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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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## Taylorseries method

C조

박재무

윤동일

장석민

### 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

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

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

WhatisTaylorseries method?

Euler’s method

Tangent linelocal approximation

Taylor series

Higher degree better approximation

### Graph of e^x

Taylortool in matlab

Taylor polynomial of degree 1 (Euler’s M)

Taylor polynomial of degree 2

### 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

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

Problem(2) has large error

### Subproblem 1 (Analytical)

Integrating factor

### 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 error?

Problem (2)

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

### 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

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

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

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:

### 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

Step size = 1

Step size = 3

Step size = 1

Step size = 3

### Problem 2 ( Taylor’s method)

Taylor polynomial

Degree 2

Taylor polynomial

Degree 5

### 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 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

### Problem 2 ( Analytical)

Integrating factor

### 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….

How can we get approximation

fast and accurate

with simple code(calculation)?

Runge-Kutta