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Finding a power series for arcsin(x)

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data
    1. Find a power series for (1-x)^(-1/2)
    2. Find a power series for (1-x^2)^(-1/2)
    3. Find a power series for arcsin(x)

    2. Relevant equations
    Binomial series, (1+x)^k= 1 + kx + k(k-1)x^2/2!+ k(k-1)(k-2)x^3/3!+....

    3. The attempt at a solution
    for 1., I have 1+ (-1/2)(-x) + (-1/2)(-1/2-1)(-x)^2/2! + (-1/2)(-1/2-1)(-1/2-2)(-x)^3/3!+....
    Simplified, 1+ x/2 +3x^2/8+ 15x^3/48 + 35x^4/128+....
    From here, I am having trouble finding the general term.
    Once I can find that, the rest is easy...just use x^2 instead of x for (2) and integrate the series to get arcsin(x) for (3).
    I know I could look up a power series for arcsin(x) in a table or website but the point is to get it using the binomial series.
     
    Last edited: Mar 23, 2009
  2. jcsd
  3. Mar 24, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi chouse12! Welcome to PF! :smile:
    Quite right! :biggrin:

    hmm :rolleyes: … a series of three questions …

    Hint: integrate :wink:
     
  4. Mar 24, 2009 #3

    lanedance

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    how about only simplifying the subtractions first to look for a pattern, then multiply out the negative signs to either + or -
    [tex]=
    1+ (-\frac{1}{2})(-x) + (-\frac{1}{2})(-\frac{3}{2})(-x)^2\frac{1}{2!} + (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-x)^3\frac{1}{3!}!+....

    [/tex]

     
  5. Mar 24, 2009 #4
    tiny-tim, Thanks for the welcome and the hint. The main problem for me seems to be getting the first series though..

    lanedance, so far I have the series expressed as (-1)^0/2^0/0!(-x)^0 + (((-1)/2^1)/1!)(-x)^1+(((-1)(-3)/2^2)/2!)(-x)^2+(((-1)(-3)(-5)/2^3)/1!)(-x)^3+....+ (((-1)(-3)(-5)...(2n-1)/2^n)/n!) (-x)^n+....
    The trouble is I don't know how to express the (-1)(-3)(-5)...(2n-1) as a single term containing "n's".

    Sorry for the newbformatting, I need to learn how to use latex
     
  6. Mar 24, 2009 #5
    Ok, I messed around with it a little more and ended up with this:
    1+ (1/2)(x) + ((1)(3)/(2)(4))x^2 + ((1)(3)(5)/(2)(4)(6))x^3+...
    So the coefficients on each term is with consecutive odd integers in the numerator, always multiplying by the next integer in the next term, and consecutive even integers in the denominator, always multiplying by the next integer in the next term (I multiplied all the negative signs to give all positive terms). Now, from this, how would I find the general term of the series?
     
  7. Mar 24, 2009 #6

    Dick

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    Look up the definition of the 'double factorial', e.g. (2n-1)!!.
     
  8. Mar 24, 2009 #7

    tiny-tim

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    Hi chouse12! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    How about ∑(2n)!/(n!)222n, = 2nCn/22n ? :smile:
     
  9. Mar 24, 2009 #8
    Dick, I did not know about double factorials or other multiple factorials before so thanks for that. However, how do I make the top odd and the bottom even? On one website I looked at, it just has n!! as a piecewise term, making the terms all even for n>0 and even and all odd for n>0 and odd. Or would it just be, as you stated, (2n-1)!!?

    Tiny-Tim, I believe that is the correct answer but could you explain how you arrived there?

    Thanks to everyone for all the help.
     
  10. Mar 24, 2009 #9
    Dick, from Tiny-Tim's answer, I notice that it is just (2n-1)!! simplified into terms with single factorials (i.e. (2n-1)!!=(2n)!/2nn!) but instead of in the denominator 2nn!, it is that quantity squared (22n(n!)2). Does this make the terms in the denominator the even integers?
     
