# Thread: Proof by use of Maclaurin's Expansion or Binomial Expansion

1. ## Proof by use of Maclaurin's Expansion or Binomial Expansion

Prove that if x is so small that x^3 and higher powers of x may be neglected that

{[1-x]/[1+x]}^1/2 = 1 - x + 0.5x^2

I just can't get it out. Help please! I have A levels beginning next week!!!!!!!!

2. Originally Posted by presando
Prove that if x is so small that x^3 and higher powers of x may be neglected that

$f(x)=\sqrt{\frac{1-x}{1+x}} = 1 - x + \frac{x^2}{2}$

I just can't get it out. Help please! I have A levels beginning next week!!!!!!!!
Yes, use MacLaurin. The first term is $\frac{f(0)}{0!}$. The next term is $\frac{f'(0)}{1!}x$. The next term is $\frac{f''(0)}{2!}x^2$.

So $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2 = 1-x+\frac{x^2}{2}$

I verified on my calculator that this is correct, so I'm not going to do out all the derivatives as it would take too long.

3. Hi

You can also separate like this

$\sqrt{\frac{1-x}{1+x}} = \sqrt{(1-x)(1-x+x^2+o(x^2))} = \sqrt{1+(-2x+2x^2+o(x^2))}$

$\sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" alt="\sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" />

$\sqrt{\frac{1-x}{1+x}}= 1-x+x^2-\frac12\:x^2 + o(x^2)= 1-x+\frac12\:x^2+o(x^2)$

4. Originally Posted by running-gag
Hi

You can also separate like this

$\sqrt{\frac{1-x}{1+x}} = \sqrt{(1-x)(1-x+x^2+o(x^2))} = \sqrt{1+(-2x+2x^2+o(x^2))}$

$\sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" alt="\sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" />

$\sqrt{\frac{1-x}{1+x}}= 1-x+x^2-\frac12\:x^2 + o(x^2)= 1-x+\frac12\:x^2+o(x^2)$
Thanks!

However, I have never seen that method used before. I'm going to write it out and see if I understand it.

5. Originally Posted by redsoxfan325
Yes, use MacLaurin. The first term is $\frac{f(0)}{0!}$. The next term is $\frac{f'(0)}{1!}x$. The next term is $\frac{f''(0)}{2!}x^2$.

So $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2 = 1-x+\frac{x^2}{2}$

I verified on my calculator that this is correct, so I'm not going to do out all the derivatives as it would take too long.
It's the simplification of f'(x) and f''(x) that's killing me.

6. You don't need to simplify them, just evaluate them. Even if it's some crazy expression, plugging in $x=0$ shouldn't be too hard.

$f'(0)=-1$
$f''(0)=1$

7. Originally Posted by redsoxfan325
You don't need to simplify them, just evaluate them. Even if it's some crazy expression, plugging in $x=0$ shouldn't be too hard.

$f'(0)=-1$
$f''(0)=1$
I'll try.