Taylor series — derivation from power series: Worked examples

This is a drill page. The parent note proved the one formula that runs everything here:

Before the notation runs off, three reminders in plain words:

$f^{(n)} (a)$ means "take the derivative $n$ times, then set $x$ equal to the centre $a$ " — a single number.
$n!$ (read " $n$ factorial") means $n \times (n - 1) \times \dots \times 2 \times 1$ , with the convention $0! = 1$ and $1! = 1$ . It is the number the $n$ differentiations pile up.
$(x - a)$ is the displacement from the centre: how far you have walked away from the safe point $a$ .

The scenario matrix

Every Taylor problem you will meet falls into one of these cells. The examples below are labelled with the cell they cover, and together they hit every row.

Cell	What makes it different	Example
A. Centre $a = 0$ , all derivatives equal	derivatives repeat forever	Ex 1 ( $e^{x}$ variant $e^{2 x}$ )
B. Centre $a = 0$ , derivatives cycle with signs	only some powers survive	Ex 2 ( $cos x$ built, sign bookkeeping)
C. Centre $a \neq = 0$	powers of $(x - a)$ , not $x$	Ex 3 ( $x$ about $a = 4$ )
D. Polynomial input (degenerate: series terminates)	finite, exact, remainder $= 0$	Ex 4
E. Negative / all-sign coefficients	geometric-type, sign of $(x - a)$ matters	Ex 5 ( $\frac{1}{1 - x}$ and $\frac{1}{1 + x}$ )
F. Limiting behaviour / convergence edge	what happens at the boundary and outside	Ex 6
G. Real-world word problem	build, truncate, estimate error size	Ex 7 (pendulum $sin θ \approx θ$ )
H. Exam twist: reuse a known series	substitute instead of differentiating	Ex 8 ( $e^{- x^{2}}$ from $e^{x}$ )

Cell A — derivatives that repeat

Worked example Ex 1. Maclaurin series of

$e^{2 x}$ (centre $a = 0$ ) Forecast: You know $e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots$ . Guess: does the " $2$ " land inside each power, or multiply out front? Write your guess before reading.

Step 1 — differentiate. $f (x) = e^{2 x}$ , so $f^{'} (x) = 2 e^{2 x}$ , $f^{''} (x) = 4 e^{2 x} = 2^{2} e^{2 x}$ , and in general $f^{(n)} (x) = 2^{n} e^{2 x}$ . Why this step? The recipe needs $f^{(n)}$ . The chain rule drops a factor $2$ each time, so after $n$ derivatives we carry $2^{n}$ .

Step 2 — plug in the centre $a = 0$ . $f^{(n)} (0) = 2^{n} e^{0} = 2^{n}$ . Why this step? Coefficients use the derivative evaluated at the centre, and $e^{0} = 1$ .

Step 3 — divide by $n!$ . $c_{n} = \frac{2 ^{n}}{n !} \Rightarrow e^{2 x} = \sum_{n = 0}^{\infty} \frac{2 ^{n}}{n !} x^{n} = 1 + 2 x + 2 x^{2} + \frac{4 x ^{3}}{3} + \dots$ Why this step? That is literally $c_{n} = f^{(n)} (a) / n!$ .

Verify: The whole series equals $e^{x}$ with $x$ replaced by $2 x$ : $\sum (2 x)^{n} / n!$ . So the " $2$ " lives inside every power as $(2 x)^{n}$ — matching the shortcut of Ex 8. First few terms: $1 + 2 x + \frac{4 x ^{2}}{2} + \frac{8 x ^{3}}{6} = 1 + 2 x + 2 x^{2} + \frac{4 x ^{3}}{3}$ . ✓

Cell B — cycling derivatives and sign bookkeeping

Worked example Ex 2. Maclaurin series of

$cos x$ — carefully tracking which terms die Forecast: Which powers of $x$ appear — even, odd, or all? And what is the sign of the $x^{4}$ term?

Step 1 — list the derivative cycle. $cos x \to - sin x \to - cos x \to sin x \to cos x$ (period 4). Why this step? Trig derivatives cycle, so instead of computing forever we spot the repeat.

