Help with proving deffferentiable

450081592 · #1 $Old$ October 28th, 2009, 09:39 AM

Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational

1. Prove that h is dierentiable at 0.

2. Prove that h is not continuous at c not = 0. You may use the fact that any
interval in R contains both rational and irrational points.
(Hint: Split the proof into two cases: one for c rational, the other for c irrational.)

3. Prove that h' is a function whose domain is {0} and that h" does not
exist.

tonio · #2 $Old$ October 28th, 2009, 10:17 AM

Quote:

Originally Posted by 450081592 $View Post$

Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational

1. Prove that h is di erentiable at 0.

2. Prove that h is not continuous at c not = 0. You may use the fact that any
interval in R contains both rational and irrational points.
(Hint: Split the proof into two cases: one for c rational, the other for c irrational.)

3. Prove that h' is a function whose domain is {0} and that h" does not
exist.

Ok, what've you done so far? Did you try the definition of h'(0), taking limits when x --> 0 and all x's are rational, or all are irrational? What happened? Where are you stuck in the rest?

Tonio

450081592 · #3 $Old$ November 2nd, 2009, 03:49 PM

ok, now I have proved part 1 then I know h is continuous and differentiable at 0, now hoe do I prove that h is not continuous at c not = 0, how do I define the interval in R? I know I need to prove that exist a |f(x) - L| > absola, how do I start this.

tonio · #4 $Old$ November 2nd, 2009, 04:11 PM

Quote:

Originally Posted by 450081592 $View Post$

ok, now I have proved part 1 then I know h is continuous and differentiable at 0, now hoe do I prove that h is not continuous at c not = 0, how do I define the interval in R? I know I need to prove that exist a |f(x) - L| > absola, how do I start this.

Do as they tell you: split in tow cases, rational and irrational: if $c\neq 0$ then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.

Tonio

450081592 · #5 $Old$ November 2nd, 2009, 06:39 PM

Quote:

Originally Posted by tonio $View Post$

Do as they tell you: split in tow cases, rational and irrational: if $c\neq 0$ then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.

Tonio

I still dont understand what that means, can you tell me how to do it please?

HallsofIvy · #6 $Old$ November 3rd, 2009, 04:10 AM

Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if $\lim_{h\to 0}\frac{f(h)- f(0)}{h}$ exists. And that will be true if and only if $\lim_{n\to \infty}\frac{f(h_n)- f(0)}{h_n}$ exists and is the same for all sequences $\{h_n\}$ that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.

450081592 · #7 $Old$ November 3rd, 2009, 11:38 AM

Quote:

Originally Posted by HallsofIvy $View Post$

Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if $\lim_{h\to 0}\frac{f(h)- f(0)}{h}$ exists. And that will be true if and only if $\lim_{n\to \infty}\frac{f(h_n)- f(0)}{h_n}$ exists and is the same for all sequences $\{h_n\}$ that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.

so should I use the absola and delta argument to prove it, my friend hinted me to take absola = to c^2/2, will that work, how do I implement it?

Can you show me the solution please, I am really confused now