Official TSR Mathematical Society

Original post by harr

Thanks for your reply. I was thinking of arc length in terms of the sup of the lengths of polygonal (?) approximations to the curveThere's a problem with thinking like that...

Mod: but thank you for the opportunity to post a completely relevant 'troll' image!

Reply 3202

That's quite an interesting result to be honest. I would have thought doing something as 'sensible' as this wouldn't have returned as bad an approximation as 4.

Reply 3203

Original post by Oh I Really Don't Care

That's quite an interesting result to be honest. I would have thought doing something as 'sensible' as this wouldn't have returned as bad an approximation as 4.

It's even worse in 3 dimensions. See http://mathdl.maa.org/images/upload_library/22/Polya/00494925.di020678.02p0385w.pdf

Reply 3204

Zhen Lin

12

Original post by Oh I Really Don't Care

That's quite an interesting result to be honest. I would have thought doing something as 'sensible' as this wouldn't have returned as bad an approximation as 4.

A stark reminder of how far geometric intuition can be from analytic reality.

As for the problem of defining lengths / surface area / etc., I think we should stick with differentiable manifolds. At least the geometrically obvious definition for that case hasn't been shown to defy expectations...

Reply 3205

Original post by Zhen Lin

A stark reminder of how far geometric intuition can be from analytic reality.

As for the problem of defining lengths / surface area / etc., I think we should stick with differentiable manifolds. At least the geometrically obvious definition for that case hasn't been shown to defy expectations...

I find the whole thing confusing to be honest, and I think the concept of considering R^n where R is some complete ordered field just doesn't quite represent space the same way R represents a line.

I know that sounds ridiculous when you consider using Cartesian coordinates, but things just go too wrong, the majority of the time (by this, I mean when you go looking, things just turn out ridiculous, like the above). Can you imagine that on the real line? Yes, there are some weird things when you consider measure on the line (say with Cantor sets) but even then, it's not too bizarre. Perhaps completeness is just slightly too strong (although it's equivalent statements clearly beg to differ, representing it as almost self evident).

I would be interested to see a novel approach to considering the plane as something other than R^2, but I don't know of anything. Better yet, considering the real line as something other than a complete ordered field.

Reply 3206

Original post by RichE

...

What's your take on the axioms for the real line and n-dimensional space?

Reply 3207

RichE

Original post by Oh I Really Don't Care

What's your take on the axioms for the real line and n-dimensional space?

That foundational aspect I've never really studied. I'm happy that you can get the real number system out of axioms for geometry like Hilbert's and vice versa.

If you're interested though in other attempts to describe geometry without set theory you might want to look at

http://en.wikipedia.org/wiki/Tarski%27s_axioms

(edited 12 years ago)

Reply 3208

harr

Original post by DFranklin

There's a problem with thinking like that...

Spoiler

Apologies for being unclear. I meant polygonal approximations where the end points of all the line segments are on the curve.

Each line segment can then be replaced by a horizontal and a vertical line segment as in your example to give the upper bound of 3.

(edited 12 years ago)

Reply 3209

twig

Original post by DFranklin

There's a problem with thinking like that...

Mod: but thank you for the opportunity to post a completely relevant 'troll' image!

Is there a pre-uni explanation to why the above is wrong? Other than the "repeat to infinity" part looking odd, I can not understand which wrong assumption is made in order to lead the result: pi=4 :frown:

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Reply 3210

Original post by twig

Is there a pre-uni explanation to why the above is wrong? Other than the "repeat to infinity" part looking odd, I can not understand which wrong assumption is made in order to lead the result: pi=4 :frown:

.

The problem is that the 'curve' you're making by removing corners never gives a good approximation to the circumference.

[Which is kind of just restating your question, to be honest. But in essence, that's the invalid assumption - that just because one curve approximates another in a "they more or less cover the same points" sense, it doesn't mean the length of that curve approximates the length of the other one].

Reply 3211

RichE

Original post by twig

Is there a pre-uni explanation to why the above is wrong? Other than the "repeat to infinity" part looking odd, I can not understand which wrong assumption is made in order to lead the result: pi=4 :frown:

.

To put it in uni speak, one would say that arc-length is not continuous in the supremum norm.

More informally, what this means is that you can have two graphs that are very close in y-values but differ enormously in arc-length. This can be achieved by having one curve have many wiggles of small-amplitude.

Consider y = 0 and y = (1/n) sin(n^2x) say on 0<x<pi.

