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

Step 1 — What is a curve, and what is a polynomial?

WHAT. A function $f$ is a rule that takes a number $x$ and gives back a height $f (x)$ . Draw all those heights and you get a curve. A polynomial is a special, tame curve made only by adding powers of $x$ : $P (x) = constant a_{0} + a line a_{1} x + a parabola a_{2} x^{2} + an S-bend a_{3} x^{3} + \dots$

$a_{0}$ is a plain number — it just lifts the whole curve up or down.
$a_{1} x$ tilts it (a straight ramp).
$a_{2} x^{2}$ bends it into a bowl.
$a_{3} x^{3}$ adds an S-shape, and so on.

Each $a_{k}$ is a dial. Turning a dial changes one feature of the shape. Our whole job is to find the dial settings that make the tame polynomial imitate a wild function.

WHY start here. Polynomials are the only curves we can fully control with a handful of numbers. Sine, cosine, logs — these are hard. So we ask: can we fake a hard curve using dials we understand?

PICTURE. Watch each power add its own personality to the shape.

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

Step 2 — Pick the meeting point: $x = 0$

WHAT. We choose one special input, $x = 0$ , and demand our polynomial agree with the real function there — not just in height, but in every layer of behaviour.

WHY $x = 0$ ? Because $x = 0$ makes powers of $x$ vanish in a beautifully clean way: $0^{1} = 0, 0^{2} = 0, 0^{3} = 0, \dots$ When we plug in $x = 0$ , every term with an $x$ disappears. That collapse is the whole trick — it lets us peel off one dial at a time. (Meeting at some other point $a \neq = 0$ is allowed too — that is the Taylor series (general centre a) — but $0$ is the friendliest.)

PICTURE. The real curve $f$ and our polynomial $P$ are forced to touch at the single point $(0, f (0))$ .

Step 3 — Match the height: find $a_{0}$

WHAT. Put $x = 0$ into the polynomial: $P (0) = a_{0} + a_{1} \cdot 0 + a_{2} \cdot 0^{2} + \dots = a_{0} .$

Every dial except $a_{0}$ is multiplied by a power of $0$ , so it drops out. Setting $P (0) = f (0)$ gives: $a_{0} = f (0)$

$a_{0}$ — the height dial, now locked to the true height of $f$ at the meeting point.
$f (0)$ — the actual value of the curve at $x = 0$ .

WHY. This is the crudest possible fake: a flat horizontal line at exactly the right height. It touches $f$ at one point but ignores whether $f$ is climbing or falling.

PICTURE. The green flat line matches height only — it will drift away instantly.

Step 4 — Match the slope: what a derivative is, and finding $a_{1}$

WHAT is a derivative? The slope of a curve at a point is how steep it is — how much height you gain per step sideways. The tool that measures this is the derivative, written $f^{'} (x)$ . Picture zooming in on the curve until it looks straight; that tiny straight tilt is $f^{'} (x)$ .

WHY this tool and not another? Height alone (Step 3) gives a flat line. To also capture which way the curve is heading, we need the one tool that reports steepness — that is exactly the derivative. No other operation answers "how tilted is it here?".

HOW. Differentiating a power lowers its exponent by one and multiplies by the old exponent: $\frac{d}{d x} x^{k} = k x^{k - 1} .$ Apply it to the whole polynomial: $P^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \dots$ Now the $a_{0}$ term vanished (the derivative of a constant is $0$ — a flat line has no tilt). Put $x = 0$ : every remaining $x$ dies, leaving $a_{1} = f^{'} (0)$

$a_{1}$ — the tilt dial.
$f^{'} (0)$ — the true slope of the curve at $x = 0$ .

Now our fake is a tangent line: right height and right tilt.

PICTURE. The line now leans exactly along the curve at $0$ .

Step 5 — Match the bend: where the $n!$ is born, finding $a_{2}$

WHAT. The tangent line is straight; the real curve bends away from it. Bending is measured by the derivative of the derivative, written $f^{''} (x)$ — "how fast the slope itself is changing".

WHY. A line can match height and tilt but never bend. To hug a curved shape we must match curvature, and the second derivative is precisely the curvature-reporter.

HOW — and watch the factorial appear. Differentiate $P^{'} (x)$ again: $P^{''} (x) = 2 a_{2} + (3 \cdot 2) a_{3} x + (4 \cdot 3) a_{4} x^{2} + \dots$ Notice the $x^{2}$ term got hit twice: first $x^{2} \to 2 x$ , then $2 x \to 2$ . Two derivatives multiplied the numbers $2$ and $1$ together — that is $2! = 2$ . Put $x = 0$ : $P^{''} (0) = 2 a_{2} = f^{''} (0) ⟹ a_{2} = \frac{f ^{''} ( 0 )}{2 !}$

The $2!$ is not a decoration — it is the leftover from differentiating $x^{2}$ twice.
In general, differentiating $x^{n}$ a full $n$ times gives $n \cdot (n - 1) \dots 2 \cdot 1 = n!$ , and every other term is either gone (lower powers) or still carrying an $x$ (higher powers, which die at $0$ ). So: $a_{n} = \frac{f ^{(n)} ( 0 )}{n !}$

This is the master coefficient rule — the whole engine, now seen, not just stated.

