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Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is there any easy way to calculate this?
Please someone help.
| Hint. First observe that
$$
\frac{1}{2i(2i+1)(2i+2)}=\frac{1}{2}\left(\frac{1}{2i(2i+1)}-\frac{1}{(2i+1)(2i+2)}\right)=\frac{1}{2}\left(\frac{1}{2i}-\frac{2}{2i+1}+\frac{1}{2i+2}\right)
$$
Then
$$
\sum_{i=1}^n\frac{1}{2i(2i+1)(2i+2)}=\sum_{i=1}^n\left(\frac{1}{2i}-\frac{1}{2i+1}\right)-\frac{1}{4}+\frac{1}{2(2n+2)}\\=\sum_{i=1}^n\frac{1}{2i(2i+1)}-\frac{1}{4}+\frac{1}{2n(2n+2)}
$$
Clearly
$$
\sum_{i=1}^n\frac{1}{2i(2i+1)}=\int_0^1\int_0^x(t+t^3+t^5+\cdots+t^{2n-1})\,dt\,dx\longrightarrow\int_0^1\int_0^x\frac{t\,dt}{1-t^2}\,dx\\=\frac{1}{2}\int_0^1\int_0^x\left(\frac{1}{1-t}-\frac{1}{1+t}\right)dt\,dx=-\frac{1}{2}\int_0^1\left(\log(1-x)+\log(1+x)\right)\,dx=\cdots
$$
We have
$$
\int_0^1 \log(1-x)\,dx=\int_0^1 \log x\,dx=\left.x\log x-x\right|_0^1=-1, \\
\int_0^1\log(1+x)\,dx=\int_1^2\log x\,dx=\left.x\log x-x\right|_1^2=2\log 2-1.
$$
Hence
$$
\sum_{i=1}^\infty\frac{1}{2i(2i+1)}=1-\log 2
$$
and finally
$$
\sum_{i=1}^\infty\frac{1}{2i(2i+1)(2i+2)}=\frac{3}{4}-\log 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\sqrt2}$
But correct answer is $\pm\frac{1}{\sqrt2}$
Where am I going wrong?
| Okay so expand the double angle to get $\cos^2(\sin^{-1}(-x))-\sin^2(\sin^{-1}(-x)=0$.
This should give you $(1-(-x)^2)-(-x)^2=0$.
Finally you have $1-2x^2=0$.
Solutions are $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Green's Theorem in the plane When calculating $\oint _{ C }^{ }{ \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } dx+\left( \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 } \right) dy } $ where $C$ Is the boundary of the region $D=\left\{ \left( x,y \right) \in \mathbb{R}^{ 2 }: 1\le { x }^{ 2 }+{ y }^{ 2 }\le 4,x\ge 0,y\ge 0 \right\} $,
${ C }_{ 1 }:\alpha \left( t \right) =\left( t,0 \right) ,t\in \left[ 1,2 \right] $,
${ C }_{ 2 }:\alpha \left( t \right) =\left( 2\cos { t } ,2\sin { t } \right) ,t\in \left[ 0,\frac { \pi }{ 2 } \right] $,
${ C }_{ 3 }:\alpha \left( t \right) =\left( 0,4-2t \right) ,t\in \left[ 0,1 \right] $, and
${ C }_{ 4 }:\alpha \left( t \right) =\left( \sin { t } ,\cos { t } \right) ,t\in \left[ 0,\frac { \pi }{ 2 } \right] $
are the curves bordering the region $D$. Differs by using Green's theorem, which is the reason for it, unless it calculates the integrals badly. Thanks for your help.
| Let $P(x,y)= \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 }$ and $Q(x,y)= \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 }$.
Then
$$
\begin{align*}
\oint _{ C }^{ }{ \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } dx+\left( \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 } \right) dy }
&= \iint_D ( Q_x-P_y ) dA \\
&= \iint_D x+y \: dA \\
&= \int_{0}^{\pi/2} \int_{1}^{2} (r \cos \theta + r\sin\theta)\; rdrd\theta \\
&= \int_{0}^{\pi/2} \int_{1}^{2} r^2 (\cos \theta + \sin\theta) drd\theta \\
&= \int_{0}^{\pi/2} \dfrac{r^3}{3} (\cos \theta + \sin\theta) \Bigg|_{1}^{2} d\theta \\
&= \int_{0}^{\pi/2} \dfrac{7}{3} (\cos \theta + \sin\theta) d\theta \\
&= \dfrac{7}{3} (\sin \theta - \cos\theta) \Bigg|_{0}^{\pi/2} \\
&= \dfrac{7}{3}(2) \\
&= \frac{14}{3}. \\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
| The limit at $\infty$ does not exist: consider the sequence
$$
a_n=\frac{\pi}{3}+2n\pi
$$
Then the numerator evaluated at $a_n$ is
$$
1-\cos\frac{\pi}{3}\cos\frac{2\pi}{3}\cos\pi=1-\frac{1}{4}=\frac{3}{4}
$$
and the denominator is $\frac{3}{4}$, so the quotient is $1$.
On the other hand, for the sequence
$$
b_n=\frac{\pi}{6}+2n\pi
$$
the numerator is $1$ and the denominator is $1/4$. So
$$
1=\lim_{n\to\infty}\frac{1-\cos a_n\cos2a_n\cos3a_n}{\sin^2a_n}
\ne
\lim_{n\to\infty}\frac{1-\cos b_n\cos2b_n\cos3b_n}{\sin^2b_n}=4
$$
For the limit at $0$, which is probably what you have to compute, the easiest way is to use l’Hôpital: after the first run you have
$$
\lim_{x\to0}\frac{\sin x\cos2x\cos3x+2\cos x\sin 2x\cos3x+3\cos x\cos2x\sin3x}{2\sin x\cos x}
$$
that you can rewrite, using $2\sin x\cos x=\sin2x$ and that
$$
\lim_{x\to0}\frac{\sin3x}{\sin2x}=\frac{3}{2}
$$
as
$$
\lim_{x\to0}
\left(
\frac{\cos2x\cos3x}{2\cos x}
+2\cos x\cos 3x
+3\cos x\cos2x\frac{\sin3x}{\sin2x}
\right)=
\frac{1}{2}+2+3\cdot\frac{3}{2}=7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Find the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$ I've been asked to find, if it exists, the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$
To start, I've tried to find the points of the ellipsoid where the distance is minimum and maximum using the method of Lagrange multipliers.
First, I considered the sphere centered on the point $P$ as if it were a level surface and I construct:
$f(x,y,z)=(x-4)^2+(y-2)^2+z^2-a^2$
And I did the same with the ellipsoid, getting the following function:
$g(x,y,z)=x^2+y^2+2z^2-6$
Then, I tried to get the points where $\nabla$$f$ and $\nabla$$g$ are paralell:
$\nabla$$f=\lambda$$\nabla$$g$ $\rightarrow$ $(2x-8,2y-4,2z)=\lambda(2x,2y,4z)$
So, solving the system I got that $\lambda=1/2$, so in consquence $x=8$ and $y=4$
But when I replaced the values of $x,y$ obtained in $g=0$ to get the coordinate $z$ I get a complex number. I suspect there is something on my reasoning which is incorrect, can someone help me?
| Since by C-S $$2x+y\leq\sqrt{(x^2+y^2)(2^2+1^2)}=\sqrt{5(x^2+y^2)},$$
we obtain
$$(x-4)^2+(y-2)^2+z^2=x^2+y^2+z^2-4(2x+y)+20\geq $$
$$\geq x^2+y^2+z^2-4\sqrt{5(x^2+y^2)}+20=6-2z^2+z^2-4\sqrt{5(6-2z^2)}+20=$$
$$=26-z^2-4\sqrt{5(6-2z^2)}\geq26-4\sqrt{30}.$$
It's obvious that the equality occurs, which gives the minimal value:
$$\sqrt{26-4\sqrt{30}}.$$
Since $-2x-y\leq\sqrt{5(x^2+y^2)}$, by the same way we can get a maximal value: $$\sqrt{26+4\sqrt{30}}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
If 25 divides $a^5 + b^5 + c^5 + d^5 + e^5$ then prove that 5 divides $abcde$. I solved the following problem and giving the outline of my solution. I'm curious to know if a shorter and more elegant solution exists.
Problem
Let $a,b,c,d,e \in \mathbb{N}$ such that $25 | (a^5 + b^5 + c^5 + d^5 + e^5)$. Prove that $5 | abcde$.
Solution Outline
Use contradiction.
$5 \nmid abcde \implies 5 \nmid a,b,c,d,e$.
If $n = 5q + r$ then $n^5 = 25k + r^5$ which can be proved by Binomial Expansion. Thus $n \equiv r (mod \; 5) \implies n^5 \equiv r^5 (mod \; 25)$.
Now possible values of $r$ are $1,2,3,4$. Hence possible values of $r^5 (mod \; 25)$ are $1,7,-7,-1$. (We assume that the set of remainders modulo 25 is $\{-12, -11, \dots, -1, 0, 1, \dots, 11, 12 \}$). Thus $a^5, b^5, c^5, d^5, e^5 \equiv 1,7,-7,-1 (mod \; 25)$ in some order.
Then I spent a major chunk of my solution proving that $x_1 + x_2 + x_3 + x_4 + x_5 = 0 (mod \; 25)$ has no solutions if $x_1, x_2, x_3, x_4, x_5 \in \{1, 7, -7, -1 \}$. This part of the proof was essentially a brute force attack, and I'm not fully satisfied with it.
Question
I'm curious to know if this last part has a shorter/more elegant solution. Also, I'd be interested in any other solution overall.
Thanks!!
| Since $5^1\binom{5}{1}=25$, we have $(n+5)^5\equiv n^5\pmod{25}$. Thus, we can generate the table
$$
\begin{array}{c|c}
n\pmod{5}&n^5\pmod{25}\\
0&0\\
1&1\\
2&7\\
3&-7\\
4&-1\\
\end{array}
$$
Since there are no solutions to $7x+y=25$ with $|x|+|y|\le5$, we must be looking for $7x+y=0$ with $|x|+|y|\le5$, which has only one solution: $x=y=0$. This corresponds to an equal number of $7\pmod{25}$ and $-7\pmod{25}$ and an equal number of $1\pmod{25}$ and $-1\pmod{25}$. Since $5$ is odd, there must be at least one $0\pmod{25}$ among $a^5,b^5,c^5,d^5,e^5$. This means at least one $0\pmod{5}$ among $a,b,c,d,e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Can we identify the limit of this arithmetic/geometric mean like iteration? Let $a_0 = 1$ and $b_0 = x \ge 1$. Let
$$
a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n+1} b_n})/2.
$$
Numeric computation suggests that regardless of the choice of $x$, $a_n$ and $b_n$ always converge to the same value. Can we prove this?
Moreover, if we assume this is true and define $f(x) = \lim_{n \to \infty} a_n$, then numeric computation shows that
$$
f'(1) \approx 0.57142857142857 \approx 4/7 \\
f''(1) \approx -8/49 \\
f^{(3)}(1) \approx 1056/4459 \\
f^{(4)}(1) \approx -65664/111475.
$$
There seems to be some patterns here.
Can we actually find the limit with these clues?
I was actually trying to check this recursion
$$
a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n} b_n})/2.
$$
I made a mistaked in my code and computed the one at the top. See The Computer as Crucible, pp 130.
| We have $1\le a_0\le b_0\le x$.
We can prove that $1\le a_n\le a_{n+1}\le b_{n+1}\le b_n\le x$ by mathematical induction.
$\displaystyle a_1=\frac{1+\sqrt{x}}{2}$ and $\displaystyle b_1=\frac{x+\sqrt{a_1x}}{2}$.
Obviously, $a_1\ge 1$. Hence $\displaystyle b_1-a_1=\frac{(x-1)+(\sqrt{a_1}-1)\sqrt{x}}{2}\ge0$
Note that
$$a_1-x=\frac{1+\sqrt{x}-2x}{2}=\frac{(1-\sqrt{x})(1+2\sqrt{x})}{2}\le0$$
and
$$b_1\le\frac{x+\sqrt{x\cdot x}}{2}=x$$
Therefore, $1\le a_0\le a_{1}\le b_{1}\le b_0\le x$.
Suppose that $1\le a_k\le a_{k+1}\le b_{k+1}\le b_k\le x$. Then
$$a_{k+2}=\frac{a_{k+1}+\sqrt{a_{k+1}b_{k+1}}}{2}\ge\frac{a_{k+1}+\sqrt{a_{k+1}a_{k+1}}}{2}=a_{k+1}$$
$$b_{k+2}=\frac{b_{k+1}+\sqrt{a_{k+2}b_{k+1}}}{2}\ge\frac{a_{k+1}+\sqrt{a_{k+1}b_{k+1}}}{2}=a_{k+2}$$
and
$$b_{k+2}=\frac{b_{k+1}+\sqrt{a_{k+2}b_{k+1}}}{2}\le\frac{b_{k+1}+\sqrt{b_{k+2}b_{k+1}}}{2}$$
So,
\begin{align}
b_{k+2}&\le\frac{b_{k+1}+\frac{b_{k+2}+b_{k+1}}{2}}{2}\\
4b_{k+2}&\le3b_{k+2}+b_{k+1}\\
b_{k+2}&\le b_{k+1}
\end{align}
This complete the induction proof.
So, $\{a_n\}$ is increasing and bounded above by $x$. $\{b_n\}$ is decreasing and bounded below by $1$. Both the sequences converge.
Let $\displaystyle \lim_{n\in\infty} a_n=a$ and $\displaystyle \lim_{n\in\infty} b_n=b$. Then we have
$$a=\frac{a+\sqrt{ab}}{2}$$
This implies that $a=b$.
Unsolved: How to find the value of the common limit?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Integrating the Fourier series to find the Fourier series of $\frac{1}{2}x^2$
The function $\phi(x) = x$ on the interval $[-l,l]$ has the Fourier series
$$x = \frac{2 l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l} \right) = \frac{2 l}{\pi}\left(\sin\left(\frac{\pi x}{l} \right) - \frac{1}{2}\sin\left(\frac{2\pi x}{l} \right) + \frac{1}{3}\sin\left(\frac{3\pi x}{l} \right) - \ldots \right)$$
Integrate the series term-by-term to find the Fourier series for $\frac{1}{2}x^2$, up tp a constant of integration (which is then the $\frac{1}{2}A_0$ term in the cosine series). Find the $A_0$ using the standard formula to completely determine the series.
Attempted solution - We have
\begin{align*}
&\sum_{m=1}^{\infty}\int_{0}^{l}\frac{2l}{\pi}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l}\right)dx\\
&= \frac{2l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\int_{0}^{l}\sin\left(\frac{m\pi x}{l}\right)dx\\
&= \frac{2l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\left[-\frac{l}{\pi m}\left(\cos\left(\frac{m\pi x}{l}\right)\Big|_0^l\right)\right]\\
&= -\frac{2 l^2}{\pi^2}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2}\left(\cos\left(\frac{m \pi l}{l}\right) - 1\right)\\
\end{align*}
I am not sure where to go from here, any suggestions are greatly appreciated.
| What has been given is
$$\frac{x^2}{2} = -\frac{2 l^2}{\pi^2}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2}\left(\cos\left(\frac{m \pi l}{l}\right) - 1\right).$$
Now consider:
$$\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
and
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} &= \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots \\
&= \sum_{n=1}^{\infty} \frac{1}{n^2} - 2 \, \sum_{n=1}^{\infty} \frac{1}{(2 n)^2} \\
&= \frac{1}{2} \, \zeta(2) = \frac{\pi^2}{12}.
\end{align}
This then leads to
$$\frac{x^2}{2} = \frac{l^2}{6} -\frac{2 l^2}{\pi^2}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2} \, \cos\left(\frac{m \pi l}{l}\right)$$
or
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \, \cos\left(\frac{n \pi x}{l}\right) = \frac{\pi^2}{12} - \left(\frac{\pi x}{2 \, l}\right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What are the diagonals in the matrix called? How correctly to designate these diagonals? They are highlighted in different colors.
\begin{pmatrix}\color{green}1&\color{orange}2&\color{red}3&\color{blue}4\\
\color{orange}5&\color{red}6&\color{blue}7&\color{green}8\\
\color{red}9&\color{blue}{10}&\color{green}{11}&\color{orange}{12}\\
\color{blue}{13}&\color{green}{14}&\color{orange}{15}&\color{red}{16}\end{pmatrix}
$\color{green}{1, 14, 11, 8}$
$\color{orange}{5, 2, 15, 12}$
$\color{red}{9, 6, 3, 16}$
$\color{blue}{13,10, 7, 4}$ - secondary diagonal
| A cyclic generator matrix ( a generator for the cyclic group of order 4 ):
$$ {\bf C} =\left[\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{array}\right]$$
if we multiply to the left of
$${\bf M} = \left[\begin{array}{cccc}0&0&0&\color{blue} 4\\0&0&\color{blue}3&0\\0&\color{blue}2&0&0\\\color{blue}1&0&0&0\end{array}\right]$$
It generates ${\bf CM} = \left[\begin{array}{cccc}0&0&\color{red}3&0\\0&\color{red}2&0&0\\\color{red}1&0&0&0\\0&0&0&\color{red}4\end{array}\right], {\bf C}^2{\bf M} = \left[\begin{array}{cccc}0&\color{orange}2&0&0\\\color{orange}1&0&0&0\\0&0&0&\color{orange}4\\0&0&\color{orange}3&0\end{array}\right]$ , ${\bf C}^3{\bf M} = \left[\begin{array}{cccc}\color{green}1&0&0&0\\0&0&0&\color{green}4\\0&0&\color{green}3&0\\0&\color{green}2&0&0\end{array}\right]$
These are the ones you color, therefore we could call it something with cyclic: cyclic diagonals or cyclically generated antidiagonals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Finding values of $t$ I have this equation -
$3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0 $
I'm struggling on how to find the values of $t$
First, I power both sides by $2$ to make the power be a whole number .
I get
$3t - \frac{2}{5} t^{-3} = 0 $
From here I'm stunned and stuck . Can I get a hint ! Thanks !
| The equation
$$
3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0
$$
is equivalent to
$$
t^{\frac{1}{2}} = \frac{2}{15} t^{\frac{-3}{2}} \;.
$$
Since $t$ cannot be zero$\color{green}{^*}$, you are allowed to multiply both sides by $t^{\frac{3}{2}}$:
$$
t^{2} = \frac{2}{15} \;.
$$
So, the solution is
$$
t = \pm\sqrt{\frac{2}{15}} \;.
$$
$\small\color{green}{^*\text{: If $t$ were zero, we would have $0^{\frac{-3}{2}} = \dfrac{1}{0}$.}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
PA = LU Decomposition with Row Exchange I am not sure how to deal with the L with we do row exchange in PA = LU decomposition. Here's my example:
$
A =
\left[ {\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 1\\
2 & 3 & 4\\
\end{array} } \right]
$
Step 1: Swap row 1 and row 3
$
U = \left[ \begin{array}{ccc}
2 & 3 & 4\\
0 & 0 & 1\\
1 & 1 & 1\\
\end{array} \right]
$
$
P = \left[ \begin{array}{ccc}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
\end{array} \right]
$
Step 2: row 3 <-- (-1/2) * row 1 + row 3
$
U = \left[ \begin{array}{ccc}
2 & 3 & 4\\
0 & 0 & 1\\
0 & 1 & 2\\
\end{array} \right]
$
$
P = \left[ \begin{array}{ccc}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
\end{array} \right]
$
Step 3: swap row 2 and row 3
$
U = \left[ \begin{array}{ccc}
2 & 3 & 4\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{array} \right]
$
$
P = \left[ \begin{array}{ccc}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{array} \right]
$
Therefore:
$
L = \left[ \begin{array}{ccc}
1 & 0 & 0\\
1/2 & 1 & 0\\
0 & 0 & 1\\
\end{array} \right]
$
The reason why I build L at the last step, is because we are sure the order of the rows after the row swaps so we can plug the low diagonal elements back to L. Element (2,1) is 1/2 because of step 2, the rest of lower diagonals are 0 because in row 3 elements (3,1) and (3,2) are originally 0s.
But I tried to use L * U but it doesn't equal to L*A. Could someone explain why I am wrong? Thanks!
| When you check you answer you have to check it by PA = LU.
What I did first was put A in an upper triangle.
$
A =
\left[ {\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 1\\
2 & 3 & 4\\
\end{array} } \right]
$
Step 1: $r_3$-2$r_1$→$r_3$
$
= \left[ \begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 1\\
0 & 1 & 2\\
\end{array} \right]
$
Step 2: $r_3$⟷$r_2$
$
= \left[ \begin{array}{ccc}
1 & 1 & 1\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{array} \right]
$ $$\\$$
Now, every row swap you did on matrix A you to that to a 3x3 identity matrix because A is a 3x3 matrix.
$
P = \left[ \begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array} \right]
$
Step 3: $r_3$⟷$r_2$
$
= \left[ \begin{array}{ccc}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{array} \right]
$ $$\\$$
Now form C=PA because it will have an LU decomposition.
