- #1
copperback
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Hi, I need help with the following numerical analysis questions. The textbook I'm using (Intro to Numerical Analysis by Stoer) doesn't have many examples so I'm having a lot of trouble figuring out what exactly I should be doing.
"Evaluating the summation as i goes from 1 to n of a sub i in floating point arithemetic may lead to an arbirarily large error. If however, all summands a sub i are of the same sign, then this relative error is bounded. Derive a crude bound for this error, disregarding terms of higher order."
What I did was to expand the n terms in the form [(a+b)(1+esub1)+c](1+esub2)... and then reduce it using the formula for relative error: [fl(y)-y]/y. But the upper bound is given by 5*10^(-t), so what am i supposed to do next?
Another question is:
"Show how to evaluate the following expression in a numerically stable fashion:"
1/(1+2x) - (1-x)/(1+x)
I combined the above to get 2x^2/[(x+1)(2x+1)]. What am I supposed to do next? Am I supposed to come up with an algorithm for solving it then calculate the relative error? The book doesn't really have any worked examples so I'm hopelessly lost.
Thanks for looking.
"Evaluating the summation as i goes from 1 to n of a sub i in floating point arithemetic may lead to an arbirarily large error. If however, all summands a sub i are of the same sign, then this relative error is bounded. Derive a crude bound for this error, disregarding terms of higher order."
What I did was to expand the n terms in the form [(a+b)(1+esub1)+c](1+esub2)... and then reduce it using the formula for relative error: [fl(y)-y]/y. But the upper bound is given by 5*10^(-t), so what am i supposed to do next?
Another question is:
"Show how to evaluate the following expression in a numerically stable fashion:"
1/(1+2x) - (1-x)/(1+x)
I combined the above to get 2x^2/[(x+1)(2x+1)]. What am I supposed to do next? Am I supposed to come up with an algorithm for solving it then calculate the relative error? The book doesn't really have any worked examples so I'm hopelessly lost.
Thanks for looking.