Is vectorizing this even possible?
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Payton Brown
il 17 Set 2020
Commentato: Payton Brown
il 18 Set 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
il 17 Set 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 Commenti
Walter Roberson
il 18 Set 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.
Più risposte (1)
madhan ravi
il 17 Set 2020
Modificato: madhan ravi
il 17 Set 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 Commenti
Stephen23
il 17 Set 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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