Connection between polynomials and Pascal's triangle

AI Thread Summary
The discussion reveals a connection between polynomials and Pascal's Triangle through specific polynomial evaluations yielding zero when using coefficients from the triangle. For a 3rd degree polynomial, the equation f(5) - 4f(4) + 6f(3) - 4f(2) + f(1) = 0 aligns with the 5th row of Pascal's Triangle, while similar patterns emerge for 2nd and 1st degree polynomials. The participant notes that these equations express the value of the polynomial at a point in terms of its values at previous points, suggesting a deeper relationship. Despite attempts to understand the underlying reasoning, clarity remains elusive. Further exploration of the Binomial Theorem and finite differences may provide insights into this phenomenon.
LightningB0LT
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I recently discovered that for a 3rd degree polynomial I was studying, f(5) - 4f(4) + 6f(3) - 4f(2) + f(1) = 0. At first I just though it was coincidental that the coefficients were the 5th row of Pascal's Triangle, but then I tried a 2nd degree polynomial and found that f(4) - 3f(3) + 3f(2) - f(1) = 0, which is the 4th row. The same thing worked for 1st and 4th degree polynomials I tried, using the 3rd and 6th row as coefficients. I've tried to reason through why this might be the case, but without success. Can someone explain this to me? Thanks in advance!
 
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LightningB0LT said:
I recently discovered that for a 3rd degree polynomial I was studying, f(5) - 4f(4) + 6f(3) - 4f(2) + f(1) = 0.

Another way of looking at the same equation is that expresses f(5) in terms of the value of the polynomial at previous consecutive values: f(5) = 4f(4) - 6f(3) + 4f(2) - f(1) and so this link might be relevant: http://ckrao.wordpress.com/2012/02/28/finite-differences-for-polynomial-extrapolation/
 
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