Differentiation of a Power Series

[DivGradCurl]

Hi there,

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

\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}

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

\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}

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

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

\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}}

and

\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}}

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.

 

[shmoe]

Hi, notice in your last example that the term with n=0 is actually zero, so you could start the index at 1 difficulty. They should have actually done this, as it's written you have a x^{-1} term, albeit with a zero coefficient.

Just look at the lowest power of x in your power series. If it's x^0, you'll lose this term when you differentiate and have to change the index.

 

[DivGradCurl]

Oh... I see. I just didn't notice that before... thank you.