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Minimum value of $M$ such that $\exists a, b, c \in \mathbb{R}$ and $$ \left|4 x^{3}+a x^{2}+b x+c\right| \leq M ,\quad \forall|x| \leq 1 $$

What i considered was that putting x= 0, 1 and -1 we get c should be between M and -M and similarily $|\pm4 + a \pm b +c| \leq M$ , from this i was not able to conclude anything about M , Or is it possible to solve it using the derivative to be 0 at minima ?

Update : i made a graph so that to try doing from that . I can make a observation that minimum will be when the M ,-M line is touching the both maxima , minima occuring there . (Not sure though if its correct ) does it help ? enter image description here

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2 Answers 2

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Let $P[x]$ be the set of all cubic polynomials with leading coefficient $4$, and let $P_M[x]$ be the subset of $P[x]$ consisting of polynomials $p(x)$ satisfying the required property that $$-M\leq p(x) \leq +M,$$ for all $x\in [-1,+1]$.

Clearly, if $p(x) \in P_M[x]$, then also $q(x)={p(x)-p(-x)\over 2} \in P_M[x]$. This implies that if $P_M[x]$ is nonempty, then so is $Q_M[x] \subset P_M[x]$ consisting only of odd functions. Hence we can restrict our attention to polynomials of the form $f_b(x)=4x^3+bx$ over the range $x\in[0,1]$.

In other words, the problem to solve becomes to compute

$$M^* = \min_b \max_{x\in[0,1]}\left|f_b(x)\right|$$

where $f_b(x)=4x^3+bx$.

Note $f'_b(x) = 12x^2+b$. If $b\geq0$, then $f_b(x)$ is nondecreasing over the range $[0,1]$. While if $b<0$, then $f_b(x)$ has a single local extremum in the range $(0, \infty)$ at $x=\sqrt{{-b \over 12}}$. This local extremum is a minimum that falls in the range $[0,1]$ iff $-12\leq b\leq 0$.

For $b\not \in (-12,0)$ and $x\in[0,1]$, then, $\left|f_b(x)\right|$ is maximized at one of the endpoints of the interval $[0,1]$, and so we have:

$$\begin{eqnarray} \min_{b\not\in(-12,0)} \max_{x\in[0,1]}\left|f_b(x)\right| &=& \min_{b\not\in(-12,0)} \left|f_b(1)\right| \\ &=& \min_{b\not\in(-12,0)} \left|4+b\right| \\ &=& 4 \end{eqnarray}$$

For $b\in[-12,0]$, on the other hand, $\left|f_b(x)\right|$ is maximized at either $x=\sqrt{{-b \over 12}}$ or $x=1$, so

$$\begin{eqnarray} \min_{b\in[-12,0]} \max_{x\in[0,1]}\left|f_b(x)\right| &=& \min_{b\in[-12,0]} \max_{x\in\bigl\{\sqrt{{-b \over 12}}, 1\bigr\}}\left|f_b(x)\right| \\ &=& \min_{b\in[-12,0]} \max\left(\left|\sqrt{{-b^3 \over 27}}\right|, \left|4+b\right|\right) \end{eqnarray}$$

The $\max$ is minimized at $b=-3$; this can be seen by solving the equation

$$\left|\sqrt{{-b^3 \over 27}}\right| = \left|4+b\right|,$$

which yields a cubic equation when squaring both sides, and by factoring that cubic.

The final answer then is $M^*=1$, attained by the cubic polynomial $4x^3-3x$.

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  • $\begingroup$ Sir i understood most part of it , few queries : 1. we restricted the domain to [0,1] because odd functions are symmetric about origin , thats why when we take modulus it will not effect anyhow if we consider just the x lying in [0,1]? 2. We can say for any M we are sure to be having odd functions in the subset , so thats why we restricted right ? 3. why cannot we consider even functions too that will also be surely there we can just restrict ourselves too ax^2 +b too Sir right ? And find min a,b of max |fa,b (x)| right?We just ignored it because its two variable minimization,compared to one? $\endgroup$
    – Orion_Pax
    Apr 19 at 15:52
  • $\begingroup$ Odd functions are symmetric about the origin, so the max of their absolute value over $[-a, +a]$ is equal to the max of their absolute value over $[0, +a]$. $\endgroup$
    – dshin
    Apr 19 at 15:59
  • $\begingroup$ In the beginning of my solution, I proved that if $P_M[x]$ is nonempty, then $P_M[x]$ contains at least one odd function. So the minimum $M$ we seek can be attained by some odd function. This is why the restriction to odd functions is justified. $\endgroup$
    – dshin
    Apr 19 at 16:04
  • $\begingroup$ Thanks a lot understood Sir , finally can this method of odd function/even function restriction can be done in case of this too : math.stackexchange.com/q/288152/922054 , as such |ax^2 + bx + c| <= 3/2 means this set contains both odd and even functions so we can focus either on odd or even whichever better for convenience to solve Sir ? Or it will not work $\endgroup$
    – Orion_Pax
    Apr 19 at 16:15
  • $\begingroup$ @Orion_Pax I constructed a function $q(x)$ that is an element of $P_M[x]$. Take a look at that function. Is it clear to you that if you changed a minus sign to a plus sign, or if you changed the $2$ to a different value, that the resultant $q(x)$ does not work? $\endgroup$
    – dshin
    Apr 19 at 16:22
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Remarks: It is related to the Chebyshev polynomials of the first kind. I used it before e.g. in Minimizing the maximum of $|x^2 - x - k|$


The condition is written as $$|x^3 + (a/4)x^2 + (b/4)x + c/4| \le M/4, \quad \forall -1 \le x \le 1.$$

Recall the well-known result:

For any given $n \ge 1$, among the polynomials of degree $n$ with leading coefficient $1$, $$f(x) = \frac{1}{2^{n - 1}}T_n(x)$$ is the one of which the maximal absolute value on the interval $[-1, 1]$ is minimal, where $T_n(x)$ is the Chebyshev polynomials of the first kind. This maximal absolute is $\frac{1}{2^{n - 1}}$, and $|f(x)|$ reaches this maximum exactly $n + 1$ times at $$x = \cos \frac{k\pi}{n}, \quad 0 \le k \le n.$$ See: see "Minimal $\infty$-norm" https://en.wikipedia.org/wiki/Chebyshev_polynomials, or https://handwiki.org/wiki/Chebyshev_polynomials

Here, $n = 3$, $T_3(x) = 4x^3 - 3x$, and $\frac{M^\ast}{4} = \frac{1}{2^2}$ i.e. $M^\ast = 1$.

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  • $\begingroup$ Sir that well known result high school level proof is there anywhere? $\endgroup$
    – Orion_Pax
    Apr 20 at 4:00
  • $\begingroup$ @Orion_Pax Perhaps Chebyshev polynomials are unkind to high school students. A proof is given in math.stackexchange.com/questions/259428/… $\endgroup$
    – River Li
    Apr 20 at 6:11
  • $\begingroup$ Understood sir thanks $\endgroup$
    – Orion_Pax
    Apr 20 at 8:27

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