Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why is $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ rational? For this question,
Show the following irrational-looking expressions are actually rational numbers.
(a) $\sqrt{4+2\sqrt{3}}-\sqrt{3}$, and
(b) ...
I solved it as follows:
$$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\
x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\
(x+\sqrt{3})^{2} &= (\sqrt{4+2\sqrt{3}})^{2},\\
\end{align}\\
x^{2}+2\sqrt{3}x-(1+2\sqrt{3}) = 0,\\
(x-1)(x+(1+2\sqrt{3}))=0.$$
My question is that, there are two numbers satisfying $x = \sqrt{4+2\sqrt{3}}-\sqrt{3}$, but one of them is irrational. Then, how can we say $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational as a whole?
| With respect to your written
note that $$\bf{x = \sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{4+\sqrt{12}}-\sqrt{3} > 0}$$
$$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\
x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\
(x+\sqrt{3})^{2} &= (\sqrt{4+2\sqrt{3}})^{2},\\
\end{align}\\
x^{2}+2\sqrt{3}x-(1+2\sqrt{3}) = 0,\\
(x-1)(x+(1+2\sqrt{3}))=0.$$ now you have two roots ,but $\bf{x >0}$
so
$$x-1=0 \to x=1 \checkmark \\x=-1-2\sqrt 3 <0 \times$$
| {
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"url": "https://math.stackexchange.com/questions/2457032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trying to solve $2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$ without seeing the obvious solution
Determine all real values of $x$ such that $$\log_{5x+9}(x^2+6x+9)+\log_{x+3}(5x^2+24x+27)=4$$
Taken from Waterloo 2012: Link
What I tried:
$$2\log_{5x+9}(x+3)+\log_{x+3}\left((x+3)(5x+9)\right)=4$$
$$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$$
Then I was stuck and I graphed it:
It was clearly $0$:
$$2\log_{9}(3)+\log_{3}(9)=3$$
But how could I do it without seeing the "zero"? Is there a way?
| To solve
$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$
first of all we note that
$$\log _a b=\frac{1}{\log _b a}$$
the equation becomes
$$2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9}(x+3)}=3$$
And then set $w=\log_{5x+9}(x+3)$
$2w+\dfrac{1}{w}=3\to 2w^2-3w+1=0\to w_1=\dfrac{1}{2};\;w_2=1$
If $w=1$ we have $\log_{5x+9}(x+3)=1$ which means $5x+9=x+3\to \color{red}{x=-\dfrac{3}{2}}$
If $w=\dfrac{1}{2}$ then $\log_{5x+9}(x+3)=\dfrac{1}{2}$
that is $(5x+9)^{1/2}=x+3$
$5x+9=(x+3)^2\to x^2+x=0\to \color{red}{x=0;\;x=-1}$
Let's verify these solutions in the given equation
for $x=0$ we have $2\log_{9}(3)+\log_{3}(9)=3\to 2\cdot \frac12+2=3$ true
for $x=-1$ equation becomes $2\log_{9-5}(3-1)+\log_{3-1}(9-5)=3\to 2\log_4 2+\log_2 4=3$ true
and for $x=-\dfrac{3}{2}$ we have $2\log_{9-\frac{15}{2}}\left(3-\frac32\right)+\log_{3-\frac32}\left(9-\frac{15}{2}\right)=\log_{\frac{3}{2}}\left(\frac12\right)^2+\log_{\frac12}\left(\frac{3}{2}\right)=3$
Hope this is useful
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$ Find the value of $$I=\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$$
My Try: I used Integration by parts
So
$$I=\frac{1}{2}\log x \times \log (1+x^2) \biggr\rvert_{0}^{1}-\int_{0}^{1}\frac{\log(1+x^2)}{2x}dx$$
So
$$I=\frac{-1}{2}\int_{0}^{1}\frac{\log(1+x^2)}{x}$$
Can we proceed from here?
| $\begin{align}I=\int_{0}^{1} \frac{x \log x }{1+x^2}\:dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}I=\dfrac{1}{4}\int_{0}^{1} \frac{\log x }{1+x}\:dx\end{align}$
And it is well known that,
$\begin{align}J=\int_{0}^{1} \frac{\log x }{1+x}\:dx=-\frac{\pi^2}{12}\end{align}$
Addendum:
Actually,
$\displaystyle K=\int_0^1 \frac{\ln x}{1-x}\,dx=-\zeta(2)$
(Taylor series expansion)
$\begin{align}K-J=\int_0^1 \dfrac{2x\ln x}{1-x^2}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}K-J&=\dfrac{1}{2}\int_0^1 \dfrac{\ln x}{1-x}\,dx\\
&=\dfrac{1}{2}K
\end{align}$
Therefore,
$J=\dfrac{1}{2}K=-\dfrac{1}{2}\zeta(2)$
NB:
One assumes that $\displaystyle \zeta(2)=\dfrac{\pi^2}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation
| By C-S
$$\sum_{cyc}\frac{x}{yz}=\sum_{cyc}\frac{x^2}{xyz}\geq\frac{(x+y+z)^2}{3xyz}\geq\frac{3(xy+xz+yz)}{3xyz}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is general solution of the PDE $(x^2+y^2)u_x+2xyu_y=-u^2$? What is the general solution of the following PDE?
$$(x^2+y^2)u_x+2xyu_y=-u^2.$$
If we write the characteristic equation
$$\frac{dx}{x^2+y^2}=\frac{dy}{2xy}=\frac{-du}{u^2}$$
then, we find $\frac{d(x+y)}{(x+y)^2}=\frac{-du}{u^2}$ and we have $c_1=1/u+1/(x+y)$.
And then?
Best regards.
| HINT :
$$\frac{dx}{x^2+y^2}=\frac{dy}{2xy}=\frac{dx+dy}{(x+y)^2}=\frac{dx-dy}{(x-y)^2}
$$
$$\frac{1}{x+y}-\frac{1}{x-y}=c_2$$
With the first characteristic equation that you alredy found, this second equation (or any other equivalent equation) allows to express the general solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466324",
"timestamp": "2023-03-29T00:00:00",
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Solving systems of equation with unknown How do i solve this system?
$$
\left\{
\begin{array}{c}
ax+y+z=1 \\
x+ay+z=a \\
x+y+az=a^2
\end{array}
\right.
$$
Ive reduced to this form. How should i continue to get infinitely many solutions, no solution and unique?
$$ \left[
\begin{array}{ccc|c}
1&a&1&a\\
0&(a^2)-1&a-1&(a^2)-1\\
0&(a-1)(a-2)&0&a-1
\end{array}
\right] $$
| We have to obtain the reduced row echelon form of the augmented matri, starting from
$$\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&a&1&a\\0&a^2-1&a-1&a^2-1\\0&(a-1)(a-2)&0&a-1
\end{array}\!\!\end{bmatrix}$$
*
*If $a=1$, this matrix becomes
$\;\smash[b]{\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&1&1\\0&0&0&0\\0&0&0&0
\end{array}\!\!\end{bmatrix}}$, which corresponds to the plane with equation:
$$\color{red}{x+y+z=1}.$$
*If $a\ne 1$, we can factor out $x-1$, and thus obtaining the matrix
$$\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&a&1&a\\0&a+1&1&a+1\\0&a-2&0&1
\end{array}\!\!\end{bmatrix},\enspace\text{and swapping columns $2$ and $3$}:\quad \begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&a&a\\0&1&a+1&a+1\\0&0&a-2&1
\end{array}\!\!\end{bmatrix}$$
(Swapping these columns amounts to exchange unknowns $y$ and $z$).
*If $a=2$, the linear system is inconsistent.
*If $a\ne2$, we can proceed with row reduction:
\begin{align}
\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&a&a\\0&1&a+1&a+1\\0&0&a-2&1
\end{array}\!\!\end{bmatrix}&\rightsquigarrow\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&a&a\\0&1&a+1&a+1\\0&0&1&\frac1{a-2}
\end{array}\!\!\end{bmatrix}\rightsquigarrow\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&1&0&\frac{a(a-3)}{a-2}\\0&1&0&\frac{(a+1)(a-3)}{a-2}\\0&0&1&\frac1{a-2}
\end{array}\!\!\end{bmatrix}\\
&\rightsquigarrow
\begin{bmatrix}\!\!\begin{array}{@{}ccc|c}
1&0&0&-\frac{(a-3)}{a-2}\\0&1&0&\\0&0&1&\frac1{a-2}
\end{array}\!\!\end{bmatrix}
\end{align}
So the solutions are
$$\color{red}{x=-\frac{(a-3)}{a-2}},\quad \color{red}{y=\frac1{a-2}},\quad \color{red}{z=\frac{(a+1)(a-3)}{a-2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Computing the binomial series: $\binom{n}{2}\cdot2^2+\binom{n}{4}\cdot2^4+\binom{n}{6}\cdot2^6+\cdots+ \binom{n}{n}\cdot2^n$ How to find the tight bound for the binomial series
$\binom{n}{2}\cdot2^2+\binom{n}{4}\cdot2^4+\binom{n}{6}\cdot2^6\ldots,+ \binom{n}{n}\cdot2^n$
I have found $3^n$ for series
$(1+2)^n=1+\binom{n}{1}\cdot2^1+\binom{n}{2}\cdot2^2+\binom{n}{3}\cdot2^3+\cdots+ \binom{n}{n}\cdot2^n$
Is there any way to prove the bound of above series.
| $$3^n = (1+2)^n=1+\binom{n}{1}*2^1+\binom{n}{2}*2^2+\binom{n}{3}*2^3\ldots,+ \binom{n}{n}*2^{n}$$
$$ (-1)^n =(1-2)^n=1-\binom{n}{1}*2^1+\binom{n}{2}*2^2-\binom{n}{3}*2^3\ldots,+ \binom{n}{n}*2^{n}$$
then $$3^n +(-1)^n = (1+2)^n+(1-2)^n=2+\binom{n}{2}*2^2+\binom{n}{4}*2^4+\binom{n}{6}*2^6\ldots, $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If $a+b+c=0$ prove that $ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $$
I made this question as a more difficult (higher degree) version of this question. My idea was that algebraic brute force methods are easy to distinguish from more sophisticated ones if the degree of the terms in the question is higher.
The question was specifically made using the method from my answer to the linked question.
| plug in the term $$c=-a-b$$ in the left-hand side of the equation we get
$$-30 a b (a+b) \left(a^2+a b+b^2\right)^2$$ and so is the right-hand side
a remark: if we compute the left-hand side minus the right-hand side and factorize, we obtain
$$-(a+b+c) \left(2 a^6-2 a^5 b-2 a^5 c-a^4 b^2+4 a^4 b c-a^4 c^2+6 a^3 b^3-3 a^3 b^2 c-3 a^3 b
c^2+6 a^3 c^3-a^2 b^4-3 a^2 b^3 c+6 a^2 b^2 c^2-3 a^2 b c^3-a^2 c^4-2 a b^5+4 a b^4 c-3 a
b^3 c^2-3 a b^2 c^3+4 a b c^4-2 a c^5+2 b^6-2 b^5 c-b^4 c^2+6 b^3 c^3-b^2 c^4-2 b c^5+2
c^6\right)$$ and it is clear that we get zero
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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3 normals to the parabola
Suppose there exists three normal lines from the point $(r, 0)$ to the
parabola $x = y^2$, and $r > \frac{1}{2}$, one of which is the
x axis. Determine the value of r where the two other normal lines would be
perpendicular?
I found the $\frac{dy}{dx} (parabola)= \frac{1}{2\sqrt{x}} $ .
Thus gradient of normal, is $-2\sqrt{x}$, and the equation of normal is $y=-2\sqrt{x} (x-r) $.
If I equate the equation of normal with the parabola,
I get $ \pm \sqrt{x}=-2\sqrt{x} (x-r) $,
which solving will result in $x=\pm\frac{1}{2} +r$,
however I am stuck and I cannot find an exact value of r. Where did I go wrong and how should I have approached this question?
| In the diagram below, if the two normals are to be perpendicular, then we have a right triangle whose vertices are the three marked points, so that
$$(2a)^2 = 2((a-0)^2+(a^2-r)^2) = 2 a^2+2 a^4-4 a^2 r+2 r^2.$$
Also, since the lines are normal to the parabola, we see that the slope of the upper normal, which is $-2a$, must be equal to the slope of the line, so that
$$-2a = \frac{a}{a^2-r}.$$
Solving these two equations for $a$ and $r$ gives $r = \frac{3}{4}$ and $a = \pm\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472271",
"timestamp": "2023-03-29T00:00:00",
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Showing that $a_n= 3^{n}+ 4^{n+1}+ 5^{n+2}$ is a solution of the recurrence relation $a_n = 12a_{n−1} − 47a_{n−2}+ 60a_{n−3}$
Let $a_n$ be given recursively by
$$a_n = \begin{cases} 30 & \text{if } n = 0\\\\ 144 & \text{if } n = 1\\\\ 698 & \text{if } n = 2\\\\ 12 a_{n−1} − 47 a_{n−2} + 60 a_{n−3} & \text{if } n \geq 3\\\end{cases}$$
Prove that $a_n= 3^{n}+ 4^{n+1}+ 5^{n+2}$. For which $n$ is $a_n$ divisible by $3$?
I honestly have no idea how to approach this. I started with brute force to obtain $a_3$ and $a_4$ as $3,408$ and $16,730$ but I do not see any pattern in the sequence. I think induction could be one approach. Right now, I'm learning about congruence classes so that might be another route but I don't know how I'd apply it.
| First part
From characteristic polynomials point of view, the recurrence
$$a_n=12a_{n-1}-47a_{n-2}+60a_{n-3}$$
has the following characteristic polynomial (as it was pointed by Jyrki Lahtonen in the comments)
$$x^3-12x^2+47x-60$$
which has the following roots $r_1=3, r_2=4,r_5=5$. Thus the general term of the recurrence is
$$a_n=Ar_1^n+Br_2^n+Cr_3^n=A\cdot3^n+B\cdot4^n+C\cdot5^n$$
Given the initial values, we have the following system of linear equations
$$\left\{\begin{matrix}
30=A+B+C
\\
144=A\cdot3+B\cdot4+C\cdot5
\\
698=A\cdot3^2+B\cdot4^2+C\cdot5^2
\end{matrix}\right.$$
which yields the following solution $A=1, B=4, C=25$. As a result:
$$a_n=3^n+4^{n+1}+5^{n+2}$$
Second part
$$3 \mid a_n \Leftrightarrow 3 \mid 4^{n+1}+5^{n+2}$$
But
$$4 \equiv 1 \pmod{3} \text{ then } 4^{n+1} \equiv 1 \pmod{3}$$
$$5 \equiv -1 \pmod{3} \text{ then } 5^{n+2} \equiv (-1)^{n+2} \pmod{3} $$
Thus
$$3 \mid 4^{n+1}+5^{n+2} \Leftrightarrow 3 \mid 1 + (-1)^{n+2}$$
which says for odd $n$, $3 \mid a_n$, but it also works for $n=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find $AC$ WITHOUT using Cosine law...
In the following figure find side $AC$ WITHOUT using Cosine law:
There is a simple solution involving Cosine law,but is it possible to calculate $AC$ without using it??
| Let $BC=x$. Then $AC=\sqrt{1+x^2}$. Use the law of sines in triangle BCD to obtain
$$\frac{x}{\sin\angle D}=\frac{1}{\sin 30^\circ}=2.$$
Notice that $\angle ABD=120^\circ$, whose supplementary angle is $60^\circ$. So we have
$$ AB\times\sin 60^\circ=AD\times\sin\angle D.$$
We may plug in numbers to obtain
$$(1+\sqrt{1+x^2})\,\frac{x}{2}=\frac{\sqrt{3}}{2}.$$
Here's the tricky part. In terms of $y=\sqrt{1+x^2}$, the equation becomes
\begin{align}
(1+y)\sqrt{y^2-1}&=\sqrt{3},\\
y^4+2y^3-2y-1&=3,\\
(y^3-2)(y+2)&=0.
\end{align}
The answer is $AC=2^{1/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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(1) Sum of two factorials in two ways; (2) Value of $a^{2010}+a^{2010}+1$ given $a^4+a^3+a^2+a+1=0$. Question $1$:
Does there exist an integer $z$ that can be written in two different ways as $z=x!+y!$,where $x,y\in \mathbb N$ and $x\leq y$?
Answer: $0!=1!$ so $0!+2!=3=1!+2!$
Question $2$:
If $a^4+a^3+a^2+a+1=0$, find the value of $a^{2010}+a^{2010}+1$.
Answer: If $a^4+a^3+a^2+a+1=0$ then $$(a-1)(a^4+a^3+a^2+a+1)=0\implies a^5-1=0\implies a=1$$So the value of the required expression is $3$
I think that it is wrong as at $a=1$,$\frac{a^5-1}{a-1}$ is not defined.
Please give me some hints to solve thes problems?
| The second solution is not so wrong
$a^5-1=0$ has five solutions, the five complex fifth roots of unity. Namely
$$\left(a_0=1,\;a_1 = -\frac{1}{4}-\frac{\sqrt{5}}{4}-i \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}},\;a_2 = -\frac{1}{4}+\frac{\sqrt{5}}{4}+i \sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}},\\a_3 = -\frac{1}{4}+\frac{\sqrt{5}}{4}-i \sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}},\;a_4 = -\frac{1}{4}-\frac{\sqrt{5}}{4}+i \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}\right)$$
One of them, $a=1$, is not a root of the fourth degree equation
$$a^4+a^3+a^2+a+1=0$$
but the other four actually are because $$a^5-1=(a-1)(a^4+a^3+a^2+a+1)$$
So plugging $a^5=1$ in the $a^{2000}+a^{2010}+1$ is perfectly legal and the actual result is $3$
Hope this is useful
| {
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"url": "https://math.stackexchange.com/questions/2474724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.
| If you need to prove it for odd $n$, then here is an elegant proof
$$2^n=(3-1)^n$$
$$=3k+(-1)^n$$
Since $n$ is odd , $(-1)^n=-1$
Also proving $2^n+1$ is divisible by $3$ is same as proving $2^n-2$ is divisible by $3$
Therefore
$$ 2^n-2= (3-1)^n-2$$
$$=3k+(-1)^n-2$$
$$=3k-1-2$$
$$3k-3 $$
And this is divisible by $3$ for all $n\in \mathbb N(odd)$
And of course it is not true for every even $n$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of sixth power of roots of $x^3-x-1=0$ Question:
Find the sum of sixth power of roots of the equation $x^3-x-1=0$
My First approach:
Let $S_i$ denote the sum of $i^{th}$ power of roots of the given equation.
Now, multiplying given equation by $x^3$ , putting each of the three roots and adding the three formed equations
we get, $S_6=S_4+S_3$
repeating same above procedure to obtain,
$S_4=S_2+S_1$ , $S_3=S_1+1$
hence, $S_6=S_2+2S_1+1=2+0+1=3$
My Second Approach:
Let $a,b,c$ be the roots of $f(x)=x^3-x-1$
Clearly, $f^/(x)/f(x)=1/(x-a)+1/(x-b)
+1/(x-c)$
=$\sum(1/x+a/x^2+a^2/x^3+a^3/x^4+\cdots)=3/x+S_1/x^2+S_2/x^3+\cdots$
hence we get, $S_6=5$
$\rule{200px}{0.5px}$
I am getting wrong answer through First Approach, Please point out my mistake or post a better solution.
Thank You
|
$S_6=S_2+2S_1+1$
This is not correct.
We have
$$S_6=S_2+2S_1+\color{red}{3}$$
since $$S_6=x_1^6+x_2^6+x_3^6=(x_1^2+2x_1+1)+(x_2^2+2x_2+1)+(x_3^2+2x_3+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Induction for a Double Sided Inequality For the question below I have split the inequalities into two and currently have
LS
$$\frac{1}{4} k^4 + k^3 < \frac{1}{4} (k+1)^4$$
RS
$$\frac{1}{4} (k+1)^4 < \frac{1}{4} k^4 + (k+1)^3$$
I am unsure of what I am solving for after this. When one side equals the other it makes sense that I have to make the two sides the same. With inequalities however how can I show one is less then the other? What am I trying to show?
| We should prove that
$${ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }<\frac { { \left( n+1 \right) }^{ 4 } }{ 4 } <{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+\left( n+1 \right) ^{ 3 }$$
$$L.H.\quad \underset { <\frac { { n }^{ 4 } }{ 4 } }{ \underbrace { { 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ \left( n-1 \right) }^{ 3 } } +{ n }^{ 3 } } <\frac { { n }^{ 4 } }{ 4 } +{ n }^{ 3 }=\frac { { n }^{ 4 }+4{ n }^{ 3 } }{ 4 } <\frac { { n }^{ 4 }+4{ n }^{ 3 }+6{ n }^{ 2 }+4n+1 }{ 4 } =\frac { { \left( n+1 \right) }^{ 4 } }{ 4 } \\ R.H.{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }+\left( n+1 \right) ^{ 3 }>\frac { { n }^{ 4 } }{ 4 } +\left( n+1 \right) ^{ 3 }=\frac { { n }^{ 4 }+4\left( n+1 \right) ^{ 3 } }{ 4 } =\frac { { n }^{ 4 }+4{ n }^{ 3 }+12{ n }^{ 2 }+12n+4 }{ 4 } >\frac { { n }^{ 3 }+4{ n }^{ 3 }+6{ n }^{ 2 }+4n+1 }{ 4 } =\frac { { \left( n+1 \right) }^{ 4 } }{ 4 } \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find pdf given transformation I'm given that the random variable $X$ is distributed as $UNIF(0,1)$.
