Questions on Maclaurin series

Question 1

Write down the general Maclaurin series of a function f in terms of its derivatives at zero.

Accepted Answer

f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + …: the coefficient of x^r is the rth derivative at zero divided by r factorial. B1 for the expansion, B1 for the general coefficient.

Question 2

Derive the Maclaurin series of cos x up to the x⁴ term, showing the derivative values you use.

Accepted Answer

The derivatives at 0 cycle 1, 0, −1, 0, 1, so cos x = 1 − x²/2! + x⁴/4! + … = 1 − x²/2 + x⁴/24 + …. M1 for the derivative values at 0, A1 for the x² term, A1 for the x⁴ term. Only even powers survive, matching cos being an even function.

Question 3

Derive the Maclaurin series of sin x up to the x⁵ term, showing the derivative values you use.

Accepted Answer

The derivatives at 0 cycle 0, 1, 0, −1, 0, 1, so only odd powers survive: sin x = x − x³/3! + x⁵/5! + … = x − x³/6 + x⁵/120 + …. M1 for the derivative values at 0, A1 for the x³ term, A1 for the x⁵ term. Odd powers only, matching sin being an odd function.

Question 4

Write down the Maclaurin series of e^x up to the x³ term, and state the range of x for which it is valid.

Accepted Answer

e^x = 1 + x + x²/2! + x³/3! + … = 1 + x + x²/2 + x³/6 + …, since every derivative of e^x at 0 equals 1. B1 for the series, B1 for valid for all real x.

Question 5

Use the first three terms of the cosine series to estimate cos 0.2, and comment on the accuracy.

Accepted Answer

1 − 0.02 + 0.0000667 = 0.980067, and the true value is 0.980067 to six decimal places. The next term is 0.2⁶/720, about 9 × 10⁻⁸. M1 for substituting into the three terms, A1 for 0.980067, B1 for the comment on accuracy. Three terms are tight here only because 0.2 is small; at x = 2 the same three terms give −0.333, against a true value of −0.416.

Question 6

Write down the series for e^3x up to the x³ term, and state the coefficient of x³ exactly.

Accepted Answer

Substitute 3x into the exponential series: 1 + 3x + 9x²/2 + 27x³/6 = 1 + 3x + (9/2)x² + (9/2)x³ + …. The x³ coefficient is 27/3! = 9/2. M1 for substituting 3x, A1 for the x² term, A1 for the x³ term, A1 for the coefficient 9/2.

Question 7

State the range of validity of the series for ln(1 + x), and deduce the range of validity for the series of ln(1 + 3x).

Accepted Answer

ln(1 + x) converges for −1 < x ≤ 1. Substituting 3x means the substituted quantity must stay in that window: −1 < 3x ≤ 1, so −1/3 < x ≤ 1/3. B1 for the original window, M1 for substituting 3x, A1 for the new window.

Question 8

Use the first four terms of the Maclaurin series for e^x to estimate e^−0.5 to 4 decimal places, and comment on the accuracy of the estimate by considering the size of the next term.

Accepted Answer

1 − 0.5 + 0.5²/2 − 0.5³/6 = 1 − 0.5 + 0.125 − 0.020833… = 0.6042. M1 for substituting x = −0.5 into four terms, A1 for 0.6042. The true value is 0.6065, so the estimate is 0.0024 low. A1 for the comparison with the true value. The next term is 0.5⁴/24 = 0.0026, and because x is negative the series alternates in sign here, so the error after truncating is no bigger than that next term, which is consistent with the 0.0024 found. B1 for the comment on accuracy. With x positive instead, every term of the e^x series is positive, so the error exceeds the next term rather than being bounded by it.

Question 9

Find the Maclaurin series of e^x cos x up to and including the x³ term, by multiplying the two standard series.

Accepted Answer

e^x = 1 + x + x²/2 + x³/6 + … and cos x = 1 − x²/2 + …. Multiplying and keeping terms up to x³: the x⁰ term is 1, the x¹ term is x, the x² term is −x²/2 + x²/2 = 0, and the x³ term is −x³/2 (from x × (−x²/2)) + x³/6 (from the e^x series alone) = −x³/3. So e^x cos x = 1 + x − x³/3 + …. M1 for both series to the required order, M1 for multiplying and collecting by power, A1 for the x² term vanishing, A1 for the x³ coefficient.

Question 10

State the range of validity of the series for ln(1 + x), and deduce the range of validity for the series of ln(1 − 2x).

