Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate
$\lim_{n \rightarrow \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know that $1 - \frac {2}{k(k+1)} = \frac{(k+2)(k-1)}{k(k+1)}$
After that I'm not able to proceed further.
Any hints/solution will be apprecaited.
Thanks you and thanks in advance.
| A formal proof
with product notation.
$lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})
$
$\begin{array}\\
\prod_{k=2}^n (1-\dfrac{2}{k(k+1)})
&=\prod_{k=2}^n \dfrac{k(k+1)-2}{k(k+1)}\\
&=\prod_{k=2}^n \dfrac{k^2+k-2}{k(k+1)}\\
&=\prod_{k=2}^n \dfrac{(k+2)(k-1)}{k(k+1)}\\
&=\dfrac{\prod_{k=2}^n (k+2)\prod_{k=2}^n (k-1)}{\prod_{k=2}^n k\prod_{k=2}^n (k+1)}\\
&=\dfrac{\prod_{k=4}^{n+2}k\prod_{k=1}^{n-1}k}{\prod_{k=2}^n k\prod_{k=3}^{n+1}k}\\
&=\dfrac{\prod_{k=4}^{n+2}k\prod_{k=1}^{n-1}k}{\prod_{k=3}^{n+1}k\prod_{k=2}^n k}\\
&=\dfrac{(n+2)(1)}{3n}\\
&=\dfrac{n+2}{3n}\\
&=\dfrac13+\dfrac{2}{3n}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$ Evaluate:
$$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$$
We let
$$\alpha = \operatorname{arccos} \frac{2}{\sqrt5} \qquad \beta = \operatorname{arccos}\frac{3}{\sqrt{10}}$$
Then we have:
$$\begin{align}
\cos(\alpha) = \frac{2}{\sqrt5} &\qquad \cos(\beta) = \frac{3}{\sqrt{10}} \\[4pt]
\sin(\alpha) = \frac{1}{\sqrt5} &\qquad \sin(\beta) = \frac{1}{\sqrt{10}}
\end{align}$$
In order to evaluate, we are told, we first determine $\sin(\alpha + \beta)$; we wind up with $1/\sqrt2$, thus we have $\pi/4$.
What I am confused about is why we have to use sin($\alpha + \beta$). For example, if I were to use $\cos(\alpha + \beta)$, I would get the answer $7/(\sqrt{10}\sqrt5)$, which I do not know what to do with. I am having trouble finding out whether there is some kind of pattern to this kind of thing, or did the author just know to use $\sin(\alpha + \beta)$ since he/she checked cos and saw nothing comes out of this?
Any help is much appreciated, thank you
| From $\sin(\alpha+\beta)=\frac1{\sqrt2}$, we should expect $|\cos(\alpha+\beta)|=\frac1{\sqrt2}$ from the formula $\sin^2 \theta + \cos^2 \theta = 1$.
\begin{align}
\cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)\color{blue}-\sin(\alpha)\sin(\beta)\\
&=\frac{2}{\sqrt5}\cdot \frac{3}{\sqrt{10}}-\frac1{\sqrt5}\cdot \frac{1}{\sqrt{10}}\\
&=\frac{6}{5\sqrt{2}}-\frac{1}{5\sqrt2}\\
&=\frac{5}{5\sqrt2}\\
&=\frac{1}{\sqrt2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the inverse of $A$
Find the inverse of $$A =\left[\begin{matrix}0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{matrix}\right]$$
My attempts:
$$A^{-1} = \frac {\operatorname{adj}A}{\det A}$$
As I can find the $\det A$ that is $\det A = -b$. Here, how can I find the inverse? Is there any easy method/ or easy procedure to find the inverse of $A$?
| Let
$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} := \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right]$$
Let $\mathrm r^\top := \begin{bmatrix} b & c & d\end{bmatrix}$. From
$$\left[\begin{array}{c|c} 0_3 & \mathrm I_3\\ \hline a & \mathrm r^\top\end{array}\right] \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right] = \left[\begin{array}{c|c} \mathrm I_3 & 0_3\\ \hline 0_3^\top & 1\end{array}\right]$$
we obtain $\mathrm P_{21} = \mathrm I_3$, then $\mathrm p_{22} = 0_3$, then $\mathrm p_{11}^\top = -\frac{1}{a} \mathrm r^\top$ and, lastly, $p_{12} = \frac 1a$. Thus,
$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} = \color{blue}{\left[\begin{array}{ccc|c} - \frac{b}{a} & - \frac{c}{a} & - \frac{d}{a} & \frac{1}{a}\\ \hline1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\end{array}\right]}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Are $1+p^3+p^6$ and $1+p^4+p^8$ coprime? What are the primes $p$ for which $1+p^3+p^6$ and $1+p^4+p^8$ are coprime?
I know it is true for $p=2$ and $p=3$ and not true for any $p \equiv 1 \mod 6$. I conjecture that it true for all primes $p \equiv 5 \mod 6$.
Any counterexample $> 10^8$.
This is relevant to OEIS sequence A046685.
| From comments, therefore CW:
$$ \left( x^{8} + x^{4} + 1 \right) $$
$$ \left( x^{6} + x^{3} + 1 \right) $$
$$ \left( x^{8} + x^{4} + 1 \right) = \left( x^{6} + x^{3} + 1 \right) \cdot \color{magenta}{ \left( x^{2} \right) } + \left( - x^{5} + x^{4} - x^{2} + 1 \right) $$
$$ \left( x^{6} + x^{3} + 1 \right) = \left( - x^{5} + x^{4} - x^{2} + 1 \right) \cdot \color{magenta}{ \left( - x - 1 \right) } + \left( x^{4} - x^{2} + x + 2 \right) $$
$$ \left( - x^{5} + x^{4} - x^{2} + 1 \right) = \left( x^{4} - x^{2} + x + 2 \right) \cdot \color{magenta}{ \left( - x + 1 \right) } + \left( - x^{3} + x^{2} + x - 1 \right) $$
$$ \left( x^{4} - x^{2} + x + 2 \right) = \left( - x^{3} + x^{2} + x - 1 \right) \cdot \color{magenta}{ \left( - x - 1 \right) } + \left( x^{2} + x + 1 \right) $$
$$ \left( - x^{3} + x^{2} + x - 1 \right) = \left( x^{2} + x + 1 \right) \cdot \color{magenta}{ \left( - x + 2 \right) } + \left( -3 \right) $$
$$ \left( x^{2} + x + 1 \right) = \left( -3 \right) \cdot \color{magenta}{ \left( \frac{ - x^{2} - x - 1 }{ 3 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x^{2} \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( - x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - x^{3} - x^{2} + 1 \right) }{ \left( - x - 1 \right) } $$
$$ \color{magenta}{ \left( - x + 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{4} - x + 1 \right) }{ \left( x^{2} \right) } $$
$$ \color{magenta}{ \left( - x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - x^{5} - x^{4} - x^{3} \right) }{ \left( - x^{3} - x^{2} - x - 1 \right) } $$
$$ \color{magenta}{ \left( - x + 2 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{6} - x^{5} - 2 x^{3} - x + 1 \right) }{ \left( x^{4} - x^{3} - x - 2 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x^{2} - x - 1 }{ 3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{8} - x^{4} - 1 }{ 3 } \right) }{ \left( \frac{ - x^{6} - x^{3} - 1 }{ 3 } \right) } $$
$$ \left( x^{8} + x^{4} + 1 \right) \left( \frac{ x^{4} - x^{3} - x - 2 }{ 3 } \right) - \left( x^{6} + x^{3} + 1 \right) \left( \frac{ x^{6} - x^{5} - 2 x^{3} - x + 1 }{ 3 } \right) = \left( -1 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
$$ 2y + x = ±1 $$ can't be deduced.
Instead another erroneous line $ y = 0 $ is outputted.
Some help will be appreciated.Alternate methods of solution would work too:)
| If the slant asymptote exists, we may denote
$$k=\lim_{x \to \infty}\dfrac{y}{x}.$$
Then $$0=\lim_{x \to \infty}\left(1+3\cdot \frac{y}{x}-4 \cdot \frac{y^3}{x^3}-\frac{1}{x^2}+\frac{y}{x^3}+\frac{3}{x^3}\right)=1+3k-4k^3=-(k-1)(2k+1)^2.$$
Thus,$$k=1,-\frac{1}{2}.$$
*
*If $k=1$, we denote $y=x+b$ and put it into the equation, we have $$-4b^3-12b^2x-9bx^2+b+3=0.$$
Then $$0=\lim_{x \to \infty} \frac{-4b^3-12b^2x-9bx^2+b+3}{x^2}=-9b.$$
Thus,$$b=0.$$
*
*If $k=-\dfrac{1}{2}$, we denote $y=-\dfrac{x}{2}+b$ and put it into the equation, we have $$-8b^3+12b^2x+2b-3x+6=0.$$
Then $$0=\lim_{x \to \infty} \dfrac{-8b^3+12b^2x+2b-3x+6}{x}=12b^2-3.$$
Thus, $$b=\pm \frac{1}{2}.$$
As a result, the slant asymptotes are $$y=x, y=-\frac{x}{2}\pm \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Linear Algebra, Least squares I'm going through the Linear Algebra course on Brilliant.org, but can't figure out how I should know that $\frac{17}{3}$ is the best value for $c$ to fill in here.
My two questions are:
1) Why do we want to minimize the distance of
$$3c^2-34c+101$$
2) How can you derive where this distance is minimised?
| You wish to minimize the distance of $c\pmatrix{1 \\ 1 \\ 1} -2 \pmatrix{1 \\ 2 \\ 3}$ to the vector $\pmatrix{5 \\ 0 \\ 0}$.
We have
$$\left\|c\pmatrix{1 \\ 1 \\ 1} -2 \pmatrix{1 \\ 2 \\ 3} - \pmatrix{5 \\ 0 \\ 0}\right\|^2 = \left\|\pmatrix{c-7 \\ c-4 \\ c-6}\right\|^2 = (c-7)^2+(c-4)^2+(c-6)^2 = 3c^2-34c+101$$
To minimize the polynomial $3c^2-34c+101$, write it as
$$3c^2-34c+101 = 3\underbrace{\left(c - \frac{17}3\right)^2}_{\ge 0} + \frac{14}3$$
so the minimum is attained for $c = \frac{17}3$ and it is equal to $\frac{14}3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
$$\frac{1}{x+1}-\frac{1}{x+5}=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
| HINT
We have that
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=$$
$$=\frac{1}{(x+1)}\color{red}{-\frac{1}{(x+2)}+\frac{1}{(x+2)}-\frac{1}{(x+3)}+\frac{1}{(x+3)}-\frac{1}{(x+4)}+\frac{1}{(x+4)}}-\frac{1}{(x+5)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Limit superior of a sequence of oscillating functions related to Chebyshev polynomials Let $n \in \mathbb N$ and consider the polynomial function $f_n \colon \mathbb R \to \mathbb R$ defined by
$$f_n(x) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} (1 - x^2)^{n-k} x^{2k}$$
for any $x \in \mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)
By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of $f_3, f_5, f_7$:
As $n \to \infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_{50}$:
Let $f \colon D \to \mathbb R$ be defined by
$$f(x) = \limsup_{n \to \infty} f_n(x)$$
whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.
So far, I have only been able to show the following (see the update below):
*
*$f$ is an even function, since all of the $f_n$'s are even.
*$0 \notin D$. Indeed, $f_n(0) = 2n + 1 \to \infty$ as $n \to \infty$.
*$f(\pm 1) = 1$, because $f_n(\pm 1) = (-1)^n$ for any $n \in \mathbb N$.
*$f \left (\pm \frac {\sqrt 2} 2 \right ) = 1$. This is because:
$$f_n \left ( \pm \frac {\sqrt 2} 2 \right ) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} \left ( \frac 1 2 \right )^n = (-1)^{\left \lfloor \frac n 2 \right \rfloor} 2^n \left ( \frac 1 2 \right )^n = (-1)^{\left \lfloor \frac n 2 \right \rfloor} \le 1$$
In particular, $f_{4m} \left (\pm \frac {\sqrt 2} 2 \right ) = 1$ for any $m \in \mathbb N$, so $\limsup_{n \to \infty} f_n \left ( \pm \frac {\sqrt 2} 2 \right ) = 1$.
By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.
What else can we say about $f$? Is it possible to find a "simple" expression?
Thank you in advance for any reply.
Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that
$$f_n(\sin \alpha) = \frac {\sin ((2n+1) \alpha)}{\sin \alpha}$$
for any $\alpha \in \mathbb R \smallsetminus \pi \mathbb Z$, which immediately provides us with many other values of $f$. For instance,
$$f_n \left (\sin \frac \pi 6 \right ) = \frac{\sin \left ( (2n+1) \frac \pi 6 \right )}{\sin \frac \pi 6} \le \frac 1 {\frac 1 2} = 2$$
In particular,
$$f_{6m+1} \left (\sin \frac \pi 6 \right ) = \frac{\sin \left ( (12 m + 3) \frac \pi 6 \right )}{\sin \frac \pi 6} = \frac{\sin \left ( 2 m \pi + \frac \pi 2 \right )}{\sin \frac \pi 6} = \frac 1 {\frac 1 2} = 2$$
for any $m \in \mathbb N$, and thus $f \left (\pm \frac 1 2 \right ) = 2$.
How can we get a simple expression for $f$ using this information?
| This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 \sqrt{x^2 \left(x^2-1\right)} \qquad \text{and}\qquad v=1-2 x^2+2 \sqrt{x^2 \left(x^2-1\right)}$$ a CAS produced
$$f_n(x)=\frac{ \left(u^n+v^n\right)}{2 }+\frac{ \left(u^n-v^n\right)}{2 }\,\frac{\sqrt{x^2 \left(x^2-1\right)}
}{ x^2}$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_n\left(\sin \left(\frac{\pi k}{12}\right)\right)$ and obtained the (may be) interesting values
$$\left(
\begin{array}{cc}
k & f_n\left(\sin \left(\frac{\pi k}{12}\right)\right) \\
0 & 2 n+1 \\
1 & \cos \left(\frac{n \pi }{6}\right)+\left(2+\sqrt{3}\right) \sin \left(\frac{n
\pi }{6}\right) \\
2 & \cos \left(\frac{n \pi }{3}\right)+\sqrt{3} \sin \left(\frac{n \pi }{3}\right)
\\
3 & \cos \left(\frac{n \pi }{2}\right)+\sin \left(\frac{n \pi }{2}\right) \\
4 & \cos \left(\frac{2 n \pi }{3}\right)+\frac{1}{\sqrt{3}}\sin \left(\frac{2 n \pi
}{3}\right) \\
5 & \cos \left(\frac{5 n \pi }{6}\right)+\left(2-\sqrt{3}\right) \sin
\left(\frac{5 n \pi }{6}\right) \\
6 & (-1)^n
\end{array}
\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$ Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$
I have done it until the the divisor is of second degree.
But here the degree of the remainder is $4$
This means the remainder will be of the form. : $ax^3+bx^2+cx+d$ which makes it difficult to solve. Is it possible to solve $4$ variants with $4$ equations?
| Using Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd,
$$x^{10}+1\equiv0\pmod{x^2+1}$$
Again $x^3-1=(x-1)(x^2+x+1)$
$$x^{10}+1= x(x^9-1)+x+1\equiv x+1\pmod{x^2+x+1}$$
Now apply CRT
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots of $f$ and $f'$ for $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$ I have this question from an admission exams. Given that $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,\ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+\frac{1}{2!}2(2-1)=0$$
$$f(3)=1-3+\frac{1}{2!}3(3-1)-\frac{1}{3!}3(3-2)(3-1)=\binom{3}{0}-\binom{3}{1}+\binom{3}{2}-\binom{3}{3}$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0\,$ and indeed that $f(101)=(1-1)^{101}=0$
Since the polynomial is also of order $101$, the roots are $x=1,2, \ldots, 101$ giving:$$S(f(x))=\sum_{j=1}^{101}j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-\frac{1}{101!}(x-1)(x-2)\cdots(x-101)$$
| According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_{n-1}x^{n-1}+\cdots + a_0\\
p'_n(x) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2}+\cdots+a_1
$$
now
$$
S_{100}(x) = -\frac{a_{100}}{a_{101}}\\
S'_{100}(x) = -\frac{(101-1)a_{100}}{101 a_{101}}
$$
Here
$$
a_{101} = \frac{(-1)^{101}}{101!}\\
a_{100} = -a_{101}\frac{100(100+1)}{2}+\frac{(-1)^{100}}{100!}
$$
then
$$
S_n(x) = 5151\\
S'_n(x) = 5100
$$
and finally
$$
S_{100}(x)-S'_{100}(x) = 5151-5100 = 51
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to calculate the indefinite integral of $\frac{1}{x^{2/3}(1+x^{2/3})}$?
$$\int \frac{dx}{x^{2/3}(1+x^{2/3})}.$$
I substituted,
$$t=\frac{1}{x^{1/3}}$$
$$\frac{dt}{dx} = -\frac{1}{3x^{4/3}}$$
$$\frac{dt}{dx} = -\frac{t^4}{3}$$
Rewriting the question,
$$\int \frac{dx}{x^{2/3}+x^{4/3}}$$
$$-\frac{1}{3} \int \frac{dt}{t^4\Bigl(\frac{1}{t^2}+\frac{1}{t^4}\Bigr)}$$
We have,
$$-\frac{1}{3} \int \frac{dt}{t^2 + 1}$$
$$-\frac{1}{3}\tan^{-1}t+C$$
$$-\frac{1}{3}\tan^{-1}\Biggl(\frac{1}{x^{1/3}}\Biggr)+C$$
But the answer given is $$3\tan^{-1}x^{1/3}+C$$
Where am I wrong?
Any help would be appreciated.
| $t=\sqrt[3]{x},\qquad dt=\frac{1}{x^{2/3}}dx$
$$\int \frac{1}{x^{\frac{2}{3}}+x^{\frac{4}{3}}}dx=3\int \frac{1}{t^2+1}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$.
Let $f(x)=x^3+ax^2+bx+c$ be a cubic polynomial with real coefficients and all real roots, also $|f(i)|=1$ where $i=\sqrt{-1}$. Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$.
As $f(i)=-i-a+ib+c=1$ and $f(i)=-i-a+ib+c=-1$
I don't know how to solve further.
| Alternatively, note that for $z=a+bi\in \mathbb Z, \bar{z}=a-bi \in \mathbb Z$, the norm is:
$$|z|=\sqrt{z\cdot \bar{z}}=\sqrt{a^2+b^2}.$$
So:
$$|f(i)|=|c-a+(b-1)i|=\sqrt{(c-a)^2+(b-1)^2}=1 \Rightarrow (c-a)^2+b^2-2b=0 \tag{1}$$
Let $x_1,x_2,x_3$ be the roots of $f(x)=0 \iff x^3+ax^2+bx+c=0.$
By the Vieta's formulas:
$$\begin{align}\begin{cases}
x_1+x_2+x_3&=-a\\
x_1x_2+x_1x_3+x_2x_3&=b\\
x_1x_2x_3&=-c\end{cases} \tag{2}\end{align}$$
Plug $(2)$ to $(1)$:
$$(x_1+x_2+x_3-x_1x_2x_3)^2+(x_1x_2+x_1x_2+x_2x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0 \iff \\
x_1^2+x_2^2+x_3^2+(x_1x_2)^2+(x_1x_3)^2+(x_2x_3)^2+(x_1x_2x_3)^2=0 \iff \\
x_1=x_2=x_3=0.$$
| {
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Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
| You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $\frac{1}{2}\sqrt{2}(1-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}-1}{2}$.
So the area of the polygon is $2\sqrt{2}- \frac{\sqrt{2}-1}{2}= \frac{3\sqrt{2}+1}{2}$.
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.
Some condition must be missing, check the problem again.
| {
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"timestamp": "2023-03-29T00:00:00",
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Locus of a vertex of a triangle inside an equilateral one, under an integer constraint Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $F$ and $G$ on the side $AB$.
These two points define the segment $AF$ (blue), $FG$ (green), and $GB$ (red). We denote $\alpha=\overline{AF}$, $\gamma=\overline{FG}$, and $\beta=\overline{GB}$.
So far, I was able to prove that if $\frac{\gamma^2}{2\alpha\beta}=1$, then $D$ lies on the arc of the circle with center in the midpoint $M$ of the side $AB$, and with diameter equal to $\overline{AB}$ (in these conditions, $ABD$ is a right triangle).
On which curves lies the point $D$ if
$$
\frac{\gamma^2}{2\alpha\beta}=n,
$$
where $n\in\{2,3,4\ldots\}$?
Thanks for your suggestions!
| I don't think you will find a nice description of the curve without using coordinates.
Let $A = (-1, 0)$, $B = (1, 0)$ and $D = (x, y)$. Then it is easy to prove that
$$\alpha = 2 - \sqrt{(x - 1)^2 + y^2} \qquad \beta = 2 - \sqrt{(x + 1)^2 + y^2}$$
$$\gamma = \sqrt{(x + 1)^2 + y^2} + \sqrt{(x - 1)^2 + y^2} - 2$$
Therefore, the equation $\gamma^2 = 2 n \alpha \beta$ amounts to:
$$\left [\sqrt{(x + 1)^2 + y^2} + \sqrt{(x - 1)^2 + y^2} - 2 \right ]^2 = 2 n \left [2 - \sqrt{(x - 1)^2 + y^2} \right ] \left [ 2 - \sqrt{(x + 1)^2 + y^2} \right ]$$
Indeed, if $n = 1$ then the equation reduces to $x^2 + y^2 = 1$, which represents the circle having diameter $AB$. If $n > 1$, the equation represents a curve having a branch shaped like an oval which becomes "pointy" as $n$ increases.
