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Integrate $\int\frac{x}{x^3-8}dx$ Integrate
$$\int\frac{x}{x^3-8}dx$$
I solved this integral by dividing it in partial fractions, than i came in two integrals $I_1,I_2$. Then, partioning $I_2$ into $I_3,I_4$ and partioning $I_4$ into $I_5,I_6$. But it took me so much work even though i got correct answer.
Is there any simple way to solve it?
Thank you in advance :)
| Recognize
$$\frac{4x}{x^3-8}
= \frac{(x^2+2x+4) -x^2 +2(x-2)}{x^3-8}
= \frac1{x-2} - \frac{x^2}{x^3-8}+\frac2{x^2+2x+4}
$$
and integrate to obtain
$$4\int \frac x{x^3-8}dx=\ln|x-2|-\frac13\ln|x^3-8|+ \frac2{\sqrt3}\tan^{-1}\frac{x+1}{\sqrt3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{\sin x}{x^2}$ is uniformly continuous at $N(0;r)^c$ for any $r >0$ I tried $\left \vert \frac{\sin x}{x^2} - \frac{\sin c}{c^2}\right \vert \leq \frac{1}{x^2} + \frac{1}{c^2} < \epsilon$, but it doesn't help me much with $\vert x - c \vert < \delta$. How can I prove this?
| Note that for $x,y \neq 0$,
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| = \left|\frac{\sin x}{x^2} - \frac{\sin x}{xy} + \frac{\sin x}{xy}- \frac{\sin y}{y^2}\right| \\\leqslant \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{1}{|y|}\left| \frac{\sin x}{x} - \frac{\sin y}{y}\right|\\ = \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{1}{|y|}\left| \frac{\sin x}{x} - \frac{\sin x }{y} + \frac{\sin x}{y}-\frac{\sin y}{y}\right|\\ \leqslant \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{|\sin x|}{|y|}\left| \frac{1}{x} - \frac{1 }{y}\right| + \frac{1}{|y|^2}\left|\sin x-\sin y\right|$$
Since $\left| \frac{\sin x}{x}\right|= \frac{|\sin x|}{|x|} \leqslant 1$ and $|\sin x | \leqslant 1$, it follows that
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| \leqslant \left(1 + \frac{1}{|y|} \right)\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{1}{|y|^2}\left|\sin x-\sin y\right| \\ =\left(1 + \frac{1}{|y|} \right)\frac{|x-y|}{|x||y|} + \frac{1}{|y|^2}\left|\sin x-\sin y\right|,$$
and for all $x,y \in [r,\infty)\cup (-\infty,-r],$
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| \leqslant \frac{1+r}{r^3}|x-y| + \frac{1}{r^2}|\sin x - \sin y|$$
Using the prosthaphaeresis formula , we get $|\sin x - \sin y | \leqslant |x - y|$ and it follows that
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| \leqslant \frac{1+r}{r^3}|x-y| + \frac{1}{r^2}| x - y| = \frac{1+2r}{r^3}|x-y|,$$
directly proving uniform continuity on $[r,\infty)\cup (-\infty,-r]$ where $r > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106933",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Convergence of $\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$. My try. Examine the convergence of:
$$\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$$
Pointwise convergence (for every $ x \in \mathbb{R}$ ):
$\displaystyle \lim_{n \to \infty} x\frac{\sin(n^2x)}{n^2} = 0$ , because $\sin(n^2x) \in [-1, 1]$
Uniform convergence (for every $ x \in \mathbb{R}$ and $n \in \mathbb{N}$ ):
form dirichlet criterion:
$$\displaystyle \sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2} = \displaystyle \sum_{n=1}^{\infty} f_n(x) \cdot g_n(x) $$ where:
*
*$f_n(x) = \frac{\sin(n^2x)}{n} \leq 2$ which works for: $ \forall x \in \mathbb{R}, n \in \mathbb{N}$
*$g_n(x) = \frac{x}{n} \implies \displaystyle \lim_{n \to \infty} g_n(x) = 0$ (monotonously)
Therefore the series is uniformly convergent for $x \in \mathbb{R}$.
Is that correct?
| In the comments you can see the issues with your solution. Here I present a possible approach to this problem. The pointwise convergence follows from the estimate
$$
\left| {x\sum\limits_{n = 1}^\infty {\frac{{\sin (n^2 x)}}{{n^2 }}} } \right| \le \left| x \right|\sum\limits_{n = 1}^\infty {\frac{{\left| {\sin (n^2 x)} \right|}}{{n^2 }}} \le \left| x \right|\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}} = \frac{{\pi ^2 }}{6}\left| x \right|.
$$
Note that
\begin{align*}
\mathop {\sup }\limits_{x \in \mathbb{R}} \left| {x\sum\limits_{n = 2N}^\infty {\frac{{\sin (n^2 x)}}{{n^2 }}} } \right| & \ge \left| {\frac{\pi }{2}(2N + 1)\sum\limits_{n = 2N}^\infty {\frac{{\sin \left( {n^2 \frac{\pi }{2}(2N + 1)} \right)}}{{n^2 }}} } \right|
\\ &
= \frac{\pi }{2}(2N + 1)\sum\limits_{k = 0}^\infty {\frac{1}{{(2N + 2k + 1)^2 }}} \\ & \ge \frac{\pi }{2}(2N + 1)\sum\limits_{k = 0}^\infty {\frac{1}{{(2N + 2k + 1)(2N + 2k + 3)}}}
\\ &
= \frac{\pi }{4}(2N + 1)\sum\limits_{k = 0}^\infty {\left( {\frac{1}{{2N + 2k + 1}} - \frac{1}{{2N + 2k + 3}}} \right)} = \frac{\pi }{4}
\end{align*}
for any positive integer $N$. Thus the series cannot be uniformly convergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove $\sum_{n=1}^{\infty}(\sqrt{n^2+3}-\sqrt{n^2-1})=\infty$ I want to prove $$\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty$$
If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\neq0$, I can prove it.
In fact, however, $$\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=0$$
As a solution to finding a sum of series, I only know one that uses a partial sum or a progression of differences. How can I apply it to this proof?
I would appreciate it if you could tell me the solution.
| Also:
\begin{align}
\sqrt{n^2+1} &= n\left(1+\frac{1}{n^2}\right)^{1/2} = n\left(1+\frac{1}{2n^2} + O(n^{-4})\right)
\\ &= n + \frac{1}{2n} + O(n^{-3})
\\
\sqrt{n^2-3} &= n - \frac{3}{2n} + O(n^{-3})\qquad\text{similarly}
\\
\sqrt{n^2+1} - \sqrt{n^2-3} &= \frac{2}{n} + O(n^{-3})
\end{align}
Thus the series diverges by comparison with $\sum \frac{1}{n}$.
Sorry, I did $+1$ and $-3$ instead of $+3$ and $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ I am trying to find the area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ using Double integrl in Polar coordinates.
Now as shown in the figure the angle $\theta$ takes values from $0$ to $\frac{\pi}{2}$.
Now $r$ depends on $\theta$.
From $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$,substituting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we get
$$r=\sin(\theta)$$ and like-wise substituting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ we get
$$r=\cos(\theta)$$
So the double integral which gives the area is:
$$A=\int_{\theta=0}^{\pi / 2} \int_{r=\sin \theta}^{\cos \theta} r d r d \theta \text { . }$$
But i am getting answer as ZERO.
| The polar equation of the circles are $r=\sin\theta=f(\theta)$ and $r=\cos\theta=g(\theta)$. They intersect at the pole and when $\theta=\frac{\pi}{4}$. Looking at the figure, the area is
$$A=\int_0^{\pi/4} \frac{\sin^2\theta}{2}\; d\theta + \int_{\pi/4}^{\pi/2}\frac{\cos^2 \theta}{2} = \frac{1}{2}\left( \int_0^{\pi/4}\frac{1-\cos (2\theta)}{2}\; d\theta + \int_{\pi/4}^{\pi/2}\frac{1+\cos(2\theta)}{2} \right)
=\frac{1}{4}\left( \left[\theta-\frac{\sin(2\theta)}{2} \right]_0^{\pi/4} + \left[ \theta+\frac{\sin(2\theta)}{2}\right]_{\pi/4}^{\pi/2} \right)
= \frac{\pi}{8} - \frac{1}{4}
$$
Note: using the symmetry about $\theta=\frac{\pi}{2}$, the area simplify to $A=\int_0^{\pi/4} \sin^2\theta\; d\theta$.
Other method: the line $y=x$ separates the region of same area. This area is the difference of the area of a quarter of circle of radius $\frac{1}{2}$ and a right triangle with two sides $\frac{1}{2}$. Then
$$A=2\left( \frac{1}{4}\frac{\pi}{4} - \frac{1}{2}\frac{1}{4}\right) = \frac{\pi}{8}-\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Converting equation into short Weierstrass form I have the curve $y^2 + 4y = x^3 + 3x^2 −x + 1$. I need to find a transformation of the form $X=x+a$ and $Y=y+b$
that turns this curve into the standard form of the elliptic curve: $$Y^2=X^3+AX+B.$$
I completed the square for the left side and now have $(y+2)^2= x^3 + 3x^2 −x + 5$, but I'm not sure what to do next to fix the right side. Any help would be great!
| You need to complete the cube in terms of $x$, this is very similar to completing the square, except that we have
$$(x+a)^3 = x^3 + 3ax^2 + 3a^2x + a^3$$
so if the coefficient of $x^2$ is $c$ then $(x + c/3)$ is the variable you should rewrite
$x^3 + cx^2 +dx + e$ in terms of so that you get
$$x^3 + cx^2 +dx + e =(x+c/3)^3 + A(x+c/3) + B$$
for some coefficients $A,B$ that you can solve for using this equation,
in your case $c/3=1$ so this should work out nicely!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$N^2$ leaves a remainder of $1$ when divided by $24$. What are the possible remainders we can get if we divide $N$ by $12$? My solution approach is as below :-
$Rem[\frac{N^2}{24}] = 1$
$\Rightarrow Rem[\frac{N \times N}{24}] = 1$
$\Rightarrow Rem[\frac{N}{24}] \times Rem[\frac{N}{24}] = 1$
$\Rightarrow \text{This can only happen when both are 1 i.e. } Rem[\frac{N}{24}]=1$
So we can say that $N=24k+1$ and when we divide this $N$ by $12$ we should always get a remainder of $1$ but the answer provided for this problem is different and looks like I am wrong. Please help me on this !
Thanks in advance !
| Let's suppose the remainder of $N$ when divided by $12$ is some integer $r \in \{0, 1, 2, \ldots, 11\}$. That is to say, $$N = 12q + r,$$ where $q$ is the quotient. Then $$N^2 = (12q + r)^2 = 144q^2 + 24qr + r^2,$$ and we are told that this has a remainder of $1$ when divided by $24$. Since $144 = 6(24)$, the first two terms on the right hand side are always divisible by $24$ and leave a remainder of $0$. So $r^2$, when divided by $24$, must leave a remainder of $1$.
Among the choices $r \in \{0, 1, 2, \ldots, 11\}$, clearly if $r$ is even, $r^2$ is also even, and cannot leave an odd remainder. This reduces our search to $r \in \{1, 3, 5, 7, 9, 11\}$, for which the squares are $$r^2 \in \{1, 9, 25, 49, 81, 121\}.$$ Then we see
$$\begin{align}
1 &= 24(0) + 1, \\
9 &= 24(0) + 9, \\
25 &= 24(1) + 1, \\
49 &= 24(2) + 1, \\
81 &= 24(3) + 9, \\
121 &= 24(5) + 1,
\end{align}$$
so the permissible remainders for $N$ upon division by $12$ are $$r \in \{1, 5, 7, 11\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of the locus of the mid point of AB as m varies I am working through a pure maths book as a hobby. This question puzzles me.
The line y=mx intersects the curve $y=x^2-1$ at the points A and B. Find the equation of the locus of the mid point of AB as m varies.
I have said at intersection:
$mx = x^2-1 \implies x^2 - mx - 1=0$
Completing the square:
$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$
$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$
$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$
x-coordinates for points of intersection are
$x= \frac{-\sqrt(4+m^2) +m}{2}$ and $x= \frac{\sqrt(4+m^2)+m}{2}$
$\implies$ x-co-ordinate of P is mid-way between the two above points, namely $\frac{2m}{2} = m$
So $x=m, y = mx = m^2\implies y = x^2$
But my book says $y=2x^2$
| The $x$-coordinate of the midpoint is the the sum of the other two divided by $2$, i.e. $$ x=\frac m2$$ And $$y=mx =\frac{m^2}{2} $$ and $$y=2x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate a missing middle digit of a multiplication between 2 big numbers using modular arithmetic? What elementary number theory methods can I use for solving this type of questions?
One of these kind of problems is to find $x$ where $985242x6565 = 172195\cdot 572167$ without multiplying the numbers again.
I tried doing, let's call $985242x6565$ $N$, $N\text{mod}10^5$ to somehow get to the $5th$ digit but this is doesn't get me anywhere and in hindsight it doesn't make any sense. How should I proceed?
| The remainder of the division by 11 is equal to the alternating sum of the digits of the number. Because
$1 \equiv 1$ (mod 11)
$10 \equiv -1$ (mod 11)
$100 \equiv 1$ (mod 11)
$1000 \equiv -1$ (mod 11)
$10000 \equiv 1$ (mod 11)
and so on...
Thus,
$172195 \equiv (-1+7-2+1-9+5) \equiv 1$ (mod 11)
$572167 \equiv (-5+7-2+1-6+7) \equiv 2$ (mod 11)
$172195⋅572167 \equiv 1\cdot2 \equiv 2$ (mod 11)
$985242x6565 \equiv (9-8+5-2+4-2+x-6+5-6+5) \equiv 4+x$ (mod 11)
and $x\equiv9$ (mod 11).
Answer: $x=9$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show $\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}^n = 2^{n-1}\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}$? Initial note: I'm interested in the combinatorics aspect of the following problem, not how to proof the relation in general.
The idea is to show the following relationship:
$$
\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}^n = 2^{n-1}{\underbrace{\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}}_{\equiv M}}
$$
by decomposing the matrix $M = A + B + C + D$ into the four individual matrices
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \qquad
B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \qquad
C = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \qquad
D = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
$$
and noting the following relationships for matrix multiplication:
$\ast$
$A$
$B$
$C$
$D$
$\bf{A}$
$A$
$B$
$0$
$0$
$\bf{B}$
$0$
$0$
$A$
$B$
$\bf{C}$
$C$
$D$
$0$
$0$
$\bf{D}$
$0$
$0$
$C$
$D$
If these matrix products were commutative, one could use the multinomial theorem in order to compute the coefficients for the resulting polynomial $(A+B+C+D)^n$. Since they are not, one needs to consider all possible combinations for the matrices $A,B,C,D$ (like for this question):
$$
\sum_{combinations} X_1X_2\dots X_n \qquad, \quad \mathrm{where} \; X_i \in \{A,B,C,D\}
$$
However due to the above relations, each pair of matrices $X_iX_{i+1}$ can be reduced to either zero or another matrix of the set $\{A,B,C,D\}$ (which means the contribution of the whole expression $X_1X_2\dots X_n$ will reduce to either $0$ or $1$). So I'm wondering whether by applying this knowledge it is possible to combine the various permutations for a given term $A^{k_1}B^{k_2}C^{k_3}D^{k_4}$ in order to determine how many nonzero permutations there are. Together with the multinomial coefficients this could then be used to calculate the result.
At this point I'm struggling to find the number of nonzero permutations for a given set of powers $\{k_1,k_2,k_3,k_4 | k_i \geq 0\}$ in the expression $A^{k_1}B^{k_2}C^{k_3}D^{k_4}$. Is there are way to determine this?
| Some more aspects regarding OPs approach. When looking at the matrix $M$ in the form
\begin{align*}
M=A+B+C+D
\end{align*}
we can associate certain walks in the graph $G(M)$ given below. The graph consists of two nodes $1$ and $2$ and directed edges corresponding to the entries of $M$. We recall Theorem 1.1. from Algebraic Combinatorics by R. P. Stanley:
*
*Theorem: For any integer $n\geq 1$, the $(i,j)$-entry of the matrix $M^n$ is equal to the number of walks from node $i$ to node $j$ of length $n$.
This means that $A,B,C,D$ can be represented as subgraphs of $G(M)$ whereby
*
*$A= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ has a path from node $1$ to $1$
*$B= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has a path from node $1$ to $2$
*$C= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ has a path from node $2$ to $1$
*$D= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ has a path from node $2$ to $2$
We want to show that
\begin{align*}
\color{blue}{M^n=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}^n=2^{n-1}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}}\tag{1}
\end{align*}
by counting paths via $A,B,C$ and $D$ of length $n$.
Paths of length $n$ from node $1$ to $1$:
*
*In order to get a path of length $n$ from node $1$ to $1$ we can take zero or more times $A$. We can also take zero or more times $BC$, noting that whenver we go via $B$ from node $1$ to $2$ we also have to go back via $C$ from $2$ to $1$. Whenever we are going from $1$ to $1$ via $BC$ we can insert zero or more times $D$ walking the loop from $2$ to $2$.
*This means we have building blocks
\begin{align*}
\color{blue}{\{A,BC,BDC,BD^2C,\ldots\}}\tag{2}
\end{align*}
which can be used to build all paths of length $n$. For instance all $2^3=8$ paths of length $4$ from node $1$ to $1$ are given by
\begin{align*}
&A^4,\\
&A^2(BC),A(BC)A,(BC)A^2,\\
&A(BCD),(BCD)A,\\
&(BC^2D),\\
&(BC)(BC)
\end{align*}
To better see the building blocks, those containing $B$ are written in parentheses. Note that no two building blocks have the same length. This is an essential property which we'll use in order to prove (1).
Paths of length $n$ from node $2$ to $2$:
The building blocks of paths from node $2$ to $2$ are due to symmetry according to (2)
\begin{align*}
\color{blue}{\{D,CB,CAB,CA^2B,\ldots\}}\tag{3}
\end{align*}
Paths of length $n$ from node $1$ to $2$:
We can use the building blocks from (2) and (3). We can either start with $B$, walk from $1$ to $2$ and append all paths of length $n-1$ from node $2$ to $2$. We can otherwise take all paths of length $n-1$ from $1$ to $1$ and append $B$ from $1$ to $2$ as last step. We obtain
\begin{align*}
\color{blue}{\{BD,BCB,BCAB,BCA^2B,\ldots\}\cup \{AB,BCB,BDCB,BD^2CB,\ldots\}}\tag{3}
\end{align*}
Paths of length $n$ from node $2$ to $1$:
Due to symmetry we have according to (3) the building blocks
\begin{align*}
\color{blue}{\{DC,CBC,CABC,CA^2BC,\ldots\}\cup \{CA,CBC,CBDC,CBD^2C,\ldots\}}\tag{4}
\end{align*}
We are now ready to prove (1). We start with counting the number of paths of length $n$ from node $1$ to $1$.We observe that the length of the building blocks (2) is unique. We consider the lengths
\begin{align*}
&\{\overline{A},\overline{BC},\overline{BDC},\overline{BD^2C},\overline{BD^3C},\ldots\}=\{1,2,3,4,5,\ldots\}\\
\end{align*}
Example: $n=4$. We consider a small example to better see what's going on. The $2^3=8$ paths of length $4$ from node $1$ to $1$ are:
\begin{align*}
\begin{array}{ll}
1+1+1+1&\qquad\qquad A^4\\
1+1+2&\qquad\qquad A^2(BC)\\
1+2+1&\qquad\qquad A(BC)A\\
2+1+1&\qquad \qquad(BC)A^2\\
2+2&\qquad\qquad (BC)^2\\
1+3&\qquad\qquad A(BDC)\\
3+1&\qquad \qquad(BDC)A\\
4&\qquad\qquad (BDDC)
\end{array}
\end{align*}
*
*We observe the number $\alpha_n$ of paths of length $n$ from node $1$ to node $1$ is equal to the number of compositions to $n$, and this number is known to be $$\color{blue}{\alpha_n=2^{n-1}}$$
*In the same way we we find the number $\delta_n$ of paths of length $n$ from node $2$ to $2$ is $$\color{blue}{\delta_n=2^{n-1}}$$
*The number $\beta_n$ of paths of length $n$ from node $1$ to $2$ is calculated according to the derivation above as
\begin{align*}
\color{blue}{\beta_n}=\alpha_{n-1}+\delta_{n-1}=2\cdot 2^{n-2}\color{blue}{=2^{n-1}}
\end{align*}
and analogously we have $$\color{blue}{\gamma_n}=\alpha_{n-1}+\delta_{n-1}\color{blue}{=2^{n-1}}$$
We conclude
\begin{align*}
\color{blue}{M^n}=\begin{pmatrix}
\alpha_n&\beta_n\\
\gamma_n&\delta_n\\
\end{pmatrix}
=\begin{pmatrix}
2^{n-1}&2^{n-1}\\
2^{n-1}&2^{n-1}\\
\end{pmatrix}
\color{blue}{=2^{n-1}\begin{pmatrix}
1&1\\
1&1\\
\end{pmatrix}}
\end{align*}
showing the claim is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
An Inequality concerning double integral
Prove $$ \frac{\pi}{8}\left(1-\cos\frac{2}{\pi}\right)\le \iint_D \sin (x^2)\cos (y^2) dxdy \le \frac{\pi}{8}(1-\cos 1),$$where the integrating region $D$ is enclosed by $x=0, y=0$ and $x+y=1$.
We can obtain
$$\iint_D \sin (x^2)\cos (y^2) dxdy=\int_0^1\int_0^{1-y}\sin(x^2)\cos(y^2)dxdy\le \int_0^1\int_0^{1-y}\sin(x^2)dxdy,$$
but which is still difficult.
| We want to bound
$$
I = \iint_D \sin (x^2)\cos (y^2) dxdy
$$
where the integrating region $D$ is enclosed by $x=0, y=0$ and $x+y=1$. Since the integrand is symmetric in $x$ and $y$, we can write this as
$$
I = \frac12 \iint_E \sin (x^2)\cos (y^2) dxdy
$$
where the integrating region $E$ is enclosed by the rectangle $x+y=0$, $x+y=1$, $x-y=-1$, $x-y=1$. Changing variables to $x+y=a$, $x-y=b$ gives
$$
I = \frac14 \int_{a=0}^1 \int_{b=-1}^1 \sin ((\frac{a+b}{2})^2)\cos ((\frac{a-b}{2})^2) dadb
$$
Using the trig theorem $\sin(A) \cos(B) = 1/2 (\sin(A - B) + \sin(A + B))$ gives
$$
I = \frac18 \int_{a=0}^1 \int_{b=-1}^1 (\sin(\frac{a^2+b^2}{2}) - \sin(a b)) dadb
$$
The integration of $\sin(a b)$ gives zero, and the trig theorem $\sin(A+B) = \cos(B) \sin(A) + \cos(A) \sin(B)$ gives us
$$
I = \frac12 \int_{a=0}^1 \sin(\frac{a^2}{2}) da \int_{b=0}^1\cos(\frac{b^2}{2}) db
$$
Note that the integrals now factorize, and that so far we have made no approximations at all.
We can now compute bounds. Since in the intervals of integration, $ \sin(\frac{a^2}{2})$ is convex and $\cos(\frac{b^2}{2})$ is concave, we have
$$
I \le \frac12 \cdot \frac{\sin(1/2)}{2} \cdot 1 = \frac{\sin(1/2)}{4} \simeq 0.12 < \frac{\pi}{8}(1-\cos 1) \simeq 0.18
$$
For the lower bound, we use the tangent slope of $f(a) = \sin(\frac{a^2}{2})$ , which is $f'(a) = a \cos(\frac{a^2}{2})$, at $a=1$, to get
$$
I \ge \frac12 \cdot \frac{(\sin(1/2))^2}{2 \cos(1/2)} \cdot \frac{1 + \cos(1/2)}{2} \simeq 0.0615
$$
Now this is reasonable bound, however it doesn't prove the conjecture since
$
0.0615 \le \frac{\pi}{8}\left(1-\cos\frac{2}{\pi}\right) \simeq 0.077
$ . It is unlikely anyway that one should be able to prove the conjectured lower bound by reasonable approximations, since numerically, $I \simeq 0.079$, hence the lower bound is very tight, it is about $0.964$ times the true value. For comparison, the conjectured upper bound is not tight at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
taking an integral without trigonometric substitution I needed to evaluate $$\int_{-\infty}^\infty\frac{1}{x^2+a^2}dx$$
I looked around and found it's solved by trigonometric substitution $x = a\tan\theta$ and the answer is $\frac{\pi}{a}$. I understand the derivation but have been curious: supposing I didn't know the "trick", are there more staightforward methods (even if more laborious) to find this integral?
For example, I figured out the power series $$\frac{1}{x^2+a^2} = \frac{1}{a^2} - \frac{1}{a^4}x^2 + \frac{1}{a^6}x^4 - ...$$
I can integrate the power series term-by-term. But what I end up with obviously diverges as $x\to \infty$ or $x \to -\infty$. Is there a way to continue with this approach and end up with the value of the integral? Or some other way, perhaps?
| Another approach is as follows:
$$\begin{align}
\int_{-\infty}^\infty\frac{1}{x^2+a^2}dx
&= 2\int_0^\infty \frac{1}{x^2+a^2}dx \\
&= \frac{2}{a^2} \int_0^\infty \frac{1}{1+(x/a)^2}dx \\
&= \frac{2}{a} \int_0^\infty \frac{1}{1+t^2}dt \\
&= \frac{2}{a} \arctan t\bigg|_0^\infty \\
&= \frac{\pi}{a}\end{align}$$
Note that if $a = 0$, the integral diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Minimal Polynomial Exercise $\newcommand{\Ker}{\operatorname{Ker}}$Let $E$ be a vector space. We are given a matrix $A$ that has a characteristic polynomial $q(t) = -(t-2)^5$. We know that $\dim\Ker(A-2I)^2 = 3$ and $\dim\Ker(A-2I)^4 = 5$. We are asked to find the minimal polynomial of $A$.
This is what I've done so far: since $\dim\Ker(A-2I)^4 = 5 = \dim(E)$, $m_A(x)$ has to divide $(x-2)^4$. Besides, $\dim*\Ker(A-2I)^2 = 3$ implies that $m_A(x)$ does not divide $(x-2)^2$. Therefore, we are left with two candidates for the minimal polynomial of $A$: $(x-2)^4$ or $(x-2)^3$. I don't know how to decide between these two.
| Recall that $$ \dim \ker (A - \lambda I)^k - \dim \ker (A - \lambda I)^{k-1}$$ gives you the number of blocks corresponding to $\lambda$ of size at least $k$. See, for example: Jordan form, number of blocks.
So in this case you know that $\dim\ker(A-2I)^2 = 3$ and $\dim\ker(A-2I)^4 = 5$. Let us consider the three possibilities for $\dim\ker(A-2I)^3$.
*
*$\dim\ker(A-2I)^3=3$ is not possible. This would imply that $\dim\ker(A-2I)^k=3$ for each $k\ge 2$. (Also, this would mean that there are no blocks of size $\ge3$, which would lead to $(A-2I)^2=0$. Which is another way to get a contradiction.)
*$\dim\ker(A-2I)^3=5$ would mean that there are $5-3=2$ blocks of size at least $3\times3$. This is not possible in a $5\times 5$ matrix.
*So only one possible case remains: $\dim\ker(A-2I)^3=4$. We get $4-3=1$ block of size at least $3\times3$, and also $5-4=1$ block of size at least $4\times 4$. The Jordan form has one block $4\times4$ and one block $1\times1$.
$$J=\begin{pmatrix}
2 & 1 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 & 0 \\
0 & 0 & 2 & 1 & 0 \\
0 & 0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 & 2 \\
\end{pmatrix}$$ This gives us $m_A(x)=m_J(x)=(x-2)^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the Largest integer that is smaller than this expression Largest integer that is smaller than
$\frac{2^{2^{2021}}}{(2^{2^1}-2^{2^0}+1)(2^{2^2}-2^{2^1}+1)...(2^{2^{2020}}-2^{2^{2019}}+1)}$
is...
My Progress
$\frac{1}{2^{2^{k+1}}-2^{2^k}+1}=\frac{2^{2^k}+1}{2^{3.2^k}+1}$
Reorder so
$\frac{2^{2^{2021}}}{(2^{2^1}-2^{2^0}+1)(2^{2^2}-2^{2^1}+1)...(2^{2^{2020}}-2^{2^{2019}}+1)}$=$\prod_{i=0}^{2018}\frac{2^{2^{k+1}}}{2^{3.2^{k}}+1}.\frac{3.2^{2^{2021}}}{2^{3.2^{2019}}}< \frac{3.2^{2^{2021}}}{2^{3.2^{2019}}}$
I'm stuck here :(
If someone was able to finish this, that would be great, I've been looking for the answer for more than 1 hour:(
| Your first step is, good.
Then use:
$$(x-1)\prod_{k=0}^n \left(x^{2^k}+1\right)=x^{2^{n+1}}-1 $$
Prove by induction.
