Need Help with Double Integral Problem?

[matt grime]

yes:

a<x and c positive means

ac<xc

c<y and x positive means

cx<xyso putting them together,

ac<cx<xy

the other bit is left as an easy exercise for the reader

 

[Claire84]

Okies, thanks for that. means one of the questions was a lot easier than was expeted.

I thought the next question was going to be something similar but it wasn't. I'll give you the question and what I've done of it, and any input would be fab.

Let 0<r<=s and take some number e with 0<e<3rs. Suppose that (xn) is a b approximation to r and (yn) is a b approximation to s, where b=e/3s. Prove that (xn,yn) is an e approximation to rs.

I started out by just showing that eventually r-b<xn<r+b and s-b<yn<s+b

I then just thought you could multiply the 2 together and do a bit of subbing in, but that has all gone very pear-shaped and I'm guessing that's not what you're supposed to do since the question says to prove this rather than to show (as the previous question did that I thought would be similar to this one). To get an idea for this proof do I need to have rs-e<xnyn<rs+e and then work backwards from that using the things they've given us?

 

[matt grime]

What does it mean to be a "b approximation" exactly?

here's how i'd start it:

|x(n)y(n)-rs| = |x(n)y(n) - sx(n) + sx(n) - rs|


so you get this is less than or equal to

|x(n)[y(n)-s]| + |s[x(n)-r]|

and possibly by this approx answer you get this is less than or equal to

|x(n)||y(n)-s| +sb

and sb=e/3

I'd need to know exactly what the terminology means though to do more

 

[Claire84]

If b>0 we can say that (xn) is a b approximation to r if (xn) is eventually confined to the interval (r-b,r+b), according to the notes anyway.

 

[matt grime]

Ok, so in my last post subs in the word 'eventually'

now if we can show that |x(n)||(y(n)-s)| is at most 2e/3 we are done, right?

well, we know y(n) is a b approx so this quantity is eventually no more than

|x(n)|b



but we know that |x(n)| < r+b eventually as well since -r-b < r-b < x(n) < r+b by definition.

so this is no more than |r+b||b|

The question is how big is r+b?

well, b is less than s, as is r, so r+b is less than 2s

so |r+b||b| < 2sb = 2e/3

so we've shown that

|x(n)y(n)-xy| < 2e/3 +e/3 = e

as required.

 

[Claire84]

Thanks for that, I'd have never thought about doing it that way! You're too clever you know. Are you working at a university now or what?

 

[matt grime]

It's just experience. Anyone can learn to do these questions after while:

I know that I've got to use e and b, and they're linked by some rule like b = e/3s

so i look and see how to tweak it so I get an s times b, cos that gives me an e/3, right?

so that tells me I want an s|x(n)-r| or something.

from more experience of other proofs I know that i can get this from (omitting the ns)

xr- rs = xy - xs +xs -rs

cos adding zero and mutliplying by 1 in some way is always good.

so I've got some more stuff to fiddle with. Now the final thing i wrote down was the third thing that i tried, but it worked out, from knowing what i was looking for.

the thing you should take away is that i was able to start solving the problem without knowing what the terms were, simply because it looked like a lot of other stuff i'd seen before.


you can do this for any analysis question pretty much.

write down the e, d definition (epsilon delta) and assume you've defined delta, now what' can you say? go back and show you can pick delta with this knowledge to get epsilon at the end.

i can show you some examples where we work out what the proof must be without doing anything at all. just tell me how much analysis you've done. do you know what it means in terms of e and d for a function to be continuous?

i finish my phd in october, god knows after that

 

[Claire84]

I've done calculus for the gand total of 3 lectures (very experienced!). I'll be doing it for the next 2 weeks (analysis, that is!), then after the easter hols I've another 2-3 weeks of it and that's me finished til the exams. So hopefully by the end of it I'll have a better idea of it all. So far we haven't met continuity in analysis- only very brifly in applied maths abnd we didn't have to prove anything. We will be covering it though, along with lots more sequence stuff. Any help during that would be great becauise as you said yourself before, it's a bit odd to get used to.

What's your phd in then- pure or applied maths or what? Not thinking about staying on with research then or have you had enough of it?

 

[matt grime]

Here's some analysis advice then: (shamelessly nicked from Tim Gowers, a very nice man who's far too clever for my own good)


Let's take f a function from R to R.

What does it mean for f to be continuous? Suppose you think of f as being some kind of measurement, such as velocity as a function of position. Suppose we have some allowable error term, e for the output. We will say f is continuous at x, if given this allowable error e in the ouput, we can find some d (d is a function of x and e, in the sense that it is allowed to depend on x and e) such that if we measure x with an error of at most d, then the error in measuring f will be no more than e.



Formally: given ANY e>0, we can find d(x,e) such that

|x-y| < d (ie measure y within d of x)

then |f(x)-f(y)|<e (f of y is no more than e away from f of x)


example, and good method for doing almost any example:

Suppose for ease we restrict to just the inteval [0,1] take f(x) = x^2

Let e be any positive number, PRETEND we've picked d.

