Removal of For Loops
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Is it possible to script the following function in MATLAB without a for loop, or any other iterative loop? I am trying to understand if it is possible to have a matrix reference itself as it is populated using a single command in MATLAB. Thanks for looking!
x(1)=0;
x(2)=1;
x(3)=2;
for n=4:51
x(n)=x(n-1) + x(n-3);
end
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Respuesta aceptada
Sean de Wolski
el 1 de Feb. de 2012
It's probably possible with filter() or similar. But I guarantee a well written for-loop will not be much slower. Just remember to preallocate x.
x = zeros(51,1);
%etc.
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Más respuestas (1)
Walter Roberson
el 1 de Feb. de 2012
I = sqrt(-1);
x = @(n) -(301/14945472)*((1+I*3^(1/2))*(93^(1/2)-93/7)*(108+12*93^(1/2))^(1/3)-744/7-(5/14)*(-1+I*3^(1/2))*(93^(1/2)-31/5)*(108+12*93^(1/2))^(2/3))*(((1/72)*(-93^(1/2)+9)*(108+12*93^(1/2))^(2/3)+(1/6)*(108+12*93^(1/2))^(1/3))^n*(((I*3^(1/2)-67/43)*93^(1/2)+279/43-((775/43)*I)*3^(1/2))*(108+12*93^(1/2))^(1/3)+((-((97/129)*I)*3^(1/2)+27/43)*93^(1/2)+((248/43)*I)*3^(1/2)-310/43)*(108+12*93^(1/2))^(2/3)+1860/43+((92/43)*I)*3^(1/2)*93^(1/2))*((1/72)*(-9+93^(1/2))*(-1+I*3^(1/2))*(108+12*93^(1/2))^(1/3)-(1/72)*(108+12*93^(1/2))^(2/3)-((1/72)*I)*(108+12*93^(1/2))^(2/3)*3^(1/2)+1/3)^n+(6696/43)*(-(1/72)*(-9+93^(1/2))*(1+I*3^(1/2))*(108+12*93^(1/2))^(1/3)-(1/72)*(108+12*93^(1/2))^(2/3)+((1/72)*I)*(108+12*93^(1/2))^(2/3)*3^(1/2)+1/3)^n*((1/72)*(-93^(1/2)+9)*(108+12*93^(1/2))^(2/3)+(1/6)*(108+12*93^(1/2))^(1/3))^n+((-((12/43)*I)*3^(1/2)+98/43)*93^(1/2)-1302/43-((248/43)*I)*3^(1/2))*(108+12*93^(1/2))^(1/3)+((-((89/129)*I)*3^(1/2)+35/43)*93^(1/2)+((279/43)*I)*3^(1/2)-217/43)*(108+12*93^(1/2))^(2/3)+1860/43-((92/43)*I)*3^(1/2)*93^(1/2))/((1/72)*(-93^(1/2)+9)*(108+12*93^(1/2))^(2/3)+(1/6)*(108+12*93^(1/2))^(1/3))^n
But watch out for floating point round-off.
(Yes, really. And yes, this is the simplified form of the expression.)
2 comentarios
Walter Roberson
el 1 de Feb. de 2012
Note: the above was produced by simplifying the output of Maple's rsolve() routine. The basic form of the answer is not very complicated, but it involves the sum of terms with the sum taken over the roots of a cubic expression, and that cubic happens to have two imaginary solutions. The expanded expression before simplification is pretty grotty.
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