False-Position Method

Summary: The chapter-22 workhorse — regula falsi. Given a transcendental equation $f(s)=0$, pick two trial values bracketing the root, compute the signed errors from log tables, and linearly interpolate. Iterate, narrowing the interval first to degrees, then minutes, then seconds, etc., gaining roughly one sexagesimal digit per pass.

Sources: chapter22, §§531–538 (used in eight of the nine problems).

Last updated: 2026-05-12.

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The technique

Suppose we want $s$ with $f(s)=0$, where $f$ is built from $s,\sin s,\cos s,\tan s$ and constants. The method has three ingredients.

Bracket. Pick two values $s_1<s_2$ such that $f(s_1)>0>f(s_2)$ (or the reverse). The errors $e_1=f(s_1)$ and $e_2=f(s_2)$ have opposite signs, so the true root lies between them.

Interpolate linearly. Treat $f$ as if it were linear on $[s_1,s_2]$. Then

$$s\approx s_1+(s_2-s_1)\cdot \frac{|e_1|}{|e_1|+|e_2|}.$$

Euler writes this as the proportion

$$|e_1|+|e_2|:|e_1|::s_2-s_1:s-s_1,$$

i.e., “the difference of errors is to the smaller error as the gap is to the excess of the true root over the smaller value.” This is the classical regula falsi statement.

Iterate. Re-bracket around the new estimate, this time at minute precision. After three or four passes the bracket is at the seconds-or-finer level. The errors shrink by roughly a factor of $60$ per pass, mirroring the sexagesimal subdivision Euler uses for angles.

Logarithmic evaluation

The errors are computed not directly from $f(s)$ but from its logarithm via log tables. A typical Problem-I computation at $s=30^{\circ }$ reads:

$$\begin{array}{rl}\log 30=& 1.4471213\\ \text{subtract }& 1.7581226\\ \log s=& 9.7189987\\ \log \cos s=& 9.9375306\\ \text{difference}& \dots \end{array}$$

The first subtraction converts $30^{\circ }$ to the corresponding arc length $\pi /6$ (using $\log (\pi /180)=-1.7581226\dots$, so subtracting $1.7581226$ from $\log n$ yields $\log (\pi n/180)$). The second line is a direct table lookup of $\log \cos 30^{\circ }$. The third line — the difference — is the (logarithm of the) error, signed by which of $s,\cos s$ is larger.

For minutes the conversion constant is $3.5362739$ (i.e., $\log (\pi /(180\cdot 60))$); for seconds $5.3144251$. So a single rule — “subtract the appropriate fixed log” — converts an angle expressed in degrees, minutes, or seconds to the logarithm of its arc length. This is what makes the iteration cheap.

Worked example — Problem I

Find $s$ with $s=\cos s$ (§531).

Pass 1 (degrees). Try $s=30^{\circ }$: $\log s=9.7189987$, $\log \cos s=9.9375306$, so $\cos s>s$, much. Try $s=40^{\circ }$: $\log s=9.8439374$, $\log \cos s=9.8842540$, still $\cos s>s$. Try $s=45^{\circ }$: $\log s=9.8950899$, $\log \cos s=9.8494850$, now $s>\cos s$. So $40^{\circ }<s<45^{\circ }$.

Pass 2 (linear interpolation gives $\approx 42^{\circ }$). Try $s=42^{\circ }$ and $s=43^{\circ }$. Errors come out $+0.0059468$ and $-0.0112184$. The proportion $171652:59468=1^{\circ }:20^{\prime },47^{\prime \prime }$ refines to $s\in (42^{\circ }20^{\prime },42^{\circ }21^{\prime })$.

Pass 3 (minutes). At $s=2140^{\prime }$ and $s=2141^{\prime }$ the errors are $+0.0002253$ and $-0.0000608$, giving $s=42^{\circ }20^{\prime }47^{\prime \prime },14^{\prime \prime \prime }$ to seconds.

The accumulated answer: $s=42^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$, arc length $0.7390847$.

Why it works on these problems

Three features of Euler’s setting make false position succeed where naive guess-and-check would not:

1. One-sided monotonicity. On the relevant bracket each $f$ here is strictly monotonic, so the signed error tells you which way to move.
2. Smooth, near-linear. Transcendental functions evaluated against a polynomial in $s$ are smooth, and over a one-degree window the linear approximation is very good — hence the interpolation step matches the true root to four extra digits.
3. Cheap evaluation. Each $f(s)$ requires only a table lookup and a subtraction, both at $\sim 7$-digit precision. No multiplication, no division, no Newton-style derivative.

Modern eyes see Newton’s method as the natural successor, replacing the secant with the tangent. Euler is comfortable doing this too (see arcs-equal-to-tangents-series where he inverts a series instead), but for one-off transcendental roots, false position with log tables is faster and needs no calculus.

Variations in the chapter

• Cascading precision. Most problems run three passes (degrees → minutes → seconds), occasionally four when sixths-of-a-second precision is required (Problem II §532 reaches $s=54^{\circ }18^{\prime }{6}^{\prime \prime }52^{\prime \prime \prime }43^{\prime \prime \prime \prime }33^{\prime \prime \prime \prime \prime }$).
• No bracket from §531-style monotonicity argument. Sometimes Euler doesn’t bother to prove uniqueness — he just observes that the trial values bracket a sign change.
• Series substitution. In Problem VI (§536) he first uses a trig identity to convert $1+\cos s+\sin s$ into $2\sqrt{2}\cos (\frac{s}{2})\cos (45^{\circ }-\frac{s}{2})$ — a product of two cosines whose logs are tabulated — before applying false position.

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — where the method appears.
• arc-equals-cosine — first and simplest application (Problem I).
• arcs-equal-to-tangents-series — the one chapter-22 problem solved by series inversion instead of false position.