Hard Maths Question.

[Sam Finch]

Sorry to ruin the forum with hard sums, but I can't find the answer to this one.

if we have 4 numbers x,y,a and b where

e^y = e^a + e^b

i'm trying to prove that

e^(x+y) = e^(x+a) + e^(x+b)

however, x,y,a and b are not normal numbers, they are hypercomplex numbers (quaternions or larger) this means that we can no longer assume that a*b = b*a, and hence e^(a+b) is no longer e^a * e^b.

I can't test this as I can't calculate e^x for these numbers and I want to know If I can assume this Identity for non commutative numbers.

does anyone know anything about this?

I am so dissapointed in you sam,
how can you not know the answer?
I am not going to tell you, you have to prove that you can figure it out..... *tsk tsk*

[Guv]

Sam, I think I can get you started for quaternions, but not for anything worse. I intend to be a bit sloppy about the use of * for multiplication, sometimes omitting it when the multiplication is implied by context. I vaguely remembered the general idea from somewhere in my past, and managed to work out the following. I left out some details that you can verify for yourself if you do not trust me (I do not advise trusting me all the time).

First, let Q = A + iB + jC + kD be a quaternion. (i^2 = j^2 = k^2 = -1) (ij = -ji = k, jk = -kj = i, ki = -ik = j) You can work out e^Q as follows.

0) (A + iB + jC + kD)*(A - iB - jC - kD) = A^2 + B^2 + C^2 + D^2, which can be used to clear denominators of quaternions in case you need to deal with division, which I do not think you will need.

1) A commutes with the other stuff, so e^Q = e^A*e^(iB + jC + kD). The following deals with e^(iB + jC + kD).

2) If N^2 = -1 and X is real, then e^(N*X) = cos(X) + N*sin(X), You can check this via the normal expansion of e^(N*X). The even powers of N are either 1 or -1, giving the series for cosine. N can be factored out of all the odd powers, giving N*(series for sine, after replacing even powers with 1 or -1).

3) Let M be magnitude of (iB + jC + kD), M = sqr(B^2 + C^2 + D^2).

4) Let L = (iB + jC + kD)/M, which results in L^2 = -1 You can work this out.

5) Then (iB + jC + kD) = L*M, where L^2 = -1 and M is real.

6) This results in e^(L*M) = cos(M) + L*sin(M)

7) Finally, e^Q = e^A*[cos(M) + L*sin(M)], where M = (B^2 + C^2 + D^2), and L = (iB + jC + kD)/M

Now that you know the exponential of a quaternion, perhaps you can prove or disprove your conjecture. Working out the above tired me out for the time being.

Let me know if the above helped. Ask questions if you want. By tomorrow I will be ready & willing to work on your conjecture if you need more help.

[Sam Finch]

Guv

Thanks, that's roughly where I'd got up to on my own, what I'm actually doing is trying to reduce square matricies into sets of hypercomplex numbers in order to extend the set of operations that can be performed on them.

so we can reduce a 2*2 matrix into something a bit like a quaternion by defining

I = [1 0]  J = [0  1]  K = [-1 0] L = [0 1]
    [0 1]      [-1 0]      [0 -1]     [1 0]


and hence J^2 = JKL = -I
          K^2 = L^2 =  I

which are incredibly similar to quaternions (it was a short documentry on Hammilton and quaternions that gave me the Idea) so you can work out pretty much all functions of 2*2 atricies in a similar way that you can with quaternions.
(there are a couple of special cases eg Q = a + bJ + cK + dL where (bJ + cK + dL)^2 = 0 which make it tricky but essentially the algebra is the same (and you have to assume a,b,c and d are complex as well otherwise things get very strange) but apart from that it's a simple and powerful way of doing algebra with 2*2 matricies.

The problem comes in extending this to larger matricies, the sensible way of doing it seems to be looking only at (2^N)*(2^n) matricies, and defining

M°I = [M 0]  M°J = [0  M]  M°K = [-M 0]   M°L = [0 M]
      [0 M]        [-M 0]        [0 -M]         [M 0]

for a (2^N)*(2^N) matrix M

which can be treated in a similar way to the above s.t.
Q = A°I + B°J + C°K + D°L where A,B,C and D are (2^N)*(2^N) matricies.

a few of Identities for the ° operator

A°B + C°B = (A+C)°B
A°B + A°C = A°(B+C)
(A°B)(C°D) = (AC°BD)
e^(M°Q) = Cos(M) + Sin(M)°Q iff  Q^2 = -I

(these are fairly easy to proove but not in ascii)

so we can still treat large matricies in a similar way to quaternions. The trouble comes because we loose comutitivity and can no longer say that for (2^N)*(2^N) matricies B,C and D (B°J + C°K + D°L) is a (2^N)*(2^N) matrix.

so this is where I'm stuck. and I'm looking for other axioms that I can work with, but as the only training I've had in hypercomplex numbers was a 10 minute documentary which explained what quaternions were and how to multily them and I havn't found any good resources on this i'm not making very much progress.

