8

Here is a similar problem posted before. I try to use this method to solve this problem.

$$f(-1)=a-b+c, f(0)=c, f(1)=a+b+c$$

So we have $|c|=|f(0)|\le 100$

$$|2a|=|f(-1)-2f(0)+f(1)|\le|f(-1)|+2|f(0)|+|f(1)|\le400$$ So we have $|a|\le200$

But how to find an upper bound for $b$?

Due to the symmetry, I make a guess for the maxima of $|a|+|b|+|c|$ occurring when $b=0$, then we have

$$y=200x^2-100~~~\text{or}~~~y=-200x^2+100$$

But is there a rigorous way to prove it?

MathFail
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4 Answers4

3

$$y=200x^2-100~~~\text{or}~~~y=-200x^2+100$$

are examples where $\vert a \vert + \vert b \vert + \vert c \vert = 300.\quad (*)$

If $\vert a \vert + \vert b \vert \leq 200,$ then $\vert a \vert + \vert b \vert + \vert c \vert \leq 300,\ $ so not an improvement over $(*).$

If $\vert a \vert + \vert b \vert > 200,\ $ then one of $\vert f(1)\vert$ or $\vert f(-1)\vert$ are equal to $ \vert a\vert + \vert b \vert + c\ \geq \vert a\vert + \vert b\vert - \vert c\vert > 100, $ which contradicts the condition "$\vert ax^2+bx+c\vert \le 100$ for all $\vert x\vert \le 1$" in the question.

Adam Rubinson
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  • Thank you, could you please explain the case 5, why $|a|+|b|>200$ implies $|f(\pm1)|>100$ – MathFail Aug 07 '22 at 20:27
  • I have actually explained that in the shortened version I just updated now. Does it make sense? – Adam Rubinson Aug 07 '22 at 20:29
  • $|f(1)|=|a+b+c|\ge|a|+|b|-|c|>100$ or $|f(-1)|=|a-b+c|\ge|a|+|b|-|c|>100$, at least one of them works depends on the sign of $a$ and $b$, right? – MathFail Aug 07 '22 at 20:34
  • that's your case. 5 – MathFail Aug 07 '22 at 20:37
  • Ah. Remember your original question said: If $|ax^2+bx+c|\le100$ for all $|x|\le 1$, What is the maxima for $|a|+|b|+|c|$." For case $5$ (and the last paragraph of my "new short version"), I show that the condition "$|ax^2+bx+c|\le100$ for all $|x|\le 1$" is not satisfied... Either $x=1$ or $x=-1$ violates this condition. – Adam Rubinson Aug 07 '22 at 20:40
  • What I mean is: For case.5 if assume $|a|+|b|>200$, since $|c|\le 100$, then it gives: $$|a|+|b|-|c|>100$$ If $a,b$ have the same sign, then $$|f(1)|=|a+b+c|\ge|a+b|-|c|=|a|+|b|-|c|>100$$ If $a,b$ have different signs, then $$|f(-1)|=|a-b+c|\ge|a-b|-|c|=|a|+|b|-|c|>100$$ So at least one of them gives contradictions. Therefore, $$|a|+|b|\le200$$ – MathFail Aug 07 '22 at 20:45
  • Yes, I think you don't need to make so many cases. Because we have got $|a|\le 200$ and $|c|\le100$. So we don't need to discuss $b$ separately. Just assume $|a|+|b|>200$ and we can reach contradictions from above steps. So we got $|a|+|b|\le 200$ and it is done. – MathFail Aug 07 '22 at 20:53
  • Yes - that's why I did the new version at the top... OK I've deleted the old version and left the new shorter one. Is the proof now OK? Yay! – Adam Rubinson Aug 07 '22 at 20:54
0

Remark: I used the approach in my answer for this question.

We have \begin{align*} |a - b + c| &\le 100, \tag{1}\\ |a + b + c| &\le 100, \tag{2}\\ |c| &\le 100. \tag{3} \end{align*}

Using (1) and (3), we have $$|a - b| \le |a - b + c| + |-c| \le 200. \tag{4}$$

Using (2) and (3), we have $$|a + b| \le |a + b + c| + |-c| \le 200. \tag{5}$$

Using (4) and (5), we have $$|a| + |b| \le 200. \tag{6}$$ (Note: If $ab \ge 0$, then $|a| + |b| = |a + b| \le 200$. If $ab < 0$, then $|a| + |b| = |a - b| \le 200$.)