Use it with $x=2,x=2^3.$
Then $$\begin{align}\frac1{\prod_{k=0}^{2019}\left(2^{2^{k+1}}-2^{2^k}+1\right)}&=\prod_{k=0}^{2019}\frac{2^{2^k}+1}{2^{3\cdot 2^k}+1} \\
&=(8-1)\frac{2^{2^{2020}}-1}{2^{3\cdot 2^{2020}}-1}
\end{align}$$
So $$\begin{align}\frac{2^{2^{2021}}}{\prod_{k=0}^{2019}\left(2^{2^{k+1}}-2^{2^k}+1\right)}&=7\cdot\frac{2^{3\cdot 2^{2020}}-2^{2\cdot 2^{2020}}}{2^{3\cdot2^{2020}}-1}\\
&=7\cdot\left(1-\frac{4^{2^{2020}}-1}{8^{ 2^{2020}}-1}\right)\end{align}$$
And that fraction is much smaller than $\frac17,$ so the greatest integer is $6.$
More generally, $$\frac{x^{2^n}}{\prod_{k=0}^{n-2}\left(x^{2^{k+1}}-x^{2^k}+1\right)}=(x^2+x+1)\left(1-\frac{x^{2^{n}}-1}{x^{3\cdot 2^{n-1}}-1}\right)$$
Also when $x>1,$ $$\frac{x^{2^n}-1}{x^{3\cdot 2^{n-1}}-1}=\frac{x^{2^{n-1}}+1}{x^{2^n}+x^{2^{n-1}}+1}<\frac{1}{x^{2^{n-1}}}$$
In particular, for $x>1$ an integer and $n>2,$ the greatest integer would always be $x^2+x.$
When $n=2$ we have:
$$\frac{x^4}{x^2-x+1}=x^2+x-\frac{x}{x^2-x+1}$$
So the greatest integer is $x^2+x-1$ in that case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5}$, then $\dfrac{a}{b} \leq \dfrac{91}{150}$ and if $\dfrac{a}{b}<\dfrac{3}{5}$, then $\dfrac{a}{b} \geq \dfrac{89}{150}$.
Hence the maximum value of $\dfrac{a}{b}$ is $\dfrac{91}{150}$. And $\dfrac{a}{b}$ is maximum when $b$ is the smallest.
So, I thought $150$ would be the smallest value of $b$.
But my friend said that the smallest value of $b$ would be $32$ as $\left\lvert \dfrac{19}{32}- \dfrac{3}{5}\right\rvert= \dfrac{1}{160}$ works actually. Though he said that he found this intuitively.
So, how to solve the problem with proper procedure?
| HINT:
If $\frac{a}{b}$ satisfies the condition then
$$|5 a - 3 b| = 1$$
the explanation being that in the parallelogram on the vectors $(3,5)$ and $(a,b)$ there should be no other integer points excepts the vertices. Now, the solutions of
$$5 a - 3 b = 1$$
are $$a = 2 + 3 t \\ b = 3 + 5 t$$
while those of
$$5 a - 3 b = -1$$
are
$$a = 1 + 3 t \\ b = 2 + 5 t $$
Now if $|5 a - 3 b| = 1$, then
$$\left|\frac{a}{b} - \frac{3}{5}\right| = \frac{1}{5 b}$$
Therefore, we need $$5 b \ge 150$$, or $$b \ge 30$$
Now, the smallest $b \equiv 2 \text{ or } 3 \mod 5$ and $\ge 30$ is $32$ ( and not $33$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Locus of point M such that its projections onto two fixed lines are always at the same distance. A point moves so that the distance between the feet of the perpendiculars drawn from it to the lines $ax^2+2hxy+by^2=0$ is a constant $c$. Prove that the equation of its locus is
$$4(x^2+y^2)(h^2-ab)=c^2(4h^2+(a-b)^2).$$
What I have done: Suppose $P$ is any point on the locus, $A$ and $B$ be the feet of the perpendiculars, and $O$ is the origin. Then I want to show that
$$OP=\frac{AB}{\sin AOB}=\frac{c}{\sin \theta}$$
where $\theta$ is the angle between the two lines.
Note that the arc $PAO$ is a semi-circle, since $\angle PAO=90$. Similarly the arc $PBO$ is a semi-circle since $\angle PBO=90$. Also $\angle APB=\angle AOB$, as it is the angle of the same arc.
Now, $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. From here I can find $\sin \theta$ and $OP=\sqrt{x^2+y^2}$. Putting all these in,
$$OP=\frac{c}{\sin \theta}$$
will give me the equation of the locus. The problem is how to show that
$$OP=\frac{AB}{\sin AOB}.$$
| Here's a non-trigonometric take on this problem.
The conic $ax^2+2hxy+by^2=0$ gives the two solutions $y=m_\pm x$ where $bm_\pm=-h\pm\sqrt{h^2-ab}$. Given $P(x_0,y_0)$, the perpendicular foot $A(x,m_+x)$ minimises $AP^2=(x-x_0)^2+(m_+x-y_0)^2$. This is minimised at $(m_+^2+1)x^*=x_0+m_+y_0$ so $$A\left(\frac{x_0+m_+y_0}{m_+^2+1},\frac{m_+(x_0+m_+y_0)}{m_+^2+1}\right)\implies B\left(\frac{x_0+m_-y_0}{m_-^2+1},\frac{m_-(x_0+m_-y_0)}{m_-^2+1}\right).$$ without loss of generality. As $AB=c$ we have $c^2=(\alpha^2+\beta^2)(x_0^2+y_0^2)$ where $\alpha=\dfrac1{m_+^2+1}-\dfrac1{m_-^2+1}$ and $\beta=\dfrac{m_+}{m_+^2+1}-\dfrac{m_-}{m_-^2+1}$. Now \begin{align}\alpha^2+\beta^2&=\frac1{m_+^2+1}+\frac1{m_-^2+1}-\frac{2(m_+m_-+1)}{(m_+^2+1)(m_-^2+1)}\\&=\frac{(m_+-m_-)^2}{(m_+m_-)^2+m_+^2+m_-^2+1}\\&=\frac{4(h^2-ab)/b^2}{a^2/b^2+2(2h^2-ab)/b^2+1}\\&=\frac{4(h^2-ab)}{4h^2+(a-b)^2}\end{align} using Vieta so the locus is a circle centred at the origin of radius $\dfrac c2\sqrt{\dfrac{4h^2+(a-b)^2}{h^2-ab}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Does $\lim_{\epsilon \to 0^+}(\int_{1/2}^{1-ε}＋\int_{1＋ε}^{3/2})\frac {\log x} {(x-1)^2} dx$ exist? Does $$\lim_{\epsilon \to 0^+} \left(\int_{1/2}^{1-ε}＋\int_{1＋ε}^{3/2}\right)\frac{\log x}{(x-1)^2} dx$$ exist?
Hint says $$\lim_{\epsilon \to 0^+} \left( \frac{\log x-a-b(x-1)-c(x-1)^2}{(x-1)^3}\right)$$ exists only if $a=0,b=1,c=-\frac12$.
I have no ideal how to apply this hint.
Another way is also appreciated. Thank you in advance.
| Look at the Taylor expansion of $\log x$ around $x = 1$:
$$\log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots.$$
Therefore, there exists a continuous function $g : [\frac{1}{2}, \frac{3}{2}] \to \mathbb{R}$ defined by
$$g(x) = -\frac{1}{2} + \frac{x-1}{3} - \frac{(x-1)^2}{4} + \cdots$$
such that $\log x = (x-1) + (x-1)^2 g(x)$.
Therefore,
$$\lim_{\varepsilon\to 0^+} \left( \int_{1/2}^{1-\varepsilon} + \int_{1+\varepsilon}^{3/2} \right) \frac{\log x}{(x-1)^2} dx =
\lim_{\varepsilon\to 0^+} \left( \int_{1/2}^{1-\varepsilon} + \int_{1+\varepsilon}^{3/2} \right) \left( \frac{1}{x-1} + g(x) \right) dx.$$
Now, if we split into the term for $\frac{1}{x-1}$ and the term for $g(x)$, in the first we get
$$\lim_{\varepsilon\to 0^+} (\log(1/2) - \log(\varepsilon)) + (\log(\varepsilon) - \log(1/2)) = \lim_{\varepsilon\to 0^+} 0 = 0.$$
In the second, setting $G$ to be an antiderivative of $g$ on $[1/2, 3/2]$ (and thus $G$ must in particular be continuous) and using the Fundamental Theorem of Calculus, we get
\begin{align*}
\lim_{\varepsilon\to 0^+} (G(3/2) - G(1 + \varepsilon)) + (G(1 - \varepsilon) - G(1/2)) = \\ (G(3/2) - G(1)) + (G(1) - G(1/2)) = G(3/2) - G(1/2).
\end{align*}
Since both components of the sum have a limit, the sum must then also have a limit. (In other words, the Cauchy principal value of the improper integral $\int_{1/2}^{3/2} \frac{\log x}{(x-1)^2} dx$ exists.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\mathbf{A}_i=\lambda_i\mathbf{a}+\mu_i\mathbf{b}+\nu_i\mathbf{c}$, then $\mathbf{A}_1\cdot\left(\mathbf{A}_2\times\mathbf{A}_3\right)=$ Here is Prob. 2.43 in the book Vector Analysis & Cartesian Tensors by D. E. Bourne and P. C. Kendall, 3rd edition:
If
$$
\begin{align}
\mathbf{A}_1 &= \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} + \nu_1 \mathbf{c}, \\
\mathbf{A}_2 &= \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} + \nu_2 \mathbf{c}, \\
\mathbf{A}_3 &= \lambda_3 \mathbf{a} + \mu_3 \mathbf{b} + \nu_3 \mathbf{c},
\end{align}
$$
show that
$$
\mathbf{A}_1 \cdot \left( \mathbf{A}_2 \times \mathbf{A}_3 \right) = \left| \begin{matrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 & \nu_3 \end{matrix} \right| \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ).
$$
My Attempt:
In what follows, we will be using the following identities:
$$
\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{b} ) = \mathbf{b} \cdot ( \mathbf{a} \times \mathbf{b} ) = \mathbf{b} \cdot ( \mathbf{b} \times \mathbf{c} ) = \mathbf{c} \cdot ( \mathbf{b} \times \mathbf{c} ) = \mathbf{c} \cdot ( \mathbf{c} \times \mathbf{a} ) = \mathbf{a} \cdot ( \mathbf{c} \times \mathbf{a} ) = 0,
$$
and also that
$$
\mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) = \mathbf{b} \cdot ( \mathbf{c} \times \mathbf{a} ) = \mathbf{c} \cdot ( \mathbf{a} \times \mathbf{b} ).
$$
Using the distributive property of the dot and the cross products, we obtain
$$
\begin{align}
& \qquad \mathbf{A}_1 \cdot \left( \mathbf{A}_2 \times \mathbf{A}_3 \right) \\
&= \left( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} + \nu_1 \mathbf{c} \right) \cdot \left[ \left( \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} + \nu_2 \mathbf{c} \right) \times \left( \lambda_3 \mathbf{a} + \mu_3 \mathbf{b} + \nu_3 \mathbf{c} \right) \right] \\
&= \left( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} + \nu_1 \mathbf{c} \right) \cdot \left[ \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) ( \mathbf{a} \times \mathbf{b} ) + \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) ( \mathbf{b} \times \mathbf{c} ) + \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) ( \mathbf{c} \times \mathbf{a} ) \right] \\
&= \lambda_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] + \lambda_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \lambda_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] \\
& \qquad + \mu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] + \mu_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \mu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] \\
& \qquad + \nu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] + \nu_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \nu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] \\
&= \lambda_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \mu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] + \nu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] \\
&= \left[ \lambda_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) + \mu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) + \nu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \right] \big[ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] \\
&= \left| \begin{matrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 & \nu_3 \end{matrix} \right| \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ),
\end{align}
$$
as required.
Is this solution correct? If so, is it clear enough? Or, do I need to include some more details?
| Your approach is correct, but there is a simpler proof using triple product formula:
$$\vec{u}.(\vec{v} \times \vec{w})=\det(u,v,w)\tag{1}$$
making the relationship to be proven equivalent to:
$$\det(A_1,A_2,A_3) \overset{?}{=} \begin{vmatrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 & \nu_3 \end{vmatrix} \det(a,b,c)\tag{2}$$
Now, your initial data with 3 relationships can be grouped into a single matrix relationship:
$$\underbrace{(A_1|A_2|A_3)}_{3 \times 3 \ \text{matrix}}=\underbrace{(a|b|c)}_{3 \times 3 \ \text{matrix}}\left(\begin{matrix} \lambda_1 & \lambda_2 & \lambda_3 \\ \mu_1 & \mu_2 & \mu_3 \\ \nu_1 & \nu_2 & \nu_3 \end{matrix} \right) \tag{3}$$
How to prove that (3) implies (2) ? Plainly by taking the determinant of both sides... (please note that in (3) we had to take the transpose of the matrix of coefficients, which is harmless for the computation of its determinant).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the line intergal: $A=\int_C \sqrt{x^2+y^2}ds$ Calculate the line intergal:
$$A=\int_C \sqrt{x^2+y^2}ds, \, \text{with} \,\, C: x^2+y^2=ax$$
Set $\left\{\begin{align*}
x&=\dfrac{a}{2}\left( \dfrac{a}{2}+\cos t\right)\\[10pt]
y&=\dfrac{a}{2}sint
\end{align*}\right., \quad 0 \leq t \leq 2\pi$
So, $$\begin{align*}
D&=\displaystyle \int_0^{2\pi}\left( \dfrac{a}{2}\right)^2\sqrt{\left( \dfrac{a}{2}+\cos t\right)^2+\sin^2t}.\sqrt{\sin^2 t + \cos^2 t}\, dt\\[10pt]
&=\dfrac{a^2}{4} \displaystyle \int_0^{2\pi} \sqrt{a\cos t+1+\dfrac{a^2}{4}}\,dt
\end{align*}$$
I have a trouble with this integral. It's not easy to calculate. Help me!
| $C: x^2+y^2=ax \implies \left(x - \dfrac{a}{2}\right)^2 + y^2 = \dfrac{a^2}{4}$
So parametrize as $r(t) = \left(\dfrac{a}{2} + \dfrac{a}{2} \cos t, \dfrac{a}{2} \sin t \right), 0 \leq t \leq 2\pi$
$|r'(t)| = \sqrt{\left( - \dfrac{a}{2}\sin t\right)^2 + \left( \dfrac{a}{2}\cos t\right)^2} = \dfrac{a}{2}$
$\sqrt{x^2+y^2} = \sqrt{ax} = \dfrac{a}{\sqrt2} \sqrt{1 + \cos t}$
So the integral is,
$I = \displaystyle \dfrac{a^2}{2\sqrt2} \int_0^{2\pi} \sqrt{1 + \cos t} \ dt = 2a^2$
Edit: there is another parametrization you can use as the circle is centered on x-axis and y-axis is tangent to it,
In polar coordinates, $x = \rho \cos t, y = \rho \sin t$ and so $x^2 + y^2 = ax \implies \rho = a \cos t$ and as we are measuring the angle from origin, the circle forms for $- \pi/2 \leq t \leq \pi/2$.
$x = \rho \cos t = a \cos^2 t, y = \rho \sin t = a \sin t \cos t$
So we have, $r(t) = (a \cos^2 t, a \sin t \cos t), - \pi/2 \leq t \leq \pi/2$
$r'(t) = (- a \sin 2t, a \cos 2t), |r'(t) = a$
$\sqrt{x^2+y^2} = a \cos t$
The integral becomes,
$I = \displaystyle a^2 \int_{-\pi/2}^{\pi/2} \cos t \ dt = 2a^2$
| {
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"source": "stackexchange",
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Collatz Conjecture: For a cycle where the maximum odd integer is $x_{max}$, does it follow that $x_{max} < 3^n$ I am working on understanding the upper limit in the case where a non-trivial cycle exists for the Collatz Conjecture.
Is the following reasoning valid for establishing that the maximum odd integer in a cycle consisting $n$ odd integers is less than $3^n$?
Let:
*
*$\nu_2(x)$ be the 2-adic valuation of $x$
*$n > 5$ be an integer
*$x_1, x_2, \dots, x_n$ be the distinct odd integers that make up a cycle of length $n$ with:
*
*$x_{i+1} = \dfrac{3x_i+1}{2^{\nu_2(3x_i+1)}}$
*$x_{i+n} = x_i$
*$x_{max}$ is the highest odd integer in the cycle
*$p_0, p_1, \dots p_n$ be integers such that:
*
*$p_0 = 0$
*$p_{i+1} = p_i + \nu_2(3x_{i+1}+1)$
*$p_n > p_{n-1} > \dots > p_1 > p_0 = 0$
(1) Since $x_1, \dots x_n$ is a cycle, we can assume that $x_1 = x_{max}$
(2) It follows from reasoning in step 2 here:
$$2^{p_n}x_{n+1} - 3^{n}x_1 = x_{max}\left(2^{p_n} - 3^n\right) = \sum\limits_{i=0}^{n-1}3^{n-1-i}2^{p_i}$$
(3) From reasoning here for a cycle:
$$p_n < 2n$$
(4) $\sum\limits_{i=0}^{n-1}3^{n-1-i}2^{p_i} < 6^n$ since:
$$\sum\limits_{i=0}^{n-1}3^{n-1-i}2^{p_i} < 3^{n-1} + 4^{n-1} + (2^{2n-n+1})(\sum\limits_{i=1}^{n-2}3^{n-i-i}2^{i}) < 3^{n-1} + 4^{n-1} + 2^{n+1}(3^{n-1}-2^{n-1}) = 3^{n-1} + 4^{n-1} + 4(6^{n-1} - 4^{n-1}) < 6^{n-1} + 6^{n-1} + 4\times6^{n-1} = 6^n$$
(5) $2^{p_n} \ge 2^{\lceil{n}\log_2{3}\rceil} > 3^n$
*
*From reasoning here, since $n > 5$, it follows that $2^{p_n} - 3^n > 2^n$
*It follows that $x_{max} < \dfrac{6^n}{2^n} = 3^n$
Did I make a mistake? Is there any step that is unclear? Is there a simpler argument to reach the same conclusion?
Edit:
There is no advantage to the 2 cases in step 5. I have updated to simplify based on suggestions in the comment.
| The start of your reasoning is quite good, although there are some issues with your later steps. Otherwise, your arguments are basically correct.
In your (4), there are several relatively minor mistakes, but your overall upper bound still applies. Using your result of $p_n \lt 2n$, then each $p_{i+1} \gt p_{i}$ for $1 \le i \le n - 1$ means each lower index $p$ value must be at least one less than the one above, with this giving that for all $0 \le j \le n - 1$
$$p_{n-j} \lt 2n - j \tag{1}\label{eq1A}$$
Thus, for example, $p_{n - 1} \lt 2n - 1$ and $p_i \lt 2n - (n - i) = n + i$ for all $1 \le i \le n$ (thus, for example, $p_1 \lt n + 1$). This then, along with $n \gt 5$, gives
$$\begin{equation}\begin{aligned}
\sum_{i=0}^{n-1}3^{n-1-i}2^{p_i} & \lt 3^{n-1} + 2^{2n-1} + (2^n)\left(\sum_{i=1}^{n-2}3^{n-1-i}2^{i}\right) \\
& = 3^{n-1} + 2(2^{2(n-1)}) + (2^n)((2)(3)(3^{n-2} - 2^{n-2})) \\
& = 3^{n-1} + 2(4^{n-1}) + (2^2)((2)(3))(6^{n-2} - 4^{n-2}) \\
& \lt 6^{n-1} + 6^{n-1} + (4)(6)(6^{n-2}) \\
& = 6^{n-1} + 6^{n-1} + 4(6^{n-1}) \\
& = 6^n
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
You made a mistake with using $4^{n-1}$ instead of $2(4^{n-1})$. Also, since the starting point for $i = 1$ has a factor of $2^i = 2$ instead the summation, the common factor outside is $2^n$, not $2^{n+1}$. Finally, assuming you tried to use the actual summation instead of performing several steps at once, then note the summation has a factor of $3$ outside, with the exponents inside being $n - 2$, not $n - 1$. To see this, using the sum of a geometric series formula gives
$$\begin{equation}\begin{aligned}
\sum_{i=1}^{n-2}3^{n-1-i}2^{i} & = 2(3^{n-2})\left(\frac{1 - \left(\frac{2}{3}\right)^{n-2}}{1 - \frac{2}{3}}\right) \\
& = 2(3^{n-2})\left(\frac{\frac{3^{n-2} - 2^{n-2}}{3^{n-2}}}{\frac{1}{3}}\right) \\
& = 2(3)(3^{n-2} - 2^{n-2})
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Regarding your step (5), here are a few more details. With $m = \left\lceil n\log_2 3 \right\rceil$ being the positive exponent $m$ which allows $2^m - 3^n \gt 0$, then since $2^{p_n} \gt 3^n$, we have $p_n \ge m$. This then gives from your linked answer for $n \gt 5$ that
$$2^{p_n} - 3^n \ge 2^{\left\lceil n\log_2 3 \right\rceil} - 3^n \gt 2^n \tag{4}\label{eq4A}$$
| {
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$\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+..+ \frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqrt{n-1}}(\sqrt{a_1}+\sqrt{a_2}+\dots+ \sqrt{a_n})$ If $a_1,a_2,\dots ,a_n\ge 0$ such that $a_1+a_2+\dots a_n=1$ show that $\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+\dots +\frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqrt{n-1}}(\sqrt{a_1}+\sqrt{a_2}+\dots+ \sqrt{a_n})$.
Attempt
WLOG assume that $a_1\leq a_2\leq \dots \leq a_n$ it implies $\frac{1}{\sqrt{1-a_1}}\leq \frac{1}{\sqrt{1-a_2}}\leq \dots \leq \frac{1}{\sqrt{1-a_n}}$ now if we denote by $S$ the LHS of the inequality then by rearrangement inequality
$$S\geq \frac{a_2}{\sqrt{1-a_1}}+\frac{a_3}{\sqrt{1-a_2}}+\dots +\frac{a_1}{\sqrt{1-a_n}}$$
$$S \geq \frac{a_3}{\sqrt{1-a_1}}+\frac{a_4}{\sqrt{1-a_2}}+\dots +\frac{a_2}{\sqrt{1-a_n}}$$
$$\vdots$$
$$S \geq \frac{a_n}{\sqrt{1-a_1}}+\frac{a_{n-1}}{\sqrt{1-a_2}}+\dots +\frac{a_1}{\sqrt{1-a_1}}$$
it is $$(n-1)S\geq \left(\frac{1}{\sqrt{1-a_1}}+\frac{1}{\sqrt{1-a_2}}+\dots +\frac{1}{\sqrt{1-a_n}}\right) \left(a_1+a_2+\dots a_n \right)\geq$$
since $\frac{1}{1-a_i}\geq \frac{a_i}{1-a_i}\geq a_i$ we get $\frac{1}{\sqrt{1-a_i}}\geq \sqrt{\frac{a_i}{1-a_i}}\geq \sqrt{a_i}$
$$\geq (\sqrt{a_1}+\sqrt{a_2}+\dots \sqrt{a_n})(a_1+a_2+\dots a_n)$$
and since $a_i>\sqrt{a_i}$
$$(n-1)S\geq (\sqrt{a_1}+\sqrt{a_2}+\dots \sqrt{a_n})^2$$
But it not look like our inequality
any advice or help was useful thanks in advice.
| Let $x \in (0,1)$, then notice that
$f(x)=\frac{x}{\sqrt{1-x}} \implies f''(x)=\frac{4-x}{4(1-x)^{5/2}}>0.$
So by Jensen's inequality, we can write that
$$S=\sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}}\ge n \frac{\sum_{k} a_k/n}{\sqrt{1-\sum a_k/n}}=\sqrt{n}\frac{\sum_k a_k}{\sqrt{n-1}}=\sqrt{n} \frac{\sqrt{\sum_k a_k}}{\sqrt{n-1}}.~~~~(1)$$
$\sum_ka_k=1$ is used partially in above. Next using RMS-AM inequality:
$\sqrt{\sum_k a_k} \ge \sum_k \sqrt{a_k}/\sqrt{n}$ in (1), we finally get
$$S=\sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}}\ge \frac{\sum_k \sqrt{a_k}}{\sqrt{n-1}}$$
Note: If f''(x)>0 for $x\in D$, then for all $x_k$ in this domain, the Jensen's inequality is $$\frac{1}{n}\sum_{k} f(x_k) \ge f(\sum_k\frac{ x_k}{n})$$
| {
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Let $x = (a^2+a+1)^{1/2}-(a^2-a+1)^{1/2}$ and $a$ belongs to real number find all possible value of $x$. Let $A = (-1/2, \sqrt{3}/2)$, $B = (1/2, \sqrt{3}/2)$, and $P = (a, 0)$. Then $P$ is a point on the $X$-axis and we are looking for all possible values of $d = PA - PB$.
By the Triangle Inequality, $PA - PB < AB = 1$. And it is clear
that all the values $-1 < d < 1$ are indeed obtainable. In fact, for such
a $d$, a half hyperbola of all points $Q$ such that $QA - QB = d$ is well
defined. (Points $A$ and $B$ are foci of the hyperbola.) Since line $AB$ is parallel to the $X$-axis, this half hyperbola intersects the
$X$- axis, i.e., $P$ is well defined.
I know that at point $A$ and $B$ the two above $\sqrt{\cdot}$ function will obtain minimum value respectively
I am not able to understand why are we looking for all the possible value of $d$, what is the relationship of $d$ with the possible value of $x$?
This is question number 14 from 101 problems in algebra (USAIMO)
| Since
$a^{1/2}-b^{1/2}
=(a^{1/2}-b^{1/2})\dfrac{a^{1/2}+b^{1/2}}{a^{1/2}+b^{1/2}}
=\dfrac{a-b}{a^{1/2}+b^{1/2}}
$,
$d(a)
=(a^2+a+1)^{1/2}-(a^2-a+1)^{1/2}
=\dfrac{2a}{(a^2+a+1)^{1/2}+(a^2-a+1)^{1/2}}
$.
Also,
by the generalized mean inequality,
$\sqrt{uv}
\le (\dfrac{\sqrt{u}+\sqrt{v}}{2})^2
\le \dfrac{u+v}{2}
$
or
$2\sqrt[4]{uv}
\le\sqrt{u}+\sqrt{v}
\le \sqrt{2(u+v)}
$
so
$d(a)
\ge \dfrac{2a}{\sqrt{4(a^2+1)}}
= \dfrac{a}{\sqrt{a^2+1}}
$
and
$\begin{array}\\
d(a)
&\le \dfrac{2a}{2\sqrt[4]{(a^2+a+1)(a^2-a+1)}}\\
&= \dfrac{a}{\sqrt[4]{(a^2+1)^2-a^2}}\\
&= \dfrac{a}{\sqrt[4]{a^4+a^2+1}}\\
&\le 1\\
\end{array}
$
So the range of values is from
$0$ to $1$.
| {
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Computing $\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$ I have this limit as my question to solve:
$$\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$$
My procedure:
$$\lim_{x\to-1}\frac{(2x+\sqrt{3-x})(2x-\sqrt{3-x})}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{4x^2+x-3}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{(x+1)(x-\frac{3}{4})}{x(x+1)(2x-\sqrt{3-x})}$$
$(x+1)$ is the zero factor and has to be eliminated.
$$\frac{-1-\frac{3}{4}}{-1(2(-1)-\sqrt{3-(-1)})}=-\frac{7}{16}$$
But when I checked my answer, I found it's wrong and the right answer is $-\frac{7}{4}$.
What was my mistake here?
| Hint: Your factorisation is not correct.
It is easier to do it if you let $t=\sqrt{3-x}$, then $x= 3-t^2$ and $t\to 2$.
| {
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Using Monotonicity to find the value of a required variable Given $f(x) = \log_c\frac{x-2}{x+2}$, is defined $\forall \; x \in [a,b]$ and the function is monotonically decreasing.
We have to find the value (or range of values) of $c$ such that there exists $a$ and $b$, where $(2 < a < b)$, and the range of the function over $[a,b]$ is $[\log_cc(b-1), \log_cc(a-1)]$.
I have concluded the following:
*
*$c \in (0,1)$
*Since the function is monotonically decreasing $$f(a) = \log_cc(a-1) \implies \frac{a-2}{a+2} = c(a-1)$$ and $$f(b) = \log_cc(b-1) \implies \frac{b-2}{b+2} = c(b-1)$$
I thought that subtracting the two equations or taking their ratio might help me find some helpful inequality, but unfortunately couldn't find anything.
Your help is appreciated. Thanks
UPDATE $1$: We know that $c \in (0,1)$ and $\frac{x-2}{x+2} \in (0,1) \; \forall \; x \in (2, \infty)$. Hence, the value attained by the function is positive. Thus,
$$\log_cc(a-1) > 0 \implies c(a-1) < 1 \implies a < \frac{1}{c} + 1$$
and a similar inequality for $b$.
| As you wrote, we have
$$\frac{a-2}{a+2} = c(a-1)$$
which is equivalent to
$$ca^2+(c-1)a-2c+2=0\tag1$$
Similarly, we have
$$cb^2+(c-1)b-2c+2=0\tag2$$
From $(1)(2)$, we want to find $c\ (0\lt c\lt 1)$ such that the equation
$$cx^2+(c-1)x-2c+2=0$$
has two distinct real roots $x$ satisfying $x\gt 2$.
