By C. B. Gupta

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**Example text**

2e x 7. Prove that ∇n yx = ∆nyx–n. 8. Evaluate (i) (2∆2 + ∆ – 1) (x2 + 2x + 1) (ii) (∆ + 1) (2∆ – 1) (x2 + 2x +1) (iii) (E + 2) (E + 1) (2x + h + x) (iv) (E2 + 3E + 2) 2x+h + x 9. Write down the polynomial of lowest degree which satisfies the following set of number 0, 7, 26, 63, 124, 215, 342, 511. 10. A third degree polynomial passes through the points (0, – 1) (1, 1) (2, 1) and (3, – 2). Find the polynomial. 11. Construct a forward difference table for x 0 5 10 15 20 25 f(x) 7 11 14 18 24 32 12.

5 − 1) 2 ∆ f (40) + ∆ f (40) 3! 2! 5 − 3) 4 ∆ f (40) 4! 5) (0. 867. Example 7. 5 Solution. 4 By Newton-Gregory forward formula f (x) = f (a) + u ∆ f (a) + u(u − 1) 2 u(u − 1) (u − 2) 3 ∆ f (a) + ∆ f (a) + .... 2! 3! 5 − 1) 2 ∆ f (1) + ∆3 f (1) 3! 2! 41) 2! 3! 4 By Newton-Gregory backward formula f (a + nh + hu) = f (a + nh) + u∇f (a + nh) + u(u + 1) 2 ∇ f (a + nh) 2! 54 + u(u + 1) (u + 2) 3 ∇ f (a + nh) + ..... 3! 41) 3! 2! 439375. Example 8. The population of a town in the decennial census were as under estimate the population for the year 1895 and 1925.

5 from the following data. 5563025 Solution. Here h = 10. 0122345 Stirling formula is u(u 2 − 1) ( ∆3 y −1 + ∆3 y−2 ) u 2 (u 2 − 1) 4 ( ∆y0 + ∆y−1 ) u 2 2 ∆ y–1 + + ∆ y– 2 + ...... + 4! 3! 2 2 2! 0003989) 2 2! 0000025) 3! 2 4! 52827374. yu = y0 + u Example 17. 309. Solution. Here h = 5. 069 39 INTERPOLATION Stirling formula is yu = y0 + 2 2 u( ∆y0 + ∆y−1 ) u 2 2 u(u 2 − 1) ( ∆3 y −1 + ∆3 y−2 ) u (u − 1) 4 + ∆ y–1 + + ∆ y–2 + ..... 2 4! 2! 3! 142) + 3! 2 2! 142) 4! 07669. 1 1. Find the missing term in the following table: x 1 2 3 4 5 y 2 5 7 — 32 2.