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Given that f is continuous and $x^3[f(x+1)-f(x)]=1$, determine $\lim_{x\rightarrow \infty}f(x)$ explicitly.

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Already answered at Delete this, those who have the Power. – TonyK Jan 19 '11 at 18:08
This is different. – Benji Jan 19 '11 at 18:09
@bobokinks, the exact same reasoning used for the other question applies. If the limit exists, then it is zero. So your question amounts to: does there exists a continuous solution? – Mariano Suárez-Alvarez Jan 19 '11 at 18:24
@Mariano Suárez-Alvarez, the same reasoning does NOT apply; in this case, the previous reasoning only gives us 0=0. – Benji Jan 19 '11 at 18:26
@Mariano: bobobinks is right. See my answer below. – Willie Wong Jan 19 '11 at 18:39
up vote 1 down vote accepted

I will assume you are working with the function only defined on the positive real numbers. As any function that is bounded at 0 and 1 (implied by continuity at 0 and 1) cannot satisfy your expression at $x = 0$.

In any case, your limit does not have any fixed value, and in fact does not necessarily exist.

  • First notice that if you define the function $g(x) = f(x) + c$, then $x^3[g(x+1) - g(x)] = x^3[f(x+1) + c - f(x) - c] = 1$ is another solution to your functional equation. This implies that should the limits exists, $\lim_{x\to\infty}f(x) + c = \lim_{x\to\infty} g(x)$.

  • Now if the limit does exists, the limit would be independent of how you approach $x\to\infty$. So in particular choose the sequence $x_k = k$ of integers, from which the functional equation $f(x+1)-f(x) = 1/x^3$ leads to a telescoping sum $$ \lim_{x\to\infty} f(x) = f(1) + \sum_{k = 1}^{\infty} \frac{1}{k^3} $$ the sum in the right hand side evaluates to roughly 1.20206.

  • But the limit does not have to exist. For any function $h(x)$ defined on $[1,2]$ satisfying $h(1) = h(2) - 1$, you can use the functional equation to generate a solution of $x^3[f(x+1) - f(x)] = 1$. You simply set $$f(x) = h(1 + \{x\}) + \sum_{k = 1}^{\lfloor x\rfloor - 1} \frac{1}{(k+ \{x\})^3}$$ where $\{x\}$ is the fractional part of $x$ and $\lfloor x\rfloor$ is the integral part of $x$. Now consider your initially chosen $h(x)$ to take the values $h(1) = 1$, $h(2) = 2$, $h(3/2) = 0$. Then we have $$\limsup_{x\to\infty} f(x) \geq h(1) + \sum_{k = 1}^{\infty} \frac{1}{k^3}$$ but also $$\liminf_{x\to\infty} f(x) \leq h(3/2) + \lim_{n\to\infty}\sum_{k = 1}^n \frac{1}{(k + 1/2)^3} < h(1) + \sum_{k=1}^\infty\frac{1}{k^3} \leq \limsup_{x\to\infty}f(x)$$ using term by term comparison of the two absolutely converging series, and that $h(3/2) < h(1)$. This shows that for such a choice, the limit doesn't exist (since the $\limsup$ is strictly larger than the $\liminf$).

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BTW, a necessary and sufficient condition for the limit to exist (by the computation performed above) is that $f(x) = C - \sum_{k = 0}^{\infty}\frac{1}{(x+k)^3}$, where $C$ will be the limit at infinity. The sum in the righthand side can be more succinctly expressed as $\zeta(3,x)$, the Hurwitz zeta function. – Willie Wong Jan 19 '11 at 18:45

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