# Numerical Methods, - PowerPoint PPT Presentation

PPT – Numerical Methods, PowerPoint presentation | free to download - id: 6e6cd0-NDRhN

The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## Numerical Methods,

Description:

### Today s class Numerical differentiation Roots of equation Bracketing methods Numerical Methods, Lecture 4 Prof. Jinbo Bi CSE, UConn* – PowerPoint PPT presentation

Number of Views:62
Avg rating:3.0/5.0
Slides: 46
Provided by: leiw154
Category:
Tags:
Transcript and Presenter's Notes

Title: Numerical Methods,

1
Todays class
• Numerical differentiation
• Roots of equation
• Bracketing methods

2
Numerical Differentiation
• Finite divided difference
• First forward difference
• First backward difference

3
Numerical Differentiation
• Centered difference approximation
• Subtract the two equations

4
Numerical Differentiation
• First forward difference

5
Numerical Differentiation
• First backward difference

6
Numerical Differentiation
• Centered difference

7
Error Propagation
• What is the effect of error in one calculation
propagating to subsequent calculations?
• Example
• Multiplying sin x with cos x
• Single variable functions

8
Error Propagation
• Use Taylor series

9
Error Propagation
10
Error Propagation
• Multivariable functions

11
Numerical stability
• Condition of a problem is a measure of its
sensitivity to changes in input values
• The condition number is defined as the ratio of
the relative function error to the relative value
error

12
Numerical stability
• Condition number lt 1 indicates a well-conditioned
function i.e. changes in the input are
attenuated
• Condition number gt 1 indicates a ill-conditioned
function i.e. changes in the input are amplified

13
Roots of equation
• Given a function f(x), the roots are those values
of x that satisfy the relation f(x) 0
• Example
• From the quadratic formula, the roots are

14
Roots of Equations
• The need to solve for roots show up in many
engineering problems
• Also, can be used to find solutions to implicit
variables

15
Example
• Find a value of R such that current is 5A at t
1s

16
Example
• It is not possible to isolate R to the left side
and thus solve for R
• R is know as an implicit variable
• Rewrite the function as a function of R set to 0

17
Roots of equations
• Still need a method to solve for this root
• Other examples of difficult to solve roots

18
Roots of equations
• Non-computer methods
• Graphical methods

19
Graphical methods
• Not exact
• Can give you a rough estimate of the root,
• Can give you insights on the number of roots and
shape of the curve
• Can use the rough estimate in more precise
numerical methods

20
Graphical methods
• Use to get an initial estimate of the root and
also to find out how many roots there are

21
Graphical methods
22
Graphical methods
23
Graphical methods
24
Roots of equation
• Non-computer/numerical method
• Exhaustive search method
• To find the root in the interval a,b, start at
xa and check if f(a) 0, then try f(a?),
f(a2?), and so on, until we get f(x)
sufficiently close to 0
• If the step value ? is sufficiently small we can
obtain an accurate result but this could take an
extremely long time. For example, if the interval
is 0,10 and the step size is ? 0.001, it will
take on average 10,000 guesses
• In addition to the inefficiency of this approach,
if f(x) is a steep function, this approach may
not produce an accurate results

25
Exhaustive search
• Example
• Find the root of the function
• Actual root is at x1.0001
• With an interval of 0.9, 1.1 and a step size of
? 0.001. The exhaustive search method will test
f(1.000) -0.01 and f(1.001) 0.086, neither of
which are that close to f(x) 0

26
Roots of equations
• More systematic methods are required
• Bracketing methods
• Open methods

27
Incremental search methods
• Locate an interval where sign changes
• Divide interval into smaller subintervals which
are then searched for sign changes
• Keep repeating until root is found with
sufficient confidence

28
Bisection method
• Also called
• Binary chopping
• Interval halving
• An incremental search method where the interval
is cut in half

29
Bisection method
• Step 1
• Choose lower xl and upper xu such that the
function changes sign over that range i.e.
f(xl) and f(xu) are different signs or f(xl)
f(xu) lt 0
• Step 2
• Estimate root to be xr(xlxu)/2

30
Bisection method
• Step 3
• Determine in which subinterval the root lies
• If f(xr)?0 is within acceptable tolerance, stop
and root equals xr
• If f(xl) f(xr) lt 0, then root is in lower
• If f(xl) f(xr) gt 0, then root is in upper

31
Bisection method
• Termination criteria
• Use approximate relative error calculation to
determine when to stop
• In general, ?a is larger than ?t

32
Bisection method
• Example
• Use range of 202204
• Root is in upper subinterval

33
Bisection method
34
Bisection method
• Use range of 203203.5
• Root is in upper subinterval

35
Error estimates
• The approximate error is upper bound estimate of
the true error
• When the root is near one of the ends of the
interval, the approximate error is fairly close
to the actual true error
• Error is fairly well-contained

36
Error estimates
• You always know that the true root is within ?x/2

37
Bisection method
38
Bisection method
• You can calculate an error estimate based just on
the initial guesses
• You can also make estimates on the error on
future iterations
• Superscripts indicates the iteration number

39
Bisection method
• Each subsequent iteration cuts the approximate
error in half
• This, allows to determine a priori exactly how
many iterations are needed to arrive at the
desired error

40
False Position Method
• The false position method works in a similar
fashion to the bisection method
and f(b) have opposite signs, which is the same
as the bisection
• Instead of choosing the initial guess xr as the
midpoint of the interval, we join the point
a,f(a) and b,f(b) with a straight line and
choose xr as the point where that straight line
crosses the x-axis.

41
False Position Method
42
False Position Method
• Algorithm is the same as bisection method with
the same three steps

43
False Position Method
• Step 1
• Choose lower xl and upper xu such that the
function changes sign over that range - i.e.
f(xl) and f(xu) are different signs - or f(xl)
f(xu) lt 0
• Step 2
• Estimate new root to be

44
False Position Method
• Step 3
• Determine in which subinterval the root lies
• If f(xr) ? 0 is within acceptable tolerance, stop
and root equals xr
• If f(xl) f(xr) lt 0, then root is in lower