Part 0 · The Toolkit — Chapter 0.3

Series, Approximation, Orders of Magnitude

How physicists actually think, and why every theory in this book is the first few terms of another one.

Tools you'll need  — Chapter 0.1: linearisation, higher derivatives, the definition of $\ee$. Chapter 0.2: the fundamental theorem of calculus and integration by parts, which is the engine of §1.

1 · Taylor's theorem, derived

Almost every book states Taylor's series and then, some pages later, mumbles about a remainder. We will do it the other way round, because the remainder is the entire point. A series without an error estimate is not a result. It is a wish.

Start from the fundamental theorem of calculus, which says nothing more than "the function at the end equals the function at the start plus the accumulated change":

That is exact. No approximation has been made and none will be. Every step below is an identity.

Our next move is to integrate by parts, and there is exactly one trick in the whole derivation. It is worth naming before we use it: the trick is the choice of antiderivative. Integrating $\dd t$ gives $t + C$ for any constant $C$, and we get to pick $C$. We pick it so that the upper boundary term vanishes. So take the antiderivative to be $t-(a+h)$, which is zero at $t=a+h$. Then

The boundary term at $t=a+h$ is zero by construction. At $t=a$ it is $-f'(a)(a-a-h)=+f'(a)h$, and it enters with a minus sign in front of the bracket's lower limit, so it contributes $+f'(a)h$. Now flip the sign inside the remaining integral, and we have

Let's stop and look at what we have. The first two terms are exactly the linear approximation of Chapter 0.1. The third term is the new thing. The vague phrase "smaller than $h$" has been replaced by an explicit integral, and that integral is the error, written down in full, with no limits taken and nothing swept anywhere.

Now do it again. The goal this time is to peel one more polynomial term off that integral, so we integrate by parts a second time. In $\int (a+h-t)f''(t)\,\dd t$, let $u=f''(t)$ and $\dd v=(a+h-t)\,\dd t$, whose antiderivative $-\tfrac12(a+h-t)^{2}$ again vanishes at the upper limit:

The underbraced boundary term is the next polynomial term, so put that whole line back into the result of the first round and we have

There is the $\tfrac12 f''(a)h^{2}$ that the zoom figure in Chapter 0.1 measured numerically. It was never a coincidence. It is the second boundary term.

The pattern is now visible. Each integration by parts peels off one more term of a polynomial and hands back an integral with one more power of $(a+h-t)$ and one more derivative on $f$. Doing it $n$ times gives Taylor's theorem with the integral form of the remainder:

(As usual $f^{(0)}=f$ and $0!=1$, so the $k=0$ term is just $f(a)$.) Setting $a=0$ and renaming $h\to x$ gives the Maclaurin form, which is what you will use nine times out of ten: $f(x)=\sum_k f^{(k)}(0)x^{k}/k!\,+\,R$.

Grind box — the induction, and the Lagrange form of the remainder

The induction. Define, for $n\ge1$,

$$R_{n} \;\equiv\; \int_{a}^{a+h}\frac{(a+h-t)^{n-1}}{(n-1)!}\,f^{(n)}(t)\,\dd t.$$

Equation (0.3.1) says $f(a+h)=f(a)+R_{1}$, which starts the induction. Integrating by parts with $u=f^{(n)}(t)$ and antiderivative $v=-\dfrac{(a+h-t)^{n}}{n!}$ (again chosen to vanish at $t=a+h$):

$$R_{n} = \Big[-\tfrac{(a+h-t)^{n}}{n!}f^{(n)}(t)\Big]_{a}^{a+h} + \int_{a}^{a+h}\frac{(a+h-t)^{n}}{n!}f^{(n+1)}(t)\,\dd t \;=\; \frac{h^{n}}{n!}f^{(n)}(a) \;+\; R_{n+1}.$$

Each application converts a remainder into "the next polynomial term plus the next remainder", so applying it $n$ times to $R_1$ gives (0.3.6) exactly. The only hypothesis used is that $f^{(n+1)}$ exists and is continuous on the closed interval, which is enough for integration by parts to be legal.

