Is vectorizing this even possible?
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Payton Brown
am 17 Sep. 2020
Kommentiert: Payton Brown
am 18 Sep. 2020
vec3(1) = 1;
i = 1;
while i<5
i = i+1;
vec3(i) = (vec3(i-1)+2)^2;
end
vec3
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Walter Roberson
am 17 Sep. 2020
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
3 Kommentare
Walter Roberson
am 18 Sep. 2020
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.
Weitere Antworten (1)
madhan ravi
am 17 Sep. 2020
Bearbeitet: madhan ravi
am 17 Sep. 2020
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
2 Kommentare
Stephen23
am 17 Sep. 2020
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
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