Abstract
Equations are solved numerically by Newton’s method, the secant method, or bisection. Asked which is fastest, one usually answers Newton. This paper asks whether the answer changes when “fast” is redefined——it changes three ways. No new mathematical theorem and no new law is claimed.
First, we actually solve $x^3-2=0$. To reach $16$ digits, Newton takes $\mathbf{5}$ iterations, the secant method $\mathbf{7}$, bisection $\mathbf{53}$ (Section 2).
Second, this is the core of the paper. Counted in total function evaluations, the secant method’s $\mathbf{7}$ beats Newton’s $\mathbf{10}$——the verdict reverses (Section 3).
Third, the efficiency index predicts it. $p^{1/m}$ is $1.414214$ for Newton and $\mathbf{1.618034}$ for the secant method——the secant method is $\mathbf{14.41\%}$ faster (Section 3).
Fourth, there is a third reversal. In evaluations per digit on the first step, bisection’s $3.3219$ beats Newton’s $3.6430$ (Section 4).
Fifth, the golden ratio stands here. The secant method’s order of convergence is the positive root of $p^2=p+1$, namely $\varphi=\mathbf{1.618034}$ (Section 5).
Sixth, the separator is what counts as one unit. Iterations, function evaluations, or the first step——it is the convention of counting, not a property of the methods, that decides (Section 6).
Scope (what is not claimed)
No new mathematical theorem and no new law is claimed——Newton’s quadratic convergence, the secant method’s order $\varphi$, bisection’s linear convergence, and the efficiency index $p^{1/m}$ are all standard. We do not build numerical analysis——all we use is one equation and three iterations. We do not prove the orders of convergence——$p=2$ and $p=\varphi$ are merely quoted, and we do not enter the derivation of the error recursions. We assert nothing about general equations——$x^3-2=0$ is simple-rooted, smooth, and well conditioned, the most favourable case. At a multiple root quadratic convergence drops to linear, and with a poor starting value Newton diverges. We assert no values for the counts——the $5$, $7$ and $53$ of Section 2 are measurements for this starting value, this equation, and this stopping criterion. Change the starting value and they move by $\pm1$. We assume equal evaluation costs——we set the cost of $f$ and $f'$ equal. In practice $f'$ may be dearer or cheaper, so the reversal of Section 3 depends on this assumption. We do not disparage bisection——bisection alone has guaranteed convergence. Comparisons of speed have meaning only on the premise that one converges at all. We assert nothing about the golden-ratio coincidence——the $\varphi$ of Section 5 and the $1/\varphi$ of Paper 269 both come from $x^2=x+1$, but we do not claim any relation as phenomena. Relation to earlier papers: Paper 257 showed that changing the speed does not change the destination——this paper is the case where the definition of “speed” itself is threefold. Paper 255 separated two things called “accuracy,” only one removed by calibration——this paper likewise treats several quantities under one word. Paper 269 showed that a crystal’s five-fold symmetry is forbidden because $2\cos72^\circ=1/\varphi$——the $\varphi$ here is a root of the same quadratic. Paper 159 showed that “ill posed” is not one word——this paper shows that “fast” is not one word. What is added is actually solving $x^3-2=0$ by three methods and counting iterations and evaluations, confirming that the verdict reverses under evaluation count, computing evaluations per digit at each stage and finding the third reversal in which bisection wins the first step, and recording that $\varphi$ and $1/\varphi$ come from the same quadratic.
1. Three Methods and Three “Speeds”
We line up three iterative methods.
| Method | Iteration | Function evaluations per step |
|---|---|---|
| Newton | $x\leftarrow x-f(x)/f'(x)$ | $\mathbf{2}$ ($f$ and $f'$) |
| Secant | $x\leftarrow x-\dfrac{f(x)(x-x_0)}{f(x)-f(x_0)}$ | $\mathbf{1}$ ($f$ only) |
| Bisection | halve the interval | $\mathbf{1}$ |
Observation 1.1 “Fast” admits three definitions.
| Definition | What counts as one unit |
|---|---|
| (a) Iteration count | one iteration |
| (b) Total function evaluations | one call to the function |
| (c) Cost per digit | one digit of accuracy |
Question: Is the winner the same under all three?
2. Actually Solving It: $5$, $7$, and $53$
Setting: $x^3-2=0$, root $2^{1/3}=1.259921049894873$. We iterate until the relative error falls below $10^{-16}$.
Check 2.1: Newton (starting at $x=1$).
| Step | $x$ | Correct digits |
|---|---|---|
| $0$ | $1.0000000000000000$ | $0.686$ |
| $1$ | $1.3333333333333333$ | $1.235$ |
| $2$ | $1.2638888888888888$ | $2.502$ |
| $3$ | $1.2599334934499771$ | $5.005$ |
| $4$ | $1.2599210500177698$ | $10.011$ |
| $\mathbf{5}$ | $1.2599210498948732$ | $\mathbf{16.000}$ |
Observation 2.2 The digits run $0.686\to1.235\to2.502\to5.005\to10.011$, roughly doubling. This is quadratic convergence——the number of correct digits doubles each step.
