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Continuity

Post Icon Posted: Submitted by 1stOrder on 29 January 2008 - 3:16am.

Joined: 2008-01-22
Posts: 2

Hey, completely took a dive on the second analysis sheet: I was wondering how you prove continuity in general- it's just I kind of didn't understand the example in the lectures ($ f(x)=x^2 $), it all went a little fast and I don't know where she got the deltas from, or why to be honest.

‹ OCTAVE Quantum Chemistry DG ›
Post Icon Posted: 29 January 2008 - 10:08am

Joined: 2006-08-31
Posts: 694

I'm going to have a stab at this one, anyone feel free to correct me.

Right, lets look at the definition of what it means for a function be continuous at a point, $ c $.

$ f $ is cts at $ c $ if $ \forall \varepsilon > 0 $, $ \exists \delta > 0 $ such that

when $ \vert x - c \vert < \delta $ we have $ \vert f(x)-f(c)\vert < \varepsilon $

The key point here is the $ \exists \delta > 0 $. In that, if someone were to give you a specific $ \varepsilon $ then you could say: "aha here's a $ \delta $ such that when...". If you construct a method for generating these $ \deltas $ then that's your proof.

So we want to say something like: take $ \delta = \frac{\varepsilon^2}{1-c} $ or whatever is need to satisfy the definition.

To get this method of finding $ \delta $ we usually perform some algebraic manipulations to get an explicit formula, or at least an upper bound on what $ \delta $ could be. We then forget that we ever made this calculation, and write out our proof and voilla it works!

For example, if $ f(x) = x^2 $ then, given some point $ c\in\mathbb{R} $ we want to show that $ f $ is cts.

Given an $ \varepsilson $ we want to find a $ \delta $ such that,

when $ \vert x - c \vert < \delta $ we have $ \vert x^2-c^2\vert < \varepsilon $

(I got bored so someone else can continue...)
Post Icon Posted: 29 January 2008 - 11:53am

Joined: 2006-10-05
Posts: 674

Think I need to brush up.

I got inequalities by taking it out of latex. I don't know how steven has done it.
Post Icon Posted: 29 January 2008 - 11:55am

Joined: 2007-10-03
Posts: 373

One thing really, it's just to find $ \delta(c,\varepsilon) $ such that $ \forall \varepsilon $ > $ 0 |x-c| $ < $ \delta(c,\varepsilon) \implies |f(x)-f(c)| $ < $ \varepsilon $ (so don't listen to Steven or Alex who gave $ \delta $ as a function of $ \epsilon $ and $ x $, you mustn't use $ x $.)

For $ x^2 $ :
$ |x^2-c^2| &lt; \varepsilon $ :
$ |x^2-c^2|=|x+c||x-c| $

Now by our definition of $ \delta $, $ |x-c| $ < $ \delta $
so $ |x+c||x-c| $ < $ \delta |x+c| $
but $ |x+c| $ < $ |x-c|+|2c| $ by the triangle inequality,
so $ |x^2-c^2|=|x+c||x-c| $ < $ \delta |x+c|&lt;\delta(|x-c|+|2c|) $ < $ \delta(\delta+|2c|) $
as a shortcut (otherwise it can get messier), also say $ \delta $ < $ c $ (so we're only considering $ c $ > $ 0 $, you can easily do the rest)
so $ |x^2-c^2| $ < $ \delta(\delta+|2c|) $ < $ 3\delta|c| $
Remember we want that to be less than epsilon, and now we got rid of everyting except $ \delta $ and $ c $, we can do that :
$ 3\delta|c| $ < $ \varepsilon \implies \delta $ < $ \frac{\varepsilon}{3|c|} $
So we've got $ \delta(c,\varepsilon)=\min(\frac{1}{2} \frac{\varepsilon}{3|c|},\frac{1}{2} |c|) $ (where the 1/2 are added just to make sure it's smaller).

Hope that helps !

(Oh and could someone tell me how to get the > and < working in LaTeX here ? This post is a mess right now :D)
Post Icon Posted: 29 January 2008 - 2:56pm

Joined: 2008-01-22
Posts: 2

So is your delta allowed to depend on x? My supervisor said it's not supposed to- was he just talking rubbish?
Post Icon Posted: 29 January 2008 - 2:58pm

Joined: 2007-10-04
Posts: 188

No, it really wouldn't make any sense. We have to pick a FIXED delta to control x, if it was allowed to move with x the definition would be meaningless. In fact it wouldn't make any sense at all to put an x in anywhere.
Post Icon Posted: 29 January 2008 - 5:55pm

Joined: 2006-11-02
Posts: 1004

Indeed, the $ \delta $ is there exactly in order to restrict the variable $ x $ to a certain interval (namely $ [c-\delta, c+\delta] $) so that the function is itself restricted to an interval ($ [f(c)-\epsilon, f(c) + \epsilon] $); $ x $ plays a totally different role compared to the other parameters.

