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

2 views (last 30 days)
Erdem Turan on 14 Nov 2022
Edited: Alan Stevens on 7 Dec 2022
% 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

Alan Stevens on 14 Nov 2022
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
Alan Stevens on 7 Dec 2022
Edited: Alan Stevens on 7 Dec 2022
Looks like it should be
fn = @(n) 2*cos(n*B).^(5/2);

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