# Differentiation of a Power Series

1. Nov 1, 2004

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.

Last edited: Nov 1, 2004
2. Nov 1, 2004

### 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.

3. Nov 1, 2004