By Gupta, C.B.,Malik, A.K. , Kumar, Vipin

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2), we get two equation in missing X1 and X2. On solving we get X1 and X2. 2. , a + nh respectively where h is the difference of the arguments. Let f (x) be a polynomial in x of degree n. So f (x) can be written as f (x) = a0 + a1(x – a) + a2 (x – a) (x – a – h) + a3 (x – a) (x – a – h) (x – a – 2h) + ...... + an (x – a) (x – a – h) ..... h 2h2 a2 = f (a + 2h) – 2ha1 – a0 a2 = = f ( a + 2h) − 2( f ( a + h) − f ( a)) − f ( a) 2 ! h2 f ( a + 2h) − 2 f ( a + h) + f ( a) ∆2 f ( a) = 2 ! h2 2 ! h2 23 INTERPOLATION Proceeding in the same way, we get a3 = ∆3 f ( a) ......

Solution. 4. = Example 12. Find the value of y at x = 5 given that: x 1 3 4 8 10 y 8 15 19 32 40 Solution. 746. Example 13. Apply Lagrange’s formula to find the cubic polynomial which includes the following values of x and y. x 0 1 4 6 yx 1 –1 1 –1 36 ADVANCED MATHEMATICS Solution. Here and x0 = 0, x1 = 1, x2 = 4, x3 = 6 y0 = 1, y1 = – 1, y2 = 1, y3 = – 1 Putting the above values in Lagrange’s formula, we get f (x) = ( x − 0) ( x − 1) ( x − 6) ( x − 1) ( x − 4) ( x − 6) ( x − 0) ( x − 4) ( x − 6) 1+ (–1) + 1 ( 4 − 0) (4 − 1) (4 − 6) ( 0 − 1) (0 − 4) (0 − 6) (1 − 0) (1 − 4) (1 − 6) + =– ( x − 0) ( x − 1) ( x − 4) (–1) (6 − 0) (6 − 1) (6 − 4) 1 3 1 1 3 [x3 – 11x2 + 34x – 24] – [x – 10x2 + 24x] – [x – 7x2 + 6x] 15 24 24 – =– 1 [x3 – 5x2 + 4x] 60 1 3 3 2 10 x + x – x + 1.

X3 – 9x2 + 17x + 6 14. 47692 15. 395 16. 32 18. 788003 19. 15 20. 875 21. 5280 22. 87 23. 2x2 – 7x + 9 24. 274. 5. 090 10. x2 + 2x + 3 17. 84 3 CHAPTER Solution of Linear Simultaneous Equations INTRODUCTION Simultaneous linear equations occur in the field of science and engineering like as analysis of a network under sinusoidal steady-state condition, determination of the output of a chemical plant and finding the cost of reaction, the analysis of electronic circuits having a number of invariant element etc.

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