If we choose $D$ on the altitude of the triangle, it is easy to see that $\frac {\gamma^2} {2 \alpha \beta} \to \infty$ as $D$ approaches $C$. So the answer to the related question is yes.
| {
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Reduction of Order Leads to Non-Elementary Integral
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
\begin{align}
\frac{dw}{dx}(x^2+x)-w(x^2+1)&=0 \\
\frac{dw}{dx}&=\frac{(x^2+x)^{-1}(x^2+1)}{w^{-1}} \\
\text{ln}(w)&=\int \frac{x^2+1}{x^2+x} \ dx \\
\text{ln}(w)&=\int 1-\frac{1}{x}+\frac{2}{x+1} \ dx \\
\text{ln}(w)&=x+\text{ln}\left(\frac{(x+1)^2}{x}\right)+C \\
w&=C_1\frac{e^x(x+1)^2}{x} \\
v&=C_1\int \frac{e^x(x+1)^2}{x} \ dx
\end{align}
Where $$C_1\int \frac{e^x}{x} \ dx$$ cannot be solved. How do I find $v$?
| Define the operators $D$ and $X$ by $(D\,h)(x):=h'(x)$ and $(X\,h)(x):=x\,h(x)$. For any function $\phi$, we also define the operator $\phi(X)$ to be $\big(\phi(X)\,h\big)(x):=\phi(x)\,h(x)$. Observe that
$$D^2-\left(\frac{X+1}{X}\right)\,D+\frac{1}{X}=\left(D-1-\frac{1}{X}+\frac{1}{X+1}\right)\,\left(D-\frac{1}{X+1}\right)\,.$$
First, let $v:=\left(D-\dfrac{1}{X+1}\right)\,u$. Then, $\left(D-1-\dfrac{1}X+\dfrac1{X+1}\right)\,v=0$. Therefore,
$$v(x)=\exp\Biggl(\int\,\left(1+\frac{1}{x}-\frac{1}{x+1}\right)\,\text{d}x\Biggr)=a\,\left(\frac{x\,\exp(x)}{x+1}\right)$$
for some constant $a$. Now, $v=\left(D-\dfrac{1}{X+1}\right)\,u$ implies that $D\,\left(\dfrac{1}{X+1}\,u\right)=\dfrac{1}{X+1}\,v$. Hence,
$$u(x)=(x+1)\,\int\,\frac{x\,\exp(x)}{(x+1)^2}\,\text{d}x=(x+1)\,\Biggl(a\left(\frac{\exp(x)}{x+1}\right)+b\Biggr)=a\,\exp(x)+b(x+1)\,,$$
for some constant $b$.
Interestingly, you can also write
$$D^2-\left(\frac{X+1}{X}\right)\,D+\frac{1}{X}=\left(D-\frac{1}{X}\right)\,(D-1)\,.$$
That is,
$$D\,\Biggl(\frac{\exp(X)}{X}\,D\,\big(\exp(-X)\,y\big)\Biggr)=0\,.$$
Consequently, for some constant $A$,
$$\big(D\,\big(\exp(-X)\,y\big)\Big)(x)=-A\,x\,\exp(-x)\,,$$
making
$$y(x)=\exp(x)\,\int\,\big(-A\,x\,\exp(-x)\big)\,\text{d}x=A\,(x+1)+B\,\exp(x)\,,$$
for some constant $B$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| Another solution
\begin{align*}
\sqrt {49 - x^2} - \sqrt {25 - x^2} &= 3\\
\sqrt {49-x^2} &= 3+ \sqrt {25-x^2}\\
49-x^2 &= 25 -x^2 + 9 + 6\sqrt {25-x^2}\\
15 &= 6 \sqrt {25 - x^2}\\
25 - x^2 &= \frac {25} 4\\
x^2 &= \frac {75}4\\
x &= \pm \frac 5 2 \sqrt 3.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $p$ and $q$ such that the maximum and minimum values of $5 +6\cos\theta + 2 \cos2\theta$ satisfy $x^2-p x+q=2$
If the maximum and minimum values of $5 +6\cos\theta + 2 \cos2\theta$ satisfy $x^2-px + q = 2$, then what are $p$ and $q$?
My thinking that, maximum valus of $\cos\theta$ is $1$ and minimum is $-1$. Now I can claim that the maximum of $5+6\cos\theta + 2 \cos2\theta)$ is $13$ and the minimum value is $9$
So $p = 13$ and $q=9$.
Is it correct or not?
Any hints/solution?
Thank u
| The minimum value is wrong. Let $f(t)=5+6t+2(2t^{2}-1)$. [Recall that $\cos (2x)=2\cos^{2}x-1$]. Therefore, $f'(t)=6+8t=0$ when $t=-3/4$ which is a possible value for $\cos x$. So the minimum value is $5+6(-3/4)+2(9/8-1)=3/4$. The quadratic with leading coefficient $1$ and roots $3/4$ and $13$ is given by $(x-\frac 3 4 ) (x-13)=0$. Comparing coefficienrs we get $p=\frac {55} 4$ adn $q=\frac {47} 4$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17
| By C-S
$$12=a^2+b^2+c^2=\frac{1}{2}(1^2+1^2)(a^2+b^2)+c^2\geq\frac{1}{2}(a+b)^2+c^2=\frac{1}{2}(2-c)^2+c^2,$$
which gives $$3c^2-4c-20\leq0$$ or
$$(3c-10)(c+2)\leq0$$ or
$$-2\leq c\leq\frac{10}{3}.$$
Now, we get $$\frac{10}{3}-(-2)=\frac{16}{3}.$$
| {
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Calculating the series $1/8+1/88+1/888+....$ I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=\dfrac18\displaystyle\sum_{k=0}^\infty\dfrac{1}{\sum_{n=0}^k10^n}$
where $$\displaystyle\sum_{n=0}^k10^n=\dfrac{10^{k+1}-1}{9}$$
then
$S=\dfrac98\displaystyle\sum_{k=0}^\infty\dfrac{1}{10^{k+1}-1}$
I have tried to calculate $\displaystyle\sum_{k=0}^K\dfrac{1}{10^{k+1}-1}$ for finite values but I failed.
What methods can we try?
| $$\frac{1}{8}+\frac{1}{88}+\frac{1}{888}+\dotsm = \frac{1}{8}\left( 1+\frac{1}{11}+\frac{1}{111}+\dotsm \right) =
\frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} \iff x= \frac{1}{10}$$
Where $\sum_{n=1}^{\infty} \frac{x^n}{1-x^n}$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
S=\frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} =\frac{9}{8}\left(\frac{\log\left(1-x\right)+\psi_{x}^{(0)}(1)}{log(x)}\right)
$$
That gives us:
$$S=\frac{9}{8}\left(\frac{\log\left(\frac{9}{10}\right)+\psi_{\frac{1}{10}}^{(0)}(1)}{log(\frac{1}{10})}\right)=0.137614773854525092047481887706797505400431...$$
Where $\psi_{x}^{(y)}(z)$ is the generalized PolyGamma function
| {
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Number of solution of $x^4-5x^3+(\lambda+2)x^2-5x+1=0$
Consider the bi-quadratic equation $E:x^4-5x^3+(\lambda+2)x^2-5x+1=0$ then, the real values of $\lambda$ so that $E$ has four different solutions is?
My attempts:
As $x=0$ is not a solution for any $\lambda$ hence we divide by $x^2$,\begin{align}x^2+\dfrac{1}{x^2}-5x-5\dfrac{1}{x}+\lambda+2&=0\\\bigg(x+\dfrac{1}{x}\bigg)^2-5\bigg(x+\dfrac{1}{x}\bigg)&=-\lambda\\t(t-5)&=-\lambda\end{align}
Here, $x+\dfrac{1}{x}=t$ cannot lie between $[-2,2]$.
Now we draw the graph of $-t(t-5)$ and $\lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+\lambda=0$, each of the roots of $Q$ gives two $x's$.
From GRAPH HERE, my solution is $\lambda\in(-\infty,-14]\cup[6,\frac{25}{4})$
But this wrong according to answer provided, please help.
| Hint: Draw a graph $$f(x)= {-x^4+5x^3-2x^2+5x-1\over x^2}$$
and see for which $\lambda$ the line $y= \lambda$ cuts the graph of $f$ four times.
| {
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Find no. of non negative integer solutions of $a+2b+3c = 200$ I want to find number of the non negative integer solutions of $a+2b+3c = 200$
(answer can contain $a,b,c = 0$)
Example one of the solutions is $a = 100, b = 50, c = 0$
I used stars and bars trick and here i get is
$\dfrac{\dbinom{200+2}{2}}{2 \times 3}$ but it is incorrect
any type of math is welcomed
what I have learnt from stars and bars technique is to find the solutions of $x+y = n , x,y,n > 0$ are positive integers but I am confused with equation of type $ax+by = n$ where $a , b> 1$
| Consider $x:=a+b+c$, $y:=b+c$, and $z:=c$. Let $n:=200$. Then, $x\geq y\geq z\geq 0$ and $x+y+z=n$. Let $$T_n:=\big\{(x,y,z)\in\mathbb{Z}_{\geq 0}^3\,|\,x+y+z=n\big\}\,.$$ Then, the symmetric group $S_3$ of order $3!=6$ acts on $T_n$ by permuting the three entries of each element of $T_n$. The number of $S_3$-orbits $N_n$ in $T_n$ is precisely the number of triples $(x,y,z)\in T_n$ with $x\geq y\geq z$.
By Burnside's Lemma,
$$N_n=\frac{1}{|S_3|}\,\sum_{g\in S_3}\,\big|\text{Fix}(g)\big|\,,$$
where $\text{Fix}(g)$ denotes the number of triples $(x,y,z)\in T_n$ fixed by $g$. The class equation of $S_3$ is $$\#\{\text{identity}\}+\#\{\text{transpositions}\}+\#\{\text{$3$-cycles}\}=1+3+2\,.$$
That is,
$$N_n=\frac{1}{6}\,\Biggl(1\cdot \binom{n+3-1}{3-1}+3\cdot\left\lfloor \frac{n+2}{2}\right\rfloor+2\cdot \left\lfloor\frac{3-(n\text{ mod }3)}{3}\right\rfloor\Biggr)\,,$$
so
$$N_n=\frac{1}{6}\,\binom{n+2}{2}+\frac{1}{2}\,\Biggl(1+\left\lfloor\frac{n}{2}\right\rfloor\Biggr)+\frac{1}{3}\,\left\lfloor\frac{3-(n\text{ mod }3)}{3}\right\rfloor\,.$$
In particular,
$$N_{200}=\frac{1}{6}\,\binom{202}{2}+\frac{1}{2}\,(1+100)+\frac{1}{3}\,(0)=3434\,.$$
From Daniel Schepler's solution, the generating function $f(t):=\sum\limits_{n=0}^\infty\,N_n\,t^n$ is given by $f(t)=\dfrac{1}{(1-t)\,(1-t^2)\,(1-t^3)}$. This answer gives a slightly more aesthetic expression for $f$:
$$f(t)=\frac{(1-t)^{-3}}{6}+\frac{(1+t)\,\left(1-t^2\right)^{-2}}{2}+\frac{\left(1-t^3\right)^{-1}}{3}\,.$$
| {
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Showing $\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$
I wish to show that if $z$ is real, then
$$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$
I have shown this result, although my inequality is the wrong way around.
I considered
\begin{align}
|z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(triangle inequality)} \\
&=|z|^2+1 \\ \\
\Rightarrow |z^2+1|&\leq |z|^2+1 \\
\frac{1}{|z^2+1|}&\geq\frac{1}{|z|^2+1} \\
\frac{|e^{iz}|}{|z^2+1|}&\geq\frac{|e^{iz}|}{|z|^2+1} \\
\left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{e^{-\text{Im}(z)}}{|z|^2+1} \\
\left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{1}{|z|^2+1} \ \ \ \ \ \ \text{(if $z$ is real $\Rightarrow$ Im$(z)=0$)} \\
\end{align}
Where did I go wrong?
Also, I wonder, would this inequality still hold if $z$ was not real?
| We have
$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{\left|e^{iz}\right|}{\left|z^2+1\right|}= \frac{1}{\left|z^2+1\right|}$$
and
$$0\le\left|{z^2+1}\right|= |z|^2+1$$
therefore the result follows.
| {
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Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$
Or $\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$
Then i can't prove $\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$
I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QM\ge AM+GM$ with $n=2$ so is that true for $n=i$?
| Thanks everyone i had the answer for my stuck.
We have : $$LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$$
Or $$-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$$
Or $$\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$$
Then i need to prove $$\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$$
Or $\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b$. Which's true by C-S $$\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)^2\le 2\left(\frac{a^2+b^2}{2}+ab\right)=\left(a+b\right)^2$$
| {
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Positive integers have the form $a^2+b^2+c^2+2^d$ For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that
$$n=a^2+b^2+c^2+2^d.$$
Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers
$$m=a^2+b^2+c^2$$
if and only if $m$ is not of the form $m=4^a(8b+7)$, for integers $a$ and $b$. If the conjecture is true, then for all $n\geq 1$ it must exist a $k$ such that $n-2^k$ is not of the form $m=4^a(8b+7)$.
I want help to prove the conjecture or to find a counter-example.
| We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $x\in\{1,2,4\}$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $\equiv 0$ or $\equiv 3\bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $\bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $n\leq 4$. We have
$$0^2+0^2+0^2+2^0=1$$
$$0^2+0^2+0^2+2^1=2$$
$$0^2+0^2+1^2+2^1=3$$
$$0^2+0^2+0^2+2^2=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many zeros does the polynomial have in the right half plane? The polynomial is $f(z) = z^4+\sqrt{2}z^3+2z^2-5z+2$
If you check the image of the imaginary axis, you see that there are no zeros, so we can use the right semicircle from $iR$ to $-iR$,and make $R$ arbitrarily big. The problem now is picking the right function to use Rouche's theorem with. I'm confused because if I were to pick something like $z^4+2$, then it would have 1 zero on the right-half plane, and two on the imaginary axis. Do we count those two as well?
But then, how do you show that $|z^4+2|>|\sqrt{2}z^3+2z^2-5z|$ on the imaginary-axis part of the semi-circle. It doesn't seem like a clean argument.
| We can use the following way.
For all real $k$ we have:
$$z^4+\sqrt2z^3+2z^2-5z+2=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\left(2k-\frac{3}{2}\right)z^2+(\sqrt2k+5)z+k^2-2\right).$$
Now, let $2k-\frac{3}{2}>0$ and $$(\sqrt2k+5)^2-4\left(2k-\frac{3}{2}\right)(k^2-2)=0.$$
We see that $k=2.65...$ is valid.
Id est, for this value of $k$ we obtain
$$z^4+\sqrt2z^3+2z^2-5z+2=$$
$$=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\left(2k-\frac{3}{2}\right)z^2+(\sqrt2k+5)z+k^2-2\right)=$$
$$=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\sqrt{2k-\frac{3}{2}}z+\sqrt{k^2-2}\right)^2=$$
$$=\left(z^2+\left(\tfrac{1}{\sqrt2}-\sqrt{2k-\tfrac{3}{2}}\right)z+k-\sqrt{k^2-2}\right)\left(z^2+\left(\tfrac{1}{\sqrt2}+\sqrt{2k-\tfrac{3}{2}}\right)z+k+\sqrt{k^2-2}\right).$$
Now, easy to see for this value of $k$ we have
$$\left(\tfrac{1}{\sqrt2}-\sqrt{2k-\tfrac{3}{2}}\right)^2-4\left(k-\sqrt{k^2-2}\right)<0$$ and
$$\left(\tfrac{1}{\sqrt2}+\sqrt{2k-\tfrac{3}{2}}\right)^2-4\left(k+\sqrt{k^2-2}\right)<0,$$ which says that our polynomial has no real roots and since
$$\tfrac{1}{\sqrt2}-\sqrt{2k-\tfrac{3}{2}}<0$$ and
$$\tfrac{1}{\sqrt2}+\sqrt{2k-\tfrac{3}{2}}>0,$$
we see that two roots exactly placed in the right half of the plane.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$ I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then:
$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac{\pi^2}{8}-\int_0^1 \frac{\arcsin^2 x}{(1+x^2)^{3/2}}dx$$ We can expand into power series the integral, we have: $\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$ and using the binomial series for $(1+x^2)^{-3/2}$ will result in: $$\sum_{n\geq1}\frac{2^{2n-1}x^{2n}}{n^2\binom {2n}n}\sum_{k\ge 0}\binom{-3/2}{k}x^{2k}$$ But I dont know how to simplify this. I tried one more thing, letting $\sin x= \sin^2 t$ gives:
$$I=2\int_0^\frac{\pi}{2}\frac{x\sin x}{\sqrt{1+\sin^2 x}}dx$$ Since $\int \frac{\sin x}{\sqrt{1+\sin^2x}}dx=-\arcsin\left(\frac{\cos x}{\sqrt 2} \right)+C$ we can integrate by parts to obtain: $$I=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\cos x}{\sqrt 2}\right)dx=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ But I am stuck, so I would appreciate some help.
Edit: By letting $\frac{\sin x}{\sqrt 2} =t $ We get: $$I=2\int_0^\frac1{\sqrt{2}} \frac{\arcsin x}{\sqrt{\frac12-x^2}}dx=2\text{Li}_2\left(\frac1{\sqrt 2}\right)-\frac{\pi^2}{24}+\frac{\ln^2 2}{4}$$ Where the latter integral was evaluated with wolfram. I would love to see a proof for that.
| Work in progress.
Since $\arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}$ for any $x\in[-1,1]$ and
$$ \int_{0}^{\pi/2}\left(\sin x\right)^{n+1/2}\,dx=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{n}{2}+\frac{3}{4}\right)}{\Gamma\left(\frac{n}{2}+\frac{5}{4}\right)} $$
we "just" need an explicit value for the series
$$ \sqrt{\frac{2}{\pi}}\sum_{n\geq 0}\frac{2^n \Gamma\left(\frac{n}{2}+\frac{3}{4}\right)^2}{(2n+1)^2 \Gamma(n+1)} $$
which is given by a linear combination of two $\phantom{}_4 F_3(\ldots;1)$ functions with quarter-integer parameters, namely $\phantom{}_4 F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},\frac{5}{4},\frac{5}{4}; 1\right)$ and $\phantom{}_4 F_3\left(\frac{3}{4},\frac{3}{4},\frac{5}{4},\frac{5}{4}; \frac{3}{2},\frac{7}{4},\frac{7}{4};1\right)$.
Fourier-Legendre expansions revealed to be extremely effective in dealing with such objects: for instance all the functions $\frac{\arcsin\sqrt{x}}{\sqrt{x}},\frac{1}{\sqrt{1-x^2}},\frac{1}{\sqrt{1-x^4}},K(x)$ have reasonably simple FL-expansions, opposed to the moderate complexity of their Maclaurin series. This observation allowed Campbell, Cantarini, Di Trani, Sondow and I to exhibit many surprising identities about $\phantom{}_3 F_2(\ldots;1)$ and $\phantom{}_4 F_3(\ldots;1)$ in terms of polylogarithms. My bet is that the same occurs here. With a step of integration by parts we have
$$ I = \int_{0}^{1}\frac{2x\arcsin x}{\sqrt{1-x^4}}\,dx = \frac{\pi^2}{4}-\int_{0}^{1}\frac{\arcsin(x^2)}{\sqrt{1-x^2}}\,dx$$
which is extremely good in simplifying the hypergeometric structure:
$$ I = \frac{\pi^2}{4}-\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}\int_{0}^{\pi/2}\left(\sin x\right)^{4n+2}\,dx $$
leads to
$$ I = \frac{\pi^2}{4}-\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{(2n+1)4^{3n+1}}=\frac{\pi^2}{4}-\frac{\pi}{4}\sum_{n\geq 0}\frac{\binom{2n}{n}\binom{4n}{2n}}{4^{3n}}\cdot\frac{4n+1}{(2n+1)^2} $$
where the last series is blatantly related to Legendre function $P_{-1/4}$. Indeed, according to Mathematica's notation for the complete elliptic integrals (i.e. the argument is the elliptic modulus)
$$ \sum_{n\geq 0}\frac{\binom{2n}{n}\binom{4n}{2n}}{4^{3n}}z^{2n} = \frac{2}{\pi\sqrt{1+z}}\,K\left(\frac{2z}{1+z}\right) $$
and the given problem boils down to computing
$$ \int_{0}^{1}\frac{1}{\sqrt{1+z}}\,K\left(\frac{2z}{1+z}\right)\,dz\quad\text{and}\quad \int_{0}^{1}\frac{\log z}{\sqrt{1+z}}\,K\left(\frac{2z}{1+z}\right)\,dz.$$
The substitution $z\mapsto\frac{x}{2-x}$ leads to three integrals which are simple to tackle through the FL machinery, namely $\int_{0}^{1}\frac{K(x)}{(2-x)^{3/2}}g(x)\,dx$ where $g(x)\in\{1,\log(x),\log(2-x)\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$
My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
| We have
$$(y^3+2y+1)-(y-1)^3=3y^2-y+2=\frac{(6y-1)^2+23}{12}>0$$
and
$$(y+2)^3-(y^3+2y+1)=6y^2+10y+7=\frac{(6y+5)^2+17}{6}>0\,.$$
That is,
$$(y-1)^3<y^3+2y+1<(y+2)^3$$
for all $y\in\mathbb{Z}$. If $y^3+2y+1$ is a cube, then either $y^3+2y+1=y^3$ or $y^3+2y+1=(y+1)^3$, which gives $y=-\dfrac12$, $y=-\dfrac13$, or $y=0$. That is, $(x,y)=(1,0)$ is the only integer solution to $x^3=y^3+2y+1$.
| {
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"url": "https://math.stackexchange.com/questions/2882685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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"answer_id": 5
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A functional equation in a TST exam Find all functions $f:\Bbb {R} \rightarrow \Bbb {R} $ such that:
$$f(f(xy-x))+f(x+y)=yf(x)+f(y)$$
Source: 2018 Hong Kong TST 2 problem 3
I recently proved that $f(x+1)=f(x)+f(1).$
| To recap your result: setting $x=y=0$ shows us that
\begin{align*}
f(f(0))+f(0)&=0\cdot f(0)+f(0) \\
f(f(0))+f(0)&=f(0) \\
f(f(0))&=0.
\end{align*}
Now from the original equation we set $y=1$ to obtain
\begin{align*}
f(f(0))+f(x+1)&=f(x)+f(1) \\
f(x+1)&=f(x)+f(1).
\end{align*}
Let us set $x=0$ in this equation to obtain
\begin{align*}
f(1)&=f(0)+f(1) \\
0&=f(0).
\end{align*}
So the function must go through the origin.
$f(x+1)=f(x)+f(1)$ is a straight-forward recurrence relation. If we assume $f(x)=b\cdot a^x$ and plug in, we obtain
\begin{align*}
b\cdot a^{x+1}&=b\cdot a^{x}+b\cdot a \\
a^{x+1}&=a^x+a \\
a^{x+1}-a^x-a&=0.
