How can I solve these recursive equations ?
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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.
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Torsten
il 13 Gen 2016
xn = a*(1-b^n)/(1-b) + b^n*x0.
Now insert n=743.
Best wishes
Torsten.
6 Commenti
Torsten
il 14 Gen 2016
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.
Più risposte (1)
Guillaume
il 14 Gen 2016
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.
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