FAC1004: Rapid-Fire Drill Pack — Leak Topics

Objective: Master the 4 leak-highlighted topic areas for the final exam.
Target: 2–3 min per problem.
Total problems: 44
Estimated time: ~110 min

Cheat Sheet (Memorize First)

Taylor & Maclaurin Series

General Formula: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n, \qquad \text{Maclaurin: } f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$

Standard Expansions (up to $x^5$):

Function Series
$e^x$ $1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} + \cdots$
$\sin x$ $x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$
$\cos x$ $1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$
$\ln(1+x)$ $x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \dfrac{x^5}{5} - \cdots$
$\tan x$ $x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots$
$\frac{1}{1-x}$ $1 + x + x^2 + x^3 + x^4 + x^5 + \cdots$
$\arctan x$ $x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$

Complex Numbers & De Moivre

Concept Formula
Cartesian $z = a + bi$
Polar $z = r(\cos\theta + i\sin\theta)$
Exponential $z = re^{i\theta}$
Euler $e^{i\theta} = \cos\theta + i\sin\theta$
De Moivre $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$
n-th roots $z^{1/n} = r^{1/n}\left[\cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right)\right]$
Complex log $\text{Log}(z) = \ln
Circle locus $|z - z_0| = r$
Perp bisector $|z - z_1| = |z - z_2|$
Half-line $\arg(z - z_0) = \theta$

Inverse Trig Derivatives

Function Derivative Domain
$\sin^{-1} x$ $\dfrac{1}{\sqrt{1-x^2}}$ $|x| < 1$
$\cos^{-1} x$ $-\dfrac{1}{\sqrt{1-x^2}}$ $|x| < 1$
$\tan^{-1} x$ $\dfrac{1}{1+x^2}$ all $x$

Hyperbolic Functions

Definitions: $$\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

Fundamental identity: $\cosh^2 x - \sinh^2 x = 1$

Derivatives:

$f(x)$ $f'(x)$
$\sinh x$ $\cosh x$
$\cosh x$ $\sinh x$
$\tanh x$ $\text{sech}^2 x$

Inverse Hyperbolic:

Function Logarithmic Form Derivative
$\sinh^{-1} x$ $\ln(x + \sqrt{x^2+1})$ $\dfrac{1}{\sqrt{1+x^2}}$
$\cosh^{-1} x$ $\ln(x + \sqrt{x^2-1})$, $x \geq 1$ $\dfrac{1}{\sqrt{x^2-1}}$, $x > 1$
$\tanh^{-1} x$ $\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$, $|x| < 1$ $\dfrac{1}{1-x^2}$, $|x| < 1$

Standard Integrals:

Integral Result
$\int \frac{dx}{\sqrt{a^2+x^2}}$ $\sinh^{-1}\left(\frac{x}{a}\right) + C$
$\int \frac{dx}{\sqrt{x^2-a^2}}$ $\cosh^{-1}\left(\frac{x}{a}\right) + C$
$\int \frac{dx}{a^2-x^2}$ $\frac{1}{a}\tanh^{-1}\left(\frac{x}{a}\right) + C$ ($|x|<a$)
$\int \sinh x , dx$ $\cosh x + C$
$\int \cosh x , dx$ $\sinh x + C$

Part A: Taylor & Maclaurin Series (Section B Focus)

Set A1 — Standard Expansions (5 problems)

Write the Maclaurin series up to $x^5$ for each function.

  1. $f(x) = e^{3x}$
  2. $f(x) = \sin(2x)$
  3. $f(x) = \cos\left(\dfrac{x}{2}\right)$
  4. $f(x) = \ln(1 + 2x)$
  5. $f(x) = (1 - x)^{-1}$

Score: ___/5

Set A2 — $\tan x$ Expansion (3 problems)

  1. Direct differentiation: Find $f'(x)$, $f''(x)$, $f'''(x)$, $f^{(4)}(x)$, $f^{(5)}(x)$ for $f(x) = \tan x$. Hence determine the Maclaurin series up to $x^5$.

  2. Division method: Use $\tan x = \dfrac{\sin x}{\cos x}$ with the known series for $\sin x$ and $\cos x$ to find the Maclaurin series for $\tan x$ up to $x^5$.

