[f(x)'^n continuous at a point (EXAM IN 4 hrs.)

[llursweetiell]

Homework Statement

Suppose f: D-->R is continuous at a. Let n >1 be a positive integer. using the epsilon-delta definition of continuity, prove g(x)=[f(x)]^n is continuous as a

Homework Equations

i know how to do it as a sequence proof; but i don't know how to use the epislon/delta definition to prove it.

The Attempt at a Solution

i tried using the composition proof too, but that didn't work. any help?

 

[snipez90]

Um, I'm assuming you mean g = f nested n times since you mentioned the "composition proof". If this is the case, I don't even think the theorem is true since if n = 2, you would need f to be continuous at f(a), but f(a) might not even be in D.

 

[llursweetiell]

the problem states:
suppose f:D--> R is continuous at a. Let n >1 be a positive integer. Using the epsilon and delta definition of continuity, prove g(x)=[f(x)]^n is continuous at a.

Does that help?

Also, the composition way did not work. which is why I'm unsure of how to go about tackling the problem.

 

[snipez90]

Sorry, the comment about the composition of functions threw me off. I should have known that the solvable interpretation is the right one.

Anyways, you need to estimate |[f(x)]^n - [f(a)]^n| knowing that you can bind |f(x) - f(a)| by any positive number. But a^n - b^n has a very nice and symmetric formal factorization, namely

$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-2}a + b^{n-1})$$

Thus you can replace a with f(x) and b with f(a) to get |[f(x)]^n - [f(a)]^n| equal to |f(x) - f(a)| times a bunch of terms within an absolute value (this is key) as a result of the factorization above. But now can you see what to bind |f(x) - f(a)| by to ensure that |[f(x)]^n - [f(a)]^n| is less than say, epsilon?

 

[llursweetiell]

So i wrote,
We know that f is continuous at a. to prove that g is continuous at a:
let epsilon be greater than 0 and let delta > 0 such that d=epsilon/$$(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})$$

then,
for |x-a| < delta, then |f(x)^n-f(a)^n|=$$f(x)^n - f(a)^n = (f(x)-f(a))(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})$$
< (f(x)-f(a)) * delta = epsilon. So g is continuous at a.

Is that what you meant?

 

[snipez90]

You have the right idea, but you don't actually get to choose delta here. Since f is continuous at a, there is already some delta for which |x-a| < delta implies
$$|f(x) - f(a)| < \frac{\varepsilon}{(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})}.$$

But this is exactly what we need.

 

[llursweetiell]

|f(x) - f(a)| < \frac{\varepsilon}{(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})}.[/tex]

QUOTE]

do you mean f(x)^n-f(a)^n < all of the factorization?

 

[snipez90]

Eh, perhaps it would be clearer if I just pieced together the proof. Let epsilon > 0 be given. By the continuity of f at a, there exists some delta > 0 such that |x-a| < delta implies
$$|f(x) - f(a)| < \frac{\varepsilon}{|f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1}|}.$$
But then
$$|f(x)^n - f(a)^n| = |f(x)-f(a)||f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1}| < \varepsilon.$$

 

[llursweetiell]

oh that makes sense. thank you so much! =)