So, we want to find $c$ satisfying
$$\begin{cases}0\lt c\lt 1
\\\\(c-1)^2-4c(-2c+2)\gt 0
\\\\\dfrac{1-c}{2c}\gt 2
\\\\c\cdot 2^2+(c-1)\times 2-2c+2\gt 0\end{cases}$$
So, the answer is $0\lt c\lt \dfrac 19$.
| {
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Finding the range of $a$ for which line $y=2x+a$ lies between circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting either
Find the range of parameter $a$ for which the variable line $y = 2x + a$ lies between the circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting or touching either circle.
Now, how I solved it was realising the line has a positive slope of $+2$, and thus if it's a tangent to circle #1 then its intercept should be negative. And so, using the condition of tangency,
$$a^2= m^2(r_1)^2+(r_1)^2= 4+1 $$
And thus, as $a$ can only be negative (otherwise a positive value of $a$ will make the line intersect the circle). Thus, making the lower bound of $a> - \sqrt 5$.
Similarly for the bigger circle $a<-\sqrt{20}=-\sqrt{(2^2)(2^2)+(2^2)}$
Hence I find that the solution should be $(-\sqrt5,-\sqrt{20})$, but the actual solution is $\left(2\sqrt 5-15,-\sqrt 5-1\right)$.
| I would draw perpendicular to the line $2x - y + a = 0$ from centers of both circles.
From circle $C_1$ centered at $(1, 1)$,
$d_1 = \dfrac{|2 -1 + a|}{\sqrt5} \gt 1$
(where $1$ is the radius of the circle $C_1$).
Similarly from $(8, 1)$, $d_2 = \dfrac{|16 - 1 + a|}{\sqrt5} \gt 2$
(where $2$ is the radius of the circle $C_2$).
Simplifying we have,
$|a + 1| \gt \sqrt5$
$|a + 15| \gt 2 \sqrt5$
If the line is in between both circles, the algebraic perpendicular distances from centers will have opposite signs. In other words, $(a+1)$ and $(a+15)$ will have opposite signs.
i) From the first (considering both negative and positive value of $a+1$),
$a \gt \sqrt5 - 1 \ $ or $ \ a \lt - \sqrt5 - 1$
ii) From the second (again considering both negative and positive values of $a+15$),
$a \gt 2 \sqrt5 - 15 \ $ or $ \ a \lt - 2 \sqrt5 - 15$.
a) When $a \gt \sqrt5 - 1$, both $(a+1)$ and $(a+15)$ have positive signs.
b) When $a \lt - 2 \sqrt5 - 15$, both $(a+1)$ and $(a+15)$ have negative signs.
c) When $2 \sqrt5 - 15 \lt a \lt - \sqrt5 - 1$, $(a+1)$ and $(a+15)$ have opposite signs.
That leads us to the answer $2 \sqrt5 - 15 \lt a \lt - \sqrt5 - 1$.
| {
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Methods for solving nonhomogeneous system of ODEs I have the system
$$ \mathbf{x}'=\begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf{x}+\begin{pmatrix}e^t\\t\end{pmatrix}. $$
The eigenvalues of the matrix are 1 and 3 with eigenvectors $(1,-1)^T$ and $(1,1)^T$ respectively. Thus, we have the fundamental matrix
$$
\Psi(t)=\begin{pmatrix}e^t & e^{3t}\\-e^t & e^{3t}\end{pmatrix}.
$$
From here I used the method of variation of parameters to assume there's a solution $\mathbf{x}=\Psi(t)\mathbf{u}(t)$, where $\mathbf{u}(t)$ satisfies $\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$. I solved this matrix equation and multiplied to find
$$
\mathbf{x}=c_1\begin{pmatrix}1\\-1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\1\end{pmatrix}e^{3t}+\frac{1}{2}\begin{pmatrix}1\\-1\end{pmatrix}te^t-\frac{5}{12}\begin{pmatrix}1\\1\end{pmatrix}e^t-\frac{1}{2}\begin{pmatrix}1\\-1\end{pmatrix}t+\frac{1}{9}\begin{pmatrix}4\\-5\end{pmatrix}.
$$
I wanted to do this same problem using the Laplace transform given $\mathbf{x}(0)=\mathbf{0}$.
When I take the Laplace transform of the system and solve for $\mathbf{X}(s)$ I get
$$
\mathbf{X}(s)=
\begin{pmatrix}
s-2&-1\\
-1&s-2
\end{pmatrix}^{-1}
\begin{pmatrix}
(s-1)^{-1}\\
s^{-2}.
\end{pmatrix}.
$$
Performing this computation and taking the inverse Laplace transform, I get
$$
\mathbf{x}=\begin{pmatrix}
\frac{1}{8}e^t+\frac{1}{8}e^{5t}-\frac{1}{4}e^t+t\\
e^t+\frac{2}{5}t+\frac{1}{2}e^{-t}+\frac{1}{50}e^{5t}-\frac{13}{25}.
\end{pmatrix}
$$
At this point I realized that there's no way this is the same answer. Does someone see where I'm going wrong or what I misunderstood about one of the methods?
| For the first part, I will use this approach. We are given
$$\mathbf{x}'=\begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf{x}+\begin{pmatrix}e^t\\t\end{pmatrix}$$
The eigenvalues/eigenvectors are
$$\lambda_1 = 1, v_1 = \begin{pmatrix}-1 \\ 1 \end{pmatrix} \\ \lambda_2 = 3, v_2 = \begin{pmatrix}1 \\ 1 \end{pmatrix} $$
The Fundamental matrix is
$$X = \begin{pmatrix}-e^t & e^{3t} \\ e^t & e^{3t} \end{pmatrix}$$
We have $$x_p = X \int X^{-1} g~dt = \begin{pmatrix}
\dfrac{e^t t}{2}+\dfrac{t}{3}-\dfrac{e^t}{4}+\dfrac{4}{9} \\
-\dfrac{e^t t}{2}-\dfrac{2 t}{3}-\dfrac{e^t}{4}-\dfrac{5}{9} \\
\end{pmatrix}$$
We can now write the solution as
$$\mathbf{x}(t) = c_1e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + x_p = c_1 e^t\begin{pmatrix}-1 \\ 1 \end{pmatrix} + c_2 e^{3t}\begin{pmatrix}1 \\ 1 \end{pmatrix} + \begin{pmatrix}
\dfrac{e^t t}{2}+\dfrac{t}{3}-\dfrac{e^t}{4}+\dfrac{4}{9} \\
-\dfrac{e^t t}{2}-\dfrac{2 t}{3}-\dfrac{e^t}{4}-\dfrac{5}{9} \\
\end{pmatrix}$$
Using the IC, $\mathbf{x}(0)=\mathbf{0}$, this reduces to
$$x(t) = \frac{e^t t}{2}+\frac{t}{3}+\frac{11 e^{3 t}}{36}-\frac{3 e^t}{4}+\frac{4}{9}\\
y(t) = \frac{e^t t}{2}-\frac{2 t}{3}+\frac{e^t}{4}+\frac{11 e^{3 t}}{36}-\frac{5}{9}$$
For the Laplace method, we have
$$\mathbf{X}(s)=
\begin{pmatrix}
s-2&-1\\
-1&s-2
\end{pmatrix}^{-1}
\begin{pmatrix}
\dfrac{1}{s-1}\\
\dfrac{1}{s^2}
\end{pmatrix} = \begin{pmatrix}
\dfrac{s-2}{(s-1) \left(s^2-4 s+3\right)}+\dfrac{1}{s^2 \left(s^2-4 s+3\right)} \\
\dfrac{s-2}{s^2 \left(s^2-4 s+3\right)}+\dfrac{1}{(s-1) \left(s^2-4 s+3\right)} \\
\end{pmatrix}$$
The final result from the inverse Laplace Transform is
$$x(t) = \frac{e^t t}{2}+\frac{t}{3}+\frac{11 e^{3 t}}{36}-\frac{3 e^t}{4}+\frac{4}{9}\\
y(t) = \frac{e^t t}{2}-\frac{2 t}{3}+\frac{e^t}{4}+\frac{11 e^{3 t}}{36}-\frac{5}{9}$$
Those two match $\Large\color{\green}{\checkmark}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$ While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried $x=\sin(u)$ then $\tan(u/2)=t$ and got
$$I=\int_0^{\pi/2}\cot(u)\ln(\sin u)\ln(1-\sin u)\ln(1+\cos u)\mathrm{d}u$$
$$=\int_0^1\frac{1-t^2}{t(1+t^2)}\ln\left(\frac{2t}{1+t^2}\right)\ln\left(\frac{(1-t)^2}{1+t^2}\right)\ln\left(\frac{2}{1+t^2}\right)\mathrm{d}t.$$
Any idea?
Thanks
| Since you are primarily interested in that series I'll evaluate it.
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
Consider the well-known expansion:
$$2\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}x^{2k}=\frac{\arcsin ^2\left(x\right)}{\sqrt{1-x^2}}$$
$$-4\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}\int _0^1x^{2k-1}\ln \left(x\right)\:dx=-2\int _0^1\frac{\ln \left(x\right)\arcsin ^2\left(x\right)}{x\sqrt{1-x^2}}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-2\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\ln \left(\sin \left(x\right)\right)\:dx+\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
You can find that integral here. I can also provide a new way to calculate it but it's lenghty.
For the remaining sum consider:
$$\int _0^1x^{k-1}\ln ^2\left(1-x\right)\:dx=\frac{H_k^2+H_k^{\left(2\right)}}{k}$$
$$\int _0^1\left(\sum _{k=1}^{\infty }\frac{1}{4^kk}\binom{2k}{k}x^k\right)\frac{\ln ^2\left(1-x\right)}{x}\:dx=\sum _{k=1}^{\infty }\frac{H_k^2}{4^kk^2}\binom{2k}{k}+\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
$$\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-16\underbrace{\int _0^1\frac{x\ln ^2\left(x\right)\ln \left(1+x\right)}{1-x^2}\:dx}_{I}+2\ln \left(2\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx-\underbrace{\sum _{k=1}^{\infty }\frac{H_k^2}{4^kk^2}\binom{2k}{k}}_{S}$$
$$\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=3\zeta \left(4\right)-3\ln \left(2\right)\zeta \left(3\right)$$
$I$ transforms into known integrals after integration by parts and $S$ can be calculated by simple means as shown in the OP's book, page $\#297$
And therefore:
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-\frac{129}{8}\zeta \left(4\right)+\frac{25}{4}\ln \left(2\right)\zeta \left(3\right)-\frac{3}{2}\ln ^2\left(2\right)\zeta \left(2\right)+8\pi \operatorname{\mathfrak{I}} \left\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Partial Fraction Decomposition of $\frac{z+4}{(z+1+2i)(z+1-2i)}$ I have the fraction $\frac{z+4}{(z+1+2i)(z+1-2i)}$. I want to partial decompose this fraction, but I am not seeing how to do it. I know that the answer is a=$\frac{1}{2}$+$\frac{3i}{4}$ and b=$\frac{1}{2}$-$\frac{3i}{4}$, where
$$\frac{z+4}{(z+1+2i)(z+1-2i)}= \frac{a}{z+1+2i}+\frac{b}{z+1-2i},\quad a,b \in \mathbb C.
$$
I tried to write $a=c+di$ and $b=e+fi$ but I didn't got the answer. I also tried $a(z+1-2i) + b(z+1+2i)$ but this is not right because it gave me $a+b=1$ and $a+b=4$.
Thank you for your help in advance!
| Let $k=1+2i$, then we want to find $a$ and $b$ such that
$$\frac{z+4}{(z+k)(z+\bar{k})}=\frac{a}{z+k}+\frac{b}{z+\bar{k}}$$
Or $$z+4=(z+\bar{k})a+(z+k)b$$
Then setting $z=-\bar{k}$ and $z=-k$ we obtain $b=\frac{-\bar{k}+4}{k-\bar{k}}=\frac{-(1-2i)+4}{(1+2i)-(1-2i)}=\frac{3+2i}{4i}=\frac{1}{2}-\frac{3}{4}i$ and $a=\frac{1}{2}+\frac{3}{4}i$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Root of the given polynomial $4x^3-3x-p$ Given the equation $$ 4x^3 - 3x-p=0 $$
In the question we were required to find the root of this equation in the interval $[1/2,1]$ and $-1\le p\le 1$. The answer is arrived at by substituting x as $\cos(\theta)$
and using the multiple angle formula. However assuming I am not able to think of this substitution in the exam(it does not appear to be a very intuitive one considering the range of $x$ is not $[-1,1]$), is there another method to arrive at this answer?
| If the solution is not rational, you can use Cardano's method to find one real root. $4x^3-3x-p=0$ is equivalent to $x^3-\frac{3}{4}x-\frac{p}{4}=0$. And now you let $t=u+v$, which only holds if $-\frac{3}{4}=-3uv$ and $-\frac{p}{4}=-(u^3+v^3)$. Therefore, $u=\frac{1}{4v}$ and $\frac{p}{4}=v^3+\frac{1}{64v^3}$. Multiply everything by $v^3$ and you get $v^6-\frac{p}{4}v^3+\frac{1}{64}=0$, which is a triquadratic equation and can be solved by the Bhaskara's formula. $u^3=\frac{p}{8}\pm\sqrt \frac{p^2-1}{64}$, so $u=\sqrt[3]{\frac{p}{8}\pm\sqrt \frac{p^2-1}{64}}$. $v$ will be its conjugated, so $t=\sqrt[3]{\frac{p}{8}+\sqrt \frac{p^2-1}{64}}+\sqrt[3]{\frac{p}{8}-\sqrt \frac{p^2-1}{64}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_{x=1}^nb_x$$
And I want an explicit formula for $c_n$. After some research, I found the explicit formula for $b_n=\frac{n(n+1)(n+2)}{6}$. Seeing the patterns from $a_n$ and $b_n$, I figured the explicit formula for $c_n$ would be $\frac{n(n+1)(n+2)(n+3)}{24}$ or $\frac{n(n+1)(n+2)(n+3)}{12}$.
Then I tried to plug in those two potential equations,
If $n=1$, $c_n=1$, $\frac{n(n+1)(n+2)(n+3)}{24}=1$, $\frac{n(n+1)(n+2)(n+3)}{12}=2$. Thus we can know for sure that the second equation is wrong.
If $n=2$, $c_n=1+4=5$, $\frac{n(n+1)(n+2)(n+3)}{24}=5$. Seems correct so far.
If $n=3$, $c_n=1+4+10=15$, $\frac{n(n+1)(n+2)(n+3)}{24}=\frac{360}{24}=15$.
Overall, from the terms that I tried, the formula above seems to have worked. However, I cannot prove, or explain, why that is. Can someone prove (or disprove) my result above?
| Hint:
$$\sum_{r=1}^n r=\frac{n(n+1)}{2}$$
$$\sum_{r=1}^n r^2=\frac {n(n+1)(2n+1)}{6}$$
$$\sum_{r=1}^n r^3=\frac {(n(n+1))^2}{4}$$
Use of these $3$ formulae is sufficient to prove the required result.
The derivation of the $3^{rd}$ formula can comes by noting:
$$(r+1)^4-r^4=4r^3+6r^2+4r+1$$
Now sum this identity over $r=1$ to $r=n$, and since $\sum r^2$ and $\sum r$ are already known, the $3^{rd}$ formula gets proven.
In general, using this process, $\sum r^n$ can be derived if $\sum r^{n-1}$ is known.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182890",
"timestamp": "2023-03-29T00:00:00",
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Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
$\textbf{My attempts} :$
From the condition, we can write $$1+2n^2+3m^2\equiv 0(\mod 5)\tag 1$$
Now, since $5$ is prime so by Fermat's little theorem, we can write $$n^4\equiv 1(\mod 5)\quad\text{and} \quad m^4\equiv 1(\mod 5).$$
So, we get $n^4-m^4\equiv 0(\mod 5)$.
Since $5$ prime so, either $5|(m^2+n^2)$ or $5|(m^2-n^2).$
Now if $5|(m^2+n^2)$ then from $(1)$ we get $$1-n^2+3n^2+3m^2\equiv 0(\mod 5)$$ So, we are done.
Now, if $5|(m^2-n^2)$ then from $(1)$ we get $$1+5n^2-3n^2+3m^2\equiv 0(\mod 5).$$ So, we shall arrive at a contradiction that $1\equiv 0(\mod 5).$
In this way, I have tried to solve this problem. I will be highly obliged if you kindly check this or correct me.
Thanks in advance.
| squares are $0,1,4 \pmod 5$ so $3m^2 \equiv 0,3,2 \pmod 5,$ next $1+3m^2 \equiv 1,4,3 \pmod 5$ Finally
$$ -(1+3m^2) \equiv 4,1,2 \pmod 5 \; , \; \; $$
$$ 2 n^2 \equiv 0,2,3 \pmod 5 $$
The overlap of these two lists, $4,1,2$ and $0,2,3$ is the single possibility $2.$ That is, we need $2n^2 \equiv 2 \pmod 5$ and $n^2 \equiv 1 \pmod 5$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:
If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
S \space$ then if $a = b\cdot c\space$ prove that $c \in S$
My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\space$ and $b = 1 + 3 \cdot y\space$ and $x = y + j\space$ for some $j >= 1$
If $a = bc \implies c \mid a \space \And \space b\mid a$
Now $b \mid a \implies (1 + 3y)\mid (1 + 3x) \implies (1 + 3y) \mid (1 + 3(y + j)) \implies (1 + 3y) | (1 + 3y + 3j) \implies (1 + 3y) | (1 + 3y) + 3j$
Since $1 + 3y \mid (1 + 3y) \implies (1 + 3y) \mid 3j$ and we keep that.
Now we know that
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = \frac{1+3(y + j)}{1 + 3y} = \frac{1 + 3y + 3j}{1 + 3y} = \frac{1 + 3y}{1 + 3y} + \frac{3j}{1+3y} = 1 + \frac{3j}{1 + 3y}$$
But we have shown that $1 + 3y \mid 3j \implies 3j = k(1 + 3y)\space$ where $1 + 3y$ is a positive integer.
Hence $$c = 1 + \frac{k(1 + 3y)}{1 + 3y} = 1 + k$$
But now I am stuck because I need to show that $k$ is a multiple of $3$ to finish the proof and I am not sure how to do that.
I know that $$k = 3\frac{j}{1 + 3y}$$ but the $\frac{j}{1 + 3y}$ is not an integer I think so I a messing up somewhere.
Update:
Following the comment of @Infinity_hunter
Let $c = r + 3k\space$ for some $r >= 1$. We have:
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = 1 + \frac{3j}{1 + 3y}$$
This was shown earlier.
But
$$c = r + 3k = 1 + \frac{3j}{1 + 3y} $$
We know that $$\frac{3j}{1 + 3y}$$ since we have shown that $1 + 3y \mid 3j$ so let $\frac{3j}{1 + 3y} = p$
So we have:
$c = r + 3k = 1 + p \implies c = r - 1 - p + 3k \implies c = (r - 1 - p) + 3k$
And I am not sure how to progress from here.
| Take $c = r + 3k_1, a = 1 + 3k_2$ and $b = 1 + 3k_3$ where$k_1,k_2,k_3 $ and $r$ are integers. By Division theorem we can chose $r$ such that $r =0,1 $ or $2$. Now we know that $a = bc$ so
$$3k_2 + 1 = (r+3k_1)(1+3k_3) = r + 3k_3r+ 3k_1 + 9k_1k_3$$
So by rearranging terms we have
$$ 1 - r = 3( k_3r + k_1+3k_1k_3 - k_2)$$ which implies that $3$ divides only $1-r$ which is possible only when $r=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the determinant of a Hessenberg matrix whose elements are the Bernoulli numbers positive? The Bernoulli numbers $B_r$ are generated by
\begin{equation}\label{Bernoullu=No-dfn-Eq}
\frac{z}{e^z-1}=\sum_{r=0}^\infty B_r\frac{z^r}{r!}
=1-\frac{z}2+\sum_{r=1}^\infty B_{2r}\frac{z^{2r}}{(2r)!}, \quad \vert z\vert<2\pi.
\end{equation}
For $r\in\mathbb{N}$, I proved that
\begin{equation*}
(-1)^r\begin{vmatrix}
B_2 & -B_0 & 0 & \dotsm & 0& 0 & 0\\
B_{4} & B_2 & -B_0 & \dotsm & 0 & 0& 0\\
B_6 & B_{4} & B_2 & \dotsm & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots &\vdots & \vdots\\
B_{2r-4} & B_{2r-6} & B_{2r-8} & \dotsm & B_2 & -B_0 & 0\\
B_{2r-2} & B_{2r-4} & B_{2r-6} & \dotsm & B_{4} & B_2 & -B_0\\
B_{2r} & B_{2r-2} & B_{2r-4} & \dotsm & B_{6} & B_{4} & B_2
\end{vmatrix}
<0.
\end{equation*}
For $r\in\mathbb{N}$, let
\begin{equation*}
\mathfrak{D}_{r}=
\begin{vmatrix}
B_2 &B_0& 0 & \dotsm & 0 & 0 & 0\\
B_4 & B_2 &B_0& \dotsm & 0 & 0 & 0\\
B_6& B_4 & B_2 & \dotsm & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
B_{2(r-2)} & B_{2(r-3)} & B_{2(r-4)} & \dotsm & B_2 & B_0& 0\\
B_{2(r-1)} & B_{2(r-2)} & B_{2(r-3)} & \dotsm & B_4 & B_2 &B_0\\
B_{2r} & B_{2(r-1)} & B_{2(r-2)} & \dotsm & B_6 & B_4 & B_2
\end{vmatrix}.
\end{equation*}
Straightforward computation gives
\begin{equation*}
\mathfrak{D}_1=\frac{1}{6}, \quad\mathfrak{D}_2=\frac{11}{180}, \quad\mathfrak{D}_3=\frac{299}{7560}, \quad\mathfrak{D}_4=\frac{10417}{226800}.
\end{equation*}
Therefore, we guess that, for $r\in\mathbb{N}$,
\begin{equation*}
\mathfrak{D}_r>0.
\end{equation*}
Prove or deny this guess.
Since $(-1)^{r+1}B_{2r}>0$ for $r\in\mathbb{N}$, it is trivial that
\begin{equation*}
\mathfrak{D}_1=B_2^2-B_0B_4>0
\end{equation*}
and
\begin{equation*}
\mathfrak{D}_2=B_2^3+B_0^2B_6-2B_0B_2B_4>0.
\end{equation*}
Does this imply that the guess $\mathfrak{D}_r>0$ is trivially true?
See also similar discussions on the ResearchGate.
| The so-called guess $\mathfrak{D}_r>0$ can be proved to be true via mathematical induction and by the following recursive relation.
Let $H_0=1$ and
\begin{equation*}
H_r=
\begin{vmatrix}
h_{1,1} & h_{1,2} & 0 & \dotsc & 0 & 0\\
h_{2,1} & h_{2,2} & h_{2,3} & \dotsc & 0 & 0\\
h_{3,1} & h_{3,2} & h_{3,3} & \dotsc & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
h_{r-2,1} & h_{r-2,2} & h_{r-2,3} & \dotsc & h_{r-2,r-1} & 0 \\
h_{r-1,1} & h_{r-1,2} & h_{r-1,3} & \dotsc & h_{r-1,r-1} & h_{r-1,r}\\
h_{r,1} & h_{r,2} & h_{r,3} & \dotsc & h_{r,r-1} & h_{r,r}
\end{vmatrix}
\end{equation*}
for $r\in\mathbb{N}$. Then the sequence $H_r$ for $r\ge0$, with $H_1=h_{1,1}$, satisfies
\begin{equation}
H_r=\sum_{\ell=1}^r(-1)^{r-\ell}h_{r,\ell} \Biggl(\prod_{j=\ell}^{r-1}h_{j,j+1}\Biggr) H_{\ell-1}
\end{equation}
for $r\ge2$, where the empty product is understood to be $1$.
This recursive relation can be found on page 222 of the paper:
N. D. Cahill, J. R. D'Errico, D. A. Narayan, and J. Y. Narayan, Fibonacci determinants, College Math. J. 33 (2002), no. 3, 221--225; available online at https://doi.org/10.2307/1559033.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a$ & $b$ , s.t. $a+b=0$ and $ab=-3$? I've a problem in this question :
In the polynomial identity $x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)$ , find $ab$ $? $
MY APPROACH
We have : $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$
Now according to the Problem : $$x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)=(x^2+1)(x^4-x^2+1)$$ Or $$x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1=x^4-x^2+1$$ $$(a+b)x^3+(ab+2)x^2+(a+b)x=-x^2$$ In order to satisfy this , $a+b=0$ and $ab+2=-1$
Hence , $$ab=-3$$
but what'll be the value of $a$ & $ b$ , which'll satisfy these conditions $?$
| The equation $t^2-(a+b)t+ab=0$ can be written $(t-a)(t-b)=0$ and thus has solutions $t=a$ and $t=b$.
In your case this equation becomes $t^2-3=0$ which has solutions $t=\pm\sqrt3,$ so $a,b=\pm\sqrt3.$
| {
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"source": "stackexchange",
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prove that $x^6=(ax)^3$ implies $x^3=e,\: \forall x\in G$ Let be $(G,\cdot)$ a group and $e$ the identity element.
Let be $a\in G$.
If $$x^6=(ax)^3,\:\forall x\in G$$
Then
$$x^3=e,\: \forall x\in G$$
What I did:
$$x^6=(ax)^3,\:\forall x\in G$$
$x \gets a^2$ then $a^3=e$.
$$x^6=(ax)^3,\:\forall x\in G \implies x^6=(xa)^3,\:\forall x\in G$$
so
$$(ax)^3=(xa)^3,\:\forall x\in G$$
How should I continue from here?
| As @ancient mathematician hinted, for all $x\in G$
$$x^{24} = \left(x^6\right)^4 = (ax)^{12} = \dots = x^3\,,$$
so $x^{21}=e$.
What's more, since you proved $(ax)^3 = (xa)^3$, it follows that
$$x^3 = (a^{-1}xa)^3 = a^{-1}x^3a\,,$$
which implies that $\forall x\in G, x^3a = ax^3$.
Thus $x^{18}=(x^3)^6 = a^3x^9 = x^9$, so $x^9=e$.
Finally $e=x^{21}=x^3$ since $9+9+3=21$.
| {
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Solving $2x + 1 = 11$: Why, when subtracting $1$, do I only do it to a single term on the left but, if dividing by $2$, I must divide both terms? Solving $2x + 1 = 11$ (for example)
Why, when subtracting $1$, do I only do it to a single term on the left but, if dividing by $2$, I must divide both terms ?
| The key thing to realise here is that multiplication is distributive over addition. This is essentially a fancy way of saying that you can expand brackets: $(3+x)\times 5=(3 \times 5)+(x\times 5)$, and $(3+7)\times 4=(3\times 4)+(7\times 4)$, and $(a+y)\times 100=(a\times 100)+(y\times 100)$. In general, the distributive property means that for any numbers $a$, $b$, and $c$, the following law is true:
$$(b+c)\times a=(b\times a)+(c\times a)$$
In your example, we are told that $2x+1=11$. If we multiply both sides by $\frac{1}{2}$ (this is the same as dividing both sides by $2$), then we get that
$$
(2x+1)\times\frac{1}{2}=11 \times \frac{1}{2} \, .
$$
Then, we can simplify the left-hand-side by using the distributive property: $(2x+1)\times\frac{1}{2}=\left(2x \times \frac{1}{2}\right)+\left(1\times \frac{1}{2}\right)=x+\frac{1}{2}$. The right-hand-side simplifies to $\frac{11}{2}$. This means that
$$
x+\frac{1}{2}=\frac{11}{2} \, .
$$
Then, if we subtract $\frac{1}{2}$ from both sides, we get that $x=\frac{11}{2}-\frac{1}{2}=\frac{10}{2}=5$. By contrast, subtraction is not distributive over addition. In other words, the following "law" is bogus:
$$
(b+c)-a=(b-a)+(c-a) \, .
$$
For example, $(5+7)-2$ does not equal $(5-2)+(7-2)$, and $(x+y)-5$ does not equal $(x-5)+(y-5)$. So, if we are told that $2x+1=11$, then we are allowed to subtract $1$ from both sides:
$$
(2x+1)-1=11-1 \, .
$$
What we can't do is "simplify" the left-hand-side by writing $(2x+1)-1$ as $(2x-1)+(1-1)$.
| {
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Probability two blocks have exactly 2 out of 4 properties the same?
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two blocks have exactly two of these four properties the same?
I ended up getting:
$${{\binom{2}{1}\binom{3}{1}\binom{4}{2}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{1}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{1}\binom{4}{1}\binom{5}{2} + \binom{2}{2}\binom{3}{1}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{2}\binom{4}{1}\binom{5}{1}}\over{\binom{120}{2}}} = {{41}\over{238}}$$Is this correct?
