TITLE: How do you evaluate an integral with an absolute value?
QUESTION [1 upvotes]: For example, how would you evaluate something like $\int_{-\pi}^{\pi} |t|\cos(kt)dt$? When I use Wolfram on the indefinite integral it gives me a $sgn(t)$ function, and I'm not sure what this is. However, if I use the $-\pi$ and $\pi$ it gives me the answer $\frac{2(\pi k \sin(\pi k)+\cos(\pi k)-1)}{\pi k^2}$.

REPLY [1 votes]: Just so this gets an answer (for future reference by anyone with the same problem):
The integrand consists of the product of two functions, lets call them $f(t)=|t|$ and $g(t)=\cos(kt)$. Both $f$ and $g$ are even functions, meaning they take the same value if we flip the sign of the argument;
\begin{align}
f(-t)=|-t|=|t|, \ \ \ \text{and} \ \ \ g(-t)=\cos(-kt)=\cos(kt).
\end{align}
The product of two even functions is itself even, meaning the entire integrand is an even function. And since even functions take the same values whether we are evaluating them at $t$ or $-t$, the integral will remain unchanged if we change the limits from $[0,\pi]\rightarrow[-\pi,0]$. That is,
\begin{align}
\int_0^\pi f(t)g(t)\,\mathrm{d}t=\int_{-\pi}^0f(t)g(t)\,\mathrm{d}t.
\end{align}
But then we can rewrite our integral as
\begin{align}
\int_{-\pi}^\pi |t|\cos(kt)\,\mathrm{d}t = 2\int_0^\pi |t|\cos(kt)\,\mathrm{d}t = 2\int_0^\pi t\cos(kt)\,\mathrm{d}t.
\end{align}
Note that since $t$ is always positive now, we might as well remove the abosulute value sign. Now, make a change of variables $kt\rightarrow x$, obtaining
\begin{align}
2\int_0^\pi t\cos(kt)\,\mathrm{d}t = 2\int_0^{\pi k} \frac{x}{k}\cos(x)\,\frac{\mathrm{d}x}{k} = \frac{2}{k^2}\int_0^{\pi k} x\cos x\,\mathrm{d}x.
\end{align}
Integrate by parts, and get
\begin{align}
\frac{2}{k^2}\int_0^{\pi k} x\cos x\,\mathrm{d}x = \left.\left(\frac{2\left[k\pi\sin(k\pi)+\cos(\pi k)\right]}{k^2}\right)\right|_0^{\pi k} = \frac{2\left(k\pi\sin(k\pi)+\cos(k\pi)-1\right)}{k^2}.
\end{align}


TITLE: Convexity of $e^{-2x}+e^{-bx^2}$, $\frac12<b<\frac32$
QUESTION [3 upvotes]: Prove that $u_b(x)=1-\frac12(e^{-2x}+e^{-bx^2})$ is concave for  $\frac12<b<\frac32$. What about b=$\frac1{20}$? $b=2$?

By disregarding the 1 and then the $\frac12$ we can turn it into the equivalent problem to show that $e^{-2x}+e^{-bx^2}$ is convex.
the second derivative is $2be^{-b x^2} (2 b x^2-1)+4 e^{-2 x}$, but showing its nonnegativity requires somehow to manipulate the different exponents, and neglecting terms has led to nothing.

REPLY [1 votes]: HINT:
Note that
$$\begin{array}{rcl}
2b\mathrm{e}^{-bx^2}>0&&\forall x\in\mathbb{R}\\
4\mathrm{e}^{-2x}>0&&\forall x\in\mathbb{R}
\end{array}$$
So you only need to verify the sign of $(2bx^2-1)$.


TITLE: Particular solution:$ (3+x) e^{-2x}$?
QUESTION [3 upvotes]: $y''+4y'+4y = (3+x) e^{-2x}$
So I'm working with undetermined coefficient and figured out solution for the left side. But what is the particular solution for the right side? I tried these but they don't work as all terms cancel to zero:
$y = (Ax+B) x e^{-2x}$
$y = (Ax+B)e^{-2x}$

REPLY [1 votes]: Considering the differential equation $$y''+4y'+4y =0$$ you surely showed that the characteristic equation has a double root corresponding to $-2$ and the solution would be $$y=c_1 e^{-2 x}+c_2 e^{-2 x} x$$
So, if I may suggest, reformulate the equation defining $$y= e^{-2x} z\quad ,\quad y'=e^{-2 x} \left(z'-2 z\right)\quad,\quad y''=e^{-2 x} \left(z''-4 z'+4 z\right)$$ Replacing, the original differential equation write $$z''e^{-2 x} =(3+x) e^{-2x}$$ that is to say $$z''=x+3$$ which explains the cubic polynomial to be taken into account.
You can easily concieve that for $$y''+4y'+4y =P_n(x)\,e^{-2x}$$ where $P_n(x)$ would be a polynomial of degree $n$, the same procedure would lead to $$z''=P_n(x)$$ and the solution would correspond to a polynomial of degree $n+2$.


