Suppose we are given several consecutive integer points at which a polynomial is evaluated. What information does this tell us about the polynomial?

We answer this question in today's post Method of Differences.

[Update] You can also check out the follow-up thread on applying this method.

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## Comments

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TopNewestVery interesting article! Thanks Calvin. I want to mention something that jumped out at me while reading the second worked exercise, where you show that \(2^{n+1}−2^n=2^n\). I recently encountered this property within the topic of

finite calculus. Finite calculus is based on the properties of the difference operator (as opposed to the derivative operator in infinite calculus). Instead of taking the limit as the difference between two points goes to zero, finite calculus considers the difference of a function evaluated at points one unit apart. Clearly then, the role of \(2^n\) within finite calculus mirrors that of \(e^x\) in infinite calculus. It seems to me that your article is an excellent demonstration of how finite calculus can be used to gain valuable information about a function/set of data. Would love to hear your thoughts on this. Thanks again!Log in to reply

Hey... I think there is a typo...if \(D_2(4)=2\) then should be \(D_1(5)=7\) not 8...

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Thanks! I've fixed it.

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Hmmm..Bt the article is quite interesting.. :-)

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Hey Clavin.. What was the meaning of \(f_i(n)\) in the equation which is after problem 1? And how should i go about proving that?

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can you explain more?

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What is it you would like him to elaborate upon?

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lets say the polynomial is f(x) if the sign changes between consecutive integers, this implies that there is atleast one root in between the integers

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