作者rainss (神是太空人....)
看板MATLAB
标题[问题] .m档的直接执行
时间Fri Feb 17 08:17:58 2012
因为我是Matlab的新新手 这个问题我实在不会处理
所以要请教板上的高手
因为必须要更精准的计算Legendre函数
所以无法使用Matlab内建的Legendre功能
我在网路上找了一个.m档是计算Legendre函数
但是这个党只有示范计算的结果
请问有没有可能把这个档案修改之後 像Matlab内建的函数一样
只要输入就可以算出值? 谢谢
function mlpmn
%This program is a direct conversion of the corresponding Fortran program in
%S. Zhang & J. Jin "Computation of Special Functions" (Wiley, 1996).
%online:
http://iris-lee3.ece.uiuc.edu/~jjin/routines/routines.html
%
%Converted by f2matlab open source project:
%online:
https://sourceforge.net/projects/f2matlab/
% written by Ben Barrowes (
[email protected])
%
% ==========================================================
% Purpose: This program computes the associated Legendre
% functions Pmn(x)and their derivatives Pmn'(x)
% using subroutine LPMN
% Input : x --- Argument of Pmn(x)
% m --- Order of Pmn(x), m = 0,1,2,...,n
% n --- Degree of Pmn(x), n = 0,1,2,...,N
% Output: PM(m,n)--- Pmn(x)
% PD(m,n)--- Pmn'(x)
% Example: x = 0.50
% Pmn(x):
% m\n 1 2 3 4
% --------------------------------------------------------
% 0 .500000 -.125000 -.437500 -.289063
% 1 -.866025 -1.299038 -.324760 1.353165
% 2 .000000 2.250000 5.625000 4.218750
% 3 .000000 .000000 -9.742786 -34.099750
% 4 .000000 .000000 .000000 59.062500
% Pmn'(x):
% m\n 1 2 3 4
% --------------------------------------------------------
% 0 1.000000 1.500000 .375000 -1.562500
% 1 .577350 -1.732051 -6.278684 -5.773503
% 2 .000000 -3.000000 3.750000 33.750000
% 3 .000000 .000000 19.485572 .000000
% 4 .000000 .000000 .000000 -157.500000
% ==========================================================
m=[];n=[];x=[];pm=[];pd=[];
pm=zeros(100+1,100+1);
pd=zeros(100+1,100+1);
fprintf(1,'%s \n',' please enter m, n and x');
% READ(*,*)M,N,X
m=2;
n=5;
x=.5;
fprintf(1,'%0.15g \n');
fprintf(1,'%s \n',' m n x pmn(x) pmn''(x)');
fprintf(1,'%s \n',' ---------------------------------------------------');
[dumvar1,m,n,x,pm,pd]=lpmn(100,m,n,x,pm,pd);
for j=0:n;
fprintf(1,[repmat(' ',1,1),'%3g',repmat(' ',1,3),'%3g',repmat('
',1,3),'%5.1g',repmat('%17.8g',1,2) ' \n'],m,j,x,pm(m+1,j+1),pd(m+1,j+1));
end; j=n+1;
%format(1x,i3,3x,i3,3x,f5.1,2e17.8);
end
function [mm,m,n,x,pm,pd]=lpmn(mm,m,n,x,pm,pd,varargin);
% =====================================================
% Purpose: Compute the associated Legendre functions
% Pmn(x)and their derivatives Pmn'(x)
% Input : x --- Argument of Pmn(x)
% m --- Order of Pmn(x), m = 0,1,2,...,n
% n --- Degree of Pmn(x), n = 0,1,2,...,N
% mm --- Physical dimension of PM and PD
% Output: PM(m,n)--- Pmn(x)
% PD(m,n)--- Pmn'(x)
% =====================================================
for i=0:n;
for j=0:m;
pm(j+1,i+1)=0.0d0;
pd(j+1,i+1)=0.0d0;
end; j=fix(m)+1;
end; i=fix(n)+1;
pm(0+1,0+1)=1.0d0;
if(abs(x)== 1.0d0);
for i=1:n;
pm(0+1,i+1)=x.^i;
pd(0+1,i+1)=0.5d0.*i.*(i+1.0d0).*x.^(i+1);
end; i=fix(n)+1;
for j=1:n;
for i=1:m;
if(i == 1);
pd(i+1,j+1)=1.0d+300;
elseif(i == 2);
pd(i+1,j+1)=-0.25d0.*(j+2).*(j+1).*j.*(j-1).*x.^(j+1);
end;
end; i=fix(m)+1;
end; j=fix(n)+1;
return;
end;
ls=1;
if(abs(x)> 1.0d0)ls=-1; end;
xq=sqrt(ls.*(1.0d0-x.*x));
xs=ls.*(1.0d0-x.*x);
for i=1:m;
pm(i+1,i+1)=-ls.*(2.0d0.*i-1.0d0).*xq.*pm(i-1+1,i-1+1);
end; i=fix(m)+1;
for i=0:m;
pm(i+1,i+1+1)=(2.0d0.*i+1.0d0).*x.*pm(i+1,i+1);
end; i=fix(m)+1;
for i=0:m;
for j=i+2:n;
pm(i+1,j+1)=((2.0d0.*j-1.0d0).*x.*pm(i+1,j-1+1)-(i+j-1.0d0).*pm(i+1,j-2+1))./(j-i);
end; j=fix(n)+1;
end; i=fix(m)+1;
pd(0+1,0+1)=0.0d0;
for j=1:n;
pd(0+1,j+1)=ls.*j.*(pm(0+1,j-1+1)-x.*pm(0+1,j+1))./xs;
end; j=fix(n)+1;
for i=1:m;
for j=i:n;
pd(i+1,j+1)=ls.*i.*x.*pm(i+1,j+1)./xs+(j+i).*(j-i+1.0d0)./xq.*pm(i-1+1,j+1);
end; j=fix(n)+1;
end; i=fix(m)+1;
return;
end
--
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