Secant method: Exercises — Maths deep dive

The one formula we lean on the whole way down: $x_{n + 1} = x_{n} - f (x_{n}) \frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )} .$

Read it as the [!mnemonic] from the parent: "New = Now minus f-Now times Run-over-Rise."

The picture above is the only geometry you need: two dots on the curve, a straight ruler (the secant line — a chord joining two curve points) through them, and the spot where that ruler crosses the horizontal axis. That crossing is the next guess.

Level 1 — Recognition

L1.1

State the Secant iteration formula and say, in one sentence, what geometric object each step builds.

Recall Solution

$x_{n + 1} = x_{n} - f (x_{n}) \frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )} .$ Each step builds the secant line (a straight chord) through the two latest points $(x_{n - 1}, f (x_{n - 1}))$ and $(x_{n}, f (x_{n}))$ , then takes that line's $x$ -intercept (where it hits the horizontal axis) as the next guess $x_{n + 1}$ .

L1.2

True or false, with one-line reasons: (a) The Secant method needs $f^{'} (x)$ . (b) It needs two starting guesses. (c) It always keeps the root bracketed like bisection.

Recall Solution

(a) False. It replaces the derivative with the chord slope $\frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}$ — no $f^{'}$ ever computed. (b) True. You need two points to define a line. (c) False. It is an open method: the new point may jump outside $[x_{n - 1}, x_{n}]$ and the sign change is not preserved. That bracketing property belongs to Regula Falsi (False Position).

L1.3

Which quantity in Newton's step does the Secant method replace, and with what?

Recall Solution

Newton uses the exact slope $f^{'} (x_{n})$ . Secant replaces it with the finite-difference slope $\frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}},$ which is the finite-difference estimate of $f^{'} (x_{n})$ built from two known heights. See Newton-Raphson method for the exact-slope version.

Level 2 — Application

L2.1

$f (x) = x^{2} - 3$ , root $3 \approx 1.73205$ . Take $x_{0} = 1, x_{1} = 2$ . Compute $x_{2}$ and $x_{3}$ .

Recall Solution

$f (1) = 1 - 3 = - 2$ , $f (2) = 4 - 3 = 1$ . $x_{2} = 2 - 1 \cdot \frac{2 - 1}{1 - ( - 2 )} = 2 - \frac{1}{3} = 1.66667.$ WHAT we did: plugged the two latest points into the boxed formula. WHY: the $x$ -intercept of the chord through them is our best straight-line guess for the root. WHAT IT LOOKS LIKE: the ruler drops from the point $(2, 1)$ toward the axis and lands left of $x = 2$ .

Now $f (1.66667) = 1.6666 7^{2} - 3 = - 0.22222$ . $x_{3} = 1.66667 - (- 0.22222) \cdot \frac{1.66667 - 2}{- 0.22222 - 1} = 1.66667 - (- 0.22222) \cdot \frac{- 0.33333}{- 1.22222} .$ $= 1.66667 - (- 0.22222) (0.27273) = 1.66667 + 0.06061 = 1.72727.$ So $x_{2} = 1.66667$ , $x_{3} = 1.72727$ — already three correct digits.

L2.2

$f (x) = x^{3} - x - 2$ , root $\approx 1.52138$ . Take $x_{0} = 1, x_{1} = 2$ . Compute $x_{2}$ .

Recall Solution

$f (1) = 1 - 1 - 2 = - 2$ , $f (2) = 8 - 2 - 2 = 4$ . $x_{2} = 2 - 4 \cdot \frac{2 - 1}{4 - ( - 2 )} = 2 - 4 \cdot \frac{1}{6} = 2 - 0.66667 = 1.33333.$ One clean step from the recipe — no derivative of $x^{3} - x - 2$ needed.

L2.3

$f (x) = e^{x} - 3 x$ , root $\approx 0.61906$ . Take $x_{0} = 0, x_{1} = 1$ . Compute $x_{2}$ .

Recall Solution

$f (0) = e^{0} - 0 = 1$ , $f (1) = e - 3 = 2.71828 - 3 = - 0.28172$ . $x_{2} = 1 - (- 0.28172) \cdot \frac{1 - 0}{- 0.28172 - 1} = 1 - (- 0.28172) \cdot \frac{1}{- 1.28172} .$ $= 1 - (- 0.28172) (- 0.78021) = 1 - 0.21980 = 0.78020.$ The method never cared that $e^{x} - 3 x$ has no closed-form root.

