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Finding the nth term of the maclaurin's expansion of sqrt(1+x)

  1. Oct 30, 2007 #1

    rock.freak667

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    [SOLVED] Finding the nth term of the maclaurin's expansion of sqrt(1+x)

    1. The problem statement, all variables and given/known data

    Find the nth term of the Maclaurin's expansion of [itex]\sqrt{1+x}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    so far I've expanded and gotten

    [tex]\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{2!4}+\frac{3x^3}{3!8}-\frac{15x^4}{4!16}+\frac{105x^5}{5!32}[/tex]

    so far I've gotten that part of the nth term should be

    [tex]\frac{x^n}{n!2^n}[/tex]

    but what I do not know is how to deal with the + and - signs as well as the numbers in th numerators...please help me
     
  2. jcsd
  3. Oct 30, 2007 #2
    (-1)^n should cover the + and -
    as for the numbers, I can't help you. I'm tired. I'll be back after I work on it.
     
  4. Oct 30, 2007 #3

    Dick

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    I think you know the pattern for the numerators. You just don't know the name for it. Look up double factorials written n!!.
     
  5. Oct 30, 2007 #4

    rock.freak667

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    But the (-1)^n doesn't work for the 2nd term or after
     
  6. Oct 30, 2007 #5

    Dick

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    You know what, I'll give you an answer. (-1)^n*(2n)!/((1-2n)*(n!)^2*4^n). I didn't work that out, I looked it up because I got annoyed with the problem. Notice the (1-2n) in the denominator changes signs between n=0 and n=1 cancelling the sign change from the (-1)^n. Getting one formula that works for all n can be more of an art than a science. I would consider it perfectly legit to write 1+x/2-x/8+... and then give a formula for the nth term after the first few. It could probably be written more simply than the one fits all version.
     
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