- Topic
- Taylor polynomials
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Taylor polynomials practice question 7
Find the coefficient of in the third-order Maclaurin polynomial of .
Differentiating three times is slow. Which two Maclaurin polynomials do you already know, and how could you combine them?
Write down the Maclaurin polynomials of (put in that of ) and of , each up to , and multiply them.
An term of the product comes from every pair of terms whose powers add to 3. As has no constant or term, there are two such pairs.
A fraction, or a decimal to three places.
Theory UBC Math 100 §9CLP-1 §3.4Key results
Worked solution
Strategy
Multiply the known Maclaurin polynomials of the two factors and keep only the terms up to x³. This is far quicker than differentiating the product three times.
Working
Substitute u = 2x into the polynomial of eᵘ.
The memorised polynomial of sin, to third order.
Multiply out, dropping every power above x³: the x³ terms are 2x² · x and 1 · (−x³/6).
Collect the x³ terms: 2 − 1/6 = 11/6.
Answer
The answer is eleven sixths.
Check
By derivatives instead, the third derivative of e2x sin x at 0 is 11, and 11/3! = 11/6.
Where this goes wrong
Keeping only 1 · (−x³/6) and answering −1/6. The x³ term of a product collects every pair of powers that add to 3, so 2x² · x contributes 2x³ as well.
Leans on
The product rule, from Differentiation rules
Try a similar question
Find the coefficient of in the third-order Maclaurin polynomial of .Saved in this browser only.