The formal definitions of linear, superlinear and quadratic convergence order are clear in the context of the convergence of the solution of nonlinear systems of equations. Recently, I have come across some works on superquadratic convergence, but I can't find the formal definition of the convergence order (superquadratic) just like we have in the rest of the existing convergence order.
NB: I need a formal definition. Definition like the ones given in this text, see the link below:
https://www.siam.org/books/textbooks/fr16_book.pdf
Super quadratic convergence means that the order of convergence is more between 2 and 3, for example, something like 1+ sqrt (2).
Dear R.C. Mittal,
Not necessarily "between 2 and 3" because a cubic convergence is still considered to be superquadratic. So, I would say that we have a superquadratic convergence when the order of convergence is greater than 2.
By definition a sequence x_k converges superquadratically to x, if
\|x_{k+1} - x\| \leq c_k \|x_k - x\|^2 for a sequence c_k \to 0
This includes cubic, quartic, and higher order rates, and also convergence, where there is no explicit rate above 2, but which is still faster than rate 2.
Dear Mahboub
1) The case p = 3 may also be called superquadratic
2) When we say superquadratic convergence, we always mean 2
I do agree with the definition given by dear @Dirk Bloemker and also that a cubic convergence is also superquadratic convergence.
Dirk Bloemer answer is just like the definition of superlinear convergence with the denominator raise to the power of two. That is x_k converges to x superquadratically if lim_k to infinty ||x_k+1-x||/||x_k-x||^2=0.
Clearly, superquadratic convergence implies superlinear convergence. Dirk Bloemer answer makes sense.
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