\end{align*}
This must hold for $x=1,$ which forces $a^2-a-a=0,$ or $a^2-2a=0,$ the solutions of which are $a=0,$ not very exciting, or $a=2,$ which is much more interesting. Indeed, we can handle the $a=0$ case by simply letting $b=0$. So we will say that the solutions are $$f(x)=b\cdot 2^x.$$
We can say something more, though: we proved that the function must go through the origin. In looking at this formula, it is evident that the trivial solution is the only solution of this type, since $2^x\not=0.$ Hence, one answer is that
$$\boxed{f(x)=0\quad\forall\,x\in\mathbb{R}.} $$
Another special answer (thanks to mfl in the comments!) we can obtain by inspection: $\boxed{f(x)=x},$ the identity function. To see this, we compute:
\begin{align*}
f(f(xy-x))+f(x+y)&=yf(x)+f(y) \\
f(xy-x)+x+y&=yx+y \\
xy-x+x+y&=yx+y \\
xy+y&=xy+y,
\end{align*}
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Linear Transformation of a Polynomial I have an operation that takes $ax^2+bx+c$ to $cx^2+bx+a$. I need to find if this corresponds to a linear transformation from $R^3$ to $R^3$, and if so, its matrix.
I know that
$$
ax^2+bx+c = \begin{bmatrix}a&b&c \end{bmatrix} \begin{bmatrix}x^2\\x\\1 \end{bmatrix}
$$
If I perform the column operation $C_1 \leftrightarrow C_3$, then I can get the desired result. However, this would mean putting the corresponding elementary matrix to the right of my coefficient matrix like so:
$$
\begin{bmatrix}a&b&c \end{bmatrix} \begin{bmatrix}0&0&1\\0&1&0\\1&0&0 \end{bmatrix} \begin{bmatrix}x^2\\x\\1 \end{bmatrix} = cx^2+bx+a
$$
Is this the answer? I have difficulty accepting that the matrix can simply be put in the middle.
| $$T((ax^2+bx+c)+(a'x^2+b'x+c'))=T((a+a')x^2+(b+b')x+(c+c'))
\\=(c+c')x^2+(b+b')x+(a+a')
\\=(cx^2+bx+a)+(c'x^2+b'x+a')
\\=T(ax^2+bx+c)+T(a'x^2+b'x+c')$$ and linearity is confirmed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What method is most preferred to find nature of the series $\sum_{n=1}^{\infty} \frac{a^n}{a^n+x^n}$? I tried to use D'Alembert's ratio test to test the convergence of the given series $\sum\limits_{n=1}^{\infty} \frac{a^n}{a^n+x^n}$ but I could only solve for $\frac{a}{x} <1$ and it is coming out to be converging but for $\frac{x}{a} <1$, I am not able to do it with ratio test.
| $$\frac{a^n}{a^n + x^n} = \frac{1}{1 + (\frac{x}{a})^n}$$
Ratio test:
$$\frac{\frac{a^{n+1}}{a^{n+1} + x^{n+1}}}{\frac{a^n}{a^n + x^n}} = a \frac{a^n + x^n}{a^{n+1} + x^{n+1}} = \frac{1 + (\frac{x}{a})^n}{1 + (\frac{x}{a})^{n+1}}$$
*
*If $\vert \frac{x}{a} \vert \leq 1$, the above limits goes to $1$. And the ratio test here is inconclusive. Note here that $\frac{a^n}{a^n + x^n} = \frac{1}{1 + (\frac{x}{a})^n}$ does not go to zero (goes to $1$). The series diverges. This test is referred to as $n^{th}$ term test.
*If $\vert \frac{x}{a} \vert > 1$, the above limit acts as
$$\frac{(\frac{x}{a})^n}{(\frac{x}{a})^{n+1}} = \frac{a}{x} < 1$$
Thanks @DanielWainfleet for the remark on absolute values.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$ Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$
I tried:
$x=\sin\alpha$ and $y=\cos\beta$
$\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}$ but if I write $\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta$, it's not true because
$\cos\alpha \sin\beta$ can be negative.
Can someone give me an idea?
| \begin{align}
& \left| xy-\sqrt{(1-x^2)(1-y^2)}\,\right| \\[10pt]
= {} & \left| \sin\alpha\cos\beta - \left| \cos\alpha\sin\beta \right| \, \right| \\[10pt]
= {} & \begin{cases} \left| \sin\alpha\cos\beta - \cos\alpha \sin\beta\right| = \left|\sin(\alpha-\beta){} \right| & \text{if } \cos\alpha>0\ \&\ \sin\beta>0, \\[5pt]
\left| \sin\alpha\cos\beta + \cos\alpha\sin\beta \right| = \left| \sin(\alpha+\beta) \right| & \text{if } \cos \alpha<0\ \&\ \sin\beta>0, \\[5pt]
\text{and similarly in the other two cases.}
\end{cases}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Taking $4$ shots or $6$?
Suppose your skill (the ability to make a shot) follows a binomial distribution (with some unknown $p$). You need to make at least half of the shots. Would you rather take $4$ or $6$? More precisely, what should be the $p$ which is the threshold of your decision?
My Try: assuming $0 < p < 1$.
$${{6}\choose{3}}p^3(1-p)^3 ={{4}\choose{2}}p^2(1-p)^2 \\
\implies p(1-p) = 6/20
$$
but apparently the answer is: $3/5$.
What's wrong with my calculation?
| Probability of making AT LEAST half shots out of 6 is
$$
\binom{6}{3}p^3(1-p)^3 +\binom{6}{4}p^4(1-p)^2 +\binom{6}{5}p^5(1-p) +\binom{6}{6}p^3
$$
and for making at least 2 out of 4 is
$$
\binom{4}{2}p^2(1-p)^2 +\binom{4}{3}p^3(1-p) +\binom{4}{4}p^4
$$
For threshold $p$, equating both, we get:
$$
10p^4-36p^3+48p^2-28p+6 = 0
$$
which reduces to
$$
2(x-1)^3(5x-3) = 0
$$
which gives $p = {\frac 35}$ since $0 \lt p \lt 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that the highest and lowest order terms will be $z^{1 + 2 + \cdots + m} = z^{\frac 1 2m(m+1)}$ and its reciprocal, respectively.
Here's the situation for low values of $m$:
$$\begin{split}
f_1(z) = z &+ \frac 1 z \\
f_2(z) = z^3 + z &+ \frac 1 z + \frac 1 {z^3} \\
f_3(z) = z^6+z^4+z^2+\ &\color{red}2 + \frac 1 {z^2}+ \frac 1 {z^4} + \frac 1 {z^6} \\
f_4(z) = z^{10} + z^8 + z^6 + \color{red}2z^4 + \color{red}2z^2 +\ &\color{red}2 + \frac {\color{red}{2}} {z^2} + \frac {\color{red}2} {z^4} + \frac 1 {z^6} + \frac 1 {z^8} + \frac{1}{z^{10}} \\
f_5(z)=z^{15} + z^{13}+ z^{11} + \color{red}2z^{9} + \color{red}2z^7 + \color{orange}3z^5 + \color{orange}3z^3 + \color{orange}3z &+ \frac{\color{orange}3} z + \frac {\color{orange}3}{z^3} + \frac {\color{orange}3} {z^5} + \frac {\color{red}2} {z^7} + \frac {\color{red}2} {z^9} + \frac 1 {z^{11}} + \frac 1 {z^{13}} + \frac 1 {z^{15}}
\end{split}$$
It seems that
*
*In each $f_m$ the exponent of $z$ falls from $\tau_m = \frac 1 2 m(m+1)$ down to $-\tau_m$, skipping every other value;
*For all $m$ the coefficient $a_{m,n}$ of the $n$-th term in the analytic part is equal to the coefficient $a_{m,-n}$ of the $n$-th term in the singular part (which makes sense intuitively).
So:
*
*Is there a closed-form expression for $a_{m,n}$?
*How can I prove fact 1. above?
| This is just a start, not a solution, but I can't fit it in a comment. Maybe it's enough for you to answer the question yourself. We can think of the problem this way. Partition $[m] = \{1,2,\dots\}$ into two sets $A$ and $B$ and let $\sum A$ and $\sum B$ be the sum of the elements of $A$ and $B$ respectively. Then $a_m,k= [z^k]f_m(z)$ is the number of such partitions with $\sum A-\sum B = k$. We have $$
a_{m,k}=\cases{
1,& $m=1,\ |k|=1$\\
0,& $m=1,\ |k|\neq 1$\\
a_{m-1,k-m}+a_{m-1,k+m},&otherwise
}
$$
This is because the partitions of $[m]$ arise from adding the element $m$ to one of the sets in a partition of $[m-1].$ Notice that I am assuming that $a_{m,k}$ is defined for all positive integers $m$ and all integers $k.$ Just make $a_{m,k}=0$ when $z^k$ does not appear in $f_m.$
I have no idea whether this can be solved for $a_{m,k}$ in closed form, but at least it makes it easy to calculate.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Show that $8$ is an eigen value of $A$ Consider the following matrix:
$$A=\begin{bmatrix}B&&&C\\D&&&F\end{bmatrix}$$ where
$$B= \begin{bmatrix} 9 &1&1&1&1\\1&9&1&1&1\\1&1&9&1&1\\1&1&1&9&1\\1&1&1&1&9\end{bmatrix}$$ and $$C=\begin{bmatrix}1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\end{bmatrix}$$
and $$D=C^T$$
and $$F=\begin{bmatrix} G&H \\I &J\end{bmatrix}$$
where $$G=\begin{bmatrix} 8 &1&1&1\\1&8&1&1\\1&1&8&1\\1&1&1&8\end{bmatrix}$$ and $$H=\begin{bmatrix} 0\\0\\0\\0\end{bmatrix}$$ and $$I=H^T$$ and
$$J=\begin{bmatrix} 5\end{bmatrix}$$
Show that $8$ is an eigen value of the matrix $A$.
How should I try to prove it?Please give some hints .I dont want a complete solution
| Hint:
*
*Examine the first two rows of $A-8I$, they are identical, what can you conclude.
Same result holds if you replace $D$ and $F$ with other matrices of the same size.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?
I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it.
Another thought: If $(a^2+b^2+c^2)^2 \geq 1$, then this statement will be true. But how can I prove it?
| Yes, that question helps a lot! Note that by AM-GM inequality $a^2+b^2+c^2\ge 3(a^2b^2c^2)^{1/3}=3$. Since $x^2\geq x$ for all $x\geq 1$, it follows that
$$(a^2+b^2+c^2)^2\geq a^2+b^2+c^2\geq a+b+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I eliminate $x$ and $y$ from the system $x^2 y= a$, $x(x+y)= b$, $2x+y=c$ to get a single equation in $a$, $b$, $c$? Alright, a homework problem.
I'm stuck at this question,
Eliminate $x$ and $y$ from the given equations to get a single
equation in terms of $a$ , $b$ and $c$
$$\begin{align}
x^2 y &= a \\
x(x+y) &= b \\
2x+y &=c
\end{align}$$
Let me tell you what I tried, I tried to get $y$ from one equation and substitute in the other two. Turns out that I'm not able to fully get rid of both $x$ and $y$. Help please.
| Combine the first and second equations to eliminate $y$:
$x(x+\frac{a}{x^2})=b$
which tidies to
$x^2+\frac{a}{x}=b$ ....... (1)
combine the second and third equations to eliminate $y$:
$x(x+c-2x)=b$
which tidies to
$x^2-cx+b=0$
solving this quadratic gives
$x=\frac{c \pm \sqrt{c^2-4b}}{2}$
which you can substitute into equation (1):
$\left(\frac{c \pm \sqrt{c^2-4b}}{2}\right)^2+\frac{a}{\left(\frac{c \pm \sqrt{c^2-4b}}{2}\right)}=b$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that $0$ has multiplicity $3$ in $M-4I$? Question from an exam:
Consider the matrix $M=$
\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3\end{bmatrix}
Show that $4$ is an eigen value of the above matrix with multiplicity $3$.
I considered $M-4I$ from which I got
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
By elementary row operations:
$$M-4I=$$
\begin{bmatrix} 1&1&1&1&1&1\\0&0&0&0&0&0\\0&0&0&0&0&0\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
So $0$ is an eigen value with multiplicity at least $2$ since $M-2I$ has two zero rows.
How to show that $0$ has multiplicity $3$ in $M-4I$?
If one can show how to proceed after this,I will be really grateful.
Please help
| Since $M$ is symmetric, the geometric and algebraic multiplicity of its eigenvalues coincide.
If you finish your row reduction, you get
$$
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
\xrightarrow{\ \ \ \ \ \ \ \ \ \ }
\begin{bmatrix} 1&1&1&0&0&-1\\0&0&0&1&0&1\\0&0&0&0&1&1\\0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\end{bmatrix}
$$
As there are three leading ones, this shows that the set of solutions of $(M-4I)x=0$ has three dependent solutions, and three independent. Thus $\dim \ker(M-4I)=3$. This means that the eigenvalue 4 has geometric and algebraic multiplicity 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $\left|\operatorname{Re}(3+i+2\bar z^2 -iz)\right| \le 6$, when $|z| \le 1$ So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y^2-ix+y)\right|$
$\left|3+2x^2-2y^2+y\right|$
$3+2x^2-2y^2+y$
And then I'm not really sure where to go. Pretty sure I took the wrong approach because not using the initial information that $|z| \le 1$
| Note that $\operatorname{Re}(3+i+2\overline z^2-iz)=\operatorname{Re}(3+2\overline z^2-iz)$ and that therefore\begin{align}\bigl\lvert\operatorname{Re}(3+i+2\overline z^2-iz)\bigr\rvert&\leqslant\bigl\lvert3+2\overline z^2-iz\bigr\rvert\\&\leqslant3+2|z|^2+|z|\\&\leqslant6,\end{align}since $|z|\leqslant1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
In how many ways can $5$ teaching posts be filled if $2$ posts are reserved for American candidates? For the post of $5$ teachers, there are $23$ applicants, $2$ posts are reserved for American candidates and there are $7$ American candidates among the applicants. In how many ways can the selection be made ?
Answer in textbook is given as
selection of $2$ American candidates out of $7$ + selection of $3$ from $16$ others left
$$\binom{7}{2} \times \binom{16}{3} $$
But, I am using combinations of $2$ Americans and $3$ others from left $16$ + combinations of $3$ Americans and $2$ others from left $16$ + combinations of $4$ Americans and $1$ other from left $16$ + combinations of $5$ Americans
$$\binom{7}{2} \times\binom{16}{3} +\binom{7}{3} \times\binom{16}{2} +\binom{7}{4} \times\binom{16}{1}+\binom{7}{5} \times\binom{16}{0}$$
Is it wrong? Why? Please explain.
| this is because the no of ways you can select 2 American candidates out of 7 is 7C2-------M
and the number of ways you can select the rest is (23-7)C(5-2)-------------N
Thus, total ways = M*N
= 7C2 * 16C3
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form solution for this integral? I'm hitting a road block in finding an expression (closed form preferably) for the following integral:
\begin{equation}
\int^{+\infty}_0 x^b \left ( 1-\frac{x}{u} \right )^c \exp(-a x^3) dx
\end{equation}
where $a,b$ are positive constants; $b>1$ is an odd multiple of $0.5$, while $c$ is a positive or negative odd multiple of $0.5$; $u$ is a (positive) parameter.
Things I have considered or tried:
*
*look up in tables (Gradshsteyn and Ryzhik): there are very few explicit results for integrals involving $\exp(-a x^3)$ (or for the other factors after transforming via $y=x^3$). Also, tabulated results involving $\exp(-a x^p)$ for more general $p$ do not include the other factors $x^b (1-x/u)^c$. One exception is (3.478.3):
\begin{equation}
\int^{u}_0 x^b (u-x)^c \exp(-a x^3) dx,
\end{equation}
but the limits of integration do not match with my case;
*there is a closed form solution (3.478.1) for the simpler integral
\begin{equation}
\int^{+\infty}_0 x^{d-1} \exp(-a x^3) dx = \frac{a^{-d/3}}{3} \Gamma(d/3).
\end{equation}
(NB: there is also an expression for the indefinite integral.)
A binomial expansion of $[1-(x/u)]^n$ for integer $n$ would produce a solution in series form. However, in my case, the exponents $b$ and $c$ are strictly half-integer. For the same reason, integration by parts does not lead to a simpler integral without the factor $[1-(x/u)]^c$;
*Wolfram Math online did not produce a result;
*the integral is an intermediate step in a longer analysis, so numerical solution (with given values for the parameter) is not practical.
Grateful for any pointers or solution.
| An integral of a product of two Meijer G-functions of rational powers of the argument gives a G-function:
$$\int_0^\infty x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx = \\
\int_0^u x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx +
(-1)^c \int_u^\infty x^b \left( \frac x u - 1 \right)^c e^{-a x^3} dx = \\
\Gamma(c + 1) \int_0^\infty x^b
G_{1, 1}^{1, 0} \left( \frac x u \middle| {c + 1 \atop 0} \right)
G_{0, 1}^{1, 0} \left( a x^3 \middle| {- \atop 0} \right) dx + \\
(-1)^c \Gamma(c + 1) \int_0^\infty x^b
G_{1, 1}^{0, 1} \left( \frac x u \middle| {c + 1 \atop 0} \right)
G_{0, 1}^{1, 0} \left( a x^3 \middle| {- \atop 0} \right) dx = \\
3^{-c - 1} \Gamma(c + 1) u^{b + 1}
G_{3, 4}^{1, 3} \left(a u^3 \middle|
{ \frac {-b} 3, \frac {-b + 1} 3, \frac {-b + 2} 3 \atop
0, \frac {-b - c - 1} 3, \frac {-b - c} 3, \frac {-b - c + 1} 3} \right) + \\
3^{-c - 1} (-1)^c \Gamma(c + 1) u^{b + 1}
G_{3, 4}^{4, 0} \left(a u^3 \middle|
{ \frac {-b} 3, \frac {-b + 1} 3, \frac {-b + 2} 3 \atop
0, \frac {-b - c - 1} 3, \frac {-b - c} 3, \frac {-b - c + 1} 3} \right).$$
When $b$ and $c$ are half-integers, one of the three numbers $(-b - c - 1)/3, (-b - c)/3, (-b - c + 1)/3$ is an integer. Therefore, the ratio of the gamma functions in the $G_{3, 4}^{4, 0}$ term always has an infinite sequence of double poles inside the integration contour, and the G-function cannot be converted to a sum of hypergeometric functions. The sum will be valid only as a limit, which will give derivatives of ${_pF_q}$ wrt parameters.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $
I understand every step until here
$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$
How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
| In order to find an explicit value for the given series, we may recall that for any $|x|<\frac{1}{e}$ we have
$$ \sum_{n\geq 1}\frac{n^{n-1}}{n!} x^n = -W(-x) $$
with $W(x)$ being the inverse function of $x e^x$ (Lambert's function), by Lagrange inversion theorem.
A rapid sequence of formal manipulations:
$$ \sum_{n\geq 1}\frac{n^n}{n!} x^{n}=\frac{-W(-x)}{1+W(-x)} $$
$$ \sum_{n\geq 1}\frac{n^n}{n!} x^{2n+1}=\frac{-x W(-x^2)}{1+W(-x^2)} $$
$$ \sum_{n\geq 1}\frac{n^n}{n!}(2n+1) x^{2n}=\frac{-x W(-x^2)}{1+W(-x^2)} $$
$$ \sum_{n\geq 1}\frac{n^n}{n!}(2n+1) x^{n}=\frac{-3W(-x)-2W(-x)^2-W(-x)^3}{(1+W(-x))^3} $$
and by replacing $x$ with $x(1-x)$, then applying $\int_{0}^{1}(\ldots)\,dx$, we get
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{n^n n!}{(2n)!} &=& \int_{0}^{1}\frac{-3W(-x(1-x))-2W(-x(1-x))^2-W(-x(1-x))^3}{(1+W(-x(1-x)))^3}\,dx\\
&=&\int_{0}^{1}\frac{-3W\left(\frac{t^2-1}{4}\right)-2W\left(\frac{t^2-1}{4}\right)^2-W\left(\frac{t^2-1}{4}\right)^3}{\left(1+W\left(\frac{t^2-1}{4}\right)\right)^3}\,dt\\&=&2\int_{-1/4}^{0}\frac{-3W(t)-2W(t)^2-W(t)^3}{\sqrt{1+4t}(1+W(t))^3}\,dt\\&=&\int_{W(-1/4)}^{0}-\frac{2 e^x x (3+2x+x^2)}{(1+x)^2 \sqrt{1+4 e^x x}}\,dx\end{eqnarray*}$$
where the last function is clearly integrable on $\left(W(-1/4),0\right)$. This integral representation is also a good way for deriving that the value of the wanted series is $\approx 1.53261$. An alternative derivation of an upper bound comes from the Cauchy-Schwarz inequality and the identities
$$ \sum_{n\geq 1}\frac{n!^2}{(2n)!}e^n = \frac{e}{4-e}+\frac{4\sqrt{e}}{(4-e)^{3/2}}\arcsin\left(\frac{\sqrt{e}}{2}\right),$$
$$ \sum_{n\geq 1}\frac{n^{2n}}{(2n)!e^n}=\frac{1}{2}\left(\frac{1}{1+W\left(\frac{1}{2\sqrt{e}}\right)}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $x^2+3x^4+o(x^4)=x^2+o(x^3)$ hold true? My textbook asks if the following holds true or false for $x\rightarrow0$:
$$x^2+3x^4+o(x^4)=x^2+o(x^3)$$
I think I understand what the little-o notation means. For $x\rightarrow x_o$ we have:
$$\lim_{x\rightarrow x_0}{\frac{f(x)}{g(x)}=0}$$
and we say $f(x)=o(g(x))$.
