Hi there,(adsbygoogle = window.adsbygoogle || []).push({});

I have a question regarding the differentiation of a power series. I understand what my book says about the index in a power series...

[tex] \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} c_n \left( x - a \right) ^n \right] = \sum _{n=1} ^{\infty} n c_n \left( x - a \right) ^{n-1} [/tex]

it changes from [tex] n=0 [/tex] to [tex] n=1 [/tex] due to the loss of a term, namely [tex] c_0 [/tex].

[tex] \frac{d}{dx} \left[ \sum _{n=1} ^{\infty} n c_n \left( x - a \right) ^{n-1} \right] = \sum _{n=2} ^{\infty} n \left( n-1 \right) c_n \left( x - a \right) ^{n-2} [/tex]

it changes from [tex] n=1 [/tex] to [tex] n=2 [/tex] due to the loss of a term, namely [tex] c_1 [/tex].

I get a bit confused when it comes down to the real stuff. For example, consider the following:

[tex] \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right] = \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n \left( 2n+1 \right) x^{2n}}{n! \left( n+1 \right) ! 2^{2n+1}} [/tex]

and

[tex] \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n \left( 2n+1 \right) x^{2n}}{n! \left( n+1 \right) ! 2^{2n+1}} \right] = \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n \left( 2n \right) \left( 2n+1 \right) x^{2n-1}}{n! \left( n+1 \right) ! 2^{2n+1}} [/tex]

The index does not change in either case. My question is: do I always need to expand the series to determine it? Or, is there a shortcut to determine the value of the index?

Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Differentiation of a Power Series

**Physics Forums | Science Articles, Homework Help, Discussion**