Recent content by razmtaz

you mentioned that h is continuous. Is this the reason that we can find a neighbourhood N around our fixed point such that the sequence of iterates of any point in the neighbourhood converged to the fixed point? And hence there is an analogous neighbourhood h(N) where the same is true for...

Sorry Sunjin I might have missed what you were getting at but here's an attempt: we have a sequence {x, f(x), f(f(x)), ...} which has x as some point in the basin of attraction of a fixed point p, but not equal to p, that converges to f(p)=p. Similarly, if we apply h to every element in this...

Homework Statement suppose f and g are conjugate show that if p is an attractive fixed point of f(x), then h(p) is an attractive fixed point of g(x). Homework Equations f and g being conjugate means there exist continuous bijections h and h^-1 so that h(f(x)) = g(h(x)) a point p...

Okay, so I ended up doing this question differently: since its a homeomorphism then it preserves the fixed points, and so h(g(x)) has a certain number of fixed points which implies that f(h(x)) has the same number, but f only has 1 fixed point and h(g(x)) has 3 i think, so that's the...

Thanks! a lot simpler than what I was attempting to do.

Homework Statement let f(x) = x3 and g(x) = x - 2x3. Show there is no homeomorphism h such that h(g(x)) = f(h(x)) Homework Equations Def let J and K be intervals. the function f:J->K is a homeomorphism of J onto K if it is one to one, onto, and both f and its inverse are...

Awesome, thanks!

Oops, guess I was thinking about g(x) = mu*e^x - x. I guess I want to say that for f(x) = mu*e^x there is some fixed point a < x_0 (x_0 is the minimum of g(x)) and some fixed point b > x_0 Since the function is positive and increasing, and at the point x_0 its derivative is equal to 1, then...

Thanks for the reply Dick. So if I am understanding you we have a minimum between the two zeroes, at ln(1/mu), and the value of the derivative of mu*e^x at this point is 1; Since we have two values on either side of this point which are higher than it, we can say the following about the...

Okay, I determined a solution using IVT, thanks for the tip morphism. There is another part of the question that asks which point is attracting and which is repelling, and to do this part I want to use a theorem that says if the value of the derivative at a fixed point is <1 then the point is...

morphism and LCKurtz thanks for the replies. Gonna try IVT in a few minutes then ill update, but for LCKurtz suggestion: Im not really sure what you mean. I found the max of x*e^-x to be 1/e at x = 1 and the min to be 0 at x = 0 (which is what we would expect, given the statement...

Homework Statement let f = \muex let 0 < \mu < 1/e Show that f has two fixed points q and p with q < p Homework Equations a fixed point p is a point such that f(p) = p The Attempt at a Solution solving f(x) = x: f(x) - x = 0 \muex - x = 0 Now I want to take logarithms...

Yeah, the problem asks for that specifically, but its restricted to the interval (-pi/3, pi/3). I wonder if there is a much easier way to solve this. We have a theorem in the text that says "suppose f is differentiable at a fixed point p. then if |f'(p)| < 1 then p is an attracting fixed...

Okay, so i tried a few times and i was unable to show algebraically that tan(x)/2 < x for (0,pi/3). To solve the problem I ended up graphing the two with maple and observing that tan(x)/2 < x is true. Then i used a corollary in my textbook which says that if a sequence is bounded and monotone...

Thanks for the reply micromass. Okay, well, if I evaluate f at 0 I get zero, which is the definition of a fixed point. Are you asking me to show that tan(x)/2 is 0 at 0, or are you asking me to try and show that f[n](0) = 0?