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A family of parabolas $p(x)$ is given for $x \in [0,1]$ by coefficients $(a,b,c)$ , everything real-valued: $$ p(x) = a x^2 + b x + c $$ The area of the parabolas is normed: $\int_0^1 p(x)\, dx = 1$ ; they have a minimum: $a > 0$ ; and they do not intersect the x-axis: $b^2 - 4ac < 0$ . So the maximum values must be at $x=0$ or $x=1$ .
What is the upper bound of these maximum values ?

Han de Bruijn
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  • Some ideas: try to show that optimal parabola touches the $x$-axis, so that you are only left with two parameters: scale $a$ and minimum location $-b/2a$ – SBF Nov 25 '14 at 10:31
  • It seems to be an easy problem, just integrate the polynomial to eliminate a variable then plug in 0 and 1. What part are you stuck on? – Suzu Hirose Nov 25 '14 at 10:35
  • @Ilya: See my answer below: you're quite right. But I wonder if there is a more intuitive way to see that the optimum is reached for a parabola that touches the $x$-axis ($M=0$ in my answer). – Han de Bruijn Nov 25 '14 at 14:38
  • Well, suppose you have some parabola - how can you increase both of its boundary values? Either increase the scale $a$, or the level $M$ (I'm using notation of @mookid). If you increase $a$ to have the same effect, it will bring up less points, hence it does not increase the area as much as the equivalent increase of $M$ would. Hence, we'd like to keep $M$ as low as possible. Also, moving the minimum point we balance the boundary maxima so it's intuitively clear that the minimum point must be $0$ or $1$ (does not really matter). – SBF Nov 25 '14 at 14:40
  • @SuzuHirose: After those hints given by mookid, indeed there is not much of a problem anymore. See my answer below. But I disagree with you that the problem is just "easy". – Han de Bruijn Nov 25 '14 at 14:43

2 Answers2

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Rewrite $p(x) = a\left( (x-m)^2 + M^2 \right)$. Then the max is, as you noticed: $$O(m,M,a) = \max (a\left( (1 - m)^2 + M^2 \right), a\left( m^2 + M^2 \right)) $$

and the domain is only constrined by : $m,M\in\Bbb R, a>0$ and as the intrgral is 1: $$ C(m,M,a) = \frac a3 ((1-m)^3 + m^3) + aM^2 = 1 $$ Now notice that $O(m,M,a) = O(1-m,M,a)$ and $C(m,M,a) = C(1-m,M,a)$ so you can get rid of the $\max$.

mookid
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The answer IMO is not at all quite trivial. Using the Ansatz by mookid we find: $$ \frac{a}{3} \left[(1-m)^3 + m^3\right] + aM^2 = 1 \quad \Longrightarrow \quad a = \frac{1}{1/3 - m + m^2 + M^2} \quad \Longrightarrow \\ p(0)(m,M) = \frac{m^2+M^2}{1/3 - m + m^2 + M^2} $$ As a function of $M$ , $p(0)(m,M)$ is maximal (or minimal) for $M=0$ : $$ \frac{\partial\, p(0)(m,M)}{\partial M} = 0 \quad \Longrightarrow \quad M = 0 $$ Hence proceeding with $M=0$ : $$ p(0)(m,0) = \frac{m^2}{1/3 - m + m^2} \\ \frac{dp(0)(m,0)}{dm} = 0 \quad \Longrightarrow \quad m \in \{ 0,2/3 \} \quad \Longrightarrow \quad p(0)(2/3,0) = 4 $$ So the upper bound is $\huge{\;4\;}$ , which means that I can pinpoint my graphics window at that value.

Notes. $$ \frac{\partial\,p(0)(m,M)}{\partial M} = \frac{2M(m-1/3)}{\left[M^2-(m-1/3)+m^2\right]^2} = 0 $$ It follows that $M=0$ is a solution, giving $\;p(0)(m,0)=m^2/(1/3-m+m^2)\;$ , leading to the upper bound as required. But there is also a solution $\;m=1/3\;$ , leading to a function value which is independent of the parabola's minimum $M$ : $\;p(0)(1/3,M) = 1$ .
The other way around: $$ \frac{\partial\,p(0)(m,M)}{\partial m} = \frac{M^2-m^2+2m/3}{\left[M^2-(m-1/3)+m^2\right]^2} = 0 $$ It follows that $m=1/3+\sqrt{1/9+M^2}$ is the solution. Substitution in $\;p(0)(m,M)\;$ yields: $$ p(0)(1/3+\sqrt{1/9+M^2},M) = 2\frac{1+\sqrt{1+9M^2}+9M^2}{18M^2-\sqrt{1+9M^2}+2} $$ And the maximum is found for derivative zero: $$ \frac{dp}{dM} = -\frac{54M\sqrt{1+9M^2}}{\left(18M^2-\sqrt{1+9M^2}+2\right)^2} = 0 \quad \Longrightarrow \quad M=0 \quad \Longrightarrow \quad m = 2/3 $$

Han de Bruijn
  • 17,070