Suppose we are given several consecutive integer points at which a polynomial is evaluated. What information does this tell us about the polynomial? To answer this question, we create the following table, known as the **Difference Table**:

$\begin{array} {l l l l l l l} n & f(n) & D_1(n) & D_2(n) & D_3(n) & \ldots \\ 1 & \\ 2 & \\ 3 & \\ \vdots \\ \end{array}$

In the first column, we fill out the points at which the polynomial is evaluated (which I will assume is $1, 2, 3\ldots$.

In the second column, we fill out the corresponding values of the polynomial at those points.

In the third column, we calculate the difference between two entries in the previous column. This is known as the **first difference** and is given by $D_1 (n) = f(n+1) - f(n)$.

In the fourth column, we calculate the difference between two entries in the previous column. This is known as the **second difference** and is given by $D_2 (n) = D_1 (n+1) - D_1 (n)$.

We continue building out subsequent columns of the table in this way, with $D_{k+1} (n) = D_k (n+1) - D_k (n)$.

For example, if we are given that $f(n)$ is a quadratic polynomial satisfying

$f(1) = 4, f(2) = 3 , f(3) = 4, f(4) = 7 \mbox{ and } f(5) = 12,$

then the difference table is as follows:

$\begin{array} {l l l l l l l} n & f(n) & D_1(n) & D_2(n) & D_3(n) & \ldots \\ 1 & 4 & -1 & 2 & 0 \\ 2 & 3 & 1 & 2 & 0\\ 3 & 4 & 3 & 2\\ 4 & 7 & 5 \\ 5 & 12 & \\ \vdots \\ \end{array}$

Note that since we are not given $f(6)$, we are unable to calculate $D_1(5), D_2(4), D_3(3)$ or $D_4(2)$. Now, there isn't (yet) any reason why this table would ever end; if we are given infinitely many values, we can always calculate the differences of terms. However, here is an interesting fact.

If a polynomial $f(x)$ has degree $k$, then the $k$th difference is constant.

Furthermore, the $k$th difference is equal to $k!$ times the leading coefficient of $f(x)$.

In the above example, we see that $D_2 (1) = D_2(2) = D_2( 3) = 2$. Since $f(x)$ is a quadratic polynomial, the above fact tells us that $D_2(n) = 2$ for all $n$, and this allows us to fill in other entries in the table. We get that $D_2(4) = 2$ so $D_1 (5) = 7$ and hence $f(6) = 19$! These values are previously unknown to us, but the difference table allows us to calculate them without knowing the polynomial $f(x)$!

How do we prove this interesting fact? Let's consider a general polynomial of degree $k$. We have

$f(n) = a_k n^k + a_{k-1} n^{k-1} + \ldots + a_1 n + a_0.$

What is the value of $D_1(n)$? By definition, we have

$D_1(n) = f(n+1) - f(n) = \sum_{i=1}^k a_i [(n+1)^i - n^i]$

By the binomial theorem, we have

$(n+1)^{i} - n^i = {i \choose 1} n^{i-1} + {i \choose 2} n^{i-2} + \ldots + {i \choose i} n^1 + {i \choose i } n^0,$

implying that there are no terms of degree $n^k$ in $D_1(n)$. Furthermore, the coefficient of $n^{k-1}$ comes solely from the term $a_k [(n+1)^k - n^k]$, and is equal to $k a_k$.

We may likewise continue by induction to consider the second difference, third difference, etc. Each time, we see that the degree of the polynomial decreases by 1. Hence, by the time we get to the $k$th difference, it is a polynomial of degree 0. Since the only polynomials of degree 0 are the constants, this implies $D_k(n)$ is a constant polynomial.

What happens to the leading coefficient at each step? We see that it gets multiplied by the degree of the (current) polynomial. Since the degree decreases by 1 at each step, we see that it gets multiplied by $k \times (k-1) \times \ldots \times 2 \times 1 = k!$. Hence, we have $D_k = a_k \times k!$, or that $a_k = \frac{D_k} { k!}$.

## 1. Construct the difference table for the function $f_k(n) = (n-1) \times (n-2) \times \ldots \times (n-k)$ for $n =1$ to $k+1$. Note that $f_k(n)$ is a polynomial of degree $k$.

Solution: This is a specially chosen function. We easily see that:

$f(i) = \begin{cases} 0 & i = 1 \mbox{ to } k \\ k! & i = k+1 \\ \end{cases} .$

This allows us to easily calculate the first difference column as:

$D_1(i) = \begin{cases} 0 & i = 1 \mbox{ to } k-1 \\ k! & i = k \\ \end{cases} .$

Similarly, for the $j$th difference column, we have that

$D_j (i) = \begin{cases} 0 & i = 1 \mbox{ to } k-j \\ k! & i = k-j+1 \\ \end{cases} .$

(*) Show that if $f(n)$ is a polynomial of degree $k$, then $f(n) = f(1) + \sum_{i=1}^k \frac{D_i (1)}{i!} \times f_i (n) .$

In particular, the quadratic polynomial $f$ in the write up above satisfying $f(1) = 4, f(2) = 3 , f(3) = 4, f(4) = 7$ and $f(5) = 12$ is given by $f(n) = n^2 - 4n +7$.

## 2. Prove that there is no polynomial (of finite degree) which satisfies $f(n) = 2^n$ for all positive integers.

Solution: Construct the difference table for $f(n) = 2^n$. Since $D_1(n) = f(n+1) - f(n) = 2^{n+1} - 2^{n} = 2^{n}=f(n)$, we see that $D_1$ is exactly equal to $f$. This remains true for each $k$th difference column, and hence there is no polynomial of degree $k$ that satisfies $f(n) = 2^n$.

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## Comments

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TopNewestCan you help me? Cos It seems that I don't understand. please

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What do you not understand? Which problem are you stuck at?

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in the first since this information is provided that the polynomial is quadratic we can assume a general quadratic plug in the given values and solve for three variables.

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Yes, that is certainly one possible approach. But what happens when there are many more equations? See Worked example 2, how would you prove that there is no polynomial which is equal to $2^n$ on the positive integers?

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Is there a typo in example 1? Should it be $f(n) = f(1) + \sum_{i=1}^k \frac{D_i(1)}{i!}\times f_i(n)$

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Thanks! I've fixed it :)

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@Calvin Lin I tried this method in solving This problem , but something went wrong. Any help please?

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@Calvin Lin Is this technique also called 'Synthetic Division'? Or is it different from it?

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Nope. Synthetic Division is just "Lazy man's long division for linear polynomials". The only thing that you save is writing out all the terms. If you don't understand Synthetic division, then just do long division.

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@Satvik Golechha - Nope. Its different I guess. But unfortunately I'm not good at that one too :(

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