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Where does this method of constructing the taylor expansion of arctanx fails?

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data
    arctan x = ∫du/(1+u2), from 0 to x


    2. Relevant equations



    3. The attempt at a solution

    I noticed that 1/(1+u2) = 1/(1+u2)1/2 × 1/(1+u2)1/2.

    I decided to take the taylor series expansion of 1/(1+u2)1/2, square the result and then integrate.

    I got 1/(1+u2)1/2 = 1 - u2 + 3u4/8 - 15u6/48...

    When squaring that result, I get 1/1+u2 = 1 - u2 +5u4/8 ...


    When I integrate, the first 2 terms are fine but the x5/8 should be x5/5.


    Did I make a mistake, or is my method of "cheating" the interval of convergence for the binomial series somehow falsifying the result?
     
  2. jcsd
  3. Jun 11, 2012 #2

    SammyS

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    I get something different when I square 1 - u2 + 3u4/8 - 5u6/16....

    1-2 u2+(7 u4)/4-(11 u6)/8 + ...

    How did you get the following?
     
  4. Jun 11, 2012 #3
    Dangit, I didn't copy my notebook properly sorry.

    I squared (1-u2/2 + 3u4/8...), which should be the right series expansion for (1+u2)-1/2.
     
  5. Jun 11, 2012 #4

    SammyS

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    For [itex]\left(1 - u^2/2 + 3u^4/8 - 15u^6/48+35u^8/128+...\right)^2[/itex]

    I get 1 - u2 + u4 - u6 +u8 + ...
     
  6. Jun 11, 2012 #5
    Sigh, I keep making stupid mistakes. I counted u4/2 twice and 3u4/8 once instead of u4/2 + 3u4/8 +3u4/8.

    Thank you for the help.
     
  7. Jun 11, 2012 #6

    HallsofIvy

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    Rather than using a square root, it would make much more sense to use that the fact that the sum of a geometric series is given by
    [tex]\sum_{n=0}^\infty r^n=1+ r+ r^2+ \cdot\cdot\cdot+ r^n+ \cdot\cdot\cdot= \frac{1}{1- r}[/tex]

    Here the function is [itex]1/(1+ u^2)[/itex] which is of the form [itex]1/(1- r)[/itex] with [itex]r= -u^2[/itex] so the sum is [itex]1- u^2+ u^4- u^6+\cdot\cdot\cdot= \sum_{n=0}^\infty (-1)^n u^{2n}[/itex]. And the integral of that is [itex]u- (1/3)u^3+ (1/5)u^5- (1/7)u^7+ \cdot\cdot\cdot= \sum_{n=0}^\infty ((-1)^n/(2n+1)!)u^{2n+1}[/itex].
     
  8. Jun 11, 2012 #7
    It is hard to cheat using power series. As long as you do your algebra correctly, the result should be valid in some interval as long as each series in the calculation has a nonzero radius of convergence.
     
    Last edited: Jun 11, 2012
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