How do you calculate the average value of a line?
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Colin Lynch
le 22 Déc 2017
Commenté : Colin Lynch
le 23 Déc 2017
I am attempting to calculate the average value of a line using the equation from this website: http://tutorial.math.lamar.edu/Classes/CalcI/AvgFcnValue.aspx
Part of this equation requires getting a definite integral from x0 to x1. I only want the positive values of these lines, and the maximum value of the y coordinate is 1. To accomplish this, I want the lower of limit of x to be where this line crosses the x-axis (x0), and I want the upper limit of x to be where y = 1 (x1). I tried to do this with the following code:
fun = @(x,c) sqrt((-.25 .*(n-(c.*n))./(100-n))./(((n-(c.*n))./(100-n))-1));
x0 = fzero(@(x)fun(x,.5), 1);
x1 = fzero(@(x)fun(x,.5)-1, 1);
q = (1/(x1-x0))*integral(@(x)fun(x,.5),x0, x1)
This code, however, spits out some strange errors that I do not know how to resolve. How can I fix this so that I can find the average value of this line?
2 commentaires
Ahmed raafat
le 23 Déc 2017
fun = @(x,c) sqrt((-.25 .*(n-(c.*n))./(100-n))./(((n-(c.*n))./(100-n))-1));
it is fun of n & c not x & c
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Walter Roberson
le 23 Déc 2017
x0 = fzero(@(x)fun(x,.5), [0 66.66666666]);
x1 = fzero(@(x)fun(x,.5)-1, [0 66.66666666]);
fun(x,1/2) has a discontinuity at 200/3 but grows fairly slowly, so it is necessary to get fairly close to 200/3 to satisfy x1
4 commentaires
Walter Roberson
le 23 Déc 2017
The discontinuity is at 100/(2 - c) . But you have to make sure you do not use exactly that number or else you will get inf there. So like 100/(2-c) * (1-10*eps)
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Ahmed raafat
le 23 Déc 2017
where is x????
fun = @(x,c) sqrt((-.25 .*(n-(c.*n))./(100-n))./(((n-(c.*n))./(100-n))-1));
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