Moreover, I'm given that:
$U = X(1-X)$
Using the CDF method (i.e. finding the CDF, then taking the derivative), how would I find the PDF of $U$? I can't isolate X in the equation, so how would I go about getting the PDF of $U$ as a result? Thank you
| Let $F(u) = \Pr(U \le u)$. The minimum and maximum values of $U$ are $0$ and $1/4$, respectively. Therefore, $F(u)=0$ when $u \le 0$, and $F(u)=1$ when $u \ge 1/4$.
Below is the calculations for $0 \le u \le 1/4$. Observe that
$$F(u) = \Pr(X(1-X)\le u) = \Pr(X-X^2 \le u) = \Pr(X-X^2-1/4 \le u-1/4).$$
But $-(X-1/2)^2 = -(X^2-X+1/4)=X-X^2-1/4$. Therefore,
$$F(u) = \Pr\left(-\left(X-\frac{1}{2}\right)^2 \le u-\frac{1}{4}\right)
\\
= \Pr\left(\left(X-\frac{1}{2}\right)^2 \ge \frac{1}{4}-u\right) \\
= \Pr\left(\left|X-\frac{1}{2}\right| \ge \sqrt{\frac{1}{4}-u}\right).$$
Consequently,
$$F(u) = \Pr\left(X \le \frac{1}{2} - \sqrt{\frac{1}{4}-u}\right) + \Pr\left(X \ge \frac{1}{2} + \sqrt{\frac{1}{4}-u}\right).$$
But $\Pr(X \le x)=x$ and $\Pr(X \ge x) = 1-x$. So
$$F(u) = 1-\sqrt{1-4u}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Proving Convergence using Monotone Convergence Theorem Prove that the sequence defined by $x_1 = 3$ and $x_{n+1} = \displaystyle\frac{1}{4-x_n}$ converges.
I know that the sequence looks like this: $\{3, 1, \displaystyle\frac{1}{3}, \frac{3}{11}...\}$ and the sequence is decreasing.
To use the Monotone Convergence Theorem I know that we have to show it is decreasing, thus show that $x_n \geq x_{n+1} (\forall n \in \mathbb{N}$) and bounded (above?).
So far I have come up with:
We need to show that $x_n \geq x_{n+1}$ $\forall n \in \mathbb{N}$ and the sequence is bounded below. We have that $x_1 = 3$. Notice $x_2 = \displaystyle\frac{1}{4-x_1}=\frac{1}{4-3}=1$ and $3\geq 1$. So the statement $x_n \geq x_{n+1}$ holds when $n=1$.
I am having trouble with the proof from this point. Can someone walk me through step by step?
There are two other parts to this problem after proving this fact:
(b) Now that we know lim$x_n$ exists, explain why lim$x_{n+1}$ exists and equal to the same value.
(c) Take the limit of each side of the recursive sequence in part (a) to explicitly compute lim$x_n$.
| I am reading "Understanding Analysis 2nd Edition" by Stephen Abbott.
This exercise is Exercise 2.4.1 on p.59 in this book.
(a)
Let $f(x):=x^2-4x+1.$
Then, $f(x_n)\leq 0$ if and only if $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3}.$
$x_{n+1}-x_n=\frac{1}{4-x_n}-x_n=\frac{x_n^2-4x_n+1}{4-x_n}=\frac{f(x_n)}{4-x_n}.$
If $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3},$ then $2-\sqrt{3}\leq 4-x_n\leq 2+\sqrt{3}.$
If $2-\sqrt{3}\leq 4-x_n\leq 2+\sqrt{3},$ then $2-\sqrt{3}\leq \frac{1}{4-x_n}\leq 2+\sqrt{3}.$
Therefore, if $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3},$ then $2-\sqrt{3}\leq x_{n+1}\leq 2+\sqrt{3}.$
And $2-\sqrt{3}\leq x_1=3\leq 2+\sqrt{3}.$
So, $2-\sqrt{3}\leq x_n\leq 2+\sqrt{3}$ for all $n\in\{1,2,\dots\}.$
So, $f(x_n)\leq 0$ for all $n\in\{1,2,\dots\}.$
So, $x_{n+1}-x_n\leq 0$ for all $n\in\{1,2,\dots\}.$
So, $\{x_n\}$ is a decreasing sequence.
And $2-\sqrt{3}$ is a lower bound of $\{x_n\}$.
So, $\{x_n\}$ converges.
(b)
Let $a:=\lim x_n$.
This means for any positive $\epsilon$, there exists a natural number $N$ such that if $n\geq N$, then $|x_n-a|<\epsilon.$
So, if $n\geq \max\{N-1, 1\}$, then $|x_{n+1}-a|<\epsilon.$
So, $\{x_{n+1}\}$ also converges to $a$.
(c)
$\{x_{n+1}\}$ converges to $a$.
By Algebraic Limit Theorem on p.50, $\{\frac{1}{4-x_n}\}$ converges to $\frac{1}{4-a}.$
Since $\{x_{n+1}\}=\{\frac{1}{4-x_n}\}$ and the limit of a sequence is unique by Theorem 2.2.7 on p.46, $a=\frac{1}{4-a}$.
So, $f(a)=0$.
So, $a=2+\sqrt{3}$ or $a=2-\sqrt{3}.$
Since $x_1=3<2+\sqrt{3}$ and $\{x_n\}$ is a decreasing sequence, $a=2-\sqrt{3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the trigonometric equation $\sin x + \cos x=\sin 2x + \cos 2x$?
Question: Solve the trigonometric equation: $\sin x + \cos x=\sin 2x + \cos 2x$.
My attempt:
$\sin x + \cos x=\sin 2x + \cos 2x$
$\implies \sin x + \cos x=2\sin x \cos x + \cos^2 x - \sin^2 x$
$\implies \sin x + \cos x=2\sin x \cos x + \cos^2 x - (1-\cos^2 x)$
$\implies \sin x + \cos x=2\sin x \cos x + 2\cos^2 x - 1$
$\implies \sin x - 2\sin x \cos x + \cos x - 2\cos^2 x= - 1$
$\implies \sin x(1-2\cos x)+\cos x(1-2\cos x)=-1$
$\implies (1-2\cos x)(\sin x+\cos x)=-1$
$\implies (1-2\cos x)=-1$ or $(\sin x +\cos x)=-1$
$\implies \cos x=1$ or $\sin^2 x +\cos^2 x + 2\sin x\cos x=1$
$\implies x=2n\pi$ or $\sin 2x=0$
$\implies x=2n\pi$ or $2x=n\pi$
$\therefore x=2n\pi$ or $x=\frac{n\pi}{2}$
But the answers given in my book are $x=2n\pi$ and $x=\frac{(4n+1)\pi}{6}$. Where have I gone wrong? Please help.
| Use Subtraction: $$\sin 2x−\sin x=\cos x−\cos 2x$$
$$2\sin\frac{x}2\cos\frac{3x}{2}=2\sin\frac{3x}{2}\sin\frac{x}{2}$$
So,
$$\sin\frac{x}{2}=0$$
OR
$$\tan\frac{3x}{2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Proving Schwarz's theorem for reversal of order of derivative
Show that the function $f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ at $(x,y) \neq (0,0)$ and equal to $0$ when $(x,y)= (0,0)$ does not satisfy conditions of Schwarz's theorem.
Here I am able to find out $f_x(x,y)$ but how to find $f_{yx}(x,y)$
| Let's take the partial derivative with respect to $x$ at $(x, y) = (0, 0)$ first. We calculate:
\begin{align*}
f_x(0, 0) &= \lim_{h \rightarrow 0} \frac{f(0 + h, 0) - f(0, 0)}{h} \\
&= \lim_{h \rightarrow 0} \frac{f(h, 0)}{h} \\
&= \lim_{h \rightarrow 0} \frac{(h)(0)(h^2 - 0^2)}{h} = 0.
\end{align*}
Similarly for $f_y(0, 0) = 0$. At other points, we have,
\begin{align*}f_x(x,y) &= \frac{\partial}{\partial x} \frac{xy(x^2 - y^2)}{x^2 + y^2} \\
&= \frac{(x^2 + y^2)\frac{\partial}{\partial x}[xy(x^2 - y^2)] - xy(x^2 - y^2)\frac{\partial}{\partial x}[x^2 + y^2]}{(x^2 + y^2)^2} \\
&= \frac{(3yx^2 - y^3)(x^2 + y^2) - xy(x^2 - y^2)(2x)}{(x^2 + y^2)^2} \\
&= y\frac{(3x^2 - y^2)(x^2 + y^2) - 2x^2(x^2 - y^2)}{(x^2 + y^2)^2} \\
&= y\frac{x^4 + 4x^2 y^2 - y^4}{(x^2 + y^2)^2}.
\end{align*}
Therefore,
\begin{align*}
f_{xy}(0, 0) &= \lim_{h \rightarrow 0} \frac{f_x(0, 0 + h) - f_x(0, 0)}{h} \\
&= \lim_{h \rightarrow 0} \frac{f_x(0, h)}{h} \\
&= \lim_{h \rightarrow 0} \frac{1}{h} \cdot h \cdot \frac{0^4 + 4 \cdot 0^2 h^2 - h^4}{(0^2 + h^2)^2} \\
&= \lim_{h \rightarrow 0} \frac{-h^4}{(h^2)^2} = -1.
\end{align*}
I'm guessing that $f_{yx}(0, 0) = 1$, but you'll have to check for yourself.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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solving $a_{n+2}-3a_{n+1}+2a_n=2n$ How can i generalize $a_n$ for $a_{n+2}-3a_{n+1}+2a_n=2n$ with $a_0=1,a_1=0$?
I can't think on any way to approach this and I will be happy if I could get help help. thanks
| The characteristic polynomial is
$x^2-3x+2
=(x-1)(x-2)
$
so the homogeneous solutions are
$1$ and $2^n$.
To find a specific solution to
$a_{n+2}-3a_{n+1}+2a_n=2n$,
try
$a_n = un^2+vn+w$.
Then
$\begin{array}\\
2n
&=u(n+2)^2+v(n+2)+w-3(u(n+1)^2+v(n+1)+w)+2)un^2+vn+w)\\
&=u(n^2+4n+4)+vn+2v+w-3(u(n^2+2n+1)+vn+v+w)+2(un^2+vn+w)\\
&=n(4u+v-6u-3v+2v)+4u+2v+w-3u-3v-3w+2w\\
&=-2un+u-v\\
\end{array}
$
Therefore
$u=-1$
and
$v=u=-1$,
so the solution is
$-n^2-n$.
Therefore the general solution is
$-n^2-n+a+b2^n$.
Choose $a$ and $b$
to fit initial conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Unclear math contest solution I'm having difficulty understanding this solution to a problem in an old math contest. I understand everything up to the word "Hence". Can anybody explain this to me?
Question
Answer
| The three roots satisfy the polynomial. So, from
$$ x^5 = \cdots = -3 + 9x - x^2 $$
we get \begin{align*}
a^5 &= -3 + 9a - a^2 \\
b^5 &= -3 + 9b - b^2 \\
c^5 &= -3 + 9c - c^2 \text{.}
\end{align*}
Summing these, we get
$$ a^5 + b^5 + c^5 = 3(-3) + 9(a+b+c) - (a^2 + b^2 + c^2) = \dots \text{.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$ Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
I already tried to use the Sqeeze theorem on it, but I just was not able to find some reasonable upper series for it, only lower:
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
$$\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right) \leq n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
Is there anyone who can give me a hint as to how to solve it?
| Consider
$$
f(x)=\sqrt[3]{1+x^3}-\sqrt{1+x^2}
$$
Then
$$
f'(x)=\frac{3x^2}{3\sqrt[3]{(1+x^3)^2)}}-\frac{2x}{2\sqrt{1+x^2}}
$$
and therefore $f'(0)=0$. Therefore
$$
0=f'(0)=\lim_{x\to0^+}\frac{\sqrt[3]{1+x^3}-\sqrt{1+x^2}}{x}=
\lim_{t\to\infty}\bigl(\sqrt[3]{t^3+1}-\sqrt{t^2+1}\bigr)
$$
with the substitution $t=1/x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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For which $n \in \mathbb{N}$ it is true that $x^2+x+1 | (x+1)^n-x^n-1$ I have to find such $n \in \mathbb{N}$ for which $x^2+x+1 | (x+1)^n-x^n-1$.
$x^2+x+1$ has 2 complex roots: $x_1=-((-1)^{(1/3)})$, $x_2=(-1)^{(2/3)}$ so I tried to divide $(x+1)^n-x^n-1$ by $(x-x_1)$ and then $(x-x_2)$ but it was too difficult. Any other hints?
| Let roots of $x^2+x+1$ be $w,w^2$. Thus for this polynomial to divide $P(x)= (x+1)^n-x^n-1$, $w,w^2$ must be roots of $P(x)$.
We must have $(w+1)^n - w^n -1 = 0$ and $(w^2+1)^n-w^{2n}-1 = 0$.
Thus we have $(-1)^nw^{2n}-w^n-1 = 0$ and $(-1)^nw^n-w^{2n}-1=0$
*
*Case $n$ is even: Then we have $w^{2n}-w^n-1=0$ and $w^n-w^{2n}-1 = 0$
*
*$n$ is multiple of $3$: These give $-1 = 0$, so no solutions in this case.
*$n$ not multiple of $3$: These give $w^n = -w^n$. So no solutions.
*Case $n$ is odd: We have $-w^{2n}-w^n - 1 = 0$ and $-w^n-w^{2n}-1 = 0$.
*
*$n$ is multiple of $3$: From second equation we have $-3 = 0$. So no solutions.
*$n$ not multiple of $3$: Now both the equations are satisfied. Hence this is the only solution.
So we have that $n$ is not a multiple of $2$ or $3$. Hence we can say $n = \boxed {6k\pm 1}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof about Fermat number without induction I want to show that $7 \mid (F_{2k + 1} + 2)$ where $k \in \mathbb{N_0}$ and $F_n := 2^{2^n} + 1$. I was able to proof this using induction, but was wondering if there is a more direct approach?
Here is a simple sketch of proof (only induction step):
$$
F_{2(k+1)+1} +2= 2^{2^{2k +3}} +3 = 2^{2^{2k + 3}} + 7q - 2^{2^{2k + 1}}
$$
for some $q \in \mathbb{Z}$.
$$
= 16^{2^{2k + 1}} + 7q - 2^{2^{2k + 1}} = 7q + 2^{2^{2k + 1}} (8^{2^{2k + 1}} -1) \\
= 7q + 2^{2^{2k + 1}} ((7 + 1)^{2^{2k + 1}} -1) \\
= 7q + 2^{2^{2k + 1}} ((\sum_{i = 0}^{2^{2k + 1}}7^i1^{2^{2k + 1}-i}) -1) \\
= 7q + 2^{2^{2k + 1}} ((\sum_{i = 1}^{2^{2k + 1}}7^i1^{2^{2k + 1}-i}))
$$
This is clearly divisible by $7$.
| $$2^{2^n}\equiv2^{2^n\pmod{\phi(7)}}\pmod7$$
Now we need $2^n\pmod6$ as $\phi(7)=6$
As $(2^n,6)=2$ for $n\ge1$
let us find $2^{n-1}\pmod{6/2}$
As $2\equiv-1\pmod3,2^{n-1}\equiv(-1)^{n-1}\pmod3$
$\implies2^n=2\cdot2^{n-1}\equiv2(-1)^{n-1}\pmod{2\cdot3}$
$\implies2^{2^n}\equiv2^{2(-1)^{n-1}\pmod6}\pmod7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the ODE $y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$ Question:
Solve the ODE given below:
$y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$
My try:
The equation is not separable because a function of $x$ is added to a function of $y$.
($y+y\sqrt{x^2y^4+1}$)
Also, it is not linear with respect to $x$ or $y$, because it has the term$\sqrt{x^2y^4+1}$.
On the other hand, it's not a complete ODE because $\frac{d}{dy}(y(1+\sqrt{x^2 y^4+1})) \neq \frac{d}{dx}(2x)$
I also tried homogenous ODE's. But this ODE is not homogenous. It doesn't seem to be a Clero DE either.
Any idea?
Thanks in advance.
| $$y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$$
$$2xyy'=-y^2(1+\sqrt{x^2 y^4+1})$$
$u(x)=xy^2 \quad\to\quad u'=y^2+2xyy'=y^2-y^2(1+\sqrt{x^2 y^4+1})=-y^2\sqrt{x^2 y^4+1})$
$$xu'=-xy^2\sqrt{x^2 y^4+1})=-u\sqrt{u^2+1}$$
$$\frac{u'}{u\sqrt{u^2+1}}=-\frac{1}{x}$$
$$\int\frac{du}{u\sqrt{u^2+1}}=-\int\frac{dx}{x}$$
$$\ln|u|-\ln|1+\sqrt{u^2+1}|=-\ln|x|+\text{constant}$$
$$\frac{u}{1+\sqrt{u^2+1}}=\frac{c}{x}$$
$$u=\frac{2cx}{x^2-c^2}$$
$$y^2=\frac{u}{x}=\frac{2c}{x^2-c^2}\quad\to\quad y=\pm\sqrt{\frac{2c}{x^2-c^2}}$$
Bringing back into the ODE shows that it agrees.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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$\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx$ How would one go about integrating $\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx$
I tried rationalizing and then doing partial fractions but got something really ugly and IBP doesn't work too well either.
| Let $x=\dfrac1u$ then
\begin{align}
I
&= \int\dfrac{-u}{\sqrt{u^2+u+1}}du \\
&= -\int\dfrac{2u+1}{2\sqrt{u^2+u+1}}du+\dfrac12\int\dfrac{1}{\sqrt{(u+\frac12)^2+\frac34}}du \\
&= -\sqrt{u^2+u+1}+\dfrac12\operatorname{arcsinh}\dfrac{2u+1}{\sqrt{3}}+C \\
&= -\dfrac{\sqrt{x^2+x+1}}{x}+\dfrac12\operatorname{arcsinh}\dfrac{x+2}{x\sqrt{3}}+C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometry Problems Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac{1}{2}$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
My student left me with a list of questions just now (he also said that he wanted to be helped at school :P)
| The first two equations, squared, allow you to solve for $\sin^2$ and $\sin^2y$ as follows:
$$\begin{cases}\sin^2x&=9\sin^2y,\\4(1-\sin^2x)&=1-\sin^2y.\end{cases}$$
Then $\sin^2x=\dfrac{27}{35},\sin^2y=\dfrac3{35}$.
From this, by the double angle formula
$$\begin{cases}\cos 2x=-\dfrac{19}{35},\\\cos 2y=\dfrac{29}{35}.\end{cases}$$
The rational ratio is $$\dfrac pq=\frac32-\frac{19}{29}=\frac{49}{58}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
| With substitution $x=u+3$ we have
\begin{align}
\lim_{x\to 3} \dfrac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
&= \lim_{u\to 0}\dfrac{1-\cos\pi u}{\sqrt[3]{u+8}-2} \\
&= \lim_{u\to 0}\dfrac{2\sin^2\frac12\pi u}{u^2}\dfrac{u^2}{\sqrt[3]{u+8}-2}\dfrac{\sqrt[3]{u+8}^2+\sqrt[3]{u+8}+4}{\sqrt[3]{u+8}^2+\sqrt[3]{u+8}+4} \\
&= \dfrac{\pi^2}{2}\lim_{u\to 0}\dfrac{u^2}{u}\lim_{u\to 0}\dfrac{\sqrt[3]{u+8}^2+\sqrt[3]{u+8}+4}{1} \\
&= \dfrac{\pi^2}{2}\times 0 \times 12 \\
&= 0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of the serie $\sum \dfrac{(\ln(n))^n}{n!}$ How to prove the convergence of the series: $$\sum_n \dfrac{(\ln(n))^n}{n!}$$
I tried to use the d'Alembert's rule:
If we write $u_n = \dfrac{(\ln(n))^n}{n!}$ then we get
$$\dfrac{u_{n+1}}{u_n} = \dfrac{1}{n+1} \times \dfrac{(\ln(n+1))^{n+1}}{(\ln(n))^n}$$
Is that possible to show that $\dfrac{u_{n+1}}{u_n} \rightarrow l <1$ ?
| Write
$$\frac{u_{n+1}}{u_n} = \frac{\ln(n+1)}{n+1}\times \left(\frac{\ln(n+1)}{\ln n}\right)^n$$
Since $\ln(n+1) = 2\ln \sqrt{n+1} < 2\sqrt{n+1}$, it follows from the squeeze theorem that
$$\lim_{n\to \infty} \frac{\ln(n+1)}{n+1} = 0$$
Now
$$\frac{\ln(n+1)}{\ln n} = 1 + \frac{\ln(n+1) - \ln n}{\ln n} = 1 + \frac{\ln(1 + \frac{1}{n})}{\ln n}$$
so by the binomial theorem,
$$\left(\frac{\ln(n+1)}{\ln n}\right)^n > 1 + n\frac{\ln(1 + \frac{1}{n})}{\ln n} = 1 + \frac{\ln(1 + \frac{1}{n})}{\frac{1}{n}}\frac{1}{\ln n}$$
The limit of $n\dfrac{\ln(1 + \frac{1}{n})}{\ln n}$ is $0$ since $\lim\limits_{n\to \infty} \dfrac{1}{\ln n} = 0$ and $\dfrac{\ln(1 + \frac{1}{n})}{\frac{1}{n}}$ is bounded by $1$. As
$$1 + n\frac{\ln(1 + \frac{1}{n})}{\ln n} < \left(\frac{\ln(n+1)}{\ln n}\right)^n < 1$$
the squeeze theorem yields $\lim\limits_{n\to \infty} \left(\dfrac{\ln(n+1)}{\ln n}\right)^n = 1$. Consequently, $\lim\limits_{n\to \infty} \dfrac{u_{n+1}}{u_n} = 0$.
| {
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Test the convergence of $\sum _{k=1}^\infty (-1)^{k-1} \frac{2^k}{2^k+k^2}$ Test the convergence of $\sum _{k=1}^\infty (-1)^{k-1} \frac{2^k}{2^k+k^2}$
my idea
$a_k=\sum _{k=1}^\infty (-1)^{k-1} \frac{2^k}{2^k+k^2}$
$a_{k+1}=\sum _{k=1}^\infty (-1)^{k} \frac{2^{k+1}}{2^{k+1}+(k+1)^2}$
$|\frac{a_{k}}{a_{k+1}}|_{k\to \infty}=\left| \frac{1}{2}\frac{2^{k+1}+(k+1)^2}{2^k+k^2}\right|_{k\to\infty}$
can any help from here please thank you
| Well,
\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1} 2^k}{2^k+k^2} &= \sum_{k=1}^{\infty} (-1)^{k-1} \frac{2^k+k^2-k^2}{2^k+k^2} \\
&= \sum_{k=1}^{\infty} (-1)^{k-1} \left [ 1 - \frac{k^2}{2^k+k^2} \right ]\\
&= \sum_{k=1}^{\infty} (-1)^{k-1} - \sum_{k=1}^{\infty} \frac{(-1)^{k-1} k^2}{2^k+k^2}
\end{align*}
The first series obviously diverges whereas the second converges. To see that split it apart. For example
$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1} k^2}{2^k+k^2} = \sum_{k=1}^{\left \lfloor \frac{2}{\log 2} \right \rfloor} \frac{(-1)^{k-1} k^2}{k^2+2^k} + \sum_{\left \lfloor \frac{2}{\log 2} \right \rfloor+1}^{\infty} \frac{(-1)^{k-1} k^2}{2^k+k^2} $$
The first sum is finite , hence convergent , and the second converges since $\displaystyle \lim_{k\rightarrow +\infty} \frac{k^2}{2^k+k^2} = 0 $ and of course the sequence $\displaystyle a_k = \frac{k^2}{2^k+k^2}$ is decreasing. Here is the invoked test.
