Fit a curve based on its area?

Hi everyone! Basically, I need to estimate a curve based on its integral. I don't know the actual function.

I know my x values will be (0:1000). I know the area when x = 0:30 is 10% of the total area, x = 0:60 is 50% and x = 0:80 is 90%.

I have no idea where to start! I have tried using curve fitting tools but didn't succeed. Can someone help me with that?

Cheers!

[David Goodmanson]

Hello Larissa, if that is all the information you have there are an unlimited number of possible functions. If the domain of the function is really 1:1000 and not 1:100, given that 90% of the integral occurs by x = 80, that would argue that the function might (might!) feature a decreasing exponential, something like (a+bx+cx^2)e^(-gx). But without imposing some kind of functional form you can't make any progress at all.

[Larissa Perez]

There will be no negative values and the curve should have a single peak, looking a bit like a normal distribution, but not symmetric. Does that help?

[Walter Roberson]

Small details like that are important...

[Larissa Perez]

So sorry! I should have said before.

[David Goodmanson]

do you know if the function is zero for x<0, or does it extend into negative values of x?

[Larissa Perez]

It is zero, does not extent into negative x.

[Larissa Perez]

Also, x=60 is the maximum of this function.

[David Goodmanson]

I'm not saying this is the function, but does its shape look somewhat like

x = 0:.01:10;
y = x.*exp(-x);
plot(x,y)

[Larissa Perez]

Yes!!! :)

[David Goodmanson]

Well, the problem now is that you have four conditions, assuming you are confident that the peak is right at x = 60. That means you have to fit an equation with four adjustable parameters. Walter has a solution technique, but so far it assumes a straight polynomial fit. The most obvious four parameter solution that has the shape you are talking about would be (ax + by^2 + cy^3)exp(-gy) [or, because of the nature of the conditions, (x + by^2 + cy^3 + dy^4))exp(-gy) might be needed, I haven't quite figured out which ].

There would be work involved to fit that, especially the coefficient g. If you knew g, which you don't, it would be a lot easier. Someone may have some tricks up his/her sleeve here. But it's a hard enough problem that it makes sense to ask, what is the context here? Are the conditions exact or pretty good but approximate? Is a not so simple many-parameter fit the way to go? Do you have any other information about where this function came from that might indicate what kind of function it is? Is the goal to do interpolation to find the area at other values of x?

[Larissa Perez]

Thank you so so much for your help! It is a particle size distribution. My range is 0:1000 micrometres. And I need to estimate this curve based on the d10, d50 and d90, which are close to the values I have mentioned (33, 60, 95). I tried using a Normal distribution, but I would need to somehow have two sigma values, so it didn't work.

[David Goodmanson]

Your best bet here may be the log normal distribution which is often used for particle sizes and a lot of other situations. You assume the log of particle size is normally distributed. For a good fit, log of the d50 size should be right in between log of the d10 size and log of the d90 size. So log(60) should be halfway between log(33) and log(95). Those three values are

ans =  3.4965    4.0943    4.5539

and not equally spaced. But fitting those three points with a straight line may be the best you can do. That fit is pretty decent, not super.

[Larissa Perez]

I really appreciate your help! Thanks for that. :)

Answer 1

[Walter Roberson]

one of the solutions is

syms a b c d e F(x) f(x)
F(x) = a*x^4+b*x^3+c*x^2+d*x+e
sol = solve([F(30)-F(0)==10,F(60)-F(0)==50,F(80)-F(0)==90,F(1000)-F(0)==100])
f(x) = subs(diff(F(x),x),{a,b,c,d},{sol.a,sol.b,sol.c,sol.d})

For each additional known point you have, add another term to F(x) and add the point to the solve() expression.

This is based upon the fact that the indefinite integral is taken and becomes an actual function, and that the definite integral is determined by the subtraction of the value of the indefinite integral at the end points. F(x) is acting as the indefinite integral. You can then create simultaneous equations out of it and solve for the coefficients. Then substitute those into the derivative of F(x) to get the original f(x)