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Author Topic: Maths thread.  (Read 57523 times)

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Haddock

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Re: Maths thread.
« Reply #850 on: June 08, 2017, 08:29:30 am »
+1

Yeah, my numbers are counting to 10, not 20, whoops  :-[

I get Polk's #s with 20, up to whatever little deviations.

I say lock the kid up and throw away the key.

Huh, interesting.  What happens when you go to 30, 40, 50?  I would guess that it would approach a stable distribution.
Hmmm.  In the limit, shouldn't it tend towards a 100% chance of 7 always getting there first?

EDIT: Which is to say it would never completely stabilise.
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Re: Maths thread.
« Reply #851 on: June 08, 2017, 08:39:09 am »
0

Well that would be stable; I mean a limit distribution.  And it's probably approaching it monotonically.  And I was thinking initially it wouldn't take too large of an N to get there.
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Re: Maths thread.
« Reply #852 on: June 08, 2017, 08:49:39 am »
0

Well that would be stable; I mean a limit distribution.  And it's probably approaching it monotonically.  And I was thinking initially it wouldn't take too large of an N to get there.
Ah sorry I see the word approach now.  Yes I agree it probably is. 
No idea how quick the convergence would be.
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Mic Qsenoch

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Re: Maths thread.
« Reply #853 on: June 08, 2017, 11:15:14 am »
+7

#1020304050601002004001000
20.00065
30.12330.003
41.596930.33280.07640.01970.00440.0017
57.366134.11742.39311.43360.89140.53790.07950.0004
620.1743819.113517.503915.787514.190312.73878.39813.1110.49550.0021
741.4785852.846360.087565.622769.880773.447183.058793.795999.015699.9947
820.1763519.131917.448415.669114.164112.738.38043.09130.48890.0032
97.366194.12212.4181.44890.86650.54390.08330.0014
101.593490.32950.07250.01850.00260.0007
110.123370.00350.0002
120.00063

Million trials each, except the 10 column (which was 10 million).
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Tables

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Re: Maths thread.
« Reply #854 on: June 08, 2017, 01:40:37 pm »
+2

First, what's the probability he rolled fairly?

Pretty much 0. This is the same kid who managed to get around 80% heads when flipping coins, and over 50% 6's when rolling a single die. It's almost certain he was either not actually rolling, or just ticking off 10's when it shouldn't have been.

Don't worry, we were pretty confident when talking about it that he didn't get the result legitimately :).

Anyway, really interesting seeing these simulation results. Out of curiousity, what are you using to simulate it - what language, and roughly what does the sim do?
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Mic Qsenoch

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Re: Maths thread.
« Reply #855 on: June 08, 2017, 01:49:15 pm »
+1

Anyway, really interesting seeing these simulation results. Out of curiousity, what are you using to simulate it - what language, and roughly what does the sim do?

Well I wrote mine in perl, but you could write it in anything with a random number generator. It just rolls 2 dice (gets a random integer from 1-6) and sums them and then adds one to a counter for that specific 2 dice sum. When one of these counters reaches whatever limit (like 20), that number is recorded as the winner. Then just do that a million times to get a million winners and keep track of how many times each number wins.
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Re: Maths thread.
« Reply #856 on: June 08, 2017, 03:14:10 pm »
+2

First, what's the probability he rolled fairly?

Pretty much 0. This is the same kid who managed to get around 80% heads when flipping coins, and over 50% 6's when rolling a single die. It's almost certain he was either not actually rolling, or just ticking off 10's when it shouldn't have been.

Statisticians hate him!
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Polk5440

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Re: Maths thread.
« Reply #857 on: June 08, 2017, 03:47:19 pm »
0

I wrote mine in STATA. Any language works. Now that I think about it, it is actually possible to do a simulation in Excel, but it's not super pretty.

