# Prove that we can't find effective bounds on the point guaranteed by the Mean Value Theorem.

I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:

For each real number $$M$$ and each real number $$\xi$$ that lies strictly between $$0$$ and $$1$$, construct a function $$f$$ such that $$f(0)=0,\; f(1)=M,\;f\text{ is continuous on }[0,1],\; f\text{ is differentiable on }(0,1),\;\text{ and }\xi\text{ is the unique point strictly between 0 and 1 such that}\;f'(\xi)=M\,.$$

For the $$M\neq 0$$ and $$\xi\neq 1/e$$ case, we can show that $$g(x)=\begin{cases} 0&\text{ if }x=0,\\ 1/e&\text{ if }x=1\\ 1&\text{ if }x=\infty,\\ \sqrt[1-x]{x}&\text{ otherwise} \end{cases}$$

is strictly increasing and continuous on $$[0,\infty]$$. Thus there is a unique positive $$\alpha$$ such that $$g(\alpha)=\xi$$. In turn, we can define $$f(x)=Mx^\alpha$$ which will satisfy the claim. For the $$M\neq 0$$ and $$\xi=1/e$$ case, take the obvious continuous extension of $$f(x)=M(x+x\ln(x))$$.

For $$M=0$$, we first choose $$\alpha\geq 1$$ and $$\beta\geq 1$$ such that $$\frac{\alpha}{\alpha+\beta}=\xi$$. We then define $$f(x)=x^\alpha(1-x)^\beta$$ which will satisfy the claim.

My question however is this:

Can we construct such an $$f$$ to be a polynomial?

An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $$\xi$$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).

• Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(\forall 0 \leq 1 \leq x) p''(x) < 0$, and $p'(\xi) = 0$ for some arbitrary (unique by construction) $\xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$. – Connor Harris Oct 30 '18 at 21:09
• @ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though. – Robert Wolfe Oct 30 '18 at 23:16
• One third of a solution: for $1/3\leq \xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<\xi\leq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=\frac{\xi(3\xi-2)}{2\xi-1}\;\text{ and }\;c_2=\frac{\xi(3\xi-2)}{1-2\xi}\,.$$ For $\xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way. – Robert Wolfe Nov 2 '18 at 3:32
• @RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(x\pm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third. – alex.jordan Nov 2 '18 at 7:29

First, imagine $$f$$ is some such polynomial for $$\xi$$. Then let $$g(x)=f(x)-Mx$$. We have $$g(0)=g(1)=0$$, and $$\xi$$ is the unique number in $$(0,1)$$ where $$g'(x)=0$$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.

Below is a proof that if you have $$\xi>\frac{1}{2}$$, take some integer $$n>\frac{1-2\xi}{\xi-1}$$, and then take $$t=\frac{(n+2)\xi^2-(n+1)\xi}{-2\xi+1}$$. Then the polynomial $$g(x)=(x+t)^nx(1-x)$$ satisfies $$g(0)=g(1)=0$$, and there is a unique number in $$(0,1)$$ where $$g'(x)=0$$, and that number is $$\xi$$.

If $$\xi<\frac{1}{2}$$, there is a symmetric construction with $$t<-1$$. And if $$\xi=\frac{1}{2}$$, just take $$g(x)=x(1-x)$$.

For example, with $$\xi=\frac{e}{\pi}$$, we can take $$n=6$$, and $$t=\frac{8(e/\pi)^2-7(e/\pi)}{-2(e/\pi)+1}\approx0.09233\ldots$$. Then $$g(x)=(x+t)^6x(1-x)$$ is such that $$g'$$ has only one zero in $$(0,1)$$, and it is located at $$\frac{e}{\pi}$$. See this demonstrated at WolframAlpha.

# Explanation

Assume $$\xi>\frac{1}{2}$$. Consider $$g(x)=(x+t)^nx(1-x)$$ for $$n\in\mathbb{N}$$ and $$t\in\mathbb{R}_{\gt0}$$. Then \begin{align} g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\\ &=(x+t)^{n-1}\big(nx(1-x)+(x+t)(1-x)-(x+t)x\big)\\ &=(x+t)^{n-1}\big(x^2(-n-2)+x(n+1-2t)+t\big)\\ \end{align} The zeros of $$g'$$ are $$-t$$ (which is not in $$(0,1)$$) and $$\frac{-(n+1-2t)\pm\sqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=\frac{A\pm B}{C}$$ Since $$n,t>0$$, it follows that $$|B|>|A|$$. It follows that one of these two roots is negative (so not in $$(0,1)$$) and the other is positive. So if the positive root is equal to $$\xi$$, then $$g$$ satisfies the Rolle's version of the proposition. We have freedom to choose $$n\in\mathbb{N}$$, $$t\in\mathbb{R}_{\gt0}$$, so maybe we can choose them well. In the following, we attempt to solve for $$t$$ in terms of $$\xi$$ and $$n$$.