Lagrange form. The integral remainder is exact but not always easy to look at. Note first that, substituting $s=a+h-t$,

$$\int_{a}^{a+h}\frac{(a+h-t)^{n}}{n!}\,\dd t \;=\; \int_{0}^{h}\frac{s^{n}}{n!}\,\dd s \;=\; \frac{h^{n+1}}{(n+1)!}.$$

For $h\gt0$ the weight $(a+h-t)^{n}/n!$ is non-negative on the whole interval, so the mean value theorem for integrals applies: there is some $\xi$ between $a$ and $a+h$ with

$$R_{n+1} \;=\; f^{(n+1)}(\xi)\int_{a}^{a+h}\frac{(a+h-t)^{n}}{n!}\,\dd t \;=\; \frac{f^{(n+1)}(\xi)}{(n+1)!}\,h^{n+1}.$$

(For $h\lt0$, reverse the orientation of the interval. The weight is then of one sign again, and the same conclusion follows.) This is the form you will actually use for estimates. You rarely know $\xi$, but you can almost always bound $\abs{f^{(n+1)}}$ on the interval.

One warning to carry into §3, since it costs nothing to state now. The Taylor series is what you get by letting $n\to\infty$ in (0.3.6). It equals $f$ precisely when $R_{n+1}\to0$. That is not the same as the series converging. Those are two different conditions, and the difference is the subject of a callout later in this chapter that you may find genuinely unsettling.

2 · The essential expansions

Five expansions carry most of the load in physics. We derive each one rather than quoting it. The derivations take one line apiece, and you should never be in the position of half-remembering whether there is a factorial in the binomial series.

2.1 · The exponential

By definition (Chapter 0.1) $\ee^{x}$ is its own derivative, so $f^{(k)}(x)=\ee^{x}$ and $f^{(k)}(0)=1$ for every $k$. Feeding that into the Maclaurin form:

Does it converge to $\ee^x$? To find out, use the Lagrange bound. On the interval between $0$ and $x$ we have $\ee^{\xi}\le \ee^{\abs{x}}$, so $\abs{R_{n+1}}\le \ee^{\abs{x}}\abs{x}^{n+1}/(n+1)!$. Once $n+1\gt 2\abs{x}$, every further step multiplies this by $\abs{x}/(n+2)\lt\tfrac12$. So the bound is eventually smaller than a constant times $2^{-n}$, and it therefore goes to zero.

The series converges to the function for every real $x$. It also converges for every complex $x$, which is something we are about to exploit.

2.2 · Sine and cosine

Derivatives of $\sin$ cycle with period four: $\sin,\cos,-\sin,-\cos,\dots$, which at $x=0$ evaluate to $0,1,0,-1,\dots$. Only odd powers survive, alternating in sign:

the second by the identical argument starting from $\cos$ (values $1,0,-1,0,\dots$ at the origin). Convergence everywhere is immediate: every derivative of $\sin$ or $\cos$ is bounded by $1$ in absolute value, so $\abs{R_{n+1}}\le\abs{x}^{n+1}/(n+1)!\to0$ by the same factorial argument.

Notice the leading term of $\sin x$. It is $x$, which is the "small-angle approximation" you have used since school. Now notice the first correction, $-x^{3}/6$, which tells you exactly when to stop using it. At $x=0.1$ rad that correction is $1.7\times10^{-4}$, a relative error of $0.17\%$. That number is not a rule of thumb. It is a computed quantity.

2.3 · The logarithm

Differentiating $\ln(1+x)$ repeatedly gives $f'=(1+x)^{-1}$, $f''=-(1+x)^{-2}$, $f'''=2(1+x)^{-3}$, and in general $f^{(k)}(x)=(-1)^{k-1}(k-1)!\,(1+x)^{-k}$. At $x=0$ that is $(-1)^{k-1}(k-1)!$, and dividing by $k!$ leaves $(-1)^{k-1}/k$:

The restriction $\abs{x}\lt1$ is not decoration, and §3 explains where it comes from. The leading behaviour $\ln(1+x)\approx x$ is one of the two most-used approximations in quantitative science. It is why small log-hazard-ratios read directly as percentages, a point we return to in the Familiar Ground box.

