Question

does anyone know a way to solve a system of five linearly independent equations in five variables?

Answer 1

Easiest way is to use a matrix of 5x5 and find its reduced row echelon form, or row echelon form and back substitute.

Answer 2

I found a way to solve a system of linear equations using multidimensional spheres

Answer 3

WOW- I found a way to that with 2, 3 and even 4 unknowns but 5 is a bit much. I used Cramer's Rule to do that but even using that for even 3 unknowns gets a little cumbersome.

Answer 4

my method involves finding the center of a multidimensional sphere given a set of points and their tangent displacements. the center is the solution of the system of equations. notice that it is very cumbersome to calculate the solution of a set of linear equations for any more variables than three using traditional methods. using my algorithm, the equaltions solve themselves

Answer 5

So, you do have a way to solve for 5 equations in 5 unknowns?

Answer 6

If your method is correct, please educate us. :)

Answer 7

the question is whether this extends to n equations with n unknowns?

Answer 8

first, I shall define the matrix representation of a sphere in n dimensions
\[
m+1\;points\;are\;needed\;to\;define\;a\;sphere\;of\;m\;dimensions
\\x_{(m+1)k}:k^{th}\;component\;of\;m^{th}\;point\;(given)
\\d_m^2:tangent\;displacement\;of\;the\;m^{th}\;point\;(given)
\\x_{0k}:k^{th}\;component\;of\;center\;(to\;be\;determined)
\\R^2:radius\;squared\;(to\;be\;determined)
\\\sum^{n}_{k=1}\left(x_{(m+1)k}-x_{0k}\right)^2=R^2+d_m^2
\\\sum^{n}_{k=1}\left(x_{(m+1)k}^2-2x_{(m+1)k}x_{0k}+x^2_{0k}\right)=R^2+d^2_m
\]
now, subtract the equation with m=0 from all the others
the result is:
\[\sum^n_{k=1}\left(\left(x^2_{(m+1)k}-x^2_{1k}\right)-2\left(x_{(m+1)k}-x_{1k}\right)x_{0k}\right)=d^2_m-d_0^2
\\this\;is\;linear\;in\;the\;components\;of\;the\;center
\\so\;a\;matrix\;can\;be\;formed\;to\;solve\;for\;the\;center\;in\;terms\;of\;the
\\given\;points\;and\;displacements
\]
however at this point, this is no simpler than the direct method of solving the original system of equations. Fortunately, the spheres can be used to build the solution geometrically
I will post the result soon

Answer 9

just commenting to follow this

Answer 10

Thank you for your interest. the final equation in my previous post is linear in the coordinates of the center, so I can rewrite it in a more familiar form:
\[\sum^{n}_{k=1}2\left(x_{(m+1)k}-x_{1k}\right)x_{0k}=\sum^{n}_{k=1}\left(x^2_{(m+1)k}-x^2_{1k}\right)-\left(d^2_m-d^2_0\right)
\\\;\\\;\\If\;A\;is\;the\;coefficient\;matrix,\;then\;the\;cell\;of\;A\;at\
\\the\;j^{th}\;row\;and\;the\;k^{th}\;column\;is:
\\A(j,k)=2(x_{(j+1)k}-x_{1k})
\\\;\\\;\\If\;B\;denotes\;the\;constants\;vector,\;then\;the\;j^{th}\;component\;of\;B\;is:\;
\\B(j)=\sum^{n}_{k=1}\left(x^2_{(j+1)k}-x^2_{1k}\right)-\left(d^2_j-d^2_0\right)
\\\;\\\;\\The\;system\;of\;equations\;can\;be\;defined\;in\;tensor\;notation\;by\:
\\AX=B
\\\;\\\;\\Solve\;for\;X\;(center\;coordinates):
\\X=A^{-1}B
\]
this is the way to find the center using matrices,
there is a more elegant way to do it with geometry

Answer 11

calculators

Answer 12

computers

Answer 13

I wonder how CAS's do it.

Answer 14

What do you mean by 'tangent displacement'.

Answer 15

the distance from the point where the tangent intersects the circle to the point itself

Answer 16

So the m+1 points are not on the actual sphere, they are off the sphere, or randomly placed, but the tangential displacements are given?

Answer 17

What's the O time of your solution?

Answer 18

the points and their displacements are given. If the displacement is 0, then the point is on the surface of the sphere.

Answer 19

linear using recursion

Answer 20

\[ \Large x_{o_k} , x_{1_k} , x_{2_k} .. x_{m_k} \]

Answer 21

Nevermind, we are solving for the center.

Answer 22

bingo

Answer 23

\[ \Large x_{o_k} , x_{1_k} , x_{2_k} .. x_{m_k} , x_{m+1_k} \]

Answer 24

\[x_{0_k}\;is\;the\;k^{th}\;component\;of\;the\;center\]

Answer 25

I have a feeling we might need a double sum.

Answer 26

no, we need a system of equations

Answer 27

$$
\Large \sum_{i=1}^{m+1} \sum^{n}_{k=1}\left(x_{i_k}-x_{0_k}\right)^2=R^2+d_m^2
$$

Answer 28

each point other than the one chosen as the origin has its own equation

Answer 29

the term origin here is a little misleading because it does not actually need to be 0

Answer 30

\[x_{1k}\;is\;the\;k^{th}\;element\;of\;the\;origin\]

Answer 31

Are you changing your problem above. you used \( \large x_{o_k} \) to denote the center.
Sorry these subscripts are tripping me up.

copy pasted: $$m+1\;points\;are\;needed\;to\;define\;a\;sphere\;of\;m\;dimensions
\\x_{(m+1)k}:k^{th}\;component\;of\;m^{th}\;point\;(given)
\\d_m^2:tangent\;displacement\;of\;the\;m^{th}\;point\;(given)
\\x_{0k}:k^{th}\;component\;of\;center\;(to\;be\;determined)
\\R^2:radius\;squared\;(to\;be\;determined)
$$

Is this a correct edit, change m to m+1
$$x_{(m+1)k}:k^{th}\;component\;of\;m+1^{th}\;point\;(given) $$

Answer 32

yeah that seems right