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}
\times
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 1 \\
2 & 3 & 4
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 1 \\
2 & 3 & 4 \\
0 & 0 & 1
\end{bmatrix}$$ $$\\$$
Now you find the LU decomposition on matrix C. And you get:
$
U =
\left[ {\begin{array}{ccc}
1 & 1 & 1\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{array} } \right]
$
$
L =
\left[ {\begin{array}{ccc}
1 & 0 & 0\\
2 & 1 & 0\\
0 & 0 & 1\\
\end{array} } \right]
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
continuity of a function series I need help finding for which $x$ the function:
$$\sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$$
is continuous.
I first need to show that the series converges uniformly, the thing is I don't know how to deal with the $(-1)^n$. I tried comparing to the series $\sum_{n=1}^{\infty}\frac{x+n}{x^2+n^2}$ but I think I hit a dead end there.
any suggestions?
| $f(x) = \sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$
We need to show that for any $\epsilon$ there exists a $\delta$ such that for all $x,y$ in the domain, when $|x-y|< \delta,$ then $|f(x) - f(y)| < \epsilon$.
The first thing we are going to need is some upper bound on $f(x)-f(y)$
$f(x) -f(y) =$$ \sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}-\frac{y+n(-1)^n}{y^2+n^2}\\
\sum_{n=1}^{\infty}\frac{x^2y - y^2x+n(-1)^n(y^2-x^2)}{x^2y^2+(x^2+y^2)n^2 + n^4}\\
(y-x)\sum_{n=1}^{\infty}\frac{xy+n(-1)^n(x+y)}{x^2y^2+(x^2+y^2)n^2 + n^4}$
Now we need to show that $\sum_{n=1}^{\infty}\frac{xy+n(-1)^n(x+y)}{x^2y^2+(x^2+y^2)n^2 + n^4}$ converges.
The denominator grow at the order of $n^4$ and numerator grow with the order of $n$, so the series converges.
What we really want to be able to say is that:
There exists an $M$ such that $\sum_{n=1}^{\infty}\frac{xy+n(-1)^n(x+y)}{x^2y^2+(x^2+y^2)n^2 + n^4} < M$
If $x,y$ are big the denominator grows more quickly than the numerator, and if $x,y$ are near $0,$ the numerator collapses to $0,$ while the denominator does not.
It seems obvious, but I suggest you tighten that up. Anyway, once that has been established you will be able to say.
$|f(x)-f(y)|< M|y-x|$
$\delta = \frac \epsilon M$
Update
You are looking for something along these lines?
let $\{f\}$ be sequence of functions such that
$f_k(x) = \sum_{n=1}^{k}\frac{x+n(-1)^n}{x^2+n^2}$
We want to show that this sequence converges uniformly to $f(x)$
for any $\epsilon>0$ there exists a $K>0$ such that $k,m>K \implies \sup|f_k(x) - f_m(x)|<\epsilon$
$f_k(x) - f_m(x) = \sum_{n=k}^{m}\frac{x+n(-1)^n}{x^2+n^2}$
Unfortunately, I have to run... and can't figure out the last bit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $5^n +5 <5^{n+1}$ $∀n∈N$ using induction Prove $5^n +5 <5^{n+1}$ $∀n∈N$
Base Case: $n=1$
$\implies 5^1 +5 <25$
$\implies 10<25$ ; holds true
Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N
Then;
$\implies 5^{k+1} +5 < 5^{k+2}$
$\implies 5\cdot 5^k +5 < 25*5^k$
I don't know how to proceed after this step.
| $5^{k+1} + 5 < 5^{k+2} = 5(5^{k} + 1 < 5^{k+1})$. As $5^{k} + 1 < 5^{k} + 5 < 5^{k+1}$(from the inductive hypothesis), you have your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
The perimeter is equal to the area The measurements on the sides of a rectangle are distinct integers. The perimeter and area of the rectangle are expressed by the same number. Determine this number.
Answer: 18
It could be $4*4$ = $4+4+4+4$ but the answer is 18.
Wait... Now that I noticed, the sides are different numbers. But I can't find a way to solve.
| Let's see, let $x$ be the length of the one side and $y$ be the length of one of the adjacent sides. Then the perimeter of the rectangle is $P(x,y)=2x+2y$ and the area is $A(x,y)=xy$. So we need $P=A$ i.e. $2x+2y=xy$. Thta is, $$2x+2y=xy \Leftrightarrow 2x-xy+2y+4=4 \Leftrightarrow (x-2)(y-2)=4.$$ Note that $4=2 \cdot 2$ or $4=4 \cdot 1=1 \cdot 4$, the case $x-2=y-2=2$ leads to $x=y=4$, an that contradicts the hipótesis. The case $x-2=4$ and $y-2=1$ leads to $x=6$ and $y=3$ (the other case, $x-2=1$ and $y-2=4$ leads to a similar case). Finally, since $(-2)(-2)=(-4)(-1)=(-1)(-4)=4$, the cases $4=(-1)(-4)=(-4)(-1)$ leads to $x=-2$ or $y=-2$, respectively. Finally, the case $(-2)(-2)=4$ gives us $x=y=0$, another contradiction.
So the answer is $(x,y)=(6,3)$ or $(x,y)=(3,6)$, and indeed $A=6\cdot 3=18=2 \cdot 6 + 2\cdot 3=P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:
$$a^2+b^2={\left(ab\right)}^2+5$$
Any ideas? Thanks!
| Inspired/ Trying to make sense of Jaideep's Answer ...
\begin{eqnarray*}
\tan(5 \theta) =\frac{5 \tan(\theta)-10 \tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta) +5\tan^4(\theta)}
\end{eqnarray*}
Let $\theta=\frac{\pi}{5}$ and $x=tan(\frac{\pi}{5})$. We have the quintic
\begin{eqnarray*}
5x-10x^3+x^5=0
\end{eqnarray*}
This has roots $tan(\frac{\pi}{5}),tan(\frac{2\pi}{5}),tan(\frac{3\pi}{5}),tan(\frac{4\pi}{5}),tan(\frac{5\pi}{5})$.
Note that $tan(\frac{3\pi}{5})=-tan(\frac{2\pi}{5})$ , $tan(\frac{4\pi}{5})=-tan(\frac{\pi}{5}) $ and $tan(\frac{5\pi}{5})=0$.
Now square this quintic & substitute $y=x^2$ and we have
\begin{eqnarray*}
25y-100y^2+110y^3-\color{red}{20}y^4+y^5=0
\end{eqnarray*}
The sum of the roots of this polynomial gives $2 (tan^2(\frac{\pi}{5})+tan^2(\frac{2\pi}{5}))=20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.
My attempt:
| Suppose that the base of our trangle is $2a$ then the coresponding height is $A/a$ and the length of the side is $b=\sqrt{a^2+A^2/a^2}$, and finally the semi perimeter of the triangle is
$$p= a+\sqrt{a^2+\frac{A^2}{a^2}}$$
Now because $pr=A$ the maximal value of $r$ corresponds to the smallest value of $p$. So, we need to find the minimum of $p$ as function of $a$.
To simpify things we introduce a new variable $x=a/\sqrt{A}$ so that
$$f(x)=\frac{p}{\sqrt{A}}= x+\sqrt{x^2+\frac{1}{x^2}}$$
Clearly,
$$\eqalign{f'(x)&=1+\frac{x^2-1/x^2}{\sqrt{x^4+1}}=\frac{\sqrt{x^4+1}+x^2-1/x^2}{\sqrt{x^4+1}}\cr
&=\frac{3x^4-1}{x^4\sqrt{x^4+1}(\sqrt{x^4+1}-x^2+1/x^2)}
}$$
Thus $f(x)$ is decreasing on $(0,1/\sqrt[4]{3})$ and increasing on $(1/\sqrt[4]{3},\infty)$. Thus $f$ attains its minimum when $x=1/\sqrt[4]{3}$. The minimum corresponds to
$$ 2a=2\sqrt{\frac{A}{\sqrt{3}}},\qquad b=\sqrt{a^2+A^2/a^2}=2a.$$
So, our triangle when $r$ is maximum is equilateral, and the desired apex is equal to $\frac{\pi}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Finding the volume of the solid generated by revolving the given curve. The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.
Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about $x-axis$ is given as $$V=2\pi\int_0^{\infty}(f(x)^2-0)dx$$
Which gives: $2\pi\int_0^\infty\dfrac{a^6}{(a^2+x^2)^2}dx$
Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form $\dfrac{1}{x^4+2x^2+1}$ for $a=1$, which is transformed into $\dfrac{\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}=2[\dfrac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}-\dfrac{(1-\frac{1}{x^2})}{x^2+2+\frac{1}{x^2}}]$ which can be integrated, but this integral is tending to infinity. Can anyone help ? IS there a simpler way of doing this problem ?
| This seems like a trig substitution problem to me, albeit a not-so-straightforward one.
Let $x = a \tan\theta$. Then $dx = a \sec^2\theta\,d\theta$, and $a^2 + x^2 = a^2 \sec^2 \theta$. Since $\theta = \tan^{-1}\left(\frac{x}{a}\right)$, $\theta = 0$ when $x=0$, and $\theta\to\frac{\pi}{2}$ as $x\to\infty$. Therefore
\begin{align*}
2\pi \int_0^\infty \frac{a^6}{(a^2 + x^2)^2}\,d\theta
&= 2\pi \int_0^{\pi/2} \frac{a^6}{a^4\sec^4\theta}\,a \sec^2\theta \,d\theta\\
&= 2\pi a^3 \int_0^{\pi/2} \cos^2\theta\,d\theta \\
&= \pi a^3 \int_0^{\pi/2}(1+\cos2\theta)\,d\theta \\
&= \pi a^3 \left[\theta + \frac{1}{2}\sin2\theta\right]^{\pi/2}_0 \\
&= \frac{\pi^2 a^3}{2}
\end{align*}
This agrees with the Wolfram Alpha answer.
I can't quite figure out how you intended to integrate $\int\frac{1 + \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx$, but it could be that you were trying to represent a convergent improper integral as the difference of two divergent improper integrals. That leads to trouble, since $\infty - \infty$ is an indeterminate limit form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Expand $\frac{z}{z^4+9}$ To Taylor Series
expand $$\frac{z}{z^4+9}$$ to taylor series
$$\frac{z}{z^4+9}=\frac{z}{9}\frac{1}{1--\frac{z^4}{9}}$$
Can we write $$\frac{z}{9}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z^4}{9}\right)^n=\sum_{n=0}^{\infty}(-1)^n\frac{z^{4n+1}}{9^{n+1}}$$?
| Your expansion as Taylor series around $0$ (i.e. as Maclaurin series) is fine. But we also have to state the validity of the series representation of $f(z)$
\begin{align*}
f(z)=\frac{z}{z^4+9}=\frac{z}{9}\cdot\frac{1}{1-\left(-\frac{z^4}{9}\right)}=\sum_{n=0}^\infty (-1)^n\frac{z^{4n+1}}{9^{n+1}}
\end{align*}
The range of validity is $\left|-\frac{z^4}{9}\right|<1$ or equivalently $|z|<\sqrt{3}$.
We now consider the Taylor expansion around other points $z_0\in\mathbb{C}$. The function
\begin{align*}
f(z)&=\frac{z}{z^4+9}\\
&=-\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(1+i\right)}
-\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(-1-i\right)}\\
&\qquad+\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(-1+i\right)}
+\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(1-i\right)}\tag{1}
\end{align*}
has four simple poles at $z_1=\sqrt{\frac{3}{2}}\left(1+i\right),z_2=\sqrt{\frac{3}{2}}\left(1-i\right),z_3=\sqrt{\frac{3}{2}}\left(-1+i\right)$ and $z_4=\sqrt{\frac{3}{2}}\left(-1-i\right)$, one in each quadrant residing on the diagonals.
According to (1) we derive an expansion at $z=z_0$ via
\begin{align*}
\frac{1}{z-a}&=\
\frac{1}{(z-z_0)-(a-z_0)}\\
&=-\frac{1}{a-z_0}\cdot\frac{1}{1-\frac{z-z_0}{a-z_0}}\\
&=-\frac{1}{a-z_0}\sum_{n=0}^\infty\left(\frac{z-z_0}{a-z_0}\right)^n\\
&=-\sum_{n=0}^\infty\frac{1}{(a-z_0)^{n+1}}(z-z_0)^n\tag{2}
\end{align*}
We obtain from (1) and (2) for $z,z_0\in\mathbb{C}\setminus\{z_1,z_2,z_3,z_4\}$:
\begin{align*}
f(z)&=\frac{z}{z^4+9}\\
&=\frac{i}{12}\sum_{n=0}^\infty\left[
\frac{1}{\left(\sqrt{\frac{3}{2}}\left(1+i\right)-z_0\right)^{n+1}}
+\frac{1}{\left(\sqrt{\frac{3}{2}}\left(-1-i\right)-z_0\right)^{n+1}}\right.\\
&\qquad\qquad\qquad\left.-\frac{1}{\left(\sqrt{\frac{3}{2}}\left(-1+i\right)-z_0\right)^{n+1}}
-\frac{1}{\left(\sqrt{\frac{3}{2}}\left(1-i\right)-z_0\right)^{n+1}}\right](z-z_0)^n
\end{align*}
The radius of convergence of the expansion around $z=z_0$ is the distance to the pole in the same quadrant of $z_0$ or the distance to the two nearest poles if $z_0$ resides on an axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all orthogonal $3\times 3$ matrices of the form... Find all orthogonal $3\times 3$ matrices of the form
\begin{bmatrix}a&b&0\\c&d&1\\e&f&0\end{bmatrix}
Using the fact that $A^TA$ = $I_n$, I set that all up and ended up with the following system of equations:
$$\left\{\begin{array}{l}a^2 + e^2 = 1\\
ab + ef = 0\\
b^2 + f^2 = 1\end{array}\right.$$
I know I can let things equal the sine and cosine of theta, but I'm not exactly sure how to write this answer down on paper. There has to be tons of possibilities, right? How many exactly?
| There are few possibilities in fact. First of all, since $c^2+d^2+1^2=1$, $c=d=0$.
Now, you know that $a^2+b^2=1$, that $e^2+f^2=1$, and that $ab+ef=0$. This means that the matrix $\left(\begin{smallmatrix}a&b\\e&f\end{smallmatrix}\right)$ is orthogonal. Therefore, there is some $\theta\in\mathbb R$ such that$$\begin{pmatrix}a&b\\e&f\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\text{ or that }\begin{pmatrix}a&b\\e&f\end{pmatrix}=\begin{pmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $a^2+b^2+2(a+b)$ minimum if $ab=2$ Let $a,b\in R$,and such
$$ab=2$$
Find the minimum of the $a^2+b^2+2(a+b)$.
I have used $a=\dfrac{2}{b}$, then
$$a^2+b^2+2(a+b)=\dfrac{4}{b^2}+b^2+\dfrac{4}{b}+2b=f'(b)$$
Let $$f'(b)=0,\,b=-\sqrt{2}$$
So $$a^2+b^2+2(a+b)\ge 4-4\sqrt{2}$$
I wanted to know if there is other way to simplify the function and find the required value without using messy methods. Can we cleanly use AM-GM inequality?
| AM-GM inequality is (at least usually) only applied to positive real numbers.
Therefore I first consider the region $a>0,b>0$.
You can also write
$$
\begin{aligned}
f(a,b)&=a^2+b^2+2(a+b)\\
&=(a+b)^2+2(a+b)-2ab\\
&=s^2+2s-4,
\end{aligned}
$$
where $s=a+b$.
By the AM-GM inequality $s\ge 2\sqrt2$ with equality only when $a=b=\sqrt2$.
Here
$$
g(s)=s^2+2s-4
$$
is an increasing function of $s$ in the interval $s\in[2\sqrt2,\infty)$ (this is kinda obvious, but you can also check that $g'(s)>0$ in this interval). Therefore its smallest value is attained at $s=2\sqrt2$.
So if $a$ and $b$ are positive, the smallest value of $f$ is
$$
f(a,b)=f(\sqrt2,\sqrt2)=g(2\sqrt2)=4+4\sqrt2.
$$
The other possibility is that $a<0,b<0$. In that case AM-GM inequality (applied to $-a$ and $-b$) gives that $a+b\le-2\sqrt2$. Because the function
$g(s)$ is decreasing in the interval $s\in(-\infty,-2\sqrt2]$ it attains its smallest value at $s=-2\sqrt2$, i.e. at the point $a=b=-\sqrt2$. So if you allow negative values for $a$ and $b$, then the minimum is
$$
f(-\sqrt2,-\sqrt2)=4-4\sqrt2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluate an indefinite integral Find the value of
$$\int{\frac{x^2e^x}{(x+2)^2}} dx$$
My Attempt: I tried to arrange the numerator as follows:
$$ e^xx^2 = e^x(x+2-2)^2 $$ but that didn't help.
Any guidance on this problem will be very helpful.
| Another approach, let $u \mapsto x+2$ and expand
\begin{align*}
\int \frac{x^2e^x}{(x+2)^2}dx &= e^{-2} \int \frac{(u-2)^2 e^u}{u^2} \,du \\
&= e^{-2} \int \left( \frac{4e^u}{u^2} - \frac{4e^u}{u} + e^u \right) \, du \\
&= e^{-2} \left( \int e^u \,du -4 \int \frac{e^u}{u} - \frac{e^u}{u^2}\, du \right) \\
\end{align*}
now for the second integral, integrate each term by parts, $f_0=e^u$, $dg_0=u^{-2}$ and $f_1 = -u^{-1}$, $dg_1 = e^u$,
$$ = e^{-2} \left( e^u - \frac{4e^u}{u} \right) +c$$
$$ = e^{u-2} - \frac{4e^{u-2}}{u} +c$$
and substitute back in
$$ = e^x - \frac{4e^x}{x+2} +c$$
$$ = \frac{(x-2)e^x}{x+2} +c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342042",
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"source": "stackexchange",
"question_score": "4",
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Find the largest constant $k$ such that $\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$
Find the largest constant $k$ such that $$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$$
My attempt,
By A.M-G.M, $$(a+b)^2+(a+b+4c)^2=(a+b)^2+(a+2c+b+2c)^2$$
$$\geq (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2$$
$$=4ab+8ac+8bc+16c\sqrt{ab}$$
$$\frac{(a+b)^2+(a+b+4c)^2}{abc}\cdot (a+b+c)\geq\frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}\cdot (a+b+c)$$
$$=(\frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{ab}})(a+b+c)$$
$$=8(\frac{1}{2c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ab}})(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c)$$
$$\geq 8(5\sqrt[5]{\frac{1}{2a^2b^2c}})(5\sqrt[5]{\frac{a^2b^2c}{2^4}})=100$$
Hence the largest constant $k$ is $100$
Is my answer correct? Is there another way to solve it? Thanks in advance.
| I think you mean that $a$, $b$ and $c$ are positives.
For positive variables your solution is true, I think.
Another way:
Let $a+b=4xc$.
Hence, by AM-GM $$\frac{((a+b)^2+(a+b+4c)^2)(a+b+c)}{abc}\geq\frac{((a+b)^2+(a+b+4c)^2)(a+b+c)}{\left(\frac{a+b}{2}\right)^2c}=$$
$$=\frac{(16x^2+(4x+4)^2)(4x+1)}{4x^2}=\frac{4(2x^2+2x+1)(4x+1)}{x^2}\geq$$
$$\geq\frac{4\cdot5\sqrt[5]{(x^2)^2\cdot x^2\cdot1}\cdot5\sqrt[5]{x^4\cdot1}}{x^2}=100.$$
The equality occurs for $x=1$, which says that the answer is $100.$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Rate of convergence of a sequence I was just playing around rediscovering calculus (trying to figure out the derivative of $\arcsin x$) when I stumbled upon this rather untrivial problem.
$\large \int _0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}$
From this formula we can apply the definition of definite integral and get the following:
$\large \lim\limits_{m \to \infty} \sum_{i=0}^{m-1} \frac{1}{\sqrt{m^2-i^2}} = \frac{\pi}{2}$
All this is fine, but what is especially interesting is how rapidly does this sequence converge to $\frac{\pi}{2}$?
Let us define sequence $N_m:$
$\large {N}_m = \left \lfloor -\log_{10} \left | \frac{\pi}{2} - \sum_{i=0}^{m-1} \frac{1}{\sqrt{m^2-i^2}} \right |\right \rfloor$
So the problem can be formulated as follows: find $m$ such that $N_m \geq k$.
Is it possible to find the exact solution using school grade math?
Is it possible to get an approximate estimate?
| Since the integrand is unbounded at $x = 1$, it is not Riemann integrable and the integral is improper:
$$\int_0^1 \frac{dx}{\sqrt{1 - x^2}} = \lim_{c \to 1}\int_0^c \frac{dx}{\sqrt{1 - x^2}} = \frac{\pi}{2}.$$
In general, there is no guarantee that a Riemann sum converges to the improper integral. Fortunately, the integrand is monotonically increasing and in this case it can be shown that the left Riemann sums will converge:
$$\lim_{m \to \infty} \sum_{k=0}^{m-1} \frac{1}{\sqrt{m^2 - k^2}} = \frac{\pi}{2}.$$
If the integrand were continuously differentiable (and Riemann integrable) then the Riemann sums would converge at the rate $O(1/m)$. In this case, the convergence will be at the slower rate $\mathbf{O(1/\sqrt{m})}$.
This convergence rate implies that the approximation error should be reduced roughly by a factor of $2$ as the number of points $m$ increases by a factor of $4$, as you should see by some numerical experimentation.
