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!!!!!!!!

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- May 1st 2009, 10:15 AMpresandoProof 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 AMredsoxfan325
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 AMrunning-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 AMpresando
- May 1st 2009, 11:30 AMpresando
- May 1st 2009, 11:33 AMredsoxfan325
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 AMpresando