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!


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


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


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.

  • $\begingroup$ I'm lazy af right now to go through such mathematical detail at the moment, so would you do me a favor and show how this can be applied to prove the stated inequality? The answer in the linked page doesn't use this form of MVT but instead some MVT for integrals. $\endgroup$ – learner Nov 22 '15 at 7:36
  • $\begingroup$ @learner Taylor's theorem is called the M.V.T. of $n^{\text{th}}$ order. $\endgroup$ – SchrodingersCat Nov 22 '15 at 19:06
  • $\begingroup$ I see, well, would you be kind enough to show how does one use Taylor's Theorem here to prove the stated inequality? I know the infinite series expansion $\sin(x)=\sum\limits_{k=0}^\infty (-1)^k\dfrac{x^{2k+1}}{(2k+1)!}$ but I'm not sure how does one rigorous prove that inequality using Taylor's Theorem (I haven't covered that part yet). $\endgroup$ – learner Nov 24 '15 at 10:51
  • 1
    $\begingroup$ You have stated the proof in your comment. It goes like this: $\sin x=\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...> x-\frac{x^3}{3!}$ since it is very obvious and can also be proven easily that $\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...>0$. Hope this helps..:-) $\endgroup$ – SchrodingersCat Nov 24 '15 at 16:00
  • $\begingroup$ it does but I was thinking there would be some rigorous proof at hand here. Nonetheless, thanks for all the help. :) $\endgroup$ – learner Nov 24 '15 at 17:03

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