So to find an answer to the problem I evaluated the following limit:
$$\lim_{x\rightarrow0}{\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}}$$
For $x\rightarrow0$, I think we have
$$\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}\sim \frac{x^2+o(x^4)}{x^2+o(x^3)}=\dots$$
Any hints on how to continue? My textbook says the answer should be true. Sorry I couldn't find the appropriate tag for this little-o notation question.
| Recall that by definition
$$f(x)=o(g(x)) \iff f(x)=g(x)\cdot \omega(x) \quad \omega(x) \to 0$$
therefore as $x \to 0$ we have
$$x^2+3x^4+o(x^4)=x^2+3x^4+x^4\cdot \omega(x)=$$$$=x^2+x^3(3x+x\cdot \omega(x)))=x^2+x^3\cdot \omega_1(x)=x^2+o(x^3)$$
but for the limit in your example the key point is that
$$o(x^{n+1})=x\cdot o(x^n) \implies \frac1x \cdot o(x^{n+1})=o(x^n)$$
Therefore we can conclude the limit as as follow
$$\lim_{x\rightarrow0}{\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}}=\lim_{x\rightarrow0}{\frac{x^2+o(x^3)}{x^2+o(x^3)}}=\lim_{x\rightarrow0}{\frac{1+o(x)}{1+o(x)}} = \frac{1+0}{1+0}=1$$
or also
$$\lim_{x\rightarrow0}\frac{x^2+3x^4+o(x^4)}{x^2+o(x^3)}=\lim_{x\rightarrow0} \frac{1+3x^2+o(x^2)}{1+o(x)}=\frac{1+0+0}{1+0}= 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$ Is there a closed form expression of
$$\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$$
, where $\{x\}=x-\lfloor x \rfloor$ denotes the fractional part of $x$? $\text{I think this limit will converge to a point between $0$ and $\log_{2}3-1$}$
by $(x+y)(x-y)=(x^2-y^2)$, but I can't go further.
| By periodicity, $\log_2 (k)$ and $\log_2 (k) - n$ has the same fractional part. Hence
\begin{align*}
\lim_n \frac 1{2^n} \sum_1^{2^n} \{\log_2(k)\} &= \lim_n \frac 1{2^n} \sum_1^{2^n} \left\{\log_2 \left(\frac k{2^n}\right) \right\} \\&= \lim_n \frac 1{2^n} \sum_1^{2^n} \log_2 \left( \frac k{2^n }\right) - \left\lfloor \log_2\left(\frac k{2^n}\right)\right\rfloor \\
&= \int_0^1 \log_2(x) \mathrm dx - \lim f(n)/2^n,
\end{align*}
where
$$
f(n) = \sum_1^{2^n} \lfloor \log_2(k/2^n)\rfloor = \sum_1^{2^n} \lfloor \log_2(k)\rfloor - 2^n \cdot n.
$$
Now calculate the sum of the integer parts. Since $\lfloor \log_2(k)\rfloor = j$ when $2^j \leq k \leq 2^{j+1}-1$, we have
$$
f(n) + 2^n \cdot n = 0 + n + \sum_0^{n-1} 2^j j.
$$
Let $S(n) = \sum_0^{n-1}j 2^j$, then $2S(n) = \sum_0^{n-1} 2^{j+1} j = \sum_1^n 2^j (j-1)$, thus
\begin{align*}
S(n) = 2S(n) - S(n) &= \sum_1^n 2^j (j-1) - \sum_0^{n-1} 2^j j= \sum_2^n 2^j (j-1) - \sum_1^{n-1} 2^j j \\
&= 2^n (n-1) + \sum_2^{n-1} 2^j (-1) -2 = 2^n (n-1) -2(2^{n-1}-1) \\
&= 2^n(n-2) +2,
\end{align*}
then
$$
\frac {f(n)}{2^n} = \frac {2^n (n-2) + 2 - 2^n n-n}{2^n} = -2 +\frac {2-n}{2^n} \to -2.
$$
Hence the result is
$$
\int_0^1 \log_2 (x)\mathrm d x + 2 = 2 - \frac 1 {\log(2)}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An optimization problem in Euclidean Geometry - finding "the smallest" inscribed triangle What is the smallest perimeter of a triangle that can be inscribed in a triangle with sides of lengths $5$, $9$, and $10$?
I have read that such a triangle has its vertices at the feet of altitudes from the three vertices of the given triangle. (I guess the given triangle must be an acute triangle.) What is an efficient computation for the location of the vertices of the inscribed triangle.
I would appreciate a link to a demonstration that the triangle with its vertices at the feet of altitudes from the three vertices of a given triangle is a triangle with the smallest perimeter that can be inscribed in it.
| Suppose we have a triangle $\Delta ABC$ with $|AB| = c$, $|BC| = a$ and $|AC| = b$. Let $E$ be the foot lying on $AB$, so that $AB \perp EC$. Without loss of generality, assume $b \ge a$. Then
\begin{align*}
|AE|^2 + |EC|^2 &= |AC|^2 = b^2 \\
|BE|^2 + |EC|^2 &= |BC|^2 = a^2 \\
|AE| + |BE| &= |AB| = c
\end{align*}
Subtracting the first two equations, we obtain
$$b^2 - a^2 = |AE|^2 - |BE|^2 = (|AE| - |BE|)(|AE| + |BE|)$$
Using this with the third equation gives us
$$b^2 - a^2 = c(|AE| - |BE|) \implies |AE| - |BE| = \frac{b^2 - a^2}{c}.$$
Using the third equation again allows us to solve for $|AE|$, in particular,
$$2|AE| = c + \frac{b^2 - a^2}{c} \implies |AE| = \frac{b^2 + c^2 - a^2}{2c}.$$
(Note the not-so-subtle or coincidental resemblance to the law of cosines!) So, for example, if we wish to find the position of the foot on the length $5$ edge, we have $c = 5$, $b = 10$, and $a = 9$ (so chosen to preserve $b \ge a$). Thus, we have
$$|AE| = \frac{10^2 + 5^2 - 9^2}{2 \cdot 5} = \frac{22}{5}.$$
The point $A$ is incident on the edges of length $5$ and length $10$. The foot along the length $5$ edge is $\frac{22}{5}$ distance from this point (and $\frac{3}{5}$ distance from the other point).
| {
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Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
| You incorrectly arrive at
$$
\frac{4(x^2-x-2)}{(x^2+2)(x+4)}\ge0
$$
which should be instead
$$
\frac{4(x^2-x-2)}{(x^2+2)(x+4)}\le0
$$
Moreover you do the “illegal” step of removing the denominator, which changes sign at $-4$.
The factors $4$ and $x^2+2$ can be removed because they're positive. Factoring the numerator we obtain
$$
\frac{(x+1)(x-2)}{x+4}\le0
$$
that's the same as
$$
\frac{(x+4)(x+1)(x-2)}{(x+4)^2}\le0
$$
The denominator is positive, so it's irrelevant as far the inequality is concerned, but we have to remember that $-4$ is a value to exclude from solutions, because it makes the expression undefined.
Thus we are reduced to
$$
(x+4)(x+1)(x-2)\le 0\qquad x\ne-4
$$
A polynomial only changes sign at odd multiplicity roots. For large values of $x$ the polynomial $(x+4)(x+1)(x-2)$ is positive. Going backwards on the real line it changes sign at $2$ (becoming negative), then at $-1$ (becoming positive) and finally at $-4$ (becoming negative). Thus it is negative over $(-\infty,-4)$ and $(-1,2)$. Since we allow values where the expression is $0$, the final solution set is
$$
(-\infty,-4)\cup[-1,2]
$$
or, in other notation,
$$
x<-4\qquad\text{or}\qquad -1\le x\le2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Proving $-\cos\alpha = \sin(\alpha-90^\circ)$ from $\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$ $$-\cos\alpha = \sin(\alpha-90^\circ)$$ but
$$\sin (\alpha-\beta) = \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta$$
How does this work out?
$$\begin{align}
\sin\alpha\cdot\cos\beta &=\phantom{-}0\\
\cos\alpha\cdot\sin(-90^\circ)&=-1
\end{align}$$
so I get
$$\sin(\alpha-90^\circ) = \cos\alpha$$
not $-\cos\alpha$
The only way to make this work is by taking the absolute value of $\beta$, but how is that logical if we are dealing with $-90^\circ$?
Thanks!
| Write
$$\sin(\alpha - 90) = \sin\alpha\cos 90 - \sin 90\cos\alpha $$ $$= \sin \alpha \cdot 0 - 1\cdot\cos \alpha = -\cos \alpha.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\int_0^{\infty} \frac{x}{\sinh(\sqrt{3}x)} dx$ I was asked, by a high school student in the UK, how to calculate the following integral:
$$\int_0^{\infty} \frac{x}{\sinh(\sqrt{3}x)} dx.$$
It has been a long time since I have done any calculus and most of my immediate thoughts used mathematics that he is unlikely to have seen before.
I know that the result is $\frac{\pi^2}{12}$ but I am interested in a proof which a (good) high school student would be satisfied by. I do not mind if it goes a little beyond the A-level further maths syllabus, but I would like to avoid having to teach complex analysis or Fourier analysis just to understand this proof.
| Smooth change of variables just to write it in a more "elegant" way.
$$\sqrt{3}x = y$$
The measure becomes $\text{d}y = \sqrt{3}\text{d}x$ and remember that $y = \sqrt{3}x$ which then combines with the other square root. Hence the integral becomes
$$\frac{1}{3}\int_0^{+\infty} \frac{y\ \text{d}y}{\sinh(y)}$$
Now:
$$\sinh(y) = \frac{e^y - e^{-y}}{2}$$
And the integral is rewritten as
$$\frac{2}{3}\int_0^{+\infty} \frac{y}{e^y - e^{-y}} \ \text{d}y$$
Collect the term $e^y$ at the denominator
$$\frac{2}{3}\int_0^{+\infty} \frac{y\ \text{d}y}{e^y(1 - e^{-2y})}$$
And make use of the geometric series for the term $\frac{1}{1 - e^{-2y}} = \sum_{k = 0}^{+\infty} e^{-2yk}$ whence
$$\frac{2}{3}\sum_{k = 0}^{+\infty} \int_0^{+\infty} y e^{-y(2k+1)}\ \text{d}y$$
Notice that the integral is now trivial:
$$\int_0^{+\infty} y e^{-y(2k+1)}\ \text{d}y = \frac{1}{(1+2k)^2}$$
What now remains is a series, which is actually trivial since:
$$\sum_{k = 0}^{+\infty} \frac{1}{(1+2k)^2} \equiv \frac{\pi^2}{8}$$
Combining with the constant and you will get eventually
$$\frac{2}{3}\frac{\pi^2}{8} = \frac{\pi^2}{12}$$
How to show the sum is $\frac{\pi}{8}$
If we write the first terms of the sum, we have:
$$1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81} + \cdots + \frac{1}{n_{\text{disp}}^2}$$
This sum is indeed the sum of all the odd squares. This is a particular sum, and we can see it as the sum of ALL the reciprocal squares, minus the sum of the EVEN reciprocal squares, indeed we can write:
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \frac{1}{64} + \frac{1}{81} + \cdots + \frac{1}{n^2}$$
if we subtract from this the sum of all even reciprocal squares, we obtain exactly our sum. Translated into mathish it's like to have (calling OUR sum $S$)
$$S = \frac{1}{n^2} - \frac{1}{(2n)^2}$$
namely again: our sum is the whole sum of reciprocal squares, minus the sum of all the EVEN reciprocal squares. We can do that simple subtraction:
$$S = \frac{3n^2}{4n^4} = \frac{3}{4n^2}$$
This means that our sum is three quarters the value of the sum of all the reciprocal squares which is a well known series (also it's the Riemann Zeta vaulted in $2$):
$$\sum_{k = 1}^{\infty} \frac{1}{n^2} = \sum_{k = 1}^{+\infty} \frac{1}{(k+1)^2} = \frac{\pi^2}{6}$$
Since our sum is three quarters of that value we get:
$$S = \frac{3}{4}\cdot \frac{\pi^2}{6} = \frac{\pi^2}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Given that $m$ is a real number not less than $-1$ The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.
Here's what I've tried:.
Using Vieta's formulas:
$rs = m^2-3m+3$
$r+s = -2m+4$
Then I set the equation as such; $(r+s)^2 - 2rs = 6$. I got $2m^2-10m+4=0$. Using the quadratic equation, I got $\frac {5+\sqrt {17}}2$ and $\frac {5-\sqrt {17}}2$. Is there anything wrong with my solution?
| It is correct:
$$(r+s)^2 - 2rs = 6 \iff \\
(-2(m-2))^2-2(m^2-3m+3)=6 \iff \\
(4m^2-16m+16)-2m^2+6m-6=6 \iff \\
2m^2-10m+4=0.$$
And the roots are:
$$m_{1,2}=\frac{5\pm \sqrt{17}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the Maximum of the Chi-Square density function Below is a problem which I did and I believe my answer is correct. I would like
somebody to confirm that (if true) and provide some additional comments about my style.
Thanks,
Bob
Problem:
Show that $\chi^2_v$ pdf has a maximum at $v - 2$ if $v > 2$.
Answer:
The $\chi^2_v$ pdf for $x >= 0$ is:
\begin{eqnarray*}
f_(x) &=& \begin{cases}
\frac{ x^{\frac{v}{2} - 1}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)}
& \text{for } x >= 0 \\
0 & \text{otherwise} \\
\end{cases} \\
\end{eqnarray*}
\newline
Now to find its maximum we compute $f'(x)$.
\begin{eqnarray*}
f'(x) &=&
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)} \\
\end{eqnarray*}
Now we set $f'(x) = 0$.
\begin{eqnarray*}
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)} &=& 0 \\
\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}} &=& 0 \\
\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2} &=& 0 \\
\big(-\frac{1}{2}\big)x + \frac{v}{2} - 1 &=& 0 \\
x &=& v - 2 \\
\end{eqnarray*}
Now we know that $x = v - 2$ is an extreme point. The question is it a maximum
or a minimum. Since $f(0) = 0$ and $f(v-2) > 0$ we conclude that $x = v - 2$ is a maximum.
| What you did makes a lot of sense, but what if it $v-2$ was a saddle point?
A better way to see it is using the double derivative test, i.e.
\begin{equation}
f'(x) =
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)}
\end{equation}
Then
\begin{equation}
f''(x)
=
\frac{1}{2^{\frac{v}{2}} \Gamma(v/2)}
\dfrac{x^{\frac{v}{2}-3}\left(x^2+\left(4-2v\right)x+v^2-6v+8\right)\mathrm{e}^{-\frac{x}{2}}}{4}
\end{equation}
At the extremum, we have
\begin{equation}
f''(v - 2)
=
\frac{(v-2)^{\frac{v}{2}-3}}{4(2^{\frac{v}{2}} \Gamma(v/2))}
( (v-2)^2 -2(v-2)(v-2) + (v-2)(v-4) )e^{-\frac{v-2}{2}}
\end{equation}
that is
\begin{equation}
f''(v - 2)
=
\frac{(v-2)^{\frac{v}{2}-3}e^{-\frac{v-2}{2}}}{4(2^{\frac{v}{2}} \Gamma(v/2))}
( -(v-2)^2 + (v-2)(v-4) )
\end{equation}
which is
\begin{equation}
f''(v - 2)
=
-2
\frac{(v-2)^{\frac{v}{2}-3}e^{-\frac{v-2}{2}}}{4(2^{\frac{v}{2}} \Gamma(v/2))}
(v-2)
\end{equation}
For $v > 2$ it is easy to see that $f''(v-2) < 0$, hence it is a maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving Homogenous First Order ODE Using Substitution Problem
Find the general solution of the following homogeneous equation.
$$ ty' = y + \sqrt { t^2 - y^2} $$
Attempt
I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, which suggests rearranging the equation into the form:
$$ y' = ... $$
where $ y = tv $ and $ y' = v + tv' $ (v is the substitution variable).
I haven't been able to get too far in the procedure to make the substitution:
$$ y' = \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{\frac{1}{t}}{\frac{1}{t}} \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{y}{t} + \frac{\sqrt { t^2 - y^2}}{t} $$
Notes
Can someone please help me reduce the RHS of the above equation to the form $ y' = f(\frac{y}{t}) $ (an equation in terms of $ (\frac{y}{t}) $), using the substitution $ y = tv $? For bonus points, please provide a solution to the ODE so myself and other viewers can check their answers!
Thanks in advance!
Solution
With the help of Isham, I was able to get the solution.
$$ ty' = y + \sqrt { t^2 - y^2 } $$
It is implied that $ t^2 - y^2 \geq 0 $, and so $ t^2 \geq y^2 \gt 0 $ or $ \lt 0 $.
$$ ty' - y = \sqrt { t^2 - y^2} $$
$$ \frac {1}{\sqrt {t^2} } (ty' - y) = \frac {1}{\sqrt {t^2} } \sqrt { t^2 - y^2} $$
$$ \frac {1}{t} (ty' - y) = \sqrt { \frac{t^2}{t^2} - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - (\frac{y}{t})^2} $$
$$ y' = \sqrt { 1 - (\frac{y}{t})^2} + \frac{y}{t} $$
Let $ v = \frac{y}{t} $. Then, $ y = tv $ and so $ y' = \frac{dy}{dt} = v + tv' $. Substituting for $ v $ in the above equation we get:
$$ v + tv' = \sqrt {1 - v^2} + v $$
$$ tv' = \sqrt {1 - v^2} $$
Note, since $ t^2 \geq y^2 $, $ \frac{y^2}{t^2} = v^2 \leq 1 $, so there are no equilibrium cases to consider.
$$ t\frac{dv}{dt} = \sqrt {1 - v^2} $$
$$ \frac{1}{\sqrt { 1 - v^2}}dv = \frac{1}{t}dt $$
$$ \int\frac{1}{\sqrt { 1 - v^2}}dv = \int\frac{1}{t}dt $$
$$ arcsin(v) = ln|t| + C_1, C_1 \in \mathbb R $$
$$ \frac {y}{t} = \sin{(ln|t| + C_1)} $$
$$ y = t \sin{(ln|t| + C_1)} $$
| Your substitution is fine.
$$ty' = y + \sqrt { t^2 - y^2}$$
$$ty' - y = \sqrt { t^2 - y^2}$$
$$\frac {ty' - y}{t^2} = \frac 1t \sqrt { 1 - (\frac yt)^2}$$
$$ \left (\frac yt \right )'= \frac 1t \sqrt { 1 - (\frac yt)^2}$$
$$ \int \frac {d\left (\frac yt \right )}{\sqrt { 1 - (\frac yt)^2}}= \int \frac {dt}t $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the relation between coefficients of quadratic equation
My attempt:
I could prove $|c| < 1$.
As it given that $f\lt 1$, so $f(0)\lt 1$.
I have solved using triangular inequality. Is there any other way?
ans is ABCD
| On the boundaries $x=0$ and $x=1$ the inequality corresponds to
*
*$x=0 \implies |c|\le 1$
*$x=1 \implies |a+b+c|\le 1 \implies 0\le|a+b|\le 2$
Let wlog $a \ge 0$, since the extrema values are also reached at $x=-\frac{b}{2a}$ we need to consider $3$ cases:
$1) \quad 0\le-\frac{b}{2a}\le1 \iff -2a\le b\le 0$
*
*$x=1 \implies |a+b+c|\le 1 \implies 0\le|a+b|\le 2 \implies 0\le a \le 2 \land -4\le b \le 0$
*$x=-\frac{b}{2a} \implies \left|\frac {b^2} {4a}-\frac{b^2}{2a}+c\right|\le 1\implies 0\le \left|-\frac{b^2}{4a}\right|\le 2\implies 0\le b^2 \le 8a\implies -2\sqrt {2a}\le b \le 0$
and since
*
*$2\sqrt {2a}=2a \iff \sqrt {2a}=a\iff a=2$
we have
*
*$0\le a \le 2$
*$-4\le b \le 0$
$2) \quad b\ge 0$
*
*$x=1 \implies |a+b+c|\le 1 \implies 0\le a+b \le 2 $
and since $a \ge 0$ we have
*
*$0\le a \le 2$
*$-2\le b \le 2$
$3) \quad b\le -2a $
*
*$x=1 \implies |a+b+c|\le 1 \implies 0\le |a+b| \le 2 \implies -2\le a+b \le 0$
and we have
*
*$0\le a \le 2$
*$-4 \le b \le 0$
Therefore all options are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using Gram-Schmidt to find the QR decomposition I'm having problems doing a QR decomposition of a matrix...
Let $A=\begin{bmatrix}
{1} & {1} & {0} \\
{0} &{1} &{1} \\
{1} & {0} &{1}
\end{bmatrix}$
Find the QR decomposition for A
Here's what I've been doing:
I choose this basis, $B=\left \{(1,0,1), (1,1,0), (0,1,1)\right \}$ (the columns of the matrix).
Now I use the Gram-Schmidt process (and this is where I'm having trouble)
$u_{1}=(1,0,1)$
$u_{2}=(1,1,0)$ (cuz $<(1,0,1), (1,1,0)>=0$)
$u_{3}=(0,1,1)-\frac{<(0,1,1), (1,1,0)>}{<(1,1,0), (1,1,0)>}(1,1,0)-\frac{<(0,1,1), (1,0,1)>}{<(1,0,1), (1,0,1)>}(1,0,1)=$ $(0,1,1)-1/2(1,1,0)-1/2(1,0,1)=(-1, 1/2, 1/2)$
And now I find the norm for all the three vectors:
$||u_{1}||=||u_{2}||=||u_{3}||=\sqrt{2}$
So the orthonormal basis must be $B'= \left \{(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0), (\frac{-1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})\right \}$ (Which it isn't orthonormal)
So $Q=\begin{bmatrix}
{\frac{1}{\sqrt{2}}} & {\frac{1}{\sqrt{2}}} & {\frac{-1}{\sqrt{2}}} \\
{0} &{\frac{1}{\sqrt{2}}} &{\frac{1}{2\sqrt{2}}} \\
{\frac{0}{\sqrt{2}}} & {0} &{\frac{1}{2\sqrt{2}}}
\end{bmatrix}$
Which $Q^{t}Q \neq I$ ($I$ being the identity matrix), so all I did was wrong...
Where's my mistake?
| You're computing $u_2$ wrongly.
I find it useful to set up a systematic way, where the information can be picked up easily.
Let $v_1$, $v_2$ and $v_3$ be the three columns of $A$.
