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Maclaurin series questions
Many standard functions can be written as an infinite polynomial. Matching the value and every derivative at zero gives the coefficients, each divided by a factorial, and that is where the expansions printed in the formulae booklet come from.
7 original questions · 27 marks · the maclaurin series notes · Further algebra and series
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Write down the general Maclaurin series of a function f in terms of its derivatives at zero.
Worked answer
f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + …: the coefficient of xr is the rth derivative at zero divided by r factorial. B1 for the expansion, B1 for the general coefficient.Derive the Maclaurin series of cos x up to the x⁴ term, showing the derivative values you use.
Worked 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.Use the first three terms of the cosine series to estimate cos 0.2, and comment on the accuracy.
Worked 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.Write down the series for e3x up to the x³ term, and state the coefficient of x³ exactly.
Worked 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. Substitution into a standard series beats fresh differentiation every time it is available.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).
Worked 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. The window transforms with the same substitution as the series.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.
Worked 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.Find the Maclaurin series of ln(1 + sin x) up to and including the term in x³.
Worked answer
Two standard series stack here. 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 four 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. Decide how far you need each series before substituting, or the x³ coefficient picks up terms that should have been discarded.
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