t -->> | 0 | ∞ |
+---------+--------+
u -->> | 1 | 0 |
+---------+--------+
$$\begin{align*}
f(s)
& = \int_0^1 \frac{u^{s-1}}{\sqrt{\frac 1{u}-1}}du\\
& = \int_0^1 u^{s-\frac 1{2}}(1-u)^{-\frac1{2}}du\\
& = \beta \left(s+\frac 1{2}, \frac 1{2} \right)\\
& = \frac{\Gamma(s+\frac 1{2})\Gamma(\frac 1{2})}{\Gamma(s+1)}\\
& \implies \color{red}{f_1(s)} = \color{blue}{f(s)}\color{green}{\left(\psi_0\left(s+\frac 1{2}\right) - \psi_0(s+1)\right)} = \color{blue}{f}\color{green}{\times g}\\
\end{align*}$$
*
*What is $\psi$-function
*Derivative of $\beta$-function
$$\begin{align*}f_1 =
& f\times g\\
& \implies f_2 = f_1g + fg_1 = (fg)g+fg_1 = fg^2+fg_1\\
& \implies f_3 = f_1g^2+2fgg_1+f_1g_1+fg_2\\
& \implies f_3(s) = fg^3 + 3fgg_1 + fg_2 = \color{blue}{f(s)\left[g^3(s)+3g(s)g_1(s) + g_2(s)\right] = f_3(s)}\text{.................#2}\\
\end{align*}$$
let's find zeros of function
$$\color{green}{f(s=0)} = \beta\left(\frac 1{2}, \frac 1{2}\right) = \frac {\sqrt\pi\sqrt\pi}{1} = \color{green}{\pi}$$
*
*$g(0)$
$$\begin{align*}
\color{green}{g(s)}
& = \left(\psi_0(s+ \frac 1{2}) - \psi_0(s+1)\right) \\
&\implies g(s=0) = \left(\color{red}{\psi_0(\frac 1{2})} - \psi(1)\right) \\
&= (\color{red}{(-\gamma-2\ln2)}+\gamma) = \color{green}{-2\ln(2)}\\
\end{align*}$$
*
*$g_1(0)$
$$\begin{align*} \color{green}{g_1(s)}
&= \left(\psi_1(s+ \frac 1{2}) - \psi_1(s+1)\right)\\
&\implies g_1(0) = \left(\color{red}{\psi_1(\frac 1{2})} - \psi_1(1)\right)\\
& = \left(\color{red}{\frac{\pi^2}{2}}-\frac{\pi^2}{6}\right) = \color{green}{\frac{\pi^2}{3}}\\
\end{align*}$$
*$g_2(0)$
$$\begin{align*}
\color{green}{g_2(s)} =
&\left(\psi_2(s+ \frac 1{2}) - \psi_2(s+1)\right)\\
&\implies g_2(0) = \left(\color{red}{\psi_2( \frac 1{2})} - \psi_2(1)\right)\\
& = \color{red}{-14\zeta(3)} + 2\zeta(3) = \color{green}{-12\zeta(3)}
\end{align*}$$
From above
$$\begin{align*}I =
& -\frac 1{8\pi^4} f_3(0) = -\frac 1{8\pi^4}\left(\pi\left((-2\ln 2)^3 + 3(-2\ln 2)\times \frac{\pi^2}{3}
- 12\zeta(3)\right)\right)\\
& = \frac {\ln^3(2)}{\pi^3} + \frac{\ln(2)}{4\pi} + \frac {3\zeta(3)}{2\pi^3} \\
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find minimum polynomial of an element of $GF(2^5)$ I am trying to find the minimum polynomial of $\alpha^5$ and using this primitive polynomial to define my operations. $p(X)=1+X^2+X^5$.
The conjugates of an element $\beta$ of $GF(2^m)$ are $\{\beta^{(2^i)},\;i\geq 0 \}$.
The minimal polynomial is
$$\phi(X)=\prod_{i=0}^{e-1} (X+\beta^{2^i}).$$
$$\phi(X) = (X+\beta)(X+\beta^2)(X+\beta^4)(X+\beta^8)(X+\beta^{16})$$
I have set $\beta = \alpha^5$ and this gives me,
$$\phi(X)=X^5+X^4+(\alpha+\alpha^2+\alpha^4)X^3+(\alpha+\alpha^2+\alpha^3+\alpha^4)X^2+X+1$$
But the answer should be, $X^5+X^4+X^2+X+1$
Edit:
I have figured it out, It was a calculation mistake and I missed some terms. Now I have got the correct answer, i.e.
$$\phi(X)=X^5+X^4+X^2+X+1$$
| Now that the OP found the mistakes in their calculation I will post my trick answer.
The element $\alpha^2$ is a conjugate of $\alpha$, so it shares with it the minimal polynomial $p(X)=X^5+X^2+1$. We can use this by observing that
$$\beta=\alpha^5=\alpha^5+p(\alpha)=1+\alpha^2.$$
Therefore $\beta$ must be a zero of
$$
q(X)=p(1+X)=(1+X)^5+(1+X)^2+1=X^5+X^4+X^2+X+1.
$$
After all,
$$
q(\beta)=q(1+\alpha^2)=p(1+(1+\alpha^2))=p(\alpha^2)=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $G=\langle\mu\rangle$ be the subgroup of $S_4$, now compute the coset of $G$ using...?
Let $G= \langle \mu \rangle$ be the subgroup $S_4$ generated by $$\mu =\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 4 & 1
\end{pmatrix}.$$
Compute all the cosets of G containing the permutation $\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}.$
$$G= \left\{\begin{pmatrix} 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4\end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\right\}$$
Coset of G is simply $ga$ $\forall g \in G$
Therefore we would have
*
*$\begin{pmatrix} 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4\end{pmatrix} \begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix} = \begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}$
*$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix}\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}=(1 3)(2 3)(4 1) (4 1)(2 3)=\begin{pmatrix} 1 & 2 & 3 & 4 \\
3 & 2 & 1 & 4\end{pmatrix}$
*$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}=(1 3) (4 2) (4 1)(2 3)=\begin{pmatrix} 1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4\end{pmatrix}$
*$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} \begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}= (2 3) (4 3)(4 1) (4 1)(2 3)=\begin{pmatrix} 1 & 2 & 3 & 4 \\
1 & 4 & 3 & 2\end{pmatrix}$
So I am not $100\%$ sure if I am thinking about this correctly. Please let me know if there are other ways or more clear ways of thinking about this.
