close the boundary in the upper half plane you need the residuum at $i$, your function is $\frac1{(z-i)^m}\cdot h(z)$ where $h(z)=\frac1{(z+i)^m}$ which is analytic at $i$
if put in a taylor expansion of $\frac1{(z+i)^m}$ at $i$ you find that the term of order $m-1$ will give you the residual
this term is $\frac1{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\frac1{(z+i)^m}$ evaluated at $i$
now the $m-1$st derivative of $(z+i)^{-m}$ is $(-m)(-m-1)\cdot...\cdot(-2m-1)(z+i)^{2m-1}$
I have a calculus text by wylie, and it does a really bad job of explaining
though, of course I don't need it for my test on monday, so I'm not worrying too much about it
but it explained it as $\frac{1}{n}[\frac{1^2}{n^2}+\frac{2^2}{n^2}+\frac{3^2}{n^2}....\frac{n-1}{n^2}]$> area > $\frac{1}{n}[\frac{1^2}{n^2} ... \frac{n^2}{n^2}]$
and then the excersises were like "use this method to determine the area under the curve between x =0 and x = 2 for x^2
do you know the second fundamental theorem of calculus?
I'll assume you don't since that was a long pause. :-)
The second fundamental theorem of calculus states that if you select any single antiderivative of $f(x)$ and call it $F(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$
@Dodsy so... what is the expression by which you describe the set of antiderivatives of $x^2$. Or for that matter, can you find any antiderivative for $x^2$?
@MARXOS $i$ and $-i$ are in fact algebraically indistinguishable quantities, which is to say that if any polynomial equation with coefficients in $\Bbb R$ has $i$ as a root, it necessarily also has $-i$ as a root.
If $C = C(t)$ is a trajectory of a time independent vector field, we define the omega limit set of $C$, denoted by $ω(C)$, to consist of all points $p$ such that there exists a sequence $t_{n} → ∞$ such that $C(t_{n}) → p.$
For me, dreams from intensive maths thinking the other day is known to create dreams that most mistaken them as me taking LSD
(Even though I never took any illicit drugs nor my kidney condition will survive it. I am dared to say if any illicit drug get into my body, I will drop dead in seconds)
I am pretty sure my 6 year of dream log is telling people that doing maths has the same effect on me as taking LSD by someone else
Maths is known to make my dreams super vivid. The mechanism is still under investigation though
Unfortunately maths that makes sense in my dreams are rare. The recent extraction of the left right middle associative tensor looking algebra is one of those rare things that might be consistent
Physics is also known to have similar effect. Perhaps jokingly speaking physics is a type of maths lol
really? I don't believe that. Couple days ago I was ill so I couldn't do math like in 4 days. And during those days I dreamed about things, weird things and I was't doing math at all. However when I do math, I never dream about anything.
Rip, I have reached almost 100 days for the fanatic badge twice now, but yesterday I missed the log in because due to timezones, it wasn't quite a day yet when I logged in yesterday.
