Polynomial interpolation using derivatives at some points

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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?

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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.

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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) = 0, 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.