Texas Instruments TI-92+, Список советов

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z9
=
a
$
x
3
2
$
y
3
2
+
b
$
x
3
2
$
y3
+
c
$
x3
$
y
3
2
+
d
$
x
3
2
+
e
$
y
3
2
+
f
$
x3
$
y3
+
g
$
x3
+
h
$
y3
+
j
In matrix form this system is written as
[3]
x
1
2
$
y
1
2 x
1
2
$
y1 x1
$
y
1
2 x
1
2 y
1
2 x
1y1 x1 y1 1
x1
2
$
y2
2 x
1
2
$
y2 x1
$
y2
2 x
1
2 y
2
2 x
1y2 x1 y2 1
x
1
2
$
y
3
2 x
1
2
$
y3 x1
$
y
3
2 x
1
2 y
3
2 x
1y3 x1 y3 1
x
2
2
$
y
1
2 x
2
2
$
y1 x2
$
y
1
2 x
2
2 y
1
2 x
2y1 x2 y1 1
x2
2
$
y2
2 x
2
2
$
y2 x2
$
y2
2 x
2
2 y
2
2 x
2y2 x2 y2 1
x2
2
$
y3
2 x
2
2
$
y3 x2
$
y3
2 x
2
2 y
3
2 x
2y3 x2 y3 1
x
3
2
$
y
1
2 x
3
2
$
y1 x3
$
y
1
2 x
3
2 y
1
2 x
3y1 x3 y1 1
x
3
2
$
y
2
2 x
3
2
$
y2 x3
$
y
2
2 x
3
2 y
2
2 x
3y2 x3 y2 1
x
3
2
$
y
3
2 x
3
2
$
y3 x3
$
y
3
2 x
3
2 y
3
2 x
3y3 x3 y3 1
$
a
b
c
d
e
f
g
h
i
=
z1
z2
z3
z4
z5
z6
z7
z8
z9
or, in terms of equivalent matrix variables
[4] X
$
a
=
z
We could solve this system for the coefficients, but that requires inverting the X matrix. We can avoid
inverting the matrix, and get the solution more quickly, if we scale the x- and y-coordinates such that
[5] x1
h
u1
=
0 x2
h
u2
=
1 x3
h
u3
=
2
[6] y1
h
v1
=
0 y2
h
u2
=
2 y3
h
v3
=
2
We scale the original x and y variables (at which to interpolate) with
and [7] u
=





x
−
x1
x2
−
x1
x
x
<
x2
x
−
x2
x3
−
x2
+
1
x
x
m
x2





v
=





y
−
y1
y2
−
y1
x
y
<
y2
y
−
y2
y3
−
y2
+
1
x
y
m
y2





With this scaling, the original 9x9 matrix becomes a matrix of constants, since all the u and v are
constants 0, 1 and 2, regardless of the actual values of x and y. Since the matrix is constant, its
inverse is constant, and we can calculate it once, in advance, and never need to calculate it again.
When we substitute the u's and v's for the x's and y's in the matrix, we have
[8] X
=
0 0 0 0 0 0 0 0 1
0 0 0 0 1 0 0 1 1
0 0 0 0 4 0 0 2 1
0 0 0 1 0 0 1 0 1
1 1 1 1 1 1 1 1 1
4 2 4 1 4 2 1 2 1
0 0 0 4 0 0 2 0 1
4 4 2 4 1 2 2 1 1
16 8 8 4 4 4 2 2 1
6 - 49