Output is coming out as a large expression

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))

Answer 1

[Walter Roberson]

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.

[AVM]

Thanks a lot for your help. But one thing I would like to know that here p, q run from 0 to n and m respectively and ''l'' also run from 0 to ''k''. Here ''k'' =2, n=6,m=5 (in first case). Then why the result (output) ultimately depends upon them? Please help me to understand this.

[Walter Roberson]

    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

[AVM]

Thanks for your clarification.