\begin{align} \xi&=\frac{-(n+1-2t)\pm\sqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\\ -2(n+2)\xi&=-(n+1-2t)\pm\sqrt{(n+1-2t)^2+4(n+2)t}\\ -2(n+2)\xi+n+1-2t&=\pm\sqrt{(n+1-2t)^2+4(n+2)t} \end{align} Squaring both sides: \begin{align} [-2(n+2)\xi+n+1]^2-4t[-2(n+2)\xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\\ [-2(n+2)\xi+n+1]^2-4t[-2(n+2)\xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\\ [-2(n+2)\xi+n+1]^2-4t[-2(n+2)\xi+n+1]&=(n+1)^2+4t\\ [-2(n+2)\xi+n+1]^2-(n+1)^2&=4t[-2(n+2)\xi+n+2]\\ 4(n+2)^2\xi^2-4(n+2)(n+1)\xi&=4t(-2(n+2)\xi+n+2)\\ \end{align} \begin{align} t&=\frac{4(n+2)^2\xi^2-4(n+2)(n+1)\xi}{4(-2(n+2)\xi+n+2)}\\ &=\frac{(n+2)\xi^2-(n+1)\xi}{-2\xi+1} \end{align} We have assumed $$\xi>\frac{1}{2}$$, so the denominator is negative. We need $$t$$ to be positive, so we need the numerator to be negative. Can we choose $$n$$ to make that happen? \begin{align} (n+2)\xi^2-(n+1)\xi&<0\\ n(\xi^2-\xi)&<\xi-2\xi^2\\ n(\xi-1)&<1-2\xi\\ n&>\frac{1-2\xi}{\xi-1} \end{align} So yes. If you have $$\xi>\frac{1}{2}$$, take some integer $$n>\frac{1-2\xi}{\xi-1}$$, and then take $$t=\frac{(n+2)\xi^2-(n+1)\xi}{-2\xi+1}$$. Then the polynomial $$g(x)=(x+t)^nx(1-x)$$ satisfies $$g(0)=g(1)=0$$, and there is a unique number in $$(0,1)$$ where $$g'(x)=0$$, and that number is $$\xi$$.

And restating from the introduction, if $$\xi<\frac{1}{2}$$, there is a symmetric construction where $$t<-1$$. And if $$\xi=\frac{1}{2}$$, just take $$g(x)=x(1-x)$$.

This approach was motivated by starting with $$x(1-x)$$, and then multiplying by some power of $$(x+t)$$ that would "warp" the parabola in between $$0$$ and $$1$$ without making it wiggle. Something that would stretch $$x(1-x)$$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $$\xi<\frac{1}{2}$$.)

• I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis. – Robert Wolfe Nov 2 '18 at 21:02
• You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $\xi>1/2$ versus $\xi<1/2$ versus $\xi=1/2$? – alex.jordan Nov 2 '18 at 21:16
• yeah. a small detail. – Robert Wolfe Nov 2 '18 at 21:29
• it seems that zhw's $(x\mapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now. – Robert Wolfe Nov 2 '18 at 21:37
• This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,\infty)$ into $\tilde{t}$ in $(-\infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $\xi<1/2$. – alex.jordan Nov 2 '18 at 22:26

Just a few words on solving the nonpolynomial problem: First take $$M=0.$$ Then $$f(x)=x(1-x)$$ is a solution for $$\xi=1/2.$$ For other values of $$\xi\in (0,1),$$ we look at $$g_p(x)=f(x^p)$$ for $$p>0.$$ We have $$g_p'(x)=px^{p-1}f'(x^p)=0$$ iff $$x=(1/2)^{1/p}.$$ Thus if $$\xi$$ is given, we take $$p=\ln(1/2)/\ln \xi,$$ and $$g_p$$ solves the problem. Finally, if $$M\ne 0$$ and $$\xi$$ is given, we take the same $$p$$ and the function $$M(x+g_p(x))$$ solves the problem.

