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Differentiation of a Power Series

  1. Nov 1, 2004 #1
    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...

    [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]


    [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?

    Last edited: Nov 1, 2004
  2. jcsd
  3. Nov 1, 2004 #2


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    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 [tex]x^{-1}[/tex] term, albeit with a zero coefficient.

    Just look at the lowest power of x in your power series. If it's [tex]x^0[/tex], you'll lose this term when you differentiate and have to change the index.
  4. Nov 1, 2004 #3
    Oh... I see. I just didn't notice that before... thank you.
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