A textbook of Computer Based Numerical and Statiscal Techniques part 8 doc
A textbook of Computer Based Numerical and Statiscal Techniques part 8 doc
... 1.79630762 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 2.6.2 Rate of Convergence of Iteration MethodLet f(x) = 0 be the equation which is being expressed as x = g(x). The iterative formula for ... (3.7 781 5) = 3. 788 63ALGEBRAIC AND TRANSCENDENTAL EQUATION65Example 5. Find the real root of equation f(x) = x3 + x2 – 1 = 0 by using iteration method.Sol. Here, f(0)= – 1 and f(1) = 1 so a root ... approximated by 3.646 correct to three decimal places.ALGEBRAIC AND TRANSCENDENTAL EQUATION63x3= g (x2) = g (3. 788 63) = 3. 789 24x4= g (x3) = g (3. 789 24) = 3. 789 27Hence, the root of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 62 docx
... the RangeLower limit a = 0Upper limit b = 6Enter the number of subintervals = 6Value of the integral is: 1.35715 98 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Enter the value of y2 ... ALGORITHM FOR TRAPEZOIDAL RULEStep 1. Start of the program for numerical integrationStep 2. Input the upper and lower limits a and bStep 3. Obtain the number of subinterval by h = (b a) /nStep 4. ... 1.10 384 6.13.30 ALGORITHM FOR SIMPSON’S 1/3 RULEStep 1. Start of the program for numerical integrationStep 2. Input the upper and lower limits a and bStep 3. Obtain the number of subinterval...
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A textbook of Computer Based Numerical and Statiscal Techniques part 9 docx
... 66 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Putting i = 2 in (1) x3 = 0.02439Therefore reciprocal of 41 is 0.0244.Example 8. Find the square root of 20 correct to 3 decimal places ... x3 up to four decimal places. So we have 12 = 3.4641.72 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = ()()100.4343 2 .81 33 1.22.41 28 log 2 .81 33 0.4343 0 .88 35+=+or x2= 2.7411Putting ... 0.6071 68 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From which we have, h = – ()()00fxfx′, where [f ′(x0)¹ 0].Hence, if x0 be the initial approximation, then next (or first) approximation...
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A textbook of Computer Based Numerical and Statiscal Techniques part 13 doc
... 0.07 38 ∇ and 4 log 50 = 0.05 08 ∇−Example 10. Given that:123456 78 1 8 27 64 125 216 343 512xyConstruct backward difference table and obtain 4 ()f8∇.1 08 COMPUTER BASED NUMERICAL AND ... 22()dfxdx and so on.The operator ∆ is an analogous to the operator D of differential calculus. In finite differences,we deal with ratio of simultaneous increments of mutually dependent quantities ... operator E is called shift operator or displacement or translationoperator. It shows the operation of increasing the argument value x by its interval of differencingh so that.CALCULUS OF FINITE...
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A textbook of Computer Based Numerical and Statiscal Techniques part 15 doc
... values of the independent variables are atequal interval.Proof: Consider the polynomial f(x) = a 0 + a 1x + a 2x2 + + a nxn (1)Where n is a positive integer and a 0, a 1, a 2, a n ... 12.Similarly we have obtained ∆3u0 = 6 and ∆4u0, ∆5u0 , are all zero as ur = r3 is a polynomial of third degree.130 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES PROBLEM ... functional values:Sol.12345672 5 10 18 26 37 50xy134 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From this table we have the third differences are quite iregular and the irregularity...
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A textbook of Computer Based Numerical and Statiscal Techniques part 21 docx
... 33.1162109]192 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 5. Apply Gauss forward formula to find a polynomial of degree three which takes the values of y as given on next page:2 4 681 021 382 0xy −2317 ... 0.256 0.0 48 26fu−−=− + × + × + × = 0.04 787 5190 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 4. Use Gauss’s forward formula to find a polynomial of degree four which takes thefollowing ... −−+−−∆−− ∆− 188 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now, let taking origin at 1.15, here h = 0.05Then, 1.17 1.150.40.05u−==On applying Gauss’s forward interpolation, we have...
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A textbook of Computer Based Numerical and Statiscal Techniques part 31 docx
... DIFFERENTIATIONThe method of obtaining the derivatives of a function using a numerical technique is known as numerical differentiation. There are essentially two situations where numerical differentiation ... Newtonforward formula, and if the same is required at a point near the end of the set of given tabular294INTERPOLATION WITH UNEQUAL INTERVAL 289 Example 1. Obtain cubic spline for every subinterval, ... + NUMERICAL DIFFERENTIATION AND INTEGRATION295values, then we use Newton’s backward interpolation formula. The central difference formula(Bessel’s and Stirling’s) used to calculate value...
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A textbook of Computer Based Numerical and Statiscal Techniques part 34 docx
... Ans.324 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 6. Evaluate the integral 6301dxx+∫by using Weddle’s rule.Sol. Divide the interval [0,6] into 6 equal parts each of ... Using the formula, the given interval of integration must be divided into an even number of sub-intervals.320 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. (i) The values of 21()1fxx=+ ... 0. 784 3) 0.1927 0. 582 0 0.9694 6(0.3777)10= + + ++++ = 0.075[1.7441 + 5(0 .83 92) +6(0.3777)]316 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.5 TRAPEZOIDAL RULEPutting n =1 in equation...
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A textbook of Computer Based Numerical and Statiscal Techniques part 36 docx
... approximation of the error, we have. 132 182 9515x ≤ .00005.Taking logarithm, we obtain 15 log x ≤ log .00005 2 182 9513afa for x ≤ . 988 .340 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example ... nnyyhfxy+=+342 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Thus, xxx371136322079++ represents y correct to 4 decimal places. In the range |x| ≤ . 988 .i.e. –0. 988 ≤ x ≤ 0. 988 .Example ... =. Ans.Here in this example, only I approximation can be obtained and so it gives that approximatevalue of y for 0.1x=Example 12. Find the series expansion that gives y as a function of...
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A textbook of Computer Based Numerical and Statiscal Techniques part 37 doc
... value of ()0.1y and ()0.2y.[Ans. 3.005, 3.020]3 48 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now, using Euler’s modified formula, we obtain()()()()( )22120.1 .82 8 2.1 .82 8 ... + +1.0 082 =()() ( ){}310.21 log 0 1 log .2 1.0 082 2y =+ + + +1.0 082 =352 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6. Solve the following initial value problem by Picard methodydyxedx= ... RUNGE-KUTTA METHODThe method is very simple. It is named after two german mathematicians Carl Runge ( 185 6-1927) and Wilhelm Kutta ( 186 7-1944). These methods are well-known as Runge-Kutta Method....
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