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(Tenth grade high school math competition):Let $$f(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4$$ where $a_{i}\in \mathbb R$, and such that for $x\in[-1,1]$ we have $$|f(x)|\le 1$$ Find the maximum of the value $|a_{2}|$.

I know Chebyshev polynomial $T_{4}(x)=8x^4-8x^2+1$ verifies such condition.

But how to prove that it generates the looked for maximum ?

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    $\begingroup$ Probably what you are looking for en.wikipedia.org/wiki/Markov_brothers%27_inequality $\endgroup$ – r9m Dec 23 '16 at 10:54
  • $\begingroup$ @r9m very interesting! (I didn't know this inequality). Maybe the OP should be reminded that $a_2$ is connected to $P''(0)$. $\endgroup$ – Jean Marie Dec 23 '16 at 11:07
  • $\begingroup$ And this is a Tenth grade problem? $\endgroup$ – Violapterin Dec 23 '16 at 11:56
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Note that we only have to consider $f$ which are even, since if $f$ is a quartic with $|f(x)|\le 1$ for $-1\le x\le 1$, then $e(x) = \frac{f(x)+f(-x)}{2}$ is also a quartic with $|e(x)|\le 1$ for $-1\le x\le 1$, and $e$ is even and has the same coefficient of $x^2$ as $f$.

As such, let $f(x) = a_0+a_2x^2+a_4x^4$. Then $f(x) = g(x^2)$, where $g(x) = a_0+a_2x+a_4x^2$, and $g$ has the property that $|g(x)|\le 1$ for $0\le x\le 1$ (since $-1\le x\le 1\iff 0\le x^2\le 1$). In particular, $|g(x)|\le 1$ for $x=0$, $\frac{1}{2}$, and $1$. Now $$ a_2 = 4\cdot\left(a_0+\frac{a_2}{2} + \frac{a_4}{4}\right) - (a_0+a_2+a_4) -3a_0 = 4g\left(\frac{1}{2}\right) - g(1) - 3g(0) $$ and so $g\left(\frac{1}{2}\right)\le 1$, $g(1)\ge -1$, and $g(0)\ge -1$ imply that $$ a_2 \le 4\cdot 1 - (-1) - 3\cdot(-1) = 8.$$ Similarly, $g\left(\frac{1}{2}\right)\ge 1$, $g(1)\le -1$, and $g(0)\le -1$ imply that $ a_2 \ge -8$. Hence, we must have $|a_2|\le 8$.

Conversely, $g(x) = 8\left(x-\frac{1}{2}\right)^2-1 = 8x^2-8x+1$ satisfies $|g(x)|\le 1$ for $0\le x\le 1$, and hence $f(x) = 8x^4-8x^2+1$ satisfies $|f(x)|\le 1$ for $-1\le x\le 1$. Hence, $8$ is the maximum possible value of $|a_2|$.

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  • $\begingroup$ Nice answer! but How to find the $a_{2}=4g(1/2)-g(1)-3g(0)$ this indentity? $\endgroup$ – communnites Dec 23 '16 at 11:51
  • $\begingroup$ @communnites we have $g(0) = a_0$, $g(1/2) = a_0+a_2/2+a_4/4$, and $g(1) = a_0+a_2+a_4$, so we have a system of three linear equations in three unknowns ($a_0$, $a_2$, $a_4$). You can then solve the system using your favorite method. $\endgroup$ – Joey Zou Dec 23 '16 at 11:54
  • $\begingroup$ Nice!Thanks.I have understand.+1 $\endgroup$ – communnites Dec 23 '16 at 11:55
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We have that

$$|f(x)+f(-x)|=2|a_{0}+a_{2}x^2+a_{4}x^4|\le |f(x)|+|f(-x)|\le 2,$$ from where

$$\frac 12|f(x)+f(-x)|\le 1.$$ That is, we have

$$|a_0+a_2x^2+a_4x^4|\le 1.$$ Now, if $p(x)=a_0+a_2x^2+a_4x^4$ satisfies $\|p\|_{\infty}=1$ we have from Remez inequality that $p=\pm T_4$ which gives $|a_2|=8.$

So, assume $0<\|p\|_{\infty}=k<1.$ Then $$p_k(x)=\frac{a_0}k+\frac{a_2}kx^2+\frac{a_4}kx^4$$ satisfies $\|p_k\|_{\infty}=1.$ That is, $p_k=\pm T_4$ from where

$$\frac{a_0}k+\frac{a_2}kx^2+\frac{a_4}kx^4=\pm(1-8x^2+8x^4).$$ In particular $$|a_2|= 8k\le 8.$$

Thus we have shown that $8$ is the maximum value for $|a_2|.$

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