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Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{1}{2}n^2-\frac{3}{4}n+\frac{9}{8}.$$
I am just wondering if someone could check my work to make sure I have the procedure correct.
| Given the homogenous solution: $a_n^{(h)} = c_03^n$ and the particular solution: $a_n^{(p)} = c_1n^2+c_2n+c_3$, and we have: $a_n = a_n^{(h)}+a_n^{(p)}$. First we find the coefficients to the particular solution:
$$
\begin{align}
&a_n^{(p)} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 3a_{n-1}^{(p)} &&= n^2+0n-3\\
\iff &c_1n^2+c_2n+c_3 -3\left(c_1(n-1)^2+c_2(n-1)+c_3\right) &&= n^2+0n-3\\
\iff &c_1n^2+c_2n+c_3 -3(c_1n^2-2c_1n+c_1+c_2n-c_2+c_3) &&= n^2+0n-3\\
\iff &\color{red}{c_1}n^2\color{green}{+c_2}n\color{blue}{+c_3}\color{red}{-3c_1}n^2\color{green}{+6c_1}n\color{blue}{-3c_1}\color{green}{-3c_2}n\color{blue}{+3c_2}\color{blue}{-3c_3} &&= n^2+0n-3\\
\iff &\color{red}{(-2c_1)}n^2+\color{green}{(6c_1-2c_2)}n+\color{blue}{(3c_1-3c_2+4c_3)} &&= \color{red}{1}n^2\color{green}{+0}n\color{blue}{-3}\\
\end{align}
$$
This gives three equations:
$$
\begin{matrix}
\color{red}{-2c_1}&&&= \color{red}{1}\\ \color{green}{6c_1}&\color{green}{-2c_2}&&= \color{green}{0}\\ \color{blue}{3c_1}&\color{blue}{-3c_2}&\color{blue}{+4c_3}&= \color{blue}{3}
\end{matrix}
$$
Solving this system yields:
$$
\begin{matrix}
c_1&&&=-\frac{1}{2}\\ &c_2&&= -\frac{3}{2}\\ &&c_3&= 0
\end{matrix}
$$
So we have $a_n^{(p)} = - \frac{1}{2}n^2-\frac{3}{2}n+0$
This means a solution is given by $a_n = c_03^n - \frac{1}{2}n^2-\frac{3}{2}n$, and given the initial condition $a_0 = 1$, we can solve this for $c_0$:
$a_0 = 1 = c_03^0 - \frac{1}{2}0^2 - \frac{3}{2}0 \iff c_0 = 1$
This means $a_n = 3^n - \frac{1}{2}n^2-\frac{3}{2}n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How can I prove the last two digits of $1+2^{2^{n}}+3^{2^n}+4^{2^n}$ always are $54$ How can I prove the last two digits of $$1+2^{2^{n}}+3^{2^n}+4^{2^n}$$ are $54$ when $n$ is a positive integer number if $n>1$
| These things go in cycles with periods of length $4$.
The last two digits of $4^{2^n}$ are $16$, $56$, $36$ or $96$.
For $n > 4$, the last two digits of $3^{2^n}$ are $41$, $81$, $61$ or $21$.
For $n > 4$, the last two digits of $2^{2^n}$ are $96$, $16$, $56$ or $36$.
Then notice that $16 + 41 + 96 + 1 = 154$, and therefore if $n > 4$ and $n \equiv 1 \pmod 4$ then $4^{2^n} + 3^{2^n} + 2^{2^n} + 1 \equiv 54 \pmod{100}$. You should be able to work out $n \equiv 2 \pmod 4$ and the other cases yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Taylor Series of $2xe^x$ I have to find the Taylor Series for $2xe^x$ centred at $x=1$. I came up with the following.
$$e^x = e^{x-1} \times e = e \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg)$$
Then consider $2xe^x$.
$$2xe^x = 2xe \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg) = \bigg( \sum_{n=0}^\infty \frac{2xe(x-1)^n}{n!}\bigg) $$
I am wondering whether this is a right Taylor Series centred at $x=1$. I do understand that the $(x -1)^n$ implies that it is centred at $1$. But, I am very not sure with my answer since there are two terms of $x$, that is $x$ and $(x-1)^n$.
Any clarification and explanation would be highly appreciated.
| Write $2xe^x=2(x-1)e^x+2e^x$ and $e^x=e\cdot e^{x-1}$ to obtain $2xe^x=2e(x-1)e^{x-1}+2e\cdot e^{x-1}$. Since
$$
e^{x-1}=\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k,
$$
we have
\begin{align*}
2xe^x&=2e(x-1)e^{x-1}+2e\cdot e^{x-1} \\ &= 2e(x-1)\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k+2e
\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k \\ &=
2e\left(\sum_{k=1}^\infty\frac{1}{(k-1)!}(x-1)^k+\sum_{k=1}^\infty\frac{1}{k!}(x-1)^k+1\right) \\ &=
2e\left(1+\sum_{k=1}^\infty\left(\frac{1}{(k-1)!}+\frac{1}{k!}\right)(x-1)^k\right) \\ &=
2e\sum_{k=0}^\infty\frac{k+1}{k!}(x-1)^k.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Using an Integral to Solve for a Variable a I am struggling to use the following equation:
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = 0
$$
where $a > 1$, to deduce that $a = \text{cosec}(\frac{\pi}{4} - \frac{\alpha}{2})$, where $\alpha$ satisfies $\alpha = \cos(\alpha)$.
I integrate the integrand, via
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = -\int_0^1 \sqrt{a^2-x^2}\, dx + \int_1^a \sqrt{a^2-x^2}\, dx
$$
But once I calculate those integrals I cannot seem to get any closer to the answer.
Any help would be great.
| First, $a>1$ and $x\in[0,a]$, then $|x|=x\ge0$
$$
\renewcommand\sgn{\operatorname{sgn}}
\renewcommand\arcsec{\operatorname{arcsec}}
\begin{array}{ll}
0\!\!\!&=\int_0^a\sqrt{a^2-x^2}\sgn(x-1)\,\mathrm dx\\
&=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx-\int_0^1\sqrt{a^2-x^2}\,\mathrm dx
\end{array}
$$
Then
$$\int_1^a\sqrt{a^2-x^2}\,\mathrm dx=\int_0^1\sqrt{a^2-x^2}\,\mathrm dx\tag{1}$$
But
$$\int_1^a\sqrt{a^2-x^2}\,\mathrm dx=\text{Area of }DAB\tag{2}$$
$$\int_0^1\sqrt{a^2-x^2}\,\mathrm dx=\text{Area of }ODBC\tag{3}$$
and area of $DAB+$ area of $ODBC=\dfrac14$area disc$(a)=\dfrac{\pi a^2}4$
Then $(1),(2),(3)$ yield
$$2\text{area of }DAB=\dfrac{\pi a^2}4\implies\text{ area of }DAB=\dfrac{\pi a^2}8=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx\tag{★}$$
Moreover: Area of sector $OAB=\pi\theta= \pi\arccos\left(\frac1a\right)= \pi\arcsec a$
On the other hand, Area of sector $OAB=$ Area of triangle $ODB+$ Area of $DAB$.
Then,
$$\boxed{\displaystyle\pi\arcsec a=\dfrac12\sqrt{a^2-1}+\int_1^a\sqrt{a^2-x^2}\,\mathrm dx}\tag{★★}$$
From $($★$)$ and $($★★$)$ it follows that: $a$ is a solution of the equation
$$\boxed{\displaystyle\pi\arcsec a=\dfrac12\sqrt{a^2-1}+\frac{\pi a^2}{8}}$$
I transformed the initial problem into a simpler one (I hope!)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Express this linear optimization problem subject to a circular disk as a semidefinite program I have to express the following problem as a semidefinite program
$$\begin{array}{ll} \text{minimize} & F(x,y) := x + y +1\\ \text{subject to} & (x-1)^2+y^2 \leq 1 \tag{1}\end{array}$$
Only affine equality conditions should be used. The hint was to examine the structure of $\mathbb{S}^2_+$, the cone of symmetric positive semidefinite matrices.
The characteristic polynomial of such a matrix is
$$C=\lambda^2-(a_{11}+a_{22})\lambda - a_{12}a_{21}$$
which has a similar form to the rewritten condition (1) $x^2-2x+y^2\leq0$. If $a_{11}=a_{22}=1,\lambda = x$ and $a_{12}=-a_{21}=y$ the characteristic polynomial would be $x^2-2x+y^2=0$. Is this useful?
My problem is, that I have no clue how to formulate a $\leq$ with equality conditions.
| First of all, let's simplify a bit:
$$(x-1)^2+y^2\leq 1 \quad\Longleftrightarrow\quad x^2-2x+1+y^2\leq 1
\quad\Longleftrightarrow\quad x^2+y^2\leq 2x$$
Note that this implies that $x\geq 0$, which makes sense since the original inequality describes a disc centered at $(1,0)$ with radius $1$.
Let's go a bit further: add $2xy$ to both sides and rewrite:
$$x^2+2xy+y^2\leq 2x+2xy
\quad\Longleftrightarrow\quad (x+y)^2 \leq 2x(y+1)$$
Now we have the implication that $2x(y+1)\geq 0$, which combined with the previous gives us $y+1\geq 0$ or $y\geq -1$. Again, this follows from the geometry of the disc, but both of these facts are important for the next step.
Now let's look at the structure of a $2\times 2$ semidefinite cone. A matrix is positive semidefinite iff all of its principal minors are nonnegative. For the $2\times 2$ case, that simply means that the diagonal elements are nonnegative, and the determinant is nonnegative:
$$\begin{bmatrix} z_{11} & z_{12} \\ z_{12} & z_{22} \end{bmatrix} \succeq 0
\quad\Longleftrightarrow\quad z_{11} \geq 0, ~ z_{12} \geq 0, ~ z_{11}z_{12} \geq z_{12}^2$$
Hmm, there is some familiar structure there. Let's do some substitution:
$$\begin{bmatrix} 2x & x+y \\ x+y & y+1 \end{bmatrix} \succeq 0
\quad\Longleftrightarrow\quad 2x \geq 0, ~ y+1 \geq 0, ~ 2x(y+1) \geq (x+y)^2$$
Bingo! Now we can write out the solution by inspection.
$$\begin{array}{ll}
\text{minimize} & x + y + 1 \\
& \begin{bmatrix} z_{11} & z_{12} \\ z_{12} & z_{22} \end{bmatrix} \succeq 0 \\
& 2x = z_{11} \\
& x+y = z_{12} \\
& y+1 = z_{22}
\end{array}$$
Another way to do it would be to eliminate $x$ and $y$ altogether, and recover them after the fact from the equations $x=z_{11}/2$, $y=z_{22}-1$:
$$\begin{array}{ll}
\text{minimize} & z_{11} / 2 + z_{22} \\
& \begin{bmatrix} z_{11} & z_{12} \\ z_{12} & z_{22} \end{bmatrix} \succeq 0 \\
& z_{11} / 2 - z_{12} + z_{22} = 1
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an integral table I could find was:
$$\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln \left|2ax + b +2\sqrt{a\left({ax^2+bx+c}\right)}\right|$$
Which would in my case give:
$$\frac{dx}{\sqrt{x^2-x}} = \ln \left|2x -1 + 2\sqrt{x^2-x}\right|$$
Which has me struggling with the absolute value signs as I need to extract x from the solution. All I know is that $x<0$ which does not seem to help me either (the square root will only be real if $x<-1$).
Is there some other formula for solving this integral which does not involve absolute value signs or which makes extracting $x$ from the solution somewhat easier? Thanks!
| $$
\displaylines{
\sqrt {x^2 - x} = t + x \cr
\Leftrightarrow x^2 - x = t^2 + 2xt + x^2 \cr
\Leftrightarrow - x = t^2 + 2xt \cr
- x = \frac{{t^2 }}{{1 + 2t}} \Rightarrow - dx = \frac{{2\left( {t^2 + 1} \right)}}{{\left( {1 + 2t} \right)^2 }}dt \cr
\frac{{dx}}{{\sqrt {x^2 - x} }} = - \frac{{\frac{{2\left( {t^2 + 1} \right)}}{{\left( {1 + 2t} \right)^2 }}dt}}{{t - \frac{{t^2 }}{{1 + 2t}}}} \cr
= \frac{{2\left( {t^2 + 1} \right)dt}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} \cr
\frac{{2\left( {t^2 + 1} \right)}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} = \frac{\alpha }{t} + \frac{\beta }{{t + 1}} + \frac{\lambda }{{1 + 2t}} \cr}
$$
$$\displaylines{
\frac{{2\left( {t^2 + 1} \right)}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} = \frac{\alpha }{t} + \frac{\beta }{{t + 1}} + \frac{\lambda }{{1 + 2t}} \cr
= \frac{{\lambda \left( {t^2 + t} \right) + \beta \left( {2t^2 + t} \right) + \alpha \left( {2t^2 + 3t + 1} \right)}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} \cr
= \frac{{t^2 \left( {\lambda + 2\beta + 2\alpha } \right) + t\left( {\lambda + \beta + 3\alpha } \right) + \alpha }}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} \cr
\Rightarrow \left\{ \begin{array}{l}
\alpha = 2 \\
\lambda + \beta + 6 = 0 \\
.\lambda + 2\beta + 4 = 2 \\
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\alpha = 2 \\
\beta = 4 \\
\lambda = - 10 \\
\end{array} \right. \cr}
$$
$$
\int {\frac{{2\left( {t^2 + 1} \right)dt}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}}} = 2\ln \left( t \right) + 4\ln \left( {1 + t} \right) - 5\ln \left( {1 + 2t} \right)
$$
$$\int {\frac{{dx}}{{\sqrt {x^2 - x} }}} = \ln \left( {\frac{{\left( { - x + \sqrt {x^2 - x} } \right)^2 \left( {1 + \left( { - x + \sqrt {x^2 - x} } \right)} \right)^4 }}{{\left( {1 + 2\left( { - x + \sqrt {x^2 - x} } \right)} \right)^5 }}} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 1
} |
Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$
I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{2^n - 2^{n-1}}{\sum_{i = 2^{n-1} + 1}^{2^n}(a + ib)}\\
\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{(2^n - 2^{n-1})^2}{\frac{2^n -2^{n-1}}{2}(2a + (2^n + 2^{n-1} + 1)b)}\\
&=\frac{2^{n+1} - 2^n}{2a + (2^n + 2^{n-1} + 1)b}\end{align}$$
after which I am more or less stuck. How can I continue on from here, or is there another method?
| Assuming $a,b\gt0$, we get
$$
\begin{align}
\sum_{i=2^{n-1}+1}^{2^n}\left(\frac1{a+ib}-\frac{2^{-n+1}}{a+2b}\right)
&\ge\sum_{i=2^{n-1}+1}^{2^n}\left(\frac1{a+2^nb}-\frac{2^{-n+1}}{a+2b}\right)\\
&=\sum_{i=2^{n-1}+1}^{2^n}\frac{a(1-2^{-n+1})}{(a+2^nb)(a+2b)}\\
&=\frac{a(2^{n-1}-1)}{(a+2^nb)(a+2b)}\\[12pt]
&\ge0
\end{align}
$$
Therefore,
$$
\sum_{i=2^{n-1}+1}^{2^n}\frac1{a+ib}\ge\frac1{a+2b}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integration involving $\sin x/x$ Let $f$ be a differentiable function satisfying:
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0$$
and $f(\pi)=0$, Considering $g(x)=f(x)\forall x\in\mathbb R_0=\mathbb R+{0}$.
If $$\int_0^{\infty}(g(x))^3dx=A$$ and $$\int_0^{\infty}\frac{1-g(x)}{x^2}dx=\frac{kA}{k+1}$$ then k is?
First I did :
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0\\\int_0^{f(x)}f^{-1}(t)dt+\int_0^xf(t)dt=\int_0^x\cos tdt\\xf(x)=\sin x$$
So $f(x)=\dfrac{\sin x}x$
But how can someone calculate?
$$\int_0^{\infty}\frac{\sin^3x}{x^3}dx$$
NB, limit to highschool level
|
But how can someone calculate $\quad\quad\quad\displaystyle\int_0^{\infty}\frac{\sin^3x}{x^3}dx$
Assuming that you know $$\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$$
Use$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$
And then Integrate by parts
For example
$$\int\frac{\sin^3xdx}{x^3}=\int\frac{3\sin xdx}{4x^3}-\int\frac{\sin xdx}{4x^3}$$
$$\int\frac{\sin xdx}{x^3}=-\frac{\sin x}{2x^2}-\int\frac{\cos x}{2x^2}dx$$
$$\int\frac{\cos x}{x^2}dx=-\frac{\cos x}{x}+\int\frac{\sin x}{x}dx$$
This method seems reasonable as $\text{Mathematica } 10.0.1.0 $ yields follwing Indefinite Integral
$$\int\frac{\sin^3x}{x^3}dx=\frac{-3 x^2 \text{Si}(x)+9 x^2 \text{Si}(3 x)-4 \sin ^2(x) (\sin (x)+3 x \cos (x))}{8 x^2}+c$$
Pat your back if you get
$$\large\int_0^{\infty}\frac{\sin^3x}{x^3}dx=\frac{3\pi}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
| Notice that $\tan x\times \cot\ x=1$, and
according to the inequality of means we get $$\tan x+\cot x \geqslant 2\sqrt{\tan x\times \cot x}=2$$
which is what we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Integral of constant divided by polynomial and another constant $$\int_{-\infty}^{-1}\frac{4}{\sqrt{x^6+2}}\,dx$$
What are the steps to integrate?
| $\int_{-\infty}^{-1}\dfrac{4}{\sqrt{x^6+2}}dx$
$=\int_\infty^1\dfrac{4}{\sqrt{(-x)^6+2}}d(-x)$
$=\int_1^\infty\dfrac{4}{\sqrt{x^6+2}}dx$
$=\int_1^\sqrt[6]2\dfrac{4}{\sqrt{x^6+2}}dx+\int_\sqrt[6]2^\infty\dfrac{4}{\sqrt{x^6+2}}dx$
$=\int_1^\sqrt[6]2\dfrac{2\sqrt2}{\sqrt{1+\dfrac{x^6}{2}}}dx+\int_\sqrt[6]2^\infty\dfrac{4}{x^3\sqrt{1+\dfrac{2}{x^6}}}dx$
$=\int_1^\sqrt[6]22\sqrt2\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{6n}}{4^n(n!)^22^n}dx+\int_\sqrt[6]2^\infty\dfrac{4}{x^3}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^n}{4^n(n!)^2x^{6n}}dx$
$=\int_1^\sqrt[6]2\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{6n}}{2^{3n-\frac{3}{2}}(n!)^2}dx+\int_\sqrt[6]2^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-6n-3}}{2^{n-2}(n!)^2}dx$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{6n+1}}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}\right]_1^\sqrt[6]2+\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-6n-2}}{2^{n-2}(n!)^2(-6n-2)}\right]_\sqrt[6]2^\infty$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+\frac{1}{6}}}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{-n-\frac{1}{3}}}{2^{n-1}(n!)^2(3n+1)}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-\frac{5}{3}}(n!)^2(6n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-\frac{2}{3}}(n!)^2(3n+1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Compute $\int \frac{\mathrm{d}x}{49x^2+1}$ So I tried solving this by taking a substitute for the integrand, $t=49x$, so its derivative is $dx = \frac {dt} {49}$. Then you insert it into the integrand and get $$\int \frac{\mathrm dt}{49(t^2 +1)} = \frac{1}{49}\int\frac{1}{t^2 +1}dt = \frac{1}{49}\arctan t + c = \frac{1}{49}\arctan 49x + c $$
Why is this not correct? Why do I have to take substitute $t=7x$ and then do the thingy? I don't know what I'm doing incorrect.
| \begin{align}
& \int\frac{dx}{49x^2+1} = \int\frac{dx}{(7x)^2+1} = \frac 1 7\int \frac{7\,dx}{(7x)^2+1} = \frac 1 7 \int\frac {dt}{t^2+1} \\[10pt]
= {} & \frac 1 7 \arctan t + C = \frac 1 7 \arctan(7x)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to prove $\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$ I want to prove the following and wanted to ask, if my proof is correct.
$$\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$$
Remark (i): For $n \in \mathbb{N}$ and $n > 1$ holds: $2 < (1+\frac{1}{n})^n$
Define $b_n := \left\{\frac{2^n n!}{n^{n}}\right\}_{n \in \mathbb{N}}$.
Then $\left| \frac{b_{n+1}}{b_n} \right| = \left| \frac{\frac{2^{n+1} (n+1)!}{(n+1)^{n+1}}}{\frac{2^n n!}{n^{n}}} \right| = 2 \left(\frac{n}{n+1}\right)^n = 2 \left(1-\frac{1}{n+1}\right)^n = \frac{2 \left(1-\frac{1}{n+1}\right)^n \cdot (1+\frac{1}{n})^n}{(1+\frac{1}{n})^n} = \frac{2}{(1+\frac{1}{n})^n} \overbrace{<}^\text{(i)} 1$ holds for $n \in \mathbb{N}$ and $n > 1$
Hence: $\lim \limits_{n \rightarrow \infty} \left|\frac{b_{n+1}}{b_n}\right| < 1$, which yields: $\lim \limits_{n \rightarrow \infty} b_n = 0$
Define $a_n := \left\{\frac{2^n n!}{n^{n+1}}\right\}_{n \in \mathbb{N}}$.
Obviously $0 \leq a_n \leq b_n$ holds for all $n \in \mathbb{N}$.
Hence with the sandwich-theorem: $\lim \limits_{n \rightarrow \infty} 0 = \lim \limits_{n \rightarrow \infty} b_n = \lim \limits_{n \rightarrow \infty} a_n = \lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$
| I present an elementary solution.
