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stgreen
Group Title
Show that the singular point of the following function is a pole. Determine the
order m of that pole and the corresponding residue B.
f(z)=(1coshz)/(z^3)
 one year ago
 one year ago
stgreen Group Title
Show that the singular point of the following function is a pole. Determine the order m of that pole and the corresponding residue B. f(z)=(1coshz)/(z^3)
 one year ago
 one year ago

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malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
I really wish my university had an applied complex variables class :/
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i think we can at least start this if i recall \(\cosh(z)=\cos(iz)\) if so, we can expand in a taylor series
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
we can try \[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
this should give \[1\cosh(z)=\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
yeah got it
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
Which would then give: \[\sum_{k=1}^{\infty}\frac{z^{2k3}}{(2k)!}\] right?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
@malevolence19 yes, that is what i get too. or shift the index either way.
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
So the the residue is just 1/2?
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
so b1 comes out to be 1/2.yeah?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
yes if my memory serves me in this context the residue is the coefficient of the \(z^{1}\) term
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, 1/2 i forgot about that in front.
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i would not place much money on this, because it has been a while, but i am pretty sure it is correct
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
and pole is??i don't have any idea what the heck is it
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
It would just be zero right?
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
It it was like: \[\frac{B}{(z\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
lambda is a singular point..not sure about pole
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
yes, the pole is at 0, because that is where the denominator is undefined. it is a simple pole because it is a zero of order one
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
^the denominator becomes undefined at singular points...so every singular point is a pole??
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
i think its pole is 1
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
a "pole" is a kind of singular point think of a simple example from pre calc class \[\frac{x^24}{x2}\] vs \[\frac{x^2}{x2}\] in the first example, 2
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
this thing implies m=1 at b1
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
in both examples, 2 is a singular point because the function is not defined at 2 but in the first example, 2 is a removable singularity, because you can remove it
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
in the second example, 2 is a pole, in pre calc a "vertical asypmtote"
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
however it is spelled
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
isolated singular point.what is it??
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
ok hold the phone
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
what you posted was about the order of the pole, not the pole itself
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
in your case above, we had \(\frac{1}{2z}\) plus stuff with positive exponents, means you have a pole of ORDER 1, not a pole at \(z=1\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
here is a nice succinct explanation (very short) with example http://www.solitaryroad.com/c612.html
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
implies order of pole=k if bk is the coefficient of z^1 term.right?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
the order of the pole is the multplicty of the zero on the denominator, to put it in pre calculus terms, example \[\frac{1}{(z5)^3}\] has a pole of order 3 at \(z=5\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{1}{z2}\] has a pole of order 1 at \(z=2\) a pole of order one is called a simple pole
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
oh *shouts got it*
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.4
thanks bro
 one year ago
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