TITLE: How to prove $\operatorname{Log}(z) = \log(|z|)+i\arg(z)$.
QUESTION [4 upvotes]: The value of the principal branch of the logarithm can be evaluated by the formula
  \begin{align*}
\operatorname{Log}(z) = \log(|z|)+i\arg(z),
\end{align*}
  where $\arg(z) \in (-\pi,\pi)$ and $\log$ denotes the usual real logarithm.

So the strategy was to represent $z \in \mathbb C \setminus (-\infty,0]$ in polar form $z = re^{i\phi}$, where $r = |z|$ and $\phi = \arg(z)$. Then, if $\phi > 0$, let $\gamma = [1,r]+\gamma_2$, where $\gamma_2 = re^{it}$ with $t \in [0,\phi]$, and prove $\int_{[1,z]} \frac{1}{z} dz = \int_\gamma \frac{1}{z} dz$. I have computed the latter integral and I got exactly the wanted formula on the right handside. Then one can do similarly the case for $\phi < 0$ and we are done.
My problem is, that I just don't see why $\int_{[1,z]} \frac{1}{z} dz = \int_\gamma \frac{1}{z} dz$ this holds.
I would really appreciate explanations here. Thanks.

REPLY [3 votes]: It follows from Cauchy's integral theorem that $\int_\gamma{dx\over z}$ will have the same value for any path $\gamma$ from $1$ to $z$ in $\mathbb{C}\setminus (-\infty,0]$. This is true since $\mathbb{C}\setminus (-\infty,0]$ is simply connected ("has no holes") and $f(z)={1\over z}$ is holomorphic in this domain. Cauchy's integral theorem is a deep result, but given this you are free to choose your $\gamma$ as you wish, and the particular $\gamma$ that you have chosen is the one which makes the computation simplest.
If you are interested in $\int_{[1,z]}{dz\over z}$, this can also be computed directly with elementary (and real) means, and will yield the same answer, but with a slightly more involved computation. (Note by the way that the notation is somewhat overloaded, in that $z$ denotes both the variable which runs along the interval, as well as the end point of the interval.)
Assume that $z=x+iy$ with $y\neq 0$, since otherwise there is no real difference between the two integrals. Parametrize $[1,z]$ by $t\mapsto t\cdot z+(1-t)\cdot 1$ with $0\leq t\leq 1$. Then
$$\int_{[1,z]}{dz\over z}=\int_0^1{z-1 \over 1+t(z-1)}dt$$
Write the integrand of the last integral on standard form $a+ib$ for complex numbers by expanding the fraction by the complex conjugate of the denominator. This yields that the integral is equal to
$$\int_0^1 \biggl({(x-1)(1+t(x-1))+ty^2 \over (1+t(x-1))^2+t^2y^2} + i{y\over (1+t(x-1))^2+t^2y^2}\biggr)dt$$
The integral of the first term is equal to
$$ {1\over 2}\log\bigl((1+t(x-1))^2+t^2y^2\bigr)\Bigr|_0^1={1\over 2}\log(x^2+y^2)=\log(|z|)$$
and the integral of the second term is (with some work) shown to be equal to
$$ i\arctan\biggl({t(x^2+y^2-2x+1)+x-1 \over y}\biggr)\bigg|_0^1$$ $$=i\arctan\biggl({x^2+y^2-x\over y}\biggr)-i \arctan\biggl({x-1\over y}\biggr)=i\arctan {y\over x}=i \arg(z)$$


TITLE: Power series for the Bessel function using Laplace transforms?
QUESTION [2 upvotes]: The Bessel's function of the first kind of order zero, $J_0$ is the solution to
$$ty''+y'+ty=0$$
which satisfies $J_0(0)=1$
The Laplace transform of this equation gives
$$\mathcal{L}[J_0(t)]=\frac{1}{\sqrt{1+s^2}}$$
The problem is to expand
$$s\mathcal{L}[J_0(t)]=\frac{s}{\sqrt{1+s^2}}$$
in a binomial series so as to express $J_0$ as a power series.

What I have tried so far:
  $$s\mathcal{L}[J_0(t)]=\frac{s}{\sqrt{1+s^2}}={\sqrt{\frac{s^2}{1+s^2}}}=\sqrt{1-\frac{1}{1+s^2}}$$
  $$\mathcal{L}[J_0(t)]=\frac{1}{s}\sum_{k=0}^\infty \binom{1/2}{k}\left(-\frac{1}{1+s^2}\right)^k$$
  But now if I apply $\mathcal{L}^{-1}$ on both sides I get a convolution and a bunch of weird sines inside the series...
  I'm guessing my problem is expressing $s/\sqrt{1+s^2}$ in terms of $1/s^k$.