Write $$\ell = \lim_{n \to \infty} \frac {2^n n!} {n^{n + 1}} = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n} \frac {2k} {n}.$$ For $k < n/2$, write $2k/n = 1 - a_k$, where $0 < a_k < 1$. Then, obviously, for $k > n/2$ we'll have $2k/n = 1 + a_k$. Hence, $$\ell = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n/2} (1 - a_k^2) = \lim_{n \to \infty} \frac {1} {n} \cdot \lim_{n \to \infty} \prod_{k = 1}^{n/2} (1 - a_k^2) = 0 \cdot 0 = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Non-linear equation system Resolution I want to prove (I am not sure it is true though) that
$$\left\{\begin{array}{c} x-y-x\left( x^2+5y^2\right)=0\\ x+y-y\left( x^2+y^2\right) =0\end{array}\right.$$
Admits no other solution in the plane than the origin.
==================================================================
I tried:
Let's consider $ \left( x,y\right)\in \mathbb{R}^2-\{ \left( 0,0\right)\} $
$y\neq0$ otherwise $x=0$
So I can change second equation as such:
$$ x^2+y^2= \frac{x+y}{y} $$
But that does not lead me anywhere actually.
| As you have tried $x^2+y^2= \frac{x+y}{y}$, and then I tried, $x^2+5y^2=\frac{x-y}{x}$.. Now, from that I 've found $y^2=-\frac{x^2+y^2}{4xy}$. But $x^2+y^2>0$[assuming $\left( x,y\right)\in \mathbb{R}^2-\{ \left( 0,0\right)\}$]. So, $xy>0$.
So, $x>0 \leftrightarrow y>0$ and $x<0 \leftrightarrow y<0$
Now, remark that $x^2+5y^2>x^2+y^2$. So, $\frac{x-y}{x}>\frac{x+y}{y}$.
Now, we have two cases : $(1)x>0$ and $y>0$ (2)$x<0$ and $y<0$
Consider first case,
$\frac{x-y}{x}>\frac{x+y}{y}$
$xy-y^2>xy+x^2$
$x^2+y^2<0$... which is not possible.
Now, consider second case, $\frac{x-y}{x}>\frac{x+y}{y}$
$[\frac{x-y}{x}]x<[\frac{x+y}{y}]x$ [because $x<0$]
$[\frac{x-y}{x}]xy>[\frac{x+y}{y}]xy$ [because $y<0$]
$y(x-y)>x(x+y)$
$xy-y^2>x^2+xy$
$x^2+y^2<0$ Which is not possible. So, We have $x=0$ and $y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use Laplace Transform to solve the following IVP: I know that this is a somewhat simple problem but I have been having trouble coming up with the little "tricks" that help with Laplace.
The problem is:
$y''+2y' +5y = e^{-t}\sin(2t)$ where $y(0) = 2, y'(0) = -1$
Attempt at Solution
$(s^2+2s+5)Y = \frac{2}{(s+1)^2+4} + (s+2)(2) + 1(-1)$
$Y = \frac{2}{(s^2+2s+5)^2}+\frac{2s+3}{s2+2s+5}$
$Y = \frac{2}{(s^2+2s+5)^2}+2(\frac{s+1}{s^2+2s+5})+\frac{1}{s^2+2s+5}$
$Y = \frac{2}{(s^2+2s+5)^2}+\frac{1}{2}\sin(2t)+2e^{-t}\cos(2t)$
And I am stuck here.
The answer given in the book is as follows:
Y = $\frac{5}{8}e^{-t}\sin(2t)+2e^{-t}\cos(2t)-\frac{1}{4}te^{-t}\cos(2t)$
Any help would be greatly appreciated.
| You need the inverse Laplace transform of
$$ \dfrac{2}{(s^2 + 2 s + 5)^2}$$
This may help:
$$ {\mathcal L} \{t e^{at} \cos(bt)\} = \dfrac{(s-a)^2 - b^2}{((s-a)^2+b^2)^2}
= \dfrac{1}{(s-a)^2 + b^2} - \dfrac{2b^2}{((s-a)^2+b^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving $\sin\theta -1=\cos\theta $
Solve$$\sin\theta -1=\cos\theta $$
Steps I took to solve this:
$$\sin^{ 2 }\theta -2\sin\theta +1=1-\sin^2\theta $$
$$2\sin^{ 2 }\theta -2\sin\theta =0$$
$$(2\sin\theta )(\sin\theta -1)=0$$
$$2\sin\theta =0, \sin\theta -1=0$$
$$\quad \sin\theta =0, \sin\theta =1$$
$$\theta =0+\pi k, \theta =\frac { \pi }{ 2 } +2\pi k$$
Why is $\theta =0+\pi k$ wrong?
| use that $$\sin(\theta)=2\,{\frac {\tan \left( \theta/2 \right) }{1+ \left( \tan \left( \theta
/2 \right) \right) ^{2}}}
$$
and $$\cos(\theta)={\frac {1- \left( \tan \left( \theta/2 \right) \right) ^{2}}{1+
\left( \tan \left( \theta/2 \right) \right) ^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Mixed Conditioning - Two Normal Distributions
Let $Z \sim \mathcal{N}(0,1)$ and $Y|Z \sim \mathcal{N}(Z, 1)$.
Show that $f_{Z|Y}(z|y)$ is a normal density, and find the parameters of this density.
What I have so far:
\begin{align*}
f_Z(z) &= \phi(z)\\
f_{Y|Z}(y|z) &= \frac{1}{\sigma} \phi\left( \frac{y - \mu}{\sigma} \right) = \phi\left( y - z \right)\\
f(y,z) &= f_Z(z) \cdot f_{Y|Z}(y|z) = \phi(z) \cdot \phi\left( y - z \right)
\end{align*}
From the above, we can use Bayes' theorem express $f_{Z|Y}(z|y)$:
\begin{align*}
f_{Z|Y}(z|y) &= \frac{ f_Z(z) \cdot f_{Y|Z}(y|z)}{ f_Y(y) }\\
&= \frac{ \phi(x) \cdot \phi(y-z) }{ \int_{-\infty}^\infty f(y,z)\,\mathrm{d}z }\\
&= \frac{ \phi(x) \cdot \phi(y-z) }{ \int_{-\infty}^\infty\phi(z) \cdot \phi\left( y - z \right)\,\mathrm{d}z }\\
\end{align*}
However, I am stuck at this point. I'm not exactly sure how to compute the integral in the denominator, and then express $f_{Z|Y}(z|y)$ as a normal distribution function. Is there a better way to approach this? Thanks!
| $$
f_{Z|Y}(z|y) =
\frac{f_Z(z)f_{Y|Z}(y|z)}{\int_{-\infty}^{+\infty}{f_Z(a)f_{Y|Z}(y|a)da}} =
\frac{\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-z)^2}{2}}}
{\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-a)^2}{2}}da}} =
\frac{e^{-z^2+yz-\frac{y^2}{2}}}{\int_{-\infty}^{+\infty}{e^{-\frac{2a^2-2ya+y^2}{2}}da}}
$$
Now we'll calculate the integral in the denominator:
$$
\int_{-\infty}^{+\infty}{e^{-\frac{2a^2-2ya+y^2}{2}}da} =
\int_{-\infty}^{+\infty}{e^{-\frac{(\sqrt{2}a - \frac{\sqrt{2}}{2}y)^2 + \frac{y^2}{2}}{2}}da} =
e^{-\frac{y^2}{4}} \int_{-\infty}^{+\infty}{e^{-\frac{(\sqrt{2}a - \frac{\sqrt{2}}{2}y)^2}{2}}da}
$$
Defining $b=\sqrt{2}a$ and multiplying and dividing by $\sqrt{\pi}$ we get:
$$
e^{-\frac{y^2}{4}}\sqrt{\pi} \int_{-\infty}^{+\infty}
{\frac{e^{-\frac{(b - \frac{\sqrt{2}}{2}y)^2}{2}}}{\sqrt{2\pi}}db}
$$
Notice that we have the pdf of the normal distribution $\mathcal{N}\left(\frac{\sqrt{2}}{2}y, 1\right)$ inside the new integral. Therefore it equals 1.
Substituting
$
\int_{-\infty}^{+\infty}{e^{-\frac{2a^2-2ya+y^2}{2}}da} = e^{-\frac{y^2}{4}}\sqrt{\pi}
$
into the initial formula we have:
$$
f_{Z|Y}(z|y) = \frac{1}{\sqrt{\pi}}e^{-z^2+yz-\frac{y^2}{2}+\frac{y^2}{4}} =
\frac{1}{\sqrt{\pi}}e^{-(z-\frac{y}{2})^2} =
\frac{1}{\sqrt{2\pi} \frac{1}{\sqrt{2}}}e^{-\frac{(z-\frac{y}{2})^2}{2(\frac{1}{\sqrt{2}})^2}}
$$
That is,
$$
Z|Y \sim \mathcal{N}\left(\frac{Y}{2}, \frac{1}{\sqrt{2}}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence of a complex function I need to proof if the following function is bounded and convergent.
$f(n)=\left(\frac{10+in}{n^{2}+2in}\right)^{n}$
Status:
This should be correct. Can anybody confirm this?
I tried it with Bernoulli now:
$\lim_{n\rightarrow\infty}q^{n}=0$ for $\left|q\right|<1,\quad q\in\mathbb{C}$
$
\begin{eqnarray*}
\left|\frac{10+in}{n^{2}+2in}\right| & = & \left|\frac{10n^{2}-20in+in^{3}-2i^{2}n^{2}}{n^{4}+4n^{2}}\right|\\
& = & \left|\frac{10n^{2}-20in+in^{3}+2n^{2}}{n^{4}+4n^{2}}\right|\\
& = & \left|\frac{12n^{2}-20in+in^{3}}{n^{4}+4n^{2}}\right|\\
& = & \left|\frac{12n^{2}+in^{3}-20in}{n^{4}+4n^{2}}\right|\\
& = & \left|\frac{12n^{2}}{n^{4}+4n^{2}}+i\frac{n^{3}-20n}{n^{4}+4n^{2}}\right|\\
& = & \sqrt{\left(\frac{12n^{2}}{n^{4}+4n^{2}}\right)^{2}+\left(\frac{n^{3}-20n}{n^{4}+4n^{2}}\right)^{2}}\\
& = & \sqrt{\frac{144n^{4}}{\left(n^{4}+4n^{2}\right)^{2}}+\frac{n^{6}-40n^{4}+400n^{2}}{\left(n^{4}+4n^{2}\right)^{2}}}\\
& = & \sqrt{\frac{n^{6}+104n^{4}+400n^{2}}{n^{8}+8n^{6}+16n^{4}}}\\
& = & \sqrt{\frac{n^{2}(n^{4}+104n^{2}+400)}{n^{4}(n^{4}+8n^{2}+16)}}\\
& = & \sqrt{\frac{(n^{2}+100)(n^{2}+4)}{n^{2}(n^{2}+4)^{2}}}\\
& = & \sqrt{\frac{n^{2}+100}{n^{2}(n^{2}+4)}}\\
& = & \sqrt{\frac{1}{n^{2}}\cdot\frac{1}{n^{2}+4}\cdot(n^{2}+100)}\\
& = & \sqrt{\frac{1}{n^{2}}}\cdot\sqrt{\frac{1}{n^{2}+4}}\cdot\sqrt{n^{2}+100}\\
& = & \frac{\sqrt{\frac{n^{2}+100}{n^{2}+4}}}{n}\\
\Rightarrow\frac{\sqrt{\frac{n^{2}+100}{n^{2}+4}}}{n} & < & 1\\
n_{1} & > & \sqrt{\frac{1}{2}(\sqrt{409}-3)}\\
n_{2} & < & -\sqrt{\frac{1}{2}(\sqrt{409}-3)}
\end{eqnarray*}
$
The function converges for $n>\sqrt{\frac{1}{2}(\sqrt{409}-3)}$. Therfore it ist also bounded.
| $$f(n)=\left(\frac{10+in}{n^{2}+2in}\right)^{n}$$
Gives:
$$f(n)=\left(\frac{\sqrt{n^2+100}e^{(arg(10+in))i}}{\sqrt{n^4+4n^2}e^{(arg(n^2+2in))i}}\right)^{n}=$$
$$f(n)=\left(\frac{\sqrt{n^2+100}}{\sqrt{n^4+4n^2}}e^{((arg(10+in))-(arg(n^2+2in)))i}\right)^{n}=$$
$$f(n)=\sqrt{\frac{{n^2+100}}{{n^4+4n^2}}}^ne^{((arg(10+in))-(arg(n^2+2in)))ni}=$$
$$f(n)=\left(\frac{n^2+100}{n^2(n^2+4)}\right)^{\frac{n}{2}}e^{in\left(-iln\left(\frac{10+in}{|10+in|}\right)+iln\left(\frac{2in+n^2}{|2in+n^2|}\right)\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $ I have the following limit to evaluate:
$$ \displaystyle \lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $$
What's the trick here?
| Using Bernoulli's Inequality,
$$
\begin{align}
\left(\frac{1+\tan(x)}{1+\sin(x)}\right)^{\large1/x^2}
&=\left(1+\frac{\tan(x)-\sin(x)}{1+\sin(x)}\right)^{\large1/x^2}\\
&=\left(1+\frac{1-\cos(x)}{x^2}\frac{\tan(x)}{x}\frac1{1+\sin(x)}x^3\right)^{\large1/x^2}\\
&\ge1+\underbrace{\vphantom{\frac1{\sin(x)}}\frac{1-\cos(x)}{x^2}}_{1/2}\underbrace{\vphantom{\frac1{\sin(x)}}\frac{\tan(x)}{x}}_1\underbrace{\frac1{1+\sin(x)}}_1\underbrace{\vphantom{\frac1{\sin(x)}}x}_0\tag{1}
\end{align}
$$
$$
\begin{align}
\left(\frac{1+\sin(x)}{1+\tan(x)}\right)^{\large1/x^2}
&=\left(1-\frac{\tan(x)-\sin(x)}{1+\tan(x)}\right)^{\large1/x^2}\\
&=\left(1-\frac{1-\cos(x)}{x^2}\frac{\tan(x)}{x}\frac1{1+\tan(x)}x^3\right)^{\large1/x^2}\\
&\ge1-\frac{1-\cos(x)}{x^2}\frac{\tan(x)}{x}\frac1{1+\tan(x)}x\tag{2}
\end{align}
$$
Therefore,
$$
\left(\frac{1+\tan(x)}{1+\sin(x)}\right)^{\large1/x^2}\le\frac1{\displaystyle1-\underbrace{\vphantom{\frac1{\sin(x)}}\frac{1-\cos(x)}{x^2}}_{1/2}\underbrace{\vphantom{\frac1{\sin(x)}}\frac{\tan(x)}{x}}_1\underbrace{\frac1{1+\tan(x)}}_1\underbrace{\vphantom{\frac1{\sin(x)}}x}_0}\tag{3}
$$
Since both the right hand sides of $(1)$ and $(3)$ tend to $1$ as $x$ tends to $0$, by the Squeeze Theorem,
$$
\lim_{x\to0}\left(\frac{1+\tan(x)}{1+\sin(x)}\right)^{\large1/x^2}=1\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Finding an interval containing the values of $ abc(a + b + c) $, given a quadratic constraint Suppose $ a, b, c $ are real numbers such that $$ (ab)^2 + (bc)^2 + (ca)^2 = k $$
then all possible values of $ abc(a + b + c) $ lies between which interval?
| Note $$x^2+y^2+z^2\ge xy+yz+zx$$
then we have
$$k=(ab)^2+(bc)^2+(ac)^2\ge abc(a+b+c)$$
other hand
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)\ge 0$$
so
$$xy+yz+xz\ge -\dfrac{1}{2}(x^2+y^2+z^2)$$
so
$$abc(a+b+c)\ge -\dfrac{1}{2}[(ab)^2+(bc)^2+(ac)^2]=-\dfrac{k}{2}$$
so
$$abc(a+b+c)\in [-\dfrac{k}{2},k]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1062995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can we prove that there is only one real solution without using the discriminant?
| Substitute $x=\frac1{t-1}$ to translate $x^3-9 x^2-15x-6 =0$ into
$$6t^3-3t^2-3t-1=0$$
and then rewrite it as $7t^3=(t+1)^3$,
which yields $t=\frac1{\sqrt[3]7-1}$ and in turn the solution
$$x= \frac1{t-1} =\frac{\sqrt[3]7-1}{2-\sqrt[3]7}=\sqrt[3]{49}+\sqrt[3]7+3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
A problem with proving using definition that $\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1$
Prove using the definition that: $$\displaystyle\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1 $$
What I did:
Let $\epsilon >0$, finding $N$: $\mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid\le\mid\frac {-2}{n^2}\mid=\frac {2}{n^2}\le\frac {2}{n}$ So $N=\frac 2 {\epsilon}$.
So for all $n>N$ we'll want to show that: $|a_n-l|<\epsilon \Rightarrow \mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid<\mid\frac {-2}{N^2} \mid=\mid\frac {-2}{\frac 1 {\epsilon^2}} \mid=\mid -2\epsilon^2 \mid=2\epsilon^2$
But that isn't smaller than $\epsilon$ so I'm doing something wrong here and I don't know what...
| Hint:-
$\dfrac{n^2-1}{n^2+1}=1-\dfrac{2}{n^2+1}<1-\dfrac{1}{n^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
prove that $\sqrt{2} \sin10^\circ+ \sqrt{3} \cos35^\circ= \sin55^\circ+ 2\cos65^\circ$ Question:
Prove that: $\sqrt{2} \sin10^\circ + \sqrt{3} \cos35^\circ = \sin55^\circ + 2\cos65^\circ$
My Efforts:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$
| $$\sqrt{2}\sin10+\sqrt{3}\cos35=2(\frac{\sqrt{2}}{2}\sin10+\frac{\sqrt{3}}{2}\cos35)=2(\sin45\sin10+\cos30\cos35)$$
using the product rule for sine and cosine yields
$$2(\sin45\sin10+\cos30\cos35)=2(\frac{\cos(45-10)-\cos(45+10)}{2}+\frac{\cos(30+35)+\cos(35-30)}{2})$$$$=\cos35-\cos55+\cos65+\cos5$$
$$=\cos(90-55)+\cos65+(\cos5-\cos55)$$
$$=\sin55+\cos65+2\sin(\frac{55+5}{2})\sin(\frac{55-5}{2})$$
$$=\sin55+\cos65+2\sin30\sin25$$
$$=\sin55+\cos65+2\frac{1}{2}\sin(90-65)$$
$$=\sin55+\cos65+\cos65$$
$$=\sin55+2\cos65$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find maximum without calculus Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?
| Putting in some more details to make the answer better:
\begin{align*}
f(x) &= 2x + 2x\sqrt{1-x^2}\\
&= 2\sin(\theta) + 2\sin(\theta)\sqrt{\smash[b]{1-\sin(\theta)^2}} &\text{via \( x = \sin(\theta)\), $\theta \in (0,\frac{\pi}{2}]$}\\
&= 2\sin(\theta)+2\sin(\theta)\cos(\theta)\\
&= 2\sin(\theta) + \sin(2\theta)\\
&= 3\left(\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(2\theta)\right)\\
&= 3\left(\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(\theta)+\frac{1}{3}\sin(\pi-2\theta)\right)\\
&\leq 3\sin\left(\frac{1}{3}\theta+\frac{1}{3}\theta+\frac{1}{3}(\pi-2\theta)\right) &\text{by Jensen's Inequality}\\
&= 3\sin\left(\frac{\pi}{3}\right)\\
&= \frac{3\sqrt{3}}{2}
\end{align*}
Note that $\sin(\cdot)$ is concave down (convex) when $\theta$ is in $\left(0, \dfrac{\pi}{2}\right)$ and $\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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} |
Solve a second order nonlinear equation I have a second order nonlinear equation:
$$-u''+ \frac{1}{4}(u')^2+au=x^2.$$
I am only interested in the solutions in $[0, \frac{x^2}{a}+\frac{1}{a^2}]$. One paper claims without proof that the unique solution should be
$$u(x)=(\sqrt{a^2+4}-a)\big(\frac{x^2}{2}+\frac{1}{a}\big).$$
Does anyone have some ideas how to calculate the solution (without guessing the form of quadratic polynomial) and how to prove the uniqueness?
Thank you!
| we can trivially see that a solution could be of the form
$$
u(x) = \alpha x^2 + \beta x + \gamma \tag{1}
$$
trialing wih Eq. (1)
we find
$$
-2\alpha + \frac{1}{4}\left(2\alpha x + \beta\right)^2 + a\left(\alpha x^2 + \beta x + \gamma\right) = x^2
$$
comparing coefficents we find
$$
-2\alpha + \frac{\beta^2}{4} + a\gamma = 0\;\;\text{$x^0$ terms}\\
\alpha\beta + a\beta = \beta(\alpha + a) = 0\;\;\text{$x$ terms}\\
\alpha^2 + a\alpha = \alpha(\alpha+a) = 1\;\;\text{$x^2$ terms}
$$
From the coefficients for $x$ and $x^2$ we require $\alpha + a\neq 0$ thus $\beta = 0$
therefore
$$
-2\alpha +a\gamma = 0 \implies \gamma = \frac{2\alpha}{a}
$$
the solution Eq.(1) is of the form with a constant
$$
u(x) = 2\alpha\left(\frac{x^2}{2} + \frac{1}{a}\right)\tag{2}
$$
now to find $\alpha$
we know that
$$
\alpha^2 + a\alpha = 1 \implies \alpha^2 + a\alpha -1 = 0
$$
solutions of the above quadratic is
$$
\alpha = \frac{-a\pm\sqrt{a^2 +4}}{2}
$$
or
$$
2\alpha = -a\pm\sqrt{a^2 +4}
$$
now we can put into Eq.(2)
$$
u(x) = \left(\pm\sqrt{a^2 +4}-a\right)\left(\frac{x^2}{2} + \frac{1}{a}\right)
$$
since we are looking for solutions $>0$ at least, we can eliminate (-) from the $\pm$ above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\sin 4x +\sqrt{3} \sin 3 x + \sin 2 x=0$ This question is from a 2012 VMK entrance exam
I was trying to solve it first by expanding $\sin 4 x = 2 \sin 2 x \cos 2x$, then by noticing that if divided by 2, one can get, e.g. $ \frac{\sqrt{3}}{2} \cos 3 x = \sin \frac{\pi}{6} \cos 3 x$ and then trying to find some way to obtain $\sin (a +b)$ or $\cos (a+b)$, but without any success.
