The 59-second version, then read on for the full derivation.
Three Lecture Rooms, One Door
You meet them in three different courses, and nobody tells you they are related. In calculus, the Taylor series rebuilds a function from a handful of local numbers at a single point. In complex analysis, Cauchy's integral formula recovers a function's value from a loop drawn around it. In a signals or physics class, the Fourier transform breaks a wave into its frequencies. Three lecture rooms, three vocabularies, three sets of homework.
This article is about the door between them. Underneath all three is a single physical act:
Spin an arrow around a circle, and average it over one full turn. Everything that spins cancels to nothing; only what stands still survives.
Taylor, Cauchy, and Fourier are three ways of pointing a flashlight at that one act. Once you see the act, the three theorems stop being separate facts to memorize and become the same sentence, read three ways.
Throughout, τ (tau) stands for one full rotation, the ratio of a circle's circumference to its radius. Wrap a thread around any circle and it is τ ≈ 6.283 radii long; one full turn is τ radians. We use τ and not 2π for one reason: every formula in this article is an average over exactly one full turn, and τ is the name of one full turn. The notation will keep reminding you what the math is actually doing.
The Spinning Arrow
The one move behind all three results: let a thing spin, average it over a full turn, and only the part that never spins survives.
Part 1, The spinning arrow, and one question
Put a unit-length arrow at the origin of the plane and let it point at angle θ. As θ grows from 0 to τ, the tip sweeps once around the unit circle. We write that tip, a point with a horizontal and a vertical part, as a single complex number:
exp(iθ) = cos θ + i sin θAn arrow spinning k times as fast is exp(ikθ): as θ goes once around, this arrow goes around k times (and for negative k, the other way). Now the one question this whole article hangs on:
If I let an arrow spin and track its average position over one full turn, where does the average land?
Drag k to change the spin rate, and press ▶ to sweep θ from 0 to τ. The gold arrow is the current position, and each faint spoke is a position the arrow has already visited, a little arrow in its own right waiting to be averaged in. To average a batch of arrows you add them tip to tail and divide by how many there are: the pink dot is exactly that average so far. Watch its dashed trail, which is the whole point: the average starts out on the rim, pinned to the very first position, then spirals steadily inward as later positions arrive to cancel the earlier ones. By the end of one full turn every spoke is matched by an opposite spoke, and the average has been dragged all the way to the centre.
The gold arrow is exp(ikθ); the faint spokes are the positions visited so far; the pink dot is their average, and its dashed trail shows it spiralling in from the rim to the centre.
The length of the running average. For any k ≠ 0 it returns to 0 after a full turn. For k = 0 the arrow never moves, so the average stays at 1.
The picture is the proof. A spinning arrow visits every direction equally, so its positions cancel in symmetric pairs, and the average collapses to the center. The single exception is k = 0: a rate-zero "spin" is an arrow frozen at 1, and the average of a constant is itself. Before compressing that into one line, let us build the average back up from the pieces we actually drew on the circle.
Part 2, Build the average from the pieces
Put the three moves together and the average reads straight off the picture:
tap a box below to lock its answer into the equation
The arrow never spins, so it stands still at full length, and its average is its whole self. This is the one we keep: the prize every extraction in this article is hunting for.
The arrow spins a whole number of loops and lands back where it began, so its positions fall into balanced pairs and cancel. Everything else vanishes, clearing the field so the survivor stands alone.
( k a whole number )
Part 3, The full-turn cancellation lemma
Why a whole number? The cancellation needs the path to close. If k is a whole number, the arrow makes a whole number of loops over one turn and lands exactly back where it started, so the positions fall into perfectly balanced pairs and sum to zero. A fractional rate would leave the arrow stopped partway around, the loop unclosed, and a leftover sliver that does not cancel. So from here on the only spins we ever worry about are the ones that complete a full cycle and close; spins that do not close we simply set aside. That is not a loss: the closing spins are exactly what each later setting is built from.
This is the only tool we need. Textbooks call it the orthogonality lemma or the averaging lemma; here we name it the full-turn cancellation lemma, because that is exactly what it says: over one full turn, everything that spins cancels, and only the still arrow survives. Sections 2, 3, and 4 are nothing but this lemma pointed at three different functions.
Fourier: Tune the Dial
Fourier, a repeating signal is a chord of pure frequencies; here we read off how loud each one is.
