Numerical methods. Examples with solution

Transcription

1 Numerical methods Examples with solution

2 CONTENTS Contents. Nonlinear Equations 3 The bisection method Newton s method Linear Systems LU-factorization Jacobi iterative method Gauss-Seidel iterative method Interpolation and approximation of functions 7 Polynom interpolation The least square method Numerical integration Trapezoid rule Composite trapezoid rule Simpson s rule Composite Simpson s rule Initial value problems for ODEs 3 Euler s method Runge-Kutta method

3 . NONLINEAR EQUATIONS. Nonlinear Equations 3

4 . NONLINEAR EQUATIONS Example.. Find the roots of equation by bisection method: x 3 = ln(0 x) with the precision ε = 0 3. Solution: x 3 ln(0 x) x f(x) It seen the root (x-intercept) lies in the interval, (from the graph,). The function values are tabulated on this interval f(x) = x 3 ln(0 x) and is found the sign of functional values are changing between. and.3. The function is continuous on a given interval. The equation has one root in the interval.,.3. 4

5 . NONLINEAR EQUATIONS The root are computed in the interval.,.3, thus a 0 =., b 0 =.3. Compute the first approximation x : x = b0 + a 0 =.3 +. =.5. Compute the functional values of the function f(x) = x 3 ln(0 x) in points a 0, x, b 0 : f(a 0 ) = 0.446, f(x ) = 0.59, f(b 0 ) = and find the interval a, b : f(a 0 )f(x ) 0 a = x 0 =.5, b = b 0 =.3. Find the approximation error b0 a 0 = 0.05 ε = 0 3. The computation continues. Compute the next approximation x = a + b = =.75 and find the interval a, b : f(a ) = 0.59, f(x ) = , f(b ) = , f(a )f(x ) 0 a = x =.75 b = b =.3. Find the approximation error b a = 0.05 ε = 0 3. The computation continues. Compute the next approximation x 3 = a + b = =.875 and find the interval a 3, b 3 : f(a ) = , f(x 3 ) = , f(b ) = , f(a ) f(x 3 ) 0 a 3 = x 3 =.875, b 3 = b =.3. 5

6 . NONLINEAR EQUATIONS Find the approximation error b a = 0.05 ε = 0 3. The computation continues. Compute the next approximation x 4 = a3 + b 3 = =.938 and find the interval a 4, b 4 : f(a 3 ) = , f(x 3 ) = 0.004, f(b 3 ) = , f(a 3 )f(x 3 ) < 0 a 4 = a 3 =.875, b 4 = x 3 =.938 Find the approximation error. b 3 a 3 = ε = 0 3. The computation continues. Compute the next approximation x 4 : x 4 = a4 + b 4 = =.906 and find the interval a 5, b 5 : f(a 4 ) = f(x 4 ) = f(b 4 ) = f(a 4 ) f(x 4 ) 0 a 5 = x 4 =.906, b 5 = b 4 =.938 Find the approximation error. b 4 a 4 = ε = 0 3. The computation continues. Compute the next approximation x 5 : x 5 = a5 + b 5 = =.9 and find the interval a 6, b 6 : f(a 5 ) = f(x 5 ) = f(b 5 ) = f(a 5 ) f(x 5 ) > 0 a 6 = x 5 =.906, b 6 = b 5 =.938 Find the approximation error b4 a 4 = ε = 0 3. The computation continues. 6

7 . NONLINEAR EQUATIONS Compute the next approximation x 4 : x 6 = a6 + b 6 = =.93 Find the approximation error b6 a 6 = < ε = 0 3. The desired precision is satisfied. All information are written to the table: i a i f(a i ) x i f(x i ) b i f(b i ) b i a i / The root of the equation is: ξ =.93 ± 0.00 Calculation in Matlab: a=., b=.3 f=inline( x^3-log(0-x) ) x=(a+b)/ while abs(b-a)/ > e-3 if f(a)*f(x)<0;b=x;else a=x;end; x=(a+b)/ end given interval definition of a function f(x) initial approximation approximations in the calculation of cycle 7