I suspect there should be some adding an extra term and then multiplying out, but can't get my head around it.
| If you don't see any trick, you can always try
\begin{align*}
&\sin(4x)=\text{Im}(\cos x+i\sin x)^4=4\cos^3\!x\sin x-4\cos x\sin^3\!x\\
&\sin(2x)=\text{Im}(\cos x+i\sin x)^3=3\cos^2\!x\sin x-\sin^3\!x\\
&\sin(3x)=\text{Im}(\cos x+i\sin x)^2=2\cos x\sin x.
\end{align*}
Plugging it in yields
$$(4\cos^3\!x-4\cos x\sin^2\!x+\sqrt3(3\cos^2\!x-\sin^2\!x)+2\cos x)\cdot\sin x=0$$
and you can solve the left factor by setting $y=\cos x$
\begin{align*}
0&=4y^3-4y(1-y^2)+\sqrt3(3y^2-(1-y^2))+2y=8y^3-4y+\sqrt3(4y^2-1)+2y\\
&=2y(4y^2-1)+\sqrt3(4y^2-1)=(2y+\sqrt3)(2y+1)(2y-1).
\end{align*}
I'm sure you could finish this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Calculating a limit of integral Computing the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$
I made the substiution $t=nx$ then, we have: $$I=\frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan t}dt$$ where $I$ is the inside integral. How would you continue?
| Since $\arctan$ is increasing we get
$$\frac{1}{ n^2}\int_{n\pi}^{2n\pi }\frac{t}{\arctan 2\pi n}dt \leq I \leq \frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan n \pi}dt$$
Now we can calculate both sides and we get
$$ \frac{4\pi ^2 - \pi^2}{2\arctan 2\pi n} \leq I \leq \frac{4\pi ^2 - \pi^2}{2\arctan \pi n} $$
So
$$ \frac{3\pi ^2 }{2\arctan 2\pi n} \leq I \leq \frac{3\pi ^2}{2\arctan \pi n} $$
At this point if you do not have to compute $I^n$ since $\lim _{x \rightarrow \infty} \arctan (x) = \pi/2$, you will get
$ \lim _{n \rightarrow \infty} I= 3\pi$.
Which implies that $\frac{I}{3\pi} \rightarrow 1$.
So the limit $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$ is of the form $1^{\infty}$. We will apply L'H rule to calculate that. We change variable and instead of $n$ we work with $t$.
By doing the standard method to apply L'H rule it suffices to calcualte
$$ \lim _{t\rightarrow \infty}t \ln ( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx ) $$
We calcualte
\begin{align*}
\lim _{t\rightarrow \infty}t \ln ( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx ) &= \lim _{t\rightarrow \infty}\frac{ \ln ( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx ) }{\frac{1}{t}} \\
&= \lim _{t\rightarrow \infty}\frac{ \frac{1} {( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx )} \frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{x}{\arctan(tx)} \right )' \ dx \ }{-\frac{1}{t^2}} \\
&= \lim _{t\rightarrow \infty}\frac{\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{x}{\arctan(tx)^2} \right ) \frac{x}{1+t^2x^2} \ dx \ }{\frac{1}{t^2}} \\
&=\lim _{t\rightarrow \infty}\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{x}{\arctan(tx)^2} \right ) \frac{t^2x}{1+t^2x^2} \ dx \\
&= \lim _{t\rightarrow \infty}\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{1}{\arctan(tx)^2} \right ) \frac{t^2x^2}{1+t^2x^2} \ dx \\
&= \lim _{t\rightarrow \infty}\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{1}{\left ( \frac{\pi}{2} \right )^2} \right )\cdot 1 \ dx \, \,\rm{by\,dominated \,convergence} \\
&=\frac{4}{3 \pi^2}
\end{align*}
So we conclude the final answer is $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n=e^{\frac{4}{3 \pi^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Prove $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $ for all $n\ge 1$ and $b,c$ are integers.
Is it possible to prove this without induction?
| Let $x_1$ and $x_2$ be the roots of the quadratic equation $$x^2 - bx + c = 0$$
Then $$x_1 = \frac{b + \sqrt{b^2 - 4c}}{2}$$ $$x_2 = \frac{b - \sqrt{b^2 - 4c}}{2} $$
The above statement is equivalent to $$2^n \,\big| \, 2^n\big({x_1}^n + {x_2}^n \big) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Integrating $\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx$ I am working on $\displaystyle\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx$
First: I use integrating by part then get
$$ x\arctan(\sin x)\Big|_0^\pi-\int_0^\pi \arctan(\sin x)\,dx $$
then I have $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$ because $x\arctan(\sin x)\Big|_0^\pi$ is equal to $0$
However, I don't know how to integrate $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$
Can someone give me a hint?
Thanks
| Split the integral at $\dfrac{\pi}{2}$, we get
$\displaystyle \begin{align} \int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx &= \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx + \int_{\pi/2}^{\pi} \frac{x\cos x}{1+\sin^2 x}\,dx \\ & = \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx - \int_0^{\pi/2} \frac{(x+\pi/2)\sin x}{1+\cos^2 x}\,dx \\ &= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx - \pi\int_0^{\pi/2} \frac{\cos x}{1+\sin^2 x}\,dx \\ &= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx - \frac{\pi^2}{4}\end{align}$
At this point we can write the above integral:
$\displaystyle \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx = \frac{\pi^2}{8} - \int_0^{\pi/2} \tan^{-1} \sin x \,dx$
The later can be evaluated in many ways. See for example Here.
See integral representations of Legendre $\chi$-function.
Also sos440 blog entry has a very nice solution to the problem.
| {
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"url": "https://math.stackexchange.com/questions/1081039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
} |
Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?
| HINT:
By taking derivative with respect to $y$ and setting it to be equal to $0$ we have :
$$\int_{0}^{y}\frac{(y-y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}=0$$
Hence:
$$\int_{0}^{y}\frac{dx}{\sqrt{x^4+(y-y^2)^2}}dx=\frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Inequality between AM-GM Prove that for $x>y>0$ $$\sqrt {xy} <\frac {x-y}{\ln x-\ln y}<\frac {x+y}2$$ Using $x=y+k$, we can turn the inequality into $$\sqrt{y^2+ky}<\frac k{\ln(1+\frac ky)}<y+\frac k2$$ Now what do i do to the $\ln\left(1+\frac ky\right) $ term? We can't approximate it using the expansion of $\ln(1+x)$ as we have no information on $k$, i.e., how big $x$ is in comparison to $y$.
Please give a hint.
| For $ 0<a<b$ note $ A(a,b)=\frac{a+b}{2}, G(a,b)=\sqrt{ab}, L(a,b)=\frac{b-a}{\ln b-\ln a}$
Proof 1.:
With $\frac{b}{a} = x > 1$ we have:
$\sqrt{x}< \frac{x - 1}{\ln x} < \frac{1}{2}( x + 1)$
Let functions $f, g : [ 1, \infty) \rightarrow \textbf{R}$:
$f( x ) = \ln x -\sqrt{x}+ \frac{1}{\sqrt{x}}, g( x ) = \ln x - \frac{2(x-1)}{x+1}$
with $f'( x ) = -\frac{(\sqrt{x}-1)^2}{2x\sqrt{x}}, g'( x ) = \frac{(x -1)^2}{x(x+1)}$ and $f(x) < 0 < g( x ),(\forall )x > 1 $.
Proof 2.:
Let $ H_{a,b}( x ): [ 0, 1 ]\rightarrow \textbf{R}$, $H_{a,b}( x ) = \frac{1}{2}( a^x.b^{1-x} + a^{1-x}.b^x )$ with $ L( a,b ) = \int^{1}_{0}H_{a,b}( x )dx, $
$H_{a,b}( 0 ) = H_{a,b}( 1 ) = A( a, b)$ and $H_{a,b}( \frac{1}{2} )= G( a, b ).$
$H^{'}_{a,b}( x ) = \frac{1}{2}( a^x.b^{1-x} - a^{1-x}.b^x )\ln \frac{b}{a}$
implies $G( a, b ) \leq H_{a,b}( x ) \leq A( a, b),(\forall) x\in [ 0, 1 ].$
In what follows we use other depictions of logarithmic mean using integral:
$\frac{1}{L( a, b )} = \frac{1}{b - a}\int^{b}_{a}\frac{dx}{x}=\frac{1}{b - a}\int^{\frac{b}{a}}_{1}\frac{dx}{x}$
$\frac{1}{L( a, b )} = \int^{1}_{0}\frac{dx}{xa + ( 1- x )b}$
$\frac{1}{L( a, b )} = \lim_{x\rightarrow 0}\int^{x}_{0}\frac{dt}{(t + a)( t + b )}$
Proof 3.:
For $(\forall)x > 0$ we have:
$\frac{2}{(x+1)^2}\leq\frac{1}{2x}\leq\frac{x+1}{4x\sqrt{x}}$
Integrating this inequality from $1$ to $\frac{b}{a}$ now obtain :
$\frac{b-a}{b+a}\leq\frac{1}{2}ln\frac{b}{a} \leq\frac{b-a}{2\sqrt{ab}}...$
Proof 4.:
For $(\forall)x \geq 0$ we have:
$x^{2} + 2x\sqrt{ab} + ab \leq x^{2} + x(a + b) + ab \leq x^{2} + x(a + b) + (\frac{a + b}{2})^2$
Reversing these inequalities and integrating them using last representation is obtained inequality.
Proof 5.:
Noting $a = e^{x}, b = e^{y}$ demonstrated inequality may be written as
$e^{\frac{x+y}{2}} \leq \frac{e^y - e^x}{y - x} \leq \frac{e^x + e^y}{2}.$
Dividing this inequality with $e^{\frac{x+y}{2}}$the first part to write:
$1\leq \frac{e^{\frac{y - x}{2}} - e^{\frac{x - y}{2}}}{y - x}\Longleftrightarrow \frac{y - x}{2} \leq \sinh\frac{y -x}{2}$ true that $t \leq \sinh t,( \forall) t\geq 0.$
Similarly, the second part is reduced to $t \geq \tanh t,( \forall) t\geq 0.$
Proof 6.:
Inequality in the previous proof can be obtained by applying Hadamard-Hermite's inequality function $f: \textbf{R}\rightarrow \textbf{R}, f( x ) = e^{x}$
(Hadamard-Hermite's inequality:
$f ( \frac{a + b}{2} )\leq\frac{1}{b - a}.\int^{b}_{a}f( x ) dx \leq \frac{f ( a ) + f ( b )}{2}.$)
Inequality of utterance can be demonstrated using CBS with integrals of convenient functions chosen.
Proof 7.:
If $f:[a, b]\rightarrow(0, \infty)$ then
$\int^{b}_{a}(\sqrt{f(x)})^2dx.\int^{b}_{a}\frac{1}{(\sqrt{f(x)})^2}dx\geq(\int^{b}_{a}(\sqrt{f(x)})^2.\frac{1}{(\sqrt{f(x)})^2}dx)^2$
and
$\int^{b}_{a}f(x)dx.\int^{b}_{a}\frac{1}{f(x)}dx\geq (b-a)^2.$
Now for $f(x)=\frac{1}{x}$ we have $(\ln b-\ln a).\frac{b^2-a^2}{2}>(b-a)^2$ ie $L ( a, b )\leq A(a, b).$
For $f(x)=e^x$ is obtained $(e^b-e^a)(e^{-a}-e^{-b})>(b-a)^2$ and noting $a=\ln x, b=\ln y$ cu $0<x<y$ result $G(x,y)< L(x,y).$
Prof 8.:
If $f:[a, b]\rightarrow(0, \infty)$ then
$\int^{b}_{a}(f(x))^2dx.\int^{b}_{a}1dx\geq(\int^{b}_{a}f(x).1dx)^2$ or $(b-a)\int^{b}_{a}(f(x))^2dx\geq(\int^{b}_{a}f(x)dx)^2.$
Now, applying this inequality for $f(x)=\frac{1}{x}$ is obtained
$(\ln b- \ln a)^2<(b-a)(\frac{1}{a}-\frac{1}{b})$
from which it follows that $G(a,b)< L(a,b).$
For $f(x)=\frac{1}{\sqrt{x}}$ rezult $4(\sqrt{b}-\sqrt{a})^2<(b-a)(\ln b-\ln a).$ Noting $\sqrt{b}=y$ and $\sqrt{a}=x$ we have
$4(y-x)<2(y^2-x^2)(\ln y-\ln x)$and finally $L(x,y)<A(x,y).$
(Paper presented at the XVIII Conference of the Mathematical Society of Romania, Iasi, 24 -26 October 2014)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$ I have these problems :
How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$
For some reason this is incorrect I'll be glad to understand why, This is what I done :
I used this formula : $(\alpha,\beta)*(\gamma,\delta)=(\alpha\gamma-\beta\delta,\alpha\delta+\beta\gamma)$
And also : $(\alpha,\beta)*(\gamma,\delta)^{-1}=(\alpha,\beta)*(\frac{\gamma}{\gamma^2+\delta^2}-\frac{\delta}{\gamma^2+\delta^2})$
*
*$$\frac{1+i}{1-i}=\\(1+i)(1-i)^{-1}=\\(1+i)(\frac{1}{2}+\frac{1}{2}i)=\\(\frac{1}{2}-\frac{i^2}{2}+\frac{1}{2}i+\frac{1}{2}i)=\\(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}i)=\\1+i\\ \\$$
*$$(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}=\\
((1+\sqrt{3i})(1-\sqrt{3i})^{-1})^{10}=\\
((1+\sqrt{3i})(\frac{1}{1+3i}+\frac{\sqrt{3i}}{1+3i}i)^{10}=\\
(\frac{1}{1+3i}-\frac{\sqrt{3i}\sqrt{3i}}{1+3i}i+\frac{\sqrt{3i}}{1+3i}i+\frac{\sqrt{3i}}{1+3i})^{10}=\\
(\frac{4}{1+3i}+\frac{\sqrt{3i}+\sqrt{3i}}{1+3i})^{10}$$
Now I want to use De-Moivre :
$$tan(args)= \frac{\frac{\sqrt{3i}+\sqrt{3i}}{1+3i})}{\frac{4}{1+3i}}=\frac{(1+3i)(\sqrt{3i}+\sqrt{3i})}{4+12i}=\frac{\sqrt{3i}-3i\sqrt{3i}+\sqrt{3i}+3i\sqrt{3i}}{4+12i}=\frac{\sqrt{3i}+\sqrt{3i}}{4+12i}$$
But I reach to math error, when trying to calculate the args.
Any help will be appreciated.
| *
*$$\left(1+i\right)\left(\frac{1}{2}+\frac{1}{2}i\right)\ne\left(\frac{1}{2}\underbrace{\color{red}{\bf -}}_{\small\text{wrong}}\frac{i^2}{2}+\frac{1}{2}i+\frac{1}{2}i\right)$$
*$$((1+\sqrt{3}i)(1-\sqrt{3}i)^{-1})^{10}=
(1+\sqrt{3}i)\left(\frac{1}{\color{red}4}+\frac{\sqrt{3}i}{\color{red}4}\right)^{10}$$
Actually both of them are easily solvable by this way:
- $$\begin{align}\frac{1+i}{1-i}&=\frac{\frac1{\sqrt2}+\frac i{\sqrt2}}{\frac1{\sqrt2}-\frac i{\sqrt2}}\\&=\frac{e^{i\pi/4}}{e^{-i\pi/4}}\\&=e^{i\pi/2}\\&=i\end{align}$$
- $$\begin{align}\left(\frac{1+\sqrt3i}{1-\sqrt3i}\right)^{10}&=\left(\frac{\frac12+\frac{\sqrt3}2i}{\frac12-\frac{\sqrt3}2i}\right)^{10}\\&=\left(\frac{e^{i\pi/3}}{e^{-i\pi/3}}\right)^{10}\\&=\left(e^{2i\pi/3}\right)^{10}\\&=e^{20i\pi/3}=e^{i(6\pi+2\pi/3)}\\&=e^{2i\pi/3}\\&=-\frac12+\frac{\sqrt3}2i\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1084725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of $\sum_{n=1}^{\infty }\frac{1}{\pi n }\sin ^k\left(\frac{2\pi n}{k}\right)$ We have: $$S_k=\sum_{n=1}^{\infty }\frac{1}{\pi n }\sin ^k\left(\frac{2\pi n}{k}\right)$$
where $k$ is an odd number greater than $1$.
I was able to find the sum of the series when $k=3,5$ as follows:
$$S_3={\sum_{n=1}^{\infty }\frac{1}{n\pi }\sin ^3\left(\frac{2\pi n}{3}\right)}=\frac{1}{8}$$
$$S_5={\sum_{n=1}^{\infty }\frac{1}{n\pi }\sin ^5\left(\frac{2\pi n}{5}\right )}=\frac{7}{32}$$
Can I get the closed-form of the other values of $S_k$ when $k=7,9,11,13,....$?
| We can treat this problem with Fourier series. For instance, since:
$$ \sin^3 x = \frac{1}{4}\left(3 \sin x - \sin(3x)\right) $$
we have:
$$\begin{eqnarray*}S_3 &=& \sum_{n\geq 1}\frac{\sin^3(2\pi n/3)}{\pi n} = \frac{1}{4}\left(3\sum_{n\geq 1}\frac{\sin(2\pi n/3)}{\pi n}-\sum_{n\geq 1}\frac{\sin(2\pi n)}{\pi n}\right)\\&=&\frac{3}{4\pi}\sum_{n\geq 1}\frac{\sin(2\pi n/3)}{n}\tag{1}\end{eqnarray*}$$
and since for any $x\not\in \pi\mathbb{Z}$ we have:
$$ \sum_{n\geq 1}\frac{\sin(nx)}{n}=\frac{1}{2}\left(\pi-(x\operatorname{mod} 2\pi)\right)\tag{2}$$
it follows that:
$$ S_3 = \frac{3}{4\pi}\cdot\frac{\pi}{6} = \color{red}{\frac{1}{8}}.\tag{3}$$
With the same approach we can compute $S_k$: we take the Fourier sine series of $\sin^k x$, then use identity $(2)$ and collect pieces. From:
$$ \sin^5 x = \frac{1}{16}\left(10\sin x-5\sin(3x)+\sin(5x)\right)$$
we get:
$$ S_5 = \frac{1}{16\pi}\left(3\pi+\frac{\pi}{2}\right)=\color{blue}{\frac{7}{32}}.$$
In general, from:
$$ \sin^{2k+1}(x) = \frac{1}{4^k}\sum_{j=0}^{k}(-1)^j\binom{2k+1}{k-j}\sin((2j+1)x)\tag{4}$$
we get:
$$ S_{2k+1}=\frac{1}{4^k \pi}\sum_{j=0}^{k-1}(-1)^j\binom{2k+1}{k-j}\frac{1}{2}\left(\pi-\frac{2(2j+1)\pi}{2k+1}\right)$$
or:
$$ S_{2k+1}=\frac{1}{2^{2k+1}}\sum_{j=1}^{k}(-1)^{k-j}\binom{2k+1}{j}\frac{4j-2k-1}{2k+1}\tag{5}$$
that further simplifies to:
$$ \color{purple}{S_{2k+1}=\sum_{n\geq 1}\frac{\sin^{2k+1}(2\pi n/(2k+1))}{n\pi}=\frac{1}{2^{2k+1}}\left((-1)^k+\binom{2k}{k}\right)}. \tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$
Then I use the partial fraction:
$$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$
So that's all. What's next step?
