Taylor series approximation: Worked examples — AI-ML deep dive

Recall Quick reminder: what does

$n!$ (factorial) mean? $n!$ means "multiply all whole numbers from $1$ up to $n$ ": $3! = 1 \cdot 2 \cdot 3 = 6$ , $4! = 24$ , $5! = 120$ . By convention $0! = 1$ . It shows up in every Taylor coefficient because differentiating $(x - a)^{n}$ exactly $n$ times spits out $n \cdot (n - 1) \dots 1 = n!$ , and we divide by it to cancel that factor. $4! =$ ::: $24$ $0! =$ ::: $1$ (by convention)

The scenario matrix

Every Taylor problem lives in one of these case classes. Each row is a distinct thing that can go right or wrong; the last column says which worked example nails it.

#	Case class	What makes it special	Example
C1	Expand at $a = 0$ , converges everywhere	all derivatives finite, series = function for all $x$	Ex 1 ( $cos x$ )
C2	Expand at $a \neq = 0$	must re-anchor derivatives at the new point	Ex 2 ( $ln x$ at $a = 1$ )
C3	Inside radius of convergence	series truly rebuilds $f$	Ex 3 (geometric, $x = 0.5$ )
C4	Outside radius — series diverges	function is fine, series explodes	Ex 3 (same, $x = 2$ )
C5	Degenerate / pathological	all derivatives $= 0$ but $f \neq = 0$	Ex 4 ( $e^{- 1/ x^{2}}$ )
C6	Error estimate (remainder bound)	how many terms for target accuracy	Ex 5 ( $e^{x}$ to 3 decimals)
C7	Real-world word problem	physical quantity + units	Ex 6 (pendulum period)
C8	Multivariable / ML core	gradient + Hessian, matrix step	Ex 7 (Newton step on a quadratic)
C9	Exam-style twist	reuse a known series cleverly instead of grinding derivatives	Ex 8 ( $lim$ via series)

Prerequisites you may want open: Derivatives and gradients, Approximation error and Big-O, Hessian matrix, Convex functions.

Example 1 — C1: a function that converges everywhere

Worked example Maclaurin series of

$cos x$ , and estimate $cos (0.3)$ Forecast: guess — will the series have only even powers, only odd, or both? And is $cos (0.3)$ closer to $0.955$ or $0.85$ ?

Step 1 — List the derivative cycle at $a = 0$ . $cos, - sin, - cos, sin, cos, \dots$ evaluated at $0$ give $1, 0, - 1, 0, 1, \dots$ Why this step? The Taylor coefficient is $f^{(n)} (0) / n!$ , so we literally need each derivative's value at $0$ — nothing else feeds the formula.

Step 2 — Kill the zero terms. Every odd derivative is $\pm sin 0 = 0$ , so only even powers survive: $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \frac{x ^{6}}{6 !} + \dots$ Why this step? A term with a zero coefficient contributes nothing; dropping them shows why $cos$ is built from even powers only (it's an even function — $cos (- x) = cos x$ ).

Step 3 — Plug in $x = 0.3$ , first two terms. $1 - \frac{0. 3 ^{2}}{2} = 1 - \frac{0.09}{2} = 1 - 0.045 = 0.955.$ Why this step? Near $0$ , $x^{4} /24 = 0.0034/24 \approx 0.00003$ is tiny, so two terms already nail 4 decimals.

Verify: true $cos (0.3) = 0.955336...$ ; our $0.955$ matches to 3 decimals. ✓ (This is C1: valid for every $x$ , we just chose a small one.)

Example 2 — C2: expanding away from zero

Worked example Taylor series of

$ln x$ around $a = 1$ , and approximate $ln (1.1)$ Forecast: why can't we use $a = 0$ here? (Look at $ln 0$ .)

Step 1 — Explain the anchor choice. $ln 0 = - \infty$ : the function isn't even defined at $0$ , so a Maclaurin ( $a = 0$ ) series is impossible. We anchor at $a = 1$ where $ln 1 = 0$ is clean. Why this step? Rule: expand near where you'll evaluate, and where $f$ and its derivatives exist. Here we evaluate at $1.1$ , close to $1$ .