8 . NONLINEAR EQUATIONS Example.. Find the roots of equation by Newton s method: x 3 = ln(0 x) with the pecision ε = 0 6. Solution: The equation has one root in the interval.,.3 (see the Example..). Compute the the first and second derivative: f(x) = x 3 ln(0 x), f (x) = 3 x + 0 x, f (x) = 6 x + (0 x), Verify assumptions of method: f(a) f (a) = = = b a and it is seen that the condition is not satisfied. Reduce the interval in which the search root is x 3 ln(0 x) It is seen that the root lies in the interval.5,.3 (from the graph) And again, verify the method conditions, this time in the interval.5,.3 : f(a) f (a) = 0.59 = < 0.05 = b a f(b) f (b) = = < 0.05 = b a

9 . NONLINEAR EQUATIONS x i f (x i ) f (x i ) It is seen from the first and second derivative values in the interval.5,.3 : Find the initial approximation x 0 = b =.3. Compute the next approximation x : f (x) > 0 na.5,.3 f (x) > 0 na.5,.3. The Newton method conditions are satisfied. x = x 0 f(x 0) =.3 f (x 0 ) = and the approximation error x x 0 = ε = 0 6. The computation continues. Compute the next approximation x : x = x f(x ) f (x ) = = and the approximation error x x = ε = 0 6. The computation continues. Compute the next approximation x 3 : x 3 = x f(x ) f (x ) = = and the approximation error x x 0 = = < ε = 0 6. The desired precision is satisfied. All information are written to the table: i x i x i x i

10 . NONLINEAR EQUATIONS The root of the equation is: ξ = ± 0 6 Calculation in Matlab: f=inline( x^3-log(0-x) ) input the function f(x) fd=inline( 3*x^+./(0-x) ) input the first derivative f (x) xs=.3 initial approximation x=xs-f(xs)/fd(xs) the next approximation computation while abs(xs-x)>e-6 approximations in the cycle computation xs=x; x=xs-f(xs)/fd(xs) end 0

11 . LINEAR SYSTEMS. Linear Systems

12 . LINEAR SYSTEMS Example.. The matrix is given. 3 A = Compute the LU-decomposition A = LU by using Gaussian elimination method with direct choice of main element. Using the LU-decomposition,compute the solution of a system of linear equations Ax = b,where b = ( 9, 59, 8), inverse matrix A and determinant deta. Solution: Compute by using Gaussian elimination method: 3 6, The matrix U is the result of the Gaussian elimination method, the matrix L we prepare of multipliers, which are written next to matrices : L = 6 0, U = In solving Ax = b first compute y from Ly = b and then from Ux = y compute x. y = 9 9 Weget : 6y +y = 59 y = 5, 3y +y 3 = 8 x 3x +x 3 = 9 x x 3 = 5 x = 3. x 3 =

13 . LINEAR SYSTEMS Inverse matrix is calculated by solving the system Aa i = e i, i =,, 3, where e i are the columns of the unit matrix I = (e,e,e 3 ). The results a i are the columns of inverse matrix, i.e. A = (a,a,a 3 ). The solution of systems is again decomposed into two steps. y =, 0, y +y = 0,, 0 Y = 6 0, 3y +y 3 = 0, 0, 3 0 x 3x +x 3 =, 0, x x 3 = 6,, 0 A = 0. x 3 =, 0, 3 0 Determinant det A be calculated as the product of the diagonal elements of the matrix L and U: deta = detldetu = ( ) (( ) ( ) ) =. Example.. Solve the system of linear equations x +x 3x 3 = 4, x +5x +x 3 = 5, 4x x +x 3 =, using Jacobi iterative method with the pecision ǫ = 0. Solution: To ensure convergence, modify the system in form with strong diagonal dominant matrix. Replace the first and third equation 4x x +x 3 =, x +5x +x 3 = 5, x +x 3x 3 = 4. 3