I think if $\mu$ as a $90$ degree rotation counter clockwise to a square that has
one of the left top corner, two in the top right corner, 3 on the bottom right corner and 4 on the bottom left corner. Though I don't think this is the right way of thinking about this, especially when applying $a$ to $ga$ (because $a$ would be a $90$ degree rotation clockwise)?
| Hint: Cosets of any group by any subgroup partition the group . . .
. . . and so each element of the desired cosets of $G$ is in exactly one coset; namely, $G\begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4298828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of multivariare function at the origin Consider the function $f:\mathbb{R}^2\to \mathbb{R}$ defined as:
$$
f(x,y)=
\begin{cases}
\frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}}.\quad&\text{ if } (x,y)\neq 0\\
0\quad& \text{ if }(x,y)= 0\,.
\end{cases}
$$
Is this function continuous at the origin?
If the limit exists it has to be $0$ since, for example, if we take the restriction $x=y$ we obtain:
$$
\lim_{x\to0}\frac{|x|^{\frac{5}{2}}x}{\sqrt{2}(x^2+x^4)|x|}=0
$$
Restrictions to any kind of powers seem to give the same result suggesting that the function has to be continuous at the origin (as the graph also seems to confirm) but I'm not able to find an estimate for $f$ to use the squeeze theorem and actually prove continuity.
Any help will be greatly appreciated.
| Alternatively:
$$\sqrt{\frac{x^2+y^2}2}\ge \sqrt{|xy|}\implies \frac{\sqrt{2}}{\sqrt{x^2+y^2}}\le\frac{1}{\sqrt{|xy|}}\tag 1$$
so $$\frac{|x|^{5/2}|y|}{(x^2+y^4)\sqrt{x^2+y^2}}=\frac{x^2\sqrt{|xy|}\sqrt{|y|}}{\sqrt 2(x^2+y^4)}\frac{\sqrt 2}{\sqrt{x^2+y^2}}\le\sqrt{|y|}\frac{x^2\sqrt{|xy|}}{\sqrt 2 x^2\sqrt{|xy|}}\le\sqrt{|y|}$$
or from $(1)$, you could've immediately written $\frac{\sqrt{|xy|}}{\sqrt{x^2+y^2}}\le\frac1{\sqrt 2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $\int_0^{2\pi} \cos(a\cos(x) + b\sin(x) + cx + d)dx$ I'm trying to estimate the following integral:
\begin{align*}
\int_0^{2\pi} \cos(a\cos(x) + b\sin(x) + cx + d)dx,
\end{align*}
where a, b, c and d are non-zero constants. Is there any way to get a closed-form solution using Bessel functions for example?
During my search to solve it, I stumble accross this equation:
\begin{align*}
\int_0^{2\pi} \cos(a\cos(x) + b\sin(x) + d)dx = 2\pi J_0(\sqrt{a^2 + b^2})\cos(d),
\end{align*}
where $J_0$ is the Bessel function of the first kind of order 0.
Edit
I was able to find a solution whenever c=1 (useful page to understand all the steps)
\begin{align*}
&\int_0^{2\pi} \cos(a\cos(x) + b\sin(x) + x + d)dx \\
&= \int_0^{2\pi}\cos(a\cos(x) + b\sin(x) + d)\cos(x)dx - \int_0^{2\pi}\sin(a\cos(x) + b\sin(x) + d)\sin(x)dx \\
&= Re \int_0^{2\pi}e^{i(a\cos(x) + b\sin(x) + d)}\cos(x)dx - Im\int_0^{2\pi}e^{i(a\cos(x) + b\sin(x) + d)}\sin(x)dx \\
&= -Re \left[ie^{id} \frac{\partial}{\partial a} \int_0^{2\pi}e^{i(a\cos(x) + b\sin(x))}dx\right] + Im\left[ie^{id} \frac{\partial}{\partial b}\int_0^{2\pi}e^{i(a\cos(x) + b\sin(x))}dx\right]\\
&= -Re \left[ie^{id} \frac{\partial}{\partial a} 2\pi J_0\left(\sqrt{a^2 + b^2}\right)\right] + Im\left[ie^{id} \frac{\partial}{\partial b}2\pi J_0\left(\sqrt{a^2 + b^2}\right)\right]\\
&=-\frac{2\pi}{\sqrt{a^2 + b^2}}J_1\left(\sqrt{a^2 + b^2}\right)\left[a\sin(d) + b\cos(d)\right]
\end{align*}
My problem is really this factor $c$ in front of the $x$ term. How can I solve this difficulty?
| So after some computations, I was able to obtain a satisfactory answer in the case where c is an integer. For simplicity, let us define:
\begin{align*}
\forall n\in\mathbb N, \qquad I_n = \int_{0}^{2\pi} \cos(x\cos(\theta) + y\sin(\theta) + n\theta + d)d\theta.
\end{align*}
I already established that:
\begin{align*}
I_0 &= 2\pi J_0(\sqrt{x^2 + y^2})\cos(d),\\
I_1 &= -\frac{2\pi}{\sqrt{x^2 + y^2}}J_1\left(\sqrt{x^2 + y^2}\right)\left[x\sin(d) + y\cos(d)\right].
