Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate an expression with assumption Let $a,b,c$ be non-zero real numbers and satisfy
\begin{equation}
\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}-\frac{a^3+b^3+c^3}{abc}=2.
\end{equation}
Evaluate
\begin{equation}
M=[(a+b)^{2018}-c^{2018}][(b+c)^{2018}-a^{2018}][(a+c)^{2018}-b^{2018}].
\end{equation}
My impression is to divide $M$ by $abc$, but I'm stuck. I have no idea what to do. I tried Mathematica, but it didn't give me an answer. Could someone please help me?
Thank you.
EDIT:
I think I got it. The assumption is equivalent to
\begin{equation}
(a+b-c)(b+c-a)(a+c-b)=0.
\end{equation}
From here, it is easy to manipulate $M$.
| For a way to factor the original condition without using hints from the second part of the question, eliminate denominators and collect the powers of $\,a\,$ to get:
$$
\begin{align}
0 &= 2abc + (a^3+b^3+c^3) - ab(a+b) - bc(b+c) - ca(c+a) \\
&= a^3 - a^2 (b + c) - a (b^2 - 2 b c + c^2) + b^3 - b c (b + c) + c^3 \\
&= \underbrace{a^3 - a^2(b+c)}_{\large{a^2\left(a-\left(b+c\right)\right)}} \;\underbrace{- a(b-c)^2+(b+c)(b-c)^2}_{\large{-(b-c)^2\left(a-\left(b+c\right)\right)}} = \ldots
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $a_n = \frac{n}{2n+1}+\frac{1}{n^3}$ is a Cauchy Sequence I want to show that
$$a_n = \frac{n}{2n+1} + \frac{1}{n^3}$$
is a Cauchy sequence.
My attempt:
$$|a_m-a_n|=|(\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3})|$$
$$\leq\frac{m}{2m+1}+\frac{n}{2n+1}+\frac{1}{m^3}+\frac{1}{n^3}$$
$$<\frac{1}{2}+\frac{1}{2}+\frac{1}{N^3}+\frac{1}{N^3}$$
$$=1+\frac{2}{N^3}$$
$$\leq \epsilon\Rightarrow N(\epsilon)\geq(\frac{2}{\epsilon-1})^{\frac{1}{3}}$$
But this seems a bit off?
Any thoughts?
| You split too much, instead:
\begin{align}\left|\frac{m}{2m+1}-\frac{n}{2n+1} + \frac{1}{m^3}-\frac{1}{n^3}\right|
&\leq \left|\frac{m}{2m+1}-\frac{n}{2n+1}\right| + \left|\frac{1}{m^3}-\frac{1}{n^3}\right|\\
&\leq \left|\frac{m(2n+1) - n(2m+1)}{(2m+1)(2n+1)}\right| + \left|\frac{n^3-m^3}{m^3n^3}\right|\\
&= \frac{|m-n|}{(2m+1)(2n+1)} + \frac{|m-n||m^2+mn+n^2|}{m^3n^3}
\end{align}
I'll let you take it from here...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$ given $4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$
$$4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$$ Solve for $x$.
I tried to put every single monomial under $x^2$ denominator, but that did not get me to anything I could solve for.
I appreciate any help.
| \begin{equation}
4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47
\end{equation}
Multiply by $x^2$, i.e.
\begin{equation}
4x^4 + 12x^3 -47x^2 + 12x + 4= 0
\end{equation}
Notice that for $x = 2$ and $x = \frac{1}{2}$, we get zero. So
\begin{equation}
4(x-2)(x-\frac{1}{2})(x - a)(x-b)
=
4x^4 + 12x^3 -47x^2 + 12x + 4
\end{equation}
Expand the left hand side, i.e.
\begin{equation}
4(x^2 - \frac{5}{2}x + 1)(x^2 - (a+b)x + ab)
\end{equation}
That is
\begin{equation}
(4x^2 - 10x + 4)(x^2 - (a+b)x + ab)
\end{equation}
Expand now
\begin{equation}
4x^4 -4(a+b)x^3 + 4abx^2
-10x^3 + 10(a+b)x^2 -10abx
+4x^2 -4(a+b)x +4ab
\end{equation}
That is
\begin{equation}
4x^4 + (-10 -4(a+b))x^3 + (4ab+10(a+b)+4)x^2 +(-10ab
-4(a+b))x +4ab
\end{equation}
Equate it to the right hand side of the third equation and we get by identification
\begin{align}
-10 -4(a+b) &= 12 \\
4ab+10(a+b)+4 &= -47 \\
-10ab-4(a+b) &= 12 \\
4ab &= 4
\end{align}
So
\begin{equation}
ab = 1
\end{equation}
and hence
\begin{equation}
(a+b) = -\frac{12 + 10(1)}{4} = -\frac{11}{2}
\end{equation}
Now let's find $a,b$
\begin{align}
a+b &= -\frac{11}{2}\\
ab &= 1
\end{align}
So, $b = -a -\frac{11}{2}$ and hence $ab = a(-a -\frac{11}{2}) = 1$ which gives a quadratic system
\begin{equation}
a^2 + \frac{11}{2}a +1 = 0
\end{equation}
Giving two roots
\begin{align}
a_1 &= \frac{1}{4}(-11 - \sqrt{105})\\
a_2 &= \frac{1}{4}(-11 + \sqrt{105})
\end{align}
It turns out that
\begin{equation}
b_1= \frac{1}{a_1} = a_2
\end{equation}
and
\begin{equation}
b_2= \frac{1}{a_2} = a_1
\end{equation}
So the roots of the initial equation are
\begin{align}
x_1 &= 2\\
x_2 &= \frac{1}{2}\\
x_3 &= \frac{1}{4}(-11 - \sqrt{105})\\
x_4 &= \frac{1}{4}(-11 +\sqrt{105})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Differential equation: $(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
Attempt:
After expanding, everything is neat except: $3x^2y dy - 3y^2 x dx$
I can't convert it to exact differential.
Also, there's weird symmetry in the equation wrt the coefficients of $x^2$ and $y^2$. But not sure how to utilise that symmetry.
As an attempt, though, I reached this:
$\dfrac{d(x^2 - y^2 -1)}{2(x^2- y^2 -1)}= \dfrac{y dy}{(2x^2 + 3y^2 -7)}$ which is not useful at all.
Answer given is:
$(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C$
| Hint:
$$(2x^2 -4 + 3y^2 -3)x dx = (3x^2 -6+ 2y^2 -2 )y dy$$
and use substitutions $x^2-2=u$ and $y^2-1=v$. You will find a homogeneous differential equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all matrices that commute with $A$
Given $$A = \begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.
Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to find other $B's$ using above Nilpotency of $A-3I$?
| With
$N = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \tag 1$
we have
$A = 3I + N; \tag 2$
then
$AB = BA \tag 3$
implies
$(3I + N)B = B(3I + N), \tag 4$
or
$3B + NB = 3B + BN, \tag 5$
whence
$NB = NB; \tag 6$
with
$B = \begin{bmatrix} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \end{bmatrix}, \tag 7$
we then have
$NB = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \end{bmatrix} = \begin{bmatrix} b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \\ 0 & 0 & 0 \end{bmatrix}, \tag 8$
and
$BN = \begin{bmatrix} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & b_1 & b_2 \\ 0 & b_4 & b_5 \\ 0 & b_7 & b_8 \end{bmatrix}; \tag 9$
comparing (8) and (9) in the light of (6) yields
$b_4 = b_7 = b_8 = 0, \; b_1 = b_5 = b_9 = 0, \; b_6 = b_2, \tag{10}$
and $b_3$ unspecified/unconstrained; therefore $B$ takes the form
$B = \begin{bmatrix} 0 & b_2 & b_3 \\ 0 & 0 & b_2 \\ 0 & 0 & 0 \end{bmatrix}; \tag{11}$
it is easy to walk these calculations back and show that every $B$ as in (11) satisfies (6) and hence (3); that $B$ take the form (11) is thus both a necessary and sufficient condtition that (3) should bind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Roots of unity and large expression
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} \\
&= \dfrac{\omega(1+\omega^3) + \omega^4(1+\omega^2)}{(1+\omega^2)(1+\omega^3)} + \dfrac{\omega^2(1+\omega) + \omega^3(1+\omega^4)}{(1+\omega^4)(1+\omega)} \\
&= \dfrac{\omega + 2\omega^4 +\omega^6}{1+\omega^2 + \omega^3 + \omega^5} + \dfrac{\omega^2 + 2\omega^3 + \omega^7}{1+\omega + \omega^4 + \omega^5} \\
&= \dfrac{2\omega + 2\omega^4}{2+\omega^2 + \omega^3} +
\dfrac{2\omega^2 + 2\omega^3}{2+\omega+\omega^4}
\end{align*}
OR
\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} \\
&= \dfrac{\omega(1+\omega) + \omega^3(1+\omega^2)}{(1+\omega)(1+\omega^2)} + \dfrac{\omega^2(1+\omega^3) + \omega^4(1+\omega^4)}{(1+\omega^3)(1+\omega^4)} \\
&= \dfrac{\omega + \omega^2 + \omega^3 + \omega^5}{1+\omega + \omega^2 + \omega^3} + \dfrac{\omega^2 + \omega^4 + \omega^5 + \omega^8}{1 + \omega^3 + \omega^4 + \omega^7} \\
&= \dfrac{2\omega+\omega^2+\omega^3}{1+\omega+\omega^2+\omega^4} + \dfrac{1+\omega+\omega^2+\omega^4}{1+2\omega^3+\omega^4}
\end{align*}
| Expand 2nd and 4th fraction with $\omega $ and $\omega ^2$ respectively: $$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}=\frac{\omega}{1 + \omega^2} + \frac{\omega^3}{\omega+ 1} + \frac{\omega^3}{1 + \omega} + \frac{\omega}{\omega^2+1}$$
$$=2\frac{\omega}{1 + \omega^2} + 2\frac{\omega^3}{\omega+ 1} $$
$$=2\frac{\omega^2+\omega + \omega^3+1}{(\omega^2+1)(\omega+ 1)}=2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determine if a system of equations is independent, dependent or inconsistent Is there a way to determine the nature of a system of equations without solving it? For example, given the system
\begin{cases}
2x + y - 4z = 6 \\[4px]
y - 2z = 2 \\[4px]
4x + 3y - 10z = -3
\end{cases}
Can I tell that this system is independent without solving it?
| Not really. You can see whether the three equations are independent by computing the determinant
$$
\det\begin{bmatrix}
2 & 1 & -4 \\
0 & 1 & -2 \\
4 & 3 & -10
\end{bmatrix}
$$
but, unfortunately, it is $0$. Besides, computing a determinant with Laplace expansion is a very expensive computation.
Much simpler is going with Gaussian elimination:
\begin{align}
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
4 & 3 & -10 & -3
\end{bmatrix}
&\to
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
0 & 1 & -2 & -15
\end{bmatrix}
&& R_3\gets R_3-2R_1
\\[6px]
&\to
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & -17
\end{bmatrix}
&& R_3\gets R_3-R_2
\end{align}
The last column is a pivot column, so the system is inconsistent.
The system would be solvable if $-10$ is changed into $14$. In this case we could go backwards
\begin{align}
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}
&\to
\begin{bmatrix}
2 & 0 & -2 & 4\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}
&& R_1\gets R_1-R_2
\\[6px]
&\to
\begin{bmatrix}
1 & 0 & -1 & 2\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}
&& R_1\gets\tfrac{1}{2}R_1
\end{align}
The third unknown is free, and the solutions are
\begin{cases}
x=2+h \\[4px]
y=2+2h \\[4px]
z=h
\end{cases}
with arbitrary $h$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 x$
Solving and re-arranging we get
$-64\sin^6 x+80\sin^4 x-20\sin^2 x+1=0$
Using the substitution $\sin^2 x=t$
$-64t^3+80t^2-20t+1=0$
I am not able to solve it from hence forth
| We need to solve
$$8\sin^2x\cos{x}=\sqrt3\sin{x}+\cos{x}$$ or
$$2\sin2x\sin{x}=\cos(x-60^{\circ})$$ or
$$\cos{x}-\cos3x=\cos(x-60^{\circ})$$ or
$$2\sin30^{\circ}\sin(30^{\circ}-x)=\cos3x$$ or
$$\cos(60^{\circ}+x)=\cos3x.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Difficulty proving integral reduction formula So I tried proving this integral reduction formula but to no avail:
If $$I_n=\int \frac{x^n}{\sqrt{ax+b}}dx$$, then
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a(2n+1)}-\frac{2nb}{a(2n+1)}I_{n-1}$$
I tried integrating by parts and my attempt went as followed:
$$\int \frac{x^n}{\sqrt{ax+b}}dx\begin{vmatrix}u=x^n\\du=nx^{n-1}dx\end{vmatrix}v=\int \frac{1}{\sqrt{ax+b}}dx\enspace v=\frac{2\sqrt{ax+b}}{a}\\
\int udv=uv-\int vdu$$
The integral then becomes
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\sqrt{ax+b}dx$$
Multiplying and dividing the integrand on the right by $\sqrt{ax+b}$ gives
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int \frac{x^{n-1}(ax+b)}{\sqrt{ax+b}}dx\\
=\frac{2x^n\sqrt{ax+b}}{a}-2n\int \frac{x^n}{\sqrt{ax+b}}dx-\frac{2nb}{a}\int \frac{x^{n-1}}{\sqrt{ax+b}}dx\\
=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}$$
So as you can see, I tried algebraic manipulation in order to yield an $I_{n-1}$ term within the integral, but this was as far as I got and I don't know how to proceed from here (or if my approach was even correct). Can I get some help with this one?
| You are almost there. Just move the $-2nI_n$ term to the other side, then divide everything by $2n+1$
$$I_n=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}\\
I_n+2nI_n=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2nb}{a}I_{n-1}\\
I_n=\frac{2x^n\sqrt{ax+b}}{a(2n+1)}-\frac{2nb}{a(2n+1)}I_{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that the highest and lowest order terms will be $z^{1 + 2 + \cdots + m} = z^{\frac 1 2m(m+1)}$ and its reciprocal, respectively.
Here's the situation for low values of $m$:
$$\begin{split}
f_1(z) = z &+ \frac 1 z \\
f_2(z) = z^3 + z &+ \frac 1 z + \frac 1 {z^3} \\
f_3(z) = z^6+z^4+z^2+\ &\color{red}2 + \frac 1 {z^2}+ \frac 1 {z^4} + \frac 1 {z^6} \\
f_4(z) = z^{10} + z^8 + z^6 + \color{red}2z^4 + \color{red}2z^2 +\ &\color{red}2 + \frac {\color{red}{2}} {z^2} + \frac {\color{red}2} {z^4} + \frac 1 {z^6} + \frac 1 {z^8} + \frac{1}{z^{10}} \\
f_5(z)=z^{15} + z^{13}+ z^{11} + \color{red}2z^{9} + \color{red}2z^7 + \color{orange}3z^5 + \color{orange}3z^3 + \color{orange}3z &+ \frac{\color{orange}3} z + \frac {\color{orange}3}{z^3} + \frac {\color{orange}3} {z^5} + \frac {\color{red}2} {z^7} + \frac {\color{red}2} {z^9} + \frac 1 {z^{11}} + \frac 1 {z^{13}} + \frac 1 {z^{15}}
\end{split}$$
It seems that
*
*In each $f_m$ the exponent of $z$ falls from $\tau_m = \frac 1 2 m(m+1)$ down to $-\tau_m$, skipping every other value;
*For all $m$ the coefficient $a_{m,n}$ of the $n$-th term in the analytic part is equal to the coefficient $a_{m,-n}$ of the $n$-th term in the singular part (which makes sense intuitively).
So:
*
*Is there a closed-form expression for $a_{m,n}$?
*How can I prove fact 1. above?
| This is not an answer, but maybe help you to find an answer.
Since the function is same as
$$
f_{m}(z) = \frac{1}{z^{m(m+1)/2}}\prod_{k=1}^{m}(1+z^{2k}),
$$
it is enough to find an explicit formula for the coefficients of the polynomial
$$
g_{m}(z) = \prod_{k=1}^{m}(1+z^{k}).
$$
Note that $f_{m}(z) = g_{m}(z^{2})/z^{m(m+1)/2}$. One can observe that the $n$-th coefficient of $g_{m}(z)$ is same as the number of ways to express $n$ as a sum of distinct numbers that are less or equal to $m$.
It seems that it is really hard to find the explicit formula for such coefficients, since the explicit formula for $q(n) = \lim_{m\to \infty}[z^{n}]g_{m}(z)$ ($n$-th coefficient of $\prod_{k=1}^{\infty} (1+z^{k})$, which is a number of ways to express $r$ as a sum of distinct numbers, without any condition on the bound of such numbers), closely related to the ordinary partition number $p(n)$:
$$
p(n) =\sum_{k=0}^{\lfloor n/2\rfloor} q(n-2k)p(k).
$$
Also, it is known that $q(n)$ has a following explicit formula
$$
q(n) = \frac{1}{\sqrt{2}}\sum_{k=1}^{n}A(2k-1, n)\frac{\partial J_{0}\left(\frac{\pi i}{2k-1}\sqrt{\frac{1}{3}\left(n+\frac{1}{24}\right)}\right)}{\partial n}
$$
where $A(k, n)$ is a generalized Kloosterman sum
$$
A(k, n) = \sum_{\substack{1\leq h\leq k \\ \gcd(h, k) = 1}}\exp\left(\pi i \sum_{j=1}^{k=1}\frac{j}{k}\left(\frac{hj}{k} - \lfloor \frac{hj}{k}\rfloor - \frac{1}{2}\right)- \frac{2\pi i n h}{k}\right)
$$
and $J_{0}$ is a first kind of Bessel function. This formula seems not to be that helpful, and at least we have a asymptotic formula
$$
q(n) \sim \frac{1}{4\sqrt[3]{3}n^{3/4}}\exp\left(\pi\sqrt{\frac{n}{3}}\right).
$$
All of these formulas are from here, and sadly I don't know proof of any of these. (Maybe the last one comes from a Hardy-Littlewood's circle method, since similar asymptotic formula for $p(n)$ is proved by such method.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Convergence and limit of recursive sequence Consider the sequence $(a_n)_{n \in \mathbb{N}}$ which has startvalue $a_0 > -1$ and recursive relation:
$$a_{n+1} = \frac{a_n}{2} + \frac{1}{1+ a_n}$$
How to prove the convergence and find the limit?
_
I think you need to show the convergence with a Cauchy sequence. It is also possible to show that the sequence is decreasing and bounded, but I found that if $-1 < a_0 < 0$ that the sequence first increases and after $a_n > 1$ it starts to decrease. So that method seems more complicated to me.
For the limit I found the values $1$ and $-2$ but don't know how to show which is the right one.
| Solution $\blacktriangleleft$. Let
$$
f(x) = \frac x 2 + \frac 1 {1+x}, \quad [x >-1],
$$
then
$$
f'(x) = \frac 12 - \frac 1{(1+x)^2} = \frac {(x+1+\sqrt 2)(x+1-\sqrt 2)} {2(1+x)^2},
$$
hence $f(x) \searrow$ on $(-1, \sqrt 2 -1)$, $\nearrow$ on $(\sqrt 2 - 1, +\infty)$, therefore $f(x) \geqslant f(\sqrt 2 -1) = \sqrt 2 -1/2$. Thus for $n \in \Bbb N^*$, $a_n \geqslant \sqrt 2-1/2.$
Now for $n \geqslant 1$,
\begin{align*}
a_{n+1}-a_n &= \frac {a_n -a_{n-1}}2 + \frac 1{1+a_n} - \frac 1{1+a_{n-1}} \\
&= (a_n -a_{n-1}) \left(\frac 12 - \frac 1 {(1+a_n) (1+a_{n-1})}\right).
\end{align*}
Since
$$
\frac 12 > \frac 12-\frac 1 {(1+a_n)(1+a_{n-1})} \geqslant \frac 12 - \frac 1 {(\sqrt 2 + 1/2)^2} = \frac 12 - \frac 1 {2 + 1/4 +\sqrt 2} >0,
$$
we have
$$
|a_{n+1} - a_n| = |a_{n-1} - a_n| \left|\frac 12 - \frac 1{(1+a_n)(1+a_{n-1})}\right| \leqslant \frac 12 |a_n - a_{n-1}|,
$$
hence $(a_n - a_{n-1})$ is contracting, thus $(a_n)$ is Cauchy. Hence $(a_n)$ converges. Since $a_n \geqslant \sqrt 2 -1/2 >0$, we have $a_n \to 1. \blacktriangleright$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
| We have
$$a+\sqrt b=b+\sqrt a\iff a-b=(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=\sqrt a-\sqrt b.$$
Then $a=b$ (which is not allowed) or $\sqrt a+\sqrt b=1$.
Now,
$$a+b=a+(1-\sqrt a)^2$$ has an extremum found by taking the derivative on $a$,
$$1-\frac{1-\sqrt a}{\sqrt a}=0$$
which gives $a=\dfrac14$. But it turns out that this is a minmum.
Then we also have to try the values at the boundaries of the domain: $a=0,b=1$ and $a=1,b=0$ both yield
$$a+b=1$$
which is the searched maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Expected number of parallel tosses of $N$ unfair coins until all coins landed head at least once I am trying to understand this answer, but it doesn't work when I plug in the numbers. Given the recurrence relation
\begin{align*}
E_n &= \dfrac{\displaystyle 1+\sum_{k=1}^{n-1}q^kp^{n-k}E_{k}}{1-q^n} \\
E_1 &= \frac{1}{p}
\end{align*}
and assuming $p = q = 1/2$, we can compute $E_5$ as:
\begin{align}
E_1 &= 2\\
E_2 &= \frac{1 + (1/2)(1/2)2}{1-(1/4)} = \frac{6}{3} = 2\\
E_3 &= \frac{1 + (1/2)(1/4)2 + (1/4)(1/2)2}{1-(1/8)} = \frac{12}{7} \approx 1.714\\
E_4 &= \frac{1 + (1/2)(1/8)2 + (1/4)(1/4)2 + (1/8)(1/2)(12/7)}{1-(1/16)} = \frac{152}{105} \approx 1.448\\
E_5 &= \frac{1 + (1/2)(1/16)2 + (1/4)(1/8)2 + (1/8)(1/4)(12/7) + (1/16)(1/2)(152/105)}{1-(1/32)} = \frac{4112}{3255} \approx 1.263,
\end{align}
which does not match the author's answer of $E_5 = 2470/651 \approx 3.79416282642$.
