# 0.16 Appendix: optimization theory

 Page 1 / 1

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

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\parallel d-\mathbf{C}h\parallel }_{2}$

which can be written as

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\left(d,-,\mathbf{C},h\right)}^{T}\left(d,-,\mathbf{C},h\right)$

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,

$\begin{array}{cc}\hfill \frac{\partial }{\partial h}\left\{{\left(d,-,\mathbf{C},h\right)}^{T},\left(d,-,\mathbf{C},h\right)\right\}& =\frac{\partial }{\partial h}\left\{{d}^{T},d,-,2,{d}^{T},\mathbf{C},h,+,{\left(\mathbf{C},h\right)}^{T},\mathbf{C},h\right\}\hfill \\ & =-2{\mathbf{C}}^{T}d+2{\mathbf{C}}^{T}\mathbf{C}h=0\hfill \\ & ⇒\phantom{\rule{0.277778em}{0ex}}{\mathbf{C}}^{T}\mathbf{C}h={\mathbf{C}}^{T}d\hfill \end{array}$

The solution of [link] is given by

$h={\left({\mathbf{C}}^{T},\mathbf{C}\right)}^{-1}{\mathbf{C}}^{T}d$

where the inverted term is referred [link] , [link] as the Moore-Pentrose pseudoinverse of ${\mathbf{C}}^{T}\mathbf{C}$ .

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

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\parallel \sqrt{w}\left(d,-,\mathbf{C},h\right)\parallel }_{2}^{2}=\sum _{k}{w}_{k}{|{d}_{k}-{C}_{k}h|}^{2}$

where ${C}_{k}$ is the $k$ -th row of $\mathbf{C}$ , one can write [link] as

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}\left(\mathbf{W},\left(,d,-,\mathbf{C},h\right){\right)}^{T}\left(\mathbf{W},\left(,d,-,\mathbf{C},h\right)\right)$

where $\mathbf{W}=\text{diag}\left(\sqrt{w}\right)$ contains the weighting vector $w$ . The solution is therefore given by

$h={\left({\mathbf{C}}^{T},{\mathbf{W}}^{T},\mathbf{W},\mathbf{C}\right)}^{-1}{\mathbf{C}}^{T}{\mathbf{W}}^{T}\mathbf{W}d$

## Newton's method and ${l}_{p}$ Linear phase systems

Consider the problem

$\underset{a}{\text{min}}\phantom{\rule{0.277778em}{0ex}}g\left(a\right)={\parallel A\left(\omega ;a\right)-D\left(\omega \right)\parallel }_{p}$

for $a\in {\mathbb{R}}^{M+1}$ . Problem [link] is equivalent to the better posed problem

$\begin{array}{ccc}\hfill \underset{a}{\text{min}}\phantom{\rule{0.277778em}{0ex}}f\left(a\right)=g{\left(a\right)}^{p}& =& {\parallel A\left(\omega ;a\right)-D\left(\omega \right)\parallel }_{p}^{p}\hfill \\ & =& \sum _{i=0}^{L}\mid {C}_{i}a-{D}_{i}{\mid }^{p}\hfill \end{array}$

where ${D}_{i}=D\left({\omega }_{i}\right)$ , ${\omega }_{i}\in \left[0,\pi \right]$ , ${C}_{i}=\left[{C}_{i,0},...,{C}_{i,M}\right]$ , and

$\mathbf{C}=\left[\begin{array}{c}{C}_{0}\\ ⋮\\ {C}_{L}\end{array}\right]$

The $ij$ -th element of $\mathbf{C}$ is given by ${C}_{i,j}=\text{cos}\phantom{\rule{1mm}{0ex}}{\omega }_{i}\left(M-j\right)$ , where $0\le i\le L$ and $0\le j\le M$ . From [link] we have that

$\nabla f\left(a\right)=\left[\begin{array}{c}\frac{\partial }{\partial {a}_{0}}f\left(a\right)\\ ⋮\\ \frac{\partial }{\partial {a}_{M}}f\left(a\right)\end{array}\right]$

where ${a}_{j}$ is the $j$ -th element of $a\in {\mathbb{R}}^{M+1}$ and

$\begin{array}{ccc}\hfill \frac{\partial }{\partial {a}_{j}}f\left(a\right)& =& \frac{\partial }{\partial {a}_{j}}\sum _{i=0}^{L}\mid {C}_{i}a-{D}_{i}{\mid }^{p}\hfill \\ & =& \sum _{i=0}^{L}\frac{\partial }{\partial {a}_{j}}\mid {C}_{i}a-{D}_{i}{\mid }^{p}\hfill \\ & =& p\sum _{i=0}^{L}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}·\frac{\partial }{\partial {a}_{j}}\mid {C}_{i}a-{D}_{i}\mid \hfill \end{array}$