1. Each row is a trial. Say you have 1000 rows. Each cell is the sum of two dice =RANDBETWEEN(1,6)+RANDBETWEEN(1,6). Do 220 columns by 1000 rows. 220 because you are guaranteed to hit 20 of some number by then. 
2. Now we need to identify the roll on which the 20th of some number is rolled. I don't know if you can do this in one step. The way I thought of doing it is the following procedure.
   2a. Copy/paste values so the numbers don't keep changing on you.
   2b. Find/replace 12 to k, 11 to j, 10 to i, 2 to a, 3 to b, 4 to c, 5 to d, 6 to e, 7 to f, 8 to g, 9 to h (in that order).
   2c. CONCATENATE each row into a single 240 character string. (Delete columns 1-220 keeping only the string which becomes the new column A.)
   2d. In columns B-L you will find the position at which each of 2-12 (coded as a-k) appears for the the 20th time. Use variations on the formula =FIND("z",SUBSTITUTE(A1,"a","z",20)). This substitutes "z" for the 20th "a" in the string then returns the position of "z". This is roll on which the 20th time "2" was rolled. Drag the formulas down the 1000 rows.
3. Now we need to identify the number that rolled 20 times first.
   3a. Copy/paste values.
   3b. Find/replace "#VALUE!" to 220.
   3c. In column M find the MIN number of rolls.
   3d. In column N find the number 2-12 associated with the minimum. I know you can do this with a 11-nested IF statement if you insert a row 1 that simply labels every column 2,3,...,12. Something like =IF(B2=M2,$B$1,IF(C2=M2,$C$1,IF(...etc...))). You can probably also do index/match or set up a vlookup or something.
4. Now you just need to count everything up! Use COUNTIF statements.
« Last Edit: June 08, 2017, 03:49:54 pm by Polk5440 »
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Re: Maths thread.
« Reply #858 on: June 10, 2017, 11:04:34 am »
0

Request: If you were to roll a pair of unbiased dice (2d6) and total the result, until you rolled the same total 20 times, what are the chances of each value being the one you reach 20 on?

This came from a lesson I was in yesterday on experimental probability. We'd gotten the pupils/groups to roll 2d6 and record their result until they got 20 of the same number, then collated results and plotted them to see the distribution and compare it to the expected result. Most of the groups got fairly reasonable results, several finished with twenty 7's, but one boy managed to get to twenty 10's before anything else, and in fact did so in a remarkably low number of rolls (73 IIRC). The teacher and I were discussing afterwards about how unlikely this was and how you'd calculate it, and neither of us were especially confident in how to do it. I suspected it would be the hypergeometric distribution, but I've honestly barely used that so I'm not sure.

Any takers? If not I'll have to go over to Reddit's theydidthemaths.
My first idea for getting the distribution was:
1) Calculate the probability that the various sums occur twenty times in a row
2) Take the sum
3) Multiply every probability with 1/sum

Does that work or am I overseeing something?
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Mic Qsenoch

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Re: Maths thread.
« Reply #859 on: June 10, 2017, 02:06:25 pm »
+1

My first idea for getting the distribution was:
1) Calculate the probability that the various sums occur twenty times in a row
2) Take the sum
3) Multiply every probability with 1/sum

Does that work or am I overseeing something?

You can't just ignore all the sequences where the number isn't rolled 20 times in a row, they aren't proportional between the different sums. Your method says reaching twenty 7s is (6/5)^20 ~= 38 times more likely than reaching twenty 6s or twenty 8s. But it's only about 2.7 times as likely.
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Re: Maths thread.
« Reply #860 on: July 07, 2017, 06:24:25 pm »
+1

Random thing I was thinking about at work: How many different triangles, up to similarity, are there with integer angles in degrees? So e.g. there's the triangle with angles 1, 1, 178, one with angles 1, 2, 177 etc.

Easier version: How many different isosceles triangles are there?

I made the answer to the main problem 2700, and the easier version 88.

Solution (feel free to check it's correct, or if you can't work out for yourself):

Let the angles in a triangle be a, b, c such that a<=b<=c. Consider two cases:

Case 1: a is odd. In this case, b can vary between a and 90 - (a + 1)/2, and of course c = 180 - a - b. b is bounded by this range due to the constraint a<=b<=c we assumed - it cannot go lower than a, and if it is equal to 90 - (a + 1)/2 then c = 180 - a - 90 + (a + 1)/2 = 90 - (a - 1)/2 which is the lowest it can go without being lower than b.

