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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

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  • $\begingroup$ Hermite interpolation requires vastly more data than you have provided. Missing: $p'(0)$, $p'(1)$, and $p(2)$. Why do you think you can apply that method? (Original version of this comment included unnecessary second derivatives.) $\endgroup$ – Eric Towers Apr 17 '20 at 22:28
  • $\begingroup$ Please use Mathjax to format math. $\endgroup$ – Micah Windsor Apr 17 '20 at 22:33
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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 $.

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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 $p$

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  • $\begingroup$ Thank you for this answer! $\endgroup$ – Fergusson Apr 17 '20 at 23:22
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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.

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  • $\begingroup$ Thank you! I forgot to add the minimization of degree constraint in the question. Do you know some way to solve my problem using divided differences? Oddly, the excercise I am working on wants me to use that method but I can't figure out how to do it $\endgroup$ – Fergusson Apr 17 '20 at 23:21
  • $\begingroup$ Also, please take a look at this: coast.nd.edu/jjwteach/www/www/30125/pdfnotes/lecture5_9v14.pdf relevant quote: "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. I guess it was wrong of me to call it hermite interpolation. More like a generalization of it maybe? $\endgroup$ – Fergusson Apr 17 '20 at 23:29

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