Comparing a series expansion to polynomial regression So I don't have a great background in mathematics but I have a quick and hopefully simple question for you guys.
I'm a graduate student and I'm doing some polynomial regression on some thermodynamic data to create a model and I'm kind of struggling a bit with understanding some of the differences between a series expansion and a regression. Maybe it's a subtlety or maybe I'm just ignorant. 
But anyway. Let's say I take a Maclaurin series expansion to third order:
$$f(x) = f(0) + f^\prime(0) x + \frac{1}{2!}f^{\prime\prime}(0) x^2 + \frac{1}{3!}f^{\prime\prime\prime}(0)x^3 + O(x^4)$$
But this is just a polynomial:
$$f(x) = a + bx + cx^2 + dx^3$$
where the coefficients are the derivatives of $x$ and their associated coefficients.
In the series expansion, the coefficients are defined to be the $n^{th}$ derivatives computed at $0$.  But when I compare this to the polynomial expansion, can I really make the identity that each of the coefficients is proportional to the derivative computed at $0$? Or are the values of the coefficients more globally defined since they have been obtained through regression?
The reason I ask is because it has important implications in the language I will use to describe the values of the coefficients in the polynomial fit.
Any help or insight would be greatly appreciated.
Edit.
I have a model where I can compute the internal energy of a system $U$ as a function of the number of particles $N$ and then some extensive property of the system $\xi$, so $U \equiv U(N,\xi)$.
I am ultimately interested in calculating the chemical potential $\mu$, which is the change in the internal energy as particles are added to they system, this is expressed as taking a derivative with respect to $N$:
$$ \mu(N,\xi) = \left(\frac{\partial U}{\partial N}\right)_{\xi}$$
So since I am able to calculate $U(N,\xi)$, I can compute $\mu(N,\xi)$ by fitting a 2D surface (polynomial) to the data and then computing its derivative with respect to $N$.
$$ U(N,\xi) = U(0,0) + U_{ ξ } \xi + U_N N + \frac{1}{2!}U_{ξξ} ξ ^2 + \frac{1}{2!}U_{NN} N^2 + U_{N ξ }N ξ  + \dots$$
This can be rewritten as 
$$ U(N,\xi) = U(0,0) + \left(\frac{\partial U(0,0)}{\partial \xi}\right)_{N} \xi + \mu(0,0) N + \frac{1}{2}\left(\frac{\partial^2 U(0,0)}{\partial ξ^2}\right)_N ξ ^2 + \frac{1}{2}\left(\frac{\partial \mu(0,0)}{\partial N}\right)_{\xi} N^2 + \left(\frac{\partial \mu(0,0)}{\partial \xi}\right) N ξ  + \dots$$
So the derivatives in this model have physical meaning, since they represent thermodynamic variables and derivatives of thermodynamic variables.
So in this case,
$$ \mu(N, ξ ) = \mu(0,0) + \left(\frac{\partial \mu(0,0)}{\partial N}\right)_{\xi} N + \left(\frac{\partial \mu(0,0)}{\partial \xi}\right) ξ + \dots $$
The main question above was if I perform a polynomial regression on my $U(N, ξ )$ data, can I reasonably identify each of the polynomial's coefficients with the $\partial_i U$ values which are supposed be computed at $N =  ξ  = 0$ in the Taylor expansion, or will these coefficients correspond to something else since they are obtained by regression.
I suppose the fitted coefficients should truly correspond to the $N =  ξ  = 0$ case.
 A: Also very late answer, but this thread did help me and might help others by adding the following explanation:
A Taylor expansion produces a best fit polynomial in the vicinity of some point $x_0$. In this regard, the fundamental difference between the polynomial obtained through regression and the one obtained through a Taylor expansion is that a regression fits the function not just at $x_0$ but over an etire domain, albeit at the expense of enough terms.
Hope this helps clarify.
A: I'm not sure if I understand your question properly but...
Take for example series of $\exp$:
$$e^x=a+b\,x+c\,x^2+d\,x^3+...$$
Plug $x=0$ to find $a$:
$$a=e^0=1$$
Differentiate both sides:
$$\frac{d}{dx}e^x=\frac{d}{dx}(a+b\,x+c\,x^2+d\,x^3+...)$$
$$e^x=b+2c\,x+3d\,x^2+...$$
And plug $x=0$ to find $b$:
$$b=e^0=1$$
Differentiate again and plug $x=0$:
$$e^x=2c+6d\,x+...$$
$$2c=e^0=1\implies c=\frac12$$
Repeat to find other coefficients.
A: Seems more than $~3~$ years has happened but maybe this may be useful for someone.
I have tested what you asked on a trial function: $~~~f(x)=x*e^x~$.
I fitted the function using a polinomial function of eight order, and then I obtained the Taylor series expansion around $~x=0~$.
If you plot both function for $~x~$ on $~[0:1]~$ you can see that the functions are not the same.