Thus your sum diverges.
| {
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Sum and product of n natural numbers is equal Sum and product of $n$ natural numbers is equal. Prove that this sum $\ge$ $n+s$ where $s$ is the smallest natural number satisfying $2^s$ $\ge$ $s+n$.
| I think the problem needs a little bit of a polishing, more specific, we can have $s \geq 0$, $n \geq 1$ and all $n$ number(s) must be strictly $>0$. Now ...
For $n=1$, we can consider $s=0$ because $2^s=2^0 \geq 1 + 0=n+s$ and every $a_1 >0 \Rightarrow a_1 \geq 1=n+s$
For $n>1$ and
$$K=a_1+a_2+...+a_n=a_1\cdot a_2 \cdot ... \cdot a_n \tag{1}$$
let's note by
*
*$S_{\color{red}{1}}=\left\{ a_k \mid a_k = \color{red}{1}, k=1..n\right\}$
*$S_{\color{red}{2}}=\left\{ a_k \mid a_k \geq \color{red}{2}, k=1..n\right\}$ and
$S_2$ is not empty, otherwise $a_1=a_2=...=a_n=1$ and then $a_1+a_2+...+a_n=n > 1=a_1\cdot a_2 \cdot ... \cdot a_n$ contradicting the equality assumption. $S_1$ may be empty, like in the following case $\color{red}{2+2=2\cdot 2}$. Both $S_1$ and $S_2$ are disjoint, i.e. $S_1 \bigcap S_2 = \varnothing $.
Then, noting $\color{red}{\left|S_2\right|=k}$, from $(1)$ and this inequality (applied to $\log_2$ translates as $x \leq 2^{x-1}, \forall x\geq 2$)
$$K=\sum\limits_{k=1}^{n}a_k = \sum\limits_{a_k \in S_1}a_k + \sum\limits_{a_k \in S_2}a_k=(n-k)+\sum\limits_{a_k \in S_2}a_k =\\ n + \sum\limits_{a_k \in S_2}(a_k - 1) \geq n + \sum\limits_{a_k \in S_2}\log_2{a_k} \tag{2}$$
and (from $(2)$!)
$$K=\prod\limits_{k=1}^{n}a_k=\prod\limits_{a_k \in S_1}a_k \cdot \prod\limits_{a_k \in S_2}a_k=\prod\limits_{a_k \in S_2}a_k=2^{\left(\sum\limits_{a_k \in S_2} \log_2{a_k}\right)}\geq n + \sum\limits_{a_k \in S_2}\log_2{a_k} \tag{3}$$
One value for $s$ is
$$s=\left \lfloor \sum\limits_{a_k \in S_2}\log_2{a_k} \right \rfloor +1 $$
And $s=k$ is not sufficient, since it doesn't work for $\color{red}{1+2+3=1\cdot 2 \cdot 3}$
A few examples
$$2+2=2\cdot 2$$
$$1+2+3=1\cdot 2 \cdot 3$$
$$1+1+2+2+2=1^2\cdot 2^3$$
$$1+1+1+1+1+1+1+1+2+2+2+2=1^8 \cdot 2^4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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easier ways of solving $\int \frac{4xdx}{(2x+1)^2}$? original integral is
$$\int \frac{4xdx}{(2x+1)^2}$$
I tried partial fractions which worked but it seems to be too long, are there easier ways?
| Here is a method which manages to avoid partial fractions by making use of two substitutions.
We begin by letting $x = \frac{1}{2} \tan^2 u, dx = \tan u \cdot \sec^2 u \, du,$ giving
\begin{align*}
\int\frac{4x}{(2x + 1)^2} \, dx &= 2 \int \frac{\tan^3 u \sec^2 u}{(1 + \tan^2 u)^2} \, du\\
&= 2 \int \frac{\tan^3 u}{\sec^2 u} \, du\\
&= 2 \int \frac{\sin^3 u}{\cos u} \, du\\
&= 2 \int \frac{(1 - \cos^2 u)}{\cos u} \sin u \, du.
\end{align*}
Now let $t = \cos u, dt = -\sin u \, du$. Thus
$$\int \frac{4x}{(2x + 1)^2} \, dx = 2 \int \frac{t^2 - 1}{t} \, dt = t^2 - 2 \ln |t| + C,$$
or
$$\int \frac{4x}{(2x + 1)^2} \, dx = \cos^2 u - 2 \ln |\cos u| + C,$$
since $t = \cos u$. Also, as $\tan^2 u = 2x$ the cosine term in terms of $x$ can be written as $\cos u = 1/\sqrt{2x + 1}$. So finally we have
$$\int \frac{4x}{(2x + 1)^2} = \ln |2x + 1| + \frac{1}{2x + 1} + C,$$
in agreement with the answer given by samjoe.
| {
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Find an f(x) that fits specific critera (Polynomial Regression) I need a function (I don't care how messy it is) that when:
$f(1) = 1$ <--this has $1!$ different terms
$f(2) = a-b$ <--this has $2!$ different terms
$f(3) = a^2b-a^2c-b^2a+b^2c+c^2a-c^2b$ <--this has $3!$ different terms
$f(4) = a^3b^2c-a^3b^2d-a^3c^2b+a^3c^2d+a^3d^2b-a^3d^2c-b^3a^2c+b^3a^2d+b^3c^2a-b^3c^2d-b^3d^2a+b^3d^2c+c^3a^2b-c^3a^2d-c^3b^2a+c^3b^2d+c^3d^2a-c^3d^2b-d^3a^2b+d^3a^2c+d^3b^2a-d^3b^2c-d^3c^2a+d^3c^2b$ <--this has $4!$ different terms
$f(5) = a^4b^3c^2d-a^4b^3c^2e-a^4b^3d^2c+a^4b^3d^2e+a^4b^3e^2c-a^4b^3e^2d-a^4c^3b^2d+a^4c^3b^2e+a^4c^3d^2b-a^4c^3d^2e-a^4c^3e^2b+a^4c^3e^2d+a^4d^3b^2c-a^4d^3b^2e-a^4d^3c^2b+a^4d^3c^2e+a^4d^3e^2b-a^4d^3e^2c-a^4e^3b^2c+a^4e^3b^2d+a^4e^3c^2b-a^4e^3c^2d-a^4e^3d^2b+a^4e^3d^2c+b^4a^3c^2d-$ <--this would have $5!$ different terms (If I typed all 120 of them)
This function lists all of the unique combinations of the letters and the powers used. Hopefully you can see the pattern between the what is stated above, I just can't find a way to describe it.
This function would be a necessary part for working out a formula for 2-dimensional polynomial regression.
I do realize that there are multiple answers for this and any help for this would be much appreciated.
| This equation works for all the values:
$$-\prod_{i=1}^{N-1}(-\prod_{j=1}^i(x_{i+1}-x_j))$$
Where $x =1,2,3,...,N-1,N$
and $N =$ length of sequence $x$
| {
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How to find the maximum value of $x^2+y^2+z^2$ after applying the technique of Lagrange multipliers (complicated set of equations)? I need to show that the maximum or the minimum of $f(x,y,z)=x^2+y^2+z^2=r^2$ given that $$(x^2+y^2+z^2)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\tag{1}$$ and $$\lambda x+ \mu y+ \gamma z=0\tag{2}$$ is given by the equation $$\frac{a^2\lambda^2}{1-a^2r^2}+\frac{b^2\mu^2}{1-b^2r^2}+\frac{c^2\gamma^2}{1-c^2r^2}=0$$
My Approach:
Say $$g(x,y,z)=(x^2+y^2+z^2)^2-(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2})$$
and
$$h(x,y,z)=\lambda x+ \mu y+ \gamma z$$
Using the method of Lagrange multipliers I get
$f_x=Ag_x+Bh_x$
$f_y=Ag_y+Bh_y$
$f_z=Ag_z+Bh_z$
Where $A$ and $B$ are constants.
Hence we get:
$$
\begin{cases}
2x=A(2(x^2+y^2+z^2)(2x)-2x/a^2)+B\lambda,\\
2y=A(2(x^2+y^2+z^2)(2y)-2y/b^2)+B\mu,\\
2z=A(2(x^2+y^2+z^2)(2z)-2z/c^2)+B\gamma.
\end{cases}\tag{*}
$$
Somehow I need to eliminate $A$ and $B$, but the direct method is too lengthy.
I'm not sure how to elegantly reach $$\frac{a^2\lambda^2}{1-a^2r^2}+\frac{b^2\mu^2}{1-b^2r^2}+\frac{c^2\gamma^2}{1-c^2r^2}=0$$
from here. Any suggestions?
| Hint: multiply the first equation in (*) by $x$, the second one by $y$ and the third one by $z$, then sum them up. You will get rid of $B$ and can find $A$ in terms of $r$. After simplification you will be able to express $x,y,z$ easily and substitute into (2). The parameter $B$ cancels out at the end.
| {
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Binomial Theorem 3Unit Australia Year 12 If X is the sum of the first, third, fifth, ... terms in the expansion of $$ (a+b)^n $$ and Y is the sum of the second, fourth, sixth, ... terms, show that $$ X^2 - Y^2 = (a^2-b^2)^n $$
So if I expand it, it is basically the sum of the odd terms would include:
$$ X=\binom{n}{0} +\binom{n}{2}x^2 + \binom{n}{4}x^4 + \binom{n}{6}x^6+ ... $$
and the sum of the even terms would include
$$ Y= \binom{n}{1}x +\binom{n}{3}x^3 + \binom{n}{5}x^5 + \binom{n}{7}x^7+ ... $$
What would you do like factorise (X-Y)(X+Y) or something but theres like infinite numbers n
| The terms of the expansion of $(a-b)^n$ are the same as those of $(a+b)^n$ except that the terms for odd powers of $b$ are multiplied by $-1$. Thus
$$ X + Y = (a+b)^n$$
while
$$ X - Y = (a-b)^n$$
So $$ X^2 - Y^2 = (X + Y)(X - Y) = (a+b)^n (a-b)^n = (a^2-b^2)^n$$
| {
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Indices: $\left(2^\left(n+3\right) - 2^n\right)/14$ The questions im working on is..
$ \left(2^\left(n+3\right) - 2^n\right)/14
$
My thought process behind solving it was to spilt the first 2 term into two separate terms then cancelling the 2^n leaving 2^3 over 14
$ (2^n \cdot 2^3-2^n)/14
$
$ (2^3)/14 = 8/14
$
however this is incorrect, how am i supposed to correctly simplify this equation?
Sorry about messing up the formatting so much
| Steps:
$2^{n+3}= 2^n2^3= 2^n×8.$
$(2^{n+3} -2^n) = 2^n(8-1)=7×2^n.$
$14=2×7$.
$\dfrac{7×2^n}{14}= 2^n/2^1= 2^{n-1}.$
| {
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Solve the equation $x^2+x+2=2^y$ in positive integers. Let $x,y$ are positive integers. Solve the equation
$$x^2+x+2=2^y$$
My work so far:
$$(2x+1)^2 + 7 = 2^{y+2}$$
If $y$ is even, then $x=1,y=2$.
| Solutions are (brute force) $$(90,13);\;(5,5);\;(2,3);\;(1,2)$$
I managed to find only the even solution $y=2$ analytically
I did in this way
Solving $x^2 + x + 2- 2^y=0$
discriminant is $1-4(2-2^y)=4\cdot 2^y-7$
It is a perfect square if
$4\cdot 2^y -7=a^2$
that is $4\cdot 2^y-a^2=7$
factoring RHS
$\left(2\cdot 2^{y/2}+a\right)\left(2\cdot 2^{y/2}-a\right)=7$
Which means that $2\cdot 2^{y/2}+a=7$ and $2\cdot 2^{y/2}-a=1$
that is $a = 2\cdot 2^{y/2}-1$ and
$2\cdot 2^{y/2}+ 2\cdot 2^{y/2}-1=7$
$ 4\cdot 2^{y/2}=8\to 2^{y/2}=2\to \color{red}{y=2;\;x=1}$
But I don't know how to find the odd solutions for $y$
Hope this helps
| {
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Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? More generally, why does a negative exponent turn positive when moved to the denominator? Would appreciate kindergarten language ;-D
| First you need to understand why
$$
7^{-1} = \frac{1}{7}
$$
In the row below, to move one to the right, multiply by $7$:
$$
7^1=7,\qquad 7^2=49,\qquad 7^3=343,\qquad\dots
$$
And consequently, to move to the left, divide by $7$. So that is how to extend it the other way: keep dividing by $7$:
$$
\dots \qquad7^{-2} = \frac{1}{49},\qquad 7^{-1} = \frac{1}{7},\qquad7^0=1,\qquad7^1=7,\qquad 7^2=49,\qquad\dots
$$
| {
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Prove this inequality $\frac{n}{3^n-2^n}<\left(\frac{2}{5}\right)^{n-1}$ Let $n\ge 3$. How to prove show that
$$\dfrac{n}{3^n-2^n}<\left(\dfrac{2}{5}\right)^{n-1}\tag1$$
or
$$3^n\cdot 2^{n-1}-2^{2n-1}-5^{n-1}\cdot n>0,n\ge 3\tag2$$
It seem right:see wolfram
when I do this problem found it:How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$
if this inequality (1) has prove it,then
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<1+\dfrac{2}{5}+\sum_{k=3}^{+\infty}\dfrac{2^{k-1}}{5^{k-1}}=\dfrac{5}{3}$$
Question: How to prove inequality (1) or (2)
| We see that $(1)$ is equivalent to
$$6^{n}\gt 4^{n}+2n\cdot 5^{n-1}\tag3$$
Let us prove by induction that $(3)$ holds for $n\ge 3$.
We see that $(3)$ holds for $n=3,4,5$ :
*
*$6^3=216\gt 214=4^3+2\cdot 3\cdot 5^{3-1}$
*$6^4=1296\gt 1256=4^4+2\cdot 4\cdot 5^{4-1}$
*$6^5=7776\gt 7274=4^5+2\cdot 5\cdot 5^{5-1}$
Supposing that $(3)$ holds for some $n\ (\ge 5)$ gives
$$\begin{align}6^{n+1}&=6\cdot 6^n\\\\&\gt 6(4^{n}+2n\cdot 5^{n-1})\\\\&=6\cdot 4^n+12n\cdot 5^{n-1}\\\\&\ge 4\cdot 4^n+(10n+10)\cdot 5^{n-1}\\\\&=4^{n+1}+2(n+1)\cdot 5^n\qquad\blacksquare\end{align}$$
| {
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Find sum of all positive real numbers $x$ such that $\sqrt[3]{2+x} + \sqrt[3]{2-x}$ becomes an integer.
Find sum of all positive real numbers $x$ such that $\sqrt[3]{2+x} + \sqrt[3]{2-x}$ becomes an integer.
I thought Euler's identity may help,so letting $a=\sqrt[3]{2+x}, b=\sqrt[3]{2-x},c=-y$ ($y \in \Bbb Z)$ we have : $a+b+c=0$ leading to :$(2+x)+(2-x)+(-y^3)=3\sqrt[3]{2+x} \sqrt[3]{2-x}(-y)$, but at this point I think sum of such $x's$ can't be found.
| Using Power Mean inequality,it's seen that: $\sqrt[3]{2+x}+\sqrt[3]{2-x}\leq 2\sqrt[3]2$.So the positive integer value of $\sqrt[3]{2+x}+\sqrt[3]{2-x}$ can be $1$ or $2$.
After many hours of thinking,I found out that Euler's identity really helps simplifying the equation:
$$y^3-4=3y\sqrt[3]{4-x^2}\Rightarrow$$
$$x^2=4-\frac{(y^3-4)^3}{27y^3}$$
$$y=1\Rightarrow x=\sqrt5$$
$$y=2\Rightarrow x=\frac{10}{3\sqrt3}$$
So,sum of such $x$'s:
$$\sqrt5+\frac{10}{3\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a formula for $\sum\limits_{n=0}^N(-1)^n\frac{(2n+1)^3}{(2n+1)^4+4}$ I found this sum in the mathematial induction chapter of The art of Computer Programming and i have no idea how to solve it.
$\dfrac{1^3}{1^4+4}-\dfrac{3^3}{3^4+4} + ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)^4+4} $
I tried writing it as $\dfrac{1^3}{1*1^3+4}-\dfrac{3^3}{3*3^3+4} + ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)*(2n+1)^3+4} $
and then writing it as
$\dfrac{1}{1*1^3+4}-\dfrac{3+5}{3*3^3+4} +...+\dfrac{(-1)^n(((2n+1)^2-(2n+1)+1)+...+((2n+1)^2+(2n+1)-1))}{(2n+1)*(2n+1)^3+4} $
but did not know how to continue.
I also tried writing it as
$\dfrac{1}{1+\dfrac{4}{1^3}}-\dfrac{1}{3+\dfrac{4}{3^3}} + ... +\dfrac{(-1)^n}{2n+1+\dfrac{4}{(2n+1)^3}} $ but without succes.
| The exercise asks you to find and prove a closed form. You can take a guided approach (e.g. Gosper's algorithm), but the intention is probably that you work out a few cases by hand, guess the desired form, and then prove it.
$$\begin{eqnarray}
n=1 & \implies & \frac{1}{5} \\
n=2 & \implies & \frac{1}{5} - \frac{27}{85} = \frac{-50}{425} = \frac{-2}{17} \\
n=3 & \implies & \frac{-50}{425} + \frac{125}{629} = \frac{21675}{267325} = \frac{3}{37} \\
n=4 & \implies & \frac{21675}{267325} - \frac{343}{2405} = \frac{-39564100}{642916625} = \frac{-4}{65} \\
\end{eqnarray}$$
So it's worth hazarding a guess that the desired form is $\frac{(-1)^{n+1}n}{p(n)}$. At this point you could break out the algebra to see what $p(n)$ would work, or you could guess (or know, if you understand the theory behind Gosper's algorithm) that $p(n)$, if it has a closed form, must be a polynomial and see whether the cubic fit through $(1,5), (2,17), (3,37), (4,65)$ works. If not, evaluate another term and fit a quartic...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a + b = 20$, then what is the maximum value of $ab^2$? Knowing that
$$a + b = 20$$
and $a,b \in \mathbb{Z}_{+}$
What is the maximum value that $ab^2$ can take?
If $b$ is even then so is $a$. Hence, let $b = 2k$ and $a = 20 - 2k$, where $k \in \{1, 2, 3, \cdots, 9\}$
$$ab^2 = (20 - 2k)(2k)^2 = 80k^2 - 8k^3$$
The value of $k$ which maximizes $ab^2$ is therefore $k = 7$ which gives $ab^2 = 1176$.
However this only shows that this is the maximum for even values of $a$ and $b$. It turns out that the maximum of $ab^2$ is when both are odd.
How do I solve this question since my approach is clearly not very elegant?
My solution:
$$a = 20 - b \Longrightarrow ab^2 = (20 - b)b^2$$
Differentiate to find the maximum value of $b$
$$40b - 3b^2 = b(40-3b) = 0$$
$$\Longrightarrow b = \left\lfloor \frac{40}{3} \right\rfloor$$ $$\Longrightarrow b = \left\lceil \frac{40}{3} \right\rceil$$
Therefore $a, b = 7, 13$ or $a, b = 6, 14$. Inspecting these $2$ cases yields the former.
