Secant method: Question bank — Maths deep dive

Ground rules — symbols and assumptions this page uses

Before the traps, pin down everything so no symbol appears "for free."

Definition The players

$f$ — the function whose root we chase. Assumptions: $f$ is continuous near the root, twice differentiable there (needed for the Taylor argument below), and the root is simple (i.e. $f (r) = 0$ but $f^{'} (r) \neq = 0$ ). Every failure mode on this page is a case where one of these assumptions is bent.
$r$ — the true root, the exact $x$ with $f (r) = 0$ . It is the target we never quite land on.
$x_{n}$ — our $n$ -th guess. $x_{0}, x_{1}$ are the two starting guesses.
$e_{n} = x_{n} - r$ — the error: how far the $n$ -th guess is from the true root. Small $∣ e_{n} ∣$ = good guess.
$Δ x = x_{n} - x_{n - 1}$ (the "run" between the two latest guesses) and $Δ f = f (x_{n}) - f (x_{n - 1})$ (the "rise" between their heights). The chord slope is $Δ f /Δ x$ .
$K, C$ — plain constants (fixed numbers that don't grow), used to say "roughly this size" in the error laws below.

Why the golden ratio? — deriving the error recurrence

True or false — justify

The Secant method needs the derivative

f^{'} (x)

False. It approximates the slope with the chord slope

\frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}

, so it never touches

f^{'} (x)

— that is its whole selling point over Newton-Raphson method.

The Secant method always keeps the root bracketed like Bisection method

False. It is an open method: the new iterate is the

x

-intercept of a chord and can land outside

[x_{n - 1}, x_{n}]

. No sign change is enforced, so no bracket is guaranteed.

x_{0}

and

x_{1}

straddle the root (opposite signs of

f

), convergence is guaranteed

False. Straddling helps the first step but is not preserved. Because we always keep the two latest points, both can drift to the same side and the method can diverge.

Two starting points are needed only to be safe; one would technically do

False. Two points are needed to define a straight line. With one point and no derivative you cannot draw any chord at all — the method is undefined.

The Secant method converges quadratically like Newton

False. Its order is the golden ratio

p = \frac{1 + 5}{2} \approx 1.618

— superlinear but strictly below Newton's

p = 2

Per function evaluation, Secant can beat Newton

True. Under our cost model (one unit per

f

f^{'}

), Newton spends

2

units per step; Secant reuses the old value and spends one. Two Secant steps give

1.61 8^{2} \approx 2.6 > 2

, so when

f^{'}

is costly Secant often wins.

The "weighted average" form

\frac{x _{n - 1} f ( x _{n} ) - x _{n} f ( x _{n - 1} )}{f ( x _{n} ) - f ( x _{n - 1} )}

is a different method from the boxed form

False. It is the same iteration, algebraically rearranged. The boxed subtraction form is just numerically safer (less catastrophic cancellation) in code.

Regula Falsi (False Position) and Secant use the same intercept formula, so they behave the same

False. Same formula, different bookkeeping. Regula Falsi keeps a bracketing endpoint (bounded, but only linear); Secant keeps the two most recent points (order 1.618, no bracket).

Spot the error

x_{n + 1} = x_{n} - f (x_{n}) \frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}

The fraction is upside-down. We divide

f

by the slope, so it must be

\frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )}

(

Δ x

over

Δ f

, i.e. run over rise). As written this multiplies by the slope and blows up.

"After computing

x_{n + 1}

, replace

x_{n - 1}

with

x_{n + 1}

and keep

x_{n}

Wrong window slide. The new pair must be the two latest points:

x_{n}

becomes

x_{n - 1}

and

x_{n + 1}

becomes

x_{n}

. Keeping the stale

x_{n}

paired with

x_{n + 1}

destroys the superlinear order.

"I recompute

f (x_{n - 1})

and

f (x_{n})

fresh every iteration."

Wasteful and self-defeating. The efficiency advantage comes from reusing the already-computed

f (x_{n})

as the next step's

f (x_{n - 1})

. Recomputing throws away the one-evaluation-per-step benefit.

"The denominator

f (x_{n}) - f (x_{n - 1})

can never be zero, so no guarding is needed."

Dangerous. Near a flat region — or right at convergence when both

f

values are tiny and nearly equal — the denominator

Δ f \to 0

and

x_{n + 1}

explodes. Always guard with a tolerance and an iteration cap.