EDIT: The answer given here is ${{35}\over{119}} = {5\over{17}}$:
https://web2.0calc.com/questions/probability-question-blocks
Who is correct?
| You have made a simple arithmetic error. I evaluated your expression and I got $\dfrac{5}{17}$.
| {
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Input of a function must give specific output, find number of positive integers that satisfy. Question: The prime factorization of an integer $n \geq 2$ is of the form $({p_1})^{a_1}(p_2)^{a_2} … (p_k)^{a_k}$ where $p_1$, $p_2$, …, $p_k$ are different prime numbers and $a_1, a_2, …, a_k$ are positive integers. Given an input of an integer $n \geq 2$, the Barbeau process outputs the number equal to $n(\frac{a_1}{p_1} + \frac{a_2}{p_2} … + \frac{a_k}{p_k})$. Determine the number of triples (a, b, c) of positive integers such that the Barbeau Process with input $2^a3^b5^c$, outputs $4(2^a3^b5^c)$.
My solution:
First of all, $2, 3$, and $5$ are all primes, which means that we can immediatly use the Barbeau Process.
We know that the input is $2^a3^b5^c$ which means that the output should be equal to $(2^a3^b5^c)(\frac{a}{2} + \frac{b}{3} + \frac{c}{5})$.
So, $(2^a3^b5^c)(\frac{a}{2} + \frac{b}{3} + \frac{c}{5}) = 4(2^a3^b5^c)$.
The $(2^a3^b5^c)$ cancels out on both sides, leaving that $\frac{a}{2} + \frac{b}{3} + \frac{c}{5} = 4$.
This gives that $\frac{15a + 10b + 6c}{30} = 4 \Longrightarrow 15a + 10b + 6c = 120$.
We first realize that since $a, b$, and $c$ are positive integers, and $10b + 6c$ and $120$ are even, $15a$ must also be even.
We also know that $15a < 120$, because otherwise, $b$ and $c$ would not be positive integers.
Testing out the cases of the even values of $a$:
First of all, $a \leq 6$ because of the constraint that $15a < 120$.
$a = 6 \Longrightarrow$ Does not give any values for $b$ and $c$.
$a = 4 \Longrightarrow$ One case, $b = 3, c = 5$.
$a = 2 \Longrightarrow$ Two cases, $b = 6, c = 5$ or $b = 3, c = 10$.
So, there are $\boxed{3}$ total triples $(a, b, c)$; $(4, 3, 5)$, $(2, 6, 5)$, and $(2, 3, 10)$.
Please help me decide if my solution is correct or wrong. Thanks in advance!
| It seems like your solution is valid. However, you should state that $a < 8$ instead of $a \leq 6$ because $120/15 = 8$.
| {
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Given $x+y+z=3$, how do I maximize $xy+yz+zx-xyz$? Given non-negative numbers $x,y,z$ such that $x+y+z=3$. How do I maximize $xy+yz+zx-xyz$. I found out that the maximum is 2 and equality holds when one of the numbers is 1 and the other two sum up to 2 but had no idea how to prove it. Can someone help please?
| We can equivalently want to maximize:
$$
\begin{aligned}
(1-x)(1-y)(1-z)
&=1-\underbrace{(x+y+z)}_{=3}+(xy+yz+zx)-xyz\\
&=-2 + (xy+yz+zx)-xyz\\[3mm]
&\qquad\text{ so let us substitute }\\
X&=1-x\ ,\\
Y&=1-y\ , \\
Z&=1-z\ .\qquad\text{ These variables satisfy:}\\
X+Y+Z &= 3-(x+y+z)=0\ ,\text{ and}\\
(*)\qquad 1&\ge X,Y,Z\ .
\end{aligned}
$$
The maximum of $XYZ$ is of course positive, obtained when we arrange for two of the variables to be negative, and one positive. We will assume (w.l.o.g) for this $X,Y\le 0$, so $Z\ge 0$.
The positive variable $Z$ is then the only one suffering from the restriction $(*)$, and we have only the condition $Z\le 1$. If $Z< 1$, we can change multiplicatively $X,Y,Z$ by multiplying them each by $1/Z$, the sum remains zero, the product gets bigger by the factor $(1/Z)^3>1$. So we may and do assume $Z=1$ while maximizing. This implies $X+Y=-1$, and of course the product $XY$ is maximal for $X=Y=-\frac12$.
We come back to the $(x,y,z)$ world and obtain a maximal value for $x=y=1-\left(-\frac 12 \right)=\frac 32$ and $z=1-1=0$. Since $z=0$, the only term surviving is $xy$, the maximal value is thus $\left(\frac 32\right)^2=\frac 94>2$.
| {
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Solving $\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$ My attempt:
Given,
$$\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$$
Taking $t=|\log_2x|$:
$$\sqrt{t^2-5t+6} \le 2\sqrt{5}$$
I know that squaring both sides will lead to extraneous solutions. In fact, I solved the rest of the problem by squaring sides and got $x \in \left[\dfrac{1}{128},128\right]$ whereas the correct answer according to Wolfram Alpha is $x \in \left[\dfrac{1}{128},\dfrac{1}{8}\right] \cup \left[\dfrac{1}{4}, 4\right] \cup [8,128]$.
What is the correct way of solving this problem that avoids extraneous solutions?
| This inequality is quite fiddly, but with care we can get to the solution. First, note that any inequality of the form $\sqrt{a}\le b$ is equivalent to $a\le b^2$ and $a\ge0$, meaning
\begin{align}
\sqrt{t^2-5t+6}\le2\sqrt{5} &\iff t^2-5t+6\le20 \quad\text{and}\quad t^2-5t+6\ge0 \\[5pt]
&\iff (t+2)(t-7)\le0\quad\text{and}\quad(t-2)(t-3)\ge0 \\[5pt]
&\iff t\in[-2,7]\quad\text{and}\quad t\not\in(2,3) \\[5pt]
&\iff t\in\left([-2,7]\setminus(2,3)\right)\\[5pt]
&\iff t\in[-2,2]\cup[3,7] \, .
\end{align}
Hence, $-2\le|\log_2(x)|\le 2$ or $3\le|\log_2(x)|\le 7$.
If $x\ge1$, then these inequalities become $-2\le\log_2(x)\le2$ (which has the solution $\frac{1}{4}<x\le4$) and $3\le\log_2(x)\le7$ (which has the solution $8\le x\le 128$). So if $x\ge1$, then $\frac{1}{4}\le x\le 4$ or $8\le x\le 128$. This boils down to $1\le x\le 4$ or $8\le x\le128$.
If $0<x<1$, then the inequalities become $-2\le-\log_2(x)\le 2$ (which has solution $\frac{1}{4}\le x\le2$) and $3\le -\log_2(x) \le 7$ (which has solution $\frac{1}{128}\le x\le\frac{1}{8}$). This boils down to $\frac{1}{128}\le x\le\frac{1}{8}$ or $\frac{1}{4}\le x\le1$.
Combining all of these solutions, we find that $\frac{1}{128}\le x\le\frac{1}{8}$ or $\frac{1}{4}\le x\le 4$ or $8\le x\le 128$.
I will give an example of how I solved one of the above inequalities. The rest of them are left as exercises:
\begin{align}
3\le -\log_2(x)\le 7 &\iff -3 \ge \log_2(x) \ge - 7 \\[5pt]
&\iff -7 \le \log_2(x) \le -3 \\[5pt]
&\iff 2^{-7} \le x \le 2^{-3} \\[5pt]
&\iff \frac{1}{128} \le x \le \frac{1}{8} \, .
\end{align}
Finally, there is a simpler way to solve $-2\le|\log_2(x)|\le 2$ or $3\le|\log_2(x)|\le 7$, but it is not obvious. A number $y$ satisfies these inequalities if and only if $\frac{1}{y}$ satisfies them. So once we have the solutions $1\le x \le 4$ or $8 \le x \le 128$, we can easily find the others. Admittedly, I didn't realise this until I solved this inequality the long-winded way.
| {
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Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer I tried to prove this using induction.
Let $k=1$; then the equation gives
$$1/7 + 1/5 +2/3 – 1/105 = 105/105 = 1,$$
which is an integer. So it is true for $k=1$. Now let it be true for $n>k$. This gives
$$105|(15n^7 + 21n^5 + 70n^3 – n).$$
For $(n+1)$ we have
$$105|(15(n+1)^7 + 21(n+1)^5 + 70(n+1)^3 – (n+1)),$$
which is the same as
$$105|(15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175)$$
since $105|(15n^7 + 21n^5 + 70n^3 – n)$.
If $105|(105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175),$
which is the same as if $$105|((105n^6 + 315n^5 + 630n^4 +945n^2 + 630n) + 805n^3 + 175),$$
which is the same as if $$105|((\text{multiple of } 105) + 805n^3 + 175).$$
I’m stuck at this part, because neither $805$ nor $175$ is a multiple of $105$, so how to prove that they are multiples of $105$?
| I think you made a mistake, and it should have been
$15n^7+105n^6+336n^5+630n^4+805n^3+735n^2+419n+105$
where you typed $15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175$,
so it should have been
$105n^6 + 315n^5 + 630n^4 + 735n^3 + 735n^2 + 420n + 105$
where you typed $105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175$.
Can you take it from here?
| {
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Computing $\sum_{k = 1}^{n} \dfrac{k}{2^k} $ in different ways I am trying to compute the following sum \begin{align} S = \sum_{k = 1}^{n} \dfrac{k}{2^k} \end{align}
I have computed this sum and found that
$$ \bbox[5px,border:2px solid red]
{
S = 2 - \dfrac{n+2}{2^{n}}
}
$$
For prooving this let's rewrite the sum in the following way:
\begin{align} S = \dfrac{1}{2} + \dfrac{1 + 1}{2^2} + \dfrac{1+1+1}{2^3} + \ldots + \dfrac{1+ \ldots+1}{2^n} = \bbox[5px,border:2px solid yellow]
{\dfrac{1}{2} + \dfrac{1}{2^2} + \ldots + \dfrac{1}{2^n}} + \bbox[5px,border:2px solid yellow]
{ \dfrac{1}{2^2} + \dfrac{1}{2^3} + \ldots+\dfrac{1}{2^n}} + \ldots + \bbox[5px,border:2px solid yellow]{
\sum_{j = k}^{n}\dfrac{1}{2^j}} + \ldots +\bbox[5px,border:2px solid yellow]{
\dfrac{1}{2^n}} = \sum_{j = 1}^{n}\left( \dfrac{1}{2^{k-1}} - \dfrac{1}{2^n} \right) = 2 - \dfrac{1}{2^{n-1}} - \dfrac{n}{2^n} =2 - \dfrac{n+2}{2^{n}} \end{align}
This computations are rather awfull and i made them just to find the answer, and i am looking for another interesting ways how to compute this sum (especcialy not elementary, using calculus, or number theory, or theory of functions of complex variable, or something else), any ways will be very appriciated!
| First, compute the value of a geometric series
$$
T_n = \sum_{k=1}^n (1/2)^k .
\tag1$$
We can multiply by $1/2$ to get
$$
\frac{1}{2} T_n = \sum_{k=1}^n (1/2)^{k+1} = \sum_{k=2}^{n+1} (1/2)^k
\tag2$$
Compute $(1) - (2)$:
$$
T_n - \frac{1}{2}T_n = \sum_{k=1}^{n} (1/2)^k- \sum_{k=2}^{n+1} (1/2)^k
\\
\frac{1}{2} T_n = (1/2) - (1/2)^{n+1}
$$
so
$$
T_n = 1 - (1/2)^{n}
\tag3$$
Now let's go on to compute
$$
S_n = \sum_{k=0}^n k (1/2)^k
\tag4$$
(Starting at $k=0$ makes no difference.)
As before, multiply by $1/2$
$$
\frac{1}{2} S_n = \sum_{k=0}^n k (1/2)^{k+1} = \sum_{k=1}^{n+1} (k-1) (1/2)^k
\tag5$$
Subtract $(4)-(5)$
$$
S_n - \frac{1}{2} S_n = \sum_{k=0}^n k (1/2)^k - \sum_{k=1}^{n+1} (k-1) (1/2)^k
\\
\frac{1}{2} S_n = 0 + \sum_{k=1}^n (1/2)^k - n(1/2)^{n+1}
\\
\frac{1}{2} S_n = 0 + T_n - n(1/2)^{n+1}
\\
\frac{1}{2} S_n = 0 + 1-(1/2)^n - n(1/2)^{n+1}
$$
so
$$
S_n = 2 -(2+n)(1/2)^n
$$
| {
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Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^2\ge a^2$ also $b+c\ge c+a \ge a+b$. Thus $\frac{1}{b+c}\le \frac{1}{c+a}\le \frac{1}{a+b}$. Hence by Rearrangement Inequality $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is the greatest permutation however I am not able to express $\frac{a^2+b^2+c^2}{2}$ as another permutation of the same, hence I require assistance.
Thank You
| I'm not sure how to use Rearrangement Inequality but I can do it with tangent line method.
Using $a+b+c=3$ we can rewrite OP as: $$\underbrace{{a^2\over 3-a}-{a^2\over 2}}_{f(a)} + {b^2\over 3-b}- {b^2\over 2}+{c^2\over 3-c} -{c^2\over 2} \geq 0$$ If you calculate a tangent at point $x=1$ on $f(x)$ we get $y={1\over 4} (x-1)$ and it is not hard to see that for $x\in (0,3)$ we have $$f(x)\geq {1\over 4}(x-1)$$
Using this we get $$f(a) +f(b)+f(c)\geq {1\over 4}(a-1)+ {1\over 4}(b-1)+{1\over 4}(c-1)=0$$
| {
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Asymptotic approximation for $\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\, dx$ as $a \rightarrow 0$
Asymptotic approximation for $$\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\, dx$$ as $a \rightarrow 0$.
In such an integral, when $a$ approaches 0, the term $a^2x^2$ is always << 1 because x is bounded from $0$ to $1$. Thus, can I directly expand this term $\frac{1}{\sqrt{1-a^2x^2}} \sim (1+\frac{a^2x^2}{2}+O(a^4x^4))$ and integrate term by term multiplying the $\sqrt{1-x^2}$. Would doing this give me a good approximation for the integral? Thanks
| Yes, with such approximation we find that as $a\to 0$,
$$\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\,dx=
\frac{\pi}{4}+\frac{\pi a^2}{32}+O(a^4).$$
You may also obtain a more precise approximation, by using Wallis' integral,
$$\int^1_0 \sqrt{1-x^2}\, x^{2n}\,dx=\int^{\pi/2}_0 \cos^2(t)\, \sin^{2n}(t)\,dt=\frac{\pi}{4^{n+1}(n+1)}\binom{2n}{n}
.$$
Therefore, as $a\to 0$,
$$\begin{align}
\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\,dx
&=\int^1_0 \sqrt{1-x^2}\sum_{k=0}^{2n}\binom{-\frac{1}{2}}{k}(-a^2x^2)^{k}\,dx+O(a^{2n+2})
\\
&=
\frac{\pi}{4}\sum_{k=0}^{2n}\frac{\binom{2n}{n}^2}{16^{n}(n+1)}\, a^{2k}+O(a^{2n+2}).
\end{align}$$
| {
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$(1+A)^n \ge (1+a_1)\cdot(1+a_2)\cdots(1+a_n) \ge (1+G)^n$ $A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that
*
*$(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
*if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$
My try (Ques -1): To prove $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)=1+\sum{a_1}+\sum{a_1}\cdot{a_2}+\sum{a_1}\cdot{a_2}\cdot{a_3}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$
Consider $a_1,a_2,\ldots,a_n$ be $n$ positive real numbers and then apply AM$\ge$GM
$\frac{(a_1+a_2+\cdots+a_n)}{n}\ge(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$ imples $\sum{a_1}\ge n\cdot{G}$ because $G=(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$
again consider $(a_1\cdot{a_2}),(a_1\cdot{a_3}),\ldots({a_1}\cdot{a_n}),(a_2\cdot{a_3}),(a_2\cdot{a_4})\ldots,(a_2\cdot{a_n}),(a_3\cdot{a_4})\ldots(a_{n-1}\cdot{a_n})$ be $\frac{n(n-1)}{2!}$ positive real numbers.
Then Applying AM$\ge$GM , we get,
$\frac{\sum{a_1}\cdot{a_2}}{\frac{n(n-1)}{2!}}$ $\ge$ $\bigl(a_{1}^{n-1}\cdot{a_{2}^{n-1}}\cdots{a_{n}^{n-1}}\bigl)^\frac{2!}{n(n-1)}$ implies $\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$
Similary , $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.
Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge 1+nG+\frac{n(n-1)}{2!}G^2+\frac{n(n-1)(n-2)}{3!}G^3+\cdots+G^n$
Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge (1+G)^n$
To prove $(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$
Consider $(1+a_1),(1+a_2),\ldots,(1+a_n)$ be $n$ positive real numbers and then applying AM$\ge$GM
$\frac{(1+a_1)+(1+a_2)+\cdots+(1+a_n)}{n} \ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$
$\frac{n+a_1+a_2+\cdots{a_n}}{n}\ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$
$(1+A)^n\ge(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)$
My try (Ques -2): To prove $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$
Consider $(k+a_1),(k+a_2)\ldots(k+a_n)$ be $n$ positive real numbers and applying AM$\ge$GM
$\frac{(k+a_1)+(k+a_2)+\cdots(k+a_n)}{n}\ge[(k+a_1)\cdot(k+a_2)\cdots(k+a_n)]^\frac{1}{n}$
$(k+A)^n\ge (k+a_1)\cdot(k+a_2)\cdots(k+a_n)$
To prove $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$
$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)=k^n+k^{n-1}\sum{a_1}+k^{n-2}\sum{a_1\cdot{a_2}}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$
now
$\sum{a_1}\ge n\cdot{G}$,
$\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$
$\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.
Therefore
$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge k^n+nk^{n-1}G+\frac{n(n-1)}{2!}k^{n-2}G^2+\cdots+G^n$
Therefore, $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge (k+G)^n$
My Ques :
*
*Have I solved the questions correctly?
*How to prove $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$
Thanks
| Your approach works fine. I feel I should point out that the second result follows trivially from the first one by thinking about what happens when you multiply all the $a_i$ by $k$. Also, there are a few places you say "integer" when you mean real number.
For the missing step in your method, you can use the same approach you did for the sum of products of $2$ variables for the sum of products of $m$ variables.
To do this, you simply need to count how many terms there are in the sum, which as you have noticed is $n \choose m$, and then (in order to compute the GM of these terms) you need to know how many of these terms the variable $a_i$ appears in. How could you count these? Hint: What are the other variables appearing in a term containing $a_i$?
Then you can simplify the exponent to get the answer you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Question on partial answer of the generalized Catalan's conjecture (case $n =2$) Edited question.
Let $n,m\in\Bbb N$ be integers greater or equal to $2$. If $3\leq m<n$ then there is no $m,n$ such that $m^n -n^m =2$. If $n<m$ and $m$ is prime then there is no $m,n$ such that $m^n-n^m =2$. If $m = 2$ then there is no $n$ such that $m^n- n^m = 2$.
Original question.
Let $n\geq 2$ be an integer and $p$ be a prime number. Then there is no $p,n$ such that $p^n - n^p =2$.
False Proof : Consider the case when $p$ is odd prime. Then
by FLT, $n^{p-1}\equiv 1\bmod p$ so $n^p = p^n-2 \equiv n\bmod p$. $\color{#C00}{\text{Then $n^{p+1} = (p+1)^n -2\equiv n\bmod p+1$ so $n^p\equiv 1\bmod p+1$}}$. $p^n - 2 = (p+1)m+1$ for some $m$ so that $p^n =(p+1)m+3 = pm+m+3$. Hence, $p(p^{n-1}-m) = m+3 = pk_0$ for some integer $k_0$ so that $m = pk_0-3$. Then $p(p^{n-1}-pk_0+3) = pk_0$ so $p^{n-1}-pk_0= k_0-3$. Since, $p(p^{n-2}-k_0) = k_0-3$ and letting $k_0 = pk_1+3$, we get $p(p^{n-2}-pk_1-3) = pk_1$ i.e. $p^{n-2}-pk_1 = k_1+3$. Hence inductively, we conclude that $p(1-k_N) = k_N\pm 3$ for some large $N$. This gives the contradiction.
Rough idea given by @robjohn ♦: We want to show that $p^n-n^p=2$ is impossible, so if $n=kp-2$, consider $p^{kp-2}-(kp-2)^p$
Dividing both sides by $(kp)^p$, we have $\frac{p^{(k-1)p}}{k^pp^2}-\left(1-\frac2{kp}\right)^p$.
If $k=1$, this is $\frac1{p^2}-\left(1-\frac2{p}\right)^p$ which is less than $0$. If $k\ge2$, then $\left(\frac{p^{k-1}}{kp^{2/p}}\right)^p-\left(1-\frac2{kp}\right)^p$ is greater than $0$ and multiplied by $(kp)^p$ is much bigger than $2$.
| If $p$ is prime, then looking at $p^n-n^p=2$ mod $p$ shows that $n\equiv-2\pmod{p}$.
We don't need to consider $p=2$, since then $n$ must be even and then $p^n$ and $n^p$ are divisible by $4$.
$\boldsymbol{n\lt p}$
Since $n\equiv-2\pmod{p}$, if $n\lt p$, we must have that $n=p-2$
$$
\begin{align}
p^n-n^p
&=p^{p-2}-(p-2)^p\tag{1a}\\
&=p^p\left(\frac1{p^2}-\left(1-\frac2p\right)^p\right)\tag{1b}
\end{align}
$$
Explanation:
$\text{(1a)}$: $n=p-2$
$\text{(1b)}$: factor $p^p$ out front
For $p=3$, $\text{(1a)}$ gives $3^1-1^3=2$ (which is the only solution).
$\left(1-\frac2p\right)^p$ is an increasing function:
$$
\begin{align}
\frac{\left(1-\frac2{p+1}\right)^{p+1}}{\left(1-\frac2p\right)^p}
&=\frac{p-2}p\left(\frac{p-1}{p+1}\frac{p}{p-2}\right)^{p+1}\tag{2a}\\
&=\frac{p-2}p\left(1+\frac2{(p+1)(p-2)}\right)^{p+1}\tag{2b}\\[3pt]
&\ge\frac{p-2}p\left(1+\frac2{p-2}\right)\tag{2c}\\[9pt]
&=1\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: $1-\frac2{p+1}=\frac{p-1}{p+1}$ and $1-\frac2p=\frac{p-2}p$
$\text{(2b)}$: $\frac{p-1}{p+1}\frac{p}{p-2}=1+\frac2{(p+1)(p-2)}$
$\text{(2c)}$: Bernoulli's Inequality
$\text{(2d)}$: simplify
Since $\frac1{p^2}$ is a decreasing function, for $p\ge4$, we have that $\frac1{p^2}-\left(1-\frac2p\right)^p\le0$, which implies $p^n-n^p\le0$
Thus, for $n\lt p$, the only solution is $3^1-1^3=2$.
$\boldsymbol{n\gt p}$
Since $n\gt p\ge3$, for $k\ge1$,
$$
\begin{align}
p^{p+k}-(p+k)^p
&\ge p^p\left(p^k-e^k\right)\tag{3a}\\[1pt]
&\ge p^p(p-e)ke^{k-1}\tag{3b}\\[3pt]
&\ge 27(3-e)\tag{3c}\\[3pt]
&\gt7\tag{3d}
\end{align}
$$
Explanation:
$\text{(3a)}$: $\left(1+\frac kp\right)^{p/k}\le e$
$\text{(3b)}$: $p^k-e^k=(p-e)\sum\limits_{j=0}^{k-1}p^{k-j-1}e^j\ge(p-e)ke^{k-1}$
$\text{(3c)}$: $p\ge3$ and $k\ge1$
$\text{(3d)}$: $27(3-e)\approx7.60639$
Therefore, $p^n-n^p\gt7$, and so no solution exists.
| {
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"url": "https://math.stackexchange.com/questions/4216663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that $T(n)= 2^{n+1} - 1$ Show that $T(n)= 2^{n+1} - 1$, for the following,
$$
T(n) = \cases{ 1 & if, $n=0$ \cr
T(n-1) + 2^n & otherwise}
$$
I did it as follows,
\begin{align*}
T(n) = T(n-2) + 2^{n-1} + 2^n \\
T(n) = T(n-3) + 2^{n-2} + 2^{n-1} + 2^n\\
&\vdots\\
T(n) = T(n-(n-1)) + 2^{n-n}+ \dots+ 2^{n-1} + 2^n
\end{align*}
From the last call, I can see that,
\begin{align*}
T(0) = T(1) + 2^{0}+ \dots+ 2^{n-1} + 2^n \\
1 = T(1) + 2^{0}+ \dots+ 2^{n-1} + 2^n\\
1 = T(1) + \sum_{i=1}^{n}2^i\\
1 = T(1) + \frac{2^{n+1}-1}{2-1}\\
\end{align*}
How we can get rid of 1 and T(1) as they appeared in my solution please?
| Your work contains a number of errors.
First of all, your equation $$T(n-1) = T(n-2) + 2^{n-1} + 2^n$$ makes no sense, since $$T(n) = T(n-1) + 2^n \tag{1}$$ from the given definition implies $$T(n-1) = T(n-2) + 2^{n-1}, \tag{2} $$ simply by replacing $n$ with $n-1$.
What you probably intended to write is
$$T(n) = T(n-1) + 2^n = \left(T(n-2) + 2^{n-1}\right) + 2^n,$$ where we "unfolded" the recursion by substituting Equation $(2)$ into Equation $(1)$. If you do it again, we get
$$T(n) = T(n-3) + 2^{n-2} + 2^{n-1} + 2^n.$$
This is what you want to write. The left hand side does not change.
Then a complete unfolding of the recursion gives
$$T(n) = T(0) + 2^1 + 2^2 + \cdots + 2^n,$$ and since $T(0) = 1$, we get $$T(n) = \sum_{k=0}^n 2^k = \frac{2^{n+1} - 1}{2-1} = 2^{n+1} - 1,$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the system of differential equations and plot the curves given the initial conditions. We are given the following system of ordinary differential equations:
$$\dot{x} = x - 4y \quad \dot{y} = x - 2y -4.$$
Thus, the slope of the trajectories is given by
$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{x - 2y -4}{x-4y}.$$
We have the initial conditions $x(0) = 15$ and $y(0) = 5$.
How do I solve this system? The book does not give the solution but claims that we can solve for the trajectories. It does not suggest a method. It also asks what happens for other initial conditions.
I tried treating each separate equation as a linear equation and solved them via integrating factors, but that of course resulted in an integral equation on the right-hand side. I was not sure if that was the right way to go about it.
| Mathematica:
$$\left\{x(t)\to \frac{1}{7} e^{-t/2} \left(56 e^{t/2} \sin ^2\left(\frac{\sqrt{7}
t}{2}\right)-3 \sqrt{7} \sin \left(\frac{\sqrt{7} t}{2}\right)+56 e^{t/2} \cos^2\left(\frac{\sqrt{7} t}{2}\right)+49 \cos \left(\frac{\sqrt{7}
t}{2}\right)\right), y(t) \to \frac{1}{7} e^{-t/2} \left(14 e^{t/2} \sin
^2\left(\frac{\sqrt{7} t}{2}\right)+5 \sqrt{7} \sin \left(\frac{\sqrt{7}
t}{2}\right)+14 e^{t/2} \cos ^2\left(\frac{\sqrt{7} t}{2}\right)+21 \cos
\left(\frac{\sqrt{7} t}{2}\right)\right)\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Stuck solving $\iint \frac{x^2}{(x^2+y^2)}dA$ over the region $R$, $R$ lies between $x^2+y^2=4$ and $x^2+y^2=49$ by changing to polar coordinates. I plugged in for $x=r \cos\theta$ and $y=r \sin\theta$ to change to polar coordinates
I set my bounds as $2 \leqq r\leqq 7$ and $0\leqq \Theta \leqq 2\pi$.
I simplified my integral from
$$\displaystyle\iint\frac{r^2 \sin^2\theta}{(r^2 \sin^2(\theta)+r^2 \cos^2(\theta))}r dA$$ to $$\iint r \sin^2 (\theta) d\theta dr$$. I run into my issue with my first integral, because I get $\frac{-r}{3} \cos^3(\Theta$) and I have to plug in 2$\pi$ and $0$, and I end up with $\frac{-r}{3}+\frac{r}{3}=0$.
| You have a mistake in your integral. First of all, it should have been $x^2 = r^2 \cos^2\theta$.
Also $ \int \displaystyle \sin^2\theta \ d\theta \ne \cos^3\theta$.
Write $\sin^2\theta$ as $\cfrac{1 - \cos2\theta}{2} \ $ or $ \ \cos^2\theta \ $ as $ \ \cfrac{1 + \cos2\theta}{2}$
But there is an easier way to do this. Due to symmetry of the region,
$\displaystyle \iint \cfrac{x^2}{x^2+y^2} \ dA = \iint \cfrac{y^2}{x^2+y^2} \ dA = \cfrac{1}{2} \iint dA$
$\displaystyle = \cfrac{1}{2} \int_0^{2\pi} \int_2^7 r \ dr \ d\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to compute $\int_0^\infty \frac{\log(x)}{\sqrt{x}(x+1)} \ dx$ using the Residue Theorem? I have made the following attempt (I won't put every detail, but if necessary, I will edit the question).