So the maximum value $ab^2$ can take is $1183$.
| By AM-GM
\begin{eqnarray*}
\frac{1}{3} \left( a + \frac{b}{2} +\frac{b}{2} \right) \geq \sqrt[3]{\frac{ab^2}{4}}.
\end{eqnarray*}
So
\begin{eqnarray*}
4 \left( \frac{20}{3} \right)^3 \geq ab^2.
\end{eqnarray*}
Now the restriction that $a$ and $b$ are integers suggest $(a,b)=(6,14)$ or $(7,13)$ computing these gives $(a,b)=(7,13)$ and a largest value of $1183$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506589",
"timestamp": "2023-03-29T00:00:00",
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Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)$.
Thus, $$z^{1/3} = (1351)^{2/3} \left(\cos \left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3}+ \frac{2k\pi}{3} \right] + i\sin\left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3} + \frac{2k\pi} {3}\right]\right)$$ for $k = 0,1,2.$
This does not seem like I am heading in the right direction. Any clues or hints would be greatly appreciated. Thank you!
| Let $\,a = \sqrt[3]{26 + 15 \sqrt{3}}\,$, $\,b=\sqrt[3]{26 - 15 \sqrt{3}}\,$, then $\require{cancel}\,a^3+b^3=26 + \cancel{15 \sqrt{3}} + 26 - \cancel{15 \sqrt{3}} = 52$ and $\,ab=\sqrt[3]{26^2 - 15^2 \cdot 3} = 1\,$.
It follows that $\,(a+b)^3 = a^3 + b^3 + 3ab(a+b) = 52 + 3(a+b)\,$, so $\,t=a+b\,$ satisfies the equation $\,t^3 - 3t -52 =0 \iff (t-4)(t^2+4t+13) = 0\,$ with the only real root $\,t = 4\,$.
Then $\,a+b=4\,$, $\,ab=1\,$ so $\,a,b\,$ are the roots of $\,x^2 - 4x + 1=0 \iff x = 2 \pm \sqrt{3}\,$ with $\,a\,$ being the larger one i.e. $\,a = 2 + \sqrt{3}\,$, $\,b = 2 - \sqrt{3}\,$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use a triple integral to compute the volume of a solid bounded by yhe surface $(x^2+y^2+z^2)^3=3xyz$ Use a triple integral to compute the volume of a solid bounded by yhe surface $$(x^2+y^2+z^2)^3=3xyz$$
I use spherical coordinates
\begin{cases}
x=r \cos \varphi \cos \theta \\
y=r \sin \varphi \cos \theta \\
z=r\sin \theta
\end{cases}
but what I get is $(x^2+y^2+z^2)^3=3xyz \Rightarrow$
$r^6=3r^3\cos \varphi \cos \theta \sin \varphi \cos \theta\sin \theta \Rightarrow$
$r^3=3 \cos \varphi \sin \varphi \cos^2 \theta \sin \theta$
and it doesn't look helpful to me. Can anyone explain?
| You can graph the parametric surface on a function grapher. I did the correction because there are some mistakes in the answers.
If you had graphed the surface, you will see that the volumen is divided in four
polygons in diferents quadrants. I will find the volumen when $x,y,z\geq0$.
$$(x^2+y^2+z^2)^3=3xyz$$
Use de Jacobian for change to spherical coordinates
\begin{cases}
x=r \cos \varphi \cos \theta \\
y=r \sin \varphi \sin \theta \\
z=r\cos \varphi
\end{cases}
Then, $(x^2+y^2+z^2)^3=3xyz \Rightarrow$
$r^3=3 \sin^2 \varphi \cos \theta \sin \theta \cos\varphi$
We will obtain the following when calculating the Jacobian.
$$r^2\sin\varphi$$
We define the volume to be integrated.
$$D:={(x,y,z)/0\leq\theta\leq\pi/2;0\leq\varphi\leq\pi/2;0\leq r\leq\sqrt[3]{3\sin^2 \varphi \cos \theta \sin \theta \cos\varphi}}$$
Because I will take one of four polygons in the firts cuadrant.
$$ 4\int_{0}^{\pi/2}d\theta\int_{0}^{\pi/2}d\varphi\int_{0}^{\sqrt[3]{3\sin^2 \varphi \cos \theta \sin \theta \cos\varphi}}r^2\sin\varphi dr =4\int_{0}^{\pi/2}d\theta\int_{0}^{\pi/2}\sin^3 \varphi \cos \theta \sin \theta \cos\varphi=\int_{0}^{\pi/2}cos \theta \sin \theta=\color{red}{\frac{1}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507417",
"timestamp": "2023-03-29T00:00:00",
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How to solve this ( and general )logarithmic Equation $$ 4^{x+1} - 6^x - 2*9^{x+1} = 0$$
I recently stumbled across this logarithmic Equation and really I have no clue how to solve this. Also , please provide a tactical way to approach such questions
| $$4^{x+1} - 6^x - 2 \cdot 9^{x+1} = 0\\
2^{2x+2} - 2^x \cdot 3^x - 2 \cdot 3^{2x+2} = 0$$
Now, divide everything by $2^{2x}$:
$$4^{x+1} - 6^x - 2 \cdot 9^{x+1} = 0\\
2^{2} - \frac{3^x}{2^x} - \frac{2 \cdot 3^{2x+2}}{2^{2x}} = 0 \\
4 - \left(\frac{3}{2}\right)^x - 2\cdot9 \left(\frac{3}{2}\right)^{2x} = 0$$
Let's pose $y = \left(\frac{3}{2}\right)^x.$ Then, your equation can be rewritten as follows:
$$4 - y - 18y^2 = 0 \Rightarrow y = -\frac{1}{2} ~\text{or}~ y = \frac{4}{9}.$$
Of course, we need to discard the negative solution, since $y = \left(\frac{3}{2}\right)^x$ must be positive by definition.
Finally:
$$\left(\frac{3}{2}\right)^x = \frac{4}{9} \Rightarrow
x\log\left(\frac{3}{2}\right) = \log\left(\frac{4}{9}\right) \Rightarrow \\
x = \frac{\log\left(\frac{4}{9}\right)}{\log\left(\frac{3}{2}\right)} \Rightarrow
x = -\frac{\log\left(\frac{9}{4}\right)}{\log\left(\frac{3}{2}\right)} \Rightarrow\\
x = -2\frac{\log\left(\frac{3}{2}\right)}{\log\left(\frac{3}{2}\right)} \Rightarrow
x = -2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Equation of tangents from external point to a circle I have point $(p,q)$ and circle $ x^2 + y^2 + 2gx + 2fy + c = 0$.
I'm aware you could do substitute in $y = mx+c$ and solve a quadratic or you could use $ y-(-f) = m(x-(-g)) \sqrt{1+m^2}$ where m is the slope of the line and $(-g, - f)$ is the coordinates of the centre. But are there other ways? perhaps using calculus/vectors/complex numbers
maybe take point $(4,-5)$ and circle $ x^2 + y^2 -6x + -4y + 4 = 0$. as example
| By Joachimsthal $$s_1^2=s \cdot s_{11}$$ defines the tangents from $(p,q)$ as a line pair with $$s=x^2 + y^2 + 2gx + 2fy + c = 0,$$ $$s_{11}=p^2 + q^2 + 2gp + 2fq + c,$$ $$s_1=xp + yq + g(x+p) + f(y+q) + c.$$
I.e.
$(xp + yq + g(x+p) + f(y+q) + c)^2-(x^2 + y^2 + 2gx + 2fy + c)(p^2 + q^2 + 2gp + 2fq + c)=0$
or
${\small (-g^2+2fq+q^2+c)(x-p)^2+(-2fg-2fp-2gq-2pq)(x-p)(y-q)+(-f^2+2gp+p^2+c)(y-q)^2=0}$
which factorises as:
$${\Tiny \frac{((-g^2+2fq+q^2+c)(x-p) -(y-q) (pq+gq+fp+fg-\sqrt{D}))
((-g^2+2fq+q^2+c)(x-p)-(y-q)(pq+gq+fp+fg+\sqrt{D}))}{-g^2+2fq+q^2+c}},$$ where $$D=(g^2+f^2-c)(p^2+q^2+2gp+2fq+c)=(g^2+f^2-c)s_{11},$$
or
$${\Tiny \frac{((p^2+2gp-f^2+c)(y-q) -(x-p) (pq+gq+fp+fg-\sqrt{D}))
((p^2+2gp-f^2+c)(y-q)-(x-p)(pq+gq+fp+fg+\sqrt{D}))}{p^2+2gp-f^2+c}},$$
making the actual lines
$$(y-q)=(x-p)\frac{pq+gq+fp+fg\pm \sqrt{D}}{p^2+2gp-f^2+c}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find k so that this rank is true? $$
\begin{bmatrix}
3 & 2 & -2 \\
1 & 1 & -1 \\
-1 & 2 & k \\
\end{bmatrix}
$$
Find the values k for which rank(A) = 3
From what I have tried myself, I have brought it down to row-echelon form where it looks something like this:
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & k+2 \\
\end{bmatrix}
$$
Through various row operations, but I seem to be stuck on whether or not I'm doing the right thing. Thanks!
To start:
$$
\begin{bmatrix}
3 & 2 & -2 \\
1 & 1 & -1 \\
-1 & 2 & k \\
\end{bmatrix}
$$
Interchanging Rows 1 and 2.
$$
\begin{bmatrix}
1 & 1 & -1 \\
3 & 2 & -2 \\
-1 & 2 & k \\
\end{bmatrix}
$$
then, Row 2 minus 3 times Row 1 and Row 3 plus Row 1.
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & -1 & 1 \\
0 & 3 & k-1 \\
\end{bmatrix}
$$
then, multiply Row 2 by (-1)
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 3 & k-1 \\
\end{bmatrix}
$$
Finally, Row 3 minus 3 times Row 2.
$$
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & k+2 \\
\end{bmatrix}
$$
| Hint:-
Rank$=3$ iff $\det(A)\ne 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by definition $\epsilon-\delta$ the limit $\lim_{(x,y)\rightarrow(0,0)}\frac{1-\cos(xy)}{(xy)^2}=\frac{1}{2}$
Prove by definition $\epsilon-\delta$
$$\lim_{(x,y)\rightarrow(0,0)}\frac{1-\cos(xy)}{(xy)^2}=\frac{1}{2}$$
My work:
Let $\epsilon >0$, and $\delta =...$
If $\sqrt{x^2+y^2}<\delta$ then $|\frac{1-cos(xy)}{(xy)^2}-\frac{1}{2}|=|\frac{2-2cos(xy)-(xy)^2}{2(xy)^2}|=\frac{|2-2cos(xy)-(xy)^2|}{2(xy)^2}\leq\frac{|2-2-(xy)^2|}{2(xy)^2}=\frac{|-(xy)^2|}{2(xy)^2}=\frac{(xy)^2}{2(xy)^2}=\frac{1}{2}$
In this step i'm stuck. Can someone help me?
| With $|\sin t|\leq|t|$ or $-t^2\leq-\sin^2t$ we write
$$2-2\cos t-t^2\leq2-2\cos t-\sin^2t=4\sin^2\dfrac12t-4\sin^2\dfrac12t\cos^2\dfrac12t=4\sin^4\dfrac12t$$
then
$$\Big|\dfrac{1-\cos t}{t^2}-\dfrac12\Big|=\Big|\dfrac{2-2\cos t-t^2}{2t^2}\Big|\leq\Big|\dfrac{4\sin^4\dfrac12t}{2t^2}\Big|\leq\Big|\dfrac{\dfrac14t^4}{2t^2}\Big|=\dfrac18t^2$$
also $|xy|\leq\dfrac12(x^2+y^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show convergence of recursive sequence and find limit value Let $(a_n)_{n \in \mathbb N}$ be a recursive sequence. It is defined as $a_1=1, \quad a_{n + 1} = \frac{4a_n}{3a_n+3}$.
I have to show that the sequences converges and find a limit value.
To show convergence I was about to use the Cauchy criterum. Unfortunately I am quite confused here because of the recursive definition.
Question: How can I show show that the sequences converges and how can I find a limit value?
| If we write $a_n$ as $p_n/q_n$, we get:
$$a_{n + 1} = \frac{4(p_n/q_n)}{3(p_n/q_n)+3} = \frac{4p_n}{3p_n+3q_n} = \frac{p_{n+1}}{q_{n+1}}$$
In other words:
$$p_{n+1} = 4p_n \longrightarrow p_n = 4^{n-1}$$
$$q_{n+1} = 3p_n + 3q_n = 3\cdot 4^{n-1} +3 q_n$$
We can find $q_n$ with generating function $Q(x) = \sum_{n=1}^\infty q_nx^n$:
$$Q(x) = x + x\sum_{n=1}^\infty q_{n+1}x^n$$
$$Q(x) = x + 3x\sum_{n=1}^\infty (q_n + 4^{n-1})x^n$$
$$Q(x) = x + 3xQ(x) + 3x\sum_{n=1}^\infty 4^{n-1}x^n$$
$$(1 - 3x)Q(x) = x + 3x\frac{x}{1-4x}$$
$$Q(x) = \frac{x}{1-3x} + \frac{3x^2}{(1-3x)(1-4x)} = \frac{1}{4} +\frac{x}{1-3x} - \frac{1}{1-3x} + \frac{3}{4(1-4x)}$$
$$Q(x) = \frac{1}{4} + \sum_{n=1}^\infty3^{n-1}x^n - \sum_{n=1}^\infty3^{n}x^n + \frac{3}{4}\sum_{n=1}^\infty 4^nx^n$$
$$Q(x) = \frac{1}{4} + \sum_{n=1}^\infty(3^{n-1} - 3^n + \frac{3}{4}4^n)x^n = \sum_{n=1}^\infty q_nx^n$$
$$q_n = 3^{n-1} - 3^n + \frac{3}{4}4^n = \frac{1}{12}(9\cdot4^{n} - 8\cdot 3^{n})$$
Thus $a_n = \dfrac{3\cdot4^{n}}{9\cdot4^{n} - 8\cdot 3^{n}}$ giving limit $\dfrac{1}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a sequence of reciprocals of square of odd natural numbers If $$a_{n}=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{(2n+1)^2}$$where $n\in N$.
Then prove that
$a_{n}<\frac{1}{4}$
| adding and subtracting $1 + \frac{1}{2^2} + \frac{1}{4^2} + \ldots + \frac{1}{(2n)^2}$, and using Basel problem we get
$\lim_{n\to \infty}a_n = \sum_{k=1}^{2n+1} \frac{1}{k^2} - \frac{1}{4}\left(\sum_{k=1}^{n} \frac{1}{k^2}\right) - 1 = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} - 1 = \frac{\pi^2}{8} - 1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Show that $\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$ is convergent.
Show that $$\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent.
$$\int_{0}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent by comparison test with $\int_{0}^{\infty} \dfrac{1}{x^4} dx$.
How do i show that $$\int_{-\infty}^{0} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent?
| Observe that for every $x\in\mathbb{R}$ you have:
$$1\geq 0 \Longrightarrow x^2+1\geq x^2\Longrightarrow \frac{x^{2}}{x^2+1}\leq 1$$
$$x^2\geq 0 \Longrightarrow 1+x^2\geq 1\Longrightarrow \frac{1}{x^2+1}\leq 1$$
multiplying both inequalities together we get
$$\frac{x^2}{(x^2+1)^2}\leq 1$$
this implies ( by multiplying both sides by ${\frac{1}{(x^2+2x+2)}}$):
$$0\leq \frac{x^2}{(x^2+1)^2(x^2+2x+2)}\leq \frac{1}{(x^2+2x+2)}=\frac{1}{(x+1)^2+1}$$
because $x^2+2x+2=(x+1)^2+1>0$
and $$\int_{\mathbb{R}}{\frac{1}{(x+1)^2+1}}dx=\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify nth Roots $6\sqrt[3]{9000} + 7\sqrt[3]{576}$ I'm having some concerns about how I am going about simplifying this radical expression. I wanted to know if this would be an accurate method of solving.
Simplify:
$$6\sqrt[3]{9000} + 7\sqrt[3]{576}$$
Radical expression product rule & greatest common factor:
$$6\sqrt[3]{125}\sqrt[3]{72} + 7\sqrt[3]{8}\sqrt[3]{72}$$
Principle of nth root:
$$6 \cdot 5\sqrt[3]{72} + 7 \cdot 2\sqrt[3]{72}$$
Simplify:
$$30\sqrt[3]{72} + 14\sqrt[3]{72}$$
Addition of radical rule:
$$30 + 14\sqrt[3]{72}$$
Simplify:
$$44\sqrt[3]{72}$$
Is $44\sqrt[3]{72}$ completely simplified?
I'm positive that 72 has no whole number $a$ that satisfies $72 = a^{3}.$
| You write "I'm positive that 72 has no whole number $a$ that satisfies $72 = a^{3}$". However, simplifying also works if there $72$ has a cubic factor. And indeed, $72$ contains a cube $a^3=8$. So we get $88\cdot \sqrt[3]{9}$.
| {
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"answer_id": 0
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Are the boundaries of the integral correct? I want to calculate the area of the triangle with vertices $(1,1,0)$, $(2,1,2)$, $(2,3,3)$.
We can parametrize the triangle using the function $\Sigma (x,y)=\left (x, y, 2x+\frac{y}{2}-\frac{5}{2}\right )$, right?
For the boundaries of x and y I have done the following:
From the verices we see that $1\leq x\leq 2$ and $$0\leq z\leq 3\Rightarrow 0\leq 2x+\frac{y}{2}-\frac{5}{2}\leq 3\Rightarrow \frac{5}{2}-2x\leq \frac{y}{2}\leq 3+\frac{5}{2}-2x\\ \Rightarrow \frac{5}{2}-2x\leq \frac{y}{2}\leq \frac{11}{2}-2x\Rightarrow 5-4x\leq y\leq 11-4x$$ Since from the vertices the smaller value of $y$ is $1$ we get that $1\leq y\leq 11-4x$.
So, do we calculate the following?
$$A(\Sigma (D))=\iint_D\|\Sigma_x\times\Sigma_y\|dxdy=\int_1^2\int_1^{11-4x}\frac{\sqrt{21}}{2}dydx$$
| To find your bounds on $x$ and $y$, project down to the $xy$-plane, where our vertices are $(1,1), (2,1)$ and $(2,3)$. Drawing this triangle helps.
That tells us we should have $x$ going from $1$ to $2$, as you noted, and $y$ should run from $y=1$ up to the line joining $(1,1)$ and $(2,3)$. That's the line $y=2x-1$.
The rest of your set-up appears to be fine.
As a double-check, you should get an area equal to $\frac12\|\langle1,0,2\rangle\times\langle1,2,3\rangle\|$, which is just $\frac12$ the magnitude of the cross-product of two legs of the triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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rewrite $\sum_{n=0}^\infty \frac{(-1)^nx^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$ Rewrite $\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$.
These two series are equal because they can both be derived from $\frac{1}{x^2+2x}$. To get the first one, you split it into $\frac{1}{x}\cdot \frac{1}{x+2}$ and then you use the power series. To get the second series, you complete the square with $\frac{1}{x^2+2x+1-1}$.
The point is those two are obviously equal but I can't figure out how to manipulate them while leaving them in series form to make them both equal each other (without using $\frac{1}{x^2+2x}$ as an in between equals). This was a problem that came up while my teacher was lecturing and he couldn't figure it out.
| One tricky part is that you have the singularity at $x=0$ built into the left side, but not the right.
So you'll need to use that $\frac{1}{x}=\frac{-1}{1-(1+x)}=-\sum (1+x)^k$.
These two are equal only on the interval $(-2,0)$ since the left side converges on $(-2,2)\setminus\{0\}$ and the right side converges on $(-2,0)$.
Writing $$f(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$$
We rewrite it as: $$\begin{align}f(x)
&=\frac{1}{2x}+\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2^{n+1}}\\
&=\frac{1}{2x}+\sum_{m=0}^{\infty}\frac{(-1)^{m+1}x^m}{2^{m+2}}\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^mx^m\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^m\left(1+x-1\right)^m\\
&=\frac1{2x}-\frac14\sum_{m=0}^{\infty}\left(\frac{-1}{2}\right)^m\sum_{k=0}^{m}\binom{m}{k}(1+x)^k(-1)^{m-k}\\
&=\frac{1}{2x}-\frac14\sum_{k=0}^{\infty}\sum_{m=k}^{\infty}\binom{m}{k}(1+x)^k(-1)^k\left(\frac 12\right)^m\\
&=\frac{1}{2x}-\frac{1}{4}\sum_{k=0}^{\infty}(-1-x)^k\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}\\
\end{align}$$
I'll prove below: $$\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}=2.$$
So you get:
$$f(x)=\frac{1}{2x}-\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k(1+x)^k$$
Also, $$\frac{1}{2x}=\frac{-1}2\frac1{1-(1+x)}=\frac{-1}{2}\sum_{k=0}^{\infty}(1+x)^k$$
Now $$(1+x)^k+(-1)^k(1+x)^k=\begin{cases}2(1+x)^k&k\text{ even}\\0&k\text{odd}\end{cases}$$
So you get: $$f(x)=-\sum_{k=0}^{\infty}(1+x)^{2k}$$
Lemma: For integer $k\geq0$, we have $$\sum_{m=k}^{\infty}\binom{m}{k}\frac{1}{2^m}=2.$$
Proof: We'll use a probability proof.
