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Differentiation

If I am differentiating an infinite power series eg the power series expansion for the exp function, do I need to assume the axiom of choice to make term by term differentiation valid?

Reply 1

Oh I Really Don't Care

No there is no need to explicitly use it in the proof.

*the proof that the derivitive of the complex series;

\displaystyle \sum a_nx^n

\displaystyle \displaystyle \sum na_nx^{n-1}

when x is within the radius of convergence of the original series [both the derivitive and series will have same radius of convergence).

Reply 2

coldfusion

DeanK22

No there is no need to explicitly use it in the proof.

*the proof that the derivitive of the complex series;

\displaystyle \sum a_nx^n

\displaystyle \displaystyle \sum na_nx^{n-1}

when x is within the radius of convergence of the original series [both the derivitive and series will have same radius of convergence).

Well actually I'm trying to prove product rule in unconventional circumstances (I'm working with operators in quantum mechanics). So if term by term differentiation in the first power series never ends then I'll never get to differentiating the second power series so the product rule fails which is very bad for me :wink:

Reply 3

Oh I Really Don't Care

Well it clearly exists - if you except that you can add infinitely many terms of the series you are essentially doing infinitely many operations here so I don't see how you can question that you get in trouble when performing the operation of finding the derivative of infinitely many terms of the series.