I have experimental data as seen below and I need to find an equation that fits it.
X Y
1.27 0.00073
6.37 0.00683
12.73 0.01261
31.83 0.02681
63.66 0.03914
127.32 0.04840
254.65 0.05412
509.30 0.05730
999.49 0.05896
My best bet so far was to assume that it was an exponential equation of the form :
y=(1-a^(-x/b))*.06
I know that the last term is .06 since as per the theory, I have an asymptote at y = .06. I also know that at x = 0, I have y = 0. I have included a graph with the experimental data (blue) and the fit with the equation (purple) below with a = 1.6 and b = 40. It looks like it but it is not good enough for my purposes.
My question is : Is this an exponential equation or should I look for something else? If it is an exponential equation, how can I adjust the coefficient to fit the curve? Bear in mind I am an engineer, not a mathematician!
Thanks.
I think that there is problem with the model itself $$y=0.06(1-a^{-\frac xb})\implies y=0.06(1-e^{-x\frac{\log(a)}b})=0.06(1-e^{-\alpha x})$$ and there is only one parameter to be adjusted, not two.
As Hans Engler commented, let us give more flexibility to the model considering $$y=0.06-\alpha e^{-\beta x}$$
A nonlinear regression leads to $\alpha=0.0584266$ and $\beta=-0.0153205$ corresponding to an $R^2=0.996263$ which looks good at first glance. However, as you noticed, the function goes too fast to the asymptote (just as you observed) and the calculated values are really overestimated for low values of $x$. This is shown in the table below. $$\left( \begin{array}{cccc} i & x_i & y_i & \text{calc}\\ 1 & 1.27 & 0.00073 & 0.00267 \\ 2 & 6.37 & 0.00683 & 0.00701 \\ 3 & 12.73 & 0.01261 & 0.01193 \\ 4 & 31.83 & 0.02681 & 0.02412 \\ 5 & 63.66 & 0.03914 & 0.03797 \\ 6 & 127.32 & 0.04840 & 0.05233 \\ 7 & 254.65 & 0.05412 & 0.05882 \\ 8 & 509.30 & 0.05730 & 0.05998 \\ 9 & 999.49 & 0.05896 & 0.06000 \end{array} \right)$$
The model inadequacy is revealed from a scatter plot of $\log(0.06-y)$ as a function of $x$. It is not close to a straight line.
Since the model "looks" to be exponential, we can think about the sum of two exponential functions $$y=0.06-\alpha e^{-\beta x}-\gamma e^{-\delta x}$$ and I think that I found a good one (I think) $$y=0.06-0.0147534 e^{-0.00337241 x} -0.0459384 e^{-0.0258599 x}$$ which gives $$\left( \begin{array}{cccc} i & x_i & y_i & \text{calc}\\ 1 & 1.27 & 0.00073 & 0.00086 \\ 2 & 6.37 & 0.00683 & 0.00660 \\ 3 & 12.73 & 0.01261 & 0.01281 \\ 4 & 31.83 & 0.02681 & 0.02658 \\ 5 & 63.66 & 0.03914 & 0.03924 \\ 6 & 127.32 & 0.04840 & 0.04869 \\ 7 & 254.65 & 0.05412 & 0.05369 \\ 8 & 509.30 & 0.05730 & 0.05735 \\ 9 & 999.49 & 0.05896 & 0.05949 \end{array} \right)$$ to which correspond $R^2=0.999949$ which is not bad. $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.0147534 & 0.0016224 & \{0.010249,0.019258\} \\ b & 0.0033724 & 0.0004319 & \{0.002173,0.004571\} \\ c & 0.0459384 & 0.0015420 & \{0.041657,0.050220\} \\ d & 0.0258599 & 0.0012963 & \{0.022261,0.029459\} \\ \end{array}$$