By C. B. Gupta
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College of CaliforniaEngineering and actual sciences extension sequence. comprises bibliographies. in line with the Statewide lecture sequence on combinatorial arithmetic provided via the collage of California, college Extension, Engineering and actual Sciences department, in 1962.
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From the following table, find the form of the function f (x). x 3 5 7 9 11 f (x) 6 24 58 108 174 24. If lx represents the number of persons living at age x in a life table, find as accurately as the data will permit the value of l47. Given that l20 = 512, l30 = 439, l40 = 346, l50 = 243. ANSWERS 1. 14 2. 31 3. 0086 4. 609 6. 67 7. 25 8. 7 lbs 9. 1295 11. 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.
5), we get an = f (x) = f (a + nh) + (x – a – nh) ∇ 2 f ( a + nh) ∇f ( a + nh) + (x – a – nh) (x – a – (n – 1)h) 2 ! h2 h n + ...... + (x – a – nh) (x – a – (n – 1)h) ...... (x – a – h) ∇ f ( a + nh) n ! 6), we get f (a + nh + hu) = f (a + nh) + u∇f (a + nh) + + ..... + u(u + 1) 2 ∇ f (a + nh) 2! u(u + 1) (u + 2) ...... (u + n − 1) n ∇ f (a + nh). n! 4. , xn not necessarily equally spaced. Since there are (n + 1) values of f (x) so (n + 1)th difference is zero. Thus f (x) is supposed to be polynomial in x of degree n.
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) ......