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

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Erdem Turan
Erdem Turan 2022 年 11 月 14 日
編集済み: Alan Stevens 2022 年 12 月 7 日
% 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

回答 (1 件)

Alan Stevens
Alan Stevens 2022 年 11 月 14 日
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
  4 件のコメント
Alan Stevens
Alan Stevens 2022 年 12 月 7 日
編集済み: Alan Stevens 2022 年 12 月 7 日
Looks like it should be
fn = @(n) 2*cos(n*B).^(5/2);

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