| $$x^4 + x^2 + 1 = 0 \implies x = \frac{1 \pm i\sqrt{3}}{2}, \frac{-1 \pm i\sqrt{3}}{2}$$
Only consider the two positive roots,
$$a, b = \frac{1 + i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2}$$
Consider a semi circle contour $C$, with radius $R$ and upper semi-circle $\Gamma$
In other words, $C = \text{line} + \Gamma$
The integral around the whole contour $C$ is given by:
$$\oint_{C} f(z) dz = (2\pi i)(\sum Res)$$
The sum of the residues is as follows: (ask if you dont understand)
$$\sum \text{Res}f(z) = \frac{-i}{2\sqrt{3}} $$
Then,
$$\oint_{C} f(z) dz = (2\pi i)(\sum Res) = (2\pi i)\cdot \frac{-i}{2\sqrt{3}} = \frac{\pi}{\sqrt{3}} $$
But realize that:
$$\oint_{C} f(z) dz = \int_{-R}^{R} f(x) dx + \int_{\Gamma} f(z) dz = \frac{\pi}{\sqrt{3}}$$
Using the M-L estimation lemma
You see that:
$$\left| \int_{\Gamma} \frac{1}{z^4 + z^2 + 1} dz \right| \le \int_{\Gamma} \left| \frac{1}{z^4 + z^2 + 1} \right|$$
You see that:
The point in polar representation is $z = Re^{i\theta}$
$$\left| \frac{1}{z^4 + z^2 + 1} \right| = \left| \frac{1}{R^4e^{4i\theta} + (R^2e^{2i\theta}) + 1} \right| $$
Since $\theta > 0$ we have:
$$\left| \frac{1}{R^4e^{4i\theta} + (R^2e^{2i\theta}) + 1} \right| = \left| \frac{1}{R^4(1) + R^2(1)) + 1} \right| = M$$
The perimeter along the semi-circle is $L(\Gamma) = \frac{1}{2} (2\pi R) = \pi R$
The Ml inequality states:
$$\left| \int_{\Gamma} \frac{1}{z^4 + z^2 + 1} dz \right| \le \int_{\Gamma} \left| \frac{1}{z^4 + z^2 + 1} \right| \le ML(\Gamma) = \frac{\pi R}{R^4(1) + R^2(1)) + 1}$$
$$\lim_{R \to \infty} \frac{\pi R}{R^4(1) + R^2(1)) + 1} = 0$$
Back to:
$$\oint_{C} f(z) dz = \int_{-R}^{R} f(x) dx + \int_{\Gamma} f(z) dz = \frac{\pi}{\sqrt{3}}$$
Take the limit as $R \to \infty$
$$\frac{\pi}{\sqrt{3}} = \int_{-\infty}^{\infty} f(x) dx + \lim_{R \to \infty} \int_{\Gamma} f(z) dz = \int_{-\infty}^{\infty} f(x) dx + 0$$
Thus,
$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^2 + 1} dx = \frac{\pi}{\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Sum $S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$ We know that
$$\sum _{k=1}^{\infty } \frac{H_k}{k^3} = \frac{\pi^4}{72}.$$
Is there a closed form for the sum
$$S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$$
Mathematica doesn't give anything resembling a closed form and I have no idea if one exists.
| Since:
$$ \sum_{k\geq 1} H_k x^k = \frac{-\log(1-x)}{1-x} $$
we have:
$$ \sum_{k\geq 1} \frac{H_k}{k+1} x^{k+1} = \frac{1}{2}\log^2(1-x), $$
$$ \sum_{k\geq 1} \frac{(-1)^{k+1}H_k}{k+1} x^{k} = \frac{1}{2x}\log^2(1+x), $$
and since $\int_{0}^{1} x^k\log(x) = -\frac{1}{(k+1)^2}$, it follows that:
$$ S=\sum_{k\geq 1} \frac{(-1)^{k}H_k}{(k+1)^3}=\int_{0}^{1}\frac{\log^2(1+x)\log x}{2x}\,dx \tag{1}$$
for the last integral, Mathematica returns:
$$ S = \frac{\pi ^4}{48}-\frac{\log^4 2}{12}+\frac{\pi^2\log^2 2}{12} -2 \operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{7}{4}\zeta(3)\log 2 \tag{2}$$
and since:
$$ \sum_{k\geq 1}\frac{(-1)^{k}\frac{1}{k+1}}{(k+1)^3}=-1+\frac{7\pi^4}{720}\tag{3} $$
it follows that:
$$ \sum_{k\geq 1}\frac{(-1)^k H_k}{k^3}=-\frac{11\pi ^4}{360}+\frac{\log^4 2}{12}-\frac{\pi^2\log^2 2}{12} +2 \operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4}\zeta(3)\log 2.\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum of infinite series $1+\frac22+\frac3{2^2}+\frac4{2^3}+\cdots$ How do I find the sum of $\displaystyle 1+{2\over2} + {3\over2^2} + {4\over2^3} +\cdots$
I know the sum is $\sum_{n=0}^\infty (\frac{n+1}{2^n})$ and the common ratio is $(n+2)\over2(n+1)$ but i dont know how to continue from here
| Once you've determined that the sum converges (you can do this by the ratio test or by integration as other users have pointed out), you can find the value quite nicely. Let
$$S=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\cdots$$
Then
$$2S=2+\frac{2}{1}+\frac{3}{2}+\frac{4}{4}+\frac{5}{8}+\frac{6}{16}+\cdots$$
From the second term on, this shares denominators with $S$ itself, so we can write
$$2S-S=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$$
But this is just a geometric series, so we get
$$S=\boxed{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
My attempt:
I have used a hit and trial approach. I found out that only for x=2 and x=3 this is happening. But how can I be sure of it? Is there any general approach?
| It's obvious that $x$ can only be the form $2^a,3^b,p^c,2^a3^b,3^bp^c$ or $2^ap^c$. I solved the case when $x=2^a3^b$ (and hope the others are similar).
First, if $a\ge 1$ and $b\ge 1$, then $2^a3^b+1$ must be the power of a prime $p$ different from $2$ and $3$. Also $2^a3^b+2=2(2^{a-1}3^b+1)$, and $2^{a-1}3^b+1$ can not be divisible by $p$ (otherwise $p|2^a3^b-2^{a-1}3^b$ or $p|2$, contradiction) and by $3$, thus $2(2^{a-1}3^b+1)$ is a power of $2$, which means $a=1$. Consider $x+3=3(2\cdot 3^{b-1}+1)$, and in a similar way we can conclude $b=1$. Then $x=6$.
Cases $a=0$ or $b=0$ is similarly solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Simplifying fractions with exponents I'm revising for an exam, which I have the solutions to. One of the questions asks me to prove that a sequence is a Cauchy sequence, sequence is written as:
$ a_n = \frac{2^{n+2}+1}{2^n},$
and then in the solutions, it has:
$\left | a_n - a_m \right | = \left | \frac{2^{n+2}+1}{2^n} - \frac{2^{m+2}+1}{2^m} \right | = \left | \frac {1}{2^n} - \frac {1}{2^m} \right |$
I am extremely rusty on my math skills, but could someone explain how the $2^{nd}$ and $3^{rd}$ absolute values are equal?
| Notice that
$$
\frac{2^{n+2}+1}{2^n} = \frac{2^{n+2}}{2^n} + \frac{1}{2^n} = 2^2 + \frac{1}{2^n} = 4 + \frac{1}{2^n}.
$$
Doing the same with the second term (the one involving $m$), we find that
$$
\left| \frac{2^{n+2}+1}{2^n} - \frac{2^{m+2}+1}{2^m} \right|
= \left| 4 + \frac{1}{2^n} - \left( 4 + \frac{1}{2^m} \right) \right|
= \left| \frac{1}{2^n} - \frac{1}{2^m} \right|.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
integration of $\large\int \frac{u^2}{(1-u^2)^2}$ $ du$ Is there a way to integrate $$\large\int \frac{u^2}{(1-u^2)^2} du$$ without using partial fraction decomposition?
| For the diversity of the solution set, I present an exotic method that works almost exclusively for this integral:
Decompose the integrand as
\begin{align*}
\frac{u^{2}}{(u^{2}-1)^{2}}
&= \frac{1}{u^{2} + u^{-2} - 2} \\
&= \frac{1}{2}\frac{1+u^{-2}}{(u-u^{-1})^{2}} + \frac{1}{2}\frac{1-u^{-2}}{(u+u^{-1})^{2} - 4} \\
&= \frac{1}{2}\frac{1+u^{-2}}{(u-u^{-1})^{2}} + \frac{1}{8} \frac{1-u^{-2}}{(u+u^{-1}) - 2} - \frac{1}{8} \frac{1-u^{-2}}{(u+u^{-1}) + 2}.
\end{align*}
By noting that
$$ \frac{d}{du}(u-u^{-1}) = 1+u^{-2}
\qquad \text{and} \qquad
\frac{d}{du}(u+u^{-1}) = 1-u^{-2}, $$
it follows that
\begin{align*}
\int \frac{u^{2}}{(u^{2}-1)^{2}} \, du
&= - \frac{1}{2}\frac{1}{u-u^{-1}} + \frac{1}{8} \log \left|(u+u^{-1}) - 2\right| - \frac{1}{8} \log \left|(u+u^{-1}) + 2\right| \\
&= \frac{u}{2(1-u^{2})} + \frac{1}{4} \log \left| \frac{1-u}{1+u} \right|.
\end{align*}
Indeed,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Pythagorean triple problem I am doing research on perfect cuboids, and I'm looking for values $a,b,c$ such that the following is integer, and I'm not sure how to continue this. Any suggestions are appreciated!
$PED$ is a very large constant=$899231100768000$
$$
\begin{align}
&\exp\left(\sigma_1+\sigma_2+\frac{\ln(a^2+b^2+c^2)}{2}-\ln(PED)
\right)\in\mathbb Z\\
&\sigma_1=\ln a+\ln b+\ln c\\
&\sigma_2=\frac{\ln(a^2+b^2)}{2}+\frac{\ln(b^2+c^2)}{2}+\frac{\ln(a^2+c^2)}{2}
\end{align}
$$
Another way to write this is:
$$
\begin{align}
&\frac {abc*\sqrt{a^2+b^2}*\sqrt{a^2+c^2}*\sqrt{b^2+c^2}*\sqrt{a^2+b^2+c^2}}{899231100768000}\in\mathbb Z\\
\end{align}
$$
| If $a=b=1$, $c=4$, then $$(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)=2\times17\times17\times18=102^2,$$ so $$abc\sqrt{a^2+b^2}\sqrt{a^2+c^2}\sqrt{b^2+c^2}\sqrt{a^2+b^2+c^2}=408$$ is an integer. Multiply each of $a,b,c$ by $899231100768000$, and you have an example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Calculated $f(2004)$ where the function $f:N \rightarrow Q^*$ It is considered that the function $f:N \rightarrow Q^*$ which has properties:
$a) f(7) = 4;$
$b) f(8013) = 8015;$
$c) f(n+2).f(n) = 1+ f(n+1).$
Calculated $f(2004)$.
All my attempts to find the required number had no result. Can anyone help me? Thanks so much!
| Using (c) to express a simple recurrence $f(n+2)=\frac{f(n+1)+1}{f(n)}$, it's easy to demonstrate the sequence operates a cycle of length 5 (provided zeros are avoided):
$$ f(1)=x $$
$$f(2)=y$$
$$ f(3) = \frac{y+1}{x}$$
$$ f(4) = \frac{x+y+1}{xy}$$
$$f(5) = \frac{x(xy+x+y+1)}{xy(y+1)} = \frac{(x+1)(y+1)}{y(y+1)} = \frac{x+1}{y}$$
$$f(6) = \frac{xy(y+x+1)}{y(x+y+1)} = x$$
$$f(7) = \frac{y(x+1)}{(x+1)} = y$$
Therefore $f(2)= f(7)=4$ and $f(3)=f(8013)= 8015$
So $f(2004)=f(4) = \frac{8016}{4} = 2004$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Surface area of a sphere above the plane $z=1$ The question is: Find the surface area of the part of the sphere $x^2+y^2+z^2=4$ that lies above the plane $z=1$. I got $4\pi(\sqrt3-1)$ but the answer key says $4\pi(\sqrt2-1)$. Am I doing something wrong or is the answer key wrong?
Thanks in advance.
Edit: This is my work thus far:
We can write the surface in terms of z as $z = \sqrt{4-x^2-y^2}$ (where we take the positive root since $z\geq1 \rightarrow z\geq0$), which can then be parameterized to $\mathbf r(x,y) = <x,y,\sqrt{4-x^2-y^2}>$ with the condition that $x^2+y^2\leq3$ (because $z$ ranges from 1 to 2).
$\mathbf r_x = <1,0,\frac{-x}{\sqrt{4-x^2-y^2}}>$
$\mathbf r_y = <0,1,\frac{-y}{\sqrt{4-x^2-y^2}}>$
$\mathbf r_x \times \mathbf r_y = <\frac{x}{\sqrt{4-x^2-y^2}},\frac{y}{\sqrt{4-x^2-y^2}},1>$
and $\| \mathbf{\mathbf r_x \times \mathbf r_y} \| = \frac{2}{\sqrt{4-x^2-y^2}}$
So the surface area $A$ equals:
$A = \iint_S1dS = \iint_R\frac{2}{\sqrt{4-x^2-y^2}}dA = \int_0^{2\pi} \int_0^{\sqrt3}\frac{2r}{\sqrt{4-r^2}}drd\theta$
$= \int_0^{2\pi}[-2\sqrt{4-r^2}]_1^\sqrt3d\theta$
$= \int_0^{2\pi}2(\sqrt3-1)d\theta$
= $4\pi(\sqrt3-1)$
| For those who want to know why the answer is $4\pi$, let's do it!
First, plot the figure:
Since the goal is to find that area, the integral is just:
$$\iint_S dS,$$
with the appropriate limits. It will be easier if we use spherical coordinates:
$$x = 2 \sin \phi \cos \theta, \quad y = 2 \sin \phi \sin \theta, \quad z = 2 \cos \phi.$$
We know that $0\le \theta \le 2 \pi$, so we have to find the limits for $\phi$. How? That surface area is for $z \ge 1$, so let's find the angle $\phi$ for $z = 1$:
$$
\begin{align}
x^2 + y^2 &= 3, \\
4 \sin^2 \phi \, \cos^2 \theta + 4 \sin^2 \phi \, \sin^2 \theta &= 3, \\
4 \sin^2 \phi &= 3, \\
\sin \phi &= \dfrac{\sqrt 3}{2}, \\
\phi &= \dfrac{\pi}{3}.
\end{align}
$$
Therefore, the limits for $\phi$ are $0$ and $\pi/3$. The integral is:
$$4 \int_0^{2\pi} \! \int_0^{\pi/3} \sin \phi \, d\phi \, d\theta = 4\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that $\frac{1}{90} \sum_{n=1}^{90} 2n \cdot \sin((2n)^\circ) = \cot (1^{\circ})$ Show that $$\frac{(2\sin(2^\circ)) + (4\sin(4^\circ))+ (6\sin(6^\circ)) + \ldots +(180\sin(180^\circ))}{90} = \cot(1^\circ).$$
I used a lot of steps, and typing it all down on here would take me an hour, but here are my last few steps up to the point where I got stuck:
$$180 (\sin(2^\circ) + \sin(4^\circ) + \sin(6^\circ) +.....+ \sin(88^\circ)) + 90$$
Using the product-to-sum formulas, I then reduced it down to
$$180(2\sin(45^\circ) [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90$$
Simplifying a bit more gives me:
$$\frac{180\sqrt{2} [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90}{90}$$ $$= 2\sqrt{2} [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90.$$
Now I am stuck. What can I do next to achieve the final result?
| $$S=\sum_{n=1}^{90}2(91-n)\sin(2n^\circ) = \sum_{n=1}^{90}2\sum_{k=1}^{n}\sin(2k^\circ).$$
Since:
$$2(\sin 1^{\circ})\sum_{k=1}^{n}\sin(2k^\circ)=\sum_{k=1}^{n}\left(\cos((2k-1)^{\circ})-\cos((2k+1)^{\circ})\right)=\cos 1^\circ-\cos((2n+1)^\circ)$$
it follows that:
$$ S = \frac{1}{\sin 1^\circ}\left(90\cos 1^\circ-\sum_{n=1}^{90}\cos((2n+1)^{\circ})\right),$$
but with the same trick shown in this other question we have that:
$$\sum_{n=1}^{90}\cos((2n+1)^{\circ})=-2\cot 1^\circ,$$
hence $\color{red}{S=92\cot 1^{\circ}}$. On the other hand, since:
$$ \sum_{n=1}^{90}2\sin(2n^\circ) = 2\cot 1^{\circ} $$
always by the same trick, it follows that:
$$\sum_{n=1}^{90}2n\sin(2n^\circ)=(91\cdot 2-92)\cot 1^\circ= \color{blue}{90\cot 1^{\circ}} $$
as wanted.
For a complex-analytic derivation, set $t=\frac{\pi}{180}, z=e^{2it}$ and consider that:
$$\sum_{n=1}^{90}2n \sin(2nt) = \Im\sum_{n=1}^{90}2n z^n=2\Im\left(\frac{z-91z^{91}+90 z^{92}}{(1-z)^2}\right).$$
Since $z^{90}=-1$, the last expression simplifies to:
$$\begin{eqnarray*}2\Im\left(\frac{92 z-90 z^{2}}{(1-z)^2}\right)&=&2\Im\left(90\frac{z}{1-z}+\frac{2z}{(1-z)^2}\right)=90\cdot\Im\left(\frac{2e^{2it}}{1-e^{2it}}\right)\\&=&90\cdot\frac{\Im(ie^{it})}{\sin t}=90\cot t.\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To show sequence $a_{n+1}= \frac{a_n^2+1}{2 (a_n+1)}$ is convergent Let $a_1=0$
and $$a_{n+1}= \dfrac{a_n^2+1}{2 (a_n+1)}$$ $\forall n> 1.$
Show that sequence $a_n$ convergent.
I tried to prove $a_n$ is less than 1 by looking at few terms. But i failed to prove my claim.
| since
$$a_{n+1}-(\sqrt{2}-1)=\dfrac{a^2_{n}+1-2(\sqrt{2}-1)(a_{n}+1)}{2a_{n}+2}=\dfrac{[a_{n}-(\sqrt{2}-1)]^2}{2a_{n}+2}\tag{1}$$
simaler
$$a_{n+1}+\sqrt{2}+1=\dfrac{(a_{n}+\sqrt{2}+1)^2}{2a_{n}+2}\tag{2}$$
$\dfrac{(1)}{(2)}$ we have
$$\dfrac{a_{n+1}-(\sqrt{2}-1)}{a_{n+1}+\sqrt{2}+1}=\left(\dfrac{a_{n}-(\sqrt{2}-1)}{a_{n}+\sqrt{2}+1}\right)^2=\cdots=\left(\dfrac{a_{0}-(\sqrt{2}-1)}{a_{0}+\sqrt{2}+1}\right)^{2^n}\to 0$$
beause
$$\dfrac{a_{0}-(\sqrt{2}-1)}{a_{0}+\sqrt{2}+1}<1$$
so
$$\lim_{n\to\infty}a_{n}=\sqrt{2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve the pde $(x^2-y^2-yz)p+ (x^2-y^2-zx)q=z(x-y)$ I'm solving this by Lagrange's method.
Lagrange's auxiliary equation is:
$\frac{dx}{x^2-y^2-yz}=\frac {dy}{x^2-y^2-zx}= \frac{dz}{z(x-y)}$
From the first two ratio and the last ratio:
$\frac{dx-dy}{z(x-y)}=\frac {dz}{z(x-y)}$
Solving which I get $z-x+y=c_1$
I'm unable to find another solution. Please help!
| since
$$\dfrac{dx}{x^2-y^2-yz}=\dfrac{dy}{x^2-y^2-xz}=\dfrac{dz}{z(x-y)}$$
$$\Longrightarrow \dfrac{xdx}{x^3-y^3-xyz}=\dfrac{ydy}{x^2y-y^3-xyz}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow\dfrac{xdx-ydy}{(x^3+y^3)-xy(x+y)}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow \dfrac{xdx-ydy}{(x+y)(x-y)^2}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow \dfrac{xdx-ydy}{x^2-y^2}=\dfrac{dz}{z}$$
$$\Longrightarrow\dfrac{d(x^2-y^2)}{2(x^2-y^2)}=\dfrac{dz}{z}$$
$$\Longrightarrow z^2=x^2-y^2+C_{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$
Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$.
Every time I work this problem out I have $2|z|^2$ but I can not get the other piece.
Here is what I got so far. Let $z = a+ bi$ and $w = c + di$, then
\begin{align*}
|z+w|^2 + |z-w|^2 &= a^2 + b^2 + c^2 + d^2 + a^2 + b^2 - c^2 - d^2\\
&= 2a^2 + 2b^2\\
&= 2(a^2 + b^2)\\
&=2|z|^2.
\end{align*}
I do not know if I have the wrong approach or not.
| You seem to be using $|z + w|^2 = a^2 + b^2 + c^2 + d^2$ and $|z - w|^2 = a^2 + b^2 - c^2 - d^2$. These are both false. Instead we have
$$|z + w|^2 = |(a+bi) + (c+di)|^2 = |(a + c) + (b + d)i|^2 = (a + c)^2 + (b + d)^2$$
and a similar calculation gives $|z - w|^2 = (a-c)^2 + (b - d)^2$. If you use these, you will find that their sum is $2(a^2 + b^2 + c^2 + d^2) = 2|z|^2 + 2|w|^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
extracting the middle term of $ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $ Is there a systematic way to extract the middle term of the following expression?
$$ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $$
This is homogeneous polynomial of degree $2m$, so I am looking for the $(zw)^m$ term.
Example $m = 2$:
$$ (z \cos \theta + w\sin \theta )^2(- z\sin \theta + w\sin \theta )^2
= \dots + (zw)^2(\cos^4 \theta - 4 \cos^2 \theta \sin^2 \theta + \sin^4 \theta) + \dots $$
The middle terms can be simplified using the cosine double angle formulas.
\begin{eqnarray}\cos^4 \theta - 4 \cos^2 \theta \sin^2 \theta + \sin^4 \theta
&=& (\cos^2 \theta - \sin^2 \theta)^2 - 2 \cos^2 \theta \sin^2 \theta \\
&=& \cos^2 2\theta - \tfrac{1}{2}\sin^2 2\theta \\
&=& \boxed{\tfrac{3}{2}\cos^2 2\theta - \tfrac{1}{2}}
\end{eqnarray}
Is it true this middle term can always be expanded as a polynomial in $x=\cos 2\theta$?
I learned of this trick while studying the Wigner D-matrices. It says the Legendre polynomials are certain matrix elements in the representation theory of $SU(2)$. Checking this fact was surprisingly hard.
| Continuing mickep's answer, these are, indeed, the Legendre Polynomials
$$
\sum_{k=0}^m (-1)^k{m\choose k}^2 \bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^{m-k}\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^k
= P_m(\cos(2\theta))
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\
c &
\end{array} ; 1\right] = \dfrac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$
I give negative parameters to the $\Gamma$ function.
What other identity can I use?
