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Power Series - Finding x values for which the series equals a certain number

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Given [itex]x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... = k[/itex], are there any values of x for the values [itex]k = -100, \frac{1}{2}, 100[/itex]?

    2. Relevant equations


    3. The attempt at a solution
    I started by finding that the series is [itex]\Sigma^{\infty}_{n=1} \frac{(-1)^{n+1}x^n}{n} = k[/itex]. I then used the ratio test to find that the radius of convergence = R = 1. I'm not really sure where to go from here though. I'm hoping someone can point me in the right direction as to how I can begin to find out if there are any values of x so that the series equals either -100, [itex]\frac{1}{2}[/itex], or 100.
     
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  3. Feb 5, 2013 #2

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    Welcome to PF, Illania! :smile:

    Easiest would be to find the sum first.
    Afterward we would need to consider in which cases that sum is valid, depending on the radius of convergence.

    Suppose we define the function ##f(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...##.

    Can you say what f'(x) is?
    Can you sum it?
    And integrate it?
     
  4. Feb 5, 2013 #3
    Thanks for the welcome. Everyone here seems very helpful so far.

    I've found that [itex]f'(x) = \Sigma^{∞}_{n=1}(-1)^{n+1}x^{n-1}[/itex]
    However, I'm not sure what you mean by summing it.
     
  5. Feb 5, 2013 #4

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    Let's write it as

    ##f'(x)=1-x+x^2-x^3+...##

    This is a geometric series.
    Do you know what the sum is of a geometric series?
     
  6. Feb 5, 2013 #5
    The sum of a geometric series is [itex]\frac{a}{1-r}[/itex], so would the sum here just be [itex]\frac{1}{1+x}[/itex]? How do you account for the [itex](-1)^{n+1}[/itex] in the sum?
     
  7. Feb 5, 2013 #6

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    Yep.

    The [itex](-1)^{n+1}[/itex] is included in the geometric series.
    It is actually:
    $$f'(x)=1+(-x)+(-x)^2+(-x)^3+...=\frac{1}{1-(-x)}=\frac{1}{1+x}$$
     
  8. Feb 5, 2013 #7
    Now, I know that the radius of convergence is the same for the series and its derivative, but how does the sum of the derivative of a series relate to the sum of the series itself?
     
  9. Feb 5, 2013 #8

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    Integrate?
     
  10. Feb 5, 2013 #9
    Integrating the original series, I got [itex]\Sigma^{∞}_{n=1}\frac{(-1)^{n+1}x^{n+1}}{n^2+n}[/itex]

    I'm still not quite sure I see the connection between everything.
     
  11. Feb 5, 2013 #10

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    f(x) represents the sum that you are looking for.

    We know now that:

    $$f'(x) = \frac {1} {1+x}$$

    What do you get if you integrate this equality?
     
  12. Feb 5, 2013 #11
    I believe it should be [itex]ln|1+x|[/itex].
     
  13. Feb 5, 2013 #12

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    Yes. Good!

    So we have ##f(x)=\ln|1+x|=k##.
    (Btw, we should also have an integration constant, but that turns out to be zero.)

    Can you find an x if ##k=-100,\frac 12,100##?
    And is that x within the radius of convergence?
     
  14. Feb 5, 2013 #13
    I'm a bit rusty on my rules for ln's and e's, but I'll give it a shot. I think I understand the general premise now. I take the power series, differentiate it and then find a formula for it. If I integrate that formula, that will go back and give me the formula for the original series, right?

    Now I can set that formula equal to whatever value of k I want and find x. If that x is within the interval of convergence, which I believe is (-1, 1] here, then that exists as a solution. If the x value lies outside of that interval, then the solution does not exist for that value of k.

    Does this sound right?
     
  15. Feb 5, 2013 #14

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    Yes. :approve:
     
  16. Feb 5, 2013 #15
    Great! Thanks so much for the help. It's really important to me that I understand exactly why something is being done and you've been a huge help with that. Now I just have to see if I can find these values of x :tongue:.
     
  17. Feb 5, 2013 #16

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    Sounds like a good plan!
     
  18. Feb 5, 2013 #17
    Ok, ok, one more question!

    After using the ratio test to find the radius of convergence, I say that [itex]|x| < 1[/itex], so [itex]-1 < |x| < 1[/itex]. For the interval of convergence, I found that the series diverges for -1 and converges for 1. For k = -100, I found that [itex] x = \frac{1}{e^{100}} - 1 = -1[/itex]. Now, -1 is not in the interval of convergence since it diverges there, but do I take the absolute value of x there and say that it is within the interval?
     
  19. Feb 5, 2013 #18

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    You are slightly off.

    From ##|x| < 1## you should get that ##-1<x<1##.
    Furthermore ##x=\frac 1{e^{100}}-1>-1##, so for k=-100 the corresponding value of x is within the interval.
     
    Last edited: Feb 5, 2013
  20. Feb 5, 2013 #19
    Can you explain why [itex]\frac{1}{e^{100}} - 1 \neq -1[/itex]?
     
  21. Feb 5, 2013 #20

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    Because ##\frac 1{e^{100}}## is not zero. It is a very small positive value.

    Suppose it would be something like 0.0001.
    Would you say that ##0.0001 - 1 = -1##?
     
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