This document discusses Joseph-Louis Lagrange and interpolation. It provides:
1) A brief biography of Joseph-Louis Lagrange, an Italian mathematician who made significant contributions to calculus and probability.
2) A definition of interpolation as producing a function that matches given data points exactly and can be used to approximate values between points.
3) An explanation of Lagrange's interpolation formula for finding a polynomial that fits a set of data points, including an example of applying the formula.
Presentation introduction by Mukunda Madhav Changmai from Jorhat Institute of Science and Technology.
Joseph-Louis Lagrange, Italian mathematician, known for variational calculus, differential equations, and proving every number can be expressed as a sum of four squares.
Interpolation, a method to create a function matching given data, allows for estimating values between known data points.
Mathematical proof of interpolation using Nth degree polynomial approximation and divided differences.
Presentation of Lagrange's interpolation formula applicable for both equal and unequal intervals.
Example using Lagrange's formula to find function values from discrete data points (0, 1, 3) and their corresponding function values (1, 3, 5).
About Joseph-Louis Lagrange
Joseph-LouisLagrange was an Italian
mathematician and astronomer. Lagrange was one of
the creators of the calculus of variations, deriving the
Euler–Lagrange equations for extrema of functionals
. Lagrange invented the method of solving
differential equations known as variation of
parameters, applied differential calculus to the
theory of probabilities and attained notable work on
the solution of equations. He proved that every
natural number is a sum of four squares .
3.
What is Interpolation?
Interpolation produces a
function that matches the given
data exactly. The function then can
be utilized to approximate the data
values at intermediate points.
Interpolation may also be used to
produce a smooth graph of a
function for which values are
known only at discrete points,
either from measurements or
calculations.
4.
Proof :
Let thegiven function be y=f(x). Let corresponding the
values x0,x1,x2 ……., xn of the argument x, the values of the
function f(x) be f(x0), f(x1), f(x2),……., f(xn-1), f(xn), where the
intervals x1-x0, x2-x1, ….., xn-xn-1 are not necessary equal. 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, xn-1] = 0
5.
Let,
f(x) = A0(x-x1)(x-x2)…..(x-xn-1)(x-xn)+ A1(x-x0)(x-x2)…..(x-xn)+
………. An(x-x0)(x-x1)…..(x-xn-1), ……….(i)
where A’s are constant.
To find A0, A1, A2,……., An We put x=x0,x1, x2 , …xn respectively in
(i). Thus putting x=x0 in (i) we get
f(x0) = A0(x0-x1)(x0-x2)…..(x0-xn)+0+0…
Or
A0 =
)())((
)(
01010
0
nxxxxxx
xf
6.
Similarly by puttingx=x1, we get
A1 = and so on.
Thus,
An =
Substituting these values of A0 ,A1, A2 ,......., An in (i) , we get
)())((
)(
12101
1
nxxxxxx
xf
)())((
)(
110 nnnn
n
xxxxxx
xf
7.
This is calledLagrange’s interpolation formula and can be used
both equal and unequal intervals .
.........
)())((
)())((
)(
)())((
)())((
)()(
12101
20
1
01010
21
0
n
n
n
n
xxxxxx
xxxxxx
xf
xxxxxx
xxxxxx
xfxf
)())((
)())((
)(
110
110
nnnn
n
n
xxxxxx
xxxxxx
xf