I'm trying to find a closed form to this: $\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
Wolfram Mathematica answered this as a closed form: $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
But I would like to have a manual solution with proof.
This is how I got that hypergeometric series:
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
$\dfrac{t_{i+1}}{t_{i}} = \frac{\binom{a+i+1-1}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \frac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \frac{(a+i) (i-n)}{ (i-a-n+1)(i+1)}$
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
UPDATE
Thanks to David H, I got closer to the solution.
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
$\lim\limits_{\epsilon \to0} \frac{\Gamma (-2 a-2 \epsilon +1) \Gamma (-a-n-\epsilon +1)}{\Gamma (-a-\epsilon +1) \Gamma (-2 a-n-2 \epsilon +1)} = \frac{4^{-a} \Gamma \left(\frac{1}{2}-a\right) \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
As you can see this result is close to the expected $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$ formula. But the $\binom{a+n-1}{n}$ factor is still missing and I don't really understand why.
| Here I post the full solution
The problem: we are looking for the closed form of this sum:
$\sum\limits_{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
The first term of the sum for $i=0$ is $\binom{a+n-1}{n}$
The ratio two consecutive terms:
$\dfrac{t_{i+1}}{t_i} = \dfrac{P(i)}{Q(i)} = \dfrac{\binom{a+i}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \dfrac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \dfrac{(i+a)(i-n)}{(i-a-n+1)(i+1)}$
So our hypergeometric series can be described this way:
$\, _2F_1(a,-n;-a-n+1;1)$
Using this and this identities:
$\begin{align}
{_2F_1}{\left(a,-n;1-a-n;1\right)}
&=\frac{n!}{(a)_{n}}C_{n}^{a}{\left(1\right)}\\
&=\frac{\Gamma{\left(n+1\right)}\,\Gamma{\left(a\right)}}{\Gamma{\left(a+n\right)}}\cdot\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(2a\right)}\,\Gamma{\left(n+1\right)}}\\
&=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(2a+n\right)}}{\Gamma{\left(a+n\right)}\,\Gamma{\left(2a\right)}}\\
&=\frac{\binom{2a+n-1}{2a-1}}{\binom{a+n-1}{a-1}}.\\
\end{align}$
The final solution:
$\begin{align}\sum\limits_{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} &= \binom{a+n-1}{n} \cdot \, _2F_1(a,-n;-a-n+1;1)\\
&= \binom{a+n-1}{n} \cdot \dfrac{\binom{2a+n-1}{2a-1}}{\binom{a+n-1}{a-1}}\\
&=\binom{2a+n-1}{n}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Combination questions Q: A team of $11$ is to be chosen out of $15$ cricketers of whom $5$ are bowlers and $2$ others are wicket keepers. In how many ways can this be done so that the team contains at least $4$ bowlers and at least $1$ wicket keeper?
| The number of ways to choose exactly $4$ bowlers and exactly $1$ keeper:
$$\binom{5}{4}\cdot\binom{2}{1}\cdot\binom{15-5-2}{11-4-1}=280$$
The number of ways to choose exactly $5$ bowlers and exactly $1$ keeper:
$$\binom{5}{5}\cdot\binom{2}{1}\cdot\binom{15-5-2}{11-5-1}=112$$
The number of ways to choose exactly $4$ bowlers and exactly $2$ keepers:
$$\binom{5}{4}\cdot\binom{2}{2}\cdot\binom{15-5-2}{11-4-2}=280$$
The number of ways to choose exactly $5$ bowlers and exactly $2$ keepers:
$$\binom{5}{5}\cdot\binom{2}{2}\cdot\binom{15-5-2}{11-5-2}=70$$
The number of ways to choose at least $4$ bowlers and at least $1$ keeper:
$$280+112+280+70=742$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$? The problem is to find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$.
I tried $2^{2 \cdot 5^6}+3^{5^4} \cdot 2^{5^4}= 2^{5^4} \cdot ( 2^{2 \cdot 5^6 -5^4}+ 3^{5^4} )$.
| Use the following lemma, which can be proven by induction 'on $b$':
For any odd prime $p$ and $a,b \in \mathbb{N}_{>0}$ such that $p^a \mid\mid x-y$ and $p^b \mid\mid k$:
$p^{a+b} \mid\mid x^k-y^k$.
For this question we don't need the "strictly divides" in the conclusion, but in general this lemma is very useful.
$4^{5^6}+6^{5^4} = (4^{5^2})^{5^4}+6^{5^4} = (4^{5^2})^{5^4}-(-6)^{5^4}$ and $5 \mid 4^{5^2}-(-6)$ by Fermat's little theorem.
Therefore by the lemma $5^5 \mid 4^{5^6}+6^{5^4}$.
Combining with what you already have gives the number of zeros.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to solve this linear equations using gauss jordan method? How to solve the linear equations in Gauss elimination method
$$2x+3y+z=1\\
x+y+z=3\\
3x+4y+2z=4$$
| \begin{align*}
\left(\begin{array}{ccc|c}
1&1&1&3\\
2&3&1&1\\
3&4&2&4
\end{array}\right)\\\\
\left(\begin{array}{ccc|c}
1&1&1&3\\
0&1&-1&-5\\
0&1&-1&-5
\end{array}\right)\\\\
\begin{array}{ccc}
x&y&z
\end{array}\qquad\qquad\\
\left(\begin{array}{ccc|c}
1&1&1&3\\
0&1&-1&-5\\
0&0&0&0
\end{array}\right)
\end{align*}
Z is your free variable .continue from here .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
| HINT: Use $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also the fact that if $x^2+x+1=0$, then $x+\frac1x$=$-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$ How can we prove the following
$$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$
I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \dfrac{1}{n^2}=\frac{\pi^2}{6}$$
But I'm stuck at this problem.
| First consider the following expansion of $\pi \cot(\pi z)$:
$$\pi \cot(\pi z) = \frac{1}{z} + \sum_{n = 1}^\infty \frac{2z}{z^2 - n^2} \quad (z \neq 0, \pm 1, \pm 2,\ldots)$$
Replacing $z$ by $iz$, we have
$$-i\pi \coth(\pi z) = \frac{1}{iz} - \sum_{n = 1}^\infty \frac{2iz}{z^2 + n^2} = -i\left(\frac{1}{z} + 2\sum_{n = 1}^\infty \frac{z}{z^2 + n^2}\right)$$
Thus
$$\pi \coth(\pi z) = \frac{1}{z} + 2\sum_{n = 1}^\infty \frac{z}{z^2 + n^2} \quad (z \neq 0, \pm i, \pm 2i,\ldots)$$
Evaluting at $z = 1$ results in
$$\pi \coth(\pi) = 1 + 2\sum_{n = 1}^\infty \frac{1}{1 + n^2},$$
or
$$\sum_{n = 1}^\infty \frac{1}{1 + n^2} = \frac{\pi \coth(\pi) - 1}{2}.$$
Therefore
\begin{align}\sum_{n = 0}^\infty \frac{1}{1 + n^2} &= \frac{\pi\coth(\pi)+ 1}{2}\\
&= \frac{1}{2}\left(\frac{\pi(e^{\pi} + e^{-\pi})}{e^{\pi} - e^{-\pi}} + 1\right)\\
&= \frac{1}{2}\left(\frac{\pi(e^{2\pi} + 1)}{e^{2\pi} - 1} + 1\right)\\
&= \frac{(\pi + 1)e^{2\pi} + (\pi - 1)}{2(e^{2\pi} - 1)}\\
&= \frac{(\pi + 1)(e^{2\pi} - 1) + 2\pi}{2(e^{2\pi} - 1)}\\
&= \frac{\pi + 1}{2} + \frac{\pi}{e^{2\pi} - 1}.\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{x\to\infty}(x(\log(1+\sqrt{1+x^2}-\log(x)))$ can someone give me a hint/solution for:
$$\lim_{x\to\infty}\left(x(\log(1+\sqrt{1+x^2}-\log(x))\right)$$
Shall I do a derivative ?
But there's no L'Hospital to use here..
Shall I change its form ?
$x(\log(1+(1+x^2)^{1/2})-\log(x))$ (minimal change)
Or what shall I do ?
Thanks
| L'Hospital's rule:
$$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} =\lim_{x \to \infty} \frac{(\ln(1+\sqrt{x^2 + 1}) - \ln x)' }{(\frac{1}{x})'} =$$$$= \lim_{x \to \infty} \frac{\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}} - \frac{1} {x} }{-\frac{1}{x^2}} = \lim_{x \to \infty}(x-\frac{x^2}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}})=$$$$=\lim_{x \to \infty}\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}}[(1+\sqrt{x^2 + 1})\cdot\sqrt{x^2+1}-x^2]=$$$$=\lim_{x \to \infty}\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}}(\sqrt{x^2 + 1}+x^2+1-x^2)=\lim_{x \to \infty}\frac{x}{\sqrt{x^2 + 1}}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to Solve Trigonometric Equations? How are you supposed to go about solving equations such as:
$$-\sqrt{3} = \frac{\sin{4\theta}}{\sin{7\theta}}.$$
I know that $\theta = 30^{\circ}$ is one such solution, but how do I find all solutions using algebra?
Thanks
Edit: I figured out one possible method of reasoning. For $-\sqrt{3} = \frac{\sin{4\theta}}{\sin{7\theta}}$ to be true, taking the "special values" of the $\sin$ function on the unit circle, one way to achieve the value of $-\sqrt{3}$ is to have either $$\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$$ or $$\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}.$$ Solving for the first case, to achieve a negative value in the denominator, $0\leq\theta\leq\frac{\pi}{4}$ (since you want $4\theta\leq180^{\circ}$ and $\sin{7\theta}\lt0$). Then the only value for $\sin\theta=\frac{\sqrt{3}}{2}$ in the first quadrant is $\theta = \frac{\pi}{6}$.
Using similar reasoning, you can deduce a symmetrical value in the case where the numerator is negative. This method to me, however, feels unprofessional and "weak." So again, is there a more definitive, algebraic solution?
| Use the identities
$$\sin4\theta = 4\cos\theta\left (\sin\theta - 2\sin^3\theta\right)$$
$$ \sin7\theta = -64\sin^7\theta + 112\sin^5\theta -56\sin^3\theta+7\sin\theta,$$ obtained from De Moivre's formula and by taking the substitution $\cos^2\theta
=1-\sin^2\theta$.
Hence,
$$ \frac{4\cos\theta\left (\sin\theta - 2\sin^3\theta\right)}{-64\sin^7\theta + 112\sin^5\theta -56\sin^3\theta+7\sin\theta} = -\sqrt{3} $$$$ \implies \cos\theta = \frac{\sqrt{3}\left(64\sin^7\theta - 112\sin^5\theta +56\sin^3\theta-7\sin\theta\right)}{4\left(\sin\theta - 2\sin^3\theta\right)}.$$
Square the equation (we keep the solutions) and do $y = \sin\theta$. Thus,
$$1-y^2 = \frac{3\left(64y^7-112y^5+56y^3-7y\right)^2}{16\left(y-2y^3\right)^2}$$ $$$$ $$\implies 12288y^{12} -43008y^{10}+59136y^8-40256y^6+13984y^4 -2272y^2 +131 = 0.$$
Now, take $z = y^2$. Hence,
$$12288z^6 -43008z^5+59136z^4-40256z^3+13984z^2 -2272z +131 = 0,$$ which has $6$ positive solutions (found numerically). So, from $z$, you can easily find $y$ and $\theta$.
Remark: Since the equation was squared, you should test the solutions $\theta$ in the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
| $x+\sqrt{x^2+1}$ is a strictly increasing, continuous function so that a single $x$ corresponds to a single $y$, and $y=-x$ is the one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$ Convergence of $$\sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n })$$
Attempt:
Initial Check : $\lim_{n \rightarrow \infty } \log~ ( n ~\sin \dfrac {1 }{ n }) = 0$
$\log~ ( n ~\sin \dfrac {1 }{ n }) < n ~\sin \dfrac {1 }{ n }$
$\implies \sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n }) < \sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{ n }$
But, $\sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{ n }$ is itself a divergent sequence. Hence, I don't think the above step is of any particular use.
Could anyone give me a direction on how to move ahead.
EDIT: I did think of trying to use the power series of $\sin$. Here's what I attempted :
$\sin \dfrac {1}{n} = \dfrac {1}{n}-\dfrac {1}{n^3.3!}+\dfrac {1}{n^5.5!}-\cdots$
$\implies n \sin \dfrac {1}{n} = 1- \dfrac {1}{n^2.3!}+\dfrac {1}{n^4.5!}-\cdots$
I couldn't proceed further due to the $\log$.
Thank you for your help.
| Since $$\log\left(n\sin \frac{1}{n}\right) = \log\left(n\left(\frac{1}{n} + O\left(\frac{1}{n^3}\right)\right)\right) = \log\left(1 + O\left(\frac{1}{n^2}\right)\right) = O\left(\frac{1}{n^2}\right)$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges, so does $\sum_{n = 1}^\infty \log(n\sin\frac{1}{n})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Finding an order of a coset in $A/B$ where $A$ is a free abelian group and $B$ is a subgroup. Let $A$ be a free abelian group with basis $x_1,x_2,x_3$ and let $B$ be a subgroup of A generated by $x_1+x_2+4x_3, 2x_1-x_1+2x_3$. In the group $A/B$ find the order of the coset $(x_1+2x_3)+B$.
How can I find this order? Just so you can reply more directly, I shall find a basis $f_1,f_2,f_3$ such that $d_1\cdot f_1,d_2\cdot f_2,d_3\cdot f_3$ is a basis of $B$ where $d_i|d_{i+1}$.
$Attempt$: so by $Smith$ algorithm, $\begin{pmatrix}1 & 2 \\1 & -1 \\4 & 2 \\\end{pmatrix}$ $\underrightarrow{R_2\to R_2-R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\4 & 2 \\\end{pmatrix}$ $\underrightarrow{R_3\to R_3-4R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\0 & -6 \\\end{pmatrix}$ $\underrightarrow{R_3\to R_3-2R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\0 & 0 \\\end{pmatrix}$ $\underrightarrow{C_2\to C_2-2C_1} \begin{pmatrix}1 & 0 \\0 & -3 \\0 & 0 \\\end{pmatrix}$$\underrightarrow{C_2\to -C_2} \begin{pmatrix}1 & 0 \\0 & 3 \\0 & 0 \\\end{pmatrix}$ and therefore, $d_1=1, \space d_2=3, \space d_3=0.$
Therefore, while $M= \begin{pmatrix}1 & 2 \\1 & -1 \\4 & 2 \\\end{pmatrix}$, there exist $Q,P\in GL_n(\Bbb{Z})$ such that $QMP^{-1}= \begin{pmatrix}1 & 0& 0 \\0 & 3 & 0\\0 & 0 & 0\\\end{pmatrix}$ and $P^{-1}$ columns are a basis of $A$ that fulfills the aforementioned requirements. One way to find $P^{-1}$ is to classify the identity element where the classification is the inverse of the row operations I used above in the opposite direction: from the end to the start: $\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\0 & 0 & 1\\\end{pmatrix}$ $\underrightarrow{R_3\to R_3 + 2R_1}\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\0 & 2 & 1\\\end{pmatrix}$ $\underrightarrow{R_3\to R_3+4R_1}\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\4 & 2 & 1\\\end{pmatrix}$ $\underrightarrow{R_2\to R_2+R_1}\begin{pmatrix}1 & 0& 0 \\1 & 1 & 0\\4 & 2 & 1\\\end{pmatrix}$ $\Rightarrow \{x_1+x_2+4x_3, x_2+2x_3 , x_3\}$ is the desired basis, where $\Rightarrow \{x_1+x_2+4x_3, 3\cdot (x_2+2x_3
)\}$ is a basis of $B$. Furthermore: $A/B\cong \Bbb{Z}_{f_1}\times \Bbb{Z}_{f_2}\times \Bbb{Z}_{f_3}/1\cdot \Bbb{Z}_{f_1}\times 3\cdot \Bbb{Z}_{f_2}\times 0\cdot \Bbb{Z}_{f_3} \cong \Bbb{Z}/\Bbb{Z}\times \Bbb{Z}/3\Bbb{Z}\cong \Bbb{Z}/3\Bbb{Z}$. How shall I continue?
| $Proof$: since no one has yet answered, I shall make an attempt and would appreciate your correction: Let us denote the basis $\{x_1+x_2+4x_3=a,\ 3 \cdot (x_2+2x_3)=b\}$ then $3a-b=3x_2+6x_6$. Then obviously, since $3x_2+6x_6\in B$ then the order of $(x_1+2x_3)+B$ is $3$, as $(x_1+2x_3)+B\ne B$, $2(x_1+2x_3)+B\ne B$ but $3(x_1+2x_3)+B=B= identity_{A/B}$.
| {
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Prove, inequality ,positive numbers $$\frac{a}{e+a+b}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{c+d+e}+\frac{e}{d+e+a}<2$$
Prove that for positive numbers $a,b,c,d,e$ there is such inequality
| For convenience of notation I'll use $a_i$'s ($1 \le i \le 5$) instead of $a,b,c,d,e$.
Wlog assume $a_1+a_2+a_3 = \min\limits_{1 \le i \le 5}\{a_{i-1}+a_i+a_{i+1}\}$
(where, the indexing is cyclic $a_{0} = a_{5}$ and $a_6 = a_1$ and so on)
Then, $\displaystyle \sum\limits_{i=1}^{3} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} \le \frac{a_1}{a_1+a_2+a_3}+\frac{a_2}{a_1+a_2+a_3}+\frac{a_3}{a_1+a_2+a_3} = 1$
And, $\displaystyle \frac{a_4}{a_3+a_4+a_5}+\frac{a_5}{a_4+a_5+a_1} < \frac{a_4}{a_4+a_5} + \frac{a_5}{a_4+a_5} = 1$.
Adding together we get:
$$\displaystyle \sum\limits_{i=1}^{5} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} < 2$$
Infact using a $5$-tuple, $(a_1,a_2,a_3,a_4,a_5) = (x,y,x,y,x)$ and letting $y \to \infty$,
$$\lim\limits_{y \to \infty} \sum\limits_{i=1}^{5} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} =\lim\limits_{y \to \infty} \frac{3x}{x+2y}+\frac{2y}{2x+y} = 2$$
Hence, $2$ is the best possible upper-bound here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx$, where $0I am trying to evaluate the following integral:
$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx,$$
where $0<c<1$.
I can't really think of a way to find it so please give me a hint.
| THIS IS ONLY A PARTIAL ANSWER
First we apply DUIS $$I'(a)=\int_{-\pi/2}^{\pi/2} \frac{\sin(x)}{a\sin(x)+1}$$
This can be integrated with a Weierstrass substitution, which I'll leave for you to verify:
$$\large\left.I'(a)=\frac{x-\frac{2\tan^{-1}\left(\frac{\tan(x/2)+y}{\sqrt{1-y^2}}\right) }{\sqrt{1-y^2}}}{y}\right|_{-\pi/2}^{\pi/2}$$
After plugging in the bounds we get:
$$\large I'(a)=\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}+\frac{\pi/2+\frac{2\tan^{-1}\left(\frac{-1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}=I_1+I_2$$
Now we have to integrate that huge monster.
$$\large I_1=\int\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+y}{\sqrt{1-y^2}}\right)}{\sqrt{1-y^2}}}{y}$$
$$\large=\int\cos\theta\frac{\pi/2-\frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\cos\theta}}{\sin\theta}(y=\sin\theta)$$
$$\large=\int\cot\theta\left({\pi/2-\frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\cos\theta}}\right)$$
$$\large=\int\frac{\pi\cot\theta}2 +\int \frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\sin\theta}$$
$$\large=\frac \pi 2 \log \sin\theta +\int \frac{2\tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right)}{\sin\theta}$$
Which is not expressible in terms of elementary functions.
| {
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Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$? I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with
$n\le 12\ 000$.
*
*If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.
*If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pmod 5$.
So, $n$ must have the form $10k+5$.
For $n=35$ and $n=55$, the number $4^n+n^4$ splits into two primes
with almost the same size.
So, is there an obvious reason (like algebraic factors) that there is no prime I am looking for ?
| If $n$ is even, so will be $4^n+n^4$ and the later will definitely be $>2$ and hence composite
Else
$$4^n+n^4=(2^n)^2+(n^2)^2=(n^2+2^n)^2-2\cdot2^n\cdot n^2$$
$$=(n^2+2^n)^2-(n2^{\frac{n+1}2})^2$$
As $n$ is odd $\iff n+1$ is even $\implies\dfrac{n+1}2$ is an integer
$$4^n+n^4=(n^2+2^n+n2^{\frac{n+1}2})(n^2+2^n-n2^{\frac{n+1}2}) $$
Establish that both factors are $>1$
| {
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Probability of Seven (Distinct) car accidents occurred on the same day Seven (Distinct) car accidents occurred in a week. What is probability that they all occurred on the same day?