Using (3) and (6), we have $$|a| + |b| + |c| \le 300.$$

On the other hand, when $a = 200, b = 0, c = -100$ (so $|a| + |b| + |c| = 300$), we have, for all $|x| \le 1$, $$|200x^2 - 100|\le \max(|200\cdot 1^2 - 100|, |200\cdot (-1)^2 - 100|, |200\cdot 0^2 - 100|) = 100.$$

Thus, the maximum of $|a| + |b| + |c|$ is $300$.

River Li
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  • How did you provide a proof for your inequalities must equivalent to the statement "$|ax^2+bx+c|≤100$ holds $\forall |x|≤1$"? – User Aug 08 '22 at 07:21
  • @User We only need (1)(2)(3) which implies $|a| + |b| + |c| \le 300$. – River Li Aug 08 '22 at 07:24
  • I cann't see how the equivalence of the result of these inequalities and the statement $|ax^2+bx+c|≤100,\forall |x|≤1$ is proven – User Aug 08 '22 at 07:28
  • @User We don't need the equivalence. We have: |ax^2+bx+c|,\forall |x|≤1 =====> (1)(2)(3) =====> $|a| + |b| + |c| \le 300$. On the other hand, $|a| + |b| + |c| = 300$ is attainable e.g. when $a = 200, b = 0, c = -100$. – River Li Aug 08 '22 at 07:31
  • Don't your steps need to include $\iff$ for the exact existence of $a,b,c$ values ​​that provide the result $|a|+|b|+|c|=300$ we found here? Because for example $...\implies x^2≥-3$ but $x^2=-3$ does not happen. – User Aug 08 '22 at 07:36
  • @User I don't get your point. First part, we prove that for all $a, b, c$ such that $|ax^2 + bx + c| \le 100$ on $[-1, 1]$, it must satisfy $|a| + |b| + |c| \le 300$. Second part, we give an example in which $|a| + |b| + |c| = 300$ and $|ax^2 + bx + c| \le 100$ on $[-1, 1]$. – River Li Aug 08 '22 at 07:40
  • What I mean, if we found $|a|+|b|+|c|≤300$, from some inequalities ,this doesnt prove alone $|a|+|b|+|c|=300$ is possible. Therofore I thought, after showing $|a|+|b|+|c|≤300$ , we need one exact example or $\iff$ between our steps with the original statement of problem. – User Aug 08 '22 at 07:59
  • @User I gave such an example from the first version of my answer, see the last 2nd paragraph. That is: On the other hand, when a=200,b=0,c=−100, it holds that |ax^2+bx+c|=|200x^2−100|≤⋯=100. – River Li Aug 08 '22 at 09:04
-1

Let $|ax^2+bx+c|\le100$ for all $|x|\le 1$
Firstly if $x=0, |c|\le 100(1)$
And if $x = \dfrac{1}{2}, then \left| \dfrac{a}{4}+\dfrac{b}{2}+c \right| < 100\Longrightarrow| a + 2b + 4c | < 400$
$| a | = | 2a + 2b + 2c - (a + 2b + 4c) + 2c | < 2 | a + b + c | + | a + 2b + 4c | + 2 | c | < 200+ 400 + 200< 800$ (if $x=1$ then $|a+b+c|\le 100$)\ then $|a|\le 800(2)$
$ | b | = | a + 2b + 4c - (a + b + c) - 3c | < | a + 2b + 4c | + | a + b + c | + 3 | c | < 400 + 100 + 300< 800$\ then $| b | < 800 (3)$
Finally $| a | + | b | + | c | \le 100+800+800\le 1700$ i.e $| a | + | b | + | c | \le1700$

MosikasikaC19
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-2

Since $300$ is the optimum value for the linear programming problems $$ \eqalign{\text{maximize}\ & a + b - c \cr \text{subject to}\ & a + b - c \le 100 \cr & a + b - c \ge -100\cr & a - b - c \le 100\cr & a - b - c \ge -100\cr & -c \le 100\cr & -c \ge -100\cr}$$ and the same with objective $a - b - c$, you certainly can't do better than that in your problem with $a \ge 0$ and $c \le 0$.
By symmetry the same holds for $a \le 0$ and $c \ge 0$. Now consider the other cases.

Robert Israel
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