Taylor series and improper integration
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 1 (Total: 20 marks)
(a) State Taylor’s theorem with local remainder. [2 marks]
(b) Find the first four non-zero terms of the power series of
log(1 + x)
1 + x
about x = 0, and state its radius of convergence. [4 marks]
(c) Define
g(x) = log(1 + x) − x + x2/2
x3 , x > −1, x ̸= 0.
(i) Calculate the limits as x → −1+, x → 0 and x → ∞. [3 marks]
(ii) Extend g at 0 and give its power series near 0. [2 marks]
(iii) Use a power series to give a rational approximation, correct to five decimal places, to
A =
∫ 1/4
0
log(1 + x)
1 + x dx.
[4 marks]
(iv) Determine whether
∫ ∞
0 g(x)/(1 + x) dx exists. [5 marks]
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Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 1 - Complete solution (Total: 20 marks)
(a) Standard Taylor formula with Rn = f (n+1)(ξ)(x − a)n+1/(n + 1)!. [2 marks] Suggested level: A
(b) Multiplying log(1 + x) = x − x2/2 + x3/3 − · · · by (1 + x)−1 = 1 − x + x2 − · · · gives
log(1 + x)
1 + x = x − 3
2 x2 + 11
6 x3 − 25
12 x4 + · · · ,
with radius 1. [4 marks] Suggested level: B
(c) (i) The limits are +∞, 1/3 and 0. [3 marks] Suggested level: A
(ii)
g(x) = 1
3 − x
4 + x2
5 − x3
6 + · · · , |x| < 1,
so g(0) = 1 /3. [2 marks] Suggested level: B
(iii) In general
log(1 + x)
1 + x =
∞X
n=1
(−1)n+1Hnxn,
where Hn = 1 + · · · + 1/n. Integrating through n = 7 at x = 1/4 gives the rational number
A7 =
7X
n=1
(−1)n+1 Hn
n + 1 4−(n+1) = 16447331
660602880 = 0.024897455 . . . .
The next term is < 1.2 × 10−6, so A = 0.02490 to five decimal places. [4 marks] Suggested
level: C/D
(iv) At 0 the integrand tends to 1/3. For large x, the numerator is x2/2 + O(x), so g(x) =
1/(2x) + O(x−2) and g(x)/(1 + x) = O(x−2). Thus the integral exists. [5 marks] Suggested
level: C
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Cycloid geometry and Fourier series
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 2 (Total: 20 marks)
(a) One arch of a cycloid is parametrised by
x = t − sin t, y = 1 − cos t, 0 ≤ t ≤ 2π.
It is made from a uniform thin wire.
x
y
0 2π
(i) Mark the cusps and highest point and describe the orientation. [2 marks]
(ii) Find the area under the arch. [3 marks]
(iii) Find the length of the wire. [3 marks]
(iv) Find the centre of mass of the wire. [4 marks]
(b) Find the Fourier series of f (x) = x2 on [−π, π], extended periodically. State its convergence at
x = ±π and use the series to prove ∑∞
n=1 n−2 = π2/6. [8 marks]
3
Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 2 - Complete solution (Total: 20 marks)
(a) (i) Cusps are (0, 0) and (2π, 0); the highest point is (π, 2). The curve is traversed left to right. [2
marks] Suggested level: A
(ii) Since dx/dt = 1 − cos t = y,
A =
Z 2π
0
(1 − cos t)2dt = 3π.
[3 marks] Suggested level: B
(iii)
ds =
q
(1 − cos t)2 + sin2 t dt = 2 sin(t/2)dt,
so L = 8. [3 marks] Suggested level: B
(iv) Symmetry gives ¯x = π. Moreover
¯y = 1
8
Z 2π
0
(1 − cos t)2 sin(t/2)dt = 1
8 · 32
3 = 4
3 .
