MA2213 Lecture 1 Rootfindi ng
Dec 25, 2015
MA2213 Lecture 1
Rootfinding
Class ScheduleLectures:
Tuesday 2-4 in LT31 (building S16)
Groups: Class will be divided into 5 groups
Each group (of about 25 students) will:
Tutorials: will meet in weeks 3,5,7,9,11
Labs: will meet in weeks 4,6,8,10,12
do tutorials and labs together
Friday 2-4 in LT31 (building S16)
be assigned a person who will supervise both
tutorials and labs, and grade written homework
(more information will be announced soon)
SuggestionsDo:
Attend lectures, tutorials, and labs
Complete assigned homework and give
to your grader during tutorials and labs
Do Not:
Send or give me assigned homework
Study the textbook and work problems
Use the Module IVLE Website
Email me questions whose answers
are on the Module IVLE Website
Purchase the required textbook
Objectives and PrerequisitesLearn Numerical Methods
Accuracy and Errors
Computer Algorithms (MATLAB)
CalculusReal Numbers and Complex NumbersFundamental Theorem of AlgebraFundamental Theorems of Calculus
Linear Equations: Geometry and Matrix Algebra
Vector Spaces and Linear Transformations
Linear Algebra
Prerequisites
Required Textbook
Elementary Numerical Methods by Atkinson and Han
About 20 books are now in the Science COOP(due to limited shelve space) but more bookswill be transferred from the warehouse)
Homework (Tutorial and Lab) may be assigned from this textbook
Suggested Reading from this textbookAppendix A&B Math Rev, D MATLAB
Lecture Schedule
1 Rootfinding 14, 17 Aug2 Interpolation 21, 24 Aug3 Approximation 28, 31 Aug4 Numerical Integration 4, 7 Sept5 Linear Equations (Direct) 11, 14 Sept 6 Linear Equations (Iterative) 18, 21 Sept
Recess Week 24-28 SeptReview and Midterm Test 2, 5 Oct
Lecture # Topic Dates
Lecture Schedule
7 Eigenvalues 9, 12 Oct 8 Nonlinear Systems 16, 19 Oct9 ODE (Explicit) 23, 26 Oct10 ODE (Implicit) 30 Oct, 2 Nov11 PDE 6, 9 Nov12 Review 13, 16 Nov
Lecture # Topic Dates
EXAM : Friday, 30 Nov 200(MORNING)
Retirement Planning (pages 71-72)
inPdeposited at beginning of each ofAmount
rinN time periods. Interest is paid per period.
End of Period Amount of Money in the Account
1 inPrA )1()1( inPrArA )1()1()1()2( 2inin PrPr )1()1( 2
1k inPrkArkA )1()()1()1(
Retirement Planning (pages 71-72)
Specifying and the value of
in terms of the value of suffices to write
a computer program to compute
)1(A
A
)1( kA)(kA
The figure shows thatif you deposited $1000 monthy at r = 0.0025you would have $584,194 after 30 years
Retirement Planning (pages 71-72)Here is the smallest MATLAB code required
function A = amount(Pin,Nin,r)A(1) = (1+r)*Pin;for k = 1:Nin-1 A(k+1) = (1+r)*A(k) + (1+r)*Pin;endyou would write and store it in a m-file amount.m and then run it interactively using the commands: A = amount(1000,360,0.0025)plot(A); grid; xlabel(‘End of Period (Month)’)ylabel(‘Amount at end of Period’)
Retirement Planning (pages 71-72)But your grader will demand documented codefunction A = amount(Pin,Nin,r)% function A = amount(Pin,Nin,r)% % Wayne Lawton, Student Number: XXXXXX% MA2213, Homework Assignment 1% Date: Tuesday 14 August 2007% % Computes Amount of Money in an Interest Bearing Account%% Inputs:% Pin = input at beginning of each period% Nin = number of input periods% r = interest earned during each period% Outputs:% A = array of length Nin% A(k) = amounts at the end of the k-th period%A(1) = (1+r)*Pin;for k = 1:Nin-1 A(k+1) = (1+r)*A(k) + (1+r)*Pin;end
Retirement Planning (pages 71-72)Comments (start with %) are useful in interactive mode
>> help amount
function A = amount(Pin,Nin,r) Wayne Lawton, Student Number: XXXXXX MA2213, Homework Assignment 1 Date: Tuesday 14 August 2007 Computes Amount of Money in an Interest Bearing Account Inputs: Pin = input at beginning of each period Nin = number of input periods r = interest earned during each period Outputs: A = array of length Nin A(k) = amounts at the end of the k-th period
Retirement Planning (pages 71-72)
An algebraic calculation gives
We now derive a closed expression for
Case 1.
ininN PrPr in )1()1( 1
ininin PNSr 0
ininN
inN PrPrPr inin )1()1()1( 1
)( inin NAS
inS
Case 2. inin SrSrr ]1)1[(0
in
N
in Pr
rrS
in )1()1( 1
Retirement Planning (pages 71-72)
At the end of period
we withdraw the first of payments each ofoutNinN
to leave 0 after the last withdrawal.
