Lagrange Polynomial - Proof

Proof

The function L(x) being sought is a polynomial in of the least degree that interpolates the given data set; that is, assumes value at the corresponding for all data points :

Observe that:

  1. In there are k terms in the product and each term contains one x, so L(x) (which is a sum of these k-degree polynomials) must also be a k-degree polynomial.
  2. \ell_j(x_i) = \prod_{m=0,\, m\neq j}^{k} \frac{x_i-x_m}{x_j-x_m}

We consider what happens when this product is expanded. Because the product skips, if then all terms are (except where, but that case is impossible, as pointed out in the definition section—in that term, and since, contrary to ). Also if then since does not preclude it, one term in the product will be for, i.e., zeroing the entire product. So

  1. \ell_j(x_i) = \delta_{ji} = \begin{cases} 1, & \text{if } j=i \\ 0, & \text{if } j \ne i \end{cases}

where is the Kronecker delta. So:

Thus the function L(x) is a polynomial with degree at most k and where .

Additionally, the interpolating polynomial is unique, as shown by the unisolvence theorem at Polynomial interpolation.

Read more about this topic:  Lagrange Polynomial

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