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 x1 and x2 in Secantmethod. Similarly, other approximations can be obtained by using two recent values of approximations. These arex3= ... f(x5) = – 0.00002ALGEBRAIC AND TRANSCENDENTAL EQUATION77Since, imaginary roots occur in conjugate pairs roots are 1.915 ± 1.908i upto 3 places of decimal. Assuming other pair of roots to be α ... root can be taken as 2.0945 correct up to four decimals.ALGEBRAIC AND TRANSCENDENTAL EQUATION796. Using N–R method, obtain formula for N and find 20 correct to two decimal places.[Ans....
- 10
- 713
- 0
A textbook of Computer Based Numerical and Statiscal Techniques part 1 ppt
... This pageintentionally leftblankThis pageintentionally leftblank A TEXTBOOK OF COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Anju Khandelwal M.Sc., Ph.D.Department of MathematicsSRMS ... Based Numerical and Statistical Techniques is primarily writtenaccording to the unified syllabus of Mathematics for B. Tech. II year and M.C .A. I year students of all Engineering colleges affiliated ... affiliated to U.P. Technical University, Lucknow.The subject matter is presented in a very systematic and logical manner. In each chapter, allconcepts, definitions and large number of examples...
- 10
- 425
- 1
A textbook of Computer Based Numerical and Statiscal Techniques part 4 ppt
... 0.10.0542ddd·Example 32. In a ∆ABC, b = 9.5 cm, c = 8.5 cm and A = 45o, 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 notalways handled the same way. It turns out that many manufacturers believe (sometimes rightly and sometimes wrongly) that ... Given δb = 2 mm = 0.2 cm a = 1 mm = 0.1 cmsin A = −⇒=1sin a a A bb∂∂ A a= 222211 11babab⋅=−−∂∂ A b= 2222211aababbab⋅− =−−− A < AAabaa∂∂δ+δ∂∂22 2212.30.1...
- 10
- 580
- 0
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 EQUATION37Here ei and ei + 1 are the errors in ith and (i + 1)th iterations respectively. ... Root of the Equation f(x) = 0Step 1: Choose two initial guess values (approximation) a and b (where (a > b)) such thatf (a) . f(b) < 0.Step 2: Evaluate the mid point x1 of a and...
- 10
- 469
- 1
A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx
... the argument half waybetween the arguments at q and r is +B A, 24 where A is the arithmetic mean of q and r and B is arithmeticmean of 3q – 2p – s and 3r – 2s – p.Sol. Given A is the arithmetic ... 0.7880Example 7. Following are the marks obtained by 492 candidates in a certain examination:04040454550505555606065. 2104 354743279MarksNo of Candidates−−−−−−Find out (a) No. of candidates, ... r and s corresponds to argument a, a + h, a + 2h and a + 3h respectivelythen the value of the argument lying half way between a + h and a + 2h will be a + h + 2h i.e., a + 32hHence,a...
- 10
- 411
- 0
A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx
... gives bestestimation when 11.44u−< This formula is obtained by taken mean of Gauss forward and Gaussbackward 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 limitsare as follows:MarksNo of candidates−−−−−0192039405960798099. ... ×−= 235 – 24.65 – 10. 76625 – 1.076625= 235 – 36.492875= 198.507 { 198∴ Total no. of candidates who obtained fewer than 70 marks are 198.Example 6. The area A of a circle of diameter d is given...
- 10
- 363
- 0
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 nth degree are constant.Let f(x) = A 0xn + A 1xn–1 + + A n–1 x + A n by a polynomial of degree n provided A 0 ≠ 0 and arguments be equally spaced so thatx1 ... of 20th Jan. and its value can be obtained similarly.[f(x)]min = [f(x)]1.184 = 63.647 °F approximately.Example 8. The mode of a certain frequency Curve y = f (x) is very near to x = 9 and...
- 10
- 231
- 0
A textbook of Computer Based Numerical and Statiscal Techniques part 30 pptx
... BASED NUMERICAL AND STATISTICAL TECHNIQUES after simplication, we get() () () () ()+++ =∫∫ ∫∫01 bb b bnnaa a a a wxdx a xwxdx a xwxdx wxfxdx() () () () ()2101 bb b bnnaa a a a xw ... dxa xdxa xdxa xdx 111135/2 2 40120000x dxa xdxa xdxa xdx=++∫∫∫∫or Simplifying above equations, we get12001201222332234523457aa a aaaaaa++=++ =++=284 COMPUTER BASED NUMERICAL ... NUMERICAL AND STATISTICAL TECHNIQUES Minimax polynomial approximation: Let f(x) be continuous on [a, b] and it isapproximated by the polynomial Pn (x) = a 0 + a 1x + +a nxn , then the minimax...