To see why analytically, observe that
$$\frac{1}{\sqrt{2}}\int_0^1 \frac{dx}{\sqrt{1 - x}} < \int_0^1 \frac{dx}{\sqrt{1 - x^2}} = \int_0^1 \frac{dx}{\sqrt{1 +x}\sqrt{1 - x}} < \int_0^1 \frac{dx}{\sqrt{1 - x}},$$
and we expect the behavior to be similar in terms of convergence to
$$\int_0^1 \frac{dx}{\sqrt{1 -x}} = \int_0^1 \frac{dx}{\sqrt{x}} = 2.$$
In this case, the right Riemann sums are convergent as
$$\lim_{m \to \infty} \frac{1}{m} \sum_{k = 1}^m \frac{1}{\sqrt{k/m}} = \lim_{m \to \infty} \frac{1}{\sqrt{m}} \sum_{k = 1}^m \frac{1}{\sqrt{k}} = 2.$$
Since,
$$2(\sqrt{k+1} - \sqrt{k}) = \frac{2}{\sqrt{k+1} + \sqrt{k}} < \frac{1}{\sqrt{k}} < \frac{2}{\sqrt{k} + \sqrt{k-1}} = 2(\sqrt{k} - \sqrt{k-1}), $$
we can bound the sum as
$$2(\sqrt{m+1} - 1) < \sum_{k=1}^m \frac{1}{\sqrt{k}} < 2 \sqrt{m},$$
and
$$2\left(\sqrt{1 + 1/m} - 1/\sqrt{m}\right) < \frac{1}{\sqrt{m}}\sum_{k=1}^m \frac{1}{\sqrt{k}} < 2.$$
Expanding $\sqrt{1 + 1/m}$ in a Taylor series we find the error behaves as
$$\left|\frac{1}{\sqrt{m}}\sum_{k=1}^m \frac{1}{\sqrt{k}} - 2 \right| = O(1/\sqrt{m})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometric series with complex numbers In a geometric series
$a_{1}=1-\sqrt{3}i$ and $a_{2}=2$
Prove that for every natural number $n$, the numbers at the $3n+2$ location in the series are real numbers.
I have started by finding the ratio of the series, which is:
$q=1-\sqrt{3}i$
Now I am stuck. The elements of the series are $5,8,11,14,\ldots$, i.e., jumps of $3$. I tried looking at the first one, using De Moivre's formula, and got that it is indeed real. But how do I prove it for every $n$?
Thank you
| The geometric progression has ratio
$$\frac{a_2}{a_1}=\frac{2}{1-i \sqrt{3}}=\frac{1}{2} \left(1+i \sqrt{3}\right)$$
Therefore the sequence can be written as
$$\{a_n\}=\left\{\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{n-1}\right\};\;n\geq 1$$
Observing that $\frac{1}{2} \left(1+i \sqrt{3}\right)$ is a cubic root of $-1$ we have
$$a_{3n+2}=\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{3n+1}=$$
$$=\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{3n}\left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)=\left(1-i \sqrt{3}\right) (-1)^{n}\left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)=$$
$$=\frac{1}{2}(-1)^n(1-i\sqrt{3})(1+i\sqrt{3})=(-1)^n \cdot 2$$
Therefore the terms indexed $(3n+2)$ are $2$ or $-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded:
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x + 7) + 8] + 15\\
& = (x^2 + 8x + 7)^2 + 8(x^2 + 8x + 7) + 15
\end{align*}
If we make the substitution $y = x^2 + 8x + 7$, we get
\begin{align*}
y^2 + 8y + 15 = (y^2 + 3y) + (5y + 15) = y(y+3) + 5(y+3) = (y+5)(y+3) = 0
\end{align*}
From whence we obtain that:
\begin{align*}
y + 5 = 0\Leftrightarrow x^2 + 8x + 12 = 0 \Leftrightarrow (x+4)^2 - 4 = 0\Leftrightarrow x\in\{-6,-2\}\\
\end{align*}
Analogously, we have that
\begin{align*}
y + 3 = 0\Leftrightarrow x^2 + 8x + 10 = 0\Leftrightarrow (x+4)^2 - 6 = 0\Leftrightarrow x\in\{-4-\sqrt{6},-4+\sqrt{6}\}
\end{align*}
Finally, the solution set is given by $S = \{-6,-2,-4-\sqrt{6},-4+\sqrt{6}\}$.
Differently from this approach, could someone provide me an alternative way of solving this problem? Any contribution is appreciated. Thanks in advance.
| Hint: let $x+4=y$ then the equation writes as:
$$0 = (y-3)(y-1)(y+1)(y+3)+15=(y^2-1)(y^2-9)+15=y^4-10y^2+24$$
The latter is a biquadratic with solutions $y^2 \in \{4, 6\}\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Laplace transform of $x^{3/2}$ Solve the Laplace transform of $x^{3/2}$
$L[x^{3/2}]=L[x (x^{1/2})]=-\frac{d}{dp}L[x^{1/2}]=-\frac{d}{dp}L[x(x^{-1/2})]=-\frac{d}{dp}(-\frac{d}{dp})\sqrt{\frac{\pi}{p}}=\frac{1}{2p}\frac{1}{2p}\sqrt{\frac{\pi}{p}}=\frac{1}{4p^2}\sqrt{\frac{\pi}{p}}$
The correct answer is $\frac{3}{4p^2}\sqrt{\frac{\pi}{p}}$, could someone tell me where I missed the $3$ please?
| Here's the correct way to do the integral.
\begin{align*}
\int_0 ^\infty e^{-st} t^{3/2} \ dt &= \int_0 ^\infty e^{-u} \left(\frac{u}{s} \right)^{3/2} \ \frac{du}{s} \\
&= \frac{1}{s^{5/2}} \int_0 ^\infty e^{-u} u^{\frac{5}{2}-1} \ du \\
&= \frac{1}{s^{5/2}} \Gamma \left( \frac{5}{2} \right) \\
&= \frac{1}{s^{5/2}} \cdot \frac{3\sqrt{\pi}}{4}
\end{align*}
The change of variables $u = st$ was used, and tables were used to evaluate the particular value of the gamma function.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find $x-\sqrt{7x}$ given that $x - \sqrt{\frac7x}=8$
If $ x - \sqrt{\frac{7}{x}}=8$ then $x-\sqrt{7x}=\text{?}$
I used some ways, but couldn't get the right form :) by the way, the answer is $1$.
Thanks in advance.
| \begin{eqnarray*}
x-\sqrt{\frac{7}{x}}=8 \\
x-8 =\sqrt{\frac{7}{x}}
\end{eqnarray*}
Now square both sides and multiply by $x$
\begin{eqnarray*}
x^2-16x+64=\frac{7}{x} \\
x^3-16x^2+64x-7=0 \\
(x-7)(x^2-9x+1)=0
\end{eqnarray*}
Now assume $x$ does not equal $7$ and taking the positive square root will give
\begin{eqnarray*}
x^2-9x+1=0 \\
x^2-2x+1= 7x \\
(x-1)^2 =7x \\
x-1 = \sqrt{7x} \\
x-\sqrt{7x}= \color{red}{1}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$$
Now, how can I prove this conjecture? I've tried a lot, but still couldn't have any idea.
| Here is a combinatorial proof: consider the set $F$ of functions $\{ 1, 2, \ldots, n \} \to \{ 1, 2, \ldots, n, n + 1 \}$, and for $k = 1, \ldots, n$, let $S_k$ be the set of such functions such that $k$ is not in the range. Then, by inclusion-exclusion:
$$\left|F \setminus \bigcup_{k=1}^n S_k \right| = |F| - \sum_{1 \le i \le n} |S_i| + \sum_{1 \le i < j \le n} |S_i \cap S_j| - \sum_{1 \le i < j < k \le n} |S_i \cap S_j \cap S_k| + \cdots + (-1)^n |S_1 \cap \cdots \cap S_n|.$$
Now, $|F| = (n+1)^n$; for each $i$, $|S_i| = n^n$, so the second term is $-\binom{n}{1} n^n$; for each $i < j$, $|S_i \cap S_j| = (n-1)^n$, so the third term is $\binom{n}{2} (n-1)^n$; and so on. Therefore, the sum on the right hand side is equal to the left hand side of your equation (using that $\binom{n}{n-x} = \binom{n}{x}$). On the other hand, $F \setminus \bigcup_{k=1}^n S_k$ is exactly the set of functions such that each of $1, 2, \ldots, n$ is in the range, which is precisely the set of permutations of $\{ 1, 2, \ldots, n \}$; therefore, the left hand side is equal to $n!$.
In fact, if we let $F$ be the set of functions $\{ 1, 2, \ldots, n \} \to \{ 1, 2, \ldots, n + r \}$ instead, this argument easily generalizes to show that for $n$ a positive integer and $r$ a nonnegative integer:
$$\sum_{x=0}^n (-1)^k \binom{n}{x} (n+r-x)^n = n!.$$
| {
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Finding a Closed Form for $\int_0^\infty A\sin(\frac{2\pi}{T}x) \exp(-bx)dx$ I'm struggling in finding a closed form for,
$$ \int_0^\infty Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right) \,dx $$
My Attempt: Let $H = \int Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx$. First I integrate by parts, splitting $f=e^{-bx}$ and $dg=\sin\left(\frac{2\pi}{T}x\right)\,dx$. Then another IBP using $u=e^{-bx}$ and $dv = \cos\left(\frac{2\pi}{T}x\right)\,dx$ and simplify,
\begin{align}
\frac{2\pi}{A\cdot T} H &= -e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac 1 b \int \cos\left(\frac{2\pi}{T}x\right)e^{-bx}\,dx \\
\frac{2\pi}{A\cdot T} H &= -e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac 1 b \left[ e^{-bx}\frac{T}{2\pi}\sin\left(\frac{2\pi}{T}x\right) + \frac{T}{2\pi b}\int e^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx \right] \\
\left(\frac{2\pi}{A\cdot T} - \frac{T}{2\pi b^2}\right)H &= e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac{T}{2\pi b} e^{-bx} \sin\left(\frac{2\pi}{T}x\right) \\
H &= -e^{-bx} \left(\cos\left(\frac{2\pi}{T}x\right) + \frac{T}{2\pi b}\sin\left(\frac{2\pi}{T}x\right) \right) \cdot \left( \frac{A\cdot T 2\pi b^2}{(2\pi b)^2 - A\cdot T^2}\right) + c\\
\end{align}
First off, is this horrid looking expression correct? If so, can it be simplified further?
Now I come to evaluate $H|_0^\infty$. $H$ appears to be defined at $x=0$, so I really just need to find $\lim_{x\to \infty^+} H$. Here is where I'm stuck. The denominator grows exponentially for $b\ge 1$, and $T$ and $A$ don't seem to have any effect on the limit. How should I evaluate this limit?
| Even without Laplace transform, consider the two antiderivatives
$$I=\int \sin(ax) e^{-bx}\,dx \qquad \text{and} \qquad J=\int \cos(ax) e^{-bx}\,dx$$ $$J+iI=\int e^{iax} e^{-bx}\,dx=\int e^{-(b-ia)x}\,dx=-\frac{e^{ -(b-ia)x}}{b-ia}\tag 1$$
$$J-iI=\int e^{-iax} e^{-bx}\,dx=\int e^{-(b+ia)x}\,dx=-\frac{e^{-(b+ia)x}}{b+ia}\tag 2$$ $$(J+iI)-(J-iI)=2iI=-\frac{e^{ -(b-ia)x}}{b-ia}+\frac{e^{-(b+ia)x}}{b+ia}\tag 3$$ Manipulate the complex to get $$2iI=-\frac{2 i e^{-b x} (b \sin (a x)+a \cos (a x))}{a^2+b^2}$$ that is to say $$I=-\frac{ b \sin (a x)+a \cos (a x)}{a^2+b^2}e^{-b x}$$ So,
$$K=\int_0^\infty \sin(ax) e^{-bx}\,dx=\frac{a}{a^2+b^2}$$ provided that $|\Im(a)|<\Re(b)$.
Replace $a$ by $\frac {2\pi}T$ to get the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is evident that $-1$ is not a root of the second equation, but how can I find out whether the other two roots are common or not?
| $$x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1=(x+1)(x^2+x+1).$$
We see that $-1$ it's not common root and
$$x^{1990}+x^{200}+1=x^{1990}-x+x^{200}-x^2+x^2+x+1=$$
$$=x((x^3)^{663}-1)+x^2((x^3)^{66}-1)+x^2+x+1\equiv0(\mod(x^2+x+1)),$$
which says that roots of $x^2+x+1$ are all common roots and we get the answer:
$$\left\{-\frac{1}{2}+\frac{\sqrt3}{2}i,-\frac{1}{2}-\frac{\sqrt3}{2}i\right\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrix and sequences Given the following matrix : $$
A= \begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix}
$$ and the sequences $a_n$, $b_n$, $ c_n$, $d_n$ such that $ A^n$= \begin{bmatrix}
a_n& b_n \\
c_n & d_n \\
\end{bmatrix}
I need to find the pair $(x,y)$ such that $a_{n+2}=xa_{n+1}+ya_n$ for every integer $n$.
What do I need to do? Do I have to find a form for the matrix raised to the power $n$? Can I use Hamilton Cayley here? Could someone please explain in detail because I haven't worked with matrices and sequences until now. The problem also asks me the number of values for $n$ such that $b_n=c_n$ and the limit as $n$ goes to infinity of $\frac{a_n}{b_n}$.
| Note that
$$\begin{pmatrix} a_{n+1}&b_{n+1}\\c_{n+1}&d_{n+1} \end{pmatrix} = \begin{pmatrix} a_n&b_n\\c_n&d_n \end{pmatrix} \begin{pmatrix} 1&1\\1&0 \end{pmatrix} = \begin{pmatrix} a_n + b_n & a_n \\ c_n + d_n & c_n \end{pmatrix} $$
and
$$\begin{pmatrix} a_{n+2}&b_{n+2}\\c_{n+2}&d_{n+2} \end{pmatrix} = \begin{pmatrix} a_{n+1}&b_{n+1}\\c_{n+1}&d_{n+1} \end{pmatrix} \begin{pmatrix} 1&1\\1&0 \end{pmatrix} = \begin{pmatrix} 2a_n + b_n & a_n + b_n \\ 2c_n + d_n & c_n + d_n \end{pmatrix} $$
So,
$$ a_{n+2} = 2a_n + b_n$$
$$ a_{n+1} = a_n + b_n $$
Can you solve for $x$ and $y$ from here?
| {
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If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$
If $a$ and$ b$ are two distinct real values such that
$$F (x)= x^2+ax+b $$
And given that $F(a)=F(b)$ ; find the value of $F(2)$
My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F(a)=F(b)$ implies $2a^2=b^2+ab$, and thus
$F(2)= 4+2a+b $
What Now?
Any help would be appreciated , thank you
| Due to the symmetry of parabolas of that form, you can see $\frac{|a+b|}{2}$ is a minimum or maximum.
This gives you $f'\left(\frac{|a+b|}{2}\right)=|a+b|+a=0$ which implies $b=0\lor 2a+b=0$.
$b=0 \implies F(b)=0 \land F(a)=2a^2$
But $F(a)=F(b)\implies a=0$ which leads to a contradiction (as $a$ and $b$ are distinct).
Hence $2a+b=0$ as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$ , and $a \leq b \leq c $?
What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$, and $a \leq b \leq c$?
My try:
The prime factorization of $1000$ is $2^3\cdot 5^3$
$a\cdot b \cdot c = 2^3\cdot 5^3$
$a=2^{a_1}\cdot 5^{a_2}$
$b=2^{b_1}\cdot 5^{b_2}$
$c=2^{c_1}\cdot 5^{c_2}$
$abc=2^{a_1+b_1+c_1}\cdot 5^{a_2+b_2+c_2}=2^3\cdot 5^3 $
$a_1+b_1+c_1=3$
How many ways are there such that $a_1+b_1+c_1=3$
Star's and Bar's method: -
Number of ways to chose $2$ separators($0s$) in a string of $5 $$ = {5\choose 2 }=10$
$N(a_1+b_1+c_1=3)=10$
Similarly, $N(a_2+b_2+c_2=3)=10$
$N(abc=1000)=10\cdot 10=\boxed{100}$
Is that okay ? Please write down any notes
| For all $a,b,c$ you got 100 but that didn't take into account $a \le b \le c$.
So count $a=b=c$ that is $a=b=c=10$ you counted that once. The remaining $99$ were overcounted.
Consider $a=b, c\ne a$. Then $a=b= 2^j5^k; c=2^{3-2j}3^{3-2j}$ so there are $3$ such cases ($j = 0,1; k = 0,1$ but not $j=k=3-2j=3-2k= 1$). You counted each of those $3$. So that accounts for $9$ when you should have only counted $3$. You have $90$ more to account for.
These are $a,b,c$ distinct. You counted each of these $6$ times when you should have counted them once. So you counted $90$ when you should have counted $15$.
So the number should be $15 + 3 + 1 = 19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.
How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$?
| Because $$n^4+22n^3+71n^2+218n+384=n(n+1)(n+2)(n+3)+16n^3+60n^2+212n+384=$$
$$=n(n+1)(n+2)(n+3)+384+16(n^3-n)+60n^2+228n=$$
$$=n(n+1)(n+2)(n+3)+384+16(n-1)n(n+1)+60n(n+1)+168n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to calculate $\lim_{x\to 0^+} \frac{x^x- (\sin x)^x}{x^3}$ As I asked, I don't know how to deal with $x^x- (\sin x)^x$.
Please give me a hint. Thanks!
| Note that
$$\frac{x^x - (\sin x)^x}{x^3} = x^x\frac{1 - \left(\frac{\sin x}{x}\right)^x}{x^3}$$
We can find the Taylor expansion
$$\left(\frac{ \sin x}{x} \right)^x = \exp\left(x \log \left(\frac{\sin x}{x} \right) \right) = 1 - \frac{x^3}{6} + O(x^5),$$
using
$$\frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4)\\ \log (1 - y) = -y + O(y^2)\\ \exp(z) = 1 + z + O(z^2)$$
Thus,
$$\frac{1 - \left(\frac{\sin x}{x}\right)^x}{x^3} = \frac{1}{6} + O(x^2)$$
Since $\lim_{x \to 0+} x^x = \lim_{x \to 0+} \exp(x \log x) = 1$ we find the desired limit is $1/6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Checking divisibility of $C^{20}_7-C^{20}_8 +C^{20}_9 -C^{20}_{10}+\dots-C^{20}_{20}$ Let $N=C^{20}_7-C^{20}_8 +C^{20}_9 -C^{20}_{10}+\dots-C^{20}_{20}$. Prove that it is divisible by $3,4,7,19$.
| Hint. Since by the binomial theorem
$$0=(1-1)^{20}=\sum_{n=0}^{20}\binom{20}{n}(-1)^n=\sum_{n=0}^{6}\binom{20}{n}(-1)^n+\sum_{n=7}^{20}\binom{20}{n}(-1)^n=\sum_{n=0}^{6}\binom{20}{n}(-1)^n-N$$
it follows that
\begin{align*}
N&=\sum_{n=0}^6\binom{20}{n}(-1)^n\\
&=1-20+190-\frac{20\cdot 19\cdot 18}{3!}+\frac{20\cdot 19\cdot 18\cdot 17}{4!}-\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{5!}+\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{6!}\\
&=-19+19\cdot 10-20\cdot 19\cdot 3+5\cdot 19\cdot 3\cdot 17- 19\cdot 3\cdot 17\cdot 16+19\cdot 17\cdot 8\cdot 15.
\end{align*}
Is it $N$ divisible by $19$ (without evaluating all these product)?
What about $3$, $4$, and $7$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\sin^2 \theta}{ \sin^2 \theta } + \frac{\cos^2 \theta }{ \sin^2 \theta }= \frac 1 { \sin^2 \theta } $$
b) $$1 + \cot^2 \theta = \csc^2 \theta
$$
How did the $ \tan^2 \theta + 1 = \sec^2 \theta$ comes into the picture?
| It is taken from Pythagoras theorem,
Let
Know that
let
c= Hypothenuse
a= Opposite
b=Adjacent
length of the triangle can be found by
$$a^2+b^2=c^2$$
Now divide everything by b
We have
$$\frac{a^2}{b^2}+1=\frac{c^2}{b^2}$$
Notice that
$$tan \theta=\frac{a}{b}$$
$$sec \theta=\frac{1}{\cos \theta}=\frac{c}{b}$$
There you go
$$1+\tan^2 \theta=\sec^2 \theta$$
There rest of the two identities are obtained by dividing c&a!