\end{align*}
Let us rewrite $I_n$ as:
\begin{align*}
I_n &= \int_{0}^{2\pi} \cos(x\cos(\theta) + y\sin(\theta) + d)\cos(n\theta)d\theta - \int_{0}^{2\pi} \sin(x\cos(\theta) + y\sin(\theta) + d)\sin(n\theta)d\theta\\
&= Re\left[\int_{0}^{2\pi} e^{i(x\cos(\theta) + y\sin(\theta) + d)}\cos(n\theta)d\theta\right] - Im\left[\int_{0}^{2\pi} e^{i(x\cos(\theta) + y\sin(\theta) + d)}\sin(n\theta)d\theta\right]\\
&= Re\left[C_n\right] - Im\left[S_n\right],
\end{align*}
where I have defined the two sequences $C_n$ and $S_n$ that I am going to compute. We already know that:
\begin{align*}
C_0 &= 2\pi e^{id} J_0(\sqrt{x^2 + y^2})\\
C_1 &= 2\pi \frac{e^{id}}{i} \frac{\partial}{\partial x}[J_0(\sqrt{x^2 + y^2})].
\end{align*}
Moreover, in the same manner, it can be shown that:
\begin{align*}
S_0 &= 0\\
S_1 &= 2\pi \frac{e^{id}}{i} \frac{\partial}{\partial y}[J_0(\sqrt{x^2 + y^2})].
\end{align*}
Using the fact that $\cos(2x) = 2\cos^2(x) - 1$ and $\sin(2x) = 2\sin(x)\cos(x)$, it can be demonstrated that:
\begin{align*}
C_2 &= \frac{2}{i}\frac{\partial C_1}{\partial x} - C_0\\
S_2 &= \frac{2}{i}\frac{\partial S_1}{\partial x} - S_0.
\end{align*}
Using Chebyshev polynomials, we can demonstrate that:
\begin{align*}
\cos[(n+1)x] = 2\cos(x)\cos(nx) - \cos[(n-1)x]\\
\sin[(n+1)x] = 2\cos(x)\sin(nx) - \sin[(n-1)x].
\end{align*}
Using induction reasoning alongside with these two relations, it is straightforwards to prove that:
\begin{align*}
C_{n+1} &= \frac{2}{i}\frac{\partial C_n}{\partial x} - C_{n-1}\\
S_{n+1} &= \frac{2}{i}\frac{\partial S_n}{\partial x} - S_{n-1}.
\end{align*}
To finish and in order to find an inductive relation for $I_n$, let us define:
\begin{align*}
\forall n\in\mathbb N, \qquad J_n = \int_{0}^{2\pi} \sin(x\cos(\theta) + y\sin(\theta) + n\theta + d)d\theta = Im\left[C_n\right] + Re\left[S_n\right],
\end{align*}
we can write:
\begin{align*}
I_{n+1} &= Re\left[C_{n+1}\right] - Im\left[S_{n+1}\right]\\
&= Re\left[\frac{2}{i}\frac{\partial C_n}{\partial x} - C_{n-1}\right] - Im\left[\frac{2}{i}\frac{\partial S_n}{\partial x} - S_{n-1}\right]\\
&= 2\frac{\partial}{\partial x}\left[Re\left[\frac{C_n}{i}\right] - Im\left[\frac{S_n}{i}\right]\right] - I_{n-1}\\
&= 2\frac{\partial}{\partial x}\left[Im\left[C_n\right] + Re\left[S_n\right]\right] - I_{n-1}\\
&= 2\frac{\partial J_n}{\partial x} - I_{n-1}.
\end{align*}
Similarly, we have:
\begin{align*}
J_{n+1} = -2\frac{\partial I_n}{\partial x} - J_{n-1}.
\end{align*}
By defining $Z_n = I_n + iJ_n$ we can use the last two recurrence relations to obtain:
\begin{align*}
Z_{n+1} = -2i \frac{\partial Z_n}{\partial x} - Z_{n-1},
\end{align*}
where $Z_n$ can also be defined using integral equation as:
\begin{align*}
\forall n\in\mathbb N, \qquad Z_n = \int_{0}^{2\pi} e^{i(x\cos(\theta) + y\sin(\theta) + n\theta + d)}d\theta.
\end{align*}
Using brute force computation (see this post for the full answer), the solution is:
\begin{align*}
\quad Z_{\pm n} = 2\pi e^{id} \left(\frac{x\pm iy}{x^2+y^2}\right)^{n}\left[r^2 J_0(r) \sum_{k=1}^{u_n}a_k^{n} \left(x^2 + y^2\right)^{k-1} + i r J_1(r) \sum_{k=1}^{u_{n+1}}b_k^{n} \left(x^2 + y^2\right)^{k-1}\right]
\end{align*}
where $n\geq 1$ and $u_n$ is defined as $u_n = \left(2n+(-1)^n -1\right)/4$. The two coefficients can be computed using:
\begin{equation*}
\forall k\geq 1, n\geq 2k,\quad a_k^{n} = (2i)^{n-2k}\frac{(n-k)!(n-(k+1))!}{(k-1)!k!(n-2k)!}
\end{equation*}
\begin{equation*}
\forall k\geq 1, n\geq 2k-1,\quad b_k^{n} = (2i)^{n-(2k-1)}\frac{(n-k)!(n-k)!}{(k-1)!(k-1)!(n-(2k-1))!)}
\end{equation*}
Remarks
(1) For negative integers, it is just a matter of sign:
\begin{align*}
\forall n\in\mathbb N, \qquad I_{-n} &= Re\left[C_n\right] + Im\left[S_n\right] \\
\forall n\in\mathbb N, \qquad J_{-n} &= Im\left[C_n\right] - Re\left[S_n\right].
\end{align*}
(2) For non integers number, considering the developments and all the computations that I have done, I suspect that factorial derivatives are involved but I have no idea how to do it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What's the maximum integer value of the segment (IG) joining the incenter to the centroid in the triangle below? For reference:
Given the obtuse triangle $ABC$, obtuse at $B$,
where $IG \parallel BC$, $I$ being the incenter and $G$ the centroid of this triangle and the perimeter of the triangle is $144$.
