Lagrange’s interpolation formula
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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.
![Proof :
Let the given 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](https://image.slidesharecdn.com/lagrangesinterpolationformula-150406095101-conversion-gate01/85/Lagrange-s-interpolation-formula-4-320.jpg)
Lagrange’s interpolation formula
- 1. Presented by- Mukunda Madhav Changmai Roll No: MTHM-22/13 Jorhat Institute of Science and Technology
- 2. About Joseph-Louis Lagrange Joseph-Louis Lagrange 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 the given 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 putting x=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 called Lagrange’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
- 8. Example: find f(x) for the data- Using Lagrange’s interpolation formula : x 0 1 3 f(x)=y 1 3 5 ))(( ))(( )( ))(( ))(( )()( 2101 20 1 1010 21 0 xxxx xxxx xf xxxx xxxx xfxf ))(( ))(( )( 1202 10 2 xxxx xxxx xf 5,3,1,3,1,0 210210 yyyxxx,here )31)(01( )3)(0( 3 )30)(10( )3)(1( 1)( xxxx xf )13)(03( )1)(0( 5 xx )6142( 6 1 2 xx
- 9. Thank You