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
𝑑𝑦
𝑑𝑥
=f(x,y) …….(1)
Given that Y(x0)=y0 its curve of solution through
P(x0,y0) 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
LL1
So Y1=L1P1=LP+R1P1
=y0+PR1 tan𝜃 =y0+hf(x0,y0)

4. Euler’s Method
Repeating this process n times
yn+1=yn+hf(xn,yn) ……….(2)
Where xn=x0+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 x0+h
And y1 .
Than the Euler’s Method become y2=y1+hf(x0 +h,y1)
where y1 is the value obtain from first approximation .
3.Similerly approximations are taken out until x0 +h=xn (Required value of y at any
point xn )

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
yn+1=yn+hf(xn,yn) …………..(1)
Putting n=0, for finding first approximation than we have,
y1=y0+hf(x0,y0)
y1=0+0.2×0 =0
Putting n=1, for finding second approximation than we have,
y2=y1+hf(x1,y1)

7. Example
y2=0+0.2× 0.2 =0.4
Putting n=2, for finding third approximation than we have,
y3=y2+hf(x2,y2)
y3=0.04+0.2× .44=0.128
Putting n=3, for finding forth approximation than we have,
y4=y3+hf(x3,y3)
y4=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 LL1 is
approximates by the tangent at P such as at P1 we have,
Y1=y0+hf(x0,y0)
Than the slope of the curve of the solution through P1 is computed at the tangent at P1
to P1Q1 is drawn meeting the ordinate through L2 in P2(x0+2h,y2)……1
Now we find better approximation y1’ of y(x0+h) by taking the slope of the curve as the
mean of the slope of the tangent at P and P1 i.e.
Y1’=y0+
ℎ
2
{f(x0+y0)+f(X0+h,Y1)}…..2
As the slope of the tangent at P1 is not known, We take Y1 is found in equation (1)

9. Modified Euler’s Method
Euler method
yn+1 =yn+hf(xn,yn)
Where xn=x0+nh
Euler’s Modified Method is
yn+1=yn+
ℎ
2
{f(xn , yn)+f(xn+1 ,yn+1)}

10. Working of Modified Euler’s Method
1. First we find the first approximation using Euler’s Method.
2. The approximated value of y1 is than modified using Euler modified method.
3. The approximated value of y1 from Euler modified method is again approximated
until the equal value of y1 is found.
4. The value of y1 is taken for the approximation of y2 using Euler method.
5. 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,
yn+1=yn+hf(xn,yn) …………..(1)
Putting n=0,
y1=y0+hf(x0,y0) ; y1=1+0.1(1)=1.1
Than using Euler’s Modified Method,
yn+1=yn+
ℎ
2
{f(xn , yn)+f(xn+1 ,yn+1)}……(2)

12. Example
y1=y0+
ℎ
2
{f(x0 , y0)+f(x1 ,y1)}
y1 =1+
0.1
2
(1+1.2)=1.11
Repeating process will give
y1 =1.1105
Than using Euler’s Method, Putting n=1;
y2=y1+hf(x1,y1)
y2=1.1105+0.1(0.1+1.1105)=1.23155
Than using Euler’s Modified Method,Putting n=1;
y2=y1+
ℎ
2
{f(x1 , y1)+f(x2 ,y2)}

13. Example
y2=1.1105+
0.1
2
(1.2105+1.43155)=1.2426
Repeating process will give
y2=1.2432
At y(0.2)=1.2432 Answer.

14. Fitting of Nonlinear Curve
1. y=a𝒙 𝒃
Taking logarithms,𝑙𝑜𝑔10 𝑦 = 𝑙𝑜𝑔10 𝑎 + 𝑏𝑙𝑜𝑔10 𝑥
Y=A+bX ……(1) Where A= 𝑙𝑜𝑔10 𝑎, X=𝑙𝑜𝑔10 𝑥,Y= 𝑙𝑜𝑔10 𝑦
Normal equation for (1) are,
𝑌=nA+b 𝑋
𝑋𝑌=A 𝑋+b 𝑋2
From A & b can be determine .Than a can be calculated from A= 𝑙𝑜𝑔10 𝑎.

15. Fitting of Nonlinear Curve
2. y=a𝒆 𝒃𝒙
Taking logarithms,𝑙𝑜𝑔10 𝑦 = 𝑙𝑜𝑔10 𝑎 + 𝑏𝑥𝑙𝑜𝑔10 𝒆
Y=A+bX ……(1) Where A= 𝑙𝑜𝑔10 𝑎 , B=𝑏 𝑙𝑜𝑔10e ,Y= 𝑙𝑜𝑔10 𝑦
Normal equation for (1) are,
𝑌=nA+B 𝑥
𝑥𝑌=A 𝑥+B 𝑥2
From A & B can be determine .Than “a” and “b” can be calculated from A= 𝑙𝑜𝑔10 𝑎
, B=𝑏 𝑙𝑜𝑔10e .

16. Fitting of Nonlinear Curve
3. xya=b
Taking logarithms,𝑙𝑜𝑔10 𝑥 + 𝑎𝑙𝑜𝑔10 𝑦 = 𝑙𝑜𝑔10 𝑏
𝑙𝑜𝑔10 𝑦=
1
𝑎
𝑙𝑜𝑔10 𝑏 -
1
𝑎
𝑙𝑜𝑔10 𝑥
Y=A+BX ……(1) Where X= 𝑙𝑜𝑔10 𝑥 , B=-
1
𝑎
, A =
1
𝑎
𝑙𝑜𝑔10 𝑏 ,Y= 𝑙𝑜𝑔10 𝑦
Normal equation for (1) are,
𝑌=nA+B 𝑋
𝑋𝑌=A 𝑋+B 𝑋2
From A & B can be determine .Than “a” and “b” can be calculated from A =
1
𝑎
𝑙𝑜𝑔10 𝑏
, B=-
1
𝑎
.

17. Fitting of Nonlinear Curve-Working
1.Given relation eg-xya=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. 𝑋=nA+B 𝑋 , 𝑋𝑌=A 𝑋+b 𝑋2)
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 : An experiment gave the following value:
It is known that v and t are connected by a relation v=atb .Find the best possible
values of a and b.
Solution : We have v=atb .
Taking log10 both side will give,
log10v=log10a+blog10t …..(1)
Equation 1 can be written as;
Y=A+bX ……..(2)
Where X=log10t ,Y=log10v, A=log10a
v(ft/min) 350 400 500 600
t(min) 61 26 7 2.6

19. Example
Than the Normal equations are ,
𝑌=nA+b 𝑋 …… (3)
𝑋𝑌=A 𝑋+b 𝑋2 ……..(4)
Now 𝑋, 𝑌, 𝑋𝑌, 𝑋2 are calculated in the following table;
v t X=log10t Y=log10v XY X2
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/)