GS1
$u_1=v_1$
$\langle u_1,u_1\rangle=2$
GS2
$\alpha_{12}=\dfrac{\langle u_1,v_2\rangle}{\langle u_1,u_1\rangle}=\dfrac{1}{2}$
$u_2=v_2-\alpha_{12}u_1=\begin{bmatrix}1/2\\1\\-1/2\end{bmatrix}$
$\langle u_2,u_2\rangle=\dfrac{3}{2}$
GS3
$\alpha_{13}=\dfrac{\langle u_1,v_3\rangle}{\langle u_1,u_1\rangle}=\dfrac{1}{2}$
$\alpha_{23}=\dfrac{\langle u_2,v_3\rangle}{\langle u_2,u_2\rangle}=\dfrac{1}{3}$
$u_3=v_3-\alpha_{13}u_1-\alpha_{23}u_2=\begin{bmatrix}-2/3\\2/3\\2/3\end{bmatrix}$
$\langle u_3,u_3\rangle=\dfrac{4}{3}$
Matrix $Q$
The matrix $Q$ has as columns the vectors $u_1$, $u_2$ and $u_3$ divided by their norms:
$$
Q=\begin{bmatrix}
1/\sqrt{2} & 1/\sqrt{6} & -1/\sqrt{3} \\
0 & 2/\sqrt{6} & 1/\sqrt{3} \\
1/\sqrt{2} & -1/\sqrt{6} & 1/\sqrt{3}
\end{bmatrix}
$$
Matrix $R$
The matrix $R$ is obtained by multiplying each row of the upper unitriangular with the entries $\alpha_{ij}$ by the norm of the corresponding $u$ vector.
$$
R=\begin{bmatrix}
1 & 1/2 & 1/2 \\
0 & 1 & 1/3 \\
0 & 0 & 1
\end{bmatrix}
\begin{array}{l}\cdot\sqrt{2}\\\cdot \sqrt{6}/2\\\cdot 2/\sqrt{3}\end{array}=
\begin{bmatrix}
\sqrt{2} & \sqrt{2}/2 & \sqrt{2}/2 \\
0 & \sqrt{6}/2 & \sqrt{6}/6 \\
0 & 0 & 2/\sqrt{3}
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ r^2 = a^2\cos^2(\theta) + b^2\sin^2(\theta) $ [Using pythagoras theorem]
$$ Area = \int_{0}^{\pi/2} \int_{0}^{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}} rdrd\theta $$
$$ = 1/2 \int_{0}^{\pi/2} r^2 \Big|_{0}^{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}} d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} (a^2\cos^2(\theta)+b^2\sin^2(\theta)) d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} ((a^2 - b^2)\cos^2(\theta)+b^2) d\theta $$
$$ = 1/4 \int_{0}^{\pi/2} (a^2 - b^2)(1+ \cos(2\theta)) d\theta +2b^2 d\theta $$
I am getting $$ \pi/8 (a^2 + b^2).$$ But the correct answer is $ \pi ab/4 $
| Your mistake is to believe that $\theta$ is the polar argument. It is not, because
$$\tan\phi=\frac yx=\frac ba\tan \theta\ne\tan\theta.$$
You can fix by taking the differential
$$(\tan^2\phi+1)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$
and substituting
$$\left(\frac{b^2}{a^2}\tan^2\theta+1\right)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$
so that the integral becomes
$$\frac12\int_0^{\pi/2}\frac{\dfrac ba(\tan^2\theta+1)}{\dfrac{b^2}{a^2}\tan^2\theta+1}(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta
\\=\frac{ab}2\int_0^{\pi/2}\frac{\sin^2\theta+\cos^2\theta}{b^2\sin^2\theta+a^2\cos^2\theta}(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$ Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$
My working:
Let $f(x)=\frac{ax^2+x-2}{a+x-2x^2}=y$
therefore,
$$(a+2y)x^2+(1-y)x-(2+ay)=0$$
Now, for $x$ to have real values, the discriminant of the above equation must not be less than zero.
So,
$$(1+8a)y^2+(14+4a^2)y+(1+8a)\ge 0$$
Now, somehow from this, we have to land on the condition for $a$ which I am unable to do.
| So, we need $$(14+4a^2)^2\ge4(1+8a)^2$$
$$\iff(4a^2+14-2-8a)(4a^2+14+2+8a)\ge0$$
$4a^2+14+2+16a=4(a^2+4a+4)=4(a+2)^2\ge0$
So,the sufficient condition is $4a^2+14-2-16a=4(a^2-4a+3)=4(a-1)(a-3)\ge0$
$\implies$ either $a\ge$max$(1,3)$ or $x\le$min$(1,3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$ To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is
\begin{align}
\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T
\end{align}
in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In addition, $(\cdot)^T$ is transposition operation. $\mathbf{G}$ is $u\times u$. We use $\mathbf{G}_u$ to underline the dimension of $\mathbf{G}$.
The question is to determine $\det(\mathbf{G})$.
As I know: We rewrite $\mathbf{G}_u$ as
\begin{align}
\mathbf{G}_u=\left[\begin{array}{ccc}
\mathbf{G}_{u-1} & b\\
b & a+b
\end{array}
\right]
\end{align}
Using the determinant of block matrix lemma
\begin{align}
\det\left[\begin{array}{ccc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}
\end{array}
\right]=\det(\mathbf{A})\det(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})
\end{align}
We then have
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})
\end{align}
It still needs to get $\mathbf{G}_{u-1}^{-1}$ via matrix inverse lemma
\begin{align}
(\mathbf{A}+\mathbf{BC})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1}\mathbf{B}(\mathbf{I}+\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1}
\end{align}
We then have
\begin{align}
\mathbf{G}_{u}
&=\frac{1}{a}\mathbf{I}-\frac{1}{a^2}b\boldsymbol{e}\left(1+\frac{b}{a}\boldsymbol{e}^T\boldsymbol{e}\right)^{-1}\boldsymbol{e}^T\\
&=\frac{1}{a}\mathbf{I}-\frac{b}{a(a+bu)}\boldsymbol{ee}^T
\end{align}
where $\boldsymbol{e}^T\boldsymbol{e}=u$ is used. Plugging it into
\begin{align}
&\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})\\
=&a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)
\end{align}
Then
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\left[{a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)}\right]
\end{align}
Although, the connection between $\mathbf{G}_u$ and $\mathbf{G}_{u-1}$ is found, I can't give the expression of $\mathbf{G}_u$. Please give me hand, thanks a lot!
| Assume $a\ne 0$, We know that
$$\det(A+uv^T)=(1+v^TA^{-1}u)\det(A)$$
Here $A=aI, u = be, v=e$
\begin{align}\det(aI+bee^T)&=(1+be^T(aI)^{-1}e)\det(aI) \\
&=\left(1+\frac{bu}a\right)a^u\\
&=a^{u}+bua^{u-1}\end{align}
If $a=0$ and $u>1$, then the determinant is $0$.
If $a=0$ and $u=1$, then the determinant is $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
$ \frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2}$. Where can I find some more series of this class? After reading Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ and noodling around on wolfram alpha, I discovered
$$ \begin{align}
&\coth(x \pi)=\frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2+n^2}
& \cot(x \pi)= \frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2-n^2} \\
& \text{csch}(x \pi )=\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2+n^2}}
&\csc(x \pi)= \frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2-n^2}} \\
& \tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}}
&\tan(x \pi) = \frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2-4x^2}} \end{align} $$
I suspect (but don't know for sure) that these can all be justified by milking the techniques from the link above.
Go Go Gadget Calculus
We should be able to derive a few more identities. For example,
$$ \Big(\cot(x \pi) \Big)'= \Big(\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{1}{x^2-n^2}} \Big )'$$
$$ \Big(\pi csc(\pi x)\Big)^2= \sum_{n=-\infty}^\infty\frac{x^2+n^2}{(x^2-n^2)^2}$$
And after some manipulations we can find
$$ \frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2} $$
Lovely. We can write then
$$\frac{\pi^2}{9}=\sum_{n=-\infty}^\infty {\frac{1+(6n)^2}{(1-(6n)^2)^2}}$$
I have a feeling at this point that this must be a well-studied subject and I wonder where I can find some more identities of this class. Does anyone have a link/resource where I can read more on these. I don't really need their derivations if they are just the techniques of the link above + elementary calculus techniques. I am just looking for a well organized list that I can refer to.
| There is some nice things to see here:
$$\tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}}$$
$$\Big(\tanh(x \pi) \Big)'= \Big( \frac{1}{\pi}\sum_{n=-\infty}^\infty{\frac{4x}{(2n+1)^2+4x^2}} \Big)'$$
$$ \pi\operatorname{sech}^2(\pi x)=\frac{1}{\pi}\sum_{n=-\infty}^{\infty}\frac{4\left(4n^2+4n-4x^2+1\right)}{\left(4n^2+4n+4x^2+1\right)^2}$$
Which is nice: taking $x=0$ we have
$$\pi^2=\sum_{n=-\infty}^{\infty}\frac{4}{4n^2+4n+1}=8\Big(1+\sum_{n=1}^\infty{\frac{1}{4n^2+4n+1}}\Big)$$
Interesting. I wonder if we can use this to demonstrate Takebe Kenko's somewhat similar looking (this can be found- with a tiny typo in denominator- on the last page of this):
$$\pi^2=4\Big(1+\sum_{n=1}^{\infty} \frac{2^{2n+1}(n!)^2}{(2n+2)
!} \Big) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Coefficient of a generating function I have the following generating function,
$$f(x)=(1+x+x^2+x^3)^4$$
I want to find the $[x^5]$ coefficient, to do so I wrote,
$$[x^5]f(x)=[x^5](1+x+x^2+x^3)^4=[x^5]\left(\frac{1-x^4}{1-x}\right)^4=
[x^5](1-x^4)^4(1-x)^{-4}
$$
I am wondering how to proceed from here?
|
We obtain
\begin{align*}
\color{blue}{[x^5]}&\color{blue}{(1-x^4)^4(1-x)^{-4}}\\
&=[x^5]\sum_{j=0}^\infty\binom{-4}{j}(-x)^{j}(1-x^4)^4\tag{1}\\
&=\sum_{j=0}^5\binom{3+j}{3}[x^{5-j}](1-x^4)^4\tag{2}\\
&=\sum_{j=0}^5\binom{8-j}{3}[x^{j}]\sum_{k=0}^4\binom{4}{k}(-1)^kx^{4k}\tag{3}\\
&=\binom{8}{3}\binom{4}{0}-\binom{4}{3}\binom{4}{1}\tag{4}\\
&\,\,\color{blue}{=40}
\end{align*}
Comment:
*
*In (1) we apply the binomial series expansion to $(1-x)^{-4}$.
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We set the upper limit of the series to $5$ since other terms do not contribute to $[x^5]$.
*In (3) we change the order of summation $j\to 5-j$ and expand the binomial.
*In (4) we select the coefficients of $x^0$ and $x^4$. Other terms do not contribute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Consider the equation $a_n +b_n\sqrt{2} = (1+\sqrt{2})^n$ where $a_n, b_n \in \mathbb{Z} \ge 1$. Prove that $\gcd(a_n, b_n) = 1$. Consider the equation $a_n +b_n\sqrt{2} = (1+\sqrt{2})^n$ where $a_n, b_n \in \mathbb{Z} \ge 1$. Prove that $\gcd(a_n, b_n) = 1$.
I know that $(1+2\sqrt{3}) = 1^3 + 3(1)^2(\sqrt{2})^2 + 3(1)(\sqrt{2})^2 + (\sqrt{2})^3 = 7+5\sqrt{2}$
so $a_n = 7, b_n = 5$ by binomial expansion.
So my strategy is to prove this via mathematical induction. The base case is easy. What I need for the inductive step is to get the relation between $(a_n+1, bn+1)$ to $(a_n, b_n)$. So we get:
$$a_{n+1}, b_{n+1}\sqrt{2} = (1+\sqrt{2})(1+\sqrt{2})^n = (1+\sqrt{2})(a_n +b_n\sqrt{2}) = (a_n + 2b_n)+(a_n+b_n)\sqrt{2}$$
So we have $a_{n+1} = a_n + 2b_n, b_{n+1} = a_n + b_n$
So now I'm trying to use the euclidean algorithm to show that $\gcd(a_{n+1}, b_{n+1}) = 1$. So we get:
$\gcd(a_{n+1}, b_{n+1}) = (a_n + 2b_n,a_n + b_n)=$
$a_n+2b_n= 1 *(a_n + b_n) + b_n =$
$a_n + b_n = 1 * (b_n) + a_n$
$b_n = 0 * (a_n) + bn$
But, where do I go from here?
Any help would be appreciated!
-IdleMathGuy
| We have
$$
\pmatrix{a_{n+1} \\ b_{n+1}}
=
\pmatrix{1 & 2 \\ 1 & 1}
\pmatrix{a_n \\ b_n}
$$
and so
$$
\pmatrix{a_n \\ b_n}
=
\pmatrix{1 & 2 \\ 1 & 1}^n
\pmatrix{a_0 \\ b_0}
=
\pmatrix{1 & 2 \\ 1 & 1}^n
\pmatrix{1 \\ 0}
$$
Therefore, $\pmatrix{a_n \\ b_n}$ is the first column of $\pmatrix{1 & 2 \\ 1 & 1}^n$. Write
$$
\pmatrix{1 & 2 \\ 1 & 1}^n
=
\pmatrix{a_n & c_n \\ b_n & d_n}
$$
Taking determinants, we get
$$
(-1)^n = a_n d_n - b_n c_n
$$
which implies $\gcd(a_n, b_n) = 1$.
(It turns out and easily proved by induction that $d_n=a_n$ and $c_n=2b_n$, which is nice but not relevant here.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16 \\
9^2 &= \phantom{1}8\cdot10+1 = 81 \\
15^2 &= 14\cdot16+1 = 225
\end{align}$$
and so on.
Then, to know any number raised to the power of $2$, you can multiply the previous number $(x-1)$ by the next one $(x+1)$, and add $1$.
So, to know the squared root of a number like $64$, you have to substract $1$ ($63$) and look for two numbers multiplied are $63$ and subtracted are $2$:
$$x \cdot y = 63 \qquad x - y = 2$$
Solving the equation you get $9$ and $7$ (or $-7$ and $-9$). The number between these is the square root ($8$).
I don't know if this apply for power of $3$.
Does this fact/theorem/relation has a name or something?
| A similar formula for $x^3$ is $$x^3=(x-1)(x^2+x+1)+1$$
You can extend this to higher powers
$$x^n= (x-1)(x^{n-1}+x^{n-2}+...+1)+1$$
Which is the basis for the sum of geometric series formula.
$$1+x+x^2+...+x^{n} = \frac {x^{n+1}-1}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
A specific homogeneous polar differential equation In an assignment of our school, we are asked to solve
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y - 3}{x - y - 1}$$
by turning it into a homogeneous polar differential equation (equation of the form $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = F\left(\frac{y}{x}\right)$) using substitutions $x = X + a$, $y = Y + b$. My solution was:
Firstly, I determined substitutions $x = X + 2$, $y = Y + 1$, such that
$$\frac{\mathrm{d}Y}{\mathrm{d}X} = \frac{X + Y}{X - Y}$$
Then, let $Y = Xv$, thus
$$\begin{aligned}
v + X\frac{\mathrm{d}v}{\mathrm{d}X} &= \frac{X + Xv}{X - Xv} \\
\int \frac{1 - v}{1 + v^2}\ \mathrm{d}v &= \int \frac{\mathrm{d}X}{X}
\end{aligned}$$
The left-hand side, specifically, gives
$$\int \frac{\mathrm{d}v}{1 + v^2} - \int \frac{v}{1 + v^2}\ \mathrm{d}v = \tan^{-1} v - \frac{\ln \left(1 + v^2\right)}{2} + \mathrm{constant}$$
Therefore,
$$\tan^{-1} v - \frac{\ln \left(1 + v^2\right)}{2} = \ln \left|X\right| + \mathrm{constant}$$
i.e.
$$2\tan^{-1} \frac{y - 1}{x - 2} = \ln \left[1 + \frac{\left(y - 1\right)^2}{\left(x - 2\right)^2}\right] + \ln \left(x - 2\right)^2 + \mathrm{constant}$$
However, our assignment didn't come with a standard solution, so I verified my answer with Wolfram Alpha, which gives
$$
2 \tan^{-1}\left(\frac{y(x) + x - 3}{-y(x) + x - 1}\right) = c_1 + \ln\left(\frac{x^2 + y(x)^2 - 2 y(x) - 4 x + 5}{2 \left(x - 2\right)^2}\right) + 2 \ln\left(x - 2\right)$$
which is different from my solution in
*
*the fraction inside function $\tan^{-1}$ is vastly different
*the denominator given by Wolfram Alpha inside the first $\ln$ is twice the denominator I gave
*the $x - 2$ in the last $\ln$ has no absolute value sign around it, but this seems a common problem of Wolfram Alpha solutions, so we can overlook it for the second
May I know whether I'm wrong, or that this is a problem of the Wolfram Alpha solution? (or that the two solutions are actually equivalent, though seemingly very unlikely?)
| Attached a plot showing the solutions
$$
S_1\to 2 \tan ^{-1}\left(\frac{x+y-3}{x-y-1}\right)=\log \left(\frac{x^2-4 x+y^2-2 y+5}{2 (x-2)^2}\right)+2 \log (x-2)+C_0\\
S_2\to \tan ^{-1}\left(\frac{y-1}{x-2}\right)-\frac{1}{2} \log \left(\frac{(y-1)^2}{(x-2)^2}+1\right)=\log (x-2)+C_1
$$
with $C_0=0, C_1 = -1.13$
I hope this helps.
In red $S_1$ and in blue $S_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$ I was solving questions on beta and gamma functions and then I came across this question.
$$ \int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx $$
Generally in questions of beta functions integrals of the type $\int x^n(1−x^m) \, dx$ can be solved by substituting $x^m = t$, but in this case substitution is unconventional because if I put $1+x^4=t$, then I will not get the required form of beta function
I tried substituting $x^2 = \tan t$ and also tried substituting $x^2 = \cos t$ but on simplifying each of them further I could not reduce them to standard form of beta function.
| Apply the substitution $x^4 = \frac{1-u}{1+u}$. Then
$$ I := \int_{0}^{1} \frac{(1-x^4)^{3/4}}{(1+x^4)^2} \, dx
= \frac{1}{2^{9/4}} \int_{0}^{1} u^{3/4}(1-u)^{-3/4} \, du
= \frac{1}{2^{9/4}} B(\tfrac{7}{4},\tfrac{1}{4})
= \frac{3\pi}{2^{15/4}}.$$
Explanations. First substitute $x^4 = t$ to obtain $ I = \frac{1}{4} \int_{0}^{1} \frac{(1-t)^{3/4}t^{-3/4}}{(1+t)^2} \, dx$. To transform this integral into the form of beta integral, we notice that
*
*$-1$, $0$ and $1$ are the only branch points/poles of the integrand,
*$0$, $1$, $\infty$ are the only branch points/poles of the integrand of $\int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1} \, du$.
This leads to consider a Möbius transformation $t = f(u) = \frac{au+b}{cu+d}$ satisfying either
*
*$f(\infty) = -1$, $f(0) = 0$ and $f(1) = 1$ and $ad-bc > 0$, or
*$f(\infty) = -1$, $f(0) = 1$ and $f(1) = 0$ and $ad-bc < 0$.
The first condition yields $f(u) = \frac{u}{2-u}$ and the second condition yields $f(u) = \frac{1-u}{1+u}$. Both can be used, since one condition is simply transformed to the other by $z \mapsto 1-z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by
$$\left(-17\right)^2\geq \left(x\sqrt{8-x^2}+x\sqrt{25-x^2}\right)^2.$$
But I think this is not the best approach.
| First $x^2 \le 8$. Second, continuing from what you have written, i.e.,
$$\frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x,$$
we have
$$\frac{-17}{\sqrt{8-0^2}+\sqrt{25-0^2}} \ge \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}} \ge x \implies x \le \frac{-17}{\sqrt{8}+5} \implies x^2 > 8.$$
Contradiction! So no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $n^3+(n+1)^3>(n+2)^3$ Here is the question: for which natural numbers do the following inequalities hold true? State a claim and prove
it:
$n^3+(n+1)^3>(n+2)^3$.
I think this statement holds when $n\ge6$, where $n^3+(n+1)^3=559$, and $(n+2)^3=512$.
I thought about using induction, but I couldn't express $(n+1)^3+(n+2)^3$ in terms of the inductive hypothesis- $n^3+(n+1)^3$.
So I'm wondering if there's a better way than induction to prove it or how to use the inductive hypothesis in a useful way given that induction is a good way to prove this.
| Need to show:
$n^3\gt \dfrac{(n+2)^3-(n+1)3}{1}=$
$=3(t^2)$, $t \in (n+1,n+2)$
(MVT)
Hence:
$3(n+2)^2 \gt (n+2)^3-(n+1)^3$.
For $n \ge 6:$
$n^3 > 3(n+2)^2$.
Since:
The inequality is true for $n=6$.
$f(x):= x^3 -3(x+2)^2$.
$f'(x) = 3x^2 -6(x+2) =$
$3(x^2-2x -4) = $
$3[( x-1)^2 -5] >0$, for $x \ge 6$,
hence $f(x)$ is strictly increasing
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What are the strict local maximisers of $x^T A x$ in the simplex? I am dealing with a matrix in this form.
$A=\begin{pmatrix}
0&a & a & a & c & c &c & c\\
a& 0& a &a & c& c& c& c\\
a& a &0 &a & c& c & c & c\\
a& a& a &0 & c &c &c &c \\
c& c & c & c & 0& b& b &b \\
c& c & c & c& b&0 & b & b\\
c& c & c & c & b & b &0 & b\\
c& c & c & c & b& b& b &0
\end{pmatrix}$
I have $a >b >0.5$ and $c <0.5$.
Want to maximize $x^T A x$, where $\sum_{i=1}^{8} x_i=1$ and all $x_i \ge 0$ (.i.e, x belongs to the simplex). How to prove that strict local maximizers of the quadratic form are only $x_1^*=(1/4,1/4,1/4,1/4, 0, 0, 0,0)$ and
$x_2^*=(0,0,0,0, 1/4, 1/4, 1/4,1/4)$. I can see it intutively as there are 4 blocks in the matrix, but how to prove it.
| Consider the gradient of $x^{T}Ax$. Since $A$ is symmetric, the gradient is $2Ax$. Let $x = [x_1\ x_2\ x_3\ldots x_7\ x_8]^{T}$, $l = x_1 + x_2 + x_3 + x_4$ and $r = x_5 + x_6 +x_7 + x_8$, then the gradient is actually (stupid notation..
$$[2(a(l - x_i) + cr)_{i=1, 2, 3, 4} \ 2(cl + b(r - x_i))_{i = 5, 6, 7, 8}]^T$$
Since $a, b > 0$, we can show that any local maximizer $x$ must have $x_1 = x_2 = x_3 = x_4$ and $x_5 = x_6 = x_7 = x_8$. To see this, suppose for example that $x_3 > x_2$. Then $2(a(l - x_3) + cr) < 2(a(l - x_2) + cr)$. This means we should trade $x_3$ for $x_2$ locally (without changing other $x_i$'s) and get larger $x^{T}Ax$. (To show the formally, you can choose to project the gradient to the line where $x_2 + x_3 = 0$ and other dimension being $0$ and see that the gradient is pointing into the simplex.)
