[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.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![[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)
_ = — | ≤ 1 42²
M
<
-h²
2h
6
where M = maxx=[a,b] |ƒ"" (x).
Assume now that h can be represented exactly (no floating point error),
but, in contrast, that f(x+h) and f(x−h) have some error in their floating
point representation. Denote these floating point approximations
f(xo+h): Fl(f(xo + h)), f(xoh): Fl(f(xo - h))
and suppose that €1₁, €2 are the floating
f(xo + h) = f(xo + h) + €₁,
(1)
ƒ'(xo)
point errors in each value; that is:
f(xo - h) = f(xo - h) + €2.
(a) Use the triangle inequality and (1) to show that the error in the
centered difference approximation in the presence of round error satisfies:
М
f(xo+h)-f(xo – h)
-
2
| S M N² + 1/
< -h²
2h
6
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fd47556-f3ce-4f39-818d-be563d9523c8%2Fbf45b478-7f8d-4395-b47e-718e14c97aa0%2Ftvhj5ab_processed.png&w=3840&q=75)
Transcribed Image Text:[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)
_ = — | ≤ 1 42²
M
<
-h²
2h
6
where M = maxx=[a,b] |ƒ"" (x).
Assume now that h can be represented exactly (no floating point error),
but, in contrast, that f(x+h) and f(x−h) have some error in their floating
point representation. Denote these floating point approximations
f(xo+h): Fl(f(xo + h)), f(xoh): Fl(f(xo - h))
and suppose that €1₁, €2 are the floating
f(xo + h) = f(xo + h) + €₁,
(1)
ƒ'(xo)
point errors in each value; that is:
f(xo - h) = f(xo - h) + €2.
(a) Use the triangle inequality and (1) to show that the error in the
centered difference approximation in the presence of round error satisfies:
М
f(xo+h)-f(xo – h)
-
2
| S M N² + 1/
< -h²
2h
6
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.
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