I have no experience with absorbing Markov chains so I have no idea where the mistake is. Am I doing something wrong or is the answer wrong?
(Apologies if this is not the right place. I would have posted this as a comment to the answer, but my reputation is not high enough to do that...)
| The recurrence relation given by @saulspatz in the comments is correct. Solved for $E_n$, it is:
$$ E_n = \dfrac{1+\sum\limits_{k=1}^{n-1}\binom{n}{k}p^{n-k}q^{k}E_k}{1 - q^n} $$
with $ E_1 = \frac{1}{p} $. This form correctly predicts $ E_2 = \frac{8}{3} $ and $ E_3 = \frac{22}{7} $, as well as $ E_5 = \frac{2470}{651}$.
| {
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"url": "https://math.stackexchange.com/questions/2892447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?
I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it.
Another thought: If $(a^2+b^2+c^2)^2 \geq 1$, then this statement will be true. But how can I prove it?
| I'll suppose $a$, $b$, $c>0$ as that always seems to be assumed in these
sorts of problems. Then your inequality is equivalent to
$$(a^2+b^2+c^2)^2\ge abc(a+b+c)=a^2bc+b^2ac+c^2ab.$$
As $bc\le\frac12(b^2+c^2)$ etc. we get
$$a^2bc+b^2ac+c^2ab\le a^2b^2+a^2c^2+b^2c^2$$
and that is clearly less than $(a^2+b^2+c^2)^2$.
We can do better:
$$\frac{a^4+b^4}2+2a^2b^2\ge3a^2b^2$$
etc. Therefore
$$a^2bc+b^2ac+c^2ab\le a^2b^2+a^2c^2+b^2c^2
\le\frac13(a^2+b^2+c^2)^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does recurrence relation works with system of equations? There are $$ a+b+c+d = 2\\2a+2^2b+2^3c+2^4d = 5\\ 3a+3^2b+3^3c+3^4d = 6\\4a+4^2b+4^3c+4^4d = 1$$
then I'm given
$$C_{n}= a+bn+cn^2+dn^3$$
from linear recurrence relation with repeated roots said that
$$(x-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^{4-k}x^{4-k}$$
and then
$$C_{n}=\sum_{k=1}^{4}\binom{4}{k}(-1)^{5-k}C_{n-k} = 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4}$$ $\forall n \geq 5$
I want to find $C_{6}$. Then i've plugged $C_{5}=-3$ into that relation which finally gave $C_{6}=-8$ but wolfram alpha told me that $C_{6} =-48
Question: How do I use recurrence relations to find out the correct $C_6$
|
How do I use recurrence relations to find out the correct $C_6$
You did find the correct $\,C_6 = -8\,$.
$\;C_{n}== 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4} \quad\quad\forall n \geq 5$
Correct, thus far.
but wolfram alpha told me that $C_{6} =-48$
What did you ask WA?
WA's answer to $\,c_{n} = 4 c_{n-1} - 6 c_{n-2} + 4 c_{n-3} - c_{n-4}\,$$\;\style{font-family:inherit}{\text{with}}\;$$\,c_1 = 2\,,$$\,c_2 = 5/2\,,$$\,c_3 = 2\,,$$\,c_4 = 1/4\,$:
$$
C_n = \frac{3}{4} - \frac{1}{24} n \big(n \left(n + 6\right) - 37\big) = \frac{-n^3 - 6 n^2 + 37 n + 18}{24}
$$
Unsurprisingly, this matches $\,\frac{p(n)}{n}\,$ from Jack D'Aurizio's answer, and $\,C_5=-3, C_6=-8\,$ indeed.
However, if you are maybe looking for $\,6a+6^2b+6^3c+6^4d\,$, instead, then that's $\;6 \cdot C_6 = -48\,$.
| {
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"url": "https://math.stackexchange.com/questions/2893452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Understanding a particular method of solving generalized version of Pell's equation So, I have understood how to solve Diophantine equations of the form
$$x^2-Dy^2=1.$$
However, when I was reading the solution of the generalized Pell's equation
$$x^2-Dy^2=c,$$
I got stuck. I knew how to solve it using continued fraction, but the book used an intriguing method, but skips over so much detail that I can't understand it. It seems to be using a method similar to Euclid's algorithm, which reduces the $c$ step by step, finally reaching $c=1$.
Here is the translated text:
Example 5. Find all integer solutions of the equation $x^2-15y^2=61$.
Solution: This is not a standard Pell's equation, therefore we need to transform it.
First, solve for the equation $l^2\equiv15(\mathrm{mod}\;61)\;(0\le l\le\frac{61}2)$, which is equivalent to $l^2=15+61h\;(l^2\le900)$. Since $0\le h\le \left[\frac{900}{61}\right]=14$, trying $h=0,1,\dots,14$ gives us $h=10, l=25$.
Next, find the solutions of $x^2_1-15y^2_2=10$, which reduces to to solving $l^2=15+10h\;(l\le\frac{10}{2}=5)$. We can see that $l=5,h=1$. Therefore $x_1=5, y_1=1$.
Next, solve the equation $x^2_2-15y^2_2=1$. This is a Pell's equation, whose fundamental solution is $x_2=4, y_2=1$.
Therefore, $x_2+\sqrt{15}y_2=\pm(4+\sqrt{15})^n$,
$x_1+\sqrt{15}y_1=\pm(4+\sqrt{15})^n(5\pm\sqrt{15})$,
$x+\sqrt{15}y=\pm(4+\sqrt{15})^n(5\pm\sqrt{15})(\frac{25\pm\sqrt{15}}{10})$.
As the three $\pm$ signs are mutually independent, the solution to the equation is therefore
$$x+\sqrt{15}y=\pm(4+\sqrt{15})^n(14+3\sqrt{15})$$ and $$x+\sqrt{15}y=\pm(4+\sqrt{15})^n(11+2\sqrt{15})$$
| The point is to first find a single solution to $x^2-15y^2=61$. Then every other solution is obtained by multiplying by units. I won't prove this here.
First, solving $l^2\equiv15\pmod{61}$ yields the identity
$$25^2=15+61\cdot10\qquad\text{ so }\qquad \frac{(25+\sqrt{15})(25-\sqrt{15})}{10}=61.$$
Now suppose we have integers $z,w\in\Bbb{Z}$ such that $z^2-15w^2=10$. Then $x,y\in\Bbb{Z}$ defined by
$$x+y\sqrt{15}:=(25\pm\sqrt{15})(z\pm w\sqrt{15}),$$
will satisfy $x^2-15y^2=61$. So we've reduced the problem to finding a solution to $z^2-15w^2=10$.
By the exact same method, solving $k^2\equiv15\pmod{10}$ yields the identity
$$5^2=15+10\qquad\text{ so }\qquad(5+\sqrt{15})(5-\sqrt{15})=10.$$
Putting these together shows that
$$61=\frac{(25+\sqrt{15})(25-\sqrt{15})}{(5+\sqrt{15})(5-\sqrt{15})}
=\frac{1}{10^2}(5+\sqrt{15})(5-\sqrt{15})(25+\sqrt{15})(25-\sqrt{15}).$$
So all $x,y\in\Bbb{Z}$ with $x^2-15y^2=61$ are of the form
$$x+y\sqrt{15}=u\cdot\frac{1}{10}(5\mp\sqrt{15})(25\pm\sqrt{15}),$$
where $u\in\Bbb{Z}[\sqrt{15}]^{\times}$ is any unit.
For values other than $61$ the process might be longer or shorter; here we were reduce to the classical Pell equation in two steps, consecutively solving the diophantine equations
$$x^2-15y^2=61,\qquad x^2-15y^2=10,\qquad x^2-15y^2=1.$$
What can be said is that the right hand side decreases with every step, so this process does eventually terminate.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How do I eliminate $x$ and $y$ from the system $x^2 y= a$, $x(x+y)= b$, $2x+y=c$ to get a single equation in $a$, $b$, $c$? Alright, a homework problem.
I'm stuck at this question,
Eliminate $x$ and $y$ from the given equations to get a single
equation in terms of $a$ , $b$ and $c$
$$\begin{align}
x^2 y &= a \\
x(x+y) &= b \\
2x+y &=c
\end{align}$$
Let me tell you what I tried, I tried to get $y$ from one equation and substitute in the other two. Turns out that I'm not able to fully get rid of both $x$ and $y$. Help please.
| I will work with 2nd and 3rd equations to solve for $x$ and $y$.
From (3) we get,
$$y=c-2x$$
Substituting the value in (2) we get,
$$x(x+(c-2x))=b$$
$$\implies x^2-cx+b=0$$
$$x=\dfrac{1}{2}(c+\sqrt{c^2-4b}) \text{ or, }\dfrac{1}{2}(c+\sqrt{c^2-4b})$$
If $x=\dfrac{1}{2}(c+\sqrt{c^2-4b})$,then,
$$y=-\sqrt{c^2-4b}$$
Now,we plug the values of x and y in equation in (1),
We get,
$$-\dfrac{1}{4}(c^2+2c\sqrt{c^2-4b}+c^2-4b)\sqrt{c^2-4b}=a$$
And this is the required equation.
[If you use another value of $x$ then,we will get another equation like it.]
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now
| Hint: This hint also handles complex roots. Let $a,b,c\in\mathbb{C}$. Prove that, for a complex number $z$, $x:=z$ is a solution to $$x^4-a\,x^2\,\sin^2(bx)+c=0$$ if and only if $x:=-z$ is a solution of the above equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Regarding bijectivity of $f(x)=\frac{x}{1-x^2}$ Consider the map $f:(-1,1)\to \mathbb R,$ $$f(x)=\frac{x}{1-x^2}$$
Munkres claims that it is an order preserving bijection.
To see it's order preserving, assume $x< y$. Then $x^2<y^2, 1-x^2 >1-y^2$, so $\frac{1}{1-x^2}<\frac{1}{1-y^2}$.
To show it is a bijection, I guess it's easiest to construct the inverse.
We have $y=\frac{x}{1-x^2}\iff yx^2+x-y=0$. But this equation has two solutions for $x$; how to see which one we need?
| Suppose that $x \neq y$ but $f(x)=f(y)$ for some $x$ and $y$.
It follows that $\frac{x}{1-x^2}=\frac{y}{1-y^2}$, that is, $y(1+xy)=x(1+xy)$ and since $1+xy\neq 0$ it follows that $x=y$, but we assumed $x \neq y$ so when $x\neq y$ we have $f(x)\neq f(y)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Power series solutions
Construct two linearly independent, power series solutions to the ODE
$$u''+zu'+u=0.$$
Hence find the solution which satisfies $u(0)=1$ and $u'(0)=1.$
I have come up with the solution for the coeffecient, however I am not sure why we multiply them with $z^2$ instead of $z.$
| Let us consider your differential equation:
$$\frac{d^2u(z)}{dz^2}+z\cdot\frac{du(z)}{dz}+u(z)=0$$
Apply the reverse product rule:
$$\frac{d}{dz}\frac{du(z)}{dz}+\frac{d}{dz}(z\cdot u(z))=0$$
Integrate with respect to $z$:
$$\int\frac{d}{dz}\left(\frac{du(z)}{dz}+z\cdot u(z)\right) dz=0$$
$$\frac{du(z)}{dz}+z\cdot u(z)=C_1$$
Multiply on both sides by $e^{\frac{z^2}{2}}$:
$$e^{\frac{z^2}{2}}\cdot\frac{du(z)}{dz}+z\cdot e^{\frac{z^2}{2}}\cdot u(z)=C_1\cdot e^{\frac{z^2}{2}}$$
Recognize that $z\cdot e^{\frac{z^2}{2}}=\frac{d}{dz}\left(e^{\frac{z^2}{2}}\right)$ and apply the reverse product rule:
$$\frac{d}{dz}\left(e^{\frac{z^2}{2}}\cdot u(z)\right)=C_1\cdot e^{\frac{z^2}{2}}$$
Integrate with respect to $z$:
$$\int\frac{d}{dz}\left(e^{\frac{z^2}{2}}\cdot u(z)\right) dz=\int C_1\cdot e^{\frac{z^2}{2}} dz$$
$$e^{\frac{z^2}{2}}\cdot u(z)=C_1\cdot\sqrt{\frac{\pi}
{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)+C_2$$
Multiply on both sides by $e^{-\frac{z^2}{2}}$:
$$u(z)=C_1\cdot\sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot e^{-\frac{z^2}{2}}+C_2\cdot e^{-\frac{z^2}{2}}$$
The first derivative is given by
$$\frac{du(z)}{dz}=C_1-C_1\cdot \sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot z \cdot e^{-\frac{z^2}{2}}-C_2\cdot z \cdot e^{-\frac{z^2}{2}} $$
Apply conditions:
$$u(0)=1\rightarrow u(0)=C_2\rightarrow C_2 =1$$
$$u'(0)=1\rightarrow u'(0)=C_1\rightarrow C_1=1$$
Thus, a solution is given by
$$u(z)=\sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot e^{-\frac{z^2}{2}}+e^{-\frac{z^2}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\left|\operatorname{Re}(3+i+2\bar z^2 -iz)\right| \le 6$, when $|z| \le 1$ So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y^2-ix+y)\right|$
$\left|3+2x^2-2y^2+y\right|$
$3+2x^2-2y^2+y$
And then I'm not really sure where to go. Pretty sure I took the wrong approach because not using the initial information that $|z| \le 1$
| Note that $3+2x^2-2y^2+y\leq5+y-4y^2$ if $x^2+y^2\leq 1$. Therefore,
$$3+2x^2-2y^2+y\leq 5+y-4y^2=5+\frac{4y(1-4y)}{4}\leq 5+\frac{1}{4}\,\left(\frac{4y+(1-4y)}{2}\right)^2=\frac{81}{16}$$
for $y\in[-1,+1]$, where we have applied the AM-GM Inequality on the right-hand inequality. We also have $3+2x^2-2y^2+y\geq 3+y-2y^2$, so
$$3+2x^2-2y^2+y\geq 3+y-2y^2=(3-2y)(1+y)\geq 0\text{ for all }y\in[-1,+1]$$ This shows that
$$0\leq \text{Re}\big(3+\text{i}+2\,\bar{z}^2-\text{i}\,z\big)\leq \frac{81}{16}$$
for all complex numbers $z$ such that $|z|\leq 1$. The left-hand side is an equality iff $z=-\text{i}$, and the right-hand side is an equality iff $z=\dfrac{\pm3\sqrt{7}+\text{i}}{8}$
| {
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $
I understand every step until here
$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$
How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
| Divide both the numerator and denominator on the left side by $n^2$.
As n approaches $\infty$, the 1/n and 2/n tend to zero, giving $\frac{(1)(1)}{(2)(2)}$ = $\frac{1}{4}$.
The limit $\lim_{n\to\infty}(\frac{n+1}{n})^n$ = $\lim_{n\to\infty}(1+\frac{1}{n})^n$ = e.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $f(x) = \frac{-2x+1}{(2x-1)^2-1}$ is one-to-one on $(0,1)$? How do I prove that the function
$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one?
| $f(x) = \frac{-2x+1}{(2x-1)^2-1}$
$f'(x) = \frac {(-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'}{[(2x-1)^2 - 1]^2}$
$=\frac {-2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2}{[(2x-1)^2 - 1]^2}=\frac {2[1-(2x-1)^2] +4(2x -1)^2}{{[(2x-1)^2 - 1]^2}}$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0\le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 \le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
| {
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"source": "stackexchange",
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Try to find eigenvalues, orthogonal vectors when you have big matrix Let
$$A =\begin{bmatrix}
1& 1& 1& 1& 1& 1& 1& 1\\
1& 1& -1& -1& 1& -1& -1& 1\\
1& -1& 1& -1& -1& 1& -1& 1\\
1& -1& -1& 1& -1& -1& 1& 1\\
1& 1& -1& -1& -1& 1& 1& -1\\
1& -1& 1& -1& 1& -1& 1& -1\\
1& -1& -1& 1& 1& 1& -1& -1\\
1& 1& 1& 1& -1& -1& -1& -1
\end{bmatrix}$$
a) Find matrix equality on which you can conclude that column of matrix A make orthogonal vector system then find norm of that vectors. Find matrix equality on which you can conclude all eigenvalue then find their algebraic multiplicity.
b) From result of a) what you can say about rank of matrix A
c) Can you diagonalize this matrix? And what you can say about geometric multiplicity ?
I know matrix equality for eigenvalue,since $A=A^T$ then
$$AA^T=A^2=\begin{bmatrix}
8& 0& 0& 0& 0& 0& 0& 0\\
0& 8& 0& 0& 0& 0& 0& 0\\
0& 0& 8& 0& 0& 0& 0& 0\\
0& 0& 0& 8& 0& 0& 0& 0\\
0& 0& 0& 0& 8& 0& 0& 0\\
0& 0& 0& 0& 0& 8& 0& 0\\
0& 0& 0& 0& 0& 0& 8& 0\\
0& 0& 0& 0& 0& 0& 0& 8
\end{bmatrix}$$
so spectrum of $A^2=\{8\}$, then spectrum of $A=\{-2\sqrt2,2\sqrt2\}$ and their algebraic multiplicity is 4 for $\lambda1$ and $\lambda2$. But I think that matrix multiplicity that column is orthogonal is that $AA^T=A^TA=8I$, and norm is $\|x_i\|=\sqrt8, i=1,\ldots,8$
b) $\operatorname{rank}(A)=8$
c) Yes I can diagonalize matrix since $A=A^T$ then if I use spectral theorem $A=Q\Lambda Q^T$ so geometric multiplicity is the same as algebraic multiplicity. What you say is this good?
| Since $A^TA = 8I$, clearly the canonical vectors $\{e_1, \ldots, e_8\}$ are eigenvectors for $A^TA$. Your previous question then implies that $\{Ae_1, \ldots, Ae_8\}$ are orthogonal vectors and these are precisely the columns of $A$.
Since $\{Ae_1, \ldots, Ae_8\}$ is orthogonal, in particular it is linearly independent so the rank of $A$ is $$r(A) = \dim \operatorname{span} \{Ae_1, \ldots, Ae_8\} = 8$$
Direct calculation gives $\|Ae_i\| = \sqrt{8}$ for all $i = 1, \ldots, 8$.
Hence the matrix $\frac{1}{\sqrt{8}}A$ has orthonormal columns, so we conclude that $\frac{1}{\sqrt{8}}A$ is an orthogonal matrix, so its eigenvalues are on the unit circle. It is also symmetric so its eigenvalues are real. Therefore $$\sigma(A) = \sqrt{8}\sigma\left(\frac{1}{\sqrt{8}}A\right) \subseteq \sqrt{8} \cdot \{-1,1\} = \{-\sqrt8,\sqrt8\}$$
Let $a(\pm\sqrt{8})$ be the algebraic multiplicities of $\pm\sqrt8$ respectively. We have
$$0 = \operatorname{Tr} A = \sum_{\lambda \in \sigma(A)} \lambda = a(\sqrt{8})\sqrt{8} - a(-\sqrt{8})\sqrt{8}$$
Therefore $a(\sqrt{8}) = a(-\sqrt{8}) = 4$, because they are equal and $a(\sqrt{8}) + a(-\sqrt{8}) = 8$.
$A$ is of course diagonalizable because it is symmetric. Hence, geometric multiplicities of $\pm\sqrt{8}$ are equal to algebraic multiplicities, which are $4$.
| {
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The limit of $\lim_{n \to \infty}\left(1 + \frac{1}{\sqrt{n}}\right)^n$ I am approaching the question from an inequality perspective. In other words, I just want to see if the equation has an upper bound or lower bound.
After expanding the equation using binomial expansion, I get the term that $$\left(1 + \frac{1}{\sqrt{n}}\right)^n \leq 1 + n^{1/2} + \frac{n}{2} + \frac{n^{3/2}}{2^{2}}$$
For $n^{1/2} + \frac{n}{2} + ...$ $$a = \sqrt{n}, r = \frac{\sqrt{n}}{2}$$I then use the sum to infinity formula $S_{\infty} = \frac{a}{1 - r}$to get $\frac{4{\sqrt{n}} + 2n}{4 - n}$, add back in the 1 and simplify to get $\frac{\frac{4}{n} + \frac{4}{\sqrt{n}} + 1}{\frac{4}{n} - 1}$.
Finally, when I apply the limit of n to infinity, I get back -1. But, this does not seem right to me. Looking back at the equation, if n is positive, the sum to infinity should be a positive number instead.
My guess is that the ratio that I used when calculating sum to infinity is wrong. The ratio should be less than 1, but my ratio is more than 1 if n tends to infinity.
How do I go about solving this?
| You can use the binomial expansion of the equation:
$$\left(1 + \frac{1}{\sqrt{n}}\right)^n = \sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$
Since all terms are positives, you can fix a lower bound:
$$\sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} \ge \sum_{k=n-1}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$
Then note that:
$$\sum_{k=n-1}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} =$$$$ \begin{pmatrix} n\\n-1 \end{pmatrix} 1^{n-1}\times\left( \frac{1}{\sqrt{n}}\right)^1 + \begin{pmatrix} n\\n \end{pmatrix} 1^{n} \times\left(\frac{1}{\sqrt{n}}\right)^0 \\
$$$$=n\times\frac{1}{\sqrt{n}} + 1$$
So, you have
$$\left(1 + \frac{1}{\sqrt{n}}\right)^n \ge 1+\sqrt{n}$$
From there, you can deduce your limit!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
complex numbers equation deg 4 Find the sum of squares of elements of set $A$ if: $$A=\Big{\{}\big|z^n+\frac{1}{z^n}\big|;\:n\in\mathbb{N},\:z\in\mathbb{C},\: z^4+z^3+z^2+z+1=0\Big{\}}.$$
| Since $z^5=1$ (so $|z|=1$) it is not difficult to find $$A=\{ 2,|z+1|,|z^2+1|\}$$
If we devide $z^4+z^3+z^2+z+1=0$ with $z^2$ and put $t=z+{1\over z}$( so $t$ is a solution of $t^2+t-1=0$), then
$$|z+1|^2 = |z^2+2z+1| = |z||z+2+{1\over z}| = |t+2|$$
and $$|z^2+1|^2 = |t|^2$$
so we have $$S= 4+|z+1|^2+|z^2+1|^2= 4+|t+2|+|t|^2 $$
1. case If $t={\sqrt{5}-1\over 2}$ we get $S= 7$
2. case If $t={-\sqrt{5}-1\over 2}$ we also get $S= 7$
So $$\boxed{S=7}$$
| {
"language": "en",
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"question_score": "3",
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Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
| You obtained the equivalent form $$\frac{4x^2-4x-8}{(x^2+2)(x+4)}\geq 0$$
Here, $x^2+2$ in the denominator is positive for all $x\in \mathbb{R}$, so you can cancel it.