My Solution:
All 7 accident occurs in 1 day in $\binom{7}{1}$ ways
All 7 accident occurs in 2 days in $\binom{7}{2}$ ways
All 7 accident occurs in 3 days in $\binom{7}{3}$ ways
All 7 accident occurs in 4 days in $\binom{7}{4}$ ways
All 7 accident occurs in 5 days in $\binom{7}{5}$ ways
All 7 accident occurs in 6 days in $\binom{7}{6}$ ways
All 7 accident occurs in 7 days in $\binom{7}{7}$ ways
Hence P(They all occurred on the Same day)$=\frac{\binom{7}{1}}{\binom{7}{1}+\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}}$
$=\frac{7}{2^7}$
Is my analysis correct or I've assumed something wrong
| There are seven days and seven accidents (which are distinguishabla from one another), so the number of possible outcomes is $7^7$. However, only seven of them fulfil our criteria: all accidents on Monday, all on Tuesday, ..., all on Saturday. So the answer is $1/7^6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ for $a,b,c \in\mathbb{R}$ Find biggest constans k such that $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ is true for any $a,b,c \in\mathbb{R}$
Could you check up my solution? I'm not sure it's ok -
$(a+b)^2 + (a+b+4c)^2 \ge 0$
and
$0\ge \frac{kabc}{a+b+c}$
so
$0 \ge k$
| Using $\text{arithmetic mean} \geq \text{geometric mean}$, we get
$$
\begin{eqnarray}
(a+b)^2+(a+b+4c)^2&=&(a+b)^2+(a+2c+b+2c)^2 \\
&\geq& \big(2\sqrt{ab}\big)^2+\big(2\sqrt{2ac}+2\sqrt{2b}\big)^2 \\
&=& 4ab+8ac+8bc+16c\sqrt{ab}
\end{eqnarray}
$$
Therefore
$$
\begin{eqnarray}
\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c)&\geq&\frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}(a+b+c)\\
&=&\Bigg(\frac4c+\frac8b+\frac8a+\frac{16}{\sqrt{ab}}\Bigg)(a+b+c)\Bigg(\frac a 2+\frac a 2+\frac b 2+\frac b 2+c\Bigg)\\
&\geq&8\Bigg(5\sqrt[5]{\frac1{2a^2b^2c}}\Bigg)\Bigg(5\sqrt[5]{\frac{a^2b^2c}{2^4}}\Bigg)\\
&=& 100
\end{eqnarray}
$$
Hence,
$$
\begin{eqnarray}
k &\leq& \frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c)\\
&\leq& 100\tag*{$\square$}
\end{eqnarray}
$$
| {
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Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| From Cauchy-Schwarz Inequality: $$\sum\limits_{cyc} \frac{a}{b}\ge \frac{(a+b+c)^2}{ab+bc+ca}$$
Hence, it suffices to prove: $\displaystyle \frac{(a+b+c)^3}{ab+bc+ca} \ge 9$
Squaring both sides and using $a^2+b^2+c^2 = 3$,
$$\begin{align} & (a+b+c)^6 \ge 81(ab+bc+ca)^2 = 27(a^2+b^2+c^2)(ab+bc+ca)^2 \end{align}$$
which follows from the Am-Gm Inequality with $3$ terms, $$a^2+b^2+c^2,ab+bc+ca,ab+bc+ca$$
as,
$\displaystyle \frac{(a+b+c)^6}{27} = \left(\frac{a^2+b^2+c^2+2(ab+bc+ca)}{3}\right)^3 \ge (a^2+b^2+c^2)(ab+bc+ca)^2$
| {
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Why does solving the spherical Bessel equation using Frobenius series produce two quadratic equations for the exponents at the singularity? The spherical Bessel equation is:
$$x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0$$
If I seek a Frobenius series solution, I will have:
\begin{align*}
&\quad y = \sum_{n = 0}^{\infty} a_nx^{n + r} \\
&\implies y' = \sum_{n = 0}^{\infty} (n + r)a_nx^{n + r - 1} \\
&\implies y'' = \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r - 2}
\end{align*}
Substituting into the ODE of interest:
\begin{align*}
&\quad x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0 \\
&\equiv \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r} + \sum_{n = 0}^{\infty} 2(n + r)a_nx^{n + r} + \sum_{n = 0}^{\infty} a_nx^{n + r + 2} + \sum_{n = 0}^{\infty} \frac{-5}{16}a_nx^{n + r} = 0 \\
&\equiv \sum_{n = 0}^{\infty} [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_nx^{n + r} + \sum_{n = 2}^{\infty} a_{n - 2}x^{n + r} = 0 \\
&\equiv [r(r-1) + 2r - (5/16)]a_0 + [(r+1)(r) + 2(r+1) - (5/16)]a_1 + \\
&\quad \sum_{n = 2}^{\infty} ([(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2})x^{n + r} = 0 \\
&\implies [r(r-1) + 2r - (5/16)] = 0 \wedge \\
&\quad [(r+1)(r) + 2(r+1) - (5/16)] = 0 \wedge \\
&\quad [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2} = 0
\end{align*}
The first conjunct is the standard indicial equation specifying $r$. The second conjunct is yet another quadratic that $r$ has to satisfy. Is there a mistake? Or should $a_1 = 0$?
| You set the value of $r$ found in the first equation into the second equation. It then defines $a_n$ in terms of $a_{n-2}$ for all $n \geq 2$. So, once we know $a_0$, $a_1$ we get $a_2$ from the second equation and so on.
| {
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Defining the equation of an ellipse in the complex plane
Usually the equation for an ellipse in the complex plane is defined as $\lvert z-a\rvert + \lvert z-b\rvert = c$ where $c>\lvert a-b\rvert$. If we start with a real ellipse, can we define it in the manner below?
For $x,y,h,k,a,b\in\mathbb{R}$ such that $a,b\neq 0$, we define a real ellipse as
$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1.
$$
Let $z = \frac{x}{a} + i\frac{y}{b}$ and $z_0 = \frac{h}{a} + i\frac{k}{b}$. If we expand the equation for an ellipse, we have
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{h^2}{a^2} + \frac{k^2}{b^2} -
\frac{2xh}{a^2} - \frac{2yk}{b^2} = 1.
$$
Notice that $\lvert z\rvert^2 = \frac{x^2}{a^2} + \frac{y^2}{b^2}$ and $\lvert z_0\rvert^2 = \frac{h^2}{a^2} + \frac{k^2}{b^2}$. Now, let's write the ellipse as
$$
\lvert z\rvert^2 + \lvert z_0\rvert^2 - \frac{2xh}{a^2} - \frac{2yk}{b^2} +
\frac{yh}{ab}i - \frac{yh}{ab}i + \frac{xk}{ab}i - \frac{xk}{ab}i =
\lvert z\rvert^2 + \lvert z_0\rvert^2 - \bar{z}z_0 - z\bar{z}_0 = 1.
$$
Thus, the equation of an ellipse in the complex plane is
$$
(z - z_0)(\bar{z} - \bar{z}_0) = \lvert z - z_0\rvert^2 = 1\Rightarrow
\lvert z - z_0\rvert = 1
$$
where $z$ and $z_0$ are defined above.
| Setting $z_0=h+ik$, you can't do much better than
$$\frac{\left(\Re(z-z_0)\right)^2}{a^2}+\frac{\left(\Im(z-z_0)\right)^2}{b^2}=1.$$
The function in the LHS isn't analytic, so taking the real and imaginary parts isn't worse nor better than using the modulus.
You could also try starting from the parametric form $(h+a\cos \theta,k+b\sin\theta)$ with the hope to match it to the polar form $r\cos\theta+ir\sin\theta$ but you will end up with an anisotropic transform $(x,y)\to(\lambda x,\mu y)$, that doesn't have a nice form by complex functions.
The classical polar form
$$r=\frac p{1-e\cos\theta}$$
leads to the unsatisfactory equation
$$|z-z_0|=e\,\Re(z-z_0)+p,$$
that can easily be related to the first.
In all above cases, these representations are missing a degree of freedom (axis orientation).
| {
"language": "en",
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"question_score": "10",
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Elementary number theory , when is $12n^2 + 1$ a square Prove that if
$$k = 2 + 2\sqrt{12n^2 + 1}$$
is an integer then it is a square.
Can anyone help me with this? All I know is that k is an integer if and only if ${12n^2 + 1}$ is a square. What do I do next?
| let
$$\sqrt{12n^2+1}=m\Longrightarrow 12n^2+1=m^2\Longrightarrow \dfrac{m-1}{2}\cdot\dfrac{m+1}{2}=3n^2$$
because $m$ is odd numbers,so $\dfrac{m-1}{2},\dfrac{m+1}{2}\in N^{+}$
since
$$\gcd\left(\dfrac{m+1}{2},\dfrac{m-1}{2}\right)=1$$
case 1:
$$\dfrac{m-1}{2}=3u^2,\dfrac{m+1}{2}=v^2,uv=n$$
$$\Longrightarrow 2+2\sqrt{12n^2+1}=2m+2=4v^2$$
case 2:
$$\dfrac{m-1}{2}=u^2,\dfrac{m+1}{2}=3v^2\Longrightarrow 3v^2=u^2+1$$
since $3v^2\equiv 0,3\pmod 4$,and
$u^2+1\equiv 1,2\pmod 4$
so that's impossible
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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System of exponential equations If $x,y,z \in \mathbb{R}$ and
$$
\begin{cases}
2^x+3^y=5^z \\
2^y+3^z=5^x \\
2^z+3^x=5^y
\end{cases}
$$
does it imply that $x=y=z=1$?
| I think so.
Let $f(x) = 5^x - 2^x - 3^x$, you have that
$$x < 1 \Leftrightarrow f(x) < 0$$
$$x = 1 \Leftrightarrow f(x) = 0$$
$$x > 1 \Leftrightarrow f(x) > 0$$
Suppose that $x \geq y \geq z$:
If $x > 1$, then $5^x > 2^x + 3^x \geq 2^y + 3^z$, therefore $5^x \neq 2^x + 3^y$.
This implies that $1 \geq x \geq y \geq z$.
If $z < 1$, then $5^z < 2^z + 3^z \leq 2^x + 3^y$, this implies that $1 \geq x \geq y \geq z \geq 1$ so $x = y = z = 1$.
By cyclicity of the equations you get the same result if $y \geq z \geq x$ and $z \geq x \geq y$.
Suppose $x \geq z \geq y$:
If $x > 1$ then $5^x > 2^x + 3^x \geq 2^y + 3^z$ therefore $1 \geq x \geq y \geq z$.
If $y < 1$ then $5^y < 2^y + 3^y \leq 2^z + 3^x$ therefore $1 \geq x \geq z \geq y \geq 1$ which implies that $x = y = z = 1$.
By cyclicity we have covered the other 3 orders, so the only solution should be $x = y = z = 1$.
| {
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Expected first return time of Markov Chain Given the following Markov Chain:
$$M = \left(
\begin{array}{cccccc}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\
\frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 & 0 \\
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & 0 & 0 \\
\frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} & 0 & \frac{1}{4} \\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\end{array}
\right)$$
I need to find $E(T_1 | X_0 = 1)$, with $T_1 = \inf\{ n \geq 1 : X_n = 1\}$, i.e. the expected first arrival time of M.
I know that I can recursively calculate the probability of arriving back at 1 after exactly n steps:
$$f_1(n) = P(X_n = 1, X_{n-1} \neq 1,...,X_1 \neq 1, X_0 = 1) $$
This can be done the following way:
$$f_1(n) = p_{i,i}(n)-\sum_{k=1}^{n-1}f_1(k)p_{i,i}(n-k)$$
where $p_{i,i}(n) = (M^n)_{i,i}$ is the probability of going from state i to state i in n steps.
So I would say that $$E(T_1 | X_0 = 1)= \sum_{n=1}^\infty nf_1(n)$$
is the expected return time to step 1. However I have no idea how to actually evaluate the series, can anybody help me please?
| I think Zoe's approach is the easiest but that she has the equations wrong.
Let
$$T_{11} = E(T_1\mid X_0=1) \\
T_{21} = E(T_1\mid X_0=2).$$
We want to find $T_{11}$. Considering the possible transitions between states $1$ and $2$ and their probabilities, we get equations:
$$T_{11} = 1+\dfrac{1}{2}T_{21}\qquad\qquad\qquad (1) \\
T_{21} = 1+\dfrac{3}{4}T_{21}\qquad\qquad\qquad (2)$$
Solving these simultaneously gives $T_{11} = 3$.
Note: Derivation of Equation $(1)$:
*
*the $1$ is to count $1$ for the current state transition;
*we have probability $1/2$ to move from State $1$ to State $2$, from where the remaining steps to return to State $1$ is $T_{21}$.
Equation $(2)$ is derived similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation $\frac{(x-y)^2}{(y-3)(3-x)} = 1$
That was my attempt, I can't think of anything else here. I'd prefer a hint
| Just did this yesterday: What's so special about the form $ax^2+2bxy+cy^2$?
Given some $g$ that is going to be set equal to $0,$ with
$$ g = a x^2 + b x y + c y^2 + d x + e y + f $$
and taking the usual
$$ \Delta = b^2 - 4 a c, $$
we get, assuming both $a, \Delta \neq 0$ for this amount of work to be necessary,
$$ \color{red}{ -4a \Delta g} \; = \; \color{blue}{ (\Delta y + bd - 2ae)^2 - \Delta (2 a x + b y + d)^2} + \color{green}{ \left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)} $$
I put extra parentheses to emphasize that $\left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)$ is a single constant term, no $x,y$ involved.
New variables to be called:
$$ u = \Delta y + bd - 2ae, \; \; \; v = 2 a x + b y + d $$
For you,
$$g= x^2 - xy + y^2 - 3 x - 3 y + 9 $$
is being set to $0,$ and
$$ a=1, b=-1,c=1,d=-3,e=-3,f=9, \Delta = -3, $$
Go through it all,
$$ -4ag = (-3y+9)^2 + 2 (2x-y-3)^2 $$
and this is being set to $0,$ the constant term came out to $0.$
So we have a single point, $y=3,$ then $2x-y-3 = 0$ says $x=3$ also.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The pedantic function $\frac{y \cdot \sin(x^5y^3+x^3)}{(x^4y^8+x^6+3y^2)\cos(x^2y)^2}$ I was shown the following "pedantic" function.
$$
f(x,y) := \frac{y \cdot \sin(x^5y^3+x^3)}{(x^4y^8+x^6+3y^2)\cos(x^2y)^2}
$$
The question is what happens as the function approaches origo. So in order for the limit to exists we need to check that the function is approaching the same limit for all directions.
So along the coordinate axes we have
$$
\lim_{x\to 0} f(x,0) = \lim_{y\to 0} f(y,0) =0
$$
Hmmm. Okay, lets test all straight lines through origo $y = k x$, some calculations again give
$$
\lim_{x \to 0} f(x,k x) = 0
$$
So it seems the limit exists, further studies show that
$$
\lim_{x \to 0} f(x,k x^n) = 0
$$
For all integers $n$, except $3$. Is this true?
Conjecture
Let $f(x,y)$ be defined as above then
$$
\lim_{x\to 0} f(x,k x^n) = 0
$$
for all $n \in\mathbb{N}$ except $n=3$. Here we have
$$
\lim_{x\to 0} f(x, k x^3) = \frac{k}{3k^2+1}
$$
where $k \in \mathbb{R}$.
Is the conjecture true? I tested $n$ up to 2000. Also are
there other lines approaching origo that leads to a different answer than zero, except $y = x^3$? Can this be proven rigorously?
| First, notice that
$$f(x,kx^n) = \frac{kx^n \cdot \sin(x^5(kx^n)^3+x^3)}{(x^4(kx^n)^8+x^6+3(kx^n)^2)\cos(x^2(kx^n))^2} = \frac{kx^n\sin(x^3+k^3x^{n+5})}{(x^6+3k^2x^{2n}+k^8x^{8n+4})[\cos(kx^{n+2})]^2}$$
As $x\to 0$, lower powers of $x$ converge to $0$ faster than higher powers. So only the lowest order powers of $x$ actually matter in the limit (this is dual to the fact that as $x\to \infty$ only the highest powers matter).
Now $\sin(Z) = Z-\frac{Z^3}{3!}+\frac{Z^5}{5!}+\cdots$ so $\sin(x^3+k^3x^{n+5})=(x^3+k^3x^{n+5})+\cdots = x^3+O(x^4)$
If you're not familiar with "big-O" notation, $O(x^4)$ means that all of the rest of the terms are of order $x^4$ or greater. They go to $0$ faster than $x^3$, so they won't matter in the limit.
Likewise, $\cos(Z) = 1-\frac{Z^2}{2!}+\cdots$ so $\cos(kx^{n+2}) = 1-\frac{k^2x^{2n+4}}{2!}+\cdots = 1+O(x)$.
Next, assuming that $n>3$, we have that $k^8x^{8n+4}$ and $3k^2x^{2n}$ are of higher order than $x^6$. So we get
$$f(x,kx^n) = \frac{kx^n(x^3+O(x^4))}{(x^6+O(x^7))(1+O(x))^2} = \frac{kx^{n+3}+O(x^{n+4})}{x^6+O(x^7)} \to 0$$
as $x \to 0$ since $x^{n+3}$ is a higher power than $x^6$ (since $x>3$).
So, yes, your conjecture does hold. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{dx}=u+x\frac{du}{dx}$.
Hence I get:
$u+x\frac{du}{dx}=-\frac{u(2-u^3)}{2u^3-1}$
This gives:
$2u^4-u+(2u^3-1)x\frac{du}{dx}=-2u+u^4$
$(2u^3-1)x\frac{du}{dx}=-(u^4+u)$
$\frac{(2u^3-1)}{(u^4+u)}\frac{du}{dx}=-\frac{1}{x}$
So I can atleast reduce the problem to a seperable differential equation, but I am not able to integrate the left side up. Do you have any tips?
| here is a cheap way of doing this knowing that $x^3 + y^3 = Cxy$ is a solution. we will make the change of variables $$u = x^3+y^3, t = xy$$
the deifferential equation satisfied by $u$ and $t$ are:
$$\frac{du}{dx} = 3x^2 + 3y^2 \frac{dy}{dx} = \frac{3(y^6-x^6)}{x(2y^3 - x^3}, \frac{dt}{dx} = x\frac{dy}{dx} + y = \frac{3y(y^3-x^3)}{2y^3 - x^3} $$
we can divide it out and get a separable equation $$\frac{du}{dt} = \frac{u}{t} $$ which has solution of the form $$u = Ct, x^3 + y^3 = Cxy$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
Proof: $n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$ I've edited the post in order to add at the end what, I think, is the complete proof of these inequalities. I want to apologize by not having given a reply as early as those given by the users who gave me hints about this exercise. It took me a little of time but your answers were very helpful (:.
This is actually part (b) of the exercise. I was able to prove these inequalities by applying the Binomial Theorem, but I have no idea about how to do it using part (a), which I'll post here too.
(a) Let $b$ be a positive integer. Prove that:
$$b^p - a^p = (b-a)(b^{p-1}+ b^{p-2}a+b^{p-3}a^{2}+...+ba^{p-2}+a^{p-1})$$
(b) Let $p$ and $n$ denote positive integers. Use part (a) to show that
$$n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$$
(a) Proof:
$$\begin{split}b^p - a^p &= (b-a)(b^{p-1}+ b^{p-2}a+b^{p-3}a^{2}+...+ba^{p-2}+a^{p-1})\\
&= (b-a)\sum_{k=0}^{p-1}\big[b^{p-(k+1)}a^k\big]\\
&=\sum_{k=0}^{p-1}\big[b^{p-(k+1)+1}a^k- b^{p-(k+1)}a^{k+1}\big]\quad \Leftarrow \small{\text{Distribute }(b-a).}\\
&=\sum_{k=0}^{p-1}\big[b^{p-k}a^k- b^{p-(k+1)}a^{k+1}\big]\\
&=b^{p-0}a^0- b^{p-[(p-1)+1]}a^{(p-1)+1}\quad\Leftarrow \small{\text{Apply the telescoping property for sums.}}\\
&= b^p - a^p\end{split}$$
(b) This is my attempt:
$$n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$$
By multiplying by $p+1$ we have
$$(p+1)n^p < (n+1)^{p+1}-n^{p+1} < (p+1)(n+1)^p$$
which can also be written as
$$(p+1)n^p < [(n+1)-n]\sum_{k=0}^{p}(n+1)^{p-(k+1)}n^{k} < (p+1)(n+1)^p\\
(p+1)n^p < \sum_{k=0}^{p}(n+1)^{p-(k+1)}n^{k} < (p+1)(n+1)^p$$
By dividing the inequalities by $(n+1)^p$ we get
$$(p+1)\left(\frac{n}{n+1}\right)^p < \sum_{k=0}^{p}\frac{n^k}{(n+1)^{k+1}} < (p+1)$$
Here I'm stuck. I guess the last step wasn't necessary.
Edit:
This is my last attempt. Hopefully it's not flawed.
We will prove each bound separately.
To prove: $(p+1)n^p < (n+1)^{p+1}-n^{p+1}$
Proof (direct):
Let the numbers $a$ and $b$ be defined as
$$a = n \quad \text{and}\quad b = n+1\qquad \text{For }n \in \mathbb{N}.$$
Then we have
$$a < b$$
And it follows
$$\frac{a^p}{a^k} < \frac{b^p}{b^k}\quad \text{For } p\ \text{and }k\in\mathbb{N}, p \neq k. \qquad (1)$$
By multiplying both sides by $a^k$ we get
$$a^p < b^{p-k}a^k$$
By taking the sum of both sides we have
$$\sum_{k=0}^pa^p < \sum_{k=0}^pb^{p-k}a^k$$
On the LHS $a^p$ is summed $p+1$ times (from $0$ to $p$). So it can also be written as
$$(p+1)a^p < \sum_{k=0}^pb^{p-k}a^k$$
Since $b-a = 1$, let multiply the RHS by $b-a$
$$(p+1)a^p < (b-a)\sum_{k=0}^pb^{p-k}a^k$$
By distributing $b-a$ inside the sum we have
$$(p+1)a^p <\sum_{k=0}^p[b^{p-(k-1)}a^k - b^{p-k}a^{k+1}]$$
And by applying the telescoping property for sums we get
$$\begin{align*}(p+1)a^p &< b^{p-(0-1)}a^0 - b^{p-p}a^{p+1}\\
(p+1)a^p &< b^{p+1} - a^{p+1}\\
(p+1)n^p &< (n+1)^{p+1} - n^{p+1}\end{align*}$$
which completes the proof.
