Curve to fit points I'm trying to find the equation that fits the following points:
$1,0.0105;
2,0.181;
3,0.47;
4,0.755;
5,1.01;
6,1.22;
7,1.39;
8,1.54;
9,1.67;
10,1.79;
11,1.88;
12,1.96;
13,2.03;
14,2.10;
15,2.15;
16,2.21;
17,2.26;
18,2.30;
19,2.34;
20,2.37;
21,2.40;
40,2.73$
To me it looks like tan(theta) except sideways, but how do I get the function to go on its side? See:
http://i421.photobucket.com/albums/pp292/SharkD2161/Support/flowlines_test_plots_zps60e790dc.png
 A: Turning a function "on its side" roughly means exchanging the roles of the $x$ and $y$ axes, which is to say swapping the domain and the codomain. The input becomes the output and the other way 'round. That is, you want (something related to) the inverse of the tangent function, which is called the arctangent and written $\arctan$, $\operatorname{atan}$, or $\tan^{-1}$.
Caution 1: the tangent function is periodic, so it doesn't have an inverse over its entire domain, but if you restrict the domain to $(-\pi/2,\pi/2)$ things should work out well.
Caution 2: Although $\tan^2 x = (\tan x)^2$ by convention, $\tan^{-1} x$ has nothing to do with $(\tan x)^{-1}$, the latter being the cotangent of $x$.
Caution 3: I haven't actually looked at the points, and, more importantly, I don't know where they came from, but just because something looks like an inverse tangent (or any other sort of curve) doesn't mean that that's actually an appropriate curve to fit. Ideally, you would fit a curve that you have some theoretical reason to suspect is the right sort for your data based on what they mean.
A: You could do a polynomial regression
Set up a matrix of any degree you want, see example below
 
Then solve the system of equations if it's manageable or have a CAS solve the system of equations for \begin{align}a0,a1,a2,a3...\end{align}
See here for details
OR you could pop the values into excel, scatter plot the points then add the trend line, exponential or polynomial and the equation as well as it's R^2 value to chose the best fit.
For a logarithmic regression you should get \begin{align}y=0.9023Ln(x)-0.3294\end{align} and an \begin{align}R^2=0.9735\end{align}
For a polynomial regression of degree 3 I get \begin{align}0.0003x^3-0.0162x^2+0.3534x-0.996\end{align} and an \begin{align}R^2=0.983\end{align}
You could do a transformation for a square root fit
A: Many functions are likely to be optimized in order to abtain a good fit with the data set.
For example a polynomial, such as it was already explained.
Another example with an exponential function (3 parameters) is shown in attachment. 
Usually the non-linear Least Squares Fitting requires an initial guess and iterative process for numerical computation. The result given in attachement was obtain thanks to an unusual straightforward method without initial guess and without iteration. The computation method is given page 17 in the paper "Régressions et équations intégrales" published on Scribd :
http://www.scribd.com/JJacquelin/documents
Note : the fitting should be much better if the last point (40 , 2.73) was suppressed