TITLE: Proposition regarding inverse operators and their norm
QUESTION [0 upvotes]: I'm trying to prove the following assertion.

Hypothesis:
$1)$ $X$ is a Banach space and $Y$ is a normed space.
$2)$ $A : X → Y$ is a bounded bijective operator and $A^{-1}$ is bounded.
$3)$ $B : X → Y$ is a bounded operator and $∥A^{-1}(B-A)∥<1$.
Thesis:
$B$ is bijective and $\Vert B^{-1}∥ ≤ \frac{∥A^{-1}∥}{1-∥A^{-1}(B-A)∥}$

Now, if $\Vert L \Vert < 1$ then $(I-L)$ is bijective, so $ I – A^{-1}(B-A) = -A^{-1}B$ is bijective. But $(-A^{-1})$ is obviously bijective, so $B$ is also bijective. I need a little help for the bound.

REPLY [1 votes]: Let $T=A^{-1}(B-A)$. Then $(I-T)^{-1}=\sum\limits_{n=0}^{\infty} T^{n}$, so $\|(I-T)^{-1}\| \leq \sum\limits_{n=0}^{\infty} \|T^{n}\| =\frac  1 {1-\|T\|}$ and the result follows from the fact that $B^{-1}=-(I-T)^{-1}A^{-1}$


TITLE: Brouwer fixed point theorem problem
QUESTION [0 upvotes]: the proof for Brouwer fixed point theorem involved creating a function g that mapped the disk $D_n$ into its boundary, where $g(x)$ was obtained by extending a segment from $f(x)$ through $x$ and into the boundary of $D_n$.
Define such $g:D_n\to D_n$ with following properties

$g$ is continuous
$g$ maps all points of $D_n$ to boundary $D_n$
$g$ leaves boundary points fixed

which shows that there is no such $g$ exists.
Now suppose that, instead, we create $g$ in a different way: suppose that $g(x)$ is obtained by extending a segment from $x$ through $f(x)$ and into the boundary of $D_n$. Explain why the argument will fail if this "new" $g$ is used in place of the "original" $g$.
How can I explain this?

REPLY [1 votes]: If $x$ is in the boundary, it should be mapped to itself. This is not so for your function, since $x \mapsto f(x)$ and extended to the boundary, but the resulting point cannot be $x$.


TITLE: show that the equation $3z^3+(2-3ai)z^2+(6+2bi)z+4=0$has exactly one real root
QUESTION [1 upvotes]: show that :
the equation $3z^3+(2-3ai)z^2+(6+2bi)z+4=0$ (where both $a$ and $b$ are real numbers) has exactly one real root.

let $x_{1},x_{2},x_{3}$ be root,and $$x_{1}x_{2}x_{3}=-\frac{4}{3},x_{1}+x_{2}+x_{3}=\dfrac{3ai-2}{3},x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=\dfrac{2bi+6}{3}$$

REPLY [1 votes]: If we had exactly 3 real roots, then the sum of all roots would be real, but it is not.
If we had exactly 2 real roots, then the product of all roots would not be real, but it is.


TITLE: Epsilon delta proof for the following limit
QUESTION [0 upvotes]: How do  write an epsilon-delta proof or the following limit assuming the limit is -2,
$$\lim_{x \to -1} \frac{x^2-1}{x+1}$$

REPLY [1 votes]: Take any $x\neq -1$. Then it is simple to show that $$\frac{x^2-1}{x+1} = (x-1)$$.
Let $\epsilon > 0$. Now take $\delta = \epsilon$ and any $x$ such that $0<|x+1|<\epsilon$. You can then see that $$|\frac{x^2-1}{x+1}+2| = |x-1+2| = |x+1| < \epsilon$$ proving your point.


text:
\subsection{Enclosure of an attractor.}
In this section we will describe an algorithm that we used to enclose an attractor. There are already existing packages for rigorous enclosures of invariant objects, see for example \cite{GAIO}.
We used, however, our own implementation dedicated for this purpose which seems to be easier and makes the software independent on the other libraries. We will use the fact that what we want to enclose is an attractor.

