[Math] Higher Order Derivative Proof .

analysisderivativesproof-verificationtaylor expansion

I would appreciate if someone could check over my proof for this question and advise me if it is correct.

Question

My attempt so far;
Now as $f$ is k times differentiable , it taylor series about $x_{0}$ can be written as follows,
$$f(x)=f(x_{0})+f'(x_{0})(x-x_{0})+f''(x_{0})\frac{(x-x_{0})^2}{2!}+…+f^{(k)}\frac{(x-x_{0})^k}{n!}$$
We are given that; $$f'(x_0)=…=f^{(k-1)}(x_0)=0 $$ and $$f^{(k)} \neq 0$$
Sow we can conclude that we end up with something looking like this… $$f(x)=f(x_{0})+f^{(k)}\frac{(x-x_{0})^k}{n!}$$.
And I also know the fact in the neighbourhood of $x_0$

1) if $x_0$ is a local minimum, then in a neighbourhood of $x_0$, $f(x)-f(x_0)>0,$
2) if $x_0$ is a local maximum, then in a neighbourhood of $x_0$, $f(x)-f(x_0)<0.$

And from what we have : $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}$$.

Now I just have to put this together;

  • CASE 1: $k$ is Even

Subcase 1: If $f^{(k)}<0$. Then looking at the interval $x \in (x_0 – \delta,x_0 + \delta)$
We have

  • $x<x_0$ implies that $(x-x_0)^{k}>0$ as $k$ is even and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}<0$$ Implying $f(x)$ is increasing on $(x,x_0)$.

  • $x>x_0$ implies that $(x-x_0)^{k}>0$ and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}<0$$ Implying $f(x)$ is decreasing on $(x_0,x)$.
    So this implies a maximum, as in the neighbourhood of $x_0$, $f(x)-f(x_0)<0.$.

Subcase 2:

If $f^{(k)}>0$. Then looking at the interval $x \in (x_0 – \delta,x_0 + \delta)$
We have ;

  • $x<x_0$ implies that $(x-x_0)^{k}>0$ as $k$ is even and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}>0$$ Implying $f(x)$ is decreasing on $(x,x_0)$.

    • $x>x_0$ implies that $(x-x_0)^{k}>0$ and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}>0$$ Implying $f(x)$ is Increasing on $(x_0,x)$.
      So this implies a minimum, as in the neighbourhood of $x_0$, $f(x)-f(x_0)>0.$.

And so this proves $(i)$;
CASE 2: $k$ is odd;

Subcase 1: If $f^{(k)}<0$. Then looking at the interval $x \in (x_0 – \delta,x_0 + \delta)$
We have

  • $x<x_0$ implies that $(x-x_0)^{k}<0$ as $k$ is odd and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}>0$$ Implying $f(x)$ is decreasing on $(x,x_0)$.

    • $x>x_0$ implies that $(x-x_0)^{k}>0$ and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}<0$$ Implying $f(x)$ is decreasing on $(x_0,x)$.

So $f$ is strictly decreasing on neighbourhood $x_0$.

Subcase 2:

If $f^{(k)}>0$. Then looking at the interval $x \in (x_0 – \delta,x_0 + \delta)$
We have ;

  • $x<x_0$ implies that $(x-x_0)^{k}<0$ as $k$ is even and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}<0$$ Implying $f(x)$ increasing on $(x,x_0)$.

    • $x>x_0$ implies that $(x-x_0)^{k}>0$ and so $$f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}>0$$ Implying $f(x)$ is Increasing on $(x_0,x)$.

So $f$ is strictly increasing on neighbourhood $x_0$.
So this Proves part $(ii)$. $\blacksquare$

EDIT: I know this proof is tedious and long but is it correct… I also realise induction would have been much quicker but felt that this strengthens my understanding better.

Best Answer

Question statement transcribed:

Exercise 11. Suppose that $f:(a, b) \rightarrow \mathbb{R}$ is $k$ times differentiable, and $f^{\prime}\left(x_{0}\right)=\cdots=$ $f^{(k-1)}\left(x_{0}\right)=0, f^{(k)}\left(x_{0}\right) \neq 0 .$ Show that

(i). if $k$ is even then $f$ has a local max at $x_{0}$ if $f^{(k)}\left(x_{0}\right)<0$ and a local min if $f^{(k)}\left(x_{0}\right)>0$, and

(ii). if $k$ is odd then $f$ is strictly increasing in a neighbourhood of $x_{0}$ if $f^{(k)}\left(x_{0}\right)>0$ and strictly decreasing if $f^{(k)}\left(x_{0}\right)<0$

Hint: Use induction on $k$, applying the induction hypothesis to $f^{\prime}$, to prove both statements simultaneously.

Comment from Vityôk (who set a bounty):

@CalvinKhor, i wanted to know how to incorporate the remainder, and i don't understand the case 2. Why in the subcase 1 of case 2 we have: $f(x)-f(x_{0})=f^{(k)}\frac{(x-x_{0})^k}{n!}>0$? I don't understand why it is $>0$. – Vityôk

  1. It is worth noting that the question (ii) is wrong. Differentiable at a point with positive derivative implies increasing in neighborhood of point? gives a counterexample (everywhere differentiable function with $f'(0)>0$) for $k=1$ (odd) which is not monotonic near $0$. What you can prove is what the OP did (modulo incorporating the remainder).

  2. To "incorporate the remainder", one should apply a correct version of Taylor's theorem. There are many variants; since we only have $k$ times differentiability, we have to use the Peano remainder form which is the weakest kind*. It reads: for every $x_0$, and for every $k$, if $f$ is $k$ times differentiable, then there exists a function $r$ with $r(x)\to 0$ as $x\to x_0$ such that $$ f(x)= \sum_{n=0}^k \frac{f^{(n)}(x_0)(x-x_0)^n}{n!} + r(x)(x-x_0)^k$$

  3. Once you have this correctly stated, the rest of the proof is easily modified to be correct (after the question is corrected). Let me write explicitly write the subcase indicated by Vityôk, which is when $k$ is odd and $f^{(k)}(x_0)<0$. Taylor's theorem gives for any $x$, $$ f(x)-f(x_0)= \left[\frac{f^{(k)}(x_0)}{k!} + r(x)\right](x-x_0)^k$$ For $x$ sufficiently close to $x_0$, $\frac{f^{(k)}(x_0)}{k!} + r(x)<0$. When $x>x_0$, then $(x-x_0)^k>0$; product of negative and positive is negative. So $f(x)<f(x_0)$. Similarly for $x<x_0$. But note we cannot get monotonicity on an interval since we know nothing about $f^{(k)}$ at other points than $x_0$.

*$k+1$ times differentiable lets you use Cauchy and Lagrange forms; absolute continuity of $f^{(k)}$---which also follows from $k+1$ times differentiability---allows you the integral form. These imply continuity of the $k$th derivative which then in fact gives a result like (ii).

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