Questions on Taylor series

Question 1

Write down the Taylor series of a function f about x = a, as far as the term in (x − a)².

Accepted 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.

Question 2

Find the Taylor series of e^x about x = 1, up to the term in (x − 1)².

Accepted 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.

Question 3

Find the Taylor series of cos x in ascending powers of (x − π/2), up to and including the term in (x − π/2)³.

Accepted Answer

The derivatives of cos x are −sin x, −cos x and sin x, so at π/2 the values of the function and its first three derivatives are 0, −1, 0 and 1. B1 for the four values. The series is cos x = −(x − π/2) + (x − π/2)³/6 + …. B1. Every even power is missing because cos x is an odd function of (x − π/2), which is the statement cos x = −sin(x − π/2) read term by term.

Question 4

Explain the difference between a Taylor polynomial and a Taylor series, and state what must be true of the remainder before the infinite series can be claimed to equal the function.

Accepted Answer

A Taylor polynomial is the sum of finitely many terms, up to some power of (x − a), and is a different function from f whose difference from f is the remainder. The Taylor series is the infinite sum. B1 for that distinction. The series equals f(x) only where the remainder after n terms tends to zero as n increases without limit; that is what has to be shown, and it can fail even when every derivative exists. B1.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

Expand √x about x = 9 up to the term in (x − 9)², and use it to estimate √9.6 to 3 decimal places.

Accepted Answer

Derivatives at 9: √9 = 3, 1/(2√x) = 1/6, and −1/(4x^3/2) = −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.

Question 9

The quadratic Taylor polynomial of sin x about x = π/6 is used to estimate sin 0.6. Find the estimate to four decimal places. Taylor's theorem gives the error as f'''(c)(x − a)³/3! for some c between a and x; use this to find a bound on the error, and compare the bound with the true value sin 0.6 = 0.564642.

Accepted Answer

At π/6: sin = 1/2, cos = √3/2 and −sin = −1/2, so the polynomial is 1/2 + (√3/2)(x − π/6) − (1/4)(x − π/6)². With x − π/6 = 0.6 − 0.523599 = 0.076401: 0.5 + 0.066165 − 0.001459 = 0.5647. M1 for the three values at π/6, A1 for the estimate.
The third derivative is −cos x, of modulus at most 1 whatever c is, so the error is at most (0.076401)³/6 = 7.4 × 10⁻⁵. M1 for bounding |f'''| by 1, A1 for the bound. The actual error is 0.564706 − 0.564642 = 6.4 × 10⁻⁵, inside the bound as it must be. The bound, 7.4 × 10⁻⁵, exceeds 0.00005, so it cannot guarantee the fourth decimal place, and indeed the true value rounds to 0.5646. The bound comes from the size of the neglected term, not from the smoothness of the sine.

Question 10

The Taylor series of ln x about x = 1 is (x − 1) − (x − 1)²/2 + (x − 1)³/3 − …. Explain why it cannot be used to find ln 3 by substituting x = 3, and describe a way of using the same series to reach ln 3.

Accepted Answer

At x = 3 the terms are 2, −2, 8/3, −4, 32/5, …, growing in size from the third term on, so the partial sums swing ever more wildly and never settle: the series diverges there. B1 for the terms growing. The series converges only for −1 < x − 1 ≤ 1, that is 0 < x ≤ 2, and 3 is outside. B1 for the interval. Inside it, write ln 3 = −ln(1/3) and put x = 1/3, where x − 1 = −2/3 and the terms shrink, or ln 3 = 2 ln √3 with x = √3 = 1.732, which also lies inside the interval. B1 for a valid detour.

Question 11

Find the Taylor series of 1/x about x = 2, up to the term in (x − 2)³. Identify the series as geometric, and state the values of x for which the infinite series converges to 1/x.

Accepted Answer

The derivatives of 1/x are −1/x², 2/x³ and −6/x⁴, which at x = 2 give 1/2, −1/4, 1/4 and −3/8. Dividing by 0!, 1!, 2! and 3!: 1/x = 1/2 − (x − 2)/4 + (x − 2)²/8 − (x − 2)³/16 + …. M1 for the derivatives at 2, A1 for the series. Each term is the term before it multiplied by −(x − 2)/2, so this is a geometric series with first term 1/2 and common ratio −(x − 2)/2. B1. It converges when the ratio has modulus less than 1, that is |x − 2| < 2, so 0 < x < 4, and there its sum (1/2)/(1 + (x − 2)/2) = 1/x is the function itself. A1. Outside that interval the polynomials are still perfectly good numbers; they simply stop approaching 1/x.

