Triangles, parallelograms and Shimura curves: Maths!

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Martijn
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Triangles, parallelograms and Shimura curves: Maths!

Post by Martijn » Thu Dec 04, 2008 16:02

I did maths at uni. And then I went on to do a PhD and then... well, anyway, there is no maths thread (or, for those accross the pond, a math thread) on Anorak and after I discovered I share a mathematical background with a fellow Anorak, I thought I'd start one. Not that I have anything to say, but that might come now that there is a thread.

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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by soft revolution » Thu Dec 04, 2008 16:42

Yes! I dabbled in maths. I'm conviced most of it is entirely made up. Square root of minus one indeed.

Div, Grad and Curl I found a bit enjoyable because it was good to know how shapes worked. Please can someone give me a practical application for Fourier Transforms or Laplace though? They never made sense to me.
And by me, I mean, Flexo.

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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by Martijn » Thu Dec 04, 2008 16:48

soft revolution wrote:Please can someone give me a practical application for Fourier Transforms
MP3s. Or digital music in general. Is that practical enough?

Here is an article that tries to explain it all.

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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by soft revolution » Thu Dec 04, 2008 16:54

Wow, that's totally brilliant. I may have paid more attention if someone had said that, equations always seem so abstract without a practical example.
And by me, I mean, Flexo.

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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by lynsosaurus » Thu Dec 04, 2008 16:59

don't they use it in the processing of remotely-sensed images, too? don't ask me how, like.

i have this on the wall above my desk.
Image

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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by Big Nose » Fri Dec 05, 2008 14:24

142857 is a cyclic number — you can find its multiples simply by rotating its digits:

* 142857 × 1 = 142857
* 142857 × 2 = 285714
* 142857 × 3 = 428571
* 142857 × 4 = 571428
* 142857 × 5 = 714285
* 142857 × 6 = 857142

http://www.futilitycloset.com/category/science-math/

I love numbers me.
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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by lynsosaurus » Fri Dec 05, 2008 15:03

that is fucking cool.

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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by SophieC » Fri Dec 05, 2008 15:10

i've now got a handy list of everyone on anorak i need to apply a chinese burn to if i ever meet them. carry on.
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Re: Triangles, parallellograms and Shimura curves: Maths!

Post by gloom button » Fri Dec 05, 2008 15:31

assuming you're not playing championship manager at the time.

i may be the fellow anorak mentioned in the first post. my brother and i are doing this thing at the moment where we solve the problems on this website - http://www.projecteuler.net - where he does the computer-programming parts and i (try to) do the maths part so that the program isn't just brute-forcing its way to a solution but doing something a wee bit more elegant. it's great, actually, it's the first time i've done any problem-solving in forever and the problems are pretty accessible and graded quite nicely as you go along. plus you get to laugh at other people's rubbish solutions once you solve one.
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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by Nicole Diver » Fri Dec 05, 2008 15:54

The project Euler website looks great. On paper I should be good at that with a degree in maths and a job that involves computer programming, I bet I'm not though...

I shall return to this thread once I've remembered (asked my boyfriend to repeat to me) my favourite maths joke.

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by Big Nose » Fri Dec 05, 2008 16:03

I became extremely moist when I learned that 0.999... is actually 1

http://en.wikipedia.org/wiki/0.999

The proofs are brilliant aren't they?

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by gloom button » Fri Dec 05, 2008 21:08

Nicole Diver wrote:The project Euler website looks great. On paper I should be good at that with a degree in maths and a job that involves computer programming, I bet I'm not though...

I shall return to this thread once I've remembered (asked my boyfriend to repeat to me) my favourite maths joke.
I have a degree in it as well, and feel guilty if I can't solve those things quickly! I'm quite pleased today though because I managed to solve a quite nice problem about pentagonal numbers and (I think) one about an arithmetic sequence of primes.

(The only good maths joke I know is the one about the two cats on an inclined plane..)
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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by lynsosaurus » Sat Dec 06, 2008 13:06

gloom button wrote: (The only good maths joke I know is the one about the two cats on an inclined plane..)
that is one of my favourite jokes ever.

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by gloom button » Mon Dec 08, 2008 11:49

gloom button wrote:and (I think) one about an arithmetic sequence of primes.
So actually in solving this problem I think I have something at least mildly interesting (or something I didn't know about before, at any rate). The problem is about arithmetic sequences of 4-digit primes whose digits are permutations of each other, and given one (1487, 4817, 8147) you're asked to find the (only) other 4-digit ascending sequence that there is.

