H = 0.1;
I = 5;
Y = 0;
BI = zeros(I,1);
for i = 1:I
b = fzero(@(b)(b*tan(b)-H),pi*i);
BI(i) = b ;
end
X = linspace(0,1,I);
Theta = zeros(I,1);
for k = 1:length(X)
for m = 1:BI
Ts = sum(((((1/BI).*(1-cos(BI)))/(0.5-(1/(4.*BI)))).*sin(BI.*Y).*(cosh(2.*BI.*(X))-tanh(2.*BI).*sinh(2.*BI.*(X)))));
end
end
I'd like to:
  • Start at X =0
  • Sum Ts for all values of BI
  • Step to next value of X
  • Sum Ts for all values of BI
  • etc.

 Akzeptierte Antwort

Edoardo_a
Edoardo_a am 7 Mär. 2023
Bearbeitet: Edoardo_a am 7 Mär. 2023

0 Stimmen

Hi, do you mean something like that?
In this case I sum all the values in the same Ts variable for all the X entry.
If you want to save a different Ts sum for each X entry then you should preallocate and store each Ts from the for loop.
H = 0.1;
I = 5;
Y = 1;
BI = zeros(I,1);
for i = 1:I
b = fzero(@(b)(b*tan(b)-H),pi*i);
BI(i) = b ;
end
X = linspace(0,1,I);
Theta = zeros(I,1);
Ts = zeros(1);
for k = 1:length(X)
for m = 1:length(BI)
Ts =Ts + ((((1/BI(m)).*(1-cos(BI(m))))/(0.5-(1/(4.*BI(m))))).*sin(BI(m).*Y).*(cosh(2.*BI(m).*(X(k)))-tanh(2.*BI(m)).*sinh(2.*BI(m).*(X(k)))));
end
end

1 Kommentar

H = 0.1;
I = 5;
Y = 1;
BI = zeros(I,1);
for i = 1:I
b = fzero(@(b)(b*tan(b)-H),pi*i);
BI(i) = b ;
end
X = linspace(0,1,I);
Theta = zeros(I,1);
Ts = zeros(1);
for k = 1:length(X)
for m = 1:length(BI)
Ts(m) = ((((1/BI(m)).*(1-cos(BI(m)*pi/180)))/(0.5-(1/(4.*BI(m))))).*sin(BI(m)*(pi/180).*Y).*(cosh(2.*BI(m)*(pi/180).*(X(k)))-tanh(2.*BI(m)*pi/180).*sinh(2.*BI(m)*(pi/180).*(X(k)))));
%->>
end
TS(k) = sum(Ts);
end
TS
TS = 1×5
0.0030 0.0029 0.0028 0.0027 0.0027
You can do store the Ts values for each BI iteration using its index, and sum the Ts variable later. Also, input values to trigonometric functions, need to be in radians, for which you can multply with pi/180

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Numerical Integration and Differential Equations finden Sie in Hilfe-Center und File Exchange

Gefragt:

am 7 Mär. 2023

Kommentiert:

am 7 Mär. 2023

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by