1. Not finding help here? Sign up for a free 30min 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!

Steps to differentiate a geometric sum

  1. Sep 5, 2014 #1
    Can someone guide me with the steps to differentiate a geometric sum, x?

    [itex]^{n}_{i=0}[/itex][itex]\sum[/itex]x[STRIKE][itex]^{i}[/itex][/STRIKE]=[itex]\frac{1-x^{n+i}}{1-x}[/itex]

    If i'm not wrong, the summation means:

    [itex]= x^0 + x^1 + x^2 + x^3 + ..... + n^i[/itex]

    Problem is:
    I have basic knowledge on differentiating a normal numbers but how do I apply differentiate on a geometric sum?
     
    Last edited: Sep 5, 2014
  2. jcsd
  3. Sep 5, 2014 #2
    Its been a while since I looked at these kinds of problems, so I checked out Paul's Online Calculus Notes, which is a good reference.

    I would think that...

    [itex] \frac {d}{dx} \sum_{n=0}^{i} x^{i} = \sum_{n=0}^{i} \frac {d}{dx} x^{i} [/itex].

    From there, I would see the what the series actually looks like, then express the summation as a formula.
     
  4. Sep 5, 2014 #3
    And I think that this is valid for the derivative of a Power Series.
     
  5. Sep 5, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\frac{d}{dx}(\sum a_n(x))= \sum \frac{da_n(x)}{dx}[/tex]

    if and only if the initial sum is "uniformly convergent". All power series are uniformly convergent inside the radius of convergence.
     
  6. Sep 5, 2014 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You are wrong: it means
    [tex]x^0 + x^1 + x^2 + \cdots + x^n = 1 + x + x^2 + \cdots + x^n.[/tex]
    Note how the last term differs from yours. Also, this is equal to
    [tex] \frac{1-x^{n+1}}{1-x}[/tex]
    if ##x \neq 1##. You wrote something different, with an ##i## in it (but there is no "##i##" outside of the summation---that is, after you do the sum, the "##i##' no longer 'exists').

    Anyway, your last sentence makes no sense: you cannot differentiate a normal number (unless you want to get 0 always), but you can differentiate functions of x, like your series.

    There are two ways to proceed here: (1) differentiate the series directly; and (2) differentiate the formula you have obtained for the sum. In other words:
    [tex] (1) \;\text{derivative } = \frac{d}{dx} (1 + x + x^2 + \cdots + x^n) = 1 + 2x + 3x^2 + \cdots + n x^{n-1}\\
    (2)\; \text{derivative } = \frac{d}{dx} \frac{1-x^{n+1}}{1-x} = \cdots \cdots[/tex]
    I will let you do the derivative here.
     
  7. Sep 5, 2014 #6
    Thank you guys for your reply.

    I did the (2) derivative by Ray Vickson and this is the draft solution I got.

    Is this correct to say that the derivatives both sides of the geometric sum , with respect of x is:

    [itex]\frac{d}{dx}[/itex][itex]\frac{1-x^{n+1}}{1-x}[/itex]
    =[itex]\frac{x^{n}(n(x-1)-1)+1}{(x-1)(x-1)}[/itex]
    =[itex]\frac{x^{n}(n(x-1)-1)+1}{(x-1)^2}[/itex] (simplified final answer)


    I got confused with the question saying "derivatives both sides of the geometric sum". I thought I will need to do the derivatives for both LHS and RHS?
     
    Last edited: Sep 6, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Steps to differentiate a geometric sum
  1. Sum of geometric series (Replies: 10)

Loading...