Recent content by shebbbbo

oh whoops, stupid error by me. that gets me to the result though so that great! thanks heaps for all your help with this question.

yeah i realized after i wrote it that it is a little ambiguous to say it like that. but i felt like you would know what i meant. so i think when you combine those two integrals (by addition) i come out with something like this... \int e-rt*\frac{1}{r}*(2 - \frac{4}{r+1})dr which is an...

thanks jackmell. that helps a heap. i have done the analysis of all the other contours and unless i have made a mathematical mistake I am pretty confident they are all 0. but your suggestion is a great way to check it. the only thing I am confused about is how you arrived at your expression...

Homework Statement Find the Inverse Laplace Transform of \frac{1}{s}*\frac{\sqrt{s}-1}{\sqrt{s}+1} The Attempt at a Solution for this question i found the singularities to be at 0 and when s = 1. (as the sqrt of 1 is ± 1) there is also a branch point that runs from 0→-∞. so if you...

Let f: ℂ→ ℂ be an entire function. If there is some nonnegative integer m and positive constants M,R such that |f(z)| ≤ M|z|m, for all z such that |z|≥ R, show that f is a polynomial of degree less that or equal to m. im really lost on this question. i feel like because...

oh yeah... that makes sense... your right we can't get i and -i in on a path integral across the real axis maybe i need to build a circle or something? i guess i have less of an idea now? thanks for your help? any idea where i should go from here?

The question asks to show using the residue theorem that \intcos(x) / (x2 +1)2 dx = \pi / e (the terminals of the integral are -∞ to ∞ but i didnt know the code to write that) I found the singularities at -i and +i so i think we change the function inside the integral to cos(z) / (z2...

ive looked on that page and stared at it for ages... will both methods work? and am i correct with what i have done so far? (ie poles at +i and -i)?

The question asks to show using the residue theorem that \int cos(x)/(x2+1)2 dx = \pi/e (the terminals of the integral are -\infty to \infty but i didnt know the code to write that) I found the singularities at -i and +i so i think we can then say \intcos (z) / (z+i)2(z-i)2 dz...

im now more confused... are you saying i don't need the ML inequality? I don't understand where the R in the first line comes from? and where do you go from the last line to get closer to the solution? im sorry i havnt understood

Having trouble with this question: The question is: establish the inequality |\inteizzdz| \leq \pi(1-e-R2)/4R on C {z(t) = Reit, t \in [0,\pi/4, R>0 When i saw the modulus of an integral i thought ML inequality. I think the length will be R\pi/4 but I am struggling with...

yeah good point. thanks for all your help!

Once i expanded them i realized they looked exactly like the triangle inequality where the modulus of the summation of terms was less than or equal to the modulus of each term summed. i didnt try to conclude |e^z| = e^|z| from what i was reading i think they are equal when z is real and...

thanks!

great thanks... quick question: does |ez| = e|z| ?