In this case then the range of b is 90 - (3a + 1)/2. This is also therefore the number of triangles with smallest angle a. So as a varies between 1 and 59 (and odd) we get the total number of triangles being sum from i=1 to 30 of [ 90 - (3(2i - 1) + 1)/2 = sum from i=1 to 30 of [ 91 - 3i ] = 1335

Case 2: a is even. Honestly, this proceeds very similarly to above if you follow the reasoning through, except b can vary between a and 90 - a/2. So we get that the total number of triangles being sum from i=1 to 30 of [90 - 3*2i/2 + 2] = sum from i=1 to 30 of [92 - 3i] = 1365

Add these two cases up, and you have all triangles with the smallest angle between 1 and 60, which covers all triangles, and there's a total of 2700 of them
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liopoil

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Re: Maths thread.
« Reply #861 on: July 07, 2017, 06:58:24 pm »
+1

Alternate solution:

We are counting unordered partitions of 180 into three parts. First we'll count the number of ordered partitions, adjust a few times to just count distinct ones.

Line up 180 dots in a line. To partition into three parts, draw two vertical lines in two of 179 spots to separate them. There are 89*179 ways to do this.

The equilateral triangle was counted exactly once.

Isosceles triangles can be made in three different ways. How many of these are there that aren't equilateral? 88, one for every even integer between 2 and 178 inclusive, except for 60.

Therefore there were 89*179 - 88*3 - 1 drawings that resulted in scalene triangles. Each of these was drawn 6 times, so after dividing by six we get that there were 2611 distinct scalene triangles. Add back the 89 other triangles to get 2700 total triangles.

I suspect that there may be an even simpler way still to count this...
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Re: Maths thread.
« Reply #862 on: July 08, 2017, 03:18:24 pm »
0

A fun (and tricky) problem I came up with at work about triangles is:

How many triangles satisfy the following properties:
1. All of the side lengths are integers.
2. The area of the triangle is equal to its perimeter.

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Re: Maths thread.
« Reply #863 on: July 08, 2017, 03:35:25 pm »
0

A fun (and tricky) problem I came up with at work about triangles is:

How many triangles satisfy the following properties:
1. All of the side lengths are integers.
2. The area of the triangle is equal to its perimeter.
Funny you would ask... my first inclination is to use... Heron's formula! Let S be half the perimeter. Heron's formula states that the area of the triangle is sqrt(S(S-A)(S-B)(S-C)) where the sides are A, B, and C. That's equal to 2S, apparently. Then:

4S^2 = S(S-A)(S-B)(S-C)
4S = (S-A)(S-B)(S-C)
16(A + B + C) = (-A + B + C)(A - B + C)(A + B - C)

Uh, then we need all integer solutions to this, and I suppose we also need to be able to make a triangle out of it (so if A < B < C, we need A + B > C). Not really sure where to go from here though...
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Re: Maths thread.
« Reply #864 on: July 08, 2017, 07:12:44 pm »
+1

No proof here, by my hypothesis would be that there are no solutions.

Some observations following from Liopoil's work:

Exactly one or three of A, B and C must be even. If zero or two are even, then the entire RHS of the last line is odd (odd + odd + odd = odd, odd + even + even = odd, odd * odd * odd = odd), and so definitely not a multiple of 16. From here, we also observe (A + B + C) must be even, and so the RHS must be a multiple of 32, or 2^5.

I don't know if considering powers of two leads anywhere but it at least feels a little promising.

At first I was thinking there might be no solutions, but there certainly is at least one: (6, 8, 10) = (A, B, C)

Considering the above point about powers of 2, we have:
2^4 | 16
2^3 | (A + B + C)

2^2 | (-A + B + C)
2^3 | (A - B + C)
2^2 | (A + B - C)

So it all checks out (as it must).

