0.5 Sampling with automatic gain control  (Page 12/19)

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To apply steepest descent to the minimization of the polynomial $J\left(x\right)$ in [link] , suppose that a current estimate of $x$ is available at time $k$ , which is denoted $x\left[k\right]$ . A new estimate of $x$ at time $k+1$ can be made using

$x\left[k+1\right]=x\left[k\right]-\mu {\left(\frac{dJ\left(x\right)}{dx}|}_{x=x\left[k\right]},$

where $\mu$ is a small positive number called the stepsize, and where the gradient (derivative) of $J\left(x\right)$ is evaluated at the current point $x\left[k\right]$ . This is then repeated again and again as $k$ increments. This procedure isshown in [link] . When the current estimate $x\left[k\right]$ is to the right of the minimum, the negative of the gradient points left. When the current estimate is to the left of the minimum, thenegative gradient points to the right. In either case, as long as the stepsize is suitably small, the newestimate $x\left[k+1\right]$ is closer to the minimum than the old estimate $x\left[k\right]$ ; that is, $J\left(x\left[k+1\right]\right)$ is less than $J\left(x\left[k\right]\right)$ .

To make this explicit, the iteration defined by [link] is

$x\left[k+1\right]=x\left[k\right]-\mu \left(2x\left[k\right]-4\right),$

or, rearranging,

$x\left[k+1\right]=\left(1-2\mu \right)x\left[k\right]+4\mu .$

In principle, if [link] is iterated over and over, the sequence $x\left[k\right]$ should approach the minimum value $x=2$ . Does this actually happen?

There are two ways to answer this question. It is straightforward to simulate the process. Here is some M atlab code that takes an initial estimate of $x$ called x(1) and iterates [link] for N=500 steps.

N=500;                          % number of iterations mu=.01;                         % algorithm stepsizex=zeros(1,N);                   % initialize x to zero x(1)=3;                         % starting point x(1)for k=1:N-1   x(k+1)=(1-2*mu)*x(k)+4*mu;    % update equationend polyconverge.m find the minimum of $J\left(x\right)={x}^{2}-4x+4$ via steepest descent (download file) 

[link] shows the output of polyconverge.m for 50 different x(1) starting values superimposed; all converge smoothly to the minimum at $x=2$ .

Explore the behavior of steepest descent by running polyconverge.m with different parameters.

1. Try mu = -.01, 0, .0001, .02, .03, .05, 1.0, 10.0. Can mu be too large or too small?
2. Try N= 5, 40, 100, 5000. Can N be too large or too small?
3. Try a variety of values of x(1) . Can x(1) be too large or too small?

As an alternative to simulation, observe that the process [link] is itself a linear time invariant system, of the general form

$x\left[k+1\right]=ax\left[k\right]+b,$

which is stable as long as $|a|<1$ . For a constant input, the final value theorem of z-Transforms (see [link] ) can be used to show that the asymptotic (convergent)output value is ${lim}_{k\to \infty }{x}_{k}=\frac{b}{1-a}$ . To see this withoutreference to arcane theory, observe that if ${x}_{k}$ is to converge, then it must converge to some value, say ${x}^{*}$ . At convergence, $x\left[k+1\right]=x\left[k\right]={x}^{*}$ , and so [link] implies that ${x}^{*}=a{x}^{*}+b$ , which implies that ${x}^{*}=\frac{b}{1-a}$ . (This holds assuming $|a|<1$ .) For example, for [link] , ${x}^{*}=\frac{4\mu }{1-\left(1-2\mu \right)}=2$ , which is indeed the minimum.

Thus, both simulation and analysis suggest that the iteration [link] is a viable way to find the minimum of the function $J\left(x\right)$ , as long as $\mu$ is suitably small. As will become clearer in later sections, suchsolutions to optimization problems are almost always possible—as long as the function $J\left(x\right)$ is differentiable. Similarly, it is usually quite straightforward to simulate thealgorithm to examine its behavior in specific cases, though it is not always so easy to carry out a theoretical analysis.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
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can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
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I don't understand what the A with approx sign and the boxed x mean
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
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
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Abhi
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Abhi
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salma
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Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Porter
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Stotaw
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Azam
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silver nanoparticles could handle the job?
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Azam
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Uday
I'm interested in Nanotube
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