NMC PresentationBY:-

Mukul Dev Khunte

Guided by:-

Dr. G.K.Singh

Head Dept. of Mathematics

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

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)

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.

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 )

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)

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.

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)

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)}

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.

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)

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)}

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.

Fitting of Nonlinear Curve

1. y=a𝒙𝒃

Taking logarithms,𝑙𝑜𝑔10𝑦 = 𝑙𝑜𝑔10𝑎 + 𝑏𝑙𝑜𝑔10𝑥

Y=A+bX ……(1) Where A= 𝑙𝑜𝑔10𝑎, X=𝑙𝑜𝑔10𝑥,Y= 𝑙𝑜𝑔10𝑦

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= 𝑙𝑜𝑔10𝑎.

Fitting of Nonlinear Curve

2. y=a𝒆𝒃𝒙

Taking logarithms,𝑙𝑜𝑔10𝑦 = 𝑙𝑜𝑔10𝑎 + 𝑏𝑥𝑙𝑜𝑔10𝒆

Y=A+bX ……(1) Where A= 𝑙𝑜𝑔10𝑎 , B=𝑏 𝑙𝑜𝑔10e ,Y= 𝑙𝑜𝑔10𝑦

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= 𝑙𝑜𝑔10𝑎

, B=𝑏 𝑙𝑜𝑔10e .

Fitting of Nonlinear Curve

3. xya=b

Taking logarithms,𝑙𝑜𝑔10𝑥 + 𝑎𝑙𝑜𝑔10𝑦 = 𝑙𝑜𝑔10𝑏

𝑙𝑜𝑔10𝑦=1

𝑎𝑙𝑜𝑔10𝑏 -

1

𝑎𝑙𝑜𝑔10𝑥

Y=A+BX ……(1) Where X= 𝑙𝑜𝑔10𝑥 , B=-1

𝑎, A =

1

𝑎𝑙𝑜𝑔10𝑏 ,Y= 𝑙𝑜𝑔10𝑦

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

𝑎𝑙𝑜𝑔10𝑏

, B=-1

𝑎.

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.

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

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

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

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/)