1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Write the Maclaurin series for (1+x)^(-1/2) as a sum

  1. May 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Write the Maclaurin series for ##\frac{1}{(1+x)^{1/2}} ## in ##\sum## form using the binomial coefficient notation. Then find a formula for the binomial coefficients in terms of n.

    2. Relevant equations
    3. The attempt at a solution

    $$\frac{1}{(1+x)^{1/2}}=1-\frac{x}{2}+\frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128} $$
    This is where I get stuck, I can't seem to figure out the formula for how the terms are increasing. Is there a way to figure out the formula without using a guess and check method?
     
  2. jcsd
  3. May 25, 2015 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do you know how to find a Maclaurin series in general ?
     
  4. May 25, 2015 #3
    As in ##(1+x)^p=1+\frac{p\cdot x}{1!}+\frac{p(p-1)\cdot x^2}{2!}+...##?
     
  5. May 25, 2015 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I was thinking more general, but that's perfectly adequate here and a good place to start.

    Of course p = -1/2 in your case.
     
  6. May 25, 2015 #5
    Yea I have derived quite a few Maclaurin series for practice but I am having trouble rewriting the terms as a power series. I think it looks something like this ##\sum_{n=0}^{\infty} \frac{x^n}{2^n}## but I'm struggling to find the rest
     
  7. May 25, 2015 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Write out a few terms.

    Write out the general term with xn in it.

    Look for patterns.
     
  8. May 25, 2015 #7
    $$(1+x)^p=1+\frac{p}{1!}x+\frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\frac{p(p-1)(p-2)(p-3)}{4!}x^4 \\
    (1+x)^{-1/2}=1+\frac{(\frac{-1}{2})}{1!}x +\frac{(\frac{-1}{2})(\frac{-3}{2})}{2!}x^2+\frac{(\frac{-1}{2})(\frac{-3}{2})(\frac{-5}{2})}{3!}x^3\\
    (1+x)^{-1/2}=1+\frac{-1}{2*1!}x+\frac{(-1)(-3)}{2*2*2!}x^2+\frac{(-1)(-3)(-5)}{2*2*2*3!}x^3 \\
    (1+x)^{-1/2}=1+\frac{(-1)(1)}{2*1!}x+\frac{(-1)^2(1)(3)}{2*2*2!}x^2+\frac{(-1)^3(1)(3)(5)}{2*2*2*3!}x^3 $$ So I think the formula is ## \sum_{0}^{\infty} \frac{(-1)^n(2n-1)!!}{2^n\cdot n!} x^n## My only concern is that when n=0 (2n-1)!! goes to (-1)!.
     
  9. May 25, 2015 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Oh, you mean (-1)!! .

    Believe it or not, (-1)!! is defined & is equal to 1.

    Similar to the way you can get ##\ 0!=1\,,\ ## the recursion relation for double factorial is ##\displaystyle \ n!! = n\cdot(n-2)!! \,,\ ## which gives you the result if n = 1 .

    Alternatively, ##\displaystyle \ (2n-1)!! = \frac{(2n)!}{2^n(n!)}\ .##

    See http://en.wikipedia.org/wiki/Double_factorial .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Write the Maclaurin series for (1+x)^(-1/2) as a sum
  1. Maclaurin Series for e^x (Replies: 11)

Loading...