And it turns out that the constant between the elements in that first sequence (3330) is the same as the constant between the elements in the other sequence, which seemed completely counterintuitive to me. Why should two more-or-less unrelated sequences of primes have the same difference between each member. So we extended the program to look at sequences of more than four digits, and it turns out that that same difference, 3330, or multiples of it, keeps turning up as the constant between elements of sequences of primes. You get other constants as well - 1450, 2970, etc - whose digits always sum to a multiple of 9, and I think I understand that part at least, because, taking a simple case, if you have a 2-digit number 10a+b and you reverse the digits (10b+a) and subtract, you get 9(b-a), and you get versions of that for other numbers.

And that's true of all numbers, actually, not just primes, but I still think that this is all a bit mysterious, and that there ought to be some interesting application to it somehow (so that, for instance, if you have a very large prime in an arithmetic sequence with a 3330 constant difference, you can tell where the next prime is - or something).

I did do other things this weekend too.
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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by indiehorse » Mon Dec 08, 2008 11:57

A few weekends ago, I started trying to explain a mathematical concept in the pub to a selection of other anorakers. I did a diagram and all. At no point did JamieC attempt to give me a Chinese burn.

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by Big Nose » Wed Dec 10, 2008 13:48

gloom button wrote:
gloom button wrote:and (I think) one about an arithmetic sequence of primes.
So actually in solving this problem I think I have something at least mildly interesting (or something I didn't know about before, at any rate). The problem is about arithmetic sequences of 4-digit primes whose digits are permutations of each other, and given one (1487, 4817, 8147) you're asked to find the (only) other 4-digit ascending sequence that there is.

And it turns out that the constant between the elements in that first sequence (3330) is the same as the constant between the elements in the other sequence, which seemed completely counterintuitive to me. Why should two more-or-less unrelated sequences of primes have the same difference between each member. So we extended the program to look at sequences of more than four digits, and it turns out that that same difference, 3330, or multiples of it, keeps turning up as the constant between elements of sequences of primes. You get other constants as well - 1450, 2970, etc - whose digits always sum to a multiple of 9, and I think I understand that part at least, because, taking a simple case, if you have a 2-digit number 10a+b and you reverse the digits (10b+a) and subtract, you get 9(b-a), and you get versions of that for other numbers.

And that's true of all numbers, actually, not just primes, but I still think that this is all a bit mysterious, and that there ought to be some interesting application to it somehow (so that, for instance, if you have a very large prime in an arithmetic sequence with a 3330 constant difference, you can tell where the next prime is - or something).

I did do other things this weekend too.
And in any base.

Octal :-

71 - 17 = 52 (7)

Hex :-

C1 - 1C = A5 (F)

Binary :-

10 - 1 = 1
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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by Gordon » Wed Dec 10, 2008 14:40

gloom button wrote: whose digits always sum to a multiple of 9
Can be simplifed to "who are divisible by 9", surely.
Toot toot.

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by soft revolution » Wed Dec 10, 2008 15:12

That depends on whether your mathsglass is half full or half empty.
And by me, I mean, Flexo.

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by Gordon » Wed Dec 10, 2008 16:03

No, I meant if the digits of x add up to a multiple of 9, then x is a multiple of 9.
Toot toot.

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Re: Triangles, parallelograms and Shimura curves: Maths!

Post by Martijn » Wed Dec 10, 2008 16:12

gloom button wrote:And it turns out that the constant between the elements in that first sequence (3330) is the same as the constant between the elements in the other sequence, which seemed completely counterintuitive to me.
It did so to me at first, but then, if you look at length-three four-digit arithmetic sequences, whose members are permutations of each other, then there is only a small number of possibilities for the difference. Knowing that the numbers all have to be prime leaves you with even fewer possibilities (because, for instance, the final digits either have to be all different, or all the same). So the fact both sequences have a difference of 3330 (and not, say, 450, which a priori is possibly too) still surpises me, but it's not as strange as it seems at first.

Random maths fact of the day: the only three arithmetic sequences whose members are cyclic permutations of each other and who are of maximal length are the ones starting with 148, 259 and 012345679.

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