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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.

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# Differentiation of a Power Series

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