Step 2 — evaluate each at $a = 0$ . $f (0) = 1, f^{'} (0) = 0, f^{''} (0) = - 1, f^{'''} (0) = 0, f^{(4)} (0) = 1, \dots$ Why this step? $sin 0 = 0$ kills every odd-order derivative; the even ones alternate $+ 1, - 1$ .

Step 3 — build coefficients. Odd $n$ : $c_{n} = 0$ . Even $n = 2 k$ : $c_{2 k} = \frac{( - 1 ) ^{k}}{( 2 k )!}$ . $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \frac{x ^{6}}{6 !} + \dots$ Why this step? Only even derivatives survive, so only even powers appear; the alternating $\pm 1$ gives $(- 1)^{k}$ .

Verify: The $x^{4}$ term is $+ \frac{1}{4 !} = + \frac{1}{24}$ (positive, as predicted by $(- 1)^{2} = + 1$ ). Differentiate term-by-term: $\frac{d}{d x} cos x = - sin x$ , and indeed $\frac{d}{d x} (1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}) = - x + \frac{x ^{3}}{6} = - (x - \frac{x ^{3}}{6}) = - sin x$ . ✓ See Maclaurin series — common expansions.

Cell C — a centre that is not zero

Here the picture matters: we approximate $x$ near the safe point $x = 4$ (where the answer is the clean number $2$ ).

Figure — Taylor series — derivation from power series

Worked example Ex 3. Taylor series of

$x$ about $a = 4$ , first three terms; estimate $4.2$ Forecast: Look at the figure. The red curve is $x$ . Will the parabola (2 terms + curvature) sit above or below the curve just right of $x = 4$ ? Guess the sign of the $(x - 4)^{2}$ coefficient.

Step 1 — differentiate. $f (x) = x^{1/2}$ , $f^{'} (x) = \frac{1}{2} x^{- 1/2}$ , $f^{''} (x) = - \frac{1}{4} x^{- 3/2}$ . Why this step? We need $f, f^{'}, f^{''}$ at the centre for the first three coefficients.

Step 2 — plug in $a = 4$ . $4 = 2$ , $4^{- 1/2} = \frac{1}{2}$ , $4^{- 3/2} = \frac{1}{8}$ . $f (4) = 2, f^{'} (4) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}, f^{''} (4) = - \frac{1}{4} \cdot \frac{1}{8} = - \frac{1}{32}$ Why this step? Coefficients use derivatives at the centre $a = 4$ , not at $0$ .

Step 3 — divide by $n!$ and use powers of $(x - 4)$ . $x \approx 2 + \frac{1}{4} (x - 4) - \frac{1}{32 \cdot 2 !} (x - 4)^{2} = 2 + \frac{1}{4} (x - 4) - \frac{1}{64} (x - 4)^{2}$ Why this step? The basis is $(x - a)^{n} = (x - 4)^{n}$ — not powers of $x$ (that mistake is warned about in the parent note).

Step 4 — estimate $4.2$ . Here $x - 4 = 0.2$ : $2 + \frac{1}{4} (0.2) - \frac{1}{64} (0.2)^{2} = 2 + 0.05 - 0.000625 = 2.049375$ Why this step? Plugging the displacement into the truncated polynomial is linear-plus-curvature approximation (Linear approximation & differentials is just the first two terms).

Verify: True value $4.2 = 2.049390 \dots$ ; our estimate $2.049375$ is off by $\approx 1.5 \times 1 0^{- 5}$ — excellent this close to the centre. The $(x - 4)^{2}$ coefficient is negative, so the parabola bends below the curve — matching the concave-down red curve in the figure. ✓

Cell D — degenerate: a polynomial (the series stops)

Worked example Ex 4. Taylor series of

$p (x) = x^{3} - 2 x + 1$ about $a = 1$ Forecast: $p$ is already a polynomial. How many terms will its Taylor series have — finite or infinite?

Step 1 — differentiate down to zero. $p^{'} (x) = 3 x^{2} - 2$ , $p^{''} (x) = 6 x$ , $p^{'''} (x) = 6$ , $p^{(4)} (x) = 0$ . Why this step? A degree-3 polynomial has a zero 4th derivative — this is the degenerate case where the series must terminate.

