I'm trying to curve fit 3 data points. The system has the characteristics of the general function:

Y(x) = C*(1-e^(-b*x))

I have 3 ordered pairs (0,0),(y1,x1),(y2,x2).

I can solve for the undetermined constants (C & b) numerically from;

y_1 = C*(1-e^(-b*x_1))

y_2 = C*(1-e^(-b*x_2))

The solution (0,0) is naturally a consequence of the equations form.

However, after I do that the curve fits (0,0) and (x_1,y_1), but falls short on matching (x_2,y_2). What is the explanation for this?

Originally posted by joe shmo
I'm trying to curve fit 3 data points. The system has the characteristics of the general function:

Y(x) = C*(1-e^(-b*x))

I have 3 ordered pairs (0,0),(y1,x1),(y2,x2).

I can solve for the undetermined constants (C & b) numerically from;

y_1 = C*(1-e^(-b*x_1))

y_2 = C*(1-e^(-b*x_2))

The solution (0,0) is naturally a consequence of the eq ...[text shortened]... s (0,0) and (x_1,y_1), but falls short on matching (x_2,y_2). What is the explanation for this?

Originally posted by iamatiger
That is a hard equation to solve!

how did you do it?!

Any chance of your initial numbers? It is an interesting conundrum, but I think the solution does depend quite strongly on the numbers because of the powers (and can the values be complex numbers, by the way?)

Originally posted by iamatiger
Any chance of your initial numbers? It is an interesting conundrum, but I think the solution does depend quite strongly on the numbers because of the powers (and can the values be complex numbers, by the way?)

Looks like logarithms might help

Originally posted by joe shmo
Sure,

(y,x)
(0,0)
(3430,6.417)
(4250,16.917)

The solution for "b" I found was between 0.26 - 0.27, C = 4201.78

If you mean the values of the constants being imaginary, I would say probably not.

I found that eliminating B worked

B = -ln(1-y1/C)/x1

which then gives

x2/x1 = ln(1-y2/C)/ln(1-y1/C)

ln(1-y2/C)/ln(1-y1/C) - x2/x1 crosses zero when C is between 4316.498599 and 4316.4986

so I estimated C to be 4316.4985995 and calculated B to be 0.24667606

These values fit your input data very well indeed

However eliminating C also worked:

C = y1/(1-exp(1-Bx1)
y2/y1 = (1-exp(-Bx2))/(1-exp(-Bx1))

{where exp(x) means e^x}

these gave exactly the same values for B and C, and look the same as your equations, so perhaps there was a typo somewhere when you did you lathering,, rinsing and repeating? Perhaps did your function exp() mean 10^x or something like that?

looking at what you said again, I don't suppose you wrote b down wrongly Joe?
b is between 0.246 and 0.247 ....

Originally posted by iamatiger
looking at what you said again, I don't suppose you wrote b down wrongly Joe?
b is between 0.2[b]46 and 0.247 ....[/b]