2.4 · The binomial series, for arbitrary real exponent

This is the workhorse. Let $f(x)=(1+x)^{\alpha}$, with $\alpha$ any real number at all, not necessarily a positive integer. Then $f'=\alpha(1+x)^{\alpha-1}$, $f''=\alpha(\alpha-1)(1+x)^{\alpha-2}$, and in general

Dividing each of those by $k!$ is what defines the generalised binomial coefficient, and doing so gives us

When $\alpha$ is a non-negative integer the product eventually hits a factor of zero, the series terminates, and you recover the finite binomial theorem you already know. For every other $\alpha$ it runs forever. Two cases deserve to be committed to memory, because between them they cover essentially every square root you will ever have to linearise:

The second is $\alpha=-\tfrac12$ with $x\to-x$, and it is the relativistic $\gamma$ factor. Put $x=v^{2}/c^{2}$ and you have, in one line, the expansion that Chapter 0.1 obtained by grinding out two derivatives. Now you also get every higher term for free, which is exactly what Worked example 2 will cash in.

Grind box — where the radius $\abs{x}\lt1$ comes from, without complex analysis

For the logarithm you can bypass Taylor's theorem entirely and get an exact remainder from the finite geometric sum. For any $t\neq-1$ and any $n$, multiply out to verify

$$\frac{1}{1+t} \;=\; 1 - t + t^{2} - \cdots + (-t)^{n-1} + \frac{(-t)^{n}}{1+t}.$$

This is an algebraic identity, not an approximation: the finite sum telescopes when you multiply both sides by $(1+t)$. Now integrate from $0$ to $x$, using $\int_0^x t^k\,\dd t = x^{k+1}/(k+1)$ and $\int_0^x \dd t/(1+t)=\ln(1+x)$:

$$\ln(1+x) = x - \frac{x^{2}}{2}+\cdots+\frac{(-1)^{n-1}x^{n}}{n} \;+\; \underbrace{(-1)^{n}\int_{0}^{x}\frac{t^{n}}{1+t}\,\dd t}_{\textstyle R_{n}}.$$

For $0\le x\lt1$ the integrand is at most $t^{n}$, so $\abs{R_{n}}\le x^{n+1}/(n+1)\to0$ and the series converges to the function. For $-1\lt x\lt0$ the factor $1/(1+t)$ is no longer bounded by $1$, so that argument does not run again unchanged. Substitute $t=-s$ and bound $1/(1-s)\le 1/(1-\abs x)$ over the whole interval, which gives $\abs{R_{n}}\le \abs{x}^{n+1}/\big[(n+1)(1-\abs x)\big]\to0$ as well. The bound degrades as $x\to-1$, and that degradation is the analytic shadow of $\ln(1+x)\to-\infty$ there. For $x\gt1$ the terms $x^{n}/n$ themselves blow up, so the series cannot converge at all, since a convergent series must have terms tending to zero. The radius is exactly $1$, and here the reason is visible on the real line, because $\ln(1+x)$ runs to $-\infty$ at $x=-1$.

The binomial radius. Take the ratio of consecutive coefficients in (0.3.11):

$$\left|\frac{\binom{\alpha}{k+1}x^{k+1}}{\binom{\alpha}{k}x^{k}}\right| = \left|\frac{\alpha-k}{k+1}\right|\abs{x} \;\longrightarrow\; \abs{x} \quad\text{as } k\to\infty.$$

By the ratio test of §3 the series converges for $\abs{x}\lt1$ and diverges for $\abs{x}\gt1$, for every non-integer $\alpha$. Again the reason is visible: $(1+x)^{\alpha}$ misbehaves at $x=-1$, one unit from the origin.

2.5 · Reference table

Here are all five in one place, together with the radius each one is good out to and the chapter where this book spends it.

Function	Series about $0$	Radius	Where it is spent in this book
$\ee^{x}$	$1+x+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+\cdots$	$\infty$	Everywhere. Time evolution, partition functions, propagators
$\sin x$	$x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\cdots$	$\infty$	Small oscillations (0.8), string modes (7.4)
$\cos x$	$1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\cdots$	$\infty$	Pendulum potential (below), lattice dispersion
$\ln(1+x)$	$x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\cdots$	$1$	Entropy, log-likelihoods, running couplings (5.11)
$\dfrac{1}{1-x}$	$1+x+x^{2}+x^{3}+\cdots$	$1$	Geometric resummation, and the propagator in 5.6
$(1+x)^{\alpha}$	$1+\alpha x+\dfrac{\alpha(\alpha-1)}{2}x^{2}+\cdots$	$1$	Every square root
$\sqrt{1+x\,}$	$1+\dfrac{x}{2}-\dfrac{x^{2}}{8}+\dfrac{x^{3}}{16}-\cdots$	$1$	$E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}$ (2.5)
$\left(1-x\right)^{-1/2}$	$1+\dfrac{x}{2}+\dfrac{3x^{2}}{8}+\dfrac{5x^{3}}{16}+\cdots$	$1$	The Lorentz factor $\gamma$ (2.5)

