Let use examine Eq. (1-1) carefully : (χ-χo)(χ-χ₁)…(χ-χk-₁)(χ-χk+₁)…(χ-χn) Lk = ---------------------------------------------------- ........ (1 - 1) (χk-χo) (χk-χ₁)…(χk-χ-₁)(χk-χ-₁).....

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다국어 초록 (Multilingual Abstract)
Let use examine Eq. (1-1) carefully : (χ-χo)(χ-χ₁)…(χ-χk-₁)(χ-χk+₁)…(χ-χn) Lk = ---------------------------------------------------- ........ (1 - 1) (χk-χo) (χk-χ₁)…(χk-χ-₁)(χk-χ-₁).....
Let use examine Eq. (1-1) carefully :
(χ-χo)(χ-χ₁)…(χ-χk-₁)(χ-χk+₁)…(χ-χn)
Lk = ---------------------------------------------------- ........ (1 - 1)
(χk-χo) (χk-χ₁)…(χk-χ-₁)(χk-χ-₁)....(χk-χn)
At first glance, this expression appears to be very complicated.
However, it is not difficult to see that Eq. (1- 1) is equivalent to the following shortand version :
n χ-χi
Lk(k)= ┰ --------- .........................(1 - 2)
i=0, i≠k χk-χ-i
n
y = Pn(χ) = Lo(χ)yo + L₁(χ)y₁+ ............... + Ln(χ)yn = Σ Lk(χ)yk ........(1 - 3)
k=0
where the pi notation has the meaning
┰n = (χ-χo) (χ-χ₁) ... (χ-χn)
d
┰n' = ---- n ( χ)
dχ
Substituting Eq. (1 - 2) into Eq. (1 - 3), We obtain
n ┰n(χ)
y - Σ ---------------- yi (i=0, 1, 2, .........., n)
i=0 (χ-χi) ┰n (χi)
This equatin is known as the Lagrange interpolation formula for unequally spaced data.