Check 2.3: The secant method (starting at $x_0=1$, $x_1=1.5$).
| Step | $x$ | Correct digits |
|---|---|---|
| $1$ | $1.2105263157894737$ | $1.407$ |
| $2$ | $1.2514085388756730$ | $2.170$ |
| $3$ | $1.2602652758390120$ | $3.563$ |
| $4$ | $1.2599187140887995$ | $5.732$ |
| $5$ | $1.2599210492568176$ | $9.295$ |
| $6$ | $1.2599210498948743$ | $15.055$ |
| $\mathbf{7}$ | $1.2599210498948732$ | $\mathbf{16.000}$ |
Observation 2.4 The digits run $1.407\to2.170\to3.563\to5.732\to9.295\to15.055$, growing by about $1.6$ each time. $2.170/1.407=1.542$, $3.563/2.170=1.642$, $9.295/5.732=1.622$——oscillating about $\varphi=1.618$.
Check 2.5: Bisection (interval $[1,2]$).
| Quantity | Value |
|---|---|
| Digits gained per step, $\log_{10}2$ | $0.301030$ |
| Steps to reach $10^{-16}$, $\log_2(1/(10^{-16}\cdot2^{1/3}))$ | $52.8175\to\mathbf{53}$ |
Observation 2.6 Under definition (a), iteration count, Newton wins with $5$. Secant $7$, bisection $53$——this is the ground for saying “Newton is fast”.
3. The Core: Counted in Evaluations, It Reverses
Setting: Newton evaluates two things per step, $f$ and $f'$. Secant and bisection evaluate one.
Check 3.1:
| Method | Iterations | Evaluations per step | Total evaluations |
|---|---|---|---|
| Newton | $\mathbf{5}$ | $2$ | $\mathbf{10}$ |
| Secant | $7$ | $1$ | $\mathbf{7}$ |
| Bisection | $53$ | $1$ | $53$ |
Observation 3.2 This is the core of the paper. In iterations $5<7$ and Newton wins; in total evaluations $7<10$ and the secant method wins——on the same computation and the same run, the verdict swaps.
Check 3.3: We confirm with the efficiency index $p^{1/m}$ ($p$ the order, $m$ the evaluations per step).
| Method | Order $p$ | $m$ | Efficiency index $p^{1/m}$ |
|---|---|---|---|
| Newton | $2.000000$ | $2$ | $1.414214$ |
| Secant | $\mathbf{1.618034}$ | $1$ | $\mathbf{1.618034}$ |
| Bisection | $1.000000$ | $1$ | $1.000000$ |
Observation 3.4 The efficiency index predicts the observed reversal. $1.618034/1.414214=1.144123$, so the secant method is $\mathbf{14.41\%}$ faster——the method with the lower order wins on efficiency.
Observation 3.5 A point to write honestly——this reversal depends on the assumption that $f$ and $f'$ cost the same. If $f'$ is cheap analytically Newton wins; if $f'$ is available only by numerical differentiation the secant method wins by more——the answer is decided not by the problem but by the cost table.
4. The Third Reversal: Bisection Wins the First Step
Setting: We compute definition (c), evaluations per digit, stage by stage.
Check 4.1: Newton ($1$ iteration $=2$ evaluations).
| Stage | Digits gained | Evaluations per digit |
|---|---|---|
| $0\to1$ | $+0.549$ | $\mathbf{3.6430}$ |
| $1\to2$ | $+1.267$ | $1.5785$ |
| $2\to3$ | $+2.503$ | $0.7990$ |
| $3\to4$ | $+5.006$ | $0.3995$ |
| $4\to5$ | $+5.989$ | $\mathbf{0.3339}$ |
Check 4.2: The secant method ($1$ iteration $=1$ evaluation).
| Stage | Digits gained | Evaluations per digit |
|---|---|---|
| $0\to1$ | $+0.687$ | $\mathbf{1.4556}$ |
| $1\to2$ | $+0.763$ | $1.3106$ |
| $2\to3$ | $+1.393$ | $0.7179$ |
| $3\to4$ | $+2.169$ | $0.4610$ |
| $4\to5$ | $+3.563$ | $0.2807$ |
| $5\to6$ | $+5.760$ | $\mathbf{0.1736}$ |
Check 4.3: Bisection is the same at every stage.
| Stage | Digits gained | Evaluations per digit |
|---|---|---|
| every one | $+0.3010$ | $\mathbf{3.3219}$ |
Observation 4.4 We compare the first step.
| First step | Final stage | |
|---|---|---|
| Newton | $3.6430$ | $0.3339$ |
| Secant | $1.4556$ | $0.1736$ |
| Bisection | $\mathbf{3.3219}$ | $3.3219$ |
Observation 4.5 Looking only at the first step, bisection’s $3.3219$ is cheaper than Newton’s $3.6430$. Superlinear methods are slow at first and abruptly fast later——the third reversal.
Observation 4.6 Bisection’s column holds the same number top and bottom. Linear convergence means the cost per digit never changes, favourable at first and hopeless at the end——at the final stage it costs $\mathbf{9.95}$ times Newton and $\mathbf{19.14}$ times the secant method.
