Can we use Taylor expansions to find the series representation of sin(x^(3/2))?

[Undoubtedly0]

It is well-known that the Taylor series of \sin x about x=0 is \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}
By extension, one might presume that the Taylor series of \sin x^{3/2} about x=0 is

\sin x^{3/2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{\frac{3}{2}(2n+1)}

This is indeed the case, but why? Finding the Taylor series of \sin x^{3/2} from the definition does not give a convergent series. In general, my question is, if

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n
why is
f(g(x)) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}g(x)^n
It seems like this last expression would involve chain differentiation, but does not. Thanks all - I hope I have explained my question properly.

 

[arildno]

Why do you think your last expression is not in conformity with the chain rule??

 

[Vargo]

Your series is a valid absolutely convergent series representation of sin(x^(3/2)). However, it is not a power series because it has terms like x^(3/2). Therefore, it cannot be a Taylor series.

The fact is that sin(x^3/2) is not twice differentiable at x=0, so it does not have a Taylor series representation at 0.

In general, you can substitute g(x) into the power series representation of f as long as g(x) remains inside the interval of convergence of the power series. You will then have a series representation for f(g(x)). However, the resulting formula won't necessarily be a power series.