Intuitive Derivation: Lagrange Interpolation

Normally, finding a curve that passes through specific points involves solving a complex system of linear equations. Lagrange Interpolation offers a smarter way: instead of solving equations, we construct the answer piece by piece using a concept similar to a "Switchboard."

The Core Idea:
We want to build a separate "sub-polynomial" for each data point that acts like a switch:

It is ON (value = 1) at its own specific time $t$.
It is OFF (value = 0) at every other data point's time.

1. Simple Example: 2 Points

Suppose we have two data points: $(t_0, y_0)$ and $(t_1, y_1)$.

We need a line $P(t)$ that passes through both.

Step A: The First Switch $L_0(t)$

We want a function $L_0(t)$ that is 1 at $t_0$ and 0 at $t_1$.

Make it Zero: To force it to be zero at $t_1$, we include the term $(t - t_1)$.
Check: If $t = t_1$, then $(t_1 - t_1) = 0$. Success.

Make it One: Currently, at $t_0$, the value is $(t_0 - t_1)$, which is just some number. To make it equal 1, we divide by that exact number.
$$L_0(t) = \frac{t - t_1}{t_0 - t_1}$$

Step B: The Second Switch $L_1(t)$

Now we want a function $L_1(t)$ that is 0 at $t_0$ and 1 at $t_1$.

Using the same logic, we flip the terms:

$$L_1(t) = \frac{t - t_0}{t_1 - t_0}$$

Step C: Combine with Heights

Now we just scale each switch by the desired height ($y$) and add them up:

$$P(t) = y_0 \cdot L_0(t) + y_1 \cdot L_1(t)$$ $$P(t) = y_0 \frac{t - t_1}{t_0 - t_1} + y_1 \frac{t - t_0}{t_1 - t_0}$$

2. Generalizing to $n$ Points

Imagine we have many points: $(t_0, y_0), (t_1, y_1), \dots, (t_n, y_n)$.

For any specific point $j$, we want to build a switch $L_j(t)$ that is:

0 at all other $t_k$ (where $k \neq j$)
1 at exactly $t_j$

Constructing the Numerator (The "Zero" Maker)

To make the polynomial zero at all other points, we just multiply linear terms for every other point:

$$(t - t_0)(t - t_1)\dots(t - t_{j-1})(t - t_{j+1})\dots(t - t_n)$$

We write this compactly using the product symbol $\Pi$:

$$\text{Numerator} = \prod_{k \neq j} (t - t_k)$$

Constructing the Denominator (The "One" Maker)

To ensure the value is 1 when $t = t_j$, we divide the numerator by itself, but evaluated at $t_j$:

$$\text{Denominator} = \prod_{k \neq j} (t_j - t_k)$$

The Basis Polynomial

Putting it together, the basis polynomial for point $j$ is:

$$L_j(t) = \prod_{k \neq j} \frac{t - t_k}{t_j - t_k}$$

3. The Final Formula

Since each switch $L_j(t)$ handles the value for point $j$ independently without affecting the others, the final interpolation polynomial is simply the weighted sum of all switches:

$$P(t) = \sum_{j=0}^{n} y_j L_j(t)$$