Step 2 — Derivatives at $a = 1$ . $f = ln x$ , $f^{'} = 1/ x$ , $f^{''} = - 1/ x^{2}$ , $f^{'''} = 2/ x^{3}$ , $f^{(4)} = - 6/ x^{4}$ . At $x = 1$ : $0, 1, - 1, 2, - 6$ . Why this step? These plug straight into $f^{(n)} (1) / n!$ .

Step 3 — Deduce the general coefficient (don't just guess it). Differentiating $1/ x$ repeatedly gives $f^{(n)} (x) = (- 1)^{n - 1} (n - 1)! x^{- n}$ for $n \geq 1$ , so at $x = 1$ : $f^{(n)} (1) = (- 1)^{n - 1} (n - 1)!$ . The Taylor coefficient is therefore $c_{n} = \frac{f ^{(n)} ( 1 )}{n !} = \frac{( - 1 ) ^{n - 1} ( n - 1 )!}{n !} = \frac{( - 1 ) ^{n - 1}}{n} .$ Why this step? This is where the alternating sign $(- 1)^{n - 1}$ and the $1/ n$ come from — the $(n - 1)!$ from repeated differentiation cancels almost all of the $n!$ , leaving just $1/ n$ . Now the pattern is derived, not inferred. Check: $c_{1} = 1$ , $c_{2} = - \frac{1}{2}$ , $c_{3} = + \frac{1}{3}$ , $c_{4} = - \frac{1}{4}$ — matching Step 2's raw derivatives.

Step 4 — Assemble in powers of $(x - 1)$ . $ln x = (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - \frac{( x - 1 ) ^{4}}{4} + \dots = \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1}}{n} (x - 1)^{n} .$ Why this step? Because we anchored at $1$ , the natural variable is $(x - 1)$ (distance from the anchor), not $x$ .

Step 5 — State the radius of convergence. This series only equals $ln x$ for $∣ x - 1∣ < 1$ , i.e. $0 < x < 2$ . Beyond $x = 2$ the series diverges even though $ln x$ stays finite. Why this step? The nearest "trouble spot" of $ln x$ is $x = 0$ (where $ln$ blows up), a distance $1$ from the anchor $a = 1$ — and that distance is the radius. Naming it tells the reader how far from $1$ they're allowed to trust the formula (our target $1.1$ is comfortably inside).

Step 6 — Evaluate at $x = 1.1$ , so $(x - 1) = 0.1$ . $0.1 - \frac{0.01}{2} + \frac{0.001}{3} = 0.1 - 0.005 + 0.000333 = 0.095333.$ Why this step? With $∣ x - 1∣ = 0.1$ well inside the radius of $1$ , three terms already give strong accuracy; each higher power of $0.1$ shrinks fast.

Verify: true $ln (1.1) = 0.0953102...$ ; our $0.095333$ is right to 4 decimals. ✓

Example 3 — C3 & C4: inside vs outside the radius

This one example shows both the safe and the broken case, because they use the same series.

Worked example Geometric series

$\frac{1}{1 - x} = 1 + x + x^{2} + \dots$ at $x = 0.5$ (inside) and $x = 2$ (outside) Forecast: the function $1/ (1 - x)$ is perfectly finite at $x = 2$ (it equals $- 1$ ). Will its series also give $- 1$ ? Guess yes/no before reading.

Step 1 — Radius of convergence. The ratio of consecutive terms is $x$ ; the sum only settles when $∣ x ∣ < 1$ . So the radius is $1$ — see the red dashed boundary in the figure above, with the safe zone shaded red. Why this step? A geometric sum $\sum x^{n}$ converges iff $∣ x ∣ < 1$ ; outside that the partial sums blow up regardless of what $f$ does.