However, you need to make a case distinction in terms of the sign of $x+4$.
| {
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"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x=\sqrt[3]{n+x}\Rightarrow x^3=n+x\Rightarrow x^3-x-n=0$
could some help me how to solve it, Thanks
| $n = 0$ would fit. Now let's assume $n > 0$.
I think you are on the right track. If the given expression, call it $y$, is a natural number, then $x = y^2 - n$ must also be an integer - and it is strictly positive, so it must be a natural number.
Then from $x^3-x-n = 0$ and $x$ being an integer, we conclude that $x$ must be a divisor of $n$, and in particular $x \le n$.
Since $n \le 50$, we get $x^3 = x + n \le 100$, which means $x \le 4$. Try $x = 1, 2, 3, 4$, compute the respective $n$, and see if $y$ is still an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solving $2^x=x^3$ Algebraically
How can I solve $2^x=x^3$ algebraically?
I could take $\log_2(\cdot)$ on both sides, but I'd still be stuck.
| This equation follows the general equation
$$a^x = x^{b}$$
and can be solved, for $a \neq 1$, as follows:
\begin{align}
a^{x} &= x^{b} \\
e^{x \, \ln a} &= x^{b} \\
x^{b} \, e^{- x \, \ln a} &= 1 \\
x \, e^{- x \, \ln a/b} &= 1 \\
- \frac{x \, \ln a}{b} \, e^{- x \, \ln a/b} &= - \frac{\ln a}{b} \\
u \, e^{u} &= - \frac{\ln a}{b} \, \hspace{5mm} u = - \frac{x \, \ln a}{b} \\
u = - \frac{x \ln a}{b} &= W_{0}\left(- \frac{\ln a}{b}\right) \\
x &= - \frac{b}{\ln a} \, W_{0}\left(- \frac{\ln a}{b}\right),
\end{align}
where $W_{0}(x)$ is the primary branch of the Lambert W-function which is defined as the solution to the equation $W(x) \, e^{W(x)} = x$. Note that in the case $a=1$ the solution is $x = e^{2 \pi i/b}$.
For the problem proposed it is determined that:
$$2^{x} = x^{3} \, \text{ leads to } \, x = - \frac{3}{\ln 2} \, W\left(- \frac{\ln 2}{3}\right).$$
| {
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"source": "stackexchange",
"question_score": "2",
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Find all natural solutions $(a,b,c) $ such that $a^2−b, b^2−c, $ and $c^2−a $ are all perfect squares Find all natural solutions $(a,b,c)$ such that
$a^2-b$,
$b^2-c$,
$c^2-a$
are all perfect squares.
| Say $a>b$
Since $b<a\leq 2a-1$ we have: $$(a-1)^2<\underbrace{a^2-b}_{=x^2}<a^2\implies a-1 <x<a$$
which is impossible.
So $a\leq b$. With the same procedure we see that $b\leq c$ and $c\leq a$, so $a=b=c$
So we have $$(a-1)^2\leq a^2-a =x^2<a^2$$ and thus $x=a-1$ so $a^2-2a+1= a^2-a \implies a=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find solutions of $2x^3\ge 1-6x^2$ The inital function I want to study is
$$\frac{x^3+1}{x+2}$$
Its derivative is
$$\frac{2x^3+6x^2-1}{(x+2)^2}$$
I want to solve the following inequality graphically:
$$2x^3\ge 1-6x^2$$
I tried drawing the graphs of both the functions. But couldn’t find a way to find the minimum $(0,1)$ and maximum $(-1,0)$. Any hints on how to find those points?
| Exact solutions with CAS Maxima:
$$2 \cos{\left( \frac{\operatorname{atan}\left( \frac{\sqrt{7}}{3}\right) -3 {\pi} }{3}\right) }-1\le x \le \cos{\left( \frac{\operatorname{atan}\left( \frac{\sqrt{7}}{3}\right) +{\pi} }{3}\right) }-1,$$
$$x \ge \cos{\left( \frac{\operatorname{atan}\left( \frac{\sqrt{7}}{3}\right) -{\pi} }{3}\right) }-1$$
or
$$-2.942241850969666\le x\le -0.4421253016684752,$$
$$x\ge 0.3843671526381418$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913038",
"timestamp": "2023-03-29T00:00:00",
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Using Gram-Schmidt to find the QR decomposition I'm having problems doing a QR decomposition of a matrix...
Let $A=\begin{bmatrix}
{1} & {1} & {0} \\
{0} &{1} &{1} \\
{1} & {0} &{1}
\end{bmatrix}$
Find the QR decomposition for A
Here's what I've been doing:
I choose this basis, $B=\left \{(1,0,1), (1,1,0), (0,1,1)\right \}$ (the columns of the matrix).
Now I use the Gram-Schmidt process (and this is where I'm having trouble)
$u_{1}=(1,0,1)$
$u_{2}=(1,1,0)$ (cuz $<(1,0,1), (1,1,0)>=0$)
$u_{3}=(0,1,1)-\frac{<(0,1,1), (1,1,0)>}{<(1,1,0), (1,1,0)>}(1,1,0)-\frac{<(0,1,1), (1,0,1)>}{<(1,0,1), (1,0,1)>}(1,0,1)=$ $(0,1,1)-1/2(1,1,0)-1/2(1,0,1)=(-1, 1/2, 1/2)$
And now I find the norm for all the three vectors:
$||u_{1}||=||u_{2}||=||u_{3}||=\sqrt{2}$
So the orthonormal basis must be $B'= \left \{(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0), (\frac{-1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})\right \}$ (Which it isn't orthonormal)
So $Q=\begin{bmatrix}
{\frac{1}{\sqrt{2}}} & {\frac{1}{\sqrt{2}}} & {\frac{-1}{\sqrt{2}}} \\
{0} &{\frac{1}{\sqrt{2}}} &{\frac{1}{2\sqrt{2}}} \\
{\frac{0}{\sqrt{2}}} & {0} &{\frac{1}{2\sqrt{2}}}
\end{bmatrix}$
Which $Q^{t}Q \neq I$ ($I$ being the identity matrix), so all I did was wrong...
Where's my mistake?
| Suppose that we have $A$ as the following matrix
$$ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \tag{1} $$
modified Gram Schmidt is
$$ v_{1} =a_{1} \tag{2}$$
$$ r_{11} = \| v_{1} \| = \sqrt{ 2}\tag{3}$$
$$ q_{1} = \frac{v_{1}}{r_{11}} = \langle \frac{1}{\sqrt{2}} , 0 , \frac{1}{\sqrt{2}} \rangle \tag{4}$$
$$ r_{12} = q_{1}^{*}v_{2} \tag{5} $$
they only share one vector entry in common
$$ r_{12} = \frac{1}{\sqrt{2}} \tag{6}$$
$$ v_{2} = \langle 1 , 1, 0 \rangle - \frac{1}{\sqrt{2}}\langle \frac{1}{\sqrt{2}} , 0 , \frac{1}{\sqrt{2}} \rangle \tag{7} $$
$$ v_{2} = \langle \frac{1}{2} , 1, \frac{-1}{2} \rangle \tag{8} $$
$$ r_{13} = q_{1}^{*}v_{3} \tag{9} $$
$$ r_{13} = \frac{1}{\sqrt{2}} \tag{10} $$
$$ v_{3} = \langle 0 , 1, 1 \rangle - \frac{1}{\sqrt{2}} \langle 0 , \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \rangle \tag{11} $$
$$ v_{3} = \langle 0 , 1, 1 \rangle - \langle 0 , \frac{1}{2} , \frac{1}{2} \rangle
= \langle 0 , \frac{1}{2} , \frac{1}{2} \rangle \tag{12} $$
Now we repeat
$$ r_{22} = \| v_{2}\| =\sqrt{(\frac{1}{2})^{2} + 1 + (\frac{1}{2})^{2} }\tag{13}$$
$$ r_{22} = \| v_{2}\| = \frac{\sqrt{6}}{2}\tag{13}$$
$$ q_{2} = \frac{v_{2}}{r_{22} } \tag{14} $$
$$ q_{2} = \langle \frac{\frac{1}{2}}{\frac{\sqrt{6}}{2}}, \frac{1}{\frac{\sqrt{6}}{2}} ,\frac{\frac{-1}{2}}{\frac{\sqrt{6}}{2}} \rangle \tag{15} $$
$$ q_{2} = \langle \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} ,\frac{-1}{\sqrt{6}} \rangle \tag{16} $$
Go on from there..
| {
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Find the generating function for sequence $1,2,4,0,8,24,120,184,312,56,568,1592,...$ I'm having trouble finding a generating function for the sequence that has a closed form. The sequence can be deduced using two powers, with alternating negative and multiples of three as shown:
first term:
$1=1$
second term:
$2=1+2^{0}$
third term:
$4=1+2^{0}+2^{1}$
fourth term:
$0=1+2^{0}+2^{1}-2^{2}$
...
n-th term:
$n = 1+2^{0}+2^{1}-2^{2}+2^{3}+2^{4}+3(2^{5})+2^{6}+2^{7}-2^{8}+...$
Therefore, the -1 appears at every $4(mod6)$ term and the 3 appears at every $1(mod6)$ term greater than 1.
So I can construct the n-th ($n>1$) term as follows:
$f(n) = 1+\sum_{i=2}^{n}2^{i-2}u(i)$
where
$u(i)=-1$, if $i\equiv4(mod6)$
$u(i)=3$, if $i\equiv1(mod6)$
$u(i)=1$, otherwise
Is there a way to get a closed form of the function $f(n)$?
| Here is an idea:$$
-2^2 = 2^2-2\times 2^2$$ and $$3\times 2^5 = 2^5 + 2 \times 2^5.$$
So, for $n \ge 7$, we can split your sum as
$$f(n) = 1 + \sum_{i=0}^{n-2}2^i - 2\sum_{i=0}^{\lfloor\frac{n-4}{6}\rfloor}2^{6i+2} + 2\sum_{i=0}^{\lfloor\frac{n-7}{6}\rfloor}2^{6i+5}.$$
The last piece should be a matter of calculating these simpler sums.
| {
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How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is.
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$, but it was not fruitful.
Could people give me some hints on how to approach this problem?
To view the general formula, please visit The value of $\sum_{n=0}^{\infty}\frac{(n+1)^k}{(2n+1)!}$, where $k\in\mathbb{W}$
| $$\sinh x=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$$
differentiating and multiplying by $x$:
$$x\cosh x=\sum_{n=0}^\infty\frac{(2n+1)x^{2n+1}}{(2n+1)!}.$$
Again:
$$x^2\sinh x+x\cosh x=\sum_{n=0}^\infty\frac{(2n+1)^2x^{2n+1}}{(2n+1)!}.$$
Find $a$, $b$ and $c$ such that
$$a+b(2n+1)+c(2n+1)^2=(n+1)^2.$$
Then
$$a\sinh x+bx\cosh x+c(x^2\sinh x+x\cosh x)
=\sum_{n=0}^\infty\frac{(n+1)^2x^{2n+1}}{(2n+1)!}.$$
Now set $x=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the complex equation $ω^2=-11/4+15i$ The question is stated as following:
"First, solve the equation:
$ω^2=-11/4+15i$
and after, with the help of that, solve:
$z^2-(3-2i)z+(4-18i)=0$"
The problem for me lies in solving the system of equations for ω;
$Re:a^2-b^2=-11/4$ and $Im:2ab=15$
Where I eventually end up with the fourth degree equation: $4b^4+11b^2+15^2=0$ which I don't know how to solve, is this solveable or am I on the wrong track? I would very much appreciate some help, thank you for your time.
| As you have noted, equating the real parts gives $a^2 - b^2 = -\frac{11}{4}$. Now since the modulus of $\omega^2$ is the squared modulus of $\omega$ (think in polar form), $|\omega|^2 = a^2 + b^2 = \sqrt{(-11/4)^2 + 15^2} = \frac{61}{4}$.
Thus adding and subtracting, $a^2 = \frac{1}{2} \left(\frac{61}{4} + -\frac{11}{4} \right) = \frac{25}{4}$ and $b^2 = \frac{1}{2} \left(\frac{61}{4} - \frac{11}{4}\right) = 9$. But since the argument of $\omega^2$ is either in the second quadrant or the second quadrant + $360º$, the argument of $\omega$ is half that of $\omega^2$, which results in the first quadrant or the third quadrant. Hence $a+bi = \frac{5}{2} + 3i, -\frac{5}{2} - 3i$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{|z-a|}{|z-b|}=c$ for a constant $c$ not equal to $1$, then prove that the expression is a circle I saw a similar question which asked to show that the locus traced out by $arg(\frac{z-a}{z-b})=c$ is a circle. I know how to prove that but what is the connection between the $2$ questions. Can I describe the equation $\frac{z-a}{z-b}=c$ by its argument?
Also, how do I find the radius and centre of the equation?
Thanks
| Let $$\frac{|z-a|}{|z-b|}=c.$$
Let $z=x+iy$ and $a=a_1+ia_2$, $b=b_1+ib_2.$
We have $$(x-a_1)^2+(y-a_2)^2= c^2(x-b_1)^2+c^2(y-b_2)^2$$
$$ (1-c^2)x^2 + ( 1-c^2)y^2 -2(a_1-c^2 b_1)x-2(a_2-c^2 b_2)y+a_1^2+b_2^2-c^2b_1^2-c^2b_2^2=0$$
Which is the equation of a circle if the constant $ a_1^2+b_2^2-c^2 b_1^2-c^2 b_2^2$ is negative.
The center and the radius could easily be found.
| {
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A ring such that $(a+b)^2=a^2+b^2$ and $(a+b)^3=a^3+b^3$
Let $(A,+,\cdot)$ be a ring such that there are $a,b \in A$ which satisfy $$(a+b)^2=a^2+b^2, \quad (a+b)^3=a^3+b^3$$
Prove that $(a+b)^n=a^n+b^n,$ for all positive integers $n.$
I have found the following solution, but I am not quite satisfied with it.
From the hypothesis we get $ab+ba=0$ and $ab^2+ba^2=0.$ We will prove the identity using induction. Suppose that it is true for $1,2,...,n-1, \: n \geq 4.$ We can write $$(a+b)^n=(a+b)^{n-1}(a+b)=(a^{n-1}+b^{n-1})(a+b)=a^n+a^{n-1}b+b^{n-1}a+b^n$$
It is left to prove that $a^{n-1}b+b^{n-1}a=0.$ We can write $$a^{n-1}b+b^{n-1}a=a^{n-2}ab+b^{n-2}ba=-a^{n-2}ba-b^{n-2}ab \quad (*)$$
But $(a+b)^{n-1}=(a+b)^{n-2}(a+b)=(a^{n-2}+b^{n-2})(a+b)=a^{n-1}+b^{n-1},$ so $$a^{n-2}b+b^{n-2}a=0 \Rightarrow b^{n-2}a=-a^{n-2}b$$
Plugging this back in $(*)$ gives $$a^{n-1}b+b^{n-1}a=-a^{n-2}ba+a^{n-2}b^2=a^{n-2}(-ba+b^2)=a^{n-3}ab(-a+b)$$
But $0=ab^2+ba^2=ab^2-aba=ab(b-a),$ so $$a^{n-1}b+b^{n-1}a=a^{n-3}\cdot 0 = 0$$ and this completes the indution.
Is there any other solution, maybe quicker or more beautiful?
| It is no hard to see by using induction $a^nb=-b^na$ and $a^nb^n=-b^na^n$
Assume that $n$ is even.
$$(a+b)^n= \underbrace{(a+b)^2(a+b)^2(a+b)^2(a+b) ^2\cdots(a+b)^2}_{\text{$\frac{n}{2}$}\ \ \text{times}}$$
$$=(a^2+b^2)(a^2+b^2)\cdots(a^2+b^2)$$
$$=(a^4+a^2b^2+b^2a^2+b^4)\cdots (a^4+a^2b^2+b^2a^2+b^4)$$
Using $a^nb^n=-b^na^n$
$$=(a^4+b^4)(a^4+b^4)\cdots(a^4+b^4)$$
$$\vdots$$
*
*Finally, we reach two components by using $a^nb^n=-b^na^n$, We get $a^nb=-b^na$ if used as much as necessary we obtain
$$=a^n+b^n$$
$\textbf{Note:}$ When we reach the odd number of components, we do not handle the final component.
$\textbf{Example:}$ The logic is as follows:
$$(a+b)^6=(a+b)^2(a+b)^2(a+b)^2$$
$$=(a^2+b^2)(a^2+b^2)(a^2+b^2)$$
$$=(a^4+a^2b^2+b^2a^2+b^4)(a^2+b^2)$$
$$=(a^4+b^4)(a^2+b^2)=(a^6+a^4b^2+b^4a^2+b^6)$$
$$=a^6-b^4ab-a^4ba+b^6$$
$$=a^6+b^5a+a^5b+b^6=a^6+b^6$$
Similarly, we can think when $n$ is odd as following:(Actually, we encounterd above the situation below.)
$$(a+b)^n= \underbrace{(a+b)^2(a+b)^2(a+b)^2(a+b)^2\cdots (a+b)^2}_{\text{$\frac{n-1}{2},
$}\ \
\text{times}}
\underbrace{(a+b)}_{\text{$1$ times}}=a^n+b^n $$
| {
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Limiting Behavior of a Rational Function at Infinity Consider the following function:
$$f(x) = \frac{2x^2+x}{x^2-1}$$
I know there's a horizontal asymtote at $y = 2$. Without graphing, is there a way to tell that the $x\rightarrow \infty$ part is approaching from above and the $x \rightarrow -\infty$ is approaching from below?
| For $x > 1$,
$$\frac{2x^2+x}{x^2-1} > \frac{2x^2+1}{x^2-1}>\frac{2x^2-2}{x^2-1}=2$$
Since the denominator is positive and $2x^2+x > 2x^2+1$. So $f(x) \to 2$ from above as $x \to \infty$.
Similarly
For $x< -2$,
$$\frac{2x^2+x}{x^2-1} < \frac{2x^2-2}{x^2-1}=2$$
Since the denominator is positive and $2x^2+x < 2x^2-2$.
| {
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Dimension of a vector space consisting of two $2$ by $2$ matrices that commute with all $2$ by $2$ matrices. What is the dimension of the vector space that consists of $2$ by $2$ matrices that commute with all $2$ by $2$ matrices?
Let $A$ be a matrix that commutes with all $2$ by $2$ matrices and let $B$ be any $2$ by $2$ matrix.
So if I pick $B$ to be $\begin{pmatrix}1&2\\ 3&4\end{pmatrix}$ and generalize $A$ to be $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, and I multiply them together, I get that:
$AB$ $=$ $\begin{pmatrix}a+2c&b+2d\\ 3a+4c&3b+4d\end{pmatrix}$ and $BA$ $=$ $\begin{pmatrix}a+3b&2a+4b\\ c+3d&2c+4d\end{pmatrix}$.
After setting the respective elements equal to each other and subtracting them from each other so that I got an equation that equalled $0$, I got the matrix ($a$ is the first column, $b$ is the second column, $c$ is the third column, $d$ is the fourth column):
\begin{pmatrix}0&-3&2&0\\ -2&-3&0&2\\ 3&0&3&-3\\ 0&3&-2&0\end{pmatrix}
Which row reduced to:
\begin{pmatrix}1&0&1&-1\\ 0&1&\frac{-2}{3}&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}
Since the rank of this matrix is $2$ that would mean that the dimension of the space is also $2$, but I am not sure if the way I have done it is correct. Especially since I have only picked one specific matrix to represent $B$.
Any help?
| It appears that you’re looking for all $2\times2$ that commute with every $2\times2$ matrix, that is, matrices $A$ such that $[A,X]=AX-XA=0$ for all $X$. Let $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}.$$ Since this has to commute with every $2\times2$ matrix, it has to commute with the elements of the standard basis, so we compute $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} - \begin{bmatrix}1&0\\0&0\end{bmatrix} \begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}0&-b\\c&0\end{bmatrix},$$ therefore $b=c=0$. Continuing on to the next basis element, $$\begin{bmatrix}a&0\\0&d\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix} - \begin{bmatrix}0&1\\0&0\end{bmatrix} \begin{bmatrix}a&0\\0&d\end{bmatrix} = \begin{bmatrix}0&a-d\\0&0\end{bmatrix},$$ so $a=d$. I’ll leave it to you to verify that the other two standard basis matrices don’t introduce any other constraints. Thus, the matrices that commute with every standard basis matrix are of the form $aI$, which clearly commutes with every $2\times2$ matrix as well. It should be obvious that these matrices do indeed form a subspace of the space of $2\times2$ matrices and what the dimension of that subspace is.
| {
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Calculating the sum of the infinite series $\frac{1}{5} + \frac{1}{3}\frac{1}{5^3} + \frac{1}{5} \frac{1}{5^5} +......$ How do I calculate the sum of the infinite series?
$$\frac{1}{5} + \frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......$$
My attempt :
I know that $$\log (\frac{1+x}{1-x}) = 2 \, \left(x + \frac{x^3}{3} + \frac{x^5}{5} +.....+\frac{x^{2r-1}} {2r-1}+..\right)$$
Now
\begin{align}\frac{1}{5} + (\frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......) &= \frac{1}{5} + \log \frac{(1 +1/5)} {(1-1/5)} - \frac{5}{2} \\ &=\frac{1}{5} + \log (\frac{6}{4})-10= -\frac{49}{5} + \log \frac{3}{2}
\end{align}
Please verify my answer, thank you!
| Let $$f(x):=\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}.$$ The radius of convergence is $1$.
Then
$$f'(x):=\sum_{k=0}^\infty{x^{2k}}=\sum_{k=0}^\infty{(x^2)^k}=\frac1{1-x^2}.$$
By integration,
$$f(x)=\int_0^x\frac{dx}{1-x^2}=\frac12\int_0^x\left(\frac1{1+x}+\frac1{1-x}\right)dx=\frac12\log\left|\frac{1+x}{1-x}\right|$$ which you evaluate at $x=\dfrac15$.
| {
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Cylotomic polynomial $\Phi_n(x) $: if $a$ is a root then so is $a^{k}, (k, n) =1$ It is well known that the $n$'th cylotomic polynomial $\Phi_n(x) \in\mathbb {Q} [x] $ has its roots as $\cos(2k\pi/n)+i\sin(2k\pi/n),(k,n)=1$.
