Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all: Worked examples

The scenario matrix

Each example below is tagged with the cell it fills. The goal: no reader ever meets a situation we skipped.

Cell	What makes it different	Example
A · positive $x$ , fast series	plug small $x > 0$ into $e^{x}$	Ex 1
B · negative $x$ , sign bookkeeping	$e^{- x}$ , watch alternating signs	Ex 2
C · substitution	replace $x$ by $x^{2}$ / $2 x$ inside a known series	Ex 3
D · trig, odd input & parity check	$sin$ of a scaled angle	Ex 4
E · edge of convergence	$ln (1 + x)$ at $x = 1$ (allowed) vs $x = - 1$ (forbidden)	Ex 5
F · non-integer power (binomial)	$(1 + x)^{1/3}$ , an infinite series	Ex 6
G · limit via series (the payoff)	$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$	Ex 7
H · combine two series	multiply $e^{x} sin x$ , collect terms	Ex 8
I · real-world word problem	pendulum small-angle approximation	Ex 9
J · exam twist / degenerate	$lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}$ mixing cells	Ex 10

Ex 1 — Cell A · positive $x$ , fast-shrinking terms

Worked example Estimate

$e^{0.2}$ using terms up to $x^{3}$ . Forecast: guess the answer will be a bit above $1.2$ — write your guess down.

Steps.

Write the series: $e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$ . Why this step? $e^{x}$ has every derivative equal to $1$ at $0$ , so all coefficients are $\frac{1}{n !}$ — no sign or parity worries here.
Put $x = 0.2$ : $1 + 0.2 + \frac{0.04}{2} + \frac{0.008}{6}$ . Why this step? We just substitute; the factorials in the denominator make later terms tiny, so truncating is safe.
Add: $1 + 0.2 + 0.02 + 0.0013 \overline{3} = 1.2213 \overline{3}$ . Why this step? We sum the four numbers to collapse the truncated series into a single estimate; the running total stabilises fast because each new term is roughly $10 \times$ smaller than the last, so four terms already fix three decimals.

Verify: true $e^{0.2} = 1.221402 \dots$ ; our error is about $0.00007$ — smaller than the first dropped term $\frac{0. 2 ^{4}}{24} \approx 0.0000 \overline{6}$ . ✓

Ex 2 — Cell B · negative $x$ , sign bookkeeping

Ex 3 — Cell C · substitution ( $x \to x^{2}$ )

Ex 4 — Cell D · trig, scaled angle & parity check

Worked example Series for

$sin (3 x)$ to three nonzero terms; estimate $sin (0.3)$ from it (so $3 x = 0.3$ , $x = 0.1$ ). Forecast: only odd powers survive (sin is odd); guess $sin (0.3) \approx 0.2955$ .

Steps.

$sin u = u - \frac{u ^{3}}{3 !} + \frac{u ^{5}}{5 !} - \dots$ . Why this step? Derivatives of $sin$ cycle $0, 1, 0, - 1$ at $0$ — killing every even power.
Substitute $u = 3 x$ : $sin (3 x) = 3 x - \frac{( 3 x ) ^{3}}{6} + \frac{( 3 x ) ^{5}}{120} - \dots = 3 x - \frac{27 x ^{3}}{6} + \frac{243 x ^{5}}{120} - \dots$ . Why this step? $(3 x)^{3} = 27 x^{3}$ ; the scale factor gets amplified by the power — a common exam trap.
Tidy: $sin (3 x) = 3 x - 4.5 x^{3} + 2.025 x^{5} - \dots$ . Why this step? We divide out the factorials to get clean decimal coefficients: $\frac{27}{6} = 4.5$ and $\frac{243}{120} = 2.025$ . Simplifying now makes the numeric substitution in the next step arithmetic-free of fractions.
Put $x = 0.1$ : $0.3 - 4.5 (0.001) + 2.025 (0.00001) = 0.3 - 0.0045 + 0.00002025 = 0.29552$ .

Verify: true $sin (0.3) = 0.295520 \dots$ . ✓ Parity check: only odd powers of $x$ appear, as an odd function demands. ✓

The figure below plots the true $sin (3 x)$ (teal) against our odd-power polynomial (dashed orange): the two curves are indistinguishable near $x = 0$ and only peel apart out toward $∣ x ∣ \approx 1$ , which is exactly why truncating at $x^{5}$ is safe for the small input $x = 0.1$ (the plum dot). It also makes the parity visible — the polynomial is antisymmetric about the origin, mirroring the odd function.

Figure — Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all

Ex 5 — Cell E · the edge of convergence

Worked example Evaluate

$ln 2$ from the $ln (1 + x)$ series (set $x = 1$ ). Then explain why $x = - 1$ is forbidden. Forecast: guess $ln 2 \approx 0.69$ ; guess $x = - 1$ blows up.

Steps.

$ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ , valid for $- 1 < x \leq 1$ . Why this step? This came from integrating the geometric series $\frac{1}{1 + x} = 1 - x + x^{2} - \dots$ , which only converges for $∣ x ∣ < 1$ ; the right endpoint $x = 1$ survives (alternating series), the left one does not.
Put $x = 1$ : $ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ (the alternating harmonic series). Why this step? $x = 1$ is included ( $x \leq 1$ ), and alternating shrinking terms converge.
Ten terms give $1 - 0.5 + 0.3 \overline{3} - 0.25 + 0.2 - 0.1 \overline{6} + 0.142857 - 0.125 + 0.1 \overline{1} - 0.1 = 0.645635$ . Why this step? This series converges slowly — expect only ~1 digit right, and error below the next term $\frac{1}{11} \approx 0.091$ .
Left endpoint $x = - 1$ : the series becomes $- 1 - \frac{1}{2} - \frac{1}{3} - \dots$ , the negative harmonic series, which diverges — matching $ln (1 + (- 1)) = ln 0 = - \infty$ . Why this step? The interval of convergence mirrors the function's own domain trouble.

Verify: true $ln 2 = 0.693147 \dots$ ; our 10-term partial sum $0.645635$ is within one term's size of it (slow convergence, but on the right track). ✓

Ex 6 — Cell F · non-integer power (binomial series)

Ex 7 — Cell G · limit via series (the classic payoff)

Worked example Evaluate

$x \to 0 lim \frac{1 - cos x}{x ^{2}}$ . Forecast: at $x = 0$ this is $\frac{0}{0}$ — indeterminate. Guess the answer is $\frac{1}{2}$ .

Steps.

Replace $cos x$ by its series: $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ . Why this step? Series turn a mysterious $\frac{0}{0}$ into honest algebra — this is the series route to limits.
Then $1 - cos x = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots$ . Why this step? The $1$ 's cancel; the leading surviving term is $\frac{x ^{2}}{2}$ , which matches the $x^{2}$ downstairs.
Divide by $x^{2}$ : $\frac{1 - cos x}{x ^{2}} = \frac{1}{2} - \frac{x ^{2}}{24} + \dots$ . Why this step? Cancelling $x^{2}$ removes the $\frac{0}{0}$ — the whole point.
Let $x \to 0$ : every remaining term with $x$ dies, leaving $\frac{1}{2}$ . Why this step? Once the indeterminacy is gone the expression is an ordinary polynomial, so the limit is just its value at $0$ — the constant term $\frac{1}{2}$ , since $- \frac{x ^{2}}{24} \to 0$ .

Verify: L'Hôpital twice gives $lim \frac{s i n x}{2 x} = lim \frac{c o s x}{2} = \frac{1}{2}$ . ✓ Numeric: at $x = 0.01$ , $\frac{1 - c o s 0.01}{0.0001} = 0.499996$ . ✓

Ex 8 — Cell H · multiplying two series

Ex 9 — Cell I · real-world word problem (pendulum)

Worked example A pendulum's true period involves

$sin θ$ ; the "small-angle" model replaces $sin θ$ by $θ$ . For $θ = 1 0^{\circ}$ , what is the relative error of that swap? Forecast: $1 0^{\circ}$ is small — guess the error is a fraction of a percent.

Steps.

Convert: $1 0^{\circ} = \frac{10 π}{180} = 0.174533$ rad. Why this step? Series for $sin$ require the angle in radians — its derivative rules assume radians.
Series: $sin θ = θ - \frac{θ ^{3}}{6} + \dots$ , so the small-angle model drops $\frac{θ ^{3}}{6}$ . Why this step? Replacing $sin θ$ by $θ$ literally throws away the $- \frac{θ ^{3}}{6}$ term and beyond.
Relative error $\approx \frac{θ - sin θ}{sin θ} \approx \frac{θ ^{3} /6}{θ} = \frac{θ ^{2}}{6}$ . Why this step? The leading error over the leading value gives a clean $\frac{θ ^{2}}{6}$ rule of thumb.
Plug in: $\frac{0.17453 3 ^{2}}{6} = \frac{0.030462}{6} = 0.005077$ , i.e. about $0.51%$ .

Verify: direct: $sin (0.174533) = 0.173648$ ; $\frac{0.174533 - 0.173648}{0.173648} = 0.005097$ , i.e. $\approx 0.51%$ . ✓ The $\frac{θ ^{2}}{6}$ estimate nails it — small angle, tiny error.

Ex 10 — Cell J · exam twist / mixed indeterminate

Recall Which cell was which? (self-test)

Positive small- $x$ estimate ::: Ex 1 (Cell A, $e^{0.2}$ ) Alternating signs from $e^{- x}$ ::: Ex 2 (Cell B) Substituting $x^{2}$ inside cos ::: Ex 3 (Cell C) Scaled trig angle & parity ::: Ex 4 (Cell D) Endpoint $x = 1$ works, $x = - 1$ fails ::: Ex 5 (Cell E) Non-integer power infinite series ::: Ex 6 (Cell F) $\frac{1 - c o s x}{x ^{2}} \to \frac{1}{2}$ ::: Ex 7 (Cell G) Multiplying $e^{x} sin x$ ::: Ex 8 (Cell H) Pendulum small-angle error ::: Ex 9 (Cell I) $\frac{e ^{x} - 1 - x}{x ^{2}} \to \frac{1}{2}$ ::: Ex 10 (Cell J)

Connections

Parent topic (Hinglish) — the derivations these examples exercise
L'Hôpital & limits via series — Ex 7 & Ex 10 are the series route
Radius & interval of convergence — Ex 5's endpoint drama
Binomial theorem (integer n) — Ex 6 with non-integer $n$
Geometric series — the engine behind $ln (1 + x)$ in Ex 5
Euler's formula — cousin of Ex 8's series-multiplication idea