Question

In order to get the value of a function at a removable singularity, could we use the residue theorem?

Answer 1

the function does not exist when there is singularity. however you can define (extend) it as it is equal to it's limit.

Answer 2

I meant a removable singularity, so that the function would have a value.

Answer 3

for this particular problem, the integrability should imply continuity but function is not continuous at x=5 (removable discontinuity) yet it is integrated. The function is defined as zero at this point.
http://www.wolframalpha.com/input/?i=integrate+e^%28-1%2F%28x-5%29^2%29+from+4+to+6

Answer 4

I could put down an example, \[(e ^{z}-1)/z\], has a removable singulatiry at 0 apparently and its value is 1 at that location.

Answer 5

hmm, I'm just not sure how to determine whether a singularity is removable or not, do I use L'hospitals to figure that out?

Answer 6

limits exist (finite) but functional value is indeterminate => removable singularity

Answer 7

for this particular case, you can use L'Hopital rule to find out the limit as it is 1.

Answer 8

I see, so in general is that the proper way to approach whether a singularity is removable or not?

Answer 9

yep ... where there is removable singularity, you can integrate it. (Lesbegue integrability) i have little knowledge on Analysis. but for this particular integral, it's almost same as x/x. just expand e^z and find the Laurent series.

Answer 10

I see thanks so much, just one more question though. Is the residue of a function basically defined to be the value of the line integral at the residue?

Answer 11

I think the residue is the value of coefficient of 1/z of the Laurent series. Let me check it again. I seem to have forgotten.

Answer 12

Or I'm probably not stating it right, but if the function has no residue, than the value around a closed loop is zero? Other wise it exists and is as you stated the coeffiencet at -1

Answer 13

yes, the residue must be zero.

Answer 14

Allright thanks so much for the help, much appreciated.

Answer 15

yep ... in Laurent series, the \( a_{-1} \) term is called residue. If there is no 1/z term in the Laurent expansion then i guess no residue.

Answer 16

And is the Laurent expansion also known as the power series of the function, or is just one form of a power series expansion? Sorry for all the questions, I'm just trying to get my head around all this stuff.

Answer 17

this is just like Taylor series usually expanded at the point of singularity
http://en.wikipedia.org/wiki/Laurent_series

Answer 18

Okay I see, thanks, which book is that btw?

Answer 19

that is solution of my past exams questions.

Answer 20

Is it an A or O level book by any chance ?

Answer 21

no it's university level.

Answer 22

I'm guessing there isnt a PDF version on the web somewhere?

Answer 23

no ... i had stored photograph of some of few problems. There are excellent resources available.

Answer 24

Allright, thats too bad. Thanks for all the help again!

Answer 25

try this ... i haven't tried though
http://ocw.mit.edu/courses/mathematics/18-04-complex-variables-with-applications-fall-2003/

Answer 26

True, I've tried them for some of there other courses. Pretty useful. I'll give it a look through.