I consider $\Gamma_{R,\varepsilon}$ as the following path:
and consider computing $\int_\Gamma \frac{\log^2(x)}{\sqrt{x}(x+1)} \ dx$:
\begin{equation}
\begin{aligned}
2\pi i\operatorname{Res}\left(\frac{\log^2(x)}{\sqrt{x}(x+1)},-1\right) & = \lim_{\substack{R\to\infty \\ \varepsilon\to0^+}}\int_\Gamma \frac{\log^2(x)}{\sqrt{x}(x+1)} \ dx = \int_0^\infty \frac{\log^2(x)}{\sqrt{x}(x+1)} \ dx - \int_0^\infty \frac{(\log(x)+2\pi i)^2}{\sqrt{x}(x+1)} = \\
& = \int_0^\infty \frac{\log^2(x)-\log^2(x)-4\pi i \log(x) + 4\pi^2}{\sqrt x(x+1)}\ dx = \int_0^\infty \frac{-4\pi i \log(x) + 4\pi^2}{\sqrt x(x+1)}\ dx
\end{aligned}
\end{equation}
Therefore,
\begin{equation}
\begin{aligned}
\int_0^\infty \frac{\log(x)}{\sqrt{x}(x+1)} \ dx & = -\frac{1}{4\pi i}\left( 2\pi i\operatorname{Res}\left(\frac{\log^2(x)}{\sqrt{x}(x+1)},-1\right) - \int_0^\infty \frac{4\pi^2}{\sqrt{x}(x+1)\ dx} \right) \\
& = -\frac 12 \operatorname{Res}\left(\frac{\log^2(x)}{\sqrt{x}(x+1)},-1\right)-\pi i\int_0^\infty \frac{1}{\sqrt{x}(x+1)}\ dx = \\
& = \frac 12 \pi^2i - \pi^2i = -\frac 12\pi^2i
\end{aligned}
\end{equation}
If I compute the original integral in a calculator, the result is $0$. Could anyone please help me out telling me what is wrong with my reasoning?
| On the "lower" part of the branch cut we have
$$\begin{align}
\int_{\gamma_3} \frac{\log^2(z)}{\sqrt{z}(z+1)}\,dz&=-\int_0^\infty\frac{(\log(x)+i2\pi)^2}{\sqrt{xe^{i2\pi}}(x+1)}\,dx\\\\
&=\int_0^\infty \frac{(\log(x)+i2\pi)^2}{\sqrt{x}(x+1)}\,dx
\end{align}$$
This might not be fit for purpose here.
But we also have on the lower branch cut that
$$\begin{align}
\int_{\gamma_3} \frac{\log(z)}{\sqrt{z}(z+1)}\,dz&=-\int_0^\infty\frac{(\log(x)+i2\pi)}{\sqrt{xe^{i2\pi}}(x+1)}\,dx\\\\
&=\int_0^\infty \frac{(\log(x)+i2\pi)}{\sqrt{x}(x+1)}\,dx
\end{align}$$
Therefore, we find that
$$\begin{align}
2\int_0^\infty \frac{\log(x)}{\sqrt{x}(x+1)}\,dx+2\pi i \int_0^\infty \frac{1}{\sqrt{x}(x+1)\,dx}&=2\pi i \text{Res}\left(\frac{\log(z)}{\sqrt{z}(z+1)}, z=-1\right)\\\\
&=2\pi i \frac{\log(e^{i\pi})}{\sqrt{e^{i\pi}}}\\\\
&=i2\pi^2\tag1
\end{align}$$
Equating the real and imaginary parts of the right and left hand sides of $(1)$, we find that
$$\int_0^\infty \frac{\log(x)}{\sqrt{x}(x+1)}\,dx=0$$
and that
$$\int_0^\infty \frac{1}{\sqrt{x}(x+1)}\,dx=\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;(u-1)^2\;du\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u^2-2u+1)\;du\\
=\frac{1}{2}\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\;du\\
=\frac{1}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{1}{3}u^{\frac{3}{2}}+C
$$
So far so good. But when I try this method on the given integral, I get the following:
$$
\int x^2\sqrt{x^2+1}\;dx\\
=\frac{1}{2}\int \sqrt{x^2+1}\;x\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;(u=x^2+1)\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u-1)^\frac{1}{2}\;du
$$
Here is where it falls down. I can't expand the $(u-1)^\frac{1}{2}$ factor like the $(u-1)^2$ factor above was, because it results in an infinite series. I couldn't prove, but I think any even exponent for the $x$ factor outside the square root will cause an infinite series to result. Odd exponents for $x$ will work, since it will cause the $(u-1)$ term to have a positive integer exponent.
How should I proceed? I don't necessarily want an answer. I just want to know if I'm missing something obvious or if it is indeed above first year calculus level and probably a typo on the question.
| One way to proceed is:
\begin{align}
\int x^2 \sqrt{x^2+1} \, dx
&= \frac{1}{2} \int \sqrt{u} \sqrt{u-1} \, du \qquad (u = x^2 + 1) \\
&= - \frac{1}{16} \int \frac{y^8 - 2 y^4 + 1}{y^5} \, dy \qquad (y = \sqrt{u} - \sqrt{u-1}) \\
&= - \frac{y^4}{64} + \frac{1}{64 y^4} + \frac{\log(y)}{8} + C.
\end{align}
Here, to change the integration variable from $u$ to $y$, one needs
\begin{equation}
\frac{dy}{du} = \frac{1}{2} \left( \frac{1}{\sqrt{u}} - \frac{1}{\sqrt{u-1}} \right) = - \frac{y}{2 \sqrt{u} \sqrt{u-1}},
\end{equation}
and also the expression of $u$ in terms of $y$, which can be obtained by squaring the both sides of $\sqrt{u-1} = \sqrt{u}-y$ and then solving it for $\sqrt{u}$:
\begin{equation}
u = \frac{(y^2+1)^2}{4y^2}.
\end{equation}
So, after the variable transformation from $u$ to $y$, one gets a factor $u(u-1)$, which is rewritten as
\begin{equation}
u(u-1) = \frac{(y^2+1)^2(y^2-1)^2}{16 y^4}.
\end{equation}
Now, after performing the integration, one needs the following substitutions
\begin{equation}
y = \sqrt{x^2 + 1} - x , \qquad
y^4 - \frac{1}{y^4} = - 8 x (2 x^2 + 1) \sqrt{x^2 + 1},
\end{equation}
to get the final result in $x$:
\begin{equation}
\int x^2 \sqrt{x^2+1} \, dx
= \frac{1}{8} \left[
x (2x^2+1) \sqrt{x^2+1} + \log(\sqrt{x^2+1} - 1)
\right] + C.
\end{equation}
Note that one can rewrite the log term as:
\begin{equation}
\log(\sqrt{x^2+1}-x) = - \log(\sqrt{x^2+1}+x) = - \sinh^{-1}(x).
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$ has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking for a simialr series solution. I suspect is is of the form
$$x = \sum_{k=0}^\infty \binom{3k+2}{k} \frac{b^{3k+2} }{(3)k+2} $$
| $x^3 - \frac {x}{b} =-1$
Let $x = 2\frac {\sqrt {3b}}{3b}\cos\theta$
$\frac {8\sqrt {3b}}{9b^2} \cos^3\theta - \frac {2\sqrt{3b}}{3b^2}\cos\theta = 1\\
\frac {2\sqrt{3b}}{9b^2}(4\cos^3\theta - 3\cos\theta) = 1\\
\cos3\theta = \frac {3b\sqrt {3b}}{2}\\
\theta = \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right), \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right) + \frac {2\pi}{3}, \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right) + \frac {4\pi}{3} \\
x = 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)\right), 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)+\frac {2\pi}{3}\right), 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)+\frac {4\pi}{3}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integration By Substitution： Why are the two results different? The sphere: $x^2+y^2+z^2\leq a^2$ is intercepted by the cylindrical surface $x^2+y^2=ax$. Calculate the intercepted volume.
Consider the intercepted volume of the upper hemisphere, and then multiply it by 2:
$$D=\{(x,y):x^2+y^2\leq ax\}$$
Now calculate $A$:
$$A=\iint\limits_{D}\sqrt{a^2-x^2-y^2}\text{d}x\mathrm{d}y=\int_{\color{magenta}{-\pi/2}}^{\pi/2}{\text{d}\theta\int^{a\cos\theta}_{0}}r\sqrt{a^2-r^2}\text{d}r
\\= \color{magenta}{2}\int_{\color{magenta}{0}}^{\pi/2}\text{d}\theta\int^{a\cos\theta}_{0}r\sqrt{a^2-r^2}\text{d}r\\
=\frac{2}{3}a^3\int^{\pi/2}_{\color{magenta}{0}}(1-\sin^3\theta)\text{d}\theta=\color{red}{\frac{a^3}{3}(\pi-\frac{4}{3})} $$ if I don't change the lower limit of integral, the result would be:
$$
\frac{1}{3}a^3\int^{\pi/2}_{\color{magenta}{-\pi/2}}(1-\sin^3\theta)\text{d}\theta= \color{red}{\frac{a^3}{3}\pi}
$$
Why does the change in limit affect the result?(The two red parts are different)
| Thanks @Benjamin_Gal.
In the comment of this question, he pointed out:
$$
(\sin^2\theta)^{3/2}=|\sin^3\theta|\ne\sin^3\theta
$$
And the correction would be:
$$
\begin{eqnarray}
A=\iint\limits_{D}\sqrt{a^2-x^2-y^2}\text{d}x\mathrm{d}y&=&\int_{\color{black}{-\pi/2}}^{\pi/2}{\text{d}\theta\int^{a\cos\theta}_{0}}r\sqrt{a^2-r^2}\text{d}r
\\
&=&\frac{1}{3}a^3\int^{\pi/2}_{\color{black}{-\pi/2}}(1-|\sin^3\theta|)\text{d}\theta=\color{black}{\frac{a^3}{3}(\pi-\frac{4}{3})}
\end{eqnarray}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $2^{2n+7} \equiv 2 (mod 3)$ for all integer and non-negative number n. Originally, I want to prove this problem:
for all integer and non-negative number $n$,
$$\frac{1}{3}\times2^{2n+7}+\frac{37}{3} \equiv 0\quad(\text{mod}\space3)$$
I tried like this:
$$\frac{1}{3}\times2^{2n+7}+\frac{37}{3} = \frac{1}{3}(2^{2n+7}+37)$$
therefore $2^{2n+7}+37$ must be $\equiv0\space(\text{mod}\space3)$ and greater than or equal to $3$.
Since $2^{2n+7}+37$ is bigger than $37$(for all $n$), we can ignore the latter.
By the way, $37 \equiv 1(\text{mod}\space3)$. Therefore $2^{2n+7} \equiv 2(\text{mod}\space3)$.
I stuck here. is there any way to prove this?
and following calculation correct?
$$2^{2n+7} = 2^7\times 2^{2n}\quad \text{and}\quad 2^7 \equiv 2 (\text{mod}\space3)$$Therefore,
$$2^{2n+7} =2^7\times 2^{2n} \equiv 2\times 2^{2n} (\text{mod}\space3)$$
| Thanks Will199 and Evariste.
for all integer and non-negative number $n$,
$2^{n}\equiv (-1)^{n} \equiv -1($when n is odd$)$ or $1$(when n is even) $(mod \quad3)$
$2n+7$ is always odd because $2n$ is even and $7$ is odd and even+odd=odd.
So, $2^{2n+7}\equiv (-1)^{2n+7} \equiv -1 \equiv 2 \quad(mod \quad3)$.
| {
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Which solution of this integral $\int{\frac{x^2+1}{x^4-x^2+1}}dx$ is correct? $$\begin{align}
\int{\frac{x^2+1}{x^4-x^2+1}}dx&=\int{\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}}dx\\
&=\int{\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+1}}dx\\
&=\int{\frac{1}{u^2+1}}du \quad(u=x-\frac{1}{x})\\
&=\arctan(u)+C\\
&=\arctan(x-\frac{1}{x})+C
\end{align}$$
The image of the integral is:
When I check it using wolfram, the result is $\arctan(\frac{x}{1-x^2})+C$
https://www.wolframalpha.com/input/?i2d=true&i=%5Cint%5C%28%7B%29Divide%5BPower%5Bx%2C2%5D%2B1%2CPower%5Bx%2C4%5D-Power%5Bx%2C2%5D%2B1%5D%5C%28%7D%29
It seems Wolfram calculates this integral like this:
$$\begin{align}
\int{\frac{x^2+1}{x^4-x^2+1}}dx&=\int\frac{x^2+1}{(x^2-1)^2+x^2}dx\\
&=\int\frac{\frac{x^2+1}{(x^2-1)^2}}{1+\frac{x^2}{(x^2-1)^2}}dx\\
&=\int{\frac{1}{1+u^2}}du \quad(u=\frac{x}{1-x^2})\\
&=\arctan(u)+C\\
&=\arctan(\frac{x}{1-x^2})+C
\end{align}$$
Which solution is correct?
| Both are. Remember that antiderivatives may differ by a constant (hence the $+ C$).
| {
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"answer_id": 1
} |
Given $B=AA^T$ and $C=A^TA$, can we calculate $C$ if we know $B$ but not $A$? Consider $B=AA^T$ and $C=A^TA$ with second-order tensor $A$. With knowledge of $B$, but no knowledge of $A$, can we calculate C?
I suspect it should not be possible since different $A$ probably can result in the same $B$, but on the other hand, by the related formulas for $B$ and $C$, maybe all these $A$ also happen to result in the same $C$? How would I go about proving/disproving something like this?
| Permute the columns of $A$ doesn't change $B$ but may change $C$, so you can't establish a function $C=f(B)$.
The reason that this is true is that $B_{ij}$ is the dot product of the $i$th and $j$th rows whereas $C_{ij}$ is the dot product of the columns. Permutations on columns have no effect on dot product of rows.
MaxD edited: Starting with
$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
we get
$$B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
\begin{pmatrix} a & c \\ b & d \end{pmatrix}=
\begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2 \end{pmatrix}$$
and
$$C=\begin{pmatrix} a & c \\ b & d \end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} =
\begin{pmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{pmatrix}.$$
Switching the two columns and starting with
$$A = \begin{pmatrix} b & a \\ d & c \end{pmatrix}$$
results in
$$B = \begin{pmatrix} b & a \\ d & c \end{pmatrix}
\begin{pmatrix} b & d \\ a & c \end{pmatrix}=
\begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2 \end{pmatrix}$$
and
$$C=\begin{pmatrix} b & d \\ a & c \end{pmatrix}
\begin{pmatrix} b & a \\ d & c \end{pmatrix} =
\begin{pmatrix} b^2+d^2 & ab+cd \\ ab+cd & a^2+c^2 \end{pmatrix}.$$
| {
"language": "en",
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"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Pairs of integers $ (x,m)$ for which $\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$ hold?
Find all pairs of integers $(x,m)$ for which $$\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$$ hold.
I have used this property :
Property:
if $$a+b+c=0 \implies a^3+b^3+c^3=3abc, $$ I come up to the following equation:
$(2m-8)^3=-216(m^2-(x-2)^{2/3})$ , such that $a=-2, b=(\sqrt[3]{x-2}+m)^{1/3}, c=(-\sqrt[3]{x-2}+m)^{1/3}$, I can't solve the last equation however i tried $x$ as a paramater instead of $m$ , The solution from wolfram alpha are $(x,m)=(2,1),(66,4)$, Any Help ?
| Let $y=\sqrt[3]{x-2}$, then we have $$2 = \sqrt[3]{m+y} + \sqrt[3]{m-y}\;\;\;\;\;\;|^3$$ $$8 = m+y +3\sqrt[3]{m^2-y^2}\cdot 2 +m-y$$ and thus $$4-m = 3\sqrt[3]{m^2-y^2} $$
so $$ 64 -48m+12m^2-m^3 = 27m^2-27y^2$$ or $$27y^2 =m^3+15m^2+48m-64$$ and finally $$\boxed{27y^2 = (m-1)(m+8)^2}$$
Since $\gcd(m-1,m+8) \mid 9$ we have $3\mid m+8$ and $3\mid m-1$ and so
$m+8 = 3b$ and $m-1 = 3a$ so $$y^2 = ab^2\implies a=c^2$$ for some $c\in\mathbb{N}$.
Now $\boxed{m = 3c^2+1}$ and $ y= bc = c(c^2+3)$ so $\boxed{x= 2+c^3(c^2+3)^3}$
*
*For $c=0$ we get $m=1$ and $x= 2$
*For $c=1$ we get $m=4$ and $x= 66$
*For $c=2$ we get $m=13$ and $x= 14^3+2$
*...
| {
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"source": "stackexchange",
"question_score": "3",
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Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$$
I tried factoring as $${(a^2-x^2)(a^2+x^2)}$$ $$\text{and}$$ $$(x^2+a^2+ax)(x^2+a^2-ax)$$ but didn't get much after this. I sew that the expression is symmetric in a,x but I don't know how to use that.
| Rewrite the integrand as
$$\frac{a^4 - x^4}{x^4 + a^2x^2 + a^4}
=-1+ \frac{\frac {3a}2(\frac 1a +\frac{a}{x^2})-\frac a2(\frac 1a -\frac{a}{x^2})}{\frac{x^2 }{a^2}+ \frac{a^2}{x^2}+1}
$$
and then integrate as follows
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx
= -x + \frac{3a}2\int \frac{d(\frac xa-\frac ax)}{(\frac xa-\frac ax)^2+3} -\frac{a}2\int \frac{d(\frac xa+\frac ax)}{(\frac xa+\frac ax)^2-1}\\
=-x +\frac{\sqrt3 a}2\tan^{-1}\frac{\frac xa-\frac ax}{\sqrt3}
+\frac a2\coth^{-1}(\frac xa+\frac ax)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231586",
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"source": "stackexchange",
"question_score": "1",
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Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$
$=>S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)}$
$\leq \frac{a^2}{4}.(\frac{1}{2a+b}+\frac{1}{2a+c})+\frac{b^2}{4}.(\frac{1}{2b+c}+\frac{1}{2b+a})+\frac{c^2}{4}.(\frac{1}{2c+a}+\frac{1}{2c+b})$
$=\frac{1}{4}.[a^2.(\frac{1}{2a+b}+\frac{1}{2a+c})+b^2.(\frac{1}{2b+a}+\frac{1}{2b+c})+c^2.(\frac{1}{2c+a}+\frac{1}{2c+b})]$
$\leq \frac{1}{4}.[\frac{a^2}{9}.(\frac{2}{a}+\frac{1}{b}+\frac{2}{a}+\frac{1}{c}) +\frac{b^2}{9}.(\frac{2}{b}+\frac{1}{c}+\frac{2}{b}+\frac{1}{a}) +\frac{c^2}{9}.(\frac{2}{c}+\frac{1}{a}+\frac{2}{c}+\frac{1}{b})]$
$=\frac{1}{36}.[a^2.(\frac{4}{a}+\frac{1}{b}+\frac{1}{c})+b^2.(\frac{1}{a}+\frac{4}{b}+\frac{1}{c})+c^2.(\frac{1}{a}+\frac{1}{b}+\frac{4}{c})]$
$= \frac{1}{36}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) $
We need prove that: $S \le \frac{1}{3}$
$=> S \le \frac{1}{12}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) \le 1 $ <- in there, I don't know how to do that :<
I'm trying to find a solution to continue for me or another solution, Can you help me?
| Another way.
After full expanding we need to prove that:
$$\sum_{cyc}(2a^4b^2+2a^4c^2-4a^3b^3+5a^4bc+4a^3b^2c+4a^3c^2b-13a^2b^2c^2)\geq0,$$ which is true by Muirhead because
$(4,2,0)\succ(3,3,0),$ $(4,1,1)\succ(2,2,2)$ and $(3,2,1)\succ(2,2,2).$
Also, we can see it by AM-GM.
| {
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"source": "stackexchange",
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"answer_id": 2
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Calculating value of $\delta$ for a function $y=\frac1{(1-x)^2}$ unbounded at $x\rightarrow 1$ The function $y=\frac 1{(1-x)^2}$ is unbounded at $x\rightarrow 1$.
How can I calculate the value of $\delta$ such that $y> 10^6$ if $\left | x-1 \right |< \delta$ ?
Taking $y=f(x)=10^6$, putting in ${y=\frac1{(1-x)^2}}$, we get
$10^6 = \frac1{(1-x)^2}\\
\frac1{(1-x)^2} = \frac1{10^6}\\
x=1-\sqrt{(10^{-6})}$ meaning $x= 0.999$.
As, $\left |x-1\right |< \delta,$ hence $ \left|0.999 -1\right| < \delta,$ or
$10^{-3} < \delta$.
It can imply that,
$ \delta$ must be at least $10^{-3}$ or greater $ \delta \geq 10^{-3}$
(After correction by user:505767)
| Pick $\delta$ such that
$$y=\frac{1}{\delta^2}=10^6$$
then at $\delta=±0.001$, $y=10^6$.
So$$|x-1|<\delta=0.001$$
in order to have $$y=\frac{1}{(1-x)^2}>\frac{1}{\delta^2}=10^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)$
$x^4+\dfrac{1}{x^4}=x^4+2+\dfrac{1}{x^4}-2=\left(x^2+\dfrac{1}{x^2}\right)^2-1-1=\left(x^2+\dfrac{1}{x^2}-1\right)\left(x^2+\dfrac{1}{x^2}+1\right)-1$
Because $x^4+\dfrac{1}{x^4}$ is rational number so
$\left(x^2+\dfrac{1}{x^2}-1\right)$ is rational number too and
$x^3+\dfrac{1}{x^3}$ is rational number so
$\left(x+\dfrac{1}{x}\right)$ is rational number.
Am I wrong? Please check my solution, thank you.
| I constructed a possible elementary algebraic way:
Let,
$$\begin{cases}x+\frac 1x=v,\thinspace v \not \in \mathbb Q\\
x^2+\frac{1}{x^2}=u,\thinspace u\in\mathbb R\\
x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\
x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{cases}$$
This implies,
$$pv=u+q\not\in \mathbb Q\implies u\not\in\mathbb Q\\$$
Then we have,
$$\begin{align}&uq=p^2-2+u\\
\implies &uq-u=p^2-2\\
\implies &u=\frac{p^2-2}{q-1}\in\mathbb Q\\
&\thinspace \text{A contradiction.}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}
I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?
| You have shown that $10^n + 3 \cdot 4^{n+2} + 5=3(2+2^{n+4})$ modulo $9$, so it should be enough to show that $2+2^{n+4}=3k$ for some $k$. You can do this by induction.
The base case is obvious, so we will assume $2+2^{2n+4}=3k$ for some n, then:
$$
2+2^{2(n+1)+4}
=
2+4\cdot 2^{2n+4}
=
(2+4(2^{2n+4}+2-2))=
12k-6
=
3(4k-2)
$$
And so, by induction $2+2^{2n+4}=3k$ always holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \right)} \right)^4} + {\left( {f\left( x \right)} \right)^8} + ..$
Let $T = f\left( x \right);U = {\left( {f\left( x \right)} \right)^2};V = {\left( {f\left( x \right)} \right)^4};W = {\left( {f\left( x \right)} \right)^8}$
Let the coefficient of $x^{10}$ in $T $ is zero
Taking U
${\left( {f\left( x \right)} \right)^2} = {\left( {x + {x^2} + {x^4} + {x^8} + ..} \right)^2}$
Hence the coefficient of $x^{10}$ in $U$ is $2x^{10}$ hence $2$
For $V$ and $W$ it is getting complicated hence any short cut or easy method to solve it
| To cut it short, for $x^{10}$ to appear in $f(x)^n$, you just need to have some $2^k$'s add up to 10. It doesn't necessarily require binary expansions.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Why is this approach wrong?
If $\alpha$ and $\beta$ are the roots of $x^{2} - 4 x - 3$, then find $$\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}$$
Solution
This is the approach that gives wrong answer.
\begin{array}{l}
x^{2}-4 x-3=0 \\
(x-4)^{2}=19 \\
So, {(\alpha-4)^{2}} {\&(\beta-4)^{2}}=19 \\
\text { Hence, } \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}=\frac{2}{19}
\end{array}
This is how it is solved :
\begin{array}{l}
x^{2}-4 x-3=0 \\
\alpha^{2}-4 \alpha-3=0 \\
\alpha(\alpha-4)=3 \\
\alpha-4=\frac{3}{\alpha} \\
\& \beta-4=\frac{3}{\beta}
\end{array}
So , $$
\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}
$$
\begin{array}{l}
=\frac{\alpha^{2}}{9}+\frac{\beta^{2}}{9} \\
=\frac{1}{9}\left(\alpha^{2}+\beta^{2}\right) \\
=\frac{22}{9}
\end{array}
Please tell me why the first approach is wrong .
| $x^2-4x-3=(x-2)^2-7$
This is the mistake in your analysis.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding three unknowns from three equations Let $a$,$b$ and $c$ be three positive real numbers such that
$$\begin{cases}3a^2+3ab+b^2&=&75\\
b^2+3c^2&=&27\\c^2+ca+a^2&=&16\end{cases}$$
Find the value of $ ab+2bc+3ca$.
My attempt: I observed that $3 . 16+27=75$. Then on replacing $16$ by $c^2+ca+a^2$, $27$ by $b^2+3c^2$ and $75$ by $3a^2+3ab+b^2$, I got $2c^2+ca=ab$.
But after this I am unable to proceed.
Is there a way to proceed from here?
Any constructive hint is appreciated.
| There are two values of the expression.
The value of
$ ab+2bc+3ca=-24\sqrt{3}$
and
$ ab+2bc+3ca=+24\sqrt{3}$.
The respective values of $a$, $b$ and $c$ are:
$a=+\sqrt{\frac{24576\sqrt{3}}{6553}+\frac{93184}{6553}}$,
$b=+\sqrt{\frac{58995}{6553}-\frac{31104\sqrt{3}}{6553}}$,
$c=-\sqrt{\frac{10368\sqrt{3}}{6553}+\frac{39312}{6553}}$;
and
$a=-\sqrt{\frac{93184}{6553}-\frac{24576\sqrt{3}}{6553}}$,
$b=-\sqrt{\frac{58995}{6553}+\frac{31104\sqrt{3}}{6553}}$,
$c=-\sqrt{\frac{-10368\sqrt{3}}{6553}+\frac{39312}{6553}}$.
| {
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"url": "https://math.stackexchange.com/questions/4241459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differential equation, tricky How would I solve the following:
$(x^2-y^2)dy+(y^2+x^2y^2)dx=0$?
Here is what I did:
$\frac{dy}{dx}(x^2-y^2)=-(y^2+x^2y^2)$
Dividing through, to leave differential on one side:
$\frac{dy}{dx}=\frac{-(y^2+x^2y^2)}{x^2-y^2}$
I then proved this is a homogeneous differential equation.
This is how I Proved it:
Let $A(x,y)=-y^2-x^2y^2$. $A(tx,ty)=-t^2y^2-t^2x^2t^2y^2$ When you factorize, you get:
$t^2A(x,y)$
Do the same for the denominator, and make the conclusion that these are homogeneous.
Is this right?
I let $y=vx$ so that $\frac{dy}{dx}=x\frac{dv}{dx}+v$
However I get stuck from here, as the equation seems to not make sense for.
Anyone to please guide me
| $$
\left(x^2 - y^2\right)dy = \left(y^2 + x^2y^2\right)dx
$$
This becomes
$$
\left(\frac{x^2}{y^2} - 1\right)dy = (1+x^2)dx
$$
$$
\frac{dx}{dy} = \frac{\frac{x^2}{y^2} - 1}{1+x^2}
$$
If we assume that $y = xt^a$ we have
$$
\frac{d}{dy} = \frac{dt}{dy}\frac{d}{dt} = \frac{1}{at^{a-1}x}\frac{d}{dt}
$$
Subbing into the original ode.
$$
\frac{1}{at^{a-1}x}\frac{dx}{dt} = \frac{t^{-2a} -1}{1+x^2}
$$
This becomes separable
$$
\frac{1+x^2}{x}dx = at^{a-1}(t^{-2a} - 1)dt
$$
$$
\ln x + \frac{x^2}{2} = \int at^{-a - 1} -at^{a-1}dt
$$
setting $a = 1$
$$
\int t^{-2} -1dt = -t^{-1} - t + C
$$
Thus we have
$$
\ln x + \frac{x^2}{2} = - \frac{t^2 + 1}{t} + C
$$
converting back we have $y = xt$
$$
\ln x + \frac{x^2}{2} = -\frac{\frac{y^2}{x^2} + 1}{y/x} + C
$$
Shorter approach
$y = xt$
We then have
$$
(x^2 - x^2t^2)dy = (x^2t^2 + x^2x^2t^2)dx\\
x^2(1- t^2)dy = x^2t^2(1 + x^2)dx
$$
We then have
$$
\frac{1-t^2}{t^2}dy = (1+x^2)dx
$$
$$dy = \frac{dy}{dt}dt = xdt$$
Then we have
$$
\frac{1-t^2}{t^2}xdt = (1+x^2)dx \implies \frac{1-t^2}{t^2}dt = \frac{1+x^2}{x}dx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Ben and Jordan have three coins between them. Two of them are fair but one of them has a 4/7 chance of showing heads. Now, Ben and Jordan both flip each coin once and write down the outcomes.
