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Maclaurin series for square root (1+x)

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Maclaurin series for square root (1+x)


    2. Relevant equations



    3. The attempt at a solution
    I attempted to find the maclaurin series for the function Square root of 1+x.

    F(0)=1 first term= 1
    F'(0)=1/2 second term= (1/2)x
    F''(0)=-1/4 Third term (-1/4)x^2
    F'''(0)=3/8 fourth term (3/8*3!) x^3
    F''''(0)=15/16 fifth (-15/16*4!) x^4
    F'''''(0)105/32 six (105/32*5!) x^5
    Therefore,
    f(x)= 1+ (1/2)x + (-1/4)x^2 + (3/8*3!) x^3 + (-15/16*4!) x^4 + (105/32*5!) x^5

    The problem is to find generalize term .
    I have ( ((-1)^(n-1)) * something *x^n) / ((2^n) * (n!))

    I cannot find that "something". because it exists as 1 for the first, second, and the third term , but then it increases to 3,15, 105
    so the previous term increases by factor of 3,5,7.... (somewhat recursive?)
    Help..
     
  2. jcsd
  3. Mar 14, 2010 #2
    1=1*1
    3=3*1
    15=5*3
    105=7*15

    notice a pattern?
     
  4. Mar 14, 2010 #3
    yeah that part is easy but it goes like this
    1,1,1,3,15,105...
    how do you account for the first two 1s
     
  5. Mar 14, 2010 #4

    vela

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    Two of the ones you'd figure would come from the n=0 and n=1 cases. It's the other one that's a bit difficult. What general form did you get for the "something" so far?

    By the way, the third term in your original post is missing the 2! in the denominator.
     
    Last edited: Mar 14, 2010
  6. Mar 14, 2010 #5
    Try following the sequence backwards:
    105/7=15
    15/5=3
    3/3=1
    1/1=1
    1/-1=-1

    Not surprising, since the function is proportional to (1+x)^1/2, and the derivatives are proportional to (1+x)^(-1/2), (1+x)^(-3/2), (1+x)^(-5/2),etc.
     
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