In the last section we looked at functions of one variable, in this section we are looking at functions of more than one variable, usually denoted x, y, z.
If we have one equation with one unknown then, in many cases, we can solve it. If we have two unknowns then we need at least 2 equations which apply to the same two unknowns.
For example, perhaps we know by some means that:
2x + 3y = 13
and at the same time we also know that:
4x + y = 11
Then we can calculate the values of x and y. There are several ways to do this, perhaps the most straightforward is by substituting one equation in the other. For instance from the second we can get:
y = 11 - 4x
We can now substitute this in the first to give:
2x + 3*(11 - 4x) = 13
multiplying out the terms gives:
2x + 33 - 12x = 13
20 = 10x
so x =2
We can then substitute this back into either equation, say:
4*2 + y = 11
so
y=3
and we have found the two variables x =2 and y=3.
Linear Functions
As explained on the polynomial page, there are two ways to define linear functions, some definitions allow a constant offset and some don't.
f(x) = a * x + b
or
f(x) = a * x
The second definition is an example of a linear map or linear operator. We can extend this for several variables:
f1(x,y) = a0 * x + a1 * y
f2(x,y) = a2 * x + a3 * y
This can be represented with vectors and matricies:
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= |
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as explained on these pages.
Bilinear Functions
A Bilinear function is a function of two variables which is linear when each variable is taken on its own (making the other variables constant).
A linear function could be represented like this:
f(x) = a * x
where:
- a = constant
- x = variable
So a bilinear function could be represented like this:
B(x,y) = x * y
because if x and y are made constant, in turn, then the equation is linear (like f(x) above).
This appears so simple that its hardly worth using a special name for it, however, when we are working with quantities made up of several scalars such as vectors then it becomes useful to define the properties of such functions.
A more mathematical way to define a bilinear operator in terms of vector spaces:
B: V x W -> X
where:
- B is the bilinear operator being defined
- V, W and X are vector spaces over the same field F
Which has the property of being linear when each variable is taken on its own (making the other variables constant).
A special case is the 'bilinear form', where X is the complete field F.
The following applies for the bilinear form:
for any a,b and c in V & W:
B(fa+b,c) = f B(a,c) + B(b,c)
B(a,f b + c) = f B(a,b) + B(a,c)
Possible additional assumptions:
- symmetric: B(x,y) = B(y,x)
- antisymmetric: B(x,y) = -B(y,x)
- alternating: B(x,x) = 0
Examples:
The bilinear operator can be used to analyse different types of multiplication, for example:
matrix multiplication
Is a bilinear map: M(m,n) x M(n,p) -> M(m,p)
x00 = v 00*w00 + v 01*w10
x01= v 00*w01 + v 01*w11
x10 = v 10*w00+ v 11*w10
x11 = v 10*w01 + v 11*w11
The dot product
Is a bilinear operator V x V -> F
x = v0 * w0 + v1 * w1 + v2 * w2
The cross product
In R3 is a bilinear operator R3 x R3 -> R3
x0 = v1 * w2 - w1 * v2
x1 = v2 * w0 - w2 * v0
x2 = v0 * w1 - w0 * v1
Quaternions
So is quaternion multiplication bilinear? the equations for multiplication are:
x0 = v0*w0 - v1*w1 - v2*w2- v3*w3
x1 = v1*w0 + v0*w1 + v2*w3 - v3*w2
x2 = v0*w2 - v1*w3 + v2*w0 + v3*w1
x3 = v0*w3 + v1*w2 - v2*w1 + v3*w0
When we are using quaternions for transforms we use two multiplications, P2=q * P1 * q', can we represent this as a bilinear function:
x0 = v3*v3*w0 + 2*v1*v3*w2 - 2*v2*v3*w1 + v0*v0*w0 + 2*v1*v0*w1 + 2*v2*v0*w2 - v2*v2*w0 - v1*v1*w0
x1 = 2*v0*v1*w0 + v1*v1*w1 + 2*v2*v1*w2 + 2*v3*v2*w0 - v2*v2*w1 + v3*v3*w1 - 2*v0*v3*w2 - v0*v0*w1
x2 = 2*v0*v2*w0 + 2*v1*v2*w1 + v2*v2*w2 - 2*v3*v1*w0 - v1*v1*w2 + 2*v3*v0*w1 - v0*v0*w2 + v3*v3*w2
Simultanous Quadratic Equations
One linear and one quadratic
Using the linear expression express one unknown in terms of the other and then substitute in the quadratic. This will give a quadratic with one variable only, which is solved in the usual way, then the other variable can be found by substiution as for linear simultanous equations.
Example 1
(1) 4x + 5y = 0
(2) 2 x2 + y x - y2 = 14
from (1) x = -(5/4)y
substitute in (2)
2 (-5/4 y)2 + (-5/4 y) y - y 2 = 14
(25/8) y2 - (5/4) y2 - y 2 = 14
25 y2 - 10 y2 - 8 y 2 = 112
7 y2 = 112
y = ±√112/7
y = ±√16
y = ±4
now substitute in (1)
4x + 5(±4) = 0
4x = ±20
x = ±5
Example 2
(1) x - y = 1
(2) x2 + y2 = 61
x = y + 1
(y + 1)2 + y2 = 61
y2 + 2y + 1 = 61
y2 + 2y - 60 = 0
y = (-2 ±√(22))/2
y = -6 or 5
x = -5 or 6
Simultanous Equations Involving Three Unknowns
The currents x,y,z amps in a circuit are given by the following equations, find the currents:
(1) 3x + 2y - 2z = 16
(2) 4x + 3y + 3z = 2
(3) -2x + y - z = 1
To eliminate z from (1) & (2) multiply (1)*3 & (2)*2 to give:
9x + 6y - 6z = 48
8x + 6y + 6z = 2
Subtracting these equations gives:
(4) 17x + 12y = 52
To eliminate z from (2) & (3) multiply (3)*3 to give:
4x + 3y + 3z = 2
-6x + 3y - 3z = 3
Subtracting these equations gives:
(5) -2x + 6y = 5
x can be found from (4) & (5)*2
17x + 12y = 52
-4x + 12y = 10
Subtracting gives:
x = 42/21
x = 2
substitute in (4) gives:
34 + 12 y = 52
y = 18/12
y = 1.5
substitute these values for x and y in (1)
6 + 3 - 2z = 16
-2z = 7
z = -7/2
z = -3.5
