Quadratic Coefficients Inequality

Let $\alpha = p(0)$ , $\beta = p(\frac{1}{2})$ , and $\gamma = p(1)$ . We know that $|\alpha|$ , $|\beta|$ , and $|\gamma|$ are all less than or equal to 1. We will use this information by first writing a, b, and c in terms of $\alpha$ , $\beta$ , and $\gamma$ . By substituting the definition of p(x), we have: $\begin{aligned} c &=& \alpha \\ a+2b+4c &=& 4\beta \\ a+b+c &=& \gamma \end{aligned}$ We then substitute $\alpha$ for c, obtaining: $\begin{aligned} a+2b &=& 4\beta-4\alpha \\ a+b &=& \gamma-\alpha \end{aligned}$ Subtracting these equations, we obtain: $b = -3\alpha+4\beta-\gamma$ We can then solve for a in the following manner: $\begin{aligned} a+b+c &=& \gamma \\ a+(-3\alpha+4\beta-\gamma)+\alpha &=& \gamma \\ a &=& 2\alpha-4\beta+2\gamma \end{aligned}$ In summary, we now have the following relationships: $\begin{aligned} a &=& 2\alpha-4\beta+2\gamma \\ b &=& -3\alpha+4\beta-\gamma \\ c &=& \alpha \end{aligned}$ It is clear from these relations, and from the fact that $|\alpha|$ , $|\beta|$ , and $|\gamma|$ are all less than or equal to 1, that $|a| \le 8$ , $|b| \le 8$ , and $|c| \le 1$ . Furthermore, all those maximum values can be obtained by letting $|\alpha|$ , $|\beta|$ , and $|\gamma|$ equal 1, and by letting the sign of $\beta$ be the opposite of the signs of $\alpha$ and $\gamma$ . These stipulations result in either $p(x) = 8x^2-8x+1$ or $p(x) = -8x^2+8x-1$ . In either case, $|a| + |b| + |c| = \textbf{17}$ .

It is easy to verify that the following symmetries hold:

1. If $p(x)=ax^2+bx+c$ satisfies the given conditions and maximizes $|a| + |b| + |c|$ , then so is $-p(x)=-ax^2-bx-c$ .

2. If $p(x)=ax^2+bx+c$ satisfies the given conditions and maximizes $|a| + |b| + |c|$ , then so is $p(1-x)=a(1-x)^2+b(1-x)+c=ax^2-(2a+b)x+a+b+c$ .

Because of 1. we can assume that $a \ge 0$ . That means $p(x)$ is a parabola opening upwards.

Because of 2. we can choose whether to have the vertex of the parabola to the left or to the right of $x=0.5$ (or exactly at that point of course). We will choose the vertex $(x_0, y_0)$ to be to the right of $x=0.5$ , i.e. $x_0=-\frac{b}{2a}\ge 0.5$ , which also means that $b < 0$ .

With these assumptions: $p(x) \le p(0) = c$ for all $0\le x \le 1 (\le 2x_0)$ . That means, if we have a solution satisfying the conditions of the problem, we can always shift the parabola upwards until we reach $p(0)=c=1$ without violating the conditions and increasing $|a|+|b|+|c|$ . That means, when we have values $a, b$ and $c$ that maximize $|a|+|b|+|c|$ , $c$ must be equal to 1.

We have now 2 cases:

a. The vertex of the parabola is $x_0=-\frac{b}{2a} \ge 1$ , or $-b \ge 2a$ . In this case the parabola is monotonically decreasing for $x$ between 0 and 1. As long as $p(1) = a + b + c > -1$ we can always decrease $b$ (which is negative, so that $|a|+|b|+|c|$ increases) until we reach $p(1)=-1$ (the condition $-b>2a$ will remain true, since $-b$ is increasing). That means, that for the optimal solution in this case we have: $p(0)=c=1$ , $p(1)=a+b+1=-1$ or $-b=a+2\ge 2a$ which gives $a\le 2$ , and finally $|a|+|b|+|c|=a+a+2+1=2a+3\le 7$ .

b. The vertex of the parabola is $0.5\le x_0=-\frac{b}{2a}< 1$ . In this case, the minimum of the parabola is at $x_0=-\frac{b}{2a}, y_0=c-\frac{b^2}{4a}$ , which is within the limits of the conditions of the problem statement ( $0\le x_0\le 1$ , so that we must also have $-1\le y_0 \le 1$ . Again, we can decrease b (increase its absolute value, which also decreases $y_0$ ), until we reach $y_0=-1$ or we reach $-\frac{b}{2a}=1$ . The second limit is part of case a., so we will check here reaching the first limit: $y_0=c-\frac{b^2}{4a}=-1$ or, with $c=1$ : $b^2=8a$ . The condition $0.5\le x_0<1$ changes now to: $0.5\le -\frac{4}{b} < 1$ , and because $b$ is negative: $-8\le b < -4$ . Finally, we get: $|a|+|b|+|c|=\frac{b^2}{8}-b+1$ which is maximized for $b=-8$ giving the value $|a|+|b|+|c|=17$ .

From the cases a. and b. we see that the best solution is 17, and can be reached for $b=-8$ , $a=\frac{b^2}{8}=8$ , c=1, with: $p(x)=8x^2-8x+1$ . We can verify: the vertex of this parabola is at $x_0=0.5, y_0=-1$ , and $p(0)=1$ , $p(1)=1$ .

Of course, because of the symmetries mentioned in the beginning, also $p(x)=-8x^2+8x-1$ satisfy the conditions and reach the maximum $|a|+|b|+|c|=17$ .