Step 2 — evaluate at $a = 1$ . $p (1) = 0$ , $p^{'} (1) = 1$ , $p^{''} (1) = 6$ , $p^{'''} (1) = 6$ . Why this step? Same recipe, centre $a = 1$ .

Step 3 — assemble. $p (x) = 0 + 1 (x - 1) + \frac{6}{2 !} (x - 1)^{2} + \frac{6}{3 !} (x - 1)^{3} = (x - 1) + 3 (x - 1)^{2} + (x - 1)^{3}$ Why this step? Divide each $p^{(n)} (1)$ by $n!$ ; every term $n \geq 4$ is $0$ .

Verify: Expand: $(x - 1) + 3 (x - 1)^{2} + (x - 1)^{3}$ . Let $u = x - 1$ : $u + 3 u^{2} + u^{3}$ ; substitute back and collect — must reproduce $x^{3} - 2 x + 1$ exactly (remainder is genuinely $0$ here). Sympy confirms below. This is the one case where "series $= f$ everywhere" is literally true. ✓

Cell E — negative / all-sign coefficients (geometric flavour)

Worked example Ex 5. Maclaurin series of

$\frac{1}{1 - x}$ and $\frac{1}{1 + x}$ Forecast: Geometric series says $\frac{1}{1 - x} = 1 + x + x^{2} + \dots$ . Guess the series for $\frac{1}{1 + x}$ — what changes?

Step 1 — differentiate $f (x) = (1 - x)^{- 1}$ . $f^{'} (x) = (1 - x)^{- 2}$ , $f^{''} (x) = 2 (1 - x)^{- 3}$ , and generally $f^{(n)} (x) = n! (1 - x)^{- (n + 1)}$ . Why this step? Each derivative drops the exponent as a positive factor (chain rule gives $+ 1$ from $- (1 - x)$ ), building $n!$ .

Step 2 — at $a = 0$ : $f^{(n)} (0) = n!$ , so $c_{n} = \frac{n !}{n !} = 1$ . $\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + \dots$ Why this step? All coefficients are $1$ — this is exactly the geometric series.

Step 3 — replace $x \to - x$ for $\frac{1}{1 + x}$ . $\frac{1}{1 + x} = \sum_{n = 0}^{\infty} (- x)^{n} = 1 - x + x^{2} - x^{3} + \dots$ Why this step? Substitution beats re-differentiating (the Cell H trick); the sign of $(x - a)^{n}$ now alternates.

Verify: At $x = 0.5$ : $\frac{1}{1 + 0.5} = 0.6667$ . Partial sum $1 - 0.5 + 0.25 - 0.125 + 0.0625 = 0.6875$ ; four more terms tighten toward $0.6667$ . Sign pattern $+, -, +, -$ confirmed. ✓

Cell F — limiting behaviour at and beyond the boundary

Worked example Ex 6. Where does

$\frac{1}{1 - x} = \sum x^{n}$ actually hold? Forecast: The formula gave clean coefficients $c_{n} = 1$ . Does that mean it equals $\frac{1}{1 - x}$ for every $x$ ? Guess before reading.

Step 1 — test inside: $x = 0.5$ . $\sum (0.5)^{n} = \frac{1}{1 - 0.5} = 2$ . ✓ Why this step? Inside the radius the terms shrink, the sum converges to the true value.

Step 2 — test the boundary $x = 1$ . Terms are $1 + 1 + 1 + \dots$ — never settles. And $\frac{1}{1 - 1}$ is undefined. Why this step? At the edge the terms do not shrink to $0$ , so the series diverges — the boundary is a genuine wall.

Step 3 — test outside: $x = 2$ . Terms $1, 2, 4, 8, \dots$ blow up, yet $\frac{1}{1 - 2} = - 1$ is a perfectly finite number. Why this step? This is the parent note's third mistake made concrete: the derivation fixes coefficients but does not promise equality outside the radius of convergence.

