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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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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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  • $\begingroup$ Funnily enough I was using this function so I could prove the sum of cubes formula! $\endgroup$
    – user85798
    Jul 10, 2013 at 21:18
  • $\begingroup$ Ah, for that multiply the sum by $(1-r)$ and you will get the sum of the squares. $\endgroup$
    – OR.
    Jul 10, 2013 at 22:47
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Easy to show that $f(n)=f(0)+\frac{n^2(n-1)^2}{4}, \forall n\in Z$ and $f(x)=f([x]+a)=f(0)+\frac{[x]^2([x]-1)^2}{4}+f(a)$. It means that there is infinity many function. If $g(x):[0;1) \to R$ is a arbitary function then $f(x)=f(0)+\frac{[x]^2([x]-1)^2}{4}+g(x-[x])$ is general solution of your functional equation. If we choose $g(x)=\frac{x^2(x-1)^2}{4}$ then we get $f(x)=f(0)+\frac{x^2(x-1)^2}{4}$. Let $h(x):=g(x)-\frac{x^2(x-1)^2}{4}$

Answer is $f(x)=C+\frac{x^2(x-1)^2}{4}+h(x-[x])$, where $C=f(0)$


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