2.6 · Euler's formula, and why complex numbers are not a convenience

The exponential series (0.3.7) converges for every real $x$, and its convergence proof used only $\abs{x}$. So it converges just as well for complex arguments.

We therefore define $\ee^{z}$ for complex $z$ by that series. There is no other sensible choice, since any definition agreeing with $\ee^{x}$ on the real line and having a power series must be this one.

Now substitute $z=\ii\theta$ with $\theta$ real, and use the four-fold cycle $\ii^{0}=1,\ \ii^{1}=\ii,\ \ii^{2}=-1,\ \ii^{3}=-\ii$, which then repeats:

Splitting the sum into even and odd $k$ is legitimate because the series converges absolutely, meaning that the sum of $\abs{\theta}^{k}/k!$ converges on its own. Absolutely convergent series may be rearranged freely without changing their value. So now compare the two brackets with (0.3.8) and read off

which is not an identity between three unrelated functions but a statement that they were the same object all along, sorted by parity. Three consequences, each of which we will use repeatedly.

1. It has unit modulus. $\abs{\ee^{\ii\theta}}^{2}=\cos^{2}\theta+\sin^{2}\theta=1$. So $\ee^{\ii\theta}$ is a point on the unit circle at angle $\theta$, and $\theta$ is arc length along that circle. The identity $\ee^{\ii\pi}=-1$ is the special case "walk half way round".

2. Multiplying by $\ee^{\ii\theta}$ is a rotation. Take any complex number $z=x+\ii y$, thought of as the point $(x,y)$ in the plane, and multiply:

Those are precisely the components of the two-dimensional rotation matrix acting on $(x,y)$. Rotation, one of the central objects of the rest of this book, is multiplication by a complex exponential.

That fact does not stay small. In Chapter 0.4 it becomes the statement that $\mathrm{U}(1)$ and $\mathrm{SO}(2)$ are the same group. In Chapter 6.3, the fact that the phase of a quantum field lives on a circle is what generates electromagnetism.

3. It converts decay into oscillation. Chapter 0.1 left you holding a comparison. The clearance equation $\dot C=-kC$ gives $C=C_{0}\ee^{-kt}$, while quantum time evolution $\ii\hbar\,\partial_{t}\psi = E\psi$ gives $\psi = \psi_{0}\ee^{-\ii Et/\hbar}$. Structurally the two are identical. The only difference is the $\ii$. Euler's formula now tells you exactly what that $\ii$ does:

A decaying exponential loses amplitude. An imaginary exponential keeps its length and changes only its direction. That is the whole reason quantum mechanics conserves probability: $\abs{\psi}^{2}$ is the squared length of a vector that is being rotated, and rotations preserve length.

This is also why complex numbers are not an accounting convenience in quantum mechanics. Real exponentials can only grow or decay. You need the circle, and the circle is $\ee^{\ii\theta}$.

One free bonus, to show the machinery paying rent immediately. From $\ee^{\ii(\alpha+\beta)}=\ee^{\ii\alpha}\ee^{\ii\beta}$, expand both sides with (0.3.14) and equate real and imaginary parts:

The trigonometric identities you were made to memorise are the statement that $\ee^{\ii\theta}$ obeys the exponential law. There is nothing else to them.

3 · Convergence, radius, and a genuine surprise

A power series $\sum_{k} a_{k}x^{k}$ is an infinite sum, and infinite sums are promises. We need a test for when the promise is kept.

3.1 · The ratio test

Suppose $\abs{c_{k+1}/c_{k}}\to L$. There are two cases, and we take them in turn.