To prove: $(n+1)^{p+1}-n^{p+1} < (n+1)^p$
Proof (direct):
By the definition of $a$ and $b$ previously given and from inequality $(1)$ we have
$$\frac{a^p}{a^k} < \frac{b^p}{b^k}$$
Let multiplying both sides by $b^k$
$$a^{p-k}b^k < b^p$$
Let take the sum of both sides
$$\sum_{k=0}^pa^{p-k}b^k < \sum_{k=0}^pb^p$$
On the RHS $b^p$ is summed $p+1$ times. Then
$$\sum_{k=0}^pa^{p-k}b^k < (p+1)b^p$$
Let multiply the LHS by $1 = b-a$
$$\begin{align*}(b-a)\sum_{k=0}^pa^{p-k}b^k < (p+1)b^p\\
\sum_{k=0}^p[a^{p-k}b^{k+1}-a^{p-(k-1)}b^k] < (p+1)b^p\end{align*}$$
And by applying the telescoping property for sums we get
$$\begin{align*}a^{p-p}b^{p+1}-a^{p-(0-1)}b^0 &< (p+1)b^p\\
b^{p+1}-a^{p+1} &< (p+1)b^p\\
(n+1)^{p+1}-n^{p+1} &< (p+1)(n+1)^p\end{align*}$$
which completes the proof.
It looks like I can go to sleep without remorse, doesn't it? (:
| $$b^{p+1} - a^{p+1} = (b-a)(b^{p}+ b^{p-1}a+b^{p-2}a^{2}+ \cdots+a^{p})$$ Substitute $$b=n+1,\,\,\,\,\,\,\,a=n$$ Then $$(n+1)^{p+1}-n^{p+1}=(n+1)^p+n(n+1)^{p-1}+n^2(n+1)^{p-2}+\cdots+n^p$$ Since $$n<n+1$$ we have $$(p+1)n^p<(n+1)^p+n(n+1)^{p-1}+n^2(n+1)^{p-2}+\cdots+n^p<(p+1)(n+1)^p.$$ Hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Problem for system of equations Find the solution of the following set of equations
\begin{equation}
\begin{cases}
yz-2z+x-1=0 \nonumber \\
zx+y-z-2=0 \nonumber \\
xy-2x-y+z+2=0
\end{cases}
\end{equation}
| By summing (2) and (3) you end up with the equation:
$$
x(z+y-2)=0.
$$
Which has two solutions, either $x=0$ or $z+y-2=0$.
For the first case you end up with the following system:
\begin{align}
z(y-2)&=1\\
y-z&=2,
\end{align}
which is equivalent to $(y-2)^2=1$ and whose solutions are $(0,1,-1)$ and $(0,3,1)$.
For the other case you end up with the following system, given that $z=2-y$:
\begin{align}
y(2-y)-2(2-y) +x = -y^2 +x+4y -4 &=1\\
x(2-y)+y-(2-y)=-xy+2x +2y -2 &=2\\
xy-2x-2y+4&=0.
\end{align}
Observe that the second and third equation are the same.
Geomtrically the first equation is the horizontal parabola with vertex $(2,1)$:
$$
(y-2)^2 = (x-1).
$$
And the second is either a vertical or an horizontal line:
$$
x(y-2)-2(y-2) = (x-2)(y-2)=0.
$$
So the solutions are the possible intersections of them! To find them explicitly consider each possible case, $x=2$ or $y=2$, from which we obtain:
If $x=2$, then $y=1$ or $y=3$, with solutions $(2,1,1)$ and $(2,3,-1)$, respectively.
If $y=2$, then $x=1$ and $z=0$, with solution $(1,2,0)$.
These are all the possible solutions.
Plot: http://www.wolframalpha.com/input/?i=plot++%28y-2%29%5E2-%28x-1%29%3D0+and+x%3D2+and+y%3D2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the following systems of equations ( Matrices) Solve the following systems of equations:
\begin{Bmatrix}
x_1 & -x_2 & -x_3 & +0x_4 & = 2 \\
-x_1 & +2x_2 & +0x_3 & +3x_4 &= 1 \\
x_1 & +0x_2 & +x_3 & +0x_4 & =5\\
\end{Bmatrix}
I try to make a zero triangle but always reach a step where I'm in a cycle trying to do the same thing over and over. I once got $x_3=1/3$
Working:
1-
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
-1 & +2 & +0 & +3 & 1 \\
1 & +0 & +1 & +0 & 5\\
\end{Bmatrix}
2- Eq2 to Eq2 + Eq1
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
1 & +0 & +1 & +0 & 5\\
\end{Bmatrix}
3- Eq3 to Eq3 + Eq1
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
0 & 1 & 2 & 0 & 3\\
\end{Bmatrix}
4- Eq3 to Eq3 - Eq2
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
0 & 0 & 3 & 0 & 1\\
\end{Bmatrix}
Now I can say $3x_3=1$
Thus, $x_3=1/3$
After that i tried but couldn't complete...
Any suggestions would be appreciated
EDIT: The last step is incorrect.
It should be :
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
0 & 0 & 3 & -3 & 0\\
\end{Bmatrix}
So what I did next is:
From Third Row: $x_3+x_4=0$
so, $x_4=-x_3$
Then, From the Second row:
$x_2-x_3+3x_4=3$
$x_2-x_3+3(-x_3)=3$
$x_2-x_3=3$
$x_2=3+x_3$
And I stopped
| when i row reduce your matrix $\pmatrix{1&-1&-1&0&2\\-1&2&0&3&1\\1&0&1&0&5}$
i get $\pmatrix{1&0&0&1&5\\0&1&0&2&3\\0&0&1&-1&0}.$
let us call the variables $x_1, x_2, x_3$ and $x_4$ corresponding to the four columns. the fourth variable $x_4$ is called the free variable. to find a particular solution we set it to zero. you get a particular solution $x_1 = 5, x_2 = 3, x_3 = 0, x_4 = 0.$
to find the homogeneous solution we the free variable to one and delete the last column. so we $x_1 = -1, x_2 = -2, x_3 = 1, x_4 = 1.$
then general solution is $x_g = x_p + tx_h.$ in component form you have $$x_1 = 5-t, x_2 = 3-2t, x_3 = t, x_4 = t \text{ where $t$ is any real number.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine matrix A from linear transformations Let $T:R^3\rightarrow R^3$ be the linear transformation such that
$$
T
\left[
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right]
=\left[
\begin{array}{c}
-2\\
5\\
-2\\
\end{array}
\right],\,\,
T
\left[
\begin{array}{c}
1\\
1\\
0\\
\end{array}
\right]
=\left[
\begin{array}{c}
4\\
1\\
4\\
\end{array}
\right],\,\,
T
\left[
\begin{array}{c}
1\\
0\\
0\\
\end{array}
\right]
=\left[
\begin{array}{c}
1\\
-1\\
1\\
\end{array}
\right]
$$
a) Find a matrix $A$ such that $T(x)=Ax$ for every $x \in R^3$
b) Find a linearly independent set of vectors in $R^3$ that spans the range of $T$
I have absolutely no idea how to do these two questions and my textbook does not provide any similar examples. Could someone be kind enough to explain the logic behind answering these questions? Your help is appreciated!
| You should be able to see that the set of vectors
$$ u_1=\left[
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right], u_2=\,\left[
\begin{array}{c}
1\\
1\\
0\\
\end{array}
\right], u_3 =\, \left[
\begin{array}{c}
1\\
0\\
0\\
\end{array}
\right] $$
spans $\mathbb{R^3}$. Let $X\in \mathbb{R^3}$ so we can write it as
$$ X = a \,u_1 + b \,u_2 + c\, u_3 .$$
where $a,b,c$ are the coordinates of the vector $X$ with respect the above basis. Apply the operator $T$ to the vector $X$ gives
$$ T(X) = aT(u_1)+ bT(u_2)+ c T(u_3) = a\left[
\begin{array}{c}
-2\\
5\\
-2\\
\end{array}
\right] + b\left[
\begin{array}{c}
4\\
1\\
4\\
\end{array}
\right]+ c \left[
\begin{array}{c}
1\\
-1\\
1\\
\end{array}
\right]= \left[ \begin{array}{ccc} -2 & 4 & 1 \\ 5 & 1 & -1 \\ -2 & 4 & 1 \end{array}\right] \left[
\begin{array}{c}
a\\
b\\
c\\
\end{array}
\right] = A X.$$
I let you finish the problem.
| {
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"url": "https://math.stackexchange.com/questions/1146112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $a$, $r$, and $s$ are odd and $b$ is even Question: Suppose that $a$, $b$, $r$, $s$ are pairwise relatively prime and that $a^2 + b^2 = r^2$ and $a^2 - b^2 = s^2$. Prove that $a$, $r$, and $s$ are odd and $b$ is even.
My incomplete answer: Suppose $r$ is even then $a$ and $b$ are both even or both odd; then $s$ even, which is a contradiction with the assumption that $\gcd(r, s) = 1$. Therefore, $r$ is odd, then one of $a$ and $b$ is odd and the other is even, thus $s$ is odd. Since $\gcd(r, s) = 1$, then $|r - s| \ge 2$ (which doesn't help to solve!). How to know which of $a$ or $b$ is odd (then the other will be even)?
Thank you.
| Use the fact that if $x$ is odd, then $x^2 \equiv 1 \pmod{4}$. This is because $(2t + 1)^2 = 4(t^2 + t) + 1$.
Now first look at $a^2 + b^2 = r^2$. If $r$ is even, then $r^2 \equiv 0 \pmod{4}$. Then either (i) $a$ and $b$ are both even or (ii) $a$ and $b$ are both odd. In Case (i), we violate the gcd condition. In Case (ii), $a^2 + b^2 \equiv 2 \pmod{4}$, contradicting the fact that $r^2 \equiv 0 \pmod{4}$.
So $r$ is odd, and one of $a$ or $b$ is odd, and the other is even. We use $a^2 - b^2 = s^2$ to show $a$ must be odd. For if $a$ is even, then $a^2 - b^2 \equiv -1 \pmod{4}$, contradicting the fact that $s^2 \equiv 1 \pmod{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$? Question:
Are there $a,b,c,d \in \mathbb N$ such that $$a^2 + b^2 = c^2 \ \ \text{and} \ \ b^2 + c^2 = d^2$$
I'm a bit lost here.
| Suppose there exist such $a,b,c,d$ with $a\le b\le c\le d,$ then
$$c^2-a^2=d^2-c^2=b^2$$
$$2c^2=a^2+d^2$$
Suppose $$a=x-y,\,\,\,\,d=x+y.$$ Then $$c^2=x^2+y^2.$$
Therefore there exist $A,B$ such that $$c=A^2+B^2,\,\,\,x=A^2-B^2,\,\,\,y=2AB.$$
Now we have $$a=A^2-2AB-B^2\le(A-B)^2$$
$$d=A^2+2AB-B^2\le(A+B)^2$$
Also $$c^2-a^2=(A^2+B^2)^2-(A^2-2AB-B^2)^2=4AB(A-B)(A+B)$$ should be a PERFECT SQUARE.
That is all $A,B,A+B,A-B$ are should be perfect squares.
Therefore $$A=a_1^2,\,\,\,B=b_1^2$$
This happens only if $B$ is a perfect square and $A=0.$ Hence the only solution is $$a=c=d,\,\,\,\,\,b=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to factorize $n^5+n+1$? How to factorize $n^5+n+1$ ?
I think I should break $a^5$ and use a factorization formula. but how is it done?
| For completeness sake, the brute force approach:
Suppose there is a linear factor. Then
$$n^5+n+1=(n+a)(n^4+bn^3+cn^2+dn+e),$$
for some integers $a$, $b$, $c$, $d$ and $e$. In particular $ae=1$ so $a=\pm1$. Then plugging in $n=-a=\mp1$ yields
$$(-a)^5+(-a)+1=((-a)+a)((-a)^4+b(-a)^3+c(-a)^2+d(-a)+e)=0.$$
But this does not hold for $a=1$ or $a=-1$, a contradiction. So there is no linear factor.
Suppose there is a quadratic factor. Then
$$n^5+n+1=(n^2+an+b)(n^3+cn^2+dn+e).$$
As before $be=1$ so $b=e=\pm1$. Further comparing coefficients shows that
\begin{eqnarray*}
a+c&=&0,\\
b+ac+d&=&0,\\
bc+ad+e&=&0,\\
bd+ae&=&1.
\end{eqnarray*}
As $b=e=\pm1$ the latter shows that $d+a=\pm1$, and the second to last equation shows that
$$ad=\mp(c+1).$$
Then from $a+c=0$ it follows that $ad=\mp(-a+1)=\pm(1-a)$, which shows that $a$ divides $1-a$. This means that $a=1$ or $a=-1$, and $d=0$ or $d=2$ correspondingly (recall that $a+d=\pm1$). Either way we find that $a+d=1$ and hence $b=e=1$. Then the second equation above becomes
$$1-a^2+d=0,$$
because $a+c=0$, and we see that only $a=1$ and $d=0$ is a valid solution. This yields
$$n^5+n+1=(n^2+n+1)(n^3-n^2+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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} |
When is $(12x+5)/(12y+2)$ not in lowest terms? I am struggling to solve this problem and would appreciate any help:
When is $\frac{12x+5}{12y+2}$ NOT in lowest terms? ($x$,$y$ are nonnegative integers)
I have found that it is not in lowest terms for $x=6$ and $y=9$ because numerator and denominator are divisible by $11$, but I'm stuck here.
EDIT: Apparently "lowest terms" isn't in common usage in maths, so I will have to explain what it means. A fraction $p/q$ with $p,q\in \mathbb{Z}$ and $q\ne 0$ is in lowest terms when $\gcd(p,q)=1$. Otherwise, it is not in lowest terms.
For example, $\frac{3}{5}$ and $\frac{9}{2}$ are in lowest terms, but $\frac{15}{3}$ and $\frac{17}{34}$ are not.
| When is $\dfrac{12x+5}{12y+2} $ irreducible?
This is really just a partial answer.
So, when does $12x + 5 = A$ and $12y + 2 = B$ where $\gcd(A, B)=1$
Note that
\begin{align}
12 &\mid 2A - 5B \\
5B &\equiv 2A \pmod{12} \\
B &\equiv 10A \pmod{12} \\
B &= 12n + 10A
\end{align}
So $\gcd(A,B)=1$ becomes $\gcd(12n, A) = 1$.
\begin{align}
12y + 2 &= B \\
12y + 2 &= 12n + 10A \\
12y &= 12n + 10A - 2 \\
\hline
10A &\equiv 2 \pmod{12} \\
5A &\equiv 1 \pmod 6 \\
A &\equiv - 1 \pmod 6 \\
A &= 6\alpha - 1 \\
\hline
12y &= 12n + 10(6\alpha - 1) - 2 \\
12y &= 12n + 60\alpha - 12 \\
y &= n + 5\alpha - 1 \\
\hline
B &= 12n + 60\alpha - 10
\end{align}
Finally, we solve for $\alpha$.
\begin{align}
12x + 5 &= A \\
12x &= 6\alpha - 6 \\
2x &= \alpha - 1 \\
\alpha &= 2x + 1
\end{align}
Pick any integer value for $x$.
$ \alpha = 2x + 1 $
$ A = 12x + 5 $
Pick any $n$ such that $\gcd(12n, A) = 1$.
$B = 12n + 10A$
$y = n + 10x + 4$ implies
$$ \dfrac{12x+5}{12y+2} = \dfrac{12x+5}{12n + 120x + 50}$$
which is irreducible as long as $\gcd(12n, 12x+5) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
| The sum $\sum_{k=2}^n\frac{1}{k^2}$ is a lower Riemann sum for the function $1/x^2$ in the interval $[1,n]$, so
$$
\sum_{k=2}^n\frac{1}{k^2}\le\int_{1}^n\frac{1}{x^2}\,dx=
\left[-\frac{1}{x}\right]_1^n=1-\frac{1}{n}
$$
Therefore
$$
\sum_{k=1}^n\frac{1}{k^2}\le 1+1-\frac{1}{n}<2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:
*
*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y - xy^{m-2} + x^{m-3}y^2 + x^2y^{m-3} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + y^{m-1}$
*$\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-3} - x^{m-4}y - xy^{m-4} + x^{m-5}y^2 + x^2y^{m-5} + \dots + x^{\frac{m-3}{2}}y^{\frac{m-3}{2}} + y^{m-3}$
*$(xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-2}y - x^{m-3}y^2 - x^2y^{m-3} + x^{m-4}y^3 + x^3y^{m-4} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + xy^{m-2}$
*So that, $\dfrac{x^m + y^m}{x+y} + (xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$
*I am figuring that $\dfrac{x^{m-2} + y^{m-2}}{x+y}$ has exactly 2 terms less than $\dfrac{x^m + y^m}{x+y}$
Is this reasoning correct? For each value that I test, it seems correct.
Thanks,
-Larry
| Your argument is fine, possibly a bit more complicated than necessary. And the theorem is true for $m$ even, too, so a better proof wouldn't use the closed form for $\frac{x^m+y^m}{x+y}$, which is only true for $m$ odd.
A more advanced way to prove this uses linear recurrences. Show that $a_n=x^n+y^n$ satisfies:
$$a_{n}=(x+y)a_{n-1} - (xy)a_{n-2}$$
It is easy to show this linear recurrence is true for $b_n=x^n$ and $c_n=y^n$ because $z^2-(x+y)z+xy=0$ has roots $z=x,y$. So it is true for $a_n=b_n+c_n$.
The recurrence is also true if $a_n=\alpha\cdot x^n + \beta \cdot y^n$ for any $\alpha,\beta,x,y$. So:
$$\frac{\alpha x^m + \beta y^m}{x+y} + (xy)\frac{\alpha x^{m-2} + \beta y^{m-2}}{x+y} = \alpha x^{m-1} +\beta y^{m-1}$$
That shows that the $x+y$ in the denominator is not part of the original pattern, $a_1$, but related to the $x,y$ independent of the $\alpha,\beta$.
This can then be generalized to more variables. If $a_n=\alpha x^n+\beta y^n+\gamma z^n$ then:
$$a_n = (x+y+z)a_{n-1} -(xy+yz+xz)a_{n-2} + (xyz)a_{n-3}$$
You can go to more variables, with the coefficients of the linear recurrence being the (alternating) basic symmetric polynomials.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Argument of $3+4i$ for $z = \sqrt{3 + 4i}$, I am trying to put this in Standard form, where z is complex. I let $w = 3+4i$ and find that the modulus, $|w|=r$, is 5. I am having trouble solving for arg(w).
I find that $\tan^{-1}{\theta} = \frac{4}{3}$. However, this is not an angle well known. How do I find it?
| Suppose $\sqrt{3+4i}$ were in standard form, say $x+yi$. Then we would have $$\begin{align}
x+yi & = \sqrt{3+4i}\\
(x+yi)^2 & = 3+4i\\
(x^2-y^2) + 2xyi & = 3+4i
\end{align}
$$
so
$$\begin{align}
x^2 -y^2 &= 3 \\
2xy &= 4 \\
\end{align}
$$
From the second equation we have $y = \frac2x$. Putting this into the first equation we obtain $$x^2 - \frac4{x^2} = 3.$$ Multiplying through by $x^2$, then setting $z=x^2$ we obtain the quadratic equation $$z^2 -3z -4 = 0$$ which we can easily solve to obtain $z=4$. (The other root, $z=-1$, is spurious since $z = x^2$ and $x$ is real.) Then since $x^2=z$ and $y=\frac2x$ we get $\color{darkblue}{x=2, y=1}$ and $\color{darkred}{x=-2, y=-1}$.
| {
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"url": "https://math.stackexchange.com/questions/1151831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Math Induction $N$ greater than or equal to $2$ $(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{n})$=$(\frac{1}{n})$ So I'm trying to make it equal.
so $n$ is equal or greater than 2.
When i substitute $2$ to n, $(1-\frac{1}{2})$ = $(\frac{1}{2})$
then
$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{k})$=$(\frac{1}{k})$
$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{k})+(1-\frac{1}{K+1})$=$(\frac{1}{k+1})$
I can't make it equal
I got an answer of
$\frac{1}{k}+\frac{k}{k+1}=\frac{1}{k+1}$ I don't know if N greater than or equal to 2 has something to do with it. The usual problems that I solve just prove n is equal to 1. I did the algebra and still can't make my LHS equal to $\frac{1}{k+1}$
| As yashg mentions, you are adding and not multiplying. This might help:
$$\left(\prod_{k=2}^n\left(1-\frac1k\right)\right)\left(1-\frac1{n+1}\right)=\prod_{k=2}^{n+1}\left(1-\frac1k\right)$$
And
$$\begin{align}\left(\frac1n\right)\left(1-\frac1{n+1}\right)&=\frac1n-\frac1{n(n+1)}\\
&=\frac{n+1}{n(n+1)}-\frac1{n(n+1)}\\&
=\frac{n+1-1}{n(n+1)}\\
&=\frac{1}{n+1}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
| $\sum\limits_{cyc}\frac{a}{b+c-a}-3=\sum\limits_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{b+c-a}=$
$=\sum\limits_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{a+c-b}\right)=\sum\limits_{cyc}\frac{2(a-b)^2}{(b+c-a)(a+c-b)}\geq0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $m+\frac{4}{m^2}\ge3$ How to prove that $m+\frac{4}{m^2}\geq3$ for every $m>0$? I multiplied both sides by $m^2$ and finally got $m^3+4-3m^2\geq0$, yet don't know how to move further. Hint?
| Using $A.M\geq G.M$
$$\frac{\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{2}{m^2}+\frac{2}{m^2}}{6}\geq\frac{1}{2}$$
$$m+\frac{4}{m^2}\geq3$$
Equality occurs if
$$\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{2}{m^2}=\frac{2}{m^2}$$
$i.e.$ for $m=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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Evaluate $\int{\frac{x^3}{x^3-3x^2+9x-27}}dx$ So for the equation below I used long division first, and got $\int{(1+\frac{-3x^2+9x-27}{x^3-3x^2+9x-27}})dx$ (So my signs should be the other way around, thank you to Alijah Ahmed for pointing it out, see answer below). Now I'm not sure how to proceed, completing the square would give me something weird, I think maybe using substitution?
$$\int{\frac{x^3}{x^3-3x^2+9x-27}}dx$$
Update: using partial fractions, $A=1.5, B=1.5, C=-4.5$ and I obtained $x+ \frac{3}{2}ln(x-3)+ \frac{3}{4}ln(x^2+9)-\frac{3}{2}arctan(\frac{x}{3})+C$ as an answer.
| As the comments have mentioned, use partial fractions.
