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- Tue Nov 19, 2013 11:46 am
- Forum: Component Pascal
- Topic: Numerical Analysis
- Replies: 13
- Views: 32428
Re: Numerical Analysis
Robert, I'd like to think that, whenever a <= 1, then 1 - a >= 0. So, if you replace a by u * u + v * v , then 1 - ( u * u + v * v ) >= 0. I've changed 1 - u * u - v * v by 1 - ( u * u + v * v ) in manumart1's example, and there is no more hit. Cheers. Gérard