Solving an Elusive Problem: Seeking Inspiration from MHB

[anemone]

Hi MHB,(Wave)

I've come across this delicious problem days ago, and upon trying my very best to solve it for multiple times, I failed each time.

I finally realized this problem is beyond my ability to do it. But I've not given up hope yet, so I want to give it one last try, and MHB is the best place to get for inspiration for solving hard problem such as this one. Hence, I would appreciate it if anyone could take a look at this problem and give me some idea.

Thanks in advance for your help!

Problem:

Find all $a,\,b,\,c\in R$ such that $|ax^2+bx+c|\le1$ for every $-0.5\le x\le 0.5$ and $a^2+b^2+c^2$ is maximum.

 

[kaliprasad]

Hi MHB,(Wave)

I've come across this delicious problem days ago, and upon trying my very best to solve it for multiple times, I failed each time.

I finally realized this problem is beyond my ability to do it. But I've not given up hope yet, so I want to give it one last try, and MHB is the best place to get for inspiration for solving hard problem such as this one. Hence, I would appreciate it if anyone could take a look at this problem and give me some idea.

Thanks in advance for your help!

Problem:

Find all $a,\,b,\,c\in R$ such that $|ax^2+bx+c|\le1$ for every $-0.5\le x\le 0.5$ and $a^2+b^2+c^2$ is maximum.

Just a wild guess

try $x=\frac{1}{2}\sin\left({t}\right)$

 

[I like Serena]

Hi anemone! ;)

Note that for each solution, we have another solution by taking all values negative.

We can rewrite $y=ax^2+bx+c$ as:
$$y=a\left(x+\frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)$$
This is a parabola with its top at:
$$(x,y)=\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$$

The first case of boundary conditions that we have, is:
$$\begin{cases}
\frac 1 4 a + \frac 1 2 b + c = 1 \\
\frac 1 4 a - \frac 1 2 b + c = 1 \\
\left|{c-\frac{b^2}{4a}}\right| \le 1 &\text{ if } \left|{-\frac{b}{2a}}\right| \le \frac 1 2
\end{cases}\Rightarrow$$
\begin{cases}
b = 0 \\
a = 4-4c \\
|c| \le 1 &\qquad&\text{ if } |0| \le \frac 1 2
\end{cases}
It follows that:
$$a^2+b^2+c^2 = (4-4c)^2 + 0^2 + c^2 = 17c^2-32c+1$$
Maximum value is if $c=-1$, meaning we have the potential solution:
$$[1]\quad a=8,\ b=0,\ c=-1,\quad a^2+b^2+c^2=65$$

The opposite boundary case yields the potential solution
$$[2]\quad a=-8,\ b=0,\ c=1,\quad a^2+b^2+c^2=65$$The other boundary condition (with its opposite) is:
\begin{cases}
\frac 1 4 a + \frac 1 2 b + c = 1 \\
\frac 1 4 a - \frac 1 2 b + c = -1 \\
\left|{c-\frac{b^2}{4a}}\right| \le 1 &\text{ if } \left|{-\frac{b}{2a}}\right| \le \frac 1 2
\end{cases}

In this case we have to consider sub cases whether the top is in range or not.

 

[anemone]

Thanks for your reply, I like Serena!;)

The other boundary condition (with its opposite) is:
\begin{cases}
\frac 1 4 a + \frac 1 2 b + c = 1 \\

I also believe the sum should be -1. Right, I like Serena?

In this case we have to consider sub cases whether the top is in range or not.

I followed (only after thinking hard on it for quite some time) but I really don't get it about this last remark...what sub cases do we need to consider whether the maximum value for $a^2+b^2+c^2=65$ is within the range or not? Sorry for being a bit blur...:(

 

[I like Serena]

I also believe the sum should be -1. Right, I like Serena?

Correct.
That is the "opposite" case where $a,b,c$ take the opposite (negative) values.
The other sum would then also be opposite.

