Having issues with fprintf statement. It gives me "error using vertcat"

4 views (last 30 days)
Stephanie Mendoza on 2 Oct 2022
clear
clc
a = 1.5;
b = 4.2;%lower and upper bounds
f = @(x) 2*x.^5 -x.^4 + 0.7*x.^3 - 2*x + 13;
I_exact = integral (f,a,b)
I_exact = 1.6393e+03
MPRTE = @ (est) abs ((I_exact -est)/I_exact)*100;
MPRAE = @(current, previous) abs((current-previous)/current)*100;
%first iteration
I11 = trap(f,a,b,1);
I12 = trap(f,a,b,2);
I21 = 4/3 * I12 - 1/3* I11;
MPRAE_I21 = MPRAE(I21,I12);
%2nd iteration
I13 = trap(f,a,b,4);
I22 = 4/3*I13 - 1/3*I12;
I31 = 16/15 *I22-1/15*I21;
MPRAE_I31 = MPRAE (I31, I22);
%3rd iteration
I14 = trap(f,a,b,8);
I23 = 4/3* I14 - 1/3*I13;
I32 = 16/15 * I23 - 1/15*I22;
I41 = 64/63 * I32 - 1/63 * I31;
MPRAE_I41 = MPRAE(I41,I32);
fprintf(['j \t k =1 \t k=2 \t k=3 \t k=4 \n' ...
'- ---------- ---------- ---------------\n',
'1 %4.5f %-4.5f %4.5f %4.10f\n', ...
'2 %4.5f %-4.5f %4.5f\n ', ...
'3 %4.5f %-4.5f\n', ...
'4 %4.5f\n'], I11,I12,I13,I14,I21,I22,I23, I31,I32,I41)
Error using vertcat
Dimensions of arrays being concatenated are not consistent.
function I = trap(func,a,b,n,varargin)
% trap: composite trapezoidal rule quadrature
% I = trap(function,a,b,n,p1,p2,...):
% composite trapezoidal rule
%
% Inputs:
% func = function to be integrated
% a,b = integration limits
% n = number of segments (default = 100)
% p1, p2,... = additional parameters used by func
% Output:
% integral estimate
% Feb 08, 2017
% Updated on Feb 13, 2018
if nargin<3, error('At least 3 input arguments required'), end %Error check to ensure 3 inputs are included
if ~(b>a), error('Upper bound must be greater than lower'), end % Ensure that b is greater than a. If not exit the function
if nargin<4 || isempty(n), n = 100; end %Ensure n is specified. If not set to 100
x = a;
h = (b-a)/n; %Compute step size h
s = func(a,varargin{:});
for i = 1:n-1
x = x+h; %Increment location of x
s = s + 2*func(x,varargin{:}); %Include the summation of all interior segments into s
end
s = s + func(b,varargin{:}); %Add the last term to s
I = (b-a)*s/(2*n); %Compute the integral approximation using the summation term.
end

Simon Chan on 2 Oct 2022
Missing dots in the second line, try the following
fprintf(['j \t k =1 \t k=2 \t k=3 \t k=4 \n'...
'- ---------- ---------- ---------------\n',... % <--- Missing dots
'1 %4.5f %-4.5f %4.5f %4.10f\n', ...
'2 %4.5f %-4.5f %4.5f\n ', ...
'3 %4.5f %-4.5f\n', ...
'4 %4.5f\n'], I11,I12,I13,I14,I21,I22,I23, I31,I32,I41)
Stephanie Mendoza on 2 Oct 2022
Thank you so much!!!

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