How to make a local function that returns a vector

I found a mistake in my question. I edited a part and the part is now underlined. Sorry for confusing way to ask.

Hello,

I wnat to make a local function that returns a vector or vectors.

For example, I will give the initial value of a vector, a(1,1), then the local function calculates a(1,2) based on the former value, a(1,1) until the last slot a(1,N), which will be calculated based on a(1,N-1).

I made a local function using function [y1, y2, ... , yN]=myfunction(x1, x2, ... , xM), but it did not work correctly. Is there any function or expression that can return a vector or vectors as the output?

I will attach my code.

% parameters
N=100;
div=0.1;
p=sqrt(3);
q=0.5;
r=2;

% vectors and initial values
a=zeros(1,N);
b=zeros(1,N);
a(1,1)=5;
b(1,1)=0.1;

% function
[a,b]=fcn2(N,p,q,r);

% arb. function
function [a,b]=fcn2(N,p,q,r)
for i=1:N
    ka1=p+q*r;
    kb1=p-q*r;
    ka2=ka1+p*q;
    kb2=kb1-q/r;
    a(1,i+1)=a(1,i)+ka1*ka2;
    b(1,i+1)=b(1,i)+kb1/(kb2+1);
end
end

Answer 1

[Torsten]

% parameters
N=100;
div=0.1;
p=sqrt(3);
q=0.5;
r=2;

% Initial values
astart=5;
bstart=0.1;

% function
[a,b]=fcn2(N,p,q,r,astart,bstart);

plot(1:numel(a),[a;b])
grid on

% arb. function
function [a,b]=fcn2(N,p,q,r,astart,bstart)
  ka1=p+q*r;
  kb1=p-q*r;
  ka2=ka1+p*q;
  kb2=kb1-q/r;
  a = astart + (0:N)*ka1*ka2;
  b = bstart + (0:N)*kb1/(kb2+1);
end

[Kosuke]

Thank you for your suggestion, I made a correction in my question. But anyway, some part will be helpful, so I will try doing that way.

[Torsten]

You can also use a loop, but this will be less efficient:

% parameters
N=100;
div=0.1;
p=sqrt(3);
q=0.5;
r=2;
% Initial values
astart=5;
bstart=0.1;
% function
[a,b]=fcn2(N,p,q,r,astart,bstart);
plot(1:numel(a),[a;b])
grid on

% arb. function
function [a,b]=fcn2(N,p,q,r,astart,bstart)
  a = zeros(1,N+1);
  b = zeros(1,N+1);
  a(1) = astart;
  b(1) = bstart;
  ka1=p+q*r;
  kb1=p-q*r;
  ka2=ka1+p*q;
  kb2=kb1-q/r;
  for i = 1:N
    a(i+1) = a(i)+ka1*ka2;
    b(i+1) = b(i)+kb1/(kb2+1);
  end
end

[Kosuke]

Thank you for the additional information.

[Dyuman Joshi]

Hello @Kosuke, if this answer solved your problem, please consider accepting the answer.