Part 1, A signal as a sum of spinning arrows
A signal is just a quantity that changes: a value at every moment, or at every position. The air pressure at your eardrum while a note plays; the voltage on a wire; the brightness along one line of a photograph; the height of the tide through a day. Plot that value against time, or angle, or distance, and you get a wiggly curve. That curve is the signal.
We will work with signals that repeat. Start with a signal in time, f(t), and suppose it returns to where it began after a period T, so that f(t + T) = f(t) for every t. Then nothing new ever happens after one period: the whole signal is already contained in a single stretch of length T, repeated forever. So we may as well wrap that one stretch around a circle, call one full lap τ, and track the angle θ around the circle instead of the running clock. We write the signal f(θ): its value when the dial sits at angle θ. Same signal, just read off a circle instead of an endless line.
Here is the move that makes everything else work. We picture each ingredient of the signal as a small arrow of fixed length, pinned at the origin and spinning at its own steady rate. The arrow that comes around once per turn is exp(iθ); the one that comes around twice is exp(2iθ); three times, exp(3iθ); and so on. The strength ck simply sets how long the k-th arrow is.
Now stack them tip to tail: start at the origin, lay down the first arrow, begin the next one from its tip, then the next from that tip. Each arrow turns at its own rate, so the whole chain swings and wobbles, and its final tip traces a path as θ advances. That traced path is the signal. A signal "carried by exp items" means exactly this, a sum of spinning arrows whose last tip draws the curve:
f(θ) = … + c0 + c1exp(iθ) + c2exp(2iθ) + c3exp(3iθ) + …(The c0 term is the arrow that does not spin, a fixed offset, and terms with negative rates spin the other way. Together they can build any repeating signal.)
And here is the goal of this section. Reading that equation left to right is easy: hand me the strengths ck and I can add up the arrows and draw the signal f. But in real life we get the equation the other way round, we are handed the combined result, f(θ) at every angle, and we do not know its composition, the individual ck that went into it. Fourier analysis is the inverse problem: recover each ck from the finished f, and so reveal exactly which frequencies, and how much of each, the signal is made of. The full-turn cancellation lemma from §1 does it in one line, as we are about to see.
One catch, full cycles only. This is the place to cash in the promise from §1: that we only ever worry about spins which complete a full cycle. Here is what that means for a signal. To repeat is to close: after one full turn the signal is exactly back where it began, f(θ + τ) = f(θ). That single demand is what forces every rate to be whole: a rate-k arrow comes back to its start after one turn only when k is a whole number, while a rate-3.5 arrow is left pointing halfway around, so a signal carrying it would not return after one turn. A signal that closes on a single turn is therefore built from full-cycle arrows and nothing else, which is precisely the case the full-turn cancellation lemma reads cleanly.
And a signal that does not close on one turn? Either it still repeats, just over a longer stretch (two turns, when a half-integer like 3.5 = 7/2 is present); then we simply call that longer stretch one turn and the rates come out whole again. Or it never repeats at all, a lone pulse or a signal running down the endless line; then there is no period to make whole, and the sum of arrows must open into a continuum of them, the Fourier transform, which we will not touch in this article.
Part 2, The two shadows: real and imaginary
Each spinning arrow lives in a plane, so it casts two shadows. Drop a perpendicular onto the horizontal axis and you read off its real part; drop one onto the vertical axis and you read off its imaginary part. For the unit arrow that is exactly the relation from §1:
shadow (real) ↑ the invisible
part (imaginary)
So as the arrow turns steadily, its horizontal shadow slides back and forth as a cosine and its vertical shadow as a sine: the same oscillation, a quarter-turn apart.
Now here is the part worth slowing down for. When you measure a real signal (the pressure at your eardrum, the voltage on a wire), you record exactly one number at each instant: the horizontal shadow, the real part. Your microphone never picks up the vertical, imaginary part. No instrument does. It is, in the most literal sense, invisible to us: there is no dial in our world that reads it off.
And yet it has to be there. Take the imaginary part away and the arrow can no longer turn; it can only slide back and forth along a single line, and a thing trapped on a line cannot "go around" anything. The imaginary part is the second dimension the arrow needs in order to spin at all. Think of it as the hidden half of the machine: a quantity we cannot measure in the one dimension our instruments live in, yet without which there is no rotation, and without rotation there is no Taylor, no Cauchy, no Fourier. The invisible part is not decoration or bookkeeping; it is the gear that makes the visible part move.
If the idea of a number you can never put on a ruler unsettles you, that is the right instinct, and it deserves its own story. We will cover it properly in a future article in this series.
