The derivative of a Taylor series?

[badamann]

I took my first calculus class over the last two semesters, and my teacher and I privately worked on some harder material together. Toward the end of the school year he gave me a question that I never answered and never found an answer for. It asked me to find the derivative of a Taylor series. I don't remember the specific problem, but if someone could shed some light, that'd be lovely!

 

[mathwonk]

you are kidding right? i.e. what do you think?

 

[badamann]

I'm not kidding at all. At the time I received the problem I'd never heard of a Taylor series, and I haven't seen a problem with one since. It's something I've never had the opportunity to learn at school thus far.

 

[A. Bahat]

I'm not sure if this is exactly what you're looking for, but if f(x)=Ʃa_nxⁿ in some interval, then f'(x)=Ʃna_nx^n-1 in the interior of that interval; that is, you can differentiate the series termwise. This is because of uniform convergence, which can be shown with the Weierstrass M-test.

 

[A. Bahat]

Really, it's a very plausible generalization of what you already know for finite polynomials.

 

[badamann]

I'm really just caught up in looking at the sigma notation.. It throws me out of whack. I'm assuming I'm making a much bigger deal out of this than I should be.

This may be a better question:

If f(x) = f(a) + [f'(a)/1!](x-a) + [f"(a)/2!](x-a)^2 + ...
Then f'(x) = ?

 

[pwsnafu]

This may be a better question:

If f(x) = f(a) + [f'(a)/1!](x-a) + [f"(a)/2!](x-a)^2 + ...
Then f'(x) = ?

Exactly what you would get if you differentiated each term individually.
Provided it converges.

 

[lugita15]

badamann, just use the power rule.

 

[badamann]

Thank you. I'm not sure why I made it out to be quite so difficult... :(