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Removal of For Loops

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Ben
Ben 2012-2-1
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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采纳的回答

Sean de Wolski
Sean de Wolski 2012-2-1
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.

更多回答（1 个）

Walter Roberson
Walter Roberson 2012-2-1
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 个评论
Sean de Wolski
Sean de Wolski 2012-2-1
nice! +1 for sure.
Walter Roberson
Walter Roberson 2012-2-1
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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