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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

what is the stm
Brian Reply
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Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
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LITNING Reply
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LITNING Reply
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LITNING
scanning tunneling microscope
Sahil
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Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
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Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
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Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
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Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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Damian Reply
absolutely yes
Daniel
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Maciej
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Abigail
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Anassong
How can I make nanorobot?
Lily
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s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
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Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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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
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