product of two continous functions


by im2fastfouru
Tags: continous, functions, product
im2fastfouru
im2fastfouru is offline
#1
Feb10-08, 07:34 PM
P: 6
There is a proof in my book that asks us to prove that the product of two continuous functions is continuous. If anyone could help please reply back, thanks!
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SticksandStones
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#2
Feb10-08, 08:12 PM
P: 104
If h(x) is undefined at some point c (and thus not continuous), will it ever have a product that is defined at that point?
HallsofIvy
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#3
Feb11-08, 07:18 AM
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It is not clear to me what SticksandStones response has to do with the question!

im2fastfouru, you know, of course, that the product, fg, will be continuous at a as long as the lim, as x goes to a, of f(x)g(x) is equal to f(a)g(a). Therefore, you must prove that, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex], then |f(x)g(x)- f(a)g(a)|< [itex]\epsilon[/itex]. Using, of course, the fact that f and g are both continuous at a, that is, that given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex] such that |f(x)- f(a)|< [itex]\epsilon[/itex] and similarly for g.

I would recommend you look at |f(x)g(x)- f(x)g(a)+ f(x)g(a)- f(a)g(a)|.

After all, your book asked you to prove this, not us!

SticksandStones
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#4
Feb11-08, 10:43 AM
P: 104

product of two continous functions


I thought he was asking something entirely different; I apologize.
lurflurf
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#5
Feb11-08, 06:03 PM
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hint
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
note
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
|f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
also recall
|a+b+c|<|a|+|b|+|c|
sutupidmath
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#6
Feb12-08, 01:11 AM
P: 1,635
Quote Quote by lurflurf View Post
hint
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
note
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
|f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
also recall
|a+b+c|<|a|+|b|+|c|
Why on earth would he do so? Halls hints are quite straightforward, he actually almost did all of what was needed to do.>< All he needs to do is show that If(x)I does not get too large as x-->a(or sth), in other words the op just needs to show that f(x) is bounded, which is easy to show since lim(x-->a)f(x)=f(a), so any function who has a limit is bounded at that limit point. all he needs to do is let e-eplylon= 1, and it is easy to show that f(x) is bounded. If(x)I=If(x)-f(a)+f(a)I<=If(x)-f(a)I+If(a)I< 1 +If(a)I

For the OP: do you see how to go now? Like Halls said, you should give some effort on your own next time, if you want to recieve more help. People here won't just give us answers(including me since i often ask for help).
HallsofIvy
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#7
Feb12-08, 05:10 AM
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lurflurf may has misread "continuous" as "differentiable".
mathwonk
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#8
Feb12-08, 08:41 AM
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the point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.

to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
lurflurf
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#9
Feb18-08, 05:51 PM
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Quote Quote by sutupidmath View Post
Why on earth would he do so? Halls hints are quite straightforward.
To solve the problem and because it is fun. I agree, I provided a different (though very slightly) view.
Quote Quote by HallsofIvy View Post
lurflurf may has misread "continuous" as "differentiable".
theorem f is continuous if and only if
lim_h->0 (f(x+h)-f(x))=0
where (for picking nits) we accept as implicit f(x) exist and limit_h->0 f(x+h) exist
Quote Quote by mathwonk View Post
the point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.

to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
A useful framework

we desire to show that

if
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
then
|f(x+h)g(x+h)-f(x)g(x)|<eps
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
thus
|f(x+h)g(x+h)-f(x)g(x)|=|[f(x+h)-f(x)]g(x)+f(x)[g(x+h)-g(x)]+[f(x+h)-f(x)][g(x+h)-g(x)]|
<|[f(x+h)-f(x)||g(x)|+|f(x)||g(x+h)-g(x)|+|f(x+h)-f(x)||g(x+h)-g(x)]|
<eps1|g(x)|+|f(x)|eps2+eps1*eps2
letting
eps1=(eps/3)/|g(x)| if |g(x)|>0
eps2=(eps/3)/max(eps1,|f(x)|) if |f(x)|>0
we easily see the result is true


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