Question 12

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.

Accepted 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.

Question 13

Show that the Taylor series of sin x about x = a is sin a + cos a (x − a) − sin a (x − a)²/2! − cos a (x − a)³/3! + …. By writing h = x − a and using the addition formula for sin(a + h) together with the Maclaurin series of sin h and cos h, show that the two approaches give the same terms, and explain what this says about a Taylor series in general.

Accepted Answer

The derivatives of sin x cycle: sin x, cos x, −sin x, −cos x, and back to sin x. At x = a they take the values sin a, cos a, −sin a, −cos a. M1 for the cycle, A1 for the four values. Taylor's form f(a) + f'(a)(x − a) + f''(a)(x − a)²/2! + f'''(a)(x − a)³/3! + … gives the printed series. A1.
With h = x − a, sin(a + h) = sin a cos h + cos a sin h. M1 for the addition formula. Substituting cos h = 1 − h²/2! + … and sin h = h − h³/3! + …: sin a (1 − h²/2!) + cos a (h − h³/3!) + … = sin a + cos a h − sin a h²/2! − cos a h³/3! + …, which is the Taylor series term for term. M1 for substituting both Maclaurin series, A1 for the agreement.
The Taylor series of f about a is the Maclaurin series of the shifted function f(a + h) in the variable h. A Taylor series about a is a Maclaurin series in x − a, which is why the coefficients are the derivatives at a divided by the same factorials.

Question 14

The Taylor series of ln x about x = 1 is (x − 1) − (x − 1)²/2 + (x − 1)³/3 − …, and for 1 < x ≤ 2 its terms alternate in sign and decrease in size, so that the error after any number of terms is less than the first term omitted. Use this to decide how many terms are needed to find ln 1.2 correct to four decimal places, and find it.

Accepted Answer

At x = 1.2 the terms have sizes 0.2, 0.02, 0.002667, 0.0004, 0.000064, 0.0000107, …. M1 for the term sizes. Four decimal places need an error below 0.00005, and the first term smaller than that is the sixth, 0.2⁶/6 = 0.0000107. So five terms are enough. A1 for identifying the sixth term as the first omitted, B1 for stating the remainder condition being used: the error is less than the first omitted term, 0.0000107, which is below 0.00005.
Sum: 0.2 − 0.02 + 0.002667 − 0.0004 + 0.000064 = 0.182331, so ln 1.2 = 0.1823. M1 for the sum, A1 for 0.1823, which agrees with the true value 0.182322. Four terms would give 0.182267, an error of 0.000055, just outside the target.

Question 15

Find the Taylor series of arctan x about x = 1, up to and including the term in (x − 1)³, and use it to estimate arctan 1.1 to five decimal places.

Accepted Answer

f(x) = arctan x, so f'(x) = 1/(1 + x²) and f'(1) = 1/2. M1 for the first derivative.
f''(x) = −2x/(1 + x²)² by the chain rule, so f''(1) = −2/4 = −1/2. M1 for differentiating the reciprocal, A1 for −1/2.
f'''(x) = −2/(1 + x²)² + 8x²/(1 + x²)³ = (6x² − 2)/(1 + x²)³ by the product rule, so f'''(1) = 4/8 = 1/2. A1.
With f(1) = π/4 the series is arctan x = π/4 + (x − 1)/2 − (x − 1)²/4 + (x − 1)³/12 + …. A1.
At x = 1.1: 0.785398 + 0.05 − 0.0025 + 0.000083 = 0.83298, and the true value is 0.832981. B1. The (x − 1)⁴ term happens to vanish here, since the fourth derivative (24x − 24x³)/(1 + x²)⁴ is zero at x = 1, which is why three terms land so close. A Maclaurin series cannot be used at x = 1.1, which lies outside its interval of convergence, |x| ≤ 1; anchoring at 1, where every derivative is a simple fraction, is what keeps the arithmetic this short.