Which is
Division by c
$$\sin^2 \theta+\cos^2 \theta=1$$
Division by a
$$1+\cot^2\theta=\csc^2\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$
Find $$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$$
$$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)
= \lim\limits_{x \rightarrow \infty} \frac{e^{-\frac{1}{x}}-1}{x^{-0.5}}
= \lim\limits_{x \rightarrow \infty} \frac{[e^{-x^{-1}}-1]'}{[x^{-0.5}]'}
= \lim\limits_{x \rightarrow \infty} \frac{x^{-2}e^{-x^{-1}}}{-0.5x^{-1.5}}
= 2 \cdot \lim\limits_{x \rightarrow \infty} \frac{1}{e^{x^-1}x^{0.5}}
$$
So as $x \rightarrow \infty$, $\frac{1}{e^{x^-1}x^{0.5}} \rightarrow \frac{1}{1 \cdot \infty} \rightarrow 0$
Is this correct ? any input is much appreciated
| Yes your way is correct,but you can do it simply as $$\lim _{ x\rightarrow \infty } \sqrt { x } \left( e^{ -\frac { 1 }{ x } }-1 \right) =\lim _{ x\rightarrow \infty } \sqrt { x } \frac { \left( e^{ -\frac { 1 }{ x } }-1 \right) }{ -\frac { 1 }{ x } } \left( -\frac { 1 }{ x } \right) =\lim _{ x\rightarrow \infty }{ -\frac { \sqrt { x } }{ x } } =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Lengthy integration problem $$ \int\frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx$$
I managed to solve the problem using partial fraction decomposition.
But that approach is pretty long as it creates five variables. Is there any other shorter method to solve this problem(other than partial fractions)?
I also tried trigonometric substitution and creating the derivative of the denominator in the numerator... but it becomes even longer. Thanks in advance!!
Just to clarify: (My partial fraction decomposition)
$$\frac{x^3+3x+2}{(x^2+1)^2(x+1)}= \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{E}{x+1}$$
| HINT:
$$x^3+3x+2=x^3+x+2x+2=x(x^2+1)+2(x+1)$$
$$\dfrac{x^3+3x+2}{(x^2+1)^2(x+1)}=\dfrac x{(x^2+1)(x+1)}+\dfrac2{(x^2+1)^2}$$
For the second set $x=\tan y$
For the first
Method$\#1:$
write the numerator as $$x\cdot\dfrac{(x^2+1)-(x^2-1)}2$$
Method$\#2:$
set $x=\tan y$ to find $$\int\dfrac{\sin y}{\sin y+\cos y}dx$$
Now express $\sin y=A(\sin y+\cos y)+B\cdot\dfrac{d(\sin y+\cos y)}{dy}$
Can you derive the values of the arbitrary constants $A,B?$
| {
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"url": "https://math.stackexchange.com/questions/2363482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find the limit $ \lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} $ Question
$$
\lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x}
$$
I'm not sure how to go about this limit, I've tried to apply L'Hopital's rule
(as shown).
It seems that the form is going to be forever indeterminate? Unless I'm missing
something, but the limit is (apparently) zero.
Working
As this is in the indeterminate form of $- \infty / + \infty$, apply L'Hopital's rule.
Let $\frac{\ln(1 - 2x)}{\tan \pi x} = f/g$, then
$f' = \frac{-2}{1 - 2x}$ , and $f'' = \frac{-4}{(1 - 2x)^2}$
$g' = \pi \sec^2 \pi x$ (which is equal to $\pi (\tan^2 (\pi x )+ 1)$ ) and
$g'' = 2 \pi^2 \sec^2 (\pi x) \tan \pi x $ ( or
$2 \pi^2 (\tan^2 (\pi x) + 1) \tan \pi x$
)
Using the first derivatives gives
$$
\lim_{x \to (\frac{1}{2})^{-}}
\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}
$$
Which is in the form
$$
\frac{- \infty}{+ \infty}
$$
So that I would now use the second derivative, which is
$$
\lim_{x \to (\frac{1}{2})^{-}}
\frac{
\frac{-4}{(1 - 2x)^2}
}{
2 \pi^2 \sec^2 (\pi x) \tan \pi x
}
$$
Or
$$
\lim_{x \to (\frac{1}{2})^{-}}
\frac{
\frac{-4}{(1 - 2x)^2}
}{
2 \pi^2 (\tan^2 (\pi x) + 1) \tan \pi x
}
$$
But this is still in an indeterminate form?
| You were doing fine:
Using the first derivatives gives
$$ \lim_{x \to (\frac{1}{2})^{-}}
\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}$$
Now simplify a bit first:
$$\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x} = \frac{2\cos^2\pi x}{\pi \left(2x-1\right)}$$
Applying l'Hôpital to the fraction in this form gives the easy $-\sin\left( 2\pi x\right)$, which evaluates to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \theta}{4}+\frac{ir \sin \theta}{4}+\frac{6 \cos \theta}{r}-\frac{6 i \sin \theta}{r} \right|$$
So
$$\left|\frac{z}{4}+\frac{6}{z}\right|=\sqrt{\frac{r^2}{16}+\frac{36}{r^2}+3 \cos (2\theta)}$$
any clue from here?
| We want to find the minimum and maximum of
$\left| \dfrac{z}{4}+\dfrac{6}{z}\right|$
When $1\le |z|\le 7$
Let $z=x+iy$ and plug it in the given expression
$$\left| \dfrac{x+iy}{4}+\dfrac{6}{x+iy}\right|=\left| \dfrac{x+iy}{4}+\dfrac{6(x-iy}{x^2+y^2}\right|=\left|\frac{6 x}{x^2+y^2}+\frac{x}{4}+i\left(\frac{y}{4}-\frac{6 y}{x^2+y^2}\right)\right|$$
Which is
$$r=\sqrt{\left(\frac{6 x}{x^2+y^2}+\frac{x}{4}\right)^2+\left(\frac{y}{4}-\frac{6 y}{x^2+y^2}\right)^2}=\frac{\sqrt{r^4-48 r^2+96 x^2+576}}{4 r}$$
where $r^2=x^2+y^2$
As we have $1\le r\le 7$ let's plug $r=1$ and $r=7$ in the previous expression
for $r=1$ we get $\dfrac{ \sqrt{96 x^2+529}}{4}$ which is $\dfrac{25}{4}$ for $x=1$, maximum
for $r=7$ we get $\dfrac{\sqrt{96 x^2+625}}{28} $ which is $\dfrac{73}{28}$ for $x=7$, minimum
Hope this helps
Edit
Actually $\dfrac{z}{4}+\dfrac{6}{z}=0$ has a solution at $z=\pm 2 i \sqrt{6}$ whose module satisfies the condition $|z|\in[1,\;7]$ therefore that is the absolute minimum, zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
On understanding the use of binomial theorem to find asymptotes of a real valued function. I have the function
$$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$
Wolfram Alpha says that it has a non linear asymptote at
$y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256}$.
I have been told to expand using the binomial theorem to obtain the asymptote at predicting-non-linear-asymptotes-of-a-real-valued-function ; however, I do not understand this. How does expanding binomially give me the asymptote ?
| hint
Let $$f (x)=(x^3+2x+9)(4x^2+3x+2)^\frac{-1}{2}$$
$$g (x)=f (1/x)=$$
$$(1+\frac {9x}{2}+\frac {1}{2x^2})(1+\frac {3x}{4}+\frac {x^2}{2})^\frac {-1}{2} $$
expand $g (x) $ near $x=0$ using binomial formula.
$$(1+\frac {3x}{4}+\frac {x^2}{2})^\frac {-1}{2} $$
$$=1-\frac {3x}{8}-\frac {x^2}{4}+\frac {27x^2}{112}+x^2\epsilon (x)$$
You will get $f (x)=g (1/x) $ and the asymptote.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
| Let $\arctan\frac{1+x}{4+x}=\frac{\pi}{4}+y$.
Hence, $y\rightarrow0$ and $$\frac{1+x}{4+x}=\tan\left(\frac{\pi}{4}+y\right)$$
or $$x=\frac{1-4\tan\left(\frac{\pi}{4}+y\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ and we need to calculate
$$\lim_{y\rightarrow0}\frac{y\left(1-4\tan\left(\frac{\pi}{4}+y\right)\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ or
$$-\frac{3}{\sqrt2}\lim_{y\rightarrow0}\frac{y}{\sin\left(\frac{\pi}{4}+y\right)-\cos\left(\frac{\pi}{4}+y\right)}$$ or
$$-\frac{3}{\sqrt2}\lim_{y\rightarrow0}\frac{y}{\frac{2}{\sqrt2}\sin{y}},$$
which is $-\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How can I solve this limit without L'Hopital rule? I have found this interesting limit and I'm trying to solve it without use L'Hopital's Rule.
$$\lim\limits_{x\rightarrow 0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}$$
I solved it with L'Hopital's rule and I found that the solution is $1$. But if I try without this rule, I can't solve it. Any ideas?
| You can use equivalents and expansion in power series.
Let's begin with the denominator:
$$\sinh x-\sin x=x+\frac{x^3}{3!}+o(x^3)-\Bigl(x-\frac{x^3}{3!}+o(x^3)\Bigr)=\frac{x^3}3+o(x^3)\sim_0\frac{x^3}3.$$
Now for the numerator:
First, by definition, $\;\operatorname{arsinh}(\sinh(x))=x$.
Next,
$$\operatorname{arsinh} x=x-\frac12\frac{x^3}3+\frac{1\cdot 3}{2\cdot4}\frac{x^5}5-\frac{1\cdot 3\cdot5}{2\cdot4\cdot6}\frac{x^7}7+\dotsm $$
We'll deduce the expansion of $\operatorname{arsinh} (\sin x)$ at order $3$. Remember asymptotic expansions can be composed:
\begin{align}
\operatorname{arsinh} (\sin x)&=\operatorname{arsinh}\Bigl(x-\frac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\frac{x^3}6\Bigr)-\frac16\Bigl(x-\frac{x^3}6\Bigr)^3+o(x^3)\\
&=x-\frac{x^3}6-\frac16x^3+o(x^3)=x-\frac{x^3}3+o(x^3),
\end{align}
so that
$$\operatorname{arsinh}(\sinh(x))-\operatorname{arsinh}(\sin(x))=x-x+\frac{x^3}3+o(x^3)=\frac{x^3}3+o(x^3)\sim_0\frac{x^3}3.$$
Ultimately, we obtain (if the computation is correct):
$$\frac{\operatorname{arsinh}(\sinh(x))-\operatorname{arsinh}(\sin(x))}{\sinh x-\sin x}\sim_0\frac{\dfrac{x^3}3}{\dfrac{x^3}3}=1. $$
| {
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"url": "https://math.stackexchange.com/questions/2366608",
"timestamp": "2023-03-29T00:00:00",
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How to solve $\frac{3}{2}x-\frac{4}{5}=\sqrt{\sin(30^\circ)+\sin\frac{7\pi}{4}}$ I did it as: $$\Bigl(\frac{3}{2}x-\frac{4}{5}\Bigr)^2=\Biggl(\sqrt{\frac{1}{2}-\frac{\sqrt{2}}{2}}\Biggr)^2.$$ But after, got big numbers $$225x^2-63-\sqrt{2}=0.$$ It would be very good, if someone showed the correct way of solving this. It must be solved in $\Bbb R.$
| $$\frac{3}{2}x-\frac{4}{5}=\sqrt{\frac{1}{2}-\frac{1}{\sqrt2}},$$
which is impossible because $\frac{1}{2}-\frac{1}{\sqrt2}<0$.
The answer is $\oslash$
| {
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"timestamp": "2023-03-29T00:00:00",
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Logarithmic series as $\sum_{n=3}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)$ Inspired by this question, I've designed the following series.
$$
\begin{align}
S&=\sum_{n=3}^{\infty}\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right) \tag1
\\\\
T&=\sum_{n=3}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right) \tag2
\end{align}
$$
Each one admits a nice closed form.
Q1. What are their closed forms?
Q2. To which family would you tell these series belong to ?
| Noting that
$$ 2n^2+n-6=(n+2)(2n-3),2n^2+n-10=(n-2)(2n+5)$$
one has
\begin{eqnarray}
S&=&\sum_{n=3}^{\infty}\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)\\
&=&\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\ln \!\left(\frac{(n+2)(2n-3)}{(n-2)(2n+5)}\!\right)\\
&=&-\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\ln \!\left(\frac{(2n+5)(2n-4)}{(2n-3)(2n+4)}\!\right)\\
&=&-\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\left(\ln\frac{2n+5}{2n+4}-\ln\frac{2n-3}{2n-4}\right)\\
&=&-\sum_{n=3}^{\infty}\left[\ln\left(\frac{2n+1}{2n}\right) \!\left(\ln\left(\frac{2n+5}{2n+4}\right)-\ln\left(\frac{2n-3}{2n-4}\right)\right)\right]\\
&=&-\sum_{n=3}^{\infty}\left[\ln\left(\frac{2n+5}{2n+4}\right)\ln\left(\frac{2n+1}{2n}\right) -\ln\left(\frac{2n+1}{2n}\right)ln\left(\frac{2n-3}{2n-4}\right)\right]\\
&=&-\bigg[\ln\left(\frac{11}{10}\right)\ln\left(\frac{7}{6}\right)-\ln\left(\frac{7}{6}\right)\ln\left(\frac{3}{2}\right)+\ln\left(\frac{13}{12}\right)\ln\left(\frac{9}{8}\right)-\ln\left(\frac{9}{8}\right)\ln\left(\frac{5}{4}\right)\\
&&+\ln\left(\frac{15}{14}\right)\ln\left(\frac{11}{10}\right)-\ln\left(\frac{11}{10}\right)\ln\left(\frac{7}{6}\right)+\ln\left(\frac{19}{18}\right)\ln\left(\frac{15}{14}\right)-\ln\left(\frac{15}{14}\right)\ln\left(\frac{11}{10}\right)\\
&&+\ln\left(\frac{23}{22}\right)\ln\left(\frac{19}{18}\right)-\ln\left(\frac{19}{18}\right)\ln\left(\frac{15}{14}\right)+\cdots\bigg]\\
&=&\ln\left(\frac{7}{6}\right)\ln\left(\frac{3}{2}\right)+\ln\left(\frac{9}{8}\right)\ln\left(\frac{5}{4}\right).
\end{eqnarray}
One can use the same way to handle $T$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality in sum of fractions
Is there a constant $C\ge 0$ such that for any $a,b,c,d>0$ the inequality holds:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge C+\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}
$$
My attempt:
$a=b=c=d$ gives $1\ge C$.
Now, take $b=c=d$; this leads to inequality with two variables ($a$ and, let's say, $b$), which for $C=1$ is equivalent to $a^3+2b^3\ge 3ab^2$. This one is true by Cauchy.
The case $a=b$, $c=d$ leads to the same thing.
Is $C=1$ good?
| For $C=1$ it's wrong. Try $a=10$, $b=15$, $c=10$ and $d=11$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.
Where is my mistake? Please explain very intuitively.
| With integrals:
$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)=\lim_{n\rightarrow\infty} \frac{1}{n}\biggl(\frac{1}{n} + \frac{2}{n} + \frac{3}{n}+\dots+\frac{n}{n}\biggr)= \int_{0}^1 x dx =\frac{1}{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "24",
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Intersection of rectangular and polar equations Find the points of intersection for the following relations:
$x^2+y^2-20 = 0$ and $\theta = -3\pi / 4$
So the first one is obviously a circle, and the second one, converting into rectangular form, is y/x = 1, and thus y=x. Substituting y as x in the first equation, I then get $x^2 = 10$, which yields $+- \sqrt {10}$, and thus the points of intersection should then be $(\sqrt{10},\sqrt{10})$ and $(-\sqrt{10},-\sqrt{10})$. However, the answer says that it is $(\sqrt{10},-\sqrt{10})$ and $(-\sqrt{10},\sqrt{10})$. What did I do wrong?
Also, for another problem, I need to find the intersection between these two:
$4x^2+16y^2-64=0$ and $x^2+y^2=9$
Following a similar approach of substitution, I was able to get 4 points: $(20/3, 7/3), (-20/3, 7/3), (-20/3,-7/3), (20/3, -7/3)$. But the answer says $(3,0) and (-3,0)$...
| How about $x=\sqrt{20}\cos \phi$, and $y=\sqrt{20}\sin \phi$. Putting $\phi = -3 \frac{\pi}{4}$ and $\phi = -3 \frac{\pi}{4}+\pi$ gives you the answer that you mentioned $(\sqrt{10}$,$\sqrt{10})$ and $(-\sqrt{10}$,$-\sqrt{10})$.
Also, for the other question $(3,0)$ cannot be the answer as it does not satisfy the first equation.
| {
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There exists rational number $a_n$, $(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$ Let $n\in \mathbb{N}$. Prove that there exists a rational number $a_n$ for which $$(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$$
My attempt :
I try $n=2$,
$(x^2+\frac{1}{2}x+1)(x^2-\frac{1}{2}x+1)=x^4+\frac{7}{4}x^2+1$
$a_2 = \frac{7}{4}$
| We define $$p_n(x)=x^{2n}+a_n\,x^n+1,$$ so we have to prove
$$\left.x^2+\frac12\,x+1\right|p_n(x).$$
This is clearly true for $n=1$ with $a_1=1/2$, and it was shown for $n=2$ with $a_2=7/4$. Induction step: we assume it's true for $n=k-1$ and $n=k$. Then, as can be seen by expanding the products,
$$p_{k+1}(x)=(x^{2k}+1)\left(x^2+\frac12\,x+1\right)-\frac12\,x\,p_k(x)-x^2\,p_{k-1}(x),$$ if only $$a_{k+1}=-\frac12\,a_k-a_{k-1},$$ so $$\left.x^2+\frac12\,x+1\right|p_{k+1}(x).$$
| {
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Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle. The question is:
If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$
My solution:
$(\cot A + \cot B + \cot C)^2\ge0$ // square of a real number
$\implies \cot^2A +\cot^2B + \cot^2 +2 \ge0 $ //Conditional identity used: $\cot A \cot B + \cot B \cot C + \cot A \cot C =1$
$\implies \cot^2A +\cot^2B + \cot^2 C \ge -2$
Thus according to me the answer should be $-2$. However, the answer key states that the answer is $1$. Where have I gone wrong?
| Since $\cot^2$ is a convex function, we can use Jensen.
$$\sum_{cyc}\cot^2\alpha\geq3\cot^2\frac{\alpha+\beta+\gamma}{3}=1.$$
The equality occurs for $\alpha=\beta=\gamma=\frac{\pi}{3},$ which says that $1$ is a minimal value.
Actually, $(\cot^2x)''=\frac{2(2+\cos2x)}{\sin^4x}>0$.
Also, we can use the following way.
Let $a$, $b$ and $c$ be sides-lengths and $S$ be an area of the triangle.
Thus, we need to prove that
$$\sum_{cyc}\frac{\cos^2\alpha}{\sin^2\alpha}\geq1$$ or
$$\sum_{cyc}\frac{\frac{(b^2+c^2-a^2)^2}{4b^2c^2}}{\frac{4S^2}{b^2c^2}}\geq1$$ or
$$\sum_{cyc}(b^2+c^2-a^2)^2\geq16S^2$$ or
$$\sum_{cyc}(b^2+c^2-a^2)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or
$$\sum_{cyc}(a^4-a^2b^2)\geq0$$ or
$$\sum_{cyc}(a^2-b^2)^2\geq0.$$
Done!
Also we can make the following. It's what you wish. I think.
$$\sum_{cyc}\cot^2\alpha=1+\sum_{cyc}(\cot^2\alpha-\cot\alpha\cot\beta)=1+\frac{1}{2}\sum_{cyc}(\cot\alpha-\cot\beta)^2\geq1$$
It's
$$\cot^2\alpha+\cot^2\beta+\cos^2\gamma=$$
$$=1+\cot^2\alpha+\cot^2\beta+\cos^2\gamma-\cot\alpha\cot\beta-\cot\alpha\cot\gamma-\cot\beta\cot\gamma=$$
$$=1+$$
$$+\frac{1}{2}\left(\cot^2\alpha-2\cot\alpha\cot\beta+\cot^2\beta+\cot^2\alpha-2\cot\alpha\cot\gamma+\cos^2\gamma+\cot^2\beta-2\cot\beta\cot\gamma+\cot^2\gamma\right)=$$
$$=1+\frac{1}{2}\left((\cot\alpha-\cot\beta)^2+(\cot\alpha-\cot\gamma)^2+(\cot\beta-\cot\gamma)^2\right)\geq1.$$
| {
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Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If $n$ is even then exists $\,k\in\mathbb{N}-\left\{0\right\}\,$ such that $n=2k.\;$ so:
$n^4(n-1)(n+1)^2=2^4 k^4(2k-1)(1+2k)^2$
If $n$ is odd then exists $k\in\mathbb{N}$ such that $n=2k+1$ so
$$n^4(n-1)(n+1)^3=2^3(2k+1)^4(k+1)^2$$
Can I conclude that the maximum positive integer that divides all these numbers is $N=2^3?$ (Please, help me to improve my english too, thanks!)
Note: I correct my "solution" after a correction... I made a mistake :\
| Newton's interpolation formula
gives
$$
n^7+n^6-n^5-n^4
=144 \binom{n}{2}+ 2160 \binom{n}{3}+ 9696 \binom{n}{4}+ 18480 \binom{n}{5}+ 15840 \binom{n}{6}+ 5040 \binom{n}{7}
$$
The gcd of the coefficients in the binomial representation is $48$ and so $n^7+n^6-n^5-n^4$ is always a multiple of $48$.
