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what the hell does this equal to? 0? infinity? undefined? z

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Thread images: 2

what the hell does this equal to?
0? infinity? undefined?

z is complex
>>
http://www.wolframalpha.com/input/?i=lim+z-%3E+infinity+e^%281%2Fz%29

http://www.wolframalpha.com/input/?i=lim+z-%3E+0+e^%281%2Fz%29
>>
>>7648337
i mean only the z->0 limit
it is 0 from one side and infinity from the other
what gives?
>>
>>7648344

doesn't a function have to meet certain criteria to be differentiable or have a limit? (continuity or something?) i don't remember

http://www.wolframalpha.com/input/?i=plot%20e^%281%2Fz%29&lk=2
>>
>>7648308

fuck you for breaking my brain!

https://en.wikipedia.org/wiki/Essential_singularity
>>
>>7648308
Undefined, because you can get 0 and infinity from the limit depending on what line you approach it from. So: it is undefined. You could determine the residue though, which is 1.
>>
>>7648355
well, I guess that settles it
>>
>>7648360
>the residue
can you explain how it is 1?
>>
>>7648367
look up the power series of the exponential function: exp(z). Replace z by 1/z and you've got the power series for the function we're discussing. According to Cauchy's residue theorem, every function that is differentiable in a closed domain C has a net integral value of 0. So every analytical function nets 0 when you integrate it over a closed region, EXCEPT for 1/(z-z0), where z0 is 0 for this case. If you look at the power series of this function, you can see that it contains exactly one "copy" of 1/z; the rest is 1/z^2, 1/z^3... : all differentiable! So if you were to integrate this function over a circle around 0, the answer would be 2*pi*i times the residu, which is the coefficient of the 1/z term: 1.
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>>7648374
>the beauty of french mathematics
>>
>>7648386
Considering how boring real analysis is, complx analysis is a blessing.
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>>7648374
thanks a lot
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>>7648374
Whoops, mistake by me.
>1/z^2, 1/z^3: all differentiable
I meant to say: all integratable, which implies that the integral is 0
>>
Check out the Grand Picard Theorem too. Basically, exp(1/z) takes every complex value except zero infinitely often in the vicinity of z=0.
>>
>>7648374
Thank you, complex analysis is really interesting.
I love how you can use it to solve crazy rational function real integrals from -inf to inf by using jordans lemma or Ml inequality then using residue theorem or something else on the contour integral.
>>
>>7648418
so, the residue is 0 or 1?
>>
>>7648489
The residue is 1. The closed integral that includes 0 is 2πi.
>>
The limit isn't defined because there isn't one single path to move a complex number to a single point as is the case in the reals.
Thread posts: 18
Thread images: 2


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