Is vectorizing this even possible?

vec3(1) = 1;
i = 1;
while i<5
    i = i+1;
    vec3(i) = (vec3(i-1)+2)^2;
end
vec3

Answer 1

[Walter Roberson]

I posted the complete vectorization several days ago; . Unfortunately the person deleted the question.

syms v v0 c
f(v) = (v+c)^2;
f5 = f(f(f(f(f(v0)))));
vec3_5 = expand(subs(f5, c, 2));

v0^32 + 64*v0^31 + 2016*v0^30 + 41600*v0^29 + 631536*v0^28 + 7511168*v0^27 + 72782528*v0^26 + 590011136*v0^25 + 4077667064*v0^24 + 24363708032*v0^23 + 127184607424*v0^22 + 584772138240*v0^21 + 2382767071968*v0^20 + 8644745151232*v0^19 + 28021844462720*v0^18 + 81349497514496*v0^17 + 211814884610908*v0^16 + 494935571753856*v0^15 + 1037540400943680*v0^14 + 1949025086827264*v0^13 + 3273934344609568*v0^12 + 4902203714779904*v0^11 + 6514485357242496*v0^10 + 7638211784159744*v0^9 + 7840967227104336*v0^8 + 6975721989473536*v0^7 + 5305860461727104*v0^6 + 3387252771621376*v0^5 + 1768336935606208*v0^4 + 726328276999680*v0^3 + 220554340195584*v0^2 + 44118436709376*v0 + 4371938082724

Where v0 = 1

[Bruno Luong]

Very uggly, but admitly it answers the question.

[Walter Roberson]

Yup ;-)

Using a for loop is not vectorizing . This solution is not vectorized in terms of the number of iterations, but it is vectorized in terms of different initial conditions.

I think it should be possible to calculate what all the terms should be, in terms of binomial coefficients and number of iterations, but the form is not coming to mind immediately.

[Payton Brown]

Thank you so much!

Answer 2

[madhan ravi]

A simple for loop is the best and easier to understand:

    vec3 = zeros(5,1);
    vec3(1) = 1;
    for k = 2:5 % edited after Stephen’s comment
        vec3(k) = (vec3(k-1)+2)^2;
    end
    vec3

[Stephen23]

Starting the for loop from one will throw an error. Better to start from two:

vec3 = ones(5,1);
for k = 2:5
    vec3(k) = (vec3(k-1)+2)^2;
end

[madhan ravi]

Ah thanks Stephen!