Is there a unique polynomial function $f(x)$ of degree $\lt n$ such that $f(n) = a_n$ where $\{a_n\}$ is a sequence?

Is it true that for every sequence $a$ of $n$ numbers there is exactly one polynomial function $f(x)$ of degree $\leq n$ such that all $f(1)=a_1,f(2)=a_2,\dots f(n)=a_{n}$? If so, is there an algorithm to, given the sequence, generate the coefficients of this function?

Intuitively, I feel like this is true, because:

• Given $a$ and $b$, you can find a polynomial function $f$ that has degree $1$ such that $f(0)=a$ and $f(1)=b$.
• By induction: given coefficients $a_1,a_2\dots a_n$, you can find a polynomial function of degree $n$ such that $f(x)$ yields a constant value for all $x$ in $\{1,2\dots n-1\}$.

Mostly, the reason I want to be able to find such a function is so when my math teacher says "find the function rule" and presents us with an obviously linear function I can give her some strange polynomial that just happens to give the correct answers for those values.

A polynomial degree $\leq n$ modeling $a_1,a_2, \cdots a_n$ would not be unique. Consider $f(1)=1$, $f(2)=4$. Obviously $f(x)=3(x-1)+1$ works, but we can let $f(x)=x^2$.

A polynomial of lowest degree that models $\{a_1,a_2,a_3,...a_n\}$ is unique. A polynomial of lowest degree that fits a set of points is called the Lagrange Polynomial for that set of points. Here we want the Lagrange Polynomial for a very special set of points. A "closed form" is possible:

$$f(x)=\sum_{i=0}^{n-1} {x-1 \choose i} \Delta^i(1)$$

Here $\Delta^0(1)=f(1)$. Also $\Delta$ is defined as the operation mapping $f(x)$ to $f(x+1)-f(x)$. This operation is called the forward difference.Then $\Delta^i(1)$ is the operation iterated $i$ times then evaluated at $1$.

Ex: Find the Lagrange polynomial for $1,3,5$.

$$\color{red}{1},3,5$$

Taking forward differences once gets.

$$\color{red}{2},2$$

Twice,

$$\color{red}{0},$$

This gives a Lagrange polynomial,

$$f(x)=\color{red}{1}{x-1 \choose 0}+\color{red}{2}{x-1 \choose 1}+\color{red}{0}{x-1 \choose 2}$$

$$=2x-1$$

Using this formula, you can troll your teacher. For example if your teacher asks find the rule for $1,3,5,$ then you can use the formula to find a rule for a sequence that goes $1,3,5,\pi$.

$$1,3,5,\pi$$

$$2,2,\pi-5$$

$$0,\pi-7$$

$$\pi-7$$

This gives,

$$f(x)=1+2{x-1 \choose 1}+0+(\pi-7){x-1 \choose 3}$$

$$=1+2(x-1)+(\pi-7)\frac{(x-1)(x-2)(x-3)}{3!}$$

Regarding trolling your math teacher, if you have a linear function $f$ such that $f(1)=a_1,f(2)=a_2,\cdots, f(n)=a_n$ you can create a polynomial function of higher degree with the same property by adding $g(x)=(x-1)(x-2)\cdots(x-n)$ to $f(x)$. Then $f(i)+g(i)=a_i+0=a_i$.

Trolling here is far easier than you're making it out to be. If all you're given is a finite set of points and your teacher asks simply for a function -- not necessarily polynomial -- that fits them, you can just set $f(x_i)=y_i$ for $(x_i,y_i)$ in your list and $f(x)=0$ otherwise.