Texas Instruments TI-89 Tip List page 253

s(x) is the splice function
In general, the splice function is
and the first derivative is
x is the point between x
2
There are several choices for s(x
the point of intersection, and set s(x
choosing a value for x
interval, we may set s(x
We solve for the five coefficients of the splice polynomial by imposing five conditions:
f 1 x 1 = s x 1
1.
f 2 x 3 = s x 3
2.
s x 2 = y 2
3.
d f 1 x 1 = d
4.
dx
d f 2 x 3 = d
5.
dx
so we solve these five equations for a, b, c, d and e:
a $ x 1 4 + b $ x 1 3 + c $ x 1 2 + d $ x 1 + e = f 1 x 1
a $ x 3 4 + b $ x 3 3 + c $ x 3 2 + d $ x 3 + e = f 2 x 3
a $ x 2 4 + b $ x 2 3 + c $ x 2 2 + d $ x 2 + e = y 2
4 $ a $ x 1 3 + 3 $ b $ x 1 2 + 2 $ c $ x 1 + d = f 1 ∏ x 1
4 $ a $ x 3 3 + 3 $ b $ x 3 2 + 2 $ c $ x 3 + d = f 2 ∏ x 3
This set of five equations in five unknowns can sometimes be solved, but just as often they cannot. As
we typically fit the splice over a narrow range of x, a singular matrix error often results from loss of
precision in solving for the coefficients. Even if the singular matrix error does not occur, the solved
coefficients are large and alternating in sign, which means that the polynomial is numerically unstable.
This problem is avoided by finding the splice polynomial as a function of a scaled value of x
x itself. As usual, proper scaling simplifies the solution. For example, suppose we set x
and x = 1, where x , x
3s 1s
polynomial is now s(x
s
s ( x s ) = a $ x s 4 + b $ x s 3 + c $ x s 2 + d $ x s + e
Above, we defined h = x
s ( x ) = a $ x 4 + b $ x 3 + c $ x 2 + d $ x + e
s ∏ ( x ) = 4 $ a $ x 3 + 3 $ b $ x 2 + 2 $ c $ x + d
and x where we specify some desired value for the splice function s(x).
1 3
). For example, if f
2
) = f (x
2 1 2
slightly away from the intersection. If the functions do not intersect on the splice
2
) between the functions so the splice passes smoothly from one to the other.
2
The splice matches f
The splice matches f
We set the splice value at x
Set the first derivatives equal at x
s x 1
dx
Set the first derivatives equal at x
s x 3
dx
and x are the scaled values of x
2s 3s
):
- x ; now we define a scaled interval half-width h
2 1
(x) and f (x) intersect, we may want to set x
1 2
) = f (x ). Alternatively, the splice may work better by
2 2
at x ; there is no discontinuity
1 1
at x ; there is no discontinuity
2 3
2
(condition 1)
(condition 2)
(condition 3)
(condition 4)
(condition 5)
, x and x
1 0
to get some desired behavior
for a smooth transition
1
for a smooth transition
3
1s
, respectively. The splice
3
:
s
[1]
[2]
as
2
[3]
[4]
[5]
[6]
[7]
, instead of
s
= -1, x = 0
2s
[9]
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