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We Know that To minimize the sum of error (objective Function)

$\ J = (y(t)-\theta (t) u(t))^2 $ (eq. 1)

is done by using least square :

$\theta (t) = \theta (t-1) + \gamma y(\theta u -y) $ (eq.2)

Where $u=input ; $ $y=output; $ $\theta=Gain Input; $ $t=time; $

But the question is how to prove that eq.2 is minimizing the eq 1 respect to $\theta$?

and what the terms that shows $J$ is minimized ?

*Lets say all variables is scalar

Thanks before

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  • $\begingroup$ You need to define your notation. What are $y$, $u$, $\theta$, $\gamma$, $n$, $t$? What is the choice variable in the minimization problem? $\endgroup$ – smcc Jul 22 '16 at 14:11
  • $\begingroup$ Thanks for the replay and corection, i've edit the equation $\endgroup$ – user3556482 Jul 22 '16 at 15:16
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It's a bit difficult to untangle this question. It should have a form where we start with a sequence on $m$ measurements of the form $$\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$$ with an input trial function $$ y(x). $$ which has parameters $a_{1}, a_{2}, \dots, a_{n}$. The method of least squares finds the vector $a$ which minimizes the total error $$ r^{2}(a) = \sum_{k=1}^{m} \left( y_{k} - y(x_{k}) \right)^{2} $$ In fact, the least squares solution is defined as $$ a_{LS} = \left\{ a \in \mathbb{R}^{n} \colon r^{2}(a) \text{ is minimized} \right\} $$ This sets up the following $n$ equations: $$ \begin{align} \frac{\partial} {\partial a_{1}} r^{2}(a) & = 0 \\ \frac{\partial} {\partial a_{2}} r^{2}(a) & = 0 \\ \vdots \\ \frac{\partial} {\partial a_{n}} r^{2}(a) & = 0 \end{align} $$

It would help to have your question nudged into something close to a format such as this.

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