presentation on Euler and Modified Euler method ,and Fitting of curve

presentation on Euler and Modified Euler method with working and example ,and Fitting of Nonlinear curve using Method of least square

1. NMC Presentation BY:- Mukul Dev Khunte Guided by:- Dr. G.K.Singh Head Dept. of Mathematics

2. History-Sir Leonhard Euler Name: Sir Leonhard Euler Born :  15 April 1707, Switzerland Died :  18 September 1783,  Saint Petersburg Russia Education:  University of Basel ( 1720–1723) Contributions to mathematics and physics: Euler worked in almost all areas of mathematics, such as geometry, infinitesimal calculus, trigonometry, algebra, and number theory, as well as continuum physics, lunar theory and other areas of physics

3. Euler’s Method Consider the equation =f(x,y) …….(1) Given that Y(x )=y its curve of solution through P(x ,y ) is shown dotted in figure. Now we have to find ordinate of any other point at Q on this curve. Dividing the curve into ‘n’ equal sub-interval each of width ‘h’. So we approximate the tangent for LL 1 So Y 1 =L 1 P 1 =LP+R 1 P 1 =y +PR 1 tan =y +hf(x ,y )  

4. Euler’s Method Repeating this process n times y n+1 =y n +hf(x n ,y n ) …… ….(2) Where x n =x +nh Equation 2 is called as Euler’s Method for finding an approximate solution .

5. Working of Euler’s Method 1.Given function is taken for the first approximation. Using initial boundary condition and value of ‘h’. **If the value of ‘h’ is not given than the initial and final value(required value) is divided into ‘n’ sub-intervals for finding value of ‘h’. 2.After getting the first approximation the second approximation is taken for the function of x +h And y 1 . Than the Euler’s Method become y 2 =y 1 +hf(x +h , y 1 ) where y 1 is the value obtain from first approximation . 3.Similerly approximations are taken out until x + h=x n (Required value of y at any point x n )

6. Example Question : Apply Euler’s Method to solve y’=x+y .Given y(0)=0.Find y at x=0.8 using step length 0.2 . Solution : Given, y’=x+y , y(0)=0 ,h=0.2 Than using Euler’s Method y n+1 =y n +hf(x n ,y n ) …………..(1) Putting n=0, for finding first approximation than we have, y 1 =y +hf(x ,y ) y 1 =0+0.2 0 =0 Putting n=1, for finding second approximation than we have , y 2 =y 1 +hf(x 1 ,y 1 )  

7. Example y 2 = + .2 =0.4 Putting n=2, for finding third approximation than we have , y 3 =y 2 +hf(x 2 ,y 2 ) y 3 =0.04+0.2 .44=0.128 Putting n=3, for finding forth approximation than we have , y 4 =y 3 +hf(x 3 ,y 3 ) y 4 =0.128+0.2 0.728=0.2736 At y(0.8)=0.2736 Answer.  

8. Modified Euler’s Method In the Euler’s Modified method , The curve of the solution in the interval LL 1 is approximates by the tangent at P such as at P 1 we have, Y 1 =y +hf(x ,y ) Than the slope of the curve of the solution through P 1 is computed at the tangent at P 1 to P 1 Q 1 is drawn meeting the ordinate through L 2 in P 2 (x +2h,y 2 )……1 Now we find better approximation y 1 ’ of y(x +h ) by taking the slope of the curve as the mean of the slope of the tangent at P and P 1 i.e. Y 1 ’= y + f(x +y )+f(X +h,Y 1 )}….. 2 As the slope of the tangent at P 1 is not known, We take Y 1 is found in equation (1 )  

9. Modified Euler’s Method Euler method y n+1 =y n +hf(x n ,y n ) Where x n =x +nh Euler’s Modified Method is y n+1 = y n + {f(x n , y n )+f(x n+1 ,y n+1 )}  

10. Working of Modified Euler’s Method First we find the first approximation using Euler’s Method. The approximated value of y 1 is than modified using E uler modified method. The approximated value of y 1 from Euler modified method is again approximated until the equal value of y 1 is found. The value of y 1 is taken for the approximation of y 2 using E uler method. And the process continues.