14 . LINEAR SYSTEMS The modified system is rewrite to iterative form x = 4 ( + x x 3 ), x = 5 (5 x x 3 ), x 3 = 3 ( 4 x x ) The recurrent formulas for Jacobi method are obtained by iteration indices. x (k+) = ( + x(k) 4 x (k) 3 ), x (k+) = (5 x(k) 5 x (k) 3 ), x (k+) 3 = ( 4 x(k) 3 x (k) ). The computation converges for any initial approximation x (0) = (x (0), x (0), x (0) 3 ), so the following initial approximation is choosen x (0) = ( 0, 0, 0 ). Substitute x (0) into the right side of the recurrent formulas. Compute continue equally x () = ( 3,, 4 3 ). x () = ( 4, 9 5, 3 ), x (3) = (.8500,.333, ), atd. Find that the true is x (k) x (k ) R 0 3, compute thus x () x (0) R = max{ 3 0, 0, 4 0 } = 3, 3 x () x () R = max{ 4 + 3, 9, } =, x (3) x () R = , etc. The results are written to the table: Stop after the eleventh iteration, because x () x (0) R = , and the result is x = 3.00 ± 0, x =.00 ± 0, x 3 =.00 ± 0. 4

15 . LINEAR SYSTEMS k x (k) x (k) x (k) 3 x (k) x (k ) R Example.3. Solve the system of linear equations x +x 3x 3 = 4, x +5x +x 3 = 5, 4x x +x 3 =, using the Gauss-Seidel iterative methods with the pecision ǫ = 0. Solution: To ensure convergence, modify the system in form with strong diagonal dominant matrix. Replace the first and third equation 4x x +x 3 =, x +5x +x 3 = 5, x +x 3x 3 = 4. The modified system is rewrite to iterative form x = ( + x 4 x 3 ), x = (5 x 5 x 3 ), x 3 = ( 4 x 3 x ) 5

16 . LINEAR SYSTEMS The recurrent formulas for Jacobi method are obtained by iteration indices. x (k+) = ( + x(k) 4 x (k) 3 ), x (k+) = (5 x(k+) 5 x (k) 3 ), x (k+) 3 = ( 4 x(k+) 3 x (k+) ). The computation converges for any initial approximation x (0) = (x (0), x (0), x (0) 3 ), so the following initial approximation is choosen x (0) = ( 0, 0, 0 ). Substitute x (0) into the right side of the recurrent formulas. Compute x () = ( ,.000,.0667 ). continue equally x () = (.9833,.9800, ), atd. Find that the true is x (k) x (k ) R 0, compute thus x () x (0) R = max{ 3 0,. 0, } = 3, x () x () R = max{ ,.98., } = 0.00, atd. The results are written to the table: k x (k) x (k) x (k) 3 x (k) x (k ) R Stop after the four iteration, because x (4) x (3) R = , and the result is x = 3.00 ± 0, x =.00 ± 0, x 3 =.00 ± 0. 6

17 3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS 3. Interpolation and approximation of functions 7

18 3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS Example 3.. The nodes x i and function values f i are given below. i 0 x i 3 f i Find the interpolation polynom for theese datas: a) in basic form, b) in Lagrang form, c) in Newton form. Solution: a)interpolating polynomial is finding in form p(x) = a 0 + a x + a x. Replace the interpolation conditions p(x i ) = f i, i = 0,, p( 3) = = a 0 3a + 9a =, p( ) = = a 0 a + a =, p() = = a 0 + a + a =. These equalities represent system of linear equations, which can be written in the form 3 9 a 0 a = a. The results are a 0 =, a 8 = 3, a = 5, the interpolation polynomial has the 8 form p(x) = x x. b) Interpolation polynomial in Lagrange form is prescribed p(x) = f 0 ϕ 0 (x) + f ϕ (x) + f ϕ (x), where ϕ 0 (x), ϕ (x), ϕ (x) are Lagrange basis polynomials: ϕ 0 (x) = (x + )(x ) ( 3 + )( 3 ) = (x + )(x ), 8 8