Flip a fair coin repeatedly, and stop when you have exactly $k+1$ heads. Let $X$ be the random variable equal to the number of tosses you needed.
Then $P(X=m+1)=\binom{m}{k}\frac{1}{2^{k+1}}$, because you must get exactly $k$ heads in the first $m$ tosses, and then the next toss must be a heads.
But then $\sum_{m=0}^{\infty}P(X=m+1)=1$. Multiplying by $2$ gives us our lemma.
[To deduce that last step, you technically also have to prove that $P(X<\infty)=1$ - that is, that is, there is zero probability that you never get $k+1$ heads.]
If you instead want to prove that:
$$\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^{n+1}}=-x\sum_{n=0}^{\infty}(1+x)^{2n}$$ when both are defined, you can do so fairly easily by noting that:
$$\begin{align}-x\sum_{n=0}^\infty(1+x)^{2n}&=(1-(1+x))\sum_{n=0}^{\infty}(1+x)^{2n}\\
&=\sum_{n=0}^{\infty}(1+x)^{2n}-\sum_{n=0}^{\infty}(1+x)^{2n+1}\\
&=\sum_{m=0}^{\infty}(-1)^m(1+x)^{m}
\end{align}$$
Then a variant of the proof above proof converts $xf(x)$ to this same series:
$$\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^{n+1}}&=
\sum_{n=0}^{\infty}\frac{(-1)^n(1+x-1)^n}{2^{n+1}}\\
&=\sum_{k=0}^{\infty}(1+x)^k(-1)^k\sum_{n=k}^{\infty}\binom{n}{k}\frac{1}{2^{n+1}}\\
&=\sum_{k=0}^{\infty}(-1)^k(1+x)^k
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So
$$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty
$$
which evaluates to be $\infty$.
| First of all:
$$\frac{1}{x(x^2+1)} = \frac{1}{x}-\frac{1}{2}\left(\frac{1}{x+i} -\frac{1}{x-i}\right)$$
Second of all:
You have to evaluate the integral $\int_1 ^ R\frac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $\infty -\infty$ is undetermined.
$$\int_1 ^ R\frac{1}{x(x^2+1)} = \int_1 ^ R \left[\frac{1}{x}-\frac{1}{2}\left(\frac{1}{x+i} -\frac{1}{x-i}\right)\right]dx$$
$$=[\log |x| - \frac{1}2{}\log(|x+i|) - \frac{1}{2}\log(|x-i|)]_1^R$$
$$=[\log |R| - \frac{1}2{}\log(|R+i|) - \frac{1}{2}\log(|R-i|)] + \frac{1}2{}\log(|1+i|) + \frac{1}{2}\log(|1-i|)]$$
$$=\log\left|\frac{R}{(R+i)^{\frac{1}{2}}(R-i)^\frac{1}{2}}\right|+\frac{1}{2}\log(1+i)(1-i)$$
$$=\log\left|\frac{R}{(R^2+1)^\frac{1}{2}}\right|+\frac{1}{2}\log(2)$$
Now taking the limit yields $\frac{1}{2}\log(2)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of differential equation in Trigonometric functions of $x$ and $y$ If $$\frac{dy}{dx} +\frac{\cos x(3\cos y-7\sin x-3)}{\sin y(3\sin x-7\cos y+7)}=0$$
Attempt: $(3\sin x\sin y-7\cos y \sin y+7\sin y)dy+(3\cos x\cos y-7\sin x\cos -3\cos x)dx =0$
could some help me how to solve it , thanks
| Hint
Let $a=\sin x$ and $b=\sin y$, then substitute it in the original equation. For $\cos x$ use $\pm\sqrt{1-a^2}$ and similarly for $b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate the integral $\int_{-1}^1\exp\left(ax+b\sqrt{1-x^2}\right)\,dx$ I wonder if there is a closed form expression for the following integral with respect to $x$:
\begin{equation}
\int_{-1}^1\exp\left(ax+b\sqrt{1-x^2}\right)\,dx
\end{equation}
where $a,b$ are some constant. Thanks so much! Any help would be greatly appreciated!
| $\int_{-1}^1e^{ax+b\sqrt{1-x^2}}~dx$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}e^{a\sin x+b\sqrt{1-\sin^2x}}~d(\sin x)$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}e^{a\sin x+b\cos x}\cos x~dx$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(a\sin x+b\cos x)^n\cos x}{n!}~dx$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^na^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{n!}~dx$
$=\int_{-\frac{\pi}{2}}^0\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\int_\frac{\pi}{2}^0\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}(-x)\cos^{k+1}(-x)}{k!(n-k)!}~d(-x)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{((-1)^{n-k}+1)a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{((-1)^{n-k}+1)a^{n-k}b^kB\left(\dfrac{n-k+1}{2},\dfrac{k}{2}+1\right)}{2k!(n-k)!}$ (according to http://mathworld.wolfram.com/BetaFunction.html)
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{((-1)^{n-k}+1)a^{n-k}b^k\Gamma\left(\dfrac{n-k+1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{2k!(n-k)!\Gamma\left(\dfrac{n+3}{2}\right)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{((-1)^{n-k}+1)a^{n-k}b^k\Gamma\left(\dfrac{n-k+1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{2k!(n-k)!\Gamma\left(\dfrac{n+3}{2}\right)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{((-1)^n+1)a^nb^k\Gamma\left(\dfrac{n+1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{2k!n!\Gamma\left(\dfrac{n+k+3}{2}\right)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^k\Gamma\left(n+\dfrac{1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{(2n)!k!\Gamma\left(n+\dfrac{k+3}{2}\right)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k}\Gamma\left(n+\dfrac{1}{2}\right)\Gamma(k+1)}{(2n)!(2k)!\Gamma\left(n+k+\dfrac{3}{2}\right)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k+1}\Gamma\left(n+\dfrac{1}{2}\right)\Gamma\left(k+\dfrac{3}{2}\right)}{(2n)!(2k+1)!\Gamma(n+k+2)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k}\pi}{4^{n+k}n!\Gamma\left(k+\dfrac{1}{2}\right)\Gamma\left(n+k+\dfrac{3}{2}\right)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k+1}\pi}{2^{2n+2k+1}n!k!(n+k+1)!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k}\pi}{4^{n+k}n!\Gamma\left(k+\dfrac{1}{2}\right)\Gamma\left(n+k+\dfrac{3}{2}\right)}+\dfrac{b\pi}{2}{_0F_1}\left(-;2;\dfrac{a^2+b^2}{4}\right)$ (according to Some properties about the Kampé de Fériet function)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difference between Fractional Exponents and Fractions? how would you explain the difference between exponential multiplication and fractional multiplication? $${x^{1/3}}{^{}{}} * {x^{1/3}}{^{}{}} *{x^{1/3}}{^{}{}} * {x^{1/3}}{^{}{}} = {x^{4/3}}$$ Why is this the same as $$4 * {^{1/3}}$$ On the other hand $$1/3 * 1/3 * 1/3 * 1/3 = 1/81$$ So in this case, this is not the same as $$4 * {{1/3}}$$
| First, remember that $$b^x\cdot b^y=b^{x+y}.$$ That’s why $$x^{1/3}\cdot x^{1/3}\cdot x^{1/3}\cdot x^{1/3}=x^{1/3+1/3+1/3+1/3}=x^{4\cdot 1/3}=x^{4/3}.$$ One the other hand $$\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}=\frac{1\cdot 1\cdot 1\cdot 1}{3\cdot 3\cdot 3\cdot 3}=\frac{1}{81}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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AM>HM Problem $\frac{1}{n+1}+...+\frac{1}{3n+1}>1$ I am having difficulty solving one of the problems from "Problems in Mathematical Analysis I" - W. J. Kaczor;M. T. Nowak .
It's a problem 1.2.5 b), and it goes like this:
1.2.5. For $n \in \mathbb{N}$, verify the following claims:
$$\tag{b} \qquad \dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3} + \ldots + \dfrac{1}{3n + 1} \, > \, 1$$
In solutions it says: "Use the arithmetic-harmonic mean inequality!"
I tried to apply it on whole inequality but got:
\begin{align}
& \dfrac{\frac{1}{n+1}+\ldots+\frac{1}{3n+1}}{2n}\, >\, \dfrac{2n}{n+1+\ldots+3n+1} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{8n^2}{2n(n+1+3n+1)} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{2n}{2n+1}
\end{align}
| The HM-AM (or Jensen's Inequality applied to $\frac1x$ and a discrete measure) says
$$
\begin{align}
\frac1{2n+1}\left(\frac1{n+1}+\cdots+\frac1{3n+1}\right)
&\ge\frac1{\frac1{2n+1}\left((n+1)+\cdots+(3n+1)\right)}\\
&=\frac1{2n+1}
\end{align}
$$
Therefore,
$$
\frac1{n+1}+\cdots+\frac1{3n+1}\ge1
$$
We can also prove this by the Cauchy-Schwarz Inequality
$$
\begin{align}
(\overbrace{1+\cdots+1}^{2n+1\text{ terms}})^2&
\le\left(\frac1{n+1}+\cdots+\frac1{3n+1}\right)\left((n+1)+\cdots+(3n+1)\right)\\
&=\left(\frac1{n+1}+\cdots+\frac1{3n+1}\right)(2n+1)^2
\end{align}
$$
which also implies
$$
1\le\frac1{n+1}+\cdots+\frac1{3n+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Identity of Fibonacci sequence Let $F_n$ be a Fibonacci sequence with initial terms $F_0=0, F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\geqslant 1$.
Prove that $F_n^2+F_{n+1}^2=F_{2n+1}$ for $n\geqslant 0$ (with mathematical induction).
My efforts: For $n=0$ it is true.
Suppose that our statement holds for $0\leqslant k \leqslant n$ i.e. $F_k^2+F_{k+1}^2=F_{2k+1}$
Let's try to prove it for $k=n+1$. $$F_{2n+3}=F_{2n+1}+F_{2n+2}=F_{n+1}^2+(F_{n}^2+F_{2n+2})= ?$$
Here I'm stuck and I have applied different methods but none of them brings a positive result.
Can anyone help to complete this?
| We can actually prove two different identities (1) and (2), of which you were asked to prove (1):
$$F_{n+1}^2+F_{n+2}^2=F_{2n+3} \quad \text{(1)}$$
$$F_{n+2}^2-F_{n}^2=F_{2n+2} \quad \text{(2)}$$
Note that the Fibonacci numbers are numbered starting from $F_1$.
We can proceed by induction. Suppose that we have the two equations
$$F_n^2+F_{n+1}^2=F_{2n+1} \quad \text{(3)}$$
$$F_{n+1}^2-F_{n-1}^2=F_{2n} \quad \text{(4)}$$
The base cases are trivial.
First Step: Prove (2).
Now, adding (3) and (4), we obtain :
\begin{align}
2F_{n+1}^2+F_n^2-F_{n-1}^2 &= F_{2n+1}+F_{2n}\\\\
(F_{n+1}-F_n)^2+F_{n+1}^2 + 2F_{n+1}F_n - F_{n-1}^2&= F_{2n+2}\\\\
(F_{n-1})^2+F_{n+1}^2 + 2F_{n+1}F_n - F_{n-1}^2&= F_{2n+2}\\\\
(F_{n+1}+F_{n})^2-F_{n}^2&= F_{2n+2}\\\\
F_{n+2}^2-F_{n}^2&= F_{2n+2} \quad \text{(2)}
\end{align}
As desired. $\square$
Second Step: Prove (1). Now that we have (2), we are home free!
Adding (2) and (3), we obtain :
\begin{align}
F_{n+2}^2-F_n^2+F_{n}^2+F_{n+1}^2 &= F_{2n+2}+F_{2n+1}\\\\
F_{n+2}^2+F_{n+1}^2 &= F_{2n+3} \quad \text{(1)}
\end{align}
And we are done. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Show that $\tan^{-1} (\tfrac{1}{8} ) \geq \frac{1}{\sqrt{65}} $ Here I'd like to check that:
$$ \frac{1}{8} \geq \tan^{-1} (\tfrac{1}{8} ) \geq \frac{1}{\sqrt{65}} $$
The numbers do check out if one uses a calculator. They were using the Taylor series:
*
*$\frac{1}{\sqrt{65}} \;\;\;\;\;= 0.12403 $
*$\tan^{-1} \frac{1}{8} = 0.12435 $
*$\frac{1}{8} = 0.125$
The Taylor series for tangent could lead to an approximation. If $0 < x < \frac{\pi}{4}$ I believe we have:
$$ x - \frac{x^3}{3} < \tan^{-1} x < x $$
I do not recommend either of these, as a first recourse, but it came up in discussion last time.
These numbers might be somewhat mysterious (and they are). I got these using complex numbers. I said that:
\begin{eqnarray*}
5 &=& 1^2 + 2^2 \\
13 &=& 2^2 + 3^2
\end{eqnarray*}
And there are angles associated to these sum of squares $\theta_5 = \tan^{-1} \frac{1}{2}$ and $\theta_{13} = \tan^{-1} \frac{2}{3} $
then, using the difference of arctangents formula:
$$ \tan^{-1} \frac{2}{3} - \tan^{-1} \frac{1}{2} = \tan^{-1} \left[
\frac{ \frac{2}{3} - \frac{1}{2}}{1 + \frac{2}{3} \times \frac{1}{2}} \right] = \tan^{-1} \tfrac{1}{8}$$
and this leads to the left side inequalty above.
| $\arctan\frac{1}{8}<\frac{1}{8}$ is trivial, and by the Shafer-Fink inequality
$$ \arctan\frac{1}{8}>\frac{3\cdot\frac{1}{8}}{1+2\sqrt{1+\frac{1}{64}}}=\frac{3}{8+2\sqrt{65}}>\frac{1}{\sqrt{65}}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the real number $x$ such that $\det A=0$ $$A= \begin{bmatrix} 1 & x & x^2 & x^3\\ x & x^2 & x^3 & 1 \\ x^2 & x^3 & 1 & x \\ x^3 & 1 & x&x^2 \end{bmatrix}$$
I don't know how to approach this. I'd like to figure this out step-by-step so any suggestions would be greatly appreciated. Thanks in advance!
| With row operations that don't change the determinant (sum to a row another row multiplied by some factor):
\begin{align}
\begin{bmatrix}
1 & x & x^2 & x^3 \\
x & x^2 & x^3 & 1 \\
x^2 & x^3 & 1 & x \\
x^3 & 1 & x & x^2
\end{bmatrix}
&\to
\begin{bmatrix}
1 & x & x^2 & x^3 \\
0 & 0 & 0 & 1-x^4 \\
0 & 0 & 1-x^4 & x-x^5 \\
0 & 1-x^4 & x-x^5 & x^2-x^6
\end{bmatrix}
&&\begin{aligned}
R_2&\gets R_2-xR_1\\
R_3&\gets R_3-x^2R_1\\
R_4&\gets R_4-x^3R_1
\end{aligned}
\end{align}
Now we can expand the determinant with respect to the first column:
$$
\det A=
\det\begin{bmatrix}
0 & 0 & 1-x^4 \\
0 & 1-x^4 & x-x^5 \\
1-x^4 & x-x^5 & x^2-x^6
\end{bmatrix}
=-(1-x^4)^3=(x^4-1)^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
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The Banach spaces $(\mathbb{R^2}, \Vert\cdot\Vert_1)$ and $(\mathbb{R^2}, \Vert\cdot\Vert_{\infty})$ are linearly isometric.(T/F) The Banach spaces $(\mathbb{R^2}, \Vert\cdot\Vert_1)$ and $(\mathbb{R^2}, \Vert\cdot\Vert_{\infty})$ are linearly isometric.(T/F)
i.e $\Vert Tx\Vert_{\infty}=\Vert x\Vert_{1}$ but after that I can't think , please help.
| The statement is true.
Let $\ell^1(2) = (\mathbb{R}^2,\|\cdot\|_1)$ and $\ell^\infty(2)=(\mathbb{R}^2,\|\cdot\|_\infty).$
Define $T:\ell^\infty(2) \to \ell^1(2)$ by
$$T \begin{pmatrix}
x \\
y
\end{pmatrix} = \frac{1}{2} \begin{pmatrix}
x-y\\
x+y
\end{pmatrix}.$$
We claim that $T$ is an onto linear isometry.
For any $\begin{pmatrix}
x \\
y
\end{pmatrix} \in \ell^\infty(2),$ we have
$$\left\Vert T \begin{pmatrix}
x \\
y
\end{pmatrix} \right\Vert_1 = \left\Vert\frac{1}{2} \begin{pmatrix}
x-y\\
x+y
\end{pmatrix}\right\Vert_1 = \frac{|x-y| + |x+y|}{2} = \max\{|x|,|y| \} = \left\Vert \begin{pmatrix}
x \\
y
\end{pmatrix} \right\Vert_\infty$$
where the second last identity is standard in any real analysis text.
Therefore, $T$ is an isometry.
Clearly $T$ is linear.
We claim that $S:\ell^1(2)\to\ell^\infty(2)$ given by
$$S\begin{pmatrix}
x \\
y
\end{pmatrix} = \begin{pmatrix}
x+y\\
-x+y
\end{pmatrix}$$ is the inverse of $T.$
Indeed,
$$(S \circ T)\begin{pmatrix}
x \\
y
\end{pmatrix} = S \left[ \frac{1}{2} \begin{pmatrix}
x-y\\
x+y
\end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix}
2x\\
2y
\end{pmatrix} = \begin{pmatrix}
x\\
y
\end{pmatrix}, $$
$$(T \circ S) \begin{pmatrix}
x\\
y
\end{pmatrix} = T \left[ \begin{pmatrix}
x+y\\
-x+y
\end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix}
2x\\
2y
\end{pmatrix} = \begin{pmatrix}
x\\
y
\end{pmatrix}.$$
Therefore, $T$ is a bijection.
Hence, $\ell^\infty(2) \cong \ell^1(2).$
| {
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Evaluating $\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$ The question is to evaluate $$\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$$
I tried using leibnitz rule
$$F'(a)=\int_0^{\pi /2} \frac{ -\sin a}{(1+\cos a \cos x)}dx$$ Now I used the substitution $\tan(x/2)=t$ to get $$-2 \sin a \int_0^1 \frac{ dt}{1+t^2 +\cos a (1-t^2)} $$ which can be rewritten as
$$-2\frac{\sin a} {1- \cos a}\int_0^{1} \frac{ dt}{t^2 +\frac{1+ \cos a}{1-\cos a}} $$ which evaluates to $-a$.i am not sure where I went wrong.Any ideas?
| Hint:
Put $b= \sqrt{\frac{1+ \cos a}{1-\cos a} }$
Then,
$$\int_0^{1} \frac{ dt}{t^2 +\frac{1+ \cos a}{1-\cos a}} =\int_0^{1} \frac{ dt}{t^2 +b^2} = \frac1b\arctan{\frac{t}{b}}\bigg|_{t=0}^{t=1}= \frac1b\arctan{\frac{1}{b}} $$
Therefore, $$ F'(a) =-2\frac{\sin a} {\sqrt{(1-\cos a)(1+ \cos a)}} \arctan{\left(\sqrt{\frac{1- \cos a}{1+\cos a} }\right)} \\= -2\frac{\sin a} {|\sin a|} \arctan{\left(\sqrt{\frac{1- \cos a}{1+\cos a} }\right)}$$
That is $$ F'(a)= \color{blue}{ -2\frac{\sin a} {|\sin a|} \arctan{\left(\sqrt{\frac{1- \cos a}{1+\cos a} }\right)}}$$
| {
"language": "en",
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Proving that the set enclosed by an ellipse is convex I'm trying to prove that the following set ($B$) defined by an ellipse is convex, but I am getting stuck.
$$B = \left\{(x_1,x_2):3(x_1-3)^2+2(x_2-3)^2 \leq 1 \right\} $$
My sad attempt:
$$ \textrm{Let }x,y \: \in \: B \:, \: \lambda \: \in \: [0,1]$$
$$ \lambda x+(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\: \lambda x_2 +(1-\lambda)y_2 )$$
$$3(\lambda x_1 + (1-\lambda)y_1 -3)^2+2(\lambda x_2+(1-\lambda)y_2-3)^2=3(\lambda ^2x_1^2 +(1-\lambda)^2y_1^2+9+2\lambda x_1(1-\lambda)y_1-6\lambda x_1 -6(1-\lambda)y_1)+2(\lambda ^2x_2^2 +(1-\lambda)^2y_2^2+9+2\lambda x_2(1-\lambda)y_2-6\lambda x_2 -6(1-\lambda)y_2)$$
and this is where I have lost all confidence in proving it. I was able to factor some of the terms into the form $(\lambda x_1 -3)^2$, $((1-\lambda) y_1 -3)^2$, $(\lambda x_2 -3)^2$, and $((1-\lambda) x_2 -3)^2$, but I don't think that helps the situation any.
| We have to prove that if $(a_1,b_1) \in B$, and $(a_2,b_2) \in B$ then $$(a,b)=(\lambda a_1+(1-\lambda)a_2,\lambda b_1+(1-\lambda)b_2) \in B.$$
The trick is to observe that $a-3 = \lambda(a_1-3)+(1-\lambda)(a_2-3)$ and $b-3 = \lambda(b_1-3)+(1-\lambda)(b_2-3)$.
