Recent content by mtayab1994

To show that ##(U_n)## is monotonic decreasing I thought of the following: ##U_{n+1}-U_{n}=\frac{1}{k}(|f(x_{n})-f(s)|-|f(x_{n-1})-f(s)|)##, but I don't think I can remove the absolute value signs because we don't know if the difference is positive or not.

X0 isn't a it's alpha α

From part three we can see that ##g## is strictly decreasing so that means ##(U_{n})## is monotonic decreasing and bounded below by a>0 so therefore it converges to a number greater than 0.

Well X0 is the first value Xn takes and if g is monotonic increasing then that means: ##g(x_{n})=f(x_{n})-kx_{n}^{3}## is monotonic decreasing and since g is continuous that means:##g(x_{n})=0## at a unique point which is s. So that means that ##(x_{n})## is bounded below by s hence it's...

So g is monotonic decreasing and ##x_0>s## so that means that it's decreasing and bounded below by s which means it's convergent?

This is what I've come up with: if g(x) is strictly monotone increasing and ##x_0##=a. Then Xn will converge because it's bounded above by b.

For number 4 i was able to transform Un to the following: U_{n}=|X_{n}^{3}-s^{3}|=\frac{1}{k}|f(X_{n-1})-f(s)| . Now do I have to use the Lipschitz function that I had proved is continuous?

I don't think so because (x^2+xy+y^2)k=3ka^2<4ka^2 that is when plugging in a for x and y.

Yes I did that part i got k|x^2+xy+y^2||x-y|<k4a^2|x-y| then i chose η=ε/(4ka^2) and the proof is done. Can i get a hint on how to start number 4?

#3 is correct I believe and #2 I derived the function but I couldn't conclude anything. While for number one I couldn't really come up with anything besides of thinking of the fact that : |x^{3}-y^{3}|=|(x-y)(x^{2}+y^{2})|

And for number 4 do I have the right thought?

Homework Statement Let f be a real function defined on the interval [a,b]/0<a<b:\forall x,y\in[a,b],x\neq y/|f(x)-f(y)|<k|x^{3}-y^{3}| where k is a positive real constant. Homework Equations 1- Prove that f is uniformly continuous on [a,b] 2- We define a function g on [a,b] such that...

Yes i know, what I meant to say is that 0 and 1 or the two missing limit points so the closed interval [0,1] is the smallest closed interval that contains all of the limit points of (0,1).

Number 2- the closure of B is [0,1] because it's the smallest closed set that contains B and it's limit points which are 0 and 1. Number 4- There's a little error. Solving for x: 1-x^{2}=0\Longleftrightarrow1=x^{2}\Longleftrightarrow x=\pm1 . So in your case you'll have to choose \mathbb{R}{-1,1}

Wow I'm really sorry that's another typo the problem statement says to show that B has a and inferior bound not just a lower bound. Hence why I proceeded like that. Is my reasoning correct?