# Polynomial with Certain Conditions

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 $$b \in [0.1,0.9]$$

• 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. Commented Sep 11, 2020 at 17:22
• @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 $a,b,c, \epsilon$. Commented Sep 11, 2020 at 17:24
• "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 ... ? Commented Sep 11, 2020 at 17:27
• @JeanMarie Yes. Commented Sep 11, 2020 at 17:28

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:

$$t_k(x):=\dfrac{1}{T_k(-2)}(x+1)T_k(x-2)$$

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.

• 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! Commented Sep 11, 2020 at 20:04
• 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 ? Commented Sep 12, 2020 at 3:06
• Yes, I'm referring to a specific case, please see the EDIT to the question. Commented Sep 12, 2020 at 14:28
• Good enough for me. Accepted! Commented Sep 12, 2020 at 16:05
• @DUO Thanks. Besides, I propose you to erase the different comments/exchanges we have had that have no interest for future readers. Commented Sep 12, 2020 at 18:31