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Taylor series questions
Maclaurin series are anchored at zero. Taylor series anchor anywhere. Move the anchor to the point you care about and the same factorial formula approximates a smooth function near whichever point you choose.
7 original questions · 30 marks · the taylor series notes · Further Pure 1
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Write down the Taylor series of a function f about x = a, as far as the term in (x − a)².
Worked answer
f(x) = f(a) + f'(a)(x − a) + f''(a)(x − a)²/2! + …. B1 for the linear term, B1 for the quadratic term with its factorial. Each coefficient is the matching derivative evaluated at a, divided by the factorial of that power.Find the Taylor series of ex about x = 1, up to the term in (x − 1)².
Worked answer
Every derivative of the exponential equals itself, so all the values at x = 1 are e. The series is e + e(x − 1) + e(x − 1)²/2 + …, which is e times the standard exponential series shifted along. B1 for every derivative being e at x = 1, M1 for substituting into the Taylor form, A1 for the series.Expand ln x in ascending powers of (x − 1), up to the term in (x − 1)³, and use it to estimate ln 1.1.
Worked answer
Derivatives at 1: ln 1 = 0, 1/x = 1, −1/x² = −1, 2/x³ = 2. So ln x = (x − 1) − (x − 1)²/2 + (x − 1)³/3 − …. At x = 1.1: 0.1 − 0.005 + 0.000333 = 0.09533, against the true 0.09531. M1 for differentiating three times, A1 A1 for the derivative values, A1 for the series, B1 for 0.09533. Three terms give four decimal places, because the anchor sits so close.Find the Taylor series of tan x about x = π/4 up to the term in (x − π/4)², and estimate tan(π/4 + 0.1) to 2 decimal places.
Worked answer
At π/4: tan = 1, sec²= 2, and the second derivative 2 sec²x tan x = 4. So tan x = 1 + 2(x − π/4) + 2(x − π/4)² + …. At the stated point: 1 + 0.2 + 0.02 = 1.22, against the true 1.2230. M1 for differentiating twice, A1 for 2, A1 for 4, A1 for the series, B1 for 1.22. Correct to 2 decimal places from a quadratic.Explain why a Taylor series anchored at x = a is generally more accurate near a than a Maclaurin series would be, when a is far from the origin.
Worked answer
Every term after the first carries a power of (x − a), which is small near the anchor, so the neglected terms are tiny there. A Maclaurin series carries powers of x instead, which are large when x is far from the origin, so far more terms are needed for the same accuracy. B1 for the powers of (x − a) being small near the anchor, B1 for the neglected terms being tiny there, B1 for powers of x being large far from the origin.Expand √x about x = 9 up to the term in (x − 9)², and use it to estimate √9.6 to 3 decimal places.
Worked answer
Derivatives at 9: √9 = 3, 1/(2√x) = 1/6, and −1/(4x3/2) = −1/108. So √x = 3 + (x − 9)/6 − (x − 9)²/216 + …. At x = 9.6: 3 + 0.1 − 0.001667 = 3.098, against the true 3.0984. M1 for differentiating twice, A1 for 1/6, A1 for −1/108, A1 for the series, B1 for 3.098. Anchoring at the nearest perfect square is what makes the arithmetic this light.y satisfies dy/dx = x + y² with y = 1 at x = 1. Find the Taylor series solution for y in ascending powers of (x − 1), up to and including the term in (x − 1)³.
Worked answer
Differentiate the equation repeatedly, substituting the known values as you go. At x = 1, y = 1 and dy/dx = 1 + 1 = 2. Differentiating dy/dx = x + y² gives d²y/dx² = 1 + 2y(dy/dx), which at x = 1 is 1 + 2(1)(2) = 5. Differentiating again, and using the product rule on the 2y(dy/dx) term, gives d³y/dx³ = 2(dy/dx)² + 2y(d²y/dx²), which at x = 1 is 2(4) + 2(1)(5) = 18. The Taylor series is y = 1 + 2(x − 1) + 5(x − 1)²/2! + 18(x − 1)³/3!, that is y = 1 + 2(x − 1) + (5/2)(x − 1)² + 3(x − 1)³ + …. B1 for dy/dx = 2, M1 for differentiating the equation, A1 for 5, M1 for the product rule at the next stage, A1 for 18, M1 for the Taylor form with its factorials, A1 for the series. Two errors dominate here. Forgetting the factorial denominators turns 18 into the coefficient rather than 3, and differentiating y² as 2y instead of 2y(dy/dx) loses the chain rule that makes the recursion work.
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