[8] [Numerical differentiation and round-off error] In this problem we'll show that numerical differentiation can be unstable in the presence of round off error. Consider the 2nd order centered difference approximation to the first derivative; in exact arithmetic we know that |ƒ'(xo) — ƒ (xo + h) — ƒ(xo — h) 2h M < -h² (1) where M = maxä[a,b] |ƒ'''(x)|. Assume now that h can be represented exactly (no floating point error), but, in contrast, that f(x₁+h) and f(xo-h) have some error in their floating point representation. Denote these floating point approximations f(xo + h) := Fl(f(x + h)), ƒ(xo – h) := Fl(f(x − h)) and suppose that €1, €2 are the floating point errors in each value; that is: ƒ (xo – h) = f(xo — h) + €2. ƒ (xo + h) = f(xo + h) + €₁, - (a) Use the triangle inequality and (1) to show that the error in the centered difference approximation in the presence of round error satisfies: M f(xo + h) — f(xo – h) | P (16) - - | ≤ 11 1² + 1 ƒ'(xo) -h² 2h h where = €2- €₁/2. (b) As h gets smaller, notice that part of the error from (a) will vanish, while the other part will increase. Use differential calculus to find h such that the error derived in part (a) will be minimized.
[8] [Numerical differentiation and round-off error] In this problem we'll show that numerical differentiation can be unstable in the presence of round off error. Consider the 2nd order centered difference approximation to the first derivative; in exact arithmetic we know that |ƒ'(xo) — ƒ (xo + h) — ƒ(xo — h) 2h M < -h² (1) where M = maxä[a,b] |ƒ'''(x)|. Assume now that h can be represented exactly (no floating point error), but, in contrast, that f(x₁+h) and f(xo-h) have some error in their floating point representation. Denote these floating point approximations f(xo + h) := Fl(f(x + h)), ƒ(xo – h) := Fl(f(x − h)) and suppose that €1, €2 are the floating point errors in each value; that is: ƒ (xo – h) = f(xo — h) + €2. ƒ (xo + h) = f(xo + h) + €₁, - (a) Use the triangle inequality and (1) to show that the error in the centered difference approximation in the presence of round error satisfies: M f(xo + h) — f(xo – h) | P (16) - - | ≤ 11 1² + 1 ƒ'(xo) -h² 2h h where = €2- €₁/2. (b) As h gets smaller, notice that part of the error from (a) will vanish, while the other part will increase. Use differential calculus to find h such that the error derived in part (a) will be minimized.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 92E
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