,beginning of period 1inN
outPamount Withdrawal #
1Amount After Withdrawal
outin PSA )1(
1k outPkArkA )()1()1(
2outoutin PPSrA ))(1()2(
outN )()1()( 1outin
Nout PSrNA out
outoutoutN PPrPr out )1()1( 2
3 outoutoutin PPrPSrA )1()()1()3( 2
Retirement Planning (pages 71-72)
Therefore, the method for geometric series gives
substituting for
outN
inN
out PrSrNA outout ]1)1[()1()( 11
out
N
inN
out Pr
rSrNAr
out
out1)1(
)1()(1 1
inS then gives
outN
inNN PrPrrr outinout ]1)1[()]1()1[()1( 11
)( outNAr
therefore if 1r and
0)( outNA then r1
is a root of the retirement polynomial
outN
outinNN
in PxPPxPxR outoutin )()(
outoutininoutoutinout PNPNPNSNAr )(1
Retirement Polynomial
If thenNNN outin is a root of Nr)1(
therefore, the quadratic formula gives
inoutinoutinoutin PPPPPPP 2/4)( 2
since .1
the polynomial
outN
outinNN
in PxPPxPxR outoutin )()(
outoutinin PxPPxP )(2
in
outininoutoutin P
PPPPPP 2/||
Retirement Polynomial
Result: If
0r
then there exists so that
is a root of the retirement poly
We first factorize
outN
outinNN
in PxPPxPxR outoutin )()(
0 outoutinin PNPN
r1
)()1()( xQxxR
)1()( 1 xxxPxQ inout NNin
)1( 1 xxP outNout
Then we observe that 0)1( outoutinin PNPNQ
and 0)( inNinoutin PPPQ
Retirement PolynomialExample What is the minimal monthly interest rate to support the following retirement dream
then
4000,240,1000,360 outoutinin PNPN0045.1)( inN
inoutin PPP
Ancient Rootfinding
• Problem : Given a function f(x), find x* such that f(x*) = 0. Example. Compute
2)(f 2 xxThe Babylonian clay tablet YBC 7289 (c. 2000–1650 BC) gives an approximation of x* = in four sexagesimal figures, which is about six decimal figures
• Example :
2
Bisection Method (pages 72-79)Theorem: If f(x) is continuous in the interval [a,b]and , then there exists such that f(x*) = 0. Hence
2
bam
a b
)(f a
)(f b
midpoint
x
y
)(f xy
there exists a x* in (a,m) such that f(x*) = 0
Observation: If
0)(f)(f ba
0)(f)(f mathen (1)or (2)or (3)
0)(f)(f bm
(2)
(1)
there exists a x* in (m,b) such that f(x*) = 0
0)(f)(f ba),[],( bmxormax
),( bax
0)(f m
(3) a root is found
Bisection Method (pages 72-79)If nathen compute sequence0)(f)(f ba
For k = 2:N
2/)(,, 11111 bambbaa
0)(f)(f 11 kk maIf
else 2,, 11
kkkkkkk
bammbaa
2,, 11
kkkkkkk
bambbma
Nmx *
Bisection Method (pages 72-79)At the algorithm executes the “k = N” instructionit computes
)(2||2|| 1* ababrx NNN
Therefore, the computed root Nmx *
],[],,[ babbaa NN there exists
such that
],[ NN bar such that .0)(f rsatisfies
Question: How large should N be chosen to be ?
Question: How large will the residual |f(x*)| be ?
Bisection Method (pages 72-79)Find root of
5.1,2,1 111 mba025.0)(f)(f 11 ma
25.1,5.1,1 222 mba
3750.1,5.1,25.1 333 mba
2)(f 2 xx ]2,1[in the interval
04375.0)(f)(f 22 ma
00820.0)(f)(f 33 ma4375.1,5.1,3750.1 444 mba
4375.1x 0664.0)(f x
Questions for Thought and Discussion
What are advantages of the bisection method?
What are disadvantages of the bisection method?
How could the bisection method be used to compute 1/y on a computer that could only add, subtract, and multiply ?
