Why does the numeric error in my implementation of Runge-Kutta not decrease like N^a?

Question

I'm trying to determine how many steps it takes for the Runge-Kutta Method ("RK4") to be within 0.01% of the exact solution of an ordinary differential equation. I'm comparing this to the Euler method. Both should result in a straight line on a loglog plot. My Euler solution seems to be correct but I am getting a curved solution for RK. They are based on the same code so I am completely confused about the problem.

EDIT: Sorry for removing the wikipedia links. It wouldn't let me keep more than one link since I'm a new user. However, both methods are detailed on Wikipedia similarly to my implementation.

Should someone wish to tackle my problem, the code is below and graphs are in this Word file on dropbox.com. Yes this is a homework problem; I'm posting this because I'd like to understand what is wrong in my thought process.

f = @(x,y) x+y; %this is the eqn (the part after the @(t,y)

This is my RK4 code:

k1=@(x,y) h*f(x,y);
k2=@(x,y) h*f(x+1/2*h,y+1/2*k1(x,y));
k3=@(x,y) h*f(x+1/2*h,y+1/2*k2(x,y));
k4=@(x,y) h*f(x+h,y+k3(x,y));

clear y x exact i
x(1)=0;
y(1)=2;
xn=1;
x0=0;

exact=3*exp(xn)-xn-1; %exact solution at x=1
%# Evaluate RK4 with 1 step for x=0...1
N=1; %# number of steps
h=(xn-x0)/N; %# step size
i=1;
y(i+1)=y(i)+1/6*k1(x(i),y(i))+1/3*k2(x(i),y(i))+1/3*k3(x(i),y(i))+1/6*k4(x(i),y(i));
RK4(N)=y(i+1);  %# store result of RK4 in vector RK4(# of steps)
E_RK4(N)=-(RK4(N)-exact)/exact*100;%keep track of %error for each N
Nsteps_RK4(N)=N;


%# repeat for increasing N until error is less than 0.01%
while -(RK4(N)-exact)/exact > 0.0001
    i=1;
    N=N+1;
    h=(xn-x0)/N;
    for i=1:N
        y(i+1)=y(i)+1/6*k1(x(i),y(i))+1/3*k2(x(i),y(i))+1/3*k3(x(i),y(i))+1/6*k4(x(i),y(i));
        x(i+1)=x(i)+h;
    end
    RK4(N)=y(i+1);
    Nsteps_RK4(N)=N;
    E_RK4(N)=-(RK4(N)-exact)/exact*100; %# keep track of %error for each N
end

Nsteps_RK4(end);

This is my Euler code:

%# Evaluate Euler with 1 step for x=0...1
clear y x i
x(1)=0;
y(1)=2;
N=1; %# number of steps
h=(xn-x0)/N; %# step size
i=1;
y(i+1)= y(i)+h*f(x(i),y(i));
Euler(N)=y(i+1); %# store result of Euler in vector Euler(# of steps)
E_Euler(N)=-(Euler(N)-exact)/exact*100;%# keep track of %error for each N
Nsteps_Euler(N)=N;
%# repeat for increasing N until error is less than 0.01%
while -(Euler(N)-exact)/exact > 0.0001
    i=1;
    N=N+1;
    h=(xn-x0)/N;
    for i=1:N
        y(i+1)= y(i)+h*f(x(i),y(i));   
        x(i+1)=x(i)+h;
    end
    Euler(N)=y(i+1);
    Nsteps_Euler(N)=N;
    E_Euler(N)=-(Euler(N)-exact)/exact*100; %# keep track of %error for each N
end

Answer 1

See: http://www.mathworks.com/help/techdoc/matlab_prog/f4-70115.html#f4-71621

Variables specified in the body of the expression [...] must have a value assigned to them at the time you construct an anonymous function that uses them. Upon construction, MATLAB captures the current value for each variable specified in the body of that function. The function will continue to associate this value with the variable even if the value should change in the workspace or go out of scope.

The value of h in k1...k4 remains constant even though you change h in the while loop.

One solution is to add h to the anonymous functions:

k1=@(x,y,h) h*f(x,y);