Newton's forward interpolation formula

Transcription

1 Newton's Interpolation Formulae Interpolation is the process of approximating a given function, whose values are known at N + 1 tabular points, by a suitable polynomial, P N (x) of degree N which takes the values y k at x = x k for k = 0, 1, 2,, N. Note that if the given data has errors, it will also be reflected in the polynomial so obtained. In the following, we shall use forward and backward differences to obtain polynomial function approximating y = f(x) when the tabular points x k 's are equally spaced. That is x k x k 1 = h. Newton's forward interpolation formula Let where the polynomial P N (x)is given in the following form: f(x) = P N (x) P N (x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a k (x x 0 ) (x x k 1 ) + + a N (x x 0 ) (x x N 1 ).. (1) for some constants a 0, a 1,, a N to be determined using the fact that P N (x k ) = y k for k = 0, 1, 2,, N So, for k = 0 substitute x = x 0 in (1) to get P N (x 0 ) = y 0 This gives us a 0 = y 0. Next, for k = 1 substitute x = x 1 in (1), P N (x 1 ) = y 1

2 Thus y 1 = a 0 + a 1 (x 1 x 0 ) = y 0 + a 1 h Next, for k = 2 substitute x = x 2 in (1), a 1 = y 1 y 0 h = y 0 h P N (x 2 ) = y 2 y 2 = a 0 + a 1 (x 2 x 0 ) + a 2 (x 2 x 0 )(x 2 x 1 ) = y 0 + 2a 1 h + a 2 (2h)(h) 2h 2 a 2 = y 2 y 0 2a 1 h = y 2 y 0 2h ( y 0 h ) = y 2 y 0 2 y 0 = y 2 y 0 2(y 1 y 0 ) = y 2 2y 1 + y 0 a 2 = y 2 2y 1 + y 0 2h 2 = 2 y 0 2h 2 a 2 = 2 y 0 2h 2 Now, using mathematical induction, we get

3 a k = k y 0 k! h k Thus, P N (x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a k (x x 0 ) (x x k 1 ) + + a N (x x 0 ) (x x N 1 ) P N (x) = y 0 + y 0 h (x x 0) + 2 y 0 2h 2 (x x 0)(x x 1 ) + + k y 0 k! h k (x x 0) (x x k 1 ) + + N y 0 N! h N (x x 0) (x x N 1 ).. (2) As this uses the forward differences, it is called NEWTON'S FORWARD DIFFERENCE FORMULA for interpolation, or simply, forward interpolation formula. For calculation purpose, we simplify the formula (2) as done below. Now if we define Then So x p x 0 h = p x p x 0 = ph (x p x 0 )(x p x 1 ) = (x p x 0 )(x p x 0 h) = (ph)(ph h) = h 2 p(p 1)

4 And (x p x 0 )(x p x 1 ) h 2 = p(p 1) (x p x 0 )(x p x 1 ) (x p x k 1 ) h k = p(p 1) (p k + 1) Thus the NEWTON'S FORWARD DIFFERENCE FORMULA for interpolation formula (2) becomes P N (x p ) = y 0 + p y y 0 2! + N y 0 N! p(p 1) + + k y 0 k! p(p 1) (p N + 1).. (3) p(p 1) (p k + 1) + EXERCISE Show that and in general, and If N=1, we have a linear interpolation given by f(x p ) = f(x 0 ) + p f(x 0 )

5 For N=2, we get a quadratic interpolating polynomial: and so on. f(x p ) = f(x 0 ) + p f(x 0 ) + 2 f(x 0 ) p(p 1) 2! It may be pointed out here that if f(x) is a polynomial function of degree then P N (x) coincides with f(x) on the given interval. Otherwise, this gives only an approximation to the true values of f(x).i f we are given additional point x N+1 also, then the error, denoted by is estimated by R N (x) = P N (x) f(x) Newton's backward interpolation formula Similarly, if we assume, P N (x) is of the form then using the fact that P N (x i ) = y i, we have b 0 = y N, b 1 = 1 h y N, b 2 = 1 2! h 2 2 y N, b k = 1 k! h k k y N