What is the probability they both get the same number of heads?
My approach:
4 possible outcomes: 3H, 3T, 2H&1T, 2T&1H.
$P(3H) = (\frac{1}{2})^2\frac{4}{7} *\binom{3}{1} = \frac {36}{84} $
$P(3T) = (\frac{1}{2})^2\frac{3}{7} *\binom{3}{1} = \frac{27}{84}$
$P(2H,1T) = \frac{2}{3}(\frac{1}{2})^2\frac{4}{7} + \frac{1}{3}(\frac{1}{2})^2\frac{3}{7} = \frac{11}{84}$
$P(2T,1H) = \frac{2}{3}(\frac{1}{2})^2\frac{3}{7} + \frac{1}{3}(\frac{1}{2})^2\frac{4}{7} = \frac{10}{84}$
P(same number of heads) = $(\frac {36}{84})^2 + (\frac{27}{84})^2 + (\frac{11}{84})^2 + (\frac{10}{84})^2 = \frac{1123}{3528} \approx 32\% $
What do you think?
| You don't need to multiply by binomial coefficient or in any other way consider different orders of coins. We can simply say that first and second coins are fair, and the last is weighted.
For example, $3H$ means simply that both fair coins and weighted came heads, thus $P(3H) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{4}{7} = \frac{1}{7}$.
Similarly, $2H, 1T$ means that either both fair coins came heads and weighted came tail, or first fair came tail and the rest came heads, or second fair came tail and the rest came tail, so $$P(2H, 1T) = \frac{1}{2}\cdot \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{4}{7} = \frac{11}{28}$$
After finding correct probabilities for all possible numbers of heads, you need to sum their squarees, as you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{x \to 0} \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}$ without L'Hopital Could you help me with this limit please:
$$
\lim_{x \to 0} \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}
$$
I know that with L'Hopital is easy but I want to do it without that theorem. I have already tried converting it to:
$$
\frac{2\sin^2(x^2/2+x)}{2\sin^2(x^2/2+x/2)}
$$
or expanding the sin of the sum of two angles:
$$
\frac{\sin(x^2/2)\cos(x)+\sin(x)\cos(x^2/2)}{\sin(x^2/2)\cos(x/2)+\sin(x/2)\cos(x^2/2)}
$$
but I can not advance further than that and eliminate the indetermination 0/0. I would really appreciate if you could help me please.
| \begin{align}
\lim_{x\to0}\frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)} &= \lim_{x\to0}\frac{2\sin^2(\frac{x^2+2x}{2})}{2\sin^2(\frac{x^2+x}{2})} \\
&= \lim_{x\to0}\frac{\sin^2(\frac{x^2+2x}{2})}{\left(\frac{x^2+2x}{2}\right)^2}\cdot\frac{\left(\frac{x^2+x}{2}\right)^2}{\sin^2(\frac{x^2+x}{2})}\cdot\frac{\left(\frac{x^2+2x}{2}\right)^2}{\left(\frac{x^2+x}{2}\right)^2} \\
&= 1\cdot1\cdot\lim_{x\to0}{\frac{\left(\frac{x^2+2x}{2}\right)^2}{\left(\frac{x^2+x}{2}\right)^2}}\\
&= \lim_{x\to0}{\frac{x^2(x+2)^2}{x^2(x+1)^2}}\\
\\
&=4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find polynomial with the lowest degree
Find all polynomial $f\in \mathbb{Z}[x]$ with the lowest degree
satisfying the condition: there exists $g,h\in \mathbb{Z}[x]$ such that
$$\big(f(x)\big)^4+2f(x)+2=(x^4+2x^2+2)g(x)+3h(x)$$
This question in Korea Winter Program 2021, a program that selects students to participate IMO, it means just a question for student without field knowledge.
I considered the remainder $\pmod{x^4+2x^2+2}$ in $\mathbb{Z}_3[x]$ and had an ugly solution with a lot of case. I wonder if there is a nice and not too cumbersome solution.
| You probably are allowed to quote Eisenstein criterion here, so $p(x)=x^4+2x+2$ and $q(x)=x^4+2x^2+2$ are irreducible polynomials in $\mathbb{Z}[x]$.
If you have seen solving two polynomials $P(x,y)=Q(x,y)=0$ using resultants, then it is all routine. You can do it without resultants but it will just be messier (though doable) to eliminate $x$.
Clearly $\deg f\leq 3$ since you can always reduce $f$ mod $x^4+2x^2+2$. Also clear that $\deg f=0,1$ don't work. So reduce to either $f(x)=ax^2+bx+c$ or $f(x)=ax^3+bx^2+cx+d$.
If $f(x)=ax^2+bx+c$, then eliminating $x$ from $x^4+2x^2+2=0$, $ax^2+bx+c-y=0$ as equations with mod 3 coefficients gives
$$
\begin{vmatrix}
1&&2&&2\\
&1&&2&&2\\
a&b&c-y\\
&a&b&c-y\\
&&a&b&c-y\\
&&&a&b&c-y\\
\end{vmatrix}=0
$$
or:
$$
y^4
+4(a-c)y^3
+2(4 a^2 + b^2 - 6 a c + 3 c^2)y^2
+4(2 a^3 + 2 a b^2 - 4 a^2 c - b^2 c + 3 a c^2 - c^3)y
+(4 a^4 + 4 a^2 b^2 + 2 b^4 - 8 a^3 c - 8 a b^2 c + 8 a^2 c^2 + 2 b^2 c^2 - 4 a c^3 + c^4)
=0
$$
(Yes, I know this is supposed to be mod 3 so a lot of simplifications are possible here, but for demonstration sake I will not simplify beyond what usual determinant for just this one time.) Since $a,b,c\in\mathbb{Z}/3$, that simplifies to
$$
y^4
+(a-c)y^3
+2(a^2 + b^2)y^2
+\dots
=0
$$
But we want $y$ to also satisfy $y^4+2y+2=0$. This is only possible if $a=c$ (coefficient of $y^3$) and $a^2+b^2=0$ (coefficient of $y^2$) which therefore give $a=b=c=0$ which doesn't work.
So the minimal degree $f$ must be 3, and same procedure gives the 4 possible $f$ mod 3 that field theory says there must be. So we start with
$$
\begin{vmatrix}
1&&2&&2\\
&1&&2&&2\\
&&1&&2&&2\\
a&b&c&d-y\\
&a&b&c&d-y\\
&&a&b&c&d-y\\
&&&a&b&c&d-y\\
\end{vmatrix}=0
$$
or
$$
y^4+(b-d) y^3+\dots=0
$$
as equation mod 3. Coefficient of $y^3$ gives $b=d$, now continue further:
$$
y^4 - (a^2 + b^2 + c^2) y^2 +\dots = 0
$$
Equating coefficient of $y^2$ to 0 says $a,b,c$ are either all zero or all nonzero. So $b=d\neq 0$, $a^2=b^2=c^2=1$, and we have
$$
y^4 + 2 a b c y + 2 = 0
$$
So we need to impose $abc=1$, i.e., an even number of $-1$ there. Lifting the result back to $\mathbb{Z}[x]$, the solutions are $f(x)=ax^3+bx^2+cx+d$ for $a,b,c,d\in\mathbb{Z}$ such that
$$
(a,b,c,d)\equiv (1,-1,-1,-1),(-1,1,-1,1),(-1,-1,1,-1),(1,1,1,1)\pmod{3}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does Diophantine equation $1+n+n^2+\dots+n^k=2m^2$ have a solution for $n,k \geq 2$? When studying properties of perfect numbers (specifically this post), I ran into the Diophantine equation
$$
1+n+n^2+\dots+n^k=2m^2, n\geq 2, k \geq 2.
$$
Searching in range $n \leq 10^6$, $k \leq 10^2$ yields no solution. So I wonder if there are any solutions, and if not, is there some elementary reason for that? Or if it can be converted to some known open problem, that would do too...
Some thoughts: Of course if we allow $n=1$ or $k=1$ we could get trivial solutions. We can also see by mod $2$ that both $n$ and $k$ must be odd. Now we could try to solve some small cases such as $k=3$ or $k=5$... So set $k=3$ and try to solve $$1+n+n^2+n^3=2m^2.$$ Left side factors and hence we want to solve $(n+1)(n^2+1)=2m^2$. Now this imples $4 \mid 2m^2$ and so $m$ is even. Also we can see that $4 \nmid n^2+1$ for any integer $n$ (becase $n^2\equiv 0,1 \pmod 4 $). So all powers of $2$ in $2m^2$ except one will divide $n+1$. So let $m=2^t r$ with $2 \nmid r$, then $2m^2=2^{2t+1}r^2$, $2^{2t} \mid n+1$, $2 \mid n^2+1$. So we can put $n=2^{2t}s-1$ with $2 \nmid s$, substitute it back, divide all powers of $2$ and we have the equation
$$
s(2^{4t-1}s^2-2^{2t}s+1)=r^2.
$$
Now the two expressions in the product are coprime, so they both have to be square, and that is farthest I got so far. Also the expression $2^{4t-1}s^2-2^{2t}s+1$ being square has a solution $s=15,t=2$, and I am yet to find another (but of course $s$ is not a square in this case so it does nothing for the original problem).
| $1+n+n^2+n^3=2m^2$ can be transformed to $y^2 = x^3+2x^2+4x+8$ with $x=2n$ and $y=4m.$
The equation $y^2 = x^3+2x^2+4x+8$ is an elliptic curve which has one torsion point $(x,y)=(-2,0)$ and the rank of curve is $0$.
Since rank is $0$, so there are no rational points of infinite order on the curve.
The only integral point is torsion point $(x,y)=(-2,0)$ .
Thus, there are no positive integral solution.
sage: E = EllipticCurve([0,2,0,4,8])
sage: E.rank()
0
sage: E.torsion_points()
[(-2 : 0 : 1), (0 : 1 : 0)]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Calculate the diagonal $d$ of three equal squares inscribed into a triangle whose sides $a$, $b$ and $c$ (and interior angles) are given We have a triangle, whose sides $a$, $b$ and $c$ (and interior angles $\alpha$, $\beta$ and $\gamma$) are given. Into this triangle three equally sized squares with a side length $r$ and a diagonal $d$ are inscribed as shown by the figure below.
I am searching for a formula depending on the triangle's properties ($a$, $b$, $c$, $\alpha$, $\beta$, $\gamma$) that calculates $d$.
A few basic thoughts about angular relationships have come to my mind:
*
*$360=2\cdot90+\beta+\frac{\gamma'}{2}+90+\frac{\alpha'}{2}=2\cdot90+\gamma+\frac{\alpha'}{2}+90+\frac{\beta'}{2}=2\cdot90+\alpha+\frac{\beta'}{2}+90+\frac{\gamma'}{2}$
*$90=\alpha+\frac{\beta'}{2}+\frac{\gamma'}{2}=\beta+\frac{\gamma'}{2}+\frac{\alpha'}{2}=\gamma+\frac{\alpha'}{2}+\frac{\beta'}{2}$
*$180=2\alpha+\beta'+\gamma'=2\beta+\gamma'+\alpha'=2\gamma+\alpha'+\beta'$
and
*
*$\alpha'+\beta'+\gamma'=90$
*$\alpha+\beta+\gamma=180$
I decompose the triangle sides as follows (see the illustration below):
*
*$c=c_a+c'+c_b$
*$a=a_b+a'+a_c$
*$b=b_c+b'+b_a$
Diagonal and side length of the three (equally sized) squares satisfy $d=r\sqrt{2}$.
Maybe using trigonometric relationships we can deduce an elegant formula that yields $d$ directly when inputting the triangle's parameters.
I kept trying yesterday and attached my sketches here: PDF
| To establish the existence of such a construct and also find a way towards a solution it is useful to consider the inverse construction, starting from a central point $M$, and drawing three squares around it, with angles between the adjacent edges of adjacent squares being $\delta, \epsilon,\zeta$ and the length of the square being $r$. It is easy to see that a triangle with non-zero angles can always be constructed from that configuration, since there are always 3 pairs of adjacent vertices that can be joined by lines. These lines cannot be parallel to one another since $\delta+\epsilon+\zeta=\pi/2$, and therefore the triangles are always non-trivial when the angles are positive.
Now it is easy business to find the properties of the triangle $(a,b,c,\alpha, \beta,\gamma)$, especially the angles. Consider the triangle defined by the line segments $a_1,b_3$ and their endpoints. As noted in the drawing, the three angles are $(\pi/4+\delta/2, \pi/4+\epsilon/2, \gamma)$ and two of them are known, so the third can be computed easily. Then we find consider the other two triangles as well that
$$\gamma=\pi/4+\zeta/2~~,~~\beta=\pi/4+\epsilon/2~~,~~\alpha=\pi/4+\delta/2$$
Now to determine the lengths $a_2,b_2,c_2$ apply the cosine law in the isosceles triangles defined by $(a_2, M)$, $(b_2, M)$, $(c_2, M)$ respectively to obtain
$$a_2=2r\sin\frac{\delta}{2}~~,~~b_2=2r\sin\frac{\epsilon}{2}~~,~~c_2=2r\sin\frac{\zeta}{2}$$
For all other lengths apply the sine law to the triangles defined by $(a_3, c_1), (a_1, b_3), (b_1, c_3)$ and solve to get
$$\begin{align}&a_1=\frac{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}r\sqrt{2}~~,~~ a_3=\frac{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}r\sqrt{2}\\ &b_1=\frac{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}r\sqrt{2}~~,~~ b_3=\frac{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}r\sqrt{2}\\ &c_1=\frac{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}r\sqrt{2}~~,~~ c_3=\frac{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}r\sqrt{2}\end{align}$$
Notably, $a_1a_3=b_1b_3=c_1c_3=2r^2$ (maybe this can be derived using power theorems). Finally putting everything together we get
$$a=r\left(2\sin\frac{\delta}{2}+\frac{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}\sqrt{2}+\frac{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}\sqrt{2}\right)$$
$$b=r\left(2\sin\frac{\epsilon}{2}+\frac{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}\sqrt{2}+\frac{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\zeta}{2}\right)}\sqrt{2}\right)$$
$$c=r\left(2\sin\frac{\zeta}{2}+\frac{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}\sqrt{2}+\frac{\sin\left(\frac{\pi}{4}+\frac{\delta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\epsilon}{2}\right)}\sqrt{2}\right)$$
In the case that the lengths of the triangle $(a,b,c)$ are known instead, one can easily find the angles $(\alpha, \beta, \gamma)$ since they are known to be acute
$$\alpha=\text{arccos}\left(\frac{b^2+c^2-a^2}{2bc}\right)$$
$$\beta=\text{arccos}\left(\frac{c^2+a^2-b^2}{2ac}\right)$$
$$\gamma=\text{arccos}\left(\frac{b^2+a^2-c^2}{2ab}\right)$$
and thus compute $\delta, \epsilon, \zeta$ and then solve for $r$ or $d$. For example
$$d=\frac{a}{\sin\alpha-\cos\alpha+\frac{\sin^2\beta+\sin^2\gamma}{\sin\beta\sin\gamma}}$$
The final formula for the diagonal can be expressed in terms of the lengths explicitly but due to lack of elegance I will avoid writing it down.
Importantly, note that not all triangle side lengths are admissible! Clearly $\alpha,\beta, \gamma \in [\pi/4,\pi/2]$ and that imposes various constraints on the possible values of $a,b,c$. These can be expressed as
$$0\leq b^2+c^2-a^2\leq \sqrt{2}bc$$
plus cyclic permutations of the above relation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the equation $- x^4 - 2 x^3 + 14 x^2 - 2 x - 1=0$ This equation:
$- x^4 - 2 x^3 + 14 x^2 - 2 x - 1 = 0$
has 4 real solutions (I saw it on GeoGebra), but I can't calculate them analytically.
I know the solving formula of the fourth degree equations (https://it.wikipedia.org/wiki/Equazione_di_quarto_grado) :
${\displaystyle x_{1,2}=-{\frac {b}{4a}}-Q\pm {\frac {1}{2}}{\sqrt {-4Q^{2}-2p+{\frac {S}{Q}}}}}$
${\displaystyle x_{3,4}=-{\frac {b}{4a}}+Q\pm {\frac {1}{2}}{\sqrt {-4Q^{2}-2p-{\frac {S}{Q}}}}}$
with
$p={\frac{8ac-3b^{2}}{8a^{2}}}\qquad \qquad {\color {white}.}$
${\displaystyle S={\frac {8a^{2}d-4abc+b^{3}}{8a^{3}}}}$
and
${\displaystyle Q={\frac {1}{2}}{\sqrt {-{\frac {2}{3}}\ p+{\frac {1}{3a}}\left(\Delta _{0}+{\frac {q}{\Delta _{0}}}\right)}}}$
${\displaystyle \Delta _{0}={\sqrt[{3}]{\frac {s+{\sqrt {s^{2}-4q^{3}}}}{2}}}}$
and
${\displaystyle q=12ae-3bd+c^{2}}$
${\displaystyle s=27ad^{2}-72ace+27b^{2}e-9bcd+2c^{3}}$
I calculate "q" and comes "196"
I calculate "s" and comes "3760"
But when I try to calculate Δ₀, in the inner root $ \sqrt{s ^ {2} -4q ^ {3}} $ I get a negative value, precisely "- 15980544"
| First note that
\begin{eqnarray*}
-x^{4}-2x^{3}+14x^{2}-2x-1=0 &\implies& x^{4}+2x^{3}-14x^{2}+2x+1=0\\
&\implies& \left(x+\frac{1}{x}\right)^{2}+2\left(x+\frac{1}{x}\right)-16=0\\
&\implies& u^{2}+2u-16=0.
\end{eqnarray*}
Then we have a quadratic equation with solutions $$u=-1\pm\sqrt{17}.$$
Therefore all the solutions are given by
$$x = \frac{1}{2}\left(-1 \pm \sqrt{17} \pm \sqrt{14 - 2 \sqrt{17}}\right).$$
| {
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Calculating the coefficients of $\left( 1 + x^1 + x^3 + \dots + x^{2k+1} \right)^n$
Let $1 \leq m \leq n(2k+1)$. Calculate the coefficient of $x^m$ in $$\left( 1 + x^1 + x^3 + \dots + x^{2k+1} \right)^n$$
So, what we actually need to compute here is the number of solutions to $a_1 \cdot 1 + a_3 \cdot 3 + \dots + a_{2k+1} \cdot (2k+1) = m$ with nonnegative integers $a_i$ such that $a_1 + a_3 + \dots a_{2k+1} \leq n$.
I was trying to connect this to the number of solutions with nonnegative $y_i$ to $y_1 + y_2 + \dots + y_r = t$ for some $r,t$ which is equal to ${t+r-1 \choose r-1}$but not sure if and how these relate.
How can we proceed from here?
| Complete rewrite, because I didn’t realize it was only the odd terms.
Your version with the $a_i$ does not count the same thing. For example, when $k=1,n=2,m=3,$ there is only one $(a_1,a_3)=(0,1)$, but the coefficient of $x^3$ in $(1+x+x^3)^2$ is two.
This is because your reformulation ignores the order of values, while the the original formulation are ordered solutions.
$$\begin{align}1+x+x^3+\cdots+x^{2k+1}&=1+x(1+x^2+\cdots +x^{2k})\\&=1+x\frac{1-x^{2k+2}}{1-x^2}
\end{align}
$$
Then:
$$\begin{align}(1+x+x^3+\cdots+x^{2k+1})^n&=\sum_{j=0}^n \binom nj\frac{(x-x^{2k+3})^j}{(1-x^2)^j}\\
\end{align}
$$
Then when $j>0,$ $$\frac{1}{(1-x^2)^j}=\sum_{i=0}^{\infty}\binom{i+j-1}{j-1}x^{2i}$$
So: $$\frac{(x-x^{2k+3})^j}{(1-x^2)^j}=\sum_{p=0}^{j}(-1)^p\binom jp x^{j+p(2k+2)}\sum_{i=0}^{\infty}\binom {i+j-1}{j-1}x^{2i}\tag1$$
The coefficient of $x^m$ in $(1)$ is $$\sum_{p,i} (-1)^p\binom jp \binom{i+j-1}{j-1}$$
where the sum runs over the pairs $(p,i)$ such that $j+2(i +p(k+1))=m.$
So $j$ only contributes to $x^m$ if $j\equiv m\pmod 2.$
Given $m-j$ even and $p,$ then $i=\frac{m-j}2-p(k+1).$
Then when $m>0,$ the coefficient of $x^m$ in your original expression is:
$$\sum_{0<j\equiv m\pmod2}\binom nj \sum_p(-1)^p\binom jp \binom{\frac{m+j}{2}-p(k+1)-1}{j-1}
$$
And $$\binom{\frac{m+j}2-p(k+1)-1}{j-1}$$ is the number of non-negative integer solutions to $$y_1+y_2+\cdots+y_j=\frac{m-j}2-p(k+1)$$ which is the number of solutions to $$(2y_1+1)+(2y_2+1)+\cdots +(2y_j+1)=m-p(2k+2)$$
$j$ is the number of non-zero values of the original sum, and $p$ is the number of $x_i$ known to have $x_i>2k+1.$ So the $(-1)^p$ factor can be seen as an inclusion-exclusion “adjustment,” requires to rule out the cases when the $x_i$ are too large.
A simple calculation: $n=3,k=2,m=5$ the:
$$\binom{3}{1}\sum_{p=0}^1(-1)^p\binom{2-3p}{0}=3\tag{j=1}$$
$$\binom33\sum_{p=0}^3(-1)^p\binom 3p \binom{3-3p}{2}=3\tag{j=2}$$
giving a total of $6.$ Corresponding to $$0+0+5,0+5+0,5+0+0,\\1+1+3,1+3+1,3+1+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$P_n$ is the maximum prime factor of $1 + P_1P_2P_3…P_{n-1}$ for any integer n ≥ 2, where $P_1$ = 2. Is there a $P_n =11$? The definition of a sequence ${P_n}$ is as following: $P_1$ = 2, $P_n$ is the maximum prime factor of $1 + P_1P_2P_3…P_{n-1}$ for any integer n ≥ 2. Is there one term equal to 11 in this sequence?
I tried the following
We know that $P_1$ = 2, $P_2$ = 3, and $P_3$ = 7. If there is a $P_n$ = 11, then $1 + P_1P_2P_3…P_{n-1}$ has a maximum prime factor of 11.
Since two consecutive integers will not have common prime factors, $1 + P_1P_2P_3…P_{n-1}$ does not have prime factors of 2, 3, and 7. Therefore, the only possible prime factorization for $1 + P_1P_2P_3…P_{n-1}$ is $5^a11^b$.How should I start from here? Thanks.
| You have verified the claim for $n \leqslant 3$. Now, let $n > 3$ and assume the contrary. As you rightly pointed out, we have:
$$P_1P_2\cdots P_{n-1} + 1 = 5^a \cdot 11^b$$
where $a$ is a non-negative integer and $b$ is a positive integer. Considering the equation modulo $4$, the LHS is $3 \bmod{4}$ since the first term is divisible by $2$ but not $4$. Hence, the RHS must also be $3 \bmod{4}$, which forces $b$ to be an odd integer. Next, consider the equation modulo $3$. The LHS is $1 \bmod{3}$, which forces $a+b$ to be even in the LHS. Since $b$ is odd, $a$ must be odd as well.
Finally, consider the equation modulo $7$. We have:
$$5^a \cdot 11^b \equiv 1 \pmod{7}$$
$$(-2)^a\cdot 4^b \equiv 1 \pmod{7}$$
$$(-1)^a \cdot 2^{a+2b} \equiv 1 \pmod{7}$$
Here, note that since $a$ is odd, we have $2^{a+2b} \equiv -1 \pmod{7}$. However, this has no solutions. This is a contradiction. Hence, $11$ will not appear in the sequence of primes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How can I derive $~\text{opposite}\cdot\sin^{}\left(\theta_{}\right)+\text{adjacent}\cdot\cos^{}\left(\theta_{}\right)=\tan^{}\left(\theta_{}\right)$ Given the below equation .
$$ b \cos^{}\left(\theta_{} \right) = a \cdot \sin^{}\left(\theta_{} \right) $$
I have to derive the below equation .
$$ b \sin^{}\left(\theta_{} \right) + a \cdot \cos^{}\left(\theta_{} \right) = \sqrt{ a ^2 + b^2 } $$
My tries are as below .
$$ \frac{ b }{ a } = \frac{ \sin^{}\left(\theta_{} \right) }{ \cos^{}\left(\theta_{} \right) } $$
$$ \frac{ b }{ a } = \tan^{}\left( \theta_{} \right) $$
$$ \text{adjacent}= a $$
$$ \text{opposite}= b $$
$$ \text{hypotenuse}=\sqrt{ a^2+ b ^2 } $$
$$ b \sin^{}\left(\theta_{} \right) + a \cdot \cos^{}\left(\theta_{} \right) $$
$$ = b \left( \frac{ b \cos^{}\left(\theta_{} \right) }{ a } \right) +a \left( \frac{ a \sin^{}\left(\theta_{} \right) }{ b } \right) $$
$$ = \frac{ b^2\cos^{}\left(\theta_{} \right) }{ a } + \frac{ a ^2 \sin^{}\left(\theta_{} \right) }{ b } $$
I've been got stucked from here .
| Of the last line .
$$ \frac{ b^2\cos^{}\left(\theta_{} \right) }{ a } + \frac{ a ^2 \sin^{}\left(\theta_{} \right) }{ b } $$
Just plug the below ones to the above formula and you can get the RHS of the problem .
$$ \sin^{}\left(\theta_{} \right) = \frac{ b }{ \sqrt{ a^2+b^2 } } $$
$$ \cos^{}\left(\theta_{} \right) = \frac{ a }{ \sqrt{ a^2+b^2 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $ \lim_{x\to \infty} (f(x+1)-f(x))=1$, then $ \lim_{x\to \infty} \frac{f(x)}{x}=1$? Let $f:\mathbb{R} \to \mathbb{R}$ be a fuction such that $\displaystyle \lim_{x\to \infty} (f(x+1)-f(x))=1$. Is it true then that $\displaystyle \lim_{x\to \infty} \frac{f(x)}{x}=1$?
I think it is and here is how I went about it. Let $\varepsilon>0$. Then there is some $\delta_\varepsilon>0$ such that $$1-\varepsilon < f(x)-f(x-1)<1+\varepsilon, \forall x>\delta_\varepsilon.$$
Thus, we may write the following inequalities for an $x> \delta_\epsilon$:
$$1-\varepsilon < f(x)-f(x-1)<1+\varepsilon\\
1-\varepsilon < f(x-1)-f(x-2)<1+\varepsilon\\
\vdots\\
1-\varepsilon < f(\delta_\varepsilon+1)-f(\delta_\varepsilon)<1+\varepsilon$$
and after we sum these up we get that $$(x-\delta_\epsilon)(1-\epsilon)<f(x)-f(\delta_\varepsilon)<(x-\delta_\varepsilon)(1+\varepsilon), \forall x>\delta_\varepsilon.$$
This implies that $$\frac{f(\delta_\varepsilon)+(x-\delta_\varepsilon)(1-\varepsilon)}{x}<\frac{f(x)}{x}<\frac{f(\delta_\varepsilon)+(x-\delta_\varepsilon)(1+\varepsilon)}{x}, \forall x>\delta_\varepsilon.$$
If we take $\displaystyle\limsup_{x\to\infty}$, we get that $1-\varepsilon < \displaystyle\limsup_{x\to\infty} \frac{f(x)}{x}< 1+\varepsilon$, $\forall \varepsilon > 0$, so $\displaystyle\limsup_{x\to\infty} \frac{f(x)}{x}=1$. In the same way we get that $\displaystyle\liminf_{x\to\infty} \frac{f(x)}{x}=1$, so $\displaystyle\lim_{x\to\infty} \frac{f(x)}{x}=1$ as desired.
Is this proof correct? I am a bit unsure that I am taking that $\limsup$ correctly, even though I can't see why it could be wrong.
| Let
$
f(x)=
-\sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)
\cdot
\cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right).
$
Then
\begin{align}
f(x+1)-f(x)
=&
\sin \left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right)
\cdot
\cos\left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right)
-
\sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)
\cdot
\cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)
\\
=&+\sin\left((\lfloor x+1\rfloor - \lfloor x\rfloor)\cdot \dfrac{\pi}{2}\right)
\\
=& +\sin\left( 1\cdot \frac{\pi}{2} \right)
\\
=& 1
\end{align}
and by $-1\leq f(x)=-
\sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)
\cdot
\cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)\leq +1$ we have
$$
\lim_{x\to 0}\frac{f(x)}{x}=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Simplify summation over collection Consider a collection $V_t$ of vectors of dimension $n$ created as follows:
When $n= 3, t=0$
$$V_0=\left\{\begin{pmatrix}0\\0\\0\end{pmatrix}\right\}$$
When $n=3, t=1$
$$V_1=\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}$$
When $n=3,t=2$
$$V_2=\left\{\begin{pmatrix}2\\0\\0\end{pmatrix},\begin{pmatrix}0\\2\\0\end{pmatrix},\begin{pmatrix}0\\0\\2\end{pmatrix},\begin{pmatrix}1\\1\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}1\\1\\0\end{pmatrix},\begin{pmatrix}0\\1\\1\end{pmatrix},\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}0\\1\\1\end{pmatrix}\right\}$$
Basically every the sum of every element within a vector is equal to $t$ and to create $V_{t+1} $ you have to loop through every element of every vector of $V_{t}$ and create a new vector with that element incremented by $1$.
