Regula Falsi (False Position)

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2023)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~25 min
• Difficulty mix: medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Regula Falsi sits at the intersection of bracketing methods and interpolation — it replaces the mechanical midpoint of bisection with the chord intersection, making it faster in practice while preserving the sign-change guarantee. UPSC 2023 tested it as a Section B computation: carry out two or three iterations by hand with clean arithmetic, showing the updated bracket after each step.

Minimum Theory

Setup. Given $f$ continuous on $[a, b]$ with $f(a)f(b) < 0$, the false-position approximation is the $x$-intercept of the straight line joining $(a, f(a))$ and $(b, f(b))$:

$c = a - f(a)\cdot\frac{b - a}{f(b) - f(a)} = \frac{a\,f(b) - b\,f(a)}{f(b) - f(a)}$

Bracket update. Evaluate $f(c)$:

• If $f(a)f(c) < 0$: root is in $(a, c)$; set $b \leftarrow c$.
• If $f(a)f(c) > 0$: root is in $(c, b)$; set $a \leftarrow c$.
• If $f(c) = 0$: stop.

Key distinction from bisection. In bisection the bracket always halves. In regula falsi one endpoint may be fixed for many iterations (the chord keeps rotating about the same anchor), so the bracket does not shrink as uniformly.

Convergence. Linear (order 1), with an asymptotic constant that depends on $f''$ and $f'''$. It is generally faster than bisection on smooth functions but can stagnate when $f$ is highly curved.

Modified Regula Falsi (Illinois method). If the same endpoint is retained twice in a row, halve its $f$-value before the next step. This breaks stagnation and achieves superlinear convergence. UPSC has not tested this variant but awareness is useful context.

Comparison with secant method. Both use a chord through two recent points, but:

Formula mnemonic. Think of the chord $y - f(a) = \frac{f(b)-f(a)}{b-a}(x-a)$; set $y = 0$ and solve for $x = c$.

Question Archetypes

iterate-k-steps (1 question; 2023)

Recognition Cues

• “Apply the method of false position” or “use regula falsi.”
• A specific equation and a specific number of iterations (usually 2–3).
• Marks are awarded step-by-step, so every bracket update must be shown.

Solution Template

1. Verify the root bracket: compute $f(a)$ and $f(b)$, confirm opposite signs.
2. Compute $c_1$ using the formula; evaluate $f(c_1)$.
3. Determine new bracket $[a_1, b_1]$ by comparing signs of $f(a)$, $f(c_1)$, $f(b)$.
4. Repeat for the requested number of iterations.
5. State the final approximation and, if asked, the error relative to the exact root.

Worked Example

2023 Paper 2, 2023-P2-Q5b (15 marks)

Using the method of false position, find a real root of $f(x) = x^3 - 4x - 9 = 0$ correct to two decimal places (perform at least 3 iterations).

Step 0 — locate the bracket.

$f(2) = 8 - 8 - 9 = -9 < 0, \qquad f(3) = 27 - 12 - 9 = 6 > 0$

Root lies in $(2, 3)$. Set $a_0 = 2$, $b_0 = 3$.

Iteration 1.

$c_1 = \frac{a_0 f(b_0) - b_0 f(a_0)}{f(b_0) - f(a_0)} = \frac{2(6) - 3(-9)}{6 - (-9)} = \frac{12 + 27}{15} = \frac{39}{15} = 2.6$

$f(2.6) = (2.6)^3 - 4(2.6) - 9 = 17.576 - 10.4 - 9 = -1.824 < 0$

Since $f(a_0) = f(2) < 0$ and $f(c_1) < 0$: root is in $(c_1, b_0) = (2.6, 3)$.

Set $a_1 = 2.6$, $b_1 = 3$.

Iteration 2.

$c_2 = \frac{2.6(6) - 3(-1.824)}{6 - (-1.824)} = \frac{15.6 + 5.472}{7.824} = \frac{21.072}{7.824} \approx 2.6937$

$f(2.6937) = (2.6937)^3 - 4(2.6937) - 9$

$(2.6937)^3 \approx 19.537$, $4(2.6937) \approx 10.775$

$f(2.6937) \approx 19.537 - 10.775 - 9 = -0.238 < 0$

Root is in $(2.6937, 3)$. Set $a_2 = 2.6937$, $b_2 = 3$.

Iteration 3.

$c_3 = \frac{2.6937(6) - 3(-0.238)}{6 - (-0.238)} = \frac{16.1622 + 0.714}{6.238} = \frac{16.876}{6.238} \approx 2.7062$

$f(2.7062) = (2.7062)^3 - 4(2.7062) - 9$

$(2.7062)^3 \approx 19.828$, $4(2.7062) \approx 10.825$

$f(2.7062) \approx 19.828 - 10.825 - 9 = 0.003 \approx 0^+$

Root is in $(2.6937, 2.7062)$.

Conclusion. After 3 iterations the root is approximately $\mathbf{x^* \approx 2.71}$, correct to two decimal places. (The true root is $x^* \approx 2.7065$.)

$\boxed{x^* \approx 2.71}$

Common Traps

• Using the midpoint formula $(a+b)/2$ instead of the chord-intercept formula — that is bisection, not regula falsi.
• Forgetting to update the bracket: both $a$ and $f(a)$ must be updated when the left endpoint changes (and similarly for $b$).
• Sign error in $f(b) - f(a)$: when $f(a) < 0$ and $f(b) > 0$, the denominator is $f(b) - f(a) > 0$; do not reverse the sign.
• Rounding intermediate values too aggressively — carry at least 4 decimal places through the iteration.

Marks-Aware Writing

Section B questions at 15 marks expect three full iterations with every intermediate value shown. Allocate roughly 4 marks per iteration (bracket check + formula + $f(c)$ evaluation + update) plus 3 marks for the conclusion and any discussion of convergence. Never skip directly to the answer — the working is where the marks live.

Practice Set

Only one historical question on this atom (shown above).