Removal of For Loops
조회 수: 1 (최근 30일)
이전 댓글 표시
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
댓글 수: 0
채택된 답변
Sean de Wolski
2012년 2월 1일
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.
댓글 수: 0
추가 답변 (1개)
Walter Roberson
2012년 2월 1일
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.)
댓글 수: 2
Walter Roberson
2012년 2월 1일
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.
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!