I would like to establish the same fact without using any knowledge of complex numbers or circular functions by working entirely in the field $K=\mathbb{Q} [x] /(\Phi_n(x)) $.
More formally
Let $\Phi_n(x) $ be defined as the irreducible factor $f(x) $ of $x^n-1$ such that $f(x) $ is not a factor of $x^r-1$ for any $r<n$ and let $K=\mathbb{Q} [x] /(\Phi_n(x))$. If $a\in K$ is a root of $\Phi_n(x) $ then so is $a^{k}, (k, n) =1$.
I demonstrate the easy case for $n=3$. Here $\Phi_{3}(x)=x^2+x+1$ and clearly if $a^2+a+1=0$ and $b=a^2$ then $b^2=a^4=a^3a=a$ and hence $b^2+b+1=a+a^2+1=0$ so that both $a$ and $b=a^2$ are roots of the polynomial $x^2+x+1$.
I think (but not sure) this is same as saying that if the polynomial $\Phi_n(x^k) $ is divided by $x^n-1$ then the remainder is $\Phi_n(x) $.
| Your statement about the remainder after polynomial long division is false. For instance, $$\Phi_{15}(x) = x^8 -x^7 + x^5-x^4+x^3-x+1$$
$$\Phi_{15}(x^2) = x^{16} -x^{14} + x^{10}-x^{8}+x^6-x^2+1$$
which has remainder $-x^{14} +x^{10} -x^8+x^6-x^2+x+1$ after division by $x^{15}-1$.
Instead, let us proceed via the definition of $\Phi_n(x)$. The roots of $\Phi_n(x)$ are exactly those $\alpha$ which satisfy $\alpha^n=1$ but not $\alpha^d=1$ for any $1\leq d<n$. So if $k$ is an integer relatively prime to $n$, we have two things: $$(\alpha^k)^n = \alpha^{kn} = (\alpha^n)^k = 1$$ $$(\alpha^k)^d = \alpha^{kd} = \alpha^{k'} \neq 1$$ where in the final line, $1 \leq k' < n$ is an integer equal to $kd$ modulo $n$ - we know that $kd$ cannot be divisible by $n$ because $d$ is smaller than $n$ and $k$ is relatively prime to $n$. Therefore $\alpha^{k'}$ is a root of $\Phi_n(x)$.
| {
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Transforming a matrix to diagonal matrix Show that the matrix $$A = \begin{bmatrix}a&h\\h&b\end{bmatrix} ,\quad a \ne b$$ is transformed to diagonal matrix $D = P^{-1}AP$, where $$P = \begin{bmatrix}\cos x& -\sin x\\\sin x& \cos x\end{bmatrix}$$ and $$\tan2x=\frac{2h}{(a-b)}$$
I understand that a $n \times n$ matrix $A$ is diagonalizable if there is a diagonal matrix $D$ such that $A$ is similar to $D$, that is, if there is an invertible matrix $P$ such that $P^{-1}AP= D$. Also, columns of $P$ are $n $ linearly independent eigenvectors of $A$ and the diagonal entries of $D$ are the eigenvalues of $A$ corresponding to the eigenvectors in $P$ in the same order. I am unable to proceed solving the above problem with these leads.
| You seem to know the theory behind the problem. So there is no reason you cannot find the eigenvalues of $A$ and the corresponding eigenvectors required for the diagonalisation. Note that since $A$ is symmetric, it is orthogonally diagonalisable.
To do this in a slightly different setting but essentially the same method, consider the quadratic form
$$Q(X,Y)=(X\quad Y)\begin{pmatrix}a&h\\h&b\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}=aX^2+2hXY+bY^2\quad,\,\,a\ne b$$
Let $x$ be the angle by which the coordinate axes $X,Y$ should be rotated about the origin so that $Q$ is transformed to another quadratic form in which the product term is absent. The vanishing of the product term is equivalent to diagonalising the matrix $A$ or the quadratic form $Q$.
Suppose $u,v$ is our set of coordinate axes.
The required transformation is given by
$$\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$$
Or, $$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}\cos x&\sin x\\-\sin x&\cos x\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}$$
So, $$Q(X,Y)=Q'(u,v)=Au^2+2Huv+Bv^2$$
, where
$$A=a\cos^2x+2h\sin x\cos x+b\sin^2 x,$$
$$B=a\sin^2x-2h\sin x\cos x+b\cos^2 x,$$
$$H=(b-a)\sin x\cos x+h(\cos^2x-\sin^2 x)$$
The product term $uv$ in $Q'$ vanishes iff
\begin{align}
H=0&\implies(b-a)\sin x\cos x+h(\cos^2 x-\sin^2 x)=0
\\&\implies (b-a)\sin 2x+2h\cos 2x=0
\\&\implies \tan 2x=\frac{2h}{a-b}
\end{align}
Thus using the rotation matrix $P$, we have transformed the matrix $A$ to a diagonal matrix $D$ which is nothing but the matrix associated with the quadratic form $Au^2+Bv^2$.
| {
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Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by
$$\left(-17\right)^2\geq \left(x\sqrt{8-x^2}+x\sqrt{25-x^2}\right)^2.$$
But I think this is not the best approach.
| Since $>,<$ are not defined concepts over the complex numbers, we can assume that $\sqrt{8-x^2}, \sqrt{25-x^2}$ are real
And $\sqrt{25-x^2} > \sqrt{8-x^2}$
Which implies $x\in [-2\sqrt 2,0]$
but for all $x$ in this interval, $\sqrt{8-x^2} - \sqrt{25-x^2} > x$
there is no solution.
| {
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What are all of the negative integral solutions of $y^2+6xy-8x=0$? I got the answer as $(0,0)$ by making $D\ge 0$ (quadratic in $y$).
However, how do I know this is the only possible answer?
| $y^2+6xy-8x=0$ iff $y^2=2x(4-3y)$. Therefore, $\dfrac{y^2}{4-3y}=2x$ is an integer.
Computing the gcd of $y^2$ and $4-3y$ gives
$$
\frac{16}{4-3y} = 9 \frac{y^2}{4-3y} + (3y+4) = 18x + (3y+4)
$$
Therefore, $4-3y$ divides $16$ and so $4-3y \in \{ \pm 1, \pm 2, \pm 4, \pm 8,\pm 1 6 \}$. The only integer solutions $y$ for which $x$ is also an integer are:
$$
\begin{array}{r|rrr}
y & 2 & 0 & 4
\\
x & -1 & 0 & -1
\end{array}
$$
| {
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Using complex numbers prove that $\sin(\frac {\pi}{m})\sin(\frac {2\pi}{m})\ldots\sin(\frac {(m-1)\pi}{m})= \frac{m}{2^{m -1}}$ Prove that for $m=2,3,\dots$
$$
\sin\left(\frac{\vphantom{1}\pi}m\right)\sin\left(\frac{2\pi}m\right)\sin\left(\frac{3\pi}m\right)\cdots\sin\left(\frac{(m-1)\pi}m\right)=\frac{m}{2^{m-1}}
$$
I have no idea how to begin at all, I'm trying to think of a way using de Moivre's theorem but I can't seem to figure it out.
I'm sorry for not showing effort but I'm completely stuck.
| $|\cos \theta + i\sin \theta - 1|\\
\sqrt {1 - 2\cos \theta} = 2\sin \frac {\theta}{2}\\
2\sin\theta = |e^{2\theta i} - 1|$
$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1|$
The set $\{e^{\frac{2n\pi}{m} i}\}$ make up the roots of $z^m - 1 = 0$ excluding the root at $z= 1$ or the roots of $(z^{m-1} + z^{m-2} + z^{m-3} + \cdots + 1)$
$(z - e^{\frac{2\pi}{m}i})(z - e^{\frac{4\pi}{m}i})(z - e^{\frac{6\pi}{m}i})\cdots(z - e^{\frac{(m-2)\pi}{m}i})= (z^{m-1} + z^{m-2} + z^{m-3} + \cdots + 1)\\
|z - e^{\frac{2\pi}{m}i}||z - e^{\frac{4\pi}{m}i}||z - e^{\frac{6\pi}{m}i}|\cdots|z - e^{\frac{(m-2)\pi}{m}i}|= |z^{m-1} + z^{m-2} + z^{m-3} + \cdots + 1| $
and now set z = 1
$|1 - e^{\frac{2\pi}{m}i}||1 - e^{\frac{4\pi}{m}i}||1 - e^{\frac{6\pi}{m}i}|\cdots|1 - e^{\frac{(m-2)\pi}{m}i}| = |1^{m-1} + 1^{m-2} + 1^{m-3} + \cdots + 1| = m$
$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1| = \frac {m}{2^{m-1}}$
| {
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Finding a particular coefficient in a polynomial I'm trying to get the coefficient of $x^6$ of this polynomial product:
$$x^2(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4).$$
I know with infinite series, you can use the closed form solution of the series to calculate the coefficient, but I haven't been able to figure out a way to calculate the coefficient for a finite series.
I hope there is a better way than doing the algebra.
| What is being asked is to find the coefficient of $x^4$ of
$$(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4),$$
which, when multiplied by $x^2$ leads to the coefficient of $x^6$ overall.
This can be obtained by expansion:
\begin{align}
P &= (1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4) \\
&= (1+x+x^2+x^3+x^4+x^5)(1+x+2 x^2 + x^3 + 2 x^4 + x^5 + x^6) \\
&= 1 + 2 x + 4 x^2 + 5 x^3 + 7 x^4 + \mathcal{O}(x^5)
\end{align}
Now, $x^2 \, P = x^2 + 2 x^3 + 4 x^4 + 5 x^5 + 7 x^6 + \mathcal{O}(x^7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $A = x - y + z$ if $3x + 5y + 7z = 29$ and $x , y , z \in \mathbb{Z}^+$ I've found the answer by trial and error $(x = 4 , y=2,z=1) \Rightarrow$ $A = 3$. I tried to solve it using modular arithmetic but it didn't work .
$$3x + 5y + 7z = 29 $$
$$ 3x + 5y+7z \equiv 29 \mod 3$$
$$ 2y +z \equiv 2 \mod 3 $$
$$ y= -k , z=2k + 2$$
putting $ y= -k , z=2k + 2$ to original equation leads to $x = -3k + 5$ . Then $A = -3k+5 +k +2k+2 = 7$ . What's wrong about my answer?
| You don't get $z = 2k + 2$. you get $z \equiv 2k + 2 \mod 3$.
Or $z = 2k + 2 + 3M$.
Plugging that into the equation we get
$3x -5k + 2k + 14k + 14 + 21M = 29$ so
$3x = -9k + 15 + 21M$ or
$x = -3k + 5 + M$ for some integer $M$. As $M$ can be any integer that is pretty useless but let's continue.
Plugging $x = -3k + 5 + M; y = -k; z = 2k + 2 + 3M$ we get:
$A = -3k + 5 + M +k + 2k +2 + 3M = 7 + 4M$
In this case it turns out that $M = -1$.
$A \equiv 7 \mod 4$ is not entirely useless however.
But solving $\mod 3$ can only give you at best a solution $\mod 3$. That's not good enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Miscalculating the determinant I am learning linear algebra and am getting stuck when trying to calculate the determinant using elementary row operations. Consider the matrix A.
\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 1 & 1 & 1 \\
-2 & -2 & 3 & 3 \\
1 & 2 & -2 & -3 \\
\end{vmatrix}
According to the solution in my textbook and Matlab the determinant should be 10. I however find -20. Here is what I did.
I first interchanged row 1 and row 3.
\begin{vmatrix}
1 & 2 & -2 & -3 \\
1 & 1 & 1 & 1 \\
-2 & -2 & 3 & 3 \\
0 & 1 & 2 & 3
\end{vmatrix}
I then substracted row 1 from row two. I also added the first row twice to the third row.
\begin{vmatrix}
1 & 2 & -2 & -3 \\
0 & -1 & 3 & -2 \\
0 & 2 & -1 & 9 \\
0 & 1 & 2 & 3
\end{vmatrix}
Then, I added the second row twice to the third row and once to the fourth row.
\begin{vmatrix}
1 & 2 & -2 & -3 \\
0 & -1 & 3 & -2 \\
0 & 0 & 5 & 5 \\
0 & 0 & 5 & 1
\end{vmatrix}
My final operation was to substract the third row from the fourth row, which gave:
\begin{vmatrix}
1 & 2 & -2 & -3 \\
0 & -1 & 3 & -2 \\
0 & 0 & 5 & 5 \\
0 & 0 & 0 & -4
\end{vmatrix}
Finally, I calculated the determinant: $(-1)^1 \cdot 1 \cdot -1 \cdot 5 \cdot -4 = -20$ The $(-1)^1$ is there since I did one operation in which I interchanged two rows.
I would really appreciate if you could tell me what I did wrong.
Martijn
| You have a few mistakes in calculations. In step $2$ when you subtract row $1$ from row $2$ it should be $1-(-3)=4$ in the fourth column. In the same step when you add row $1$ twice to row $3$ it should be $3+2\times(-3)=-3$ in the fourth column.
| {
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"question_score": "6",
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"answer_id": 0
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Solution for $\frac{3}{x-9} \gt \frac{2}{x+2}$ What I did:
$\frac{3}{x-9} \gt \frac{2}{x+2}$
$3(x+2) \gt 2(x-9)$
$3x+6 \gt 2x-18$
$x \gt -24$
When typing this in in symbolab, it showed me that the solution is $-24 \lt x \lt -2$ or $x \gt 9$
What did i do wrong ?
How come i didnt get the correct solution ?
| A shorter way
The domain of the inequality is $\mathbf R\smallsetminus\{-2,9\}$.
You can multiply both sides of the inequality by a positive number. To remove the denominators, we'll multiply by $(x+2)^2(x-9)^2$ and obtain
\begin{align}
3(x+2)^2(x-9)>2(x+2)(x-9)^2&\iff(x+2)(x-9)\bigl(3(x+2)-2(x-9)\bigr) >0\\
&\iff (x+2)(x-9)(x+24)>0.
\end{align}
Now , a cubic polynomial with simple roots has alternating signs on the successive intervals determined by the roots, and this particular polynomial tends to $+\infty$ when $x$ tends to $+\infty$, so the solutions are
$$-24<x<-2\enspace\text{ or }\enspace x>9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the relationship of the length of triangle's sides.
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
Now determine what kind of triangle $\triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $b\ge4$ and $c\ge 1$. Also $10a+2\sqrt{b-4}-22\ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
| HINT:
Write the equality as $$a^2-10a+22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}=0$$ and since we know that $a$ is real, $$22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}\le25\\b-2\sqrt{b-4}\le3-|\sqrt{c-1}-2|\le3.$$ But if $f(b)=b-2\sqrt{b-4}$, $f'(b)=1-\dfrac1{\sqrt{b-4}}=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=\cdots\,\,?$
Spoiler:
The triangle is equilateral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Constructing a field with exactly $81$ elements I was thinking $\frac{\mathbb{Z_3}[x]}{(x^3+x+1)} \times \frac{\mathbb{Z_3}[x]}{(x^3+x+1)}$.
$(x^3+x+1)$ is irreducible in $\mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!
How would I do this if working over $\mathbb{Z_9}$ and $\mathbb{Z_2}$? in $\mathbb{Z_9}$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!
| Let's see. What are the irreducible quadratics: $x^2+x+2\,,x^2+2x+2\,,x^2+1$.
I count $3$ (thanks @Lubin).
So, there are ${3\choose 2}+3=6$ combinations to check.
These are the $6$ quartics I get: $x^4+1\,,x^4+x^3+x+1\,,x^4+2x^3+2x+2 \,,x^4+2x^3+x^2+x+1\,,x^4+x^3+x^2+2x+1\,,x^4+2x^2+1$.
But there are $2\cdot 3^4=162$ quartics to choose from... $81$ of them monic.
So, if I choose a quartic that doesn't have a root and isn't among the $6$ products, it will be irreducible.
So, how about $x^4+2x^3+2$?
Finally, $\mathbb F_{3^4}\cong\frac{\mathbb Z_3[X]}{x^4+2x^3+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Divergence of infinite series $(\frac{3k-2}{4k+2})^{2k-3}$ I would like to ask if my solution for testing the divergence of the infinite series below is correct.
$$\sum_{k=1}^{\infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3}$$
I used the Cauchy ratio test.
$$
\lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3} =
\lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k} \cdot \left(\frac{3k-2}{4k+2}\right)^{-3} =
\lim_{k \to \infty} \left(\frac{\left(3k-2\right)^{2}}{\left(4k+2\right)^{2}}\right)^{k} \cdot \left(\frac{3k-2}{4k+2}\right)^{-3} =
\lim_{k \to \infty}\left(\frac{9k^{2}-12k+4}{16k^2+16k+4}\right)^{k} \cdot \left(\frac{k\left(3-\frac{2}{k}\right)}{k\left(4+\frac{2}{k}\right)}\right)^{-3} =
\lim_{k \to \infty} \left(\frac{k^{2}\left(9-\frac{12}{k}+\frac{4}{k^{2}}\right)}{k^{2}\left(16+\frac{16}{k}+\frac{4}{k^{2}}\right)}\right)^{k} \cdot \left(\frac{3}{4}\right)^{-3} =
\lim_{k \to \infty} \frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k}
$$
Therefore:
$$
\left\lvert \frac{\frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k}}{\frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k+1}} \right\rvert =
\left\lvert \frac{\frac{9^{k}}{16^{k}}}{\frac{9^{k+1}}{16^{k+1}}} \right\rvert =
\left\lvert \frac{9^{k}\cdot 16^{k+1}}{16^{k}\cdot 9^{k+1}} \right\rvert =
\left\lvert \frac{9^{k}}{9^{k+1}} \cdot \frac{16^{k+1}}{16^{k}} \right\rvert =
\left\lvert 9^{k-k+1}\cdot16^{k+1-k} \right\rvert =
\left\lvert 9\cdot16 \right\rvert =
144
$$
As $144 > 1$, the series is divergent.
Is this a correct solution?
| $$
\left\lvert \frac{9^{k}}{9^{k+1}} \cdot \frac{16^{k+1}}{16^{k}} \right\rvert =
\left\lvert 9^{k-\color{red}{(k+1)}}\cdot16^{k+1-k} \right\rvert =
\left\lvert 9^{k-k-1}\cdot16^{k+1-k} \right\rvert =
\left\lvert \frac {16} 9 \right\rvert $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a Jordan Basis after finding the Jordan Canonical Form The question asked to find the Jordan Canonical Form and Jordan Basis of $\begin{bmatrix}1 & 1 & 0 & -1\\0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$, and after finding the characteristic polynomial, I used its eigenvalues and their multiplicities to find the JCF.
I found the JCF to be $\begin{bmatrix}0 & 0 & 0 & 0\\0 & 1 & 1 & 0\\0&0&1&1\\ 0 & 0 & 0 & 1\end{bmatrix}$, but I am unsure of how to find a Jordan Basis for this.
Intuitively, I'd think the Jordan Basis for this would be the vectors that make up the three linearly independent columns, but my textbook doesn't really explicitly state what the basis is when they define the JCF. Is my intuition correct or am I way off? If so how can I find the Jordan Basis for this JCF?
| let your matrix be $A$ and name $B = A - I.$ the characteristic polynomial says $A B^3 = 0.$ This is also the minimal polynomial.
The method I like is to find a column vector $w$ with $B^3 w = 0$ but $B^2 w \neq 0.$ I like zeros and ones, so I am choosing
$$
w =
\left(
\begin{array}{r}
0 \\
0 \\
0 \\
1 \\
\end{array}
\right)
$$
This is going to be the far right column, number 4. Next $v = B w$ with
$$
v =
\left(
\begin{array}{r}
-1 \\
1 \\
0 \\
0 \\
\end{array}
\right)
$$
We reach a genuine eigenvector finally with $u = Bv$
$$
u =
\left(
\begin{array}{r}
1 \\
0 \\
0 \\
0 \\
\end{array}
\right)
$$
because $Bu = B^2 v = B^3 w = 0,$ so $Au = u.$
The 0 eigenvector is the first column. Call the whole thing $R$
$$
R =
\left(
\begin{array}{rrrr}
0&1&-1&0 \\
0&0&1&0 \\
1&0&0&0 \\
0&0&0&1 \\
\end{array}
\right)
$$
Next calculate $\det R = 1$ and
$$
R^{-1} =
\left(
\begin{array}{rrrr}
0&0&1&0 \\
1&1&0&0 \\
0&1&0&0 \\
0&0&0&1 \\
\end{array}
\right)
$$
finally $J = R^{-1} A R$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by the limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$ I was reviewing $\mathbb{R}-$analisys with a friend and I'm thinking about one of the questions...
Prove by the $\epsilon-\delta$ limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$.
My answer was very long, someone could do a better answer? Does is it OK in this way? Thanks very much.
