<s>[INST] Writing a function $f$ when $x$ and $f(x)$ are known I'm trying to write a function. For each possible input, I know what I want for output. The domain of possible inputs is small:
$$\begin{vmatrix}
    x &f(x)\\
    0      & 2\\
    1      & 0\\
    2      & 0\\
    3       &0\\
    4       &0\\
    5       &0\\
    6       &1\\
\end{vmatrix}$$
My thought is to start with a function $g(x)$ that transforms $x$ to a value from which I can subtract a function $h(g(x))$ and receive my desired $f(x)$:
$$\begin{vmatrix}
    x &g(x)& h(g(x)) &f(x)\\

    0 &   7&       5  &  2\\
    6 &   6 &      5 &   1\\
    5  &  5  &     5&    0\\
    4   & 4   &    4   & 0\\
    3 &   3    &   3  &  0\\
    2  &  2     &  2 &   0\\
    1   & 1      & 1&    0
\end{vmatrix}$$
But I'm not sure where to go from there, or even whether I'm heading in the right direction. How should I approach creating this function? Is there a methodical way to go about it, or is it trial and error (and knowledge)? Feel free to suggest modifications to how I'm stating my problem, too. Thanks.
 [/INST] You can generalise the problem: suppose you know the value of $f(x)$ for a particular finite set of values of $x$. (Here, you know the value of $f(x)$ when $x=0,1,2,3,4,5,6$.) Then you can find a possible polynomial function $f$ which takes the given values using the following method.
Suppose you know the value of $f(x)$ when $x=x_1, x_2, \dots, x_n$, and that $f(x_j) = a_j$ for $1 \le j \le n$.
Let
$$P_i(x) = \lambda (x-x_1)(x-x_2) \cdots (x-x_{i-1})(x-x_{i+1}) \cdots (x-x_n)$$
Where $\lambda$ is some constant. That is, it's $\lambda$ times the product of all the $(x-x_k)$ terms with $x-x_i$ left out. Then $P_i(x_k) = 0$ whenever $k \ne i$.
We'd like $P_i(x_i) = 1$: then if we let
$$f(x) = a_1 P_1(x) + a_2 P_2(x) + \cdots + a_n P_n(x)$$
then setting $x=x_j$ sends all the $P_i(x)$ terms to zero except $P_j(x)$, leaving you with $f(x_j) = a_jP_j(x_j) = a_j$, which is exactly what we wanted.
Well we can set $\lambda$ to be equal to $1$ divided by what we get by setting $x=x_j$ in the product: this is never zero, so we can definitely divide by it. So we get
$$P_i(x) = \dfrac{(x-x_1)(x-x_2) \dots (x-x_{i-1})(x-x_{i+1}) \dots (x-x_n)}{(x_i-x_1)(x_i-x_2) \dots (x_i-x_{i-1})(x_i-x_{i+1}) \dots (x_i-x_n)}$$
Then $P_i(x_k) = 0$ if $k \ne i$ and $1$ if $k=i$, which is just dandy.
More concisely, if $f$ is to satisfy $f(x_j)=a_j$ for $1 \le j \le n$ then
$$f(x) = \sum_{j=1}^n \left[ a_j \prod_{i=1}_{i \ne j}^n \frac{x-x_i}{x_j-x_i} \right]$$
This method is called Lagrange interpolation.

So in this case, your $x_1, x_2, \dots, x_7$ are the numbers $0, 1, \dots, 6$ and your $a_1, a_2, \dots, a_7$ are, respectively, $2,0,0,0,0,0,1$. Substituting these into the above formula, we get:
$$\begin{align} f(x) &= 2 \times \dfrac{(x-1)(x-2) \dots (x-6)}{(0-1) (0-2) \dots (0-6)} + 0 \times (\text{stuff}) + 1 \times \dfrac{x(x-1) \dots (x-5)}{6(6-1)(6-2) \dots (6-5)}\\
&= 2 \dfrac{(x-1) (x-2) \dots (x-6)}{720} + \dfrac{x(x-1) \dots (x-5)}{720} \\
&= \dfrac{(x-1)(x-2)(x-3)(x-4)(x-5)}{720} \left[ 2(x-6) + x \right] \\
&= \boxed{\dfrac{(x-1)(x-2)(x-3)(x-4)^2(x-5)}{240}} \end{align}$$
You can check easily that this polynomial satisfies the values in your table.

In fact, in this particular case, all the above machinery wasn't necessary. It's plain that $f(x)=0$ when $x=1,2,3,4,5$, and so $x-j$ must divide $f(x)$ for $j=1,2,3,4,5$, and so
$$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)g(x)$$
for some polynomial $g(x)$. Since we're only worried about $x=0,6$ beyond this, i.e. $2$ values of $x$, it suggests we have $2$ free parameters in $g(x)$ and hence $g(x)=ax+b$ is linear. That is, we have
$$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(ax+b)$$
Substituting $x=0$ and $x=6$, respectively, gives
$$\begin{align}2 &= -5! \cdot b \\ 1 &= 5! \cdot (6a+b) \end{align}$$
and solving simultaneously gives $b=-\dfrac{2}{120}$ and $a = \dfrac{3}{720}$, which (after simplification) yields the desired result.
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