Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

find the taylor expansion around a singular point from a function f(z) = [exp(z)/(z-1)^{3}]!

Linear Algebra
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
find the taylor expansion around a singular point from a function \[f(z) = \frac{ \exp (z) }{ (z-1)^{3} } !\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Do you know what a singular point is?
@gerryliyana the point is that your radius of convergence is the distance your function is from its nearest singular point, where the function has a singularity (see: http://en.wikipedia.org/wiki/Mathematical_singularity#Complex_analysis ). I'm going to presume that stray \(!\) is a typo.
then, the question is wrong?
@gerryliyana no, but it won't converge I think.
with a radius of convergence of 0
ah ok.., then What form of expansion ??
Watch mate... we know \(\exp(z)\) is entire, but \(\frac1{(z-1)^3}\) is not -- it has a pole at \(z=1\). So we've found a singularity. Now let's try constructing a Taylor expansion around \(z=1\):$$\exp(z)=\sum_{n=0}^\infty\frac{e}{n!}(z-1)^n\\\frac{\exp(z)}{(z-1)^3}=\sum_{n=0}^\infty\frac{e}{n!}(z-1)^{n-3}$$oops, that's not a Taylor series, though -- it's a Laurent series.
yeah..., it's laurent.., i'm sorry, paper told us laurent., :). Thank you oldrin
@gerryliyana if you want it int he proper Laurent form rewrite:$$\frac{\exp(z)}{(z-1)^3}=\sum_{n=-3}^\infty\frac{e}{(n+3)!}(z-1)^n$$
ok.., thank you.., hei have a facebook ?? add me on fb at https://www.facebook.com/gerryliyana

Not the answer you are looking for?

Search for more explanations.

Ask your own question