So I'm just making sure if my deduction is correct. Let $f$ be a smooth function, we Taylor/Maclaurin expand it around its unique, strict, local minima $a=0.$
So we get this Taylor expansion:
$$f(x)=f(0) + \sum_{j=1}^{r}\frac{f^j(0)}{j!}x^j+o(x^{r})$$
Now let's assume that $r$ is the order of the first non-vanishing derivative at zero (so yes, we're not considering the case of smooth not real analytic functions pointed out in the first comment!), i.e. $f^j(0)=0 \forall 1\le j\le r-1.$ Therefore, the above expansion becomes:
$$f(x)=f(0) + \frac{f^r(0)}{r!}x^r+o(x^{r})$$
I wanted to show that $r$ must be even.
Proof by contraction: if not, assume $r$ is odd.
I just wanted to check if the following immediately implies that $r$ is even? For if $r$ is odd, then we can say for all small enough $x$ of either sign:
$f(x)-f(0)>0\implies \frac{f^r(0)}{r!}x^r+o(x^{r}) > 0 \implies x^r(\frac{f^r(0)}{r!} + o(1)) >0$
Assume now, if possible that $r$ is odd.
So when $x$ is positive, so is $x^r,$ so this implies $\frac{f^r(0)}{r!} + o(1) >0,$ which, passing to the limit to zero, implies: $\frac{f^r(0)}{r!} \ge 0,$ and when $x$ is negative, this will imply that $\frac{f^r(0)}{r!} \le 0,$ (because $x^r<0$ when $r$ is odd and $x<0.$). So put together, we'll have ${f^r(0)}=0,$ which will contradict our assumption that " $r$ is the odd order of the first non-vanishing derivative at zero" . is my deduction/reasoning correct?
One more thing: similar argument and conclusion also hold for higher dimension than one, correct? Thank you in advance!