Consequently,
$$3(a-3)^2 + 2(b-3)^2 = 3(\lambda (a_1-3)+(1-\lambda)(a_2-3))^2 + 2(\lambda (b_1-3)+(1-\lambda)(b_2-3))^2.$$
Simplifying the RHS, we get
$$3(a-3)^2 + 2(b-3)^2 = \lambda^2(3(a_1-3)^2+2(b_1-3)^2) + (1-\lambda)^2(3(a_2-3)^2+2(b_2-3)^2) \\+ 2\lambda(1-\lambda)(3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3)).$$
Since $3(a_1-3)^2+2(b_1-3)^2 \le 1$ and $3(a_2-3)^2+2(b_2-3)^2 \le 1$, we get
$$3(a-3)^2 + 2(b-3)^2 \le \lambda^2 + (1-\lambda)^2+ 2\lambda(1-\lambda)(\color{red}{3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3)}).$$
By Cauchy-Schwarz inequality, the red expression can be bounded as
$$3(a_1-3)(a_2-3)+2(b_1-3)(b_2-3) \le \sqrt{(3(a_1-3)^2+2(b_1-3)^2)(3(a_2-3)^2+2(b_2-3)^2)}=1.$$
Therefore,
$$3(a-3)^2 + 2(b-3)^2 \le \lambda^2 + (1-\lambda)^2+ 2\lambda(1-\lambda) =1.$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546650",
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} |
laurent series of $\frac{z^2-2z+2}{(z-1)(z^2-2z-3)}$ around $z-1$ for $0<|z-1|<2$ I want to find the Laurent series of $L=\frac{z^2-2z+2}{(z-1)(z^2-2z+3}$ around $(z-1)$
Here, I used the fact that
$L=\frac{5}{8(z-3)}+\frac{5}{8(z+1)}-\frac{1}{4(z-1)}$
When $|z-1|<2$, we have
$L=\frac{-1}{4(z-1)}+\frac{5}{8} \frac{1}{(z-1)-2} +\frac{5}{8( (z-1)+2)}$
$=\frac{-1}{4(z-1)} +\frac{5}{-16}\frac{1}{1- \frac{(z-1)}{2}} + \frac{5}{16} \frac{1}{1+ \frac{(z-1)}{2}} =-\frac{1}{4(z-1)}+ \frac{5}{16} \sum_{k=0}^\infty \frac{(z-1)^k}{2^k}(-1+(-1)^k)=-\frac{1}{4(z-1)}+\frac{5}{16} \sum_{n=0}^\infty \frac{(z-1)^{2n+1}}{2^{2n+1}} (-2)$
However,in my book, Elements d'analyse complexe by Real Gelinas,
the answer for 0<|z-1|<2 is
$\frac{-1}{4} (z-1 + \frac{1}{z-1} )\sum_{n=0}^\infty (\frac{z-1}{2n})^{2n}.$
Hence, I am missing a term.
Where is my mistake?
| Note : Since you've made a mistake in your expressions while solving (the one mentioned in the comments) I'll go over a more straight-forward and common approach, by demonstrating its all steps.
The trick is to form the expression that is asked to form the Laurent series around, which in that specific case is : $z-1$. This is what I'm going to demonstrate down below :
$$L(z) = \frac{z^2-2z+2}{(z-1)(z^2-2z-3)}= \frac{(z-1)^2+1}{(z-1)(z+1)(z-3)}$$
$$=$$
$$\frac{(z-1)^2+1}{(z-1)(z-1+2)(z-1-2)} = \frac{(z-1)^2+2}{(z-1)(z-1)(z-1)(1+\frac{2}{z-1})(1-\frac{2}{z-1})}$$
$$=$$
$$\frac{(z-1)^2+1}{(z-1)^3(1+\frac{2}{z-1})(1-\frac{2}{z-1})}$$
Now, we know the following simple geometric series :
$$\frac{1}{1-w}\sum_{n=0}^\infty w^n \quad |w|<1$$
$$\frac{1}{1+w} = \sum_{n=0}^\infty (-1)^nw^n \quad |w|<1$$
Applying these for $\frac{2}{z-1} = w$ :
$$[(z-1)^2+1] \frac{1}{(z-1)^3}\sum_{n=0}^\infty(-1)^n2^n(z-1)^{-n}\sum_{n=0}^\infty 2^n(z-1)^{-n} $$
$$=$$
$$\bigg[\frac{1}{z-1}+\frac{1}{(z-1)^3}\bigg]\sum_{n=0}^\infty(-1)^n2^n(z-1)^{-n}\sum_{n=0}^\infty 2^n(z-1)^{-n} $$
Can you now combine all of these into one expression and yield your results ? For the relation of $|z-1|$ given, it's easy to spot since you'd want $|\frac{2}{z-1}| < 1$ and on the same time $z-1\neq 0$ which means given your first condition that $|z-1| > 0$. Combining these you get : $0<|z-1|<2$.
| {
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Two different answers by applying AM GM Inequality Find minimum value of $f(x)=\sec^2 x+\csc ^2x$
We have $$f(x)=2+\tan^2 x+\cot ^2 x$$
Method $1.$ we have $$\frac{\tan^2 x+\cot^2 x}{2} \ge 1$$ $\implies$
$$f(x) \ge 4$$
Method $2.$
We have $$\frac{2+\tan^2 x+\cot^2 x}{3} \ge 2^{\frac{1}{3}}$$ $\implies$
$$f(x) \ge 3(2^{\frac{1}{3}})-2$$
But whats wrong in method $2$?
| Because in your second AM-GM the equality should be occur for $2=\tan^2x=\cot^2x,$ which is impossible.
The right way is the following.
By AM-GM
$$f(x)\geq2+2\sqrt{\tan^2x\cot^2x}=2+2=4.$$
The equality occurs for $x=45^{\circ},$ which says that $4$ is a minimal value.
| {
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Prove that $\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$ (without calculus) I'm trying to prove the following proposition (I'm not supposed to use calculus):
$$\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$$
(I'm assuming that $0 \notin \Bbb{N}$)
This is what I've tried so far:
$\sum_{i=1}^{n} \frac{n+i}{i+1} = \sum_{i=1}^{n}[ \frac{n}{i+1} + \frac{i}{i+1} ] = [\sum_{i=1}^{n} \frac{n}{i+1}] + \sum_{i=1}^{n} \frac{i}{i+1} = n[\sum_{i=1}^{n} \frac{1}{i+1}] + \sum_{i=1}^{n} \frac{i}{i+1}$
Let $j= i +1$ . So now we have:
$\sum_{i=1}^{n} \frac{n+i}{i+1} = n[\sum_{j=2}^{n+1} \frac{1}{j}] + \sum_{j=2}^{n+1} \frac{j-1}{j}$
$= n[\sum_{j=2}^{n+1} \frac{1}{j}] + \sum_{j=2}^{n+1} \frac{j}{j} - \sum_{j=2}^{n+1} \frac{1}{j}$
$= n[\sum_{j=2}^{n+1} \frac{1}{j}]- \sum_{j=2}^{n+1} \frac{1}{j} + \sum_{j=2}^{n+1} 1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + [\sum_{j=1}^{n+1} 1 ] -1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n+1 -1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n$
Therefore proving the following inequality is the same as proving the original:
$$ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n(n-1)$$
If $n=1$ then $ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n = 1$ and $1 + n(n-1) = 1$
Now I assume that $n>1$:
$ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n(n-1) \iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n^2 - n$
$\iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] \le 1 + n^2 - 2n$
$\iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] \le (n-1)^2$
$\iff \sum_{j=2}^{n+1} \frac{1}{j} \le \frac{(n-1)^2}{n-1}$
$\iff \sum_{j=2}^{n+1} \frac{1}{j} \le n-1$
$\iff 1 + \sum_{j=2}^{n+1} \frac{1}{j} \le n$
$\iff \sum_{j=1}^{n+1} \frac{1}{j} \le n$
And here is where I'm stuck. I know that an upper bound for the harmonic sum can be found using an integral test but I'm not supposed to use calculus. Is there a discrete way of proving that $\sum_{j=1}^{n+1} \frac{1}{j} \le n$ for all natural $n \ge 2$ ? Or maybe another way of proving the proposition without ending up with the harmonic sum?
| $$\sum_{j=1}^{n+1} \frac{1}{j} \le n$$ is easy to prove of $n \geq 2$. Note that the inequality fails for $n=1$.
Indeed
$$\sum_{j=1}^{n+1} \frac{1}{j}=\sum_{j=1}^{n-1} \frac{1}{j}+\left( \frac{1}{n}+\frac{1}{n+1}\right)<\sum_{j=1}^{n-1} 1+\left( \frac{1}{2}+\frac{1}{2}\right)=n $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integrate $\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \,dx$
Integrate $$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}dx$$
I solved the question by using the identity $\cos^4(x)+\sin^4(x) = \frac{1}{4}(\cos4x+3)$ and the substitution $u=\cos4x +3$, which turned it into a relatively familiar integral (see my answer below).
However, I'm pretty sure there are easier ways I am missing, so please feel free to post alternative answers.
There is a similar question here.
Problem Source: James Stewart Calculus, 6E
| First, some preliminary manipulation. $$\frac{\sin x \cos x}{\sin^4x + \cos^4x} = \frac{\sin x \cos x}{(1-\cos^2x)^2 + \cos^4x}\\=\frac{\sin x \cos x}{2\cos^4x -2\cos^2x+1}=\frac{4\sin x \cos x}{8\cos^4x -8\cos^2x+1 + 3}\\=\frac{4\sin x \cos x}{\cos4x+ 3}$$
The last step uses the quadruple angle formula for cosine. Now $\sin 2x = 2\sin x \cos x$, using this twice yields:$$\frac{4\sin x \cos x}{\cos4x+ 3}=\frac{2\sin 2x }{\cos4x+ 3} =\frac{2\sin 2x \cos 2x}{(\cos4x+ 3)\cos 2x} = \frac{\sin 4x }{(\cos4x+ 3)\cos 2x} $$
We can now use the half-angle formula for cosine, which is $\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$. $$\frac{\sin 4x }{(\cos4x+ 3)\cos 2x}=\frac{\sin 4x }{(\cos4x+ 3)\sqrt{\frac{1+\cos 4x}{2}}} \\=\frac{\sqrt{2} \times \sin 4x }{(\cos4x+ 3)\sqrt{\cos 4x + 3 - 2}}$$
The time is ripe to substitute with $u=\cos 4x +3$. Then $du = -4\sin 4x \ dx$ and $$\int \frac{\sqrt{2} \times \sin 4x }{(\cos4x+ 3)\sqrt{\cos 4x + 3 - 2}} \ dx\\ = \frac{\sqrt{2}}{-4}\int \frac{1}{u\sqrt{u - 2}} \ du$$
To finish this relatively simple integral, I did another substitution, this time with $t=\sqrt{u-2}$ and $du =2t \ dt$. $$\frac{\sqrt{2}}{-4}\int \frac{1}{u\sqrt{u - 2}} \ du = \frac{\sqrt{2}}{-4}\int \frac{2t}{ut} \ dt\\=\frac{\sqrt{2}}{-2}\int \frac{1}{t^2 +2} \ dt$$
This is an integral that involves $\arctan$:$$\frac{\sqrt{2}}{-2}\int \frac{1}{t^2 +2} \ dt=-\frac{1}{2}\arctan(\frac{t}{\sqrt{2}})=-\frac{1}{2}\arctan(\frac{\sqrt{u-2}}{\sqrt{2}})\\=-\frac{1}{2}\arctan(\sqrt{\frac{\cos 4x +1}{2}})=-\frac{1}{2} \arctan(\cos 2x)$$
Checking with wolframalpha, this differentiates to the correct result.
| {
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"question_score": "5",
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How do you find the values that $x$ can take by squaring? $$|x+4| \cdot |x-4| = x+4$$
How do you find the values that $x$ can take by squaring?
How I've tried
$$|x+4|^2 \cdot |x-4|^2 = (x+4)^2$$
and
$$|x^2+16| \cdot |x^2+16| = x^2+16$$
I think I went wrong.
| Do squaring!
$$ (\vert x- 4 \vert \cdot \vert x + 4 \vert)^2 = (x + 4)^2 $$
$$ (\vert x- 4 \vert)^2 \cdot (\vert x + 4 \vert)^2 = (x + 4)^2 $$
$$ (x- 4)^2 \cdot ( x + 4)^2 = (x + 4)^2 $$
$$ (x- 4)^2 \cdot ( x + 4)^2 - (x + 4)^2=0 $$
$$ (x + 4)^2 \left[(x - 4)^2 - 1\right] =0 $$
$$ (x+4)^2=0, (x-4)^2 - 1 = 0$$
$$\implies x= -4, x = 4±\sqrt{1}$$
$$\implies x= -4, x = 5,x= 3$$
Njoy!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565037",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to calculate last four digits of $2^{2017}$? can u suggest me any short trick of solving these kind of problems. How to find last four digits of any number raised to some power ?
| Like How to find last two digits of $2^{2016}$,
I shall find $2^{2017-4}\pmod{10^4/2^4}$
As $2^2=5-1$
$\displaystyle2^{2012}=(5-1)^{1006}\equiv1-\binom{1006}15+\binom{1006}25^2-\binom{1006}35^3\pmod{5^4}$
Now $\displaystyle1006\equiv6\pmod{5^3}\implies\binom{1006}15\equiv5\cdot6\pmod{5^4}$
$\displaystyle\binom{1006}2\equiv15\pmod{5^2}\implies\binom{1006}25^2\equiv5^2\cdot15\pmod{5^4}$
and $\displaystyle\binom{1006}3\equiv0\pmod5\implies\binom{1006}35^3\equiv0\pmod{5^4}$
$\displaystyle\implies2^{2012+1}\equiv2(1-5\cdot6+5^2\cdot15)\equiv346\cdot2\pmod{5^4}$
Multiply out by $2^4$
| {
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Simplify: $\arctan \sqrt{\frac{1-\cos x}{1+\cos x}};$ $0$$\arctan \sqrt{ \frac{1-\cos x}{1+\cos x} } $$
There are two ways you can procced with this:
1.$$=\arctan{\sqrt{\left(\frac{2\sin^2{\frac{x}{2}}}{2\cos^2{\frac{x}{2}}}\right)}}$$
$$=\arctan{\left(\tan\frac{x}{2}\right)}$$
$$=\frac{x}{2}$$
2. $$=\arctan{\sqrt{\left(\frac{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}-2\cos\frac{x}{2}\sin \frac{x}{2}}{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\cos{{\frac{x}{2}}}\sin{\frac{x}{2}}}\right)}}$$
$$=\arctan{
\sqrt{
\left(\frac
{\cos \frac{x}{2} -\sin \frac{x}{2} }{\cos \frac{x}{2} +\sin \frac{x}{2} } \right) ^2
}
}$$
$$=\arctan \left(\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\right)$$
Dividing by $\cos \frac{x}{2}$
$$=\arctan \left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right) $$
$$=\arctan\left[ \tan \left( \frac{\pi}{4}-\frac{x}{2} \right)\right] $$
$$=\frac{\pi}{4}-\frac{x}{2}$$
So, are both these answers correct? or one of them is incorrect?
|
1.$$=\arctan{\sqrt{\left(\frac{2\sin^2{\frac{x}{2}}}{2\cos^2{\frac{x}{2}}}\right)}}$$
$$=\arctan{\left(\tan\frac{x}{2}\right)}$$
Careful here:
$$\sqrt{\tan^2\tfrac{x}{2}} = \left| \tan\tfrac{x}{2} \right|$$
and you only have $\left| \tan\tfrac{x}{2} \right|=\tan\tfrac{x}{2}$ whenever $\tan\tfrac{x}{2} \ge 0$; while for $\tan\tfrac{x}{2} \le 0$ you would get $\left| \tan\tfrac{x}{2} \right|=-\tan\tfrac{x}{2}$.
This explains the difference and you'll need to know which interval $x$ is in to further simplify; perhaps $\color{blue}{0 <} x < \pi$ was meant instead of only $x<\pi$?
| {
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} |
How can I prove, that this formula is related to the binomial series? I am trying to solve the following formula:
$$\sum^{\lfloor (n-1)/2\rfloor}_{m=k}\binom{m}{k}\binom{n}{2m+1} = \binom{n-k-1}{k}2^{n-2k-1}$$
However, I haven't got a clue. Can anybody shed some light on this?
| We seek to verify for $n-k-1\ge k$ the identity
$$\sum_{m=k}^{\lfloor (n-1)/2 \rfloor} {m\choose k} {n\choose 2m+1}
= {n-k-1\choose k} 2^{n-2k-1}.$$
The LHS is
$$\sum_{m=k}^{\lfloor (n-1)/2 \rfloor}
{m\choose k} {n\choose n-2m-1}
= \sum_{m=k}^{\lfloor (n-1)/2 \rfloor}
{m\choose k} [z^{n-2m-1}] (1+z)^n
\\ = \sum_{m=k}^{\lfloor (n-1)/2 \rfloor}
{m\choose k} [z^{n}] z^{2m+1} (1+z)^n.$$
Now observe that when $2m+1\gt n$ we get zero from the coefficient
extractor so in fact it encodes the upper limit of the summation,
which we may extend to infinity since there is no contribution from
those values of $m$. We obtain
$$[z^n] (1+z)^n \sum_{m\ge k}
{m\choose k} z^{2m+1}
= [z^{n-1}] (1+z)^n \sum_{m\ge k}
{m\choose k} z^{2m}
\\ = [z^{n-1}] (1+z)^n \sum_{m\ge 0}
{m+k\choose k} z^{2m+2k}
= [z^{n-2k-1}] (1+z)^n \sum_{m\ge 0}
{m+k\choose k} z^{2m}
\\ = [z^{n-2k-1}] (1+z)^n \frac{1}{(1-z^2)^{k+1}}
= [z^{n-2k-1}] (1+z)^{n-k-1} \frac{1}{(1-z)^{k+1}}.$$
Extracting the coefficient we find
$$\sum_{q=0}^{n-2k-1} {n-k-1\choose q}
{n-2k-1-q+k\choose k}
\\ = \sum_{q=0}^{n-2k-1} {n-k-1\choose q}
{n-k-1-q\choose k}.$$
The product of binomials is
$$\frac{(n-k-1)!}{q! \times k! \times (n-2k-1-q)!}
= {n-k-1\choose k} {n-2k-1\choose q}$$
and we get
$${n-k-1\choose k} \sum_{q=0}^{n-2k-1} {n-2k-1\choose q}
= {n-k-1\choose k} 2^{n-2k-1}$$
as claimed.
| {
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"url": "https://math.stackexchange.com/questions/2567436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Sum of fifth power of the roots of equation $x^3-x^2+1=0$ The equation $x^3-x^2+1=0$ has three roots $\alpha$, $\beta$ and $\gamma$. Find the value of $\alpha^5 + \beta^5 + \gamma^5$
I tried it this way:
$x^3=x^2-1$
$\alpha + \beta + \gamma = 1$
$\alpha \cdot \beta \cdot \gamma = -1$
$\alpha \cdot \beta + \beta \cdot \gamma + \alpha \cdot \gamma = 0$
So, $\alpha^3=\alpha^2-1$
$\alpha^5=\alpha^4-\alpha^2$
And similarly for $\beta$ and $\gamma$
Now I did add them but I am unable to find something useful in it.
| \begin{align}\alpha^5+\beta^5+\gamma^5&=\alpha^4-\alpha^2+\beta^4-\beta^2+\gamma^4-\gamma^2\\&=\alpha^3-\alpha-\alpha^2+\beta^3-\beta-\beta^2+\gamma^3-\gamma-\gamma^2\\&=\alpha^2+1-\alpha-\alpha^2+\beta^2+1-\beta-\beta^2+\gamma^2+1-\gamma-\gamma^2\\&=3-(\alpha+\beta+\gamma).\end{align}Can you take it from here?
| {
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"source": "stackexchange",
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Proving that $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for every natural $n$
Prove that $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for every natural $n$
Of course, should be done by induction. Base case ($n=1$) is easy.
I got stuck with the step:
Let's assume that it's right for $n$. Then,
$$2^{2\cdot 3^{n-1}} = 1 + 3^n \pmod{3^{n+1}}$$
We want to evaluate $$2^{2 \cdot 3^n} \pmod {3^{n+2}}$$
I think we can somehow utilize the fact that $2,3$ are coprime and reduce $3^{n+2}$ to $3^{n+1}$.
I'd be glad for help on that.
Thanks!
| $\begin{align}2^{2\cdot3^{n-1}}&=(3-1)^{2\cdot3^{n-1}} \\\\ &=3^{n+1}\cdot k-2\cdot3^n+1 \\\\ &=3^{n+1}\cdot k-(3-1)\cdot3^n+1 \\\\ &=3^{n+1}\cdot (k-1)+3^n+1 \\\\ &\equiv3^n+1\pmod{3^{n+1}}\end{align}$
| {
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} |
Finding the values that $x$ can take $\left|\frac{-10}{x-3}\right|>\:5$ $$\left|\frac{-10}{x-3}\right|>\:5$$
*
*Find the values that $x$ can take.
I know that
$$\left|\frac{-10}{x-3}\right|>\:5$$
and
$$\left|\frac{-10}{x-3}\right|<\:-5$$
| $$\left| \frac { -10 }{ x-3 } \right| >\: 5\\ \frac { 10 }{ \left| x-3 \right| } >5\\ \left| x-3 \right| <2\\ -2<x-3<2\\ 1<x<5\\ \left( 1;5 \right) -\left\{ 3 \right\} \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$. I'm looking for help with (b) and (c) specifically. I'm posting (a) for completeness.
