arcs-equal-to-tangents-series

Arcs Equal to Their Tangents (Series Solution)

Summary: Problem IX of chapter 22 (§§539–540). Find all arcs $s$ with $s=\tan s$. The first is $s=0$; the next sits in the third quadrant just below $270^{\circ }$; further ones populate the fifth, seventh, ninth quadrants and so on. The countably-infinite family is parametrized by an integer $n$; Euler inverts the arctangent series to derive an explicit asymptotic expansion in $1/(2n+1)$, then tabulates the first ten arcs.

Sources: chapter22, §§539–540.

Last updated: 2026-05-12.

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Setup

The equation $s=\tan s$ has trivial solution $s=0$. The function $\tan s$ has periodic vertical asymptotes at $s=(2k+1)\pi /2$ for $k=0,1,2,\dots$, between which it sweeps $(-\infty ,+\infty )$. So on every interval $((2k-1)\pi /2,(2k+1)\pi /2)$ there is exactly one real $s$ with $s=\tan s$, except the interval centered at $\pi$ (where $\tan s$ is negative on the left half and positive on the right, never quite passing through the still-larger linear $s$).

Concretely:

• 2nd quadrant $(\pi /2,\pi )$: $\tan s<0$, $s>0$, no solution.
• 3rd quadrant $(\pi ,3\pi /2)$: $\tan s$ runs $0\to +\infty$ as $s\to 3\pi /2^{-}$, crossing $s$ once just before $3\pi /2=270^{\circ }$.
• 5th, 7th, $\dots$ quadrants: one solution per “odd quadrant,” each just below $(2n+1)\pi /2$.

So the solutions are indexed by $n=0,1,2,\dots$ with the $n$-th solution slightly less than $(2n+1)\pi /2$.

Series inversion

Let $q=\pi /2$ and write the unknown as

$$s_n=(2n+1)q-\epsilon ,\epsilon >0\text{small for large }n.$$

Then $\tan s_n=\tan ((2n+1)q-\epsilon )=\cot \epsilon$ (because $\tan$ has period $\pi$ and $\tan (\pi /2-\epsilon )=\cot \epsilon$). The equation $s_n=\tan s_n$ becomes

$$(2n+1)q-\epsilon =\cot \epsilon =\frac{1}{\tan \epsilon }.$$

Letting $x=\tan \epsilon$, the arctangent series gives $\epsilon =x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5-\cdots$. The equation is $1/x=(2n+1)q-\epsilon$, so $x\approx 1/((2n+1)q)$ to leading order, decreasing as $n$ grows.

Euler iterates the substitution. The cleaner statement: $1/x=(2n+1)q-\epsilon$ where $\epsilon =\arctan x$. Expanding both sides in $1/((2n+1)q)$:

$$\frac{1}{x}=(2n+1)q-\frac{1}{(2n+1)q}-\frac{2}{3(2n+1{)}^3{q}^3}-\frac{13}{15(2n+1{)}^5{q}^5}-\frac{146}{105(2n+1{)}^7{q}^7}-\frac{2343}{945(2n+1{)}^9{q}^9}-\cdots .$$

Multiplying through and rearranging (Euler does this implicitly), the arcs themselves admit the asymptotic series

$$s_n=(2n+1)\cdot 1.5707963268-\frac{0.6366197700}{2n+1}-\frac{0.1720081700}{(2n+1{)}^3}-\frac{0.0906259600}{(2n+1{)}^5}-\frac{0.0589283400}{(2n+1{)}^7}-\frac{0.0425854300}{(2n+1{)}^9}-\cdots .$$

The first coefficient $0.6366\dots =2/\pi$ recovers the leading correction. The expansion is an asymptotic series in $1/(2n+1)$; each term comes from one more reciprocal-power in the arctangent inversion.

Converted to degree-minute-second units (multiply each coefficient by $180/\pi \cdot 3600=206264.8\dots$ to get seconds and rescale), Euler’s form is

$$s_n=(2n+1)\cdot 90^{\circ }-\frac{131313^{\prime \prime }}{2n+1}-\frac{35479^{\prime \prime }}{(2n+1{)}^3}-\frac{18692^{\prime \prime }}{(2n+1{)}^5}-\frac{12155^{\prime \prime }}{(2n+1{)}^7}-\frac{8784^{\prime \prime }}{(2n+1{)}^9}.$$

Tabulated values

Substituting $n=0,1,2,\dots$ Euler gives the ten smallest non-trivial solutions, expressed as ”$(2n+1)\cdot 90^{\circ }$ minus a correction”:

$n$	$s_n$	Correction
0	$1\cdot 90^{\circ }-90^{\circ }=0$	(the trivial root)
1	$3\cdot 90^{\circ }-12^{\prime }32^{\prime \prime },48^{\prime \prime \prime }$
2	$5\cdot 90^{\circ }-7^{\prime }22^{\prime \prime },32^{\prime \prime \prime }$
3	$7\cdot 90^{\circ }-5^{\prime }14^{\prime \prime },22^{\prime \prime \prime }$
4	$9\cdot 90^{\circ }-4^{\prime }{3}^{\prime \prime },59^{\prime \prime \prime }$
5	$11\cdot 90^{\circ }-3^{\prime }19^{\prime \prime },24^{\prime \prime \prime }$
6	$13\cdot 90^{\circ }-2^{\prime }48^{\prime \prime },37^{\prime \prime \prime }$
7	$15\cdot 90^{\circ }-2^{\prime }26^{\prime \prime },5^{\prime \prime \prime }$
8	$17\cdot 90^{\circ }-2^{\prime }{8}^{\prime \prime },51^{\prime \prime \prime }$
9	$19\cdot 90^{\circ }-1^{\prime }55^{\prime \prime },16^{\prime \prime \prime }$

The correction shrinks roughly as $1/(2n+1)$ — i.e., the roots cling tighter and tighter to the asymptote $s=(2n+1)\pi /2$ as $n$ grows.

Why series rather than false position

Two reasons:

1. Infinitely many roots. False position finds one root at a time; the family wants a uniform parametrization. Inverting the series gives every root at once.
2. Numerical stability. Near each asymptote $\tan s$ varies very fast — at $s=(2n+1)\pi /2-\epsilon$, $\tan s=1/\epsilon$ — so the linearization that false position relies on (see false-position-method) becomes very poor. The natural small parameter is $1/(2n+1)$, not $s$, and the arctangent series is its native expansion.

The asymptotic series itself diverges as a power series in $1/(2n+1)$ for any fixed $n$ — the coefficients $1/3,2/15,13/315,\dots$ grow factorially. For $n\ge 1$ the truncated five-term form recovers $s_n$ to seconds. The connection between $s=\tan s$ and the Cotes spiral / Bessel-function zeros is left implicit; here Euler is content to display the family.

Connection to Problem VII

Problem VII (arc-equals-half-tangent) solved the related $2s=\tan s$ at $s=66^{\circ }46^{\prime }54^{\prime \prime }$ in the first quadrant. The “general” version $s=\tan s$ here has no first-quadrant solution other than $s=0$ — the slopes don’t match.

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• arc-equals-half-tangent — first-quadrant cousin solved by false position.
• false-position-method — contrasted as the technique-of-choice when a single root is wanted at moderate magnitude.
• transcendental-curves — these are points where the sine-line / tangent-line of tangent-and-secant-lines crosses the diagonal $y=x$.