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I want to shift the index of $$\sum_{n=2}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}$$ to start at $n=0$. If I expand this series I get:

$$a_2\cdot 2\cdot 1\cdot x^0+a_3\cdot 3\cdot 2\cdot x^1+\cdots$$

But if I start simply at $n=0$ without doing any modification, I get:

$$a_0\cdot 0\cdot (-1)\cdot x^{-2}+a_1\cdot 1\cdot0\cdot x^{-1}+a_2\cdot2\cdot1\cdot x^0+\cdots$$

which is the same as the series above, because the first $2$ terms are $0$, so I tought I could just write it like this:

$$\sum_{n=0}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}$$

but My book writes it as the following:

$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$$

what's happening?

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  • $\begingroup$ Both sums are the same because of the reason that you name, that the first two terms are equal to $0$. $\endgroup$ – Jimmy R. Nov 25 '16 at 10:52
  • $\begingroup$ Don't start at $0$, just apply directly the shift $n\mapsto n+2$ and the result will follow. (Note that with this shift the new index stars at $0$, as wished) $\endgroup$ – b00n heT Nov 25 '16 at 10:53
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Both are correct. In the book, they just "shifted" the whole expression.

They just replaced $n$ with $n+2$. So instead of starting with $n=2$ they start with $n=0$, but they use $n+2$ in the expression, so it is the same as just using $n=2$ in the original formula.

If you expand both series, you will see it is the same.

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Nothing happens !

$\sum_{n=0}^{\infty}a_n\cdot n\cdot (n-1)\cdot x^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$

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  • 2
    $\begingroup$ Please improve this answer (-1) $\endgroup$ – b00n heT Nov 25 '16 at 10:55

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