A textbook of Computer Based Numerical and Statiscal Techniques part 10 ppt
A textbook of Computer Based Numerical and Statiscal Techniques part 10 ppt
... next approximation can be obtained by using the values of x 1 and x 2 in Secant method. Similarly, other approximations can be obtained by using two recent values of approximations. These are x 3 = ... f(x 5 ) = – 0.00002 ALGEBRAIC AND TRANSCENDENTAL EQUATION 77 Since, imaginary roots occur in conjugate pairs roots are 1.915 ± 1.908i upto 3 places of decimal. Assuming other pair...
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A textbook of Computer Based Numerical and Statiscal Techniques part 1 ppt
... This page intentionally left blank This page intentionally left blank A TEXTBOOK OF COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Anju Khandelwal M.Sc., Ph.D. Department of Mathematics SRMS ... Based Numerical and Statistical Techniques is primarily written according to the unified syllabus of Mathematics for B. Tech. II year and M.C .A. I year students of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 4 ppt
... 0.1 0.05 42 d dd · Example 32. In a ∆ABC, b = 9.5 cm, c = 8.5 cm and A = 45 o , find allowable errors in b, c, and A such that the area of ∆ABC may be determined nearest to a square centimeter. Sol. Let area of ... semantics are also specified in detail by the IEEE standards, are not always handled the same way. It turns out that many manufacturers believe (sometimes rightly...
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A textbook of Computer Based Numerical and Statiscal Techniques part 6 pptx
... negative. Thus f(0.5) is negative and f(1) is positive. Then the root lies between 0.5 and 1. 40 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The second approximation to the root ... f(0) is negative and f(1) is positive, therefore, a root lies between 0 and 1. ALGEBRAIC AND TRANSCENDENTAL EQUATION 37 Here e i and e i + 1 are the errors in...
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A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx
... the argument half way between the arguments at q and r is + B A, 24 where A is the arithmetic mean of q and r and B is arithmetic mean of 3q – 2p – s and 3r – 2s – p. Sol. Given A is the arithmetic ... 0.7880 Example 7. Following are the marks obtained by 492 candidates in a certain examination: 04040454550505555606065 . 2104 354743279 Marks No of Candidates −−−−−− Fin...
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A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx
... gives best estimation when 11 . 44 u−< This formula is obtained by taken mean of Gauss forward and Gauss backward difference formula. Gauss forward formula for interpolating central difference ... 4! −− −− +−+ = 147.2251 Approx. Example 5. In an examination, the number of candidates who obtained marks between certain limits are as follows: Marks No of candidates −−−−−0192039405...
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A textbook of Computer Based Numerical and Statiscal Techniques part 27 ppt
... result. Example 7. The following are the mean temperatures (°F) on three days, 30 days apart round the pds. of summer and winter. Estimate the app. dates and values of max. and min. temperature. INTERPOLATION ... differences of a polynomial of n th degree are constant. Let f(x) = A 0 x n + A 1 x n–1 + + A n–1 x + A n by a polynomial of degree n provided A 0...
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A textbook of Computer Based Numerical and Statiscal Techniques part 30 pptx
... BASED NUMERICAL AND STATISTICAL TECHNIQUES after simplication, we get () () () () () +++ = ∫∫ ∫∫ 01 bb b b n n aa a a a wxdx a xwxdx a xwxdx wxfxdx () () () () () 21 01 bb b b n n aa a a a xw ... dxa xdxa xdxa xdx 1111 3 5/2 2 4 012 0000 x dxa xdxa xdxa xdx=++ ∫∫∫∫ or Simplifying above equations, we get 12 0 012 0 12 2 233 2 2345 2 3457 aa a a aa aaa ++= ++ = ++= 2...
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A textbook of Computer Based Numerical and Statiscal Techniques part 35 pptx
... numerical solutions of differential equations. In researches, especially after the advent of computer, the numerical solutions of the differential equations have become easy for manipulations. Hence, ... the values of 1 y and its derivativies in Taylor series expansion, we obtain () () () () 23 10 10 1101 1 1 2! 3! xx xx yx y x x y y y −− ′ ′′ ′′′ =+ − + + + 330 COMPU...
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A textbook of Computer Based Numerical and Statiscal Techniques part 40 pptx
... sensitive to small changes in A and B i.e., small change in A or B causes a large change in the solution of the system. On the other hand if small changes in A and B give small changes in the solution, the ... find that 12 iteration are necessary in Gauss-Jacobi Method to get the same accuracy as achieved by 7 iterations in Gauss-Seidel method. SOLUTION OF SIMULTANEOUS LINEAR...
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