$48$ is largest factor of all values of $n^7+n^6-n^5-n^4$ because it is the gcd of the values for $n=2$ and $n=4$.
| {
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If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\sin x} = 1.$$
$$3\sin x-2\cos x = \sqrt{2}\sin x\cos x$$
$$(3\sin x-2\cos x)^2 = 2\sin^2 x\cos^2 x$$
$$9\sin^2 x+4\cos^2 x-12 \sin x\cos x = 2\sin^2 x\cos^2 x$$
Could some help me to solve it, thanks
| To take advantage of $\sqrt2$ in simplification restructure the equation to:
$$\frac{3 \cdot \dfrac{1}{\sqrt{2}}}{\cos x}-\frac{2 \cdot \dfrac{1}{\sqrt{2}}}{\sin x} = 1,$$
and so by inspection it is obviously $ x= \pi/4.$
| {
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"source": "stackexchange",
"question_score": "4",
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What are the steps for deriving a complicated generalization of a partial sum of a taylor series? I looked at the Taylor series for $$-\frac{x}{x-2}$$ and found it to be $$ \sum_{k=1}^{\infty}\frac{x^k}{2^k}$$
but then I also found that this series' partial sum is a bit more complicated in the form of
$$\frac{x 2^{-k}(2^{k}+x^{k})}{x+2} $$
My question is: how was such a partial sum derived, and how would you derive it for something more complicated, like for instance
$$\sum_{k=1}^{\infty}\frac{x^{k^2}}{k!}?$$
| The Taylor series has a strict formula. Suppose $f(x)$ has any derivative at $x_0$, then
$$f(x)=\sum _{n=0}^{\infty } \dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
where $f^{(n)}$ are the derivatives of the function, with the convention that $f^{(0)}=f,\;f^{(1)}=f',\ldots$
If $x_0=0$ then the formula simplifies in the MacLaurin version
$$f(x)=\sum _{n=0}^{\infty } \dfrac{f^{(n)}(0)}{n!}x^n\quad(*)$$
You have to compute all derivatives at $x=0$ and plug them in the formula.
In your example $f(x)=-\dfrac{x}{x-2}$ the first derivatives from first to sixth are
$$\frac{2}{(x-2)^2},-\frac{4}{(x-2)^3},\frac{12}{(x-2)^4},-\frac{48}{(x-2)^5},\frac{240}{(x-2)^6},-\frac{1440}{(x-2)^7}$$
If you plug $x=0$ you get
$$f^{1}(0)=\frac{1}{2},f^{2}(0)=\frac{1}{2},f^{3}(0)=\frac{3}{4},f^{4}(0)=\frac{3}{2},f^{5}(0)=\frac{15}{4},f^{6}(0)=\frac{45}{4}$$
and now plug in the MacLaurin formula $(*)$
$$-\dfrac{x}{x-2}=f(0)+f^{1}(0)x+\frac{f^{2}(0)}{2!}x^2+\frac{f^{3}(0)}{3!}x^3+\frac{f^{4}(0)}{4!}x^4+\frac{f^{5}(0)}{5!}x^5+\frac{f^{6}(0)}{6!}x^6+\ldots$$
and then
$$-\dfrac{x}{x-2}=\frac{1}{2}x+\frac{\frac{1}{2}}{2!}x^2+\frac{\frac{3}{2}}{4!}x^3+\frac{\frac{3}{2}}{4!}x^4+\frac{\frac{15}{4}}{5!}x^5+\frac{\frac{45}{4}}{6!}x^6+\ldots$$
and finally
$$-\dfrac{x}{x-2}=\frac{1}{2}x+\frac{1}{4}x^2+\frac{1}{8}x^3+\frac{1}{16}x^4+\frac{1}{32}x^5+\frac{1}{64}x^6+\ldots$$
It's not a "choice", is the formula
Hope it helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the value of a given trigonometric series.
Find the value of $\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}.$
My attempts:
I converted the given series to a simpler form:
$\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\cot^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+1.$
Then I found the following values because I already knew the values of $\sin22.5^{\circ}$ and $\cos22.5^{\circ}$:
$\cos^2(\frac{\pi}{16})= \dfrac{2+\sqrt{2+\sqrt2}}{4}$
$\sin^2(\frac{\pi}{16})= \dfrac{2-\sqrt{2+\sqrt2}}{4}$
$\sin^2(\frac{\pi}{8})= \dfrac{2-\sqrt2}{4}$
$\cos^2(\frac{\pi}{8})= \dfrac{2+\sqrt2}{4}$
However, at this stage I feel that my method of solving this problem is unnecessarily long and complicated. Could you guide me with a simpler approach to this question?
| Use the following identity:
$$ \cot x - \tan x = 2 \cot 2x$$
Square this
$$ \cot^2 x + \tan^2 x = 2 + 4 \cot^2 2x$$
Use this again with your idea of grouping terms as $\tan^2 x + \cot^2x$, and you will see that all you need to know is the value of $$\cot \frac{4\pi}{16} = \cot \frac{\pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$
If
$$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of $$\cos2\theta+\cos2\phi$$
My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2\cos\theta\cos\phi= b^2$$
Multiplying by 2 and subtracting 2 from both sides we obtain,
$$\cos2\theta+ \cos2\phi = 2b^2-2 - 4\cos\theta\cos\phi$$
How do I continue from here?
PS: I also found the value of $\sin(\theta+\phi)= \dfrac{2ab}{a^2+b^2}$
Edit: I had also tried to use $\cos2\theta + \cos2\phi= \cos(\theta+\phi)\cos(\theta-\phi)$ but that didn't seem to be of much use
| $$\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi),$$
$$a^2+b^2=2+2\cos(\theta-\phi)$$ and
$$b^2-a^2=\cos2\theta+\cos2\phi+2\cos(\theta+\phi).$$
Thus, $$\cos2\theta+\cos2\phi=2\cdot\frac{b^2-a^2-(\cos2\theta+\cos2\phi)}{2}\cdot\frac{a^2+b^2-2}{2},$$
which gives
$$\cos2\theta+\cos2\phi=\frac{(a^2+b^2-2)(b^2-a^2)}{a^2+b^2}.$$
| {
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Find three numbers such that the sum of all three is a square and the sum of any two is a square It seems like it should be a well known problem. Ive read that Diophantine himself first posed it. I couldnt find a solution when i researched for one.
It asks to find ordered triples $(x,y,z) $ such that
$$x+y+z=a^2 $$
$$x+y=b^2$$
$$x+z=c^2$$
$$y+z=d^2$$
As an example (41, 80, 320). Any guidance is appreciated. The ideal would be some kind of parametric solution for x, y, and z
| For the General case of formula there. https://artofproblemsolving.com/community/c3046h1172008_combinations_of_numbers_in_squares
The system of equations:
$$\left\{\begin{aligned}&a+b=x^2\\&a+c=y^2\\&b+c=z^2\\&a+b+c=q^2\end{aligned}\right.$$
Solutions have the form:
$$a=4t((2t-p)k^2+2(2t-p)^2k-2p^3+9tp^2-14pt^2+8t^3)$$
$$b=4(p^2-3pt+2t^2)k^2+8(4t^3-8pt^2+5tp^2-p^3)k+$$
$$+4(p^4-6tp^3+15p^2t^2-18pt^3+8t^4)$$
$$c=k^4+4(2t-p)k^3+4(p^2-3pt+3t^2)k^2-8(p^2-3pt+2t^2)tk+$$
$$+4t(2p^3-9tp^2+14pt^2-7t^3)$$
$$x=2(2t-p)(k+2t-p)$$
$$y=k^2+2(2t-p)k+2t^2$$
$$z=k^2+2(2t-p)k+2(t-p)^2$$
$$q=k^2+2(2t-p)k+6t^2-6tp+2p^2$$
$k,t,p$ - integers asked us.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculation of a partial derivative I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right) = -\frac{x}{\Vert (x,y) \Vert^3}$.
2) Thinking of that as $\frac{\partial}{\partial x}\left( (\sqrt{x^2+y^2})^{-1}\right)$ and $\frac{\partial}{\partial a}\sqrt{a}=\frac{1}{2\sqrt{a}}$ I get $x \cdot (x^2+y^2)$.
3) Thinking of that as $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$ and $\frac{\partial}{\partial a}\frac{1}{a}=-\frac{1}{a^2}$ I get $-\frac{2x}{x^2+y^2}$.
Why 2) and 3) are not correct?
| In your calculations you should use these 3 formulas:
*
*$\frac{\partial}{\partial x}\frac{1}{a(x,y)} = \frac{-1}{a^2(x,y)} \cdot \frac{\partial}{\partial x} a(x,y)$
*$\frac{\partial}{\partial x}\sqrt{b(x,y)} = \frac{1}{2\sqrt{b(x,y)}} \cdot \frac{\partial}{\partial x} b(x,y)$
*$\frac{\partial}{\partial x}(x^2+c(y)) = 2x$
In point 2:
*
*$c(y)=y^2$
*$a(x,y) = x^2 + c(y)$
*$b(x,y) = \frac{1}{a(x,y)}$
*Compute $\frac{\partial}{\partial x} \sqrt{b(x,y)}$
In point 3 :
*
*$c(y)=y^2$
*$b(x,y) = x^2+c(y) $
*$a(x,y) = \sqrt{b(x,y)} $
*Compute $\frac{\partial}{\partial x} \frac{1}{a(x,y)}$
| {
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Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\overline{AH}=\overline{OA}-\overline{OH}=1-\cos x$$ when $x$ is very small
w.r.t. * we can conclude $$x^2 \approx 2(1-\cos x) \\\to \cos x \sim 1- \frac{x^2}{2}$$
Now I have two question :
$\bf{1}:$ Is there other idea(s) to prove (except Taylor series) $x\to 0 , \space \cos x\sim 1-\frac{x^2}{2}\\$
$\bf{2}:$ How can show $\sin x \sim x- \frac{x^3}{6}$ with a geometric concept ?
Thanks in advance.
| Now a full answer (assuming we have a result slightly stronger than yours, that $\cos x = 1-\frac{x^2}{2}+o(x^2)$, which is the true meaning of $\cos x\sim 1-\frac{x^2}{2}$, but which you haven't proven - you haven't come up with a bound for your error.)
Let $D$ be the midpoint of the arc in $MA$. The area of the quadrilateral $OMDA$ is $\frac{\sin(x)}{2}$.
The area of circular wedge between $M$ and $A$ is $\frac{x}{2}$.
Taking the tangents to the circle at $M$ and $A$ and finding their intersection, $C$, you get that the are of $OMCA$ is $2\tan(x/2)$. So you have:
$$\sin\frac{x}{2}<\frac{x}{2}<\tan\frac{x}{2}$$
Setting $u=\frac{x}{2}$ you get:
$$\sin u< u < \frac{\sin u}{\cos u}$$
Or $\cos u<\frac{\sin u}{u}<1$
This is enough to see that $\sin u = u+O(u^3)$ (using that $\cos(u)=1+O(u^2)$,) but not enough to show what you want.
The goal is to find and approximation of the difference $u-\sin u$, twice the area of the region between the line segment $AD$ and the arc $AD$.
Letting $$A_n=\frac{1}2\sin \frac{x}{2^{n}},$$ is the area of $2^{n}$ copies of the triangle $MOA$ with angle $\frac{x}{2^n}$. From this, we see that $2^{n}A_n<\frac{x}{2}$, since the $2^n$ triangles fit inside the wedge.
Now, $A_{n+1}$ is gotten by taking half of the triangle for $A_{n+1}$ and adding a triangle with base $\sin(x/2^{n+1})$ and height $1-\cos(x/2^{n+1})$.
So: $$A_{n+1}=\frac{A_n}{2}+\frac{1}{2}\sin(x/2^{n+1})(1-\cos(x/2^{n+1})).$$
So if $B_{n}=2^{n}A_n$, then $B_0=\frac{\sin x}{2}$ and $$B_{n+1}=B_n+2^{n}\sin(x/2^{n+1})(1-\cos(x/2^{n+1})).$$
Writing $f(x)=x-\sin x$ and $g(x)=\cos x-1+\frac{x^2}{2}$ as our error functions. We get:
$$\begin{align}B_{n+1}-B_{n}&=2^{n}\sin(x/2^{n+1})(1-\cos(x/2^{n+1}))\\&=2^{n}\left(\frac{x}{2^{n+1}}+f\left(\frac{x}{2^{n+1}}\right)\right) \left(\frac{x^2}{2^{2n+3}}+g\left(\frac{x}{2^{n+1}}\right)\right)\\
&=\frac{x^3}{2^{2n+4}}+\frac{x}{2}g\left(\frac{x}{2^{n+1}}\right)+\frac{x^2}{2^{n+3}}f\left(\frac{x}{2^{n+1}}\right)+f\left(\frac{x}{2^{n+1}}\right)g\left(\frac{x}{2^{n+1}}\right)
\end{align}$$
Now, we know that $f(x)=O(x^3)$ and $g(x)=o(x^2)$, so we get:
$$\begin{align}\lim_{n\to\infty} B_n &= B_0+\sum_{n=0}^{\infty}\left(B_{n+1}-B_n\right)\\
&=\frac{\sin x}{2}+\left(x^3\sum_{n=0}^{\infty}\frac{1}{2^{2n+4}}\right)+o(x^3)+O(x^5)+o(x^5)\\
&=\frac{\sin x}{2}+\frac{x^3}{12}+o(x^3)
\end{align}$$
But $\lim_{n\to\infty} B_n =\frac{x}{2}$, because this is the area of the circular wedge.
So we have: $$x-\sin x = \frac{x^3}{6} + o(x^3)$$
You can actually prove that $B_n\to \frac{x}{2}$ by noting that $\frac{B_{n}}{\cos(x/2^n)}$ is the area of $2^n$ triangles that cover the circular wedge, and thus you have:
$$B_n<\frac{x}{2}<\frac{B_n}{\cos(x/2^n)}$$
Simce $B_n$ is increasing, and $\cos(x/2^n)\to 1$, we see that both values converge to $\frac{x}{2}$.
The full proof first prove that $\sin x = x+o(x)$.
From this, though, we can show that:
$$\cos(x)=1-2\sin^2 \frac{x}{2} = 1-\frac{x^2}{2}+o(x^2).$$
Then the rest of the argument follows, since $\cos(x)=1+O(u^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 3
} |
How to change the appearence of the correct answer of $\cos55^\circ\cdot\cos65^\circ\cdot\cos175^\circ$ I represented the problem in the following view and solved it: $$\begin{align}-\sin35^\circ\cdot\sin25^\circ\cdot\sin85^\circ\cdot\sin45^\circ&=A\cdot\sin45^\circ\\ -\frac{1}{2}(\cos20^\circ-\cos70^\circ)\cdot\frac{1}{2}(\cos50^\circ-\cos120^\circ)&=A\cdot\sin45^\circ\\ \cos20^\circ\cdot\cos50^\circ-\cos50^\circ\cdot\cos70^\circ+\frac{\cos20^\circ}{2}-\frac{\cos70^\circ}{2}&=A\cdot(-4)\cdot \sin45^\circ\\ \frac{1}{2}(\cos30^\circ+\cos70^\circ)-\frac{1}{2}(\cos20^\circ+\cos120^\circ)&=A\cdot(-4)\cdot \sin45^\circ\\ \frac{\sqrt{3}}{4}+\frac{\cos70^\circ}{2}-\frac{\cos20^\circ}{2}+\frac{1}{4}+\frac{\cos20^\circ}{2}-\frac{\cos70^\circ}{2}&=-2\sqrt{2}A\\ A&=-\frac{\sqrt{6}+\sqrt{2}}{16} \end{align}$$ I believe that the above answer is true. But that didn't match a variant below:
A) $-\frac{1}{8}$
B)$-\frac{\sqrt{3}}{8}$
C) $\frac{\sqrt{3}}{8}$
D) $-\frac{1}{8}\sqrt{2-\sqrt{3}}$
E) $-\frac{1}{8}\sqrt{2+\sqrt{3}}$
I did the problem over again. After getting the same result, I thought that the apperance of my answer could be changed to match one above, so I tried to implement one of formulae involving radical numbers: all to no avail. How to change that?
| \begin{align}
A &= - \sqrt{\left(\frac{\sqrt{6}+\sqrt{2}}{16}\right)^2}\\
&=- \sqrt{\left(\frac{\sqrt{3}+1}{8\sqrt{2}}\right)^2}\\
&=- \frac18\sqrt{\left(\frac{3+1+2\sqrt3}{2}\right)}\\
&=- \frac18\sqrt{2+\sqrt{3}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the determinant of the following $4 \times 4$ matrix
Use a cofactor expansion across a row or column to find the determinant of the following matrix
$$B=\begin{pmatrix}1 &c&0&0\\0&1&c&0\\0&0&1&c\\c&0&0&1\end{pmatrix}$$
Clearly indicate the steps you take.
I have tried
$$\det B = 1 \det \begin{pmatrix}1 &c&0\\0&1&c\\0&0&1\end{pmatrix}+(-c) \det \begin{pmatrix}c &0&0\\1&c&0\\0&1&c\end{pmatrix}$$
$$ = \det\begin{pmatrix}1 &c\\0&1\end{pmatrix}+c(-c)det\begin{pmatrix}c &0\\1&c\end{pmatrix}$$
$$=1-c^4$$
| Viewing $\rm B$ as a block matrix,
$$\det \left[\begin{array}{ccc|c} 1 & c & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & c\\ \hline c & 0 & 0 & 1\end{array}\right] = \det \left( \begin{bmatrix} 1 & c & 0\\ 0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} - \begin{bmatrix} 0\\ 0\\ c\end{bmatrix} \begin{bmatrix} c\\ 0\\ 0\end{bmatrix}^\top \right) = \det \begin{bmatrix} 1 & c & 0\\ 0 & 1 & c\\ -c^2 & 0 & 1\end{bmatrix} = \color{blue}{1 - c^4}$$
Viewing $\rm B$ as a circulant matrix, its eigenvalues are given by
$$\lambda_k = 1 + c \, \exp \left( i \frac{k \pi}{2} \right)$$
for $k \in \{0, 1, 2, 3\}$. Hence, the determinant of $\rm B$ is
$$\det \begin{bmatrix} 1 & c & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & c\\ c & 0 & 0 & 1\end{bmatrix} = (1 + c) (1 + i c) (1 - c) (1 - i c) = (1 - c^2) (1 + c^2) = \color{blue}{1 - c^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges
Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that, for every $n\ge1$,
$$\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2} \tag 2$$ Prove that $x_n=a_1 + a_2 + .. + a_n$ diverges.
It is clear that $x_n$ is increasing, so it has to have a limit. I tried to prove the limit is $+\infty$ but without success. No divergence criteria from series seems to work here.
UPDATE
Attempt:
Suppose a stronger inequality holds, namely that, for every $n\ge1$, $$\frac{a_{n+1}}{a_n}\geqslant1-\frac1n \tag 1$$ Then:
$$\frac {a_3}{a_2} \ge \frac 1 2\qquad
\frac {a_4}{a_3} \ge \frac 2 3\qquad
\ldots\qquad
\frac {a_{n-1}}{a_{n-2}} \ge \frac {n-3}{n-2}\qquad
\frac {a_n}{a_{n-1}} \ge \frac {n-2}{n-1}$$
Multiplying all the above yields
$$\frac {a_n}{a_2} \ge \frac 1 {n-1}$$
The last inequality proves the divergence.
| You may notice that
$$ \frac{a_{n+1}}{a_n} \geq \left(1-\frac{1}{n}\right)\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{1}$$
hence:
$$ \frac{a_{N+1}}{a_2}\geq \frac{1}{N}\prod_{n=2}^{N}\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{2}$$
but the infinite product $\prod_{n\geq 2}\left(1+\frac{1}{n^2-n-1}\right)^{-1}$ is convergent to a positive number ($-\frac{1}{\pi}\cos\frac{\sqrt{5}}{2}\approx 0.296675134743591$). It follows that $a_{N+1}\geq \frac{C}{N}$ so $\sum_{n\geq 2}a_n$ is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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By using the definition of limit only, prove that $\lim_{x\rightarrow 0} \frac1{3x+1} = 1$
By using the definition of a limit only, prove that
$\lim_{x\rightarrow 0} \dfrac{1}{3x+1} = 1$
We need to find $$0<\left|x\right|<\delta\quad\implies\quad\left|\dfrac{1}{3x+1}-1\right|<\epsilon.$$
I have simplified $\left|\dfrac{1}{3x+1}-1\right|$ down to $\left|\dfrac{-3x}{3x+1}\right|$
Then since ${x\rightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$\left|\dfrac{1}{3x+1}-1\right|=\left|\dfrac{-3x}{3x+1}\right|<\left|\dfrac{-3x}{4}\right|<\left|\dfrac{-3\delta}{4}\right|=\epsilon$$
No sure if the solution is correct
| If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $\frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < \frac14$. Hence $\delta < \frac14$.
if $-\frac14 < x < \frac14$, $$-\frac34+1 < 3x+1< \frac34+1$$.
$$\frac14 < 3x+1< \frac74$$
$$\left|\frac{1}{3x+4} \right| < 4$$
Hence $$12\delta < \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$
My attempt :
Substitute $a=x^2, b=y^2, c=z^2$, we have to prove that
$\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2} \geq \frac{3}{5}$
By C-S,
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(2y^2x+3y^2z^2)\frac{x}{z}\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(x^2y^2+3y^2xz\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$
| Let $a = x^2, b = y^2, c=z^2$, where $x, y, z > 0$. By C-S, we have
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \sum_{cyc} \frac{2yx + 3z^2}{zx} \geq \left( \sum_{cyc} \frac{x}{y} \right)^2.$$
But
$$ \sum_{cyc} \frac{2yx + 3z^2}{zx} = 5 \sum_{cyc} \frac{x}{y}.$$
Combining them we get
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \geq \frac{1}{5} \sum_{cyc} \frac{x}{y}.$$
It remains to apply AM-GM.
| {
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"url": "https://math.stackexchange.com/questions/2394297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Taylor Series of $\frac{1}{3z+1}$ at $a=-2$
Find the Taylor Series of $\frac{1}{3z+1}$ at $a=-2$
When we are asked to find the taylor series at a point $p$ the taylor series should be at the form $\sum a_n(z-p)^n$?
