
#1
Dec1409, 12:25 PM

P: 11

1. The problem statement, all variables and given/known data
Suppose f: D>R is continuous at a. Let n >1 be a positive integer. using the epsilondelta definition of continuity, prove g(x)=[f(x)]^n is continuous as a 2. Relevant 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. 3. The attempt at a solution i tried using the composition proof too, but that didn't work. any help? 



#2
Dec1409, 01:33 PM

P: 1,106

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.




#3
Dec1409, 01:47 PM

P: 11

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. 



#4
Dec1409, 02:04 PM

P: 1,106

[f(x)'^n continuous at a point (EXAM IN 4 hrs.)
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 [tex]a^n  b^n = (ab)(a^{n1} + a^{n2}b + ... + b^{n2}a + b^{n1})[/tex] 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? 



#5
Dec1409, 06:51 PM

P: 11

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/[tex](f(x)^{n1} + f(x)^{n2}f(a) + ... + f(a)^{n2}f(x) + f(a)^{n1})[/tex] then, for xa < delta, then f(x)^nf(a)^n=[tex]f(x)^n  f(a)^n = (f(x)f(a))(f(x)^{n1} + f(x)^{n2}f(a) + ... + f(a)^{n2}f(x) + f(a)^{n1})[/tex] < (f(x)f(a)) * delta = epsilon. So g is continuous at a. Is that what you meant? 



#6
Dec1409, 11:02 PM

P: 1,106

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 xa < delta implies
[tex]f(x)  f(a) < \frac{\varepsilon}{(f(x)^{n1} + f(x)^{n2}f(a) + ... + f(a)^{n2}f(x) + f(a)^{n1})}.[/tex] But this is exactly what we need. 



#7
Dec1409, 11:37 PM

P: 11

[QUOTE=
[tex]f(x)  f(a) < \frac{\varepsilon}{(f(x)^{n1} + f(x)^{n2}f(a) + ... + f(a)^{n2}f(x) + f(a)^{n1})}.[/tex] QUOTE] do you mean f(x)^nf(a)^n < all of the factorization? 



#8
Dec1509, 12:46 AM

P: 1,106

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 xa < delta implies
[tex] f(x)  f(a) < \frac{\varepsilon}{f(x)^{n1} + f(x)^{n2}f(a) + ... + f(a)^{n2}f(x) + f(a)^{n1}}. [/tex] But then [tex] f(x)^n  f(a)^n = f(x)f(a)f(x)^{n1} + f(x)^{n2}f(a) + ... + f(a)^{n2}f(x) + f(a)^{n1} < \varepsilon. [/tex] 



#9
Dec1509, 09:46 AM

P: 11

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



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