<< Chapter < Page Chapter >> Page >

Optimization theory is the branch of applied mathematics whose purpose is to consider a mathematical expression in order to find a set of parameters that either maximize or minimize it. Being an applied discipline, problems usually arise from real-life situations including areas like science, engineering and finance (among many other). This section presents some basic concepts for completeness and is not meant to replace a treaty on the subject. The reader is encouraged to consult further references for more information.

Solution of linear weighted least squares problems

Consider the quadratic problem

min h d - C h 2

which can be written as

min h d - C h T d - C h

omitting the square root since this problem is a strictly convex one. Therefore its unique (and thus global) solution is found at the point where the partial derivatives with respect to the optimization variable are equal to zero. That is,

h d - C h T d - C h = h d T d - 2 d T C h + C h T C h = - 2 C T d + 2 C T C h = 0 C T C h = C T d

The solution of [link] is given by

h = C T C - 1 C T d

where the inverted term is referred [link] , [link] as the Moore-Pentrose pseudoinverse of C T C .

In the case of a weighted version of [link] ,

min h w d - C h 2 2 = k w k | d k - C k h | 2

where C k is the k -th row of C , one can write [link] as

min h W ( d - C h ) T W ( d - C h )

where W = diag ( w ) contains the weighting vector w . The solution is therefore given by

h = C T W T W C - 1 C T W T W d

Newton's method and the approximation of linear systems in an l p Sense

Newton's method and l p Linear phase systems

Consider the problem

min a g ( a ) = A ( ω ; a ) - D ( ω ) p

for a R M + 1 . Problem [link] is equivalent to the better posed problem

min a f ( a ) = g ( a ) p = A ( ω ; a ) - D ( ω ) p p = i = 0 L C i a - D i p

where D i = D ( ω i ) , ω i [ 0 , π ] , C i = [ C i , 0 , ... , C i , M ] , and

C = C 0 C L

The i j -th element of C is given by C i , j = cos ω i ( M - j ) , where 0 i L and 0 j M . From [link] we have that

f ( a ) = a 0 f ( a ) a M f ( a )

where a j is the j -th element of a R M + 1 and

a j f ( a ) = a j i = 0 L C i a - D i p = i = 0 L a j C i a - D i p = p i = 0 L C i a - D i p - 1 · a j C i a - D i

Now,

a j C i a - D i = sign ( C i a - D i ) · a j ( C i a - D i ) = C i , j sign ( C i a - D i )

where Note that

lim u ( a ) 0 + a j u ( a ) p = lim u ( a ) 0 - a j u ( a ) p = 0

sign ( x ) = 1 x > 0 0 x = 0 - 1 x < 0

Therefore the Jacobian of f ( a ) is given by

f ( a ) = p i = 0 L C i , 0 C i a - D i p - 1 sign ( C i a - D i ) p i = 0 L C i , M - 1 C i a - D i p - 1 sign ( C i a - D i )

The Hessian of f ( a ) is the matrix 2 f ( a ) whose j m -th element ( 0 j , m M ) is given by

j , m 2 f ( a ) = a 2 a j a m f ( a ) = a m a j f ( a ) = i = 0 L p C i , j a m D i - C i a p - 1 sign ( D i - C i a ) = i = 0 L α a m b ( a ) d ( a )

where adequate substitutions have been made for the sake of simplicity. We have

a m b ( a ) = a m C i a - D i p - 1 = ( p - 1 ) C i , m C i a - D i p - 2 sign ( C i a - D i ) a m d ( a ) = a m sign ( D i - C i a ) = 0

Note that the partial derivative of d ( a ) at D i - C i a = 0 is not defined. Therefore

a m b ( a ) d ( a ) = b ( a ) a m d ( a ) + d ( a ) a m b ( a ) = ( p - 1 ) C i , m C i a - D i p - 2 sign 2 ( C i a - D i )

Note that sign 2 ( C i a - D i ) = 1 for all D i - C i a 0 where it is not defined. Then

j , m 2 f ( a ) = p ( p - 1 ) i = 0 L C i , j C i , m C i a - D i p - 2

except at D i - C i a = 0 where it is not defined.

Based on [link] and [link] , one can apply Newton's method to problem [link] as follows,

  • Given a 0 R M + 1 , D R L + 1 , C R L + 1 × M + 1
  • For i = 0 , 1 , ...
    1. Find f ( a i ) .
    2. Find 2 f ( a i ) .
    3. Solve 2 f ( a i ) s = - f ( a i ) for s .
    4. Let a + = a i + s .
    5. Check for convergence and iterate if necessary.

