Suppose we want to interpolate an infinite number of points on the Cartesian plane using a continuous and differentiable function \(f\). How can this be done?

**Solution**

Given \(n\) points on the Cartesian plane, the set of points can be interpolated using a polynomial of at least degree \(n-1\). Given an infinite number of points to interpolate, we need an infinite polynomial:

\[f(x) = {a}_{0} + {a}_{1}(x-{x}_{0}) + {a}_{2}{(x-{x}_{0})}^{2} +...\] where \(\left|x-{x}_{0}\right|\) is within the radius of convergence.

Observation: \[f({x}_{0}) = {a}_{0}\] \[f'({x}_{0}) = {a}_{1}\] \[f''({x}_{0}) = 2{a}_{2}\] \[f'''({x}_{0}) = 6{a}_{3}\] \[{f}^{(4)}({x}_{0}) = 24{a}_{4}\] \[{f}^{(n)}({x}_{0}) = n!{a}_{n}\]

Solving for each constant term expands the original function into the infinite polynomial: \[f(x) = \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } { f }^{ (n) }({ x }_{ 0 } } ){ (x-{ x }_{ 0 }) }^{ n }.\]

Check out my other notes at Proof, Disproof, and Derivation

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## Comments

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TopNewestI find this interesting. I use Taylor Series a lot but I had never thought of what actually gives rise to them. Great note!

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There is probably a more rigorous proof out there. This note is more of an intuitive derivation than a proof.

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I still thought it was pretty informative.

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