find a polynomial $p$ with $p(0), p(1), p'(2)$ known

Question

I would like to find a polynomial $p$ so that $p(0) = 1, p(1) = 2, p'(2) = -1/2$, using Hermite interpolation, preferably with the divided differences method in the wikipedia page for Hermite interpolation.

(More generally, when we are given some functional values at some points and derivatives of varying order at other points, find the polynomial satisfying those conditions. Surely such missing-information interpolation is possible?)

Thanks in advance!

EDIT: In the slides I am reading it says

It is also possible to set up specialized Hermite interpolation functions which do not include all functional and/or derivative values at all nodes • There may be some missing functional or derivative values at certain nodes • This lowers the degree of the interpolating function

this is what I am looking for, not actual Hermite interpolation, which the kind answerers have explained to me is actually impossible

Answer 1

Applying method of differences to $p(0), p(1)$, we see that the polynomial must be of the form $ p(x) = (x+ 1) + x(x-1) \times A(x)$.

Checking the condition at $p'(2)$, we get $ - \frac{1}{2} = p'(2) = 1 + ( 4 - 1 ) \times A(x) + 2(2-1) \times A'(x) $.

So, we could have $A(x) = - \frac{1}{2}$, which yields $ - \frac{x^2}{2} + \frac{3x}{2} + 1 $.

Answer 2

The information you've provided are too small, $p(0)=1$ meaning the constant term is $1$, $p(1)=2$

$p'(2)=-1/2$, because of this lowest the degree $p$ can take is $2$

$p_2(x)= ax^2+bx+1$

$a+b+1=2$, $a+b=1$

$p'(x) = 2ax^2+b$ $2a4+b=8a+b=-1/2$

You can finish off to write out $p$

Answer 3

You only have half the data you need to apply Hermite interpolation. So that can't work.

Instead, let's find all the polynomials that meet the constraints you give. Let $p(x) = \sum_{i=0}^\infty a_i x^i$. We will restrict $p$ to have finite degree later. \begin{align*} 1 &= p(0) = a_0 + \sum_{i=1}^\infty a_i 0^i = a_0 \text{,} \\ 2 &= p(1) = \sum_{i=0}^\infty a_i 1^i = \sum_{i=0}^\infty a_i \text{, and } \\ -1/2 &= p'(2) = \sum_{i=1}^\infty i a_i 2^{i-1} \text{.} \end{align*}

From the first, we find that the constant term coefficient is $1$. From the second, that the sum of all the coefficients is $2$ (so the sum of the non-constant term coefficients is $1$), and the third tells us $(a_i)_{i \geq 1}$ include negative values and eventually decrease faster than $1/2^i$ to ensure convergence. (Notice that having only finitely many $a_i$ be nonzero meets the "decrease faster" requirement.

Is there a degree $0$ solution? No: $p(0) \neq p(1)$.

Is there a degree $1$ solution? No: the line through $(0,p(0))$ and $(1,p(1))$ has slope $1 \neq -1/2$.

Is there a degree $2$ solution? Yes. The third equation requires the (constant) slope of $p'$ is $(-1/2)/2 = -1/4$, so $$1 + \frac{3}{2}x - \frac{2!}{4}x^2$$ is such a polynomial.

Is there a degree $3$ solution? Yes, infinitely many. For any $a_3 \in \Bbb{R}$ (including $a_3 = 0$, which recovers the degree $2$ solution), $$ 1 + \frac{16 a_3 + 9}{6} x - \frac{22a_3+3}{6} x^2 + a_x x^3 $$ is such a polynomial.

Is there a degree $4$ solution? Yes, infinitely many. We solve the linear system \begin{align*} \left\{ \begin{matrix} a_0 & = & 1 \\ a_0 + a_1 + a_2 + a_3 + a_4 & = & 2 \\ a_1 + 4a_2 + 12a_3 + 32a_4 & = & -1/2 \end{matrix} \right. \end{align*} For any $a_3, a_4 \in \Bbb{R}$, $$ 1 + \frac{56 a_4 + 16 a_3 + 9}{6} x - \frac{62 a_4 + 22 a_3 + 3}{6} x^2 + a_3 x^3 + a_4 x^4 $$ is such a polynomial.

And we can keep going, setting up the same linear system, solving for $a_0$, $a_1$, $a_2$ in terms of the subsequent coefficients for each degree, and getting a succession of infinite families of solutions.

To summarize: if you want a unique solution, either you need to add the constraint to minimize degree, or you need some way to decide which of the endless sea of polynomials which meet the constraints is the one that solves your problem.