Function that calls functions

1 次查看(过去 30 天)
These first two codes are codes for an integration...
Integral 1....
function I=Integral_Numerical_1_potential(f,M,a,b)
h = (b-a)/2;
I1 = f(1); % First term of approximation
I2 = 0; % Initialization variable for second term of approximation
I3 = 0; % Initialization variable for third term of approximation
I4 = f(M); % Last term of approximation
if(length(f)>1)
for i = 2:2:M
I2 = I2 + f(i)+f(i+1);
end
I2 = 4*I2;
for i = 3:2:M-1
I3 = I3 + f(i);
end
I3 = 2*I2;
end
I = (h/3)*(I1+I2+I3+I4);
end
Integral 2.....
function I=Integral_Numerical_2(f,M,a,b)
h = (b-a)/2;
I1 = f(1); % First term of approximation
I2 = 0; % Initialization variable for second term of approximation
I3 = 0; % Initialization variable for third term of approximation
I4 = f(M); % Last term of approximation
if(length(f)>1)
for i = 2:3:M-1
I2 = I2 + f(i)+f(i+1);
end
I2 = 3*I2;
for i = 4:3:M-1
I3 = I3 + f(i);
end
I3 = 2*I2;
end
I = (3*h/8)*(I1+I2+I3+I4);
end
They are both called by the following function on the basis of if the value of m is even integral 1 is used, if the value of m is divisible by 3 it uses integral 2 and if neither of those apply it combines both integrals to find the answer.....
function nfin=FinEfficiency(R,T)
M=length(R)-1;
a=T;
b=1;
nfin=0;
if mod(M,2)==0 && M>2
nfin=2*Integral_Numerical_1_potential(R,M,a,b)/(1-a^2)
elseif mod(M,3)==0 && M>3
nfin=2*Integral_Numerical_2(R,M,a,b)/(1-a^2)
else
nfin=2*(Integral_Numerical_1_potential(R(1:4),3,a,b)+Integral_Numerical_2(R(4:end),M-3,a,b))/(1-a^2)
end
end
I keep getting an error in this last function saying index exceeds the number of array elements (1) in line 12 whenever I try to run it and I'm not sure how to fix it.
line 12 = nfin=2*(Integral_Numerical_1_potential(R(1:4),3,a,b)+Integral_Numerical_2(R(4:end),M-3,a,b))/(1-a^2)

采纳的回答

per isakson
per isakson 2020-4-25
Caveat: I'm guessing.
Replace
nfin=2*(Integral_Numerical_1_potential(R(1:4),3,a,b)+Integral_Numerical_2(R(4:end),M-3,a,b))/(1-a^2)
by
i1 = Integral_Numerical_1_potential(R(1:4),3,a,b);
i2 = Integral_Numerical_2(R(4:end),M-3,a,b);
nfin = 2*( i1 + i2 )./(1-a.^2);
What values of R and T do you use when calling FinEfficiency() ?
  3 个评论
per isakson
per isakson 2020-4-25
编辑:per isakson 2020-4-25
"I was just testing with R = .36 and T = 1"
With these input values the error should not come as a surprise. R(1:4) will throw the error. And on the following line R(4:end) will return an empty double.
This returns a numerical result
>> nfin=FinEfficiency( 0.36*ones(50,1),(301:350)')'
nfin =
Columns 1 through 10
0.13192 0.13149 0.13105 0.13062 ...
Ryan Mulligan
Ryan Mulligan 2020-4-25
Ah that would make sense, plugged in vectors and it all worked fine haha, thank you for the help.

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Sparse Matrices 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by