"In an assignment A(I) = B, the number of elements in B and I must be the same" and cant work out why

I understand that it means that the matrices involved must be the same size. Apologies if i'm being stupid! Here is the main code:

function [] = BH();
mp=0.75;
ms=0.25;
mbh=1e6;
f0=-1.982;
f(1)=f0;
t0=(sqrt(6)*tan(f0/2)*(3+(tan(f0/2)^2)));
t(1)=t0;
R=((6*(mbh.^(1/3)))/(1+cos(f0)));
fdot=(sqrt(6)/36)*(1+cos(f0)).^2;
Rdot=((6*(mbh.^(1/3)))/((1+cos(f0)).^2))*fdot*sin(f0);

%initial conditions
xBH=0;
yBH=0;
vxBH=0;
vyBH=0;
xp(1)=(6*mbh.^(1/3)*cos(f0)/(1+cos(f0)));
yp(1)=(6*mbh.^(1/3)*sin(f0)/(1+cos(f0)))-ms;
vxp(1)=((Rdot*cos(f0))-(R*fdot*sin(f0)))+ms;
vyp(1)=((Rdot*sin(f0))+(R*fdot*cos(f0)));
xs(1)=(6*mbh.^(1/3)*cos(f0)/(1+cos(f0)));
ys(1)=(6*mbh.^(1/3)*sin(f0)/(1+cos(f0)))+mp;
vxs(1)=((Rdot*cos(f0))-(R*fdot*sin(f0)))-mp;
vys(1)=((Rdot*sin(f0))+(R*fdot*cos(f0)));
h=0.5;
nsteps=(f0)/-h;
for i=1:nsteps;
    k1=h.*fun(f(i),xp(i));
    k2=h.*fun(f(i)+k1/2,xp(i)+k1/2);
    k3=h.*fun(f(i)+k2/2,xp(i)+k2/2);
    k4=h.*fun(f(i)+k3,xp(i)+k3);
    f(i+1)=f(i)-(k1./6)-(k2./3)-(k3./3)-(k4./6);
    xp(i+1)=xp(i)+(k1(1,2)./6)+(k2(1,2)./3)+(k3(1,2)./3)+(k4(1,2)./6)
    t(i+1)=t(i)+h;
end

It uses a separate function called "fun":

function a = fun(f,xp)
a(1) = f;
a(2) = xp;

this returns the error:

In an assignment A(I) = B, the number of elements in B and I must be the same.

Error in fun (line 2) a(1) = f;

Error in BH (line 30) k2=h.*fun(f(i)+k1/2,xp(i)+k1/2);

I don't see why though as there are only two variables specified in the "k1" line of the for loop. Any help much appriciated!

Answer 1

[Marta Salas]

k1 is the output of fun so it's an array, it has 2 values. So, when you call

fun(f(i)+k1/2,xp(i)+k1/2)

you are trying to assign an array k1 (dim=1x2) to a(1) (dim=1x1) . The dimension doesn't agree.

[Simon Williams]

Ok, thanks

So what do I need to do? Makes the array in fun (1x2)?

[Marta Salas]

What's the purpose for the function fun?

[Simon Williams]

It's ok now. Its working now oddly, I don't know what I changed!

If you are interested however I am writing a simulation of a three body system, two of which are a binary star system and the third a supermassive black hole. The binary is to pass by the black hole on a parabolic orbit.

This is now the working code:

function [] = BH();
mp=0.75;
ms=0.25;
mbh=1e6;
f0=-1.982;
f(1)=f0;
t0=(sqrt(6)*tan(f0/2)*(3+(tan(f0/2)^2)));
t(1)=t0;
R=((6*(mbh.^(1/3)))/(1+cos(f0)));
fdot=(sqrt(6)/36)*(1+cos(f0)).^2;
Rdot=((6*(mbh.^(1/3)))/((1+cos(f0)).^2))*fdot*sin(f0);

%initial conditions
xBH=0;
yBH=0;
vxBH=0;
vyBH=0;
xp(1)=(6*mbh.^(1/3)*cos(f0)/(1+cos(f0)));
yp(1)=(6*mbh.^(1/3)*sin(f0)/(1+cos(f0)))-ms;
vxp(1)=((Rdot*cos(f0))-(R*fdot*sin(f0)))+ms;
vyp(1)=((Rdot*sin(f0))+(R*fdot*cos(f0)));
xs(1)=(6*mbh.^(1/3)*cos(f0)/(1+cos(f0)));
ys(1)=(6*mbh.^(1/3)*sin(f0)/(1+cos(f0)))+mp;
vxs(1)=((Rdot*cos(f0))-(R*fdot*sin(f0)))-mp;
vys(1)=((Rdot*sin(f0))+(R*fdot*cos(f0)));
h=0.1;
nsteps=(f0)/-h;

for i=1:nsteps
    k1=h.*fun(f(i),xp(i));
    k2=h.*fun(f(i)+k1(1,1)/2,xp(i)+k1(1,2)/2);
    k3=h.*fun(f(i)+k2(1,1)/2,xp(i)+k2(1,2)/2);
    k4=h.*fun(f(i)+k3(1,1),xp(i)+k3(1,2));
    t(i+1)=t(i)+h;
    f(i+1)=f(i)-(k1(1,1)./6)-(k2(1,1)./3)-(k3(1,1)./3)-(k4(1,1)./6);
    xp(i+1)=xp(i)+(k1(1,2)./6)+(k2(1,2)./3)+(k3(1,2)./3)+(k4(1,2)./6)
end

And the function:

function xi=func(f,xp)
xi(1)=f;                              %outputting new xp
xi(2)=xp;                              %outputting new f