11. Example Question : Apply Euler’s Modified Method to solve y’=x+y .Given y(0)=1.Find y at x=0.2 using step length 0.1 . Solution : Given, y’=x+y , y(0)=1 ,h=0.1 Than using Euler’s Method, y n+1 =y n +hf(x n ,y n ) …………..(1) Putting n=0, y 1 =y +hf(x ,y ) ; y 1 =1+0.1(1)=1.1 Than using Euler’s Modified Method, y n+1 =y n + {f(x n , y n )+f(x n+1 ,y n+1 )}……(2) Putting n=1, for finding second approximation than we have , y 2 =y 1 +hf(x 1 ,y 1 )  

12. Example y 1 =y + {f(x , y )+f(x 1 , y 1 )} y 1 =1+ ( 1+1.2)=1.11 Repeating process will give y 1 =1.1105 Than using Euler’s Method, Putting n=1; y 2 =y 1 +hf(x 1 ,y 1 ) y 2 =1.1105+0.1(0.1+1.1105)=1.23155 Than using Euler’s Modified Method,Putting n=1; y 2 =y 1 + { f(x 1 , y 1 )+f(x 2 , y 2 )}  

13. Example y 2 =1.1105+ (1.2105+1.43155)=1.2426 Repeating process will give y 2 =1.2432 At y(0.2)=1.2432 Answer.  

14. Fitting of Nonlinear Curve y=a Taking logarithms, Y= A+bX ……(1) Where A= , X= ,Y= Normal equation for (1) are, = nA+b =A +b From A & b can be determine .Than a can be calculated from A= .  

15. Fitting of Nonlinear Curve 2. y=a Taking logarithms, Y= A+bX ……(1) Where A= , B= e ,Y= Normal equation for (1) are, = nA+B =A +B From A & B can be determine .Than “a” and “b” can be calculated from A= , B= e .  

16. Fitting of Nonlinear Curve 3 . xy a =b Taking logarithms, = - Y=A+BX ……(1) Where X= , B= - , A = ,Y= Normal equation for (1) are, = nA+B =A +B From A & B can be determine .Than “a” and “b” can be calculated from A = , B= - .  

17. Fitting of Nonlinear Curve-Working 1.Given relation eg- xy a =b is first arranged in the form of linear law using logarithm. 2.Observed set of “n” values is substituted in equation. 3.Normal equation for each constant is formed.(i.e. = nA+B =A +b ) 4.Solve these normal equations as simultaneous equation for obtain the values of A,B. 5.Substitute these values of A,B, in relation with a,b . Which will give the values of “a”, “b”. 6.Substituting the values of “a” and “b” in given equation will give curve of best fit.  

18. Example Question : A n experiment gave the following value: It is known that v and t are connected by a relation v= at b .Find the best possible values of a and b. Solution : We have v= at b . Taking log 10 both side will give, log 10 v=log 10 a+blog 10 t …..(1) Equation 1 can be written as; Y= A+bX ……..(2) Where X=log 10 t , Y=log 10 v , A=log 10 a v( ft /min) 350 400 500 600 t(min) 61 26 7 2.6

19. Example Than the Normal equations are , = nA+b …… (3) =A +b ……..(4) Now , , , are calculated in the following table;   v t X=log 10 t Y=log 10 v XY X 2 350 61 1.7853 2.5441 4.542 3.187 400 26 1.415 2.6021 3.682 2.002 500 7 0.8451 2.6990 2.281 0.714 600 2.6 0.4150 2.7782 1.153 0.172 4.4604 10.6234 11.658 6.075

20. Example Therefor equation 3 and 4 become 4A+4.46b=10.623 4.46A+6.075b=11.075 Solving these ,A=2.845,b=-0.1697 and a=Antilog(2.845)=699.8

21. Refrences Higher Engineering Mathematics by B.S. Grewal (40th edition)-Khanna Publishers. Numerical Methods in Engineering and Science by Dr. B.S. Grewal, Khanna Publishers . The Minitab Blog ( http://blog.minitab.com/ )

22. Thank you