So
$$\frac{1}{3z+1}=\frac{1}{2z-1+z+2}=\frac{1}{2z-1}\frac{1}{1+\frac{z+2}{2z-1}}=\frac{1}{2z-1}\frac{1}{1-(-\frac{z+2}{2z-1})}=\frac{1}{2z-1}\sum_{n=0}^{\infty}(-\frac{z+2}{2z-1})^n=\frac{1}{2z-1}\sum_{n=0}^{\infty}(-1)^n\frac{(z+2)^n}{(2z-1)^n}=\sum_{n=0}^{\infty}(-1)^n\frac{(z+2)^n}{(2z-1)^{n+1}}$$?
| No, your expansion is not a power series centered at $-2$.
Let $w=z+2$ then for $|3w/5|<1$, or $|z+2|<5/3$,
\begin{align*}\frac{1}{3z+1}&=\frac{1}{3(w-2)+1}=\frac{1}{3w-5}=\frac{-1/5}{1-3w/5}=
-\frac{1}{5}\sum_{n=0}^{\infty}\left(\frac{3w}{5}\right)^n\\
&=-\frac{1}{5}\sum_{n=0}^{\infty}\left(\frac{3(z+2)}{5}\right)^n.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Taylor expansion of $\cos^2(\frac{iz}{2})$
Expand $\cos^2(\frac{iz}{2})$ around $a=0$
We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$
So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$
We have $t=\frac{iz}{2}$
$$\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({zi})^{4n}}{2^{4n}{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({z})^{4n}}{2^{4n}{4n^2!}}$$
But the answer in the book is $$1+\frac{1}{2}\sum_{n=1}^{\infty}\frac{({z})^{2n}}{{2n!}}$$.
| If $f(x) = \sum_{n=0}^{\infty} a_n x^n$ then
$$ f^2(x) = \left( \sum_{n=0}^{\infty} a_n x^n \right) \left( \sum_{m=0}^{\infty} a_m x^m \right) = \sum_{k = 0}^{\infty} \left( \sum_{l = 0}^k a_l a_{k - l} \right) x^k \neq \sum_{n=0}^{\infty} a_n^2 x^{2n}.$$
In your case,
$$ \cos^2(x) = \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} \right) \left( \sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!} x^{2m} \right) = \sum_{k=0}^{\infty} \left( \sum_{l=0}^k \frac{(-1)^l}{(2l)!} \frac{(-1)^{k - l}}{(2(k-l))!} \right) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} \left( \sum_{l=0}^k { 2k \choose 2l} \right) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k 2^{2k-1}}{(2k)!} x^{2k}. $$
Plug in $x = \frac{iz}{2}$ and see what you get.
| {
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"url": "https://math.stackexchange.com/questions/2396742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Polynomial $x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ is reducible over $\mathbb{Q}$? Clearly, there are no roots, but how can I find factors of higher degree?
| To see how to get the answer, know the factorization of a finite geometric series:
$$ \begin{array}{ll} 1+x+\cdots+x^8 & \displaystyle=\frac{x^9-1}{x-1} \\[5pt] & \displaystyle =\frac{(x^3)^3-1}{x-1} \\[5pt] & \displaystyle =\frac{(x^3-1)\big((x^3)^2+x^3+1\big)}{x-1} \\[5pt] & = (x^2+x+1)(x^6+x^3+1) \end{array} $$
As Cauchy mentions in the comments, this generalizes with cyclotomic polynommials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right)<1 $ $$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right) $$
Which is the solution of$\ f(x)<1 $ ? I did:
$$\ \frac{2x+3}{7-x}>0 $$
$\ x<7 $ results in $\ x> \frac{-3}{2} $ whereas $\ x>7 $ results in $\ x< \frac{-3}{2} $ so x=($\ \frac{-3}{2} $,7) and also
$\ x-2>0 $ so $\ x>2 $
$$\ x=(2,7)-\{3\} $$
So after finding out where $\ f $ is defined, I did this:
$$\log _{x-2}\left(\frac{2x+3}{7-x}\right) <\log \:_{x-2}\left(x-2\right) $$
$$\ \frac{2x+3}{7-x} <x-2 $$
$$\ x^2-7x+17<0 $$
And I figured I must have done a mistake somewhere as $\ d<0 $ and $\ x^2 $'s coefficient is $\ >0 $ so $\ f $ would be $\ >0$. Could you let me know where is my mistake? The correct answer is $\ x=(2,3) $
| The domain gives $2<x<7$, $x\neq3$ and since $\frac{2x+3}{7-x}\neq x-2$,
we get the answer without any cases immediately by the intervals method:
$$(2,3).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.
| This is a supplement to the answer by knm to show that $k\ge n+2$ (for large enough $n$).
Let $S_n=1^n+\cdots+n^n$. First, we show that $S_{n+1}/S_n>2(n+1)$. To do that, it suffices to show that $(r+1)^{r+1}/r^r>2(n+1)$ for all $r=0,\ldots,n$. Since the LHS is an increasing function, it has it's maximum at $r=n$, so we actually get
$$
\frac{(r+1)^{r+1}}{r^r} \ge \frac{(n+1)^{n+1}}{n^n}
> (n+1)\cdot\left(1+\frac{1}{n}\right)^n \ge 2(n+1).
$$
From this follows that
$$
\frac{S_{n+1}}{S_n} = \frac{\sum_{r=0}^n (r+1)^{r+1}}{\sum_{r=0}^n r^{r}}
> (n+1)\cdot\left(1+\frac{1}{n}\right)^n \ge 2(n+1).
$$
If $S_n>(n+1)!$ for some $n$, then if $S_n=k!$ we must have $k\ge n+2$.
However, if $S_n>(n+1)!$, then $S_{n+1}>2(n+1)S_n>(n+2)!$, so once we find one $n$ with $S_n>(n+1)!$, this is also the case for all larger $n$.
In particular, $S_3=36>4!$, so for all $n\ge3$ we have $S_n>(n+1)!$, which implies that $k\ge n+2$ if $k!=S_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.
| There's actually a general formula for these kinds of expressions. Namely$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}2}$$
Where $X,Y$ are real numbers. Simply substituting $X=6$ and $Y=\sqrt{20}$ gives the proper denesting. The proof of this is quite simple. Assume that$$X\pm Y=\left(\sqrt A\pm \sqrt B\right)^2$$and expand via binomial theorem. Collecting terms, you will end up with two equations from which you can solve for $A$ and $B$ in terms of $X$ and $Y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Compute the determinant The following problem is taken from here exercise $2:$
Question: Evaluate the determinant:
\begin{vmatrix}
0 & x & x & \dots & x \\
y & 0 & x & \dots & x \\
y & y & 0 & \dots & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & y & y & \dots & 0
\end{vmatrix}
My attempt:
I tried to use first row substract second row to obtain
\begin{pmatrix}
y & -x & 0 \dots & 0
\end{pmatrix}
and also first row subtracts remaining rows.
However, I have no idea how to proceed.
| If you continue subtracting the $(k+1)$th row from the $k$th, you end up with
$$ \Delta_n = \begin{vmatrix} -y & x \\
& -y & x \\
&& -y & x \\
&&&\ddots & \ddots \\
&&&& -y & x \\
y & y & y & \cdots & y & 0 \end{vmatrix}, $$
where blank spaces are zeros. One can also pull out a factor of $xy$ now, but there's not a lot of point. Expanding along the first column gives
$$ \Delta_n = -y\Delta_{n-1} +(-1)^{n-1} y \begin{vmatrix} x \\
-y & x \\
& -y & x \\
&&\ddots & \ddots \\
&&& -y & x \end{vmatrix} = -y \Delta_{n-1} + (-1)^{n-1} yx^{n-1}. $$
Iterating this and using $\Delta_2=-xy$ gives
$$ \Delta_n = (-1)^{n-1} (xy^{n-1}+x^2y^{n-2}+\dotsb+x^{n-1}y) = (-1)^{n-1}xy \frac{x^{n-1}-y^{n-1}}{x-y}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding the quadratic equation from its given roots.
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c
=0$ , then form an equation whose roots are:
$\alpha+\dfrac{1}{\beta},\beta+\dfrac{1}{\alpha}$
Now, using Vieta's formula,
For new equation,
Product of roots ($P$) = $\dfrac{a^2+c^2+2ac}{ac}$
Sum of roots ($S$) = $\dfrac{2c-b}{c}$
Hence,
the required equation is:
$acx^2 - ax(2c-b)+(a+c)^2=0$ // as quadratic equation = $x^2-Sx +Px=0$
But the answer key states that the answer is:
$acx^2 +b(a+c)x
+(a+c)^2=0$
I am really doubtful of this answer. Where have I gone wrong (or is the answer in the key wrong?)?
| $$\alpha +\beta =-\frac { b }{ a } \\ \alpha \beta =\frac { c }{ a } \\ a\left( x-\left( \alpha +\frac { 1 }{ \beta } \right) \right) \left( x-\left( \beta +{ \frac { 1 }{ \alpha } } \right) \right) =0\\ a\left( x-\frac { \alpha \beta +1 }{ \beta } \right) \left( x-\frac { \beta \alpha +1 }{ \alpha } \right) =0\\ a\left( { x }^{ 2 }-\frac { \left( \alpha \beta +1 \right) \left( \beta +\alpha \right) }{ \alpha \beta } x+\frac { { \left( \beta \alpha +1 \right) }^{ 2 } }{ \beta \alpha } \right) =0\\ a\left( { x }^{ 2 }-\frac { \left( \frac { c }{ a } +1 \right) \left( -\frac { b }{ a } \right) }{ \frac { c }{ a } } x+\frac { { \left( \frac { c }{ a } +1 \right) }^{ 2 } }{ \frac { c }{ a } } \right) =0\\ a{ x }^{ 2 }+\frac { \left( c+a \right) b }{ c } x+\frac { { \left( c+a \right) }^{ 2 } }{ c } =0\\ c{ a }{ x }^{ 2 }+\left( bc+ba \right) x+{ \left( c+a \right) }^{ 2 }=0$$
| {
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} |
proof that the pattern exists for all n Explain the pattern
$$(\sqrt2-1)^1= \sqrt2 - \sqrt1$$
$$(\sqrt2-1)^2 = \sqrt9 - \sqrt8$$
$$ (\sqrt2-1)^3 = \sqrt{50} - \sqrt{49} $$
that is $(\sqrt2-1)^n$ is equal to difference of two consecutive numbers one of which are squares.
| We will prove a stronger result that $(\sqrt{2}-1)^{2n}=\sqrt{x_n^2}-\sqrt{2y_n^2}$ for some $x_n, y_n$ where $x_n^2-2y_n^2=1$, and $(\sqrt{2}-1)^{2n-1}=\sqrt{2z_n^2}-\sqrt{w_n^2}$ for some $z_n, w_n$ where $2z_n^2-w_n^2=1$ (here $\{x_n\}, \{y_n\}, \{z_n\}, \{w_n\}$ are all increasing sequences of positive integers such that $x_1=3, y_1=2, z_1=1, w_1=1$.)
$n=1$ case is trivial, so suppose that the result holds for $n$. Then we have $(\sqrt{2}-1)^{2n+1}=(\sqrt{2z_n^2}-\sqrt{w_n^2})(3-2\sqrt{2})=(\sqrt{2(3z_n+2w_n)^2}-\sqrt{(4z_n+3w_n)^2})$. It is enough to prove $2(3z_n+2w_n)^2-(4z_n+3w_n)^2=1$, which is equivalent to $2z_n^2-w_n^2=1$, and this holds due to induction hypothesis. That is, we can set $z_{n+1}=3z_n+2w_n, w_{n+1}=4z_n+3w_n$.
Similarly, we have $(\sqrt{2}-1)^{2n+2}=(\sqrt{x_n^2}-\sqrt{2y_n^2})(3-2\sqrt{2})=\sqrt{(3x_n+4y_n)^2}-\sqrt{2(2x_n+3y_n)^2}$. It is enough to prove $(3x_n+4y_n)^2-2(2x_n+3y_n)^2=1$, which is equivalent to $x_n^2-2y_n^2=1$, and this holds due to induction hypothesis. That is, we can set $x_{n+1}=3x_n+4y_n, y_{n+1}=2x_n+3y_n$. This proves the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find all positive integers $n > 1$ such that the polynomial $P(x)$ belongs to the ideal generated by the polynomial $x^2 +x +1$ in $\Bbb Z_n[x]$
Find all positive integers $n > 1$ such that the polynomial $x^4 + 3x^3 + x^2 + 6x + 10$ belong to the ideal generated by the polynomial $x^2 + x + 1$ in $\Bbb Z_n[x]$.
My attempt: I was using the Division Algorithm
$$P(X)= X^4 + 3x^3 + X^2 +6X + 10 = (X^2 + x + 1)( x^2 + 2x -2) + (6x + 12).$$
Here I got the remainder $6x + 12$ not equal to $0$, so $P(x)$ is irreducible over $\Bbb Z_n[x]$ because it cannot be factored into the product of two non-constant polynomials.
My thinking is that $\Bbb Z_3[x]$ is only the satisfied $x^2 + x + 1$
$$(1)^2 + 1 + 1 =3$$ and we if we divide $3/3 =1$ and remainder $= 0$.
Therefore $3$ is the only positive integer $n > 1$ such that the polynomial $P(x)$ belong to the ideal generated by the polynomial $x^2 + x + 1$ in $\Bbb Z_n[x]$.
Is my answer is correct or not? I would be more thankful to rectifying my mistake.
| Because $x^4+3x^3+x^2+6x+10$ is in the ideal,
$p(x)=x^4+3x^3+x^2+6x+10-(x^2 + x + 1)( x^2 + 2x -2)=6x+12 $ is also in the ideal.
So $6x+12=q(x)*(x^2+x+1)=q(x)x^2+(x+1)q(x)$.
As $deg(6x+12)=1$, $q(x)x^2=0$. Also $deg q(x)=1$, which leads to $6x+12=0$.
Therefore, n=2,3 or 6.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a clever one? Quadratic completion?)
\begin{align}
\frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}} & = \\
& = (x^2+1)^{-1/2} -x^2*(x^2+1)^{-3/2} \\
& = (x^2+1)^{-1/2} * ( 1 - x^2 *(x^2+1)^{-1}) \\
& = ...
\end{align}
To give a bit more context, I was calculating the derivative of $\frac{x}{\sqrt{x^2+1}}$ in order to use newtons method for approximating the roots.
| Hint:$$term_a = \frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}=\\
\frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(\sqrt{x^2+1})^{3}}=\\
\frac{x^2+1}{(\sqrt{x^2+1})^{3}} - \frac{x^2}{(\sqrt{x^2+1})^{3}}=\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| Here is a layout of the conversion algorithm (successive Euclidean divisions)
$$\begin{array}{ccrcrc*{10}{c}}
&&44&&4 \\
9&\Bigl)&397&\Bigl)&44&\Bigl)& \color{red}{\mathbf 4}\\
&&\underline{36}\phantom{7}&&\underline{36}\\
&&37&&\color{red}{\mathbf 8} \\
&&\underline{36} \\
&&\color{red}{\mathbf 1}
\end{array}$$
This is based on Horner's scheme:
\begin{align}
[397]_{10}&=9\cdot 44+ \color{red}1=9(9\cdot 4+\color{red}8)+\color{red}1=9^2\cdot\color{red}4+9\cdot\color{red}8+1\cdot \color{red}1.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
How can I prove the following sequence converges Given the following n-th term sequence:
$$a_{n} = \sqrt[n]{1^2+2^2+...+n^2}$$
You're asked to evaluate the limit of the given sequence, justifying your operations.
What strategy should I take on this? I have considered taking some inequality in order to, eventually, be able to use the Squeeze Theorem.
I've tried exploring some initial terms, viz:
$$
a_{1} = \sqrt[n]{1^2} = 1^\frac{1}{n}\\
a_{2} = \sqrt[n]{1^2 + 2^2} = 5^\frac{1}{n}\\
a_{3} = \sqrt[n]{1^2 + 2^2 + 3^2} = 14^\frac{1}{n}\\
\vdots\\
a_{n} = \sqrt[n]{1^2+2^2+...+n^2} = \sqrt[n]{k + n^2} = (k + n^2)^\frac{1}{n}
$$
Supposing $k$ is the sum of all the $n-1$ terms of the sequence, rightly before $n²$. We can see that $a_{n}$ is always smaller than $a_{n+1}$ for any $n$ strictly positive.
I'm not sure, though, what else I can try. I would appreciate a hand here.
Answer is:
\begin{align} 1 \end{align}
| $$1 = \sqrt[n]{1^2} \leq \sqrt[n]{1^2 + 2^2 + \ldots + n^2} \leq \sqrt[n]{n^2 + n^2 + \ldots + n^2} = \sqrt[n]{n^3} \xrightarrow{n\to\infty}1$$
By the squeeze theorem, the sequence converges to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Reduction formula for Integral I am trying to find a reduction formula for $$\int\frac{\sin ^n(x)}{x} dx .$$
Some numerical calculations seem to indicate a reduction formula should exist, but the usual tricks (break up the $\sin ^n(x)$ etc) don't seem to help.
Any pointers/solutions would be great.
| This is not an answer, but can lead you to the answer
$\text{Si}=\int\frac{\sin (x)}{x} \,dx;\;\text{Ci}=\int\frac{\cos (x)}{x} \,dx$
Some solutions of
$$\int\frac{\sin ^n(x)}{x} \,dx$$
for $n=1..10$
$$
\begin{array}{l|l}
n & integral\\
\hline
1 & \text{Si}(x) \\
2 & \frac{\log (x)}{2}-\frac{\text{Ci}(2 x)}{2} \\
3 & \frac{1}{4} (3 \text{Si}(x)-\text{Si}(3 x)) \\
4 & \frac{1}{8} (-4 \text{Ci}(2 x)+\text{Ci}(4 x)+3 \log (x)) \\
5 & \frac{1}{16} (10 \text{Si}(x)-5 \text{Si}(3 x)+\text{Si}(5 x)) \\
6 & \frac{1}{32} (-15 \text{Ci}(2 x)+6 \text{Ci}(4 x)-\text{Ci}(6 x)+10 \log (x)) \\
7 & \frac{1}{64} (35 \text{Si}(x)-21 \text{Si}(3 x)+7 \text{Si}(5 x)-\text{Si}(7 x)) \\
8 & \frac{1}{128} (-56 \text{Ci}(2 x)+28 \text{Ci}(4 x)-8 \text{Ci}(6 x)+\text{Ci}(8 x)+35 \log (x)) \\
9 & \frac{1}{256} (126 \text{Si}(x)-84 \text{Si}(3 x)+36 \text{Si}(5 x)-9 \text{Si}(7 x)+\text{Si}(9 x)) \\
10 & \frac{1}{512} (-210 \text{Ci}(2 x)+120 \text{Ci}(4 x)-45 \text{Ci}(6 x)+10 \text{Ci}(8 x)-\text{Ci}(10 x)+126 \log (x)) \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mathematical Induction - Is this correct? The question is :
Prove that
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{n-1} = 3 \cdot 2^n - 3$$
for all integers $n ≥ 1$
My Solution (I'll skip the basis step and go straight to the inductive step):
II.
