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

Mathematics Asked by weymar andres on November 16, 2021

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.

One Answer

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?

Answered by angryavian on November 16, 2021

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