Now consider a sub-collection $F_t$ of $V_t$ that just contains the vectors where every element is at least $1$, evidently when $n=3$, $F_0,F_1,F_2=\emptyset$.
I am trying to find a simpler expression for: $$\sum_{r=\text{Element of }F_t}\prod_{k=1}^nx_k^{r_{k}}$$
for example if $n=2,t=3$:
$$V_3=\left\{\begin{pmatrix}3\\0\end{pmatrix},\begin{pmatrix}0\\3\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix}\right\}$$
$$\implies F_3=\left\{\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix},\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\2\end{pmatrix}\right\}$$
$$\implies \sum_{r=\text{Element of }F_3}\prod_{k=1}^2x_k^{r_{k}}= x_1^2x_2^1 + x_1^1x_2^2 + x_1^2x_2^1 + x_1^1x_2^2+x_1^2x_2^1+x_1^1x_2^2=3x_1^1x_2^2+3x_1^2x_2^1$$
I tried doing it using an expression similar to the multinomial summation however this didn't work since it's not taking into count repeated elements. However I imagine if there is a way to write it with a sum it'll look similar.
| You can find a simple expression for this using the principle of inclusion exclusion to subtract the monomials where some variable has an exponent of zero.
$n=2$: $$(x_1+x_2)^t-x_1^t-x_2^t$$
$n=3$: $$(x_1+x_2+x_3)^t-(x_1+x_2)^t-(x_1+x_3)^t-(x_2+x_3)^t+x_1^t+x_2^t+x_3^t$$
etc. In general, the expression for $n$ is
$$
\sum_{S\subseteq \{1,\dots,n\}}(-1)^{n-|S|} \left(\sum_{i\in S}x_i\right)^t
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability of One Event Less Than Probability of Second Event I am having a bit of trouble proving some cases when one probability is smaller than the other probability for all positive integers $a, b$, so some suggestions would be appreciated.
Here is the problem:
For all $a, b \in \mathbb{Z}^+$, if $P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}$ and $P(B) = \dfrac{a^2 + b^2}{a^2 + 2ab + b^2}$, prove that $P(A) < P(B)$.
So I attempted using cases.
Case 1: If $a = b \neq 0$, then we want to show that $P(A) < P(B)$
\begin{align*}
P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b} = \dfrac{2a^2 - 2a}{4a^2 - 2a} < \dfrac{a^2 + b^2}{a^2 + 2ab + b^2} = \dfrac{2a^2}{4a^2} = \dfrac{1}{2} = P(B)
\end{align*}
Thus, this implies that $P(A) < P(B)$ for all $a, b \in \mathbb{Z}^+$
The next two cases are the cases I am having trouble with.
Case 2: If $a > b > 0$, then we want to show that $P(A) < P(B)$.
Case 3: If $b > a > 0$, then we want to show that $P(A) < P(B)$.
I am not sure how to approach cases 2 and 3. But case 3 should follow from case 2. So some assistance would be appreciated. Thanks
| $$P(A) < P(B)\\
\iff \frac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}<\frac{a^2 + b^2}{a^2 + 2ab + b^2}\\
\iff \frac{a^2+b^2-(a+b)}{(a+b)^2-(a+b)}-\frac{a^2+b^2}{(a+b)^2}<0\\
\iff \frac{(a^2+b^2-(a+b))(a+b)-(a^2+b^2)((a+b)-1)}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{(a^2+b^2)(a+b)-(a+b)^2-(a^2+b^2)(a+b)+(a^2+b^2)}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{-2ab}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{ab}{(a+b)^2((a+b)-1)}>0.$$
For $a,b\in\mathbb Z^+,$
$ab,\,(a+b)^2,\,((a+b)-1)$ are each positive (noting that $a+b\geq2$),
so that last inequality is true,
so $P(A)<P(B).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Limit in two variables using epsilon-delta In a class assignment, I was asked to find the following limit:
$$\underset{(x,y)\to (0,0)}{\lim}\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}$$
I computed it to be $0$ numerically and as such, I wanted to prove it using the $\varepsilon-\delta$ definition, but I have not been able to find a $\delta$ such that $\left\vert{\sqrt{x^2+y^2}}\right\vert < \delta$ implies $\left\vert{\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}}\right\vert < \varepsilon$. I also tried using the squeeze theorem, but I have not been able to find an upper-bound for $\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}$ that does not go to infinity as $(x,y)$ goes to $(0,0)$.
Any help or clue would be greatly appreciated. Thank you in advance!
| We can use that
$$\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}=\frac{1-\cos(x^2\cdot y)}{(x^2\cdot y)^2}\frac{(x^2\cdot y)^2}{x^6+ y^4}$$
with
$$\frac{1-\cos(x^2\cdot y)}{(x^2\cdot y)^2} \to \frac12$$
then for a proof by $\varepsilon-\delta$ it suffices to consider
$$\underset{(x,y)\to (0,0)}{\lim}\frac{(x^2\cdot y)^2}{x^6+ y^4}$$
and we can use that by AM-GM
$$\frac{(x^2\cdot y)^2}{x^6+ y^4}\le \frac{(x^2\cdot y)^2}{2|x|^3 y^2}=\frac{|x|}2 \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $\log_{1/3} \left(x^3 - 1\right) > \log_{1/3} \left(|3x^2 - 3x|\right)$ I tried over and over again but I simply can't solve:
$$\log_{1/3} \left(x^3 - 1\right) > \log_{1/3} \left(|3x^2 - 3x|\right)$$
So, if someone could please explain to me how to solve this step by step, I would be very greatful. By the way, the solution for this problem is $x \in \varnothing$.
Edit
This is what I tried to do:
$$\log_\frac{1}{3} \left(x^3 - 1\right) > \log_\frac{1}{3} \left(|3x^2 - 3x|\right) $$
to find the domain
$$\begin{cases}
x^3-1>0 \implies x^3>1 \implies x>\sqrt[3] 1 \implies x>1 \\
3x^2 - 3x>0 \implies \ldots \implies x\in(-\infty,0)\cup(1,\infty)\\
-3x^2 + 3x<0 \implies \ldots \implies x\in(-\infty,0)\cup(1,\infty)
\end{cases}$$
then domain is $x>1$.
We need to solve
$$x^3 - 1 <|3x^2 - 3x| \implies
\begin{cases}
x^3 - 1 <3x^2 - 3x \quad (1)\\
x^3 - 1 <-3x^2 + 3x \quad (2)
\end{cases}$$
from $(1)$ we otain
$$x^3 -3x^2 + 3x-1<0 \implies x<1$$
from $(2)$ we otain
$$x^3 +3x^2 - 3x-1<0 \implies x\in(-\infty,-2-\sqrt 3)\cup(-2-\sqrt 3,1)$$
| We have that $\log_{\frac 13}x$ function is strictly decreasing then
$$\log_\frac{1}{3} \left(x^3 - 1\right) > \log_\frac{1}{3} \left(|3x^2 - 3x|\right) \iff x^3 - 1 <|3x^2 - 3x|$$
under the conditions
*
*$x^3 - 1>0 \iff x>1$
*$|3x^2 - 3x| \neq 0 \iff x\neq 0 \quad x \neq 1$
that is $x>1$ therefore
$$x^3 - 1 <|3x^2 - 3x| \iff x^3 - 1 <3x^2 - 3x \iff x^3-3x^2+3x - 1=(x-1)^3<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Zeros of a polynomial in a simple form Let $m\geq 3$ be a natural number and define
$$f(x)=1-x^m-(1-x)(1+x)^{m-1}.$$
Then, if $m$ is even, the only real roots of $f$ are $x=-1,0,1$ and if $m$ is odd, the roots are $x=0,1$.
My work: for the case where $m$ is even, i proved that
$$f(x)=\sum_{k=0}^{m-2}\left(\begin{pmatrix}
m-1\\
k+1
\end{pmatrix}-
\begin{pmatrix}
m-1\\
k
\end{pmatrix}\right)x^{m-k-1}$$
on one hand. Decomposing it as $f(x)=(x-1)x(x+1)P(x)$, where
\begin{align*}
P(x)=\sum_{k=0}^{m-4}b_kx^{m-k-4},
\end{align*}
I get an alternative representation
\begin{align*}
f(x)=\sum_{k=0}^{m-2}(b_k-b_{k-2})x^{m-k-1},
\end{align*}
with $b_{-2}=b_{-1}=b_{m-2}=b_{m-1}=0$. Combining both I get
\begin{align*}
b_k-b_{k-2}=\begin{pmatrix}
m-1\\
k+1
\end{pmatrix}-
\begin{pmatrix}
m-1\\
k
\end{pmatrix}.
\end{align*}
From here I can clearly see that $b_k\geq 0$, however this does not imply that $P>0$ as intended. I don't know how to proceed, any suggestions please? It is entirely possible that I approached the problem from the wrong perspective...
| I was able to prove that $f$ has no more positive roots. Indeed, I needed also to study the sign of this polynomial, so I end up proving a bit more. I used the following reasoning:
\begin{align*}
f(l)=&1-l^m-(1-l)\sum_{k=0}^{m-1}
\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k-1}\\
=&1-l^m-\left[\sum_{k=0}^{m-1}\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k-1}-\sum_{k=0}^{m-1}\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k}\right]\\
=&1-l^m-\left[1+\sum_{k=0}^{m-2}\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k-1}-\left(\sum_{k=1}^{m-1}\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k}+l^m\right)\right]\\
=&\sum_{k=0}^{m-2}\left(\begin{pmatrix}
m-1\\
k+1
\end{pmatrix}-
\begin{pmatrix}
m-1\\
k
\end{pmatrix}\right)l^{m-k-1}\\
=&\sum_{k=0}^{m-2}\frac{m-2k-2}{k+1}
\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k-1}
\end{align*}
We will study the sign of these coefficients. It suffices to study the sign of $a_k:=m-2k-2$ for $k=1, ..., m-2$. If $m$ is even, then $a_k>0$ for $k=0, ..., m/2-2$, $a_{m/2-1}=0$ and $a_k<0$ for $k=m/2, ..., m-2$. On the other hand, if $m$ is odd, then $a_k>0$ for $k=0, ..., (m-3)/2$ and $a_k<0$ for $k=(m-1)/2, ..., m-2$.
The important property is that these coefficients only change sign once, hence by Descartes' rule of signs, $f$ has, at most, one positive root. Since clearly $f(1)=0$ and $f(0)=0$, $f$ does not change sign in the interval $(0,1)$. We will now proceed to prove it is negative in this interval.
Going back to the equality
\begin{align}\label{Equation_expression_polynomial}
f(l)=\sum_{k=0}^{m-2}\left(\begin{pmatrix}
m-1\\
k+1
\end{pmatrix}-
\begin{pmatrix}
m-1\\
k
\end{pmatrix}\right)l^{m-k-1}.
\end{align}
Now we observe the following. If $m\geq 4$ is even, then $f(-1)=f(0)=f(1)$, hence we can write
\begin{align*}
f(l)=(l-1)l(l+1)P(l).
\end{align*}
we will prove that $P(l)>0$, which will imply $f(l)\leq 0$ for $0\leq l \leq 1$. Let
\begin{align*}
P(x)=\sum_{k=0}^{m-4}b_kx^{m-k-4},
\end{align*}
then one can write $f$ in the alternative form
\begin{align*}
f(x)=\sum_{k=0}^{m-2}(b_k-b_{k-2})x^{m-k-1},
\end{align*}
with $b_{-2}=b_{-1}=b_{m-2}=b_{m-3}=0$. Comparing with \eqref{Equation_expression_polynomial}, we get
\begin{align*}
b_k-b_{k-2}=\begin{pmatrix}
m-1\\
k+1
\end{pmatrix}-
\begin{pmatrix}
m-1\\
k
\end{pmatrix}.
\end{align*}
Satisfying the symmetry $b_k=b_{m-k-4}$. Indeed, recalling that
\begin{align*}
c_k:&=\begin{pmatrix}
m-1\\
k+1
\end{pmatrix}-
\begin{pmatrix}
m-1\\
k
\end{pmatrix}=\frac{m-2k-2}{k+1}
\begin{pmatrix}
m-1\\
k
\end{pmatrix}\\
&=\frac{a_k}{k+1}
\begin{pmatrix}
m-1\\
k
\end{pmatrix},
\end{align*}
we can use the obvious anti-symmetry $c_k=-c_{m-k-2}$ and an inductive reasoning to imply $b_k=b_{m-k-4}$.
Now we can prove again by inductive reasoning that $b_k>0$, hence $P(l)>0$ for $l\in(0,1)$. This finally implies that $f(l)\leq 0$ for $l\in[0,1]$, as intended.
| {
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$10$ students ordering $10$ different combinations of $5$ dishes so that each dish appears in at least $1$ combination
In how many ways can $10$ students in the queue for sandwiches order $10$ different combinations of $5$ side dishes so that every dish is in at least $1$ combination? Every combination can consist of $0$ to $5$ (included) dishes.
I tried to find some more general set of examples, so to speak. I viewed different combinations as different sets with at most $5$ different elements/dishes. Those sets are different if they either have different cardinality or equal cardinality, but different elements.
*
*$\binom{5}5=1\implies$ at most $1$ student could order all $5$ dishes.
*$\binom{5}4=5\implies$ at most $5$ students can order $4$ dishes.
*$\binom{5}3=10\implies$ all $10$ students can order $3$ dishes.
*$\binom{5}2=10\implies$ all $10$ students can order $2$ dishes.
*$\binom{5}1=5\implies$ at most $5$ students can order only $1$ dish.
*$\binom{5}0=1\implies$ at most $1$ student can order nothing
Let $x_i\ge 0$ be the number of students who ordered $i$ dishes, $i\in\{0,1,\ldots,5\}.$ Then, we have $$x_0+x_1+x_2+x_3+x_4+x_5=10$$ and the following restrictions $$x_0\le 1, x_1\le5,x_2\le10,x_3\le10,x_4\le 5,x_5\le1.$$
Since $x_i\le\text{something},$ I couldn't use the substitution $t_0=x_0-\text{something},$ so, I instead tried to use the method of complements, hence, for example, instead of $x_3\le 10$ I would take $x_3\ge 11,$ analogously for all $i$ sum the solutions an substract them from the solutions of $x_0+x_1+x_2+x_3+x_4+x_5=10,$ with no restrictions other than $x_i\ge 0$ and make use of the result:
There are $\binom{n+k-1}k\quad n$ tuples of nonnegative integers satisfying $x_i\ge0\forall i\in\{1,\ldots,n\}.$
However, something went wrong here and I have problems with every dish being in at least $1$ combination and I don't know how to proceed. How should I deal with this problem?
| To be honest , as i wrote in comment section , i could not clearly understand the question.Maybe ,it is because of my english. Anyway , it seems that OP understood their question and produced a solution way ,but he suffer from calculating the equation of $$x_0 +x_1 +x_2 +x_3 +x_4 +x_5 =10$$ where $x_i \geq 0 , x_0 \leq 1 , x_1 \leq 5 ,x_2 \leq 10 , x_3 \leq 10 ,x_4 \leq 5 , x_5 \leq 1$
He tried "complemet " method , but he cannot achieve it , so i recommend him to use generating functions such that
*
*Generating function of $x_0$ is equal to $$x^0 +x^1 = \frac{x^2-1}{x-1}$$
*Generating function of $x_1$ is equal to $$x^0 +x^1 +x^2 +x^3 +x^4 +x^5= \frac{x^6-1}{x-1}$$
*Generating function of $x_2$ is equal to $$x^0 +x^1+..+x^9 +x^{10} = \frac{x^{11}-1}{x-1}$$
*Generating function of $x_3$ is equal to $$x^0 +x^1+..+x^9 +x^{10} = \frac{x^{11}-1}{x-1}$$
*Generating function of $x_4$ is equal to $$x^0 +x^1 +x^2 +x^3 +x^4 +x^5= \frac{x^6-1}{x-1}$$
*Generating function of $x_5$ is equal to $$x^0 +x^1 = \frac{x^2-1}{x-1}$$
Now , find the coefficient of $x^{10}$ in the expansion of $$\frac{x^2-1}{x-1} \times \frac{x^6-1}{x-1} \times \frac{x^{11}-1}{x-1} \times \frac{x^{11}-1}{x-1} \times \frac{x^6-1}{x-1} \times \frac{x^2-1}{x-1} $$
CALCULATION
So , answer is $721$
$\mathbf{\text{EDITION=}}$ I now understand that question ask that how many ways are there to distribute $5$ different dishes to $10$ students such that a student can take at least $0$ and at most $5$ dishes. It is classical distributing distinct objects into distinct boxes problem. To solve it , the most suitable method is exponential generating functions.
Because of the restriction , we know that the exponential generating function of each student is equal to $$\bigg(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\bigg)$$
Now , because of there are $10$ students , find the expansion of $$\bigg(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\bigg)^{10}$$
After you find the expansion , find the coefficient of $\frac{x^5}{5!}$ or find the coefficient of $x^5$ and multiply it by $5!$
LOOK AT FOR COMPUTATION
Then , the answer is $$\frac{2500}{3} \times 5! = 100,000$$
| {
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Find the partial fraction: $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ I'm trying to find the partial fraction for the following $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ by the process of long-division, here's what I have tried:
Leaving $(x-3)$ as it is in the denominator and aiming for $(x-1)$ and $(x-2)$.
First step:
$\frac{1^2}{(x-1)^2(1-2)^2(1-3)} = \frac{1}{-2(x-1)^2}, \frac{2^2}{(2-1)^2(x-2)^2(2-3)} = \frac{4}{-(x-2)^2}$
We should then substract this from the first equation to get:
$\frac{x^2}{(x-1)^2(x-2)^2(x-3)} - \frac{1}{-2(x-1)^2} - \frac{4}{-(x-2)^2} = \frac{9x^3-36-45x^2+72x}{2\left(x-1\right)^2\left(x-2\right)^2\left(x-3\right)}$
Taking synthetic division on the numerator I find that 2 and 3 is not a root as there exists remainders. Have I approached this step in the wrong way?
| Hint: $f(x) = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x-2}+\dfrac{D}{(x-2)^2}+\dfrac{E}{x-3}$. And you have to find these constants $A,B,C,D,E$.
| {
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Prove that $|x-2y|$ is the squaring number with $x^2-4y+1=2(x-2y)(1-2y)$ For $x,y$ are positive integer, statisfying $x^2-4y+1=2(x-2y)(1-2y) \,\,\ (1)$. Prove that $|x-2y|$ is a square number.
I try to compute, from $(1)$ We have $(1) \Leftrightarrow x^2+4xy-2x-8y^2+1=0$ and $x=\dfrac{1}{4} \left(x+\sqrt{3x^2-4x+2}\right)$ or $x=\dfrac{1}{4} \left(x-\sqrt{3x^2-4x+2}\right)$. I can't show $|x-2y|$ is a square number.
Another way, I see $x=1$ and $y=0$ is a solution of $(1)$.
| Continuing from what you've already stated,
$$\begin{equation}\begin{aligned}
x^2 + 4xy - 2x - 8y^2 + 1 & = 0 \\
x^2 - 2x + 1 & = 8y^2 - 4xy \\
(x - 1)^2 & = (2^2)y(2y - x)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This shows $y \mid (x-1)^2$, so $\gcd(x, y) = 1$. Furthermore,
$$\gcd(y, 2y - x) = \gcd(y, (2y - x) - 2y) = \gcd(y, -x) = 1 \tag{2}\label{eq2A}$$
Can you finish the rest (note $y$ must also be a square number)?
| {
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Find all $f$ which satisfies the condition: $f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy$ $$ f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy $$
$$
P(x, 0): f(0)=0 \\
P(0, y): f(f(y)-f(-y))=-y^2\\
P(x, x): f(f(2x))=x^2 \\
P(x, -x): f(-f(2x))=-x^2 \\
P(-x, x): f(-f(-2x))=-x^2 \\
P(-x, -x): f(f(-2x))=x^2 \\
$$
I don't have an idea of the function $f$, but I think we can show that $f$ is an even function.
$$
P(f(x), f(y)): f(f(f(x)+f(y))-\frac{x^2-y^2}{4})=f(x)f(y) \\
P(f(y), f(x)): f(f(f(x)+f(y))+\frac{x^2-y^2}{4})=f(x)f(y) \\
P(f(x), -f(y)): f(\frac {x^2-y^2}{4} - f(f(x)+f(y)))=-f(x)f(y) \\
P(f(y), -f(x)): f(-\frac {x^2-y^2}{4} - f(f(x)+f(y)))=-f(x)f(y) \\
$$
Please give a perfect process of getting the functions.
| Let us rewrite the problem as $f(f(a)-f(b)) = \frac{a^2-b^2}{4}$.
Suppose that there exists a solution;
Claim $1)$ If $f(u) = f(v)$ then $u = \pm v$.
Proof; As you noted (taking $a=b$) we have that $f(0) = 0$. Now
$$0 = f(0) = f(f(u)-f(v)) = \frac{u^2-v^2}{4} \Rightarrow u = \pm v.$$
Claim $2)$ $f(u) = f(-u)$.
Proof; Set $s = f(u)-f(-u)$. We have
$$f(s) = f(f(u)-f(-u)) = 0$$
From $f(0) = 0$ and Claim $1)$ we must have that $s = \pm 0 = 0$.
Now for all $a \in \mathbb{R}$ we have
$$\frac{a^2}{4} = \frac{a^2-0^2}{4} = f(f(a)-f(0)) = f(f(a))$$
$$-\frac{a^2}{4} = \frac{0^2-a^2}{4} = f(f(0)-f(a)) = f(-f(a)) $$
which contradicts claim 2) when we consider $a \neq 0$.
| {
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Counting solutions to $x^2+y^2 \equiv d \pmod{p}$ for a prime $p \equiv 3 \pmod 4$ For any given $p \equiv 3 \pmod{4}$ and $d=1, 2, \dots, p-1$, we would like to show that there are always exactly $p+1$ solutions to $x^2 + y^2 \equiv d \pmod{p}$.
This conjecture comes from some numerical experiments, and certainly holds for all small primes $p \equiv 3 \pmod{4}$.
We can show this in the special case when $d$ is a quadratic residue $\pmod{p}$, but are stuck on the general case.
| alright, there are $p^2$ ordered pairs. The only pair giving $x^2 + y^2 = 0$ is $(0,0)$ because $-1$ is not a square.
For each of $(p-1)/2$ residues, there are $p+1$ pairs, total is $ \frac{p^2 - 1}{2}$
Thus $0$ and residues total to $ \frac{p^2 + 1}{2}$
Finally, nonresidues total to $ \frac{p^2 - 1}{2},$ and there are $(p-1)/2$ non-residues. The count for each non-residue is also $p+1$
P.S. the reason the counts are all the same for nonresidues: let $n_1, n_2$ be nonresidues, so there is some nonzero $a$ with $n_2 = n_1 a^2.$ Then we have a bijection of $(x,y)$ pairs: $x^2 + y^2 = n_1 $ maps to $(ax)^2 + ( ay)^2 = n_1 a^2 = n_2$
| {
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Find the range of the function $f(x) = \cos(x)[\sin(x) + \sqrt{\sin^2(x) + \frac{1}{2}}]$
$f(x) = \cos(x)[\sin(x) + \sqrt{\sin^2(x) + \frac{1}{2}}]$
First of all, I tried to convert $\sqrt{\sin^2(x) + \frac{1}{2}}$ into perfect square, I tried many attempts but I failed to do so.
Next, I thought of differentiating it, and the find the maxima/minima, but the expression I got after differentiation was pretty horrible
and in the last I thought of a step wise process:
*
*$0 \le \sin^2(x) \le 1$
*$\frac{1}{2} \le \sin^2(x)+\frac{1}{2} \le
\frac{3}{2}$
*$\sqrt{\frac{1}{2}} \le \sqrt{\sin^2(x)+\frac{1}{2}} \le
\sqrt{\frac{3}{2}}$
and I was stuck here. Please try to provide hint, rather than full solution. I would like to try it myself.
Thanks in advanced.
| $$f(x)-\cos x\sin x=\cos x\sqrt{\frac{1}{2}+\sin^2 x}\Rightarrow
f(x)^2-2f(x)\cos x\sin x=\frac{1}{2}\cos^2 x\Rightarrow\\ 4f(x)\cos x\sin x=2f(x)^2-\cos^2 x\Rightarrow 16f(x)^2\cos^2 x(1-\cos^2 x)=(2f(x)^2-\cos^2 x)^2$$
$$f(x)^2=y,\ \cos^2 x=t:$$
$$16yt(1-t)=(2y-t)^2\Rightarrow 4y^2+(16t^2-20t)y+t^2=0\Rightarrow\\ 8yy'+(16t^2-20t)y'+(32t-20)y+2t=0$$
Here $y'=\frac{dy}{dt}$. In critical points $\frac{df(x)}{dx}=0 \Rightarrow \frac{dy}{dt}=0$
$$(32t-20)y+2t=0\Rightarrow y=-\frac{t}{16t-10} \Rightarrow \frac{4t^2}{(16t-10)^2}-\frac{t(16t^2-20t)}{16t-10}+t^2=0\Rightarrow\\ 4t^2-t(16t^2-20t)(16t-10)+t^2(16t-10)^2=0 \Rightarrow 32t^2(5t-3)=0$$
$t=0$ gives minimum $y=0$. $t=\frac{3}{5}$ gives maximum $y=\frac{3}{2}$.
Minimum $f(x)=-\sqrt{\frac{3}{2}}$ is obtained at $\cos x=-\sqrt{\frac{3}{5}}$, $\sin x=\sqrt{\frac{2}{5}}$.
Maximum $f(x)=\sqrt{\frac{3}{2}}$ is obtained at $\cos x=\sqrt{\frac{3}{5}}$, $\sin x=\sqrt{\frac{2}{5}}$.
| {
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Determining the period of $ \frac{\sin(2x)}{\cos(3x)}$ I would like to compute the period of this function which is a fraction of two trigonometric functions. $$ \frac{\sin(2x)}{\cos(3x)}$$
Is there a theorem for this? what trick to use to easily find the period?
I started by reducing the fraction but I'm stuck on the rest.
For example, let $T$ be the period to be calculated: $$\frac{\sin(2x)}{\cos(3x)} =\frac{\sin(2x + 2 T)}{\cos(3x+3T)} = \frac{\sin(2x) \cos(2T)+\sin(2T) \cos(2x)}{\cos(3x) \cos(3T)-\sin(3T)\sin(3x)}$$
Thanks for your help.
| $$\frac{\sin(2x)}{\cos(3x)}=\frac{2\sin(x)\cos(x)}{4\cos^3(x)-3\cos x}=\frac{2\sin(x)}{4\cos^2(x)-3}$$
$\sin x$ and $\cos x$ has a period of $2\pi$. Therefore $\frac{2\sin(x)}{4\cos^2(x)-3}$ has a period of $2\pi$.
The fundamental period must be $2\pi/n$ where $n\in\mathbb{N}$. We have
$$\frac{2\sin(x)}{4\cos^2(x)-3}=\frac{2\sin(x+2\pi/n)}{4\cos^2(x+2\pi/n)-3}$$
Substituting $x=\frac{(n-2)\pi}{n}$ we get$$\frac{2\sin(\frac{(n-2)\pi}{n})}{4\cos^2(\frac{(n-2)\pi}{n})-3}=\frac{2\sin(\pi)}{4\cos^2(\pi)-3}=0$$
$$\implies \sin\bigg(\frac{(n-2)\pi}{n}\bigg)=0$$
This is true only if $n=1$ or $2$. If $n=2$ then
$$\frac{2\sin(x)}{4\cos^2(x)-3}=\frac{2\sin(x+\pi)}{4\cos^2(x+\pi)-3}=-\frac{2\sin(x)}{4\cos^2(x)-3}$$
Therefore $n\ne2$. So the fundamental period is $2\pi$.
| {
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If $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$, then find the value of $\frac{(1+x)^3}{1+x^3}$
Let $x$ be a real number such that $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$. What is the value of $\frac{(1+x)^3}{1+x^3}$?
Of course, one way is that to solve for $x$ from the quadratic $37(1+x)^2=13(1+x^2)$ which gives the value $x=\frac{-37\pm\sqrt{793}}{24}$. Thus we can compute $\frac{(1+x)^3}{1+x^3}=\frac{13}{49}$.
But I am trying to find the result from the equation $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$ without solving for $x$ using some algebra. Here are my attempts to do that:
We have $$\begin{align} & \frac{(1+x)^2}{1+x^2}=\frac{13}{37}\\ &\implies \frac{(1+x)^2}{(1+x)^2-2x}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{(1+x)^3-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+3x(2+x)-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+x(4+x)}=\frac{13}{37}
\end{align}$$
I can't seem to proceed from here.
| To go a little further, let's calculate all $f(n)=\dfrac{1+x^n}{(1+x)^n}$
I take the inverse of the desired quantity because it makes it easier to have a factorizable expression on denominator.
