# Changing variable in summation

1. Feb 10, 2014

### Jhenrique

Like in the integration, exist a formula to compute the summation by parts, that is: $$\frac{\Delta }{\Delta x}(f(x)g(x))=\frac{\Delta f}{\Delta x}g+f\frac{\Delta g}{\Delta x}+\frac{\Delta f}{\Delta x}\frac{\Delta g}{\Delta x}$$$$\sum \frac{\Delta }{\Delta x}(f(x)g(x))\Delta x = \sum \frac{\Delta f}{\Delta x}g\Delta x + \sum f\frac{\Delta g}{\Delta x}\Delta x + \sum \frac{\Delta f}{\Delta x}\frac{\Delta g}{\Delta x}\Delta x$$$$\sum f(x) \frac{\Delta g}{\Delta x}(x)\Delta x = f g - \sum \frac{\Delta f}{\Delta x} g \Delta x - \sum \frac{\Delta f}{\Delta x} \frac{\Delta g}{\Delta x}\Delta x$$http://en.wikipedia.org/wiki/Indefinite_sum#Summation_by_parts

But in the integration is possbile make a change the variable (integration by substitution). So, analogously, is possible to make a change the variable in the summation too?

I tried some like:$$\sum f(x) \Delta x =\sum f(x(u)) \frac{\Delta x}{\Delta u} \Delta u$$But, I think that this identity is wrong, because my calculus are wrong using this formula.

2. Feb 10, 2014

### Simon Bridge

I don't think you've done the substitution correctly.

If you want:
$$\sum_{i=1}^N f(u(x_i))\Delta x = \sum_{i=1}^N g(u_i)\Delta u : u_i=u(x_i)$$

Then you want $$g(u_i) = f(x_i)\frac{\Delta x}{\Delta u} : x_i = h(u_i)=u^{-1}(u_i)$$

i.e. $f(u(x))dx \rightarrow f(u)(dx/du)du$