Also, note that the denominator $x^3-3x^2+9x-27$ factorises to $(x-3)(x^2+9)$ which helps, so you want to find $A$, $B$ and $C$ such that
$$\frac{A}{x-3}+\frac{Bx+c}{x^2+9}=\frac{\color{red}{+}3x^2\color{red}{-}9x\color{red}{+}27}{(x-3)(x^2+9)}$$
(you had the signs for the numerator after long division wrong - corrections in red in the equation above, i.e. you should have had $\int{(1+\frac{3x^2-9x+27}{x^3-3x^2+9x-27}})dx$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the summation of series $a^{i}(x+i)$ where $i$ is from $1$ to $N$ I have the following series and I am unable to figure out which series it belongs to and how to solve it
$a(x+1)+a^{2}(x+2)+…+a^{N}(x+N)$
Above series is a generalization of my actual series
$\dfrac{1}{2}(x+1)+\dfrac{1}{4}(x+2) +\cdots +\dfrac{1}{2^{N}}(x+N)$
| Read all sums as sums from $0$ to $n$
Using the identity $\sum_k a^k=\frac{1-a^{N+1}}{1-a}$ we can calculate $\sum_k ka^k$
$\sum_k (k+1)a^k=1+a+a^2+...+a^N+a+a^2+....+a^N+a^2+a^3+...+a^N+...+a^N=\frac{1-a^{N+1}}{1-a}+a\frac{1-a^{N}}{1-a}+a^2\frac{1-a^{N-1}}{1-a}+....a^N\frac{1-a}{1-a}=\frac{1+a+a^2+...+a^N - Na^{N+1}}{1-a}=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}}{1-a}$
So $\sum_k ka^k=\sum_k (k+1)a^k-\sum_k a^k=\frac{\frac{1-a^{N+1}}{1-a}-Na^{N+1}-(1-a^{N+1})}{1-a}=\frac{1-Na^{N+1}+(N-1)a^{N+2}}{(1-a)^2}-\frac{1}{1-a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$ it is not giving telescooping series.
$=\frac{1}{3}+\frac{3}{14}+\frac{2}{13}+0-\frac{1}{6}-\frac{1}{7}$
| An observation:
We can reduce this series to another one. Note that $$\frac{1}{n(n\pm 1)+1}=\frac{|n\pm 1 -n|}{n(n\pm 1)+1}=|\tan(\arctan (n\pm 1)-\arctan n)|$$ Thus given summation is $$S=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\arctan(n)-\arctan(n-1))-\tan(\arctan(n+1)-\arctan n)\right)\\=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\beta_n-\beta_{n-1})-\tan(\beta_{n+1}-\beta_n)\right)$$ where $\beta_n:=\arctan n,\ n\ge 1$. I am trying to think if there is a way to find this summation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration by De Moivre Theorem For example , integrate $\sin ^4(x) dx$.
I solved this question by reduction formula which is fairly easy. But my senior said that it would be easier if you expand the $\sin ^4(x) $by using De Moivre Theorem then integrate it. I've no idea of how De Moivre's Theorem would apply. Can anyone show me ? Thanks
| In order to evaluate the integral we have by De Moivre
$$(\cos x+i\sin x)^4=\cos4x+i\sin4x$$
Expanding the left hand side results in
$$\cos^4x+4i\cos^3x\sin x-6\cos^2x\sin^2x-4i\cos x\sin^3x+\sin 4x=\cos4x+i\sin4x$$
Equating the real component on both sides of the equation, we have
$$\cos^4x-6\cos^2x\sin^2x+\sin^4x=\cos4x$$
Rearranging this (and using some trig identities, including the double angle formula for $\cos$) results in
$$\begin{align}\sin^4x&=\cos4x+6\cos^2x\sin^2x-\cos^4x\\\sin^4x&=\cos4x-6(1-\sin^2x)\sin^2x-(1-\sin^2x)^2\\8\sin^4x &=\cos4x-4\color{blue}{(1-2\sin^2x)}+3\\\sin^4x &=\frac{1}{8}\cos4x-\frac{1}{2}\color{blue}{\cos2x}+\frac{3}{8}\end{align}$$
Thus the integral simplifies to
$$\int\sin^4x\ dx =\int\frac{1}{8}\cos4x-\frac{1}{2}\cos2x+\frac{3}{8}\ dx\\=\frac{1}{32}\sin4x-\frac{1}{4}\sin2x+\frac{3}{8}x+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the initial expression by any obvious means (squaring, multiplying by conjugate etc). Any idea how to do it? Can it be done?
| \begin{align}
\Big(\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}\Big)^2&=14+16-4\sqrt{7}-2\sqrt{14}\sqrt{16 - 4 \sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{14}\sqrt{4 - \sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{56 - 14\sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{49 - 2\times 7 \sqrt{7}+(\sqrt{7})^2}\\
&=30-4\sqrt{7}-4\sqrt{(7- \sqrt{7})^2}\\
&=30-4\sqrt{7}-4(7- \sqrt{7})\\
&=2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue? If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue ?
I notice quadratic residues of odd prime $13$ :
$$1^2\equiv 12^2\equiv 1\\2^2\equiv 11^2\equiv 4\\3^2\equiv 10^2\equiv 9\\4^2\equiv 9^2\equiv 3\\5^2\equiv 8^2\equiv 12\\6^2\equiv 7^2\equiv 10$$
$4^{-1}=10\\9^{-1}=3\\1^{-1}=1\\12^{-1}=12$
Is this true always ? How to go about proving ? thanks!
| Maps like $\,x\mapsto x^{-1}$ that are multiplicative $\,h(ab) = h(a)h(b)$ preserve squares $\,h(a^2)= (h(a))^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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What am i doing wrong when solving this differential equation $$
f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}
$$
Given that $f(x) = cx$
$$
\frac{c}{2}x^2 + k_1 = \ln(e^y y')
$$
$$
k_2\int e^{\frac{c}{2}x^2} dx = e^y
$$
$$
y = \ln(k_2\int e^{\frac{c}{2}x^2} dx)
$$
Therefore
$$
y'^2 - cxy' + y'' = 0
$$
$$
y' = \frac{c}{2}x \pm \sqrt{\frac{c^2}{4}- y''}
$$
$$
\frac{e^{\frac{c}{2}x^2}}{\int e^{\frac{c}{2}x^2} dx} = \frac{c}{2}x \pm \sqrt{\frac{c^2}{4}x^2- y''}
$$
Yet when $c=4$
$$
\frac{2\sqrt{\frac{2}{\pi}}e^{2x^2}}{\operatorname{erfi}(\sqrt{2}x)+c} = 2x \pm\sqrt{4x^2 + 8e^{2x^2}\frac{e^{2x^2}-\sqrt{2\pi}x\operatorname{erfi}(\sqrt{2}x)}{\pi \operatorname{erfi}(\sqrt{2}x)^2}}
$$
Is not correct, what am i doing wrong?
| $(e^{y(x)})'=k_2e^{\frac12Cx^2}$ implies $e^{y(x)}=k_2\int_a^xe^{\frac12Ct^2}dt+k_3$, where $k_3=e^{y(a)}$ and which for $C>0$ does not involve the error function.
Then,
$$y(x)=\ln\left(k_2\int_a^xe^{\frac12Ct^2}dt+k_3\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
| Recall that, as $u \to 0$, by the Taylor series expansion, we readily have
$$
\begin{align}
e^u& =1+u+\mathcal{O}(u^2)\\
\ln (1+u)&=u+\mathcal{O}(u^2)
\end{align}
$$ giving, as $x \to \infty$,
$$
x^2\ln \left(1+\frac {2}{x^2-1}\right)=x^2 \left(\frac {2}{x^2-1}+\mathcal{O}\left(\frac {1}{x^4}\right)\right)=2+\mathcal{O}\left(\frac {1}{x^2}\right)
$$ and
$$
\begin{align}
\Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}&=\Big(\frac{x^2-1+2}{x^2-1}\Big)^{x^2}\\\\
&= \left(1+\frac {2}{x^2-1}\right)^{x^2}\\\\
&=e^{\large x^2\ln \left(1+\frac {2}{x^2-1}\right)}\\\\
&=e^{\large 2+\mathcal{O}(\frac {1}{x^2})}\\\\
&\to e^2
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9.
Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise say a 9 if the question required us to prove that the expression is divisible by 9. However in this case, I do not end up with such a thing.
My work so far below:
Inductive Hypothesis: $7^k(3k+1)-1=9N$ where $N\in\mathbb{N}$
Inductive Step:
$7^{k+1}(3k+4)-1 \\ =7\times 7^k(3k+1+3)-1 \\ =7\times \left [ 7^k(3k+1)+3\times 7^k \right ] -1 \\ = 7 \times \left [ 9N+1 + 3 \times 7^k \right ] -1 \\ = 63N+21\times 7^k+6 \\ = 3 \left [ 21N+7^{k+1}+2 \right ]$
So now I need to somehow prove that $21N+7^{k+1}+2$ is divisible by 3, but I'm not quite sure how to proceed from here...
| Here's the inductive step. Assume the claim is true for $n=k$. Then we know that
$$9\mid 7^k(3k+1)-1.$$
Consider the case where $n=k+1$. In this case, the expression is
$$7^{k+1}(3(k+1)+1)-1.$$
Now, let's simplify this expression to
$$7[7^k(3k+1)+7^k\cdot 3]-1.$$
We observe that $7^k(3k+1)$ appears in this expression, which is almost the inductive hypothesis. By adding and subtracting 1, we get
$$7[7^k(3k+1)-1+1+7^k\cdot 3]-1.$$
The expression $7^k(3k+1)-1$ is divisible by 9 by the inductive hypothesis, so it can be ignored. This leaves that we'd like to have
$$7^{k+1}\cdot 3+6$$ being divisible by $9$.
To prove this, you can do another induction: $7^n\cdot 3+6$ is divisible by $9$ for all $n\geq 1$. When $n=1$, then this expands to $27$, which is divisible by $9$.
For the inductive case, assume that for $n=k$,
$$9\mid 7^k\cdot 3+6$$ and consider $n=k+1$.
In this case, you have $$7^{k+1}\cdot 3+6=7[7^k\cdot 3]+6.$$
Noticing that $7^k\cdot 3$ is almost the inductive hypothesis, this can be simplified to
$$7[7^k\cdot 3+6-6]+6.$$
The $7^k\cdot 3+6$ is divisible by $9$ by the inductive hypothesis, and this leaves $7\cdot 6-6=36$, which is divisible by $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Number of distinct terms in the expansion of $\big(x+\frac{1}{x}+x^2+\frac{1}{x^2}\big)^{15}$
Number of distinct terms in the expansion of $\bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15}$ is equal to ?
We can write the above as,
$$ \bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15} = \dfrac{1}{x^{30}}(1+x+x^3+x^4)^{15} $$
Now my teacher says that we can expand the polynomial $(1+x+x^3+x^4)^{15} $ as,
$$ (1+x+x^3+x^4)^{15} = a_0 + a_1x + a_2x^2 + a_3x^3.....a_{60}x^{60} $$
Hence, as each term is divided by $x^{30}$, the number of distinct terms would be equal to 61.
But my question is how do we know that the expansion of the polynomial $(1+x+x^3+x^4)^{15} $ will contain all powers of $x$ from $x^0$ to $x^{60}$ ?
| A general term (without the coefficient) of the polynomial $(1+x+x^3+x^4)^{15}$ will look like
$$x^{a+3b+4c} \qquad \text{ where } 0 \leq a,b,c, \leq 15.$$
So your question is now to show that $a+3b+4c$ can take on all integer values between $0$ and $60$.
For numbers in $0 \to 15$ let $a$ vary and have $b=c=0$.
Now try doing the same for other ranges to convince yourself that all powers will be obtained for the right combinations of $a,b$ and $c$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Remainder of division by $23$ Find
$$(2014^{16}+2016^{32})^9 \ \ \ (\text{mod }23)$$
I know that
$$2014=2\times 19 \times 53$$
$$2016=2^5\times 3^2\times 7$$
This means that $2014 \equiv 2 \times (-4)\times (-3)$. I've tried solving this in multiple ways but I can't get the same answer as Wolframalpha, which is $(2014^{16}+2016^{32})^9 \equiv 15$ (mod $23$).
I could use some help.
| This can be done simply and quickly with purely mental arithmetic mod $\,23.\,$
We transform powers to powers of $\,2\,$ and use $\,\color{#c00}{a^{22}\equiv 1}\,$ if $\,23\nmid a\,$ by little Fermat.
$2016 = 20(100)+16 \equiv -3(8)+16 \equiv -8 \ \Rightarrow\ 2014\equiv 2016-2\equiv -10$
$2016^{32}\equiv (-8)^{32} \color{#c00}\equiv 8^{10}\equiv \color{#0a0}{2^{30}}\color{#c00}\equiv 2^8 = (2^4)^2 = (-7)^2 = 3$
$2014^{16}\equiv (-10)^{16} = 100^8 \equiv 8^8\equiv 2^{24}\color{#c00}\equiv 2^2\equiv 4$
$(2016^{32}\! + 2014^{16})^9\equiv (3\!+\!4)^9\equiv (-16)^9 \equiv -(2^4)^9\equiv -\color{#0a0}{2^{30}} 2^6 \equiv -3(-5) \equiv 15$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
When is $20q^4-40q^3+30q^2-10q$ a square for positive integer $q$?
For what $q$ is the following polynomial a square?
$$
\begin{align}
&20q^4-40q^3+30q^2-10q\\
=\:&10q(q - 1)(2q^2 - 2q + 1)
&q\in\mathbb N
\end{align}
$$
I know of two single cases, $q=1$ gives $0$ and $q = 2$ gives $100$. I tested $3\le q\le12$ and found none.
This is what I've tried so far:
I found that
$$
\begin{align}
\gcd(q,q - 1) &= 1\\
\gcd(q, 2q^2 - 2q + 1) &= 1\\
\gcd(q - 1, 2q^2 - 2q + 1) &= 1
\end{align}
$$
The problem is the same as determining whether there exists a $t\in\mathbb N$ such that
$$
q(q - 1)(2q^2 - 2q + 1) = 10t^2
$$
If $q$ is even, write $q=2p$ and say
$$
p(2p-1)(8p^2-4p+1) = 5t^2
$$
The residues modulo $5$ are as follows
$$
\begin{align}
p\equiv0\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv1\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv2\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv3\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv4\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv4\pmod{5}
\end{align}
$$
Which means that $p\not\equiv4\pmod5$, and we now need to test if it's a square in the rest of the cases.
Case r = 5p
In this case we are solving $r\cdot(10r - 1)(200r^2 - 20r + 1) = t^2$ and since the factors are coprime, they must all be squares, however modulo $7$ atleast one factor is not one of the quadratic residues for all cases.
Case r = 5p + 1
In this case we are solving $(5r+1)(10r+1)(40r^2+12 r+1) = t^2$ and since the factors are coprime, they must all be squares, however modulo $3$ atleast one factor is not one of the quadratic residues for cases $1$ and $2$, and the last case $0$ we have an observed solution at $q=2$, however for $q=12$ we don't get a square, so it remains to prove whether there are other than $r=0$ which are square.
At this point I stopped because there are many cases, and I don't know how to prove that it is only for $r=0$ when $r=5p+1$ that the polynomial is square, or if there are other such $r$.
So the thing I'm asking specifically is: Is there a more elegant way, and if not, how do I prove that the observed cases are the only solutions (or if not, what is the set of solutions).
| A similar Pell approach.
We want solutions to
$$10q(q-1)(2q^2-2q+1)=n^2$$
So $q\in{0,1}$ with $n=0$ are trivial examples. We might get solutions as linear combinations of these.
Let $2p=q(q-1)$.
Then $$5(4p)(4p+1)=n^2$$
We know $n$ must be a multiple of 5, so let n=5k.
$$5(4p)(4p+1)=25k^2$$
$$(4p)^2+4p=5k^2$$
Let x=4p and Complete the square:
$$(x)^2+x+\frac{1}{4}=5k^2+\frac{1}{4}$$
$$ (x+\frac{1}{2})^2-5k^2=\frac{1}{4} $$
Then multiply by 5:
$$(2x+1)^2-5(2k)^2=1$$
Since $x=4p$, and $2p=q(q-1)$,
$$[4q(q-1)+1]^2-5(2k)^2=1$$
With factoring:
$$ (2q-1)^4-5(2k)^2=1$$
We get a Pell equation: $m^2-5n^2=1$
Initial values for $(m,n)=(9,4)$ implying $q=2$ and $k=2$, so $n=10$
Then you can apply the Brahmagupta Solution to Pell's equation which allows one to generate infinitely many solutions from a single one.
Given $x^2-py^2=1$ and solution $(x_0,y_0)$:
$$x_{n+1}=x_0x_n+py_0y_n$$
$$y_{n+1}=y_0x_n+x_0y_n$$
So you want $x_n$ that are odd perfect squares.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integral solutions to $a+b+c=abc$. Find all integral solutions of the equation $a+b+c=abc$.
Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I've tried by taking $a,b,c=1,2,3$.
| Of course, if one of $a,b,c$ are zero, then you can only have $\{a,b,c\}=\{0,k,-k\}$ for some integer $k$. So suppose none of $a,b,c$ are zero. Since $a,b,c$ satisfies $a+b+c=abc$ if and only if $-a,-b,-c$ satisfies $a+b+c=abc$, we can assume that either exactly one of $a,b,c$ is negative or none are negative.
Now dividing $a+b+c=abc$ by $abc$ results in the equation $$\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1.$$
*
*If one of $a,b,c$ are negative, so say $a<0$, (and so $b,c\geq 1$) then we have $\frac{1}{bc}\leq 1< 1-\frac{1}{ab}-\frac{1}{ac}=\frac{1}{bc}$, an impossibility.
*If $a,b,c>0$, then at least one of $ab, ac$ or $bc$ is at most $3$, and all are greater than $1$. Without loss of generality, we can say $1<ab\leq 3$. This can only be the case if $a,b\in \{1,2,3\}$, with exactly one of them equal to $1$. So say $a=1$. Hence $1+b+c=bc$ with $b=2$ or $b=3$. If $b=2$ we get $c=3$; if $b=3$, then $c=2$.
In summary, the integral solutions of $a+b+c=abc$ are $\{0,k,-k\}$ for any integer $k$, and $\{1,2,3\}$ and $\{-1,-2,-3\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2)$. Given $a,b,c,d \in \Bbb R$, I want to prove that: $$(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2).$$
But the bound I'm getting is just too weak. I know that: $$|a| = \sqrt{a^2} \leq \sqrt{a^2+b^2} \implies |a|+|b| \leq 2\sqrt{a^2+b^2}\implies (|a|+|b|)^2\leq 4(a^2+b^2).$$ This gives me: $(|a|+|b|)^2(|c|+|d|)^2 \leq 16(a^2+b^2)(c^2+d^2)$. Not good. I tried reverse engineering it: $$\begin{align} (|a|+|b|)^2 &\leq 2(a^2+b^2) \\ |a|+|b| &\leq \sqrt{2}\sqrt{a^2+b^2} \\ |a|+|b| &\leq \sqrt{2a^2+2b^2} \\ |a|+|b|&\leq \sqrt{(a\sqrt{2})^2+(b\sqrt{2})^2},\end{align}$$ which suggests that I may try to prove that: $$|a| \leq \frac{\sqrt{2}}{2}\sqrt{a^2+b^2}.$$
I'm failing to see the step needed here. I just need a little push. Thanks.
| Note that
$$
(|a|+|b|)^2=a^2+b^2+2|a||b|\leq a^2+b^2+a^2+b^2=2(a^2+b^2)
$$
and similarly for $(|c|+|d|)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.
I feel that this may be too elementary, is this the correct way to solve this?
| This isn't elegant, but it doesn't take that long. Suppose, for contradiction, that there is a solution modulo $49$. In order for there to be a solution modulo $49$, there must be a solution modulo $7$.
$$x^2+4x+4\equiv \begin{cases}
4 \pmod{7} &\mbox{if } x \equiv 0 \pmod{7}\\
2 \pmod{7} &\mbox{if } x \equiv 1 \pmod{7}\\
2 \pmod{7} &\mbox{if } x \equiv 2 \pmod{7}\\
4 \pmod{7} &\mbox{if } x \equiv 3 \pmod{7}\\
1 \pmod{7} &\mbox{if } x \equiv 4 \pmod{7}\\
0 \pmod{7} &\mbox{if } x \equiv 5 \pmod{7}\\
1 \pmod{7} &\mbox{if } x \equiv 6 \pmod{7}\\
\end{cases}
$$
Now, following Paolo Leonetti, if $x \equiv 5 \equiv -2 \pmod{7}$, then $$x^2+4x+4 \equiv 0 \pmod{49}$$
But this, along with our original equivalence, gives the ridiculous conclusion
$$14 \equiv (x^2+4x+18)-(x^2+4x+4)\equiv 0 \pmod{49}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
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