## limit of a function defined in terms of another function

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user#471729
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### limit of a function defined in terms of another function

Hello all.

I've got a mathematics problem which I am not intelligent enough to know the answer to, so I would appreciate any help that you give.

Suppose I have a function f such that 0 < f(x) < 1 for all real x on the interval [0, 1]. Is it possible to find limn -> infinityfn(0). If so, how?

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### Re: limit of a function defined in terms of another function

I haven't really looked at this problem, but I would probably start by looking at the limit definition of a derivative.
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Korandder
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### Re: limit of a function defined in terms of another function

No, you can not find such a limit in all cases.
For example f(x) = sin(x). f(1)(x) = cos(x), f(2)(x) = -sin(x), f(3)(x) = -cos(x), f(4)(x) = sin(x) so f(1)(0) = 1, f(2)(x) = 0, f(3)(x) = -1, f(4)(x) = 0 and so on. This sequence does not have a limit.

However complex theory provides a formula for the nth derivative of a analytic function - f(n)(z) = n!/(2 pi i) intC (f(s) ds)/(s - z)n+1 where C is a simple closed path exterior to z. Calculating the limit of this as n goes to infinity is not simple. This falls into the class of problem which are quite easier with complex numbers than with real numbers.

Mathmagic
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### Re: limit of a function defined in terms of another function

Korandder wrote:No, you can not find such a limit in all cases.

Except cos(x) and sin(x) don't satisfy the conditions. The interval is [0,1], and the range is 0 < f(x) < 1. But sin(0) = 0. This satisfies the domain conditions, but not the range.
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user#471729
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### Re: limit of a function defined in terms of another function

I'm not looking for limn -> infinityf(n)(x), but limn -> infinityfn(x), i.e. f1(x) = f(x), f2(x) = f(f(x)), f3(x) = f(f(f(x))), etc.

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### Re: limit of a function defined in terms of another function

That very much depends on the function. For instance, given 0<c<1 you could define f(x)=x+c for 0<=x<1-c and f(x)=x-1+c otherwise. You can see that if c is rational, then you've got only a finite collection of points that fk0) will be. On the other hand, if c is irrational, then the points fk(0) form a dense subset of [0,1]. For fk(0) to have some kind of destination would require a specific kind of function.

What sorts of functions do you have in mind?
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user#471729
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### Re: limit of a function defined in terms of another function

Let's say a quadratic function. I would like to minimise loss of generality but if that's not possible, then I'll settle for losing some.

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### Re: limit of a function defined in terms of another function

I am guessing this an area where you want to start off with linear functions before you held off into the chaotic world of non linear. I what we have here is a recurrence relation. For simple linear functions you can easily find a closed form solution in terms of n which you could take the limit of. For non linear functions the map says 'here be dragons'.

jestingrabbit
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### Re: limit of a function defined in terms of another function

If you're going to restrict to continuous functions then you want to start by finding fixed points ie points where f(x)=x. You are guaranteed at least one of these just by continuity.

If we limit ourselves to quadratics, like you said, whose image is inside [0,1] then you can see that, regardless of the coefficient of x2 you definitely get one point that crosses the line f(x)=x (ie a stationary point) and you possibly get another too. If 0 is stationary, then the limit you want is 0. If is its not stationary, then it will head off to a stationary point.

There are then two possibilities. There is one and only one stationary point, in which case the limit will be that stationary point. Otherwise, you need to work out which of the two stationary points you end up at. You can work out that a second stationary point can only exist at x=1 (or else there will be 0<=x<=1 such that f(x)>1). In that case, only when the coeff of x2 is positive. Notice that if you start a little less than 1 in that case, you move towards the other stationary point. So, if f(0)=1 and 1 is a stationary point, then the answer is 1. Otherwise its the other other stationary point that its heading towards.
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user#471729
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### Re: limit of a function defined in terms of another function

thank you very much. you've been very helpful. i imagine that it is similar but more complicated for higher order continuous functions (e.g., ax3 + bx2 + cx + d may have up to three stationary points). Is this to do with the multiple complex roots of the real numbers?

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### Re: limit of a function defined in terms of another function

user#471729 wrote:thank you very much. you've been very helpful. i imagine that it is similar but more complicated for higher order continuous functions (e.g., ax3 + bx2 + cx + d may have up to three stationary points). Is this to do with the multiple complex roots of the real numbers?

