Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
$x^3[f(x+1)-f(x-1)]=1$

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

share|improve this question
    
Already answered at math.stackexchange.com/questions/18172/x3fx1-fx-1-1. Delete this, those who have the Power. –  TonyK Jan 19 '11 at 18:08
    
This is different. –  bobobinks 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. –  bobobinks Jan 19 '11 at 18:26
    
@Mariano: bobobinks is right. See my answer below. –  Willie Wong Jan 19 '11 at 18:39
add comment

1 Answer

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 $\liminf$).

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.