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!!!!!!!!
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!!!!!!!!
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.
Hi
You can also separate like this
$\displaystyle \sqrt{\frac{1-x}{1+x}} = \sqrt{(1-x)(1-x+x^2+o(x^2))} = \sqrt{1+(-2x+2x^2+o(x^2))}$
$\displaystyle \sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2$
$\displaystyle \sqrt{\frac{1-x}{1+x}}= 1-x+x^2-\frac12\:x^2 + o(x^2)= 1-x+\frac12\:x^2+o(x^2)$