Now,

$\frac{\partial }{\partial {a}_{j}}\mid {C}_{i}a-{D}_{i}\mid =\text{sign}\left({C}_{i}a-{D}_{i}\right)·\frac{\partial }{\partial {a}_{j}}\left({C}_{i}a-{D}_{i}\right)={C}_{i,j}\phantom{\rule{0.277778em}{0ex}}\text{sign}\left({C}_{i}a-{D}_{i}\right)$

where Note that

$\underset{u\left(a\right)\to {0}^{+}}{lim}\frac{\partial }{\partial {a}_{j}}\mid u\left(a\right){\mid }^{p}=\underset{u\left(a\right)\to {0}^{-}}{lim}\frac{\partial }{\partial {a}_{j}}\mid u\left(a\right){\mid }^{p}=0$

$\text{sign}\left(x\right)=\left\{\begin{array}{cc}1& x>0\hfill \\ 0& x=0\hfill \\ -1& x<0\hfill \end{array}\right)$

Therefore the Jacobian of $f\left(a\right)$ is given by

$\nabla f\left(a\right)=\left[\begin{array}{c}p\sum _{i=0}^{L}{C}_{i,0}\phantom{\rule{0.277778em}{0ex}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}\text{sign}\left({C}_{i}a-{D}_{i}\right)\\ ⋮\\ p\sum _{i=0}^{L}{C}_{i,M-1}\phantom{\rule{0.277778em}{0ex}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}\text{sign}\left({C}_{i}a-{D}_{i}\right)\end{array}\right]$

The Hessian of $f\left(a\right)$ is the matrix ${\nabla }^{2}f\left(a\right)$ whose $jm$ -th element ( $0\le j,m\le M$ ) is given by

$\begin{array}{ccc}\hfill {\nabla }_{j,m}^{2}f\left(a\right)=\frac{\partial {a}^{2}}{\partial {a}_{j}\partial {a}_{m}}f\left(a\right)& =& \frac{\partial }{\partial {a}_{m}}\frac{\partial }{\partial {a}_{j}}f\left(a\right)\hfill \\ & =& \sum _{i=0}^{L}p\phantom{\rule{0.222222em}{0ex}}{C}_{i,j}\phantom{\rule{0.277778em}{0ex}}\frac{\partial }{\partial {a}_{m}}\mid {D}_{i}-{C}_{i}a{\mid }^{p-1}\text{sign}\left({D}_{i}-{C}_{i}a\right)\hfill \\ & =& \sum _{i=0}^{L}\alpha \frac{\partial }{\partial {a}_{m}}b\left(a\right)d\left(a\right)\hfill \end{array}$

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

$\begin{array}{ccc}\hfill \frac{\partial }{\partial {a}_{m}}b\left(a\right)& =& \frac{\partial }{\partial {a}_{m}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}\hfill \\ & =& \left(p-1\right){C}_{i,m}\mid {C}_{i}a-{D}_{i}{\mid }^{p-2}\text{sign}\left({C}_{i}a-{D}_{i}\right)\hfill \\ \hfill \frac{\partial }{\partial {a}_{m}}d\left(a\right)& =& \frac{\partial }{\partial {a}_{m}}\text{sign}\left({D}_{i}-{C}_{i}a\right)=0\hfill \end{array}$

Note that the partial derivative of $d\left(a\right)$ at ${D}_{i}-{C}_{i}a=0$ is not defined. Therefore

$\begin{array}{ccc}\hfill \frac{\partial }{\partial {a}_{m}}b\left(a\right)d\left(a\right)& =& b\left(a\right)\frac{\partial }{\partial {a}_{m}}d\left(a\right)+d\left(a\right)\frac{\partial }{\partial {a}_{m}}b\left(a\right)\hfill \\ & =& \left(p-1\right){C}_{i,m}\mid {C}_{i}a-{D}_{i}{\mid }^{p-2}{\text{sign}}^{2}\left({C}_{i}a-{D}_{i}\right)\hfill \end{array}$