19 3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS ϕ (x) = ϕ (x) = (x + 3)(x ) ( + 3)( ) = (x + 3)(x ), 4 (x + 3)(x + ) = (x + 3)(x + ). ( + 3)( + ) 8 It is obtained p(x) = 8 (x + )(x ) + 4 (x + 3)(x ) + (x + 3)(x + ). 4 c) Interpolation polynomial in Newton form is prescribed p(x) = f 0 + f[x, x 0 ](x x 0 ) + f[x, x, x 0 ](x x 0 )(x x ), where f[x, x 0 ] a f[x, x, x 0 ] are the relative difference st and nd order. The results are written to the table: i x i f i st order nd order Using the computed values in the first row of the table, find interpolation polynomial: p(x) = (x + 3) + 5 (x + 3)(x + ). 8 9

20 3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS Example 3.. Aproximate the following data x i y i by the line ϕ(x) = c x + c by the least squares method. Solution: Find the system of linear equations for the unknown coefficients c, c. 5 5 c x i + c x i = i= i= 5 5 c x i + c = i= i= 5 y i x i i= 5 y i i= Compute individual totals: 5 x i = = 90.5 i= 5 x i = = 8.5 i= 5 y i x i = = 57 i= 5 y i = = 8.5 i= Replace the system of linear equations: 90.5c + 8.5c = c + 5c = 8.5 Solve the system and the coefficients c =.3647, c = are obtained. The line ϕ(x) =.3647x is the result. 0

21 3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS 6 4 zadane body nalezena funkce Calculation in Matlab: x=[,.5,3,5,7] y=[0,,6.5,6,5] n=length(x) imput x i imput y i number of points A=[sum(x.^) sum(x);sum(x) n] matrix system b=[sum(y.*x);sum(y)] c=a\b f=c()*x+c() plot(x,y, o,x,f) vector of right-hand sides calculation of the coefficients line values in points x i graph

22 4. NUMERICAL INTEGRATION 4. Numerical integration

23 4. NUMERICAL INTEGRATION Example 4.. Compute the integral by composite trapezoid tule for n = 8: 3 e x dx Solution: Compute the step h in the interval a, b, h = b a n = 3 ( ) 8 = 0.5. Write to the table division points x 0,...,x 8 and the values of integrated functions in these points. i x i y i = e x i Finally, compute the approximate value of the integral: I 0.5 ( ) ( ) n y0 + y n y0 + y 8 7 = h + y i = y i = i= i= ( = ( ) ) = = The result is: I 0.5 =

24 4. NUMERICAL INTEGRATION Calculation in Matlab: a=-,b=3 interval bounds a, b n=8 number of segments on dividing n h=(b-a)/n x=a:h:b computation of step h computation of x i y=exp(x.^) computation of y i I=h*((y()+y(n+))/+sum(y(:n))) computation of I = h ( y 0 +y n + ) n i= y i In Matlab is indexed from, and we index points from 0, so the indexes in the formula in Matlab are shifted by. 4

25 4. NUMERICAL INTEGRATION Example 4.. Compute the integral by composite trapezoid rule with the precision ε = 0 3 : 3 e x dx Solution: Compute the integral composite trapezoid formula for n =, 4, 8, 6,..., until the error is less than the precision, i.e. I h I h < ε. n h I h I h I h ε = 0 3 = < 0.00 The result is: 3 e x = ±

