# Polynomial interpolation using derivatives at some points

Given $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4), (x_5, y_5)$, we can interpolate a polynomial of degree 4 using Lagrange method.

But, when we are given $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4), (x_5, y'''_5)$, how can we interpolate the same degree-4 polynomial?

Will Birkhoff interpolation be a good method to solve this or some modified version of Lagrange can be used?

First, note that a 4th degree polynomial might not exist, for a simple example, consider finding a quadratic which has points $f(0) = 0, f(1) = 1, f'( \frac{1}{2} ) = 1$. The first 2 conditions naturally imply that $f'( \frac{1}{2}) = 0$, which contradicts the third.

But otherwise, we can use a similar idea to the building blocks of Lagrange Interpolation.

Use $f_1 = B(x-x_2)(x-x_3)(x-x_4)(x-A)$, where $A$ is chosen such that $f''' (x_5) = 0$ and $B$ is chosen such that $f_1 (x_1) = 1$. There are no issues here. Define $f_2, f_3, f_4$ similarly.

Use $f_5 = C (x-x_1)(x-x_2)(x-x_3)(x-x_4)$, where $C$ is chosen such that $f'''(x_5) = 1$. The only issue here, is if $[ (x-x_1)(x-x_2)(x-x_3)(x-x_4)]''' (x_5) = 0$, in which case such a constant does not exist. In this case, we need to tag on another linear term, to make a polynomial of degree 6.

• Well, a 4th degree polynomial does exist, at least in the problem I'm trying to solve. In fact, the functional values themselves are obtained from a 4th degree polynomial. May 11, 2014 at 17:58
• @saura as explained, I've demonstrated where the issue occurs, which is actually only for a small number of cases (3 to be exact, where the derivative of the quartic is equal to 0) in most cases, there are no issues. May 11, 2014 at 20:31

One method, though I am unsure of the accuracy would be:

$y'''_5\approx \frac{y_5-3y_4-3y_3+y_2}{(x_5-x_4)(x_4-x_3)(x_3-x_2)}$ You can rearrange this to find an approximate $y_5$, and then you can use lagrange polynomials as normal, either using all data points for a degree $5$ approximation, or just use $4$ of them for a degree $4$ approximation.

• Nice idea, but it would be pretty inaccurate esp if $x_i$ are spaced far apart. May 11, 2014 at 13:11
• The result is close but unacceptable. The $(x_i)$'s are integers and the spacing between them is at most 4. Please clarify that using 5 $(x_i, y_i)$'s, a 5th degree polynomial cannot be interpolated using Lagrange. May 11, 2014 at 18:05
• @Sauradyuti You could interpolate a 5th order polynomial, with idea I have put, but since it would be 5th order, it would not be the same as the original 4th order polynomial. May 17, 2014 at 20:05