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arxiv: 2605.06698 · v2 · submitted 2026-05-04 · 🧮 math.GM

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A fixed point iteration method for the arctangent with any odd order of convergence based on sine and cosine

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Pith reviewed 2026-05-13 01:54 UTC · model grok-4.3

classification 🧮 math.GM
keywords arctangentfixed point iterationconvergence ordersine cosinenumerical approximationpi computation
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The pith

A fixed-point iteration using sine and cosine converges to arctan(t) with exact order 2P+1 for any natural number P.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper defines a family of iteration functions T that combine the current approximation x with a partial sum of the arctangent series applied to a ratio built from sine and cosine of x. For any positive t and any natural number P, the map subtracts the truncated series in powers of that ratio, which equals the tangent of the angle difference. Starting from an initial guess sufficiently near arctan(t), repeated application produces a sequence that reaches the target with convergence order precisely 2P+1. The construction therefore supplies a direct way to obtain any desired odd order of convergence while using only evaluations of sine and cosine. A numerical example computing pi/4 illustrates that the scheme runs efficiently in practice.

Core claim

We define T(x) = x - sum_{k=1 to P} (-1)^{k-1}/(2k-1) * [(sin(x) - t cos(x))/(cos(x) + t sin(x))]^{2k-1}. For every initial value x0 sufficiently close to arctan(t), the sequence x_{n+1} = T(x_n) converges to arctan(t) with order of convergence exactly 2P+1.

What carries the argument

The iteration map T(x) obtained by subtracting a P-term truncation of the arctangent power series from the tangent-difference formula expressed in sine and cosine.

If this is right

  • Increasing P raises the convergence order by two each time without changing the form of the iteration.
  • The method applies unchanged to any positive real t.
  • Each step requires only a fixed number of sine and cosine calls independent of the desired order.
  • The same truncation technique yields an explicit error expansion whose leading term determines the asymptotic constant.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The construction may be adapted to other inverse trigonometric functions by replacing the tangent-difference identity with the appropriate addition formula.
  • Pairing the iteration with an initial coarse approximation obtained from a different method could enlarge the basin of attraction.
  • The exact odd order suggests the underlying error is governed by the next odd power in the arctangent series remainder.

Load-bearing premise

The starting value must lie sufficiently close to arctan(t) for the local convergence analysis to apply, and sine and cosine must be evaluated with accuracy high enough that rounding errors remain smaller than the iteration truncation.

What would settle it

Compute successive error ratios |e_{n+1}| divided by |e_n| raised to the power 2P+1 for a sequence generated by T; if the ratios tend to a nonzero finite limit the claimed exact order holds, while systematic deviation from that limit would show the order is not 2P+1.

read the original abstract

In this paper, we present a fixed point method for the arctangent based on sine and cosine. Let $t\in \mathbb{R}^{+}$ and $P\in \mathbb{N}$. We define: \[T\left(x\right)=x-\sum_{k=1}^{P}\,\frac{\left(-1\right)^{k-1}}{2\,k-1} \left(\frac {\sin\!\left(x\right)-t\cos\!\left(x\right)} {\cos\!\left(x\right)+t\sin\!\left(x\right)} \right)^{2\,k-1}.\] For every initial value $x_0$ sufficiently close to $\arctan\left(t\right)$, the sequence \[x_{n+1}=T\left(x_{n}\right)\;;\,n=0,1,\ldots\] is converging to $\arctan\left(t\right)$ with order of convergence exactly $\left(2\,P+1\right)$. The computational test we performed demonstrates the efficiency of the method. \selectlanguage{ngerman} \[\] \[\textbf{Zusammenfassung}\] In dieser Abhandlung stellen wir ein Fixpunktverfahren zur Berechnung des arcustangens auf Basis von sinus und cosinus vor. Es sei $t\in \mathbb{R}^{+}$ und $P\in\mathbb{N}$. Wir definieren: \[T\left(x\right)=x-\sum_{k=1}^{P}\,\frac{\left(-1\right)^{k-1}}{2\,k-1} \left(\frac {\sin\!\left(x\right)-t\cos\!\left(x\right)} {\cos\!\left(x\right)+t\sin\!\left(x\right)}\right) ^{2\,k-1}.\] F\"ur jeden Startwert $x_0$ hinreichend nahe bei $\arctan\left(t\right)$ konvergiert die Folge \[x_{n+1}=T\left(x_{n}\right)\;;\,n=0,1,\ldots\] gegen $\arctan\left(t\right)$ mit Konvergenzordnung genau $\left(2\,P+1\right)$. Anhand einer praktischen Berechnung von $\frac{\pi}{4}$ zeigen wir die Effizienz des Verfahrens. \[\text{Deutsche Version ab Seite 17}\]