26 4. NUMERICAL INTEGRATION Example 4.3. Compute the integral by composite Simpson rule for n = 6: 0 arctg (x) + x + dx Solution: Compute the step h in the interval a, b, h = b a n = 0 = Write to the table division points x 0,...,x 6 and the value of integrated functions in these points. i x i y i = arctg (x i ) + x i i x i y i = arctg (x i ) + x i Finally, compute the approximate value of the integral: I 0.5 = h 3 (y 0 + y n + 4 S L + S S ) = (y 0 + y (y + y 3 + y 5 + +y 7 + y 9 + y + y 3 + y 5 ) + (y + y 4 + y 6 + y 8 + y 0 + y + y 4 ) = = ( ) =

27 4. NUMERICAL INTEGRATION The result is: I 0.5 = 5.00 Calculation in Matlab: a=0,b= interval bounds a, b n=6 number of segments on dividing n h=(b-a)/n computation of steph x=a:h:b computation of x i y=atan(x).^+x+ computation of y i SL=sum(y(::n)) computation of S L SS=sum(y(3::n-)) computation of S S I=h/3*(y()+y(n+)+4*SL+*SS) computation of I In Matlab is indexed from, and we index points from 0, so the indexes in the formula in Matlab are shifted by. 7

28 4. NUMERICAL INTEGRATION Example 4.4. For the function f(x) = sin x + arctanx + x 4 compute approximately the first derivative on the interval 0, by the formula f (x) f(x + h) f(x h) h with a step h = 0.5 a h = 0.. Compare the results with values of exact calculated derivative. Solution: The exact derivative is found by the formula: f (x) = x cosx + +x 4 +x 4x 3 arctan x ( + x 4 ). Find the nodes x i = ih, i = 0,..., 8 and compute the functional values in them f i = f(x i ) to compute approximate derivative with step h = 0.5. There are the first three columns in the table: i x i f i f[x i+,x i ] f (x i ) e i The formula for calculation of approximate derivative is written in form f (x i ) f i+ f i h = f[x i+, x i ], 8

29 4. NUMERICAL INTEGRATION. Compute the approximate derivative values in all internal nodes, eg f (0.75) =.008;. This is the fourth column of the table. The fifth column contains the exact values of derivatives that are in the last column compared with the approximate values: e i = f[x i+, x i ] f (x i ). For h = 0. the procedure is analogous, we obtain more accurate values, as shown in the following images: h= derivace aproximace derivace h= derivace aproximace derivace

30 4. NUMERICAL INTEGRATION Calculation in Matlab: a=0,b= interval bounds a, b h=0.5 step h x=[a:h:b] computation of nodes x i n=length(x) number of nodes n + df= *x.*cos(x.^) +((+x.^4)./(+x.^)-... 4*x.^3.*atan(x))./(+x.^4).^ computation of f (x i ) f=sin(x.^)+atan(x)./(+x.^4) computation of f i = f(x i ) adf=[nan;(f(3:n)-f(:n-))/(*h);nan] approximate derivative tab=[[0:n-],x,f,adf,df,abs(adf-df)] table plot(x,df, :,x,adf, - ) graph 30

31 5. INITIAL VALUE PROBLEMS FOR ODES 5. Initial value problems for ODEs 3

32 5. INITIAL VALUE PROBLEMS FOR ODES Example 5.. Solve the differential equation with the initial condition: y = 0. y, y( ) =, x + by the Euler method in the interval, 3 with the step h = 0.5. Solution: First, compute the number of dividing segments in the interval a, b : n = b a h = 3 ( ) 0.5 = 0. The initial condition is x 0 =,, y 0 =. Compute x, y by the Euler metod: x = x 0 + h = =.5 ( ) ( ) y = y 0 + h x y 0 = ( ) = 0.85 ( ) + Compute x, y, x 3, y 3 : x y = x + h = = ( ) ( ) = y + h x + 0. y = ( 0.85) = (.5) + = x 3 = x + h = = 0.5 ( ) ( ) y 3 = y + h x + 0. y = ( ) = ( ) + = 0.37 Other approximate values of solution x 4, y 4,..., x 0, y 0 compute accordingly. 3