What happens if |x-y| < d?

|x^2 - y^2| = |(x-y)(x+y)| < d |x+y|

now, we've restricted x and y to be in the inteval [0,1], so |x+y| is no more than 2 isn't it?


so |x-y| < d implies |x^2-y^2| < 2d

so if i can go back ab tweak it so that 2d<=e I've cracked it. So I just set d = e/2, and we're there.


This pattern almost always works. Ok, here I made it an easier example because I chose a smaller interval, and so d is just a function of e. Show that if we don't have this restriction it is still continuous. d will need to depend on x as well.


Not sure about the future. Am having midlife crisis about that as we speak.

Poor Juan Pablo, too.

 

[Claire84]

In Analysis do you mainly have to work with thins that don't have a nice wee restriction like the example you put down? Gahh! I mean I think it'd sometimes be hard enough, let alone having something infinite! For the part where you don't have the restrictions, how far through that method can you go before you have to take into account rhe lack of restrictions, or does it cal for you to do things a little differently rom the start.

Anyhoo, as for your future, you should definitely come to Queen's and associate with the barmy army (Pure Maths department). Loadsa fun! The post-doctoral guys look at least 10 years older than they atually are.

Oh and yeah, not such ana amazing race- and to think I emerged from my bed at 6.45 a.m on a Sunday just to see it. Ah well, second place ain't all bad. Could have ended up like Rafie boy with popped engine.

 

[matt grime]

For the full real line, remember that x is 'fixed',
so |x-y| <d tells us that y is in the region

x-d to x+d right? so x+y is in the region

2x-d to 2x+d

just adding on x to it.

so |x+y| < |2x|+d

so we get |x^2-y^2| < d(d+2|x|)

now we just choose d such that

d(d+2|x|)< e

this defines d as a function of x and e.

to get even more finicky, if d =1 works for us, then stop there. if not and we have to pick d smaller than 1, then d+2|x| must be less than 1+|2x| so let d be e/(1+2|x|)


what i showed in the first example is a special case of continuity called uniform continuity - which tells us we can chose d independently of x. for the full real line we can't do this.

we also learn other tricks to make life easier: if f and g are continuous at the right points so is f +g, f*g and f composed g, and f/g (as long as g is not zero there)


so we can show things like (2x-1)^2 must be continuous cos obviously 2x-1 is continuous and we showed squaring as continuous.

at least your early morning was spent with coffee and duvet, tomorrow I'm supposed to be on a badminton court by 9am! come on, I'm a lazy arsed pure mathematician.

 

[Claire84]

How was your badminton? Did you actually make it out of bed in time or just lie there and pretend you thought it was a Sunday? I think my 9a/m Physics lectures are worse though, esp. with v.bad hangover!

Anyhoo, we had a pure tutorial thing today and the turor guy was talking to me about that question I asked you about before-

Let 0<r<=s and take some number e with 0<e<3rs. Suppose that (xn) is a b approximation to r and (yn) is a b approximation to s, where b=e/3s. Prove that (xn,yn) is an e approximation to rs.

and he said an easier way to do it (well, for us level one numpties) would be to take r-b<xn<r+b and s-b<yn<s+b and then multiply the two together so that you'd get
rs-rb-sb+(b^2)<xnyn<rs+rb+sb+(b^2)

So if you wanted to show that (xnyn) is an e-approximation to rs then you'd have to have e>rb+sb+(b^2) and e>rb+sb-(b^2). So would you then just have to show that e>rb+sb+(b^2) because then this would imply that the other was correct? How would I go about doing this then using the info that was given in the question? Do you take

rb+sb+(b^2)
and then sub in b+e/3s so you end up with re/3s + e/3 + (e/3s)^2? Is there any obvious way of doing this? Thanks.

 

[matt grime]

rb+sb+b^2?

ok,

well if i recall correctly r<s

so this is less than

sb+sb+b^2

b was e/3s

so we get 2s*3/3s +b^2

so we 've got to get rid of that b^2

well, what do we know?

e<3rs

and b = e/3s

so b <r

and r <s again

so we can say b^2 < br <bs =e/3 againso the whole thing is less than
2e/3 +e/3 = e

this is exactly what I did. I just used the explicitly what a "b approximation" is more that was all, which I felt more illuminating as to why things are true, and reminiscent of the proof about the product of convergent sequences. I quite like the methods your course is adopting; they aren't standard, but they are quite interesting, and might be 'better' pedagogically. I suspect soon he'll whip out the "ta da! a sequence is convergent to x if it is a b approximation for all b" line.Badminton was interesting. I took most of the rest of the day off to recover. Oh, and they're painting my office too, so I couldn't go into work.

 

[Claire84]

Thank you! I'd been faffing about with it for an hour trying to get various terms to cancel and guess what? I was being a numpty.

Anyway, hopefully our course will continue to be relatively interesting. I haven't been too horrified by analysis yet but I expect the worst is still to come. We've been doing convergent sequences for a couple of days now and I'm understanding them okay and stuff, but I'll know how much I know about it when the homework appears on Friday(day of dread for everyone doing level one pure, we're such babies ).