# Evaluation of slope at iteration ith - Newton-Raphson method

I'd like to know how Ansys computes the slope (=stiffness matrix) at point x1 in figure. I'm studying the way in which Ansys uses the Newton-Raphson method when there are nonlinearities.

In the slide there is the specification that we don't know the real curve F-x, thus I have some problem to understand which slope is taken at x1 by Ansys.

Thank you so much for your time.

I can't speak for ANSYS specifically, but one of the most popular families of Newton's method are Newton-Krylov methods. In higher dimensions, one has to solve a linear system using the Jacobian, or stiffness matrix in this case. Krylov-based methods are a popular method for Newton's method because Krylov methods actually don't require you to know then entire stiffness matrix, you only have to be able to compute the product of the stiffness matrix with a vector. This can be approximated with a finite difference like so: $$J(x)v\approx \frac{F(x+hv)-F(x)}{h},$$ where $$J(x)$$ is the Jacobian matrix of $$F$$ at $$x$$. This is great for these solvers because by using a Krylov method to solve the linear problem in Newton, you don't ever have to explicitly compute a Jacobian matrix.
The slope taken is linearized version of equations presented in the model. Linearization usually is implemented by removing non-linear parts: instead of those their differential expressions were to be used $$f(x) \approx f(a) + f'(a)(x-a)$$. That is shown on the picture. That differential $$f'(a)$$ in question could be computed by interpolation techniques, which is separate (complex) task. Then you solve the equation, with usual methods (linear algebra approach). Then after solution is found, again you recompute all differentials. And so on.