REPLY [3 votes]: Sorry, I don't know how to use LaTex well; there may be a few errors. One can use the following verion of the binomial series $$(1+1/s^2)^{-1/2}=\sum_{k=1}^{\infty}\frac{(-1)^k{(\frac{1}{2})^\left({k}\right)}}{k!s^{2k}}$$ where ${\frac{1}{2}^\left({k}\right)}$ is the Pochhammer symbol (in this case, the rising factorial). Hence taking the Laplace transform of $\frac{1}{s(1+1/s^2)^1/2}$ gives: $$y(t)=\sum_{k=1}^{\infty}\frac{(-1)^k{(\frac{1}{2})^\left({k}\right)}t^{2k}}{k!(2k)!}$$
By using the definition of the rising factorial, and factorial, you should be able to show that $$\frac{{(\frac{1}{2})^\left({k}\right)}}{(2k)!}=\frac{1}{4^kk!} $$ which gives us a series for $J_0(t)$: $$J_0(t)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k!^2}(\frac{t}{2})^{2k}$$


TITLE: Integer solutions of $x^3-19 = y^2$
QUESTION [6 upvotes]: I am asked to find all integer solutions of the elliptic curve $x^3-19 = y^2$.
I've noticed so far that $x=7$ and $y=18$ is a solution and I've also noticed that $x$ and $y$ are coprime and that $x$ must be odd.
I've been trying to work in $K=\mathbb{Q}(\sqrt{-19})$, but since $19 \equiv 1 \pmod{4}$, I haven't found a way to relate this equation with the ring of integers of $K$.

REPLY [5 votes]: So $x^3=(y+\sqrt{-19})(y-\sqrt{-19})$. One shows in the usual way that
$y+\sqrt{-19}$ and $y-\sqrt{-19}$ are coprime in $R=\Bbb Z[\frac12(1+\sqrt{-19})]$. Then $R$ is a PID ($\Bbb Q(\sqrt{-19})$ has class number $1$) and its only units are $\pm1$. Then we get $y+\sqrt{-19}=\alpha^3$
with $\alpha\in R$. This means that
$$\left(\frac{a+b\sqrt{-19}}2\right)^3=y+\sqrt{-19}$$
where $a$ and $b$ have the same parity. Then $a^3-57ab^2=8y$
and $3a^2b-19b^3=8$. This last equation narrows down $a$ and $b$ to
a finite set.


TITLE: Whether the indefinite integral of a differentiable function is differentiable or not
QUESTION [0 upvotes]: Let $f$ be a differentiable function on $[0,1]$.Is the indefinite integral of $f$ differentiable on $[0,1]$?
By the fundamental theorem of calculus i know that the integral will be differentiable in $(0,1)$.But what about differentiability at the points 0 and 1?

REPLY [0 votes]: Unless I'm missing something I'd say
If $f$ is differentiable in $[0,1]$ it is also continuous in $[0,1]$, therefore the family of integral functions $F(x) + C$ does exist and the derivative of such functions is $f(x)$ which is continuous, hence $F(x)+C$ is differentiable.


TITLE: Multiplying significant figures and decimal places
QUESTION [0 upvotes]: One of my lecturers today said that multiplying a number leaves the number of significant figures to which that number is correct unchanged, however it doesn't leave the number of decimal places correct.
He gave the following example: $128\times 0.99687$.
According to him, if 0.99687 is correct to 4 dp then the product is not correct to 4dp. However if 0.99687 is correct to 4sf then the product is also correct to 4sf.
I am struggling to see why this is the case. I have tried thinking about how decimal places differ from significant figures but I never really learned this in detail, only the "mechanics" of how to work dp's and sf's out, so any help would be appreciated.

REPLY [1 votes]: "Correct to 4 decimal places" means that the 4th digit after the decimal point remains the same after rounding. But if we multiply this by 1000 then that 4th digit is now the 1st one after the decimal point. So the number is no longer correct in the 4 decimal place - it is now only correct in the 1st decimal place.
Conversely, dividing by 1000 the number is now correct to the 7th decimal place!
This happens because multiplication (or division) changes the positions of all digits.
However, if as with 0.9969, there are only 4 significant digits in the number, then after multiplication (or division) by 1000 there will still be only 4 significant digits.


TITLE: How to solve this functional equation involving hyperbolic functions?
QUESTION [2 upvotes]: I'm reading this (physics) book. They have the recurrence relation (book eq. 14.2.14)
$$f(K_1,0)=-\frac{1}{2}\ln\{2\sqrt{\cosh(2K_1)}\}+\frac{1}{2}f(\ln\sqrt{\cosh(2K_1)},0).\qquad(1)$$
They give the following solution (book eq. 14.2.15)
$$f(K_1,0)=-\ln(2\cosh K_1).\qquad(2)$$
I tried to check the solution was correct simply plugging (2) into (1) but end up with a $\ln(\cosh(\ln(\cosh(K_1)))$ thing.
Questions:

How do I obtain the solution (2) from (1)?
Is there an easy way to check the solution is correct?

REPLY [3 votes]: Note that
\begin{align}
\cosh x = \frac{e^x+e^{-x}}{2}
\end{align}
then you see that
\begin{align}
2 \cosh( \ln( \sqrt{\cosh(2K_1)}) =&\ 2\frac{\exp\left(\ln \sqrt{\cosh(2K_1)}\right)+\exp\left(-\ln \sqrt{\cosh(2K_1)} \right)}{2}\\
=&\ 2\frac{\sqrt{\cosh 2K_1}+ (\sqrt{\cosh 2K_1})^{-1}}{2} =  \frac{\cosh 2K_1 +1}{\sqrt{\cosh 2K_1}}.
\end{align}
Then it follows
\begin{align}
-\frac{1}{2}\ln\left(2\sqrt{\cosh 2K_1} \right)-\frac{1}{2}\ln\left(\frac{\cosh 2K_1 +1}{\sqrt{\cosh 2K_1}} \right)=&\ -\frac{1}{2}\ln\left( 2\cosh(2K_1)+2\right)\\
=&\ -\frac{1}{2}\ln(4\cosh^2 K_1).
\end{align}