Taylor polynomials practice question 13
A function has third derivative . If is approximated by its second-order Maclaurin polynomial , find the best bound on the error that the Taylor remainder formula gives.
The error after is set by the next derivative at some unknown point between 0 and 0.5. Since you cannot know the point, what can you do with that derivative instead?
The remainder after is for some between 0 and 0.5. Bound on that interval.
and , so .
An exact fraction, or a decimal to three decimal places.
Theory UBC Math 100 §9.6CLP-1 §3.4Key resultsFlavour A only
Worked solution
Strategy
The Lagrange remainder says the error is the next derivative, at some unknown point, times the step cubed over 3!. You cannot know the point, so bound the derivative over the whole interval.
Working
Taylor remainder for a second-order polynomial about 0.
Numerator at most 1, denominator at least 4.
(0.5)³ = 1/8.
Answer
The answer is at most one over 192, about 0.0052.
Check
If f‴ were 1/4 throughout, its largest value here, the error would be exactly 0.5³/24 = 1/192, the bound found.
Where this goes wrong
Dividing by 2! instead of 3!, or using the step squared. After a second-order polynomial the remainder involves the third derivative, 3!, and the cube of the step.
The idea behind it
The Taylor polynomials lesson, a figure to move, in a few steps
Leans on
Bounding a function on an interval, from Functions, graphs and inverses
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