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Expansion of infinite series

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Use Maclaurin’s theorem to derive the first five terms of the series expansion for ##(1+x)^{r}##, where -1<x<1. Assuming the series, obtained above, continues with the same pattern, sum the following infinite series

    ##1 + \frac{1}{6} - \frac{(1)(2)}{(6)(12)} + \frac{(1)(2)(5)}{(6)(12)(18} - \frac{(1)(2)(5)(8}{(6)(12)(18)(24)}+...##

    2. Relevant equations
    Maclaurin series

    3. The attempt at a solution

    I have taken the derivative of ##(1+x)^{r}## several times and obtained the power series
    ##1+rx+\frac{r(r-1)}{2!} x^{2} +\frac{r(r-1)(r-2)}{3!} x^{3} + \frac{r(r-1)(r-2)(r-3}{4!} x^4+...##

    Now, the problem is, how do I relate with the infinite series above?
     
  2. jcsd
  3. Jan 25, 2014 #2

    haruspex

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    The sequence 1, 2, 5, 8 is not obvious. The 1 doesn't seem to fit. What would you have to have instead of the 1 to fit with the remaining terms?
     
  4. Jan 25, 2014 #3
    I don't get it... Just rewrite them as 2, 5,8? :confused:
     
  5. Jan 25, 2014 #4

    haruspex

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    You are given an infinite series, but only by example of the first four terms. The first thing you need to do is figure out what the general term looks like. The pattern of the denominators is straightforward, but it's not obvious what the pattern is in the numerators. Looks like you need to see a pattern in the sequence 1, 2, 5, 8 .... but what is it?
    To answer that, try throwing away the 1. What would you put there instead of 1 to make a clear pattern?
     
  6. Jan 25, 2014 #5
    -1? :confused:
     
  7. Jan 25, 2014 #6

    haruspex

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    Right. So, how can you rewrite the series (without changing the actual values of the terms) so that the sequence goes -1, 2, 3, 5, 8, ...
     
  8. Jan 25, 2014 #7
    (-3+2)?
     
  9. Jan 25, 2014 #8

    haruspex

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    No. Here's the original series:
    ##1 + \frac{1}{6} - \frac{(1)(2)}{(6)(12)} + \frac{(1)(2)(5)}{(6)(12)(18} - \frac{(1)(2)(5)(8}{(6)(12)(18)(24)}+...##

    I'm asking you to rewrite that so that in the numerator you see the sequence -1, 2, 5, 8 instead of 1, 2, 5, 8.. You need to adjust something else so that the values of the terms don't change.
     
  10. Jan 25, 2014 #9
    I think I can rewrite the signs...

    ##1 - \frac{-1}{6} + \frac{(-1)(2)}{(6)(12)} - \frac{(-1)(2)(5)}{(6)(12)(18)} + \frac{(-1)(2)(5)(8)}{(6)(12)(18)(24)}+...##
     
  11. Jan 25, 2014 #10

    haruspex

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    Good. (And note that the signs in front of the terms now alternate nicely, +-+-.., whereas in the OP it started ++-+..)
    Next, the Maclaurin series has r(r-1)(r-2)... in the numerators, i.e. the product of a sequence decreasing by 1 at each step. You have (-1)(2)(5)(8)..., an sequence increasing in 3s. What can you do to make those match?
    (If you can't think how to fix up both of those differences, just try to fix up one: i.e. either get the sequence to be decreasing, or get it to increase by 1s.)
     
  12. Jan 25, 2014 #11
    I don't get it... Do I need to rewrite the series in reverse?
     
  13. Jan 25, 2014 #12

    haruspex

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    No. The common difference in the arithmetic sequence is +3. You need a sequence with common difference -1. What simple arithmetic operation can you apply to every term in the sequence to make that change? Remember here I'm just talking about the sequence -1, 2, 5, 8..., not the series.
     
  14. Jan 25, 2014 #13

    Dick

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    No, don't do that. Just take the direct approach. Equate the first terms of your two series. r*x=1/6 and r*(r-1)*x^2/2=(-1*2/(6*12)). You should be able to solve for r and x.
     
  15. Jan 26, 2014 #14
    Well, I got r=1/3 and x=1/2...
     
  16. Jan 26, 2014 #15

    haruspex

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    That's right.
    I should have thought of suggesting Dick's way. I'm sure that was much easier for you. I went the way I did because that was the easiest for me - I could see the answer straight away by making those adjustments.
     
  17. Jan 26, 2014 #16
    Alright, what should I do afterwards? :P
     
  18. Jan 26, 2014 #17

    Dick

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    To get the sum of the series?? Your series is the infinite series expansion of (1+x)^(1/3) when x=1/2, right? You might want to check a few other terms to make sure. Don't make me have to say the obvious about what the sum is. Just think about it. If you are thinking for over 5 seconds then you are thinking about it wrong.
     
  19. Jan 26, 2014 #18
    Is it cube root of 3/2?
     
  20. Jan 26, 2014 #19

    Dick

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    Yes, it's (1+1/2)^(1/3).
     
  21. Jan 26, 2014 #20
    That's it? :shock:
     
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