A few days ago, I posted a problem (Question about maximum and minimum of a function of C^(k+1))
My problem can be found on the given hyperlink, but for your convenience: "Let $f: \Bbb R \to \Bbb R$ be of class $C^{k+1}$. Suppose $f^{(j)}(a) = 0$ for $0 \leq j < k$ and $f^{(k)}(a) \neq 0$. If $k$ is even, show that $f$ has a maximum or minimum at $a$ depending on the sign of $f^{(k)}(a)$."
I THOUGHT that I could solve this through induction hypothesis, but I couldn't. Here is the reason I realized why this can't be proven by induction.
(for the base case $k=2$, the case is proven using the single variable Second Derivative test. )
Assume (for $k=l-2$) that $f(a)=f'(a)=...=f^{(l-3)}(a)=0$ and $f^{(l-2)}(a) \neq 0$ implies that $f$ has maximum at $a$ if $f^{(l-3)}(a) < 0$ and that $f$ has minimum at $a$ if $f^{(l-3)}(a) > 0$.
Then, we have to use our inductive hypothesis to prove that if $f(a)=f'(a)=...=f^{(l-2)}(a)=f^{(l-1)}(a)=0$ and $f^l(a) \neq 0$ then f has an extremum depending on the sign of $f^l(a)$. However, we cannot use our inductive hypothesis because we must prove that IF $f(a)=f'(a)=...=f^{(l-2)}(a)=f^{(l-1)}(a)=0$ then the condition is satisfied; Since our inductive hypothesis has $f^{(l-2)}(a)\neq 0$, we cannot apply our inductive hypothesis in any way.
I am really stuck on this problem. Since I have been learning about Taylor polynomials and partial derivatives, I think that my solution has to be somehow related to this problem. Any help would be great.
Use Taylor's Theorem:
We suppose that $f^{(k)}(a)>0$. Since $f^{(k)}$ is continuous, there is a neighborhood $U$ of $a$ such that $f^{(k)}(t)>0$ for all $t \in U$.
Now let $h \in \Bbb R $ such that $a+h \in U$. Taylor says: there is $t \in U$ with
$$f(a+h)=f(a)+\frac{f^{(k)}(t)}{k!}h^k$$.
Since $k$ is even, we have $h^k>0$ and therfore
$$f(a+h) \ge f(a)$$