Let z be a function of a finite number of variables i.e. z=f(a,b,c,...). If we have the mathematical formula connecting z and the variables, we can determine how the value of z varies with a change in any of the input variables a,b,.... so on. It is clear that if in the formula some variable is raised to some high power (exponentiation), z will change steeply with respect to that variable. So numerical stability of z is less with respect to this particular variable because a small change (error) in this variable can lead to a large error in z. Viewed in this way, determinant of a matrix is a function of its elements. Many computations use determinant in the calculation as in Cramer's rule. So numerical stability is essentially dependent on the powers of the variables involved in the formula/relation. All the condition numbers, singular values etc are actually an index/summary of these powers. Is my understanding right? Or is there something more?

  • $\begingroup$ I think in practice Cramer's rule is never used. It's rare to use determinants also. If you want to numerically solve a linear system of equations, usually you use a method such as Gaussian elimination. $\endgroup$ – littleO Apr 25 '16 at 22:21
  • $\begingroup$ I am not talking about solution of linear system of equations or any thing in particular. I am talking about numerical stability in general and have taken determinant just as an example. $\endgroup$ – Seetha Rama Raju Sanapala Apr 25 '16 at 22:24
  • $\begingroup$ It's not so clear to me that if the input variable $a$ (for example) is raised to a high power then $z$ will change steeply when $a$ changes. If $a$ is small then $a^n$ is almost a constant $0$ for large $n$; also, the formula might multiply $a^n$ by a very small coefficient. As an example, Taylor series often give good results even when they include terms with very high powers of the variable. $\endgroup$ – David K Apr 25 '16 at 22:27
  • $\begingroup$ Suppose a=$b^{n}$. The derivative a with respect to b is $n b^{n-1}$. So considering each of b and n greater than 1, the steepness increases as n goes up. That is my reasoning. @littleO: We are talking about the final and ideal solutions, not about the methods used. How final results change when some inputs have errors. $\endgroup$ – Seetha Rama Raju Sanapala Apr 25 '16 at 22:37
  • $\begingroup$ I should have clarified, my comment was only tangential to the question. $\endgroup$ – littleO Apr 25 '16 at 22:39

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