My attempt
We have $|f(x)-(-3)|=|f(x)+3|=|\frac{x^2-1}{x-3}+3|=|\frac{x^2-1+3x-9}{x-3}|=|\frac{x^2+3x-10}{x-3}|=|\frac{(x-2)(x+5)}{x-3}| \qquad (i)$
and
$|x+5|=|x+7-2|<|x-2|+|7|=|x-2|+7 \qquad (ii)$
$ |x-3|=|x-2-1|>|x-2|+|-1|=|x-2|+1\Longrightarrow \dfrac{1}{|x-3|}<\dfrac{1}{|x-2|+1} \qquad(iii)$
So, by (ii) and (iii)
$|\frac{(x-2)(x+5)}{x-3}|=\dfrac{|x-2||x+5|}{|x-3|}<\dfrac{|x-2|(|x-2|+7)}{|x-2|+1} \qquad (iv)$
But also, if $|x-2|<\delta \qquad (v)$, we have
$|\frac{(x-2)(x+5)}{x-3}|<^{(iv)}\dfrac{|x-2|(|x-2|+7)}{|x-2|+1}<^{(v)}\dfrac{\delta(\delta+7)}{\delta+1} \qquad(vi)$
And more than that
$\dfrac{\delta(\delta+7)}{\delta+1}<\dfrac{\delta(\delta+1)}{\delta+1}=\delta \qquad (vii)$
So, if $\delta<\epsilon$, we have
$|f(x)-(-3)|=^{(i)}|\frac{(x-2)(x+5)}{x-3}|<^{(vi)}\dfrac{\delta(\delta+7)}{\delta+1}<^{(vii)}\delta<\epsilon$, Q.E.D.
| I think (i) part is correct, and assume we choose $x$ from $|x-2|<\dfrac12$ then
$$\dfrac32<x<\dfrac52$$
$$-\dfrac32<x-3<-\dfrac12$$
$$\dfrac{13}{2}<x+5<\dfrac{15}{2}$$
these show
$$|\frac{(x-2)(x+5)}{x-3}|<\dfrac{15}{2}|x-2|.2<15\delta$$
so it is sufficient to have $\delta\leq\min\{\dfrac{1}{15}\varepsilon,\dfrac12\}$.
| {
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Which is greater, $\left(\frac{e}{2}\right)^\sqrt{3}$ or $(\sqrt{2})^{\pi/2}$? (no calculators) From a math contest in 1985:
Determine which of the following is greater: (no calculators)
$$\left(\frac{e}{2}\right)^\sqrt{3} \, \hspace{3mm} \text{or} \hspace{3mm} \, (\sqrt{2})^{\pi/2}$$
Hints are welcome, but I'm totally lost and would really appreciate a solution.
| This comparison is potentially pretty tricky inasmuch as the numerical values are pretty close, within $2\%$:
$$\left(\frac{e}{2}\right)^{\sqrt{3}} = 1.701\ldots, \qquad (\sqrt 2)^{\pi / 2} = 1.723\ldots$$
The following method is pretty quick-and-dirty, and I doubt that it's the most elegant way. But it doesn't use anything more advanced than knowledge of a few well-known Taylor series, and it doesn't involve integers with more than two digits.
Hint Since $\log$ is an increasing function, we may as well take the logarithms of both sides and compare quantities with the awkward exponents in more tractable places, so we're comparing $\sqrt{3}(1 - \log 2)$ and $\frac{\pi}{4} \log 2$. Subtracting these and collecting terms in $\log 2$ gives
$$\left(\frac{\pi}{4} + \sqrt{3}\right) \log 2 - \sqrt{3},$$
and the claim that $(\sqrt 2)^{\pi / 2} > \left(\frac{e}{2}\right)^{\sqrt{3}}$ is equivalent to showing that this quantity is positive. At this point we can see that this amounts to establishing sufficiently tight lower bounds for $\pi, \log 2$.
One efficient way to produce such bounds is to use the first terms of the Maclaurin series
$$\operatorname{arctanh} x \sim x + \frac{1}{3} x^3 + \cdots \qquad \textrm{and} \qquad \arcsin x \sim x + \frac{1}{6} x^3 + \cdots,$$ the identities
$$\log 2 = \log\left(\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}\right) = 2 \operatorname{arctanh} \frac{1}{3} \qquad \textrm{and} \qquad \pi = 6 \arcsin \frac{1}{2},$$ and the facts that all of the coefficients of both of these series are nonnegative. These are good choices in part because both series converge relatively quickly at the indicated values.
Solution Using the first two terms of each series gives \begin{align*}\log 2 &= 2 \operatorname{arctanh} \frac{1}{3} > 2 \left[\left(\frac{1}{3}\right) + \frac{1}{3}\left(\frac{1}{3}\right)^3\right] = \frac{56}{81}, \\ \pi &= 6 \arcsin \frac{1}{2} > 6 \left[\left(\frac{1}{2}\right) + \frac{1}{6}\left(\frac{1}{2}\right)^3\right] = \frac{25}{8} ,\end{align*} giving the bound $$\left(\frac{\pi}{4} + \sqrt{3}\right) \log 2 - \sqrt{3} > \left[\frac{1}{4}\left(\frac{25}{8}\right) + \sqrt{3}\right] \frac{56}{81} - \sqrt{3} = \frac{25}{4 \cdot 81}(7 - 4 \sqrt{3}).$$ So, it suffices to show that $7 > 4 \sqrt{3}$, but squaring both sides shows that this is equivalent to $49 > 48$.
| {
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Simplify the Dot Product in terms of $a$ and $b$ Where $a$ and $b$ are arbitrary vectors
$(a+2b) \cdot (2a-b)$
$$a\cdot(2a-b)+2b\cdot(2a-b) = 2(a\cdot a)-a\cdot b+1(b\cdot a)-2(b\cdot b)$$
$$=2(a)-ab+4ab-2(b)^2$$
$$=2a^2-2b^2$$
$$=2(a^2-b^2)$$
Where did i go wrong in simplifying this?
| If you denote dot product as $\cdot$, you get
$$
(a+2b) \cdot (2a - b)
= 2 a \cdot a + 4 b \cdot a - a \cdot b - 2 b \cdot b
= 2 a \cdot a + 3 b \cdot a - 2 b \cdot b
$$
and you can further simplify $a\cdot a = |a|$ if you like.
Your error is the 2nd step (3rd line), where you assumed $$2b\cdot(2a-b) = 1 b\cdot a - 2 b \cdot b,$$ the the correct coefficient of $b \cdot a$ is 4, not 1.
| {
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Proving that the series with the general term $u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$ diverges.
$$u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$$ for $ n \in \mathbb{N}^*$.
*
*Prove that $(u_n)$ is convergent toward $0$.
*Prove that the series with the general term $(-1)^n u_n$ converges.
*Prove that $$\sum_{n=0}^{\infty} (-1)^n u_n = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$
*Compute $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$
Hint: You can start by proving $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos(x)}dx$$
*Prove that the series with the general term $u_n$ diverges.
Hint: You can start by proving $$u_n \geq \frac{1}{n + 1}$$
*
*For $ 0 \leq x \leq \frac{\pi}{2}$ we have: $0 \leq u_n \leq \frac{\pi}{2}$ wich I can't derive the convergence from it. How can I prove it is convergent toward 0?
I am stuck with question 4. and 5, I could not see how to prove the hints.
| You can see the convergence of $u_n$ by induction:
$$\int^{\pi/2}_0 \cos x=[ \sin x]^{\pi/2}_0=1$$
$$\int^{\pi/2}_0 \cos^2 x=[ x/2+(1/4)\sin 2x]^{\pi/2}_0=\frac{\pi}{4}$$
$$\int^{\pi/2}_0 \cos^3 x=[ \sin x-(1/3)\sin^3 x]^{\pi/2}_0=1-\frac{1}{3}$$
$$\int^{\pi/2}_0 \cos^4 x=[(3/8)x+(1/32)\sin 4x+(1/16) \sin 2x]^{\pi/2}_0=\frac{3\pi}{16}$$
$$\int^{\pi/2}_0 \cos^5 x=[ \sin x -(1/5)\sin ^5x+..]^{\pi/2}_0=1-\frac{1}{5}. . .$$
$$\int^{\pi/2}_0 \cos^6 x=\frac{1}{192}[60x+48\sin 2x +9\sin 4x+4\sin^3 2x]^{\pi/2}_0=\frac{15\pi}{48}$$
Now if you sum these terms you can see the sum diverges.
Or we may use hint for 4; we have:
$$\int^{\pi/2}_0 \frac{dx}{1+\sin x}=\int^{\pi/2}_0(1-\sin x +\sin^2 x-\sin^3 x+ . . .= \Sigma\int^{\pi/2}_0(-1)^n \sin^n x=1 $$
Because:
$\sin x=\cos (\frac{\pi}{2}-x)=\cos 2(\frac{\pi}{4}-\frac{x}{2})$
$1+\sin x= 1+\cos 2(\frac{\pi}{4}-\frac{x}{2})= 2\cos^2 (\frac{\pi}{4}-\frac{x}{2})$
$$\int^{\pi/2}_0 \frac{dx}{1+\sin x}=\frac{1}{2}\int^{\pi/2}_0 \frac{dx}{\cos^2 (\frac{\pi}{4}-\frac{x}{2})}=[\tan (\frac{\pi}{4}-\frac{x}{2})]^{\pi/2}_0=1$$
That is the sum of $(-1)^n u_n$ converges to 1 and $u_n$ diverges.
$\int^{\pi/2}_0 \frac{dx}{1+\cos x}$
$\cos x= \cos 2(\frac{x}{2})$
$1+\cos x=1+\cos 2(\frac{x}{2})=2\cos^2 (\frac{x}{2})$
$$\int^{\pi/2}_0 \frac{dx}{1+\cos x}=\frac{1}{2}\int^{\pi/2}_0 \frac{dx}{\cos^2 x/2}=[\tan x/2]^{\pi/2}_0=1$$
| {
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} |
$a = \lim_{n \rightarrow \infty} ( 1+ \frac{1}{n^2})^n$ and $b =\lim_{n \rightarrow \infty}( 1+\frac{1}{n})^{n^2}$
If $$a = \lim_{n \rightarrow \infty} \Big( 1+ \frac{1}{n^2}\Big)^n\;\;\;\; \text{and}\;\;\;\; b =\lim_{n \rightarrow \infty}\Big( 1+\frac{1}{n}\Big)^{n^2}$$
then choose the correct option
*
*$a= 1, b=\infty$
*$a=0 ,b=1 $
*$a=\infty, b=1$
*$a =1,b=0$
My works : First option will correct
$a = \lim_{n \rightarrow \infty} ( 1+ \frac{1}{n^2})^n= \lim_{n \rightarrow \infty}e^\frac{1}{n}=1$
$b =\lim_{n \rightarrow \infty}( 1+\frac{1}{n})^{n^2}=\lim_{n \rightarrow \infty}e^n= \infty$
Is my answer correct or not ?
| Generalize:
$$
a_{k,l} = \lim (1+\frac{1}{n^k})^{n^l}\\
$$
So $a=a_{2,1}$ and $b=a_{1,2}$
$$
\ln a_{k,l} = \lim n^l \ln (1+\frac{1}{n^k})\\
= \lim \frac{\ln (1+\frac{1}{n^k})}{n^{-l}}\\
= \lim \frac{\frac{-k/n^{k+1}}{(1+\frac{1}{n^k})}}{-l n^{-l-1}}\\
= \lim \frac{k}{l} \frac{n^{l+1}}{n^{k+1} (1+\frac{1}{n^k})}\\
= \lim \frac{k}{l} \frac{n^{l-k}}{(1+\frac{1}{n^k})}\\
$$
If $l>k$ then $\ln a_{k,l} \to + \infty$. If $l<k$, then $\ln a_{k,l} \to 0$ and therefore $a_{k,l} \to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How evaluate $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ How prove
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$
$\mathbf {My\,Attempt:}$
I put $$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right) = \int_0^1 \frac{x^n}{1+x}dx$$ and hence
$$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^2=\int_0^1\int_0^1 \frac{(xy)^n}{(1+x)(1+y)}dxdy$$
So, the sum equals $$\int_0^1\int_0^1 \frac{1}{(1+x)(1+y)}\left(\sum\limits_{n=1}^{\infty} \frac{(xy)^n}{n}\right)dxdy=-\int_0^1\int_0^1 \frac{\ln(1-xy)}{(1+x)(1+y)}dxdy$$
The inner integral is
$$\int_0^1 \frac{\ln(1-xy)}{1+y}dy=\mathrm{Li}_2 \left(\frac{1-x}{1+x}\right)-\mathrm{Li}_2\left(\frac{1}{1+x}\right)+\ln(1-x)\ln\left(\frac{2x}{1+x}\right)$$
The remaining stuff is a lot of calculations.
$\text{Any hint for a better method or idea?}$
| $$\iint_{(0,1)^2}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy = 2\iint_{0\leq y\leq x\leq 1}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy $$
equals
$$ 2\int_{0}^{1}\int_{0}^{1}\frac{-x\log(1-x^2 z)}{(1+x)(1+xz)}\,dx\,dz = 2\int_{0}^{1}\frac{\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2(1-x)-\log(x)\log(1-x^2)}{1+x}\,dx$$
and we may tackle three separate integrals.
$$ \int_{0}^{1}\text{Li}_2\left(\frac{1}{1+x}\right)\frac{dx}{1+x}=\int_{1/2}^{1}\frac{\text{Li}_2(x)}{x}\,dx=\int_{1/2}^{1}\sum_{n\geq 1}\frac{x^{n-1}}{n^2}\,dx=\zeta(3)-\text{Li}_3\left(\tfrac{1}{2}\right), $$
$$ \int_{0}^{1}\frac{\text{Li}_2(1-x)}{1+x}\,dx = \int_{0}^{1}\frac{\text{Li}_2(x)}{2-x}\,dx\stackrel{\text{IBP}}{=}-\int_{0}^{1}\frac{\log(1-x)\log(2-x)}{x}\,dx=\frac{\pi^2}{4}\log(2)-\zeta(3)$$
and the third one is similar. Interestingly, the second integral equals
$$ \sum_{n\geq 1}\frac{1}{n}\left[\frac{1}{(n+1)^2}+\ldots+\frac{1}{(2n)^2}\right]=\frac{\pi^2}{4}\log(2)-\zeta(3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
How to apply induction to this formula? I want to justificate following equation:
$$\sum_{k=0}^n \frac{(-1)^k}{k!(n-k)!}\frac{1}{2k+1} = \frac{2^n}{(2n+1)!!}$$
I calculated the both sides for $n$ from 1 to 10 and it was true. How the mathematical induction can be applied to this equation? Or is there other way to justificate it?
| There is a standard technique for this type of sum which has appeared
here several times. Introduce
$$f(z) = (-1)^n \frac{1}{2z+1} \prod_{q=0}^n \frac{1}{z-q}.$$
We use the fact that residues sum to zero and we have for the sum over
$0\le k\le n$
$$\sum_{k=0}^n \mathrm{Res}_{z=k} f(z)
= \sum_{k=0}^n (-1)^n \frac{1}{2k+1}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^{n} \frac{1}{k-q}
\\ = \sum_{k=0}^n (-1)^n \frac{1}{2k+1}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= \sum_{k=0}^n
\frac{1}{k!} \frac{(-1)^k}{(n-k)!}
\frac{1}{2k+1}$$
so this is the target sum. The residue at infinity is zero since
$\lim_{R\to\infty} 2\pi R/R/R^{n+1} = 0$ so the sum is minus the
residue at $z=-1/2.$ We find
$$- \mathrm{Res}_{z=-1/2} f(z)
= \frac{(-1)^{n+1}}{2} \prod_{q=0}^n \frac{1}{-1/2-q}
= \frac{1}{2} \prod_{q=0}^n \frac{1}{1/2+q}
\\ = 2^n \prod_{q=0}^n \frac{1}{2q+1}
= \frac{2^n}{(2n+1)!!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Is there an elegant way to determine $Av$ given $Au_1, Au_2$, and $Au_3$ for a $3\times3$ matrix $A$? Let A be a 3x3 matrix such that ${A} \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ -13 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \\ 4 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -11 \end{pmatrix}.$
Find $\ {A} \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$
My original approach to this problem was to let $\ {A} $ be some 3x3 matrix with variable elements, and solve for each one, But i was just wondering if there was a more elegant way to solve it.
Thanks in advance!
| Try writing
$$\begin{pmatrix} 3\\ -11\\ -1\end{pmatrix}=a\begin{pmatrix} 3\\ 4\\ 5\end{pmatrix}+b\begin{pmatrix} 4\\ 5\\ 6\end{pmatrix}+c\begin{pmatrix} 5\\ -9\\ 1\end{pmatrix}$$
for some real numbers $a$, $b$ and $c$. Then
$$A\begin{pmatrix} 3\\ -11\\ -1\end{pmatrix}=aA\begin{pmatrix} 3\\ 4\\ 5\end{pmatrix}+bA\begin{pmatrix} 4\\ 5\\ 6\end{pmatrix}+cA\begin{pmatrix} 5\\ -9\\ 1\end{pmatrix}=a\begin{pmatrix} 2\\ 7\\ -13\end{pmatrix}+b\begin{pmatrix} -6\\ 0\\ 4\end{pmatrix}+c\begin{pmatrix} 3\\ 3\\ -11\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$
My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\tan x.\tan y}\implies \tan^2x+\tan^2y+2\tan x\tan y\geq 4\tan x\tan y\\
2\tan x\tan y\leq\tan^2x+\tan^2y\implies \color{red}{?}
$$
or
$$
1-\tan x\tan y=\frac{\tan x+\tan y}{\tan(x+y)}\implies\tan x\tan y=1-\frac{\tan x+\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}\\
\tan x\tan y=1-\frac{\frac{\tan x+\sqrt{3}\tan^2x+\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}=1-\frac{1+\tan^2x}{1+\sqrt{3}\tan x}\leq\color{red}{?}\\
=\frac{1+\sqrt{3}\tan x-1-\tan^2x}{1+\sqrt{3}\tan x}=\frac{\tan x(\sqrt{3}-\tan x)}{1+\sqrt{3}\tan x}=\color{red}{?}
$$
Note: I prefer not to do differentiation
| $$F(x)=\tan x\tan y=\dfrac{2\sin x\sin y}{2\cos x\cos y}=\dfrac{\cos(x-y)-\cos(x+y)}{\cos(x-y)+\cos(x+y)}=1-\dfrac{2\cos(x+y)}{\cos(x-y)+\cos(x+y)}$$
Method$\#1:$
For $x+y=\dfrac\pi3$
$$F(x)=1-\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$$ which will be maximum
if $-\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$ is maximum
if $\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$ is minimum positive
if $2\cos\left(2x-\dfrac\pi3\right)+1$ is maximum positive
Method$\#2:$
For $x+y=\dfrac\pi3,$
$F(x)=\dfrac{2\cos\left(2x-\dfrac\pi3\right)-1}{2\cos\left(2x-\dfrac\pi3\right)+1}$
$\iff2\cos\left(2x-\dfrac\pi3\right)=\dfrac{1+F(x)}{1-F(x)}$
$0<x<\dfrac\pi3,2\ge2\cos\left(2x-\dfrac\pi3\right)\ge\dfrac12$
$2\ge\dfrac{1+F(x)}{1-F(x)}\ge\dfrac12$
$2\ge\dfrac{1+F(x)}{1-F(x)}\iff\dfrac{1-3F(x)}{1-F(x)}\ge0\implies$ either $F(x)\le\dfrac13$ or $F(x)>1$
$\dfrac{1+F(x)}{1-F(x)}\ge\dfrac12\iff\dfrac{1+3F(x)}{1-F(x)}\ge0\iff\dfrac{F(x)+\dfrac13}{F(x)-1}\le0\iff-\dfrac13\le F(x)<1$
$\implies F(x)\le\dfrac13$ which occurs if $\cos\left(2x-\dfrac\pi3\right)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first part, I tried realising LHS so I got $$LHS=\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2+\cos^2\theta}=\frac{1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta}{2+2\sin\theta}.$$
but now I am stuck :( . Any help would be greatly appreciated, thanks!
| From numerator of your solution
$$1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta$$
$$(1+2\sin\theta+\sin^2\theta-\cos^2\theta)+(2i\cos\theta+2i\cos\theta\sin\theta)$$
$$(1+2\sin\theta+\sin^2\theta-1+\sin^2\theta)+2i\cos\theta(1+\sin\theta)$$
$$(2\sin\theta)(1+\sin\theta)+2i\cos\theta(1+\sin\theta)$$
$$2(1+\sin\theta)(\sin\theta+i\cos\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Help with this proof by induction with inequalities. Show that mathematical induction can be used to
prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker inequality we originally tried to prove using mathematical induction.
Base Case: P(2)
\begin{aligned}
\frac{1}{2}\cdot\frac{2(2)-1}{2(2)} &< \frac{1}{\sqrt{3(2) + 1}}\\
\frac{3}{8} &< \frac{1}{\sqrt{7}}\\
\frac{1}{8} &< \frac{1}{3\sqrt{7}}\\
\end{aligned}
This is true as $8 > 3\sqrt{7}$.
Inductive Hypothesis: $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}}$
In the inductive step, we want to show that $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n+4}}$.
Using the inductive hypothesis, we can get to the following:
\begin{aligned}
\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} &< \frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
&< \frac{1}{\sqrt{3n+1}}\cdot 1\\
\end{aligned}
I am not sure how to get to $< \frac{1}{\sqrt{3n+4}}$ from here because i know that if the denominator would get bigger by adding 3 to it so the inequality wouldn't follow...
| Assuming $n\ge0$, we can show that
$$
\frac{2n+1}{2n+2}\le\frac{\sqrt{3n+1}}{\sqrt{3n+4}}\tag1
$$
by squaring both sides to get the equivalent
$$
\frac{4n^2+4n+1}{4n^2+8n+4}\le\frac{3n+1}{3n+4}\tag2
$$
and cross-multiplying to get the equivalent
$$
12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4\tag3
$$
which is true since $n\ge0$.