You can get the second graph arbitrarily "uniformly" close to the first by choosing n suitably large but their arc lengths won't be close.

Reply 3212

harr

Original post by twig

Is there a pre-uni explanation to why the above is wrong? Other than the "repeat to infinity" part looking odd, I can not understand which wrong assumption is made in order to lead the result: pi=4 :frown:

.

Here's my (not very good) explanation:
Consider the sequence 1, 1/2, 1/3, 1/4, etc. Each term is non-zero, but the limit is zero.

Say I have a function L which gives the length of a curve. We have some sequence c_n of curves converging to a curve c (in this case a circle). L(c_n)=4 and lim(c_n)=c.
Then lim(L(c_n))=lim(4)=4.
And L(lim(c_n))=L(c).
But why should L(lim(c_n))=lim(L(c_n))? We don't expect the limit of 1, 1/2, 1/3, 1/4, ... to be non-zero just because each term is non-zero. I could create a function iszero which takes one value for 0 and another for all other numbers. But then you can see that lim(iszero(1/n))=/=iszero(lim(1/n)). You can't just interchange a function and a limit.

Reply 3213

Original post by RichE

..

Thanks. (Or in other words, I just got "Please Rate Some Other Member" for about the 4th time today. It's somewhat awesome/scary that if you google PRSOM, the first link credits it as a TSRism).

Reply 3214

Original post by DFranklin

Thanks. (Or in other words, I just got "Please Rate Some Other Member" for about the 4th time today. It's somewhat awesome/scary that if you google PRSOM, the first link credits it as a TSRism).

Best description of the student room;

2. The Student Room
A large British student forum that is 99% inhabited by geeks and nerds who spend the best part of their days talking about schools and universities whilst also pretending they're cool and get laid at parties every weekend (whilst having 300,000+ post counts), whilst also discussing how they did in their latest exams, as well as bitching about how they ONLY got an A on (insert GCSE here) or ONLY 598/600 marks in their AS level Physics exam and how they are demanding a remark for 600/600, because obviously it doesn't matter about the fact they have an A. MUST GET THAT REMARK.

For some reason, they're all obsessed with Oxbridge and the other top 5 universities, and don't seem to acknowledge that getting less than A* in your GCSE's or A's in your A-levels doesn't make you thick and doesn't doom you to ex-polytechnic universities (what's wrong with them anyway?)

Politically, the forum is very left-wing, so if you mention ANY negative point about immigration, islam, etc. prepare to be bashed to smitherines in the debate topics.
- Typical discussion on The Student Room -

Geek Girl: I got my GCSE's back today. I got 3 A*, 5 A's and 4 B's. I'm so happy. I'm doing four A levels.

Geek Boy: OMFG. HAHAHAHA. YOU FAILED YOUR GCSES AND YOU FAILED AT LIFE. YOU'LL NEVER GET INTO OXBRIDGE UNLIKE ME, I GOT 12 A* AND I'M DOING 6 A LEVELS .

Geek Girl 2: Just 12 A*?! I GOT 15 A*
YOU'RE BOTH THICK. I'M CLEARLY THE BEST OBBRIDGE CANDIDATE HERE

.. and you know the rest/

Reply 3215

SimonM

Seems accurate to me.

Reply 3216

twig

Thanks people for the explanations.

Agreed

Original post by Oh I Really Don't Care

I find the whole thing confusing to be honest, and I think the concept of considering R^n where R is some complete ordered field just doesn't quite represent space the same way R represents a line.

I know that sounds ridiculous when you consider using Cartesian coordinates, but things just go too wrong, the majority of the time (by this, I mean when you go looking, things just turn out ridiculous, like the above). Can you imagine that on the real line? Yes, there are some weird things when you consider measure on the line (say with Cantor sets) but even then, it's not too bizarre. Perhaps completeness is just slightly too strong (although it's equivalent statements clearly beg to differ, representing it as almost self evident).

I would be interested to see a novel approach to considering the plane as something other than R^2, but I don't know of anything. Better yet, considering the real line as something other than a complete ordered field.

Cantor sets aren't so bad. It's the unmeasurable monstrosities like the Vitali set that are bizarre.

There is the suggestion that the axiom of choice is too strong for mathematics that is ‘intuitive’. Certainly the existence of non-measurable sets depends on some form of the axiom of choice: there are models of set theory which have no non-measurable sets and do not satisfy the axiom of choice.

Reply 3219