Question: help!!! due today 4, Why is Richardson extrapolation more useful and effective as a method for estimating the derivative of a function at a given

help!!! due today
 help!!! due today 4, Why is Richardson extrapolation more useful and

4, Why is Richardson extrapolation more useful and effective as a method for estimating the derivative of a function at a given point (when compa method of Assignment #1)? a. Because the method employs smaller stepsizes b. Because the method quickly develops higher order, more accur without taking such small stepsizes (h), thus avoiding major rounding errors c. Because the method is a lower order technique. d. Because the truncation errors are larger 5. Rounding errors can be caused by Problems with computer representability Problems with loss of significance Both a. and b. None of the above a. b. c. d. 6. The value 1/3 is approximately represented as a floating point number in single precision as: b. c. d. 7Truncation errors can often be reduced by a. Using a smaller stepsize in an approximation b. Using a higher order method (viz., using more terms in a Taylor series) c. Both a. and b. d. None of the above 8 The naive Gauss elimination method is a. b. c. d. good for when the determinant of the system is zero difficult to use for linear equations never convergent faster than Gauss elimination with scaled partial pivoting 9 The Lagrange form of the interpolating polynomial is: always faster to compute than the Newton form less useful for theorem proving more useful when f' (x) is available easier to use when the f values might change a. b. c. d

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