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Power series expansion

  1. Jan 24, 2015 #1
    1. The problem statement, all variables and given/known data
    "Use power series to evaluate the function at the given point"
    ## ln (x+ \sqrt{1+{x^2}}) - sin x ##

    at ## x = 0.001 ##

    2. Relevant equations
    Relevant power series:
    A: ## ln (1+x) = \Big( \sum_{n=0}^\infty\frac{({(-1)^{n+1}}{x^n})}{n} \Big) ##

    B: ## {(1+x)^p} = \sum_{n=0}^{\infty }\binom{p}{n}{x^n} ##

    C: ## sin x = \Big ( \sum_{n=0}^\infty\frac{({(-1)^{n}{x}^{2n+1}})}{(2n+1)!} \Big) ##

    3. The attempt at a solution

    Starting out with the logarithmic expression, there are two power series.

    If ## \sqrt{1+{x^2}} = y ##

    then ## ln (x+y) = \Big( \sum_{n=0}^\infty\frac{({(-x)^{n+1}}{y^n})}{n} \Big) ##

    and ## y = \sum_{n=0}^{\infty }\binom{(\frac{1}{2})}{n}{({(x^2)^n)}} ##

    I hope I am correct this far? Of course, I can now expand the series and multiply term by term, but this is quite tedious - especially since (using wolfram alpha), the first terms are in the power of 5 and 7. I also tried making a general expression, but then I end up with something like..
    ## ln (x+y) = \Big( \sum_{n=0}^\infty\frac{({(-x)^{n+1}})}{n} \Big) * (\sum_{n=0}^{\infty }\binom{(\frac{1}{2})}{n}{({(x^2)^n)}}) ##

    .. and I don't know how to simplify this.

    There must be some obvious mistake, or a simpler way to do this?
     
  2. jcsd
  3. Jan 24, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    That does not look right and I don't see where this comes from. You cannot simply replace occurences of 1 by a variable. In particular, the two occurences of "1" in the expansion given above have a completely different meaning.

    What is ##\binom{(\frac{1}{2})}{n}##?
    There is a power series for ##\sqrt{1+x}## that you can use.

    With the correct power series, it is sufficient to consider the first few terms - I guess you just need the first non-vanishing order of x.
     
  4. Jan 24, 2015 #3
    "In particular, the two occurrences of "1" in the expansion given above have a completely different meaning."
    Yes, I was unsure of this.

    And maybe I was unclear - I used the binomial series to expand ## \sqrt{1+{x^2}} ##
    ## \sqrt{1+{x^2}} = \sum_{0}^{\infty }\binom{\frac{1}{2}}{n}{(x^2)^n} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} ... ## where ## \binom{\frac{1}{2}}{n} ## is the binomial expression 0.5 choose n.
    I think this is correct?

    I don't fully understand, however - how do I make a power series of:
    ## ln (1 + x + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16}...) ## ... ?
     
  5. Jan 24, 2015 #4

    mfb

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    Staff: Mentor

    Okay, I found the appropriate generalization of the standard binomial coefficients needed for that.

    ##\ln(1+x+\frac{x^2}{2}+...) = \ln(1+c)= ...## with c coming from the sum. You won't need many terms to find one that does not cancel.
     
  6. Jan 24, 2015 #5
    Of course! Thanks!
     
  7. Jan 24, 2015 #6

    Ray Vickson

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    Write ##1 + x + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} + \cdots## as ##1+z##, and use the series for ##\ln(1+z)## in powers of ##z##.
     
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