Accepted Answer

ln(1 + x) converges for −1 < x ≤ 1. Substituting −2x means the substituted quantity must stay in that window: −1 < −2x ≤ 1. Dividing by −2 reverses both inequalities, so x = −1/2 lands on the closed end and x = 1/2 on the open one: −1/2 ≤ x < 1/2. M1 for substituting −2x, M1 for dividing by a negative and reversing the inequalities, A1 for the new window with the closed end correctly placed at −1/2, A1 for open at 1/2.

Question 11

Using cos 2x = 1 − 2sin²x, or otherwise, find the Maclaurin series of cos²x up to and including the x⁴ term.

Accepted Answer

cos²x = (1 + cos 2x)/2. The series for cos 2x, from substituting 2x into the cosine series, is 1 − (2x)²/2! + (2x)⁴/4! + … = 1 − 2x² + (2/3)x⁴ + …. So cos²x = (1 + 1 − 2x² + (2/3)x⁴)/2 = 1 − x² + (1/3)x⁴ + …. M1 for the identity connecting cos²x and cos 2x, M1 for substituting 2x into the cosine series, A1 for the cos 2x series to x⁴, A1 for the final series. Halving after substituting is faster here than multiplying the cosine series by itself, since squaring a series to x⁴ needs every cross term up to that power.

Question 12

Using ln((1 + x)/(1 − x)) = ln(1 + x) − ln(1 − x), show that the series begins 2(x + x³/3 + …), and use x = 1/3 to estimate ln 2 from two terms.

Accepted Answer

Subtracting the two log series cancels the even powers and doubles the odd ones: 2(x + x³/3 + x⁵/5 + …). At x = 1/3 the argument is (4/3)/(2/3) = 2, and two terms give 2(1/3 + 1/81) = 0.6914, against ln 2 = 0.6931. M1 for subtracting the two series, A1 for the even powers cancelling, A1 for the odd-power form, M1 for putting x = 1/3, A1 for the argument being 2, A1 for 0.6914. Far faster than the plain ln(1 + x) series at x = 1, which needs hundreds of terms for the same accuracy.

Question 13

Find the Maclaurin series of ln(1 + sin x) up to and including the term in x³.

Accepted Answer

Write u = sin x = x − x³/6 + …, then ln(1 + u) = u − u²/2 + u³/3 − …. Work to x³ throughout, so u² = x² + … with the next correction at x⁴, and u³ = x³ + …. Substituting, ln(1 + sin x) = (x − x³/6) − x²/2 + x³/3 + … = x − x²/2 + x³/6 + …. M1 for putting u = sin x, M1 for the ln(1 + u) expansion, A1 for u² to the required order, A1 for u³, M1 for substituting and collecting, A1 for the final series. Differentiating three times instead is legitimate but long; the second derivative of ln(1 + sin x) alone is a quotient with sin x and cos²x in it.

Question 14

Find the coefficient of x⁴ in the Maclaurin series of ln(1 + x²), and state the full range of x for which the series is valid.

Accepted Answer

Substitute u = x² into ln(1 + u) = u − u²/2 + u³/3 − …, keeping terms up to x⁴: ln(1 + x²) = x² − x⁴/2 + …. M1 for the substitution u = x², A1 for the x² term. So the coefficient of x⁴ is −1/2. A1 for the coefficient. The window on u is −1 < u ≤ 1, that is −1 < x² ≤ 1. Since x² ≥ 0 for every real x, the left inequality never binds, and x² ≤ 1 alone gives −1 ≤ x ≤ 1, closed at both ends. M1 for stating the window on u, A1 for recognising x² ≥ 0 makes the lower bound automatic, A1 for the closed interval on x. This is a case where the substituted quantity can never reach the excluded end of the original window, and x = −1 and x = 1 both land on the closed end u = 1, so both endpoints of the new interval are included.

Question 15

Using repeated differentiation, find the Maclaurin series of tan x up to and including the x³ term.

Accepted Answer

f(x) = tan x, f(0) = 0. f'(x) = sec²x = 1 + tan²x, using the identity to simplify before evaluating, so f'(0) = 1. f''(x) = 2 sec²x tan x, and tan 0 = 0 makes f''(0) = 0 without needing sec²0. f'''(x) = 4 sec²x tan²x + 2 sec⁴x, so f'''(0) = 0 + 2 = 2. Assembling, tan x = 0 + x + 0 + (2/3!)x³ + … = x + x³/3 + …. M1 for f'(0) using the identity sec²x = 1 + tan²x, A1 for f''(0) = 0, M1 for f'''(x) differentiated and simplified using tan 0 = 0, A1 for f'''(0) = 2, A1 for the x coefficient, A1 for the x³ coefficient.