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The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.

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  • 8
    $\begingroup$ What math contest is this from? $\endgroup$
    – Potato
    Apr 20, 2013 at 17:38
  • 1
    $\begingroup$ @Potato: KöMaL, Problem B. 4538. $\endgroup$
    – Martin
    Apr 28, 2013 at 14:44
  • 5
    $\begingroup$ Another example of a question that should have been closed before anybody answered it! $\endgroup$ Apr 30, 2013 at 11:03

4 Answers 4

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Hint 1: $3=(x+1)-(x-2)$

$(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$


Hint 2: Is there a function closely related to $f$ that would verify a simpler equation?

$g(x)=f(x)+1$

$ $

$(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \Leftrightarrow g(x)=\cfrac{x+1}{x-2}g(x-1)$


$g(x)=\cfrac{x+1}{x-2}g(x-1)=\cfrac{x+1}{x-2}\cfrac{x+1-1}{x-2-1}g(x-1-1) = \left(\prod\limits_{k=0}^{n-1} \cfrac{x+1-k}{x-2-k}\right) g(x-n)$

Now give the good value to $n$, simplify the product and you'll have the expression of $g$ you need.

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  • $\begingroup$ Yeah right. Some terms cancel each other >_< $\endgroup$
    – xavierm02
    Apr 20, 2013 at 20:42
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    $\begingroup$ What is "the good value"? $\endgroup$ Apr 28, 2013 at 14:31
  • $\begingroup$ The value sot hat you can evaluate $g(x-n)$ $\endgroup$
    – xavierm02
    Apr 28, 2013 at 14:39
  • $\begingroup$ Big fan of answers with this format +1 $\endgroup$
    – Alexander Gruber
    May 1, 2013 at 3:26
  • $\begingroup$ Same here, format+1. $\endgroup$
    – Shuhao Cao
    May 5, 2013 at 1:11
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Let $x-1=y$, then, $$(y-1)f(y+1)-(y+2)f(y)=3$$ $$\implies f(y+1)=\frac{3+(y+2)f(y)}{y-1} \;\;\;\;\;(1)$$

Lemma: $\forall \; n \geq 2 \in \mathbb{N}$, $f(n)=n(n-1)(n+1)-1$.

Base Case: If $n=2$ then $f(n)=1 \cdot 2 \cdot 3 -1=5$ which is true by information provided in the question.

Inductive Step: Assume for $n=k$ that $f(k)=k(k-1)(k+1)-1$.

Then by equation $(1)$, $$f(k+1)=\frac{3+(k+2)(k(k-1)(k+1)-1)}{k-1}$$ $$=\frac{k^4+2k^3-k^2-3k+1}{k-1}$$ $$=\frac{(k-1)(k^3+3k^2+2k-1)}{k-1}$$ $$=k^3+3k^2+2k-1$$ $$=k(k^2+3k+2)-1$$ $$=k(k+1)(k+2)-1$$

Thus completing the induction.

Hence our Lemma is true and $f(n)=n(n-1)(n+1)-1 \; \forall \; n \geq 2 \in \mathbb{N}$

In particular, if $n=2013$, $f(2013)=2013 \cdot 2012 \cdot 2014-1=8157014183$ (I confess I used a calculator).

P.S I found the lemma by trying small cases.

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The conditions allow you to calculate $f(x+1)$ if you know $f(x)$. Try calculating $f(3), f(4), f(5), f(6)$, and looking for a pattern.

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Running the following Mathematica program:

f = DifferenceRoot[Function[{f, x}, {(x - 2) f[x] - (x + 1) f[x - 1] == 3, f[2] == 5}]];
f[2013]

We get the answer = 8157014183.

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