The rational function $c(x) := \frac{x+1}{x-1}$ is called Cayley transformation

d) Is $c^{-1}$ continuous?

Question

The rational function $c(x) := \frac{x+1}{x-1}$ is called Cayley transformation

d) Is $c^{-1}$ continuous?

blockcolder · Answer

What definition of continuity do you know?

anonymous · Answer

a continuous function is a function for which, intuitively, "small" changes in the input result in "small" changes in the output. Otherwise, a function is said to be "discontinuous". A continuous function with a continuous inverse function is called "bicontinuous".

anonymous · Answer

is it?

blockcolder · Answer

There are many ways to prove continuity of a function. Here are 3 of them:
1. Prove that $\lim_{x \to a}f(x)=f(a)$.
2. Prove that $f'(a)$ exists for all a in the domain.
3. Prove that $\lim_{h \to 0}f(a+h)=f(a)$.

anonymous · Answer

ok...

anonymous · Answer

wich is the better one in our situation?

blockcolder · Answer

Try Method 2. It is my preferred one, actually.

anonymous · Answer

blockcolder the problem is i dont have enough time to solve it i will be very happy if you can do it for me i must give it to mentor in 2-3 hours

blockcolder · Answer

What's the inverse function of c(x)? Can you write it here?

anonymous · Answer

$$c(x)^{-1}$$ ?

blockcolder · Answer

Yeah. Write its equation here.

anonymous · Answer

$$c(x)^{-1} =c \frac{1}{x} = $$

anonymous · Answer

i am stucking now..

blockcolder · Answer

Did you understand what I said in your previous question?

anonymous · Answer

hmm, i thought it was  complete solution, do i need to write something additional??

blockcolder · Answer

I thought you worked it out on your own. What I said back then was just how to do it.

anonymous · Answer

no unfortunately for mathematics  have really less time..but in one and half month i have exam i must start to learn good analysis math

anonymous · Answer

but i am allowed to write math exam just if i have enough pooints..

blockcolder · Answer

You're lucky; the function is its own inverse, as I just discovered a while ago.
So now what you have to do is prove that $c^{-1}(x)=\frac{x+1}{x-1}$ is continuous.
For this one, method 1 gives the fastest solution.
$$\begin{align}\lim_{x \to a}f(x)&=\lim_{x \to a} \frac{x+1}{x-1}\
&=\lim_{x \to a}(x+1) \cdot \lim_{x \to a}\frac{1}{x-1}\
&=\lim_{x \to a}(x+1) \cdot \frac{1}{\lim_{x \to a}x-1}\
&=(a+1)\cdot \frac{1}{a-1}\
&=\frac{a+1}{a-1}\
&=f(a)
\end{align}$$
Hopefully you know the limit rules I used here?

anonymous · Answer

to be honest i dont know very well but in first step i need to points for exam participation thank you very much