1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Changing variable in summation

  1. Feb 10, 2014 #1
    Like in the integration, exist a formula to compute the summation by parts, that is: [tex]\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}[/tex][tex]\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[/tex][tex]\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[/tex]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:[tex]\sum f(x) \Delta x =\sum f(x(u)) \frac{\Delta x}{\Delta u} \Delta u[/tex]But, I think that this identity is wrong, because my calculus are wrong using this formula.
     
  2. jcsd
  3. Feb 10, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Changing variable in summation
  1. Variable Change (Replies: 69)

Loading...