33 5. INITIAL VALUE PROBLEMS FOR ODES The approximate values of the solution are in the table and graph: i x i y i y Priblizne reseni x Calculation in Matlab: a=-,b=3 h=0.5 n=(b-a)/h x()=-,y()=- for i=:n x(i+)=x(i)+h y(i+)=y(i)+h*(/(x(i)^+)-0.*y(i)) end [x y ] plot(x,y) interval bounds a, b input step h number of dividing segments n input initial condition computation of x i, y i in the cycle extract into the table graph 33

34 5. INITIAL VALUE PROBLEMS FOR ODES Example 5.. Solve the differential equation with the initial condition: y = sin(x) y x 3 + y, y() = 0, by the 4th order Runge Kutta method in the interval, with the step h = 0.. Solution: First, compute the number of dividing segments in the interval a, b : n = b a h The initial condition is x 0 =,, y 0 = 0. = 0. = 5. Compute x, y by the 4th order Runge Kutta method: k = h f (x 0, y 0 ) = h (sin (x 0 ) y 0 x y 0) = 0. ( sin () ) = k k 3 = 0. ( = h f x 0 + h, y 0 + k ) = ( = h sin x 0 + h ) ( y 0 + k ) ( x 0 + h ) 3 + y 0 + k = ( ( = 0. sin + 0. ) ( ) ( + 0. ) ) = ( = h f x 0 + h, y 0 + k ) = ( = h sin x 0 + h ) ( y 0 + k ) ( x 0 + h ) 3 + y 0 + k = ( ( = 0. sin + 0. ) ( ) ( + 0. ) ) = = k 4 = h f (x 0 + h, y 0 + k 3 ) = = h (sin ) (x 0 + h) (y 0 + k 3 ) (x 0 + h) 3 + y 0 + k 3 = = 0. (sin ( + 0.) ( ) ( + 0.) ) = x = x 0 + h = + 0. =. 34

35 5. INITIAL VALUE PROBLEMS FOR ODES y = y (k + k + k 3 + k 4 ) = = 0 + ( 0. + ( 0.304) + ( 0.337) ) = Compute x, y : k = h f (x, y ) = h (sin (x ) y x 3 + y ) = = 0. (sin (.) ( 0.3) ) = = ( k = h f x + h, y + k ) = ( = h sin x + h ) ( y + k ) ( x + h ) 3 + y + k = ( ( = 0. sin ) ( ) ( ) ) = ( k 3 = h f x + h, y + k ) = ( = h sin x + h ) ( y + k ) ( x + h ) 3 + y + k = ( ( = 0. sin ) ( ) ( + 0. ) ) = k 4 = h f (x + h, y + k 3 ) = h (sin ) (x + h) (y + k 3 ) (x + h) 3 + y + k 3 = = 0. (sin ( + 0.) ( ) ( + 0.) ) = 0.95 x = x + h = =.4 y = y + 6 (k + k + k 3 + k 4 ) = = ( ( ) + ( ) 0.95) = Other approximate values of solution x 3, y 3,..., x 5, y 5 compute accordingly. 35

36 5. INITIAL VALUE PROBLEMS FOR ODES The approximate values of the solution are in the table and graph: 0 Priblizne reseni i x i y i y x Calculation in Matlab: f=inline( sin(x)*y-x^3+y ) right sidef(x, y) a=,b= h=0. n=(b-a)/h x()=,y()=0 for i=:n k=h*f(x(i),y(i)) k=h*f(x(i)+h/,y(i)+k/) k3=h*f(x(i)+h/,y(i)+k/) k4=h*f(x(i)+h,y(i)+k3) x(i+)=x(i)+h y(i+)=y(i)+/6*(k+*k+*k3+k4) end [x y ] plot(x,y) interval bounds a, b input step h number of dividing segments n input initial condition computation of x i, y i in the cycle extract into the table graph 36