computing a interpolating polynomial for some function

I am trying to understand how to use the commands polyfit and polyval so i am trying to interpolate some function, I wrote something so simple but even it is simple it gives me error
x=linspace(-1,1,20);
y= @(x) 1/(1+16*(x^2));
p = polyfit (x,y,2);
plot(x,p)
and this is the error message
Error using polyfit (line 48)
The first two inputs must have the same number of elements.
Error in polyfit_polyval (line 7)
p = polyfit (x,y,2);

 採用された回答

Torsten
Torsten 2022 年 1 月 26 日
編集済み: Torsten 2022 年 1 月 26 日

0 投票

x = linspace(-1,1,20);
yfunc = @(x) 1./(1+16*x.^2);
y = yfunc(x);
...
or
x=linspace(-1,1,20);
y=1./(1+16*x.^2);
...

9 件のコメント

Anas Gharsa
Anas Gharsa 2022 年 1 月 26 日
so the problem was the "."?? because x is a vector
Torsten
Torsten 2022 年 1 月 26 日
Yes, the dot after 1./ and the dot in x.^2.
Look up elementwise multiplication and division.
Anas Gharsa
Anas Gharsa 2022 年 1 月 26 日
okay! thank you so much
Anas Gharsa
Anas Gharsa 2022 年 1 月 26 日
now i tried what you said but it gives me error the same
x=linspace(-1,1,20);
y=1./(1+16*x.^2);
p = polyfit (x,y);
y1 = polyval(p,x);
plot(x,y,'o')
hold on
plot(x,y1,'-')
Error in polyfit_polyval (line 6)
p = polyfit (x,y);
Torsten
Torsten 2022 年 1 月 26 日
You didn't specify the degree of the fitting polynomial.
Anas Gharsa
Anas Gharsa 2022 年 1 月 26 日
okay my bad !!!!!! but why the graph looks like this
Torsten
Torsten 2022 年 1 月 26 日
It's the best you can get for a quadratic polynomial as interpolant.
Anas Gharsa
Anas Gharsa 2022 年 1 月 26 日
i understand now!! Thank you once again
Steven Lord
Steven Lord 2022 年 1 月 26 日
In your original code, the main problem was that y was a function handle but polyfit requires its first two inputs to be numeric arrays of the same size. Note that Torsten defined the function not as y but as yfunc then evaluated yfunc with x as the input to generate y. So the data that got passed into polyfit as its second input was the result of evaluating that function at the points in x rather than the function.
The use of the element-wise operators allowed the function to be evaluated on an array and return an array with the same size. The matrix form of the operator would require the inputs to be square matrices.
A = [1 2; 3 4]
A = 2×2
1 2 3 4
y = A.*A % each element of y is the square of the corresponding element of A
y = 2×2
1 4 9 16
z = A*A % this perform matrix multiplication instead of element-wise
z = 2×2
7 10 15 22

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