# Jacobian Matrix

casio fx-9860GII programming numerical methods derivatives

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### #1 TOM_DLB_89

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Posted 15 February 2018 - 09:36 AM

I've made a program that solves non-linear systems (2x2 and 3x3) of equations by newton method: r[i+1]=r[i]-J(r[i])^-1*F(r[i]).

The user is able to insert the functions (that I've set the program to store in FMEM - fn1, fn2...). During the iteration the values of each function will be in the vector F.

The big deal is defining the Jacobian Matrix. I wasn't able to find any way to make it unless the user insert by itself all the derivatives (4 or 9 functions) and the program allocate them, after calculating its values, in the matrix J.

Any suggestions or tips?

### #2 pan.gejt

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Posted 15 February 2018 - 06:23 PM

Since there is no CAS inside 9860G II you have to calculate the derivatives numerically. Unfortunatelly, the numerical differentiation built in the calc is for single variable functions only, e.g. f(x)=x^3-ln(x)+ x.

I don't know how to calculate the derivative numerically for function with two or more variables, e.g. f(x)=x^2*y.

The syntax is d/dx(fx),a,tol)

### #3 piu58

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Posted 15 February 2018 - 07:53 PM

It is not difficult to differentiate numerical. You need to define your function as a subroutine. To get the derivate at a point x, simply calc f(x) and f(x+d), where d is a small (but not too small) difference of the original x. The derivate ist the difference of the values divided by d.

### #4 TOM_DLB_89

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Posted 15 February 2018 - 07:59 PM

Yes, that is my problem. I need to input manually the partial derivatives formulas so the program will calculate the value.
Most of the times I get some equations like those

and it is a hard work writing down the derivatives hahahaha.

Is there any way to solve non linear systems in casio fx-9860GII? I mean, systems like the one above. (I know that some systems are 'solvable' with the intersect function in graph mode).

*edit: K1, K2, na, nb, nc, nd, and ne are constants. The variables are E1 and E2.

Edited by TOM_DLB_89, 15 February 2018 - 08:01 PM.

### #5 slugrustle

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Posted 24 November 2018 - 06:21 PM

piu58 has it right, although I would use the central or symmetric finite difference instead of the forward finite difference.

Given a two-variable function f(x,y), ∂f/∂x ≈ (f(x+h,y) - f(x-h,y)) ÷ (2×h), and ∂f/∂y ≈ (f(x,y+h) - f(x,y-h)) ÷ (2×h). h should be small relative to the variable against which you are differentiating, but not so small that you run into truncation error during subtraction to form the numerator. This can be extended to any number of function arguments. You can see these equations printed nicely at https://en.wikipedia...veral_variables

If accuracy becomes an issue with finite differences, you can use Richardson Extrapolation to improve the accuracy.

If you have the time and the interest, I recommend using either the Levenberg–Marquardt algorithm or Trond Steihuag's Conjugated Gradient method (http://www.ii.uib.no...apers/trust.pdf) and solving the update step with a QR decomposition instead of using a straight inverse or applying Cholesky Decomposition to the normal equations. I have had great success with both of these in my previous job, which largely involved programming and optimizing nonlinear solvers.

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