How can I solve these recursive equations ?

Hello I want to take the solution of a equation and use this as a new variable and so on like in the following demonstrated.

x1=a+bx0
x2=a+bx1
x3=a+bx2 ......

How can I solve this by a loop or so because I have to do this until 743 and I need every of the x values, so in the end I want to have a x matrix with 743x1 dimension.

[Stephen23]

Whatever you do, do not create the variable names dynamically, just save the values in a vector instead. Here is an explanation of why it is very poor programming practice to create variable names dynamically:

http://www.mathworks.com/matlabcentral/answers/196952-how-to-create-new-variables-in-batches-with-strcat

Answer 1

[Torsten]

xn = a*(1-b^n)/(1-b) + b^n*x0.

Now insert n=743.

Best wishes

Torsten.

[Fox]

Thanks. However it doesn't solve my problem. I need every of the x values at the end. I tried the following,but I always get the mistake of "Index exceeds matrix dimensions."

xb(1)=a/(1-b); % xb is a scalar , a and b also scalars

   for m=1:743
   x2(m+1)=a+ b*xb(m);
   end

Do you see my mistake here or anybody else ?

[Torsten]

x = zeros(743);
x(1)=a/(1-b); % xb is a scalar , a and b also scalars

for m=1:742
   x(m+1)=a+b*x(m);
end

Best wishes

Torsten.

[Guillaume]

Well, better would have been to keep with the original formula and adjust it to work with matrices:

n = 1:743;
xn = a*(1-b.^n)/(1-b) + b.^n*x0

[Fox]

Hi thanks with the transformed formular it works. But now I've an addttional problem. In a second step I want to adjust my a values. like in the following described.

x1=a(1,:)+bx0 
x2=a(2,:)+bx1
x3=a(3,:)+bx2

a is now a vektor of 743x1 which also influences the resulting value. I tried to solve this by a loop but I get the wrong solutions.

m=1:743
  for i=1:743
   xm = a(i,:)*(1-b.^m)/(1-b) + b.^m*xx0;
   end

[Torsten]

The formula is only valid for constant a. For a depending on n you will have to refer to the loop solution from above:

x(1) = some value;
for m=1:742
   x(m+1)=a(m+1)+b*x(m);
end

Best wishes

Torsten.

[Fox]

Thank you very much.

Answer 2

[Guillaume]

The filter function allows you to compute all your elements in one go for both use cases (a constant or variable):

• constant a and b:

numelemrequired = 743;
x = filter(1, [1 -b], [x0 repmat(a, 1, numelemrequired)])

• variable a and b:

x = filter(1, [1 -b], [x0 a]) %where a is a vector of length 743

Note that your a, b, and x are not the same a, b, and x used by the documentation of filter.

[Fox]

thanks it works