Newton’s Method (pages 79-89)
Question: How should we compute
2xx
)(f
)(xf
1'
112 xxx
x
y )(f xy
1x
Observation: Clearly |||| 12 xxxx
3x ?
Question: What happens if f(x) = ax + b ?
Newton’s Method (pages 79-89)
Question: How should we compute
)(f
)(xf
1'
112 xxx
3x ?
)(f
)(xf
2'
223 xxx
1,)(f
)(xf'
n1 n
xxx
nnn
We see a GENERAL PATTERN
Newton’s Method (pages 79-89)
Question: What happens if f(x) = ax + b ?
a
b
a
bx
xxx n
nnn
n'
n1
ax
)(f
)(xf
for ANY starting value for1x
a
b
a
bx
xxx
1
11
'1
12
ax
)(f
)(xf
332 xxxa
b
Newton’s Method (pages 79-89)Find root of
11 x
n
nnn x
xxx
2
22
1
1,2)(f 12 xxx
Solution: The iteration is
5.12
2112
x
4167.15.12
25.15.1
2
3
x
nn xxe 4142.12 x
4142.01 e
0858.02 e
0025.03 e
74514142156862.14 x 000002123.04 e
Newton’s Method (pages 79-89)Find root of )2(1 nnn xbxx ,)(f 1 xbx
11 20 bx
1b
Question Why should ?
Question When would this algorithm be useful ?
12 b
by
x
y)(f xy
)(f xy
Taylor’s Polynomial (pages 2-11)Definition If derivatives on
is called the degree n Taylor Polynomial of f at a.
has
then the polynomialf
nn axaaxaaxaa ))((f))((f))((f)(f (n)!
12''21'
Example
)(xPn
2''21' ))((f))((f)2(f axaaxa
n and],[ ],[ a
],[,1)(f 3 xxxthe degree 2 Taylor Polynomial at 2 is )(2 xP
2)2(6)2(129 xx
9126 2 xx
Taylor’s Polynomial (pages 2-11)
BLUE 1+x^3 GREEN Taylor Polynomial
Error in Taylor Polynomial (pages 11-23)
Theorem If an interval
has n+1 continuous derivatives onf
fdegree n Taylor Polynomial for
and andnP|,[
at then
}],max{},,[min{)( xaxaxc
is the
)()(f)( xPxxR nn
|,[ aa
Taylor’s Remainder satisfies
],[)),((f)()( 1)(n1)!1(
1 xxcaxxR nnn
where
(this means that c(x) is between a and x)
Proof This very important result is proved in most Calculus textbooks, e.g. pages 1166-1167 inThomas’ CALCULUS, Tenth International Edition
Error Analysis (pages 83-86)Theorem Assume that
derivatives on an intervalhas 2 continuousf
1xNewton’s algorithm starting
at
and LB < 2 where0)(f x
}:|(x)fmin{|2
}:|(x)f|max{'
''
Ix
IxB
|/,[ 22LL xxI
and
whose error
nn xxe gives a sequence
|||||| 22
1 nLB
nn eeBe
Ix 1Then for every
and 0)(f ' x for every Ix
Ixn sequence satisfies
and thereforenx converges to *x
Error Analysis (pages 83-86)
Newton’s Method gives )(f/)(xf 'n1 nnn xxx
2"
' )(2
)(f))((f)(f)(f0 nnnn xxxxxxx
nxatthen the degree 2
f],[x 2
*2
*n
LL xx Proof If
Taylor Polynomial for satisfies
where }],max{},,[min{ ** xxxx nn
therefore the errors satisfy
11 nn xxe
hence
)(f/)](f))((f[)(f/)(f '''*nnnnnnn xxxxxxxxx
2'''2*''' )](f2/)(f[))]((f2/)(f[ nnnn exxxx |||||| 2
21 nn
BLnn eeeBe complete the proof!
Questions for Thought and Discussion
What are advantages of Newton’s method?
How can the Bisection Method and Newton’s Method be combined to produce a method that has advantages of both ?
What are disadvantages of Newton’s method?
Homework Due Tutorial 1Question 1. Solve problem 1. a. on page 77
Question 2. Find the real root of the polynomial (in the previous question) using Newton’s method
Suggested Reading: pages 71-89,Appendix A. Mean Value TheoremsAppendix D. MATLAB: An Introduction
Discuss with your tutor the “Questions for Thought and Discussion”
||ln2ln||ln 1 nn eBe
Extra Credit Use the following formula
to derive a ‘closed form’ upper bound for || ne