Step 2 — Inside (C3): $x = 0.5$ . Partial sums: $1, 1.5, 1.75, 1.875, 1.9375, \dots \to 2$ . True value $1/ (1 - 0.5) = 2$ . ✓ The series rebuilds the function — this is the black dot inside the red band in the figure, climbing toward $f$ . Why this step? We pick a point strictly inside the radius to see convergence actually happen: each added term halves the remaining gap to $2$ .

Step 3 — Outside (C4): $x = 2$ . Partial sums: $1, 3, 7, 15, 31, \dots \to + \infty$ . But true value $1/ (1 - 2) = - 1$ . The series diverges to $+ \infty$ while the function sits calmly at $- 1$ (the red arrow in the figure points to this clash). Why this step? This is the headline warning from the parent note made concrete: Taylor is local. Beyond the radius the series says nothing.

Verify: at $x = 0.5$ , 5 terms give $1.9375$ (heading to $2$ ✓); at $x = 2$ , 5 terms give $31$ while $f = - 1$ ✗-by-design. Both confirmed. See Approximation error and Big-O.

Example 4 — C5: the pathological function

Worked example

$f (x) = e^{- 1/ x^{2}}$ with $f (0) = 0$ — a smooth function whose Taylor series is wrong Forecast: if I compute the Maclaurin series of this function, what do I get? (Trick: think about how fast $e^{- 1/ x^{2}} \to 0$ .)

Step 1 — Behaviour near $0$ . As $x \to 0$ , $1/ x^{2} \to + \infty$ , so $e^{- 1/ x^{2}} \to e^{- \infty} = 0$ — and it does so faster than any power of $x$ . Why this step? This "flatter than every polynomial" behaviour is exactly what makes derivatives vanish.

Step 2 — Every derivative at $0$ is $0$ . Each derivative is (polynomial in $1/ x$ ) $\times e^{- 1/ x^{2}}$ ; the exponential crushes the polynomial, giving limit $0$ . So $f^{(n)} (0) = 0$ for all $n$ . Why this step? The Taylor coefficient $f^{(n)} (0) / n!$ is therefore $0$ for every $n$ .

Step 3 — The verdict. $Maclaurin series = 0 + 0 \cdot x + 0 \cdot x^{2} + \dots = 0.$ Yet $f (0.5) = e^{- 4} \approx 0.0183 \neq = 0$ . Why this step? The series says " $f$ is identically zero"; the function disagrees everywhere but $0$ . Smoothness does NOT guarantee series = function (case C5).

Verify: $f (0.5) = e^{- 1/0.25} = e^{- 4} = 0.018316...$ while its series gives $0$ . Mismatch confirmed. ✓

Example 5 — C6: how many terms for a target accuracy?

Worked example Approximate

$e^{0.5}$ to within $1 0^{- 3}$ using the remainder bound Forecast: guess how many terms — 3, 4, or 6?

Step 1 — Recall the Lagrange remainder. $R_{k} = \frac{f ^{(k + 1)} ( ξ )}{( k + 1 )!} x^{k + 1}, ξ \in (0, x) .$ For $e^{x}$ , $f^{(k + 1)} = e^{ξ}$ , and on $[0, 0.5]$ we have $e^{ξ} \leq e^{0.5} < 2$ . Why this step? We bound the unknown $ξ$ by the largest possible value so the error estimate is a guarantee, not a guess. See Approximation error and Big-O.

Step 2 — Bound the error at $x = 0.5$ . $∣ R_{k} ∣ \leq \frac{2 \cdot 0. 5 ^{k + 1}}{( k + 1 )!} .$ Why this step? Plugging the worst-case $e^{ξ} < 2$ turns a mystery into a computable ceiling.

Step 3 — Find smallest $k$ with $∣ R_{k} ∣ < 1 0^{- 3}$ . $k = 3$ : $\frac{2 \cdot 0. 5 ^{4}}{4 !} = \frac{2 \cdot 0.0625}{24} = 0.00521$ — too big. $k = 4$ : $\frac{2 \cdot 0. 5 ^{5}}{5 !} = \frac{2 \cdot 0.03125}{120} = 0.000521$ — under $1 0^{- 3}$ . ✓ So we need terms up to $n = 4$ (order-4 polynomial, 5 terms). Why this step? We test $k$ upward and stop at the first one that clears the target, because using more terms than needed wastes work — the smallest sufficient $k$ is the answer the question asks for.