In the left figure below, the solid curve is the visible real-part shadow of f (the thing an instrument would actually record), and the dashed curve is the invisible imaginary part, drawn here only so you can watch the machinery your instruments never could.
This stack is a chord played by spinning arrows. The analysis problem is the reverse: someone hands you the finished sound f, and you must recover the individual strengths ck: how much of each frequency is in the mix. It sounds hard. The full-turn cancellation lemma makes it a one-liner.
Watch the arrow emit both, we catch only one. The animation makes the point better than words. A single arrow turns at a steady rate. At every instant it casts its two shadows, and both stream outward as waves: the real part rolls out as a cosine, the imaginary part as a sine. The arrow emits them equally, but only the real wave ever reaches a detector. The imaginary wave is sent out into a dimension our instruments have no port for.
The turning arrow exp(iθ) sends out both parts. The real wave reaches the detector; the imaginary wave is emitted just the same, we simply have no instrument that registers it.
A tempting story, handled with care. It is natural to picture the spin as "really" happening in a hidden dimension, with only its shadow reaching us. The shadow half is exactly right: what any instrument records is the real part, a 1D projection of a 2D turn. But the second axis is not a hidden place where the rotation secretly happens; it is a mathematical partner we attach because turning is easier to reason about than wobbling, and for a real signal it is fully determined by the part we can see. Whether that extra dimension is "imaginary" or "invisible" in some deeper sense is a genuine question, one that goes beyond the scope of this series.
Part 3, Recovering a single strength
Why hunt for a single strength ck? That one number is the answer to "how much of this frequency is in here": the weight of one pure tone in a chord, the strength of one ripple in an image, the size of one harmonic in a vibration. Reading those weights back off the finished signal, one at a time, is exactly what analysis means, and the full-turn cancellation lemma is the tool that does it.
Let us do one concretely. Take the rate-3 arrow's strength, c3. Multiply the whole signal by exp(−3iθ), the rate-3 arrow run backwards. This drops every arrow's rate by 3, since exp(ikθ) exp(−3iθ) = exp(i(k−3)θ):
So far we have only re-tuned the arrows; nothing has been averaged yet. The averaging is the next move, and to average over one full turn is to integrate across it: apply 1τ ∫0τ( … ) dθ to the whole sum. Only now does the full-turn cancellation lemma bite: every spinning arrow (any nonzero rate) integrates to nothing, while the one arrow that stands still (rate 0, worth exp(0) = 1) integrates to its whole self. Every term but one is wiped out, and the total collapses to exactly c3.
Nothing was special about 3: the same move recovers ck for any whole number k. Multiply the signal by exp(−ikθ) and average; only the rate-k arrow is brought to a standstill, every other one cancels, and it hands you exactly its strength:
That is the Fourier coefficient. There is no cleverness in it beyond §1: freeze the arrow you want, average away the rest. Set the strengths on the left, choose which one to recover with the dial, and press ▶, the pink dot is the running average, and after one full turn it lands exactly on the strength you dialed for.
THE MOVE, END TO END
The signal f(θ) = Σ ckexp(ikθ), its real part (solid) and imaginary part (dashed) over one full turn.
Extraction: the running average of f(θ)exp(−ikθ) starts at the first sample and spirals inward along the dashed trail as θ sweeps one turn, landing on ck.
Try it: dial up k = 2 and the dot lands on whatever you set c2 to, ignoring the other three completely. The signal can be as tangled as you like; the average reads off one ingredient at a time, blind to all the others. That blindness is the full-turn cancellation lemma.
This is the Fourier series, the natural language for anything periodic, anything that lives on a circle. A coefficient is what you get when you freeze one arrow and average the rest into oblivion; written plainly, here it is:
THE FOURIER COEFFICIENT
Cauchy: The Same Average, Drawn as a Circle
Cauchy, a smooth field is rigid: its values on a loop fix everything inside. The rim knows the interior.
The same trick has a second home, and it is the one that built much of modern physics and engineering. In §2 a signal was a value at every moment. Now picture a value at every point of a plane: the temperature across a metal sheet, the height of a stretched drum-skin, the voltage across a flat plate, the flow of smooth water. Steady, source-free fields like these are what a complex function f captures, and they turn out to be strangely rigid.
That rigidity is the punchline of this section: the value at the centre of a circle is just the average of the values around its rim. Measure the edge and you know the middle. A plate cannot hide a hot-spot at its centre while its rim stays cool; the centre is pinned to the average of its surroundings. That fact is Cauchy's theorem, and it is the §2 averaging lemma in a new coat. First see it; then we watch it fall out of the spinning-arrow lemma.
