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I really wish my university had an applied complex variables class :/

we can try
\[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]

this should give
\[1-\cosh(z)=-\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]

yeah got it

Which would then give:
\[-\sum_{k=1}^{\infty}\frac{z^{2k-3}}{(2k)!}\] right?

divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator

@malevolence19 yes, that is what i get too. or shift the index either way.

So the the residue is just 1/2?

so b1 comes out to be -1/2.yeah?

yes if my memory serves me in this context the residue is the coefficient of the \(z^{-1}\) term

^right

Yeah, -1/2 i forgot about that in front.

and pole is??i don't have any idea what the heck is it

It would just be zero right?

lambda is a singular point..not sure about pole

^the denominator becomes undefined at singular points...so every singular point is a pole??

i think its pole is 1

this thing implies m=1 at b1

in the second example, 2 is a pole, in pre calc a "vertical asypmtote"

however it is spelled

isolated singular point.what is it??

ok hold the phone

what you posted was about the order of the pole, not the pole itself

here is a nice succinct explanation (very short) with example
http://www.solitaryroad.com/c612.html

implies order of pole=k if bk is the co-efficient of z^-1 term.right?

\[\frac{1}{z-2}\] has a pole of order 1 at \(z=2\)
a pole of order one is called a simple pole

oh *shouts got it*

good!

thanks bro

yw