$f^{(j)}(a) = 0$ except for the last derivative. Under which conditions $a$ can be a minimizer of $f$?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ and suppose that $f^{(j)}(a) = 0, j=0,\cdots,n-1$ and $f^{(n)}(a) \neq 0$. Under which conditions the point $x=a$ can be a minimizer of $f$? Based on your answer: $f(x) = x^{13}$ has a minimum in $x=0$? What about $x^{16}$?

Based on what I understand, I can't say that a point that has first derivative $0$ is a minimum or maximum in this case, because the second derivative is $0$.

The only function that I know that have this property of every derivative being $0$ except in the last, is the $x^n$, but only for $a=0$. If I try to think for $a=1$ and degree $n$ then I don't think there is a solution. So I think I should only think about $x^n$.

If $n$ is even, then I know that $x^n\ge 0$ and therefore $0$ would be a minium. For $n$ odd, I don't think there's a minimum. However, the exercise asks about arbitrary $a$ so that's confusing me.

UPDATE

Can we say for sure that $(x-c)^n$ is the only possible function $f$?

If that's the case, then for a point $x_0$ to be a minimizer, we must have $$(x_0-c)^n\le (x-c)^n$$ for $x$ near $x-c$.

I can prove that if $n$ is even, then $(x-c)^n = (x-c)^{2k}$ for some natural $k$. Then:

$$ 0 = (c-c)^2\le (x-c)^2$$

by multiplying the inequation above by itself $k$ times we have

$$ 0 = (c-c)^{2k}\le (x-c)^{2k} = (x-c)^{n}$$

therefore $c$ is a minimizer for $(x-c)^{n}$

If I prove that there are no other minimizers for the case of $n$ even, then a condition for $x$ to me a minimizer would be $f(x)=0$.

So, suppose there are another minimizer $x_1$ of $f$ such that $x_1\neq c$. Then $0<(x_1-c)^2$. Since $0<x_1-c$:

$$0<\left(\frac{x_1-c}{h}-c\right)^2 <(x_1-c)^2 $$

Therefore $x_1$ cannot be a minimizer unless $x_1 = c$. We can then prove $x_1$ is not a minimizer for $(x_1-c)^n$

Therefore a condition for $x$ to me a minimizer of $f$ would be $f(x) = 0$, when $n$ is even.

For $n$ odd, we have

$$(x-c)^n = (x-c)^{2k+1} = (x-c)^{2k}(x-c)$$

for any $x$ except $c$, $$\left(\frac{x-c}{h}-c\right)^{2k}\left(\frac{x-c}{h}-c\right) < (x-c)^{2k+1} < \left(\frac{x+c}{h}-c\right)^{2k}\left(\frac{x+c}{h}-c\right)$$

therefore for any neighborhood of $x$ we have elements less than $x$ and elements greater than $x$. So no minimum in any case unless $x=c$. In this case we can see that

$$(0-h)^{2k+1}<0^{2k+1}<(0+h)^{2k+1}$$

So in general, for $x$ to be a minimizer of $f(x) = (x-c)^{n}$ we must have $n$ even, and $f(x)=0$

Am I right?

UPDATE:

Why, in the answer below,

$${\rm sgn}\left({c\over n!}+r(x)\right)={\rm sgn}(c)\qquad(-h<x<h)$$

Answer 1

We cannot deal with this question without reference to Taylor's theorem. Since this "Taylor business" is translation invariant we may WLOG assume $a=0$.

Assume $$f^{(k)}(0)=0\quad(1\leq k<n),\qquad c:=f^{(n)}(0)\ne0\ .$$ Then by Taylor's theorem we have $$f(x)=f(0)+{c\over n!}x^n + o(x^n)\qquad(x\to0)\ ,$$ or, if you prefer: There is a function $x\mapsto r(x)$, continuous at $x=0$, with $r(0)=0$, such that $$f(x)=f(0)+\left({c\over n!}+r(x)\right) x^n\ .$$ (If $f$ is in fact $n+1$ times differentiable in a full neighborhood of $0$ the error term is of order $x^{n+1}$.) Since $\lim_{x\to0} r(x)=r(0)=0$ there is an $h>0$ such that $${\rm sgn}\left({c\over n!}+r(x)\right)={\rm sgn}(c)\qquad(-h<x<h)\ .$$ Hence we may conclude the following:

• If $n>1$ is odd then $f(x)-f(0)$ assumes different signs on opposite sides of $x=0$. Therefore we have no local extremum at $x=0$.
• If $n>1$ is even then $f$ assumes a local minimum at $x=0$ if $c>0$, and a local maximum, if $c<0$.