Note that ${\text{sign}}^{2}\left({C}_{i}a-{D}_{i}\right)=1$ for all ${D}_{i}-{C}_{i}a\ne 0$ where it is not defined. Then

${\nabla }_{j,m}^{2}f\left(a\right)=p\left(p-1\right)\sum _{i=0}^{L}{C}_{i,j}{C}_{i,m}\phantom{\rule{0.277778em}{0ex}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-2}$

except at ${D}_{i}-{C}_{i}a=0$ where it is not defined.

• Given ${a}_{0}\in {\mathbb{R}}^{M+1}$ , $D\in {\mathbb{R}}^{L+1}$ , $\mathbf{C}\in {\mathbb{R}}^{L+1×M+1}$
• For $i=0,1,...$
1. Find $\nabla f\left({a}_{i}\right)$ .
2. Find ${\nabla }^{2}f\left({a}_{i}\right)$ .
3. Solve ${\nabla }^{2}f\left({a}_{i}\right)s=-\nabla f\left({a}_{i}\right)$ for $s$ .
4. Let ${a}_{+}={a}_{i}+s$ .
5. Check for convergence and iterate if necessary.

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

$\nabla f\left(a\right)=p{\mathbf{C}}^{T}y$

where

$y=\mid \mathbf{C}{a}_{i}-D{\mid }^{p-1}\text{sign}\left(\mathbf{C}{a}_{i}-D\right)=\mid \mathbf{C}{a}_{i}-D{\mid }^{p-2}\left(\mathbf{C}{a}_{i}-D\right)$

Also,

${\nabla }_{j,m}^{2}f\left(a\right)=p\left(p-1\right)\phantom{\rule{0.277778em}{0ex}}{C}_{j}^{T}\mathbf{Z}{C}_{m}$

where

$\mathbf{Z}=\text{diag}\left(\mid ,\mathbf{C},{a}_{i},-,D,{\mid }^{p-2}\right)$

and

${C}_{j}=\left[\begin{array}{c}{C}_{0,j}\\ ⋮\\ {C}_{L,j}\end{array}\right]$

Therefore

${\nabla }^{2}f\left(a\right)=\left({p}^{2}-p\right){\mathbf{C}}^{T}\mathbf{Z}\mathbf{C}$

From [link] , the Hessian ${\nabla }^{2}f\left(a\right)$ can be expressed as

${\nabla }^{2}f\left(a\right)=\left({p}^{2}-p\right){\mathbf{C}}^{T}{\mathbf{W}}^{T}\mathbf{W}\mathbf{C}$

where

$\mathbf{W}=\text{diag}\left(\mid ,\mathbf{C},{a}_{i},-,D,{\mid }^{\frac{p-2}{2}}\right)$

The matrix $\mathbf{C}\in {\mathbb{R}}^{\left(L+1\right)×\left(M+1\right)}$ is given by

$\mathbf{C}=\left[\begin{array}{ccccccc}\text{cos}M{\omega }_{0}& \text{cos}\left(M-1\right){\omega }_{0}& \cdots & \text{cos}\left(M-j\right){\omega }_{0}& \cdots & \text{cos}{\omega }_{0}& 1\\ \text{cos}M{\omega }_{1}& \text{cos}\left(M-1\right){\omega }_{1}& \cdots & \text{cos}\left(M-j\right){\omega }_{1}& \cdots & \text{cos}{\omega }_{1}& 1\\ ⋮& ⋮& \ddots & ⋮& & ⋮& ⋮\\ \text{cos}M{\omega }_{i}& \text{cos}\left(M-1\right){\omega }_{i}& \cdots & \text{cos}\left(M-j\right){\omega }_{i}& \cdots & \text{cos}{\omega }_{i}& 1\\ ⋮& ⋮& & ⋮& \ddots & ⋮& ⋮\\ \text{cos}M{\omega }_{L-1}& \text{cos}\left(M-1\right){\omega }_{L-1}& \cdots & \text{cos}\left(M-j\right){\omega }_{L-1}& \cdots & \text{cos}{\omega }_{L-1}& 1\\ \text{cos}M{\omega }_{L}& \text{cos}\left(M-1\right){\omega }_{L}& \cdots & \text{cos}\left(M-j\right){\omega }_{L}& \cdots & \text{cos}{\omega }_{L}& 1\end{array}\right]$