The function $p(x) = 8x^2 - 8x + 1$ has its minimum at $x = \frac{1}{2}$ , with $p(\frac{1}{2}) = -1$ ,and so its maximum on the interval will be at $0$ or $1$ . $p(0) = p(1) = 1$ , so this function, with a sum of 17, satisfies the conditions.

Otherwise:

Note that $k = a + b + c = p(1)$ , $l = \frac{a}{4} + \frac{b}{2} + c = p(\frac{1}{2})$ and $m = c = p(0)$ satisfy $|k|, |l|, |m| \leq 1$ .

By the triangle inequality, $|a| = |2k - 4l + 2m| \leq 2|k| + 4|l| + 2|m| \leq 8$ . Also, $|b| = |4l - k - 3m| \leq 4|l| + |k| + 3|m| \leq 8$ , and $|c| = |m| \leq 1$ .

So $|a| + |b| + |c| \leq 17$ .

At $x=0$ $p(x) = c$ therefore $c = 1$

Need to contain turning point in range so by symmetry turning point at x=0.5

Differentiating gives $2ax +b = 0$ at turning point so at $x =0.5$ $a =-b$

Gives $p(x) = ax^2-ax+1$

$p(0.5)=-0.25a+1$ & $p(0.5) = -1$ to fit in constraint

so $a=8 b=-8 c=1$

Ok, so let's picture this. By completing a square, any quadratic equation can be made into a parabola. We want as large a value for a,b,c, so we want our parabola to be less-flat. Flat parabolas have smaller a,b terms.

Let's pretend we want our parabola to open upwards (a>0). It should hit the points (0,1) (.5 , -1) (1,1). Why you ask? Try to draw the parabola and you'll see that this is the "steepest" parabola we can draw and satisfy the problem's conditions.

ax^2 + bx + c (0,1) means 1 = a(0^2) + b(0) + c; c=1 (1,1) means 1 = a(1^2) + b(1) + c; a + b = 0 (.5,-1) means -1 = a(1/4) + b(1/2) + c; (1/4)a + (1/2)b = -2 If you solve the system for a and b, you'll get that a=8 and b=-8

|8| + |-8| + |1| = 17

But wait, we need to check the case with a downward facing parabola that hits (0,-1) (.5,1) (1,-1). If you do the same process as before, you'll get a = -8, b = 8, c = -1

|-8| + |8| + |-1| = 17

The largest possible value of |a| + |b| + |c| is 17.

From this problem: \left |p(x)=ax^2+bx+c \right |\leq 1), where $0\leq x\leq 1$

Choose $x=0$ , so $\left | c \right |\leq 1$ so, for maximum $\left | a \right |+\left | b \right |+\left | c \right |$ , we choose $c=\left | 1 \right |$

$\mathbf{Case}$ $\mathbf{for:}$ $\mathbf{c=1}$

$p(x)=ax^2+bx+1$ , for $x=0$ we get $p(x)=1$ and this point consider $p(x)$ maximum for this point,

For this case imposible for $a\leq 0$

$\mathbf{Prove:}$

set $b=-a$ and we get:

$ax^2-ax \geq 0$ for $0\leq x\leq 1$

it's contradiction for $p(x)\geq 1$

So, $a$ positive number, and $b$ must negative number (because if $b$ are positive number for $x=1$ , $p(x)>1$ contradiction.)

And this case result $a\geq -b$

Maximum $p(x)=1$ and happened when $x$ boundary point, $p(x)$ have minimum for $p(x)=-1$

we have axis of symetry for $x=\frac {1}{2}$ if $f(1)=1$

Minimum point happened when $x=\frac {1}{2}$ so that $a=-b$

$p(x)=a(\frac {1}{2})^2-a(\frac {1}{2})+1=-1$ and we get $a=-8$ ,

So that , $\left | a \right |+\left | b \right |+\left | c \right |$ maximum if $p(x)=8x^2-8x+1$

$\mathbf{Case}$ $\mathbf{for:}$ $\mathbf{c=-1}$

with same way we get:

$p(x)=-8x^2+8x-1$

so, maximum

$\left | a \right |+\left | b \right |+\left | c \right|=17$

Since $p(0) = c$ , we know $-1 \le c \le 1$ . Since $p(1) = a+b+c$ , we know $-1 \le a+b+c \le 1$ . Since $p \left(\frac{1}{2} \right) = \frac{1}{4}a + \frac{1}{2}b + c$ , we know $-4 \le a + 2b + 4c \le 4$ .

Since $-4 \le a + 2b + 4c \le 4$ and $-1 \le a+b+c \le 1$ , we get $-5 \le a + 2b+4c - (a + b + c) \le 5$ , or $-5 \le b + 3c \le 5$ . Combining this inequality with $-1 \le c \le 1$ yields $-8 \le b \le 8$ . Similarly, since $-4 \le a + 2b + 4c \le 4$ and $-1 \le a+b+c \le 1$ , we get $-6 \le a + 2b + 4c - 2(a+b+c) \le 6$ , which gives us $-6 \le -a + 2c \le 6$ . Combining this with $-1 \le c \le 1$ yields the inequality $-8 \le a \le 8$ . Therefore, $|a| + |b| + |c| \le 8 + 8 + 1 = 17.$

To see that we cannot get any smaller, consider $p(x) = 8x^2 - 8x + 1.$ This polynomial satisfies the condition $|p(x)| \le 1$ for $0 \le x \le 1$ , and $|a| + |b| + |c| = 8 + 8 + 1 = 17$ for this quadratic polynomial. Note also that $p(x) = 8 \left(x - \frac{1}{2} \right)^2 - 1$ , so this is a parabola with vertex at $\left(\frac{1}{2},-1 \right)$ passing through the points $(0,1)$ and $(1,1)$ .