# taylor expansion at around 0

I need to find the taylor expansion for the following function: $$f(x):=x^{3/2}\sin\bigl(\sqrt{x}\bigr)\ .$$ I tried finding a pattern using term by term diffrentiation, but it didnt help since the derivatives are not continuous at $0$. Then i thought maybe i can just represent $\sin\bigl(\sqrt{x}\bigr)$ as taylor sum and then just multiply it by $x^{3/2}$. Would that be valid? I would be happy to hear if it is, and if there are other ways to solve this. Thanks.

• Yes, that would be valid. – Kenny Lau May 6 '17 at 8:22
• If you want life to be easier, you could start using $x=y^2$ – Claude Leibovici May 6 '17 at 8:33

For all real $x>0$ you have \eqalign{f(x)&=x^{3/2}\>\sin(x^{1/2})\cr &=x^{3/2}\left(x^{1/2}-{1\over6}x^{3/2}+{1\over120}x^{5/2}-{1\over 5040}x^{7/2}+\ldots\right)\cr &=x^2-{1\over6}x^3+{1\over120}x^4-{1\over5040}x^5+\ldots\ .\cr} The resulting power series converges for all $x>0$, and even for all $x\in{\mathbb C}$. By general principles it follows that it is in fact the Taylor series of the function $f$ it represents.