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How would I go about solving $f(r+1) - f(r) = r^3$?

I know the answer is $f(r) = c + \frac{1}{4}r^2(r-1)^2$, but I have no idea what method can be used to solve it. I have another functional equations problem that I hopefully will be able to solve if I can see an example of how to do this one.

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1 Answer 1

The difference $f(r+1)-f(r)$ is pretty much like a derivative. What you can do is 'integrate'.

$$1^3+2^3+\ldots+(n-1)^3=\sum_{r=1}^{n-1}(f(r+1)-f(r))=f(n)-f(1).$$

We know the sum of the first few cubes. Then $f(n)=(\text{sum of the first few cubes})+f(1)$.

If it was a function on the reals you can do similarly but sum from $r=\{n\}$ to $n-1$. Then you get that it is equal to the sum of the cubes plus the value at $\{n\}$. So, instead of depending on $f(1)$, it depends on the values on $[0,1)$.

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Funnily enough I was using this function so I could prove the sum of cubes formula! –  LTS Jul 10 '13 at 21:18
    
Ah, for that multiply the sum by $(1-r)$ and you will get the sum of the squares. –  ABC Jul 10 '13 at 22:47

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