what's "error: x(7): out of bound 1 (dimensions are 1x1)"?

Question

Mohammad Ashralph Mohd Ashri 2023-1-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1887152-what-s-error-x-7-out-of-bound-1-dimensions-are-1x1

编辑： Dyuman Joshi 2023-1-22

clc
clear
% Define the data points
x = [1 4 6 5 3 1.5 2.5 3.5];
y = [0 1.3862944 1.7917595 1.6094379 1.0986123 0.40546411 0.91629073 1.2527630];
% Define the number of data points
n = length(x);
% Initialize the coefficient array
a = zeros(1,n);
% Set the first coefficient to the first data point
a(1) = y(1);
% Use the recursive formula to compute the coefficient
for i = 2:n
  for j = i:n
    y(j) = (y(j) - y(j-1))/(x(j) - x(j-i+1));
  end
  a(i) = y(i);
end
% Print the coefficients
disp(a)
% Define the function using the coefficients
f = @(x) a(n);
for i = n-1:-1:1
  f = @(x) a(i) + f(x).*(x - x(i));
end
% Approximate the values of ln(3), ln(5), and ln(3.256)
approx_3 = f(3);
approx_5 = f(5);
approx_3256 = f(3.256);
% Print the approximations
disp(approx_3)
disp(approx_5)
disp(approx_3256)
% Compute the errors of the approximations
error_3 = abs(approx_3 - log(3));
error_5 = abs(approx_5 - log(5));
error_3256 = abs(approx_3256 - log(3.256));
% Print the errors
disp(error_3)
disp(error_5)
disp(error_3256)

Here is an Octave script that uses the Newton's Divided Difference method to determine the parameters of an interpolating polynomial for a given set of data points and approximates the value of ln(x) for a given set of points.

and the output is as follows:

0 0.4621 0.0405 -0.2564 -0.0061 0.1040 0.1267 -0.1406

error: x(7): out of bound 1 (dimensions are 1x1)

error: called from

@<anonymous> at line 30 column 17

Project1 at line 34 column 10

I don't understand these errors and how to solve them, any help is appreciated.

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Answer 1

Dyuman Joshi 2023-1-2

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1887152-what-s-error-x-7-out-of-bound-1-dimensions-are-1x1#answer_1139557

编辑：Dyuman Joshi 2023-1-22

在 MATLAB Online 中打开

Use another letter/symbol for the function handle input variable, as x has already been defined as an array -

% Define the data points
x = [1 4 6 5 3 1.5 2.5 3.5];
y = [0 1.3862944 1.7917595 1.6094379 1.0986123 0.40546411 0.91629073 1.2527630];
% Define the number of data points
n = length(x);
% Initialize the coefficient array
a = zeros(1,n);
% Set the first coefficient to the first data point
a(1) = y(1);
% Use the recursive formula to compute the coefficient
for i = 2:n
  for j = i:n
    y(j) = (y(j) - y(j-1))/(x(j) - x(j-i+1));
  end
  a(i) = y(i);
end
% Print the coefficients
disp(a)
         0    0.4621    0.0405   -0.2564   -0.0061    0.1040    0.1267   -0.1406
% Define the function using the coefficients
f = @(k) a(n);
for i = n-1:-1:1
  f = @(k) a(i) + f(k).*(k - x(i));
                        %^ edit the bracket for k or x accordingly
end
% Approximate the values of ln(3), ln(5), and ln(3.256)
approx_3 = f(3);
approx_5 = f(5);
approx_3256 = f(3.256);
% Print the approximations
disp(approx_3)
   -0.6225
disp(approx_5)
    3.0363
disp(approx_3256)
   -0.4451
% Compute the errors of the approximations
error_3 = abs(approx_3 - log(3));
error_5 = abs(approx_5 - log(5));
error_3256 = abs(approx_3256 - log(3.256));
% Print the errors
disp(error_3)
    1.7211
disp(error_5)
    1.4268
disp(error_3256)
    1.6256