- Topic
- Taylor polynomials
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 3
Taylor polynomials practice question 12
Models of a swinging pendulum replace by for small angles in radians. Because has no term, is its second-order Maclaurin polynomial . Find the best bound on the error that the Taylor remainder for gives.
Here is , not only . Which derivative, which factorial and which power of the step does the remainder after use?
With , the remainder after is for some between 0 and .
. How large can be for in ?
An exact fraction.
Theory UBC Math 100 §9CLP-1 §3.4Key results
Worked solution
Strategy
Write the error as the remainder after T₂, which involves the third derivative at an unknown point c, and replace |f‴(c)| by the largest value it can take on the interval.
Working
The remainder formula with n = 2, about 0.
sin differentiates to cos, −sin, then −cos. On (0, 0.2), cos c is just below 1, so no smaller bound holds for every c.
About 0.00133. The actual error is also about 0.00133, so the bound is nearly exact.
Answer
The answer is at most one over 750, about 0.00133.
Check
The actual error, 0.2 − sin 0.2 ≈ 0.001331, sits just under the bound 1/750 ≈ 0.001333.
Where this goes wrong
Dividing by 2! instead of 3!, which gives 1/250. After T₂ the remainder uses the third derivative, so it divides by 3! and multiplies by the cube of the step.
Leans on
The derivatives of sine and cosine, from Exponential and trigonometric functions
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