The idea is very simple. First, we fix a cover $\mathcal X$ of the observed attracting domain. Then we choose a box $V$ from observed attracting region, and we enclose its forward trajectory using the sets from the cover $\mathcal X$ as long as the trajectory does not leave $\supp{\mathcal X}$. Since the cover $\mathcal X$ is finite by its definition, after a finite number of steps there are no new sets in the cover of the trajectory of $V$ and we can stop the procedure. This approach generates an enclosure of some invariant set, that is expected to be an attractor.

The Algorithm~\ref{alg:enclosure} summarizes the above considerations.

\begin{algorithm}[htbp]\linesnumbered\label{alg:enclosure}

 \caption{enclose forward trajectory}
 \KwIn{$V$ : element of $\mathcal X$,\newline
         $\mathcal G=\graph{V}{E}$ : graph,\newline
         $f$ : a map
         }
  \KwResult{$\mathcal G$ - graph}
  \KwData{
    $\mathcal U, \mathcal F$ : subsets of $\mathcal X$,\newline
    $\mathcal G'=(\mathcal V',\mathcal E')$ : graph
  }
  \Begin{
  $(\mathcal V,\mathcal E)\leftarrow(\emptyset,\emptyset)$\;
  $\mathcal U\leftarrow\{V\}$\;

  \Repeat{$\mathcal U=\emptyset$}
  {
    $(\mathcal V',\mathcal E')\leftarrow(\emptyset,\emptyset)$\;
   \ForEach{$U\in \mathcal U$}{
        \If{$f(U)\not\subset\supp{\mathcal X}$}{\Return \textbf{Failure};}
        $\mathcal F\leftarrow [f(U)]_{\mathcal X}$; // computed cover of $f(U)$\;
        $\mathcal V'\leftarrow \mathcal V'\cup\{U\}\cup \mathcal F$\;
        \ForEach{$F\in \mathcal F$}{
          $\mathcal E' \leftarrow\mathcal E'\cup\{(U,F)\}$\;
        }
   }
   $\mathcal U\leftarrow \mathcal V'\setminus(\mathcal V\cup\mathcal U)$\;
   $(\mathcal V,\mathcal E)\leftarrow(\mathcal V,\mathcal E)\cup(\mathcal V',\mathcal E')$\;
  }
  \Return{$\mathcal G$};
}
\end{algorithm}

\begin{lemma}
Fix a cover $\mathcal X$ of $X$. Assume Algorithm~\ref{alg:enclosure} is called with its arguments $V$, $\mathcal G=\graph{V}{E}$ and $f$.
If the algorithms stops without the \textbf{Failure} result then
\begin{enumerate}
  \item[(i)] the combinatorial map $\mathcal F\colon\mathcal V\multimap \mathcal V$ encoded by the graph $\mathcal G=\graph{V}{E}$ is combinatorial representation of $f|_{\supp{\mathcal V}}$, with the natural cover $\mathcal V$ of $\supp{\mathcal V}$,
 \item[(ii)] $\supp{\mathcal V}$ is a positive invariant set for $f$, i.e. $f(\supp{\mathcal V})\subset \supp{\mathcal V}$,
\item[(iii)] $\bigcup_{i=0}^\infty f^i(V)\subset \supp{\mathcal V}$.
\end{enumerate}
\end{lemma}
\begin{proof}
First observe that the algorithm always stops. By contradiction, assume it does not stop. This means that in each iteration of the main repeat-until loop (lines 4-15) the set $\mathcal U$ is not empty. This implies, that the number of elements in the set $\mathcal V$, which is updated in line 14, is strictly increasing. But this is a contradiction with $\mathcal V\subset \mathcal X$, which is a finite set by the definition of cover.

Let $\mathcal G$ be the result returned by the algorithm and let $\mathcal F\colon\mathcal V\multimap\mathcal V$ be the combinatorial map represented by the graph $\mathcal G$.

We are proving the assertion (i). Let $x\in\supp{\mathcal V}$ and let $V\in \mathcal V$ be such that $x\in V$. Observe, that if $\mathcal F(V)$ is not empty then $f(V)\subset\supp{\mathcal F(V)}$ - as guaranteed in lines 9, 11, 12 and 14 of the algorithm. Hence, in this case the assertion follows. To finish this step we observe, that the main repeat-until loop stops when the set $\mathcal U$ is empty. But this set contains of exactly these sets $U\in\mathcal V$ (lines 10 and 13) for which the value $\mathcal F(U)$ is not computed yet. Hence, after the algorithm stops, for each $V\in\mathcal V$, $\supp{\mathrm{out}(V,\mathcal G)}$ contains a computed enclosure of $f(V)$.

