Dear all,

Thank you for your Help.

h: stepsize;

x in [0,x0];

I give all the step of my code, but I think there is a mistake. I wait for your Help.

I would like to compute the error between Method Huen with step size h and step size 2h using the definition of epsilon given below:

## The error written epsilon(x0,h)= sqrt(1/(N+1) * sum_i=0^N (y_i^{2h}-y_(2i)^h)^2 ), where y_i^(2h) is the approximation of y(i*2*h).

## We want : loglog epsilon versus h.

epsilon:=(x0,h)->sqrt( 1/(N+1)*add( (Heun1(f,x0,i)-Heun2(f,x0,i))^2,i=0..N ) );

f:=(x,y)=1/(1+cos(y));

ode:=diff(y(x),x)=f(x,y);

ic:=y(0)=1; h:=x0/(2*N);

## Method Heun with step size 2h

> Heun1 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (2*i+2)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[i], y[i]], i = 0 .. N)];

end proc;

### Now Heun with step size h ( the same h)

> Heun2 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (i+1)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[2*i], y[2*i]], i = 0 .. N)];

end proc;

Thanks you for your help.