2
$\begingroup$

How to find a differentiable map $T : \mathbb{R} \rightarrow \mathbb{R}$ whose fixed points are exactly integers?

I have to

a)Find points $|T'(x)| > 1$ (if any)

and prove that

b)There must exist at least one $x$ such that $|T'(x)| > 1$.

I think the functions $$x - Tx = x(x^2 - 1)(x^2 - 4)(x^2 - 9)...$$ and $$x - Tx = x(e^x - e^1)(e^{-x} - e^1)(e^x - e^2)(e^{-x} - e^2)(e^x - e^3)...$$ will satistfy the criteria.

Is there any closed form expressions for $T'(x)$?

How can I proceed further?

  • 2
    Where does the first sentence of your post end? Neither of those infinite products converge except where they are $0$. Note in particular that a necessary condition for an infinite product to converge to a nonzero finite limit is that the terms in the product converge to $1$. $\pi x(1-\frac{x^2}{1^2})(1-\frac{x^2}{2^2})(1-\frac{x^2}{3^2})\cdots$ would work, but it has a simpler form.2011-02-16
  • 1
    I find the statement of the question somewhat confusing. It begins with the task of finding *one* function whose fixed points are precisely the integers, then a) is (apparently) a question about that one function, whereas part b) is (apparently) a question about **all** such functions. Could this be clarified?2011-02-16
  • 2
    By the way, it seems to me that there are solutions here which are much simpler than writing down explicit infinite products. For instance, the function $f(x) = x + \sin x$ exhibits qualitatively the correct behavior: i.e., its graph crosses the line $y = x$ at an infinite, evenly spaced set of points. So it could be touched up a bit to give the sought after example...(Of course this is not really unrelated to the other solutions proposed, but my point is that one does not have to work out this relationship to solve the problem.)2011-02-16
  • 2
    @Pete: With the slight modification needed to make your $f$ have fixed points at the integers, you get ($x$ plus) the example I gave in my comment. I agree that one shouldn't use the infinite product form for this! I was being coy, but that is why I said "it has a simpler form". (The example is included on the Wikipedia page I linked to.)2011-02-16
  • 0
    @Jonas: I know. :)2011-02-16

4 Answers 4