Assertions (ii) follows from (i) and the definition of combinatorial representation.

To prove (iii) it is enough to observe that $V\subset\supp{\mathcal V}$ as guaranteed in line 3, 10 and 14 of the algorithm. Then the assertion follows from (ii).
\end{proof}

Using Algorithm~\ref{alg:enclosure} one can compute combinatorial representation of a map restricted to some positive invariant set, which is expected to be an attractor. For further applications this graph should be as small as possible and it should still contain enclosure of the invariant set of $f$ restricted to $\supp{\mathcal V}$. Therefore it is important to choose the starting box $V$ such that it contains points from the attractor. An easy way to do that is to take a point from the observed basin of attraction and compute non-rigorously its forward trajectory for some long time. Then we can choose a set $V$ from the cover that contains the last computed point.

config: arxiv


TITLE: How many solutions are $ a+b+c+d+e = 10 $
QUESTION [1 upvotes]: I would appreciate if somebody could help me with the following problem:
Q: How many solutions are
there to the equation
$$ a+b+c+d+e = 10 $$
where $a,b,c,d,e\in \{0,1,2,3,4\}$

REPLY [1 votes]: We outline the strategy, and leave it to you to put the pieces together.
You are probably familiar with the number of solutions of $a+b+c+d+e$ in non-negative integers. This is solved by a technique often called Stars and Bars. For the record, the number is $\binom{14}{4}$.
We need to take away from this the number of solutions where one or more of the variables is $\ge 5$.
The solutions where one of the numbers is $\ge 6$ are easy to count. There are $5$ ways to choose which of $a,b,c,d,e$ is $\ge 6$. And given that (say) $a\ge 6$, the number of solutions is the number of solutions of $a+b+c+d+e$ in non-negative integers. That's a Stars and Bars problem.
Now we count the solutions where one or more of the variables is equal to $5$. There are $\binom{5}{2}$ solutions where two of the numbers are $5$.
For exactly one of the numbers equal to $5$, things are more complicated. Which one it is can be chosen in $5$ ways. Then we need to solve $x+y+z+w=5$, with none of the entries equal to $5$. This is the number of solutions in non-negative integers of $x+y+z+w=5$, minus the $4$ solutions in which one of the variables is $5$.


TITLE: Is this exercise in complex correctly done?
QUESTION [1 upvotes]: Find all $z \in \mathbb C$ such:
$$|z|=1, Re(z^4)=-Im(z^4) $$
So what I thougt was:
First, let $z$ be $$z = |z|e^{ix+2k\pi}, |z|=1$$
then,
$$z^4 = e^{i4x+8k\pi}$$
given
$$Re(z^4)=-Im(z^4)$$
this only happens if $4x+8k\pi = \pi/4$ or $4x+8k\pi = \pi5/4$, then
$$x= \pi/16-2k\pi$$
or
$$x= \pi5/16-2k\pi$$

REPLY [2 votes]: yes but it would be more correct if you provide general solution
equate angle to $$\pi/4+(2n+1)\pi/2$$ instead of $$\pi/4$$ and $$5\pi/4$$


TITLE: "Fully correlated" definition
QUESTION [5 upvotes]: Really sorry to be a noob, but I'm a programmer, not a mathematician, and all of my knowledge about statistics come from this book "Schaum's Outline of Theory and Problems of Probability, Random Variables, and Random Processes".
I'm implementing an UKF for target tracking using C++. Everything went well until an error about covariance matrix of state is not positive definite happened.
After a little research, I found this link Under what circumstance will a covariance matrix be positive semi-definite rather than positive definite? which  almost answer everything I need.
Only one thing I don't understand: The answer says "This happens if and only if some linear combination of X is ‘fully correlated’". Can anyone explain for me what does "fully correlated" mean? And example would be great. I have search Google about its definition but there is no luck at all.

REPLY [3 votes]: (I was the poster of the question to which the OP refers and I am heartily embarrassed by my distinct lack of clarity).
The intention was that the relationship is exactly linear and the correlation is 1 or -1.
That is:
$\rho_{i,j} = \frac{\sigma_{i,j}}{\sigma_i \sigma_j} = \pm 1$
where $\sigma_{i,j}$ is the covariance of elements $i$ and $j$ and  ${\sigma_i}^2$ and ${\sigma_j}^2$ is the variance of $i$ and $j$ respectively. In all cases $i \ne j$, except for the scalar case which is covered in the original post.
There is the assumption that the random variable is multivariate normal, which is the an assumption of the (Unscented) Kalman-based estimators that are the subject of both the original question and this one.
I hope this doesn't cause past answers to be wrong!