Part 1, See it first: data on a plate
Here is the whole idea in one concrete field, no complex numbers yet. Picture the temperature across a flat plate: every point of the plate carries a value, shown here as its colour. That is what we mean by data spread over a plane. Mark a centre P, draw a circle of radius r around it, and take any point z on the rim. Average the temperatures all the way round the rim and you land exactly on the temperature at the centre P, and it stays true wherever you drag the circle.
Data on a plane. Colour is the value (a temperature) at each point, on a fixed scale, 0 (deep blue) to 1 (red), so the readouts mean the same thing in every random field. The white circle has centre P and a rim point z; the dots on the rim are coloured by their own temperatures. On the right, the average of those rim colours comes out identical to the colour at the centre. That equality is the mean-value property, drag P or change r and it never breaks.
Why the centre is never the hottest spot, in the fields we study. Hit randomize as often as you like: in every one, the brightest red and deepest blue sit on the edges, never alone in the middle. This is not a law of all temperature fields; it is the mark of the source-free ones. With no flame or sink inside, each interior point settles to the average of its surroundings, so it cannot out-rank them all. Put a real source in the middle, a hotplate, a point charge, and the core can be the hottest point; but that field is no longer source-free, and describing it needs extra terms (negative powers of z−P) that fall outside the clean stack we use here. The rigidity centre = rim average is exactly the source-free case, which is what Cauchy's theorem and this article are about.
Part 2, Prove it: the rim is a bundle of spinning arrows
So much for seeing it. Why must it hold? Because the rim, run through any nice function, turns out to be nothing but a bundle of §2's spinning arrows, and we already know what one full turn does to those: everything that spins averages to nothing, only the still part survives. Everything from here on is that single observation, made precise.
Pick a point P and a radius r, and let a second point z walk the circle of radius r around P:
z = P + r exp(iθ)Both P and z are complex numbers: points in the plane, not ticks on a line. As θ runs from 0 to τ, the displacement from the centre is
z − P = r exp(iθ)which is exactly one of our spinning arrows: length r, turning once around. The full-turn cancellation lemma is poised to fire; we only have to feed the circle through a function first.
A nice function is a stack of powers. The functions complex analysis cares about, the smooth and tear-free ones, are exactly those you can build as a sum of whole-number powers of that same displacement z−P:
f(z) = a0 + a1(z−P) + a2(z−P)² + a3(z−P)³ + …Two kinds of symbol here. The plain P is the centre of the circle, one fixed point. The subscripted a0, a1, a2, … are the coefficients, the dial-settings that weight each power. The coefficients are read off at the centre, in particular a0 = f(P), as we are about to see.
(That you can always do this is the deep fact; take it as given for now, we earn it in §4.) Writing f this way pays off immediately, because on the circle every power becomes a spinning arrow. Feeding in z−P = r exp(iθ) from just above,
(z−P)k = rk exp(ikθ)so the k-th term turns at rate k (k whole turns per lap) on a circle of radius rk. Every term of f on the circle is one of our arrows; only the constant a0, the rate-0 term, does not spin.
Keeping the circle small. Throughout this section we hold the radius under one, r < 1. Two reasons. It keeps the circle safely inside the region where f really is this clean stack of powers of z−P. And it puts each higher power on a shorter leash, r > r² > r³ > …, so when we average over a turn the spinning terms fade fast and the still constant stands clear. (The cancellation works at larger radii too, as long as f stays well behaved out there; r < 1 is simply the cleanest window to watch it in.)
Now set it in motion. A complex function sends a point to a point, so we watch two planes at once. Left, the input: z walks the circle around P, with the spinning arrow z−P drawn from P out to z. Right, the output: where the function sends it, f(z), tracing its own loop in matching colours. Below: that same f split into its four terms ak(z−P)k. Drag a slider to resize a term, drag P to move the centre, press ▶ to walk z once around.
Input. z circles the centre P; the dashed arrow is the displacement z−P = r exp(iθ). Drag to move P.
Output. f(z), the four terms added tip to tail. Its loop wanders, but its average lands exactly on a0 = f(P).
The four terms. Each pink dot is that term's average over the turn: the three spinning terms (rates 1–3) average to the origin whatever their coefficient; only the rate-0 term (gold) survives, sitting at a0. All six panels share one scale.