The matrix $\mathbf{H}={\nabla }^{2}f\left(a\right)$ is positive definite (for $p>1$ ). To see this, consider $\mathbf{H}={\mathbf{K}}^{T}\mathbf{K}$ where $\mathbf{K}=\mathbf{W}\mathbf{C}$ . Let $z\in {\mathbb{R}}^{M+1}$ , $z\ne 0$ . Then

${z}^{T}\mathbf{H}z={z}^{T}{\mathbf{K}}^{T}\mathbf{K}z={\parallel \mathbf{K}z\parallel }_{2}^{2}>0$

unless $z\in N\left(\mathbf{K}\right)$ . But since $\mathbf{W}$ is diagonal and $\mathbf{C}$ is full column rank, $N\left(\mathbf{K}\right)=0$ . Thus ${z}^{T}\mathbf{H}z\ge 0$ (identity only if $z=0$ ) and so $\mathbf{H}$ is positive definite.

## Newton's method and ${l}_{p}$ Complex linear systems

Consider the problem

$\underset{x}{\text{min}}\phantom{\rule{0.277778em}{0ex}}e\left(x\right)={\parallel \mathbf{A}x-b\parallel }_{p}^{p}$

where $\mathbf{A}\in {\mathbb{C}}^{m×n}$ , $x\in {\mathbb{R}}^{n}$ and $b\in {\mathbb{C}}^{m}$ . One can write [link] in terms of the real and imaginary parts of $\mathbf{A}$ and $b$ ,

$\begin{array}{ccc}\hfill e\left(x\right)& =& \sum _{i=1}^{m}{|{A}_{i}x-{b}_{i}|}^{p}\hfill \\ & =& \sum _{i=1}^{m}{|\text{Re}\left\{{A}_{i}x-{b}_{i}\right\}+jIm\left\{{A}_{i}x-{b}_{i}\right\}|}^{p}\hfill \\ & =& \sum _{i=1}^{m}{|\left({R}_{i}x-{\alpha }_{i}\right)+\left({Z}_{i}x-{\gamma }_{i}\right)|}^{p}\hfill \\ & =& \sum _{i=1}^{m}{\left(\sqrt{{\left({R}_{i}x-{\alpha }_{i}\right)}^{2}+{\left({Z}_{i}x-{\gamma }_{i}\right)}^{2}}\right)}^{p}\hfill \\ & =& \sum _{i=1}^{m}{g}_{i}{\left(x\right)}^{p/2}\hfill \end{array}$

where $\mathbf{A}=\mathbf{R}+j\mathbf{Z}$ and $b=\alpha +j\gamma$ . The gradient $\nabla e\left(x\right)$ is the vector whose $k$ -th element is given by

$\frac{\partial }{\partial {x}_{k}}e\left(x\right)=\frac{p}{2}\sum _{i=1}^{m}\left[\frac{\partial }{\partial {x}_{k}},{g}_{i},\left(x\right)\right]{g}_{i}{\left(x\right)}^{\frac{p-2}{2}}=\frac{p}{2}{q}_{k}\left(x\right)\stackrel{^}{g}\left(x\right)$

where ${q}_{k}$ is the row vector whose $i$ -th element is

$\begin{array}{cc}\hfill {q}_{k,i}\left(x\right)=\frac{\partial }{\partial {x}_{k}}{g}_{i}\left(x\right)& =2\left({R}_{i}x-\alpha {\alpha }_{i}\right){R}_{ik}+2\left({Z}_{i}x-\gamma {\gamma }_{i}\right){Z}_{ik}\hfill \\ & =2{R}_{ik}{R}_{i}x+2{Z}_{ik}{Z}_{i}x-\left[2{\alpha }_{i}{R}_{ik}+2{\gamma }_{i}{Z}_{ik}\right]\hfill \end{array}$

Therefore one can express the gradient of $e\left(x\right)$ by $\nabla e\left(x\right)=\frac{p}{2}\mathbf{Q}\stackrel{^}{g}$ , where $\mathbf{Q}=\left[{q}_{k,i}\right]$ as above. Note that one can also write the gradient in vector form as follows