If $L\lt1$, pick any $r$ with $L\lt r\lt1$. Then beyond some index $K$ we have $\abs{c_{k+1}}\le r\abs{c_{k}}$, hence $\abs{c_{K+m}}\le r^{m}\abs{c_{K}}$. The tail of the series is therefore dominated term by term by the geometric series $\abs{c_{K}}\sum_{m}r^{m}=\abs{c_{K}}/(1-r)$, which is finite. A series of positive terms with bounded partial sums converges, because its partial sums increase and are bounded above.

If instead $L\gt1$ the terms eventually grow, so they cannot tend to zero, and a series whose terms do not tend to zero cannot converge. Hence:

Apply this to a power series by putting $c_{k}=a_{k}x^{k}$. The ratio is $\abs{a_{k+1}/a_{k}}\abs{x}$, so if $\abs{a_{k}/a_{k+1}}\to R$ the series converges for $\abs{x}\lt R$ and diverges for $\abs{x}\gt R$. That number $R$ is the radius of convergence.

Try it on two of §2's expansions. For $\ee^{x}$, $\abs{a_{k}/a_{k+1}}=(k+1)!/k!=k+1\to\infty$, so the radius is infinite. For $\ln(1+x)$, $\abs{a_{k}/a_{k+1}}=(k+1)/k\to1$, so the radius is $1$, as the grind box already showed by hand.

3.2 · The surprise

Here is a function with nothing whatsoever wrong with it:

(the series follows from $\frac{1}{1-u}=\sum u^{k}$ with $u=-x^{2}$, or from the same exact finite-sum identity used in the grind box). On the real line $f$ is smooth, positive, bounded by $1$, infinitely differentiable, and utterly featureless. It is the Lorentzian lineshape, and it does nothing interesting at $x=\pm1$ or anywhere else.

Yet by the ratio test its series converges only for $\abs{x}\lt1$. At $x=2$ the terms are $1,-4,16,-64,\dots$, and the "approximation" runs away to infinity no matter how many terms you take.

Something is setting a radius of $1$, and it is not visible on the real line. Watch it happen:

target sin x 1/(1+x²)

order N

N = 1

trust radius ≈ 0.49

3.3 · Where the boundary actually is

The series in (0.3.19) makes perfect sense with a complex variable $z$ in place of $x$, and there the answer is immediate: $1/(1+z^{2})$ blows up at $z=\pm\ii$, both at distance $1$ from the origin. In fact at $z=\ii$ every term of the series equals $(-1)^{k}\ii^{2k}=(-1)^{k}(-1)^{k}=1$, so the partial sums are $1,2,3,\dots$. The series does not merely fail to represent the function there. It diverges outright.

That is enough to pin the radius, given one small lemma: the set where a power series converges is a disc. If $\sum a_{k}z_{1}^{k}$ converges then its terms are bounded, say $\abs{a_{k}z_{1}^{k}}\le M$, and for any $z$ with $\abs{z}\lt\abs{z_{1}}$,

which is a convergent geometric series. So convergence at one point forces absolute convergence everywhere strictly closer to the origin. Contrapositive: divergence at $z=\ii$ forces divergence at every $\abs{z}\gt1$. The radius is exactly $1$, and it was determined by two points that do not lie on the real line at all.

4 · Asymptotic series — where physics really lives

Everything so far has been about series that converge. Now the uncomfortable truth. Most of the series physics actually runs on do not converge, for any value of the parameter, ever. They are still the most accurate predictions humanity has made.

That apparent contradiction is resolved by a single idea, and this section is where the chapter earns its title.

4.1 · Two ways to take a limit

A power series statement involves two variables, the argument $x$ and the number of terms $N$, and there are two different things you can mean.

Convergent: fix $x$, let $N\to\infty$, and demand the partial sums approach $f(x)$. That is what §3 was about.

Asymptotic: fix $N$, let $x\to0$, and demand the error die faster than the last term you kept. Formally, $f(x)\sim\sum_{k}a_{k}x^{k}$ as $x\to0^{+}$ means that for each fixed $N$,

These are genuinely different demands, and the two limits do not commute. A series can satisfy (0.3.21) for every $N$ while diverging for every $x\neq0$.

Put the two questions side by side. The convergent case asks "does adding terms help forever?". The asymptotic case asks "does the approximation improve as the parameter gets small?".

