Consider the quantitya^{2} - b^{2} where a and b are real numbers. (a) Under what conditions should one expect an unusually large relative error in th

Globokim8 2020-10-20 Answered

Consider the quantitya2  b2 where a and b are real numbers.
(a) Under what conditions should one expect an unusually large relative error in the computed value of a2  b2 when this expression is evaluated in finite precision arithmetic?
(b)cWs 4-digit (decimal) rounding arithmetic to evaluate both a2  b2 and (a + b)(a  b) with a =995.1 and b=995.0. Calculate th relative error in each result.
(c) The expression (a + b)(a  b) is algebraically equivalent to a2  b2, but it is a more accurate way to calculate this quantity if both a and b have exact floating point representations. Why?

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Expert Answer

Arnold Odonnell
Answered 2020-10-21 Author has 109 answers
Solution:
a.The condition is that the computed value is rounded off when this expression is evaluated in finite precision arithmetic.
b.Given: a=995.1 and b=995.0
a2  b2=(995.1)2  (995.0)2
=990224.01  990025
=199.01
4 digit rounding arithmetic we have a2  b2=199.0.
The relative error becomes |199.01  199.0|199.01=0.01199.010.525 × 104
Conclusion:
Given: a=995.1 and b=995.0
(a + b)(a  b)=(995.1 + 995.0)(995.1  995.0)
=(199.01)(0.1)
4 digit rounding arithmetic we have (a + b)(a  b)=199.0.
The relative error becomes |199.01  199.0|199.01=0.01199.010.525 × 104
c.The expression (a + b)(a  b) is more accurate than a2  b2 when a and b have exact floating point representations because the expression
(a + b)(a  b) involves simple addition and subtraction of decimals, then easy multiplication takes place,
however the expression a2  b2 involves squaring of decimals resulting in more decimals , then subtraction takes place. The latter may involves round off whereas the former may not.
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