Suppose f: [1,2] > R is continuous function such that for every natural number n, there exists x on [1,2] with f(x)<1/n. Show that there exists a number c on [1,2] such that f(c) = 0
I did the following:
f(x)<1/n implies 1/n < f(x) <1/n. By sandwich theorem as n tends to infinity, f(x) converges to 0.
Then I thought of applying the Intermediate Value Theorem to prove that f(c) = 0 exists. But for that I need to have f(1)<=f(0)<=f(2) and I am not sure if convergence of f(x) to 0 proves that f(0) lies on that interval (since it tends to 0 but how do we know that it is ever equal to 0?)
Thank you in advance!

spacewalker
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 09012017 15:52

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DFranklin
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 09012017 16:28
(Original post by spacewalker)
Suppose f: [1,2] > R is continuous function such that for every natural number n, there exists x on [1,2] with f(x)<1/n. Show that there exists a number c on [1,2] such that f(c) = 0
I did the following:
f(x)<1/n implies 1/n < f(x) <1/n. By sandwich theorem as n tends to infinity, f(x) converges to 0.
You could say something like:
For all n, we can find x_n with f(x_n) < 1/n
and then make the deduction that f(x_n) > 0.
It's not immediate how this helps you; in particular if x_n keeps changing, you've got no "obvious" value to try for 'c'.
That said, I think this is the right place to start, but you've got some work to do to finish from here.
FWIW, unless I'm missing something I don't see the IVT being useful to you here  the tricky case is when f(x) never changes sign.
Edit: this is kind of a spoiler, but have you encountered the BolzanoWeierstrauss theorem?
2nd edit: You can actually do this just using theorems you should know with very little actual work. I'm not sure it's what would be expected, but it can be done.Last edited by DFranklin; 09012017 at 19:17. 
HapaxOromenon3
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 09012017 18:14
(Original post by spacewalker)
Suppose f: [1,2] > R is continuous function such that for every natural number n, there exists x on [1,2] with f(x)<1/n. Show that there exists a number c on [1,2] such that f(c) = 0
I did the following:
f(x)<1/n implies 1/n < f(x) <1/n. By sandwich theorem as n tends to infinity, f(x) converges to 0.
Then I thought of applying the Intermediate Value Theorem to prove that f(c) = 0 exists. But for that I need to have f(1)<=f(0)<=f(2) and I am not sure if convergence of f(x) to 0 proves that f(0) lies on that interval (since it tends to 0 but how do we know that it is ever equal to 0?)
Thank you in advance! 
DFranklin
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 09012017 18:24
(Original post by HapaxOromenon3)
The given data implies that f(x) is infinitesimal for some x (due to the 1/n condition), and the only real infinitesimal is 0, so as the range is the reals rather than the hyperreals, f(x) is 0 for some x. Thus the required result is established. For more on infinitesimals and hyperreals, see https://www.math.washington.edu/~mor...ers/gianni.pdf 
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 09012017 19:03
(Original post by DFranklin)
Aside from the fact that hyperreals are definitely not the intended approach, I'm not entirely convinced you have justified that "f(x) is infinitesimal for some x". 
DFranklin
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 09012017 19:05
(Original post by HapaxOromenon3)
This follows from letting n be infinite, so that 1/n is infinitesimal, and so f(x)<1/n implies f(x) is infinitesimal. 
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 10012017 08:14
(Original post by DFranklin)
As I said, x depends on n. Because of this, I don't see how your conclusion follows. It absolutely requires more justification than you have provided.
so if we let n be infinite (which is part of the standard extension of the natural numbers to the hyperreals), there exists x with 1<=x<=2 such that f(x)<1/infinity, i.e. f(x) is infinitesimal. 
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 10012017 09:02
(Original post by HapaxOromenon3)
"for every natural number n, there exists x on [1,2] with f(x)<1/n"
so if we let n be infinite (which is part of the standard extension of the natural numbers to the hyperreals), there exists x with 1<=x<=2 such that f(x)<1/infinity, i.e. f(x) is infinitesimal.
In any case, this clearly isn't the intended approach for the given question and is likely to just to confuse the OP. The question is a lot more about the BolzanoWeierstrass Theorem and how continuous functions treat limits. 
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 10012017 09:20
(Original post by RichE)
The given hypotheses apply to the usual natural numbers, which don't include infinity, so I don't see how "letting n be infinity" is an option. Even if it can be done this way, where have you used continuity? This question is not true for discontinuous functions.
The problem I have with nonstandard analysis is that it seems very easy to convince yourself that "I don't need to worry about XXX, because ... infinitesimal" when this isn't actually the case. Pretty sure that's going on here.
In any case, this clearly isn't the intended approach for the given question and is likely to just to confuse the OP. The question is a lot more about the BolzanoWeierstrass Theorem and how continuous functions treat limits.
Spoiler:Showf attains its bounds, but the bound must be 0 from the condition given.
What do you think? 
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 10012017 09:24