Assume that
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} = 3\cdot 2^k - 3$$
Consider $n=k+1$:
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} + 3 \cdot 2^{(k+1)-1}= 3\cdot 2^{k+1} - 3$$
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} + 3 \cdot 2^k= 3\cdot 2^{k+1} - 3$$
$$3\cdot 2^k - 3 + 3\cdot 2^k = 3\cdot 2^{k+1} - 3$$
Cancelling out the $3$, I get:
$$3\cdot 2^k + 3\cdot 2^k = 3\cdot 2^{k+1}$$
Dividing both sides by $3$, I get:
$$2^k + 2^k = 2^{k+1}$$
$$2^{k+k} = 2^{k+1}$$
$$2^{2k} ≠ 2^{k+1}$$
III. Therefore, $P(n)$ is not true for all integers $n ≥ 1$
Is my solution correct? I find it weird that we were given a problem that isn't true but I guess that's intentional (unless I'm the one who's wrong.)
| This question has been answered by this user (and others) via the comment. In particular quoting the comment:
$2^k+2^k=2×2^k=2^{k+1}$. This is the part where you made mistake. Otherwise it's fine, and you have actually proved it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Volume of a solid bounded by 6 planes I need the volume of a figure delimited by planes
$$ 1 \le x + y + z \le 2 \\
0 \le x - y - z \le 3 \\
-1 \le 2x + y - z \le 4 $$
That should hold whenever this sum here :
$$ x \begin{pmatrix}
1 \\ 1 \\ 2
\end{pmatrix} +
y\begin{pmatrix}
1 \\-1 \\ 1
\end{pmatrix} +
z \begin{pmatrix}
1 \\-1 \\-1
\end{pmatrix}$$
belongs to the cube $ (1,2) × (0,3) × (-1,4) $
Thus I suppose I am to scale the cube by the inverse transform to get this volume :
$$ \left| \det \begin{pmatrix}
1 & 1 & 1 \\
1 & -1 & -1 \\
2 & 1 & -1
\end{pmatrix} ^{-1} \right| * (2-1)(3-0)(4-(-1))$$
Is this correct ?
| Apart from a typo $\bigl((2-0)$ should be replaced by $(3-0)\bigr)$ this is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show identity for $e^x$ I want to show that, using the definition $e^x := \lim_{n \to \infty} (1+\frac{x}{n})^n$, the following identity holds:
$$ e^{x+y}=e^xe^y$$
This is what I tried:
\begin{align}
e^xe^y &=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n \\
&= \lim_{n \to \infty}\left(\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)\right)^n \\
&=\lim_{n \to \infty}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n
\end{align}
However, this seems to be more than $e^{x+y}=\lim_{n \to \infty}(1+\frac{x+y}{n})^n$. How can I resolve this seeming contradiction to the upper identity?
| For $x, y \geq 0$ we have:
$$\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) = 1 + \frac{x+y}{n} + \frac{xy}{n^2}\geq 1+\frac{x+y}{n}$$
For $n \in \mathbb{N}$, using the identity $u^n-v^n = (u-v)\sum_{k=0}^{n-1}u^{n-1-k}v^k$ we have:
\begin{align}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1 + \frac{x+y}{n}\right)^n &= \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^n - \left(1+\frac{x+y}{n}\right)^n \\
&= \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k
\end{align}
Using the first inequality we get:
$$\frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1} \leq \left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1+\frac{x+y}{n}\right)^n \leq \frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1}$$
The left inequality is obtained by noticing that $\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right) = \left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) \geq \left(1+\frac{x+y}{n}\right)$ so by replacing the larger term with the smaller one in every product, we get a lower bound for the sum:
\begin{align}\frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k &\geq \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1+\frac{x+y}{n}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k \\
&= \frac{xy}{n^2}\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}\right)^{n-1} \\
&= \frac{xy}{n^2} \cdot n\,\left(1+\frac{x+y}{n}\right)^{n-1} \\
&= \frac{xy}{n} \left(1+\frac{x+y}{n}\right)^{n-1}
\end{align}
Similarly, for the right inequality we replace the smaller term with the larger one:
\begin{align}\frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k &\leq \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^k \\
&= \frac{xy}{n^2}\sum_{k=0}^{n-1}\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\
&= \frac{xy}{n^2} \cdot n\,\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\
&= \frac{xy}{n} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\
&= \frac{xy}{n} \left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1}
\end{align}
Letting $n\to\infty$ both sides converge to $0$ so by the squeeze theorem we have:
$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1+\frac{x+y}{n}\right)^n = 0$$
So
$$e^xe^y = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{x+y}{n}\right)^n = e^{x+y}$$
Just for clarification, the limits of the lower and upper bound are obtained using the limit of product and quotient theorem:
$$\lim_{n\to\infty} \frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1} = \lim_{n\to\infty} \frac{xy}{n} \frac{\left(1+\frac{x+y}{n}\right)^n}{\left(1+\frac{x+y}{n}\right)} = 0 \cdot\frac{e^{x+y}}{1} = 0$$
$$\lim_{n\to\infty} \frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1} = \lim_{n\to\infty}\frac{xy}{n}\frac{\left(1+\frac{x}{n}\right)^{n-1}}{\left(1+\frac{x}{n}\right)}\frac{\left(1+\frac{y}{n}\right)^{n-1}}{\left(1+\frac{y}{n}\right)} = 0\cdot \frac{e^x}{1}\cdot\frac{e^y}{1} = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n - L \rvert < \epsilon$
So $\lvert \frac{\sqrt {n^2 +2}}{4n+1} - \frac {1}{4} \rvert$ simplifies to $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}$
Now $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ is
simplified to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$ Then I would square everything to remove the square root and simplify fractions but I end up with $n> \sqrt{\frac{1}{8(\epsilon^2-\frac{1}{16}}}$
We can't assume $\epsilon > \frac{1}{4}$ so somewhere I went wrong. Any help would be appreciated.
| For one, we have
$$
\frac{\sqrt{n^2+2}}{4n+1} \le \frac{\sqrt{n^2+2}}{4n} = \sqrt{\frac{1}{16}+\frac{1}{8n}} \le \frac{1}{4}+ \sqrt{\frac{1}{8n}},
$$
and we also have
$$
\frac{\sqrt{n^2+2}}{4n+1} \ge \frac{n}{4n+1} = \frac{1}{4 + 1/n}.
$$
As $n\to\infty$, both of these tend to $1/4$. By the squeeze theorem, we obtain the desired result.
If you need a more formal proof, you can go the extra mile like this:
Let $\epsilon>0$ be arbitrary. By the Archimedean property of $\Bbb R$, there exists $N_1\in\Bbb N$ such that if $n\ge N_1$, then
$\sqrt{\dfrac{1}{8n}}<\epsilon$. Thus $\lim_{n\to\infty}\dfrac{1}{4}+\sqrt{\dfrac{1}{8n}} = \dfrac{1}{4}$.
Now consider the difference
$$
\left|\frac{1}{4}-\frac{1}{4+1/n}\right| = \frac{1}{4}\cdot\frac{1}{4n+1}.
$$
Again, by the Archimedean property of $\Bbb R$, there exists $N_2\in\Bbb N$ such that if $n\ge N_2$, then $\frac{1}{4}\cdot\frac{1}{4n+1} <\epsilon$. Thus $\lim_{n\to\infty}\dfrac{1}{4+1/n} = \dfrac{1}{4}$. By the squeeze theorem, the claim is proved.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the closed form of $\sum_{k=2}^{\infty}{\lambda(k)-1\over k}?$ Given that:
$$\sum_{k=2}^{\infty}{\lambda(k)-1\over k}\tag1$$
Where $\lambda(k)$ is Dirichlet Lambda Function
We are seeking to determine the closed form $(1)$ and came close to estimates it to $1-{\frac12}\left(\gamma+\ln{\pi}\right)$.
where $\gamma=0.5772156...$ is Euler-Mascheroni Constant
How can we evalauate the exact closed form for $(1)$?
| Another approach is to use the ordinary generating function of the Riemann zeta function, namely $$\sum_{n=2}^{\infty} \zeta(n) x^{n-1} = - \gamma - \psi (1-x), \quad |x| <1. $$
(This is just the Maclaurin series of the digamma function $\psi(1-x)$.)
Integrating both sides of the generating function, we get
$$\sum_{n=2}^{\infty} \frac{\zeta(n)}{n} \, x^{n} = -\gamma x + \log \Gamma(1-x) + C, $$ where the constant of integration must be zero since the value of the series is $0$ when $x=0$.
Therefore,
$$\begin{align} &\sum_{n=2}^{\infty} \frac{\lambda(n)-1}{n} \\ &= \sum_{n=2}^{\infty} \left(\frac{\zeta(n)}{n} - \frac{1}{n} \right) - \sum_{n=2}^{\infty} \frac{\zeta(n)}{2^{n}n} \tag{1}\\&= \lim_{x \to 1^{-}}\sum_{n=2}^{\infty} \left(\frac{\zeta(n)}{n} - \frac{1}{n} \right)x^{n} - \sum_{n=2}^{\infty} \frac{\zeta(n)}{2^{n}n} \tag{2} \\ &=\lim_{x \to 1^{-}}\left({\color{red}{\sum_{n=2}^{\infty} \frac{\zeta(n)}{n}x^{n}}} - {\color{green}{\sum_{n=2}^{\infty}\frac{x^{n}}{n}}} \right) - \sum_{n=2}^{\infty} \frac{\zeta(n)}{2^{n}n} \\ &= \lim_{x \to 1^{-}} \big({\color{red}{\log \Gamma(1-x)- \gamma x}} {\color{green}{ + \ln(1-x) + 1}} \big) + \frac{\gamma}{2} - \log \Gamma \left(1-\frac{1}{2} \right) \\ &=\lim_{x \to 1^{-}} \big(\log \Gamma(1-x) + \ln(1-x) \big)+1 -\frac{\gamma}{2} - \frac{1}{2} \, \log \pi \\ &= \lim_{x \to 1^{-}} \log \big(\Gamma(1-x)\cdot (1-x) \big)+1 -\frac{\gamma}{2} - \frac{1}{2} \, \log \pi \\ &= \lim_{x \to 1^{-}} \log \Gamma(2-x) +1 -\frac{\gamma}{2} - \frac{1}{2} \, \log \pi \\ &=0 +1 -\frac{\gamma}{2} - \frac{1}{2} \, \log \pi \\ &= 1 - \frac{1}{2} \left(\gamma + \log \pi \right) \end{align}$$
$(1)$ $\lambda(s) = (1-2^{-s}) \zeta(s)$
$(2)$ Abel's theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$? How do I find the closed form of this integral:
$$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$
First, I have a partial fraction of it:
$$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$
$$A=\frac{1}{(x^3-2x+4)'}|_{x=-2}=\frac{1}{(3x^2-2)}|_{x=-2}=\frac{1}{10}$$
$$Bx+C=\frac{1}{x+2}|_{x^2-2x=-2}=\frac{1}{(x+2)}\frac{(x-4)}{(x-4)}|_{x^2-2x=-2}=$$
$$=\frac{(x-4)}{(x^2-2x-8)}|_{x^2-2x=-2}=\frac{(x-4)}{(-2-8)}|_{x^2-2x=-2}=-\frac{1}{10}(x-4)$$
Thus:
$$\frac{1}{x^3-2x+4}=\frac{1}{10}\left(\frac{1}{x+2}-\frac{x-4}{x^2-2x+2}\right)$$
$$I=\frac{1}{10}\left(\int_0^2\frac{\ln x}{x+2}dx-\int_0^2\frac{(x-4)\ln x}{x^2-2x+2}dx\right)$$
What should I do next?
| Just for your curiosity.
If you enjoy special functions of complex arguments, the antiderivative can be computed.
$$\frac 1 {x^3-2x+4}=\frac{\frac1{10}}{
x+2}-\frac{\frac{1}{20}-\frac{3 i}{20}}{x-(1-i)}-\frac{\frac{1}{20}+\frac{3 i}{20}}{x-(1+i)}$$ which makes that we are left with integrals $$I_a=\int \frac{\log(x)}{x-a}\,dx=\text{Li}_2\left(\frac{x}{a}\right)+\log (x) \log \left(1-\frac{x}{a}\right)$$ where appears the polylogarithm function.
This would lead to the result Thomas Andrews gave in a comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating
$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$
My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting
$ u = a \tan \theta$. With this in mind: I let $x = \sqrt{3} \tan \theta$ .Moving on, $dx = \sqrt{3} (\sec \theta )^2 $. And.....if $\theta$ is an
acute angle, we see in the illustration below that...
$\tan \theta = \frac{x}{\sqrt{3}}$
Then...getting the equivalent form of $\sqrt{x^2 + 3}$ in terms of $\theta$:
$$\sqrt{x^2 + 3}$$
$$ =\sqrt{(\sqrt{3} \tan \theta)^2 + 3}$$
$$= \sqrt{3(\tan \theta)^2 + 3}$$
$$= \sqrt{3((\tan \theta)^2 + 1)}$$
$$= \sqrt{3} \sqrt{(\sec \theta)^2} $$
$$\sqrt{x^2 + 3} = \sqrt{3} \sec \theta $$
Getting the equivalent form of $x^4$ in terms of $\theta$:
$$x^4 = (\sqrt{3} \tan \theta)^4$$
$$ = 9 (\tan \theta)^4 $$
Getting the equivalent form of $dx$ in terms of $\theta$:
$$dx = \sqrt{3} (\sec \theta )^2 d\theta$$
Substituting these equivalent expressions to the integral above, we get:
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{1}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{\sqrt{3} (\sec \theta )^2}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } d\theta$$
$$ = \int \frac{\sec \theta}{9 (\tan \theta)^4 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\tan \theta)^2)^2 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\sec \theta)^2 - 1 )^2 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\sec \theta)^2 - 1 )^2 } d\theta$$
$$ = \frac{1}{9}\int \frac{\sec \theta}{((\sec \theta)^2 - 1 )^2 } d\theta$$
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$$
My problem is....I don't know how to evaluate $\frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$.
I can't go forward. I'm stuck. How to evaluate $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ properly?
| Hint:
$$\dfrac{\sec t}{\tan^4t}=\dfrac{\cos^3t}{\sin^2t}=\dfrac{(1-\sin^2t)\cos t}{\sin^2t}$$
Set $\sin t=u$
OR directly, $u=\dfrac{\tan t}{\sec t}=\dfrac x{\sqrt{x^2+3}}$ assuming $0\le t<\dfrac\pi2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n^2+4=7(7q^2+2rq)+r^2+4$
$7(7q^2+2rq)$ is either divisible by $7$, or it is $0$ (when $q=0$), so it is $r^2+4$ we are concerned with.
Assume that $r^2+4$ is divisible by 7. Then $r^2+4=7k$ for some integer $k$.
This is the original problem we were faced with, except whereas $n$ could be any integer, $r$ is constrained to be one of $0,1,2,3,4,5$ or $6$.
Through trial and error, we see that no valid value of $r$ satisfies $r^2+4=7k$ so we have proved our theorem by contradiction.
I'm pretty sure this is either wrong somewhere or at the very least not the proof that the question intended. Any help would be appreciated.
| Here is a proof that seems a bit roundabout, but it relates simpler methods to Fermat's Little Theorem.
First note that clearly $n^2+4$ will not be a multiple of $7$ if $n$ is one. Now for other values of $n$, raise $n^2+4$ to the power of $7-1=6$:
$(n^2+4)^6=n^{12}+(6)(4)n^{10}+(15)(4^2)n^8+...+4^6$
$\equiv n^{12}+3n^{10}+2n^8+6n^6+4n^4+5n^2+1 \bmod 7$
Note that the coefficients are in geometric progression $\bmod 7$, with common ratio $3\equiv -4$. This is because each binomial coefficient is obtained from an adjacent one by multiplying by $(7-k)/k\equiv -1$.
Now apply Fermat's Little Theorem, thus $n^6\equiv 1, n^8\equiv n^2, n^{10}\equiv n^4, n^{12}\equiv n^6\equiv 1$. After this reduction add terms with like powers and note that most of them cancel. We are left with
$(n^2+4)^6\equiv 1\bmod 7$
This completes the proof that $n^2+4$ must fail to be a multiple of $7$.
We can do a similar analysis with $(n^2-r)\bmod p$, where $p$ is an odd prime, raising the binomial to the power of $p-1$. The binomial coefficients always have the common ratio $-1\bmod p$, which causes the successive coefficients in $(n^2-r)^{(p-1)}$ to have common ratio $r\bmod p$. Then the cancellation noted above always occurs if $r^{(p-1)/2}\equiv -1\bmod p$. Thus if $r^{(p-1)/2}\equiv -1\bmod p$, we end with $(n^2-r)^{(p-1)}\equiv 1 \bmod p$ thus proving that $(n^2-r)$ cannot be a multiple of $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Finding generating function for a given sequence The task is to find the generating function for the sequence $$d_n = (\sum_{k=0}^n \binom{n}{k})*(\sum_{k=0}^n \frac{(k-1)^2}{2^k})$$
let's call this function $D(x)$.
I marked $a_n = \frac{1}{2^n}$; $b_n = \frac{-2n}{2^n}$; $c_n = \frac{n^2}{2^n}$ with respective generating functions $A(x)$, $B(x)$, $C(x)$.
So I found $$A(x) = \frac{1}{1-\frac{x}{2}}$$ and
$$B(x) = \frac{-x}{(1-\frac{x}{2})^2}$$
$$C(x) = \frac{2x+3x^2}{4(1-\frac{x}{2})^3}$$
using derivatives.
now the generating function for $a_n + b_n +c_n$ should be $A(x)+B(x)+C(x)$ and it's $$\frac{x^2}{(1-\frac{x}{2})^3}$$.
Next by using convolution (with let's say $f_n = 1$) we'll get that the function for the series $$u_n = \sum_{k=0}^n \frac{(k-1)^2}{2^k})$$ is $$\frac{x^2}{(1-\frac{x}{2})^3(1-x)}$$
Now since $h_n = \sum_{k=0}^n \binom{n}{k} = 2^n$ the appropriate function is $$H(x) = \frac{1}{1-2x}$$.
Again with using convolution with $h_n$ and $u_n$ we'll get that $$D(x) = H(x)U(x) = \frac{x^2}{(1-\frac{x}{2})^3(1-x)(1-2x)}$$
Is my answer good? or shall i keep expanding it to partial fractions.
| I think you had to compute the sums first and then multiply. I mean, the sequence is:
$$d_n = \left(\sum_{k=0}^n \binom{n}{k}\right)\cdot\left(\sum_{k=0}^n \frac{(k-1)^2}{2^k}\right)$$
Which means
$$d_n=2^n\cdot\left(\sum _{k=0}^n \frac{k^2}{2^k}-2\sum _{k=0}^n \frac{k}{2^k}+\sum _{k=0}^n \frac{1}{2^k}\right)=\\=2^n\cdot\left[(6-2^{-n} n^2-2^{2-n} n-3\cdot 2^{1-n})-2(-2^{-n} n-2^{1-n}+2)+(2-2^{-n})\right]=\\=2^n\cdot\left[2^{-n} \left(2^{n+2}-n^2-2 n-3\right)\right]=2^{n+2}-n^2-2n-3$$
$d_n=2^{n+2}-n^2-2n-3$
And its generating function is
$$F(x)=\frac{4}{1-2 x}-\frac{-x^2-x}{(x-1)^3}-\frac{2 x}{(x-1)^2}-\frac{3}{1-x}=\\=\frac{4 x^2-3 x+1}{(x-1)^3 (2 x-1)}$$
Hope this answers to your question
| {
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"url": "https://math.stackexchange.com/questions/2424357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
domain and range of function $y= 1+\sum_{n=2}^\infty x^n$ My friend give me a question to find domain and range of
$y= 1+\sum_{n=2}^\infty x^n$
there is no more description about the problem, so I think the domain of that function is all $\mathbb{R} $ and range of that function is $ \mathbb{R}$ too
but he told me that I was wrong, domain of that function is $-1 < x < 1 $ and range of that function is $y \leq -3 $ or $y \geq 1 $ that was shocked me.
I never knew whether domain of convergence or divergence function form will be an answer of this function.
Moreover, according to WFA they said $y= 1+\sum_{n=2}^\infty x^n$
equal to $y = 1 - \frac{x^2}{x-1}$ for $\left | x \right | <1$ so domain of function should be $ \left \{ x\in \mathbb{R} : x\neq 1 \right \}$, isn't it?
