Expert: Paul Klarreich - 5/24/2008

Question
Hi Paul,
I am trying to find the Maclaurin series for f(x)=x/(1+(x/2))in terms of sum n=0 to infinity nth term
I don't seem to get a pattern to go off from.
PLease help...
Thank you - Sami

Answer
Questioner:   sami
Category:  Calculus
Private:  No
 
Subject:  Maclaurin series
Question:  Hi Paul,
I am trying to find the Maclaurin series for f(x)=x/(1+(x/2))in terms of sum n=0 to infinity nth term
I don't seem to get a pattern to go off from.
PLease help...
Thank you - Sami
.....................................
Hi, Sami,

If you use the general formula:  f(n-th)(0)/n! times x^n, you will quickly go crazy with the derivatives.  You have to look for some other way to expand in powers of x.  Best way is the infinite binomial expansion:

  x
-------- =
1 + x/2

x(1 + x/2)^-1

Now expand (1 + x/2)^-1 using the binomial theorem:

Start with:

(1 + b)^n =

        n(n-1)b^2     n(n-1)(n-2)             n(n-1)...(n-(k-1))
1 + nb + ---------- +  ----------- b^3 + ....+ ------------------ b^k +
         2*1            3 2 1                      k!

The plan is to put n = -1 and b = x/2, and finally throw in that x.


If n = -1, we have:

(1 + b)^-1 =

        -1(-2)         -1(-2)(-3)               -1(-2)...(-k)
1 -  b + ------ b^2 +  ----------- b^3 + .... + --------------- b^k
         2!                3!                          k!

       (-1)^2(2!)        (-1)^3(3!)               (-1)^k(k!)
1 - b + ---------- b^2 +  ---------- b^3 + .... + ------------ b^k
         2!                   3!                      k!


1 - b + (-1)^2 b^2 +  (-1)^3 b^3 + .... + (-1)^k b^k

-- we're getting there.  Now put b = x/2:

1 - (x/2) + (-1)^2 (x/2)^2 +  (-1)^3 (x/2)^3 + .... + (-1)^k (x/2)^k + ..

         (-1)^2          (-1)^3                (-1)^k
1 - x + ---------- x^2 +  -------- x^3 + .... + ------- x^k + ...
         2^2!              2^3                   2^k!

OK, then -- all you have to do is multiply each term by an 'x' and you have all your terms.