[4 marks] Suggested level: C
(b) a0 = 2π2/3, an = 4(−1)n/n2 and bn = 0, so
x2 = π2
3 + 4
∞X
n=1
(−1)n cos nx
n2 .
The periodic extension is continuous at ±π, so it converges there to π2. Substitution x = π yields
π2 = π2/3 + 4 P n−2, hence the result. [8 marks] Suggested level: A/B
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Linear systems and integrating factors
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 3 (Total: 20 marks)
(a) Consider
˙x = x − y, ˙y = 4x − 4y.
(i) Find the general solution, all fixed points and their stability. [6 marks]
(ii) Sketch the phase portrait and find the asymptotic point for (x(0), y(0)) = (0 , 1). [4 marks]
x
y
0
(iii) For the modified system X ′ = AX + et(1, 2)T , find a particular integral. [3 marks]
(b) Consider (
1 + y2e−y cos x
)
dx +
(
x + 2ye−y sin x
)
dy = 0.
(i) Show that the equation is not exact. [2 marks]
(ii) Find an integrating factor depending only on y, and solve implicitly. [5 marks]
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Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 3 - Complete solution (Total: 20 marks)
(a) (i) The eigenvalues are 0, −3 with eigenvectors (1, 1)T and (1, 4)T . Thus
X = c1(1, 1)T + c2e−3t(1, 4)T .
Every point on y = x is fixed; the line is transversely stable. [6 marks] Suggested level: A
(ii) From (0, 1), c2 = 1 /3, c1 = −1/3. Therefore the trajectory tends to (−1/3, −1/3). Other
trajectories lie on lines parallel to (1, 4). [4 marks] Suggested level: B
(iii) Seek pet. Then (I − A)p = (1, 2)T , giving p = (3/4, 1)T . [3 marks] Suggested level: C
(b) (i) My = e−y(2y − y2) cos x and Nx = 1 + 2 ye−y cos x, so the equation is not exact. [2 marks]
Suggested level: A
(ii)
Nx − My
M = 1,
so λ(y) = ey. Multiplication gives the exact differential of
U = y2 sin x + xey.
Thus y2 sin x + xey = C . [5 marks] Suggested level: B
4
Bifurcation, Green function and Jacobian
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 4 (Total: 20 marks)
(a) Consider ˙y = y(r − y2).
(i) Find all fixed points and determine their stability. [5 marks]
(ii) Draw and classify the bifurcation. For r = 1, give the basins of attraction. [4 marks]
r
x∗
0
(b) Use Fourier transforms to solve
−y′′ + a2y = δ(x), a > 0,
subject to y(x) → 0 as |x| → ∞ . [5 marks]
(c) For the transformation
x = eu cos v, y = eu sin v,
compute the Jacobian, find the inverse Jacobian, and prove
ux = vy, u y = −vx.
[6 marks]
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Worked solution and marking guidance
MATH40004 Calculus and Applications Original 2026 Resit Training Material
Question 4 - Complete solution (Total: 20 marks)
(a) (i) y = 0 always; y = ±√r for r > 0. Since f ′(0) = r, zero is stable for r < 0 and unstable for
r > 0. At ±√r, f ′ = −2r < 0, so both are stable. [5 marks] Suggested level: A
(ii) A supercritical pitchfork occurs at r = 0 . For r = 1 , y0 > 0 tends to 1, y0 < 0 tends to −1,
and y0 = 0 remains at the unstable fixed point. [4 marks] Suggested level: D
(b) Fourier transformation gives (ω2 + a2)by = 1. Since F {e−a|x|} = 2a/(a2 + ω2),
y(x) = 1
2a e−a|x| .
[5 marks] Suggested level: B
(c)
J =
eu cos v −eu sin v
eu sin v e u cos v
, det J = e2u.
Thus
J −1 = e−u
cos v sin v
− sin v cos v
.
Therefore ux = e−u cos v = vy and uy = e−u sin v = −vx. [6 marks] Suggested level: B/D
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