Therefore, if
$$
\frac12\frac34\cdots\frac{2n-1}{2n}\le\frac1{\sqrt{3n+1}}\tag4
$$
applying $(1)$, we get
$$
\begin{align}
\frac12\frac34\cdots\frac{2n-1}{2n}\color{#C00}{\frac{2n+1}{2n+2}}
&\le\frac1{\sqrt{3n+1}}\color{#C00}{\frac{\sqrt{3n+1}}{\sqrt{3n+4}}}\\
&=\frac1{\sqrt{3n+4}}\tag5
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do I evaluate the integral $\int_0^1\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$. stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was define the S to be equal to $$ \int_0^1 (1-x^3 +x^5-x^8+x^{10}-x^{13}+........) dx $$
I simplified this and got the above integral I tried to do partial fraction but did not succeed.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{x^{2} + x + 1 \over
x^{4} + x^{3} + x^{2} + x + 1}\,\dd x} =
\int_{0}^{1}{\pars{x^{3} - 1}/\pars{x - 1} \over
\pars{x^{5} - 1}/\pars{x - 1}}\,\dd x
\\[5mm] = &\
\int_{0}^{1}{1 - x^{3} \over 1 - x^{5}}\,\dd x
\,\,\,\stackrel{x^{5}\ \mapsto\ x}{=}\,\,\,
{1 \over 5}\int_{0}^{1}{x^{-4/5} - x^{-1/5} \over 1 - x}\,\dd x
\\[5mm] = &\
{1 \over 5}\pars{\int_{0}^{1}{1 - x^{-1/5} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-4/5} \over 1 - x}\,\dd x} =
{1 \over 5}\bracks{\Psi\pars{4 \over 5} - \Psi\pars{1 \over 5}}
\\[5mm] = &\
{1 \over 5}\bracks{\pi\cot\pars{\pi{1 \over 5}}} =
\bbx{{1 \over 5}\root{1 + {2 \over \root{5}}}\,\pi} \approx 0.8648
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding minimum value of $ \frac{x^2 +y^2}{y} $
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are
real
numbers satisfying $7x^2 + 3xy + 3y^2 = 1$
Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$
in $7x^2+3xy+3y^2=1$
$$3r^2+4r^2\cos^2 \alpha+3r^2\sin \alpha \cos \alpha =1$$
$$3r^2+2r^2(1+\cos 2 \alpha)+\frac{3r^2}{2}\sin 2 \alpha =1$$
$$8r^2+r^2(4\cos 2 \alpha+3\sin \alpha)=2$$
So $$r^2=\frac{2}{8+(4\cos 2 \alpha+3\sin \alpha)}$$
$$\frac{2}{8+5}=\frac{2}{13}\leq r^2\leq \frac{2}{8-5}=\frac{2}{3}$$
we have to find minimum of $$\frac{x^2+y^2}{y}=\frac{r}{\sin \alpha}$$
How can i find it, could some help me
| Making $y = \lambda x$ we have
$$
\min f(x,\lambda) \ \ \mbox{s. t. }\ \ g(x,\lambda) = 0
$$
here
$$
\begin{cases}
f(x,\lambda) = \frac{1+\lambda^2}{\lambda}x\\
g(x,\lambda) = x^2(7+3\lambda+3\lambda^2)-1=0
\end{cases}
$$
this minimization problem is equivalent to
$$
\min F(\lambda) = \left(\frac{1+\lambda^2}{\lambda}\right)^2\frac{1}{7+3\lambda+3\lambda^2}
$$
and then
$$
F'(\lambda)= 0\to (1 + \lambda^2) (3 \lambda^3 + 2 \lambda^2- 9 \lambda -14 )=(\lambda-2) (7 + 8 \lambda + 3 \lambda^2) = 0
$$
hence $\lambda = 2\to x = \pm \sqrt{\frac{1}{7+3\times 2+3\times 2^2}} = \pm\frac{1}{5}$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculating the cross product of a cross product so I really can't see what I am doing wrong. I want to use this formula:
$a\times (b\times c) = b(a\cdot c) - c(a\cdot b)$
Calculate the rotation of $v(x,y,z)=(x,y,z)^T \times \omega$ with $\omega \in \mathbb R^3$
Solution:
$a\cdot c=\nabla\cdot \omega=0$
and
$a\cdot b = \nabla \cdot (x,y,z)^T= \partial_x x + \partial_y y + \partial_z z = 3$
so we get $-\omega 3=-3\omega$
The actual solution (which I do get by direct calculation) is: $-2\omega$
| On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
\nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A}\ (\nabla \cdot \mathbf{B}) - \mathbf{B}\ (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B}
.
$$
Applying this on your case gives
$$\begin{align}
\nabla \times (\mathbf{x} \times \mathbf{\omega})
&= \mathbf{x}\ (\nabla \cdot \mathbf{\omega}) - \mathbf{\omega}\ (\nabla \cdot \mathbf{x}) + (\mathbf{\omega} \cdot \nabla) \mathbf{x} - (\mathbf{x} \cdot \nabla) \mathbf{\omega} \\
&= \mathbf{x} \ 0 - \mathbf{\omega}\ 3 + \mathbf{\omega} - \mathbf{0}
= -2 \mathbf{\omega}
.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof: For all n > 0, if $0I need to solve this problem without using induction, only with the following properties.
(1) Trichotomy law;
(2) if $a$ and $b$ are in $P$, then $a+b$ is in $P$;
(3) if $a$ and $b$ are in $P$, then $ab$ is in $P$.
Where P is the collection of all Positive Numbers.
My idea is to use the fact that $y^n - x^n = (y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})$.
Thus, since $x \in P$, and $y \in P$, I can build, by (2) and (3), the expansion ($x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})\in P$. How $(y - x)\in P$ (Since x < y), and by (3), $(y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1}) \in P$, which implies that $y^n - x^n\in P$, so $x^n < y^n$.
I'm in the right way? Thanks in advance.
| First a comment:
You don't need the assumption that
$x^2 < y^2$.
In general,
any statement of the form
$\forall n P(n)$
can only be proved by induction,
ultimately.
In your case,
$y^n - x^n = (y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})
$
has to be proved by induction
with $P(n)$ being
$y^n-x^n
=(y-x)\sum_{k=0}^{n-1} x^ky^{n-1-k}
$.
Another proof on the factorization
could use
$\begin{array}\\
(y+x)(y^n-x^n)
&=y^{n+1}-yx^n+xy^n-x^{n+1}\\
&=y^{n+1}-x^{n+1}+xy(y^{n-1}-x^{n-1})\\
\end{array}
$
or
$y^{n+1}-x^{n+1}
=(y+x)(y^n-x^n)-xy(y^{n-1}-x^{n-1})
$.
From this,
if
$x^n-y^n = (x-y)q_n(x, y)$
for $n$ and $n-1$
then
$\begin{array}\\
y^{n+1}-x^{n+1}
&=(y+x)(y-x)q_n(y, x)-xy(y-x)q_{n-1}(y, x)\\
&=(y-x)((y+x)q_n(y, x)-xyq_{n-1}(y, x))\\
\end{array}
$
so
$q_{n+1}(y, x)
=(y+x)q_n(y, x)-xyq_{n-1}(y, x)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need someone to explain answer to question without using Markov chains. Consider the following question here:
Probability of two consecutive head or tail or any one of them in a row?
Question. Fair coins are tossed and when either four consecutive heads and tails appear the process will be stopped. What is the probability of two consecutive head or tail or any one of them in a row?
Answer. Because heads and tails play symmetric roles in the stopping criterion (and presumably have equal chances in each Bernoulli trial), it suffices to find the probability of getting two consecutive heads before the process stops.
If the process stops with four consecutive heads, obviously that means also we got two consecutive heads before the process stops. So we can focus on the probability $p$ that we get two consecutive heads before the process stops with four consecutive tails.
One way to compute this is by defining a Markov chain with two absorbing states, a) two consecutive heads and b) four consecutive tails. Finding $p$ amounts to finding the probability of reaching the first of these absorbing states (two consecutive heads).
The Wikipedia write-up of
absorbing Markov chains
may be overly concise, so here are some details. Ordering the
transient states before the absorbing states gives a
probability transition matrix having block structure:
$$ P = \begin{bmatrix} Q & R \\ 0 & I \end{bmatrix}$$
so that $Q$ gives the transition probabilities between the
transient states and $R$ the transition probabilities from
transient to absorbing states.
We define the fundamental matrix
$N = \sum_{k=0}^\infty Q^k = (I-Q)^{-1}$, and the product
$NR$ then gives the probabilities of eventually reaching an
absorbing state starting from a transient state.
In the case at hand it is convenient to use four transient
states (consecutive runs of one Head or of one to three
Tails, resp.) and two absorbing states (two consecutive
Heads or four consecutive Tails). Taking the states in
just this order gives:
$$ P = \begin{bmatrix} 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0
\\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0
\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 & 0
\\ \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}
\\ 0 & 0 & 0 & 0 & 1 & 0
\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$
$$ N = (I-Q)^{-1} = \begin{bmatrix}
\frac{16}{9} & \frac{8}{9} & \frac{4}{9} & \frac{2}{9}
\\ \frac{14}{9} & \frac{16}{9} & \frac{8}{9} & \frac{4}{9}
\\ \frac{4}{3} & \frac{2}{3} & \frac{4}{3} & \frac{2}{3}
\\ \frac{8}{9} & \frac{4}{9} & \frac{2}{9} & \frac{10}{9}
\end{bmatrix} $$
$$ NR = \begin{bmatrix} \frac{8}{9} & \frac{1}{9}
\\ \frac{7}{9} & \frac{2}{9}
\\ \frac{2}{3} & \frac{1}{3}
\\ \frac{4}{9} & \frac{5}{9} \end{bmatrix} $$
For economy I omitted an "empty" state, so we imagine our
Markov process to initialize in state one Head with probability
$1/2$ and likewise in state one Tail with probability $1/2$.
From the above computation it follows that our chance of
reaching two consecutive Heads before four consecutive Tails is:
$$ (1/2)\frac{8}{9} + (1/2)\frac{7}{9} = \frac{5}{6} $$
This is the same as our chance of reaching two consecutive
Tails before four consecutive Heads.
I was wondering if someone could give an explanation of this answer that doesn't end up using fancy stuff like Markov chains? This is a practice question for the Math GRE subject test, so there probably is a way to do this that doesn't involve fancy stuff.
| If I understand correctly, you want to compute the probability of getting 2 heads before 4 tails (ie ending the process). Let
*
*$p$ be the probability of landing heads (here $p=1/2$)
*$P(A \mid 0)$ be the probability of 2 consecutive heads at the beginning of the sequence
*$P(A\mid 1)$ be the probability of 2 consecutive heads given last toss was tails
*$P(A\mid 2)$ be the probability of 2 consecutive heads given last 2 tosses were tails
*$P(A\mid 3)$ be the probability of 2 consecutive heads given last 3 tosses were tails
*$f$ be the probability of 2 consecutive heads given last coin was heads.
Then note that
*
*$P(A\mid 3) = 0(1-p) + f p$
*$P(A\mid 2) = P(A\mid 3)(1-p) + fp$
*$P(A\mid 1) = P(A\mid 2)(1-p) + fp$
*$P(A\mid 0) = P(A\mid 1)(1-p) + fp$
*$f = P(A\mid 1) (1-p) + p$.
So
*
*$P(A\mid 2) = fp(2-p)$
*$P(A\mid 1) = fp(2-p)(1-p) + fp = fp(3-3p+p^2)$
*$f = fp(3-3p+p^2)(1-p)+p = 7f/16 + 1/2 \implies f = 8/9 \implies P(A\mid 1) = 7/9$
*$\therefore P(A\mid 0) = (7/9)/2 + (8/9)/2 = 5/6$.
| {
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"answer_count": 1,
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} |
Dimension of $W_{2}$? Let $A = \begin{bmatrix} 1 & -1 & -5 & 1 & 4\\ -1 & 2 & 8 & -3 & -4\\ 3 & -1 & -9 & 0 & 4 \\ 2 & 2 & 2 & -5 & -10\\ 0&-3&-9&5&13\end{bmatrix}$
Now we define the subspace $W_{1},W_{2}$ of $A$ as follows -
$W_{1} = \{X \in M_{5 \times 5}| AX = 0\}$
$W_{2} = \{Y \in M_{5 \times 5} | YA =0\}$
I can see that $W_{1}$ is the nullspace of $A$
using rank nullity theorem I got Nullity of $A$ as $2$ since we have the rank of matrix $A$ to be 3.
Now I am thinking about the dimension of $W_{2}$?
As from the comments and we know that row rank = column rank, hence dim$(W_{2}) = 2$
But Now I am thinking about the dimension of $W_{1} \cap W_{2}$ and $W_{1} + W_{2}$?
Any ideas?
| Let us first look at
\begin{align*}
W'&= \{X\in M_{5\times 1}; AX=0\}\\
W''&= \{Y\in M_{1\times 5}; YA=0\}
\end{align*}
In the other words we look at similar equations but with column/row vectors instead of matrices.
By a direct computation you can get that $\operatorname{rank}A=4$, which implies that $\dim(W')=\dim(W'')=1$. You can also compute that $W'$ is the span of the column vector $\vec a=(2,-3,1,0,0)^T$ and that $W''$ is the span of the row vector $\vec b^T=(5,0,-1,-1,2)$.
If we denote the columns of the matrix $X$ as $\vec c_1,\dots,\vec c_5$ then we have
$$AX = A\begin{pmatrix} \vec c_1 & \vec c_2 & \ldots & \vec{c_5} \end{pmatrix} = \begin{pmatrix} A\vec c_1 & A\vec c_2 & \ldots & A\vec{c_5} \end{pmatrix} = \begin{pmatrix} \vec 0 & \vec 0 & \ldots & \vec 0 \end{pmatrix}.$$
I.e., each of the columns fulfills the condition $A\vec c_i=\vec 0$. So we see that the matrices in $W_1$ are precisely those matrices where each column is a multiple of $\vec a$.
Similarly, we get for the rows of the matrix $X\in W''$ the condition $\vec r_i^TA=\vec 0^T$, and $W_2$ consists of those matrices where each row is multiple of $\vec b$.
We get that
\begin{align*}
W_1&=\{
\begin{pmatrix}
2a & 2b & 2c & 2d & 2e \\
-3a & -3b & -3c & -3d & -3e \\
a & b & c & d & e \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}; a,b,c,d,e \in \mathbb R\}
\\
W_2&=\{
\begin{pmatrix}
5s & 0 & -s & -s & -2s \\
5t & 0 & -t & -t & -2t \\
5u & 0 & -u & -u & -2u \\
5v & 0 & -v & -v & -2v \\
5w & 0 & -w & -w & -2w \\
\end{pmatrix}; s,t,u,v,w \in \mathbb R\}
\end{align*}
And we also see that $\dim(W_1)=\dim(W_2)=5$.
Now the matrices in the intersection $W_1\cap W_2$ are precisely the matrices which can be expressed in both ways.
$$\begin{pmatrix}
2a & 2b & 2c & 2d & 2e \\
-3a & -3b & -3c & -3d & -3e \\
a & b & c & d & e \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}=
\begin{pmatrix}
5s & 0 & -s & -s & -2s \\
5t & 0 & -t & -t & -2t \\
5u & 0 & -u & -u & -2u \\
5v & 0 & -v & -v & -2v \\
5w & 0 & -w & -w & -2w \\
\end{pmatrix}
$$
Those are precisely the multiples of
$$
\begin{pmatrix}
10& 0 &-2 &-2 &-4 \\
-15& 0 & 3 & 3 & 6 \\
5 & 0 &-1 &-1 &-2 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
This matrix generates $W_1\cap W_2$.
We see that $\dim(W_1\cap W_2)=1$.
From the equation
$$\dim W_1+\dim W_2=\dim(W_1+W_2)+\dim(W_1\cap W_2)$$
we can calculate that $\dim(W_1+W_2)=9$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
| $$\dfrac{d\arctan\sqrt{\dfrac{x+1}{x-1}}}{dx}=\dfrac1{1+\dfrac{x+1}{x-1}}\cdot\dfrac{d\sqrt{\dfrac{x+1}{x-1}}}{dx}$$
For $x-1\ne0,$
$\sqrt{\dfrac{x+1}{x-1}}=\sqrt{\dfrac{\sqrt{x^2-1}}{(x-1)^2}}=\dfrac{\sqrt{x^2-1}}{|x-1|}$
$F=\dfrac{d\sqrt{\dfrac{x+1}{x-1}}}{dx}=\dfrac{d\dfrac{\sqrt{x^2-1}}{|x-1|}}{dx}$
If $x-1>0,F=\dfrac x{(x-1)\sqrt{x^2-1}}-\dfrac{\sqrt{x^2-1}}{(x-1)^2}=\dfrac{x(x-1)-(x^2-1)}{(x-1)\sqrt{x^2-1}}=\dfrac1{\sqrt{x^2-1}}$
What if $x-1<0?$
| {
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"answer_count": 6,
"answer_id": 0
} |
The error for approximation of bessel function
The sixth degree polynomial
$1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304}$
Is sometimes used to approximate the Bessel function $J_0(x)$ of the first kind of order zero for $0 \leq x\leq 1$. Show that the error $E$ involved in this approximation is less than $0.00001$.
I know that
$J_0(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(n!)^2} ({\frac{x}{2}})^{(2n)}= 1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304}+{x^8\over 147456}+...$
I think, We want to show that
$J_0(x)-( 1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304})\leq {x^8\over 147456}$
But How can I show that, and how can I use the information $0 \leq x\leq 1$, please?
“ this question in calculus course”
| Statement:
$E(x)=J_0(x)-\sum\limits_{n=0}^3\frac{(-1)^n}{(n!)^2} (\frac{x}{2})^{2n}\leq\frac{x^8}{2^8 (4!)^2}\tag{1}$
$E(x)=\sum\limits_{n=4}^\infty\frac{(-1)^n}{(n!)^2} ({\frac{x}{2}})^{2n}=\sum\limits_{n=0}^\infty\frac{(-1)^{n+4}}{(n+4)!^2} ({\frac{x}{2}})^{2(n+4)}=\frac{x^8}{2^8}\sum\limits_{n=0}^\infty\frac{(i)^{2n}}{(n+4)!^2} ({\frac{x}{2}})^{2n}$ $\hspace{0,5cm}$ where $\hspace{0,2cm}$ $i^2=-1$
Enough to prove that $\hspace{0,2cm}$ $\sum\limits_{n=0}^\infty\frac{(i)^{2n}}{(n+4)!^2} ({\frac{x}{2}})^{2n} \le \frac{1}{(4!)^2}\tag{2}$
From other hand the following inequality is true for every $k\ge1 $ (integer) and inside the domain of $x$ is defined by $\hspace{0,2cm}$ $0\le x\le 1$:
$\prod\limits_{k=1}^{n}\frac{(k+4)^2}{(i\frac{x}{2})^4}\gt1 \tag{3}$
$\prod\limits_{k=1}^{n}\frac{(k+4)^2}{(i\frac{x}{2})^4}=\frac{(n+4)!^2}{(4!)^2 (i\frac{x}{2})^4}\gt 1$
Realign the inequality: $\hspace{0,2cm}$
$\frac{1}{(4!)^2 (i\frac{x}{2})^2}\gt\frac{(i\frac{x}{2})^2}{(n+4)!^2}\tag{4}$
Take the sum of both sides:
$\sum\limits_{n=0}^\infty\frac{1}{(4!)^2 (i\frac{x}{2})^2}\gt\sum\limits_{n=0}^\infty\frac{(i\frac{x}{2})^2}{(n+4)!^2}=\sum\limits_{n=0}^\infty\frac{i^{2n}}{(n+4)!^2}(\frac{x}{2})^{2n}$
RHS of the inequality is the same the sum then the sum in (2). LHS of the inequality is geometric series equal to:
$\sum\limits_{n=0}^\infty\frac{1}{(4!)^2 (i\frac{x}{2})^2}=\frac{1}{(4!)^2}\frac{1}{1+\frac{4}{x^2}}\lt \frac{1}{(4!)^2}$ for every $x$
So
$E(x)\lt \frac{x^8}{2^8 (4!)^2}$ regarding that $0\le x \le 1$ the maximum error value is at $x=1$ where $E(x)\lt 10^{-5}$
Similar method can be seen on the link
Prove that ${e\over {\pi}}\lt{\sqrt3\over{2}}$ without using a calculator.
from user90369.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Verify the proof that $x_n = \ln^2(n+1) - \ln^2n$ is a bounded sequence.
Let $n\ \in \mathbb N$ and:
$$
x_n = \ln^2(n+1) - \ln^2n
$$
Prove that $x_n$ is a bounded sequence.
I've taken the following steps. Consider $x_n$
$$
\begin{align}
x_n &= \ln^2(n+1) - \ln^2n = \\
&= (\ln(n+1) + \ln n)(\ln (n+1) - \ln n) = \\
&= \ln \frac{n + 1}{n}\cdot \ln (n(n+1)) = \\
&= \ln\left({1 + {1\over n}}\right)\cdot \ln(n(n+1))
\end{align}
$$
Now multiply and divide by $n$:
$$
\begin{align}
x_n &= {n \over n} \ln\left({1 + {1\over n}}\right)\cdot \ln(n(n+1)) = \\
&= \ln\left({1 + {1\over n}}\right)^n \cdot\ln \sqrt[^n]{(n(n+1))}
\end{align}
$$
Now consider $\left({1 + {1\over n}}\right)^n$. There are plenty of proofs that it is bounded. In my case I've used expansion with binomial coefficients to prove that :
$$
2< \left({1 + {1\over n}}\right)^n < 3 \implies \\
\ln2 < \ln \left({1 + {1\over n}}\right)^n < \ln3
$$
So now we want to prove that $\ln \sqrt[^n]{(n(n+1))}$ is bounded. Start with the following:
$$
\ln \sqrt[^n]{n(n+1)} < \ln \sqrt[^n]{(n+1)^2}
$$
Consider the following equation:
$$
\begin{align}
\sqrt[^n]{(n+1)^2} &= 1+a_n \iff \\
\iff (n+1)^2 &= (1+a_n)^n = \sum_{k=0}^{n}\binom{n}{k}a_n^k
\end{align}
$$
Now:
$$
\sum_{k=0}^{n}\binom{n}{k}a_n^k \ge \frac{n(n+1)}{2}a_n^2 \implies \\
\implies (n+1)^2 \ge \frac{n(n+1)}{2}a_n^2 \implies \\
\implies a_n \le \sqrt{2 + {2\over n}}
$$
So $a_k$ is clearly bounded. Which means:
$$
\sqrt[^n]{(n+1)^2} < 1 + \sup\{a_n\} = 3
$$
Also $\sqrt[^n]{(n+1)^2} > 1$. So:
$$
\ln1 < \ln \sqrt[^n]{(n(n+1))} < \ln3
$$
Now going back to initial expression:
$$
\ln1 \cdot \ln2 < \ln\left({1 + {1\over n}}\right)^n \cdot\ln \sqrt[^n]{(n(n+1))} < \ln3 \cdot \ln3
$$
Meaning $x_n$ is bounded. Have I missed something?
| Another concise option: $\ln(n+1)<\ln n+\frac{1}{n}$ so $0<\ln^2(n+1)-\ln^2 n<2\frac{\ln n}{n}+\frac{1}{n^2}<2\cdot 1+1=3$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Sequence Inequality question from RMO 2018 Define a sequence {$a_n$} of real numbers by $a_1 = 2$ and $a_{n+1} = \dfrac{a_n^2+1}{2}$, for $n\ge 1$,
Prove that for every natural number $N, \sum_{j=1}^{N} \frac{1}{1+a_j} \lt 1$
I tried mathematical induction after coming to the step where
$Sum_n = \frac{1}{2}\left(\frac{a_1-1}{a_2-1} + \frac{a_2-1}{a_3-1} +\cdots \frac{a_n-1}{a_{n+1}-1}\right)$.
