# Finite Series Expansion

1. Jul 11, 2013

### JulmaJuha

Hello

How to do expand this: $(\sum_{j=1}^{n}(X(t_j)-X(t_{j-1}))^2 - t)^2$ where $X(t_j)-X(t_{j-1}) = \Delta X_j$

to this: $(\sum_{j=1}^{n}(\Delta X_j)^4 + 2*\sum_{i=1}^{n}\sum_{j<i}^{ }(\Delta X_i)^2(\Delta X_j)^2$ $-2*t*\sum_{j=1}^{n}(\Delta X_j)^2+t^2$

I get near the North Pole.... but it seems that I've forgotten some fundamental rules of finite series to do the last part of the manipulation .I assume that the blue part has to be expaned further. This is what I've done:

$\sum_{j=1}^{n}(\Delta X_i)^2 = a,t=b$

$E[(a-b)^2]=E[a^2-2ab+b^2]=E[\color{blue} {(\sum_{j=1}^{n}(\Delta X)^2)^2} \color{black} -2*t*\sum_{j=1}^{n}(\Delta X)^2+t^2]$

Any help would be greatly appreciated.

Last edited: Jul 11, 2013
2. Jul 11, 2013

### krome

I may be misunderstanding something, but I think the second term (with the double sum in $i$ and $j$) should be multiplied by $2$. Either that or the sum in $j$ should be over $j \neq i$ rather than $j<i$.

Anyway, you are correct to say that the blue term needs to be expanded further. Just try writing out one explicit example, say for $n=2$. Often, the compact summation notation obscures otherwise obvious patterns.

Also, you can use $( \sum_j f_j )^2 = ( \sum_j f_j ) ( \sum_i f_i )$.

3. Jul 11, 2013

### JulmaJuha

krome you're correct, its supposed to be multiplied by 2

4. Jul 11, 2013

### Ray Vickson

No, it should not be multiplied by 2. If $a_i = (\Delta X_i)^2$, you have to expand the sum
$$\left(\sum_i (a_i-t)\right)^2 = \left( \sum_i a_i - nt \right)^2 = \sum_i a_i^2 + 2\sum_{i<j} a_i a_j - 2nt \sum_i a_i + n^2 t^2.$$

5. Jul 11, 2013

### krome

Good lord! I must be going mad or selectively blind. I swear when I read this last night the second term did not have a factor of 2!