PICTURE. Watch $x^{n}$ get differentiated $n$ times and count the numbers $n, n - 1, \dots, 1$ pile up into $n!$ .

Step 6 — Feed $e^{x}$ into the engine

WHAT. Now pick the friendliest possible function: $f (x) = e^{x}$ , the curve whose slope everywhere equals its own height, so $\frac{d}{d x} e^{x} = e^{x}$ .

WHY $e^{x}$ first. Because every derivative of $e^{x}$ is again $e^{x}$ , and $e^{0} = 1$ . So every numerator $f^{(n)} (0)$ is the same number, $1$ . The engine barely has to work: $a_{n} = \frac{f ^{(n)} ( 0 )}{n !} = \frac{1}{n !} .$

HOW. Stack the pieces: $e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots = n = 0 \sum \infty \frac{x ^{n}}{n !}$

$1$ — matches the height ( $e^{0} = 1$ ).
$x$ — matches the slope ( $1$ ).
$\frac{x ^{2}}{2 !}$ — matches the bend, divided by the factorial so it doesn't overpower.
Each extra term is smaller because $n!$ grows explosively, so a few terms already hug the curve near $0$ . (It in fact hugs for all $x$ — see Radius & interval of convergence.)

PICTURE. Add terms one at a time and watch the polynomial creep outward, wrapping around $e^{x}$ .

Step 7 — The same engine, four more functions

WHAT. Nothing new to invent — just turn the same crank with different derivative patterns at $0$ .

$sin x$ : derivatives cycle $sin \to cos \to - sin \to - cos$ , giving values at $0$ of $0, 1, 0, - 1, \dots$ . Only odd powers survive, signs alternate: $sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$
$cos x$ : the same cycle starting at $cos 0 = 1$ : values $1, 0, - 1, 0, \dots$ . Only even powers survive: $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$
$ln (1 + x)$ : faster to integrate a Geometric series: $\frac{1}{1 + x} = 1 - x + x^{2} - \dots$ integrates term-by-term to $ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots (- 1 < x \leq 1)$
$(1 + x)^{n}$ : derivatives peel off factors $n, n (n - 1), \dots$ , giving the binomial series $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$

WHY grouped. They differ only in the numerators $f^{(n)} (0)$ . The $n!$ downstairs, the $x^{n}$ , the "match-at- $0$ " idea — all identical.

PICTURE. One machine, four output belts.

The one-picture summary

The single figure above compresses everything: pick the point $x = 0$ → match height, then slope, then bend, then bend-of-bend → each match forces one dial $a_{n} = f^{(n)} (0) / n!$ → the growing polynomial hugs the true curve tighter with every term.

Recall Feynman retelling — the whole walkthrough in plain words

Imagine you're standing at one spot on a hilly road ( $x = 0$ ). You can't see the whole road, but you can feel four things: how high you are, how steeply you're tilted, how hard the road curves, and how fast that curve is itself changing. Turns out those feelings are enough to redraw the road near you. So we build a fake road out of simple pieces: a flat block (height), a ramp (slope), a gentle bowl (bend), an S-wiggle (bend-of-bend), and so on. We tune each piece so it matches one feeling exactly — height first, then tilt, then curve. Each new piece we add is deliberately shrunk by dividing by a factorial, so the big shapes don't stampede over the small corrections. When the function is $e^{x}$ , every feeling is the same number ( $1$ ), so the pieces are just $1, x, x^{2} /2, x^{3} /6, \dots$ . Sine and cosine feel a repeating pattern of $0, 1, 0, - 1$ , so their pieces skip powers and flip sign. That's it — a Maclaurin series is just a stack of tuned shapes copying how a curve feels at one point.

Recall

Why does putting $x = 0$ isolate one coefficient at a time? ::: Because every term still carrying an $x$ becomes $0$ , so only the constant term of that derivative survives. Where does the $n!$ in $a_{n}$ come from? ::: From differentiating $x^{n}$ exactly $n$ times: $\frac{d ^{n}}{d x ^{n}} x^{n} = n!$ . Why is $e^{x}$ the easiest series to build? ::: Every derivative of $e^{x}$ is $e^{x}$ and $e^{0} = 1$ , so every numerator $f^{(n)} (0) = 1$ .

Connections

Parent topic — the finished formulas
Taylor series (general centre a) — same machine, meeting point $a \neq = 0$
Geometric series — the shortcut behind $ln (1 + x)$
Radius & interval of convergence — how far each series stays faithful
Euler's formula — combine the $e^{x}$ , $sin$ , $cos$ series
L'Hôpital & limits via series — series settle tricky $0/0$ limits
Binomial theorem (integer n) — terminating case of $(1 + x)^{n}$