# Thread: [SOLVED] Taylor polynomial question?

1. ## [SOLVED] Taylor polynomial question?

Hi =)

Use a suitable Taylor polynomial for e^x to evaluate the 24th derivative of f(x) = exp (x^8) at x = 0 (express your answer in terms of factorials).

Can someone teach me how to do so?

2. $\exp(x) = 1+x+{x^2\over2!}+{x^3\over3!}\cdots$ so $\exp(x^8) = 1+x^8+{x^{16}\over2!}+{x^{24}\over3!}\cdots$. The coefficient of $x^n$ in this power series is equal to ${1\over n!}\;{\mathrm{d}^n\exp(x^8)\over\mathrm{d}x^n}(0)$ by Taylor's theorem. Equating these for n=24 gives ${1\over 3!} = {1\over 24!}\;{\mathrm{d}^{24}\exp(x^8)\over\mathrm{d}x^{2 4}}(0)$.
...The coefficient of $x^n$ in this power series is equal to ${1\over n!}\;{\mathrm{d}^n\exp(x^8)\over\mathrm{d}x^n}(0)$ by Taylor's theorem. Equating these for n=24 gives ${1\over 3!} = {1\over 24!}\;{\mathrm{d}^{24}\exp(x^8)\over\mathrm{d}x^{2 4}}(0)$.
4. ${\mathrm{d}^nf(x)\over\mathrm{d}x^n}(x)$ is Leibniz's notation for the nth derivative of f at the point x. Perhaps you write it $f^{(n)}(x)$?