Physics almost always wants the second question. In physics the parameter is small and handed to you, whether it is $v/c$ or $\hbar$ or the fine structure constant, and the number of terms is a choice you make.

4.2 · A completely explicit example

Consider

This integral is perfectly well defined and finite for every $x\gt0$: the integrand is positive and bounded by $\ee^{-t}$, so $0\lt F(x)\le1$.

We want a series in $x$, so expand the denominator with the exact finite-sum identity from the grind box, $\frac{1}{1+u}=\sum_{k=0}^{N}(-u)^{k}+\frac{(-u)^{N+1}}{1+u}$ with $u=xt$, and integrate term by term using $\int_{0}^{\infty}\ee^{-t}t^{k}\,\dd t=k!$ (integrate by parts $k$ times):

Everything in those two lines is exact. Our next goal is to put a bound on that remainder, and since $1/(1+xt)\le1$ for $x,t\gt0$, it obeys

The right-hand side is precisely the magnitude of the first term you dropped. That bound is beautiful, and it is the practical rule of thumb for asymptotic series: the error is no bigger than the first omitted term.

Note what has and has not been shown. Here it is a theorem, because the remainder came out as an explicit alternating integral we could bound. For a general asymptotic series it is a reliable heuristic and not a theorem, and the honest general statement is only $\abs{R_{N+1}}=O(x^{N+1})$. The bound also proves (0.3.21) immediately, since $(N+1)!\,x^{N+1}/x^{N}=(N+1)!\,x\to0$. So the series is asymptotic to $F$.

Now apply the ratio test to $\sum(-1)^{k}k!x^{k}$. The ratio is $(k+1)\abs{x}$, which tends to infinity for every $x\neq0$. The radius of convergence is zero. The series diverges everywhere except the single point $x=0$.

Both statements are true at once, and the resolution is in the shape of the terms $k!\,x^{k}$. Multiply one term by $x(k+1)$ to get the next. While $k\lt1/x$ that factor is less than one and the terms shrink. Once $k\gt1/x$ it exceeds one and they grow, forever. The series is useful up to the turning point and poisonous after it.

parameter x

x = 0.100

best at N = 9, error = 1.77e-4 (e^(−1/x) = 4.54e-5)

Grind box — optimal truncation and the $\ee^{-1/x}$ accuracy floor

We want the $N$ minimising the error bound $N!\,x^{N}$. Take logarithms and use the crude Stirling estimate, which we can derive on the spot by comparing a sum to an integral:

$$\ln N! = \sum_{k=1}^{N}\ln k \;\approx\; \int_{1}^{N}\ln t\,\dd t = N\ln N - N + 1.$$

(The comparison is legitimate because $\ln t$ is increasing, so the sum is trapped between $\int_{0}^{N}$ and $\int_{1}^{N+1}$. We need only the leading terms.) Then

$$\Phi(N) \equiv \ln\!\big(N!\,x^{N}\big) \approx N\ln N - N + N\ln x.$$

Differentiate with respect to $N$, treating it as continuous:

$$\Phi'(N) = \ln N + 1 - 1 + \ln x = \ln(Nx) \;=\; 0 \quad\Longrightarrow\quad \boxed{\,N_{\ast} = \tfrac1x\,}.$$

It is a minimum because $\Phi''=1/N\gt0$. Substituting back, at $N_{\ast}x=1$:

$$\Phi(N_{\ast}) = N_{\ast}\ln(N_{\ast}x) - N_{\ast} = -N_{\ast} = -\frac1x \quad\Longrightarrow\quad \text{best error} \;\sim\; \ee^{-1/x}.$$

Check it against the figure. At $x=0.1$ this predicts stopping at about ten terms with an accuracy near $\ee^{-10}=4.5\times10^{-5}$. The numerically computed optimum is $N=9$ with error $1.8\times10^{-4}$. At $x=0.05$ the predicted floor is $\ee^{-20}=2.1\times10^{-9}$ and the actual optimum is $N=19$ with error $1.1\times10^{-8}$. The exponential scaling is exactly right. The factor of a few comes from the subleading $\sqrt{2\pi N}$ in Stirling that we discarded.

Remember this shape. An asymptotic series in a small parameter $x$ delivers accuracy $\ee^{-1/x}$ and no better. That exponentially small leftover is not noise. It is a real physical effect that the power series is structurally incapable of seeing, as the warning box below explains.

