TAILIEUCHUNG - A textbook of Computer Based Numerical and Statiscal Techniques part 42

A textbook of Computer Based Numerical and Statiscal Techniques part 42. By joining statistical analysis with computer-based numerical methods, this book bridges the gap between theory and practice with software-based examples, flow charts, and applications. Designed for engineering students as well as practicing engineers and scientists, the book has numerous examples with in-text solutions. | 396 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Now dS o da n X2 y - axi - bxi xi 0 i 1 or n n n xf i 1 and X xy a x2 X i i i i dS o db 1 X 2 yi - axi - bx2 -x2 o i i n n n X x2 yi aX xf bX x4 i 1 i 1 i 1 2 Dropping the suffix i from 1 and 2 then the normal equations are X xy aX x2 bX x3 X x2 y aX x3 bX x4 b Fitting of the Curve y ax Error of estimate for ith point xi yi is L. b Ci I yi axi--- l xi or We have n S X e2 i 1 By the principle of least square the value of S is minimum Now n X 2 yi i 1 I axi 0 and 0 da db dS 0 xi 1 7 n n X xiyi aX xi nb i 1 i 1 1 1 0 CURVE FITTING 397 and dS 0 db n i X 2 y i i I xi b 1 or 0 n x y 1 xi na n bX 4 Z1 xi 2 Dropping the suffix i from 1 and 2 then the normal equations are X xy aX x2 nb Xiy bX 4 x x Where n is the number of pair of values of x and y. Fitting of the Curve y C c Ix Error of estimate for i th point x y is f co ei 1 y - x - c xi We have S X ei i 1 2 By the principle of least square the value of S is minimum Now dS 0 and d 0 dS o dco X 2 yi - - c1y xi -- 0 x x i 1 1 1 n 1 and ac1 0 n X21yi v ci x w xr 0 xi i 1 i 398 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES or n n i n X y c0 ci X x i 1 i 1N xi i 1 2 Dropping the suffix i from 1 and 2 then the normal equations are x y c z 4 ci x x x Nx Xy co Xx Example 8. Find the curve of bestfit of the type y aebx to the following data by the method of least sqaures x 1 5 7 9 12 y 10 15 12 15 21 Sol. The curve to be fitted is y aebx or Y A Bx where Y log10 y A log10 a and B b log10 e- Therefore the normal equations are X Y 5A BX x X xY AX x BX x2 x y Y log y x2 xY 1 10 1 1 5 15 25 7 12 49 9 15 81 12 21 144 X x 34 X Y Xx2 300 X xY Substituting the values of Xx etc. and calculated by means of above table in the normal equations. We get Therefore 5A 34B and 34A 300B On solving these equations we obtain A B B a anti log10 A b -j-------- Hence

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