Here's another: (5, 12, 13) = (A, B, C).
2^4 | 16
2^1 | (A + B + C)

2^2 | (-A + B + C)
2^1 | (A - B + C)
2^2 | (A + B - C)

It seems like not all Pythagorean Triples (or multiples of them) generate a solution to this problem. Pythagorean Triples are of the form:

A=m^2-n^2 ,B=2mn, C=m^2+n^2

Maybe someone can do something using that?
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Re: Maths thread.
« Reply #865 on: July 08, 2017, 08:31:17 pm »
+1

Well, here a few more thoughts...

If (A,B,C) are positive integers satisfying 16(A + B + C) = (-A + B + C)(A - B + C)(A + B - C), then they must make a valid triangle. For if they violated the triangle inequality somehow, then exactly one term on the RHS would be negative while the LHS is positive. So we don't need to worry about that at all.

If the area A divides the perimeter P, say P = Ak, then after scaling sides of the triangle by k we have P' = Pk and A' = Ak^2 = Pk, so P' = A' in the new triangle. This doesn't really help with the above equations though.

Building on Tables' parity argument:

Let's look at the possibilities modulo 4. At least one of the RHS terms must be divisible by 4. Then if A,B,C are all even, then either one is divisible by 4 or all 4 are. In the case that only 1 is even we can't determine anything. Hmmm, not so helpful...

What if we looked at it modulo one of the side lengths. For example we have 16(B + C) = -(B + C)(B - C)^2 modulo A. That looks nicer but probably isn't actually too helpful... if gcd(A,B+C) = 1 it gets even simpler, (B - C)^2 = -16 modulo A... for example in the case (13,12,5) = (A,B,C) we have 7^2 = 10 = -16 modulo 13. Still not sure where this would go.

I'm guessing there are only finitely many solutions and most of them aren't right triangles.
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Re: Maths thread.
« Reply #866 on: July 10, 2017, 12:50:52 am »
0

Not sure if it helps / if trying to solve the Diophantine equation is the right path, but lets suppose A = dA', B = dB', and C = dC', where d is some common divisor of all 3. Then you get

16 * d * (A' + B' + C') = d^3 * (-A' + B' + C')(A' - B' + C')(A' + B' - C')
==> 16 * (A' + B' + C') = d^2 * (-A' + B' + C')(A' - B' + C')(A' + B' - C')

So the number of solutions is equivalent to the number of solutions to the above, where d is some integer and gcd(A', B', C') = 1.

You can also apply a similar thing to original statement. Suppose we have a triangle where perim = area and all side lengths share a common factor d. To remove any worries about integer-ness I'll work with the perimeter instead of the semiperimeter, then convert back.

2 * semiperimeter = sqrt(s * (s-a) * (s-b) * (s-c))
4 * (a+b+c) = 8 * semiperimeter = sqrt(p * (p-2a) * (p-2b) * (p-2c)) = sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c))
Let a', b', c' be the side lengths of the scaled down triangle, with p' and s' defined similarly.
4 * d * (a'+b'+c') = d^2 * sqrt((a'+b'+c')(-a'+b'+c')(a'-b'+c')(a'+b'-c'))
4 * d * p' = d^2 * sqrt(p' * (p'-2a') * (p'-2b') * (p'-2c'))
(perimeter of scaled down triangle) = p' = d * sqrt(s' * (s'-a') * (s' - b') * (s' - c')) = d * (area of scaled down triangle)

This shows the problem is equivalent to finding all triangles such that
1. Their side lengths are integers
2. Their side lengths have no common factors.
3. For some integer d, perimeter = d * area.

I think you should be able to argue something about the number of solutions for a given d, but I got stuck there.
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Re: Maths thread.
« Reply #867 on: July 12, 2017, 10:44:49 pm »
+1

I asked some other people with experience with contest math and they solved it pretty quickly. There are   5  triangles.