Verify: Radius of convergence is $1$ . At $x = 2$ the true value $- 1$ is finite but the series is nonsense — the coefficients were "right" and the series still fails. See Power series — radius & interval of convergence and Taylor's theorem with remainder. ✓

Cell G — real-world word problem

Worked example Ex 7. Pendulum: why physicists write

$sin θ \approx θ$ Statement: A pendulum's motion obeys $\ddot{θ} = - \frac{g}{L} sin θ$ . To get simple harmonic motion, engineers replace $sin θ$ by $θ$ . For a swing of $θ = 1 0^{\circ}$ , how big is the error?

Forecast: $1 0^{\circ}$ feels small. Guess the percentage error of using $θ$ instead of $sin θ$ : under $1%$ , a few $%$ , or over $10%$ ?

Step 1 — write the series (radians!). From Ex 2's sibling: $sin θ = θ - \frac{θ ^{3}}{6} + \dots$ . Why this step? The first-order truncation (Linear approximation & differentials) is exactly $sin θ \approx θ$ ; the next term $- \frac{θ ^{3}}{6}$ is the leading error.

Step 2 — convert and evaluate. $1 0^{\circ} = 0.17453$ rad. True $sin (0.17453) = 0.17365$ . Why this step? Trig series are valid only with $θ$ in radians; degrees would corrupt every power.

Step 3 — relative error. $\frac{θ - s i n θ}{s i n θ} = \frac{0.17453 - 0.17365}{0.17365} = 0.00509 \approx 0.51%$ Why this step? The dropped term $\frac{θ ^{3}}{6}$ scales as $θ^{3}$ , so small angles give tiny error — that is why the approximation is trusted.

Verify: Error $\approx 0.51%$ — under $1%$ , so a $1 0^{\circ}$ swing is safely "simple harmonic". Units: both quantities dimensionless (radians), ratio is pure percent. ✓

Cell H — exam twist: reuse, don't re-derive

Worked example Ex 8. Maclaurin series of

$e^{- x^{2}}$ (no differentiation!) Forecast: Differentiating $e^{- x^{2}}$ repeatedly is a nightmare. Is there a shortcut? Guess how.

Step 1 — start from the known series. $e^{u} = \sum_{n = 0}^{\infty} \frac{u ^{n}}{n !} = 1 + u + \frac{u ^{2}}{2} + \frac{u ^{3}}{6} + \dots$ . Why this step? The recipe is legitimate but slow; substitution into a known series gives the same unique coefficients (uniqueness is guaranteed by the parent derivation).

Step 2 — substitute $u = - x^{2}$ . $e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{( - x ^{2} ) ^{n}}{n !} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} x ^{2 n}}{n !} = 1 - x^{2} + \frac{x ^{4}}{2} - \frac{x ^{6}}{6} + \dots$ Why this step? $(- x^{2})^{n} = (- 1)^{n} x^{2 n}$ — only even powers, alternating signs, factorials from the original.

Step 3 — sanity on one coefficient. Coefficient of $x^{4}$ is $\frac{( - 1 ) ^{2}}{2 !} = \frac{1}{2}$ . Why this step? A quick check that substitution kept the factorials intact.

Verify: Direct differentiation gives $f^{(4)} (0) /4!$ . Computing $f^{(4)} (0) = 12$ , so the $x^{4}$ coefficient is $12/24 = \frac{1}{2}$ — matches the substitution. Substitution wins. ✓

Recall Which cell is which? (self-test)

A polynomial's Taylor series is ::: finite — it terminates (remainder $0$ ), Cell D. For $x$ about $a = 4$ the powers are ::: powers of $(x - 4)$ , Cell C. Whether $\sum x^{n} = \frac{1}{1 - x}$ at $x = 2$ ::: no — outside the radius the series diverges though the value is finite, Cell F. Fastest route to $e^{- x^{2}}$ 's series ::: substitute $u = - x^{2}$ into $e^{u}$ , Cell H.

Connections

Taylor series — derivation from power series (index 4.3.16) (the recipe every example runs)
Maclaurin series — common expansions (Ex 1, 2, 5, 8 all live here)
Linear approximation & differentials (Ex 3, 7 use the first-order truncation)
Power series — radius & interval of convergence (Ex 6's wall at $x = 1$ )
Taylor's theorem with remainder (the error size in Ex 3, 7)
Geometric series (Ex 5's all-ones coefficients)
L'Hôpital's rule (leading Taylor terms often replace it)