Solution to Algebraic &Transcendental Equations

1 / 19

# Solution to Algebraic &Transcendental Equations - PowerPoint PPT Presentation

A. Solution to Algebraic &Transcendental Equations. Algebraic functions. The general form of an Algebraic function:. f i = an i -th order polynomial. Example :. f 3. f 2. f 0. Polynomials are a simple class of algebraic function. a i ’s are constants. Transcendental functions.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Solution to Algebraic &Transcendental Equations' - andrew

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

A

### Solution to Algebraic &Transcendental Equations

Algebraic functions

The general form of an Algebraic function:

fi = an i-th order polynomial.

Example :

f3

f2

f0

Polynomials are a simple class of algebraic function

ai’s are constants.

Transcendental functions

A transcendental function is non-algebraic.

May include trigonometric, exponential, logarithmic functions

Examples:

Equation Solving

Given an approximate location (initial value)

find a single real root

Root

Finding

non-linear

Single variable

Open

Methods

Brackting

Methods

Iterative

Newton-

Raphson

Secant

Bisection

False-

position

A.1

### Iterative methods

April 5, 2009

Problem
• Find the root of

f(x) = e-x – x

• There is no exact or analytic solution
• Numerical solution:
Iterative Solution
• Generate
• x2=e-x1= e-1= 0.368
• x3=e-x2= e-0.368 = 0.692
• x4=e-x3= e-0.692=0.500

In general:

After a few more iteration we will get

Convergence Examples

Convergent spiral pattern

Convergent staircase pattern

Divergence Example

Divergent spiral pattern

Divergent staircase pattern

Existence of Root

There exists one and only one root if

L is Lipschitz constant,

Convergence?

If x=a is a solution then,

error reduces at each step

i.e. iteration will converge

If magnitude of 1st at x=a

derivative is less than 1

A.2

### Aitken’s Process

April 5, 2009

kth Order Convergence
• Pervious iterative method has linear (1st order) convergence, since:
• For kth order convergence we have:
• Now consider a 2nd order method.

Aitken’s 2 process

Aitken’s process
• If  is a root of the equation i.e., =g() then,
• Now if we use
Algorithm

  guess_value;

while (!   g()) {

}

http://www.buet.ac.bd

/cse/users/faculty/reazahmed/cse317.php