A hint that should help unblock things: Given any triangle with side lengths a, b, c, you can find positive x, y, z such that a = x+y, b = y+z, and c = x+z. Try working in (x,y,z) space instead.
« Last Edit: July 12, 2017, 10:47:50 pm by Titandrake »
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Re: Maths thread.
« Reply #868 on: July 12, 2017, 11:49:28 pm »
0

I asked some other people with experience with contest math and they solved it pretty quickly. There are   5  triangles.

A hint that should help unblock things: Given any triangle with side lengths a, b, c, you can find positive x, y, z such that a = x+y, b = y+z, and c = x+z. Try working in (x,y,z) space instead.

Yea, that's how I did it.
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Re: Maths thread.
« Reply #869 on: July 13, 2017, 12:44:08 am »
0

I asked some other people with experience with contest math and they solved it pretty quickly. There are   5  triangles.

A hint that should help unblock things: Given any triangle with side lengths a, b, c, you can find positive x, y, z such that a = x+y, b = y+z, and c = x+z. Try working in (x,y,z) space instead.
Huh. I guess that's vaguely motivated by the triangle inequality/the appearance of (A+B+C)/2 in Heron's?

Well, it does really simplify things:

4S^2 = S(S-A)(S-B)(S-C)
4S = (S-A)(S-B)(S-C)
4(x + y + z) = xyz

At least one of x,y,z must divide 4 (and so is either 1,2, or 4). WOLOG say it is x. Now:

xyz - 4(y + z) = 4x
yz - (4/x)(y + z) = 4
(y - 4/x)(z - 4/x) = 4 + 16/x2

If x = 1:

(y - 4)(z - 4) = 20
Solution pairs are (5,24), (6,14), and (8,9)

If x = 2:

(y - 2)(z - 2) = 8
Solution pairs are (3,10) and (4,6)

If x = 4:
(y - 1)(z - 1) = 5
The only pair is (2,6) and we already have this triple above.

The 5 solutions for x,y,z given above result in the triangles:

(6,25,29),(7,15,20),(9,10,17),(5,12,13),(6,8,10)
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Re: Maths thread.
« Reply #870 on: July 13, 2017, 09:10:48 pm »
0

It's motivated by turning a constrained problem into an unconstrained one. Instead of thinking about positive A,B,C that satisfy the triangle inequality, you can think about x,y,z that are positive with no other constraints.

You have the right answer but there are some missing edge cases in the proof you gave.

x,y, and z are not necessarily integers. If the perimeter is odd, then S is a half-integer, and x = S-A is also a half integer. For example, if A=B=C=1, then the values for x,y,z are x=y=z=0.5. So you need to argue about that case as well. Additionally, from 4(x + y + z) = xyz, at least one of x,y,z has to share factors with 4, but that doesn't mean it has to divide 4. For example, if x,y,z are all even, then both sides are divisible by 8, and you could have x = 8. You need to argue something about the smallest value of x,y,z.
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Re: Maths thread.
« Reply #871 on: July 13, 2017, 11:35:17 pm »
0

Ok wow yeah that bit was just nonsense. If A,B,C satisfied the previous equation they would automatically satisfy the triangle inequality though, so we didn't need to worry about that.

Say x <= y <= z. Say x >= 4, then let x = 4 + i, y = 4 + j, and z = 4 + k for non-negative j,k. Then:

 xyz = (4 + i)(4 + j)(4 + k) >= (16 + 4j + 4i)(4 + k) >= 64 + 16i + 16j + 16k > 48+4i +4j+4k = 4(4+i + 4+j + 4+k) = 4(x + y + z)

Which means no solutions. Then we can just check all x in {0.5, 1, 1.5, 2, 2.5, 3, 3.5} like I did before and get those solutions.

Showing that x < 4 was more messy than it really needed to be. The intuition is just that (x,y,z) = (4,4,4) already has LHS < RHS and increasing x,y, or z will increase the RHS at least as much as the LHS.
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silverspawn

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Re: Maths thread.
« Reply #872 on: July 23, 2017, 01:24:25 pm »
+1

I have a few questions about the fundamentals behind vectors. Maybe someone can help?

