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I have never had to index shift a summation series before, and it seems relatively straightforward, however, I am looking at an example in my textbook that doesn't make sense. I am wondering if someone might be able to outline the steps that appear to be missing.

Apologies for the formatting issue; perhaps someone can help with that?

Here is how it appears in the textbook:

$(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n$

$= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + \sum_{n=0}^{\infty}[2n(n-1)+6n+2]a_{n}x^n$

So it looks as though the two rightmost terms were combined somehow, but I cannot figure how this took place.

Can someone help with my understanding? I've never done anything like this before, and I'm frustrated I can't understand the rest of the example because of this.

Thanks!

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  • $\begingroup$ So I think I've made some progress in terms of figuring out what was done. It looks as though the exponents of the terms were reduced to x^n, and that this had the effect of changing the start of the index to zero. It's hard to explain. Does anyone know what I mean? $\endgroup$
    – TheColonel
    Aug 31, 2014 at 3:08
  • $\begingroup$ Yep. I think I know what you mean, the lumping in the $n=0$ and $n=1$ terms is a bit slippery. $\endgroup$ Aug 31, 2014 at 3:18
  • $\begingroup$ See my reply below. The only thing I had to go off of was my intuition and the text. So I had to make some conclusions. Check out the screenshot from the text. $\endgroup$
    – TheColonel
    Aug 31, 2014 at 3:21

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Ok, just distribute to see: \begin{align} &(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n =\\ &=\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +2x^2\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2}+ \sum_{n=1}^{\infty}6na_{n}xx^{n-1}+\sum_{n=0}^{\infty}2a_{n}x^n \\ &= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty}2n(n-1)a{_{n}}x^2x^{n-2}+ \sum_{n=1}^{\infty}6na_{n}x^{n}+\sum_{n=0}^{\infty}2a_{n}x^n \\ &= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty}2n(n-1)a{_{n}}x^{n}+ \sum_{n=2}^{\infty}6na_{n}x^{n}+\sum_{n=2}^{\infty}2a_{n}x^n +6a_1x+2a_0+2a_1x \\ &= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty} \left( 2n(n-1)+ 6n+2 \right)a_nx^n +8a_1x+2a_0\\ \end{align} Now, notice the rightmost sum begins at $n=0$ in your post, can we see the exceptional terms as the $n=0$ and $n=1$ terms in the text's expression?

Added: I thought it might be useful to show how to group all the terms: continuing, \begin{align} &(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n =\\ &= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=0}^{\infty} \left( 2n(n-1)+ 6n+2 \right)a_nx^n \\ &= \sum_{j=0}^{\infty}(j+2)(j+1)a{_{j+2}}x^{j} +\sum_{n=0}^{\infty} \left( 2n(n-1)+ 6n+2 \right)a_nx^n \\ &= \sum_{j=0}^{\infty}(j+2)(j+1)a{_{j+2}}x^{j} +\sum_{n=0}^{\infty} \left( 2n(n-1)+ 6n+2 \right)a_nx^n \\ &= \sum_{n=0}^{\infty} \left( [2n(n-1)+ 6n+2]a_n+(n+2)(n+1)a{_{n+2}} \right)x^n \end{align} So, then when we have this is zero we find relations between the $n$-th and $n+2$-th coefficients.

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  • $\begingroup$ to answer my question: yes, yes we can. $\endgroup$ Aug 31, 2014 at 3:19
  • $\begingroup$ James, thanks for the reply. Here is a screenshot of the lines in question from the example. The index shows as zero. Unless it's some sort of a mistake: imgur.com/c8FAx1b $\endgroup$
    – TheColonel
    Aug 31, 2014 at 3:19
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    $\begingroup$ no it's not a mistake. The $n=1$ gives $[2(1)(0)+6(1)+2]a_1x=8a_1x$ and $n=0$ gives $[2(0)(-1)+6(0)+2]a_0=2a_0$ so those exceptional terms get put into the right sum by starting $n$ at $0$ rather than $2$. $\endgroup$ Aug 31, 2014 at 3:21
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    $\begingroup$ Yep. Think of it as the author sweeping the dirt into two piles rather than leaving 4. $\endgroup$ Aug 31, 2014 at 3:29
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    $\begingroup$ Exactly, sadly the stack exchange software is now scolding us for discussion of math here. I hope the added few lines are helpful as you study further... $\endgroup$ Aug 31, 2014 at 3:39

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