This is a blue board which means that it's for everybody (Safe For Work content only). If you see any adult content, please report it.

Thread replies: 18

Thread images: 2

Thread images: 2

File: fuck this shit, I would be a stripper if I was a girl.png (14KB, 650x469px) Image search:
[iqdb]
[SauceNao]
[Google]

14KB, 650x469px

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.

>>

>>7648374

>the beauty of french mathematics

>>

>>7648386

Considering how boring real analysis is, complx analysis is a blessing.

>>

>>7648374

thanks a lot

>>

>>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

Thread images: 2

If a post contains copyrighted or illegal content, please click on that post's

All trademarks and copyrights on this page are owned by their respective parties. Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.

This is a 4chan archive - all of the content originated from that site. This means that 4Archive shows an archive of their content. If you need information for a Poster - contact them.