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

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Charles Naegele
Charles Naegele 2018-5-18
评论： Charles Naegele 2018-5-18
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
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Charles Naegele
Charles Naegele 2018-5-18
Also, does the indexing for 'i' make sense considering my vectors contain 3000 values, or should i = 1:3001?
Charles Naegele
Charles Naegele 2018-5-18
I think the accepted answer works for my application. Thank you for your time.

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采纳的回答

Star Strider
Star Strider 2018-5-18
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.

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