Subdividing intervals from unequally spaced numbers

Hi,

i want to generate points (say 11, variable xch) for a set of intervals defined by unequal spaced numbers (variable xn_ch). It seems that the code generates the points (xch) but with error, because xch vector dimension should not exceed the dimension of the initial generated numbers (xch, size=5). How to solve? Thanks in advance.

This is the code:

n=5

for i=1:n

xn_ch(i)=cos(pi*(2*i-1)/(2*n+2));

end

for i=1:n

xch=linspace(xn_ch(i),xn_ch(i+1),11)

end

xn_ch

This is the error:

Index exceeds the number of array elements. Index must not exceed 5.

Error in code (line 6)

xch=linspace(xn_ch(i),xn_ch(i+1),11)

Answer 1

[Manan Jain]

Hi!

The error you're encountering occurs because the loop index i in the second loop exceeds the range of valid indices for the xn_ch array. In the last iteration of the loop, i becomes 6, while xn_ch has only 5 elements.

n = 5;
for i = 1:n
   xn_ch(i) = cos(pi * (2 * i - 1) / (2 * n + 2));
end 

for i = 1:n-1
    xch = linspace(xn_ch(i), xn_ch(i+1), 11);
end

xn_ch

To solve this issue, you can modify the loop to iterate from 1 to n-1 instead of n.

I hope this helps!

[Angelo]

Many thanks.

The x variable dimesion is 11, the fixed amount of points in linspace. How to accumulate in a single variable all the points generated from each subinterval? Is it possible to increase the x dimension? From the output it looks like x points come only from the last interval [-0.2588 -0.7071]. I need all together the points generated from each subinterval.

[Voss]

n = 5;
xn_ch = cos(pi * (2 * (1:n) - 1) ./ (2 * n + 2))

xn_ch = 1×5

0.9659 0.7071 0.2588 -0.2588 -0.7071

How to accumulate in a single variable all the points generated from each subinterval?

One way is to put them in a matrix:

npts = 11;
xch = zeros(npts, n-1);
for i = 1:n-1
     xch(:,i) = linspace(xn_ch(i), xn_ch(i+1), npts);
end
disp(xch)

    0.9659    0.7071    0.2588   -0.2588
    0.9400    0.6623    0.2071   -0.3036
    0.9142    0.6174    0.1553   -0.3485
    0.8883    0.5726    0.1035   -0.3933
    0.8624    0.5278    0.0518   -0.4381
    0.8365    0.4830    0.0000   -0.4830
    0.8106    0.4381   -0.0518   -0.5278
    0.7848    0.3933   -0.1035   -0.5726
    0.7589    0.3485   -0.1553   -0.6174
    0.7330    0.3036   -0.2071   -0.6623
    0.7071    0.2588   -0.2588   -0.7071

[Angelo]

It works fine. Another question, hopefully my last, there is a double counting for the last point of each interval (the first for the next), see for example the value 0.7071 in the first column, repeated in the second and so on for the other two columns. I mean it is correct the subdivision (21 point for each subinterval) but the overall number of points is 84 with three more terms (i will use the numbers as x-values for a sum containing f(x) so i have addiction terms. How can avoid this problem?

[Angelo]

I found this:

b = unique(b(:).')

But is it possible to solve without adding new code rows?