Once this is done, the question reduces to deciding the value of $l, r$ as $x_1 = x_2 = x_3 = x_4 = l/4$ and $x_5 = x_6 = x_7 = x_8 = r/4$. The gradient also becomes $[\frac{3}{4}al + cr \ cl + \frac{3}{4}br]^T$. Now discuss cases: there are two end points: $l = 0, r = 1$ and $l = 1, r = 0$, and for other points, we want the gradient to be perpendicular to the $l + r = 1$ line: the two component of the gradient should be equal.
When $l = 0, r = 1$, the gradient is $[c, \frac{3}{4}b]^T$. For this to be a local maximizer, we want $c \le \frac{3}{4}b$. In other words, $4c - 3b \le 0$.
When $l = 1, r = 0$, the gradient is $[\frac{3}{4}a, c]^T$. For this to be a local maximizer, we want $c \le \frac{3}{4}a$. In other words, $4c - 3a \le 0$.
For other points, we want $\frac{3}{4}al + cr = cl + \frac{3}{4}br$. Since $l + r = 1$, we can solve this equation and get:
\begin{align*}
l &= \frac{4c - 3b}{(4c - 3b) + (4c - 3a)} \\
r &= \frac{4c - 3a}{(4c - 3b) + (4c - 3a)}
\end{align*}
We get solutions with $l, r > 0$ exactly when $4c - 3a>0$ and $4c - 3b > 0$.
| {
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Evaluate the integral with and without Green's theorem Evaluate the line integral $\oint_C y^2dx + xdy$ when $C$ has the vector equation $\alpha(t)=(2\cos^3t)i+(2\sin^3t)j$, $0 \leq t \leq 2\pi$.
My attempt
BY USING GREEN's THEOREM,
i.e., $P=y^2$, $Q=x$
we get $\iint_c 1-2y dxdy$
Now putting $x=r\cos^3t$ and $y=r\sin^3t$
we get $dxdy=Jdrdt$, where $J=3r \sin^2t \cos^2t$
hence $dxdy=3r \sin^2t \cos^2t drdt$
Hence the required integral is,
$\int_0^{2\pi} \int_0^2 (1-2r \sin^3t)(3r \sin^2t \cos^2t)drdt$
which is equal to $\pi /4$
But the answer given in APOSTOL's is $3 \pi /2$
NOW BY NORMAL METHOD
put $x=2 \cos^3t$ and $y=2 \sin^3t$
$dx=-6 \cos^2t \sin t dt$ and $dy=6 \sin^2t cost dt$,
Hence the integral becomes $-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$ which is equal to ZERO.
WHAT MISTAKE AM I DOING AND THE CORRECT ANSWER IS $3 \pi /2$ ?
| Your second method
Use $$\int_0^{2\pi}\sin^n x\ dx=\dfrac{2\pi}{2^{2k}}{2k\choose k}$$
for even $n=2k$ and for odd $n$ it is zero, then
$$-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t (1-\sin^2t) dt + 12 \int_0^{2 \pi} (1-\sin^2t)^2\sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t\ dt + 24\int_0^{2 \pi} \sin^9t\ dt + 12\int_0^{2 \pi} \sin^2t\ dt -24 \int_0^{2 \pi} \sin^4t\ dt + 12\int_0^{2 \pi} \sin^6t\ dt $$
$$0+0+12\dfrac{2\pi}{2^{2}}{2\choose 1} -24\dfrac{2\pi}{2^{4}}{4\choose 2} + 12\dfrac{2\pi}{2^{6}}{6\choose 3}=\color{blue}{\dfrac{3\pi}{2}}$$
| {
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Solution for $\frac{3}{x-9} \gt \frac{2}{x+2}$ What I did:
$\frac{3}{x-9} \gt \frac{2}{x+2}$
$3(x+2) \gt 2(x-9)$
$3x+6 \gt 2x-18$
$x \gt -24$
When typing this in in symbolab, it showed me that the solution is $-24 \lt x \lt -2$ or $x \gt 9$
What did i do wrong ?
How come i didnt get the correct solution ?
| HINT
The first step
$$\frac{3}{x-9} \gt \frac{2}{x+2}\iff 3(x+2) \gt 2(x-9)$$
is wrong since $x-9$ and $x+2$ are not both positive (or both negative) in general.
To solve properly use that
$$\frac{3}{x-9} \gt \frac{2}{x+2}\iff \frac{3}{x-9} - \frac{2}{x+2}>0\iff\frac{3(x+2)-2(x-9)}{(x-9)(x+2)}>0$$$$\iff\frac{x+24}{(x-9)(x+2)}>0$$
and study separetely the sign for numerator and denominator.
| {
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Integral $\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}$ Greetings I saw here (among the last integrals) that: $$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi C$$
Where $C$ is Catalan's constant. Did this integral appear here before? (my quick search did not found anything).
I gave it a try and got stuck. Denoting the integral as $I$ and using that $\ln \left(\frac ab\right)=\ln a- \ln b\ $ we have: $$I=K(1,1)-K(1,-1)-K(-1,1)+K(-1,-1)$$ Where $$K(a,b)=\int_0^1\frac{\ln(1+ax)\ln(1+bx^2)}{x}dx$$ Differentiating under the integral sign: $$\frac{\partial^2}{\partial a \partial b}K(a,b)=\int_0^1 \frac{x^2}{(1+ax)(1+bx^2)}\,dx$$ By partial fractions we get: $$\frac{1}{a^2+b}\left(\int_0^1 \frac{ax}{bx^2+1}\,dx -\int_0^1 \frac{1}{bx^2+1}\,dx +\int_0^1 \frac{1}{ax+1} \,dx\right)$$
$$=\frac{1}{a^2+b}\left(\frac{a\ln(1+b)}{2b}-\frac{\arctan \left(\sqrt b\right)}{\sqrt{b}} +\frac{\ln(1+a)}{a}\right) $$ And now since $K(0,b)=K(a,0)=0$ $$K(a,b)=\frac12\int_0^a \int_0^b \frac{x\ln(1+y)}{y(x^2+y)}\,dy\,dx-\int_0^a \int_0^b \frac{\arctan \left(\sqrt y\right)}{\sqrt{y}(x^2+y)}\,dy\,dx +\int_0^a \int_0^b \frac{\ln(1+x)}{x(x^2+y)}\,dy\,dx$$ Is there a clever way to solve this?
Another way is to start by using: $$-\frac12\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^\infty \frac{x^{2n+1}}{2n+1}$$ $$I=4\sum_{n,k=1}^\infty \frac{1}{(2n+1)(2k+1)}\int_0^1 x^{4n+2k+2}\,dx=4\sum_{n,k=1}^\infty \frac{1}{(2n+1)(2k+1)(4n+2k+3)}$$ But I dont know how to deal with this series. I would appreciate some help with this integral!
| Since the integrand is an even function, we can write
$$ I
= \frac{1}{2}\int_{-1}^{1} \log\left(\frac{1-x}{1+x}\right)\log\left(\frac{1-x^2}{1+x^2}\right)\,\frac{dx}{x}. $$
Now deforming the line contour $[-1, 1]$ to the semicircular contour from $-1$ to $1$ and substituting $x = e^{i\theta}$,
$$ I = -\frac{i}{2} \int_{0}^{\pi} \log(-i\tan(\theta/2)) \log(-i\tan \theta) \, d\theta, $$
where we utilized the identity $\frac{1-e^{i\theta}}{1+e^{i\theta}} = -i\tan(\theta/2)$. Now we note that, for $\theta \in (0, \pi/2) \cup (\pi/2, \pi)$,
*
*$\log(-i\tan(\theta/2)) = \log\tan(\theta/2) - \frac{i\pi}{2}$,
*$\log(-i\tan\theta) = \log\lvert\tan\theta\rvert - \operatorname{sign}(\tan\theta)\frac{i\pi}{2}$.
*$\int_{0}^{\pi} \log\lvert\tan\theta\rvert \, d\theta = 2 \int_{0}^{\pi/2} (\log\sin\theta - \log\cos\theta) \, d\theta = 0$.
Plugging these back and taking real parts only (since we know that $I$ is real),
\begin{align*}
I
&= -\frac{\pi}{4} \int_{0}^{\pi} \left( \log\lvert\tan\theta\rvert + \operatorname{sign}(\tan\theta)\log\tan(\theta/2) \right) \, d\theta \\
&= -\frac{\pi}{2} \int_{0}^{\pi/2} \log\tan(\theta/2) \, d\theta \\
&= -\pi \int_{0}^{1} \frac{\log u}{1+u^2} \, du, \qquad (u=\tan(\theta/2)) \\
&= \pi C.
\end{align*}
Generalization. Utilizing a similar idea s in the computation above, we can prove that
Proposition. Let $p$, $q$ be positive integers. Write $g = \gcd(p,q)$ and assume that $p/g$ and $q/g$ are not simultaneously odd. Then
\begin{align*}
&\int_{0}^{1} \log\left(\frac{1-x^p}{1+x^p}\right)\log\left(\frac{1-x^q}{1+x^q}\right)\,\frac{dx}{x} \\
&\hspace{6em} = \pi \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \left( \frac{1}{p}\tan\left((2n+1)\frac{\pi p}{2q}\right) + \frac{1}{q} \tan\left((2n+1)\frac{\pi q}{2p}\right)\right)
\end{align*}
Of course, the above can be simplified further by using either Hurwitz zeta function or trigamma function
.
| {
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"timestamp": "2023-03-29T00:00:00",
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Pre-calculus Complex coordinate law of cosine help
Let $\theta = \angle BAC$. Then we can write $\cos \theta = \dfrac{x}{\sqrt{2}}$
Find x.
Currently, that is all I have. I think I should use Law of cosine. Where should I progress?
| $$a = BC = \sqrt{2^2+3^2}=\sqrt{13}$$
$$b = AC = \sqrt{3^2+1^2}=\sqrt{10}$$
$$c = AB = \sqrt{2^2+1^2}=\sqrt{5}$$
so $$\cos \theta = {b^2+c^2-a^2\over 2bc} = {1\over 5\sqrt{2}}\implies x=1/5$$
| {
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"source": "stackexchange",
"question_score": "2",
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How to do $\lvert x - 1 \rvert < \lvert x-3 \rvert$
Solve for $x$ such that $$\lvert x - 1 \rvert < \lvert x-3 \rvert$$
I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work?
| Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $\ge 0$ and $x-3$ can be $<$ or $\ge 0$.
Or in other words $x$ can be $<$ or $\ge 1$ or $x$ can be $< $ or $\ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x \ge 3$ both) or redundant (if $x\ge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases. $x < 1; 1 \le x < 3; x\ge 3$.
Case 1: $x < 1$. Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-x\implies 1 < 3$ ... which is always true. This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2: $1 \le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x \implies 2x < 4\implies x < 2$. This means if $1 \le x < 3$ then we know further than $x < 2$ and that $2\le x < 3$ is impossible.
Case 3: $x \ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3\implies -1 < -3$. That is simply impossible. So we know that $x \ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$. 2) $x$ could be any value so that $\le x < 2$ but $x$ can not be $2 \le x < 3$. And 3) $x \ge 3$ is impossible.
Combining those three results we get .... $x < 2$. And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.
If $x \ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$ and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 < 3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$\sqrt {k^2} = |k|$ and if $0 \le a$ and $0 \le b$ then $a^2 < b^2 \iff a < b$.
So $\sqrt {(x-1)^2} = |x-1| < |x - 3| = \sqrt{(x-2)^2} \iff $
$(x-1)^2 < (x-3)^2 \iff$
$x^2 -2x + 1 < x^2 - 6x + 9 \iff$
$4x < 8 \iff$
$x < 2$.
| {
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How to find the slant asymptote for $y=\frac{x^2\arctan\left(x\right)}{3x+3}$ I got stuck at the end and wonder if you can give some guide on how to proceed to find the slant asymptote?
$$y=\frac{x^2\arctan\left(x\right)}{3x+3}$$
$$y= kx+m\\
\frac{f(x)}{x} \rightarrow k\\
f(x) - kx \rightarrow m $$
Solution:
$$\lim _{x\to \infty }\left(\frac{\frac{x^2\arctan\left(x\right)}{3x+3}}{x}\:\right)=\lim _{x\to \infty }\left(\frac{x^2\arctan\left(x\right)}{x(3x+3)}\:\right)=\lim _{x\to \infty }\left(\frac{x^2\arctan\left(x\right)}{(3x^{2\:\:}+3x)}\:\right)\\
=\lim _{x\to \infty }\left(\frac{x^2}{x^2} \cdot\frac{\arctan\left(x\right)}{(3+\frac{3}{x})}\:\right)=\frac{\pi}{6}\Longrightarrow k=\frac{\pi}{6}\\
f(x)-kx \implies \frac{x^2\arctan\left(x\right)}{3x+3} - \frac{\pi}{6} \cdot {x}\\
\frac{6\cdot x^2\arctan -3\pi{x^2} - 3\pi {x}}{18(x+1)}$$
Searching for $m$.
| $\arctan(x)=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)=\frac{\pi}{2}-\frac{1}{x}+O\left(\frac{1}{x^3}\right)$ for $x\to \infty$, hence
$$ \lim_{x\to +\infty}\left[\frac{x^2\arctan(x)}{3x+3}-\frac{\pi}{6}x\right]=\lim_{x\to +\infty}-\frac{(2+\pi ) x}{6 (1+x)}=-\frac{\pi+2}{6} $$
and the equation of the slant asymptote is $y=\frac{\pi}{6}x-\frac{\pi+2}{6}$.
| {
"language": "en",
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"source": "stackexchange",
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What is $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$? I am trying to evaluate $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$. I have found that $15^{15} \equiv 1 \pmod{7}$ and that $16^{16} \equiv 2 \pmod{7}$.
To evaluate $15^{15} \pmod{17}$, I did the following:
$$15 = 2 \times 7 + 1 \equiv 1 \pmod{7}$$
$$15^{15} \equiv 1 \pmod{7}$$
Then, to evaluate $16^{16}$, I wrote:
$$16 = 15 + 1 \equiv 1 + 1 = 2 \pmod{7}$$
$$16^{16} \equiv 2^{16} \pmod{7}$$
$$2^{3} = 8 = 7+1 \equiv 1 \pmod{7}$$
$$2^{16} = 2^{3} \times 2^{13} \equiv 2^{13} = 2^{3} \times 2^{10} \equiv 2^{10} \equiv \dots \equiv 2 \pmod{7}$$
Nevertheless, I have not managed to figure out how to evaluate $17^{17}$. How should I go about this and is my overall approach for evaluating the sum in question a good one?
| Euler's Theorem states that
$$ a^{\phi(n)}\equiv1 \pmod{n}$$
For all $n$, where $\phi(n)$ is the Euler totient function, denoting the number of positive integers less than $n$ relatively prime to $n$. For primes, $\phi(n)=n-1$. Thus
$$ a^{n-1}\equiv1\pmod n. $$
This result is also known as Fermat's little theorem. Now, $7$ is prime, so anything to the power of $6$ is congruent to $1$. We have
$$\begin{split}
&15^{15}+16^{16}+17^{17}+18^{18}+19^{19}+20^{20}\\
\equiv\, & 15^3 + 16^4 + 17^5 + 18^0 + 19^1 + 20^2 \\
\equiv \, & 1^3 + 2^4 + 3^5 + 4^0 + 5^1 + 6^2\\
\equiv\, & 1+16+243+1+5+36\\
\equiv\, & 302\\
\equiv\, & 1\pmod{7}.
\end{split}$$
| {
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How To Evaluate $ \int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx $? $$
\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx
$$
What I tried was
*
*To convert $N^r$ into $sin2x$
*To Use Identity $(a+b)^3 = (a^3 + b^3)(a^2 + b^2 - ab) $
But none of them proved useful ?
How do I evaluate this Integral ?
| Hint
Start with: $\sin x=\frac{\tan x}{\sec x}$ and $\sec^2 x =\tan^2 x +1$. Then,
$$\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx = \int \sec^2x\,\frac{\tan^2x}{\left(\tan x+1\right)^2\left(\tan^2x-\tan x +1\right)^2}\,\mathrm{d}x.$$
Now use $u=\tan x$. Then use $t=\left(u+1\right)\left(u^2-u+1\right)$, sucht that
$$\int \frac{\sin^2x\cdot\cos^2x}{(\sin^3x + \cos^3x)^2}\ dx =\frac{1}{3}\int\frac{\mathrm{d}t}{t^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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what is the characteristic polynomial and minimal polynomial of A and B? What is the characteristic polynomial and minimal polynomial of $A$ and $B$ ?
$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},B=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{pmatrix}$
My attempt : for $A$) , $ch_A(x) = (x-a)^4$ , $m_A = (x-a)^2(x-a)$
for $B)$ $ch_B(x) = (x-a)^4$ , $m_B = (x-a)^2(x-a)^2$
where $ch$ and $m$ denote the characteristic and minimal polynomial.
Is my answer is correct or not ???
Any hints/solution will be apprciated
thanks u
| $A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},$ first i break them into two jordan nlock forms that $P=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$ and $Q=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$
now , the characteristic and minimal polynomial of $P = (\lambda -a)^2=f(t)$
and the characteristic and minimal polynomial of $Q = (\lambda -a)^2=g(t)$
Now the minimial polynomial of A $m_A = Lcm(f(t),g(t))=lcm\{(\lambda -a)^2,(\lambda -a)^2\}=(\lambda -a)^2$
similarly for B matrix we get $m_B = Lcm(f'(t),g'(t))=lcm\{(\lambda -a)^2,(\lambda -a)^2\}=(\lambda -a)^2$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Counting number of occurences Let $1\leq k\leq n$. We find all finite sequences of positive integers with sum $n$. Suppose that the term $k$ occurs a total of $T(n, k)$ times in all these sequences.
Find $T(n, k)$.
As an example to make things more clear: $T(3,1)=5$, $T(3,2)=2$ and $T(3,3)=1$
This on base of the sums $1+1+1=3$, $1+2=3$, $2+1=3$ and $3=3$.
In this I'm pretty sure we can do this by choosing $r$ $x$'s and letting them be equal to $k$ individually, in groups of $2$ and so on. We can then sum it over $r$. But I'm unable to write a solution and obtain an expression for $T(n,k)$.
Thanks in advance!
| With $q$ the length of the sequence and $k$ marked with $u$ we get the
generating function
$$G(z, u) = \sum_{q\ge 0} (z+z^2+\cdots+uz^k+\cdots)^q
\\ = \sum_{q\ge 0} z^q (1+z+\cdots+uz^{k-1}+\cdots)^q
\\ = \sum_{q\ge 0} z^q \left(\frac{1}{1-z} + (u-1)z^{k-1}\right)^q
\\ = \frac{1}{1-z(1/(1-z)+(u-1)z^{k-1})}
\\ = \frac{1-z}{1-z-z(1+(u-1)(1-z)z^{k-1})}.$$
The desired quantity is given by
$$[z^n] \left.
\frac{\partial}{\partial u} G(z, u)\right|_{u=1}
\\ = [z^n] \left. \frac{1-z}{(1-z-z(1+(u-1)(1-z)z^{k-1}))^2}
(z (1-z) z^{k-1})\right|_{u=1}
\\ = [z^n] \frac{1-z}{(1-2z)^2} (z (1-z) z^{k-1})
\\ = [z^n] \frac{1-2z+z^2}{(1-2z)^2} z^k
= [z^{n-k}] \frac{1-2z+z^2}{(1-2z)^2}.$$
We thus have for $k=n$ the value one, corresponding to the composition
$n=n$, for $k=n-1$ the value
$$[z^1] \frac{1-2z+z^2}{(1-2z)^2} = 4 - 2 = 2$$
corresponding to the compositions $1 + (n-1) = (n-1) + 1$ and for
$k\le n-2$
$$(n-k+1) \times 2^{n-k} - (n-k) \times 2^{n-k}
+ (n-k-1) \times 2^{n-k-2}
\\ = 2^{n-k} + (n-k-1) \times 2^{n-k-2}.$$
for a closed form of
$$\bbox[5px,border:2px solid #00A000]{
2^{n-k-2} (n-k+3).}$$
We clearly obtain zero when $k\gt n.$
| {
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Using Laplace transforms to evaluate$\int_{0}^{\infty}\frac{\sin^2(x)}{x^2(x^2 + 1)} dx$ Recently I've been playing around with Feynman's Trick to evaluate integrals. Obviously, one of it's many great features is that it allows derivatives to make expressions simpler. I was wondering whether Laplace Transforms could equally be applied.
I'm not qualified to say the following is proper or rigorous, it was just an experiment.
Consider
$$I = \int_{0}^{\infty}\frac{\sin^2(x)}{x^2(x^2 + 1)}\, \mathrm dx.$$
Let
$$I(t) = \int_{0}^{\infty}\frac{\sin^2(tx)}{x^2(x^2 + 1)} \,\mathrm dx$$
Take the Laplace Transform to yield
\begin{align*}
\mathscr L[I(t)] &= \int_{0}^{\infty}\frac{\mathscr L[\sin^2(tx)]}{x^2(x^2 + 1)}\,\mathrm dx\\
&= \int_{0}^{\infty}\frac{\mathscr 1}{x^2(x^2 + 1)}\frac{2x^2}{s(s^2 + 4x^2)}\,\mathrm dx\\
&= \frac{2}{s}\int_{0}^{\infty}\frac{1}{(x^2 + 1)(4x^2 + s^2)}\,\mathrm dx.
\end{align*}
Splitting via Partial Fraction Decomposition we arrive at
\begin{align*}
\mathscr L[I(t)] &= \frac{2}{s(s^2 - 4)}\int_{0}^{\infty}\left[ \frac{1}{x^{2} + 1} - \frac{4}{4x^{2} + s^2}\right] \,\mathrm dx\\
&= \frac{2}{s(s^2 - 4)}\left[\arctan(x) - \frac{2}{s}\arctan\left(\frac{2x}{s}\right)\right]_{0}^{\infty}\\
&= \frac{2}{s(s^2 - 4)}\left[\frac{\pi}{2} - \frac{2}{s}\frac{\pi}{2} \right]\\
&= \frac{\pi}{s^2(s + 2)}
\end{align*}
And so
$$I(t) = \mathscr L^{-1}\left[\frac{\pi}{s^2(s + 2)}\right] = \pi\left[\frac{t}{2} + \frac{e^{-2t}}{4} - \frac{1}{4}\right]$$
Hence,
$$I(1) = \pi\left[\frac{1}{2} + \frac{e^{-2}}{4} - \frac{1}{4}
\right] = \frac{\pi}{4}\left[1 + e^{-2}\right]$$
which is correct. I'm unsure if this is mere luck or whether this is a viable method.