6 Thus, using backward differences and the transformation x p = x N + ph, that is p = x p x N backward interpolation formula as follows: h we obtain the Newton's P N (x p ) = y N + p y N + 2 y N 2! + N y N N! p(p + 1) + + k y N k! p(p + 1) (p + N 1).. (4) p(p + 1) (p + k 1) +

7 Example 1: Using Newton s forward interpolation formula find the value of y at x=8 from the following table. X Y Ans. Here, x0 h xp Using Newton s forward interpolation formula x p x 0 h = p = 1.6 P N (x p ) = y 0 + p y y 0 2! + N y 0 N! p(p 1) + + k y 0 k! p(p 1) (p N + 1).. (3) p(p 1) (p k + 1) +

8 Here N = 5, so till fifth order difference we calculate in the following difference table 2 nd 3 rd n x y First y 2 y 3 y 4 th 4 y 5 th 4 y x0 h xp p= (xp-x0)/h p= 1.6 [p] p [p]^ p [p]^ p [p]^ p [p]^ f(xp)= Ans.= 12.77

9 Example 2: The following table gives corresponding values of x and y. Prepare a forward difference table and express y as a function of x. Also obtain y when x=2.5. X Y Ans. First we we form the forward difference table n x y first 2nd 3rd 4th Using the difference table, we can write the Newton forward interpolation polynomial as follows

10 P N (x) = y 0 + y 0 h (x x 0) + 2 y 0 2h 2 (x x 0)(x x 1 ) + 3 y 0 k! h k (x x 0)(x x 1 )(x x 2 ) + 4 y 0 4! h 4 (x x 0)(x x 1 )(x x 2 )(x x 3 ).. (2) = 7 + 4(x 0) (x 0)(x 1) Now putting x = 2. 5, we get = (ANS). 1 (x 0)(x 1)(x 2) + (x 0)(x 1)(x 2)(x 3) 24 P N (2.5) = 7 + 4(2.5 0) (2.5 0)(2.5 1) + 2 (2.5 0)(2.5 1)(2.5 2) (2.5 0)(2.5 1)(2.5 2)(2.5 3) 24 Instead of the polynomial if simply f(2. 5) is asked then we could solve as the following. Here, x0 h xp Using Newton s forward interpolation formula x p x 0 h = p = 2.5

11 P N (x p ) = y 0 + p y y 0 2! + N y 0 N! p(p 1) + + k y 0 k! p(p 1) (p N + 1).. (3) p(p 1) (p k + 1) + Here N = 4, so till fourth order difference we calculate in the following difference table n x y First y 4 2 nd 2 y rd 3 y th 4 y x0 h xp p= (xp-x0)/h p= 2.5 [p] 2.5 p [p]^ p [p]^ p [p]^ f(xp)= Ans.=

12 Example 3: Find the value of an annuity at 5 3 %, given the following table 8 x y Here we have used Newton Forward interpolation to get First y 2 nd 2 y 3 rd n x y 3 y 4 y x0 h xp p= (xp-x0)/h 4 th p= 2.75 [p] p [p]^ p [p]^ p [p]^ f(xp)= Ans.=

13 Example 4. Using Newton Backward Interpolation find the population in thousands in the year 1925, given the data for 5 censuses starting from 1891 till X F(x) Ans. Here, 1925 is near 1931 so we use backward interpolation. xn h xp x p x N h = p Using Newton s backward interpolation formula P N (x p ) = y N + p y N + 2 y N 2! p(p + 1) + + k y N k! = 0.6 p(p + 1) (p + k 1) + + N y N N! p(p + 1) (p + N 1)

14 x f(x) first 2nd 3rd 4th xn h xp p p= -0.6, p+1= 0.4, p+2= 1.4, p+3= 2.4, p+4= 3.4, f(xp)= Ans=