Calculate the maximum integer value of $IG$.
My progress:
My draw and the relationships I found
$IG \parallel BC \implies a = \dfrac{b+c}{2}\\
\dfrac{AI}{IJ} = \dfrac{b+c}{2p}=\dfrac{b+c}{144}\\
c^2+a^2=2BM^2+\dfrac{b^2}{2}$
From angle bissector theorem ($\triangle ABC-AN::$),
$\dfrac{BJ}{CJ}=\dfrac{c}{b}\\
AJ^2 =bc-CJ\cdot BJ \\
\triangle ADK \sim \triangle ACB \implies:\\
\dfrac{AM}{CM}=\dfrac{AK}{BK}=\dfrac{b}{c}$
but I don't see where to fit $IG$..
| As you find out $2a = b+c$ so $a= 48$ and $b+c= 96$.
By angle bisector theorem we have $$BJ = {ac\over b+c}= {c\over 2}$$
Then $$IG = {2\over 3} JN = {2\over 3} ({a\over 2}-{c\over 2})$$
So $$IG = {b-48\over 3}$$ Since $b< a+c \implies b<72 $ so $$IG < {24\over 3} =8\implies IG\leq 7$$
Value $7$ is achieved at $b = 71$ and $ c = 25$ so $IG_{\max} =7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ Let $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ find the value of $\left[ {I + \sqrt 3 } \right] $ {where [.] represent greatest integral function}
Let $T(x)={{x^3} - 3{x^2} + 5}$, $T'\left( x \right) = {x^3} - 3{x^2} + 5 = 3{x^2} - 6x = 3x\left( {x - 2} \right) > 0,x \in \left( {2,3} \right)$
$T(x)$ is increasing for $x\in(2,3)$
Not able to proceed further
| Built around $x=\frac 52$ the series expansion of the integrand is
$$2 \sqrt{\frac{2}{15}}-2 \sqrt{\frac{2}{15}} \left(x-\frac{5}{2}\right)+\frac{1}{5}
\sqrt{\frac{6}{5}} \left(x-\frac{5}{2}\right)^2+\frac{1}{3} \sqrt{\frac{10}{3}}
\left(x-\frac{5}{2}\right)^3-\frac{111}{50} \sqrt{\frac{3}{10}}
\left(x-\frac{5}{2}\right)^4+O\left(\left(x-\frac{5}{2}\right)^5\right)$$
Integrate (some terms will disappear because of the symmetry) and you have a very good approximation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that 1 is a triple root of a polynomial I'm studying for an exam and trying to prove whether 1 is a triple root for the polynomial:
$$x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$$
for every $n\geq1$.
In our math class we never solved such a problem. So far we only used horner's scheme to prove that someone is a root, double root or triple root.
Can you please help me solve this problem?
| Let $f_n(x)=x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$. We will prove the statement by induction on $n$.
For $n=1$ we get $f_1(x)=x^3-3x^2+3x-1=(x-1)^3$ and the statement holds.
Now assume $f_n(x)$ has a factor $(x-1)^3$, then take
\begin{align}
f_{n+1}(x)-xf_n(x)
&=x^{2n+3}-x^{2n+2}-2x^{n+2}+2x^{n+1}-1+x\\
&=x^{2n+2}(x-1)-2x^{n+1}(x-1)+(x-1)\\
&=(x-1)(x^{2n+2}-2x^{n+1}+1)\\
&=(x-1)(x^{n+1}-1)^2\\
&=(x-1)^3\left(\frac{x^{n+1}-1}{x-1}\right)^2\\
&=(x-1)^3\left(1+x+x^2+\dots+x^n\right)^2.
\end{align}
The RHS has a factor $(x-1)^3$, so does $f_n(x)$ (by the induction hypothesis), hence $f_{n+1}(x)$ has a factor $(x-1)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ .
My approach is as follow
$y = \sqrt {{x^2} + 4x + 4 + 9} + \sqrt {{x^2} - 8x + 16 + 25} \Rightarrow y = \sqrt {{{\left( {x + 2} \right)}^2} + 9} + \sqrt {{{\left( {x - 4} \right)}^2} + 25} $
Within the square roots the minimum values are 3 and 5 hence the minimum value should be greeter than 8 but not able to find the actual minimum value.
| I was going to do a geometrical solution, but looks like that's already been done.
As for a calculus solution, we have
$$\frac{dy}{dx}=\frac{x+2}{\sqrt{x^2+4x+13}}+\frac{x-4}{\sqrt{x^2-8x+41}}$$
As, you have already shown $\sqrt{x^2+4x+13}>0$ and $\sqrt{x^2-8x+41}>0$ from completing the square. Hence, the only critical points occur when the RHS is equal to zero.
$$\frac{x+2}{\sqrt{x^2+4x+13}}+\frac{x-4}{\sqrt{x^2-8x+41}}=0$$
$$\frac{x+2}{\sqrt{x^2+4x+13}}=-\frac{x-4}{\sqrt{x^2-8x+41}}$$
$$\frac{x^2+4x+4}{x^2+4x+13}=\frac{x^2-8x+16}{x^2-8x+41}$$
$$\frac{x^2+4x+4}{x^2+4x+13}-1=\frac{x^2-8x+16}{x^2-8x+41}-1$$
$$\frac{-9}{x^2+4x+13}=\frac{-25}{x^2-8x+41}$$
$$9x^2-72x+369=25x^2+100x+325$$
$$16x^2+172x-44=0$$
$$4x^2+43x-11=0$$
$$(4x-1)(x+11)=0$$
$$x=-11,\frac{1}{4}$$
However, since we squared both sides of the equation, we expect one of the solutions to be extranneous. Note that if $x=-11$, then $x+2$ and $x-4$ are both negative. Hence, $\frac{x+2}{\sqrt{x^2+4x+13}}$ and $\frac{x-4}{\sqrt{x^2-8x+41}}$ will have the same sign. Then of course we could not have $\frac{x+2}{\sqrt{x^2+4x+13}}=-\frac{x-4}{\sqrt{x^2-8x+41}}$. So our solution is $x=\frac{1}{4}$. It is not hard to deduce that this is a minima.
