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Suppose I have a summation of the form $\sum\limits_{n=0}^\infty n(n-1) c_n x^{n-2}$. As is done in my textbook, we can replace $n$ with $n + 2$ and begin the summation at $n = 2$ instead of $n = 0$, giving us $\sum\limits_{n=2}^{\infty} (n+2)(n+1) c_n x^{n-2}$. I'm having difficulty following this logic. Is there a general rule for rewriting the summation in this way? What if I wanted to take the summation that begins at, say, $n = 1$? How would I go backwards, in this sense?

Thanks in advance.

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1 Answer 1

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There is something wrong in the second expression.

If you shift the index $$\sum\limits_{n=0}^\infty n(n-1) c_n x^{n-2}=\sum\limits_{n=2}^\infty n(n-1) c_n x^{n-2}=\sum\limits_{n=0}^{\infty} (n+2)(n+1) c_{n+2} x^{n}$$

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  • $\begingroup$ Thanks for this. I must have written the wrong expression, which was initially taken from a textbook. Can you explain what the general rule for shifting the starting point of the summation is? $\endgroup$
    – user465188
    Feb 10, 2018 at 16:06
  • $\begingroup$ @Matt.P. One more typo in a textbook probably. May I confess that I hate to see shifitng the index for no specific reason. When you will learn (if not yet done) solutions of differential equations using series, contact me and I shall try to show you that shifitng the index makes life more difficult. $\endgroup$
    – Claude Leibovici
    Feb 10, 2018 at 16:15
  • $\begingroup$ The funny thing is, that's precisely what I'm working on now. The textbook shifted the index when taking the second derivative, which does seem to make things messier. $\endgroup$
    – user465188
    Feb 10, 2018 at 16:19
  • $\begingroup$ @Matt.P. This is exactly the point ! I never do it (but this is a personal point of view). $\endgroup$
    – Claude Leibovici
    Feb 10, 2018 at 16:45

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