Level 3 — Analysis

L3.1

Show the two formula forms agree. Starting from $x_{n + 1} = x_{n} - f (x_{n}) \frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )},$ derive the weighted-average form $x_{n + 1} = \frac{x _{n - 1} f ( x _{n} ) - x _{n} f ( x _{n - 1} )}{f ( x _{n} ) - f ( x _{n - 1} )} .$

Recall Solution

Put everything over the common denominator $D = f (x_{n}) - f (x_{n - 1})$ : $x_{n + 1} = \frac{x _{n} D - f ( x _{n} ) ( x _{n} - x _{n - 1} )}{D} .$ Expand the numerator: $x_{n} f (x_{n}) - x_{n} f (x_{n - 1}) - x_{n} f (x_{n}) + x_{n - 1} f (x_{n}) .$ The $x_{n} f (x_{n})$ terms cancel, leaving $x_{n - 1} f (x_{n}) - x_{n} f (x_{n - 1})$ . Hence $x_{n + 1} = \frac{x _{n - 1} f ( x _{n} ) - x _{n} f ( x _{n - 1} )}{f ( x _{n} ) - f ( x _{n - 1} )} . ■$ WHY the boxed form is still preferred: the weighted form subtracts two nearly-equal large products near convergence, losing precision (cancellation). The boxed form only adds a small correction to $x_{n}$ .

L3.2

Numerical blow-up. For $f (x) = x^{2} - 2$ suppose two consecutive iterates are $x_{n - 1} = 1.41421$ and $x_{n} = 1.41422$ . Explain, with numbers, why the next step is dangerous.

Recall Solution

$f (1.41421) = 1.4142 1^{2} - 2 = - 0.0000106$ , $f (1.41422) = 1.4142 2^{2} - 2 = + 0.0000177$ . Denominator $= f (x_{n}) - f (x_{n - 1}) = 0.0000177 - (- 0.0000106) = 0.0000283$ — tiny. Dividing by a near-zero denominator massively amplifies any rounding error in those two almost-equal $f$ values. Fix: guard with a tolerance if abs(denominator) < eps: stop, plus a max-iteration cap.

L3.3

Open-method jump. For $f (x) = x^{2} - 2$ with $x_{0} = 0, x_{1} = 3$ , compute $x_{2}$ and check whether it stays inside $[0, 3]$ . Comment.

Recall Solution

$f (0) = - 2$ , $f (3) = 7$ . $x_{2} = 3 - 7 \cdot \frac{3 - 0}{7 - ( - 2 )} = 3 - 7 \cdot \frac{3}{9} = 3 - 2.33333 = 0.66667.$ Here $x_{2} = 0.66667 \in [0, 3]$ , so this step stays inside. But nothing forced it to: because Secant ignores signs, a poorly chosen pair can produce an $x_{2}$ outside the starting interval — the hallmark of an open method. Bracketing methods like Bisection method can never do that.

Level 4 — Synthesis

L4.1

Derive the order of convergence. Given the error recurrence $e_{n + 1} \approx K e_{n} e_{n - 1}$ (with $e_{n} = x_{n} - r$ ), show the order $p$ solves $p^{2} - p - 1 = 0$ and equals $φ = \frac{1 + 5}{2}$ .

Recall Solution

Assume the defining relation of order- $p$ convergence: $e_{n + 1} = A e_{n}^{p}$ for some constant $A$ . Shift the index down by one: $e_{n} = A e_{n - 1}^{p}$ , so $e_{n - 1} = (e_{n} / A)^{1/ p}$ . Substitute both into $e_{n + 1} = K e_{n} e_{n - 1}$ and into $e_{n + 1} = A e_{n}^{p}$ : $A e_{n}^{p} = K e_{n} (e_{n} / A)^{1/ p} = K A^{- 1/ p} e_{n}^{1 + 1/ p} .$ Match the powers of $e_{n}$ on both sides: $p = 1 + \frac{1}{p} ⟹ p^{2} = p + 1 ⟹ p^{2} - p - 1 = 0.$ Positive root: $p = \frac{1 + 5}{2} \approx 1.61803$ . $■$

L4.2

Efficiency per evaluation. Newton does 2 evaluations/step and has order 2. Secant does 1 new evaluation/step and has order $1.618$ . Show that two Secant steps (matching Newton's evaluation budget) beat one Newton step in effective order.

Recall Solution

Effective order per pair of evaluations:

Newton: one step of order $2$ per 2 evaluations $\Rightarrow$ effective order $2$ .
Secant: two steps, each order $1.618$ , compose to order $1.61 8^{2} = 2.618$ per 2 evaluations. Since $2.618 > 2$ , Secant extracts more accuracy per function evaluation whenever an $f^{'}$ evaluation costs about as much as an $f$ evaluation. This is why Secant wins when the derivative is expensive.

L4.3

Compare methods on the same problem. For $f (x) = x^{2} - 2$ , $x_{0} = 1, x_{1} = 2$ , list what information each method uses at the first step: Secant, Newton, Regula Falsi, Bisection.

Recall Solution

Method	Info used at step 1	Keeps a bracket?
Secant	$f (1), f (2)$ — two heights	No
Newton	$f (2), f^{'} (2) = 4$ — one height + slope	No
Regula Falsi	$f (1), f (2)$ + sign check	Yes (replaces same-sign end)
Bisection	only signs of $f (1), f (2)$	Yes (halves interval)
Secant and Regula Falsi share the same $x$ -intercept formula; only the bookkeeping (whether a bracket is enforced) differs — which is exactly why one is order $1.618$ and the other only linear.

