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).

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    $\begingroup$ 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 $p$. $\endgroup$ – Connor Harris Oct 30 '18 at 21:09
  • $\begingroup$ @ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though. $\endgroup$ – Robert Wolfe Oct 30 '18 at 23:16
  • $\begingroup$ 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. $\endgroup$ – Robert Wolfe Nov 2 '18 at 3:32
  • $\begingroup$ @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. $\endgroup$ – 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.


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 $\xi<\frac{1}{2}$.)

  • $\begingroup$ 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. $\endgroup$ – Robert Wolfe Nov 2 '18 at 21:02
  • $\begingroup$ 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 $\xi=1/2$? $\endgroup$ – alex.jordan Nov 2 '18 at 21:16
  • $\begingroup$ yeah. a small detail. $\endgroup$ – Robert Wolfe Nov 2 '18 at 21:29
  • $\begingroup$ it seems that zhw's $(x\mapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now. $\endgroup$ – Robert Wolfe Nov 2 '18 at 21:37
  • $\begingroup$ 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 $\xi<1/2$. $\endgroup$ – 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 $1-\xi.$

  • $\begingroup$ 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. $\endgroup$ – Robert Wolfe Nov 2 '18 at 21:42
  • $\begingroup$ +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 $\xi\to1^{-}$. $\endgroup$ – 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 $\frac{1}{3} \leq\xi \leq \frac{2}{3}.$

  • 1
    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Nov 5 '18 at 3:46

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