I reckon most Identities for general quaternions will work for larger sets, as long as they don't involve splitting them up into component parts.


anyway, that's what I'm trying to do, I probably ran through it a bit quickly but I hape it made some sense. Basicly any kind of Identities you still remember would be a huge help.

I was going to suggest http://mathworld.com, but they seem to be having legal issues.

[Guv]

Sam, it is not clear to me exactly what you are trying to do.

First piece of advice: Quaternions must be defined in terms of real numbers and i, j, & k. For the quaternion (a + ib + jc + kd): a, b, c, & d must be real. If not, you must do algebra to reduce what you have to standard form with real values for a, b, c, & d.

One approach to dealing with quaternions would be to define a quaternion type and a set of Functions for various operations using quaternion variables as arguments. This is likely to be the most efficient way to go, both in memory usage and speed. I would expect messy brute force code to be the only more efficient technique.

One of my future projects is to learn to use VB Classes. Yonatan posted something which gives me a good clue. Meanwhile, I am not sure about this, but it seems to me you could define a Quaternion Class and a bunch of properties/operations.

If neither of the above interests you, I do remember the following from somewhere in my past.

Matrix operations using complex numbers can be implemented using 2X2 sub-matrices. Consider transforming a complex number (a + ib) into the following skew symmetric 2X2 matrix.

A11 = a A12 = b
A21 = -b A22 = a

Addition, subtraction, & multiplication of such matrices is isomorphic to complex addition, subtraction, & multiplication.

Furthermore, if you (for example) convert a 10X10 matrix with complex elements to a 20X20 real matrix using the above transformation algorithm, you can use real matrix software to do complex matrix operations. For example: If you use matrix inversion software designed for real numbers on the 20X20, the result will correspond to the inverse of the 10X10 complex matrix. Similarly for the use of simultaneous linear equation & eigenvalue/eigenvector software.

I used the above technique many years ago on a mainframe when software was required in a hurry to process a bunch of matrices with complex elements. It worked fine.

The above suggests to me that an analogous scheme for quaternion arithmetic might require the use of 4X4 matrices, instead of 2X2 matrices as suggested in your last post. I did some algebraic work and believe that the 4X4 corresponding to Quaternion (a + ib + jc + kd) is the following.

| a b c d |
| -b a -d c |
| -c d a -b |
| -d -c b a |

I believe that addition, subtraction, & multiplication of matrices transformed as indicated above is isomorphic to quaternion addition, subtraction, & multiplication. I am pretty sure that I got the above correct, but am not certain. At any rate, I expect a skew symmetric 4X4 to be the correct transform for a quaternion. There might be several 4X4 transforms that would work. There are more degrees of freedom here that for the 2X2 transform of a complex number, and I think there are two skew symmetric 2X2 transforms that would work for complex numbers.

I have never tried the above for quaternion arithmetic, but really think that it would work. The 4X4 matrices seem to be correct. I know that "partitioning" a matrix in a regular fashion and treating each sub-matrix as an element of a matrix of lesser order works. So I think you could use real matrix software to invert quaternion matrices and/or solve systems of linear equations with quaternion coefficients.

Perhaps the above suggested technique might work for whatever you have in mind. It is slick, easy to implement, and should result in well structured code, but obviously requires more memory and CPU time than other techniques.

By the way, did you prove or disprove your original conjecture yet? My guess is that it will be disproved. This is just a guess: I did not make any attempt to prove or disprove it.

Keep me up to date on your efforts.

[OB1]

What type of mathematics are you all talking about? This seems a bit higher than algebra. =)

[Guv]

OB1, following are a few clues. First this is only a little trickier than algebra. If you coped with algebra and spent some time concentrating on this stuff, it would not seem so tuff.

Mathematicians got upset well over 100 years ago because simple equations like "X^2 + 1 = 0" had no solution. So they invented something called complex numbers.

