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

• May 1st 2009, 10:15 AM
presando
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!!!!!!!!
• May 1st 2009, 10:52 AM
redsoxfan325
Quote:

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

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

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

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

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

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.
• May 1st 2009, 11:01 AM
running-gag
Hi

You can also separate like this

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

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

$\displaystyle \sqrt{\frac{1-x}{1+x}}= 1-x+x^2-\frac12\:x^2 + o(x^2)= 1-x+\frac12\:x^2+o(x^2)$
• May 1st 2009, 11:25 AM
presando
Quote:

Originally Posted by running-gag
Hi

You can also separate like this

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

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

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

Thanks!

However, I have never seen that method used before. I'm going to write it out and see if I understand it.
• May 1st 2009, 11:30 AM
presando
Quote:

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

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

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.
• May 1st 2009, 11:33 AM
redsoxfan325
You don't need to simplify them, just evaluate them. Even if it's some crazy expression, plugging in $\displaystyle x=0$ shouldn't be too hard.

$\displaystyle f'(0)=-1$
$\displaystyle f''(0)=1$
• May 1st 2009, 11:37 AM
presando
Quote:

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

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

I'll try.