Taylor series — derivation from power series: Visual walkthrough

Step 1 — What a polynomial even is, as a picture

Before we touch anything infinite, let us agree what the pieces are.

WHAT. A polynomial is a sum of power terms. A power term is a coefficient (a fixed number) multiplied by a power of a variable: $fixed number c_{2} \cdot power of the displacement (x - a)^{2}$

$x$ is the input we slide around.
$a$ is a fixed point we call the centre — the place we build our approximation around.
$(x - a)$ is the displacement: how far $x$ has moved from the centre. It is $0$ exactly when $x = a$ .
$c_{n}$ is the coefficient — a dial we are allowed to turn to shape the curve.

WHY start here. The entire derivation is about finding the right value for each dial $c_{n}$ . So we must first see what each dial physically controls.

PICTURE. Below, the same centre $a$ is fixed. Each term $c_{n} (x - a)^{n}$ is drawn on its own: a flat line ( $n = 0$ ), a straight ramp ( $n = 1$ ), a parabola ( $n = 2$ ), a cubic ( $n = 3$ ). Notice they all pass through zero-height at $x = a$ except the flat one. That single fact is the seed of the whole proof.

Figure — Taylor series — derivation from power series

Step 2 — The one move that makes displacement vanish: set $x = a$

WHAT. Substitute the centre itself, $x = a$ , into the series.

WHY this move and not another? Look at any term with $n \geq 1$ : it contains the factor $(x - a)$ . At $x = a$ that factor is $(a - a) = 0$ , and $0$ times anything is $0$ . So substituting $x = a$ is a term-killer: it deletes every term except the flat one, $c_{0}$ . We are choosing the one input value that silences everybody but a single survivor.

$f (a) = survives c_{0} + = 0 c_{1} \cdot 0 + = 0 c_{2} \cdot 0 + \dots = c_{0}$

PICTURE. Watch the vertical line $x = a$ slice through the stacked terms. Every curved/ramped term is caught exactly at its zero-height crossing; only the flat line $c_{0}$ has any height left. That height is $f (a)$ .

Step 3 — Why we need a NEW tool: the derivative

We have $c_{0}$ . But setting $x = a$ again would just give $c_{0}$ once more — we learn nothing new. We are stuck.

WHAT. We introduce the derivative $f^{'} (x)$ : the slope (steepness) of the curve at each point.

WHY the derivative specifically? We need a machine that (a) deletes the constant $c_{0}$ so it stops blocking our view, and (b) drags the next dial $c_{1}$ down into the "flat" slot so that setting $x = a$ can catch it. Differentiation does exactly both:

the derivative of a constant is $0$ (deletes $c_{0}$ ),
the derivative of $c_{1} (x - a)$ is $c_{1}$ (turns the ramp into a new flat line).

This is the only elementary tool that lowers every power by one — see Linear approximation & differentials for the $n = 1$ picture in isolation.

$f^{'} (x) = from c_{0} 0 + ramp \to flat c_{1} + still has (x - a) 2 c_{2} (x - a) + still has (x - a) 3 c_{3} (x - a)^{2} + \dots$

PICTURE. The top panel shows the ramp $c_{1} (x - a)$ tilting upward; the bottom panel shows its derivative — a flat line at height $c_{1}$ . The derivative "reads off the steepness" and reports it as a constant.

Step 4 — Fire the term-killer again: set $x = a$ in $f^{'}$

WHAT. Now substitute $x = a$ into $f^{'} (x)$ .

WHY. Same logic as Step 2, one level up. Every term in $f^{'}$ that still carries $(x - a)$ dies; only the freshly-flattened $c_{1}$ survives.

$f^{'} (a) = c_{1} + all killed 2 c_{2} \cdot 0 + 3 c_{3} \cdot 0 + \dots = c_{1}$

PICTURE. The vertical line $x = a$ slices the derivative graph. All the curved leftovers are caught at zero-height; only the flat $c_{1}$ line has height. That height is $f^{'} (a)$ , the slope of $f$ at the centre.

We now see the rhythm: differentiate to flatten the next term, then set $x = a$ to catch it. Repeat.

Step 5 — Differentiate twice: where the factorial is born

WHAT. Differentiate again to reach $c_{2}$ . Watch the number that falls out of the quadratic term.

WHY watch the number? This is the crux of the whole page. The term $c_{2} (x - a)^{2}$ does not flatten in one step — it needs two. And each step multiplies by the current exponent:

$c_{2} (x - a)^{2} \frac{d}{d x} 2 c_{2} (x - a)^{1} \frac{d}{d x} 2 \cdot 1 c_{2} = 2! c_{2}$

The first derivative peels off the exponent $2$ .
The second derivative peels off the new exponent $1$ .
Their product $2 \cdot 1 = 2! = 2$ rides in front of $c_{2}$ .

So after two derivatives: $f^{''} (a) = 2! c_{2} \Rightarrow c_{2} = \frac{f ^{''} ( a )}{2 !}$

PICTURE. A "peeling staircase": the parabola $(x - a)^{2}$ at the top, its first derivative $2 (x - a)$ in the middle (the " $2$ " now visible as a slope), its second derivative $2$ at the bottom (a flat line at height $2$ ). The falling exponents $2$ then $1$ are labelled in red as they drop.

