Efficient Summation in MATLAB for Biphasic Model: Varying n from 0-3 to 0-100

[tunk]

i need to write this into MATLAB

http://www.engin.umich.edu/class/bme456/ch10fitbiphasic/biphasfit19.gif

which i have done here:

uj = (-sig/Ha)*(xj-(((2*h)/(pi^2))*((((-1)^n)/((n+1/2)^(1/2)))*sin((n+1/2)*((pi*xj)/h))*exp(((-Ha*ko)/(h^2))*((n+1/2)^2)*(pi^2)*t))));

how do i vary n, and get the equation to sum over n=0-3? and then again from n=0-100?

 

[Feldoh]

Call everything inside the summation f(n)

for i=1:4 %n from 0 to 3 case
y(i) = f(i-1)
end

sum(y);

for i=1:101 %n from 0 to 3 case
y(i) = f(i-1)
end

sum(y);

-----

Should work

 

[yrael]

Call everything inside the summation f(n)

for i=1:4 %n from 0 to 3 case
y(i) = f(i-1)
end

sum(y);

for i=1:101 %n from 0 to 3 case
y(i) = f(i-1)
end

sum(y);

-----

Should work

I got the following error for the same equation, following your summation of f(n),
"? Subscript indices must either be real positive integers or
logicals.

Error in ==> scriptrun at 16
f(n)=sum(((abs((-1).^n))/((n+1./2).^(2))));"




My code:

for i=1:3 %n from 0 to 2 case
f(n)=sum(((abs((-1).^n))/((n+1./2).^(2))));
y(i)=f(i-1);
end
sum(y);
uj = (-sig/Ha)*(xj-(((2*h)/(pi^2))*f(n)*sin((n+1/2)*((pi*xj)/h))*exp(((-Ha*ko)/(h^2))*((n+1/2)^2)*(pi^2)*t)));

Any advise?

 

[Feldoh]

You're summing uj = (-sig/Ha)*(xj-(((2*h)/(pi^2))*f(n)*sin((n+1/2)*((pi*xj)/h))*exp(((-Ha*ko)/(h^2))*((n+1/2)^2)*(pi^2)*t))) that's what f(n) is. just calling that expression will give an error, I believe.

 

[blue_raver22]

To vary n from 0-3, you can use a for loop in MATLAB. First, define a variable n that ranges from 0 to 3. Then, within the for loop, you can calculate the summation equation for each value of n and add it to a variable (e.g. sum) that stores the total sum. The code would look something like this:

sum = 0; % initialize the sum variable
n = 0:3; % define the range of n
for i = n % loop through each value of n
% calculate the summation equation for the current value of n
uj = (-sig/Ha)*(xj-(((2*h)/(pi^2))*((((-1)^i)/((i+1/2)^(1/2)))*sin((i+1/2)*((pi*xj)/h))*exp(((-Ha*ko)/(h^2))*((i+1/2)^2)*(pi^2)*t))));
% add the current value of uj to the sum variable
sum = sum + uj;
end
% the final value of sum will be the total summation from n=0 to 3

To vary n from 0-100, you can follow the same approach but change the range of n to 0:100 in the for loop. The code would look like this:

sum = 0; % initialize the sum variable
n = 0:100; % define the range of n
for i = n % loop through each value of n
% calculate the summation equation for the current value of n
uj = (-sig/Ha)*(xj-(((2*h)/(pi^2))*((((-1)^i)/((i+1/2)^(1/2)))*sin((i+1/2)*((pi*xj)/h))*exp(((-Ha*ko)/(h^2))*((i+1/2)^2)*(pi^2)*t))));
% add the current value of uj to the sum variable
sum = sum + uj;
end
% the final value of sum will be the total summation from n=0 to 100