Has anyone used this method before?
| Noting that the iterated integral
$$I_1=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dt\right)\,dx$$
is finite, the Fubini-Tonelli Theorem guarantees that the iterated integral formed by interchanging the order of integration,
$$I_2=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dx\right)\,dt$$
is also finite with
$$I_1=I_2$$.
But $I_2$ is the Laplace Transform of $\int_0^\infty \frac{\sin^2(tx)}{x^2(x^2+1)}\,dx$. Hence, we assert that
$$\begin{align}
\mathscr{L}\left(\int_0^\infty \frac{\sin^2(tx)}{x^2(x^2+1)}\,dx\right)(s)&=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dt\right)\,dx\\\\
&=\int_0^\infty\left(\frac2s \frac{1}{(x^2+1)(4x^2+s^2)}\right)\,dx
\end{align}$$
which agrees with the development in the OP!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Find the coefficient of $x^{10}$ We have been given the following function.
$f(x)$= $x$ +$x^2$ + $x^4$ + $x^8$ + $x^{16}$ + $x^{32}$ + ...upto infinite terms
The question is as follows:
What is the coefficient of $x^{10}$ in $f(f(x))$?
I tried solving it myself and I found the answer too but the method of solving was too much time-consuming.
I had solved it manually by only considering the first four terms of $f(f(x))$. This method took me about 10 minutes.
But the problem is that this question was asked in a competitive exam called JEE which requires solving the question in max. 3-4 minutes.
So, I wanted to know if there was a faster method to solve this problem.
Thanks in advance.
| $f(x)$ contains no tenth power.
$f^2(x)$ has $x^2\cdot x^8$, with a coefficient $2$.
$f^4(x)$ has $x\cdot x\cdot x^4\cdot x^4$ with a coefficient $\dfrac{4!}{2!2!}=6$ and $x^2\cdot x^2\cdot x^2\cdot x^4$ with a coefficient $\dfrac{4!}{3!1!}=4$.
$f^8(x)$ has $x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x^2\cdot x^2$ appearing $\dfrac{8!}{6!2!}=28$ times.
Total $40.$
Using a CAS,
$$\cdots+40x^{10}+22x^9+16x^8+8x^7+8x^6+6x^5+3x^4+2x^3+2x^2+x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$4ac-b^2\leq 3a(a+b+c)$ in quadratic Suppose that the polynomial $(b+c)x^2+(a+c)x+(a+b)$ doesn't have real roots, where $a,b,c\in\mathbb{R}$. Prove that $4ac-b^2\leq 3a(a+b+c)$.
The quadratic not having real roots means that $$(a+c)^2-4(b+c)(a+b)<0$$ which translates to $(a^2+2ac+c^2)-4(b^2+ab+ac+bc)<0$, or
$$a^2+c^2-2ac-4b^2-4ab-4bc<0$$
which is still quite far from the inequality in question. We need to eliminate the $c^2$ term, which might be possible using a square form like $(c-b)^2\geq 0$, but it doesn't really get us closer.
| We have $$(a+c)^2<4(a+b)(b+c)$$ or
$$a^2+2ac+c^2<4b^2+4ab+4ac+4bc$$ or
$$c^2+a^2+4b^2-2ac-4bc+4ab<8ab+8b^2$$ or
$$(c-a-2b)^2<8b(a+b),$$ which gives
$$a^2(c-a-2b)^2\leq8a^2b(a+b).$$
In another hand, we need to prove that
$$4ac-b^2\leq 3a^2+3ab+3ac$$ or $$a(c-a-2b)\leq2a^2+ab+b^2,$$ which is obvious for $a(c-a-2b)<0.$
Let $a(c-a-2b)\geq0.$
Thus, it's enough to prove that
$$a^2(c-a-2b)^2\leq(2a^2+ab+b^2)$$ for which it's enough to prove that
$$8a^2b(a+b)\leq(2a^2+ab+b^2)^2,$$ which is true by AM-GM:
$$(2a^2+ab+b^2)^2\geq4\cdot2a^2(ab+b^2)=8a^2b(a+b).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$
The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\sqrt{1+\tan^2\theta}}=\frac{1}{\sqrt{1+\frac{81}{4}}}=\frac{2}{\sqrt{85}}
$$
$$
\cos\theta=\frac{\sin\theta}{\tan\theta}=\frac{\frac{3}{4}}{\frac{9}{2}}=\frac{3}{4}.\frac{2}{9}=\frac{1}{6}
$$
Why is it getting confused here ?
| This is an impossible situation. If you use the identity
$$\tan(\theta) = {\sin(\theta)\over\cos(\theta)},$$
you get $\cos(\theta) = 1/6$.
However, $\cos(\theta) = \sqrt{7}/4$ by the Pythagorean identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ (both arbitrary functions)
To make more clear what I'm talking about I will give a example.
Rewrite $f(x,y) = 1+x-y-xy$ as $g(x)h(y)$
If we choose $g(x) = (1+x)$ and $h(y) = (1-y)$ we have
$f(x,y) = g(x) h(y) \implies (1+x-y-xy) = (1+x)(1-y)$
I'm trying to do the same with $f(x,y) = 1-x^2y^2 = (1-xy)(1+xy)$.
New question:
Is there also a contradiction for $f(x,y) = \frac{xy}{1-x^2y^2}$ ? Or it's possible to write $f(x,y) $ as $g(x)h(y)$ ?
| Without loss of generality let $f(x) = ax^2 + b$ and $g(y) = cy^2 + d$. Now
$$
1-x^2y^2 = f(x)g(y) = acx^2y^2 +adx^2 = cby^2 +bd
$$
By comparing terms we obtain:
\begin{align}
ad&=0\\
cb&=0\\
ac&=-1\\
bd&=1
\end{align}
By $bd=1$ we obtain $a=c=0$. Hence this problem has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
How do I evaluate the integral $\int_0^1\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$. stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was define the S to be equal to $$ \int_0^1 (1-x^3 +x^5-x^8+x^{10}-x^{13}+........) dx $$
I simplified this and got the above integral I tried to do partial fraction but did not succeed.
| If your purpose is to evaluate
$$ S=\sum_{k\geq 0}\left[\frac{1}{10k+1}-\frac{1}{10k+4}+\frac{1}{10k+6}-\frac{1}{10k+9} \right]$$
you do not need an indefinite integral, just an integral over $(0,1)$: that's a huge difference, in some cases.
Actually, since $1+9=4+6=10$, you may just invoke the identities
$$ \sum_{k\geq 0}\left[\frac{1}{10k+1}-\frac{1}{10k+9}\right]=\frac{\pi}{10}\cot\frac{\pi}{10} $$
$$ \sum_{k\geq 0}\left[\frac{1}{10k+4}-\frac{1}{10k+6}\right]=\frac{\pi}{10}\cot\frac{4\pi}{10} $$
which follow from Herglotz' trick / the reflection formula for the digamma function.
In particular
$$ S = \frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I use the definition of partial derivative to get $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2 + y^2)^2}$? $\\f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2}, & \text{if $(x,y) \ne (0,0)$} \\
0, & \text{if $(x,y) = (0,0)$}
\end{cases}$
Using the definition of partial derivatives:
\begin{align*}
f_x(x,y) &= \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}
\\&= \lim_{h\to 0} \frac{\frac{(x+h)y}{(x+h)^2 + y^2}-\frac{xy}{x^2 + y^2}}{h}\\&= \lim_{h\to 0} \frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}
\end{align*}
I'm not sure how to get rid of the variable $h$. I wanted to know exactly how I could further simplify in order to get $\frac{y(y^2-x^2)}{(x^2 + y^2)^2}$?
| Notice that
\begin{align}&\phantom{ = }\frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&= \frac{\color{red}{xy(x^2 +y^2)} + hy(x^2 +y^2) \color{red}{-xy(x^2 + y^2)} - xy(2hx+ h^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&=\frac{hx^2y+hy^3 -2hx^2y-h^2xy}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&=\frac{hy(y^2 -x^2-hx)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&=\frac{y(y^2 -x^2-hx)}{(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}\\
&\to \frac{y(y^2-x^2)}{(x^2+y^2)^2},
\end{align}
as $h\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the residue with a Laurent series expansion. I have a question about the following problem:
"Detect the error in the following argument. The function $f(z)=\frac{1}{z(z-1)^2}$ has an isolated singularity at $z=0$. The Laurent series is $f(z)=\frac{1}{(z-1)^3}-\frac{1}{(z-1)^4}+\frac{1}{(z-1)^5}-...$
for $|z-1|>1$. Apparently $z=1$ is an essential singularity with residue 0."
Now I know that the error is that one has to compute the Laurent series on the annulus $0<|z-1|<1$, but why is this the case? Do you always have to take the inner annulus to compute the residue?
A clear explanation would be much appreciated!
| We consider the expansion of $f$ at $z_0=1$ since the center $1$ is indicated by the terms $\frac{1}{(z-1)^k}$.
The function
\begin{align*}
f(z)=\frac{1}{z(z-1)^2}=\frac{1}{z}-\frac{1}{z-1}+\frac{1}{(z-1)^2}\tag{1}
\end{align*}
is to expand around the center $z_0=1$. Since there are isolated singularities, namely a single pole at $z=0$ and a double pole at $z=1$ we have to distinguish two regions
\begin{align*}
D_1:&\quad 0<|z-1|<1\\
D_2:&\quad |z-1|>1
\end{align*}
*
*The first region $D_1$ is a punctured disc with center $z_0=1$, radius $1$ and the pole $0$ at the boundary of the disc.
In the interior we have a representation of the fractions with a pole at $z=1$ as principal part of a Laurent series at $z_0=1$, while the fraction with pole at $z=0$ admits a representation as power series.
*The other region $D_2$ containing all points outside the closure of $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=1$.
The expansion of $f$ as Laurent series at $z=1$ in $D_1$:
We obtain
\begin{align*}
\color{blue}{f(z)}&\color{blue}{=\frac{1}{z(z-1)^2}}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{1+(z-1)}\\
&\color{blue}{=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\sum_{n=0}^\infty (-1)^n(z-1)^n}\tag{2}\\
\end{align*}
The expansion of $f$ as Laurent series at $z=1$ in $D_2$:
We obtain
\begin{align*}
\color{blue}{f(z)}&\color{blue}{=\frac{1}{z(z-1)^2}}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{1+(z-1)}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z-1}\frac{1}{1+\frac{1}{z-1}}\\
&=\frac{1}{(z-1)^2}-\frac{1}{z-1}+\frac{1}{z-1}\sum_{n=0}^\infty (-1)^n\frac{1}{(z-1)^n}\\
&\color{blue}{=\sum_{n=3}^\infty (-1)^{n+1}\frac{1}{(z-1)^n}}\tag{3}\\
\end{align*}
Conclusion:
*
*We see two valid series expansions (2) and (3) of $f$ at $z=1$. One is in the punctured disc $D_1$ and the other in the region $D_2$.
*In order to determine the type of singularity at $z=1$ we have to consider the region near the singularity. This means we can check the Laurent series expansion in $D_1$ but not that in $D_2$. From the series expansion (2) we clearly see that $z=1$ is a pole of order $2$. This can also be immediately deduced from the representation (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help find the mistake in this problem of finding limit (using L'Hopital)
Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$
Attempt
\begin{align*}
&\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\
= &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right)\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right).
\end{align*}
Both the terms are in $\frac00$ form. So applying L'Hopital on both the limits we have,
$$= \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\sin{x}}{x\cos{x}+\sin{x}}\right).$$
The second term is in $\frac00$ form. So applying L'Hopital on the second limit we have,
\begin{align*}
= &\lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& 1
\end{align*}
The correct answer is $\dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?
| $$\lim_{x\to0}\left(\dfrac1{x^2}-\dfrac1{\tan^2x}\right)=\lim_{x\to0}\dfrac{\tan x-x}{x^3}\cdot\lim_{x\to0}\dfrac{\tan x+ x}x\cdot\left(\lim_{x\to0}\dfrac x{\tan x}\right)^2$$
Now the limit for the last two are too simple for L'hospital
For $\lim_{x\to0}\dfrac{\tan x-x}{x^3}$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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What is the coefficient of the term $x^3y^5$, as a result of the binomial expansion of the following term? We have the term
$(1+xy+y^2)^n$
If we expand it using the binomial theorem, why is the factor of the term $x^3y^5$ the following: $4{n\choose 4}$? (The binomial coefficient n choose 4 multiplied by 4)
| It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^3y^5]}&\color{blue}{(1+xy+y^2)}\\
&=[x^3y^5]\sum_{j=0}^n\binom{n}{j}(xy)^j(1+y^2)^{n-j}\tag{1}\\
&=[y^5]\binom{n}{3}y^3(1+y^2)^{n-3}\tag{2}\\
&=\binom{n}{3}[y^2]\sum_{j=0}^{n-3}\binom{n-3}{j}y^{2j}\tag{3}\\
&=\binom{n}{3}\binom{n-3}{1}\tag{4}\\
&=\frac{n(n-1)(n-2)}{3!}\cdot(n-3)\\
&=4\cdot\frac{n(n-1)(n-2)(n-3)}{4!}\\
&\,\,\color{blue}{=4\binom{n}{4}}
\end{align*}
Comment:
*
*In (1) we apply the binomial theorem to $(xy+(1+y^2))^n$. This way we can easily select the coefficient of $x^3$.
*In (2) we select the coefficient of $x^3$.
*In (3) we use the formula $[y^{p-q}]A(y)=[y^p]y^qA(y)$ and apply the binomial theorem again.
*In (4) we select the coefficient of $y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n$ Given the following series:
$$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n,$$ where $$a_n = \frac{2^2.4^2\cdots(2n)^2}{3.4.5.6\cdots (2n+2)} x^{2n+2}$$
Now we try Ratio test first: $$\frac{a_{n+1}}{a_n} = \frac{2n^2+4n+2}{n+2}x^2$$
but then I cant conclude from that, how to proceed next?
| Apply Ratio Test
$$
\begin{align}
\frac{2^2}{3\cdot4}x^4 + \frac{2^2\cdot4^2}{3\cdot4\cdot5\cdot6}x^6+\cdots
&=\sum_{k=2}^\infty\frac{(k-1)!^22^{2k-2}}{(2k)!/2!}x^{2k}\\
&=\frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}}\\
\end{align}
$$
The Ratio Test says
$$
\begin{align}
\lim_{k\to\infty}\frac{a_{k+1}}{a_k}
&=\lim_{k\to\infty}\frac{k^2}{(k+1)^2}\frac{4x^2}{\frac{(2k+2)(2k+1)}{(k+1)^2}}\\
&=\lim_{k\to\infty}\frac{4k^2}{(2k+2)(2k+1)}x^2\\[9pt]
&=x^2
\end{align}
$$
Therefore, the series converges for $|x|\lt1$.
The Case $\boldsymbol{|x|=1}$
Using the estimate $(9)$ given in this answer, we have
$$
\begin{align}
\frac12\frac{4^k}{k^2\binom{2k}{k}}
&\le\frac12\frac{4^k}{k^2\frac{4^k}{\sqrt{\pi(k+1/3)}}}\\
&=\frac{\sqrt{\pi(k+1/3)}}{2k^2}\\[9pt]
&\le\frac{\sqrt{\pi/3}}{k^{3/2}}
\end{align}
$$
So the series converges for $|x|=1$.
Applying Raabe's Test
Another approach for the case $|x|=1$ is to use Raabe's Test:
$$
\begin{align}
\lim_{k\to\infty}k\left(\frac{a_k}{a_{k+1}}-1\right)
&=\lim_{k\to\infty}k\left(\frac{(2k+2)(2k+1)}{4k^2}-1\right)\\
&=\lim_{k\to\infty}k\left(\frac{6k+2}{4k^2}\right)\\[3pt]
&=\frac32
\end{align}
$$
which is greater than $1$, so the series converges for $|x|=1$.
The Actual Sum
Applying the substitution $x\mapsto4x^2$ to this answer gives
$$
\sum_{k=0}^\infty\frac{(2x)^{2k}}{\binom{2k}{k}}
=\frac1{1-x^2}\left[1+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]
$$
Integrating once gives
$$
\frac12\sum_{k=0}^\infty\frac{(2x)^{2k+1}}{(2k+1)\binom{2k}{k}}
=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}
$$
Integrating again gives
$$
\frac14\sum_{k=0}^\infty\frac{(2x)^{2k+2}}{(2k+2)(2k+1)\binom{2k}{k}}
=\frac{\sin^{-1}(x)^2}{2}
$$
Therefore,
$$
\begin{align}
\frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}}
&=\frac12\sum_{k=1}^\infty\frac{(2x)^{2k+2}}{(2k+2)(2k+1)\binom{2k}{k}}\\[6pt]
&=\sin^{-1}(x)^2-x^2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with induction proof I need help with the following induction proof which I am not sure if I am doing correctly.
$$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$$
I check for $n=1$ (Base case)
$$\frac{1}{1\cdot3}=\frac13$$
$$\frac{1}{2\cdot1+1}=\frac13$$
Now, is this the correct next step in my proof?
$$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
I we assume it is correct for $n=k$ then it is also true for $n=k+1$ which means that the RHS must be equal to the LHS.
| Your idea is correct but $$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1))(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
Should have been $$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
Good luck finishing it up.
| {
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"url": "https://math.stackexchange.com/questions/2972591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $26^{32}\bmod 12$? What is the correct answer to this expression:
$26^{32} \pmod {12}$
When I tried in Wolfram Alpha the answer is $4$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $0.$
| First, note that $26 \equiv 2 \pmod {12}$, so $26^{32} \equiv 2^{32} \pmod {12}$.
Next, note that $2^4 \equiv 16 \equiv 4 \pmod {12}$, so $2^{32} \equiv \left(2^4\right)^8 \equiv 4 ^8 \pmod {12}$, and $4^2 \equiv 4 \pmod {12}$.
Finally, $4^8 \equiv \left(4^2\right)^4 \equiv 4^4 \equiv \left(4^2\right)^2 \equiv 4^2 \equiv 4 \pmod {12}$.
Then we get the result.
There are slicker solutions with just a few results from elementary number theory, but this is a very basic method which should be easy enough to follow.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $A=\sum_{n=1}^{3027}\sin(n\pi/2018)$ and $B=\sum_{n=1}^{3027}\cos(n\pi/2018)$, evaluate $A(1-\cos(\pi/2018))+B\sin(\pi/2018)$
Let
$$A=\sum_{n=1}^{3027}\sin\frac{n\pi}{2018} \qquad B=\sum_{n=1}^{3027}\cos\frac{n\pi}{2018}$$
Evaluate
$$A\left(1-\cos\frac{\pi}{2018}\right)+B\sin\frac{\pi}{2018}$$
I don't know how to do it. I used a smaller case. I changed $3027 \to 3$ and changed the denominator by $2$, and I noticed that the answer is $-1$. I don't know how to solve the bigger case.
| I'll generalize the problem a bit, writing $m$ for $3027$ and $2\theta$ for $\pi/2018$.
Adapting the identities proven here, we can write
$$\begin{align}
A &:= \sum_{n=1}^{m}\sin 2n\theta = \frac{\sin(m+1)\theta}{\sin\theta}\;\sin m\theta \\[4pt]
B &:= \sum_{n=1}^{m}\cos 2n\theta = \frac{\sin(m+1)\theta}{\sin\theta}\;\cos m\theta - 1
\end{align}$$
where we subtract $\cos 0=1$ from the cosine sum (and $\sin 0=0$ from the sine sum!) because our summations start at index $1$ instead of $0$. Then we have
$$\begin{align}
A \left(1-\cos 2\theta\right) + B\sin 2\theta
&= \phantom{+}A\cdot 2 \sin^2\theta + B\cdot 2\sin\theta\cos\theta \tag{1}\\[4pt]
&= \phantom{+}2\sin\theta\;\sin(m+1)\theta\;\sin m\theta \\
&\phantom{=}+ 2 \cos\theta\;\sin(m+1)\theta\;\cos m\theta - 2\sin\theta\cos\theta \tag{2}\\[4pt]
&= \phantom{+}2\sin(m+1)\theta\;\left(\sin\theta\sin m\theta+\cos\theta\cos m\theta\right)-\sin 2\theta \tag{3}\\[4pt]
&= \phantom{+}2\sin(m+1)\theta\;\cos(m-1)\theta -\sin 2\theta \tag{4} \\[4pt]
&= \phantom{+}\left( \sin 2m\theta + \sin 2\theta \right) - \sin 2\theta \tag{5} \\[4pt]
&= \phantom{+} \sin 2m\theta \tag{6}
\end{align}$$
where $(1)$ uses the double-angle identities, $(4)$ uses the angle-difference identity for cosine, and $(5)$ invokes a somewhat-lesser-known product-to-sum identity.
Restoring the specific values for the problem at hand,
$$2\theta = \frac{\pi}{2018} \qquad m = 3027 \qquad\to\qquad 2m\theta = \frac{3\pi}{2} \qquad\to\qquad \sin 2m\theta = -1 \tag{$\star$}$$
It's a bit of an identity slog, but there it is. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that for each $n \geq 1$ it holds that $2^{n-1} F_n \equiv n \pmod{5}$ I want to show that for each $n \geq 1$ it holds that:
$$2^{n-1} F_n \equiv n \pmod{5}$$
where $F_n$ is the $n$-th Fibonacci number.
Could you give a hint how we can show this?
The sequence $F_n$ of Fibonacci numbers is defined by the recurrence relation:
$$F_n=F_{n-1}+F_{n-2} \\ F_1=1, F_2=1.$$
Right? Do we use the definion in order to get to the desired result?
| Use the general term formula:
$$2^{n-1} F_n \equiv 2^{n-1}\cdot \frac{\phi^n-\psi^n}{\sqrt{5}} \equiv \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2\sqrt{5}} \equiv\\
\frac{2{n\choose 1}\sqrt{5}+2{n\choose 3}(\sqrt{5})^3+2{n\choose 5}(\sqrt{5})^5+\cdots +2{n\choose 2\lfloor{\frac{n-1}{2}\rfloor}+1}(\sqrt{5})^{2\lfloor{\frac{n-1}{2}\rfloor}+1}}{2\sqrt{5}}\equiv\\
n+{n\choose 3}\cdot 5+{n\choose 5}\cdot 5^2+\cdots +{n\choose 2\lfloor{\frac{n-1}{2}\rfloor}+1}\cdot 5^{\lfloor{\frac{n-1}{2}\rfloor}}\equiv
n \pmod{5}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Formal proof of the sequence involving double factorials $x_n = \frac{(2n)!!}{(2n-1)!!}$ is not bounded. I'm trying to prove the following:
Let $n\in \mathbb N$ and
$$
x_n = \frac{(2n)!!}{(2n-1)!!}
$$
Where:
$$
(2n)!! = 2\cdot 4 \cdot 6 \cdot \dots \cdot 2n \\
(2n - 1)!! = 1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)
$$
Prove $\{x_n\}$ is not bounded.