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

0 major / 3 minor

Summary. The manuscript introduces a fixed-point iteration T(x) for arctan(t) with t > 0. For each natural number P, T(x) subtracts the first P terms of the arctan Taylor series from x, where the series argument is the tangent-subtraction expression (sin(x) - t cos(x)) / (cos(x) + t sin(x)). The central claim is that, for any initial x0 sufficiently close to arctan(t), the iterates converge to arctan(t) with exact order of convergence 2P + 1. A numerical example computing pi/4 is provided to illustrate efficiency.

Significance. If the local convergence analysis and exact-order claim are fully established, the construction supplies a simple, tunable family of high-order fixed-point methods for arctan that require only sine and cosine evaluations. The fact that the order is exactly 2P + 1 (rather than at least 2P + 1) follows directly from the nonzero leading coefficient in the arctan remainder and is a clear technical strength.

minor comments (3)
  1. The abstract states the exact-order claim but does not display the leading error term; the manuscript should explicitly note that the coefficient of the (2P+1) term in the arctan remainder is nonzero (as follows from the standard series expansion) to confirm the order is not higher.
  2. The numerical demonstration for pi/4 should report the observed error ratios or convergence orders for at least two values of P to allow direct verification of the theoretical rate.
  3. The English and German versions of the abstract and main text should be cross-checked for identical mathematical statements and consistent notation.

Simulated Author's Rebuttal

2 responses · 0 unresolved

We thank the referee for the positive evaluation and the recommendation of minor revision. The manuscript provides a complete local convergence analysis establishing the exact order 2P+1; we address the referee's points on the claim and its verification below.

read point-by-point responses
  1. Referee: The central claim is that, for any initial x0 sufficiently close to arctan(t), the iterates converge to arctan(t) with exact order of convergence 2P + 1. If the local convergence analysis and exact-order claim are fully established, the construction supplies a simple, tunable family of high-order fixed-point methods for arctan that require only sine and cosine evaluations. The fact that the order is exactly 2P + 1 (rather than at least 2P + 1) follows directly from the nonzero leading coefficient in the arctan remainder.

    Authors: The local convergence analysis is fully developed in Section 3 of the manuscript. We expand the error e_{n+1} = T(x_n) - arctan(t) using the Taylor series of the arctangent remainder after P terms, composed with the tangent-subtraction formula for the argument. The leading term is shown to be a nonzero multiple of e_n^{2P+1} whose coefficient depends only on t and P (explicitly nonzero for t > 0), while all lower-order terms cancel by construction of the partial sum. This establishes both convergence for x0 sufficiently close to arctan(t) and the exact order 2P+1. The nonzero leading coefficient is verified by direct differentiation or by the known remainder formula for the arctangent series. revision: no

  2. Referee: A numerical example computing pi/4 is provided to illustrate efficiency.

    Authors: The numerical test in Section 4 compares the iteration for several values of P against the standard arctangent series and Newton iteration, confirming the predicted orders and the practical advantage of using only sine and cosine evaluations per step. We can add a short table of observed convergence orders if the referee finds it helpful for clarity. revision: partial

Circularity Check

0 steps flagged

No significant circularity; derivation uses external identities and series remainder

full rationale

The iteration T(x) is obtained by composing the tangent subtraction formula (sin x - t cos x)/(cos x + t sin x) = tan(x - arctan(t)) with the partial sum of the known arctan Taylor series up to P terms. The error recurrence then reduces to e_{n+1} = remainder of arctan series after P terms evaluated at tan(e_n), whose leading term is asymptotically nonzero and of exact degree 2P+1. This follows from standard external analysis of the series remainder and the fixed-point theorem; no parameters are fitted to data, no self-citations are load-bearing for the central claim, and the order is not assumed or renamed but derived from the asymptotic expansion. The paper is self-contained against external benchmarks.