  3. Use your series for $\tan x$ to estimate $\tan(0.3)$ and compare with the exact value to 4 decimal places.

Score: ___/3

Set A3 — Substitution & Multiplication (4 problems)

  1. Use substitution to find the Maclaurin series for $e^{-2x^2}$ up to $x^4$.

  2. Find the series for $\sin(x^2)$ up to $x^{10}$ by substitution into the known series for $\sin x$.

  3. Multiply series to find the expansion of $x^2\cos x$ up to $x^6$.

  4. Find the series for $\dfrac{\sin x}{x}$ up to $x^4$ (note the removable singularity at $x=0$). Hence approximate $\displaystyle\int_0^{0.5} \frac{\sin x}{x},dx$ by integrating term by term.

Score: ___/4

Part B: Complex Numbers & De Moivre

Set B1 — De Moivre's Theorem (5 problems)

  1. Express $z = 1 + i$ in polar form. Hence find $z^6$ using De Moivre's theorem and give your answer in Cartesian form $a + bi$.

  2. Let $z = \sqrt{3} - i$. Find $z^4$ in Cartesian form.

  3. Use De Moivre's theorem to express $\cos(3\theta)$ in terms of powers of $\cos\theta$.

  4. Express $\sin(4\theta)$ in terms of $\sin\theta$ and $\cos\theta$ using De Moivre's theorem.

  5. Find all values of $(-1)^{1/4}$ (the fourth roots of $-1$) and express them in Cartesian form.

Score: ___/5

Set B2 — n-th Roots (4 problems)

  1. Find all cube roots of $z = 8\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$. Express them in polar form and plot them on an Argand diagram.

  2. Find the fourth roots of $z = -16$ and express them in Cartesian form $a + bi$.

  3. Solve $z^3 = 1$ (roots of unity). Plot all three roots on an Argand diagram and state their geometric relationship.

  4. Find the fifth roots of $z = 32e^{i\pi/2}$. Express each root in exponential form.

Score: ___/4

Set B3 — Loci & Complex Equations (4 problems)

  1. Describe and sketch the locus $|z - 2i| = 3$. Write the Cartesian equation.

  2. Find and describe the locus $|z - 1| = |z + i|$. What geometric object is this?

  3. Describe the locus $\arg(z - 1) = \dfrac{\pi}{4}$ and sketch it on an Argand diagram.

  4. Find the region defined by $|z| \leq 2$ and $0 \leq \arg(z) \leq \dfrac{\pi}{3}$. Sketch and describe the shape.

Score: ___/4

Part C: Inverse Trig & Hyperbolic Functions

Set C1 — Derivatives (5 problems)

Find $\dfrac{dy}{dx}$.

  1. $y = \sin^{-1}(5x)$
  2. $y = \tan^{-1}(e^{2x})$
  3. $y = \sinh(4x^2 + 1)$
  4. $y = \cosh^{-1}(3x)$ (state the domain)
  5. $y = \tanh^{-1}(\sin x)$

Score: ___/5

Set C2 — Integrals (5 problems)

Evaluate each integral.

  1. $\displaystyle\int \frac{dx}{\sqrt{16 + x^2}}$
  2. $\displaystyle\int \frac{dx}{\sqrt{x^2 - 25}}$ for $x > 5$
  3. $\displaystyle\int \frac{dx}{9 - x^2}$
  4. $\displaystyle\int \sinh(5x),dx$
  5. $\displaystyle\int_0^1 \cosh(3x),dx$

Score: ___/5

Set C3 — Identity Verification (5 problems)

  1. Prove $\cosh^2 x - \sinh^2 x = 1$ using the exponential definitions of $\cosh x$ and $\sinh x$.

  2. Show that $\sinh(2x) = 2\sinh x \cosh x$ using exponential definitions.

  3. Verify $\cosh x + \sinh x = e^x$ directly from the definitions.

  4. Prove $\cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y$ using exponential definitions.

  5. Show that $\frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{1+x^2}}$ by differentiating the logarithmic form $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$.