So the minimum value of $y$ is
$$\sqrt{\frac{1}{16}+14}+\sqrt{\frac{1}{16}+39}$$
$$=\sqrt{\frac{225}{16}}+\sqrt{\frac{625}{16}}$$
$$=\frac{15}{4}+\frac{25}{4}$$
$$=\boxed{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find maximum of $\sin{x}+\sin{y}-\sin{(x+y)}+\sqrt{3}\left(\cos{x}+\cos{y}+\cos{(x+y)}\right)$ The trick is that we can't use derivatives.
If we look at $\cos{x}+\cos{y}+\cos{(x+y)}$, the function cos attains its maximum of $1$ at $0$ radians, so $x=y=0$ and $x+y=0$ and the maximum of $\cos{x}+\cos{y}+\cos{(x+y)}=1+1+1=3$.
What about part of the function with $\sin$? If we use the same logic: $\sin$ attains its maximum of $1$ at $\pi/2$ radians, so $x=y=\pi/2$, so $x+y=\pi$, we get the maximum of $\sin{x}+\sin{y}-\sin{(x+y)}=1+1-0=2$, but I guess it's not true, cause Wolfram says the maximum is $0$.
Any hint would help a lot!! thank you!
| $$
\sin{x}+\sin{y}-\sin{(x+y)}+\sqrt{3}[\cos{x}+\cos{y}+\cos{(x+y)}]
\\=\sin x+\sin y -\sin(x+y)+\tan \frac\pi3[\cos x+\cos y+\cos(x+y)]\\
=\frac1{\cos\dfrac\pi3}[\sin(x+\frac\pi3)+\sin(y+\frac\pi3)+\sin(\frac\pi3-x-y)]\\
=\frac1{\cos\dfrac\pi3}[\sin(x+\frac\pi3)+\sin(y+\frac\pi3)+\sin[ \pi-(\frac{\pi}3+x)-(\frac\pi3+y)]\\
$$
The expression in the square bracket is the sum of sines of 3 angles in a triangle. Its maximum is obtained when all are equal. This can be seen if we consider the sum of 2 angles in a triangle $\alpha,\beta,\gamma$. For 2 angles $\alpha,\beta$ and a fixed $\gamma$ the sum is: $\sin \alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}2 \cos\dfrac{\alpha-\beta}2$. It is maximized when $\alpha=\beta$. Then the sum of each with a $\gamma$ will be maximized when they are equal as well and we conclude that all the 3 must be equal. It follows that:
$$
x+\dfrac\pi3=y+\dfrac\pi3= \pi-(\frac{\pi}3+x)+(\frac\pi3+y)
$$
and therefore $x=y=0$ and the maximum value of the expression is: $3\tan\dfrac\pi3=3\sqrt3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Help with finding the equation of line tangent to the semi circle with the equation: $y=\sqrt{1-x^2}$
Ok so I need to show that the equation of the tangent above to the semi-circle with the equation: $y=\sqrt{1-x^2}$, is $y=-\frac{1}{\sqrt{3}}x$ + $\frac{2}{\sqrt{3}}$
What we from the question:
The tangent intersects the $x$-axis at $(2,0)$, (assume that you do not know any other points of intersection)
What I tried to do so far:
Since its a tangent, I decided to differentiate the semi-circle function using the chain rule, getting:
$\frac{dy}{dx}$ = $\frac{-x}{\sqrt{1-x^2}}$
Then using the equation of a line formula:
$y=mx+c$
--> $y=\frac{-x}{\sqrt{1-x^2}}x +c$
Substituting the point in:
-->$0=\frac{-2}{\sqrt{1-2^2}}(2) +c$
Now clearly something is wrong because I will get the square root of a negative, which does not make any sense. Need help from here.
*Note: I know that there are other ways to solve this, but I would prefer if calculus was used to solve this question.
|
$OA = (\cos t, \sin t)$ and $AB = (2 - \cos t, -\sin t)$. For $OA \perp AB$, we need the dot product $\cos t(2 - \cos t) + \sin t \cdot - \sin t = 0 \implies 2 \cos t - 1 = 0, t = \frac{\pi}{3}$.
Thus the line segment $AB = \left(\frac{3}{2}, -\frac{\sqrt3}{2} \right)$, which has slope $-\frac{\sqrt3}{3}$ and passes through $(\cos \pi/3, \sin \pi/3) = (1/2, \sqrt3/2)$.
Here is a non-calculus method for reference:
We know that $OA = 1, OB = 2$ and $\angle OAB$ is right since it is tangent to the circle.
Thus $\cos \angle AOB = \frac{1}{2} \implies \angle AOB = \pi/3$. Thus the gradient of $OA$ is $\text{rise}/\text{run} = \tan \pi/3 = \sqrt{3}$, and so $AB$ has a gradient of $-\frac{1}{\sqrt{3}}$.
$AB$ must pass through $A$ which is $(\cos \pi/3, \sin \pi/3) = (1/2, \sqrt3/2)$, therefore the equation is:
$$y - \sqrt3/2 = -\frac{1}{\sqrt3} \left(x - \frac{1}{2} \right) \implies y = -\frac{1}{\sqrt3}x + \frac{1 + (\sqrt{3})^2}{2\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the maximum value of $(x+2) \cdot (12-3x)$ , x is a Real number Question:- Find the maximum value of $(x+2) \cdot (12-3x)$. $x$ is a Real number.