Average around the circle → the centre value. That is the whole proof, and we just watched it run. In symbols: on the circle, every power (z−P)k = rk exp(ikθ) is a spinning arrow, so f is one still term plus a chorus of spinners:
Average over one full turn and, by the full-turn cancellation lemma, every spinning arrow vanishes; only the still term a0 survives:
That is Cauchy's mean-value property, the plate's rule from the top of the section, now proven for every nice function: the average of f around any circle equals its value at the centre. However big or small the circle, the wobble of f around the loop averages out perfectly and leaves only the value in the middle. So one full turn hands us a single number, a0 = f(P), the value at the centre.
Part 3, The same average is Cauchy's integral formula
That rim-average is already the formula Cauchy is famous for, the two are one line in different clothes, and one short differentiation shows it. As z walks the circle z = P + r exp(iθ), ask what one small step dθ does to it. The only moving piece is exp(iθ); write u = iθ and lean on the one fact from the exp story, that exp is its own rate of change:
dz = r d exp(iθ) = r i exp(iθ) dθ = i (z−P) dθu = iθ, du = i dθ: d exp(iθ) = exp(u) du = i exp(iθ) dθ
Why bother rewriting it in z? Because the z-form forgets the angle and even the circle: it returns f(P) from any loop that surrounds P (shrink it, stretch it, dent it, the answer doesn't move), and it is the version that generalizes, to pull out every coefficient in §4, and to crack otherwise-hopeless integrals (the residue theorem). The angle picture is the intuition; the z-form is the workhorse the rest of complex analysis runs on.
The move itself is simple: both formulas are the same integral, the rim-average, written in the point z instead of the angle θ. The average adds up f in steps of dθ; to add it up in steps of dz instead, swap each dθ for the matching dz from the line above. That swap is the whole trick:
the answer the differential the integral sign the function's input
Cauchy's integral formula: the line the swap above lands on, with the same full turn ∮
The circle: a clean quarter-turn. Stepping θ by dθ slides z along by dz, which is the radius z−P given a quarter-turn (that is the i: multiplying by i rotates the plane a quarter-turn, a story we will tell in a future series) and scaled by dθ: dz = i(z−P)dθ. Radius and step meet at a right angle, so the faint stubs are all equal length r dθ, and the loop integral reads off cleanly as f(P).
The blob: same answer, tilted steps. Bend the rim and dz tilts off the radius, so that clean quarter-turn is gone. Yet the loop integral (1/τi) ∮ f/(z−P) dz still lands on the very same f(P) (watch the readout hold steady): Cauchy's formula doesn't care what shape you draw, only that the loop encircles P. The catch is how the centre is built: now a lopsided, weighted average (rim points sitting closer, or where the boundary bends, count for more), not the flat one. That is why the heat-map picture stays on the circle: only there does “centre colour = the plain average of the rim colours” hold, with every rim point weighted equally.
That is the whole of it. But that last factor, 1z−P, is worth slowing down on: it is the weight in the weighted average we just met on the blob. It is not a new ingredient, only the price of counting in z instead of in the angle θ, and it falls out in three steps.
One. A fair rim-average counts every slice of angle equally: each dθ gets the same vote. That is what “average” means here.
Two. But a step in z is not a step in angle. From the line above, dz = i(z−P) dθ: one step in z is the angular step dθ stretched by the radius (z−P) (and turned a quarter-turn by i). So adding up raw f dz would over-count each point in proportion to how far it sits from P: a lopsided average.
Three. Dividing by (z−P) cancels exactly that stretch, putting every angle back on one equal vote:
the radius cancels, and with the 1τi out front this is the plain rim-average again
So 1z−P is the corrective weight that un-stretches each dz back into a fair dθ, and it is the very weighting from before. On the circle that weight is dz/(z−P) = i dθ, a constant quarter-turn, so it levels out to a perfectly even average. On a blob the radius changes and dz tilts off it, so the weight is no longer constant: the same 1/(z−P) still reweights every step, but it can no longer level the votes (that is the lopsided average). The weight is the heart of the formula; the circle is just the shape where it comes out uniform.
(The sign ∮ just means “once around the closed loop”, the same full turn as §2's ∫0τ … dθ.) So “centre = rim average” and Cauchy's integral formula are one statement, read two ways.