TITLE: Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$
QUESTION [7 upvotes]: I am trying to find a closed form for this infinite series:
$$ S=\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$$
Whith $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$ the harmonic numbers.
I found this integral representation of S:
$$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx$$
Sketch of a proof:
Recall the integral representation of the harmonic numbers: $H_n=\displaystyle\int_{0}^{1}\frac{1-x^n}{1-x}dx$
By plugging it into the definition of S and interchanging the order of summation between $\displaystyle\sum$ and $\displaystyle\int$ (justified by the uniform convergence of the function series $\displaystyle\sum\left(x\to\frac{4^n}{n^2{2n\choose n}}\frac{1-x^n}{1-x}\right)$, because $\forall x\in[0,1],\frac{1-x^n}{1-x}<n$), we get:
$$S=\int_{0}^{1}\frac{1}{1-x}\sum\limits_{n=1}^{\infty}\frac{4^n(1-x^n)}{n^2{2n\choose n}}dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
Using the result $\displaystyle\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}=\frac{\pi^2}{2}$.
At that point, we will rely on the taylor series expansion of $\arcsin^2$:
$$\arcsin^2(x)=\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}x^{2n}, |x|<1$$
Out of which we get $\displaystyle\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}=2\arcsin^2\left(\sqrt{x}\right)$
So,
$$S=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-2\arcsin^2\left(\sqrt{x}\right)\right)dx$$
Which, through the substitution $u=\sqrt{x}$, gives the integral representation above.
But beyond that, nothing so far. I tried to use the integral representation of $\frac{H_n}{n}$ to switch the order of summation, but it didn't lead anywhere. Any suggestion?

REPLY [2 votes]: From here, we have
$$\frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n=1}^\infty\frac{(2z)^{2n-1}}{n{2n \choose n}}$$
substitute $z=\sqrt{y}$, we get
$$\sum_{n=1}^\infty\frac{4^ny^n}{n{2n \choose n}}=2\sqrt{y}\frac{\arcsin\sqrt{y}}{\sqrt{1-y}}$$
Now multiply both sides by $-\frac{\ln(1-y)}{y}$ then integrate from $y=0$ to $1$ and using the fact that $-\int_0^1 y^{n-1}\ln(1-x)\ dy=\frac{H_n}{n}$, we get
\begin{align}
\sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n \choose 2}}&=-2\int_0^1\frac{\arcsin\sqrt{y}}{\sqrt{y}\sqrt{1-y}}\ln(1-y)\ dy\overset{\arcsin\sqrt{y}=x}{=}-8\int_0^{\pi/2}x\ln(\cos x)\ dx\\
&=-8\int_0^{\pi/2}x\left\{-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}\right\}\ dx\\
&=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x\cos(2nx) dx\\
&=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}\right)\\
&=\pi^2\ln2+2\pi\sum_{n=1}^\infty\frac{(-1)^n\sin(n\pi)}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^n\cos(n\pi)}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\\
&=\pi^2\ln2+0+2\sum_{n=1}^\infty\frac{(-1)^n(-1)^n}{n^3}-2\operatorname{Li}_3(-1)\\
&=\pi^2\ln2+2\zeta(3)-2\left(-\frac34\zeta(3)\right)\\
&=\pi^2\ln2+\frac72\zeta(3)
\end{align}


TITLE: Integral of something like Gamma Function
QUESTION [0 upvotes]: I am currently reading a paper that uses this integral as being less that one automatically. I am just wondering how do I know this to be true? Any help would be appreciated. Kindest regards, Catherine
\begin{equation}
 \int^{\varepsilon^{-2}cT}_0 e^{-\eta} \eta^{-\beta/m} \, d\eta
\end{equation}
Everything apart from $\eta$ is constant.

REPLY [0 votes]: You can take a look at the incomplete Gamma function. Indeed, your integral can be expressed in terms of the lower incomplete gamma function $\gamma(s, x)$:
$$\int_0^{\epsilon^{-2}cT} e^{-\eta}\eta^{-\beta/m}\,d\eta=\gamma\left(-\beta/m+1,\epsilon^{-2}cT\right)$$
Right now, I don't think there is a way to show that this integral will be less than one, unless contraints are placed on the parameters.


TITLE: Ugly Differentiation
QUESTION [1 upvotes]: If $$y=\frac{ax^2}{(x-a)(x-b)(x-c)} + \frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1$$
Then prove that $$\frac{dy}{dx}=\frac{y}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$$
What is the smartest way to solve this?