The solution given in the textbook applies AM-GM to solve this
$$3 \cdot (x+2) \cdot (4-x)$$
Now, $x+2 = a$ and $4-x = b$
$a+b=6$
Objective to maximize $3 \cdot a \cdot b$
by AM-GM inequality,
$$\frac{a+b}{2} \geq \sqrt{ab}$$
$$a \cdot b \leq 9$$
Therefore $3 \cdot a \cdot b \leq 27$
But is the solution correct? as we know that while applying AM-GM, the terms should be positive real but $a$ and $b$ are not positive for every $x$ as Real number , if I am correct the range of $(-\infty,-2)$ to $(4, \infty)$ should be given for $a$ and $b$ to be both positive and then apply AM-GM, please suggest if some other way is possible too to solve this.
| Technically, as you mentioned, you cannot directly use AM-GM in this case. But we can just use the inequality $$(a+b)^2 \geq 4ab$$ instead. This inequality holds for any real numbers $a$ and $b$, since
\begin{align*}
(a+b)^2 &= a^2 + 2ab + b^2 \\
&= (a^2 - 2ab + b^2) + 4ab \\
&= (a-b)^2 + 4ab \\
&\geq 0 + 4ab \\
&= 4ab.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How do we solve $x^2 + \{x\}^2 = 33$ without computer? This is a problem taken from a group on Facebook. I wonder how to solve this without numerical process.
$x^2 + \{x\}^2 = 33\tag{1}$
My unfinished attempt:
$$\begin{align}
x^2 + \{x\}^2 &= 33\\
x^2 + \left(x - \lfloor x \rfloor\right)^2 &= 33\\
x^2 + x^2 - 2x \lfloor x \rfloor + \lfloor x \rfloor^2 &= 33\\
2x\left(x - \lfloor x \rfloor\right) + \lfloor x \rfloor^2 &= 33\\
2x\{x\} + \lfloor x \rfloor^2 = 33
\end{align}$$
I'm stuck at there. Also we know that the fractional part of $x$ is bounded i.e.:
$$0\leq\{x\}<1$$
From that, I can predict where the two solutions are placed at:
$$\begin{align}
0&\leq \{x\} < 1\\
0 &\leq \{x\}^2 < 1\\
x^2 &\leq x^2 + \{x\}^2 < x^2 + 1\\
x^2 &\leq 33 < x^2 + 1 \tag{$x^2 + \{x\}^2 = 33$}\\
S &= -\sqrt{33}\leq x \leq \sqrt{33} \quad \bigcap \quad x< -\sqrt{32} \lor x > \sqrt{32}\\
\therefore S &= -\sqrt{33}\leq x < -4\sqrt2 \quad \bigcup \quad 4\sqrt{2} < x \leq \sqrt{33}\\
S &\approx -5.74456 \leq x < -5.65685 \quad \bigcup \quad 5.65685 < x \leq 5.74456
\end{align}$$
I don't know if that's even going to help. Anyway, here are the two solutions (to the original problem) that's given by Wolfram Alpha:
$$\begin{align}
x_1 &= \frac12 \left(3\sqrt{17} - 1\right) \approx -5.815\\
x_2 &= \frac12 \left(-1 - 3\sqrt{113}\right) \approx 5.685
\end{align}$$
You see $x_1$ isn't at the interval. I'm confused.
|
My Solution for the problem-Magdy Essafty
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\textstyle{\frac{1\cdot 2}{3!} +\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\frac{4\cdot2^4}{6!}+\cdots}$ up to n terms? $$
\frac{(1)2}{3!} + \frac{(2)2^2}{4!} + \frac{(3)2^3}{5!} + \frac{(4)2^4}{6!} + \cdots =\sum\limits_{k=1}^{n}\frac{k\cdot 2^k}{(k+2)!}
$$
My attempt:
$$
\begin{align}
e^x&=\sum_{n=1}^{\infty}\frac{x^n}{n!} \\
(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!} \\
x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n}}{n!} \\
x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{n(n+1)(n+2)x^{n}}{(n+2)!}
\end{align}
$$
After this attempt I realized exponential series is for infinite terms whereas the question concerns finite terms so approach may not work.
Can you please give any hints on the right approach to be tried?
| Making the problem more general, write
$$\sum_{m=1}^n \frac{m\, x^m}{(m+2)!}=\frac 1 x\sum_{m=1}^n \frac{ x^{m+1}}{(m+1)!}-\frac 2 {x^2}\sum_{m=1}^n \frac{ x^{m+2}}{(m+2)!}$$
$$\sum_{m=1}^n \frac{ x^{m+1}}{(m+1)!}=e^x\frac{ \Gamma (n+2,x)}{(n+1)!}-x-1$$
$$\sum_{m=1}^n \frac{ x^{m+2}}{(m+2)!}=e^x\frac{ \Gamma (n+3,x)}{\Gamma (n+3)}-\frac{x^2}{2}-x-1$$ Recombining everything
$$\sum_{m=1}^n \frac{m\, x^m}{(m+2)!}=\frac{x+2}{x^2}-\frac{x^{n+1}}{(n+2)!}+e^x\frac {x-2}{x^2}\frac{ \Gamma (n+3,x)}{ (n+2)!}$$
You are lucky with $x=2$
$$\sum_{m=1}^n \frac{m\, 2^m}{(m+2)!}=1-\frac{2^{n+1}}{ (n+2)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the value of ${{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right)}$ I am trying to solve:
${\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right)$
My solution is as follow:
$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right) $
Since:
$\csc{^{ - 1}}\left( {\sqrt 2 } \right) = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4};{\cos ^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$
Then:
$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + \frac{\pi }{4} + \frac{\pi }{6}} \right)$
I am not able to proceed further.