Note that for problem [link] the Jacobian of f ( a ) can be written as

f ( a ) = p C T y

where

y = C a i - D p - 1 sign ( C a i - D ) = C a i - D p - 2 ( C a i - D )

Also,

j , m 2 f ( a ) = p ( p - 1 ) C j T Z C m

where

Z = diag C a i - D p - 2

and

C j = C 0 , j C L , j

Therefore

2 f ( a ) = ( p 2 - p ) C T Z C

From [link] , the Hessian 2 f ( a ) can be expressed as

2 f ( a ) = ( p 2 - p ) C T W T W C

where

W = diag C a i - D p - 2 2

The matrix C R ( L + 1 ) × ( M + 1 ) is given by

C = cos M ω 0 cos ( M - 1 ) ω 0 cos ( M - j ) ω 0 cos ω 0 1 cos M ω 1 cos ( M - 1 ) ω 1 cos ( M - j ) ω 1 cos ω 1 1 cos M ω i cos ( M - 1 ) ω i cos ( M - j ) ω i cos ω i 1 cos M ω L - 1 cos ( M - 1 ) ω L - 1 cos ( M - j ) ω L - 1 cos ω L - 1 1 cos M ω L cos ( M - 1 ) ω L cos ( M - j ) ω L cos ω L 1

The matrix H = 2 f ( a ) is positive definite (for p > 1 ). To see this, consider H = K T K where K = W C . Let z R M + 1 , z 0 . Then

z T H z = z T K T K z = K z 2 2 > 0

unless z N ( K ) . But since W is diagonal and C is full column rank, N ( K ) = 0 . Thus z T H z 0 (identity only if z = 0 ) and so H is positive definite.

Newton's method and l p Complex linear systems

Consider the problem

min x e ( x ) = A x - b p p

where A C m × n , x R n and b C m . One can write [link] in terms of the real and imaginary parts of A and b ,

e ( x ) = i = 1 m | A i x - b i | p = i = 1 m | Re { A i x - b i } + j I m { A i x - b i } | p = i = 1 m | ( R i x - α i ) + ( Z i x - γ i ) | p = i = 1 m ( R i x - α i ) 2 + ( Z i x - γ i ) 2 p = i = 1 m g i ( x ) p / 2

where A = R + j Z and b = α + j γ . The gradient e ( x ) is the vector whose k -th element is given by

x k e ( x ) = p 2 i = 1 m x k g i ( x ) g i ( x ) p - 2 2 = p 2 q k ( x ) g ^ ( x )

where q k is the row vector whose i -th element is

q k , i ( x ) = x k g i ( x ) = 2 ( R i x - α α i ) R i k + 2 ( Z i x - γ γ i ) Z i k = 2 R i k R i x + 2 Z i k Z i x - [ 2 α i R i k + 2 γ i Z i k ]

Therefore one can express the gradient of e ( x ) by e ( x ) = p 2 Q g ^ , where Q = [ q k , i ] as above. Note that one can also write the gradient in vector form as follows

e ( x ) = p R T diag ( R x - α ) + Z T diag ( Z x - γ ) · ( R x - α ) 2 + ( Z x - γ ) 2 p - 2 2

The Hessian H ( x ) is the matrix of second derivatives whose k l -th entry is given by

H k , l ( x ) = 2 x k x l e ( x ) = x l p 2 i = 1 m q k , i ( x ) g i ( x ) p - 2 2 = p 2 i = 1 m q k , i ( x ) x l g i ( x ) p - 2 2 + g i ( x ) p - 2 2 x l q k , i ( x )

Now,

x l g i ( x ) p - 2 p = p - 2 2 x l g i ( x ) g i ( x ) p - 4 2 = p - 2 2 q l , i ( x ) g i ( x ) p - 4 2 x l q k , i ( x ) = 2 R i k R i l + 2 Z i k Z i l

Substituting [link] and [link] into [link] we obtain

H k , l ( x ) = p ( p - 2 ) 4 i = 1 m q k , i ( x ) q l , i ( x ) g i ( x ) p - 4 4 + p i = 1 m ( R i k R i l + Z i k Z i l ) g i ( x ) p - 2 2

Note that H ( x ) can be written in matrix form as

H ( x ) = p ( p - 2 ) 4 Q diag g ( x ) p - 4 2 Q T + p R T diag g ( x ) p - 2 2 R + Z T diag g ( x ) p - 2 2 Z

Therefore to solve [link] one can use Newton's method as follows: given an initial point x 0 , each iteration gives a new estimate x + according to the formulas

H ( x c ) s = - e ( x c ) x + = x c + s

where H ( x c ) and e ( x c ) correspond to the Hessian and gradient of e ( x ) as defined previously, evaluated at the current point x c . Since the p -norm is convex for 1 < p < , problem [link] is convex. Therefore Newton's method will converge to the global minimizer x as long as H ( x c ) is not ill-conditioned.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Iterative design of l_p digital filters. OpenStax CNX. Dec 07, 2011 Download for free at http://cnx.org/content/col11383/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Iterative design of l_p digital filters' conversation and receive update notifications?

Ask