- Topic
- Taylor polynomials
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Taylor polynomials practice question 6
Find the coefficient of in the third-order Maclaurin polynomial of .
Differentiate three times with the chain rule; each derivative brings down the exponent and a factor of 2 from the inside.
First derivative: . Differentiate twice more the same way.
The coefficient of is , and at 0 every power of is 1.
A number, as a fraction or a decimal.
Theory UBC Math 100 §9CLP-1 §3.4Key results
Worked solution
Strategy
Compute the third derivative at 0 and divide by 3!. Keeping the powers of (1 + 2x) unexpanded makes the evaluation at 0 immediate.
Working
Chain rule; the 2 cancels the ½.
Again.
And once more; at 0 every power of (1 + 2x) is 1.
So T₃(x) = 1 − x + (3/2)x² − (5/2)x³.
Answer
The answer is negative five halves.
Check
By the binomial series, the u³ coefficient of (1 + u)−1/2 is −5/16, and u = 2x multiplies it by 8, giving −5/2.
Where this goes wrong
Forgetting the inner factor of 2 on one of the three differentiations. Each application of the chain rule to (1 + 2x) contributes a 2; missing one changes the coefficient by a factor of 2.
Leans on
The chain rule, from The chain rule and logarithms
Try a similar question
Find the coefficient of in the third-order Maclaurin polynomial of .Saved in this browser only.