If say the vector space V is R^3, then there seem to be two different things. One is the actual elements of R^3, ordered triples, like (7, 2, 4) which would be something like the set {{7}, {7,2}, {7,2,4}}.

And the other thing is the coordinate vectors. Now, I think one of two things must be true:

A) coordinates aren't actually real mathematical elements, but are just notation for proper n-tuples. So if I have a basis B = ((7,2,0),(0,1,0),(0,0,2)) then the "vector" (1,0,2)_B (<- meaning B in index to signal that it's from the basis B) is actually just a different notation for the set {{7}, {7,2}, {7,2,4}}.
B) They are their own things and ought to have their own, distinct representation as sets.

So if it's A then, for one, isn't it totally bongus to write down basis in coordinates? If coordinates mean "a times this basis vector, b times this basis vector..." and you define a basis in terms of coordinates, you're referencing the thing you just want to define. It seems like you should always write down a basis as ordered n-tupels (I guess that's just sloppy, though why be sloppy if it takes the same amount of time?). And furthermore, writing any vector without a basis in the index would also not make sense but just be a quick notation if the basis is obvious, the same way we write 0 instead of the n-tupel with n zeros. But okay, so far no big deal.

What about matrices? I guess matrices themselves are fine, because they are mathematical things which are proven to form a ring with addition and matrix multiplication. Fine. But multiplying matrices with vectors? Given a linear function and a basis, we say there is a distinctly defined matrix which does the same as the function, and the matrix then only cares about the coordinates... which I thought weren't a real thing.  ???

But where it really gets confusing to me (and where I actually ran into real problems doing tasks) is when you do stuff like, let f : V -> V be an endomorphism that maps v -> A * v where A is a matrix. How is that legal if f takes proper n-tupels as arguments and A takes weird coordinate things? If say the matrix is ((0,0),(0,1)) and the vector is (1,1), couldn't you just say, well I choose the basis ((1,1),(0,1)) then my vector (1,1) has the coordinates (1,0), therefore Av = 0, whereas if I choose the standard basis, then A(1,1) = (0,1). Even if you "convert" your vector "back" by "applying" the coordinates, (0,0) is definitely not equal to (0,1).

And if it's B... well then what the hell are coordinates?

It seems to me that no-one ever explained this properly (fwiw I did hear people criticize the lecture for that), and we're just supposed to do what seems intuitive if the problems come up and the tasks will be designed in such a way that it'll work out. But that's super unsatisfying.
« Last Edit: July 23, 2017, 03:11:02 pm by silverspawn »
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fisherman

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Re: Maths thread.
« Reply #873 on: July 23, 2017, 02:40:54 pm »
0

First of all polar coordinates only really make sense in 2 dimensions, so there are a couple of options for what you mean by polar coordinates in R^3.

So let's stick to the plane. Basically the answer is what you call "B".

First let me take a two-dimensional real vector space V = R^2 with the standard basis whose elements I'll call x and y. Now take another two-dimensional real vector space W = R^2 with the standard basis whose elements I'll call r and theta. Now we define a map f : W --> V by f( ar + b theta) = (a cos b, a sin b).

Now for a point in V we can talk about its polar coordinates, although we are cheating slightly when we do. If we take a point p =(a,b) in V, really it IS the ordered pair (a,b). So when we say its polar coordinates are such and such, we really mean that we are giving the rectangular coordinates of a point q in W with f(q) = p. Since the map f is not injective, this is why polar coordinates are sometimes not well defined.
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silverspawn

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Re: Maths thread.
« Reply #874 on: July 23, 2017, 03:10:05 pm »
0

okay that's... not at all what I'm talking about. Maybe you don't call them polar coordinates. I'm talking about notation via base vectors. In any finite dimension, not just the plane.

I removed "polar" to make it less confusing. I definitely know them to be "coordinates" maybe I projected the "polar" from somewhere else. I know what polar coordinates you're talking about are which again, is not that.
« Last Edit: July 23, 2017, 03:12:06 pm by silverspawn »
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