Taylor Series Expansion

An option's price is some complicated curved function of the stock price. Nobody prices a hedge by recomputing that whole curved function every time the stock ticks — instead, a trading desk quotes delta and gamma, and uses them to predict the P&L of a small move without ever touching the pricing model again. That shortcut works because of one of the most useful facts in all of applied mathematics: near any point, a smooth curve can be approximated to arbitrary accuracy by a polynomial built entirely out of derivatives at that one point. That polynomial is the Taylor series.

The analogy: describing a hill to someone standing on it

Tell a friend standing on a hillside what the terrain looks like nearby, without a map — just words. First fact: how high up they are (their current altitude — zeroth-order information, no shape at all yet). Second fact: which way is downhill and how steep (the slope right where they stand — first-order information, enough to predict altitude a few steps away, assuming the hill were flat everywhere). Third fact: whether the hill is curving away underfoot, bowl-shaped or dome-shaped (second-order information — refines the flat-slope guess for a hillside that's actually curved). Each additional fact about the local shape — how it changes, how that change is changing — sharpens your prediction of what's a little further along, without you ever having to see the rest of the hill. Stack enough of these local facts and your prediction becomes a nearly perfect map of the terrain nearby. That is the Taylor series: build a global-looking approximation entirely from information available at a single point.

Writing it down

For a function $f$ that's smooth enough (has enough derivatives) at a point $a$, the Taylor series expands $f$ near $a$ as:

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

In words: the function's value near $x$ equals its value at $a$ (the zeroth-order, "current altitude" term), plus the slope at $a$ times how far you've moved (first-order, "walk downhill at this rate"), plus a correction for curvature (second-order, weighted by $1/2!=1/2$), plus ever-smaller corrections for how the curvature itself is changing, and so on. Each term uses one more derivative, evaluated only at the single point $a$, and gets divided by a growing factorial that keeps higher terms from exploding for well-behaved functions.

Truncate after the second-order term and you get the second-order Taylor approximation, the single most-used version in practice:

$$f(x) \approx f(a) + f'(a)(x-a) + \tfrac12 f''(a)(x-a)^2 .$$

In words: approximate the curve by a parabola that matches the function's value, slope, and curvature exactly at $a$. In options language, $f'(a)$ is delta and $f''(a)$ is gamma — the entire delta-gamma P&L approximation used on every trading desk is nothing but a truncated Taylor series.

Adjust the curve here and picture zooming in on any single point: locally, every smooth curve looks like some parabola — a specific choice of the coefficients this explorer lets you drag. Taylor's theorem is the statement that those local matching coefficients are always exactly the function's own derivatives at that point.

Worked example 1: approximating $\sqrt{x}$ near $x=100$, by hand

Let $f(x)=\sqrt{x}$, expand around $a=100$ (where $f(100)=10$, a convenient exact value). Derivatives: $f'(x) = \tfrac12 x^{-1/2}$, so $f'(100) = \tfrac12(0.1) = 0.05$. $f''(x) = -\tfrac14 x^{-3/2}$, so $f''(100) = -\tfrac14(0.001) = -0.00025$.

Estimate $\sqrt{103}$ using $x-a = 3$: first order alone gives $10 + 0.05(3) = 10.15$. Adding the second-order term: $10 + 0.05(3) + \tfrac12(-0.00025)(9) = 10.15 - 0.001125 = 10.148875$.

Compare to the true value: $\sqrt{103} = 10.148892\ldots$ The first-order guess is off by about $0.000117$; the second-order guess is off by only about $0.000017$ — roughly seven times more accurate from one extra term, and correct to five decimal places using nothing but two derivatives evaluated at a single, convenient point.

Worked example 2: delta-gamma P&L on an option book

An option is worth $4.50 with delta $0.55$ and gamma $0.03$ (both per $1 move in the underlying, standard desk convention). The stock jumps from $100 to $102, a move of $x-a=2$.

First-order (delta-only) estimate: $\Delta\text{P\&L} \approx 0.55 \times 2 = 1.10$, so the option should be worth about $4.50+1.10=5.60$.

Second-order (delta-gamma) estimate: add $\tfrac12 \times 0.03 \times 2^2 = \tfrac12(0.03)(4) = 0.06$. Total predicted move: $1.10+0.06=1.16$, so the option should be worth about $4.50+1.16=5.66$.

Suppose the true model price at $102 (recomputed from scratch) is $5.665. The delta-only approximation was off by $0.065; the delta-gamma approximation was off by only $0.005 — thirteen times tighter, from tracking one more derivative, exactly mirroring the square-root example. This is precisely why "gamma P&L" is a standard line item on option risk reports: it is the second-order Taylor correction, isolated and tracked on its own.

What this means in practice

• Every Greek is a Taylor coefficient. Delta, gamma, vega, theta are just $f'$, $f''$, and partial derivatives with respect to different inputs — the whole Greeks framework is Taylor's theorem applied to an option pricing function.
• It's why hedges need rebalancing. A Taylor approximation is only accurate near the expansion point $a$; as the stock moves further from where delta and gamma were last computed, the truncated series drifts from the true price, which is exactly the error a trader is managing by rehedging.
• It underlies Newton's method and finite-difference Greeks. Both tools invert or extend a Taylor expansion — Newton's method uses the first-order term to jump toward a root; bump-and-reprice Greeks estimate the derivatives themselves by comparing function values at nearby points.