Level 5 — Mastery

L5.1

Full multi-step run to a tolerance. Solve $f (x) = cos x - x = 0$ with $x_{0} = 0.5, x_{1} = 1$ . Iterate until $∣ x_{n + 1} - x_{n} ∣ < 1 0^{- 4}$ . Report every iterate. (Root $\approx 0.73909$ .)

Recall Solution

All angles in radians. $f (0.5) = cos 0.5 - 0.5 = 0.87758 - 0.5 = 0.37758$ . $f (1) = cos 1 - 1 = 0.54030 - 1 = - 0.45970$ .

Step to $x_{2}$ : $x_{2} = 1 - (- 0.45970) \cdot \frac{1 - 0.5}{- 0.45970 - 0.37758} = 1 - (- 0.45970) \cdot \frac{0.5}{- 0.83728} .$ $= 1 - (- 0.45970) (- 0.59717) = 1 - 0.27452 = 0.72548.$

Step to $x_{3}$ : $f (0.72548) = cos (0.72548) - 0.72548 = 0.74829 - 0.72548 = 0.02281$ . $x_{3} = 0.72548 - 0.02281 \cdot \frac{0.72548 - 1}{0.02281 - ( - 0.45970 )} .$ $= 0.72548 - 0.02281 \cdot \frac{- 0.27452}{0.48251} = 0.72548 - 0.02281 (- 0.56895) = 0.72548 + 0.01298 = 0.73846.$

Step to $x_{4}$ : $f (0.73846) = cos (0.73846) - 0.73846 = 0.73951 - 0.73846 = 0.00105$ . $x_{4} = 0.73846 - 0.00105 \cdot \frac{0.73846 - 0.72548}{0.00105 - 0.02281} .$ $= 0.73846 - 0.00105 \cdot \frac{0.01298}{- 0.02176} = 0.73846 - 0.00105 (- 0.59651) = 0.73846 + 0.00063 = 0.73909.$

Now $∣ x_{4} - x_{3} ∣ = ∣0.73909 - 0.73846∣ = 0.00063 > 1 0^{- 4}$ , so one more: $f (0.73909) = cos (0.73909) - 0.73909 = 0.73909 - 0.73909 \approx 0.0000005$ . The next iterate moves by well under $1 0^{- 4}$ , so we stop with $r \approx 0.73909$ .

Iterate list: $0.5, 1, 0.72548, 0.73846, 0.73909$ .

L5.2

Design a stopping rule. Write pseudocode for a robust Secant routine that (i) caps iterations, (ii) guards the denominator, (iii) stops on step size. Explain each guard.

Recall Solution

secant(f, x0, x1, tol, maxit):
    f0 = f(x0);  f1 = f(x1)
    for k in 1..maxit:
        denom = f1 - f0
        if abs(denom) < 1e-14:        # guard (ii): flat chord → blow-up
            stop "denominator too small"
        x2 = x1 - f1*(x1 - x0)/denom  # boxed form (least cancellation)
        if abs(x2 - x1) < tol:        # guard (iii): step size small enough
            return x2
        x0, f0 = x1, f1               # slide the window...
        x1, f1 = x2, f(x2)            # ...reuse f1, one new eval per step
    stop "no convergence in maxit"    # guard (i): iteration cap

Why each guard: (i) prevents infinite loops when the method diverges/oscillates; (ii) catches the near-zero denominator from L3.2 before it produces inf; (iii) the step-size test is cheap and directly measures how much progress remains. Note the single new $f$ -evaluation per loop — that is the efficiency edge from L4.2.

L5.3

Convergence-rate sanity check. Using the run in Worked Example 1 of the parent ( $f = x^{2} - 2$ ), the errors were $0.081 \to 0.014 \to 0.00042$ . Verify these are consistent with superlinear (order $\approx 1.618$ ) behaviour rather than linear.

Recall Solution

For linear convergence, $e_{n + 1} / e_{n}$ would be roughly constant. Here: $\frac{0.014}{0.081} = 0.173, \frac{0.00042}{0.014} = 0.030.$ The ratio drops ( $0.173 \to 0.030$ ), i.e. the fraction of error removed keeps growing — that is superlinear, not linear. Cross-check the order estimate from three errors: $p \approx \frac{l n ( e _{n + 1} / e _{n} )}{l n ( e _{n} / e _{n - 1} )} = \frac{l n ( 0.030 )}{l n ( 0.173 )} = \frac{- 3.507}{- 1.755} \approx 1.998.$ With only three noisy points and rounded errors this lands near $2$ ; a longer clean run settles to $φ \approx 1.618$ . The key qualitative fact — ratios shrinking, digits multiplying — is confirmed.

Recall One-line self-audit

Can you, from memory, (a) write the boxed formula, (b) say why no derivative is needed, (c) name the order $φ \approx 1.618$ , and (d) list the three guards from L5.2? If yes to all four, you own Secant method at D4 level.