Hello, I am trying to solve an equation involving two vectors.

I have vector P (3000x1) and T (1x3000). I plotted P vs. T just fine. Now, I have to solve an equation, tau = -T/ln(P/P(0)) I tried this, but it returns 3000 numbers:

if true
  % code
end
for i = 1:3000
tau(i) = -T(i)/log(P(i)/P(1));
end

[Ameer Hamza]

Why do you want one number? tau must be some property of the system which varies with T and P. This tells you that tau have different values at each point.

[Charles Naegele]

Tau is defined as a time constant which I believe should be constant throughout the experiment. I understand that the equation would yield different values depending on P and T, but the objective is to find a single value for tau.

[Abraham Boayue]

What value of tau are you expecting to obtain? Give a few examples that might help explain what the problem should do.

[Charles Naegele]

Also, does the indexing for 'i' make sense considering my vectors contain 3000 values, or should i = 1:3001?

[Charles Naegele]

I think the accepted answer works for my application. Thank you for your time.

Answer 1

[Star Strider]

No need for the loop. Just use the mrdivide,/ (link) function:

tau = -T / log(P')/P(1);

Note the transposition (') operator to create a row vector from ‘P’. That will give you a single value for your ‘tau’ parameter.

[Charles Naegele]

Thank you!

[Star Strider]

As always, my pleasure!