By A.K. Jaiswal, Anju Khandelwal
Salient good points of this paintings contain: mathematical derivation of every process is given to construct the scholars knowing of numerical research; quite a few solved examples are given; laptop courses for the majority numerical equipment mentioned were provided in 'C' language; blunders research for the majority equipment are provided; each one bankruptcy starts off with an advent of involved subject; and, workout questions provide a chance to the scholars to check their realizing of the innovations.
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Additional resources for A Textbook of Computer Based Numerical and Statistical Techniques
5. 85 × l0 in normalized floating-point mode. 6. Explain Machine Epsilon in floating point arithmetic. 6 [Ans. 04963] [Ans. 4485e8] 7. 7320e0, using normalized point arithmetic and proves that they are not the same. Compare with the value of (x2 – 2) + 2x. [Ans. 2000e0 using normalized floating point + 3! 5! arithmetic with 4-digit mantissa. [Ans. 005)] 9. 2583e1 . 8. Find the value of sin x ≈ x − Perform the following operations, and indicate the error in the result, assuming symmetric rounding: 1.
00235 radians δA < 8°5′. We have or Example 35. Find the smaller root of the equation x2 – 30x + 1 = 0 correct to three places of decimal. State different algorithm, which algorithm is better and why? Sol. 966629 Therefore second algorithm is comparatively a better one as this gives the result correct to four figures. Example 36. Find the smaller root of the equation x2 – 400x + 1 = 0 using four-digit arithmetic. Sol. 0000. 0025. 8000 × 10 3 This is the exact root of the given equation. Remark: When two nearly equal numbers are subtracted then there is a loss of significant figures.
5) − 1 = −1 , which is negative. 5) is negative and f(1) is positive. 5 and 1. 875, which is positive. 75) is positive. 75. 297, which is negative. 625) is obtained negative. 75. 625) is negative. 688. 045, which is negative. 688. 093, which is positive. 673. 023, which is positive. 665. 66. Perform five interactions of Bisection method to obtain the smallest positive root of equation f ( x) = x3 − 5x + 1 = 0. Sol. 15) = +ve. 15. 15. 1375. 13125. 13125. 129. x5 = Example 5. 2 by Bisection Method.