Fun with Newton Polynomials, Part II

by t0rajir0u, Jun 24, 2008, 6:42 AM

I'll be in Canada for the next two weeks and "structure-preserving maps, Part II" is still a work in progress, so until then have a snippet that I will return to later:


Recall my musings on Newton polynomials. I neglected to realize that there is in fact an extremely efficient computational process that lets you write a polynomial of degree $ n$ in Newton form given only $ n + 1$ of its values.

The key is Lemma 2 from the above post, which is exactly analogous to the operation that takes the Taylor series of a function

$ f(x) = \sum a_k \frac {x^k}{k!}$

to its derivative. Essentially, we remove the first coefficient and shift all of the other ones to the left; in other words,

$ f'(x) = \sum a_{k + 1} \frac {x^k}{k!}$.

The property shared here is that $ \frac {d}{dx} \frac {x^k}{k!} = \frac {x^{k - 1}}{(k - 1)!}$ just as $ \Delta {x \choose k} = {x \choose k - 1}$. Now, recall that we find a Taylor series by repeatedly differentiating and plugging in zero. In other words,

$ f^{(k)}(0) = a_k$.

The same process can be applied here. In other words, given that $ P(x) = \sum b_k {x \choose k}$ we find that $ \Delta^k P(0) = b_k$. So all we have to do is take repeated finite differences (exactly analogous to taking repeated derivatives) and look at the first row of a table!

Let's redo the example from the above post. Our first four values were $ 0, 1^4, 2^4 = 16, 3^4 = 81, 4^4 = 256$. In tabular format,

$ \begin{tabular}{ | l | c | c | c | c | r | } x & P(x) & \Delta & \Delta^2 & \Delta^3 & \Delta^4 \\ \hline 0 & 0 & 1 & 14 & 36 & 24 \\ 1 & 1 & 15 & 50 & 60 \\ 2 & 16 & 65 & 110 & \\ 3 & 81 & 175 & & \\ 4 & 256 & & & \\ \hline \end{tabular}$.

The coefficients in the top row are exactly the Newton polynomial coefficients we found earlier. Splendid! Note that this gives an extremely efficient method to interpolate a Newton polynomial through any values $ P(0) = p_0, P(1) = p_1, ...$. Finding a polynomial through those values in terms of the standard basis $ \{ x^k \}$ just requires expanding out each Newton polynomial.

Among the many benefits of the above technique, note how it trivializes Problem 2 from the original post. See Newton's forward difference formula.


Practice Problem 1 (Generalization of Problem 2): A polynomial of degree $ n$ satisfies $ P(k) = p^k, k = 0, 1, 2, ... n$. Find $ P(n + 1)$. (Note that it is not necessary that $ p$ be an integer or even real.)

Practice Problem 2: Use the results of Practice Problem 1 to solve the practice problem in the original post. (Hint: how do you decompose a periodic function into a sum of functions of the form $ P(k) = p^k$? When the period is $ 3$, what values can $ p$ take?) Compare with the solutions presented in the original thread.

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4 Comments

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Facebook wrote:
[t0rajir0u] can hopefully get across the Canadian border without his passport.

No, you can't back in the US though. :D

by n0vad3m0n, Jun 24, 2008, 12:44 PM

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That would be a bummer. XD

Edit: Which university will you be attending in the future?

by metafor, Jun 24, 2008, 6:24 PM

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MIT. (Still in Canada, but in a hotel for now!)

by t0rajir0u, Jun 30, 2008, 2:41 AM

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If your polynomial basis is the falling factorial function, then it really resembles differentiation...

$ x_{(k)} = (x)(x - 1)...(x - k + 1)$

Then $ \Delta x_{(k)} = k\cdot x_{(k - 1)}$

where $ \Delta$ is the forward difference operator.

And furthermore,

$ p(x)=\sum_{k=0}^n \frac{ x_{(k)}\cdot \Delta^k (p(x))_{x=0}}{k!}$

by Altheman, Jun 30, 2008, 5:06 AM

Various mathematical thoughts, ranging from problem-solving techniques to attempts to tie together related concepts.

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