Double summation in matlab

I would like to do this double summation . where alpha=2; beta=1.5;

I am confuse about the command availbale in online .Pl somebody help me what is way of solving this summation.

[Walter Roberson]

Because you used addition here where the original question used multiplication.

Answer 1

[Mohammad Sami]

You can define a function to calculate the expression inside the brackets.

a = @(p,param)param.^p./factorial(p); % (param^n)/factorial(n)
% param is alpha or beta, p = n or m
sum(sum(a(0:10,2)' * a(0:15,1.5)))

[AVM]

Thanks. Then shall I have to do it separtely? and then multiply the two things?

[AVM]

Pl give the complete coding if possible.

[AVM]

clc;
clear;
syms n m
alpha=2;
beta=2;
f=@(n,m)(alpha.^n)+(beta.^m);
symsum(symsum(f,n,0,3),m,0,3)

Why this coding gives the result 120 insted of 30. Pl help me.

[Mohammad Sami]

This is missing the factorial in the divisor.

[Mohammad Sami]

Its already complete. I just put the multiplication and sum in one line

[Mohammad Sami]

I think 120 is the correct answer.

a = @(p,param)param.^p; % (param^n)/factorial(n)
% param is alpha or beta, p = n or m
sum(sum(a(0:3,2)' + a(0:3,2)))

I don't have the symbolic toolbox myself so can't test it.

I have cross checked with https://www.wolframalpha.com/input/?i=sum+2%5Ek+%2B+2%5El%2C+k%3D0+to+3%2C+l+%3D0+to+3

[AVM]

@sami: Thanks. Actually I was wrong. I have done a mistake in analytic calcualtion. It's working.

[AVM]

@Sami: I am trying to solve the following case according to your code of my previous problem, I am little bit confuse how to handle it. Pl help me.

clc;
clear;
syms alpha r1 theta mu nu lambda t n1 m1 theta2

r2=1.0;
theta2=2.0;

alpha1=0.5;
alpha2=1.0;
lambda=2.0;
omega=1;
phi=pi/3;

mu=cosh(r2);
nu=exp(1i*theta2).*sinh(r2);

f=((((alpha1.*exp(-1i*omega*t))*r1.*exp(-1i*theta)).^(n1))*(((alpha1.*exp(1i*omega*t))*r1.*exp(1i*theta)).^(m1))./(factorial(m1).*factorial(n1))).*...
   1./sqrt((((mu).^2)-((abs(nu)).^2).*exp(-2*1i*lambda*t*(n1-m1)))).*...
  exp(((1./mu).*exp(-2*1i*lambda*t*(n1-m1)).*real(((alpha2).^2)*conj(nu))+((alpha2).^2).*(exp(-1i*lambda*t*(n1-m1))))./...
     (((mu).^2)-((abs(nu)).^2).*exp(-2*1i*lambda*t*(n1-m1))));
 
 

Now I would like to evaluate this double summation fucntion 'f' over n1 and m1 from 0 to 20 each of them. Pl help me to do that

[Mohammad Sami]

I believe you should be able to solve this with the symbolic toolbox. I dont have the toolbox to check if it works.

Similar to what you posted earlier. Since only m1, n1 and t don't have a fixed value, only declare those as symbols.

Declare f as a function.

% your ealier post.
syms n m
alpha=2;
beta=2;
f=@(n,m)(alpha.^n)+(beta.^m);
symsum(symsum(f,n,0,3),m,0,3)

for the current question

syms n1 m1 t
r2=1.0;
theta2=2.0;
alpha1=0.5;
alpha2=1.0;
lambda=2.0;
omega=1;
phi=pi/3;
mu=cosh(r2);
nu=exp(1i*theta2).*sinh(r2);
f=@(n1,m1)((((alpha1.*exp(-1i*omega*t))*r1.*exp(-1i*theta)).^(n1))*(((alpha1.*exp(1i*omega*t))*r1.*exp(1i*theta)).^(m1))./(factorial(m1).*factorial(n1))).*...
   1./sqrt((((mu).^2)-((abs(nu)).^2).*exp(-2*1i*lambda*t*(n1-m1)))).*...
  exp(((1./mu).*exp(-2*1i*lambda*t*(n1-m1)).*real(((alpha2).^2)*conj(nu))+((alpha2).^2).*(exp(-1i*lambda*t*(n1-m1))))./...
     (((mu).^2)-((abs(nu)).^2).*exp(-2*1i*lambda*t*(n1-m1))));
 
 symsum(symsum(f,n1,0,20),m1,0,20)

[Atrolita afra]

[Atrolita afra]