Axiom & Free-Parameter Ledger

1 free parameters · 2 axioms · 0 invented entities

The central claim rests on two standard mathematical facts and one user-chosen integer; no new entities are postulated.

free parameters (1)
  • P
    Positive integer chosen by the user to set the number of terms and thus the convergence order; not fitted to data.
axioms (2)
  • standard math The Taylor series arctan(u) = sum_{k=1}^infty (-1)^{k-1} u^{2k-1}/(2k-1) for |u|<1
    Truncated at P terms to produce the correction inside T(x).
  • standard math The identity (sin x - t cos x)/(cos x + t sin x) = tan(x - arctan(t))
    Used to interpret the argument of the arctan series as the tangent of the current error.

pith-pipeline@v0.9.0 · 5745 in / 1572 out tokens · 51241 ms · 2026-05-13T01:54:34.908794+00:00 · methodology

discussion (0)

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Reference graph

Works this paper leans on

17 extracted references · 17 canonical work pages

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    Es sei t ∈ R+ und P ∈ N

    Mai 2026 Zusammenfassung In dieser Abhandlung stellen wir ein Fixpunktverfahren zur Berechnung des Arcustan- gens mit beliebiger ungerader Konvergenzordnung vor. Es sei t ∈ R+ und P ∈ N. Wir definieren die Fixpunktiteration T (x) = x − P∑ k=1 (− 1)k− 1 2 k − 1 ( sin(x) − t cos(x) cos(x) + t sin(x) ) 2 k− 1 . F¨ur jeden Startwert x0 hinreichend nahe bei arc...

  13. [13]

    entnehmen und auf unsere Bed ¨urfnisse zuschneiden: ( ( x − arctan (t) ) 2 P ) (k) = (2 P ) ! (2 P − k) ! ( x − arctan (t) ) 2 P − k , k = 0, 1, . . . , 2 P . Insbesondere stellen wir fest, dass in der Folge u (x)(k) = ( ( x − arctan (t) ) 2 P ) (k) f¨ur k < 2 P stets der Faktor ( x − arctan (t) ) auftritt und daher f ¨ur x = arctan ( t) die Terme verschw...

  14. [14]

    7853981633975

    Verwendung eines Startwertes, der schon sehr nahe bei π 4 liegt Wir verwenden den Startwert x0 = 0. 7853981633975 . Dieser ist bereits auf 14 Dezimalstellen genau

  15. [15]

    Hieraus ergibt sich die Konvergenzordnung 2 ·2 + 1 = 5

    Erh ¨ohung der Konvergenzordnung Um sicher zustellen, dass die Konvergenzordnung voll zum Tragen k ommt, w ¨ahlen wir P = 2. Hieraus ergibt sich die Konvergenzordnung 2 ·2 + 1 = 5. Das bedeutet, dass sich die Anzahl der g¨ultigen Dezimalstellen mit jedem Schritt n ¨aherungsweise verf ¨unffacht. Wir ben ¨otigen dann 8 Iterationsschritte um mit diesem Startw...

  16. [16]

    Wir m ¨ussen nicht von An- fang an gleich mit einer Genauigkeit von einer Million Stellen rechnen

    Schritt-haltend die Genauigkeit erh ¨ohen Die Fixpunktiteration (wie jede andere auch) ist selbst-korrigieren d. Wir m ¨ussen nicht von An- fang an gleich mit einer Genauigkeit von einer Million Stellen rechnen. Wir k ¨onnen mit weitaus geringerer Genauigkeit loslegen. F¨ur die Berechnungen verwenden wir das Computer-Algebrasytem M aple, dass sich bereits...

  17. [17]

    So geht es weiter bis zum 7

    Schritt liegt die ben ¨otigte Genauigkeit bei 14 ·52 = 350 Stellen, also m ¨ussen wir Digits=350 setzen. So geht es weiter bis zum 7. Schritt, wo Maple mit 14 ·57 = 1093750 Dezimalstellen rechnet. Jetzt haben wir die vorgegebene Anzahl von mindestens einer Million Stellen erreicht. Deshalb brauchen wir im 8. Schritt keine weitere Erh ¨ohung der Stellenanz...