It is well-known that the Taylor series of [itex]\sin x[/itex] about [itex]x=0[/itex] is [tex]\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

By extension, one might presume that the Taylor series of [itex]\sin x^{3/2}[/itex] about [itex]x=0[/itex] is

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

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

[tex] f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n [/tex]

why is

[tex] f(g(x)) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}g(x)^n [/tex]

It seems like this last expression would involve chain differentiation, but does not. Thanks all - I hope I have explained my question properly.

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# Extending Taylor expansions

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