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A-Level Maclaurin series
What the A-Level syllabus expects for Maclaurin series, and how to practise it.
What the syllabus expects
- The standard expansions of (1+x)ⁿ for rational n, eˣ, sin x, cos x and ln(1+x)
- Deriving the leading terms of a Maclaurin series by repeated differentiation, by repeated implicit differentiation, or by combining standard series
- The set of x-values for which a standard series is convergent
- Treating a Maclaurin series as an approximation to a function
- Small-angle approximations: sin x ≈ x, cos x ≈ 1 − ½x², tan x ≈ x
- Working out the general term of such a series
How it's examined
Questions on this topic most often ask you to find, show, state, compare. About 4% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (5 marks)
Given 2xy + ln y = ln 3, show that (2xy^2 + y) d^2y/dx^2 + 4y^2 dy/dx - (dy/dx)^2 = 0, and hence find the Maclaurin series for y as far as the x^2 term.
Show the worked answer
Differentiate 2xy + ln y = ln 3: 2y + 2x y' + y'/y = 0 ... (1). Differentiate (1): 2y' + (2y' + 2x y'') + (y'' y - y'²)/y² = 0, i.e. 4y' + 2x y'' + y''/y - y'²/y² = 0. Multiply by y²: 4y² y' + 2x y² y'' + y y'' - y'² = 0, so (2xy² + y)y'' + 4y² y' - y'² = 0 (shown). At x=0: ln y = ln3 => y=3. From (1): 6 + y'/3 = 0 => y' = -18. From the identity at x=0: 3 y'' + 4(9)(-18) - (-18)² = 0 => 3 y'' - 648 - 324 = 0 => y'' = 324. Maclaurin: y = 3 + (-18)x + (324/2)x² = 3 - 18x + 162x².
Example 2 (3 marks)
In triangle ABC with AB = 2, angle CAB = x and angle CBA = pi/6, where AC = 2/(cos x + sqrt3 sin x), and x is small, show that AC is approximately a + bx + cx^2, determining a, b and c.
Show the worked answer
For small x: cos x ~ 1 - x²/2, sin x ~ x. Denominator = cos x + sqrt3 sin x ~ 1 + sqrt3 x - x²/2. So AC = 2(1 + sqrt3 x - x²/2)⁻¹. Let u = sqrt3 x - x²/2; (1+u)⁻¹ ~ 1 - u + u² (to x²). u² = 3x² + ... So 1 - u + u² = 1 - sqrt3 x + x²/2 + 3x² = 1 - sqrt3 x + (7/2)x². Hence AC ~ 2[1 - sqrt3 x + (7/2)x²] = 2 - 2 sqrt3 x + 7x².
Example 3 (3 marks)
Find, in terms of e, the value of Σ_{r=6}^∞ (r+1)/r!.
Show the worked answer
The sum starts at r = 6, and there is no formula for a series that starts partway. So sum the whole series from r = 0 and then subtract the terms you do not want. Step 1: split the general term so each piece is a recognisable factorial series. For r ≥ 1, (r + 1)/r! = r/r! + 1/r! and r/r! = r/(r × (r-1)!) = 1/(r - 1)!, so (r + 1)/r! = 1/(r - 1)! + 1/r! For r = 0 the term is (0 + 1)/0! = 1, which is exactly the r = 0 term of Σ 1/r!, so nothing is lost. Step 2: sum from r = 0 to ∞. The Maclaurin series e^x = Σ_{r=0}^∞ x^r/r! at x = 1 gives Σ_{r=0}^∞ 1/r! = e and the shifted series is the same sum written from a different starting index: Σ_{r=1}^∞ 1/(r - 1)! = 1/0! + 1/1! + 1/2! + ... = e Therefore Σ_{r=0}^∞ (r + 1)/r! = Σ_{r=1}^∞ 1/(r - 1)! + Σ_{r=0}^∞ 1/r! = e + e = 2e Step 3: subtract the terms r = 0 to r = 5, which the required sum leaves out. r = 0: 1/0! = 1 r = 1: 2/1! = 2 r = 2: 3/2! = 3/2 r = 3: 4/3! = 4/6 = 2/3 r = 4: 5/4! = 5/24 r = 5: 6/5! = 6/120 = 1/20 Over the common denominator 120: 120/120 + 240/120 + 180/120 + 80/120 + 25/120 + 6/120 = 651/120 = 217/40 Step 4: combine. Σ_{r=6}^∞ (r + 1)/r! = 2e - 217/40 (As a check this is about 5.4366 - 5.425 = 0.0116, a small positive number, as it must be for a tail of positive terms.)
More worked questions on this topic
- Find the first three non-zero terms of the Maclaurin expansion of e^x sin(x + π). (3 marks)
- By differentiating that result further, obtain the Maclaurin series of y up to and including th (3 marks)
- Using BC^2 approx 1 + sqrt3 theta + (3/2)theta^2 from part (b), show BC approx 1 + (sqrt3/2)the (3 marks)
- You may use MF27 expansions. Given f(θ)=1/(3-sinθ) for θ>0, show that for small θ, f(θ)≈p+qθ+rθ (4 marks)
- For y=f(x) with ln y=π/4−arctan(eˣ) and (1+e^{2x}) dy/dx + y eˣ = 0, determine the Maclaurin ex (5 marks)
- Given y=e^{π/4−arctan(eˣ)} has Maclaurin expansion y≈1−(1/2)x+(1/8)x². Deduce, with exact coeff (4 marks)
- You may quote standard expansions from MF27. With a>0, expand a/(a-x)-1 as a series in ascendin (4 marks)
- Triangle ABC has AC = 1, with BAC measuring pi/3 and ABC measuring (pi/6 + theta), both in radi (5 marks)
- (a) Expand 1/sqrt(1 + 5x) as a series as far as the term in x^2, and give the values of x for w (6 marks)
More A-Level H2 Maths topics
Functions · Graphs and their transformations · Equations and inequalities · Sequences and series · Vectors in two and three dimensions: basic properties · Scalar and vector products · all of A-Level H2 Maths