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Can you please help? Thank you very much.

$f(x+1) - f(x) = 2x + 3$

I tried differentiating two sides, but I didn't achieve anything.

Cassiel Bclamson
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4 Answers4

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Let $g\colon \Bbb R\to\Bbb R$ be any differntiable function with period $1$. Then $$f(x)=x^2+2x+g(x) $$ is a solution.

On the other hand, for any solution $f$, $g(x):=f(x)-x^2-2x$ turns out to obey the equation $g(x+1)-g(x)=0$, hence those are all solutions of the given equation.

Hagen von Eitzen
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    Is there a "proper" way getting the $x^2 + 2x$ without giving the problem and adventoruous look and guessing well? – Imago Jul 26 '16 at 14:17
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Hint:
If we assume that $k,x\in\mathbb{N}\setminus\{0\}$, then we can write it using the equivalent form, $$f(k)=f(k-1)+2k+3.$$ Now, by applying it multiple times by summing for $k$ from $1$ to $x$, we get $$\sum_{k=1}^ x f(k)=\sum_{k=1}^ x f(k-1)+\sum_{k=1}^ x (2k+3).$$ $f(1),f(2),\ldots,f(n-1)$ appear on both sides of the equation, so by cancelling things: $$f(x)=f(0)+\sum_{k=1}^x(2k+3).$$ To complete things, try to find $f(0)$, and see whether this extends nicely to $\mathbb{R}$.

Workaholic
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Try $$f(x)=ax^2+bx+c$$

Equate coefficients to obtain $a$ and $b$ ($c$ can be anything).

smcc
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  • How do you know that f is polynomial? Thanks. – Cassiel Bclamson Jul 26 '16 at 14:10
  • It doesn't have to be. See Hagen's answer. It should be obvious from inspection that a polynomial will work because $f$ is a polynomial. (You did not say whether you wanted to find all solutions or not.) – smcc Jul 26 '16 at 14:15
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You can rewrite the original equation as the recurrence

$$f(x+1)=f(x)+2x+3.$$

Then knowing $f$ for some $x$ allows you to compute it for the next values,

$$f(x+2)=f(x+1)+2(x+1)+3=f(x)+2x+3+2(x+1)+3,$$ $$f(x+3)=f(x+2)+2(x+2)+3=f(x)+2x+3+2(x+1)+3+2(x+2)+3,$$ $$\cdots$$

You should see the pattern emerge, $f(x+k)$ is related to $f(x)$ by adding $k$ terms $2(x+i)$ with increasing $i$, and $k$ terms $3$. So, using the formula for the triangular numbers,

$$f(x+k)=f(x)+2kx+(k-1)k+3k=f(x)+2kx+k^2+2k.$$

At this stage, there's nothing more that you can do. $f$ is free to vary arbitrarily in an interval of lenght one, say $[0,1)$, and the other values are constrained by the above formula.