Having gotten this how would go about proving it using induction?
| By computing $1-\sum_{k=1}^{N}\frac{1}{a_k+1}$ with the help of Mathematica it is not difficult to conjecture that
$$ 1-\sum_{k=1}^{N}\frac{1}{a_k+1} = \frac{2^{2^N-1}}{\prod_{k=1}^{N}b_k}\tag{1}$$
with $\{b_n\}_{n\geq 1}=\{3,7,37,1033,868177,701129422753,\ldots\}$.
It looks like $b_n = (a_n+1) 2^{2^{n-1}-1} $, hence if we manage to prove
$$ \frac{1}{a_N+1} = \frac{2^{2^{N-1}-1}}{\prod_{k=1}^{N-1}(a_k+1)2^{2^{k-1}-1}}-\frac{2^{2^N-1}}{\prod_{k=1}^{N}(a_k+1)2^{2^{k-1}-1}},\tag{2}$$
which is equivalent to
$$ \frac{1}{a_N+1} = \frac{1}{2^{N}\prod_{k=1}^{N-1}(a_k+1)}\left(1-\frac{2}{a_N+1}\right)\tag{3}$$
or to
$$ 1 = \frac{a_N-1}{2^{N}\prod_{k=1}^{N-1}(a_k+1)}\tag{4}$$
we are done. On the other hand $(4)$ is exactly what we get by "unpacking"
$$ a_N-1 = \frac{a_{N-1}+1}{2}(a_{N-1}-1) \tag{5}$$
through induction. Now we may remove the conjectural part. $(5)\mapsto(4)\mapsto(3)\mapsto(2)$ and from $(2)$ it follows that
$$ \sum_{k=1}^{N}\frac{1}{a_k+1}= 1-\frac{2^N}{\prod_{k=1}^{N}(a_k+1)}.\tag{6}$$
The given exercise is equivalent to the following claim: if $k_1=2$ and $k_{n+1}=k_n^2-k_n+1$, then
$$ 1 = \sum_{n\geq 1}\frac{1}{k_n} = \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1807}+\ldots $$
which is pretty reminiscent of some Machin-like formulas.
Expert problem solvers may easily recognize the Sylvester sequence A000058.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Person with seven friends invites subset of three for one week Suppose that a person with seven friends invites a different subset of three friends to dinner every night for one week (seven days). How many ways can this be done so that all friends are included at least once?
I know the number of ways to do this with no restriction is $\binom{7}{3}^7$.
I need to use Inclusion-Exclusion..however I am having trouble determining the sets to use.
| This answer is essentially the same as Thomas Bladt's but I use a different format for counting. $$\frac {\binom{7}{3}!}{(\binom{7}{3}-7)!} - 7\cdot \frac{(\binom{7}{3}-\binom{6}{2})!}{(\binom{7}{3}-\binom{6}{2}-7)!}+ \binom{7}{2}\cdot \frac{(\binom{7}{3}-\binom{6}{2}-\binom{5}{2})!}{(\binom{7}{3}-\binom{6}{2}-\binom{5}{2}-7)!}$$ Which reduces to.....
$$\frac{35!}{28!}-7\cdot \frac{20!}{13!} + 21\cdot \frac{10!}{3!}$$
| {
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"answer_count": 2,
"answer_id": 1
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What is $df^{-1}(f(\begin{smallmatrix}x\\y\end{smallmatrix}))$?
$$f:\mathbb{R}\times[0,2\pi) \to \mathbb{R}^2,f\begin{pmatrix}x \\y\end{pmatrix}=\begin{pmatrix}e^x \cos y \\e^x \sin y\end{pmatrix}$$
which is $\exp(z)$
I know that $df(x,y)=\begin{pmatrix}e^x\cos y && -e^x\sin y\\e^x\sin y&&e^x\cos y\end{pmatrix}$.
How can I determine $df^{-1}\Bigg(f\begin{pmatrix}x \\y\end{pmatrix}\Bigg)$?
| We have $df^{-1}\Bigg(f\begin{pmatrix}x \\y\end{pmatrix}\Bigg)=(df(x,y))^{-1}=\begin{pmatrix}e^x\cos y && -e^x\sin y\\e^x\sin y&&e^x\cos y\end{pmatrix}^{-1}$.
| {
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Probability of picking two marbles each from two colors when selecting $4$ marbles out of $30$ marbles I have $\ 30 $ marbles. $\ 25 $ are white, $\ 3 $ are blue and $\ 2 $ are red. same color marbles are identical.
If I pick randomly and without replacement $\ 4 $ marbles, what is the probability that I'll pick two each two of two colors?
Trying to make it easier, I assumed all marbles are different, so there are $\ 30 \cdot 29 \cdot 28 \cdot 27 $ ways to pick them and then number of options for :
Picking $2$ blue and $2$ red marbles are $\ {3 \choose 2}{25 \choose 2} \cdot 4! $ options.
Picking $2$ blue and $2$ white marbles are $\ {3 \choose 2}{2 \choose 2} \cdot 4! $ options.
Picking $2$ white and $2$ red marbles are $\ {25 \choose 2}{2 \choose 2 }\cdot 4! $.
The three events are mutually exclusive, so I should be able to just add them all together but that's the wrong answer. Any suggestions?
| Correct answer is $\frac{401}{9135}$. It is calculated as follows:$\frac{\binom{25}{2}*\binom{3}{2}}{\binom{30}{4}}+\frac{\binom{25}{2}*\binom{2}{2}}{\binom{30}{4}}+\frac{\binom{3}{2}*\binom{2}{2}}{\binom{30}{4}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that $5|n^5-n$ and $2|n^2-n$ but i can not see that somehow help me, do you have some idea?
| Method 1:We can try factoring but ... I don't want to.
Method 2: Euler's theorem. If $\gcd(n, 10) = 1$ then $n^{\phi(10)}=n^4 \equiv 1\pmod {10}$ So $n + 3n^3 + 7n^7 + 9n^9 \equiv n + 3n^3 + 7n^3 + 9n = 10n + 10n^3 \equiv 0 \mod 10$.
But if $\gcd(n,10) \ne 1$???
Well chinese remainder theorem.
$\mod 2$ we have $0^k \equiv 0$ and $1^k \equiv 1$ so $n^k \equiv n$ so
$$n + 3n^3 + 7n^7 + 9n^9 \equiv n + n^3 + n^7 + n^9 \pmod 2$$ and $a^k \equiv a \pmod 2$ so $$n + n^3 + n^7 + n^9 \equiv 4n \equiv 0 \pmod 2$$
$\mod 5$ we have $\gcd(n,5)$ means $n^{5-1} \equiv 1 \pmod {10}$.
If $\gcd(n,5)= 1$ then $$n+3n^3 + 7n^7 + 9n^9 \equiv n + 3n^3 + 2n^3 + 4n \mod 5\\
\equiv 5n +5n^3 \equiv 0 \mod 5$$ But if $\gcd(n,5) \ne 1$ the $\gcd(n,5) = 5$ so $n + 3n^3 + 7n^7 + 9n^9 \equiv 0 \mod 5$.
So $2|n+3n^3 + 7n^7 + 9n^9$ and $5|n+3n^3 + 7n^7 + 9n^9$ so $10|n+3n^3 + 7n^7 + 9n^9$. Always.
====
Actually since $\phi(10) = \phi(5) = 4$ we get $n+3n^3 + 7n^7 + 9n^9 \equiv 10n + 10n^3$ when $\gcd(n, 5) \ne 0$. so the only case to consider is if $\gcd(n, 5) = 5$.
i.e. if $5|n$.
Well if $5|n$ then obviously $5| n+3n^3 + 7n^7 + 9n^9$ and just remains to show $n+3n^3 + 7n^7 + 9n^9$ is even (when $5|n$).
Well, ... that's easy. If $n$ is even then it is the sum of four even terms. If $n$ is odd it is the sum of $4$ odd terms.
| {
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calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$ So I multiplied by the conjugate and got $$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$
and this is where I got stuck.
|
So I multiplied by the conjugate and got
What conjugate expression was that exactly...?
You want to get rid of the cube root by using:
$$a+b=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}=\frac{a^3+b^3}{a^2-ab+b^2}$$
with, in your case, $a=x$ and $b=\sqrt[3]{1-x^3}$.
| {
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Calculate the limit : $\lim_{x\rightarrow \infty}\tan ({\frac{\pi x}{2x+1}})^\frac{1}{x}$ Exercise :
Calculate the following limit
$$\lim_{x\rightarrow \infty}\tan \bigg({\frac{\pi x}{2x+1}}\bigg)^\frac{1}{x}$$
Attempt :
$$\lim_{x\rightarrow \infty} \frac{1}{x} = \frac {1}{\infty} = 0$$
$$\lim_{x\rightarrow \infty}\tan ({\frac{\pi x}{2x+1}})^\frac{1}{x}=\lim_{x\rightarrow \infty}\tan ({\frac{\pi x}{2x+1}})^0 = 1$$
Is it correct ?
| As an alternative
$$\frac{\pi x}{2x+1}=\frac{\frac{\pi}2 (2x+1)-\frac{\pi}2}{2x+1}=\frac{\pi}2-\frac{\pi}{4x+2}$$
then
$$\left[\tan \bigg({\frac{\pi x}{2x+1}}\bigg)\right]^\frac{1}{x}=\left[\cot \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}=\frac{1}{\left[\tan \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}} \to 1$$
indeed
$$\left[\tan \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}=\left[\frac{\tan \bigg(\frac{\pi}{4x+2}\bigg)}{\frac{\pi}{4x+2}}\right]^\frac{1}{x}\left(\frac{\pi}{4x+2}\right)^\frac1x\to 1^0\cdot 1=1$$
indeed
$$\left(\frac{\pi}{4x+2}\right)^\frac1x=e^{\frac{\log \left(\frac{\pi}{4x+2}\right)}{x}}=e^{\frac{\log \left(\frac{\pi}{4x+2}\right)}{\frac{\pi}{4x+2}}\cdot\frac{\pi}{x(4x+2)}}\to e^0=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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determine the error bound for the interpolation error at $x = \frac{\pi}{4}.$ I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = \sin x$, given the points $(0,0), \left(\frac{\pi}{2}, 1\right), (\pi, 0).$
Solution:
$$P_2(x) = \frac{4}{\pi ^2}x(\pi - x)$$
Here is the question below I am having trouble with
Calculate $P_2\left(\frac{\pi}{4}\right)$, an approximation for $f\left( \frac{\pi}{4}\right) = \sin \left( \frac{\pi}{4} \right)$ and determine the error bound for the interpolation error at $x = \frac{\pi}{4}.$
Calculating $P_2\left(\frac{\pi}{4}\right)$
$$P_2\left(\frac{\pi}{4}\right) = \frac{4}{\pi ^2} \times \frac{\pi}{4}\left( \pi - \frac{\pi}{4} \right)= \frac{3}{4}$$
If we plug $x = \frac{\pi}{4}$ into $f(x) = \sin x$ we get $f(\pi /4) = \sin (\pi /4) \approx 0.7071 $
Here is my attempt below at finding the error bound
Writing the error as $err(x) = \sin x - P_2(x) = \sin x - \frac{4}{\pi ^2}x(\pi - x)$
Differentiating once gives,
$$\cos (x) - \frac{4(\pi - 2x)}{\pi ^2}$$
Differentiating twice gives,
$$\frac{8}{\pi ^2}-\sin(x)$$
Finally differentiating a third time gives,
$$- \cos x $$
But how do I use this to get my error bound
| For a polynomial interpolation of order $n$, the maximum error is given by
$${\rm err}=\frac{1}{(n+1)!}\max |f^{(n+1)}(x)|\max|\prod_{p=0}^n(x-x_p)|$$
Here $x_p$ are the roots of your polynomial.
Let's suppose that you are interested in finding the maximum only in the interval from $0$ to $\pi$.
The maximum of the derivative in your case is $1$, the maximum for the product $x(\pi-x)$ occurs at $\pi/2$ so $${\rm err}=\frac{1}{3!}1\left(\frac\pi2\right)^2$$
| {
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$\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=$? Consider the convergent integral (I looked up difficult indefinite integrals on google images and then I saw this integrand and I was like hey let's see if it converges)
$$I=\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
We have the numerical approximation $$I\approx0.5553603672697931...$$
Which Wolfram alpha says is close to $\frac\pi{4\sqrt{2}}$. All my attempts at this integral have been fruitless and I need some help. Here's the only attempt of mine that actually made the integrand smaller:
$x=\tan u$: $$I=\int_{\pi/4}^{\pi/2}\frac{\tan^2u-1}{\sqrt{\tan^4u+1}}du$$
Next step:
bang head on floor in agony
Any suggestions?
| $$
\begin{aligned}
\int_{1}^{\infty} \frac{x^{2}-1}{\left(x^{2}+1\right) \sqrt{x^{4}+1}} d x =& \int_{1}^{\infty} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right) \sqrt{x^{2}+\frac{1}{x^{2}}}} d x \\
=& \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^{2}-2}} \\
=& \int_{2}^{\infty} \frac{d u}{u \sqrt{u^{2}-2}},\textrm{ where }u=x+\frac{1}{x}. \\
=& \int_{2}^{\infty} \frac{1}{u^{2}} d\left(\sqrt{u^{2}-2}\right) \\
=& \int_{2}^{\infty} \frac{d \sqrt{u^{2}-2}}{\left(\sqrt{u^{2}-2}\right)^{2}+2} \\
=& \frac{1}{\sqrt{2}}\left[\tan ^{-1} \frac{\sqrt{u^{2}-2}}{\sqrt{2}}\right]_{2}^{\infty} \\
=& \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\\
=& \frac{\pi}{8} \sqrt{2}
\end{aligned}
$$
| {
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Prove the inequality: $\left(\frac{a}{a+2b}\right)^2+\left(\frac{b}{b+2c}\right)^2+\left(\frac{c}{c+2a}\right)^2 \geq \frac{1}{3}$
Let $a,b,c$ are all positive real numbers. Prove
$$\left(\frac{a}{a+2b}\right)^2+\left(\frac{b}{b+2c}\right)^2+\left(\frac{c}{c+2a}\right)^2 \geq \frac{1}{3}$$
Can anyone give a hint?
| By C-S and by the Vasc's inequality we obtain:
$$\sum_{cyc}\frac{a^2}{(a+2b)^2}=\sum_{cyc}\frac{a^4}{a^2(a+2b)^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+4a^3b+4a^2b^2)}=$$
$$=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}4a^3b+\sum\limits_{cyc}2a^2b^2+(a^2+b^2+c^2)^2}\geq\tfrac{(a^2+b^2+c^2)^2}{\frac{4}{3}(a^2+b^2+c^2)^2+\frac{2}{3}(a^2+b^2+c^2)^2+(a^2+b^2+c^2)^2}=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the analytic form of expression for the following integral $$
\hspace{0.3cm} \large {\int_{0} ^{\infty} \frac{\frac{1}{x^4} \hspace{0.1cm} e^{- \frac{r}{x^2}}\hspace{0.1cm}e^{- \frac{r}{z^2}} }{ \frac{1}{x^2} \hspace{0.1cm} e^{- \frac{r}{x^2}}+ \frac{1}{y^2} \hspace{0.1cm} e^{- \frac{r}{y^2}}}} dr \hspace{.2cm} ; \hspace{1cm} x>0,y>0,z>0 $$ where $ x $ ,$ y $ and $z $ are constants independent of $ r $.
| The integral is pretty awkward as written. Let's rename the parameters:
$$a=\frac{1}{x^2} \\ b= \frac{1}{y^2} \\ c= \frac{1}{z^2}$$
Then we have:
$$a^2 \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}$$
So we need to find the following integral:
$$I(a,b,c)= \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}$$
I'd really prefer the integral to be symmetric in $(a,b)$, so let's change the parameter:
$$c=b+s$$
Now we have:
$$J(a,b,s)=J(b,a,s)= \int_0^\infty \frac{e^{-(a+b+s)r} ~dr}{ae^{-ar}+be^{-br}}$$
We notice that the integrand is in the form of harmonic mean:
$$2 a bJ(a,b,s)= \int_0^\infty \frac{2a e^{-ar} b e^{-br} }{ae^{-ar}+be^{-br}}~e^{-sr} ~dr \leq \int_0^\infty \sqrt{ab} e^{-\frac{a+b}{2}r} e^{-sr} ~dr$$
Using the harmonic mean - geometric mean inequality, we have obtained a very nice bound, assuming:
$$\frac{a+b}{2}+s>0 \qquad \Rightarrow \qquad \frac{a-b}{2}+c >0$$
$$2 a bJ(a,b,s) \leq \frac{2\sqrt{ab}}{a+b+2s} \\ J(a,b,s) \leq \frac{1}{(a+b+2s)\sqrt{ab}}$$
The equality holds for the case $a=b$, where we get:
$$J(a,a,s) = \frac{1}{2a(a+s)}=\frac{1}{2ac}$$
Which is confirmed by considering the original integral.
However, what about the general expression? Let's transform the integrand a little:
$$J(a,b,s)=\frac{1}{a} \int_0^\infty \frac{e^{-(b+s)r} ~dr}{1+\frac{b}{a} e^{-(b-a)r}}$$
Let's introduce a substitution:
$$e^{-(b-a)r}=t \\ -(b-a) dr= \frac{dt}{t}$$
We assume $b>a$, without the loss of generality, since the integral is symmetric in $a,b$.
We obtain:
$$J(a,b,s)=\frac{1}{a(b-a)} \int_0^1 \frac{t^{(b+s)/(b-a)-1} ~dt}{1+\frac{b}{a} t}$$
$$J(a,b,s)=\frac{1}{a(b-a)} \int_0^1 t^{\frac{b+s}{b-a}-1} \left( 1+\frac{b}{a} t\right)^{-1} ~dt$$
This integral can be expressed in terms of the hypergeometric function:
$${_2 F_1} (\alpha, \beta; \gamma; p)$$
$$\alpha=1 \\ \beta=\frac{b+s}{b-a} \\ \gamma=\beta+1=\frac{2b-a+s}{b-a} \\ p=-\frac{b}{a}$$
$$J(a,b,s)=\frac{1}{a(b+s)} {_2 F_1} \left(1,\frac{b+s}{b-a};\frac{2b-a+s}{b-a};-\frac{b}{a} \right)$$
For the original integral we get:
$$a^2 \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}=\frac{a}{c} ~{_2 F_1} \left(1,\frac{c}{b-a};1+\frac{c}{b-a};-\frac{b}{a} \right)$$
| {
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How to solve this radical expression I've been trying to solve this expression for at least two hours now... And I always get stuck towards the end, I don't know what I'm missing.
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy}{x-\sqrt{xy}})\times (\sqrt{xy} + \frac{xy}{x+\sqrt{xy}})$
My first step was to rationalize the fractions inside the parenthesis like so
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy(x+\sqrt{xy})}{(x-\sqrt{xy})(x+\sqrt{xy})})\times (\sqrt{xy} + \frac{xy(x-\sqrt{xy})}{(x+\sqrt{xy})(x-\sqrt{xy})})$
to get
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy(x+\sqrt{xy})}{(x^2-xy)})\times (\sqrt{xy} + \frac{xy(x-\sqrt{xy})}{(x^2-xy)})$
then
$\frac 1{xy} \times (\sqrt{xy} - \frac{x^2y+ xy\sqrt{xy}}{(x^2-xy)})\times (\sqrt{xy} + \frac{x^2y-xy\sqrt{xy}}{(x^2-xy)})$
and then I'm kind of lost, nothing I've tried works.
I tried grouping each fraction by x and simplyfing removing it, and then computing the lcm inside the parenthesis in order to subtract the $\sqrt{xy}$. Or the other way around, first I did the lcm and subtracted and then simplyfied. I even tried multiplying the first factor by the second and simplifying as I went on. I tried using Wolfram Alpha to help me with each step, too. I think my calculations are correct, I'm just missing some simplification or something similar. The result should be
$\frac{x-4y}{x-y}$
I was able to get the $x-y$ but not the $x-4y$. I hope I didn't mess up the equations in the question, I'm really tired.
| For simplicity, set $z=\sqrt{xy}$, with $z^2=xy$.
Then your expression becomes
\begin{align}
\frac {1}{xy}\left(z - \frac{xy}{x-z}\right)\left(z + \frac{xy}{x+z}\right)
&=\frac {1}{xy}\frac{xz-z^2-xy}{x-z}\frac{xz+z^2+xy}{x+z}\\[4px]
&=\frac {1}{xy}\frac{xz-2xy}{x-z}\frac{xz+2xy}{x+z}\\[4px]
&=\frac {1}{xy}\frac{x^2(z-2y)(z+2y)}{(x-z)(x+z)}\\[4px]
&=\frac {1}{xy}\frac{x^2(z^2-4y^2)}{x^2-z^2}\\[4px]
&=\frac {1}{y}\frac{x(xy-4y^2)}{x^2-xy}\\[4px]
&=\frac {1}{y}\frac{xy(x-4y)}{x(x-y)}\\[4px]
&=\frac{x-4y}{x-y}
\end{align}
A similar strategy might be to set $a=\sqrt{x}$ and $b=\sqrt{y}$ (assuming both are positive, but the same holds when both are negative). Then we have
$$
\sqrt{xy}-\frac{xy}{x-\sqrt{xy}}=ab-\frac{a^2b^2}{a^2-ab}=ab-\frac{ab^2}{a-b}
=ab\left(1-\frac{b}{a-b}\right)=\frac{ab(a-2b)}{a-b}
$$
Similarly
$$
\sqrt{xy}+\frac{xy}{x+\sqrt{xy}}=\frac{ab(a+2b)}{a+b}
$$
so your expression becomes
$$
\frac{1}{a^2b^2}\frac{a^2b^2(a-2b)(a+2b)}{(a-b)(a+b)}=
\frac{a^2-4b^2}{a^2-b^2}=\frac{x-4y}{x-y}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
| Let $y:=x-\dfrac{3}{2}$. The equation becomes
$$\frac{1}{\left(y+\frac{3}{2}\right)^2}+\frac{1}{\left(y-\frac{3}{2}\right)^2}=\frac{104}{25}\,.$$
This is equivalent to
$$\frac{y^2+\frac{9}{4}}{\left(y^2-\frac{9}{4}\right)^2}=\frac{52}{25}\,.$$
Let $z:=\dfrac{1}{y^2-\frac{9}{4}}$, we have
$$\frac{9}{2}z^2+z=z^2\left(\frac{1}{z}+\frac{9}{2}\right)=\frac{52}{25}\,.$$
That is,
$$\frac{9}{2}\left(z+\frac{4}{5}\right)\left(z-\frac{26}{45}\right)=0\,.$$
Thus, $z=-\dfrac45$ or $z=\dfrac{26}{45}$.
