Polynomial regression interpolation? 
Possible Duplicate:
Writing a function $f$ when $x$ and $f(x)$ are known 

I'm not versed in mathematics, so you'll have to speak slowly...
If I want to fit a curve to the points,
X  Y
1  0.5
2  5.0
3  0.5
4  2.5
5  5.0
6  0.5

Where would I begin? For my purposes, this needs to be a sixth-order fit...

Here is a step by step way of finding a polynomial $f(x)$ such that for $1 \leq i \leq 6$, $f(i)$ has the values you stated. We will "build up" $f(x)$ slowly, calling the intermediate stages $f^{(1)}(x), f^{(2)}(x), \ldots $ and so on.


*

*Set $y = f^{(1)}(x) = 0.5$ where the right side doesn't really depend on $x$ and the $0.5$ was chosen from the given data: when we set $x=1$ in $f^{(1)}(x)$, this (zero-degree) polynomial evaluates to $0.5$, and thus fits the point $(1,0.5)$ that was given to you.

*If we set $x=2$ in $f^{(1)}(x)$, we will get $0.5$ but we really need
to get $5.0$. So, let us test $$y = f^{(2)}(x) = f^{(1)}(x) + c(x-1)
= 0.5 + c(x-1)$$
where $c$ is just some number (for now) to see what happens. If we set $x=1$
in $f^{(2)}(x)$, we get $0.5$ as before (yay!) but
if we set $x=2$, we get $0.5 + c$ and so by choosing $c = 4.5$, we can get
$f^{(2)}(x)$ to give the desired values for both $x = 1$ and $x = 2$.

$f^{(2)}(x) = 0.5 + 4.5(x-1)$  is a polynomial such that $f^{(2)}(1) = 0.5$ and
  $f^{(2)}(2) = 5.0$.\
  Notice that $4.5 = 5.0 - f^{(1)}(2)$ which I will write for future reference
  as $\displaystyle\frac{5.0 - f^{(1)}(2)}{2-1}$.


*$f^{(2)}(3) = 0.5 + 4.5(3-1) = 9.5$ which is not what we want. So,
let us try
$$y = f^{(3)}(x) = f^{(2)}(x) + c(x-1)(x-2)$$
where, as before, we will choose the value of $c$ in just a bit.  Note
that if we set $x$ to either $1$ or $2$ in $f^{(3)}(x)$, that last term
evaluates to $0$ and so $f^{(3)}(1) = f^{(2)}(1) = 0.5$, 
$f^{(3)}(2) = f^{(2)}(2) = 5.0$ and so we are OK there. But, 
$$f^{(3)}(3) = f^{(2)}(3) + c(2)(1) = 9.5 + 2c$$
and so choosing $c = -4.5$ will make $f^{(3)}(3)$ evaluate to 
the desired $0.5$.

$f^{(3)}(x) = 0.5 + 4.5(x-1) - 4.5(x-1)(x-2)$ is a polynomial such that
  $f^{(3)}(1) = 0.5$, $f^{(3)}(2) = 5.0$, and $f^{(3)}(3) = 0.5$.
  Note that
  $-4.5 = \displaystyle\frac{0.5 - f^{(2)}(3)}{(3-1)(3-2)}$.

Since I am getting tired of typing, let me briefly explain how the
rest of the thing will work.  We set 
$$f^{(4)}(x) = f^{(3)}(x) 
+ \frac{2.5 - f^{(3)}(4)}{(4-1)(4-2)(4-3)}(x-1)(x-2)(x-3)$$
where the second term vanishes when we set $x = 1, 2, 3$ and so
we get the desired values of $y$, and when we set $x = 4$,
we get
$$f^{(4)}(4) = f^{(3)}(4) + \frac{2.5 - f^{(3)}(4)}{(4-1)(4-2)(4-3)}(4-1)(4-2)(4-3) = 2.5$$
which is exactly what we want. Lather, rinse, and repeat for
$f^{(5)}(x)$ and $f^{(6)}(x)$.
If you feel inclined to learn more about this idea, look for
Newtonian interpolation on the Internet.