The idea is that "i" is defined as the square root of -1. Using this definition, a complex number is (a + ib) or (x + iy). where a, b, x, & y are ordinary numbers, and i is the square root of minus one. There are definitions for addition, subtraction, multiplication, and division of complex numbers.

The above worked pretty good, and was useful for many practical purposes.

Quaternions are an extension of the above idea. A quaternion is (a + ib + jc + kd), where a, b, c, & d are ordinary numbers; i^2 = j^2 = k^2 = -1; i*j = k, j*k = i, & k*i = j. The tricky gimmick here is that j*i = -k, k*j = -i, i*k = -j. Changing the order of multiplication of i, j, k changes the sign of the product.

It seems weird at first, but it works. There are definitions for add, subtract, multiply, & divide of quaternions.

Ahhhh I see,
that was a lot easier to understand then when sam tried to explain it to me a few months ago.....

[Sam Finch]

Guv,

Sorry, I didn't make myself clear, I'm actually doing the opposite of what you suggested.

I'm trying to evaluate the exponent of a matrix, it would seem none of the current methods of doing this are satisfactory which is why I'm doing it in a crazy way.

the numbers I have defined are not quaternions, but have similar properties (for the basic set anyway) they are made up of 4 complex numbers. and take the form

Q = a + bJ + cK +dL where a,b,c and d are complex numbers
and
J^2 = JKL = -1
K^2 = L^2 = 1
these do in fact have an almost identical algebra to quaternions, although a bit more complicated in places, and we do need the coefficiants to be complex otherwise it becomes even more difficult.

The reason I defined these is that they are isomorphic to 2*2 matricies, as any 2*2 matrix can be represented by a combination of I,J,K and L above

eg

[2 9]   =   [3 0]  +   [0  4]    +   [-1 0]    +   [0 5]
[1 4]       [0 3]      [-4 0]        [0  1]        [5 0]

so we can very easily define e^M for a 2*2 matrix M because they are isomorphic to our defined set of hypercomplex numbers.

So that's all lovely, although not groundbreaking as e^M has already been defined for small matricies, so I'm trying to extend this Idea to larger matricies. All current methods of evaluating e^M involve finding the eigenvalues of M, so a better method is certianly needed (you can't find the eigenvalues of a matrix by looking at it's exponent BTW, so I don't think we need to solve the characteristic equation to find it)

So basicly I'm not trying to do hypercomplex arithmetic using matricies, I'm trying to do matrix arithmetic using hypercomplex numbers.

For larger matricies I am as you say trying to partition the matrix into Sub Matricies. So larger matricies are isomorphic to sets of hypercomplex numbers

M = a + bJ + cK + dL where a,b,c and d are sub matricies.


and then if we can compute e^M for the general hypercomplex number in terms of it's coefficiants we have an elegant recursive method of finding the exponent of a matrix and hence solving large sets of differential equations.

NB, I'm not trying to do this for any particular reason, I'm just one of those people who does maths for fun.

I havn't found anything conclusive about my conjecture yet, nor anything helpful, I've actually been looking at the problem for a few weeks without finding anything. Basicly I'm trying to find rules about what e^(Q1 + Q2) is for non comutitive multiplication, and I can't really find any.

[OB1]

Thank you so very much for the help Guv. You made it all alot clearer. You must have a Mathematics background?

[Guv]

Sam, your definitions of J, K, & L seem incomplete and/or inconsistent.

You refer to I, J, K, & L.

I assume that I^2 = -1, and is the one that is used with complex numbers. You provide the following 3 statements.

Q = a + bJ + cK + dL, where a, b, c, & d are complex.
J^2 = JKL = -1 (This implies J = KL}
K^2 = L^2 = 1 (Implies K = 1 or K = -1, and same for L)

The above do not seem consistent, and are certainly not a
complete set of definitions. Furthermore, If a, b, c, & d can be complex numbers, you have not expresses Q in a canonical form, which seems like an incorrect basis for what you are doing.

At least as a start toward a canonical form:
Why not Q = x + jI +bJ + cK + dL, where x and y are real?

What about products like JK, KJ, JL, LJ, IJ, JI, et cetera?

There seem to be 16 possible products, and you have omitted quite a few. For example to work with dL, where d is complex: You need to know what the product of I and L.

Also, what is the transform rule for your isomorphisms?

Am I missing something? Have you left something out?