In the case $z=-\dfrac{4}{5}$, we have
$$y^2-\frac{9}{4}=\frac{1}{z}=-\frac{5}{4}\,,$$
so $y^2=1$, or $y=\pm1$. In this case, $x=\dfrac{1}{2}$ or $x=\dfrac{5}{2}$.
In the case $z=\dfrac{26}{45}$, we have
$$y^2-\frac{9}{4}=\frac{1}{z}=\frac{45}{26}\,.$$
That is, $y^2=\dfrac{207}{52}$, so $y=\pm\dfrac{3\sqrt{299}}{26}$. Hence,
$$x=\frac{39\pm3\sqrt{299}}{26}\,.$$
| {
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Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tried:
$\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)}dx=\frac{2\pi}{\sqrt3}$. I've tried setting a variable as the power, ie: $I(r)=\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^r}dx$ but differenciating the inside w.r.t to $r$ doesn't seem to form any differential equation that can be solved.
Answers and hints appreciated!
| For real $x$, we have $x^2 + x + 1 = (x+\frac12)^2 + \frac34 \ge \frac34$.
This implies for any $t \in (0,\frac34)$, following expansion in $t$ converges:
$$\frac{1}{x^2+x+1 - t} = \sum_{k=0}^\infty \frac{t^k}{(x^2 + x + 1)^{k+1}}$$
Since everything on RHS is non-negative, by Tonelli, we can integrate them term by term:
$$\int_{-\infty}^\infty \frac{dx}{x^2+x+1 - t} = \sum_{k=0}^\infty t^k \int_{-\infty}^\infty \frac{dx}{(x^2+x+1)^{k+1}}
$$
Set $a = \frac12$ and $b = \sqrt{1-t}$. Notice $b > |a|$, the formula you have
tell us the integral on LHS is
$$\frac{\pi}{\sqrt{b^2 - a^2}} = \frac{\pi}{\sqrt{\frac34 - t}}
= \frac{2\pi}{\sqrt{3}\sqrt{1 - \frac{4t}{3}}}$$
Recall $\displaystyle\;\frac{1}{\sqrt{1-4s}}$ is the generating function for the central binomial coefficients:
$$\frac{1}{\sqrt{1-4s}} = \sum_{k=0}^\infty \binom{2k}{k} s^k$$
This leads to
$$\sum_{k=0}^\infty t^k \int_0^\infty \frac{dx}{(x^2+x+1)^{k+1}} =
\frac{2\pi}{\sqrt{3}}\sum_{k=0}^\infty \binom{2k}{k}\frac{t^k}{3^k}
$$
By comparing coefficients of $t^k$ on both sides, we obtain
$$\int_{-\infty}^\infty \frac{dx}{(x^2+x+1)^{k+1}}
= \frac{2\pi}{3^k\sqrt{3}}\binom{2k}{k}\quad\text{ for } k \in \mathbb{N}
$$
In particular, for $k = 2$, this give us
$$\int_{-\infty}^\infty \frac{dx}{(x^2+x+1)^3} = \frac{2\pi}{3^2\sqrt{3}}\binom{4}{2} = \frac{4\pi}{3\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Sum of a finite complex series
Let
$$C=\cos\theta+\cos(\theta+ \frac{2\pi}{n})+ \cos(\theta+ \frac{4\pi}{n})+...+\cos(\theta+ \frac{(2n-2)\pi}{n})$$
and
$$S=\sin\theta+\sin(\theta+ \frac{2\pi}{n})+ \sin(\theta+ \frac{4\pi}{n})+...+\sin(\theta+ \frac{(2n-2)\pi}{n})$$
Show that $C+iS$ forms a geometric series and hence show that $C=0$ and $S=0$
Then $C+iS$ = $(\cos\theta+\cos(\theta+ \frac{2\pi}{n})+ \cos(\theta+ \frac{4\pi}{n}))+i(\sin\theta+\sin(\theta+ \frac{2\pi}{n})+ \sin(\theta+ \frac{4\pi}{n}))$
$$e^{i\theta}+exp (i(\theta+\frac{2\pi}{n}))+...+exp (i(\theta+\frac{(2n-2)\pi}{n}))$$
This is a far as I've got as I'm a little stuck where to go from here. Feel like I'm quite close to the answer though.
| Yes, you are close indeed:
\begin{align}
&\exp(i\theta)+ \exp(i(\theta+2\pi/n)) +\ldots+ \exp(i(\theta+2\pi(n-1)/n))\\
&= \exp(i\theta) \sum_{k=0}^{n-1} \exp(2\pi ik/n) = \exp(i\theta) \sum_{k=0}^{n-1} \exp(2\pi i/n)^k\\
&= \exp(i\theta) \frac{1-\exp(2\pi i/n)^n}{1-\exp(2\pi i/n)} =
\exp(i\theta) \frac{1-\exp(2\pi i)}{1-\exp(2\pi i/n)} = 0,
\end{align}
and hence $C+iS = 0,$ implying that $C=S=0$.
EDIT: By request, we can avoid using the sigma notation as follows. Put $q=\exp(2\pi i/n)$ for clarity. Then
\begin{align}
&\exp(i\theta)+ \exp(i(\theta+2\pi/n)) +\ldots+ \exp(i(\theta+2\pi(n-1)/n))\\
&= \exp(i\theta) \big( 1+\exp(2\pi i/n) + \ldots + \exp(2\pi i(n-1)/n) \big)\\
&= \exp(i\theta) \big( 1+q + \ldots +q^{n-1} \big)
= \exp(i\theta) \frac{1-q^n}{1-q},
\end{align}
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.
I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.
| You can do it using the mean of the $m$ power for $m=\frac{3}{2}$ (or convexity of $x^\frac{3}{2}$ for that matter) :
We have that $\frac{x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^\frac{3}{2}}{3} \geq (\frac{x+y+z}{3})^\frac{3}{2}$ for any $x,y,z > 0$
Substituting $x=a^2, y=b^2, z=c^2$ we obtain $\frac{a^3+b^3+c^3}{3} \geq (\frac{a^2+b^2+c^2}{3})^\frac{3}{2} = 9^\frac{3}{2}=27$ and therefore the sought minimum is $3\cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theorem,
$$2^{4}\equiv 1\pmod5$$
Raising to the power k we get,
$$2^{4k}\equiv 1\pmod5$$
$$2^{4k+3}\equiv 8\pmod5$$
$$2^{4k+3}\equiv 3\pmod5$$
So, $5\mid 2^{n}-3\quad\forall \quad n=4k+3$ where $k$ is any non-negative integer.
Similarly, by Fermat's theorem
$$2^{12}\equiv 1\pmod{13}$$
$$2^{12k}\equiv 1\pmod{13}$$
$$2^{12k+4}\equiv 16\pmod{13}$$
$$2^{12k+4}\equiv 3\pmod{13}$$
Therefore, $13\mid 2^{n}-3\quad\forall \quad n=12k+4$
How do I show that it contains no term which is divisible by 65?
Thank you!
| Note that $4k+3$ is of the form $12k+3, 12k+7,\text { or } 12k+11,$ never of the form $12k+4,$ so there are no numbers in common in your sequences.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Domain of definition for $u_x + uu_y = 1$ How do i find the domain of definition for $u_x + uu_y = 1$
with $u = x/2$ on $y=x$ , $0 \leq x \leq 1$
I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 \leq s \leq 1$ at $t=0$
The characteristic equations are:
$dx/dt = 1$, $dy/dt = u$, $du/dt = 1$
Solving $dx/dt = 1$ gives $x=t +s$
Solving $du/dt = 1$ gives $u=t+ s/2$
Solving $dy/dt = u$ gives $y = (1/2)t^2 + st/2 + s$
This is where I get stuck,
$t=x-s$ so $y = (1/2)(x-s)^2 + (1/2)s(x-s) + s$
and $0 \leq s \leq 1$ so am i correct in saying that the domain of definition is the region between $y = (1/2)x^2 $ and $y = (1/2)(x-1)^2 + (1/2)(x-1) + 1$
| $$u_x+uu_y= \tag 1$$
Your three equations written on a equivalent form:
$$\frac{dx}{1}=\frac{dy}{u}=\frac{du}{1}=dt$$
A first family of characteristic equations comes from $\frac{dx}{1}=\frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $\frac{dy}{u}=\frac{du}{1}$
$$\frac{u^2}{2}-y=c_2$$
The general solution of the PDE expressed on the form of implicite equation is :
$$\frac{u^2}{2}-y=F(u-x) \tag 2$$
where $F$ is an arbitrary equation, to be determined according to boundary condition :
$$u(x,x)=\frac{x}{2}\quad\implies\quad\frac{x^2}{8}-x=F\left(\frac{x}{2}-x\right)$$
$X=-\frac{x}{2}\quad;\quad x=-2X$
$$F(X)=\frac{(-2X)^2}{8}-(-2X)=\frac{X^2}{2}+2X$$
So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$
$$\frac{u^2}{2}-y=\frac{(u-x)^2}{2}+2(u-x)$$
After simplification :
$$u(x,y)=\frac{y+\frac{x^2}{2}-2x}{x-2}$$
You can find the domain of definition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$ is of order $3$ mod a prime $p$, then ... The question says:
Prove that if $a$ is of order $3$ modulo a prime $p$, then $1+a+a^2\equiv 0 \pmod p$. Moreover, $a+1$ is of order $6$.
For the First Part:
The typical idea is to start with $a^3 \equiv 1 \pmod p \to a^3 -1 \equiv 0 \pmod p$. Factoring the term on the left hand side, the rest is straightforward.
However, I need to check the following idea:
$$1+a+a^2 \equiv a^3+a^2+a \equiv a(1+a+a^2)\equiv a^2(1+a+a^2)$$
$$\equiv a^3(1+a+a^2) \equiv 0 \pmod p$$
Since, by the hypothesis, $a^3$ cannot be zero modulo the prime $p$, the desired result holds.
If this idea holds true, it can be generalized to the following result:
if $a$ is of order $k$, then, modulo prime $p$, then $1+a+a^2+\cdots + a^{k-1}$ is divisible by $p$.
Is it??
For the Second Part:
I can see that:
$$1+a+a^2 \equiv 0 \to 1+a \equiv -a^2 \pmod p$$
However, does this implies anything? Given that $a$ is of order $3$, but what about $-a$??
Please Help, and Thanks in advance,,
| For the second part, taking from the first part that $1+a = -a^2\pmod p\implies (1+a)^6= (-a^2)^6 = a^{12}\pmod p= (a^3)^4 =1^4\pmod p=1\pmod p$ . Thus $1+a$ is of order $6$ as claimed. And if $a$ is of order $3$ then $(-a)^3 = -a^3 = -1 = p-1\pmod p$. And from this we have: $(-a)^6 = ((-a)^3)^2 = (-1)^2 = 1\pmod p$. So $-a$ is of order $6$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$
Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?
| Observe that:
$$(k+2)^2+\cdots+(2k+2)^2=$$$$\left[(k+1)^2+(k+2)^2+\cdots+(2k)^2\right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$
where $f(k)$ denotes the RHS of 2).
If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*\sqrt{1-\frac{3}{x}}}$
5) $\frac{3x}{x-x(\sqrt{1-\frac{3}{x}})}$
6) $\frac{3}{1-(\sqrt{1-\frac{3}{x}})}$
Now I would just take the limit, it would result in $\frac{3}{1-1}$ which would be undefined. For some reason, the $x$ in the denominator of step 5 should turn into $-(-x)$ which in turn would be positive and therefore be $\frac{3}{1+\sqrt{1=\frac{3}{x}}}$ which would equal $\frac{3}{2}$.
I really don't get it. Apparently the $-\infty$ would mean that $\sqrt{x^2}$ = $-x$. We didn't even evaluate the limit yet.. how does that turn into $-x$, just because we know the limit is negative does not mean we evaluated it yet..., why not simplify until there is no more simplification to be done, which is what I did in my steps, which would evaluate to undefined?
Would love some help, thanks!
| Using Taylor's expansion:
When x approches $-\infty$:
$x + \sqrt{x^2-3x} = x - x(1 - \frac{3}{2x}+\text{o}(x)) = x - x + \frac{3}{2}+\text{o}(1) \rightarrow \frac{3}{2}$.
So, yeah
$
\begin{align}
\lim_{x\to-\infty}x + \sqrt{x^2-3x} =\frac{3}{2}.
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle... Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1\times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?
Attempt
Possible areas are 1,4,9,16,25,36 and 49.
Found this 2,3,5,8 as the answer (@pic). But I want to know that if there exist any other solutions or not. If not, what's the reason?
| Unless one side of the big rectangle is $8$, the $8\times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing the $8\times 8$, we are left with three smaller squares of side-lengths $a<b<c<8$ and the $2\times 1$, forming a rectangle. Again, if the large $c\times c$ square has neighbours on two edges, we run into problems. We conclude that one edge of the rectangle is $c$. Hence we have
$$2+a^2+b^2+c^2=8c. $$
Numerically, we could have $c=7$, then $a^2+b^2=5$, i.e., $a=1$, $b=2$. Or $c=6$, then $a^2+b^2=10$, so $a=1$, $b=3$. Or $c=5$, then $a^2+b^2=13$, so $a=2$, $b=3$. $c=4$ is not possible, nor is $c\le 3$.
One readily sees that it is impossible to fill a $1\times 7$ when one part os $2\times 2$, or fill a $2\times 6$, when one part is $3\times 3$. The remaining case $c=5$ leads to the well-known solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the remainder of the division of polynomials $x^{2007}$ divided by $x^2-x+1$.
I consider to solve this problem, should I break the $x^{2007}$ to find the formula $x^2-x+1$?
| Write: $$\boxed{x^{2007} = k(x)(x^2-x+1)+r(x)}$$ where $r(x)$ is a linear polynomial.
Say $a$ is zero of $x^2-x+1$, then $$ a^2-a+1=0\;\;\;/\cdot (a+1)$$ we get $$a^3+1 =0 \;\;\Longrightarrow \;\;a^3 = -1$$ and if we put $x=a$ in boxed equation we get $$-1= a^{2007} = k(a)\cdot 0+r(a)$$
and the same for other zero $b =\overline{a}$.
So if $r(x) = kx+n$, then we have a system $$-1=ka+n$$ $$-1=kb+n$$
which can easly be solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explain this derivative identity: $ \frac{1}{2^n} \frac{d^n}{dy^n} \frac{(1+y)^{2n+3}(1-y)}{((1+y)^2 -2yx)^2} \bigg|_{y=0} = (n+1)! x^n $ I have the following result that I believe to be true:
$$
\frac{1}{2^n} \frac{d^n}{dy^n} \frac{(1+y)^{2n+3}(1-y)}{((1+y)^2 -2yx)^2} \bigg|_{y=0} = (n+1)! x^n
$$
The LHS is something that arose in physics research. The RHS has been inferred by checking with Mathematica for n from 0 to 100. However, proving this result has evaded me.
It's not surprising that the RHS is a polynomial in x with coefficient $(n+1)!$ (the only way to get an $x^n$ term is to operate the derivative on the denominator repeatedly). What is surprising is that the coefficients vanish for all but the leading term. This looks like it might be amenable to induction or recursion, but I was unable to make any meaningful headway with those techniques.
Good luck!
| We have
$$\frac1{(1-t)^2}=\sum_{m=0}^\infty (m+1)t^{m},\qquad (|t|<1)$$
Put $t=\left(\dfrac{2y}{(1+y)^2}\right)x$, where $0<|y|<1$,
$$\frac1{\left(1-\left(\frac{2y}{(1+y)^2}\right)x\right)^2}=\sum_{m=0}^\infty (m+1)\left(\dfrac{2y}{(1+y)^2}\right)^mx^{m}.\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Multiply $(1+y)^{2n-1}(1-y)$ on both sides,
$$\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}=\sum_{m=0}^\infty(m+1)\big((1+y)^{2n-2m-1}(y^m-y^{m+1})\big)(2x)^m.\qquad (|x|<\frac{(1+y)^2}{2|y|})\tag{*}$$
Consider the coefficient of $y^n$ in $(*)$,
*
*when $m<n$, the coefficient is given by
$$\binom{2n-2m-1}{n-m}y^{n-m}\cdot y^m-\binom{2n-2m-1}{n-m-1}y^{n-m-1}\cdot y^{m+1}=0;$$
*when $m=n$,
\begin{align*}
(1+y)^{-1}(y^n-y^{n+1})=\left(\sum_{k=0}^\infty (-y)^k\right)(y^n-y^{n+1})=y^n+\underbrace{\text{something}}_{\text{“deg of $y$” $>n$}},
\end{align*}
so the coefficient is $(n+1) (2x)^n$;
*when $m>n$, $2n-2m-1<0$,
\begin{align*}
(1+y)^{2n-2m-1}(y^m-y^{m+1})=\left(\sum_{k=0}^\infty (-y)^k\right)^{1+2m-2n}(y^m-y^{m+1})
\end{align*}
does not contain $y^n$.
Therefore,
$$\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}=\color{blue}{\underbrace{\text{something}}_{\text{“deg of $y$” $<n$}}}+\big((n+1) (2x)^n\big)y^n+\color{orange}{\underbrace{\text{something}}_{\text{“deg of $y$” $>n$}}},\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Take the $n$-th derivative with respect to $y$ and set $y\to 0$,
\begin{align*}
\left.\frac{\partial^n}{\partial y^n}\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}\right|_{y=0}&=\color{blue}{0}+\big((n+1) (2x)^n\big)(y^n)^{(n)}+\color{orange}{0},\qquad (|x|<+\infty)\\
\left.\frac{\partial^n}{\partial y^n}\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}\right|_{y=0}&=(n+1)! (2x)^n.\qquad (x\in \Bbb R)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate inverse of matrix with -1 on diagonal and 1 on the rest Calculate the inverse of the matrix
\begin{bmatrix}
-1& 1& ...& ...&1 \\
1& -1& 1& ... &1 \\
...& ...& ...& ...&1 \\
1&1 &1 & ...&1 \\
1& 1 &1 &1 &-1
\end{bmatrix}
$-1$ on the diagonal and $1$ on the rest.
The key I think is to perform a sequence of elementary transformations on
[ A | $I_{n}$ ] until we get [ $I_{n}$ | $A^{-1}$ ] but that seems to be complicated.
| The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n \geq 3$
for $n=3:$
$$
\left(
\begin{array}{rrr}
-1 & 1&1 \\
1 & -1&1 \\
1 & 1&-1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
a & b&b \\
b & a&b \\
b & b&a \\
\end{array}
\right)=
\left(
\begin{array}{rrr}
1 & 0&0 \\
0 & 1&0 \\
0 & 0&1 \\
\end{array}
\right)
$$
for $n=4,$ different $a,b:$
$$
\left(
\begin{array}{rrrr}
-1 & 1&1&1 \\
1 & -1&1&1 \\
1 & 1&-1&1 \\
1 & 1&1&-1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
a & b&b&b \\
b & a&b&b \\
b & b&a&b \\
b & b&b&a \\
\end{array}
\right)=
\left(
\begin{array}{rrrr}
1 & 0&0&0 \\
0 & 1&0&0 \\
0 & 0&1&0 \\
0 & 0&0&1 \\
\end{array}
\right)
$$
for $n=5,$ still different $a,b:$
$$
\left(
\begin{array}{rrrrr}
-1 & 1&1&1&1 \\
1 & -1&1&1&1 \\
1 & 1&-1&1&1 \\
1 & 1&1&-1&1 \\
1&1&1&1&-1 \\
\end{array}
\right)
\left(
\begin{array}{ccccc}
a & b&b&b &b\\
b & a&b&b&b \\
b & b&a&b &b \\
b & b&b&a &b \\
b&b&b&b&a \\
\end{array}
\right)=
\left(
\begin{array}{rrrrr}
1 & 0&0&0 &0 \\
0 & 1&0&0&0 \\
0 & 0&1&0 &0 \\
0 & 0&0&1 &0 \\
0&0&0&0&1
\end{array}
\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Laurent Series expansion about the point $z_0 = i$ of $\frac{z}{z^2+1}$ I am trying to construct the Laurent series expansion of $f(z) = \frac{z}{z^2+1}$ about $z_0 = i$ in the region $\{z \in \mathbb{C}: 0 < |z - i| < 2\}$ but I am stuck.
We can re-write $f(z) = \dfrac{z}{z^2+1} = \dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $\pm i$. We can use partial fractions to obtain:
$$\frac{1}{(z+i)(z-i)} = \frac{i}{2(z+i)} - \frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$\frac{1}{z+i} = \frac{1}{2i + (z - i)} = \frac{1}{2i}\left(\frac{1}{1+\frac{1}{2i}(z-i)}\right)$$
Using the geometric series this gives:
$$ = \frac{-i}{2} \left(\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right)$$
Then we have that:
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?
| Following your argument,
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right) = (z-i+i)\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)\\ = \left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^{n+1}}{(2i)^n}\right) - 1\right) -\left(\frac{1}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the Limit $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$ Find $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$
My ideas: let $ n \in \mathbb N$ be constant, looking at $\frac{k^{n}}{1+k^{n+2}}$, we know
$$\frac{k^{n}}{1+k^{n+2}}\leq\frac{k^{n}}{k^{n+2}}=\frac{1}{k^{2}}$$
but this does not help me because $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\pi^{2}/6$
any ideas?
| I would do in this way
$$
\eqalign{
& \sum\limits_{1\, \le \,k} {{{k^{\,n} } \over {1 + k^{\,n + 2} }}}
= \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {{{k^{\,n + 2} } \over {1 + k^{\,n + 2} }}} \right)}
= \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {1 - {1 \over {1 + k^{\,n + 2} }}} \right)} = \cr
& = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}} - {1 \over 2} - \sum\limits_{2\, \le \,k} {{1 \over {k^{\,2} }}\left( {{1 \over {1 + k^{\,n + 2} }}} \right)}
\;\mathop \approx \limits^{n \to \infty } \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}
- {1 \over 2} - \sum\limits_{2\, \le \,k} {{1 \over {k^{\,2} }}\left( {{1 \over {k^{\,n + 2} }}} \right)} = \cr
& = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}} - {1 \over 2} - \left( {\sum\limits_{1\, \le \,k} {{1 \over {k^{\,n + 4} }}} - 1} \right)
= \zeta (2) + {1 \over 2} - \zeta (n + 4) \to \quad \zeta (2) - {1 \over 2} \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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