DFranklin
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 10012017 09:27
(Original post by RichE)
Quite possibly  I guess it depends whether that theorem has appeared in the course yet. 
spacewalker
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 10012017 11:36
(Original post by DFranklin)
This isn't true. When it says that "for every natural number n we can find x with f(x)<1/n", it means (colloquially) "give me an n, and I'll give you an x". In other words, the value of x will depend on n. Your argument relies on x not changing with n.
You could say something like:
For all n, we can find x_n with f(x_n) < 1/n
and then make the deduction that f(x_n) > 0.
It's not immediate how this helps you; in particular if x_n keeps changing, you've got no "obvious" value to try for 'c'.
That said, I think this is the right place to start, but you've got some work to do to finish from here.
FWIW, unless I'm missing something I don't see the IVT being useful to you here  the tricky case is when f(x) never changes sign.
Edit: this is kind of a spoiler, but have you encountered the BolzanoWeierstrauss theorem?
2nd edit: You can actually do this just using theorems you should know with very little actual work. I'm not sure it's what would be expected, but it can be done.
For each x on [1,2] we can find n such that f(x)<1/n.
So we can create a sequence x_n (which is bounded) so that f(x_n)<1/n. By BolzanoWeierstrass theorem any bounded sequence has a convergent subsequence. So, we have x_t ( which a subsequence of x_n) converging to limit L for example with f(x_t)<1/t. Since f is continuous we have lim (x_t) = L (L is on (1,2)) implies lim f(x_t) = f(L) and lim (+/)1/t = 0. Thus, by sandwich theorem 0< f(L)<0. Therefore, f(L) = 0 and so such c exists and it is equal to L
Thanks 
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 10012017 11:40
(Original post by spacewalker)
For each x on [1,2] we can find n such that f(x)<1/n.
Also you want weak inequalities surrouding f(L).Last edited by RichE; 10012017 at 11:42. 
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 10012017 11:49
(Original post by RichE)
I don't think this is quite what you mean.
Also you want weak inequalities surrouding f(L).
Thank you in advance! 
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 10012017 12:28
(Original post by spacewalker)
Thank you for your reply! Could you please explain on how I can get a weak inequality? It is just the question states f(x)<1/n and so the inequality is given to be strict.
Thank you in advance!
Secondly limits only respect weak inequalities. Note
1/n > 0 does not imply 0 = lim (1/n) > 0
but does imply
and such a result holds generally for weak inequalities 
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 10012017 13:05
(Original post by RichE)
Firstly you don't want to conclude 0<f(L)<0 as *no* number satisfies this.
Secondly limits only respect weak inequalities. Note
1/n > 0 does not imply 0 = lim (1/n) > 0
but does imply
and such a result holds generally for weak inequalities
f(x_t)<= 1/t but I am not sure why we can say that since we are given f(x_n)<1/n 
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 10012017 13:08
(Original post by DFranklin)
I'm thinking continuity comes in via the "nonstandard analysis treatment of infinitesimals". However, there's something *else* he hasn't used that is also pretty crucial (the result is false without it). [Not trying to be coy to you, but I don't want to spell it out for Hapax].
The problem I have with nonstandard analysis is that it seems very easy to convince yourself that "I don't need to worry about XXX, because ... infinitesimal" when this isn't actually the case. Pretty sure that's going on here. 
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 10012017 13:09
(Original post by spacewalker)
I agree. However, as far as I understand the only way to get this is to say that
f(x_t)<= 1/t but I am not sure why we can say that since we are given f(x_n)<1/n 
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 10012017 13:25
(Original post by HapaxOromenon3)
The idea that I was trying to convey is that as n gets larger and larger, f(x) is forced into smaller and smaller intervals of the form [k, k], and since f(x) is continuous, these intervals must eventually "converge" (I don't know if that's precisely the right word) to the single point 0.
I don't think it is helpful for you to continue derailing this thread. If you want to discuss nonstandard analysis, start your own thread (but be warned that I doubt many will care to reply to it). 
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 10012017 13:31
(Original post by spacewalker)
I agree. However, as far as I understand the only way to get this is to say that
f(x_t)<= 1/t but I am not sure why we can say that since we are given f(x_n)<1/n
to .
That is, limits don't respect strong inequalities, but applying a limit to a strong inequality gives a valid weak inequality.
[This is really no different from what RichE is saying, just a slightly different way of thinking about it].
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