Notice that $f(1)=1$ therefore:
$f(n)=f(n)f(1)=\dfrac{(1+x^n)(1+x)}{(1+x)^{n+1}}=\dfrac{(1+x^{n+1})+x(1+x^{n-1})}{(1+x)^{n+1}}=f(n+1)+\dfrac x{(1+x)^2}f(n-1)$
Though notice that $\alpha=\dfrac{x}{(1+x)^2}=\dfrac{1-f(2)}{2}=-\frac{12}{13}$
You end up with a linear recurrence with constant coefficients:
$$f(n+1)-f(n)+\alpha f(n-1)=0\qquad f(0)=2\quad f(1)=1\tag{E}$$
Which solves to
$$f(n)=r^n+\bar r^n\qquad r,\bar r=\frac 12\pm\frac 1{26}\sqrt{793}$$
Note: the given roots are solutions of $x^2-x+\alpha=0$
And you are able to effectively calculate $\dfrac{(1+x)^n}{1+x^n}=\dfrac 1{f(n)}$ for any given $n$. Of course for $f(3)$ it is quicker to just report in (E).
Remark: since $|\bar r|<1$ then $\bar r^n\to 0$ quickly, so for large values of $n$ we have $f(n)\approx r^n$.
| {
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Prove $a^2+b^2+c^2=x^2+y^2+z^2$ given that $a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$ Prove $$a^2+b^2+c^2=x^2+y^2+z^2$$ given that $$a^2+x^2=b^2+y^2=c^2+z^2=\\(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$$ where $a,b,c,x,y,z \in\mathbb R$
A friend forwarded me this problem which recently appeared in a math contest at Costa Rica. I've been working on it but I can't find a definite proof, though I have discovered several facts:
Let $u$ be equal to the value of the equation, then $ab+xy=\frac{-u}{2}$ and $a^2+b^2+c^2+x^2+y^2+z^2=3u$ which implies that $(a+b+c)^2+(x+y+z)^2 = 0 \implies a+b+c=x+y+z=0$.
I also noticed that, because of the equalities, working with complex numbers or vectors would be really tempting, as if $r=a+ix$, $s=b+iy$, $t=c+iz$, then we have $|r|=|s|=|t|=|r+s|=|s+t|=|r+t|$, but I can't find a way to make it useful.
Any ideas/hints/solutions?
| Your second observation is astute! Simply for interest, I would make one further observation, which is that the question geometrically asks us to prove
Given three $2D$ vectors (equivalently, complex numbers $r,s,t$) with given equality of lengths, prove the two $3D$ vectors $$\mathbf{u}=(a \quad b \quad c) , \qquad \mathbf{v}=(x \quad y \quad z)$$ are also of equal length (see the paper linked below for the condition that three vectors in the plane are images under orthogonal projection of an orthornormal basis in $\mathbb{R}^3$).
As others have pointed out, since $$|r|=|s|=|t|=|r+s|=|s+t|=|t+r|,$$ then, by the parallelogram law of addition, $r,s,t$ form an equilateral triangle, with centre of mass on the origin, in $\mathbb{C}=\mathbb{R}^2$. We know the cube roots of unity $$1,\quad \omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}, \quad \omega^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$ form an equilateral triangle in $\mathbb{C}$. We know also that there exists a complex number $\alpha$ such that the transformation $$z\mapsto \alpha z$$ maps $\triangle rst$ onto $\triangle 1\omega \omega^2$. Under this transformation, the lengths of $r,s,t$ are all scaled by some real number $R$. Since, as you can check for yourself, the result holds in this special case, and since rotation is an isometry (distance preserving map), we see that $$R|\mathbf{u}|=R|\mathbf{v}|$$ and so the result holds in general. $\square$
Alternatively, for a more purely complex algebraic proof note that, taking $r,s,t$ as points on a circle centred at the origin, $$\tag{1}\triangle rst\space \mathrm{is \space equilateral} \iff \begin{cases} r+s+t=0\\r^2+s^2+t^2=rs+st+tr\end{cases}$$ Equating the real parts of each equation in $(1)$ gives $$\tag{2}a+b+c=0\implies ab+bc+ca=-\frac{1}{2}(a^2+b^2+c^2)$$ and $$a^2+b^2+c^2-(x^2+y^2+z^2)=ab+bc+ca-(xy+yz+zx)\tag{3}.$$ Since $(2)$ also holds for imaginary parts, we conclude $$a^2+b^2+c^2=x^2+y^2+z^2\tag{4}.\quad \square$$ This path to the result is perhaps the cleanest, but, as presented, it offers less clue as to why the symmetry occurs. The proofs of the equations in $(1)$, however, recover some of this why and point to a wider theory of orthogonal projections. Both equations can also in fact be deduced from rotations of the plane, which is a good exercise.
| {
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Finding the maximum value of $y=\frac{x^2+x-1}{x^2-x+1}$ without calculus The problem is:
Find the maximum value of
$$y=\frac{x^2+x-1}{x^2-x+1}$$ assuming $x\geq 0$. I am only interested in pre-calculus solutions.
Using calculus it is easy to see that the maximum is $y=\frac{5}{3}$ occuring at $x=2$.
Other forms are
$$y=\frac{(x+\frac{1}{2})^2-\frac{3}{4}}{(x-\frac{1}{2})^2+\frac{5}{4}}$$
$$y=\frac{(\sqrt{x}-\frac{1}{\sqrt{x}})(\sqrt{x}+\frac{1}{\sqrt{x}})+1}{(\sqrt{x}-\frac{1}{\sqrt{x}})^2+1}$$
| A simple approach is to rewrite it as a quadratic for $x$
$$ (y-1)x^2 - (y+1)x + (y+1)=0$$
And require that the discriminant be not negative
$$(y+1)^2 - 4(y+1)(y-1) \ge 0$$
$$(y - \frac{5}{3})(y+1) \le 0$$
Which gives us the range for $y$. To be thorough, it's also good to check that the maximum value for $y$ does not correspond to a zero in the denominator of the original fraction. In this case, it doesn't.
| {
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"url": "https://math.stackexchange.com/questions/4285892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all functions $f:\mathbb R\to\mathbb R$ which satisfy $f(xy)=yf(x)+x+f\bigl(f(y)-f(x)\bigr)$ -Closed: It has a solution in AOPS.-
Find all functions $f:\mathbb R\to\mathbb R$ which satisfy $$f(xy)=yf(x)+x+f\bigl(f(y)-f(x)\bigr)$$
for all $x,y\in\mathbb R$.
My attempt:
\begin{align}
&P(x, x): f\left(x^2\right)=xf(x)+x+f(0). \\
&x=1; \ f(0)=-1. \\
\\
&P(x, 0): -1=x+f\bigl(-1-f(x)\bigr). \implies f\bigl(-f(x)-1\bigr)=-x-1. \\
&x=-1; f\bigl(-f(-1)-1\bigr)=0. \implies \exists t \ \text{ s.t. } f(t)=0. \\
\\
&P(t, 0): -1=t+f(-1). \\
&P(0, t): -1=-t+f(-1). \\
&\therefore 2t=f(1)-f(-1).
\ \\
&P(-1, 1): f(-1)=-f(1)+1+f(-2t). \\
&f(-1)+f(1)-2+f(2t)-f(-2t)=0.
\end{align}
| Solution from AOPS:
\begin{align}
&\text{let } P(x, y): f(xy)=yf(x)+x+f(f(y)-f(x)). \\
\ \\
&P(x, 1): f(f(1)=f(x))=-x. \Rightarrow f: \text{ Bijective.} \\
\ \\
&P(x, 0): f(-f(x)=1)=-x-1. \\
&P(0, x): x-1=f(f(x)+1). \\
&\therefore f(f(x)+1)+f(-f(x)-1)=-2. \\
&\text{Because } f \text{ is surjective, } f(-x)+f(x)=-2. \\
\ \\
&P(-x, -y): \\
&f(xy)=-yf(-x)-x+f(f(-y)-f(-x)) \\
&\ \ \ \ \ \ \ \ \ \ =-y(-2-f(x))-x+f((-2-f(y))-(-2-f(x))) \\
&\ \ \ \ \ \ \ \ \ \ =-y+yf(x)-x+f(f(x)-f(y)) \\
&\ \ \ \ \ \ \ \ \ \ =2y+yf(x)-x-f(f(y)-f(x))-2. \\
\ \\
&\text{Comparing this to the original F.E.: } \\
&yf(x)+x+f(f(y)-f(x))=2y+yf(x)-x-f(f(y)-f(x))-2 \\
& \Rightarrow 2f(f(y)-f(x))=2y-2x-2 \Rightarrow f(f(y)-f(x))=y-x-1. \\
\ \\
&\text{Contributing this to the original F.E.: } \\
&f(xy)=yf(x)+y-1. \\
&x=1; f(y)=yf(1)+y-1. \\
& \Rightarrow f(xy)-f(y)=y(f(x)-f(1)). \\
&x=0; -1-f(y)=-y-yf(1). \\
&\Rightarrow f(y)+1=y(f(1)+1). \\
&\text{let } f(1)+1=c. \\
& \Rightarrow \dfrac {f(y)+1}{y}=c. \\
&\Rightarrow f(y)=cy-1. \\
\ \\
&\text{Contributing this to the original F.E.:} \\
&cxy-1=cxy-y+x+c^2(y-x)-1. \\
&\Rightarrow x-y+c^2(y-x)=0. \\
&\Rightarrow (y-x)(c^2-1)=0. \\
&\text{For }x \neq y: \ c^2=1, \ c=\pm 1. \\
\ \\
&\therefore f(x)=\pm x-1.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/4286833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x - \sqrt{ \frac{2}{x} } = 3 $ then $x-\sqrt{2x} = ?$ Please find the value without calculate the value of $x$. I tried to multipy it by $x$, I tried to square it but I still can't find the solution. I tried to make some equation that my help to solve this:
$$x^{2} + \frac{2}{x}-2\sqrt{2x} =9, $$
$$x^{2} - \sqrt{2x} = 3x,$$
$$x^2 - \frac{2}{x} = 3\left(x+\sqrt{\frac{2}{x}}\right).$$
| Let $\sqrt{\dfrac x2}=y\implies x=2y^2$
$$\implies2y^2-\dfrac1y=3\iff2y^3-3y-1=0$$
Let $x-\sqrt{2x}=2y^2-2y=2a,y^2-y-a=0\ \ \ \ (1)$
$2y-\dfrac1y=3-2a\iff2y^2+(2a-3)y-1=0\ \ \ \ (2)$
$(1),(2)$ will represent the same equation iff $$\dfrac12=\dfrac{-1}{2a-3}=\dfrac a1$$
Clearly $2a=?$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Compute the series $\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}$ Compute the series $$\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}.$$
How do I go about with the index notation, for example to arrange the series instead as $\sum_{n=1}^{\infty}a_n $?
I have tried to simplify the expression as:
$$\begin{align}&\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}=\sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies&\sum_{n=3}^{\infty} \frac{n^4}{n^6} - \sum_{n=3}^{\infty} \frac{1}{n^6} = \sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies& \sum_{n=3}^{\infty} \frac{1}{n^2} - \sum_{n=3}^{\infty} \frac{1}{n^6}\end{align}$$
I'm not sure what to do with the index $n=3$ as I know that I can simplify it otherwise as $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{\pi^6}{945}$.
| You have at least two options to transform your sum into a sum where the index goes from $1$ on.
One, you can use the fact that
$$\sum_{n=1}^\infty a_n = a_1 + a_2 + \cdots + a_N + \sum_{n=N+1}^\infty a_n,$$
which can be shown from the definition of the infinite sum.
Or, you you can use the fact that
$$\sum_{n=k}^\infty a_n = \sum_{m=1}^\infty a_{k-1+m}.$$
which can be seen if you introduce a substitution $m=n+1-k$ (which also menas that $n=m+k-1$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The image set of the function $f(x) = \dfrac{x -1}{x + 1}$ generates $\Bbb{Q}$ multiplicatively? Define $f(x) = \dfrac{x - 1}{x+1}$.
Compute the composition $f\circ f(x) = \dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x - 1}{x+1} +1} =\dfrac{\dfrac{x - 1 -x -1}{x+1}}{\dfrac{x-1 + x + 1}{x+1}} =\dfrac{-2}{2x} = \dfrac{-1}{x}$
$$
f^3(x) = \dfrac{-1}{f(x)} \\
f^4(x) = \dfrac{-1}{f^2(x)} = x
$$
Thus $f : Z= \Bbb{Z}\setminus\{-1\} \hookrightarrow \Bbb{Q}^{\times}$ is an injection of sets, because $f$ is invertible. Its image space is such that for all $x,y \in f(Z)$ we have that $xy = 0$ or $xy \notin f(Z)$.
I was wondering, can we generate $\Bbb{Q}$ multiplicatively using $f(Z)$?
$$
\dfrac{0}{1}, \dfrac{1}{3}, \dfrac{1}{2}, \dfrac{3}{5}, \dfrac{2}{3}, \dfrac{5}{7}, \dfrac{3}{4}, \dfrac{7}{9}, \dfrac{4}{5}, \dfrac{9}{11}, \dots
$$
are the elements for non-negative $x \in \Bbb{Z}$.
| Let $H$ be the subset of $\Bbb Q$ which can be written as a product of finitely many elements of $f(Z)$. We want to show that $H = \Bbb Q$.
We have:
*
*$0 \in H$ because $0 = f(1)$;
*$-1\in H$ because $-1 = f(0)$;
*$\frac 1 n \in H$ for any positive integer $n$ because $\frac 1 n = f(2n - 1)f(2n - 3)\cdots f(3)$.
*$n \in H$ for any positive integer $n$ because $n = f(1 - 2n)f(3 - 2n)\cdots f(-3)$.
These obviously imply the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicating at most 2 negative real roots
I end up with following 2 possibilities:-
1)1 positive, 2 negative real roots
2)1 positive, 2 imaginary roots
how to progress further ?
| Just another way - to check for negative roots, it is enough to show $x^3+x^2+9=3x$ is not possible for any positive $x$. Using AM-GM, $x^3+(x^2+4)+5 \geqslant x^3+2\sqrt{4x^2}+5=x^3+(4x)+5>3x$, so this isn't possible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A non standard way to a probability problem We have teams $A$ and $B$.Every team has equal chance to win. A team wins after they have won $4$ matches. Team $A$ won the first match, what is the probability, that team $B$ won the whole game.
My try:
So team $A$ needs $3$ more wins to win the game. Team $B$ needs $4$ more wins to win the game. So the winner will be determined in maximum of $7$ games. So now I tried to do it my own way (I dont know if it is usual to do it this way)
Let's say that we are looking at the wins of team $A$. (They need $3$ more wins).
$$...A_i,...A_j,...A_n$$,
where $3$ dots symbolize the wins of team $B$. So $$(i-1) + (j-i-1) + (n-j-i-1) \leq 3 $$
(Because there must not be more than 3 losses, because at 4 team $B$ wins the game )
we get:
$$n-i \leq 6 \quad \text{and} \quad 0<i<n $$
If we look what integer tuples can satisfy the upper inequalities we get: $$N = \{(1,6),(2,6), (3,6),(4,6),(5,6),\\(1,5),(2,5),(3,5),(4,5),(1,4),(2,4),(3,4),(1,3),(2,3),(1,2)\}$$
Where $N$...possible winnings of team $A$
So we can count the number of elements ($15$) an that would be our number in the numerator, where in the denominator is the number of elements in our sample space.
To calculate the sample space I would do:
$$|\omega| = 1 + (\binom{4}{3} - 1)\\ + (\binom{5}{3} - \binom{4}{3} - 1)\\ + (\binom{6}{3} - (1 + (\binom{4}{3} - 1) + (\binom{5}{3} - \binom{4}{3} - 1)))\\ + (\binom{7}{3} -(1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1) + (\binom{6}{3} - (1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1)))) )$$
I guess we could also write this with a recursive formula. However we get $|\omega|=35$.
(What I did here is that $1$ is for $3$ consecutive wins, $\binom{4}{3} - 1$ are other possible matches in 4 games and so on.)
So if my procedure is correct the probability is $$P(\text{team B wins the whole game}) = 1 - \frac{|N|}{|\omega|} = 1 - \frac{15}{35} = \frac{20}{35}$$
However is my process atleas partially correct (I do not know the solution).
| Your approach looks right, but I think you made a couple mistakes when calculating $N$. First, you erred with your initial inequality. It should be
$$(i-1)+(j-i-1)+(n-j-1)\leq 3$$
$$\implies n\leq 6$$
This result should make sense because if you have $7$ games then clearly $B$ would have $4$ wins since $A$'s $3$rd win is always the last game.
The second mistake you made is when calculating $|N|$. You must find all triples $(i,j,n)$ that satisfy the inequality $n\leq 6$ (not just tuples $(i,n)$). This can be done with some combinatorics (or just brute force). However, one must consider the fact that two runs of games of different length do not have the same probability of occuring. For example, a game of length $6$ would have a probability of $2^{-6}$ of occuring while a game of length $5$ would have a probability of $2^{-5}$ of occuring.
So we will not actually solve for $|N|$, but rather a weighted sum of number of triples.
We will first fix $n$ as some integer that is at least $3$ (this is because we need at least $3$ games to get $3$ wins) and at most $6$. Then there are $n-1$ slots left before the $A_n$th game. There are $\binom{n-1}{2}$ ways to choose which of these slots get the other $2$ $A$ wins. So our probability becomes $\sum_{n=3}^6 \binom{n-1}{2}2^{-n}$
To actually solve this sum, we can use a related identity from putnam 2020 A2. Our sum is equivalent to
$$=\sum_{n=0}^3 \binom{n+2}{2}2^{-n-3}$$
$$=\sum_{n=0}^3 \binom{2+n}{n}2^{-n-3}$$
$$=2^{-5}\sum_{n=0}^3 \binom{2+n}{n}2^{2-n}$$
$$=2^{-6}\binom{5}{3}+2^{-5}\sum_{n=0}^2 \binom{2+n}{n}2^{2-n}$$
$$=2^{-6}\binom{5}{3}+2^{-5}4^2$$
$$=\frac{10}{64}+\frac{1}{2}$$
$$=\frac{21}{32}$$
Hence, the probability of $B$ winning is $1-\frac{21}{32}=\boxed{\frac{11}{32}}$
On a side note, while we never used the value of $|\omega|$, I would like to point out a more elegant approach to find its value since you did compute it. Consider all permutations of $3$ $A$'s and $4$ $B$'s. For each of these cut off from the $3$rd $A$ or $4$th $B$ depending on which comes first. These will each give new unique sequences of wins, so the total is just $\binom{7}{4}=35$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f(3)$ if $f(f(x))=x^{2}+2$ Let $a,b,f(x),x$ be positive integers such that If $a>b$ then $f(a)>f(b)$ and $f(f(x))=x^{2}+2$ . Find $f(3)$
My approach:
Replacing $x$ with $f(x)$ in the equation gives $f(f(f(x))) = f(x)^2 + 2$, but $f(f(x)) = x^2 + 2$ so $$f(x^2+2) = f(x)^2 + 2$$
how do i proceed after this. Please help. Thanks alot!
| Note that $f$ is by definition strictly monotonic. So consider: If $f(3)<3$ then we can only have $f(3)=1$ or $f(3)=2$. Else if $f(3)=3$ we have $f(3)=f(f(3)) = 3^2+2=11$, so this cannot be. Finally if $f(3)>3$ we have $f(3) < f(f(3))=11$.
$f(3)=1,2$ would imply $f(1)=f(f(3)) = 11>f(3)$.
Generally we get that $f(x)$ cannot be $x$, else $x=x^2+2$. Also $f(f(1)) = 3$, $f(f(2)) = 6$, $f(f(3)) = 11$. But as $f(3)>3$ (and $f(x)>3$ for $x>3$) this means $f(2)=3$.
But then $f(3) = f(f(2)) = 6$. Thus we’re done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any formula for $S(k)=\sum_{n=1}^{\infty} \frac{1}{n^{k} k^{n}},$ where $k\in N$? I had just used the definite integral $$\int_{0}^{\frac{1}{2}} \frac{\ln (1-x)}{x} d x $$ to evaluate the infinite sum
$$ \displaystyle S=\sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}}. $$
However, I just wonder if there is a formula for a more general series $$ \displaystyle S(k)=\sum_{n=1}^{\infty} \frac{1}{n^{k} k^{n}}, \text{ where }k \in N. $$
Now let me start with the former using the infinite geometric series
$ \displaystyle \frac{1}{1-t}=\sum_{n=0}^{\infty} t^{n} \quad \text { for }|t|<1 \tag*{(1)}$
Integrating both sides of (1) w.r.t. t from 0 to x yields
$ \displaystyle \begin{array}{r} \displaystyle \int_{0}^{x} \frac{1}{1-t} d t= \displaystyle \sum_{n=0}^{\infty} \int_{0}^{x} t^{n} d t \\ \displaystyle -\ln (1-x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n}\end{array} \tag*{(2)}$
Dividing the equation (2) by x and integrating both sides from $0$ to $\frac{1}{2} $ gives
$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}}=\displaystyle -\int_{0}^{\frac{1}{2}} \frac{\ln (1-x) d x}{x} \tag*{(3)}$
Now let’s tackle the integral
$ \displaystyle J:=\int_{0}^{\frac{1}{2}} \frac{\ln (1-x) d x}{x} \tag*{} $
by the substitution $y=\ln (1-x) $.
$\displaystyle \begin{aligned}J &=\int_{-\ln 2}^{0} \frac{y}{1-e^{y}} e^{y} d y \\&=\int_{-\ln 2}^{0} y \sum_{k=0}^{\infty} e^{(k+1) y} d y \\&=\sum_{k=0}^{\infty} \int_{-\ln 2}^{0} y e^{(k+1) y} d y\\ &\stackrel{I B P}{=} \sum_{k=0}^{\infty} \frac{1}{k+1}\left(\ln 2 \cdot \frac{1}{2^{k+1}}-\left[\frac{e^{(k+1) y}}{k+1}\right]_{-\ln 2}^{0}\right) \\&= \ln 2 \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}-\sum_{k=1}^{\infty}\frac{1}{k^{2}}+\sum_{k=1}^{\infty} \frac{1}{k^{2} 2^{k}} \\&= \ln 2 \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}-\frac{\pi^{2}}{6}+S\end{aligned} \tag*{} $
$ \displaystyle \textrm{Putting }x= \dfrac{1}{2} \textrm{ in (2) yields }$
$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}=-\ln 2 \tag*{} $
Now we can conclude that
$ \displaystyle \begin{array}{c}\displaystyle S=-\left(\ln ^{2} 2-\frac{\pi^{2}}{6}+S \right) \\\boxed{S(2)=\sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}} =\frac{1}{2}\left(\frac{\pi^{2}}{6}-\ln ^{2} 2\right)}\end{array}\tag*{} $
How about the general series $S(k)$?
:|D Wish you enjoy the proof!
Your suggestion, comments and formula for evaluating the general series S(k) are warmly welcome!
| As already pointed out by @Varun Vejalla in the comments section, the polylogarithm function is the way to go.
It´s defined by
$$\text{Li}_n\left(x\right)=\sum_{k=1}^\infty \frac{x^k}{k^n} \tag{1}$$
Letting $n=2$ and $x=\frac12$ in $(1)$ we obtain
$$\text{Li}_2\left(\frac12\right)=\sum_{k=1}^\infty \frac{1}{k^2 2^k} \tag{2}$$
Now, there is a functional equation that the dilogarithm function obbeys, namelly:
$$\text{Li}_2\left(1-\frac1x \right)=-\frac{\ln^2\left(x\right)}{2}-\text{Li}_2\left(1-x\right) \tag{3}$$
Letting $x=\frac12$ in $(3)$
$$\text{Li}_2\left(-1 \right)=-\frac{\ln^2\left(\frac12\right)}{2}-\text{Li}_2\left(\frac12\right)$$
which gives us
$$\text{Li}_2\left(\frac12 \right)=\frac{\pi^2}{12}-\frac{\ln^2\left(2\right)}{2} \tag{4}$$
euqting $(2)$ and $(4)$ we have
$$\sum_{k=1}^\infty \frac{1}{k^2 2^k}=\frac{\pi^2}{12}-\frac{\ln^2\left(2\right)}{2} \qquad \blacksquare$$
Which is the desired result. We used the fact that
$$\text{Li}_2\left(-1\right)=\sum_{k=1}^\infty \frac{(-1)^k}{k^2}=-\eta(2)=-\frac{\pi^2}{12}$$
We can go further with this method. Recall the Trilogarithm function
$$
\text{Li}_{3}(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{3}}
\tag{5}$$
letting $x=\frac{1}{2}$ in $(5)$ we obtain
$$
\operatorname{Li}_{3}\left(\frac{1}{2}\right)=\sum_{n=1}^{\infty} \frac{1}{2^{n} n^{3}}
\tag{6}$$
The Trilogarithm has the following identity
$$
L i_{3}(x)+L i_{3}(1-x)+L i_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}
\tag{7}$$
plugging $x=\frac{1}{2}$ in $(7)$ we obtain
$$
\begin{gathered}
L i_{3}\left(\frac{1}{2}\right)+L i_{3}\left(1-\frac{1}{2}\right)+L i_{3}\left(1-\frac{1}{2}\right)=\zeta(3)+\frac{\ln ^{3}\left(\frac{1}{2}\right)}{6}+\frac{\pi^{2} \ln \left(\frac{1}{2}\right)}{6}-\frac{\ln ^{2}\left(\frac{1}{2}\right) \ln \left(1-\frac{1}{2}\right)}{2} \\
2 L i_{3}\left(\frac{1}{2}\right)+L i_{3}(-1)=\zeta(3)-\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{6}+\frac{\ln ^{3}(2)}{2}
\end{gathered}
$$
Note that setting $x=-1$ in $(5)$ we get that $\text{Li}_{3}(-1)=-\eta(3)=-\frac{3}{4} \zeta(3)$, then
$$
\begin{gathered}
2 L i_{3}\left(\frac{1}{2}\right)=\frac{3}{4} \zeta(3)+\zeta(3)+\frac{\ln ^{3}(2)}{3}-\frac{\pi^{2} \ln (2)}{6} \\
L i_{3}\left(\frac{1}{2}\right)=\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3)
\end{gathered}
$$
And we finally conclude that
$$
\sum_{n=1}^{\infty} \frac{1}{2^{n} n^{3}}=\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3) \qquad \blacksquare
$$
Of course higher functional equations exist for higher polylog functions and the same methodology could be apllied, but these equations get more and more complex.
| {
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How to derive easily : $|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$? ( Context : a limit problem). Source : Lebossé & Hemery, Algèbre et Analyse ( Terminales CDT ), $1966$.
$|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$
This equality is asserted in the solution of a limit problem , namely : show that the limit of $f(x)=\sqrt{x+3}$ when $x$ goes to $1$ is $2$.
I tried to understand this assertion using the fact that $a/b = c/d$ implies that $ a\times c = b\times d$.
So I said to myself that $|\sqrt{x+3} -2|\times (\sqrt {x+3} +2)$ should yield $|x+1|$, and that , in that case , the assertion would be justified.
Since $\sqrt{x+3} -2 \geq 0 \iff x\geq 1$ one has
(1) if $x\geq 1 $ , $|\sqrt{x+3} -2|= \sqrt{x+3} -2$ , hence $|\sqrt{x+3} -2|\times (\sqrt {x+3} +2)= (\sqrt{x+3} -2)\times (\sqrt {x+3} +2) = (\sqrt{x+3})^2 - ( 2^2) = (x+3) - 4 = x-1$
(2) if $x\lt 1 $ , $|\sqrt{x+3} -2|= -(\sqrt{x+3} -2) = 2- \sqrt{x+3}$ , hence $|\sqrt{x+3} -2|\times (\sqrt {x+3} +2)= 2- \sqrt{x+3} \times (2+ \sqrt {x+3}) = (2^2 - (\sqrt{x+3})^2 = 4- (x+3) = 4-x-3 = 1-x $.
My question :
(1) how do we go formally from " the product is $(x-1)$ if $x\geq 1$ and $(1-x)$ if $x\lt 1$ to " the product is equal to $|x-1|$" ? I see that $x-1$ is the ( additive ) inverse of $1-x$ but I cannot use it to draw the desired conclusion .
(2) is there a quick way to arrive at $|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$, without all these manipulations?
| Regardless of the value of $x$, $\sqrt{x+3}+2>0$, so $\sqrt{x+3}+2=|\sqrt{x+3}+2|$ and $(\sqrt{x+3}+2)^{-1}$ exists.
Multiplying $(\sqrt{x+3}+2)^{-1}|\sqrt{x+3}+2|=1$, $|\sqrt{x+3}-2|$ changes into...
$$
|\sqrt{x+3}-2|=\frac{|\sqrt{x+3}-2|\cdot|\sqrt{x+3}+2|}{\sqrt{x+3}+2}=\frac{|\sqrt{x+3}^2-2^2|}{\sqrt{x+3}+2}=\frac{|x-1|}{\sqrt{x+3}+2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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