REPLY [3 votes]: HINT:
$$\dfrac c{x-c}+1=\dfrac x{x-c}$$
$$\dfrac{bx}{(x-b)(x-c)}+\dfrac x{x-c}=x\cdot\dfrac{b+(x-b)}{(x-b)(x-c)}=\dfrac{x^2}{(x-b)(x-c)}$$
$$\frac{ax^2}{(x-a)(x-b)(x-c)}+\dfrac{x^2}{(x-b)(x-c)}=x^2\cdot\frac{a+(x-a)}{(x-a)(x-b)(x-c)}=\dfrac{x^3}{(x-a)(x-b)(x-c)}$$
Now take logarithm in both sides and differentiate

REPLY [2 votes]: HINT:
Use Partial Fraction Decomposition,
$$\frac{ax^2}{(x-a)(x-b)(x-c)} + \frac{bx}{(x-b)(x-c)}+\frac{c}{c-x}=\dfrac A{x-a}+\dfrac B{x-b}+\dfrac C{x-c}$$
$$\implies ax^2+bx(x-a)+c(x-a)(x-b)=A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b)$$
Compare the constants and the coefficients of $x,x^2$ to find $A,B,C$


TITLE: Riemann surface, residue and branch
QUESTION [0 upvotes]: Evaluate the residue of $ω(z) =\frac{\sqrt{(z−1)(z−5)}}{z} dz $ at point z = 0 for the branch $\sqrt{(z − 1)(z − 5)}$ defined by having positive limit values on the northern shore of the cut [1, 5].
The last part means : 'positive imaginary values' here. That mean that when z go to some point from (1,5) the product (z-1)(z-5)-> something negative, and square root of (z-1)(z-5) -> something purely imaginary. For example, for z->3 we have $\sqrt{(z-1)(z-5)}$-> plus or minus 2i. And by ''positive'' here we mean 2i. The northern shore in the initial formulation mean  that limit is not just z->3, but $z=3+i.\epsilon$, $\epsilon->0$, $\epsilon>0$. Or the same that argument of (z-1)(z-5) on the ''upper'' or ``nothern'' side of the segment (1,5) is $\pi$ and not $-\pi$.
Does anybody can help me, I'm totally lost on how to prove it...

REPLY [1 votes]: Your immediate task is to find which value to assume for $\sqrt{(z-1)(z-5)}$ at $z=0$ such that it comes from the right branch.
Pick a small positive real $\delta$ and imagine moving $z$ from $1+\delta$ in a small counterclockwise semicircle around $1$ to $1-\delta$ and then along the real axis to $0$. Because we're not crossing the branch cut, the select square root should be continuous along this journey.
When $z=1+\delta$ we have $(z-1)(z-5) \approx -4\delta$ since the difference between $z-5$ and $1-5$ is negligible in this context. Since we start out from the "northern shore", the square root we start our journey with is $2i\sqrt\delta$.
As we move around the semicircle, the $z-5\approx -4$ factor hardly changes, but $z-1$ factor turns counterclockwise around the origin at a constant distance of $\delta$. So $(z-1)(z-2)$ also moves counterclockwise from $-4\delta$ to $4\delta$. In order to stay continuous the square root has to move counterclockwise from $2i\sqrt\delta$ at half the angular speed -- so it ends up at $-2\sqrt\delta$.
Now moving along the real axis to $0$, the value of $(z-1)(z-5)$ stays positive all the way, so the relevant square root will not change sign during this process. It stays negative and ends up at $-\sqrt5$ when we reach $z=0$.


TITLE: The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$
QUESTION [10 upvotes]: $\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$

I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to $2$. But how can this problem be solved with proper procedure?

REPLY [4 votes]: Take $f(x,n)=x+2^n$. We can see that;
$$\begin{aligned}
 f(x,n) &= \sqrt{2^{2n}+x\left(x+2^{n+1}\right)} \\
 &= \sqrt{2^{2n}+xf(x,n+1)} \\
  &= \sqrt{2^{2n}+x\sqrt{2^{2\left(n+1\right)}+x\sqrt{2^{2\left(n+2\right)}+x\sqrt{...}}}}\\
&=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\\
\end{aligned}$$
Taking $x=1,n=0$; we get;
$$2=\sqrt{4^{0}+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}$$


TITLE: lfp property for dagger symmetric monoidal categories and their internal categories
QUESTION [1 upvotes]: We can define internal categories in a monoidal category like this.  Let $C$ be a dagger symmetric monoidal category.  Will $C$ be locally finitely presentable?  Let $C_{int}$ be the category of internal categories in $C$.  Is $C_{int}$ locally finitely presentable?