P.S. Using < in latex is like using &, you can't.
Post Icon Posted: 29 January 2008 - 6:51pm

Joined: 2006-08-31
Posts: 694

Whoops, minor (read: major) mistake on my part, my delta was supposed to depend on $ \varepsilon $ and $ c $, not $ x $ apologies for the confusion.

P.S. I can make $ &lt; $ and $ &gt; $ because I'm special (an admin).
Post Icon Posted: 29 January 2008 - 11:40pm

Joined: 2006-10-10
Posts: 518

Another example of the massive inequality between admins and reg members. Ohohoho, geddit?
Post Icon Posted: 30 January 2008 - 11:34am

Joined: 2006-10-05
Posts: 674

I just lost the will to live.
Post Icon Posted: 30 January 2008 - 2:15pm

Joined: 2006-08-31
Posts: 694

Okay, did a couple of hours work, $ &lt; $, $ &gt; $ and $ \& $ should work for all users. Give it a try! (Don't hijack this thread though!)
Post Icon Posted: 30 January 2008 - 2:29pm

Joined: 2007-10-03
Posts: 373

Wow Steven, good work, thanks !

$ \backslash\mathrm{begin\{hijack\}} $

$ \pi &gt; \psi $

$$ \[ \left( \begin{array}{lclr} 1 & 1 & 1 & 1\\ 1 & 11 & 111 & 1111\\ 1 & 111 & 111111 & 1111111111\\ 1 & 1111 & 1111111111 & 11111111111111111111\end{array} \right)\] $$

$ \backslash\mathrm{end\{hijack\}} $

Nice one.

Now implement a conjecture-checker algorithm that checks whatever conjecture you want up to large numbers using distributed computing on all of the computers of the maths building.
Then the forum will be complete.
Post Icon Posted: 30 January 2008 - 2:39pm

Joined: 2006-08-31
Posts: 694

Maths building, you kidding. There's a lovely cluster (153rd fastest in the world or something) over in the CSC.
Post Icon Posted: 30 January 2008 - 2:43pm

Joined: 2007-10-03
Posts: 373

Wow, quite impressive !
Do you know what they use it for ?
Surely we could convince them that checking our conjectures is more important :p
Post Icon Posted: 30 January 2008 - 2:51pm

Joined: 2006-08-31
Posts: 694

The system comprises of 960 × 3 GHz Intel Xeon "Woodcrest" cores arranged as 240 × 2-way dual core nodes, 1.92 TB of RAM and QLogic InfiniPath interconnect. When fully operational, the system will deliver 11.5 TFLOPS theoretical peak performance and rank among the most powerful clusters in the UK.

Yes please.
Post Icon Posted: 30 January 2008 - 3:36pm

Joined: 2006-11-02
Posts: 1004

Awesome work with the $ &lt; $s, this should be useful.

I should point out though that $ \psi(x) &gt; \pi(x) $ trivially for $ x\geqslant 4 $...
Post Icon Posted: 30 January 2008 - 3:46pm

Joined: 2007-10-03
Posts: 373

No, no :
$ \psi \in \left[-\pi,\pi \right[ $

See image :
Post Icon Posted: 30 January 2008 - 6:28pm

Joined: 2006-11-02
Posts: 1004

No.
$ \psi(x) := \sum_{p^k\leq x} \log p $

$ \pi(x) := \sum_{p\leq x} 1 $

This proves my previous statement.
Post Icon Posted: 30 January 2008 - 6:34pm

Joined: 2006-10-05
Posts: 674

$ \Gamma (x) &gt; \zeta (x)\quad \forall x \in \{What \; we\; care\; about\} $

or care to take on $ \firall x \in \mathbb{R}_{&gt;0} $?
Post Icon Posted: 2 February 2008 - 6:23pm

Joined: 2007-10-03
Posts: 373

How about just $ x &gt; 2.51 $ ?

Anyway.
Here's the absolute proof that $ \pi &gt; \psi $
Post Icon Posted: 2 February 2008 - 6:54pm

Joined: 2006-11-02
Posts: 1004

Unfortunately, the ordering relation we need to use here is "more important than" instead of ">" and in that respect $ \zeta $ kicks $ \Gamma $'s ass. Also, the 34th step of your proof is blatantly wrong. $ \psi &gt; \pi $, no matter how hard you try proving the opposite.

Am I wrong?
I simultaneously oppose, agree with, and neutralise all criticism ad infinitum.
My point is literal.
There is no point creating a theory of everything that doesn't work.

Thank you.

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