Error: Index exceeds the number of array elements. Index must not exceed 1.

Index exceeds the number of array elements. Index must not exceed 1.

Error in bisectioniterations (line 35)

xr(i)=(xu(i)+xl(i))/2;

Unsure why this error is coming up in my code for bisection method of iteration.

clc;
clear all;
close all;
%Mini Project
%Bisection Method
syms f(x);
f(x)=(-4.868e-7)*(4*x^3)+(1.496e-4)*(3*x^2)+(-0.0094)*(2*x)+(-0.3324);
i=1;
xl(i)=-200;
xu(i)=200;
xr(i)=(xl(i)+xu(i))/2;
if f(xr(i))~=0
    while i<100
        i=i+1;
        if f(xl(i-1))*f(xr(i-1))<0
            xu(i)=xr(i-1);
            fu(i)=f(xr(i-1));
        else 
            xl(i)=xr(i-1);
            fl(i)=f(xr(i-1));
        end 
        xr(i)=(xu(i)+xl(i))/2;
        if i>1
            ea=abs((xr(i)-xr(i-1))/xr(i))*100;
            if ea<0.05
                break;
            end
        end
        i=i+1;
    end
end

Index exceeds the number of array elements. Index must not exceed 1.

figure(1)
plot(i,ea)
hold on
grid on
xlabel('Iteration')
ylabel('Percent Approximation Error')

y=-200:10:200;
figure(2)
hold on
grid on
box on
for i=1:1:length(y)
    plot(y,f(y),'LineWidth',2)
end
plot(xr,f(xr),'*','MarkerSize',10);
xlabel('Temperature (K)')
ylabel('Derivative of f(T)')
yline(0);
r_bisection=(-4.868e-7)*(xr^4)+(1.496e-4)*(xr^3)+(-0.0094)*(xr^2)+(-0.3324*xr)+14.55;          
et=abs((-13.2493-xr)/-13.2493)*100;

Answer 1

[Cris LaPierre]

The error means your are trying to index an element of your array that does not exist. The error is telling you that your array only contains 1 element, so your index is >1.

Here, xu and xl both start as scalars. Your if statement adds an ith value to only one of the vectors, so at any given time, one will have i elements, and the other can only have at most i-1 elements. However, the line of code throwing the error assumes they both have i elements.

You also index i twice - once at the top of your while loop, and again at the bottom.

I had to add som '.^' to your r_bisection equation to address a new error.

You will need to update your code to fix these issues. Here's a first pass:

clc;
clear all;
close all;
%Mini Project
%Bisection Method
syms f(x);
f(x)=(-4.868e-7)*(4*x^3)+(1.496e-4)*(3*x^2)+(-0.0094)*(2*x)+(-0.3324);
i=1;
xl(i)=-200;
xu(i)=200;
xr(i)=(xl(i)+xu(i))/2;
if f(xr(i))~=0
    while i<100
        i=i+1;
        if f(xl(i-1))*f(xr(i-1))<0
            xu(i)=xr(i-1);
            fu(i)=f(xr(i-1));
            xl(i)=0;
            fl(i)=0;
        else 
            xl(i)=xr(i-1);
            fl(i)=f(xr(i-1));
            xu(i)=0;
            fu(i)=0;
        end 
        xr(i)=(xu(i)+xl(i))/2;
        if i>1
            ea=abs((xr(i)-xr(i-1))/xr(i))*100;
            if ea<0.05
                break;
            end
        end
        % i=i+1;
    end
end

figure(1)
plot(i,ea)
hold on
grid on
xlabel('Iteration')
ylabel('Percent Approximation Error')

y=-200:10:200;
figure(2)
hold on
grid on
box on
for i=1:1:length(y)
    plot(y,f(y),'LineWidth',2)
end
plot(xr,f(xr),'*','MarkerSize',10);
xlabel('Temperature (K)')
ylabel('Derivative of f(T)')
yline(0);

r_bisection=(-4.868e-7)*(xr.^4)+(1.496e-4)*(xr.^3)+(-0.0094)*(xr.^2)+(-0.3324*xr)+14.55;          
et=abs((-13.2493-xr)/-13.2493)*100;

[Lily]

Any idea why ea comes up as NaN?

[Cris LaPierre]

xr(i) and xr(i-1) are 0, and 0/0 returns NaN.

i=2;
xr = [0 0];
ea=abs((xr(i)-xr(i-1))/xr(i))*100

ea = NaN

[Voss]

@Lily: I imagine you should make xl(i) equal to xl(i-1) (not zero) where needed, similarly for xu, and make ea a vector (i.e., store ea(i) each iteration), and it doesn't seem like you need fu or fl at all. Something like this:

clc;
clear all;
close all;
%Mini Project
%Bisection Method
syms f(x);
f(x)=(-4.868e-7)*(4*x^3)+(1.496e-4)*(3*x^2)+(-0.0094)*(2*x)+(-0.3324);
i=1;
xl(i)=-200;
xu(i)=200;
xr(i)=(xl(i)+xu(i))/2;
if f(xr(i))~=0
    while i<100
        i=i+1;
        if f(xl(i-1))*f(xr(i-1))<0
            xu(i)=xr(i-1);
            xl(i)=xl(i-1); % <-- use previous xl, not 0
        else
            xl(i)=xr(i-1);
            xu(i)=xu(i-1); % <-- use previous xu, not 0
        end 
        xr(i)=(xu(i)+xl(i))/2;
        % if i>1 % <-- i is always > 1 by this point, no need to check it
            ea(i)=abs((xr(i)-xr(i-1))/xr(i))*100; % <-- make ea a vector so you can plot it properly
            if ea(i)<0.05 % <-- use ea(i) here
                break;
            end
        % end
        % i=i+1;
    end
end


figure(1)
plot(ea) % <-- plots ea against 1:numel(ea)
hold on
grid on
xlabel('Iteration')
ylabel('Percent Approximation Error')

y=-200:10:200;
figure(2)
hold on
grid on
box on
for i=1:1:length(y)
    plot(y,f(y),'LineWidth',2)
end
plot(xr,f(xr),'*','MarkerSize',10);
xlabel('Temperature (K)')
ylabel('Derivative of f(T)')
yline(0);

r_bisection=(-4.868e-7)*(xr.^4)+(1.496e-4)*(xr.^3)+(-0.0094)*(xr.^2)+(-0.3324*xr)+14.55;          
et=abs((-13.2493-xr)/-13.2493)*100;