(a) Show convergence for $a_n=\sqrt{n+1}-\sqrt{n}$ towards $0$ and test $\sqrt{n}a_n$ for convergence.
(b) Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$.
(c) For which $\alpha\in\mathbb{Q}_+$ does $n^\alpha b_n$ converge?
I'm pretty sure I solved (a). I have proven the convergence of $a_n$ by using the fact that $$\sqrt{n}<\sqrt{n+1}\leq\sqrt{n}+\frac{1}{2\sqrt{n}}$$ which holds true since $$(\sqrt{n}+\frac{1}{2\sqrt{n}})^2=n+1+\frac{1}{4n}\geq n+1\,.$$
This gives us $$0<\sqrt{n+1}-\sqrt{n}\leq\frac{1}{2\sqrt{n}}$$
and after applying the squeeze theorem with noting that $\frac{1}{2\sqrt{n}}\longrightarrow0$ we can tell that also $a_n\longrightarrow0$.
Now $x_n=\sqrt{n}a_n=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$.
We have \begin{align*}\sqrt{n}(\sqrt{n+1}-\sqrt{n})&=\sqrt{n}\sqrt{n+1}-\sqrt{n}\sqrt{n}\\&=\sqrt{n(n+1)}-n\\&=\sqrt{n^2+n}-n\\&=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\\&=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}\\&=\frac{n}{\sqrt{n^2+n}+n}\\&=\frac{n}{n\sqrt{1+\frac{1}{n}}+n}\\&=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\end{align*}
and hence since the harmonic sequence $\frac{1}{n}$ converges towards 0 we have $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$
| Hint: Use the fact that $a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+...+ab^{k-2}+b^{k-1})$ where $a=\sqrt[k]{n+1}$ and $b=\sqrt[k]{n}$
or
$$0<a-b=\frac{a^k-b^k}{(a^{k-1}+a^{k-2}b+...+ab^{k-2}+b^{k-1})}<\frac{1}{kb^{k-1}}=\frac{1}{k\sqrt[k]{n^{k-1}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$SL(2, Z_2)$ closed under multiplication How can $SL(2, Z_2)$ (i.e matrices with entries in $\mathbb{Z_2}$ and determinant 1) form a group, when:
$\begin{pmatrix} 1&0 \\ 1&1 \end{pmatrix}$ * $\begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}$ = $\begin{pmatrix} 1&1 \\ 1&2 \end{pmatrix}$ which doesn't have entries in $\mathbb{Z_2}$
and if we let $\begin{pmatrix} 1&1 \\ 1&2 \end{pmatrix} =$ $\begin{pmatrix} 1&1 \\ 1&0 \end{pmatrix}$ then $\begin{pmatrix} 1&1 \\ 1&0 \end{pmatrix}$ has determinant $-1$
So either way we have the matrices are not closed under multiplication. Where am I wrong?
| $1=-1$ in $\mathbb Z_2$, and you should reduce all entries modulo $2$, so the first matrix has a zero in the bottom right entry.
i.e: $$\begin{pmatrix}1&1\\1&2 \end{pmatrix}=\begin{pmatrix}1&1\\1&0 \end{pmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
infinity sum Bessel-Type about $\sum _{k=-\infty }^{\infty } (-i)^k \cos (k) J_k(x)$ I would like to propose a problem following the sequence of series could find the general term when Co(x)^n it seem easy but i can not find
$$\sum _{k=-\infty }^{\infty } (-i)^k \cos (k) J_k(x)=\cos (x \cos (1))-i \sin (x \cos (1))$$
$$\sum _{k=-\infty }^{\infty } (-i)^k \cos ^2(k) J_k(x)=\frac{1}{2} (\cos (x)+\cos (x \cos (2))-i (\sin (x)+\sin (x \cos (2))))$$
$$\sum _{k=-\infty }^{\infty } (-i)^k \cos ^8(k) J_k(x)=\frac{1}{128} (35 \cos (x)+56 \cos (x \cos (2))+28 \cos (x \cos (4))+8 \cos (x \cos (6))+\cos (x \cos (8))-i (35 \sin (x)+56 \sin (x \cos (2))+28 \sin (x \cos (4))+8 \sin (x \cos (6))+\sin (x \cos (8))))$$
| We wish to find an expression for
$$\sum^\infty_{k = -\infty} (-i)^k \cos^n (k) J_k (x),$$
where $J_k (x)$ is the Bessel function of the first kind of order $k$, $i$ is the imaginary unit, and $n \in \mathbb{N}$.
To do this we will make use of the Jacobi-Anger expansion
$$e^{i x \cos \phi} = \sum^\infty_{k = -\infty} i^k J_k (x) e^{ik \phi}.$$
As
$$\cos (k) = \frac{1}{2} (e^{ik} + e^{-ik}),$$
on applying the binomial theorem we have
$$\cos^n (k) = \left (\frac{e^{ik} + e^{-ik}}{2} \right )^n = \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} e^{i(2m - n)k}.$$
Thus
\begin{align*}
\sum^\infty_{k = -\infty} (-i)^k \cos^n (k) J_k (x) &= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \sum^\infty_{k = -\infty} (-i)^k J_k (x) e^{i(2m - n)k}\\
&= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \sum^\infty_{k = -\infty} i^k (-1)^k J_k (x) e^{i(2m - n)k}\\
&= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \sum^\infty_{k = -\infty} i^k J_k (-x) e^{i(2m - n)k}, \quad {\rm since} \,\, J_k(-x) = (-1)^k J_k (x)\\
&= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} e^{-ix \cos (2m - n)},
\end{align*}
where in the last line we have made use of the Jacobi-Anger expansion. Hence
$$\sum^\infty_{k = -\infty} (-i)^k \cos^n (k) J_k (x) = \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \Big{(}\cos [x \cos (2m - n)] - i \sin [x\cos (2m - n)] \Big{)},$$
and is the desired expression you seek.
| {
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"source": "stackexchange",
"question_score": "1",
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Solve the diophantine equation $5^m + n^2=3^p$
Solve the diophantine equation: $5^m + n^2=3^p$ where $m,n,p \in
\mathbb{N}-\{0\}$
One solution is $m=1,n=2,p=2$. Now, applying modulo 4:
$1 + 0 = (-1)^p \mod 4 \tag 1$ or
$1 + 1 = (-1)^p \mod 4 \tag 2$
But only (1) is possible, therefore both $n, p$ are even.
Also, applying modulo 3:
$(-1)^m + n^2 = 0 \mod 3 \tag 3$ therefore $m$ is odd and $n=3k \pm 1$
I could not get any further.
| Following lulu, to prove there are no solution with $p\geq 3$, we only need to prove the equation $$2(3^k) - 5^m = 1$$
has no solution with $k\geq 2$.
Note that $5$ is a primitive root modulo $3^2$, hence $5$ is a primitive root modulo $3^n$ for all positive integer $n$. Thus
$$5^m \equiv -1 \pmod{3^{k}} \implies \frac{\phi(3^{k})}{2}\mid m \implies 3^{k-1}\mid m $$
Therefore $(1)$ implies, for some $r\geq 1$
$$1 = 2({3^{k}}) - 5^{r(3^{k-1})} \leq 2({3^{k}}) - {5^{{3^{k-1}}}}$$
a contradiction, because RHS is negative when $k\geq 2$.
Therefore, the only positive integer solution of $5^m + n^2=3^p$ is $m=1,n=2,p=2$.
| {
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"source": "stackexchange",
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Solving $\sin x - \cos x = 1$ I have been trying to solve $\sin x - \cos x = 1$ by squaring both sides but has not been able to obtain the solution. Here is what I did:
$$\begin{align}(\sin x - \cos x)^2 &=1^2\\\sin^2x-2\sin x\cos x+\cos^2x&=1\\1-\sin 2x&=1\\\sin2x&=0\end{align}$$
Obviously $x=0$ is not a solution. May I ask why this is the case or where did things go wrong?
Thank You
| Note that : $$\sin x - \cos x = \sqrt2 \sin(x - \frac{\pi}{4})$$ In this case, $$\sin x - \cos x = \sqrt2 \sin(x-\frac{\pi}{4})=1 \implies \sin(x-\frac{\pi}{4})=\frac{1}{\sqrt2}$$
Thus, two possible solutions are:
*
*$$x - \frac{\pi}{4}=\frac{\pi}{4} \implies x = \frac{\pi}{2}+2k\pi \, ,k \in \mathbb{Z}$$
*$$x -\frac{\pi}{4}=\frac{3\pi}{4} \implies x = \pi + 2k\pi \, , k \in \mathbb{Z}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Evaluating $\left(\frac{-2}{p}\right)$
Prove that for $p\ge 3$, a prime: $$\left(\frac{-2}{p}\right) = \begin{cases}
1 & p\equiv1,3\pmod{8}\\
-1 & p\equiv-1,-3\pmod{8}
\end{cases}$$
I already now that:
$$\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p^{2}-1}{8}}=(-1)^{\frac{(p-1)(p^{2}-1)}{8}}=(-1)^{\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\frac{p-1}{4}}$$
And I'm trying to show when the exponent is even (for the $1$ case).
We have:
*
*$\frac{p-1}{4} = 2k \implies p- 1 = 8k \implies p = 8k + 1$. So we can infer that if $p\equiv 1\pmod{8}$ then $\left( \frac{-2}{p} \right)=1$
*$\frac{p-1}{2}$ - doesn't yield anything new to us (I think)
*$\frac{p+1}{2} = 2t \implies p+1 = 4t \implies p = 4t-1$
but I don't really see how to infer the other option $(p \equiv 3 \pmod{8})$
| If $p\equiv 3\pmod{8}$ then $\frac{p-1}{2}$ is even and $\frac{p^2-1}{8}$ is odd, so $\left(\frac{-2}{p}\right)=-1$.
Here you may find a proof avoiding quadratic reciprocity and exploiting some field theory.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\int\limits_0^\infty x \left \lfloor{\frac1x}\right \rfloor \, dx$ This is an integral I computed but can't find the result online or on wolfram. So here's a proof sketch, please indulge this sanity check:
$$\int_0^\infty x \left \lfloor{\frac1x}\right \rfloor \ dx = \int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx$$
$$= \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} nx \ dx =\sum_{n=1}^\infty\frac n2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) $$
$$=
\sum_{n=1}^\infty\frac n2 \left(\frac{2n+1}{n^2(n+1)^2}\right)$$
$$= \sum_{n=1}^\infty\frac{1}{(n+1)^2} + \frac12 \sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$
$$= \frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right)$$
$$=\frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1}\right) -\frac12\left(\sum_{n=1}^\infty\frac{1}{(n+1)^2}\right)$$
$$= \left(\frac{\pi^2}{6} -1\right) + \left(\frac12\cdot 1\right) - \frac12\left(\frac{\pi^2}{6} -1\right)$$
$$= \frac{\pi^2}{12}.$$
Basically, I used the Basel sum several times, and the fifth line follows from a partial sum decomposition. The seventh follows from the known result for the Basel sum, as well as the fact that the first series in the 6th line telescopes.
I hope this is all correct.
| Alternatively, one may follow the same line of thought and use summation by parts formula to calculate the infinite sum. Similarly,
$$\int_0^{+\infty} x\lfloor \frac{1}{x}\rfloor dx =\sum_{n=1}^{\infty}\frac{n}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n)$$
where $f_n = n$ and $g_n = 1/n^2$. Now, summation by parts for the last expression gives:
$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 -\sum_{n=2}^{\infty}\frac{1}{n^2}\right)$$
But it is well-known that
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
Hence,
$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 - (\frac{\pi^2}{6}-1)\right) = \frac{\pi^2}{12}$$
| {
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Fallacy in showing that $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ Given a cyclic quadrilateral $ABCD$, $AB=a, BC=b, CD=c, \angle ABC=120^\circ, \angle ABD=30^\circ$, then, show that $$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$
I tried to do it using some trig bashing.
What I did was basically assigned the angle $\angle BCA = \theta$ and used trigonometry.
It's not hard to see that the radius of the circle is $\frac{c}{2}$.
I've squared the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$.
And using $\text {Sine Rule}$, the equation can be reduced further.
I used
*
*$\frac{a}{\sin\theta}=c$
*$\frac{b}{\sin(60^\circ - \theta)}=c$
And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.
Then I continued to reduce it using the addition-subtraction formulae of trigonometry.
And at the end of the day what I get is $$\boxed{\cos(30^\circ-\theta)=\frac{1}{2}}$$ after all those addition-subtraction of trigonometric equations.
And that's not true I believe.
May I get rectified?
| You assigned the angle $\angle BCA = \theta$ and used trigonometry.
You got the radius of the circle as $\frac{c}{2}$.
Square the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$.
And using $\text {Sine Rule}$, the equation can be reduced further.
*
*$\frac{a}{\sin\theta}=c$
*$\frac{b}{\sin(60^\circ - \theta)}=c$
And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.
Reduce it using the addition-subtraction formulae of trigonometry.
$$1+2(\sin\theta+\sin(60-\theta) = 2\sqrt{(1+\sin\theta)(1+\sin(60-\theta)}$$
On reduction we get
$$2m+1 =2\sqrt{m+m^2+\frac{1}{4}}$$
Where $m=\cos(30-\theta)$
On squaring and simplifying we get
$$4m^2+4m+1=4m^2+4m+1$$ which is always true.
Hence the statement is proved.
| {
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Prove that $L\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\}=\frac{1}{s(\sqrt {s+1})}$ Question:
Prove that $L\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\}=\frac{1}{s(\sqrt {s+1})}$
Note: $L$ stands for Laplace transform.
My try:
$$L\left\{\frac{2}{\sqrt \pi} \int_0^{\sqrt t}e^{-u^2}du\right\}=\frac{2}{\sqrt \pi} L\left\{\int_0^{\sqrt t}e^{-u^2}du\right\} \tag{$*$} $$
Now, According to the rule $L\{\int_0^x f(u)du\}=\frac{1}{s}L\{f(x)\}$, I assumed $f(u)=e^{-u^2}$. From what it seems, $x$ is replaced with $\sqrt t$. So, We have $L\{\int_0^{\sqrt t}e^{-u^2}du\}=\frac{1}{s} L\{f(\sqrt t)\}$ .
Since $f(u)=e^{-u^2}$, We have $f(\sqrt t)=e^{-t}$.
So, by all of these, $(*)$ is equal to $\frac{2}{\pi} \frac{1}{s} L\{e^{-t}\}=\frac{2}{\pi} \frac{1}{s}\frac{1}{s+1}$ which is not the thing that the question claims.
What did I do wrong?
| Note that the function: $$f(t) = \frac{2}{\sqrt \pi} \int_{0}^{\sqrt t} e^{-u^2} \, du$$ is written as $\operatorname{erf}(\sqrt t)$. Now, we can use the series representation of the integrand to get: $$\begin{align} \operatorname{erf}(\sqrt t) = \frac{2}{\sqrt \pi} \int_{0}^{\sqrt t} \left[ 1- \frac{u^2}{1!}+ \frac{u^4}{4!} - \frac{u^6}{3!}+…\right]\, du\\ =\frac{2}{\sqrt \pi} \left[\frac{\sqrt t}{1}- \frac{t^{3/2}}{3\times 1!} + \frac{t^{5/2}}{5\times 2!} - \frac{t^{7/2}}{7\times 3!} + … \right]\end{align}$$ Thus, we have: $$\begin{align} L[\operatorname{erf}(\sqrt t)] = \frac{2}{\sqrt \pi} L\left[\frac{\sqrt t}{1}- \frac{t^{3/2}}{3\times 1!} + \frac{t^{5/2}}{5\times 2!} - \frac{t^{7/2}}{7\times 3!} + … \right] \\ = \frac{2}{\sqrt \pi} \left[ \frac{\Gamma(3/2)}{s^{3/2}} - \frac{\Gamma(5/2)}{s^{5/2}\times 3\times 1!} + \frac{\Gamma(7/2)}{s^{7/2}\times 5\times 2!} - …\right]\\ = \frac{1}{s^{3/2}}\left[1-\frac{1}{2s}+\frac{(-1/2)(-3/2)}{2!}\frac{1}{s^2} + \frac{(-1/2)(-3/2)(-5/2)}{3!}\frac{1}{s^3} + …\right] \\ = \frac{1}{s^{3/2}}\left(1+\frac{1}{s}\right)^{\frac{-1}{2}} = \frac{1}{s\sqrt{1+s}} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| $$\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\frac{x-1}{\sqrt{x}}\left(1-\frac{\sqrt{x}}{\sqrt{x+1}}\right)=\frac{x-1}{\sqrt{x}}\left(1-\frac{1}{\sqrt{1+\frac1x}}\right)=\frac{x-1}{\sqrt{x}}\left(1-\left({1+\frac1x}\right)^{-\frac12}\right)=\frac{x-1}{\sqrt{x}}\left(1-1+\frac1{2x}+o\left(\frac1{x}\right)\right)=\frac{x-1}{\sqrt{x}}\left(\frac1{2x}+o\left(\frac1{x}\right)\right)=\frac{x-1}{x\sqrt{x}}\left(\frac1{2}+o(1)\right)\to0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
} |
Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
By taylor polynomials we get: $e^{x^4}=1+x^4+\frac{x^8}{2}+\mathcal{O}(x^{12})$
$\sin(x^2)=x^2-\frac{x^6}{6}+\mathcal{O}(x^{10})$
$\cos(x^3)=1-\frac{x^6}{2}+\mathcal{O}(x^{12})$
so putting these together:
$$ \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)} = \frac{x^2+x^6+\frac{x^{10}}{2}-x^2+\frac{x^6}{6}-\mathcal{O}(x^{10})}{\frac{x^6}{2}-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}$$
Now I am not too familiar with the Big-Oh notation for limits so I am stuck here.
How does arithmetic work with them, can I simplify the oh's in the numerator and can I take $x$'s out?
| From here you with some adjustment conclude:
$$...=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6++\mathcal{O}(x^{10})}{\frac{1}{2}x^6+\mathcal{O}(x^{10})}=\frac73+\mathcal{O}(x^{4})\to\frac73$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve (in)equality of three variables with trigonometric solutions I'm working through a set of inequality problems and I'm stuck on the following question:
Find all sets of solutions for which $$(a^2+b^2+c^2)^2=3(a^3b+b^3c+c^3a)$$ holds. Note that $a,b,c\in\mathbb{R}$.
Firstly, one can easily see that when $a=b=c$ then the equality holds: $$\text{LHS}=(a^2+a^2+a^2)^2=(3a^2)^2=9a^4$$ and $$\text{RHS}=3(a^4+a^4+a^4)=3(3a^4)=9a^4.$$
A hint is given to use the substitutions $a=x+2ty$, $b=y+2tz$ and $c=x+2tz$ for real $t$. The LHS is rather nice in that it simplifies to $$(4t(xy+xz+yz)+(1+4t^2)(x^2+y^2+z^2))^2$$ but I can't find a similar simplification for the RHS. I guess this is a type of uvw question but I don't know where to start.
There are actually four sets of solutions: $$a=b=c,$$ $$\frac{a}{\sin^2\frac{4\pi}7}=\frac{b}{\sin^2\frac{2\pi}7}=\frac{c}{\sin^2\frac{\pi}7},$$ $$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7},$$ and $$\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$$ I have no idea how the trigonometric expressions are obtained. How could the equality be solved?
| Hint:
Take the square root of all terms in $\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$.
Now you obtain a case of a well known equation: the sine rule.
$\frac {4\pi}7 + \frac {2\pi}7 +\frac {\pi}7= \pi$, so this is a triangle with side lengths $\sqrt a, \sqrt b$ and $\sqrt c$ and angles $\frac {\pi}7, \frac {2\pi}7 $ and $ \frac {4\pi}7$.
Hope you can take it from here!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $a^4+b^4+c^4$ The problem:
The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?
Based on the above information, we have:
\begin{align}
a + b + c &= 6 \\
a^2 + b^2 + c^2 & = 8 \\
a^3 + b^3 + c^3 & = 5 \\
\end{align}
I had a feeling that this vaguely had to do with Viete's theorem, which states for a cubic polynomial $f(x) = x^3 - px^2 + qx - r$ which has roots $\alpha , \beta , \gamma$,
\begin{align}
p & = \alpha+\beta+\gamma \\
q & = \alpha \beta+\alpha\gamma+\beta\gamma \\
r & = \alpha\beta\gamma
\end{align}
Notice that we already have $p$, because $a+b+c=6=\alpha + \beta + \gamma = p$. Then to find $q$:
\begin{align}
2q& = 2\alpha\beta+2\alpha\gamma+2\beta\gamma \\
& = [(a+b+c)^2-(a^2+b^2+c^2)] \\
& = (a^2+ab+ac+b^2+ab+bc+c^2+ac+bc)-a^2-b^2-c^2 \\
& = 2ab+2ac+2bc
\end{align}
\begin{align}
q & = ab+ab+bc \\
& = \frac{1}{2}[(a+b+c)^2-(a^2+b^2+c^2)] \\
& = \frac{1}{2}[6^2-8] \\
& = 14
\end{align}
So now we have $f(x) = x^3-6x^2+14x-r$. And it follows that $f(\alpha)=f(\beta)=f(\gamma)=0$.
\begin{align}
f(\alpha) & = \alpha^3-6\alpha^2+14\alpha-r = 0\\
f(\beta) & = \beta^3-6\beta^2+14\beta-r = 0\\
f(\gamma) & = \gamma^3-6\gamma^2+14\gamma-r = 0\\
\end{align}
\begin{align}
0 & = f(\alpha) + f(\beta) + f(\gamma) \\
& = (\alpha^3+\beta^3+\gamma^3)-6(\alpha^2+\beta^2+\gamma^2)+14(\alpha+\beta+\gamma)-3r\\
& = 5-6(8)+14(6)-3r\\
3r& = 41\\
r & = \frac{41}{3} \\
\end{align}
Now we have that $f(x)=x^3-6x^2+14x-\frac{41}{3}$. Here's where I am stuck. Of course, the above working out was the culmination of hours of trying things out, and eventually we have this equation. If you are reading this now, I'd appreciate if you gave any hints as to how I should continue the problem.
