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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?

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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. –  Jonas Meyer Feb 16 '11 at 1:38
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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? –  Pete L. Clark Feb 16 '11 at 3:42
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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.) –  Pete L. Clark Feb 16 '11 at 3:46
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@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.) –  Jonas Meyer Feb 16 '11 at 3:56
    
@Jonas: I know. :) –  Pete L. Clark Feb 16 '11 at 4:02

4 Answers 4

You're on the right track but that product does not work. Find a function that has zeros exactly at the integers. Then add $x$ to it.

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Since you are interested in infinite products, I recommend taking a look at the Wikipedia article for a little orientation on these. I believe you will find it useful in finding the example you want.

(b) is a consequence of the mean value theorem. Consider $T(\frac{1}{2})$, for example. It is either less than or greater than $\frac{1}{2}$. If less, then the average rate of change from $\frac{1}{2}$ to $1$ is greater than $1$. If greater, then the average rate of change from $0$ to $\frac{1}{2}$ is greater than $1$.

The difficulty of (a) could depend on the example, and that must wait until you have an explicit example. Of course by (b) you know that there will be such points.

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Can you unpack your explanation on (b)? I think I have a rough idea, but am having a hard time grasping it. –  user43901 Mar 22 '13 at 21:54
    
@user43901: If $T(1/2)<1/2$, then by the mean value theorem there exists $x\in (1/2,1)$ such that $T'(x)>1$. If $T(1/2)>1/2$, there's such an $x$ in $(0,1/2)$. –  Jonas Meyer Mar 25 '13 at 6:57

$f(x)=\sin(2\pi x)+x$ satisfies $f(n)=n$ for all $n\in\mathbb{Z}$ I don't know what you want with the rest...

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A simple check when $\sin(\pi x) = 0$ gives that $x$ is an integer. So using $\sin(2\pi x)$ would yield a zero result when $x=\frac{n}{2}$ for any integer $n$. However using $\sin(\pi x)$ should solve the problem. –  Asaf Karagila Feb 16 '11 at 17:19
    
I don't understand the downvote unless it is over the factor 2. This seems to satisfy the request to me. –  Ross Millikan Feb 16 '11 at 19:52
    
Ross: I downvoted because this is a very bad answer. The fixed points are not exactly the integers. I added a way to correct this, and once it is edited I will unvote the downvote. (Besides, after a bit your vote is locked and you can't undo it until there's an edit.) –  Asaf Karagila Feb 16 '11 at 20:15

I think this problem is equivalent to the problem of finding a curvature on a plane which intersect with $y=x$ only at interger points .Of course ,this curvature must be the graph of a function.but it is not difficult to find.For example,you can piece some circle curves together.(I don't know how to post the picture.)

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