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I'm not great at maths so go easy.

I'm following this video on how to solve a problem in which repeated guesses are taken at the inputs, $x_{1-6}$, of a function $f(x_1,x_2,x_3,x_4,x_5,x_6)$ and the output, a vector $y = f(x_{1-6})$ is used to determine the next iteration. $y0$ the desired output is known.

To determine the next guess, $x_{1-6_2}$, the numerical Jacobian $J = (f(x_{1-6}+e) - f(x_{1-6}))/e$ is calculated. With $dy = y0 - y$, $dx =J^+\,dy$ gives the next iteration of guesses $x_{1-6}$: $x_{n+1} = x_n+dx$.

This $dx =J^+\,dy$, gives the values for $dx$ that is the least square solution for $J\,dx = dy$.

So my confusion is about the Jacobian and why it's used here. Why does finding the values of $dx$ that minimises the least square $J\,dx = dy$ give guesses that incrementally move $x$ to the correct answer?

Also would it be more accurate of those '$d$'s were partial derivatives?

Thanks, I'm trying to understand this so I can give an explanation of it to other engineers.

Michael Hardy
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Rory McDonald
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  • This comes from the Taylor expansion to the first order, $f(x+\Delta x)\approx f(x)+J(x)\Delta x$. When you set it to $y_0$, you get $\Delta x=J^{-1}(x)(y_0-f(x))$. –  Feb 27 '20 at 15:13
  • @YvesDaoust Ah ok, I kinda get it now. So why doesn't it immediately jump to y0? is it because the first order expansion is only an aproximation? Would it jump more quickly if you expanded it more? Side question: Is there a way to get a second order Jacobian? What do I need to google to find out more? – Rory McDonald Feb 27 '20 at 15:55
  • And thank you! . – Rory McDonald Feb 27 '20 at 16:03
  • The second order approximation uses the Hessian tensor. It is usually not worth the extra complexity to use it to solve equations. –  Feb 27 '20 at 16:19
  • Would be too easy to jump instantly to $y_0$. But this does occur with linear functions. –  Feb 27 '20 at 16:20

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