Summation for every value of "n" (or summation with looping)

% Hi, i need to find the "Q" variable for every instance of "n" and divide those values by "Pr"  
clear all
clc;
n=[1:1:50]
B=7.5 % angle value
%Q= (1+2*(cos(n1*B))^(5/2)+2*(cos(n2*B)^(5/2) + ... ); --> this is an
%example of how iterations should be in short; "1+2*(cos(n*B))^(5/2)"
Pr= 395
%P1= Pr/Q

Answer 1

[Alan Stevens]

Like so:

B=7.5; % angle value
fn = @(n) (1+2*cos(n*B)).^5/2;
n=1:50;
Qn = fn(n);
Q = sum(Qn);
Pr= 395;
P1= Pr/Q;
disp(P1)

    0.3323

% Or do you mean
Q(1) = fn(1);
for n = 2:50
    Q(n) = fn(n) + Q(n-1);
end
P1 = Pr./Q;

disp(P1)

  Columns 1 through 19

   56.7538   56.9083   57.8755   45.1792    3.2258    2.8098    2.8098    2.8159    2.8096    1.6914    1.4594    1.4594    1.4620    1.4619    1.1756    0.9907    0.9907    0.9918    0.9918

  Columns 20 through 38

    0.9020    0.7481    0.7477    0.7482    0.7482    0.7211    0.5987    0.5973    0.5974    0.5974    0.5904    0.4997    0.4958    0.4959    0.4959    0.4945    0.4321    0.4245    0.4245

  Columns 39 through 50

    0.4246    0.4244    0.3848    0.3723    0.3723    0.3724    0.3724    0.3497    0.3322    0.3322    0.3323    0.3323

[Erdem Turan]

The first solution is more of what i needed however, i wanted to add 1 only in the beginning, the real function is:

Pr=P1*[1+ 2(cos(B)^(5/2)+ 2(cos(2B)^(5/2) + ... + 2(cos(n*B)^(5/2)]

i need to find P1 from this function. Would the following be a correct way to do it? :

B=7.5; % angle value
fn = @(n) (2*cos(n*B)).^5/2;
n=1:50;
Qn = fn(n);
Q = (1+sum(Qn));
Pr= 395;
P1= Pr/Q;
disp(P1)
% Thank you in advance

[Alan Stevens]

Yes, that looks right.

[Erdem Turan]

I have a problem with the writing of the fn function the orgininal equation is,

Pr=P1*[1+ 2(cos(B)^(5/2)+ 2(cos(2B)^(5/2) + ... + 2(cos(n*B)^(5/2)] i'm not sure if the ^5/2 is correctly placed in the "fn" function could you please clear this up for me?

[Alan Stevens]

Looks like it should be

fn = @(n) 2*cos(n*B).^(5/2);