Originally Posted by

**Ackbeet** Hmm. We have

$\displaystyle \displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\dots,$ as you had before. We don't need more terms because we're only looking for the first four terms. According to your formula there, we have

$\displaystyle \displaystyle{(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3},$

so

$\displaystyle \displaystyle{(1+x)^{-1/2}=1+(-1/2) x+\frac{(-1/2)(-1/2-1)}{2!}x^2+\frac{(-1/2)(-1/2-1)(-1/2-2)}{3!}x^3+\dots}$

$\displaystyle \displaystyle{=1-x/2+\frac{3x^{2}}{8}-\frac{5 x^{3}}{16}+\dots}$

We don't need more terms than these, because the lowest power of x in the sin series is to the first power; hence, the cubic in this series will go to a fourth power.

Both of these are correct. Moving on, then:

$\displaystyle \displaystyle{\frac{\sin(x)}{\sqrt{1+x}}=\left(x-\frac{x^{3}}{3!}+\dots\right)\left(1-x/2+\frac{3x^{2}}{8}-\frac{5 x^{3}}{16}+\dots\right)}$

$\displaystyle \displaystyle{=x-\frac{x^{2}}{2}+\frac{3x^{3}}{8}-\frac{5x^{4}}{16}-\frac{x^{3}}{6}+\frac{x^{4}}{12}+\dots}$

$\displaystyle \displaystyle{=x-\frac{x^{2}}{2}+\frac{5x^{3}}{24}-\frac{11x^{4}}{48}+\dots}$

Can you see where your mistake is now?