See the weight do its work. Part 1 drew the circle around P, so every rim point sat the same distance away and the vote was flat. Here the loop is held fixed and you drag P around inside it, so the distances differ and the weight finally has something to do:
The weight is the whole story. Same heat-map plate as Part 1, but the loop is now held fixed while you drag the centre P around inside it. Every rim point votes for P with weight 1/|z−P|²: nearer points (taller bars, fatter spokes) count for more. On a circle that weighted vote lands exactly on P's own value, whether P sits dead centre (all weights equal, the flat average of Part 1) or far off to one side (the near rim dominates). The flat equal-vote average, by contrast, ignores where P is entirely: it stays stuck on the value at the region's geometric centre. So weighting is what lets the rim find P. Now bend the loop into a blob: the weighted vote no longer lands exactly on P (watch the small Δ appear), because off the circle the honest weight needs which way each piece of rim faces, not only how far it sits. There is no length-only weight that nails a blob, which is the precise reason the clean heat-map shortcut stays the circle's privilege.
One last worry, settled: doesn't dividing by z−P blow up at the centre? Only at the centre itself, where z−P = 0, and we never go there. We are averaging over the rim, where z−P = r exp(iθ) stays a fixed distance r from the centre, never 0. The circle keeps us safely off that one point the whole way round.
Cauchy is one fact: the boundary loop fixes the interior. So far it has handed us a single number, the centre value f(P); in §4 we re-tune the very same average to pull out every coefficient, and find they are the function's own Taylor coefficients. That one fact, written in a single line:
CAUCHY'S INTEGRAL FORMULA
Taylor: The Averages Were the Coefficients All Along
Taylor: near a point, a function is a stack of pure powers, and each power's strength is exactly the number Fourier and Cauchy hand back.
Part 1, What a single point already knows
Before any formula, the question Taylor answers is a physical one: how much can one spot tell you about its surroundings? A surprising amount: near a point a function is made of a few simple pieces, its building blocks, and that one spot fixes how much of each it carries. The pictures below are about seeing those blocks.
Watch a ball at the top of its arc for a single instant. You can read where it is, how fast it is drifting sideways, and the steady downward tug bending its path. That snapshot is enough: from those few local facts the whole arc follows, forward and back. The curve was hiding inside the point.
And there is a deeper picture still, the one that keeps §2's spinning arrows coming back: near any resting point, almost every system behaves like a spring. Look closely at the bottom of a valley and it curls like a parabola, and that one bending layer sets the rhythm. A pendulum keeping time, a plucked string sounding its note, atoms humming in a crystal: all of them oscillate because, up close, their stored energy is that single bend. It is why oscillation turns up everywhere in physics.
One instant rebuilds the arc. A thrown ball traces the green path. Freeze it at the gold dot and three facts fix everything: its place, the way it is heading, and the steady downward pull. Those are the first three building blocks: the level (its place), the lean (its heading), the bend (the constant pull). Because the pull never changes, no deeper block is needed, so these three redraw the whole arc exactly, a parabola. Press watch.
Every valley is a spring. The green curve is a stored-energy landscape; the gold dashed parabola is its bend block alone. At the very floor the lean flattens to nothing, so the bend is the only block left to drive the motion, and the ball rocks like a weight on a spring. That bend (the same block the ball's pull supplied) sets the rhythm; deeper blocks only fine-tune it.
Your calculator does the same thing, only as arithmetic. Asked for sin(0.3) it keeps no secret table; it adds a few simple terms (0.3, then a small correction, then a smaller one) and stops. Every value it shows you is a function rebuilt from how it behaves at a single point.
sin(0.3), built from a single point. The calculator knows sin only at 0, so it adds the building blocks there: 0.3, then a small correction (−0.3³/6), then a tiny one (+0.3⁵/120). The gold dot is the running total; the cyan ring is the true value. After just three blocks the total has locked onto sin(0.3) to six decimals. Press add a term and watch the estimate climb onto the curve.
Behind all three sits Taylor's idea: near a chosen point P, a function is pinned down by a few local building blocks, and you have just met them. The ball's place, heading, and pull were the level, the lean, and the bend; the valley's spring was the bend acting alone. These blocks are nothing exotic: each is one of §1's spinning arrows, the component §2 and §3 kept adding up, here held still so we can stack it. Stack them (level, lean, bend, the turn of that bend, and on) and you rebuild any function nearby. The rest of this section asks how big each block is, then finds that §2 and §3's averaging has been handing us those sizes all along.
A level, a lean, a bend. The f curve (faint cyan) is the same in all three panels, with P marked by the gold dot. Each panel's gold block is tagged with its strength ak, shown right on the chart: a0 is the height at P (the level), a1 the slope of the tilt (the lean), a2 the curl the parabola adds beyond a straight line (the bend). Press slide P to move the point along the curve and watch all three strengths change. Stacked together, these blocks are Taylor.