| \begin{align}
\theta &= \cos^{-1}\sqrt{\frac{2+\sqrt{3}}{4}}\\
\cos \theta &= \sqrt{\frac{2+\sqrt{3}}{4}}\\
\cos^2 \theta &= \frac{2+\sqrt{3}}{4}\\
\frac 12 + \frac 12 \cos 2\theta &= \frac{2+\sqrt{3}}{4}\\[0.5em]
2 + 2 \cos 2\theta &= 2+ \sqrt 3\\
\cos 2\theta &= \frac{\sqrt 3}{2}\\
2\theta &= \frac{\pi}{6}\\[1em]
\theta &= \frac{\pi}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculating the Integral for the Brachistochrone Problem I am trying to prove that the shortest possible time of descent in the Brachistochrone problem is:
$$ T = \sqrt\frac{c_1}{2g} \theta_1 $$
I have the following 2 parametric equations:
$$ x = \frac{c_1}{2} (\theta - \sin\theta) \qquad y = \frac{c_1}{2} (1 - \cos\theta) $$
Solving $$ \frac{dx}{d\theta} = \frac{c_1}{2} (1-\cos\theta) \qquad \frac{dy}{d\theta} = \frac{c_1}{2} (\sin\theta) $$
I know that $$ T = \int_0^{x_1} \frac{\sqrt{1 + (y')^2}}{\sqrt{2gy}} dx \tag{1} $$
Now solve $$y' = \frac{dy}{dx} = \frac{dy / d\theta}{ dx / d\theta } = \frac{\sin\theta}{1-\cos\theta} $$
Therefore $$ (y')^2 = \frac{\sin^2\theta}{(1-\cos\theta)^2} = \frac{1+\cos\theta}{1-\cos\theta} $$
Now plugging everything back into \ref{1} and putting the integral in terms of $\theta$:
$$T = \int_0^{\theta_1} \frac{\sqrt{1 + \frac{1+\cos\theta}{1-\cos\theta}}}{\sqrt{2g\frac{c_1}{2} (1 - \cos\theta)}} \frac{c_1}{2} (1-\cos\theta) d\theta$$
I tried solving this integral but I am having a hard time with it. I would appreciate some guidance on how to tackle this integral.
| We can work with dimensionless variable, $\large X = {2x \over c_1}\;Y = {2y \over c_1}$
$X = θ - \sin(θ)$
$Y = 1 - \cos(θ)$
${dX \over dθ} = 1 - \cos(θ) = Y$
${dY \over dθ} = \sin(θ) = ±\sqrt{1-\cos^2(θ)} = ±\sqrt{1-(1-Y)^2} = ±\sqrt{2Y-Y^2}$
$\displaystyle \left({dX \over dθ}\right)^2 + \left({dY \over dθ}\right)^2 = Y^2 + (2Y-Y^2) = 2Y$
$\displaystyle T = \int_0^{X_1} \sqrt{1+\left({dY \over dX}\right)^2 \over c_1\;g\;Y} \left( {c_1\over2} dX\right)
= \sqrt{c_1 \over 4g} \int_0^{θ_1} \sqrt{2Y\over Y} dθ
= \sqrt{c_1 \over 2g}\; θ_1
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find minimum $n$ for $x^2+3x+6=f_1(x)^2+f_2(x)^2+ \cdots +f_n(x)^2$ The problem
Find minimum $n$ for $x^2+3x+6=f_1(x)^2+f_2(x)^2+ \cdots +f_n(x)^2$Find the minimum value of $n$ such that there exists $f_i \in \mathbb{Q[i]}, i=1,2,...,n$ such that, \begin{align}x^2+3x+6=f_1(x)^2+f_2(x)^2+ \cdots +f_n(x)^2\end{align}
My attempts
What I have been thinking is that $f_i(x)$ must all be of degree $1$. The reason is, assume $f_i(x)= a_kx^k +\cdots$, then $f_i(x)^2$ has the leading coefficient of $a_k^2$, and therefore if $\text{deg}f_i= \max \{\text{deg}f_t \mid t=1,2,\cdots,n$} is not gonna cancel anyway.
So anyway $f_s(x) = (a_sx+b_s)$, but then what? How should I develop from here? Is there any theorem on how rational numbers can/cannot sum up to integers?
Any help is appreciated!
| Let $\mathbf{v} = (v_1,\ldots,v_n)$ be a vector with $\|\mathbf{v}\| = 1$ and entries in $\mathbf{Q}$.
It is a fact, not completely obvious, that $\mathbf{v}$ extends (as either a row or a column) to an orthogonal matrix with rational entries. (For example, see Is every unit vector in $\mathbb{Q}^n$ the first column of a rational orthogonal matrix?). In particular, there exists an orthogonal matrix $M \in \mathrm{O}(n)$ with $M \mathbf{e}_1 = \mathbf{v}$, or replacing $M$ by its inverse, with $M \mathbf{v} = \mathbf{e}_1$.
The polynomial $x^2 + 3 x + 6$ can clearly be written as the sum of $n$ squares if and only if the same is true for $4(x^2 + 3 x + 6) = (2x + 3)^2 + 15$, and this can clearly be written as the sum of $n$ squares if and only if the same is true for $x^2 + 15$. Now write
$$x^2 + 15 = (v_1 x + u_1)^2 + (v_2 x + u_2)^2 + (v_3 x + u_3)^2 + (v_4 x + u_4)^2.$$
Equating coefficients, we deduce, with $\mathbf{v} = (v_1,v_2,v_3,v_4)$ and $\mathbf{u} = (u_1,u_2,u_3,u_4)$, that
$$1 = v^2_1 + v^2_2 + v^2_3 + v^2_4 = \|\mathbf{v}\|,$$
$$0 = 2 v_1 u_1 + 2 v_2 u_2 + 2 v_3 u_3 + 2 v_4 u_4 = 2 \mathbf{v}.\mathbf{u},$$
$$15 = u^2_1 + u^2_2 + u^2_3 + u^2_4 = \|\mathbf{u}\|.$$
Now take a rational orthogonal matrix $M$ above, so $M \mathbf{v} = \mathbf{e}_1$, and $M \mathbf{u} = \mathbf{x}$ for some rational vector $\mathbf{x}$. Since orthogonal matrices preserve angles and lengths, we get
$$\mathbf{e}_1.\mathbf{x} = 0, \ \|\mathbf{x}\| = 15.$$
The first equality implies that $\mathbf{x} = (0,a,b,c)$ for rational $a$, $b$, and $c$, and the second implies that
$$a^2+b^2+c^2 = 15.$$
This, however, has no solutions in rational numbers because $15 \equiv 7 \bmod 8$ and no such numbers are sums of three rational squares by a standard $2$-adic argument. (When is a rational number a sum of three squares?)