  11. Mar 24, 2009 #10

    lanedance

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    how about this
    [tex]
    1.3.5.7....(2n-1)
    [/tex]
    [tex]
    =\frac{1.2.3.4.5.6.7....(2n-1)}{2.4.6...(2n-2)}
    [/tex]
    [tex]
    =\frac{(2n-1)!}{2(1).2(2).2(3)...2(n-1)}
    [/tex]
    [tex]
    =\frac{(2n-1)!}{2^{n-1}(n-1)!}
    [/tex]
     
  12. Mar 24, 2009 #11
    lanedance, that is almost correct. That gives coefficients of 1, 2/2, 24/8, 720/48...
    which simplify, to 1, 1, 3, 15...
    The denominator is correct but the numerator is not. The coefficients should be 1, 1/2, 3/8, 15/48.... so if the bottom term was squared then it would be correct. So any ideas on how to find the formula with the denominator squared (while showing work, not just squaring the bottom since it isn't right)?
     
  13. Mar 24, 2009 #12

    lanedance

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    i only gave the term for the (1.3.5...(2n-1)) part of the sequence

    your sequence is
    [tex]= 1+ \frac{1}{2}x + \frac{1}{2}\frac{3}{2}x^2\frac{1}{2!} + \frac{1}{2}\frac{3}{2}\frac{5}{2}x^3\frac{1}{3!}+....[/tex]

    so the coefficient for the nth power of x should be
    [tex] a_n = \frac{1.3.5...(2n-1)}{2^n.n!} [/tex]

    so you get the extra terms on the denominator... i've done this pretty quickly, should work, though you may have to be careful with how n is defined - ie which term it relates to
     
  14. Mar 24, 2009 #13
    Oh that clears everything up....Thank you so much for the help.
     
  15. Mar 24, 2009 #14
    Just another quick question...I need to find the interval of convergence for this series. I set up the ratio test and I think I must be doing it wrong because I am getting convergence for all real values of x but in my textbook it says the binomial series only converges from -1 to 1.

    Here is my work:
    lim n--> infinity of abs. value (((2n+1)!/(2n+1(n+1)!)2)/((2n)!/(2n(n)!)2)) (x)

    and I simplified that to lim n--> infinity of abs. value(2n+1/4(n+1)2) abs. value of (x)
    and that limit goes to 0
     
  16. Mar 24, 2009 #15

    Dick

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    If you've settled on a_n=(2n-1)!!/(2^(n-1)*n!). And I'll accept Lanedance's opinion on that, since I haven't checked it and it seems about right. Then I get that a_(n+1)/a_n->1. That makes the limit a_(n+1)*x^(n+1)/(a_n*x^n) equal to x. Note: (2(n+1)-1)!!/(2n-1)!!=(2n+1).
     
  17. Mar 24, 2009 #16

    lanedance

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    this should be (2n-1)! in the denominator (instead of (2n!)) which will give you an extra remaining factor of n in the numerator and the limit Dick gievs

     
  18. Mar 24, 2009 #17

    Dick

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    Nooo. (2(n+1)-1)!!/(2n-1)!!. The double factorials give you a (2n+1) factor in the numerator. The single factorials give you an n+1 in the denominator and the 2^n gives you a 2 in the denominator. lim (2n+1)/((n+1)*2)=1.
     
  19. Mar 24, 2009 #18

    lanedance

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    Hey Dick, I agree with the double factroial thing which makes life alot easier, by if you're working with singles

    [tex] a_n = \frac{(2n-1)!}{(2^n.n!)^2} [/tex]
    (i haven't checked this either but it should be close based on working)

    chouse wrote

    i meant it should be

    [tex] a_n = \frac{\frac{(2n+1)!}{(2^{n+1}.(n+1)!)^2}} {\frac{(2n-1)!}{(2^n.n!)^2}}[/tex]

    which reduces to the required 1 in the limit
     
  20. Mar 25, 2009 #19

    tiny-tim

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    Hi chouse12! :smile:

    the general term is (n odds)/(n evens),

    so I multiplied top and bottom to get (n odds)(n evens)/(n evens)2

    (n odds)(n evens) is just (2n)!,

    and (n evens) is n!2n (because eg 2.4.6.8 = 2.1.2.2.2.3.2.4 = 2.2.2.2.1.2.3.4) :wink:

    Incidentally, 2nCn/22n is the chance of getting exactly n heads on tossing a coin 2n times :biggrin:

    oh … and how are you doing on arcsin(x)?
     
  21. Mar 25, 2009 #20

    Dick

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    Oh. Ok, then.
     
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