arctangent-series

The Arctangent Series

Summary: §139–§141 of Chapter 8. Inverting Euler’s formula gives $z=\frac{1}{2i}\log \frac{\cos z+i\sin z}{\cos z-i\sin z}$, which simplifies via $\tan z=\sin z/\cos z$ to $z=\frac{1}{2i}\log \frac{1+i\tan z}{1-i\tan z}$. Substituting $x=i\tan z$ in the fast-converging logarithmic series of Chapter 7 yields

$\arctan t=\frac{t}{1}-\frac{t^3}{3}+\frac{t^5}{5}-\frac{t^7}{7}+\cdots .$

At $t=1$ this is Leibniz’s $\pi /4=1-1/3+1/5-\cdots$. At $t=1/\sqrt{3}$ it becomes $\pi /6=(1/\sqrt{3})(1-1/(3\cdot 3)+1/(5\cdot 9)-\cdots )$, converging at geometric rate.

Sources: chapter8 (§139, §140, §141)

Last updated: 2026-05-11

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§139 — The arc as a complex logarithm

Take §138: $\cos z+i\sin z=e^{iz}$ and $\cos z-i\sin z=e^{-iz}$. Their ratio is $e^{2iz}$, so

$\log \frac{\cos z+i\sin z}{\cos z-i\sin z}=2iz,z=\frac{1}{2i}\log \frac{\cos z+i\sin z}{\cos z-i\sin z}.$

Euler arrives at this formula not via §138 (which he derives independently a few sections earlier) but via a parallel infinitesimal/infinite calculation: with $n=1/j$ infinitely small, $\cos (z/j)=1$ and $\sin (z/j)=z/j$, then using $\log (1+x)=j((1+x{)}^{1/j}-1)$ from §125 with $1+x=\cos z+i\sin z$ and $\cos z-i\sin z$ in turn (source: chapter8, §139). The cosine equation collapses to a tautology; the sine equation yields the boxed formula above.

The interpretation: every arc $z$ is the imaginary part (up to the $1/2i$ prefactor) of a complex logarithm. This anticipates the full theory of complex logarithms — the multivaluedness, the branch cuts — but Euler stays within a real-valued reading here.

§140 — Substituting tangent

Divide numerator and denominator inside the logarithm by $\cos z$:

$\frac{\cos z+i\sin z}{\cos z-i\sin z}=\frac{1+i\tan z}{1-i\tan z},$

so

$z=\frac{1}{2i}\log \frac{1+i\tan z}{1-i\tan z}.$

But the §123 series gives

$\log \frac{1+x}{1-x}=\frac{2x}{1}+\frac{2x^3}{3}+\frac{2x^5}{5}+\frac{2x^7}{7}+\cdots ,$

so substituting $x=i\tan z$:

$\log \frac{1+i\tan z}{1-i\tan z}=2i\tan z+\frac{2i^3(\tan z{)}^3}{3}+\frac{2i^5(\tan z{)}^5}{5}+\cdots .$

Using $i^3=-i$, $i^5=i$, $i^7=-i$, $\dots$:

$\log \frac{1+i\tan z}{1-i\tan z}=2i\left(\tan z-\frac{(\tan z{)}^3}{3}+\frac{(\tan z{)}^5}{5}-\cdots \right).$

Dividing by $2i$:

$z=\tan z-\frac{(\tan z{)}^3}{3}+\frac{(\tan z{)}^5}{5}-\frac{(\tan z{)}^7}{7}+\cdots .$

(source: chapter8, §140). Setting $t=\tan z$, so $z=\arctan t$:

$\boxed{\arctan t=\frac{t}{1}-\frac{t^3}{3}+\frac{t^5}{5}-\frac{t^7}{7}+\frac{t^9}{9}-\cdots }$

The series converges for $|t|\le 1$ (an observation Euler does not formalize but uses).

§140 — Leibniz’s $\pi /4$

At $t=1$, the arc whose tangent is 1 is $\pi /4$, so

$\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots .$

(source: chapter8, §140). Euler attributes this discovery to Leibniz. This series gives $\pi$ in closed form as an alternating sum of reciprocal odd integers — beautiful, but practically useless for computation: each correct decimal digit costs about ten new terms.

Chapter 10 re-derives Leibniz’s formula as a special case of a vast family of character-style series obtained by applying Newton’s identities to the §164 arc-form products. Chapter 15 then sieves Leibniz’s series by primes to obtain the Euler-product form $\pi /4={\prod }_{p}p/(p\mp 1)$ — a Dirichlet $L$-function in disguise (see prime-sign-series-for-pi). Chapter 18 converts Leibniz’s series to Brouncker’s continued fraction $4/\pi =1+1^2/(2+3^2/(2+5^2/(2+\cdots )))$ via the §369 reciprocal-series template.

§141 — A faster series via $\arctan (1/\sqrt{3})$

For practical $\pi$ computation, choose $t<1$ to get geometric-rate convergence. Try $t=1/\sqrt{3}$, the tangent of $\pi /6$:

$\frac{\pi }{6}=\frac{1}{\sqrt{3}}-\frac{1}{3\cdot 3\sqrt{3}}+\frac{1}{5\cdot 3^2\sqrt{3}}-\frac{1}{7\cdot 3^3\sqrt{3}}+\cdots ,$

i.e.

$\pi =\frac{2\sqrt{3}}{1}-\frac{2\sqrt{3}}{3\cdot 3}+\frac{2\sqrt{3}}{5\cdot 3^2}-\frac{2\sqrt{3}}{7\cdot 3^3}+\cdots .$

(source: chapter8, §141). Each term is about a third of the previous, so a dozen terms give roughly six correct digits. Euler comments: “By means of this series the value of $\pi$ itself, which was previously exhibited, was determined with incredible labor” — the prior 113-digit decimal of §126 was computed exactly this way.

§141 first considers $t=1/10$, which converges spectacularly fast but does not correspond to any “nice” fraction of the circumference, so $\pi$ cannot be extracted from $\arctan (1/10)$ alone. The §141 lesson: for a useful arctangent identity, $t$ must be both small (for convergence) and a known fraction of $\pi$.

The remaining inconvenience of the §141 series is that every term is irrational ($\sqrt{3}$ in the denominator). §142 resolves this by splitting $\pi /4$ as a sum of arctangents of rational numbers.

Why the series exists at all

The arctangent series is not derivable by elementary trigonometry. It requires the bridge between trig and exponentials (Euler’s formula, §138) plus the logarithmic series of Chapter 7. Without that bridge, the function $\arctan$ is defined only implicitly, and there is no algebraic technique in pre-Eulerian mathematics for expanding it as a power series.

After Euler, the series is in some sense trivial: it is just $\log \frac{1+i\tan z}{1-i\tan z}$ divided by $2i$, expanded by the standard logarithm series, with $i\tan z$ in place of $x$. The whole derivation is two lines once §138–§140 are accepted.

Convergence rate at a glance

$t$	Arc	Series	Terms for 6 digits
$1$	$\pi /4$	$1-1/3+1/5-\cdots$	~$10^6$
$1/\sqrt{3}$	$\pi /6$	$(1/\sqrt{3})(1-1/9+1/(5\cdot 9)-\cdots )$	~13
$1/2$	$\arctan (1/2)\approx 26.57°$	$1/2-1/(3\cdot 8)+1/(5\cdot 32)-\cdots$	~10
$1/3$	$\arctan (1/3)\approx 18.43°$	$1/3-1/81+1/(5\cdot 243)-\cdots$	~6

The §142 Machin decomposition uses the last two rows.

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