REPLY [5 votes]: For a dagger-category $C$, we have $C^{op} \simeq C$. But the only time that a category $C$ and its opposite can both be locally finitely presentable is when $C$ is a poset (this is covered in the text by Adámek and Rosicky). So the answer to the first question is: only when $C$ is an algebraic lattice.
The category of internal categories in an lfp category is again lfp. (Because the category of internal categories will be monadic over the category of internal graphs, and the latter is lfp.) However, a poset won't have any interesting internal categories.


TITLE: combinatorics for buying a new car
QUESTION [0 upvotes]: Question: Mindy is buying a new car. for the car she has selected the following options are available: power windows, a sunroof, and leather interior. if she must select one or more of these options, how many different combinations of options can she select?
My attempt: $3x2x1$
I don't know how to express if possibilities if she can choose one option, two options, or all 3 options .

REPLY [3 votes]: It won't be $3 \times 2 \times 1$: this counts the number of permutations of $\{\text{power windows},\text{sunroof},\text{leather interior}\}$.  (Perhaps this is the number of orderings in which they could be installed?)
The number of ways of choosing one option is $\binom{3}{1}$.  The number of ways of choosing two options is $\binom{3}{2}$.  The number of ways of choosing three options is $\binom{3}{3}$.  So, the number of ways of not choosing zero options is the sum of these.
However, there's a better way:  each option is a Yes/No choice:

Power windows: Yes or No.
Sunroof: Yes or No.
Leather interior: Yes or No.

So there are $2 \times 2 \times 2$ possibilities in total, and we just have to exclude the case where she says No to all of them.


TITLE: Which Dynkin diagram is being spoken about here? Why is there a double line?
QUESTION [2 upvotes]: I'm confused about the following comment in Knapp's Lie Groups 2ed, page 397. Here, $\Delta$ is a root system associated to a complex semisimple Lie algebra, $\alpha, \beta$ are orthogonal roots and we also have that $\alpha \pm \beta \in \Delta$ (that is, they are not strongly orthogonal).

I believe that the phrase simple component here means irreducible component. Since $\alpha \pm \beta$ are roots, $\alpha$ and $\beta$ must lie in the same irreducible component of $\Delta$. Further, since $\Delta$ comes from a semisimple Lie algebra, this irreducible component is a reduced root system (and hence the comment about only two root lengths).
Question: What set of simple roots within (an irreducible component of) $\Delta$ is the author using to obtain the Cartan matrix/Dynkin diagram with double line?
Attempt: I'm guessing there might be a way to show the existence of a simple system $\Pi \subset \Delta$ containing both $\alpha$ and $\beta - \alpha$. This would show that
\begin{equation}
2\frac{(\alpha, \beta -\alpha)}{|\alpha|^2} \cdot 2 \frac{(\alpha, \beta-\alpha)}{|\beta-\alpha|^2} = 2,
\end{equation}
thereby giving the double line. But I don't see how to construct this $\Pi$.
For example, if someone can provide a proof/reference showing that one can choose a simple system containing both $\alpha$ and $\beta - \alpha$, I will be happy.

REPLY [3 votes]: To put my comments into a proper answer:
We can see that the number of lines between two nodes in the diagram is given by $$N := \langle\alpha_i, \alpha_j\rangle\langle\alpha_j, \alpha_i\rangle = 2\frac{(\alpha_i, \alpha_j)}{|\alpha_i|^2} \cdot 2 \frac{(\alpha_j, \alpha_i)}{|\alpha_j|^2} = 4\frac{(\alpha_i, \alpha_j)^2}{|\alpha_i|^2|\alpha_j|^2} = 4 cos^2 (\theta).$$
Where $\theta$ is the angle between them. Since we have $ \langle\alpha_i, \alpha_j\rangle,\langle\alpha_j, \alpha_i\rangle,N $ all integers and from the above they are between $-3$ and $3$ ($N$ is also positive) there are only a few possibilities.
Additionally we have (as long as $\alpha_i, \alpha_j$ are not orthogonal):
$$ \frac{\langle\alpha_i, \alpha_j\rangle}{\langle\alpha_j, \alpha_i\rangle} = \frac{|\alpha_i|^2}{|\alpha_j|^2}.$$
Comparing the possibilities, it is clear that $\frac{|\alpha_i|^2}{|\alpha_j|^2}$ is in fact equal to $N$ whenever $N$ is non-zero (assuming $\alpha_i$ is the longer root). So now we just need the fact that if there are two roots somewhere in our system with $\frac{|\alpha_i|^2}{|\alpha_j|^2} = 2$ then there are two non orthogonal simple roots (in any choice of simple roots) with the same ratio of lengths.
In fact it should be clear that a set of simple roots should automatically have at least one root of each length so we just need a pair that are non-orthogonal. But this is also immediate since the Dynkin diagram is connected and we have only two lengths present: At some point between our two originally chosen nodes we must have a transition from long to short.