Part 2, What Taylor says: a stack of building blocks
§3's loop handed us one number, a0 = f(P). But a function carries a whole stack of them, a0, a1, a2, …, and before we compute the rest it is worth asking what they even are.
Stand at P and zoom in. Up close, almost any smooth function looks like a flat level; pull back a little and a faint lean appears; a little more and a gentle bend; then the bend itself begins to turn, and onward. Taylor's idea is to rebuild f at a nearby (real) point x by stacking those building blocks, and on the page each block is a pure power of the displacement (x−P):
f(x) = a0 + a1(x−P) + a2(x−P)² + a3(x−P)³ + …block k is the pure power (x−P)k; the coefficient ak is how much of it f carries near P
So each coefficient is simply a strength, the amount of one building block in the mix: a0 sets the level, a1 the lean, a2 the bend, and every further block makes the rebuilt curve hug the true f over a wider stretch. Slide the count up below and watch the layers stack onto f, one block at a time.
The function f (green) and its Taylor sum at P (gold), keeping blocks up to degree N. The faint curve is just the newest block on its own, aN(x−P)N; the gold curve is the running sum of every block up to N. Block 0 is the flat level, block 1 the straight-line lean, block 2 the bend, and each new block lets the sum cling to f a little further out. Add terms to stack them, or drag to move P.
What the real line gives us, and what it withholds. So far the picture has told us what the coefficients are: each ak is the strength of one block, and stacking the blocks rebuilds f, just as the chart showed. Half the job is even free, since the shape (x−P)k is fixed the moment P and k are chosen. But the picture stops there. It shows the blocks piled together and gives us no way to reach in and read off a single strength ak on its own. Knowing what the coefficients mean is not the same as holding a method to extract one.
And that is exactly what the circle is for. We own one tool for pulling a single piece out of a pile: §1's move, freeze the one you want and average the rest away over a full turn. But a full turn needs a circle, and a straight line has none: you cannot walk around a point on the real axis. So we lift the real input x onto a circle in the plane, z = P + r exp(iθ) (the same plane, and the same reason, as Principle 3). There the still blocks start to spin, §1's move can finally bite, and one turn's averaging lifts out any ak we ask for. The coefficients are the same numbers either way: we draw them on a line and extract them on a circle.
Part 3, Reading off any coefficient: Taylor = Fourier = Cauchy
§3 read the centre value f(P) with one average: the k = 0 coefficient. Every other one is the same average, re-tuned. Watch a single equation carry it, in four moves, each line the one above with one operation done to both sides.
1 · Each block becomes a spinning arrow. Part 2 wrote f as a stack of blocks, f = a0 + a1(x−P) + a2(x−P)² + … . Now lift the input onto a circle, z = P + r exp(iθ): the displacement z−P = r exp(iθ) is a single arrow of length r pointing at angle θ. Multiplying complex numbers adds their angles, so squaring that arrow doubles its angle and cubing it triples the angle: (z−P)k is an arrow of length rk that turns k full times for every one lap of θ: a spinning arrow at rate k. So block k = ak(z−P)k is an arrow of length akrk spinning at rate k, and laying all the blocks tip-to-tail gives f on the rim.
f(P+r exp(iθ)) = a0 + a1r exp(iθ) + a2r² exp(2iθ) + a3r³ exp(3iθ) + a4r⁴ exp(4iθ) + …2 · Freeze one block, say block 3. To single out block 3, spin the whole picture backward at rate 3: multiply every value by exp(−3iθ). Block 3 loses its spin and stands still (its rate becomes 3−3 = 0), while every other block k keeps turning, now at the shifted rate k−3.
exp(−3iθ) f = a0 exp(−3iθ) + a1r exp(−2iθ) + a2r² exp(−iθ) + a3r³ + a4r⁴ exp(iθ) + …3 · Average over one full turn. By the full-turn cancellation lemma (§1) every block that is still spinning sweeps all directions equally and averages to 0; only the one standing still, block 3, survives. Average keeps its §3 meaning: the full complex value, the centroid of the arrow-tips in the plane, never their length. So the survivor a3r³ is itself a point in the plane, complex in general, not a real number. The line below is the one above, averaged term by term:
1τ∫0τexp(−3iθ) f dθ = 0 + 0 + 0 + a3r³ + 0 + …4 · Read the three names. That surviving average is §2's Fourier coefficient c3, and its value is Taylor's strength a3 (scaled by the circle's r3). Nothing singled out block 3: the same freeze-and-average pulls out any block k we name. So for every k, Taylor's number and Fourier's number are the same number, the hinge of the whole article:
ak rk = ckThe same average wears a second face. Written as a loop in z instead of an angle in θ, it becomes Cauchy's coefficient formula, the pipeline below. Two swaps do the translation: freezing a block (the exp(−ikθ) above) becomes dividing by (z−P)k+1, and averaging over the angle becomes the single loop ∮(…) dz. Run together they also clear the rk and hand back ak by itself, with the on-the-circle identity below the figure making the swap exact.