Step 4 — Compute the estimate. $1 + 0.5 + \frac{0.25}{2} + \frac{0.125}{6} + \frac{0.0625}{24} = 1 + 0.5 + 0.125 + 0.020833 + 0.002604 = 1.648437.$ Why this step? Now that Step 3 guarantees the truncation error is below $1 0^{- 3}$ , we actually sum those 5 terms to produce the number — the bound told us how many, this step delivers the value.

Verify: true $e^{0.5} = 1.648721...$ ; error $= 0.000284 < 1 0^{- 3}$ . ✓ Bound respected.

Example 6 — C7: real-world word problem (with units)

Worked example Pendulum period for a small swing

A pendulum of length $L = 1 m$ swinging with amplitude $θ_{0}$ has period $T = 4 \frac{L}{g} \int_{0}^{π /2} \frac{d ϕ}{1 - s i n ^{2} ( θ _{0} /2 ) s i n ^{2} ϕ},$ where $g = 9.81 m/s^{2}$ is the acceleration due to gravity. The messy integral hides a factor $sin (θ_{0} /2)$ ; physicists approximate using $sin ψ \approx ψ$ for small angles to collapse it to the familiar $T \approx 2 π L / g$ . Justify this small-angle move with Taylor and estimate its error at $θ_{0} = 0.2 rad$ ( $\approx 11. 5^{\circ}$ ), so $ψ = θ_{0} /2 = 0.1 rad$ . Forecast: guess the relative error of $sin ψ \approx ψ$ at $ψ = 0.1$ : about $0.02%$ , $0.2%$ , or $2%$ ?

Step 1 — The relevant series. $sin ψ = ψ - \frac{ψ ^{3}}{6} + \dots$ Here $ψ = θ_{0} /2 = 0.1 rad$ . In the figure above, the black curve is the exact $sin ψ$ and the red dashed line is the approximation $ψ$ . Why this step? The small-angle rule is the first Taylor term; the dropped $ψ^{3} /6$ is the error — the vertical red gap in the figure.

Step 2 — Size the dropped term. $\frac{ψ ^{3}}{6} = \frac{0. 1 ^{3}}{6} = \frac{0.001}{6} = 0.0001 \overline{6} .$ Why this step? This absolute error (in radians) tells us how far the straight red line overshoots the true $sin$ curve at $ψ = 0.1$ .

Step 3 — Relative error. $\frac{ψ ^{3} /6}{ψ} = \frac{ψ ^{2}}{6} = \frac{0.01}{6} = 0.001 \overline{6} = 0.167%.$ Why this step? Relative error is dimensionless — the radians cancel top and bottom — so it directly answers the forecast and tells the engineer the approximation is safe at this amplitude regardless of unit choice.

Verify: $sin (0.1) = 0.0998334...$ , and $ψ = 0.1$ , so relative error $= \frac{0.1 - 0.0998334}{0.0998334} = 0.001668 = 0.167%$ . ✓ Matches the $ψ^{2} /6$ prediction — closest to $0.2%$ . (Units sanity: $L / g$ has units $m / (m/s^{2}) = s^{2} = s$ , so $T$ is in seconds as a period must be.)

Example 7 — C8: multivariable, Newton's step on a quadratic

Worked example One Newton step lands exactly on the minimum of a quadratic

Let $L (θ) = θ_{1}^{2} + 2 θ_{2}^{2} - 2 θ_{1} - 8 θ_{2} + 9$ , starting at $θ_{0} = (0, 0)$ . Write $Δ$ for the step we take — the vector we add to $θ_{0}$ to get the next point, i.e. the change in $θ$ . Forecast: for a pure quadratic, how many Newton steps to reach the minimum — 1, 5, or "depends"?