TITLE: Multivariate differentiable function on a compact is a Lipschitz function?
QUESTION [2 upvotes]: Let $n,m\in\mathbb{N}$, such that $E=(\mathbb{R}^n,\|\cdot\|_E)$ and $F=(\mathbb{R}^m,\|\cdot\|_F)$ with $\|\cdot\|_E$ and $\|\cdot\|_F$ two norms (we do not care which one since $n,m\in\mathbb{N}$).
I have a little doubt about the "lipschitzness" of a mapping $f$. Assuming $f:E\rightarrow F$, differentiable over $E$ (thus continuous over $E$). If $C\subset E$ is compact, then $f$ is $k$-lipschitz for a given $k>0$ over $C$, right ?
If $n=m=1$, I know that is true, but I didn't find the same result in multivariate fashion (maybe because it is too obvious) and I did not go further on this question since it's been a while I do not have pratice math.
Since $\mathbb{R}^n$ has very good topological properties, I think that this result is true. Am I wrong ?
Thanks !
Edit : Nice counter examples below !
The aforementioned statement is right only if $f$ is $C^1$.

REPLY [2 votes]: It’s not true, even for $m = n = 1$. Note, for a differentiable function, Lipschitzness is equivalent to having a bounded derivative. So, we just need a differentiable function with an unbounded derivative. Let:
$$f(x) = \begin{cases}x^{5/3} \sin\left(\frac1x\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0.\end{cases}$$
The derivative exists at $x \neq 0$, as it is a product and composition of differentiable functions. Using the usual derivative rules, we get
$$f’(x) = \frac{5}{3}x^{2/3} \sin\left(\frac{1}{x}\right) - x^{-1/3}\cos\left(\frac{1}{x}\right).$$
For $x = 0$, we use the limit definition:
$$f’(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} h^{2/3} \sin\left(\frac{1}{h}\right),$$
where the above expression of $h$ is sandwiched between $\pm h^{2/3}$, both of which tend to $0$ as $h \to 0$. Thus, $f’(0) = 0$. So, the function is defined everywhere, including every compact set containing $0$ in its interior.
We just need to show that $f’$ is unbounded around $0$. If we take $x_n = \frac{1}{2n\pi}$, then $x_n \to 0$ and
$$f’(x_n) = \frac{5}{3}x_n^{2/3} \sin(2\pi n) - \sqrt[3]{2\pi n}\cos(2\pi n) = -\sqrt[3]{2\pi n} \to -\infty$$
as $n \to -\infty$. Thus, $f$ cannot be Lipschitz on any compact set containing $0$ in its interior, e.g. $[-1, 1]$.


TITLE: Non-homogeneous Differential Equation (Stuck at integration part)
QUESTION [0 upvotes]: I need a hand for solving the integration part of the differential equation $y''+4y=x^2sin2x$ .
$(D-2i)(D+2i)y=x^2sin2x$  ,   $t= \dfrac{x{^2}sin2x}{D+2i}$
$t'+2it=x^2sin2x$, $t=uv$
$v=e^{-2ix}$
$du=(e^{2ix}) x^2sin2xdx$
I am stuck at this part. Can someone help me to solve it? Thanks in advance.

REPLY [1 votes]: Let, $P=\dfrac{1}{D^{2}+4}x^{2}\sin2x$ & $Q=\dfrac{1}{D^{2}+4}x^{2}\cos2x$.
Now, $Q+iP=\dfrac{1}{D^{2}+4}x^{2}e^{2ix}=e^{2ix}\dfrac{1}{D^{2}+4iD}x^{2}=e^{2ix}\dfrac{1}{4iD}(1+\dfrac{D}{4i})^{-1}x^{2}=\dfrac{e^{2ix}}{4i}(x^{3}-\dfrac{x^{2}}{4i}-\dfrac{x}{8})$.
Then, P.I.=$P$=imaginary part=$\dfrac{1}{16}x^{2}\sin2x-\dfrac{1}{4}(x^{3}-\dfrac{x}{8})\cos2x$.
If my calculation is right then it is the answer.But in calculation may be some mistake.Check It !!