Output is coming out as a large expression

3 次查看(过去 30 天)
When I run the code below, the outputs appear as a large expression. Here, I have a upper limit in one of the sym summation min() condition. Pl somebody help me to get the output in simple numeric form.
clc;
syms n m p q s l up_lt
mu=1.0051;
nu=0.1015;
eta=0.0922;
k=2;
assume(p>=0);
assume(q>=0);
assume(l>=0);
assume(s>=0);
assume(m>=0);
assume(n>=0);
x1= 2.*q + l -m;
y1= 2.*p - n;
up_lt= min(x1,y1); %%%%%%%%% Upper limit of the sum in sGnmp3
Gnmp1= (((-1).^(k-l))-1).*nchoosek(k,l).*((eta.*(mu-nu)).^(k-l)).*((mu.*nu./2).^l);
sGnmp1= symsum(Gnmp1,l,0,k);
Gnmp2= (1./(factorial(p).*factorial(q))).*nchoosek(p,n-p).*nchoosek(q,m-q);
sGnmp2= symsum(symsum(Gnmp2,q,0,m),p,0,n);
Gnmp3= ((2.*mu./nu).^s).*factorial(s).*nchoosek(x1,s) .*nchoosek(y1,s).*hermiteH(x1-s,0).*hermiteH(y1-s, 0);
sGnmp3= symsum(Gnmp3,s,0,up_lt);
Gnmm1= (((-1).^(k-l))+1).*nchoosek(k,l).*((eta.*(mu-nu)).^(k-l)).*((mu.*nu./2).^l);
sGnmm1= symsum(Gnmm1,l,0,k);
Gnmp= symfun((sqrt(factorial(n).*factorial(m)).*((nu./(2.*mu)).^((n+m)./2)).*sGnmp1.*sGnmp2.*sGnmp3),[n,m]);
Gnmm= symfun((sqrt(factorial(n).*factorial(m)).*((nu./(2.*mu)).^((n+m)./2)).*sGnmm1.*sGnmp2.*sGnmp3),[n,m]);
outp= vpa(Gnmp(6,5)) %%% n and m can be taken as any positive integer.
outm= vpa(Gnmm(5,5))

采纳的回答

Walter Roberson
Walter Roberson 2020-6-14
It is not possible to get that output in simple numeric form. The result depends upon the unresolved variables l, p and q.
About the best you can do is produce an output under each of the two possible conditions implied by your use of min()
syms n m p q s l up_lt
Q = @(v) sym(v);
mu = Q(1.0051);
nu = Q(0.1015);
eta = Q(0.0922);
k=2;
x1 = 2.*q + l - m;
y1 = 2.*p - n;
up_lt_vals = [x1, y1];
up_lt_cond = [x1 <= y1, y1 <= x1];
n_up_lt = length(up_lt_vals);
outp = zeros(n_up_lt, 1, 'sym');
outm = zeros(n_up_lt, 1, 'sym');
for up_lt_idx = 1 : n_up_lt
assume([n, m, p, q, s, l, up_lt], 'clear')
assumeAlso(p>=0);
assumeAlso(q>=0);
assumeAlso(l>=0);
assumeAlso(s>=0);
assumeAlso(m>=0);
assumeAlso(n>=0);
assumeAlso(up_lt_cond(up_lt_idx));
up_lt = up_lt_vals(up_lt_idx); %%%%%%%%% Upper limit of the sum in sGnmp3
Gnmp1 = (((-1).^(k-l))-1).*nchoosek(k,l).*((eta.*(mu-nu)).^(k-l)).*((mu.*nu./2).^l);
sGnmp1 = symsum(Gnmp1,l,0,k);
Gnmp2 = (1./(factorial(p).*factorial(q))).*nchoosek(p,n-p).*nchoosek(q,m-q);
sGnmp2 = symsum(symsum(Gnmp2,q,0,m),p,0,n);
Gnmp3 = ((2.*mu./nu).^s).*factorial(s).*nchoosek(x1,s) .*nchoosek(y1,s).*hermiteH(x1-s,0).*hermiteH(y1-s, 0);
sGnmp3 = symsum(Gnmp3,s,0,up_lt);
Gnmm1 = (((-1).^(k-l))+1).*nchoosek(k,l).*((eta.*(mu-nu)).^(k-l)).*((mu.*nu./2).^l);
sGnmm1 = symsum(Gnmm1,l,0,k);
Gnmp = symfun((sqrt(factorial(n).*factorial(m)).*((nu./(2.*mu)).^((n+m)./2)).*sGnmp1.*sGnmp2.*sGnmp3),[n,m]);
Gnmm = symfun((sqrt(factorial(n).*factorial(m)).*((nu./(2.*mu)).^((n+m)./2)).*sGnmm1.*sGnmp2.*sGnmp3),[n,m]);
outp(up_lt_idx) = simplify(Gnmp(6,5), 'steps', 5); %%% n and m can be taken as any positive integer.
outm(up_lt_idx) = simplify(Gnmm(5,5), 'steps', 5);
end
disp(up_lt_cond(1));
disp([outp(1); outm(1)]);
disp(up_lt_cond(2));
disp([outp(2); outm(2)]);
but this will not be very different than your existing code with min(). If more information was known about p, q, or l, then perhaps there might be additional simplification.
  3 个评论
Walter Roberson
Walter Roberson 2020-6-14
Gnmp3 = ((2.*mu./nu).^s).*factorial(s).*nchoosek(x1,s) .*nchoosek(y1,s).*hermiteH(x1-s,0).*hermiteH(y1-s, 0);
That line uses x1 and y1, which depend upon l, p, and q.
The Gmnp3 is not used within a symsum over l, p, or q, unlike sGnmp2

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Assumptions 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by