$\nabla e\left(x\right)=p\phantom{\rule{0.277778em}{0ex}}\left[{\mathbf{R}}^{T},\text{diag},\left(\mathbf{R}x-\alpha \right),+,{\mathbf{Z}}^{T},\text{diag},\left(\mathbf{Z}x-\gamma \right)\right]·\left[{\left({\left(\mathbf{R}x-\alpha \right)}^{2},+,{\left(\mathbf{Z}x-\gamma \right)}^{2}\right)}^{\frac{p-2}{2}}\right]$

The Hessian $\mathbf{H}\left(x\right)$ is the matrix of second derivatives whose $kl$ -th entry is given by

$\begin{array}{cc}\hfill {\mathbf{H}}_{k,l}\left(x\right)& =\frac{{\partial }^{2}}{\partial {x}_{k}\partial {x}_{l}}e\left(x\right)\hfill \\ \hfill & =\frac{\partial }{\partial {x}_{l}}\frac{p}{2}\sum _{i=1}^{m}{q}_{k,i}\left(x\right){g}_{i}{\left(x\right)}^{\frac{p-2}{2}}\hfill \\ & =\frac{p}{2}\sum _{i=1}^{m}\left[{q}_{k,i},\left(x\right),\frac{\partial }{\partial {x}_{l}},{g}_{i},{\left(x\right)}^{\frac{p-2}{2}},+,{g}_{i},{\left(x\right)}^{\frac{p-2}{2}},\frac{\partial }{\partial {x}_{l}},{q}_{k,i},\left(x\right)\right]\hfill \end{array}$

Now,

$\begin{array}{cc}\hfill \frac{\partial }{\partial {x}_{l}}{g}_{i}{\left(x\right)}^{\frac{p-2}{p}}& =\frac{p-2}{2}\left[\frac{\partial }{\partial {x}_{l}},{g}_{i},\left(x\right)\right]{g}_{i}{\left(x\right)}^{\frac{p-4}{2}}\hfill \\ & =\frac{p-2}{2}{q}_{l,i}\left(x\right){g}_{i}{\left(x\right)}^{\frac{p-4}{2}}\hfill \\ \hfill \frac{\partial }{\partial {x}_{l}}{q}_{k,i}\left(x\right)& =2{R}_{ik}{R}_{il}+2{Z}_{ik}{Z}_{il}\hfill \end{array}$

${\mathbf{H}}_{k,l}\left(x\right)=\frac{p\left(p-2\right)}{4}\sum _{i=1}^{m}{q}_{k,i}\left(x\right){q}_{l,i}\left(x\right){g}_{i}{\left(x\right)}^{\frac{p-4}{4}}+p\sum _{i=1}^{m}\left({R}_{ik}{R}_{il}+{Z}_{ik}{Z}_{il}\right){g}_{i}{\left(x\right)}^{\frac{p-2}{2}}$

Note that $\mathbf{H}\left(x\right)$ can be written in matrix form as

$\begin{array}{cc}\hfill \mathbf{H}\left(x\right)=& \frac{p\left(p-2\right)}{4}\left(\mathbf{Q},\phantom{\rule{0.277778em}{0ex}},\text{diag},\left(g,{\left(x\right)}^{\frac{p-4}{2}}\right),{\mathbf{Q}}^{T}\right)+\hfill \\ & p\left({\mathbf{R}}^{T},\text{diag},\left(g,{\left(x\right)}^{\frac{p-2}{2}}\right),\mathbf{R},+,{\mathbf{Z}}^{T},\text{diag},\left(g,{\left(x\right)}^{\frac{p-2}{2}}\right),\mathbf{Z}\right)\hfill \end{array}$

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

$\begin{array}{ccc}\mathbf{H}\left({x}^{c}\right)s& =& -\nabla e\left({x}^{c}\right)\\ {x}^{+}& =& {x}^{c}+s\end{array}$

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

how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
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?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
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
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
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
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!

#### Get Jobilize Job Search Mobile App in your pocket Now! By Christine Zeelie By OpenStax By Robert Morris By Stephanie Redfern By Anh Dao By Sandy Yamane By Anh Dao By Abishek Devaraj By JavaChamp Team By Mistry Bhavesh