4. Construction of the functions $p_i (t)$

To keep things simple, let's start with the case where $n = 2$ and $t_0 = 3$, $t_1 = 5$, $t_2 = 8$, and let's just try to find $p_0 (t)$. Thus we want $p_0 (t)$ so that $p_0 (3) = 1$, $p_0 (5) = 0$, $p_0 (8) = 0$.

As a first goal, let's just try to make a polynomial that is zero at $t =5$ and at $t =8$ but not at $t = 3$. That is easy: The function $(t - 5)(t - 8)$ works. It is nonzero at $t = 3$ because the value there is a product of two nonzero factors.

However, at $t = 3$ the value is $10$ and not $1$ as desired. Fortunately, all we need to do is to divide the function by the 10 and we're done:

$p_0 (t) = {\frac{\displaystyle (t-5)(t-8)}{\displaystyle 10}}$.

To show where this result came from, we could write $p_0$ as

$p_0 (t) = {\frac{\displaystyle (t-5)(t-8)}{\displaystyle (3-5)(3-8)}}$.

Now we're ready for $p_1 (t)$. The same reasoning gives

$p_1 (t) = {\frac{\displaystyle (t-3)(t-8)}{\displaystyle (5-3)(5-8)}}$. You can check that $p_1$ at $t = 3, 5, 8$ has values $0, 1, 0$ respectively. Similarly, $p_2 (t) = {\frac{\displaystyle (t-3)(t-5)}{\displaystyle (8-3)(8-5)}}$.

Now it should be clear how to make $p_0 (t)$..., $p_n (t)$ for general $n$: $p_i (t)$ is a fraction whose numerator is a product of factors $(t - t_j)$ except for $j = i$; the denominator is a product of the same factors except with $(t_i - t_j)$ instead of $(t - t_j)$. To write all this more compactly, it is handy to use a sign $\prod$ (capital pi) for a product, the same way that $\sum$ is used for summation.

Theorem 4.1 . The solution to the polynomial interpolation problem can be expressed as

$P(t) = p_0 (t) P_0 + \dots + p_n (t) P_n$, where for each $i$,

$p_i (t) = {\frac{\displaystyle \textstyle \prod_{j \neq i} (t - t_j)}{\display... ...od _ {j \neq i}} {\frac{\displaystyle t - t _ j}{\displaystyle t _ i - t _ j}}$.

Proof of Theorem . Just substitute in $t = t_k$, and you will find that each $p_j$ has the right value, and consequently $P(t_k) = P_k$. Observe that each $p_i (t)$ has degree $n$. Each coordinate function of $P (t)$ is a linear combination of $p_0 (t),\dots, p_n (t)$ and so has degree at most $n$.

Proof of Theorem . The existence of the required solution is shown by Theorem . The uniqueness is shown in the Exercises.

Note. The polynomials $p_i (t)$ depend only on $n$ and the times $t_0,\dots, t_n$, and not on the data points $P_i$. In Figure are shown graphs of the polynomials $p_i (t)$ for $n = 4$ and times $0,1,2,3,4$.

Figure: Graphs of basis polynomials