Step 1 — Gradient at $θ_{0}$ . $\nabla L = (2 θ_{1} - 2, 4 θ_{2} - 8)$ . At $(0, 0)$ : $\nabla L (θ_{0}) = (- 2, - 8)$ . Why this step? The 1st-order Taylor term needs the gradient — the direction of steepest ascent. See Derivatives and gradients.

Step 2 — Hessian (curvature). $H = (2004), H^{- 1} = (\frac{1}{2} 0 0 \frac{1}{4}) .$ Why this step? The 2nd-order Taylor term $\frac{1}{2} Δ^{⊤} H Δ$ — where $Δ$ is the step defined above — is the parabola we fit; $H$ is its curvature. Since $H ⪰ 0$ , $L$ is convex. See Hessian matrix.

Step 3 — Newton step. $Δ = - H^{- 1} \nabla L = - (\frac{1}{2} 0 0 \frac{1}{4}) (- 2 - 8) = (12) .$ New point: $θ_{1} = (0, 0) + (1, 2) = (1, 2)$ . Why this step? Newton minimizes the exact 2nd-order Taylor model. For a genuine quadratic the model is the function, so one step is exact. See Newton's Method.

Verify: minimize directly — $\partial L / \partial θ_{1} = 2 θ_{1} - 2 = 0 \Rightarrow θ_{1} = 1$ ; $\partial L / \partial θ_{2} = 4 θ_{2} - 8 = 0 \Rightarrow θ_{2} = 2$ . Minimum at $(1, 2)$ , exactly where Newton landed. ✓ Answer to forecast: 1 step.

Example 8 — C9: exam twist (compose known series)

Worked example Evaluate

$x \to 0 lim \frac{1 - cos x}{x ^{2}}$ using series Forecast: guess the limit — $0$ , $\frac{1}{2}$ , or $1$ ? (Both top and bottom go to $0$ .)

Step 1 — Substitute the known $cos$ series. From Example 1, $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , so $1 - cos x = 1 - (1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots) = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots$ Why this step? The exam trap is to grind L'Hôpital twice; reusing the Maclaurin series is faster and shows why the answer appears. The two $1$ 's cancel, leaving a series that starts at $x^{2}$ .

Step 2 — Divide by $x^{2}$ . $\frac{1 - c o s x}{x ^{2}} = \frac{\frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots}{x ^{2}} = \frac{1}{2} - \frac{x ^{2}}{24} + \dots$ Why this step? Dividing the leading $x^{2}$ term reveals a constant $\frac{1}{2}$ plus terms that each still carry a positive power of $x$ .

Step 3 — Take $x \to 0$ . As $x \to 0$ every remaining term ( $- x^{2} /24$ , and higher) vanishes, leaving just the constant: $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2} .$ Why this step? Once the expression is a clean power series $\frac{1}{2} - \frac{x ^{2}}{24} + \dots$ , the $0/0$ form is gone — taking the limit is just reading off the constant term, which is the whole payoff of the series trick.

Verify: at $x = 0.01$ , $\frac{1 - cos ( 0.01 )}{0.0001} = \frac{4.99996 \times 1 0 ^{- 5}}{1 0 ^{- 4}} = 0.499996 \approx 0.5$ . ✓

Recall Which case did each example hit?

Ex1 covers ::: C1 (converges everywhere) Ex2 covers ::: C2 (anchor at $a \neq = 0$ ) Ex3 covers ::: C3 and C4 (inside vs outside radius) Ex4 covers ::: C5 (pathological, series $\neq =$ function) Ex5 covers ::: C6 (remainder / accuracy count) Ex6 covers ::: C7 (word problem with units) Ex7 covers ::: C8 (multivariable Newton step) Ex8 covers ::: C9 (exam twist via known series)

Connections

Taylor series approximation — the parent this page stress-tests.
Approximation error and Big-O — Examples 3, 5 use the remainder bound.
Newton's Method — Example 7 is one exact Newton step.
Hessian matrix — the curvature matrix in Example 7.
Convex functions — Example 7's $H ⪰ 0$ .
Gradient Descent — the 1st-order cousin of Newton.
Derivatives and gradients — every coefficient is a derivative.