prove that $f(x)=\sum_{|\alpha|\leq k}\frac{1}{\alpha !}D^{\alpha}f(0)x^{\alpha} + O(|x|^{k+1})$

Question

If $\alpha$ is multiindex and $f$ is smooth,prove that $f(x)=\sum_{|\alpha|\leq k}\frac{1}{\alpha !}D^{\alpha}f(0)x^{\alpha} + O(|x|^{k+1})$. The hint is to use taylors form for $g(t)=f(tx)$. If i do this i will found that $g(t)=\sum \frac{g^{n}(0)}{n!}t^n = \sum \frac{ D^nf(0)x^n t^n}{n!}$. How i continued? i need to do induction over size of $\alpha$. Any hint please, thank you.

Answer 1

The key is to show using the chain rule that $$g^{(n)}(0) = \sum_{|\alpha| = n} D^\alpha f(0) x^\alpha.$$ Combining with your first step and plugging in $t=1$ yields $$f(x) = g(1) = \sum_{n \ge 0} \frac{1}{|\alpha|!}\sum_{|\alpha| = n} D^\alpha f(0) x^\alpha.$$ Can you take it from here?