[Sam Finch]

This is one of the problems with maths, you run out of letters very quickly. and it's quite hard do define things properly without symbols.

I,J,K and L

I, whaen used is the Identity matrix, not i as in i^2 = -1, we use it when working with matricies.


JK and L are not real or complex, and none are equal to eack other.

a Hyprecomplex number a + bJ + cK + dL defines a point in 8D space, the basis for this space is the set of vectors

{1, i, J, Ji, K, Ki, L, Li}

(in this case i^2 is -1 as in standard complex numbers)

the algebra of these things is quite tricky, but it's similar to that of quaternions. and it is closed and complete (apart from the reciprical of 0).

using the matricies defined in my second post as J,K and L
and I as 1, we see that these hypercomplex numbers are isomorphic to 2*2 matricies.


We call this set of numbers X1

to deal with 4*4 matricies we define the ° operator (I'm sure there's one like this but I havn't seen it so I'm using °.)

consider a 2*2 matrix M = [p q] 
                          [r s]

where p,q,r and s are real or complex numbers.

for any matrix A

A°M = [pA qA]
      [rA sA]

° is defined similarly for larger M.

so we can express any 4*4 matrix as a°I + b°J + c°K + d°L where a,b,c and d are 2*2 matricies, or members of X1, we call this set of hypercomplex numbers (in 32D space) X2.

similarly we can define sets of Hypercomplex numbers X3 for 8*8 matricies, X4 for 16*16s etc.

hopefully I've sort of explained that, basicly we have sets of hypercomplex numbers defined that are isomorphic to (2^N)*(2^N) matricies. and for any q in the set Xn

q = a°I + b°J + c°K + d°L

where a,b,c and d are elements of X(n-1)

the trouble I'm having is defining exponents etc of numbers in X2 and up, what I'm trying to do is something along the lines of.

q = a°I + b°J + c°K + d°L
  = (a°I + b°J)(1°I) + (ci°I + bi°J)(-i°K)

and as (-i°K)^2 = (-i^2)°(K^2) = -1°I (which is equivilent to -1) we can calculate e^q if we can prove that the Conjecture above holds (because cos and sin are defined in terms of exponentials)


I hope that made some sense. It is right, but hard to explain, especially without by beloved symbols.

[Guv]

Sam, I cannot help thinking that I am getting an incomplete picture. What I see seems somehow strange.

Your definitions bother me.

Your definition of K is a negative 2X2 identity matrix. That is: K = -I. This suggests to me that something is basically wrong with your approach (if this is what you want, why not call it -I?).

Also, you say that JKL = I, but it looks to me like

[-1  0]
[ 0  1]

I do not think your 2X2 matrices behave much like quaternions. JL = -LJ (like quaternions), but the products are not K & -K (unlike quaternions). JK = KJ = -J (since K = -I) very unlike quaternions, for which multiplication is not commutative.

[Sam Finch]

Sorry, that was a typo that got cut and pasted around without me noticing,

I = [1 0]  J = [0  1]  K = [-1 0] L = [0 1]
    [0 1]      [-1 0]      [0  1]     [1 0]

which does work, so now JKL = -I etc. which is enough to prove everything else from.

[Kc0bfv]

I am glad this is not part of my high school AP calc class. What do I take to learn this stuff (in more detail)?

[parksie]

Torture studies probably helps

We don't do matrices as part of the Maths A-Level any more I'm only doing it because I'm taking Further Maths as well...

Will have to borrow some thick dusty books and read more...

[Ianpbaker]

My head hurts

I got an A in maths GCSE when I was at school and was really into this sort of this stuff. But now 6 years down the road, it makes as much sense to me as japaneese

[Kc0bfv]

I'm guessing you're not Japanese

Not to imply that I am, of course

[Sam Finch]

Don't be put off just because you don't understand this, Basically it was me trying to explain something I was working on without fully understanding it myself and not knowing the right terms for various bits of it. I guarantee that no matter how much maths you learn you will never have to understand the method here, because it doesn't work, If you ever did have to understand it it would be better explained.


Just to let you know that I managed to adapt this method to something that did work, but there are matrices it doesn't work for and there are much better ways of doing the same thing anyway, but it wasn't a complete waste of time.

[X-Quest]

Geeeeez, I was a math major in college, made it through Calc 3, Linear algebra, numeric analysis, etc. I'm glad I gave this crap up ! Good luck gents, I have burned the book of Applied Differential Equations, and have move on the the "...for Dummies" series.