# Uniform continuity

• Apr 8th 2010, 11:44 AM
matlabnoob
Uniform continuity
Before i posted this,i tried reading my notes again and again! (Crying) so any help....id be thankful!
here is the problem...

consider f(x) = lnx, with A element in (0,1)
confirm the continuity of f in (0,1)
Let x0 be an element in (0,1)
then |lnx - lnx0| = | ln x/x0 | = ln(1 - (x0 - x)/x0)
this is smaller than or equal to 2|(x0 - x)/x0| if |(x0 - x)/x0| < 1/2

ok so first.. why is |(x0 - x)/x0| < 1/2?? did someone decide to randomnly pick 1/2...? and why is |(x0 - x)/x0| < 1/2... where did the ln go..? wouldnt it make more sense if it was .. ln|(x0 - x)/x0| < 1/2 ?

Take delta = min (epsilon*(x0)/2 , 1/2*x0)

how was delta chosen to be these??

thank you! any help ... would save me a lot of time..effort. im looking through books..online.. believe me im trying but this is logic (and i am poor at it!)(Crying)

i havent written up the whole problem,only the stage i am stuck so far!
• Apr 8th 2010, 03:06 PM
Drexel28
Quote:

Originally Posted by matlabnoob
Before i posted this,i tried reading my notes again and again! (Crying) so any help....id be thankful!
here is the problem...

consider f(x) = lnx, with A element in (0,1)
confirm the continuity of f in (0,1)
Let x0 be an element in (0,1)
then |lnx - lnx0| = | ln x/x0 | = ln(1 - (x0 - x)/x0)
this is smaller than or equal to 2|(x0 - x)/x0| if |(x0 - x)/x0| < 1/2

ok so first.. why is |(x0 - x)/x0| < 1/2?? did someone decide to randomnly pick 1/2...? and why is |(x0 - x)/x0| < 1/2... where did the ln go..? wouldnt it make more sense if it was .. ln|(x0 - x)/x0| < 1/2 ?

Take delta = min (epsilon*(x0)/2 , 1/2*x0)

how was delta chosen to be these??

thank you! any help ... would save me a lot of time..effort. im looking through books..online.. believe me im trying but this is logic (and i am poor at it!)(Crying)

i havent written up the whole problem,only the stage i am stuck so far!

This is hard to read and follow.

You only need to prove that it's continuous at one and the rest follows. I don't know how you define the natural log though, like this $\ln(x)=\int_1^x\frac{d
\xi}{\xi}$
?
• Apr 8th 2010, 05:08 PM
xalk
Quote:

Originally Posted by matlabnoob
Before i posted this,i tried reading my notes again and again! (Crying) so any help....id be thankful!
here is the problem...

consider f(x) = lnx, with A element in (0,1)
confirm the continuity of f in (0,1)
Let x0 be an element in (0,1)
then |lnx - lnx0| = | ln x/x0 | = ln(1 - (x0 - x)/x0)
this is smaller than or equal to 2|(x0 - x)/x0| if |(x0 - x)/x0| < 1/2

ok so first.. why is |(x0 - x)/x0| < 1/2?? did someone decide to randomnly pick 1/2...? and why is |(x0 - x)/x0| < 1/2... where did the ln go..? wouldnt it make more sense if it was .. ln|(x0 - x)/x0| < 1/2 ?

Take delta = min (epsilon*(x0)/2 , 1/2*x0)

how was delta chosen to be these??

thank you! any help ... would save me a lot of time..effort. im looking through books..online.. believe me im trying but this is logic (and i am poor at it!)(Crying)

i havent written up the whole problem,only the stage i am stuck so far!

Do you mean that:

δ =min{ $\frac{\epsilon.x_{o}}{2} ,\frac{1}{2x_{o}}$}
• Apr 9th 2010, 03:58 PM
matlabnoob
Quote:

Originally Posted by xalk
Do you mean that:

δ =min{ $\frac{\epsilon.x_{o}}{2} ,\frac{1}{2x_{o}}$}

Yes this is what i mean!
(how do you use mathematical symbols?i didnt know how to use them here)

• Apr 9th 2010, 03:59 PM
matlabnoob
Quote:

Originally Posted by Drexel28
This is hard to read and follow.

You only need to prove that it's continuous at one and the rest follows. I don't know how you define the natural log though, like this $\ln(x)=\int_1^x\frac{d
\xi}{\xi}$
?

i understand the background of it..that i have to prove continuity and uniform continuity will imply continuity. but the problem confuses me so much. it looks all very random!
• Apr 9th 2010, 04:00 PM
Drexel28
Quote:

Originally Posted by matlabnoob
Yes this is what i mean!
(how do you use mathematical symbols?i didnt know how to use them here)

Ok. Well you have three choices A) rewrite in LaTeX or a word document (with equation editor) and post it up and I will review it B) leave it as it is and maybe xalk or another member will help you C) I can give an alternative answer. Which do you want?
• Apr 9th 2010, 04:06 PM
Drexel28
Quote:

Originally Posted by matlabnoob
.that i have to prove continuity and uniform continuity will imply continuity

I'm not sure what you mean by this. U.C. implies cont. but the other direction is not necessarily true (e.g. $f:(0,1]\to\mathbb{R}$, with $f:x\mapsto\frac{1}{x}$)
• Apr 9th 2010, 04:17 PM
matlabnoob
Quote:

Originally Posted by Drexel28
Ok. Well you have three choices A) rewrite in LaTeX or a word document (with equation editor) and post it up and I will review it B) leave it as it is and maybe xalk or another member will help you C) I can give an alternative answer. Which do you want?

I hopethis turns out well...i tried in word(thanks!)

Consider:
f(x) = lnx, A Є (0,1)
Confirm the continuity of f Є (0,1) by doing the following…
Let x0 Є (0,1), then |lnx - ln x0| = |lnx / x0| = ln(1 – (x0 – x)/ x0| ≤ 2|( x0 – x)/ x0| if |( x0 – x)/ x0| < ˝
Take δ = min {ε. x0/2, 1/2 x0}
• Apr 9th 2010, 04:19 PM
matlabnoob
Quote:

Originally Posted by Drexel28
I'm not sure what you mean by this. U.C. implies cont. but the other direction is not necessarily true (e.g. $f:(0,1]\to\mathbb{R}$, with $f:x\mapsto\frac{1}{x}$)

aha..of course.thanks for this! will note down for future reference