The three moves, live. Each coloured arrow is a block ak(z−P)k, laid tip-to-tail; their sum (the white dot) is f on the rim. Freeze the block you pick (here block 3): it turns gold and stands still while the rest keep spinning. Press play to walk θ once around. The pink dot is the running average; it spirals onto that block's strength (the gold ring) as the turn completes. Every spinning block averages to nothing; only the frozen one survives. The readout prints both as complex numbers, so you can watch the pink average land exactly on the gold strength akrk. Hit 🎲 new blocks to randomize the strengths and see it hold for any function.
Why trade the angle for a loop at all? Because the loop form is Cauchy's coefficient formula: the same strength ak, now wearing the third of our three names, and read off from any circle at once instead of being tied to the one we drew. That payoff is what makes the rewrite worth doing. In the four steps we multiplied by exp(−3iθ) and averaged dθ; the machine below instead divides by (z−P)k+1 and loops dz. Nothing to take on faith, though: both halves fall out of what we already have. Step 1 made the k-th power a rate-k arrow, (z−P)k = rk exp(ikθ); invert it and the freeze exp(−ikθ) stands on its own. And §3 already found dz = i(z−P) dθ, the loop's version of the angle-step. Lined up:
So the freeze turned into a division by (z−P)k, and the angle-step dθ supplied the loop's dz with one extra power. Feed the identity into the hinge (the §2 average that gave akrk): the rk the loop carries cancels the one on the left, and what stands there is Cauchy's coefficient formula, handing back ak with no radius left in it at all:
Read end to end, the whole move is one machine: feed in f, turn the crank once around the loop, read off ak.
THE MOVE, END TO END
The three lecture rooms collapse into one identity. Taylor asks what are the coefficients? Fourier computes them by averaging spinning arrows over a full turn. Cauchy is the bridge that says the two are literally equal. One move, one τ's worth of averaging, wearing three names.
One loose thread, worth pulling: ak carries no r, yet it is read off a circle of radius r. No contradiction: the loop integral comes out the same for every radius, so the coefficient cannot depend on which circle you drew. Any circle around P, so long as f stays smooth inside it, returns the same ak. That radius-blindness is the rigidity of complex analysis, stated in one line.
With every strength in hand, the blocks reassemble into the picture we started with, now exact: Taylor's series, each coefficient the number one full turn hands back.
TAYLOR'S SERIES
f(z) = a0 + a1(z−P) + a2(z−P)² + a3(z−P)³ + …each strength ak is exactly the loop coefficient derived just above, the number one full turn hands back.
Why τ, and the One Sentence Behind All Three
Why the same τ sits in all three, and the single identity each of them reads.
One thread ties the three together, and it shows in the symbol they all carry. τ appears in every one of them for the same reason: each is an average over one full turn, and τ is the name of one full turn (the 2π you usually see is that same turn in disguise). So stand back from the symbols, and a single identity has been running under §2, §3, and §4 all along. Here it is in full: spin the input once around a circle about P, freeze block k by multiplying by the k-th arrow, and average over one full turn.
the τ is one full turn; the exp(−ikθ) is the freeze; the average is everything that spins cancelling to nothing.
Three theorems are just three readings of that one line.
So here is the whole article in one sentence: to measure how much of the k-th ingredient a function holds, spin its input once around a circle and average it against the k-th arrow.
That single average is read in three lecture rooms. At the centre it is a building-block strength: that is Taylor. On the rim it is a frequency: that is Fourier. Taken as the journey itself it is a contour integral: that is Cauchy. One τ's worth of averaging, written out three times.
Three theorems turned out to be one sentence, read three ways, and the sentence was about going around a circle exactly once.
Whenever τ appears in an equation,
a circle will always emerge somewhere.
Related: Euler's Formula in its Full Elegance →