On to polynomial solutions: WLOG $$M=0$$, for we can use the same idea as above for $$M\ne 0.$$

Let $$f(x)=x(1-x)$$ as above. If $$p$$ is a polynomial with $$p(0)=0,$$ $$p(1)=1$$ and $$p'>0$$ on $$(0,1),$$ then $$f\circ p$$ is a polynomial that solves the problem for for the unique value $$\xi \in(0,1)$$ such that $$p(\xi)=1/2.$$

For $$0\le b \le 1,$$ set $$p_b(x)=(1-b)x^2 +bx.$$ Then $$p_b(0)=0,p_b(1)=1,$$ and $$p_b'>0$$ on $$(0,1).$$ Consider the equation $$p_b(x)-1/2=0.$$ If $$b=1,$$ then $$x=1/2$$ is a solution. For $$b\in [0,1)$$ we have a quadratic equation whose solution in $$[0,1]$$ is

$$x= \frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$

Verify the right side, as a function of $$b,$$ strictly decreases on $$[0,1)$$ from $$1/2^{1/2}$$ to $$1/2.$$ Thus for $$\xi\in [1/2,1/2^{1/2}],$$ there is a unique $$b_{\xi}\in [0,1]$$ such that $$p_{b_{\xi}}(\xi)=1/2.$$ Verify that $$b_{\xi}$$ is given by the formula

$$b_{\xi} = \frac{1/2-\xi^2}{\xi(1-\xi)}.$$

So we've solved the problem for $$\xi\in[1/2,1/2^{1/2}].$$ But we've also solved it for $$\xi\in [1/2^{1/2},1/2^{1/4}].$$ Just check that that for such $$\xi,$$ $$f\circ p_{b_{\xi^2}}(x^2)$$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $$\xi\in [1/2,1).$$

What about $$\xi\in (0,1/2]?$$ That's easy, now that we've handled the other side. Just check that if $$g$$ is a solution for $$\xi\in [1/2,1),$$ then $$1-g(1-x)$$ is a solution for $$1-\xi.$$

• That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer. – Robert Wolfe Nov 2 '18 at 21:42
• +1 Suppose $\xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $\xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $\xi\to1^{-}$. – alex.jordan Nov 2 '18 at 22:58

Robert's comment about cubics is the best we can do.

Proposition: If $$f$$ is a polynomial of degree at most $$3$$ that satisfies $$f(0) = f(1) = 0$$ and has exactly one value $$\xi \in (0, 1)$$ for which $$f'(\xi) = 0$$, then $$\frac{1}{3} \leq \xi \leq \frac{2}{3}$$.

Proof: Let $$f$$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $$\xi = \frac{1}{2}$$, which requires a quadratic; it's trivial to see that for no other value of $$\xi$$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $$f(x) = ax^3 + bx^2 + cx$$ (which must satisfy $$a + b + c = 0$$) without breaking any condition on $$f$$. The solutions to $$f'(\xi) = 3\xi^2 + 2k \xi - (1+k) = 0$$ are thus $$\xi = \frac{k \pm \sqrt{k^2 + 3k + 3}}{3}.$$

This can be solved for $$k$$ by rearranging and squaring to get $$(3 \xi - k)^2 = k^2 + 3k + 3$$, or $$k = \frac{1 - 3 \xi^2}{2 \xi - 1}$$ but the squaring means that $$\xi$$ could be either the upper or the lower solution for any given $$k$$. Regardless, we know that the fully general formula for a cubic that satisfies $$f(0) = f(1) = 0$$ and has a not necessarily unique stationary point at $$\xi$$, up to scaling of the coefficients, is $$f(x) = (2 \xi - 1) x^3 + (1 - 3 \xi^2) x^2 + (3 \xi^2 - 2\xi) x.$$

We now just need to see which of these cubics have two stationary points in $$(0, 1)$$. By the Vieta formulas, the solutions to $$f'(x) = 3(2 \xi - 1) x^2 + 2(1 - 3 \xi^2) x + (3 \xi^2 - 2\xi) = 0$$ add up to $$\frac{2 (3 \xi^2 - 1)}{3 (2 \xi - 1)}.$$ If $$\xi$$ is one solution, then the other solution (call it $$\xi'$$) is \begin{align*} \xi' &= \frac{2 (3 \xi^2 - 1)}{3 (2 \xi - 1)} - \xi \\ &= \frac{6 \xi^2 - 2}{6 \xi - 3} - \frac{6 \xi^2 - 3 \xi}{6 \xi - 3} \\ &= \frac{3 \xi - 2}{6 \xi -3} \\ &= \frac{3 \xi - \frac{3}{2}}{6 \xi - 3} - \frac{\frac{1}{2}}{6 \xi - 3} \\ &= \frac{1}{2} - \frac{1}{12\xi - 6}.\end{align*} Thus, $$\xi' \notin (0, 1)$$ if and only if $$|12 \xi - 6| \leq 2$$, i.e., if $$\frac{1}{3} \leq\xi \leq \frac{2}{3}.$$

• Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution. – Robert Wolfe Nov 5 '18 at 3:46