I've had the idea of multiplying $f(x)$ by $x$ to get fourth powers, but I haven't tried that yet. Perhaps that might yield some results?
| Let set $\begin{cases}S_1=a+b+c=6\\S_2=a^2+b^2+c^2=8,\\S_3=a^3+b^3+c^3=5\end{cases}$
$S_4=\frac 16\left({S_1}^4+8\,S_3\,S_1-6\,S_2\,{S_1}^2+3\,{S_2}^2\right)=0$
I found the relation by removing terms successively, if you are interested, here were my steps...
$A=S_4-S_3S_1\\
B=S_2S_1^2-S_4\\
C=2A+B\\
D=S_1^4-6C\\
E=D+3S_2^2-4S_4\\
F=E-4S_3S_1+4S_4=0$
If I develop lab bhattacharjee's answer with my notations we have:
$ab+bc+ca = \frac 12({S_1}^2-S_2)$
$abc = \frac 16({S_1}^3-3S_1S_2+2S_3)$
$a^2b^2+b^2c^2+c^2a^2=\frac 1{12}(-{S_1}^4+6{S_1}^2S_2+3{S_2}^2-8S_1S_3)$
And we arrive finally at the same relation $6{S_2}^2-6S_4=-{S_1}^4+6{S_1}^2S_2+3{S_2}^2-8S_1S_3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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For $x,y$ in G, prove that $x = y =e$ if $xy^2=y^3x $ and $yx^2=x^3y$ For $x,y$ in G, prove that $x = y =e$ if $xy^2=y^3x $ and $yx^2=x^3y$ where $e$ is the identity element in G.
I have filled pages trying to solve but unable to reach the solution.
Also, Is there any general method to approach such questions?
| Let $xy^2=y^3x$ say $(1)$ and $yx^2=x^3y$ say $(2)$
From the relation $(1)$, we have $xy^2x^{−1}=y^3$ say $(3)$.
Computing the power of $n$ of this equality yields that
$xy^{2n}x^{−1}=y^{3n}$ for any $n\in \mathbb{N}$.
In particular, we have $xy^4x^{−1}=y^6$ and $xy^6x^{−1}=y^9$..
Substituting the former into the latter, we obtain
$x^2y^4x^{−2}=y^9,$ say $(4)$
Cubing both sides gives $x^2y^{12}x^{−2}=y^{27}$.
Using the relation $(3)$ with $n=4$, we have $xy^8x^{−1}=y^{12}$
Substituting this into equality $(4)$ yields $x^3y^8x^{−1}=y^{27}$
Now we have
$$y^{27}=x^3y^8x^{−1}=(x^3y)y^8(y^{−1}x^{−3})=yx^2y^8x^{−2}y.$$
Squaring the relation $(4)$, we have $x^2y^8x^{−2}=y^{18}$.
Substituting this into the previous, we obtain $y^{27}=y^{18}$, and hence
$$y^9=e$$
Note that as we have $xy^2x^{−1}=y^3$, the elements $y^2,y^3$ are conjugate to each other.
Thus, the orders must be the same. This observation together with $y^9=e$ imply $y=e$.
It follows from the relation $(2)$ that $x=e$ as well.
| {
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"url": "https://math.stackexchange.com/questions/2589770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $x^3+x^2=1$, express the infinite product $(1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$ in the form $A+Bx+Cx^2$.
Given that $x^3+x^2=1$ and $x\in\mathbb{R}$, express the infinite product $$(1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$$
in the form $A+Bx+Cx^2$.
In the earlier parts of the question, I have already shown that
$$x^4=-1+x+x^2$$
$$x^{-1}=x+x^2$$
$$1-x+x^2-x^3+x^4-x^5+\ldots=x^2$$
$$\frac{1}{1-x}=2+2x+x^2$$
I also know that
$$1+x=\frac{1}{1-x+x^2-x^3+x^4-x^5+\ldots}.$$
Can anyone give me a hint? Perhaps there is a way to do it using the previous parts, but I cannot see how.
| Hint. One may recall that
$$
(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots=\frac1{1-x}, \qquad |x|<1,
$$ then insert
$$
x=\frac1{x+x^2}
$$ in the preceding identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Decomposing a particular fraction I am working on Problem I.4.7 in Lang's Complex Analysis. It asks to find the convergence of
\begin{equation}\sum_\limits{n=1}^\infty \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}.\end{equation} The hint says to multiply and divide each term by $1-z$ and do a partial fraction decomposition to get a telescoping sum. I tried this and got
\begin{equation}\frac{A}{(1-z^n)(1-z)} + \frac{B}{(1-z^{n+1})(1-z)} + \frac{C}{(1-z^n)(1-z^{n+1})} = \frac{z^{n-1}-z^n}{(1-z^n)(1-z^{n+1})(1-z)}.\end{equation}
By inspection, I found that
\begin{equation}
\begin{split}
A&= 0\\
B&=-\frac{1}{z}+1 \\
C&= \frac{1}{z}
\end{split}
\end{equation}
is a solution. This however, does not give me a telescoping sum. The resources I have found only talk about partial fraction decomposition after reducing to irreducible quadratics, which does not seem tenable here. Are there any more general techniques to partial fraction decomposition might help here?
Edit: Here is the decomposition the above obtains, in case it is useful:
\begin{equation}
\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}=\frac{1}{(1-z^{n+1})(1-z)} - \frac{1}{z(1-z^{n+1})(1-z)} + \frac{1}{z(1-z^{n+1})(1-z^n)}
\end{equation}
|
The hint says to multiply and divide each term by $1-z$ and do a partial fraction decomposition
Not entirely sure what the hint meant to say, but the following telescopes nicely regardless:
$$
\begin{align}
\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} &= \frac{z^{n-1}}{(1-z)^2(1+z+\ldots+z^{n-1})(1+z+\ldots+z^{n-1}+z^n)} \\[5px]
&= \frac{z^{n-1}}{(1-z)^2} \cdot \frac{1}{z^n}\left(\frac{1}{1+z+\ldots+z^{n-1}} - \frac{1}{1+z+\ldots+z^{n-1}+z^n}\right)
\end{align}
$$
Then:
$$
\begin{align}
z(1-z)^2\sum_\limits{n=1}^{N} \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} &= \sum_\limits{n=1}^{N} \left(\frac{1}{1+z+\ldots+z^{n-1}} - \frac{1}{1+z+\ldots+z^n}\right) \\[5px]
&= 1 - \frac{1}{1+z+\ldots+z^N} \\[5px]
&= 1 - \frac{1-z}{1-z^{N+1}} \\[5px]
&= \frac{z(1-z^N)}{1-z^{N+1}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find pairs of $(x, y)$ satisfying the given equation. Find all pairs $(x, y)$ of integers such that $$xy+\frac{x^3+y^3}{3}=2007$$
$\mathbf{Try:}$
The given expression can be simplified as $$x^3+y^3+3xy=6021$$
$$\Rightarrow (x+y-1)[(x-y)^2+(y+1)^2+(x+1)^2]=12040$$
I tried considering the divisors of $12040$ but seems to be a much tedious task. Now how to proceed further.
| $$\frac{x^3+y^3}{3}=\frac{(x+y)^3-3xy(x+y)}{3},$$ which says that $(x+y)^3$ is divided by $3$ and from here $x+y$ is divided by $3$,
but it gives that $xy$ is divided by $3$.
Thus, $x$ and $y$ are divided by $3$.
Let $x=3a$ and $y=3a$, where $a$ and $b$ are integers.
Hence, we obtain $$ab+a^3+b^3=223.$$
It's a bit of easier.
Also, we have
$$(x+y-1)(x^2+y^2+1-xy+x+y)=6020=4\cdot5\cdot7\cdot43$$
and $x+y-1\equiv2(\mod3),$ which gives not so many cases because $$x^2+y^2+1-xy+x+y>0.$$
Now, it's obvious that $x+y>0$ and
$$6020=(x+y-1)(x^2+y^2-xy+x+y+1)=(x+y-1)((x+y)^2-3xy+1)\geq$$
$$\geq(x+y-1)\left((x+y)^2-\frac{3}{4}(x+y)^2+x+y+1\right)=\frac{1}{4}(x+y-1)(x+y+2)^2,$$
which gives
$$x+y\leq27.$$
We see that $x+y-1=14$ or $x+y-1=20$ only.
If $y=15-x$ then $$x^2+(15-x)^2+1+15-x(15-x)=430,$$
which has no integer roots.
If $y=21-x$ then
$$x^2+(21-x)^2+1+21-x(21-x)=301$$ or
$$(x-3)(x-18)=0,$$
which gives the answer:
$$\{(3,18),(18,3)\}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem of existing matrices
Let $n,k \geq 2$ be two integers. Prove that there must exist some $n \times n$ invertible, non-diagonal matrices: $X_1,X_2, \dots , X_k$ with real entries, such that $$X_1^{-1}+X_2^{-1}+\dots+X_k^{-1}=(X_1+X_2+\dots+X_k)^{-1}$$
Since this is a problem which asks to construct the matrices, I tried to make it less general. I tried to find some matrices with $X_i^2=I_n$ and hence $X_i^{-1}=X_i$, so what would be left to take care of is the equality $(X_1+X_2+\dots+X_k)^2=I_n.$
For the case when $k$ is odd, I chose $X_1$ to be the matrix with $1$ on the antidiagonal and $0$ anywhere else; then for $X_i=(-1)^iX_1,\: i \geq 2$ the problem is solved.
However, I have problems when $k$ is even. I couldn't find such matrices even when $k=2$.
| Here's a method that uses complex numbers. Let us note the association
\begin{align}
a+ib \longleftrightarrow
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}.
\end{align}
Hence it suffices to consider solutions to
\begin{align}
\frac{1}{z_1+z_2} = \frac{1}{z_1} + \frac{1}{z_2} \implies -1 = \frac{z_1}{z_2} + \frac{z_2}{z_1} = z + \frac{1}{z}.
\end{align}
In particular, let us focus on solutions lying on the unit circle. Hence we get
\begin{align}
-1 = e^{i\theta} + e^{-i\theta} = 2\cos \theta
\end{align}
which means $\theta = \frac{2\pi}{3}$ is a solution, that is,
\begin{align}
\frac{z_1}{z_2} = e^{i\frac{2\pi}{3}} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}.
\end{align}
Let us choose $z_1 = i$ and $z_2 = \frac{\sqrt{3}}{2}-\frac{i}{2}$, then the corresponding matrices
\begin{align}
X_1 =
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\ \ \ \text{ and } \ \
X_2 =
\begin{pmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2}\\
-\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix}
\end{align}
satisfy the relation
\begin{align}
(X_1+X_2)^{-1} = X_1^{-1} + X_2^{-1}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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what are the zeroes of $f(x) = 2^x + 3^x - 5^x?$ 1) How do I find the number of zeroes of such exponential equations?
2) Is there any easier way to manually plot such sum of exponential functions?
| At $x=0$, we know that the formula becomes $3^0+2^0-5^0=1+1-1=1$, and when $x<0$ $3^x>5^x$ and $2^x>5^x$. Therefore, $2^x+3^x>5^x$, and there are no roots for $x<0$. For $x>0$, we know that $2^x<3^x<5^x$, so at some point $5^x=2^x+3^x$ and at point $x$ there will be a zero. That just happens to be $x=1$ since $2^x+3^x-5^x=2^1+3^1-5^1=2+3-5=5-5=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If 2 matrices are such that $(A+B)^k=A^k+B^k$ for $k=2,3$, show that $(A+B)^m=A^m+B^m $ for all $m \in \mathbb{N}$ Let $A,B \in M_n(C) $. The matrix $A-B$ is invertible and $(A+B)^k=A^k+B^k $, $k \in {2,3} $. Prove that $(A+B)^m=A^m+B^m $ for every $m \in N $.
PS. I obtained $AB+BA=0$ and $A^2B+B^2A=0$, but I need your help, please :(
| You already showed
$$
AB + BA = 0\\
A^2B+B^2A = 0
$$
Continuing from there,
\begin{align*}
&A^2B+B^2A = 0\\[4pt]
\implies\;&A(AB)+B(BA) = 0\\[4pt]
\implies\;&A(AB)+B(-AB) = 0\\[4pt]
\implies\;&(A-B)(AB) = 0\\[4pt]
\implies\;&AB = 0&&\text{[since $A-B$ is invertible]}\\[4pt]
\end{align*}
Note that $A-B\;$invertible implies $B-A\;$invertible. Thus, the hypothesis is symmetric in the variables $A,B.\;$Hence, by switching the variables, we get $BA=0$.
It follows that
$$(A+B)^m = A^m + B^m$$
for all $m \ge 2,\;$since in the expansion of
$$(A+B)^m = (A+B)(A+B)\cdots(A+B)$$
all the "middle terms" vanish.
Trivially, $(A+B)^m = A^m + B^m$ holds for $m=1,\;$hence
$$(A+B)^m = A^m + B^m$$
for all positive integers $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Choosing a random number from 1, 2 or 3, and summing them, what is the probability that the sum is 4 at one point? For example we get 1+1+1+2 which can't be 4 after this point or 1+3 which is 4 after the second number chosen.
So we don't need to sum more than 4 random numbers since the minimum sum equal to for is 1+1+1+1. So for the 4 random numbers that get chosen we have $3^4$ combinations in total. There are 7 ways to sum 4 numbers or less to 4 which are $$ 1+1+1+1$$ $$1+1+2$$ $$1+2+1$$ $$2+1+1$$ $$2+2$$ $$3+1$$ $$1+3$$ so there should be a probability of $\frac{7}{81}$ which for some reason isn't the right answer. What did I get wrong?
| Lord Shark is correct. The probability of $2+2$ is not $\frac{1}{81}$ but $\frac{1}{9}.$
More generally, if you select numbers from $1$ to $n$ and want the probability that the sum at some point is exactly equal to $k$, this is the coefficient of $x^k$ in:
$$\sum_{i=0}^{\infty} \left(\frac{x+x^2+\cdots +x^n}{n}\right)^i=\frac{n(1-x)}{n-(n+1)x+x^{n+1}}$$
So if $a_k$ is the coefficient of $x^k$ then $$a_{n+1+k}=\frac{1}{n}\left((n+1)a_{n+k}-a_{k}\right)$$
We also get $a_0=1,a_1=\frac{1}{n}$ and for $2\leq k\leq n$ we get $a_{k}=\frac{1}{n}\left(1+\frac{1}{n}\right)^{k-1}$.
In particular, then, $a_{n+1}=\frac{1}{n}\left(\left(1+\frac{1}{n}\right)^n-1\right)$
For $n=3$, this gives $a_4=\frac{4^3-3^3}{3^4}=\frac{37}{81}$.
When $n=3$, we get the exact formula:
$$a_{k}=\frac{1}{2}+\frac{1}{4}\left(\left(\frac{-1+\sqrt{-2}}{3}\right)^k+\left(\frac{-1-\sqrt{-2}}{3}\right)^k\right)$$
Which gives us an estimate:
$$\frac{1}{2\cdot 3^k}\leq\left|a_k-\frac{1}{2}\right|<\frac{1}{2\cdot 3^{k/2}}$$
For general $n$ we have $\lim_{k\to\infty} a_k=\frac{2}{n+1}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
On commuting matrices Consider the complex matrix $$A=\begin{pmatrix}i+1&2\\2&1\end{pmatrix}$$ and the linear map $$f:M(2,\mathbb{C})\to M(2,\mathbb{C}),\qquad X\mapsto XA-AX.$$
I want to find a basis of $\ker f$.
I already know the canonical basis $\{E_{11},E_{12},E_{21},E_{22}\}$ and computed $$f(E_{11})=\begin{pmatrix}0&2\\-2&0\end{pmatrix},f(E_{12})=\begin{pmatrix}2&0\\0&-2\end{pmatrix},f(E_{21})=\begin{pmatrix}-2&0\\0&2\end{pmatrix},f(E_{22})=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$$
Does this help to find the basis?
| You have already found$$f(E_{11})=\begin{pmatrix}0&2\\-2&0\end{pmatrix},f(E_{12})=\begin{pmatrix}2&0\\0&-2\end{pmatrix},f(E_{21})=\begin{pmatrix}-2&0\\0&2\end{pmatrix},f(E_{22})=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$$Note that $$f(E_{11})+f(E_{22})=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$ Similarly $$ f(E_{12})+f(E_{21})=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$ Thus $ E_{12}+E_{21}$and $ E_{12}+E_{21}$ belong to the Ker(f).
Therefore, $$\{E_{11}+E_{22},E_{12}+E_{21}\}$$ =$$\{ \begin{pmatrix}1&0\\0&1\end{pmatrix}, \begin{pmatrix}0&1\\1&0\end{pmatrix}\} $$is a basis for the Ker(f)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
For what values of $p$ does the series $\sum_{n=1}^{\infty}\left(1-n\sin{\frac{1}{n}}\right)^p$ converge? For what values of $p$ is the series $$\sum_{n=1}^{\infty}\left(1-n\sin{\frac{1}{n}}\right)^p \quad \text{converge}?$$
This is my professors solution: Note that
$$1-n\sin{\frac{1}{n}}=\frac{1}{6n^2}+O(1/n^4),$$
so the series is a positive serie for every $p$. Furthermore:
$$\frac{\left(1-n\sin{\frac{1}{n}}\right)^p}{\frac{1}{n^{2p}}}=\frac{\left(\frac{1}{6n^2}+O(1/n^4)\right)^p}{\frac{1}{n^{2p}}}\rightarrow\frac{1}{6^p}, \ \text{as} \ n\rightarrow\infty.$$
Since $\sum_{k=1}^{\infty}1/n^a$ converges iff $a>1$ it follows that our series converges iff $p>1/2$ according to the limit comparison test.
Questions:
Can someone elaborate on the steps here and break down things a bit?
*
*How did he know that he should choose $1/n^{2p}$ to compare?
*How does that final limit tend to $1/6^p$?
| For point 1
Note that for $n$ large $$a_n\sim \frac{1}{n^{2p}}$$
thus we choose this “tail” for the ratio test.
For point 2
$$\frac{\left(\frac{1}{6n^2}+O\left(\frac{1}{n^4}\right)\right)^p}{\frac{1}{n^{2p}}}=\left(\frac{n^2}{6n^2}+O\left(\frac{n^2}{n^4}\right)\right)^p=\left(\frac{1}{6}+O\left(\frac{1}{n^2}\right)\right)^p$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Uniform convergence of $n\sin\sqrt{4\pi^2n^2+x^2}$ on $[0,a]$ and $\mathbb{R}.$ Let $f_n(x)=n\sin\sqrt{4\pi^2n^2+x^2}$. Prove that $(f_n)$ is uniformly convergent on $[0,a]$ for every $a>0.$ Is the convergence uniform on $\mathbb{R}$?
Attempt. For $x\in \mathbb{R}$ constant we have:
$$f_n(x)=n\sin(\sqrt{4\pi^2n^2+x^2}-2\pi n)=n\sin\bigg(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\bigg)=
\frac{\sin\bigg(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\bigg)}{\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}}\,\frac{n\,x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\rightarrow 1\cdot \frac{x^2}{4\pi}=\frac{x^2}{4\pi},$$
so we are done with the pointwise convergence.
So far I am stuck on the uniform convergence, in other words how to prove $$\max_{0\leq x\leq a}|f_n(x)-x^2/(4\pi)|\rightarrow 0$$ and $$\sup_{x\in \mathbb{R}}|f_n(x)-x^2/(4\pi)|=\sup_{x\geq 0}|f_n(x)-x^2/(4\pi)|\nrightarrow 0.$$
Thanks for the help.
| I want to offer you two different ways… choose which you prefer:
1.) Take $y \ge 0$ s.t. $$y^2 = 2\pi^2n^2 + x^2$$
Then $$x \ge 0 \iff y \ge 2\pi n$$
And we get:
$$\begin{align*}\sup_{x\geq 0}\left|f_n(x)-\frac{x^2}{4\pi}\right| &= \sup_{y \ge 2\pi n} \left|n\sin(y) - \frac{y^2 - 4\pi^2n^2}{4\pi}\right| \\\\&\ge \sup_{\substack{y = 2kn\pi \\ k\in \Bbb N}} \frac{4k^2n^2\pi^2 - 4\pi^2n^2}{4\pi} \\\\ &=\sup_{\substack{y = 2kn\pi \\ k\in \Bbb N}}(k^2-1)\pi n^2 \\ &= \infty\end{align*}$$
2.) Another way: We have $$-n \le f_n(x) \le n$$ and so $$-n -\frac{x^2}{4\pi} \le f_n(x)-\frac{x^2}{4\pi} \le n -\frac{x^2}{4\pi}$$
Hence we can conclude $$\lim_{x\to \infty} f_n(x) -\frac{x^2}{4\pi} = -\infty$$
And so we get directly: $$\sup_{x\geq 0}\left|f_n(x)-\frac{x^2}{4\pi}\right| = + \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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