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Suppose we where given the following values of $P(x)$ : $P(1) = 2, P'(1) = 3, P(2) = 1, P'(2) = -1, \text{ and } P''(2) = 1.$

How can we find a polynomial of degree less than 5 that satisfies these conditions using generalized Lagrange Interpolation?

I understand how regular Lagrange Interpolation works and I have tried to derive a method in a similar fashion but to no avail.

My current idea is to use the first few terms of the Taylor series: $$P(x) = A(P(1) + P'(1)(x-1))+B(P(2) + P'(2)(x-2) + \frac{P''(2)(x-2)^2}{2!})$$ but I am struggling to determine $A$ and $B$.

Conrad Crowley
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1 Answers1

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I have solved this problem myself.

In the general case: Suppose we are given the values of $n$ derivatives at a given point $x=a$ and $m$ derivatives at a point $x=b$. That is, $$P(a) = k_1,P^{'}(a)=k_2, \dots, P^{(i)}(a)=k_i,\dots, P^{(n-1)}(a) = k_n$$ $$P(b) = l_1,P^{'}(b)=l_2, \dots, P^{(i)}(b)=l_i,\dots, P^{(n-1)}(b) = l_m$$

We have the Interpolation formula: $$P(x) = \frac{(x-a)^n}{(b-a)^n}\left( l_1 + l_2(x-a) + l_3 \frac{(x-a)^2}{2!} + \dots \right) + \frac{(x-c)^m}{(a-b)^m}\left( k_1 + k_2(x-a) + k_3 \frac{(x-a)^2}{2!} + \dots \right)$$

Where the terms inside the parentheses are the first $n$ and $m$ terms of the Taylor series respectively.

Conrad Crowley
  • 387
  • I noticed this answer is wrong beyond some apparent typos (e.g., $c$ in the second term). Perhaps a similar approach gives the correct answer, but it needs some more computations or clever ideas. – Hhan May 03 '23 at 04:26
  • @hhan can you elaborate further here? I don't see where this is failing – Conrad Crowley May 04 '23 at 20:17
  • First, I believe the formula should be of the form like $P(x)=(x-a)^n/(b-a)^n(l_1 + l_2 (x-b) + \cdots ) + (x-b)^m /(a-b)^m (k_1 + k_2 (x-a) + \cdots)$. Let us compute $P'(b)$ in your formula. Note that $P'(x)=(x-a)^n/(b-a)^n (l_2 + (x-b)\cdots) + n(x-a)^{n-1}/(b-a)^n (l_1 + (x-b)\cdots) + (x-b)^{m-1}\cdots$. Thus $P'(b)=l_2 + n l_1 /(b-a)$, which differs from what we want. I found a different method here: https://en.m.wikipedia.org/wiki/Hermite_interpolation. – Hhan May 09 '23 at 11:25