"Since Secant approximates Newton, its error also satisfies

e_{n + 1} \approx C e_{n}^{2}

No. Secant's error recurrence is

e_{n + 1} \approx K e_{n} e_{n - 1}

(it uses two points, so both old errors appear), which yields

p = 1.618

, not

2

. The mixed

e_{n} e_{n - 1}

product is exactly why the order is between linear and quadratic.

x_{n + 1}

is a weighted average, so it always lies between

x_{n - 1}

and

x_{n}

False conclusion. The "weights"

f (x_{n})

and

f (x_{n - 1})

can have the same sign, making one weight effectively negative — so it is not a convex average and

x_{n + 1}

can fall outside the interval. Only when the signs differ is it a genuine convex blend.

Why questions

Why does the Secant method need two starting guesses when Newton needs only one?

Newton pairs one point with an exact slope

f^{'} (x)

; Secant has no slope, so it substitutes a second point to define the chord. It trades a derivative for an extra evaluation.

Why is the order exactly the golden ratio and not some arbitrary number?

The error recurrence

e_{n + 1} \approx K e_{n} e_{n - 1}

forces

p = 1 + \frac{1}{p}

, i.e.

p^{2} - p - 1 = 0

, whose positive root is

\frac{1 + 5}{2}

. The two-point product is baked into the structure — see the derivation section above.

Why does Secant lose the bracketing that Regula Falsi (False Position) keeps?

Secant always discards the oldest point regardless of sign; Regula Falsi deliberately discards the endpoint with the same sign as

f (x_{n + 1})

, preserving a sign change. The choice of what to keep is the whole difference.

Why can Secant sometimes be more efficient than Newton despite lower order?

Efficiency is measured per cost unit (one

f

or one

f^{'}

), not per step. Secant's cost is one unit per step;

1.61 8^{2} \approx 2.6 > 2

over two steps means it can outrun Newton when

f^{'}

is expensive or unavailable.

Why does the boxed form avoid the cancellation that the "weighted average" form suffers?

The boxed form computes the correction

f (x_{n}) \cdot Δ x /Δ f

and subtracts it from

x_{n}

; near convergence this correction is tiny and

x_{n}

is intact. The average form subtracts two large nearly-equal products

x_{n - 1} f (x_{n})

and

x_{n} f (x_{n - 1})

, losing significant digits.

Why does

x_{n - 1} \to x_{n}

turn the Secant formula into Newton's?

The chord slope

Δ f /Δ x = \frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}

is exactly the difference quotient whose limit defines

f^{'} (x_{n})

. As the two points merge, the chord becomes the tangent — see Finite differences.

Edge cases

What happens if

f (x_{0}) = f (x_{1})

exactly?

The chord is horizontal (slope

0

, so

Δ f = 0

), and it has no

x

-intercept — the method breaks on the very first step. You must pick different starting values.

What if

x_{0} = x_{1}

Both the run

Δ x = x_{n} - x_{n - 1}

and the rise

Δ f = f (x_{n}) - f (x_{n - 1})

are zero, giving

\frac{0}{0}

. The method is undefined: two distinct points are required to form a chord.

What if

f

has a local extremum (flat spot) between the two guesses?

The chord slope

Δ f /Δ x

becomes tiny, the correction

f \cdot Δ x /Δ f

becomes huge, and the next iterate can be flung far away — possibly to a different root or to divergence. This is why flat regions are hazardous.

What if the function has multiple roots and the starting pair straddles two of them?

The chord's intercept depends on the two heights, not on which root

r

you want; Secant may converge to a root you did not intend, or oscillate. It has no mechanism to prefer a particular root.

At a double (repeated) root, what happens to convergence?

Here

f^{'} (r) = 0

, so our "simple root" assumption fails and the constant

K = f^{''} (r) /2 f^{'} (r)

blows up. Like Newton, Secant then slows dramatically; the superlinear

1.618

order degrades toward linear and many more iterations are needed.

What if the two starting guesses are both very far from the root?

The Taylor picture only holds near

r

, so far away the chord is a poor model and the first intercept can overshoot wildly. Being an open method, there is no safety net — it may diverge, unlike Bisection method which always makes progress.

What is the expected behaviour right at convergence, and why is it a trap?

x_{n} \to r

, both

f (x_{n})

and

f (x_{n - 1})

shrink toward

0

, so

Δ f = f (x_{n}) - f (x_{n - 1}) \to 0

. The very success of the method makes it numerically fragile — stop on a step-size tolerance

∣ x_{n + 1} - x_{n} ∣

, not by chasing

f = 0

forever.

Recall One-line self-test

Cover every answer, run the whole page top to bottom. If you can justify each in a full sentence — not "yes"/"no" — you own the concept. Weak spots almost always cluster in Edge cases and the bracketing items; revisit Order of convergence and Regula Falsi (False Position) there.