Well, a "stationary point" is a solution to ax3 + bx2 + cx + d = x, i.e. a root of the polynomial ax3 + bx2 + (c-1)x + d. So yes, it can have at most 3 real roots, depending on what restrictions you apply to the image.
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### Re: limit of a function defined in terms of another function

so you've got a function f:[0,1]->[0,1] and you're looking at the limit of its iterations?

Sounds like dynamics, and I believe the answer is "No."

These things get chaotic easily.
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### Re: limit of a function defined in terms of another function

If we limit ourselves to quadratics, like you said, whose image is inside [0,1] then you can see that, regardless of the coefficient of x2 you definitely get one point that crosses the line f(x)=x (ie a stationary point) and you possibly get another too. If 0 is stationary, then the limit you want is 0. If is its not stationary, then it will head off to a stationary point.

You are claiming there is no possibilities of cycles? Or other stranger beasts?

Ie f(f(x)) = x, but f(x) != x?

This might be the case, given the restriction to quadratics and the interval restriction.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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### Re: limit of a function defined in terms of another function

Can't you interpolate a quadratic onto any 3 points? In particular, I think if you find the quadratic passing through (0, .5), (.5, .8 ), and (.8, .5), you'll get a quadratic function which satisfies the given conditions, but repeated applications of the function will keep oscillating between .5 and .8, hence will have no limit.
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### Re: limit of a function defined in terms of another function

MartianInvader wrote:Can't you interpolate a quadratic onto any 3 points? In particular, I think if you find the quadratic passing through (0, .5), (.5, .8 ), and (.8, .5), you'll get a quadratic function which satisfies the given conditions, but repeated applications of the function will keep oscillating between .5 and .8, hence will have no limit.

Hm, good call. So for f(x) = -2x2+1.6x+.5, for instance, the limit doesn't exist.

user#471729
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### Re: limit of a function defined in terms of another function

Isn't (3 + sqrt(109))/20 the limit in that case? The parabola intersects with the line y = x...

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### Re: limit of a function defined in terms of another function

user#471729 wrote:Isn't (3 + sqrt(109))/20 the limit in that case? The parabola intersects with the line y = x...

Yes, that is the stationary point for that quadratic function. If you enter (3 + sqrt(109))/20 into the function and keep iterating it you'll still get (3 + sqrt(109))/20 each time, but if you enter 1/2, then after the first iteration, you'll have 4/5, and continued iteration will yield an oscillation between those two values. It's wrong to say that that function doesn't have a limit, but it's okay to say that either of those two values for an initial value won't have a limit with continued iteration.

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### Re: limit of a function defined in terms of another function

Can we find a reasonable set of functions for which the iteration will tend to a limit except on a set of measure zero?

Or a reasonable set of functions in which the iteration will tend to a finite cycle? ... Except on a set of measure zero?

If we can guarantee that ||f(a)-f(b)|| < ||a-b|| for all a, b in the compact starting set, can we get it to tend to a limit? I think we can do it if we have a constant 0<K<1 for which ||f(a)-f(b)|| <= K||a-b|| for all a,b in the starting set.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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### Re: limit of a function defined in terms of another function

I'm surprised that nobody has yet mentioned the logistic map f(x) = kx (1-x) where k in (0, 4] is a parameter that controls how chaotic the solutions are.
For k<=1 the only fixed point is 0; for 1 < k <= 3 there is another fixed point at 1 - 1/k; for 3 < k <= 1+sqrt(6) (approx. 3.4495) fn(x) converges to a 2-cycle; for successively smaller intervals we get a 4-cycle, 8-cycle, 16-cycle, etc. and enter a chaotic region above around k = 3.57. (See the Wikipedia entry at http://en.wikipedia.org/wiki/Logistic_map for more detail and a nice bifurcation diagram.)

Obviously this doesn't fit the problem as stated since f(0) = f(1) = 0, but it should be possible to twerk it slightly and still keep most of the interesting behaviour, e.g. f(x) = k(x+d) (1+d-x) where we make d small (and positive), and limit k so that f(1/2) = k(1/4 + d + d2) < 1. 0 is not a stable fixed point of the original map for k > 1, so you should be able to get the same sort of results.