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

1. May 25, 2015

### Potatochip911

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. May 25, 2015

### SammyS

Staff Emeritus
Do you know how to find a Maclaurin series in general ?

3. May 25, 2015

### Potatochip911

As in $(1+x)^p=1+\frac{p\cdot x}{1!}+\frac{p(p-1)\cdot x^2}{2!}+...$?

4. May 25, 2015

### SammyS

Staff Emeritus
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.

5. May 25, 2015

### Potatochip911

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

6. May 25, 2015

### SammyS

Staff Emeritus
Write out a few terms.

Write out the general term with xn in it.

Look for patterns.

7. May 25, 2015

### Potatochip911

$$(1+x)^p=1+\frac{p}{1!}x+\frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\frac{p(p-1)(p-2)(p-3)}{4!}x^4 \\ (1+x)^{-1/2}=1+\frac{(\frac{-1}{2})}{1!}x +\frac{(\frac{-1}{2})(\frac{-3}{2})}{2!}x^2+\frac{(\frac{-1}{2})(\frac{-3}{2})(\frac{-5}{2})}{3!}x^3\\ (1+x)^{-1/2}=1+\frac{-1}{2*1!}x+\frac{(-1)(-3)}{2*2*2!}x^2+\frac{(-1)(-3)(-5)}{2*2*2*3!}x^3 \\ (1+x)^{-1/2}=1+\frac{(-1)(1)}{2*1!}x+\frac{(-1)^2(1)(3)}{2*2*2!}x^2+\frac{(-1)^3(1)(3)(5)}{2*2*2*3!}x^3$$ 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)!.

8. May 25, 2015

### SammyS

Staff Emeritus
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 .