4.3 · The punchline

Now the statement this section exists for.

The perturbation series of quantum electrodynamics is asymptotic and divergent. It predicts the electron's magnetic moment to about twelve significant figures. That is the most accurately confirmed prediction in the history of science.

The expansion parameter is the fine structure constant $\alpha\approx1/137$, and the series for the electron's anomalous magnetic moment $a_{e}=(g-2)/2$ runs in powers of $\alpha/\pi\approx1/430$. The measured value is $a_{e}=1.15965218059(13)\times10^{-3}$, which is twelve significant figures.

The calculated value has been worked out through five orders in $\alpha$, with over twelve thousand Feynman diagrams at the fifth order. It agrees within the combined uncertainties, at the level of about one part in $10^{12}$.

That the series diverges is not a suspicion but an argument, due to Dyson in 1952, and it is worth sketching because it is pure §3 reasoning. If the series in $\alpha$ had any nonzero radius of convergence, it would also converge for small negative $\alpha$, since a radius means a disc.

But with $\alpha\lt0$ like charges attract. The vacuum could then lower its energy without bound by spontaneously separating clumps of like charge, so there is no stable ground state and no sensible theory at all. A function cannot be analytic at a point on the boundary between "fine" and "catastrophically ill-defined". Hence the radius is zero. (The mechanism, if you want one: the number of Feynman diagrams at order $n$ grows like $n!$, exactly as in our toy integral.)

So how is anyone getting twelve digits? By the grind box. With $x=\alpha/\pi\approx1/430$, optimal truncation sits near $N_{\ast}\approx430$ terms and the accuracy floor is $\ee^{-\pi/\alpha}\sim10^{-187}$. Nobody is anywhere near the turning point. The terms are still falling steeply at fifth order and will keep falling for another four hundred.

Physicists are not being sloppy with a divergent series. They are using an asymptotic expansion correctly, in the regime where the error bound is the first omitted term and that term is astronomically small. Chapter 5.11 makes this precise and shows what the leftover $10^{-187}$ actually is.

5 · Dimensional analysis and orders of magnitude

The last tool in this chapter costs almost nothing and repeatedly returns answers that look like they should have required a calculation.

5.1 · Homogeneity, and why it is not a convention

Every additive term in a physical equation must have the same dimensions. The reason is not etiquette. A physical law must hold whatever units you measure in.

Suppose an equation reads $A=B+C$ where $B$ is a length and $C$ a time. Switch from metres to feet and $B$ changes by a factor of $3.28$ while $C$ does not. An equation true in one unit system would then be false in another, which would make the choice of units a piece of physics. It is not. Hence $[B]=[C]$, always.

In mechanics we track three independent dimensions, mass $\mathsf{M}$, length $\mathsf{L}$ and time $\mathsf{T}$, and write $[v]=\mathsf{L}\mathsf{T}^{-1}$, $[F]=\mathsf{M}\mathsf{L}\mathsf{T}^{-2}$, and so on.

The same argument in its stronger form is the Buckingham π idea. If a physical relation involves $n$ dimensional quantities built from $k$ independent dimensions, it is equivalent to a relation among just $n-k$ dimensionless combinations.

The reason is simple enough to state in one sentence: you may use $k$ of the quantities to define your units, setting them to $1$, after which only $n-k$ genuinely independent numbers remain to be related. Everything below is that sentence applied.

5.2 · Demonstration: the pendulum

What can the period $T$ of a pendulum depend on? Plausibly the length $L$ ($\mathsf{L}$), the bob mass $m$ ($\mathsf{M}$), and gravity $g$ ($\mathsf{L}\mathsf{T}^{-2}$). That is $n=4$ quantities and $k=3$ dimensions, so there is $4-3=1$ dimensionless group and the physics must be that this group is a constant.

Start with the mass. It is the only quantity carrying $\mathsf{M}$, so no dimensionless group can contain it, since any power of $m$ would leave an uncancelled $\mathsf{M}$ behind. That is a conclusion before any calculation: the period does not depend on the mass. Now seek $T \propto L^{a}g^{b}$:

so $T=C\sqrt{L/g}$ for some pure number $C$. Two lines, no differential equation, and we have the $\sqrt{L}$ scaling and the independence from mass.

