# Extending Taylor expansions

1. Oct 29, 2012

### 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.

2. Oct 29, 2012

### arildno

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

3. Oct 30, 2012

### 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.