Thus the minimal possible number of squares is $5$, but clearly
$$x^2 + 15 = (x)^2 + 3^2 + 2^2 + 1^2 + 1^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coin flip streak variance Let's say I flip a coin 12 times. Let $X$ be the number of streaks of 3 heads in a row. How do I calculate $var(X)$?
Example:
Let 'H' represent head And 'T' tail;
if we flip 4 times and get 'HHHH' its 2 streaks of 3 Heads.
if we flip 7 times and get 'HHHTHHH' it's also 2 streaks.
That's what I got so far:
$var(X) = E[X^2] - (E[X])^2$.
let $x_i = 1$ if its a start of 3 heads $else$ $x_i = 0$
$E[X] = E[x_1 + x_2 + ... + x_9 + x_{10}]$ = $10E[x_i] = 10P(x_i) = \frac{10}{2^3}$.
I'm stuck on how to calculate $E[X^2]$ ?
| $$X=\sum_{i=1}^{10}X_i$$
\begin{align}
E[X^2] = E\left[\sum_{i=1}^{10}X_i\right]+2E\left[ \sum_{i<j}X_iX_j\right]
\end{align}
Note that we have $E[X_iX_j]=\begin{cases} \frac1{2^4}, &j=i+1\\ \frac1{2^5}&, j=i+2\\ \frac1{2^6} &, j> i+2\end{cases}$
Let me compute the crossed terms:
\begin{align}
E\left[\sum_{i=1}^{9}\sum_{j=i+1}^{10}X_iX_j\right] &= \sum_{i=1}^7\sum_{j=i+1}^{10}E[X_iX_j]+E[X_8X_9]+E[X_8X_{10}] + E[X_9X_{10}]\\
&=\sum_{i=1}^7\left(\frac1{2^4}+\frac1{2^5}+\sum_{j=i+3}^{10}\frac1{2^6} \right)+ \frac1{2^4} + \frac1{2^5} + \frac1{2^4}\\
&= 7\left( \frac1{2^4}+\frac1{2^5}\right)+\sum_{i=1}^7\frac{8-i}{2^6}+\frac{2}{2^4}+\frac1{2^5}\\
&=\frac9{2^4}+\frac{8}{2^5}+\frac{7}{2^3}-\frac{7(8)}{2}\cdot \frac1{2^6}\\
&=\frac9{2^4}+\frac{8}{2^5}+\frac{7}{2^3}-\frac{7}{2^4}\\
&=\frac2{2^4}+\frac{8}{2^5}+\frac{7}{2^3}\\
&= 1.25
\end{align}
I believe you can compute the variance now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using AM-GM Inequalities: For a,b,c all positive and $a+b+c=1$, prove that $ab+bc+ca\le 1/3$ Using AM-GM Inequalities.
For a,b,c all positive and $a+b+c=1$, prove that $ab+bc+ca\le 1/3$.
My attempt:
I took AM-GM of $a+b$ and $c$, and then $a$ and $b+c$ and so on.
Using that i got $\le 3/8.$
| Since $\frac{a^2}{2}+\frac{b^2}{2}\geq 2 \sqrt{\frac{a^2}{2}\times\frac{b^2}{2}}=ab$
$1=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=$
$=(\frac{a^2}{2}+\frac{b^2}{2})+(\frac{b^2}{2}+\frac{c^2}{2})+(\frac{c^2}{2}+\frac{a^2}
{2})+2ab+2bc+2ca\geq $
$\geq ab+bc+ca+2ab+2bc+2ca$
Therefore
$ab+bc+ca \leq \frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Nested sine and cosine half angle formulas I am having trouble with this problem. I know that it is related to the sine and cosine half-angle formulas. I substituted $\frac{\sqrt{3}}{2}$ with $\cos(\frac{\pi}{6})$ and got $\frac{\pi}{24}$ for the first 2 loops. I don't know what to do for the last 2 loops. Can anyone give an explanation or hint on what I should do?
Calculate $$\arccos \sqrt{\frac{\displaystyle 1 + \sqrt{\frac{1 - \sqrt{\frac{\displaystyle 1 - \sqrt{\frac{\displaystyle 1 + \frac{\displaystyle \sqrt{3}}{2}}{2}}}{2}}}{2}}}{2}}.$$
As usual, the output of an inverse trig function should be in radians.
| By evaluating each radical we have, $$\sqrt{\dfrac{1+\frac{\sqrt3}2}2}=\sqrt{\dfrac{1+\cos\frac{\pi}6}{2}}=\cos\frac{\pi}{12}$$
$$\sqrt{\dfrac{1-\cos\frac{\pi}{12}}{2}}=\sin\frac{\pi}{24}$$
$$\sqrt{\dfrac{1-\sin\frac{\pi}{24}}{2}}=\sqrt{\dfrac{\sin^2\frac{\pi}{48}+\cos^2\frac{\pi}{48}-2\sin\frac{\pi}{48}\cos\frac{\pi}{48}}{2}}=\dfrac{|\sin\frac{\pi}{48}-\cos\frac{\pi}{48}|}{\sqrt2}$$$$=\frac1{\sqrt2}.\cos\frac{\pi}{48}-\frac1{\sqrt2}\sin\frac{\pi}{48}=\cos(\frac{\pi}4+\frac{\pi}{48})=\cos\frac{13\pi}{48}$$
And finally,
$$\sqrt{\dfrac{1+\cos{\frac{13\pi}{48}}}{2}}=\cos\frac{13\pi}{96}$$
Hence, $\arccos(\cos\frac{13\pi}{96})=\frac{13\pi}{96}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4331128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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.