Normalized to unity by summation
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I have the following code where I am trying to get unity after performing a summation. But it is not giving me unity using '' symsum''. Is there any alteranative way to execute the summation with a complicated expression. Pl somebody help me to solve that. Here my code is given below. Although theoretically it is giving ''one'' for n ranging from 1 to inf.
clc;
syms n
x=1.0;
y=0.5;
g=0.2;
l=0.1;
om=sqrt(((x).^2)-(4.*(g.^2)));
mu=sqrt((x+om)./(2.*om));
nu=((x-om)./(2.*g)).*mu;
eta=(((l)./((2.*g)+x)).*(1+((x-om)./(2.*g)))).*mu;
%% n dependency start
En=((n+(1./2)).*om)-(x./2)-(((l).^2)./((2.*g)+x));
Enn=((n-(1./2)).*om)-(x./2)-(((l).^2)./((2.*g)+x));
Dn=(y./2).*(exp(-2.*((eta).^(2)))).*(laguerreL(n,(4.*((eta).^2))));
Dnn=(y./2).*(exp(-2.*((eta).^(2)))).*(laguerreL((n-1),(4.*((eta).^2))));
Em=En - Dn;
Ep=Enn + Dnn;
eps=(Ep-Em)./2;
Deln=(eta.*y./sqrt(n)).*exp(-2.*(eta.^2)).*laguerreL((n-1),1,(4*(eta^2)));
xn=sqrt(((eps).^2)+((Deln).^(2)));
zetap=sqrt(((xn.^2)+(eps.^2))./(2.*xn));
zetam=sqrt(((xn.^2)-(eps.^2))./(2.*xn));
z= 1i.*(mu-nu).*eta./sqrt(2.*mu.*nu);
a1=(zetap./sqrt(factorial(n-1))).*((- nu./(2.*mu)).^(-1./2)).* hermiteH(n-1, z);
b1=(Deln./abs(Deln)).*(zetam./sqrt(factorial(n))).* hermiteH(n, z);
a2=(zetam./sqrt(factorial(n-1))).*((-nu./(2.*mu)).^(-1./2)).* hermiteH(n-1, z);
b2= (Deln./abs(Deln)).*(zetap./sqrt(factorial(n))).*hermiteH(n, z);
c0= -(1./sqrt(2.*mu)).*exp(-((eta.^2)./2)+ ((nu.*(eta).^2)./(2.*mu)));
cp= -c0.*((-nu./(2.*mu)).^(n./2)).*(a1 - b1);
cm= -c0.*((-nu./(2.*mu)).^(n./2)).*(a2 + b2);
c=((abs(cp)).^2) + ((abs(cm)).^2);
c1=(abs(c0)).^2;
out= ((abs(c0)).^2) + symsum(c,n,1,100);
vpa(out,5)
% Actually I need to get c1 + sum_{n=1}^{inf} c = 1. Is it possible to get by using symsum??
采纳的回答
Walter Roberson
2020-6-1
编辑：Walter Roberson
2020-6-1
You can use
out= ((abs(c0)).^2) + symsum(c,n,1,inf);
and that will be faster than for upper bound 100. The symsum() will stay unevaluated, and will be converted into numeric form when you vpa().
Alternately you could use
out= ((abs(c0)).^2) + sum(subs(c,n,1:100));
This will be a bit slow. I would suggest instead
guard_digits = 10;
out= ((abs(c0)).^2) + sum(vpa(subs(c,n,1:100), guard_digits);
Any of these methods will give the same result to within round-off error -- and none of them will give 1.0 for the final result, not even the one with symsum() to infinity. I recommend that you re-check the formulaes.
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