- 10
- 228
- 0
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 havebecome easy for manipulations. Hence, ... the values of 1y and its derivativies in Taylor series expansion, we obtain ()()()()23 10 10 1101 1 1 2! 3!xx xxyx y x x y y y−−′ ′′ ′′′=+ − + + +330 COMPUTER BASED NUMERICAL AND ... below some of the methods of numerical solutions of the ordinary differential equations. No doubt, such numerical solutions are approxi-mate solutions. But, in many cases approximate solutions...
- 10
- 349
- 0
A textbook of Computer Based Numerical and Statiscal Techniques part 40 pptx
... sensitiveto 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 accuracyas achieved by 7 iterations in Gauss-Seidel method.SOLUTION OF SIMULTANEOUS LINEAR EQUATION385Example 8. ... coefficient matrix are diagonally dominant.10x + y – 2z = 7.74x + 12y + 3z = 39.663x + 4y + 15z = 54.8376 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES largest coefficient of y. We continue...
- 10
- 663
- 2
Từ khóa: the appropriate application of computer aided design and manufacture techniques in silicone facial prostheticsdistance calculations based on molecules described by a vector of 1 apos s and 0 apos sebook a textbook of assaying for the use of those connected with mines by cornelius beringer and john jacob beringerbreak the map into regions based upon a combination of management goals sources and habitatsa brief history of computer music languages and systems the synergy between technology and creativitya culture of creativity design education and the creative industriesa study of perceptual learning styles and achievementa period of slow economic growth and high inflationa period of robust economic growth and high employmenthistory of computer based control systemsstructure and interpretation of computer programs abelson and sussman mit pressa summary of the first two and a half partspath—a foundation of file based i othe morality of hiv aids a comparison of russian orthodox church and secular ngo approachessusceptibility testing protocols for antibiograms and preventive surveillance a continuum of data collection analysis and presentationNghiên cứu vật liệu biến hóa (metamaterials) hấp thụ sóng điện tử ở vùng tần số THzGiáo án Sinh học 11 bài 13: Thực hành phát hiện diệp lục và carôtenôitGiáo án Sinh học 11 bài 13: Thực hành phát hiện diệp lục và carôtenôitGiáo án Sinh học 11 bài 13: Thực hành phát hiện diệp lục và carôtenôitPhối hợp giữa phòng văn hóa và thông tin với phòng giáo dục và đào tạo trong việc tuyên truyền, giáo dục, vận động xây dựng nông thôn mới huyện thanh thủy, tỉnh phú thọTrả hồ sơ điều tra bổ sung đối với các tội xâm phạm sở hữu có tính chất chiếm đoạt theo pháp luật Tố tụng hình sự Việt Nam từ thực tiễn thành phố Hồ Chí Minh (Luận văn thạc sĩ)Nghiên cứu khả năng đo năng lượng điện bằng hệ thu thập dữ liệu 16 kênh DEWE 5000Thiết kế và chế tạo mô hình biến tần (inverter) cho máy điều hòa không khíChuong 2 nhận dạng rui roTổ chức và hoạt động của Phòng Tư pháp từ thực tiễn tỉnh Phú Thọ (Luận văn thạc sĩ)BT Tieng anh 6 UNIT 2Tăng trưởng tín dụng hộ sản xuất nông nghiệp tại Ngân hàng Nông nghiệp và Phát triển nông thôn Việt Nam chi nhánh tỉnh Bắc Giang (Luận văn thạc sĩ)chuong 1 tong quan quan tri rui roNguyên tắc phân hóa trách nhiệm hình sự đối với người dưới 18 tuổi phạm tội trong pháp luật hình sự Việt Nam (Luận văn thạc sĩ)Giáo án Sinh học 11 bài 14: Thực hành phát hiện hô hấp ở thực vậtGiáo án Sinh học 11 bài 14: Thực hành phát hiện hô hấp ở thực vậtBÀI HOÀN CHỈNH TỔNG QUAN VỀ MẠNG XÃ HỘIĐổi mới quản lý tài chính trong hoạt động khoa học xã hội trường hợp viện hàn lâm khoa học xã hội việt namTÁI CHẾ NHỰA VÀ QUẢN LÝ CHẤT THẢI Ở HOA KỲQUẢN LÝ VÀ TÁI CHẾ NHỰA Ở HOA KỲ