LAGRANGE'S INTERPOLATION FORMULA
This is again an Nth
degree polynomial approximation formula to the function f(x), which
is known at discrete points xi,
i = 0, 1, 2 . . .
Nth.
The formula can be derived from the Vandermonds determinant but a much
simpler way of deriving this is from Newton's divided difference formula.
If f(x) is approximated with an Nth
degree polynomial then the Nth
divided difference of f(x) constant and (N+1)th
divided difference is zero. That is
f [x0, x1, . . . xn, x] = 0
From the second property of divided difference we can write
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f0
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+
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fn
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fx
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= 0 |
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+ . . . +
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(x0 - x1) . . . (x0 - xn)(x0
-
x)
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(xn - x0) . . . (xn - xn-1)(xn
-
x)
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(x - x0) . . . (x - xn)
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or
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(x - x1) . . . (x - xn)
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(x - x0) . . . (x - xn-1)
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f(x) =
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f0 + . . . +
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fn
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(x0 - x1) . . . (x0 - xn)
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(xn - x0) . . . (xn
- xn-1)
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n
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(
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n
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)fi
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S
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x - xj
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j = 0
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(xi - xj)
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i = 0
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j ¹ 1
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Since Lagrange's interpolation is also an Nth
degree polynomial approximation to f(x) and the Nth
degree polynomial passing through (N+1)
points
is unique hence the Lagrange's and Newton's divided difference approximations
are one and the same. However, Lagrange's formula is more convinent to
use in computer programming and Newton's divided difference formula is
more suited for hand calculations.
Example : Compute
f(0.3)
for the data
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x
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0
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1
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3
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4
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7
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f
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1
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3
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49
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129
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813
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using Lagrange's interpolation formula (Analytic value is 1.831)
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(x - x1) (x - x2)(x-
x3)(x - x4)
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(x - x0)(x - x1)
(x - x2)(x - x3)
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f(x) =
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f0+ . . . +
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f4
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(x0 - x1) (x0 - x2)(x0
-
x3)(x0 - x4)
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(x4 - x0)(x4 - x1)(x4
-
x2)(x4 - x3)
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(0.3 - 1)(0.3 - 3)(0.3 - 4)(0.3 - 7)
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(0.3 - 0)(0.3 - 3)(0.3 - 4)(0.3 - 7)
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=
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1+
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3 +
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(-1) (-3)(-4)(-7)
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1 x (-2)(-3)(-6)
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(0.3 - 0)(0.3 - 1)(0.3 - 4)(0.3 - 7)
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(0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 7)
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49 +
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129 +
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3 x 2 x
(-1)(-4)
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4 x 3 x
1 (-3)
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(0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 4)
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813
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7 x 6 x
4 x 3
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