Normalized to unity by summation

Question

AVM 2020-6-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/539439-normalized-to-unity-by-summation

编辑： Walter Roberson 2020-6-1

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??

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Answer 1

Walter Roberson 2020-6-1

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/539439-normalized-to-unity-by-summation#answer_444103

编辑：Walter Roberson 2020-6-1

在 MATLAB Online 中打开

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.