Estimate Square Roots With Linearization

You have a number like 4.1 and need its square root without a calculator. The fast method is linearization: use a nearby perfect square, find the square root of that perfect square, then adjust. To estimate square roots with linearization, take the square root of the perfect square, add the derivative of the square root function multiplied by the difference between your number and the perfect square. The perfect square 4 gives √4 = 2. The derivative of √x is 1/(2√x), so at x = 4 it is 0.25. Multiply 0.25 by 0.1 (the distance from 4 to 4.1), add to 2, and get 2.025. The actual √4.1 is 2.02485. The estimate is an overestimate by about 0.00015. The error direction is always an overestimate for √x because the function is concave down, its second derivative is negative, so the tangent line sits above the curve.

Working Through √4.1 Step by Step

Choose the perfect square closest to your target. For 4.1, that perfect square is 4. The point of tangency a is 4. The linearization formula from the fact sheet is L(x) = √a + (x − a)/(2√a). Plug in a = 4: √4 = 2, and 2√a = 4, so L(x) = 2 + (x − 4)/4. For x = 4.1, L(4.1) = 2 + 0.1/4 = 2 + 0.025 = 2.025. The table below shows L(x) against the true value f(x) = √x for points near 4.

x	L(x)	√x	Error |E(x)|	Over/Under
3.9	1.975	1.97484	0.00016	Overestimate
4.0	2.000	2.00000	0.00000	Exact
4.1	2.025	2.02485	0.00015	Overestimate
4.5	2.125	2.12132	0.00368	Overestimate
5.0	2.250	2.23607	0.01393	Overestimate

How to Compute the Linearization Yourself

Identify the perfect square a nearest to your target x. Compute √a from memory, for 4, 9, 16, 25, 36, and so on. The derivative at a is 1/(2√a). For a = 4, that is 0.25. For a = 9, it is 1/(2*3) = 1/6 ≈ 0.1667. Then the estimate L(x) = √a + (1/(2√a)) * (x − a). The actual √x, when you need it, can be found using any square root calculator. In this case, the estimate at x = 4.1 is 2.025, the true value is 2.02485, and the error is 0.00015.

Error Direction From Concavity

The second derivative of √x is f''(x) = −1/(4x^(3/2)). For any positive x, this is negative. A negative second derivative means the function is concave down. When a function is concave down, its tangent line lies above the curve, so every linearization estimate for √x is an overestimate. This is not a guess: it is determined by the sign of f''(a). For √x, every estimate is an overestimate, without exception.

Error Bound Using Taylor's Inequality

Taylor's inequality for n = 1 gives a worst-case error bound: |E(x)| ≤ (M/2) * |x − a|², where M is the maximum of |f''| on the interval between a and x. On [4, 4.1], f''(x) = −1/(4x^(3/2)). Its absolute value is largest at the left endpoint, x = 4, because the denominator is smallest there. |f''(4)| = 1/(4 * 8) = 1/32 = 0.03125. So M = 0.03125. Then the bound is (0.03125/2) * (0.1)² = 0.015625 * 0.01 = 0.00015625. The actual error 0.00015 is inside this bound. The error grows with the square of the distance from a, so moving from |x − a| = 0.1 to 1.0 multiplies the bound by 100.

When Linearization Breaks Down

Take x = 10 and a = 4. The distance |x − a| is 6. L(10) = 2 + 0.25 * 6 = 2 + 1.5 = 3.5. The true √10 is about 3.162. The error is 0.338, more than 10% of the true value. That estimate is not useful for any practical purpose. The error bound from Taylor's inequality is (0.03125/2) * 6² = 0.015625 * 36 = 0.5625, which is large enough to warn you the estimate is poor. The method works only when a is a perfect square close to x. For √10, you would choose a = 9 instead. With a = 9, √9 = 3, f'(9) = 1/6 ≈ 0.1667, and L(10) = 3 + 0.1667 = 3.1667, with error about 0.0047, much better.

What Goes Wrong With Distance

The error grows quadratically with |x − a|. Double the distance and the error bound quadruples. For √x at a = 4, an estimate at x = 4.2 (distance 0.2) has error bound 4 times larger than at x = 4.1. At x = 4.5 (distance 0.5), the bound is 25 times larger. The linearization is exact at x = a, then degrades rapidly. This is the single most common failure: picking a perfect square too far from the target. The second most common failure is forgetting the function value at a, L(x) starts at f(a), not zero. A third is mixing up dy and Δy. dy is the change along the tangent line, Δy is the actual change on the curve. For √4.1 from 4, dy = 0.25 * 0.1 = 0.025, Δy = 2.02485 − 2 = 0.02485, they differ by the error.

Linearization suits AP Calculus AB/BC and Calculus I students who need to compute linearizations by hand and interpret whether the estimate is an overestimate or underestimate using concavity. It also suits physics students who use small-angle approximations for sin θ and tan θ in pendulum and optics derivations, provided they work in radians. Students learning differentials and how dy relates to Δy for error propagation will find the square root example direct. The single thing that most often goes wrong is picking a point of tangency too far from the target, causing error that grows quadratically into uselessness. Anyone needing exact function values, numerical root-finding with Newton's method, or higher-order Taylor polynomial approximations beyond n = 1 should skip this method. Those readers need a Taylor series calculator or a root-finding resource, not a tangent line that is only accurate near one point.