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Homework Help: Steps to differentiate a geometric sum

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


    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


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    [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

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    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{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
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