What dimensional analysis cannot give you is $C$. It happens to be $2\pi$, and getting it requires actually solving the equation of motion, which we do in Worked example 1.

Worse, dimensional analysis is blind to dimensionless parameters. The amplitude $\theta_{0}$ is already a pure number, so nothing forbids $C$ from being a function $C(\theta_{0})$. It is. The period of a real pendulum grows with amplitude, and the "constant" $C$ is only constant in the small-angle limit.

So the method tells you the structure and is silent about everything dimensionless. That is precisely why $\alpha\approx1/137$ had to be measured rather than derived.

5.3 · The payoff: the Planck scale

Now something that looks like it should be impossible. Take the three constants that define the three great theories:

Constant	Belongs to	Dimensions
$c=3.00\times10^{8}\ \mathrm{m\,s^{-1}}$	Relativity	$\mathsf{L}\mathsf{T}^{-1}$
$\hbar=1.055\times10^{-34}\ \mathrm{J\,s}$	Quantum mechanics	$\mathsf{M}\mathsf{L}^{2}\mathsf{T}^{-1}$
$G=6.674\times10^{-11}\ \mathrm{m^{3}kg^{-1}s^{-2}}$	Gravity	$\mathsf{M}^{-1}\mathsf{L}^{3}\mathsf{T}^{-2}$

Ask for a length: $\ell = G^{a}\hbar^{b}c^{d}$. Matching exponents of $\mathsf{M}$, $\mathsf{L}$, $\mathsf{T}$ gives three linear equations:

From the first, $b=a$. The third then gives $d=-3a$. Substituting both into the second, $3a+2a-3a=2a=1$, so $a=b=\tfrac12$ and $d=-\tfrac32$. Putting those exponents back gives the Planck length, and with it a time and a mass:

The mass looks unremarkable at first, about the mass of a flea's egg. Then convert it to an energy: $m_{P}c^{2}=1.22\times10^{19}\ \mathrm{GeV}$, roughly $10^{15}$ times the energy the Large Hadron Collider delivers per collision.

Now the part that matters. The three equations (0.3.26) form a $3\times3$ linear system whose coefficient matrix has determinant

A nonzero determinant means the system has exactly one solution. That is true for a length, and equally for any other target dimension. So $\ell_{P}$ is not a length you can build from $G,\hbar,c$. It is the only one.

There is no dimensionless knob to slide it, no free parameter, no alternative combination. Combine gravity with quantum mechanics and relativity, and a length $10^{-35}\,\mathrm m$ appears whether you wanted it or not.

That is the content of the sentence in Chapter 7.1 saying that quantum gravity has a built-in scale. It is worth contrasting with electromagnetism, where the analogous attempt fails.

Take $\hbar$, $c$ and the Coulomb constant $k_{e}\equiv e^{2}/4\pi\epsilon_{0}$ (dimensions $\mathsf{M}\mathsf{L}^{3}\mathsf{T}^{-2}$, i.e. energy × length). From those three you can build no length at all, because they are dimensionally dependent. The combination $k_{e}/\hbar c=\alpha$ is already a pure number, so the corresponding determinant vanishes.

That absence of an intrinsic scale is the deep reason QED behaves so differently from gravity, and it is why the gravitational coupling being dimensionful is exactly the disease diagnosed by power counting in Chapter 7.1.

5.4 · Orders of magnitude

The habit that goes with dimensional analysis is estimating first and computing later. You decide whether an effect can possibly matter before spending a week on it. Here is one example, using only what is on this page. How does gravity compare to electricity inside an atom? Both forces fall as $1/r^{2}$, so the ratio is a pure number independent of separation:

Forty orders of magnitude. That is why no chemistry textbook mentions gravity. Read the other way, it is also why quantum gravity is experimentally out of reach. To make gravity comparable to the other forces you must reach the Planck energy, which is where (0.3.27) said it would be. A single ratio, computed in one line, explains both the structure of Part IV and the predicament of Part VII.

Before spending these tools, one expansion from outside physics, where the mathematics is not analogous but identical.

6 · Worked examples

Two examples, one from each half of the chapter. The first uses Taylor's theorem to explain why the harmonic oscillator keeps reappearing. The second uses the binomial series to go one term past Newton.

7 · Your turn