The inequality $|ax^2+bx+c| \le 1$ combined with $-\frac 1 2 \le x \le \frac 1 2$ gives us 2 pairs of boundary conditions.
Within each pair the boundary conditions are mutually exclusive.
We can combine them in $2 \times 2 = 4$ ways.

However, each of those combinations has an equivalent opposite combination that corresponds to $a,b,c$ having opposite (negative) values.
As a consequence we only have to check 2 of the combinations.

I followed (only after thinking hard on it for quite some time) but I really don't get it about this last remark...what sub cases do we need to consider whether the maximum value for $a^2+b^2+c^2=65$ is within the range or not? Sorry for being a bit blur...:(

This is not about the case where $a^2+b^2+c^2=65$.
It's about the second case that I did not write out.

In the first case the top of the parabola always has an x-value between $-\frac 1 2$ and $\frac 1 2$.
In the second case there are 2 possibilities.
Either the top of the parabola is between $-\frac 1 2$ and $\frac 1 2$, giving us another boundary condition.
Or it is not, in which case we can ignore where the top is.

 

[anemone]

This is not about the case where $a^2+b^2+c^2=65$.
It's about the second case that I did not write out.

In the first case the top of the parabola always has an x-value between $-\frac 1 2$ and $\frac 1 2$.
In the second case there are 2 possibilities.
Either the top of the parabola is between $-\frac 1 2$ and $\frac 1 2$, giving us another boundary condition.
Or it is not, in which case we can ignore where the top is.

Thanks for your further reply...I can understand the first and second case, because I can picture them in a graph
View attachment 2690

...but I couldn't do picture anything useful in the other boundary condition (where x isn't between $\pm 0.5$)...hmm, isn't it the axis of symmetry of any parabola will divide the parabola into two congruent halves, in this case, the given condition where $|ax^2+bx+c|\le 1$ has strike out any other possibilities other than the first two that we needed?(Sweating)

 

[I like Serena]

Thanks for your further reply...I can understand the first and second case, because I can picture them in a graph
View attachment 2690

Good! (Happy)

...but I couldn't do picture anything useful in the other boundary condition (where x isn't between $\pm 0.5$)...hmm, isn't it the axis of symmetry of any parabola will divide the parabola into two congruent halves, in this case, the given condition where $|ax^2+bx+c|\le 1$ has strike out any other possibilities other than the first two that we needed?(Sweating)

The graphs of $ax^2+bx+c$ with $a \ge 0$, corresponding with the boundary cases, are the following:

View attachment 2691The first set of cases correspond to the parabola around zero, which is the one you already have.

The second set of cases has a subcase where the top of the parabola is left of $-\frac 1 2$.
And another subcase where the top of the parabola is at $x=\frac 1 2$.

Each of these boundary cases has an opposite with the graph flipped upside down.
There are 6 boundary cases in total.

 

[anemone]

Good! (Happy)
The graphs of $ax^2+bx+c$ with $a \ge 0$, corresponding with the boundary cases, are the following:

View attachment 2691The first set of cases correspond to the parabola around zero, which is the one you already have.

The second set of cases has a subcase where the top of the parabola is left of $-\frac 1 2$.
And another subcase where the top of the parabola is at $x=\frac 1 2$.

Each of these boundary cases has an opposite with the graph flipped upside down.
There are 6 boundary cases in total.

Thanks, I like Serena! I don't think I got your point.(Tmi) Please look at my weak argument below:

The one of the two other cases that we need to consider is like follows:

The case where the curve is at -1 when $x=-0.5$ and at 1 when $x=0.5$.

This gives $\dfrac{1}{4}a+\dfrac{1}{2}b+c=1$ $\dfrac{1}{4}a-\dfrac{1}{2}b+c=-1$ which then yields $b=2$ and $a=-4c$ (i.e. if $a>0$, $c<0$ or otherwise.)

