calculus - Orders of mean value theorem?

Q. Use Mean Value Theorem of appropriate order to prove that $\sin(x)\gt x-\dfrac{x^3}{3!}$

Now, I know the stated inequality was proved in a previous post, viz. Proof for $\sin(x)\gt x-\frac{x^3}{3!}$ but my question here is what does the problem poser (aka my calculus professor) mean by "mean value theorem of appropriate order" ?

I'm sorry if this is a naive question, I'm a beginner at differential calculus. Thanks for any help!

Answer

Quoting from this linked page,

Now let’s take an arbitrary function f that is n-times differentiable on an open interval containing $[x,x+h]$. To prove the mean value theorem, we subtracted a linear function so as to obtain a function that satisfied the hypotheses of Rolle’s theorem. Here, the obvious thing to do is to subtract a polynomial p of degree n to obtain a function that satisfies the hypotheses of our higher-order Rolle theorem.

The properties we need $p$ to have are that $p(x)=f(x)$, $p'(x)=f'(x)$, and so on all the way up to $p^{(n-1)}(x)=f^{(n-1)}(x)$, and finally $p(x+h)=f(x+h)$. It turns out that we can more or less write down such a polynomial, once we have observed that the polynomial $q_k(x)=(u-x)^k/k!$ has the convenient property that $q_k^{(j)}(x)=0$ except when $j=k$ when it is $1$. This allows us to build a polynomial that has whatever derivatives we want at $x$. So let’s do that. Define a polynomial $q$ by

$q(u)=f(x)+q_1(u)f'(x)+q_2(u)f''(x)+\dots+q_{n-1}(u)f^{(n-1)}(x)$

Then $q^{(k)}(x)=f^{(k)}(x)$ for $k=0,1,\dots,n-1$. A more explicit formula for $q(u)$ is

$\displaystyle f(x)+(u-x)f'(x)+\frac{(u-x)^2}{2!}f''(x)+\dots+\frac{(u-x)^{n-1}}{(n-1)!}f^{(n-1)}(x)$

Now $q(x+h)$ doesn’t necessarily equal $f(x+h)$, so we need to add a multiple of $(u-x)^n$ to correct for this. (Doing that won’t affect the derivatives we’ve got at $x$.) So we want our polynomial to be of the form

$p(u)=q(u)+\lambda(u-x)^n$

and we want $p(x+h)=f(x+h)$. So we want $q(x+h)+\lambda h^n$ to equal $f(x+h)$, which gives us $\lambda=h^{-n}(f(x+h)-q(x+h))$. That is,

$\displaystyle p(u)=q(u)+\frac{(u-x)^n}{h^n}(f(x+h)-q(x+h))$

A quick check: if we substitute in $x+h$ for $u$ we get $q(x+h)+(h^n/h^n)(f(x+h)-q(x+h))$, which does indeed equal $f(x+h)$.

For the moment, we can forget the formula for $p$. All that matters is its properties, which, just to remind you, are these.

$p$ is a polynomial of degree $n$.
$p^{(k)}(x)=f^{(k)}(x) for k=0,1,\dots,n-1$.
$p(x+h)=f(x+h)$.
The second and third properties tell us that if we set $g(u)=f(u)-p(u)$, then $g^{(k)}(x)=0$ for $k=0,1,\dots,n-1$ and $g(x+h)=0$. Those are the conditions needed for our higher-order Rolle theorem. Therefore, there exists $\theta\in(0,1)$ such that $g^{(n)}(x+\theta h)=0$, which implies that $f^{(n)}(x+\theta h)=p^{(n)}(x+\theta h)$.

Let us just highlight what we have proved here.

Theorem. Let f be continuous on the interval $[x,x+h]$ and n-times differentiable on an open interval that contains $[x,x+h]$. Let $p$ be the unique polynomial of degree $n$ such that $p^{(k)}(x)=f^{(k)}(x)$ for $k=0,1,\dots,n-1$ and $p(x+h)=f(x+h)$. Then there exists $\theta\in(0,1)$ such that $f^{(n)}(x+\theta h)=p^{(n)}(x+\theta h)$.

Note that since $p$ is a polynomial of degree $n$, the function $p^{(n)}$ is constant. In the case $n=1$, the constant is $\frac{f(x+h)-f(x)}{h}$, the gradient of the line joining $(x,f(x))$ to $(x+h,f(x+h))$, and the theorem is just the mean value theorem.