# Find limits of value/derivatives defining a polynomial at 2 points to bound it in between

The following properties define the polynomial $p(x)$ uniquely:

$\text{deg}(p(x))=7\\p(-1)=y_1,\ p'(-1)=d_{1,1},\ p''(-1)=d_{2,1},\ p'''(-1)=d_{3,1}\\ p(1)=y_2,\ \ \ \ p'(1)=d_{1,2},\ \ \ \ p''(1)=d_{2,2},\ \ \ \ p'''(1)=d_{3,2}$

The parameters $y_1,\ d_{1,1},\ d_{2,1},\ d_{3,1},\ y_2,\ d_{1,2},\ d_{2,2},\ d_{3,2}$ are real and chosen in such a way that the implication $-1\leq x\leq 1\Rightarrow-1\leq p(x)\leq1$ is true.

For each parameter, I need to find all real numbers that, when replacing the parameter, make the implication $-1\leq x\leq 1\Rightarrow-1\leq p(x)\leq1$ true.

There is always one solution, because, for instance, if I replace $y_1$ with a real number $r$, one solution is $r=y_1$. Moreover, the complete solution is an interval, $a\leq r\leq b$, because all points with $x\in[-1,1]$ will move the same way, when a parameter is modified monotonously.

How should this be done numerically? (I don't think there is an general exact solution)

I am not sure I understand the problem. The polynomial contains 8 coefficients and you give 8 conditions; so we have 8 linear equations for 8 unknowns. The system seems to have a unique solution; using a CAS, I have been able to generate the expressions for each coefficient as a function of the imposed conditions. These expressions are not complex at all; they are just linear combinations of the values of the imposed conditions.

Be sure I apologize not to put them here; I am almost blind and typing the formulas would be very hard for me. Please, tell me if I can do something else for helping you.

Happy New Year

• I'm not asking for p(x) explicitly. $y_1,\ d_{1,1},\ d_{2,1},\ d_{3,1},\ y_2,\ d_{1,2},\ d_{2,2},\ d_{3,2}$ are initial conditions defining p(x). Changing one of these, while not changing the others, will modify p(x). I ask for the interval of real numbers for each of the values/derivatives that will make p(x) satisfy the implication, assuming this specific value/derivative concerned to be the only one that is different from the initial conditions. Dec 31, 2013 at 13:49
• What I was trying to mean is that, for a given set of the conditions, the polynomial is unique. If you change the value of any condition, this will alter the polynomial. The next question is hard to me and I do not feel able to give any appropriate answer. Sorry for that and good luck. Dec 31, 2013 at 13:56

You could start with the convex combinations of the Chebyshev polynomials $T_n(x)=\cos(n\arccos(x))$. More general than the constants functions and also than just the convex combinations of the monomials.

More precisely, the convex combinations of $T_n$ and $-T_n$, so after collecting the coefficients for positive and negative versions,

$$\sum_{n=0}^7c_kT_n(x)\text{ with }\sum_{k=0}^7|c_k|\le 1.$$