Observation 4.7 A point to write honestly——only the secant method’s last stage ($6\to7$) is poor at $1.0582$. That is because machine precision was reached and there was no headroom left, not a property of the method——a column whose significant figures are exhausted must not be used as evidence about speed.
5. Why the Golden Ratio Stands Here
Observation 5.1 The secant method’s error satisfies $e_{n+1}\sim C e_n e_{n-1}$. Substituting $e_n\sim A e_{n-1}^p$, the exponent must satisfy $p^2=p+1$——whose positive root is $\varphi=(1+\sqrt5)/2=1.618033988749895$.
Observation 5.2 Paper 269 wrote that a crystal forbids five-fold symmetry because $2\cos72^\circ$ is not an integer. That value is $\mathbf{0.618033988749895}$, that is $\mathbf{1/\varphi}$——the numerical difference is $1.11\times10^{-16}$, agreement at machine precision.
Check 5.3:
| Source | Value | Equation satisfied |
|---|---|---|
| Order of the secant method | $\varphi=1.618033988749895$ | $p^2=p+1$ |
| Crystal five-fold (Paper 269) | $2\cos72^\circ=0.618033988749895$ | $x^2=-x+1$ ($x=1/\varphi$) |
Observation 5.4 A point to write honestly——this paper claims no significance for the coincidence. Both are roots of quadratics with integer coefficients, and such numbers are few——the same number standing in two places is not the same root (the distinction of Paper 112).
6. Separator: What Counts as One Unit
Observation 6.1 Separator——what must be moved for the winner to change? The definition of one unit.
| Definition of “fast” | Winner | Value |
|---|---|---|
| (a) Iteration count | Newton | $5$ |
| (b) Total function evaluations | Secant | $7$ |
| (c) Per digit on the first step | Bisection | $3.3219$ |
Observation 6.2 Under three definitions, three different methods win. All three are correct——what changed is not the methods but the counting.
Observation 6.3 Paper 159 showed that “ill posed” is not one word. This paper shows that “fast” is not one word——saying “fast” without naming a unit is like saying “large” without naming a number.
Observation 6.4 And there is a fourth axis——whether it converges at all. Bisection alone always converges given a sign-changing interval——Newton diverges for some starting values. Comparisons of speed have meaning only on the premise that one converges.
7. Roll Call and Closing
- Newton, $5$ iterations (Check 2.1): Digits double.
- Secant, $7$ iterations (Check 2.3): Digits grow by about $1.6$.
- Bisection, $53$ iterations (Check 2.5): $0.301$ digits per step.
- Total evaluations $10$ against $7$ (Check 3.1): It reverses.
- Efficiency $1.414214$ against $1.618034$ (Check 3.3): $14.41\%$ faster.
- First step $3.3219<3.6430$ (Observation 4.4): Bisection wins.
- Final stage $3.3219$ against $0.3339$ (Observation 4.4): $9.95$ times the cost.
- $\varphi$ is the positive root of $p^2=p+1$ (Observation 5.1): $1.618033988749895$.
- $2\cos72^\circ=1/\varphi$ (Check 5.3): Difference $1.11\times10^{-16}$.
- Three definitions, three winners (Observation 6.1): The counting decides.
Closing: “Which is fastest” has three answers, and a different method wins each. Actually solving $x^3-2=0$: in iterations Newton wins with $\mathbf{5}$, and in total function evaluations the secant method wins with $\mathbf{7}$——because Newton needs two things, $f$ and $f'$, per step. The efficiency index $p^{1/m}$ predicts it——Newton $1.414214$, secant $\mathbf{1.618034}$, so the method with the lower order is $\mathbf{14.41\%}$ faster. And there is a third reversal——on the first step alone, bisection’s $\mathbf{3.3219}$ is cheaper than Newton’s $\mathbf{3.6430}$. Superlinear methods start slow, while bisection’s cost per digit never changes. At the final stage, that constancy becomes a factor of $\mathbf{9.95}$. One thing separates them——what counts as one unit. Iterations, function evaluations, or the first step. All three are correct, and all three return different answers. Saying “fast” without naming a unit is like saying “large” without naming a number.
References
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- Yamagishi, Y. (2026). The Allotted Region and the Freedom Within [Paper 112]. Zenodo. 10.5281/zenodo.21962834
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- Yamagishi, Y. (2026). “Accuracy” Is Two Things, and Calibration Removes Only One [Paper 255]. Zenodo. 10.5281/zenodo.22168883
- Yamagishi, Y. (2026). Changing the Speed Does Not Change the Destination [Paper 257]. Zenodo. 10.5281/zenodo.22168985
- Yamagishi, Y. (2026). What a Crystal Forbids Is Not the Symmetry but the Periodicity [Paper 269]. Zenodo. 10.5281/zenodo.22169855
Reference
Yamagishi, Y. (2026). “Which Is Fastest” Has Three Answers, and a Different Method Wins Each [Paper 288]. YOI: https://www.yamagishiyuuki.com/yoi/288
DOI: 10.5281/zenodo.22171328
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 288. This paper is a preprint and has not been peer reviewed.