Score: ___/5

Set C4 — Rapid Mixed Fire (4 problems)

  1. Differentiate $y = \sin^{-1}(\tanh x)$ and simplify.

  2. Evaluate $\displaystyle\int \frac{e^x}{\sqrt{e^{2x} + 9}},dx$ (Hint: $u = e^x$).

  3. Solve the complex equation $z^2 = -5 + 12i$. (Hint: let $z = x + yi$.)

  4. Find the Maclaurin series for $e^x\cos x$ up to $x^3$ by multiplying the known series.

Score: ___/4

Final Scorecard

Part Sets Problems Raw Score
A — Taylor & Maclaurin Series A1, A2, A3 12 ___/12
B — Complex Numbers & De Moivre B1, B2, B3 13 ___/13
C — Inverse Trig & Hyperbolic C1, C2, C3, C4 19 ___/19
TOTAL 44 ___/44

Proficiency Benchmarks

  • 32/44 (73%) — Proficient. You can handle standard exam problems.
  • 37/44 (85%) — Solid. Fast and accurate.
  • 41/44 (93%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <90 min: Excellent mechanical fluency.
  • 90–120 min: Good. Review missed patterns.
  • >120 min: Drill the specific sets you scored lowest on again tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Reason
e.g. 14 sign error
e.g. 37 missing chain rule

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Set A1 — Standard Expansions

  1. $e^{3x} = 1 + 3x + \dfrac{9x^2}{2!} + \dfrac{27x^3}{3!} + \dfrac{81x^4}{4!} + \dfrac{243x^5}{5!} + \cdots = 1 + 3x + \dfrac{9x^2}{2} + \dfrac{9x^3}{2} + \dfrac{27x^4}{8} + \dfrac{81x^5}{40} + \cdots$

  2. $\sin(2x) = 2x - \dfrac{8x^3}{3!} + \dfrac{32x^5}{5!} - \cdots = 2x - \dfrac{4x^3}{3} + \dfrac{4x^5}{15} - \cdots$

  3. $\cos\left(\dfrac{x}{2}\right) = 1 - \dfrac{x^2}{2! \cdot 4} + \dfrac{x^4}{4! \cdot 16} - \cdots = 1 - \dfrac{x^2}{8} + \dfrac{x^4}{384} - \cdots$

  4. $\ln(1+2x) = 2x - \dfrac{(2x)^2}{2} + \dfrac{(2x)^3}{3} - \dfrac{(2x)^4}{4} + \dfrac{(2x)^5}{5} - \cdots = 2x - 2x^2 + \dfrac{8x^3}{3} - 4x^4 + \dfrac{32x^5}{5} - \cdots$

  5. $(1-x)^{-1} = 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots$

Set A2 — $\tan x$ Expansion

  1. $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=2$, $f^{(4)}(0)=0$, $f^{(5)}(0)=16$ $$\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \cdots$$

  2. $\dfrac{x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots}{1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots} = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$ ✓

  3. $\tan(0.3) \approx 0.3 + \frac{0.027}{3} + \frac{2(0.00243)}{15} = 0.3 + 0.009 + 0.000324 = 0.3093$ Exact: $\tan(0.3) \approx 0.3093$ (matches to 4 d.p.)

Set A3 — Substitution & Multiplication

  1. $e^{-2x^2} = 1 - 2x^2 + \dfrac{4x^4}{2!} - \cdots = 1 - 2x^2 + 2x^4 - \cdots$

  2. $\sin(x^2) = x^2 - \dfrac{x^6}{3!} + \dfrac{x^{10}}{5!} - \cdots = x^2 - \dfrac{x^6}{6} + \dfrac{x^{10}}{120} - \cdots$

  3. $x^2\cos x = x^2\left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots\right) = x^2 - \dfrac{x^4}{2} + \dfrac{x^6}{24} - \cdots$

  4. $\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \cdots = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots$ $$\displaystyle\int_0^{0.5} \frac{\sin x}{x},dx \approx \left[x - \frac{x^3}{18} + \frac{x^5}{600}\right]_0^{0.5} = 0.5 - \frac{0.125}{18} + \frac{0.03125}{600} \approx 0.5 - 0.00694 + 0.000052 = 0.4931$$

Set B1 — De Moivre's Theorem

  1. $z = 1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$ $$z^6 = (\sqrt{2})^6\left(\cos\frac{6\pi}{4} + i\sin\frac{6\pi}{4}\right) = 8\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = 8(0 - i) = -8i$$

