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Apr
13
comment Calculate $\int_\Gamma ze^{z}dz$ where $\Gamma$ is line from point $z_1=0$ to point $z_2=\frac{\pi i}{2}$
@user227317 yes. See my answer for full details, but you got the answer! You can also use Blatter's approach too.
Apr
12
comment Calculate $\int_\Gamma ze^{z}dz$ where $\Gamma$ is line from point $z_1=0$ to point $z_2=\frac{\pi i}{2}$
@user227317 no, as I said use the substitution $z=it$. Also @ JessicaK's solution will also work. See also @ ChrisrianBlatter's answer.
Apr
12
comment Calculate $\int_\Gamma ze^{z}dz$ where $\Gamma$ is line from point $z_1=0$ to point $z_2=\frac{\pi i}{2}$
You need to parametrize the curve $\Gamma$, e.g. let $z(t)=it$ where $t\in[0,\pi/2]$. Looks like integration by parts may be helpful too.
Apr
8
comment $F(x)+G(y)= e^{x+y}?$
Yes, the way it is written confused me for a moment. However, on expanding the middle equality now I clearly see it does equal $F(1)-F(0)$. I was looking at what you had written from a different perspective - I was trying to construct the middle equality from the first by rearranging the original equation. Sorted now I see clearly what's happing - that "old trick" of adding something and then taking it away, so yes maybe best read right to left. +1 for your answer!
Apr
8
comment $F(x)+G(y)= e^{x+y}?$
I could be wrong, but shouldn't the middle equality be $-G(y)+e^{1+y}-(-G(y)+e^y)$ ? Maybe it's equivalent...
Apr
8
comment How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +…+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$
Try proof by induction ?
Mar
30
comment Question on branches and $\iff$.
Thank you. So if I restrict my attention to the principal branch (and I will also assume $f$ and $g$ are continuous) then $f+ig=0\iff e^{f(x)}\cos(g(x))+ie^{f(x)}\sin(g(x))=1$ ?
Mar
27
comment What does $O\left(\frac{1}{\log\log T}\right)$ mean?
Thanks. That's what I thought. So basically what the paper says is that as $T\to\infty$ the number of zeros outside the region is some constant multiple of $1/(\log\log T)$ ?
Mar
25
comment Series with $e^{\frac{1}{n}}$
Shouldn't that be $+O(n^{-4})$ in your equality?
Mar
25
comment Series with $e^{\frac{1}{n}}$
You could first try expanding $e^{1/n}$ to some order ($O(1/n^k)$). Presumably the value of $k$ will be related to the existence of $1/(2n^2)$ in your summand. Then simplify if possible, and see what happens from there. Haven't tried it so can't be 100% sure, but that's what I'd do first.
Mar
24
comment Complex integration confusion
Try $z=e^{i\theta}$, where $\theta\in[0,2\pi]$.
Mar
21
comment Why Does $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ sum to $ (1-(1-p)^{n+1}) $?
The binomial theorem states that: $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$
Mar
19
comment Having trouble understanding generalized complex numbers
@ Vim yes, that's exactly the problem I was thinking might arise.
Mar
19
comment Having trouble understanding generalized complex numbers
@ Vim yes, as I said I'm not too sure about that part as I haven't looked into it a great deal. I think if the discriminant is negative then you still have a linear equation so you could still solve for $i$. My worry was that the $i$ resulting from the $\sqrt{\cdot}$ was a "different" $i$ from that being defined. But as I say I haven't looked into this in any depth whatsoever!
Mar
19
comment Having trouble understanding generalized complex numbers
Just a thought (and might not be applicable here), regarding (1) could you not use the quadratic formula to obtain a "non-recursive" definition, e.g. $$i=\frac{q\pm\sqrt{q^2-4p}}{2}.$$ You would probably need $q^2-4p\geq 0$, although I'm not too sure about that...
Mar
17
comment Any way to simplify integral of Confluent Hypergeometric Function of the First Kind?
@NathanMcKenzie out of interest, from what problem does this integral arise?
Mar
17
comment Any way to simplify integral of Confluent Hypergeometric Function of the First Kind?
You may have better luck using the change of variable $y=\frac{\log n}{t}+1$. This gives $dt=\frac{\log(n)dy}{y^2}$ and $t=-\log n\implies y=0$, $t=0\implies y=\infty$. Then the integral becomes: $$-z\log(n)\int_0^\infty \frac{e^{\frac{\log (n)(1-s)}{y-1}}}{(y-1)^2} {}_1F_1\left(1-z,2,\frac{\log n}{y-1}\right)dy.$$ There are plenty of tables of integrals with limits over $\mathbb{R}^+$. See e.g. Gradshteyn and Ryzhik, p.814.
Mar
11
comment Is there a name for an object with both position and velocity?
@Sebastian, so for example a point in phase space might be $p=(x,y,u,v)$, where $(x,y)$ is position and $(u,v)$ is velocity. Thank you.
Feb
26
comment Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge?
I understand you're using the Lambert W function which enables you to obtain $x=g(y)$, but could you please explain in a little more detail how you got the substituted integral?
Feb
19
comment What is known about the complex solutions to $\zeta(s)=-1$?
I suppose this is related to "level curves" of a function...