Since $\left|a\left(x+\dfrac{1}{2a} \right)^2+\left(c-\dfrac{1}{a^2} \right)\right|\le 1$ and for the case where $a>0$, $c<0$, i.e. $\left|\left(c-\dfrac{1}{a^2} \right)\right|\le 1$ we know that $a^2\ge 1$ (or $a>1$), and hence $c\le -\dfrac{1}{4}$. This means $c$ is maximum when $c=-\dfrac{1}{4}$, $a=1$, $b=2$. $\therefore a^2+b^2+c^2=5\dfrac{1}{16}$ But this doesn't sound so right, because I can tell the quadratic equation $\left|3x^2+2x-\dfrac{3}{4}\right|\le 1$ does satisfy the above condition and for this case, the maximum of $a^2+b^2+c^2=13.5625$. Something is wrong with my understanding and hence my working and assumption are all wrong with the checking this other case.

 

[I like Serena]

Thanks, I like Serena! I don't think I got your point.(Tmi) Please look at my weak argument below:

The one of the two other cases that we need to consider is like follows:

The case where the curve is at -1 when $x=-0.5$ and at 1 when $x=0.5$.This gives
$\dfrac{1}{4}a+\dfrac{1}{2}b+c=1$
$\dfrac{1}{4}a-\dfrac{1}{2}b+c=-1$ which then yields $b=2$ and $a=-4c$ (i.e. if $a>0$, $c<0$ or otherwise.)

Agreed!

Let's stick with $a>0, b=2, a=-4c$ for now.

Since $\left|a\left(x+\dfrac{1}{2a} \right)^2+\left(c-\dfrac{1}{a^2} \right)\right|\le 1$ and for the case where $a>0$, $c<0$, i.e. $\left|\left(c-\dfrac{1}{a^2} \right)\right|\le 1$ we know that $a^2\ge 1$ (or $a>1$), and hence $c\le -\dfrac{1}{4}$. This means $c$ is maximum when $c=-\dfrac{1}{4}$, $a=1$, $b=2$.

I'm afraid I've lost you here.
Either way, I believe it should be:
$$\left|a\left(x+\dfrac{1}{a} \right)^2+\left(c-\dfrac{1}{a^2} \right)\right|\le 1$$

The point I was trying to make, is that the top of the parabola is at $x=-\frac 1 a$.

And I've just realized that I have not properly distinguished the cases yet.
We've made the assumption that the function is at +1 respectively -1 at the interval boundaries. That means the top cannot be inside the interval anymore.
So we have the extra condition that:
$$|x_{top}| = \left|-\frac 1 a\right| \ge \frac 1 2$$
$$|a| \le 2$$
Since the top is outside the interval, it does not matter where it is exactly.Now we still need to set up new cases where the top defines one of the boundaries.
That means that with $a>0$, the top needs to be at $-1$, and one of the interval boundaries needs to be at $+1$ while the other needs to be between $-1$ and $+1$.

 

[anemone]

Agreed!

Let's stick with $a>0, b=2, a=-4c$ for now.I'm afraid I've lost you here.
Either way, I believe it should be:
$$\left|a\left(x+\dfrac{1}{a} \right)^2+\left(c-\dfrac{1}{a^2} \right)\right|\le 1$$

Yes, you're right. It was a typo.

The point I was trying to make, is that the top of the parabola is at $x=-\frac 1 a$.

And I've just realized that I have not properly distinguished the cases yet.
We've made the assumption that the function is at +1 respectively -1 at the interval boundaries. That means the top cannot be inside the interval anymore.
So we have the extra condition that:
$$|x_{top}| = \left|-\frac 1 a\right| \ge \frac 1 2$$
$$|a| \le 2$$
Since the top is outside the interval, it does not matter where it is exactly.

I see. That means we can eliminate this case, as this case doesn't satisfy the given condition where $|ax^2+bx+c|\le 1$.

Now we still need to set up new cases where the top defines one of the boundaries.
That means that with $a>0$, the top needs to be at $-1$, and one of the interval boundaries needs to be at $+1$ while the other needs to be between $-1$ and $+1$.

Hmm...If $a>0$, and it has the minimum value of $-1$ and let's say $x=0.5$ gives $y=1$, and due to the axis of symmetry of the quadratic function, I see that when $x=-0.5$, $y$ has to take the value of $1$, isn't it?