  2. $z = \sqrt{3} - i$: $r = \sqrt{3+1}=2$, $\theta = -\frac{\pi}{6}$ $$z^4 = 2^4\left(\cos\left(-\frac{4\pi}{6}\right) + i\sin\left(-\frac{4\pi}{6}\right)\right) = 16\left(\cos\frac{2\pi}{3} - i\sin\frac{2\pi}{3}\right) = 16\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -8 - 8\sqrt{3},i$$

  3. $(\cos\theta + i\sin\theta)^3 = \cos(3\theta) + i\sin(3\theta)$ Expanding LHS: $\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$ Equating real parts: $\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta$

  4. $(\cos\theta + i\sin\theta)^4 = \cos(4\theta) + i\sin(4\theta)$ Expanding and equating imaginary parts: $$\sin(4\theta) = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta = 4\sin\theta\cos\theta(\cos^2\theta - \sin^2\theta) = 4\sin\theta\cos\theta\cos(2\theta)$$

  5. $(-1)^{1/4}$: $z = -1 = 1(\cos\pi + i\sin\pi)$ Roots: $\cos\left(\frac{\pi + 2\pi k}{4}\right) + i\sin\left(\frac{\pi + 2\pi k}{4}\right)$, $k=0,1,2,3$ $k=0$: $\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$, $k=1$: $-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$, $k=2$: $-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$, $k=3$: $\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$

Set B2 — n-th Roots

  1. $z^{1/3} = 2\left[\cos\left(\frac{\pi/3 + 2\pi k}{3}\right) + i\sin\left(\frac{\pi/3 + 2\pi k}{3}\right)\right]$ $k=0$: $2\left(\cos\frac{\pi}{9} + i\sin\frac{\pi}{9}\right)$ $k=1$: $2\left(\cos\frac{7\pi}{9} + i\sin\frac{7\pi}{9}\right)$ $k=2$: $2\left(\cos\frac{13\pi}{9} + i\sin\frac{13\pi}{9}\right)$ Plot: three points equally spaced on a circle of radius 2, separated by $120^\circ$.

  2. $z = -16 = 16(\cos\pi + i\sin\pi)$ $$z^{1/4} = 2\left[\cos\left(\frac{\pi + 2\pi k}{4}\right) + i\sin\left(\frac{\pi + 2\pi k}{4}\right)\right]$$ $k=0$: $\sqrt{2} + i\sqrt{2}$, $k=1$: $-\sqrt{2} + i\sqrt{2}$, $k=2$: $-\sqrt{2} - i\sqrt{2}$, $k=3$: $\sqrt{2} - i\sqrt{2}$

  3. $z^3 = 1$: $z = \cos\left(\frac{2\pi k}{3}\right) + i\sin\left(\frac{2\pi k}{3}\right)$ $k=0$: $1$, $k=1$: $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$, $k=2$: $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$ These form an equilateral triangle centred at the origin.

  4. $z^{1/5} = 2e^{i(\pi/2 + 2\pi k)/5}$ $k=0$: $2e^{i\pi/10}$, $k=1$: $2e^{i\pi/2}$, $k=2$: $2e^{i9\pi/10}$, $k=3$: $2e^{i13\pi/10}$, $k=4$: $2e^{i17\pi/10}$ Five equally spaced points on a circle of radius 2.

Set B3 — Loci & Complex Equations

  1. $|z - 2i| = 3$: Circle centred at $(0, 2)$ with radius 3. Cartesian: $x^2 + (y-2)^2 = 9$

  2. $|z - 1| = |z + i|$: Perpendicular bisector of the segment joining $(1, 0)$ and $(0, -1)$. Cartesian: $(x-1)^2 + y^2 = x^2 + (y+1)^2 \Rightarrow -2x + 1 = 2y + 1 \Rightarrow y = -x$ This is the line $y = -x$.

  3. $\arg(z-1) = \pi/4$: Half-line (ray) starting at $(1, 0)$ at an angle of $45^\circ$ to the positive real axis.

  4. $|z| \leq 2$ and $0 \leq \arg(z) \leq \pi/3$: A sector of a disc — all points inside or on the circle of radius 2 centred at the origin, with argument between $0$ and $60^\circ$ inclusive.

