Prove by induction.
Let $$f(x)={ e }^{ { x }^{ 2 } }$$ Then for every $n\in \mathbb{N}$ there exists a polynomial ${ P }_{ n }(x)$ of degree $n$ such that $$\frac { { d }^{ n } }{ { dx }^{ n } } f(x)={ P }_{ n }(x){ e }^{ { x }^{ 2 } }$$
Have tried the base case:
When $n=1$,$$\frac { d }{ dx } f(x)=2x{ e }^{ { x }^{ 2 } }$$
When $n=2$,$$f''(x)=(2x{ e }^{ { x }^{ 2 } })'=4{ x }^{ 2 }{ e }^{ { x }^{ 2 } }$$
1 Answer
$\begingroup$It's actually very easy to "see" the inductive step. The tedious part is writing it out rigorously.
In the inductive step, assume the proposition is true for some $n=k$, i.e. $f^{(k)} = P_k(x)e^{x^2}$.
Then by product and chain rules, $f^{(k+1)} = P'_k(x)e^{x^2} + 2xe^{x^2}P_k(x) = e^{x^2}(P'_k(x) + 2xP_k(x))$
Represent $P_k(x) = a_0x^k + a_1x^{k-1} + ... + a_k = \sum_{i=0}^ka_ix^{k-i}$, where $a_0 \neq 0$ (lead coefficient non zero).
Then $P'_k(x) = \sum_{i=0}^k(k-i)a_ix^{k-i-1}$. The final (constant) term in the summation vanishes, but the summation can still remain indexed this way.
Also, $2xP_k(x) = 2\sum_{i=0}^ka_ix^{k-i+1}$
Hence $P'_k(x) + 2xP_k(x) = \sum_{i=0}^k(k-i)a_ix^{k-i-1} + 2\sum_{i=0}^ka_ix^{k-i+1} \\= \sum_{i=0}^k (2a_ix^{k-i+1} + (k-i)a_ix^{k-i-1})$
You can now observe that the summation is a polynomial of degree $k+1$ as the lead coefficient $a_0$ does not vanish and the power of the $x$ term of the lead term is $k+1$ (when $i=0$). Define $P_{k+1}(x) = P'_k(x) + 2xP_k(x)$. You now have an explicit transformation from $P_k(x)$ to $P_{k+1}(x)$, as you've shown that $P'_k(x) + 2xP_k(x)$ always gives a polynomial of degree $k+1$ . This completes the inductive step. After the base case is shown, you're done.
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