Numerically Stable Formulas for x >= 0 and x < 0

[happyg1]

Hi, here's my question...We have to determine whether or not the formulas down there are numerically stable for the cases where x >=0 and x<0. I say that the formulas are stable for x>=0 but not for x<0 because you are subracting numbers that could be close to each other. My problem is that I am having a really hard time writing a new formula that is numerically stable for negative x values. See if you can point me down the right path.
the formulas are:

u= +-sqrt((x+sqrt(x^2+y^2)))/2)

v=y/2u

Thanks for any help that you can give,
CC

 

[lippyka]

To make the formulas numerically stable for negative x values, you can use a slightly different approach. Instead of subtracting two numbers close to each other, you can use the absolute value of x in the formula. So the new formula would be: u= +-sqrt(|x|+sqrt(x^2+y^2))/2)v=y/2u

 

[quicknote]

I understand the importance of using numerically stable formulas in calculations. A numerically stable formula is one that produces accurate results even when small changes are made to the input values. In the case of x<0, the existing formulas may not be numerically stable because of the possibility of subtracting numbers that are close in value, leading to potential rounding errors.

To address this issue, we can use a technique called "numerical conditioning" to improve the stability of the formulas. This involves modifying the formulas in a way that minimizes the potential for rounding errors. In this case, we can rewrite the formulas as:

u= sqrt((x+sqrt(x^2+y^2)))/2) for x>=0

u= -sqrt((x+sqrt(x^2+y^2)))/2) for x<0

v=y/2u

By separating the formulas for positive and negative x values, we can avoid the possibility of subtracting numbers that are close in value. This approach should improve the numerical stability of the formulas for all values of x.

However, it is always important to carefully consider the potential for numerical instability when using mathematical formulas, particularly in cases where the input values may vary widely. It may be helpful to perform sensitivity analysis or use alternative methods, such as using a logarithmic transformation, to improve the stability of the calculations. I hope this helps guide you in finding a numerically stable solution for negative x values.