Graphics Programs Reference
In-Depth Information
It can be shown that fitting
y
i
directly (which involves the solution of a transcen-
dentalequation) results in
f
(
x
)
614
e
0
.
5442
x
. The corresponding standarddeviation
=
3
.
is
σ
=
1
.
022, which is very close to the result in Part(2).
EXAMPLE 3.10
Write aprogram that fits apolynomialofarbitrary degree
k
to the datapoints shown
below. Use the program to determine
k
that best fits this datainthe least-squares
sense.
x
−
0
.
04
0
.
93
1
.
95
2
.
90
3
.
83
5
.
00
y
−
8
.
66
−
6
.
44
−
4
.
36
−
3
.
27
−
0
.
88
0
.
87
x
5
.
98
7
.
05
8
.
21
9
.
08
10
.
09
y
3
.
31
4
.
63
6
.
19
7
.
40
8
.
85
Solution
The following program prompts for
k
.Executionisterminatedbypressing
“return.”
% Example 3.10 (Polynomial curve fitting)
xData = [-0.04,0.93,1.95,2.90,3.83,5.0,...
5.98,7.05,8.21,9.08,10.09]';
yData = [-8.66,-6.44,-4.36,-3.27,-0.88,0.87,...
3.31,4.63,6.19,7.4,8.85]';
format short e
while 1
k = input('degree of polynomial = ');
if isempty(k) % Loop is terminated
fprintf('Done') % by pressing ''return''
break
end
coeff = polynFit(xData,yData,k+1)
sigma = stdDev(coeff,xData,yData)
fprintf('\n')
end
The results are:
Degree of polynomial = 1
coeff =
1.7286e+000
-7.9453e+000
Search WWH ::
Custom Search