Set C1 — Derivatives

  1. $\frac{dy}{dx} = \frac{5}{\sqrt{1-25x^2}}$

  2. $\frac{dy}{dx} = \frac{2e^{2x}}{1+e^{4x}}$

  3. $\frac{dy}{dx} = 8x\cosh(4x^2 + 1)$

  4. $\frac{dy}{dx} = \frac{3}{\sqrt{9x^2-1}}$, domain $x > \frac{1}{3}$

  5. $\frac{dy}{dx} = \frac{\cos x}{1-\sin^2 x} = \frac{\cos x}{\cos^2 x} = \sec x$

Set C2 — Integrals

  1. $\int \frac{dx}{\sqrt{16+x^2}} = \sinh^{-1}\left(\frac{x}{4}\right) + C = \ln\left(\frac{x+\sqrt{x^2+16}}{4}\right) + C$

  2. $\int \frac{dx}{\sqrt{x^2-25}} = \cosh^{-1}\left(\frac{x}{5}\right) + C = \ln\left(x+\sqrt{x^2-25}\right) + C$ (for $x>5$)

  3. $\int \frac{dx}{9-x^2} = \frac{1}{3}\tanh^{-1}\left(\frac{x}{3}\right) + C$ (for $|x|<3$)

  4. $\int \sinh(5x),dx = \frac{1}{5}\cosh(5x) + C$

  5. $\int_0^1 \cosh(3x),dx = \left[\frac{1}{3}\sinh(3x)\right]_0^1 = \frac{1}{3}\sinh(3)$

Set C3 — Identity Verification

  1. $\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2$ $= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1$ ✓

  2. $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}$ $2\sinh x\cosh x = 2\cdot\frac{e^x - e^{-x}}{2}\cdot\frac{e^x + e^{-x}}{2} = \frac{e^{2x} - e^{-2x}}{2} = \sinh(2x)$ ✓

  3. $\cosh x + \sinh x = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} = \frac{2e^x}{2} = e^x$ ✓

  4. $\cosh(x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2}$ $$\cosh x\cosh y + \sinh x\sinh y = \frac{e^x+e^{-x}}{2}\cdot\frac{e^y+e^{-y}}{2} + \frac{e^x-e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2}$$ $$= \frac{e^{x+y} + e^{x-y} + e^{-x+y} + e^{-x-y}}{4} + \frac{e^{x+y} - e^{x-y} - e^{-x+y} + e^{-x-y}}{4}$$ $= \frac{2e^{x+y} + 2e^{-(x+y)}}{4} = \frac{e^{x+y} + e^{-(x+y)}}{2} = \cosh(x+y)$ ✓

  5. $\frac{d}{dx}\sinh^{-1}x = \frac{d}{dx}\ln(x + \sqrt{x^2+1}) = \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}} = \frac{\frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}$ ✓

Set C4 — Rapid Mixed Fire

  1. $\frac{dy}{dx} = \frac{\text{sech}^2 x}{\sqrt{1 - \tanh^2 x}} = \frac{\text{sech}^2 x}{\sqrt{\text{sech}^2 x}} = \frac{\text{sech}^2 x}{\text{sech } x} = \text{sech } x$ (for all $x$)

  2. Let $u = e^x$, $du = e^x,dx$: $$\int \frac{e^x}{\sqrt{e^{2x} + 9}},dx = \int \frac{du}{\sqrt{u^2 + 9}} = \sinh^{-1}\left(\frac{u}{3}\right) + C = \sinh^{-1}\left(\frac{e^x}{3}\right) + C$$

  3. Let $z = x + yi$. $(x+yi)^2 = x^2 - y^2 + 2xyi = -5 + 12i$ Equating: $x^2 - y^2 = -5$ and $2xy = 12 \Rightarrow y = \frac{6}{x}$ Substitute: $x^2 - \frac{36}{x^2} = -5 \Rightarrow x^4 + 5x^2 - 36 = 0 \Rightarrow (x^2 + 9)(x^2 - 4) = 0$, so $x^2 = 4$, $x = \pm 2$ Solutions: $z = 2 + 3i$ or $z = -2 - 3i$

  4. $e^x\cos x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\right)\left(1 - \frac{x^2}{2} + \cdots\right)$ $$= 1 + x + \frac{x^2}{2} - \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^3}{2} + \cdots$$ $$= 1 + x + \left(\frac{1}{2} - \frac{1}{2}\right)x^2 + \left(\frac{1}{6} - \frac{1}{2}\right)x^3 + \cdots$$ $$= 1 + x - \frac{x^3}{3} + \cdots$$

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