product of two vectors with different dimensionality

[dedanoe]

A x B = (+i Sqrt[AxB]) x (-i Sqrt[AxB])

there's no scalar or cross product... vectors multiply alike sets of numbers.

[Cartesiantheater]

Would you please do the following vector multiplication using your method?


A = i + 2j + 3k

B = i + 2j - 3k


Step by step so I can see how you do it. Do the equivalent of both the scalar product and the vector product if you don't mind. I am always interested in new methods.

[dedanoe]

[A = i + 2j + 3k] and [B = i + 2j - 3k]

assuming i, j, k are the unit vectors, not the imaginary units of complex numbers... (which ain't much of a difference though)

[A = (1, 2, 3)] and [B = (1, 2, -3)]

first traditional scalar product of them two is 1 x 1 + 2 x 2 + 3 x (-3) = 1 + 4 - 9 = 4
this is not product of vectors but a mere orthogonal projection

then the traditional cross product of them two will be (2 x (-3) - 3 x 2, 3 x 1 - 1 x (-3), 1 x 2 - 2 x 1) = (-12, 6, 0)
this is not product of vectors but intersection of two equations in a system:
[equation1: 1 x + 2 y + 3 z = 0 and equation2: 1 x + 2 y - 3 z = 0 | how much are x, y, z]
you can rewrite the system as:
[equation1: 1 x + 2 y = - 3 z and equation2: 1 x + 2 y = + 3 z | how much are x, y]
x = (-3 z times 2 - 2 times 3 z) / (1 times 2 - 2 times 1)
y = (1 times 3 z - (-3) z times 1) / (1 times 2 - 2 times 1)
simplify that until
x = -3 times 2 - 2 times 3 = -12
y = 1 times 3 + 3 times 1 = +6
z = 1 times 2 - 2 times 1 = +/- 0

with A and B as they are, product would be [(i + 2j + 3k) times (i + 2j - 3k) = (i + 2j)^2 - (3k)^2]

my objection to this kind of multiplication is that only 1 is of such kind to be same on any degree
the product AxB cannot be set equal to C as if A, B, C are all linear numbers by their powers if C is not 1
in all the other cases AxB must be product of another two numbers like the product of their geometric middle (+i Sqrt[AxB]) x (-i Sqrt[Axb])
when three numbers enter product AxBxC then on the resulting side they exit as product of three other numbers
(+1 nroot(index = 1, degree = 3, value = AxBxC)) times
(-1/2 + i Sqrt[3]/2 nroot(index = 1, degree = 3, value = AxBxC)) times
(-1/2 - i Sqrt[3]/2 nroot(index = 1, degree = 3, value = AxBxC))

so what i propose is when taking the product of (1, 2, 3) and (1, 2, -3) accepting them two to be sets of numbers and do the multiplication on sets... for every number of the first set make him pair up with every number of the seconds set (((1, 1), (1, 2), (1, -3)), ((2, 1), (2, 2), (2, -3)), ((3, 1), (3, 2), (3, -3))) now for every pair (x, y) substitute it with (+i SQRT[x y], -i SQRT[x y]) then split them on two sets

((+i Sqrt[1], +i Sqrt[2], +i Sqrt[-3]), (+i Sqrt[2], +i Sqrt[4], +i Sqrt[-6]), (+i Sqrt[3], +i Sqrt[6], +i Sqrt[-9]))

((-i Sqrt[1], -i Sqrt[2], -i Sqrt[-3]), (-i Sqrt[2], -i Sqrt[4], -i Sqrt[-6]), (-i Sqrt[3], -i Sqrt[6], -i Sqrt[-9]))

A times B exits as A' times B' exists as A'' times B'' exits as ...
this are the transformations not the differentials

the product of father A and mother B is child that has the properties of both A and B.
for the father the child is +i sqrt[AxB] for the mother it is -i Sqrt[AxB]

if vectors are lines then product of two of them is surface, product of three of them is volume and tnn, tnn...



it is like my famous equation 1^0 = A = O^1 that made me the author of all literacy... the powers of Me 1 and my mom O are different but their final value daddy A is the same. the left eye has three abilities and the right has five together united they make the 15 pairs of networking third eye. the true source of American power is networking.

[Cartesiantheater]

What is the practical use of this? The traditional vector product is essential to the physical sciences, particularly physics and engineering. Would you eliminate it or merely redefine it as something other than multiplication? (by the way, you do make an interesting argument)

[dedanoe]

has anybody ever questioned how did simple numbers came to produce complex numeric formats like matrices ??? the traditional vector product is political version of systemic solution: equaiton1 means that institution1 of the system will be willing to participate in the political project if this, this and this interest of theirs is met, equation2 means institution2 will do the similar for similar reasons... the systemic solution that is intersection of all the interest functions of the institutions of the system, is the politician's only way to pay all the parties and have the project finished in joined effort.

what lead me to think of this is the intention of mine to compose a general lever simulator "from atoms to galaxies and back and forth in all the parallel senses" and then i had to define a matrix as information type class of (numbers: array of array of complex and submatrices: array of array of matrix class). then i said to me how do i jump the vector, are not the matrices = vectors of vectors, then vectors of vectors can be of additional vectors of vectors. so i chose to work with vectors of vectors, i must define then the operations with them like addition, subtraction, multiplication, division, powering and rooting, get me the orthogonal and unitary data of this type for how do i do me the conservative rotation.

i got me a new way of multiplying and dividing vectors recently when i was working on my extension of Regula Falsi on NxN system of equations ( https://sites.google.com/site/dedano...nearSystem.zip | the code is in unit1.pas i think ) : let A be (a1, a2, a3) and B be (b1, b2, b3) i am looking for A over B as vector result... i can random define me matrix S of type array [1..2, 1..2] of same data type as A and B and project A on every vector of that matrix S to get the A' matrix [1..2, 1..2] of the projections then project B on that same S matrix of vectors to get me the B' matrix of projections for B. then i have projection_matrix A' = some_matrix M times projection_matrix B' | solve M ??? then M[1,1] x S[1,1] + M[1,2] x S[2,1] = Result of how much is A over B. i don't know if the solution is unique and should it be at all.

[dedanoe]

now that i recalled on my general regula falsi... Newton's tangent method says: x = x0 + f(x0) / f'(x0) but Tyler's expansion series say: f(x) - f(x0) = f'(x0) (x-x0)^1 / 1! + f''(x0)(x-x0)^2 / 2! + f'''(x0)(x-x0)^3 / 3! + ... + O[x0] so rounding it up to the third differential we have a cubic polynomial by (x-x0) which resolved gives three cube roots where we make f(x) = 0. am asking will this convergence be significantly faster adding to it, that much higher polynomial powers can still enter the game ??? how greater is the karatness degree of the qur'an root taken from a number that much faster it results with 1.

it works indeed... Higher_Degree_Tyler_Approximator.zip

[longyueu]

i don't know if the solution is unique and should it be at all.

[Cartesiantheater]

i don't know if the solution is unique and should it be at all.

For the cross product we use the Right-hand rule convention for that, as I'm sure you know.