Recursion Issue

Question

Andrew 2011-2-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1227-recursion-issue

Hi again everyone, I am still trying to fix my recursion... If anyone could help me with these conditions that would be great I seem to keep on getting infinite recursions.

My codes are below:

Main:

close all
clear all
clc
theta = [-pi:pi/100:pi]';
plotValues = cos(theta);
lamda = [0:(2*pi)/100:(2*pi)]';
set1 = pnm(1,1,theta,1);
set2 = pnm(3,1,theta,1);
set3 = pnm(5,1,theta,1);
set4 = pnm(7,1,theta,1);
hold on
plot(plotValues,set1,'g');
plot(plotValues,set2,'r');
plot(plotValues,set3,'m');
plot(plotValues,set4,'b');
axis([-1 1 -4 4]);
legend('n=1; m=1','n=3; m=1', 'n=5; m=1', 'n=7; m=1')
hold off
%set1non = pnm(1,1,theta,0);
%set2non = pnm(3,1,theta,0);
%set3non = pnm(5,1,theta,0);
%set4non = pnm(7,1,theta,0);
%hold off
%hold on
%figure
%plot(plotValues,set1non,'g');
%plot(plotValues,set2non,'r');
%plot(plotValues,set3non,'m');
%plot(plotValues,set4non,'b');
%axis([-1 1 -4 4]);
%legend('n=1; m=1','n=3; m=1', 'n=5; m=1', 'n=7; m=1')
RnmZ = pnm(4,0,theta,1)*cos(0*lamda);
SnmZ = pnm(4,0,theta,1)*sin(0*lamda);

PNM:

function [ values ] = pnm( n,m,theta,norm )
%What to do...
if ((n ~= 0 && m ~= 0) || (n~=0 && m==0) || (n==0 && m~=0))
      if (norm == 0 )
          if (m >= 2)
              pnmCurrentn = 2*(m-1)*((cos(theta)/sqrt(1-cos(theta)^2)))*pnm(n,m-1,theta,norm)-(n-m+2)*(n+m-1)*pnm(n,m-2,theta,norm);
              values = pnmCurrentn;
          elseif (norm == 0 && m == 0)
              pn = -((n-1)/n)*pnm(n-2,0,theta,norm)+((2*n-1)/n)*(cos(theta))*pnm(n-1,0,theta,norm);
              values = pn;
          elseif ( n >= m+2)
              pnmCurrentn = ((2*n-1)/(n-m)).*cos(theta).*pnm(n-1,m,theta,norm)-((n+m-1)/(n-m))*pnm(n-2,m,theta,norm);
              values = pnmCurrentn;
          elseif (n==1 && m==1)
              values = sin(theta);
          elseif ( n==2 && m ==1 )
              values = 3.*sin(theta).*cos(theta);
          end
      elseif ( norm == 1)
          if ( n == 1 && m == 1)
              w11 = sqrt(3);
              values = w11*sqrt((1-cos(theta).^2))*pnm(m-1,m-1,theta,norm);
          elseif (( n==0 && m==0))
              values = 1;            
          elseif ( n == m)
              wmm = sqrt((2*m-1)/(2*m));
              values = wmm.*(1-cos(theta).^2).^(1/2)*pnm(m-1,m-1,theta,norm);
          elseif (n~=m || (n~=0 && m==0))
              wnm = sqrt(((2*n+1)*(2*n-1))/((n+m)*(n-m)));
              wnmB = sqrt(((2*(n-1)+1)*(2*(n-1)-1))/(((n-1)+m)*((n-1)-m)));
              values = wnm*(cos(theta).*pnm(n-1,m,theta,norm)-inv(wnmB)*pnm(n-2,m,theta,norm));        
          end
      end
elseif ((n == 0 && m == 0))
      if (norm ~= 1)
          values = 1; 
      elseif (norm == 1)
          values = 1;
      end
end
end

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Answer 1

Oleg Komarov 2011-2-11

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1227-recursion-issue#answer_1753

在 MATLAB Online 中打开

Andrew,

1) Deleting your old questions and posting incrementing new ones with highly related content instead of editing the original post, is not a nice way to recognize others effort.

2) On line 45:

pnm(n-2,m,theta,norm)

n is decreased by 2, but you supplied n = 3 with set2 = pnm(3,1,theta,1); , i.e. on the second recursion you'll call on line 45:

pnm(-1,1,theta,norm)

which will enter an endless recursion since you don't contemplate cases with negative values of n

Oleg

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Andrew 2011-2-11

Sorry about that, I'm just getting really irritated by seeing all my code. I won't do it again though.

I was wondering for the following code segment how would I be able to make it so the symbolic t can be used to input a value rather then staying symbolic. I use symbolic in order to take the derivative but I can't use it thereafter.

syms t

differ = (cos(t)^2-1)^n

differ = diff (differ);

t = theta;

values = (1/((2^n)*factorial(n)))*(1-cos(theta)^2)^(m/2)*differ

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