Is it possible to create a polynomial $p(x)$, in terms of $a,b,c \geq0, \in\mathbb{R}$ and $\epsilon >0, \in \mathbb{R}$, has a fixed degree (aka, a degree $n$ that does not depend on $a,b,c, \epsilon$), $p(0)=a$, and on $[b,c]$, $|p(x)| \leq \epsilon$?

EDIT: I will also accept the specific case of $\epsilon=1$ and $b \in [0.1,0.9]$

  • $\begingroup$ Do you mean that $a,b,c,\epsilon, n$ (where $n$ is the degree) are defined beforehand? In this case, the answer is no in general. $\endgroup$
    – Jean Marie
    Commented Sep 11, 2020 at 17:22
  • $\begingroup$ @JeanMarie Yes, $a,b,c, \epsilon$ are all defined beforehand, but degree $n$ can be anything, as long as it doesn't change depending on $a,b,c, \epsilon$. $\endgroup$
    – DUO Labs
    Commented Sep 11, 2020 at 17:24
  • 1
    $\begingroup$ "can be anything" must be precised: to you mean that you want to know whether for any set of positive numbers $a,b,c,\epsilon$ there exists a degree $n$ such that ... ? $\endgroup$
    – Jean Marie
    Commented Sep 11, 2020 at 17:27
  • $\begingroup$ @JeanMarie Yes. $\endgroup$
    – DUO Labs
    Commented Sep 11, 2020 at 17:28

1 Answer 1


Let $n$ denote the degree of the polynomial.

First case) If the question is with quantifiers in this order $\forall a,b,c,\epsilon, \ \exists n$ etc. (therefore $n$ is function of $a,b,c,\epsilon$) the answer is "yes". Here is why.

For didactic reasons, I will consider first the particular case $a=1,b=1,c=3$.

Consider Chebyshev polynomials $T_k$ (https://en.wikipedia.org/wiki/Chebyshev_polynomials), known to be such that:

$$|T_k(x)| \le 1 \ \ \ \text{for} \ \ -1 \le x \le 1$$

Then define:


Therefore we have $t_k(0)=1$ and

$$\max_{1 \le x \le 3} t_k = \dfrac{3}{T_k(-2)}$$

The result follows because we can take $k$ such that $|T_k(-2)|$ is as large as we want.

In the general case, consider the inverse of transformation $X=\dfrac{c-b}{2}x+\dfrac{b+c}{2}$ that maps line segment $x \in [-1,1]$ onto $X \in [b,c]$.

$ t_k(x):=\underbrace{\dfrac{a (c+1)}{T_k(\tfrac{b+c}{b-c})}}_{A} * \dfrac{(x+1)}{(c+1)} * T_k\left(\dfrac{2x-(b+c)}{c-b}\right)\tag{1}$

(please note that the $(c+1)$ expressions can be cancelled. Now let us examine (1):

  • The two conditions are fulfilled.

  • As the middle expression is bounded by $1$, we just have to make sure that expression (A) can be made arbitrarily small. This will be done by playing on degree $k$: indeed Chebyshev polynomial $T_k(x)$ for $x$ outside $[-1,1]$ can be written:

$$T_k(x)=\cosh(k \ \text{arccosh}(x))\tag{2}$$

therefore can be made arbitrarily large for any $x$ outside $[-1,1]$ (which is the case here for $x=-\tfrac{b+c}{c-b}<-1$).

Second case) If, on the contrary (it was not the way I had understood the question) the question is with quantifiers in this order:

$"\exists n$ (such that) $\forall a,b,c,\epsilon$, one can find a polynomial verifying the two conditions", the answer is "no".

Edit: Here is why. Let us fix for example the degree $n$ to $4$. Do we agree that your issue is equivalent to disprove the fact that if a polynomial $p(x)$ is such that :

$$\max_{x \in [-1,1]} p(x) \in [-1,1] \ \text{and prove that} \ p(-2) \ \text{can't be arbitrarily large} ?\tag{3}$$

(it will be easier to make a reasoning on interval $[-1,1]$ and arbitrary value $x_0=-2$ instead of $[0.1,0.9]$ and $x_0=0$ in order to avoid transformations of polynomials $T_n$).

Let us expand $p(x)$ on the basis of polynomials $T_n(x)$ (for $n\leq 4$):

$$p(x)=\sum_{k=0}^4 a_kT_k(x)\tag{4}$$ Due to condition (3), and the fact that $\max_{x \in [-1,1]}|T_k(x)|=1$:

$$\sum |a_k| \leq 5 \tag{5}$$

Using now relationship (2):

$$p(-2)=\sum_{k=0}^4 a_k \cosh(k \ \text{arccosh}(-2))$$

which is a bounded quantity due to (5).

Therefore $p(-2)$ cannot achieve arbitrary large values.

  • 1
    $\begingroup$ Also, after testing it, I do say it works, so even though it's not exactly what I wanted, it's good enough, so I'll wait a bit before accepting. +1 from me! $\endgroup$
    – DUO Labs
    Commented Sep 11, 2020 at 20:04
  • $\begingroup$ If we can achieve $|f(x)| \le \epsilon$ for any positive $\epsilon$, $\epsilon = 1$ is just a particular case. But maybe you mean something else ? $\endgroup$
    – Jean Marie
    Commented Sep 12, 2020 at 3:06
  • $\begingroup$ Yes, I'm referring to a specific case, please see the EDIT to the question. $\endgroup$
    – DUO Labs
    Commented Sep 12, 2020 at 14:28
  • $\begingroup$ Good enough for me. Accepted! $\endgroup$
    – DUO Labs
    Commented Sep 12, 2020 at 16:05
  • $\begingroup$ @DUO Thanks. Besides, I propose you to erase the different comments/exchanges we have had that have no interest for future readers. $\endgroup$
    – Jean Marie
    Commented Sep 12, 2020 at 18:31

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