Taylor series method
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Taylor series method. C 조 박재무 윤동일 장석민. outline. What is Taylor series method Main Topic 1 Main Topic 2 Future work. Problem 1 Result 1 How to solve problem 1 Summary 1. Problem 2 Result 2 How to solve problem 2 Summary 2. What is Taylor series?.

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Taylor series method

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Taylor series method

Taylorseries method

C조

박재무

윤동일

장석민


Outline

outline

  • What is Taylor seriesmethod

  • Main Topic 1

  • Main Topic 2

  • Future work

Problem 1

Result 1

How to solve problem 1

Summary 1

Problem 2

Result 2

How to solve problem 2

Summary 2


Taylor series method

WhatisTaylorseries?

The Taylor series of a real or complex-valued functionƒ(x) that is infinitely differentiable at a real or complex number ais the power series which can be written in the more compact sigma notation as

When a = 0, the series is also called a Maclaurin series


Taylor series method

WhatisTaylorseries?

Taylor seriesis a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivative at a single point.

Example


Taylor series method

WhatisTaylorseries method?

Euler’s method

Tangent linelocal approximation

Taylor series

Higher degree better approximation


Graph of e x

Graph of e^x

Taylortool in matlab

Taylor polynomial of degree 1 (Euler’s M)

Taylor polynomial of degree 2


Problem 1

Problem 1

Compute the Taylor polynomials of degree 4 for the solutions to the given initial value problems. Use these Taylor polynomials to approximate the solution at x = 1.


Result 1

Result 1

Problem(1) has error that can’t be ignored

Problem(2) has large error


Subproblem 1 numerical

Subproblem1(Numerical)


Subproblem 1 analytical

Subproblem 1 (Analytical)

Integrating factor


Subproblem 2 numerical

Subproblem 2 (numerical)


Subproblem 2 analytic

Subproblem 2(Analytic)


How can we decrease the error

How can we decrease the error?

Problem (1)

: Increase the degree of Taylor polynomial

Error : 0.247498 Error : 0.000298


How can we decrease the error1

How can we decrease the error?

Problem (2)

Can not be solve the problem by increasing degree…..


How can we decrease the error2

How can we decrease the error?

Problem (2)

Euler’s method can decrease the error

When mesh size = 0.1,

y(1) = 2.094317

errer = 0.050841

x=0;

y=4;

for i=1:100

yp=y*(2-y);

y1=y+0.1*yp;

y=y1;

x=x+0.01;

end


Solve d e in matlab

Solve D.E in matlab

syms x y

f=dsolve('Dy=x-2*y', 'y(0)=1', 'x')

p=taylor(f,5,x)

f = x/2 + 5/(4*exp(2*x)) - ¼

p = (5*x^4)/6 - (5*x^3)/3 + (5*x^2)/2 - 2*x + 1


Summary 1

Summary 1

Taylor series method can find approximation

But it need higher degree

and some points in some function can not

enable to get approximation


Problem 2

Problem 2

Compare the use of Euler’s method with that of Taylor series to approximate the solution (x) to the initial value problem

Do this by completing the following table:


Result 2 x

Result 2(X)


Result 2

Result 2


Result 2 error

Result 2(Error)


Problem 2 euler s method

Problem 2 ( Euler’s method)

x=0;

y=0;

for i=1:10  for i=1:100

slope=-y+cos(x)-sin(x);

y1=y+0.1*slope;  y1=y+0.3*slope;

y=y1;

x=x+0.1;  x=x+0.3;

end


Problem 2 taylor s method x

Problem 2 ( Taylor’s method)(X)


Problem 2 taylor s method x1

Problem 2 ( Taylor’s method)(x)

Step size = 1

Step size = 3

Step size = 1

Step size = 3


Problem 2 taylor s method

Problem 2 ( Taylor’s method)


Problem 2 taylor s method1

Problem 2 ( Taylor’s method)

Taylor polynomial

Degree 2

Taylor polynomial

Degree 5


Graph degree 2

Graph(degree 2)

y=cos(x)-exp(-x)

y=cos(x)-exp(-x)

Taylor polynomial of degree 2 with step size 0.1

Taylor polynomial of degree 2 with step size 1


Problem 2 taylor s method2

Problem 2 ( Taylor’s method)

Degree: 2

Degree: 5

x=0;

y=0;

step=0.1;

for i=1:10

slope=-y+cos(x)-sin(x);

slope2=-2*cos(x)+y;

slope3=cos(x)+sin(x)-y;

slope4=y;

slope5=-y+cos(x)-sin(x);

y1=y+step*slope

+(step)^2/2*slope2

+(step)^3/(3*2*1)*slope3

+(step)^4/(4*3*2*2)*slope4

+(step)^5/(5*4*3*2*1)*slope5;

y=y1;

x=x+step;

end

x=0;

y=0;

step=0.1;

for i=1:10

slope=-y+cos(x)-sin(x);

slope2=-2*cos(x)+y;

y1=y+step*slope+

(step)^2/2*slope2;

y=y1;

x=x+step;

end


Convergence rate

Convergence rate


Problem 2 analytical

Problem 2 ( Analytical)

Integrating factor


Problem 2 analytical1

Problem 2 ( Analytical)


Summary 2

Summary 2

Good approximation

In same method

- Step size : small

In same step size

- Taylor Method is better than Euler’s Method

- Taylor polynomial of higher degree is better

than Taylor polynomial of lower degree

But Taylor polynomial of higher degree

needs many calculation


Think more

Think more….

How can we get approximation

fast and accurate

with simple code(calculation)?

Runge-Kutta


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