Taylor polynomials
A differentiable function looks like its tangent line close to a point, more like a parabola built from its second derivative, and more still like a polynomial built from as many derivatives as you care to take. Exams ask for the polynomial, for the estimate it gives, for a derivative read back off its coefficients, and, in flavour A, for a bound on how wrong the estimate can be.
In this topic
- Taught
- Before the final
- Questions
- 16
- Total marks
- 47
- Suggested time
- 76 min
- Difficulty
Introductory to Capstone
Show the 5 hardest
Key results
- The Taylor polynomial
When has derivatives at ; written out, , and with it is the Maclaurin polynomial.
- Reading a derivative off
When is the coefficient of in the Taylor polynomial of about , for .
- Standard Maclaurin polynomials
Each line lists the terms of the function’s Maclaurin polynomials, which approximate it near , with in radians for sine and cosine.
- The Lagrange remainder
For some strictly between and , when has derivatives on an interval containing and .
- Bounding the error
When has derivatives on an interval containing and , and for every between and .
Read the theoryUBC Math 100 §9 Taylor polynomialsCLP-1 §3.4 Taylor polynomials
Worked example
Find the second-order Taylor polynomial of about .
Strategy
Find the value and the first two derivatives at the centre π/6, where sine and cosine are exact, and build the polynomial in powers of x − π/6, dividing the second derivative by 2!.
Working
Differentiate as functions first.
Then evaluate at the centre, not at 0.
The square term is f″(π/6)/2! = −1/4. Expanding the brackets gives the same polynomial in another form.
Answer
The answer is T two of x equals one half, plus root 3 over 2 times (x minus pi over 6), minus one quarter of (x minus pi over 6) squared.
Check
At x = 0.5, T₂ and sin 0.5 both come to 0.4794 to four places.
Where this goes wrong
Using the Maclaurin coefficients of sin x, which are 0, 1 and 0, with powers of x − π/6, which gives x − π/6. The coefficients of a Taylor polynomial are the derivatives at its own centre, here π/6.
On the paper
Tabulate and its derivatives at the centre, then write the polynomial in powers of , not , unless .
Leans on
The derivatives of sine and cosine, from Exponential and trigonometric functions
Your turn
Find the third-order Taylor polynomial of about .
Practice questions
5 of the 16, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.
- Introductory2 marks · 3 min
The third-order Maclaurin polynomial of a function is given below. Find .
Answer this questionStrategy
Taylor coefficients are derivatives divided by factorials, so a derivative at the base point is a coefficient multiplied back by the factorial.
Working
The x³ coefficient.
3! = 6.
Answer
The answer is negative 8.
Check
Differentiating T₃ three times wipes out 2 − x + 5x² and leaves the constant −8 from −(4/3)x³.
Where this goes wrong
Answering −4/3, the coefficient itself. The factorial is built into every Taylor coefficient and has to be undone.
Leans on
Higher-order derivatives, from Differentiation rules
Try a similar question
The third-order Taylor polynomial of a function about is given below. Find . - Introductory2 marks · 3 min
Find the coefficient of in the third-order Maclaurin polynomial of .
Answer this questionStrategy
Substitute into a polynomial you already know rather than differentiating three times: replace u by 2x in the third-order Maclaurin polynomial of log(1 + u).
Working
The memorised polynomial: the uᵏ term is divided by k, and the signs alternate.
Put u = 2x; each power applies to the 2 as well as to the x.
A check by derivatives: f‴(x) = 16/(1 + 2x)³, so f‴(0) = 16.
Answer
The answer is 8/3.
Check
At x = 0.1 the cubic 2x − 2x² + (8/3)x³ gives 0.18267, close to log 1.2 ≈ 0.18232.
Where this goes wrong
Writing (2x)³ as 2x³, which gives 2/3. The substitution replaces u by the whole of 2x, so the 2 is cubed too: (2x)³/3 = 8x³/3.
Leans on
Composing functions, from Functions, graphs and inverses
Try a similar question
Find the coefficient of in the third-order Maclaurin polynomial of . - Standard3 marks · 4 min
A function has , and . Use the quadratic approximation of centred at to estimate .
Answer this questionStrategy
The quadratic approximation adds one term to the tangent line: half the second derivative times the square of the step.
Working
Second-order Taylor polynomial at 1.
The linear estimate alone would be 2.8; the curvature term adds 0.03.
Answer
The answer is 2.83.
Check
At x = 1 the polynomial gives 3, its derivative −2 + 6(x − 1) gives −2, and its second derivative is 6, matching all three given values.
Where this goes wrong
Using f″(1) = 6 as the coefficient instead of 6/2 = 3, which gives 2.86. The second-order coefficient is always the second derivative divided by 2.
Leans on
Linear approximation, from Linear approximation
- Challenging3 marks · 6 min
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.
Answer this questionStrategy
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.
Leans on
Bounding a function on an interval, from Functions, graphs and inverses
- Capstone5 marks · 10 min
Use the second-order Taylor polynomial of about to estimate , and use the Taylor remainder to give a bound on the error of your estimate.
Answer this questionStrategy
Build the quadratic approximation from exact derivatives at 16, evaluate it one unit away, and bound the third-derivative remainder using the worst case on the interval.
Working
Derivatives at the base point.
f″(16)/2 = −1/512.
The true value is 4.1231056…
f‴ is positive and decreasing, so its maximum on (16, 17) is at 16, where 165/2 = 1024. The actual error is about 5.9 × 10⁻⁵, inside the bound.
Answer
The answer is about 4.1230, with error at most one over 16384.
What earns the marks
- Computes f(16) = 4, f′(16) = 1/8 and f″(16) = −1/256, and writes T₂(x) = 4 + (x − 16)/8 − (x − 16)²/512.
- Evaluates T₂(17) = 4 + 1/8 − 1/512 ≈ 4.1230.
- Bounds the remainder: |R₂(17)| = |f‴(c)|/3! · 1³ with f‴(x) = 3/(8x5/2) and c in (16, 17), so |R₂(17)| ≤ 3/(8 · 165/2 · 6) = 1/16384 ≈ 6.1 × 10⁻⁵.
Check
Squaring back, 4.123046875² ≈ 16.9995, and the actual error, about 0.000059, is inside the bound 1/16384 ≈ 0.000061.
Where this goes wrong
Bounding the remainder with the second derivative instead of the third, or evaluating f‴ at 17 instead of finding its maximum on the interval. The remainder after T₂ uses f‴, and the bound must cover every c in (16, 17).
Leans on
Higher-order derivatives, from Differentiation rules
Related topics
Needs
Linear approximation
Tangent-line estimates, whether they run high or low, and approximating a value without a calculator.
14 questions