Which one got the right answer and what is real definition of domain and range
Sorry for my english, Thank you in advance.
| The expression
$$ 1 + x + x^2 + x^3 + \dotsb = \sum_{k=0}^{\infty} x^k $$
is a geometric series. We say that a series converges if the corresponding sequence of partial sums converges. That is
$$ \sum_{k=0}^{\infty} x^k \text{ converges}\iff \lim_{n\to\infty} \sum_{k=0}^{n} x^k =: S_n < \infty.$$
But note that
$$ S_n = 1 + x + x^2 + x^3 + \dotsb + x^n
\implies x S_n = x + x^2 + x^3 + x^4 + \dotsb + x^{n+1}.$$
Subtracting and canceling terms as appropriate, this gives us
$$ (1-x)S_n = S_n - xS_n = 1 - x^{n+1}
\implies S_n = \frac{1-x^{n+1}}{1-x}.$$
Taking the limit, we get
$$ \lim_{n\to\infty} S_n
= \begin{cases}
\frac{1}{1-x} & \text{ if $|x|<1$ (since $x^{n+1} \to 0$),} \\
\infty & \text{ if $|x|>1$ (since $|x|^{n+1} \to \infty$), and} \\
\text{indeterminate} & \text{otherwise.}
\end{cases}
$$
Now, observe that if $|x|<1$, then
$$ f(x) = 1 + \sum_{k=2}^{\infty} x^k = \left(\sum_{k=0}^{\infty} x^k\right) - x = \frac{1}{1-x} - x.$$
For other values of $x$, the series won't converge, and the function will be undefined. Hence the domain of $f$ (presuming that the domain is real, and not complex) is the interval $(-1,1)$, and on this domain it is given by
$$ f(x) = \frac{1}{1-x} - x = \frac{1 - x(1-x)}{1-x} = \frac{1 - x + x^2}{1-x} = 1 + \frac{x^2}{1-x}.$$
Finally, suppose that $y$ is in the range of $f$. Then there is some $x\in (-1,1)$ such that $y=f(x)$. Solving, we get
$$ y = \frac{ 1-x + x^2}{1-x}
\implies (1-x)y = y - yx = 1-x+x^2
\implies 0 = x^2 + (y-1)x + (1-y).
$$
Thus
$$ x = \frac{1 - y \pm \sqrt{ (y-1)^2 - 4(1-y)}}{2}, $$
and so the equation $y = f(x)$ has a real solution $x$ as long as
$$ (y-1)^2 - 4(1-y)
= y^2 - 2y + 1 - 4 + 4y
= y^2 + 2y - 3
= (y+3)(y-1)
\ge 0. $$
This happens whenever $y\le-3$ or $y\ge 1$, i.e.
$$ y \in (-\infty,-3]\cup [1,\infty). $$
However, note that we also require $-1 < x < 1$. For such $x$, we have $1-x > 0$, and so
$$ y = 1 + \frac{x^2}{1-x} \ge 0.$$
Therefore the range of $f$ is contained in the set
$$ \Big[ (-\infty,-3]\cup [1,\infty) \Big] \cap [0,\infty)
= [1,\infty).$$
Slightly more carefully, note that $f$ is continuous on $(-1,1)$, that $f(0) = 0$, and that
$$\lim_{x\to 1^{-}} f(x) = \infty.$$
This is sufficient to show that $[1,\infty)$ is contained in the range of $f$, which finally allows us to conclude that the range of $f$ is
$[1,\infty).$
Something worth noting is that the domain of $f$ is the set $(-1,1)$, since $f$ is only defined when the series is convergent. However, as we observed above, on this domain, we have the identity
$$
f(x) = 1 + \sum_{k=2}^{\infty} x^k = 1 + \frac{x^2}{1-x}.
$$
The right-most expression defines a legitimate function on $\mathbb{R} \setminus \{1\}$. However, this is not the same function as $f$, since $f$ was defined by the series, and not by the rational expression on the right. We can, however, think of the function
$$ g(x) = 1 + \frac{x^2}{1-x} $$
as an extension of $f$ to a larger domain. Indeed, in the context of complex analysis (a branch of mathematics that deals with functions that have domains in the complex plane), it can be shown that this extension is (in some sense) the "right" extension.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove an equation with summation and binomial coefficients Good day everybody,
I am looking to prove this:
$$\delta_{a,b}=\left(\frac{1}{2}\right)^{2\left(a-b\right)}\sum_{\gamma=0}^{a-b}\left(-1\right)^{a-b-\gamma}\frac{b+\gamma}{b}\left(\begin{array}{c}
2a\\
a-b-\gamma
\end{array}\right)\left(\begin{array}{c}
2b-1+\gamma\\
2b-1
\end{array}\right)$$
for $a \ge 0 $ and $b \ge 1$.
The case $b=0$ is an exception to the rule but is also of interest for me. Therefore I would actually most appreciate a proof of matrices $D$ and $C$ to be inverse:
$$D_{mn}=\begin{cases}
\left(-1\right)^{\frac{m-n}{2}}\left(\begin{array}{c}
m\\
\frac{m-n}{2}
\end{array}\right)\left(\frac{1}{2}\right)^{m} & \text{ if }n\leq m\text{ and }n-m\text{ is even}\\
0 & \text{else}
\end{cases}
$$
and
$$C_{ij}=\begin{cases}
1 & \text{if }i=j=0\\
2^{j}\left[\left(\begin{array}{c}
\frac{i+j}{2}-1\\
j-1
\end{array}\right)+2\left(\begin{array}{c}
\frac{i+j}{2}-1\\
j
\end{array}\right)\right] & \text{else if }i\geq j\text{ and }i-j\text{ is even}\\
0 & \text{else}
\end{cases}$$
where indexing starts from zero (and thus contains also $b=0$ case).
Thank you.
| We write $n$ instead of $\gamma$, focus on the essentials and skip the constant $(-1)^{a-b}2^{-2(a-b)}$.
We obtain for integers $a\geq b>0$
\begin{align*}
\color{blue}{\sum_{n=0}^{a-b}}&\color{blue}{(-1)^n\frac{b+n}{b}\binom{2a}{a-b-n}\binom{2b-1+n}{n}}\tag{1}\\
&=\sum_{n=0}^{a-b}\frac{b+n}{b}\binom{2a}{a-b-n}\binom{-2b}{n}\tag{2}\\
&=\sum_{n=0}^{a-b}\binom{2a}{a-b-n}\binom{-2b}{n}
-2\sum_{n=1}^{a-b}\binom{2a}{a-b-n}\binom{-2b-1}{n-1}\tag{3}\\
&=\binom{2a-2b}{a-b}-2\sum_{n=0}^{a-b-1}\binom{2a}{a-b-n-1}\binom{-2b-1}{n}\tag{4}\\
&=\binom{2a-2b}{a-b}-2\binom{2a-2b-1}{a-b-1}[[a>b]]\tag{5}\\
&\color{blue}{=[[a=b]]}
\end{align*}
Comment:
*
*In (1) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3) we split the sum and apply $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ to the right-hand sum.
*In (4) we apply Chu-Vandermonde's identity
to the left sum and shift the index of the right sum by one to start with $n=0$.
*In (5) we use Iverson brackets in the right expression, since the corresponding sum in the line before is zero if $a=b$. We also use the binomial identity $\binom{2p}{p}=2\binom{2p-1}{p-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Some formula related with factor of (a+b+c+d) I am looking for some math formula
For example
\begin{align}
& a^2 -b^2 = (a+b)(a-b) \\
&a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)
\end{align}
First one related with factor a+b and the second one related with factor a+b+c
then
How about some formula related with a,b,c,d
i.e., is there are some equation factors into (a+b+c+d)?
| It will be probably
$$a^4+b^4+c^4+d^4-b^2a^2-c^2a^2-d^2a^2-b^2c^2-b^2d^2-d^2c^2+16abcd=(a+b+c+d)(a^3+b^3+c^3+d^3-ba^2-ca^2-da^2-ac^2-bc^2-dc^2-ab^2-cb^2-db^2-ad^2-bd^2-cd^2+4bca+4bda+4cad+4bcd)$$
OR
$$\sum a^4 -\sum a^2b^2 +16abcd=(\sum a)(\sum a^3-\sum ab^2 +4\sum abc)$$
$$$$
$$\sum \text{ represents cyclic summation}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$
My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \dfrac{1}{6}<\sin x<1\tag*{}$
$\implies 5-2\sin x\geq36\sin^2x+1-12\sin x\implies 18\sin^2x-5\sin x-2\leq0$
$\implies(2\sin x-1)(9\sin x+2)\leq0$
$\implies\sin x\ \epsilon\ \bigg(\dfrac{1}{6},\dfrac{1}{2}\bigg]$
All of above implies $\sin x\ \epsilon\ \bigg[-1,\dfrac{1}{2}\bigg]$.
Answer is given in the form: $\bigg[\dfrac{\pi(12n-7)}{6},\dfrac{\pi(12n+1)}{6}\bigg]\ (n\epsilon Z)$
How do I reach the form given in options? I even don't know what I've is correct or not.
Please help.
| Hint:)
Other way that may be helpful. Let $y=5-2\sin x$, from $\sqrt{5-2\sin x}\geq 6\sin x-1$ you find
$$3y-\sqrt{y}-14\geq0$$
gives
$(\sqrt{y}-2)(\sqrt{y}+\dfrac73)\geq0$. You can discuss about conditions and find that $\sin x\leq\dfrac12$. This concludes
$$\color{blue}{\left[2k\pi+\dfrac{3\pi}{2}-\dfrac{2\pi}{3},2k\pi+\dfrac{3\pi}{2}+\dfrac{2\pi}{3}\right]}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\frac{3}{abc} \ge a + b + c$, prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c$ Let $a$, $b$ and $c$ be positive numbers such that $\frac{3}{abc} \ge a + b + c$. Prove that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c.$$
Any way I use - I get stuck after 2 or 3 steps...
| The condition gives
$$1\geq\sqrt{\frac{(a+b+c)abc}{3}}.$$
Thus,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{(a+b+c)abc}{3}}=$$
$$=\sqrt{\frac{(ab+ac+bc)^2(a+b+c)}{3abc}}\geq\sqrt{\frac{3abc(a+b+c)(a+b+c)}{3abc}}=a+b+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving system of equations (3 unknowns, 3 equations) So, I've been trying to solve this question but to no avail. The system of equations are as follows:
1) $x+ \frac{1}{y}=4$
2) $y+ \frac{1}{z}=1$
3) $z + \frac{1}{x}=\frac{7}{3}$
Attempt:
Using equation 1, we can rewrite it as $\frac{1}{y}=4-x \equiv y=\frac{1}{4-x}$
. Using $y=\frac{1}{4-x}$, we can substitute into equation 2 to get $\frac{1}{4-x}+\frac{1}{z}=1$ (eq. 2')
Together with equation 3, we can eliminate z by inversing equation 3 to be:
$x + \frac{1}{z}=\frac{3}{7}$
Now, we subtract both (eq. 2') and the inversed equation to get
$\frac{1}{4-x} - x = \frac{4}{7}$
However, this is going no where. The final answer should be x=$\frac{3}{2}$, y = $\frac{2}{5}$ and z = $\frac{5}{3}$.
Any tips or help would be greatly appreciated!
| Eq. (2) implies $z=1/(1-y)$, and Eq. (1) says $y = 1/(4-x)$. Consequently, Eq. (3) becomes
$$\frac{1}{x}+\frac{4-x}{3-x}=\frac{7}{3} \implies 4x^2-12x+9 = (2x-3)^2 = 0 \implies x = \frac{3}{2}.$$
Consequently, $y = 1/(4-x) = 2/5$, and $z = 1/(1-y) = 5/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $\lim_{x\to 0^+} \Biggr(\tan^{-1}{\Big(\dfrac{b}{x}\Big)} - \tan^{-1}{\Big(\dfrac{a}{x}\Big)}\Biggr)= 0$? For $$f(x) = \tan^{-1}{\Big(\frac{b}{x}\Big)} - \tan^{-1}{\Big(\frac{a}{x}\Big)}$$
where $a$ and $b$ are differently valued constants, why is $\lim_{x\to0^+}= 0$? I understand that separately both expressions tend toward $\dfrac{\pi}{2}$ for any value of $a$ or $b$, but surely the expressions cannot both give $\dfrac{\pi}{2}$ for the same value of $x$ at any given time?
| You can use the MTV to see it. For a differentiable function $g$ holds
$g(x_1)-g(x_2)=g'(\xi)(x_1-x_2)$ for some $\xi\in (x_1,x_2)$.
Now define $g(y)=\arctan(y)$ and you get
$$
f(x)=g\left(\frac{b}{x}\right)-g\left(\frac{a}{x}\right)=g'(\xi)\left(\frac{b}x-\frac{a}x\right)=\frac1{1+\xi^2}\cdot\frac{b-a}{x}
$$
where $\xi\in\left(\frac{a}{x},\frac{b}{x}\right)$. Hence
$$
\frac{(b-a)x}{x^2+b^2}=\frac{1}{1+\left(\frac{b}{x}\right)^2}\cdot\frac{b-a}x\leq \frac1{1+\xi^2}\cdot\frac{b-a}{x}\leq
\frac{1}{1+\left(\frac{a}{x}\right)^2}\cdot\frac{b-a}x=\frac{(b-a)x}{x^2+a^2}.
$$
So we can conclude
$$
0=\lim_{x\to 0^+}\frac{(b-a)x}{x^2+b^2}\leq \lim_{x\to 0^+} f(x)\leq \lim_{x\to 0^+}\frac{(b-a)x}{x^2+a^2}=0.
$$
The function must not achieve the limit for some final value. But from
$$
\frac{(b-a)x}{x^2+b^2}\leq f(x)\leq \frac{(b-a)x}{x^2+a^2}
$$
you can see that the value of $f$ becomes smaller for large $x$ and goes to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two methods of finding $\int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\sin x}$ Two methods of finding $$I=\int_{0}^{\pi/2} \frac{dx}{1+\sin x}$$
Method $1.$ I used substitution, $\tan \left(\frac{x}{2}\right)=t$ we get
$$I=2\int_{0}^{1}\frac{dt}{(t+1)^2}=\left. \frac{-2}{t+1}\right|_{0}^{1} =1$$
Method $2.$ we have $$I=\int_0^{\pi/2}\frac{1-\sin x}{\cos ^2 x} \,\mathrm dx = \left. \int_0^{\pi/2}\left(\sec^2 x-\sec x\tan x\right)\,\mathrm dx= \tan x-\sec x\right|_0^{\pi/2}$$
What went wrong in method $2$?
| \begin{align}
\tan x - \sec x & = \frac{\sin x -1}{\cos x} = \frac{(\sin x - 1)(-\sin x -1)}{(\cos x)(-\sin x - 1)} = \frac{1-\sin^2 x}{(\cos x)(-\sin x - 1)} \\[10pt]
& = \frac{\cos^2 x}{(\cos x)(-\sin x - 1)} = \frac{-\cos x}{1+\sin x} = \frac 0 2 \text{ if } x = \frac \pi 2.
\end{align}
Or you can use the identity
$$
\tan x \pm \sec x = \tan\left( \frac \pi 4 \pm \frac x 2 \right).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$
$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$.
Show that $(a-c)(b-c)<0$.
This is a question presented in the "Olimpiadas do Ceará 1987" a math contest held in Brazil. Sorry if this a duplicate.
Given the assumptions, it is easy to show that
$$0<(a-b)^2<a^2+b^2-ab=c^2.$$ But could not find a promising route to pursue.
Any hint or answer is welcomed.
| If $a<b$, then
$$
c^2 = a^2+b^2-ab = a^2+b(b-a) > a^2 \\
c^2 = a^2+b^2-ab = b^2-a(b-a) < b^2
$$
which means $a<c<b$ and $(a-c)(b-c)<0$.
If $a>b$, then
$$
c^2 = a^2+b^2-ab = a^2-b(a-b) < a^2 \\
c^2 = a^2+b^2-ab = b^2+a(a-b) > b^2
$$
which means $b<c<a$ and $(a-c)(b-c)<0$.
| {
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"url": "https://math.stackexchange.com/questions/2431928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integrating modular function $f(x,y)=|xy|$ in the area of the circle $x^2+y^2=a^2$ I need to solve the integral $$\iint_\omega |xy|\,dx\,dy$$ where $\omega:x^2+y^2=a^2$ I created the following integral and I have no idea how can I integrate modular function:
$$\int_{-a}^a dy\int_{-\sqrt{a^2-y^2}}^\sqrt{a^2-y^2} |xy|\,dx$$The only thing that I do know is that $\int |x|\,dx=\frac{x|x|}{2}+C$, but later on I face the next problem: $$\int_{-a}^a|y|\cdot|\sqrt{a^2-y^2}|\cdot \sqrt{a^2-y^2}\,dy$$
Could anyone help me?
| Why not convert to polar coordinates?
With $x=r\cos(\theta)$ and $y=r\sin(\theta)$. The area element becomes $rdrd\theta$. Then the integral over the area of the circle becomes
\begin{align}
I &=\int_0^a\int_0^{2\pi}r^2|\cos(\theta)\sin(\theta)|rdrd\theta=\int_0^ar^3dr\int_0^{2\pi}|\cos(\theta)\sin(\theta)|d\theta\\
&=\frac{a^4}{4}\frac{1}{2}\int_0^{2\pi}|\sin(2\theta)|d\theta
=\frac{a^4}{8}\int_0^{2\pi}|\sin(2\theta)|d\theta=\frac{a^4}{4}\int_0^{\pi}|\sin(2\theta)|d\theta\\
&=\frac{a^4}{2}\int_0^{\pi/2}\sin(2\theta)d\theta=\frac{a^4}{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of $ \frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) } $ without L'Hospital technique I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.
$$ f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} $$
This limit will go to zero (I had tried using calculator manually).
I have tried to open the trigonometric identities and the fact that $ \lim\limits_{x \rightarrow 0}\frac{\sin (ax)}{x} = a $ gives :
$$ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos (3x)( \cos (3x) - 1)}{6 x^{2}} $$
$$ \cos (3x) = \cos (2x) \cos(x) - \sin (2x) \sin(x) = \cos^{3}(x) - 3\sin^{2}(x) \cos(x) $$
$$ \lim_{x \rightarrow 0 } \frac{ \cos (3x) }{x^{2}} = \lim_{x \rightarrow 0} \frac{\cos^{3}(x)}{x^{2}} - 3 $$
Any view on this and more efficient way to solve this? Thanks before.
| Just another way using the standard Taylor expansions.
$$\cos(3x)=1-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$
$$\sin(2x)=2 x-\frac{4 x^3}{3}+O\left(x^5\right)$$
$$\tan(3x)=3 x+9 x^3+O\left(x^5\right)$$
makes
$$\cos(3x)-1=-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$
$$\sin(2x)\tan(3x)=6 x^2+14 x^4+O\left(x^6\right)$$ Now, long division leads to
$$\frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) }=-\frac{3}{4}+\frac{37 x^2}{16}+O\left(x^3\right)$$ showing the limit and how it is appraoched.
Just for illustration, use $x=\frac \pi {12}$ (quite far away from $0$); the exact value of the expression is $\sqrt{2}-2\approx -0.585786$ while the above approximation would give $\frac{37 \pi ^2}{2304}-\frac{3}{4}\approx -0.591504$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help converting spherical to cartesian?
Convert the following equation from spherical to coordinate:
sin2(ϕ) + cos2(ϕ)⁄4 = 1/ρ2;
I have converted the right side of the equation to 1/x2+y2+z2, but the fraction on the left side of the equation is throwing me off. Any advice/help/solutions?
| $x = \rho\cos\theta\cos\phi\\
y = \rho\sin\theta\cos\phi\\
z = \rho\sin\phi$
$\sin^2\phi + \frac 14 \cos^2\phi = \frac 1{\rho^2}\\
\rho^2\sin^2\phi + \frac 14 \rho^2 \cos^2\phi = 1\\
z^2 + \frac 14 \rho^2 \cos^2\phi = 1$
$x^2+y^2 = \rho^2 \cos^2\phi$
$z^2 + \frac 14 x^2 + \frac 14 y^2 = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality on tangent and secant function
Let $\{\alpha, \beta, \gamma, \delta\} \subset \left (0,\frac {\pi}{2}\right)$ and $ \alpha + \beta+\gamma+\delta = {\pi}$. Prove that $\sqrt {2} \left(\tan \alpha +\tan \beta +\tan\gamma+\tan \delta\right)\ge \sec \alpha +\sec \beta +\sec \gamma +\sec \delta$
| Let $f(x)=\sqrt2\tan{x}-\frac{1}{\cos{x}}$.
Thus, $$f''(x)=\frac{(1+\sqrt2-\sin{x})(\sin{x}-\sqrt2+1)}{\cos^3x}>0$$
for $x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right)$.
Thus, by Vasc's RCF Theorem it's enough to prove our inequality for
$\beta=\gamma=\alpha$ and $\delta=\pi-3\alpha$, where $\alpha\in\left[\frac{\pi}{4},\frac{\pi}{3}\right)$ and the rest is smooth.
Indeed, we need to prove that
$$\frac{3(\sqrt2\sin{\alpha}-1)}{\cos{\alpha}}+\frac{\sqrt{2}\sin\left(180^{\circ}-3\alpha\right)-1}{\cos\left(180^{\circ}-3\alpha\right)}\geq0.$$
Let $\sin\alpha=t$.
Thus, $\frac{1}{\sqrt2}\leq t<\frac{\sqrt3}{2}$ and we need to prove that
$$\frac{3(\sqrt2t-1)}{\cos\alpha}-\frac{\sqrt2\sin3\alpha-1}{\cos3\alpha}\geq0$$ or
$$3(\sqrt2t-1)-\frac{\sqrt2(3t-4t^3)-1}{4\cos^2\alpha-3}\geq0$$ or
$$3(\sqrt2t-1)-\frac{\sqrt2(3t-4t^3)-1}{4(1-t^2)-3}\geq0$$ or
$$3(\sqrt2t-1)+\frac{\sqrt2(3t-4t^3)-1}{4t^2-1}\geq0$$ or
$$4\sqrt2t^3-6t^2+1\geq0,$$
which is true by AM-GM:
$$4\sqrt2t^3+1=2\sqrt2t^3+2\sqrt2t^3+1\geq3\sqrt[3]{\left(2\sqrt2t^3\right)^2\cdot1}=6t^2.$$
Done!
We can use also the Tangent Line method.
$$\sum_{cyc}\left(\sqrt2\tan\alpha-\sec\alpha\right)=\sum_{cyc}\left(\sqrt2\tan\alpha-\sec\alpha-\sqrt2\left(\alpha-\frac{\pi}{4}\right)\right)\geq0$$
because $f(x)\geq0$ for all $0<x<\frac{\pi}{2}$, where
$$f(x)=\sqrt2\tan{x}-\sec{x}-\sqrt2\left(x-\frac{\pi}{4}\right).$$
Indeed, $$f'(x)=\frac{\sqrt2}{\cos^2x}-\frac{\sin{x}}{\cos^2x}-\sqrt2=\frac{\sin{x}(\sqrt2\sin{x}-1)}{\cos^2x},$$
which gives $x_{min}=\frac{\pi}{4}.$
Id est, $$f(x)\geq f\left(\frac{\pi}{4}\right)=0$$
and we are done!
| {
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Subsets and Splits