Intuitively it feels like it is unbounded. Let's consider the following fraction:
$$
x_n = \frac{2\cdot 4 \cdot 6 \cdot \dots \cdot (2n)}{1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) }
$$
Now if we consequently take a number from nominator and denominator we get:
$$
\frac{2}{1} > 1 \\
\frac{4}{3} > 1 \\
\frac{6}{5} > 1 \\
\dots \\
\frac{2n}{2n-1} > 1
$$
So the fraction is a product of rational numbers each of which is greater than $1$ and the product of rational numbers greater than $1$ is increasing.
I've tried to formalize that by expanding $(2n)!!$ and $(2n-1)!!$:
$$
(2n)!! = (2n)(2n-2)(2n-4)\cdots(4)(2) = 2^kn!\\
(2n-1)!! = (2n-1)(2n-3)\cdots(5)(3)(1) = \\
= \frac{(2n-1)(2n-2)(2n-3)\cdots(5)(4)(3)(2)(1)}{(2n-2)(2n-4)\dots(4)(2)} = \\
\frac{(2n-1)!}{2^{n-1}(n-1)!}
$$
So using the above:
$$
x_n = \frac{2^nn!2^{n-1}(n-1)!}{(2n-1)!} = \frac{2^{2n-1}n!(n-1!)}{(2n-1)!}
$$
Here is where I got stuck. How do i proceed with the proof using some constant $M$ and some number $N > n$ such that $x_N > M$? Should i introduce some inequality?
I've seen a similar question, but the sequence there is proven to be divergent which i guess is more a calculus concept (yet very similar to (un)boundedness), and i'm in search of a precalculus solution.
| As simple as it gets:
LEMMA: For $k\in N$ and $k>1$
$$\frac{2k}{2k-1}\gt\frac{\sqrt{3k+1}}{\sqrt{3k-2}}\tag{1}$$
Proof: It's trivial. Both sides are positive, square them:
$$\frac{4k^2}{4k^2-4k+1}\gt\frac{3k+1}{3k-2}$$
..which is equivalent to:
$$4k^2(3k-2)\gt(4k^2-4k+1)(3k+1)$$
$$12k^3-8k^2\gt12k^3+4k^2-12k^2-4k+3k+1$$
$$12k^3-8k^2\gt12k^3-8k^2-k+1$$
$$0\gt-k+1$$
...which is obviously true.
End of lemma proof.
By using the inequality (1) we have:
$$\frac{2}{1}=\frac{\sqrt{4}}{\sqrt{1}}$$
$$\frac{4}{3}\gt\frac{\sqrt{7}}{\sqrt{4}}$$
$$\frac{6}{5}\gt\frac{\sqrt{10}}{\sqrt{7}}$$
$$...$$
$$\frac{2n-2}{2n-3}\gt\frac{\sqrt{3n-2}}{\sqrt{3n-5}}\tag{1}$$
$$\frac{2n}{2n-1}\gt\frac{\sqrt{3n+1}}{\sqrt{3n-2}}\tag{1}$$
Multiply all this and you get:
$$a_n=\frac{(2n)!!}{(2n-1)!!} > \sqrt{3n+1}$$
So $a_n$ is obviously unbounded. Cute, isn't it? :)
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$
Given a sequence
$$
\begin{cases}
x_{n+1} = \frac{x_n^4 + 1}{5x_n} \\
x_1 = 2 \\
n \in \mathbb N
\end{cases}
$$
Prove it has lower bound at $1\over 5$ and upper bound at $2$
I've tried to find a closed form for the recurrence relation, but couldn't arrive at anything. Also:
$$
x_{n+1} = \frac{x_n^4+1}{5x_n}=\frac{2(x_n^4 +1)}{2\cdot5x_n}=\frac{2}{5x_n}\cdot\frac{x_n^4+1}{2} \ge\frac{2}{5x_n}\sqrt{x_n^4\cdot1} =\frac{2x_n}{5}
$$
So I got:
$$
x_{n+1} \ge \frac{2x_n}{5} \tag1
$$
I have no ideas how to proceed. I'm not even sure it's valid to use AM-GM here.
So my main questions are:
*
*Does this recurrence have a closed form?
*What else should I try to solve the problem?
Please note this is precalculus. I'm not allowed to use calculus.
Update
Using $(x_n^2 - 1)^2 > 0$ i get the same result as in $(1)$. Expanding the terms only shows that the sequence is greater than $0$:
$$
x_{n+1} \ge 2\cdot \left(2\over 5\right)^n
$$
which is tending to $0$ with growing $n$.
Update 2
Consider the following expressions:
$$
x_1 = 2 \\
x_2 = \frac{x_1^3}{5} + \frac{1}{5x_1} \\
\dots \\
x_{n+1} = \frac{x_n^3}{5} + \frac{1}{5x_n} \\
$$
Multiply both sides of each expression by some $z$ in the power of $n$:
$$
z\cdot x_1 = 2\cdot z \\
z^2\cdot x_2 = \left(\frac{x_1^3}{5} + \frac{1}{5x_1}\right)z^2 \\
\dots \\
z^{n+1}\cdot x_{n+1} = \left(\frac{x_n^3}{5} + \frac{1}{5x_n}\right) \cdot z^{n+1} \\
$$
Now sum them up:
$$
\sum_{k=1}^{n+1}x_k\cdot z^k = 2z + {1 \over 5}\left( \sum_{k=2}^{n+1}x_{k-1}^3z^k + \sum_{k=2}^{n+1}{z^k\over 5x_{k-1}} \right) = \\
= 2z + {1\over 5z} \left(\sum_{k=1}^{n}x_k^3z^k + \sum_{k=1}^{n}{z^k\over x_k}\right)
$$
Now define:
$$
G(z) = \sum_{k=1}^{n+1}x_k\cdot z^k
$$
From this point there may be a way to express RHS in terms of $G(z)$ but i couldn't handle that.
Update 3
This goes beyond precalculus level but anyway here is another observation inspired by @amam_Abdallah.
Define $x_{n+1} = f(x_n)$ if this function have fixed points then:
$$
\overline{x} = \frac{\overline{x}^3}{5} + \frac{1}{\overline{x}} \iff \\
\iff \overline{x} = \sqrt[^3]{5\overline{x} - {1\over \overline{x}}}
$$
This equation has two solutions:
$$
\overline{x} = \sqrt{{5\over 2} \pm {\sqrt{21} \over 2}}
$$
Perhaps this will lead someone to ideas on how to use that fact.
Update 4
Some more thoughts on the sequence:
$$
x_{n+1} = \frac{1}{5}x_n^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}x_{n-1}^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}\left(\frac{1}{5}x_{n-2}^3 + \frac{1}{5x_{n-2}} \right)^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \dots
$$
Can this somehow be wrapped into something in the form of $\prod \dots$ or $\sum \dots$?
| Hint:
This is not a detailed answer, but this plot says all.
| {
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Prove that we can't find effective bounds on the point guaranteed by the Mean Value Theorem. I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:
For each real number $M$ and each real number $\xi$ that lies strictly between $0$ and $1$, construct a function $f$ such that
$$f(0)=0,\; f(1)=M,\;f\text{ is continuous on }[0,1],\; f\text{ is differentiable on }(0,1),\;\text{ and }\xi\text{ is the unique point strictly between 0 and 1 such that}\;f'(\xi)=M\,.$$
For the $M\neq 0$ and $\xi\neq 1/e$ case, we can show that
$$g(x)=\begin{cases}
0&\text{ if }x=0,\\
1/e&\text{ if }x=1\\
1&\text{ if }x=\infty,\\
\sqrt[1-x]{x}&\text{ otherwise}
\end{cases}$$
is strictly increasing and continuous on $[0,\infty]$. Thus there is a unique positive $\alpha$ such that $g(\alpha)=\xi$. In turn, we can define $f(x)=Mx^\alpha$ which will satisfy the claim. For the $M\neq 0$ and $\xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+x\ln(x))$.
For $M=0$, we first choose $\alpha\geq 1$ and $\beta\geq 1$ such that $\frac{\alpha}{\alpha+\beta}=\xi$. We then define $f(x)=x^\alpha(1-x)^\beta$ which will satisfy the claim.
My question however is this:
Can we construct such an $f$ to be a polynomial?
An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $\xi$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).
| Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $\xi \in (0, 1)$ for which $f'(\xi) = 0$, then $\frac{1}{3} \leq \xi \leq \frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $\xi = \frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $\xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(\xi) = 3\xi^2 + 2k \xi - (1+k) = 0$ are thus $$\xi = \frac{k \pm \sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 \xi - k)^2 = k^2 + 3k + 3$, or $$k = \frac{1 - 3 \xi^2}{2 \xi - 1}$$ but the squaring means that $\xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $\xi$, up to scaling of the coefficients, is $$f(x) = (2 \xi - 1) x^3 + (1 - 3 \xi^2) x^2 + (3 \xi^2 - 2\xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 \xi - 1) x^2 + 2(1 - 3 \xi^2) x + (3 \xi^2 - 2\xi) = 0$ add up to $\frac{2 (3 \xi^2 - 1)}{3 (2 \xi - 1)}.$ If $\xi$ is one solution, then the other solution (call it $\xi'$) is \begin{align*} \xi' &= \frac{2 (3 \xi^2 - 1)}{3 (2 \xi - 1)} - \xi \\ &= \frac{6 \xi^2 - 2}{6 \xi - 3} - \frac{6 \xi^2 - 3 \xi}{6 \xi - 3} \\ &= \frac{3 \xi - 2}{6 \xi -3} \\
&= \frac{3 \xi - \frac{3}{2}}{6 \xi - 3} - \frac{\frac{1}{2}}{6 \xi - 3} \\
&= \frac{1}{2} - \frac{1}{12\xi - 6}.\end{align*}
Thus, $\xi' \notin (0, 1)$ if and only if $|12 \xi - 6| \leq 2$, i.e., if $\frac{1}{3} \leq\xi \leq \frac{2}{3}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$ If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$
$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$
Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$
I am stuck here.
| Hint.- Just to give another way. Developing the product and simplify you have the equivalent inequality
$$7xyz\le xy+xz+yz$$ and you have too $$\frac13\ge\sqrt[3]{xyz}$$
| {
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How many ordered pairs satisfy $\log(x^3+\frac{1}{3}y^3+\frac{1}{9})=\log x+\log y$? How many ordered pairs satisfy $\log(x^3+\frac{1}{3}y^3+\frac{1}{9})=\log x+\log y$?
It simplifies to $x^3+\frac{1}{3}y^3+\frac{1}{9}=xy$
I dont know how to proceed further.
| $$\log\left(x^3+\frac{1}{3}y^3+\frac{1}{9}\right)=\log x+\log y\tag1$$
First of all, we have to have
$$x^3+\frac{1}{3}y^3+\frac{1}{9}\gt 0\quad\text{and}\quad x\gt 0\quad\text{and}\quad y\gt 0\tag2$$
Under $(2)$, we have
$$(1)\implies x^3+\frac{1}{3}y^3+\frac{1}{9}-xy=0\tag3$$
As lab bhattacharjee suggests in the comments, the LHS of $(3)$ is of the form $$A^3+B^3+C^3-3ABC$$
Letting
$$a=x,\qquad b=\frac{y}{\sqrt[3]3},\qquad c=\frac{1}{\sqrt[3]{9}}$$
we have
$$\begin{align}(3)&\implies x^3+\left(\frac{y}{\sqrt[3]3}\right)^3+\left(\frac{1}{\sqrt[3]{9}}\right)^3-3\cdot x\cdot\frac{y}{\sqrt[3]{3}}\cdot\frac{1}{\sqrt[3]{9}}=0
\\\\&\implies a^3+b^3+c^3-3abc=0
\\\\&\implies (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0
\\\\&\implies (a+b+c)\times \frac 12((a-b)^2+(b-c)^2+(c-a)^2)=0
\\\\&\implies a+b+c=0\quad\text{or}\quad a=b=c
\\\\&\implies x+\frac{y}{\sqrt[3]3}+\frac{1}{\sqrt[3]{9}}=0\qquad\text{or}\qquad (x,y)=\left(\frac{1}{\sqrt[3]{9}},\frac{1}{\sqrt[3]3}\right)\end{align}$$
Since $x\gt 0$ and $y\gt 0$ from $(2)$, we see that
$$x+\frac{y}{\sqrt[3]3}+\frac{1}{\sqrt[3]{9}}=0$$
does not hold.
Since $(x,y)=\left(\frac{1}{\sqrt[3]{9}},\frac{1}{\sqrt[3]3}\right)$ satisfies $(2)$, we see that the only solution for $(1)$ is
$$x=\frac{1}{\sqrt[3]{9}},\qquad y=\frac{1}{\sqrt[3]3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=$? Consider the convergent integral (I looked up difficult indefinite integrals on google images and then I saw this integrand and I was like hey let's see if it converges)
$$I=\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
We have the numerical approximation $$I\approx0.5553603672697931...$$
Which Wolfram alpha says is close to $\frac\pi{4\sqrt{2}}$. All my attempts at this integral have been fruitless and I need some help. Here's the only attempt of mine that actually made the integrand smaller:
$x=\tan u$: $$I=\int_{\pi/4}^{\pi/2}\frac{\tan^2u-1}{\sqrt{\tan^4u+1}}du$$
Next step:
bang head on floor in agony
Any suggestions?
| Since $$x^4 + 1 = x^2 (x^2 + x^{-2}) = x^2 ((x + x^{-1})^2 - 2),$$ we have for $x > 0$
$$\frac{x^2-1}{(x^2+1)\sqrt{x^4 + 1}} = \frac{x^2 (1 - x^{-2})}{x^2 (x + x^{-1}) \sqrt{(x + x^{-1})^2 - 2}} = \frac{1-x^{-2}}{(x+x^{-1})\sqrt{(x+x^{-1})^2 - 2}}.$$ Now with the substitution $$u = x + x^{-1}, \quad du = (1 - x^{-2}) \, dx,$$ we obtain $$\int \frac{x^2-1}{(x^2+1)\sqrt{x^4 + 1}} \, dx = \int \frac{du}{u\sqrt{u^2-2}} = \frac{1}{2} \int \frac{2u \, du }{u^2 \sqrt{u^2 - 2}}.$$ One more substitution of the form $$u^2 = v^2 + 2, \quad 2u \, du = 2v \, dv$$ gives $$\frac{1}{2} \int \frac{2v \, dv}{(v^2 + 2)v} = \int \frac{dv}{v^2 + 2}.$$ This demonstrates that an elementary antiderivative exists, the computation of which I have left to you as a straightforward exercise.
| {
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Knowing when a pattern is a linear homogeneous recurrence relation I'm quite stuck on finding a pattern here:
Suppose we have:
*
*1 inch letters: f, i, t
*2 inch letters: a, c, d, e, g, n, o, p, s, u
And we want to find a recurrence relation for making a banner of length $n$ inches. We can repeat letters and order matters. So "ff" would count as 1 way and "cs" and "sc" are both valid ways too.
Essentially I can write things in the form:
(ways to make with just 1s) + (ways to make with just 2s) + (ways to make with both)
So:
*
*n = 0: 1 way
*n = 1: 3 ways
*n = 2: $3^2 + 10 = 19$ ways
*n = 3: $3^3 + 2*(3*10)=87$ ways
*etc.
Somebody suggested the form $A_{n} = xA_{n-1} + yA_{n-2}$ and I solved a system of equations using this form to find x and y. This turned out to be the correct answer but he did not explain where he got this from. The answer was:
$$A_0 = 1\\A_1 = 3\\A_n = 3A_{n-1} + 10A_{n-2}, n\geq2$$
I can see it is the form of a linear homogeneous recurrence relation but I don't know how he spotted the pattern in doing calculations of different banners of length $n$. How do we know to use the form of a linear homogeneous recurrence relation? I'm not interested in solving the recurrence relation, only in finding it.
| To complete @weee's solution, use generating functions. Note that $A_0 = 1$ (just one empty banner) and $A_1 = 3$ (number of one-inch banners). Define $g(z) = \sum_{n \ge 0} A_n z^n$, write your recurrence as:
$\begin{equation*}
A_{n + 2} = A_{n + 1} + 10 A_n
\end{equation*}$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} A_{n + 2} z^n
&= 3 \sum_{n \ge 0} A_{n + 1} z^n + 10 \sum_{n \ge 0} A_n z^n \\
\frac{g(z) - A_0 - A_1 z}{z^2}
&= 3 \frac{g(z) - A_0}{z}
+ 10 g(z) \\
\frac{g(z) - 1 - 3 z}{z^2}
&= 3 \frac{g(z) - 1}{z}
+ 10 g(z)
\end{align*}$
Solve for $g(z)$, as partial fractions:
$\begin{align*}
g(z)
&= \frac{1}{1 - 3 z - 10 z^2} \\
&= \frac{5}{7} \cdot \frac{1}{1 - 5 z} + \frac{2}{7} \cdot \frac{1}{1 + 2 z}
\end{align*}$
From here we read the coefficients directly:
$\begin{align*}
A_n
&= [z^n] g(z) \\
&= \frac{5}{7} \cdot 5^n + \frac{2}{7} \cdot (-2)^n \\
&= \frac{5^{n + 1} - (-2)^{n + 1}}{7}
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Concerning the product of all unique positive divisors
If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
(A) $n^3$
(B) $n^4$
(C) $n^6$
(D) $n^8$
(E) $n^9$
I tried using the formula: $\frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.
| Since the product of positive divisors of $n$ is $n^{\frac{d(n)}{2}}$. We want
$$n^{\frac{d(n)}{2}}=n^2.$$
This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is
$$(n^2)^{\frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding a quadratic equation using roots
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_{1/2}$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_{1/2}=\large{-b\pm{\sqrt{4ac}}\over2a}$
and from here:
$$\begin{align}x_1^3&=\bigg({-b+{\sqrt{4ac}}\over2a}\bigg)^3\\&={(\sqrt{4ac}-b)^2(\sqrt{4ac}-b)\over8a^3}\\&={(4ac-2b\sqrt{4ac}+b^2)(\sqrt{4ac}-b)\over8a^3}\\&={4ac\sqrt{4ac}-4abc-8abc-2b^2\sqrt{4ac}+b^2\sqrt{4ac}-b^3\over8a^3}\\&={4ac\sqrt{4ac}-12abc-b^2\sqrt{4ac}-b^3\over8a^3}\end{align}$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.
| Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convert Polynomial to Sum I have to convert the polynomial: $P_4=x^4+7x^3-13x^2-103x-84$ into the form: $$P_4(x)=\sum_{i=0}^4 a_i(x-1)^i$$ using Horner's Method.
I can evaluate both of them using Horner's Method, but I can't figure out how to convert them into one another.
| Note that:
$$a_0+a_1(x-1)+a_2(x-1)^2+a_3(x-1)^3+a_4(x-1)^4=x^4+7x^3-13x^2-103x-84 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_0}=1+7-13-103-84=\color{red}{-192};\\
(x-1)[a_1+a_2(x-1)+a_3(x-1)^2+a_4(x-1)^3]=x^4+7x^3-13x^2-103x+108 \stackrel{\div (x-1)}{\Rightarrow} \\
a_1+a_2(x-1)+a_3(x-1)^2+a_4(x-1)^3=x^3+8x^2-5x-108 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_1}=1+8-5-108=\color{red}{-104};\\
(x-1)[a_2+a_3(x-1)+a_4(x-1)^2]=x^3+8x^2-5x-4 \stackrel{\div (x-1)}{\Rightarrow} \\
a_2+a_3(x-1)+a_4(x-1)=x^2+9x+4 \stackrel{x=1}{\Rightarrow} \\
\color{red}{a_2}=1+9+4=\color{red}{14};\\
\vdots$$
Can you continue and finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove if $f$ is convex on $(a,b)$, then $f$ is bounded on every closed subinterval of $(a,b)$ Prove :
$f : (a,b) \to \mathbb{R} $ is convex, then $f$ is bounded on every closed subinterval of $(a,b)$
where $f$ is convex if $f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda)f(y), \forall x,y \in (a,b), \forall \lambda \in [0,1]$
Try
$f$ is bounded above
Let $J = [\alpha, \beta] \subset (a,b)$.
$\forall c \in [\alpha, \beta]$, $\exists \lambda_0 \in [0,1]$ s.t. $c = \lambda_0 \alpha + (1-\lambda_0) \beta$, and
$$
f(c) = f(\lambda_0 \alpha + (1-\lambda_0) \beta) \le \lambda_0 f(\alpha) + (1-\lambda_0) f(\beta)
$$
thus, $f$ is bounded above on $[\alpha, \beta]$
But I'm stuck at how I should proceed to prove that $f$ is bounded below.
| Here is an analytic proof that a convex function $f:[\alpha,\beta] \to \mathbb{R}$ is bounded on the closed interval.
Take $M = \max (f(\alpha),f(\beta))$. Note that any $x \in [\alpha,\beta]$ is of the form $x = \lambda\alpha + (1-\lambda)\beta$ where $0 \leqslant \lambda \leqslant 1$.
Hence, we have for all $x \in [\alpha, \beta]$, the upper bound
$$f(x) \leqslant \lambda f(\alpha) + (1-\lambda)f(\beta) \leqslant \lambda M + (1-\lambda)M = M$$
To find a lower bound, write $x = \frac{a+b}{2} + \theta$. Since $\frac{a+b}{2} = \frac{1}{2} \left(\frac{a+b}{2} + \theta \right) + \frac{1}{2} \left(\frac{a+b}{2} - \theta \right) $, we have by convexity
$$f\left(\frac{a+b}{2}\right) \leqslant \frac{1}{2}f \left(\frac{a+b}{2} + \theta \right) + \frac{1}{2}f \left(\frac{a+b}{2} - \theta \right),$$
and
$$f(x) = f\left(\frac{a+b}{2} + \theta \right) \geqslant 2f\left(\frac{a+b}{2}\right)- f \left(\frac{a+b}{2} - \theta \right)$$
But from the upper bound we have $-f \left(\frac{a+b}{2} - \theta \right) \geqslant -M$, and so we have the lower bound
$$f(x) \geqslant 2f\left(\frac{a+b}{2}\right)-M = m$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.