Step 6 — The general term: differentiate $n$ times, catch $n! c_{n}$

WHAT. Do the peeling for a general term $c_{n} (x - a)^{n}$ .

WHY. We want a single formula covering every dial at once, so we track all the peeled exponents.

Each differentiation lowers the exponent by one and multiplies by the old exponent: $(x - a)^{n} \to n (x - a)^{n - 1} \to n (n - 1) (x - a)^{n - 2} \to \dots \to = n! n (n - 1) (n - 2) \dots 1$

After exactly $n$ derivatives, all $n$ exponents $n, n - 1, \dots, 1$ have been peeled and multiplied together.
Their product is the factorial $n! = n \cdot (n - 1) \dots 2 \cdot 1$ .
Lower terms (powers $< n$ ) have differentiated down to $0$ and vanished; higher terms still carry $(x - a)$ and will be killed by $x = a$ .

Set $x = a$ to catch the lone survivor: $f^{(n)} (a) = n! c_{n} \Rightarrow c_{n} = \frac{f ^{(n)} ( a )}{n !}$

Here $f^{(n)}$ means " $f$ differentiated $n$ times."

PICTURE. A ladder of exponents falling: $n \to n - 1 \to \dots \to 1$ , each rung labelled with the factor it contributes, and the running product $n!$ accumulating on the right. The division bar $\div n!$ at the bottom shows the factorial getting cancelled out of the dial.

Step 7 — Edge case: what if $f^{(n)} (a) = 0$ ?

WHAT. Suppose some derivative is zero at the centre, e.g. for $cos x$ at $a = 0$ the odd derivatives all give $0$ .

WHY cover this. A dial being zero is not an error — it means that term is simply absent, and the curve has no ramp/kink of that order at the centre. The formula handles it automatically: $c_{n} = 0/ n! = 0$ .

$cos x : f^{'} (0) = 0, f^{'''} (0) = 0, \dots \Rightarrow c_{1} = c_{3} = \dots = 0$ $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots (only even powers)$

PICTURE. $cos x$ near $0$ is perfectly flat-topped and symmetric — no left/right tilt — so no odd (asymmetric) term is allowed. The even terms build the symmetric dome; the odd dials are switched off.

Step 8 — Edge case: centre not at zero ( $a \neq = 0$ )

WHAT. Redo the peeling when the centre is, say, $a = 1$ for $ln x$ .

WHY. The killer move needs the factor $(x - a)$ to vanish at the centre. If we lazily wrote powers of $x$ instead of $(x - 1)$ , plugging $x = 1$ would not zero them, and the whole term-killing machine breaks. The basis must be $(x - a)^{n}$ .

$f (x) = ln x : f (1) = 0, f^{'} (1) = 1, f^{''} (1) = - 1, f^{'''} (1) = 2$ $c_{0} = 0, c_{1} = \frac{1}{1 !} = 1, c_{2} = \frac{- 1}{2 !} = - \frac{1}{2}, c_{3} = \frac{2}{3 !} = \frac{1}{3}$ $ln x = (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - \dots$

PICTURE. The approximating polynomial hugs $ln x$ tightly near $x = 1$ and drifts away far from it — the accuracy is anchored at the centre, exactly where $(x - a)$ is tiny.

The one-picture summary

Everything above compresses into a single loop, repeated for $n = 0, 1, 2, \dots$ :

$flattens c_{n}, builds n! differentiate n times ⟶ kills all but one term set x = a ⟶ f^{(n)} (a) = n! c_{n} ⟶ c_{n} = \frac{f ^{(n)} ( a )}{n !}$

Recall Feynman: retell the whole walkthrough in plain words

We guessed our curvy function is secretly a giant polynomial with a row of adjustable dials. To find the first dial, we plugged in the centre point — this is like reading the height of the slide right where you're standing; every wobbly part is momentarily flat there, so only the base height shows. To find the next dial we measured the slope (the derivative), which erases the base height and turns the ramp into a new readable height — then plugged in the centre again. To reach dial number three we measured how the slope changes (differentiate twice), and so on. Each time we differentiate, the powers "peel off" as multiplying numbers that pile up into $n!$ ; so we divide by $n!$ to keep the dials honest. If a dial comes out zero, that shape simply isn't part of the curve. And we always measure powers of "distance from the centre," $(x - a)$ , so that plugging in the centre truly silences everyone but the one dial we're after.

Connections

Parent: full algebraic derivation
Maclaurin series — common expansions — the $a = 0$ pictures for $e^{x}, sin, cos$
Linear approximation & differentials — Steps 3–4 in isolation ( $n = 1$ )
Taylor's theorem with remainder — how wrong the truncated picture is far from $a$
Power series — radius & interval of convergence — where these pictures stop being valid
Geometric series — the all-dials-equal- $1$ special case
L'Hôpital's rule — leading terms of these series replace it