# Combine partial sums

$$\begin{eqnarray*} 2xu''_1 + (1+x) u'_1 + u_1 &=& 2 \sum_{n=0}^{\infty} \Big( n + \frac12 \Big) \Big( n-\frac12 \Big) a_n x^{n-\frac12} + \sum_{n=0}^{\infty} \Big( n + \frac12 \Big) a_n x^{n-\frac12} + \\ && + \sum_{n=0}^{\infty} \Big( n + \frac12 \Big) a_n x^{n+\frac12} + \sum_{n=0}^{\infty} a_n x^{n+\frac12} \\ &=& \sum_{n=1}^{\infty} \Big( n + \frac12 \Big) 2n a_n x^{n-\frac12} + \sum_{n=0}^{\infty} \Big( n + \frac32 \Big) a_n x^{n+\frac12} \\ &=& \sum_{n=0}^{\infty} \left( \Big( n + \frac32 \Big) 2(n+1) a_{n+1} + \Big( n + \frac32 \Big) 2n a_n \right) x^{n+\frac12} \\ &=& 0. \end{eqnarray*}$$

It is part of a Frobenius problem, that I am looking at. Can someone explain what is happening in this? For example, why becomes it $$\Big(n+\frac{1}{2} \Big)2n?$$ I have tried to do $$\left( \Big(n+\frac{1}{2}\Big)\Big(n-\frac{1}{2}\Big) + \Big(n+\frac{1}{2}\Big)\right)a_{n},$$ but i am not getting it right.

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## 2 Answers

\begin{align*} &2\sum_{n=0}^\infty\left(n+\frac12\right)\left(n-\frac12\right)a_nx^{n-\frac12}+\sum_{n=0}^\infty\left(n+\frac12\right)a_nx^{n-\frac12}=\\ &\sum_{n=0}^\infty\left(2\left(n^2-\frac14\right)+\left(n+\frac12\right)\right)a_nx^{n-\frac12}=\\ &\sum_{n=0}^\infty(2n^2+n)a_nx^{n-\frac12}=\\ &\sum_{n=0}^\infty2n\left(n+\frac12\right)a_nx^{n-\frac12}=\\ &\sum_{n=1}^\infty\left(n+\frac12\right)2na_nx^{n-\frac12}, \end{align*}

and

\begin{align*} \sum_{n=0}^\infty\left(n+\frac12\right)a_nx^{n+\frac12}+\sum_{n=0}^\infty a_nx^{n+\frac12} &= \sum_{n=0}^\infty \left(n+\frac12+1\right)a_nx^{n+\frac12}\\ &=\sum_{n=0}^\infty\left(n+\frac32\right)a_nx^{n+\frac12}; \end{align*}

this yields the second equality. The third is just an index shift so that the two summations can be combined into one: \begin{align*} \sum_{n=1}^\infty\left(n+\frac12\right)2na_nx^{n-\frac12}&=\sum_{n=0}^\infty\left((n+1)+\frac12\right)2(n+1)a_{n+1}x^{(n+1)-\frac12}\\ &=\sum_{n=0}^\infty\left(n+\frac32\right)2(n+1)a_{n+1}x^{n+\frac12}\;. \end{align*}

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Thank you, made a silly mistake to forget multiply 2 to get -1/2 :S – krunarsson Nov 16 '11 at 15:29

If you put $u= \sum_{n=0}^\infty a_n x^{n+\frac{1}{2}}$ then $u'$ becomes $u'= \sum_{n=0}^\infty a_n (n+\frac{1}{2}) x^{n-\frac{1}{2}}$ and $u''$ becomes $u''=\sum_{n=0}^\infty a_n (n+\frac{1}{2})(n-\frac{1}{2})x^{n-\frac{3}{2}}$. So putting all this into the formula $2xu''+(1+x)u'+u$ you get the righthand side.

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I know, But what happens between line 3 and 4. When (n+1/2)(n-1/2) becomes (n+1/2)2n? – krunarsson Nov 16 '11 at 15:03