Linearizing Square Roots and Rational Powers

Why Linearize Powers?

Some numerical calculations are awkward even when the underlying function is familiar. A square root such as \(\sqrt{50}\) is not an integer, but it is close to \(\sqrt{49}=7\). A rational power such as \(26^{1/3}\) is close to \(27^{1/3}=3\). In each case, a nearby input has a function value and derivative that are easy to calculate.

The idea is to replace the function near that convenient input with its tangent line. This does not produce an exact identity for nearby values. It produces a local approximation whose accuracy is best when the input is close to the base point.

Linearizing Square Roots

Write the square-root function as \(f(x)=x^{1/2}\), for \(x>0\). The power rule gives \(f'(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\). Thus, at any positive base point \(a\),

The base \(a\) should make both \(\sqrt a\) and the slope \(1/(2\sqrt a)\) easy to compute, and should be close to the input \(x\). Perfect-square base points are especially useful. If \(x=a+h\), the same formula is

This says that a small change \(h\) in the input produces an estimated change of \(h/(2\sqrt a)\) in the square root. The approximation is linear in \(h\); it is not an exact formula for \(\sqrt{a+h}\).

Linearizing Rational Powers

For a rational exponent \(r\), the same method applies to \(f(x)=x^r\). In this discussion, take \(a>0\) and inputs \(x>0\), so the powers and derivatives are defined as real numbers. The power rule gives \(f'(x)=r x^{r-1}\). Therefore the linearization at \(a\) is

For a root, \(r=1/n\) for a positive integer \(n\); for example, a cube root uses \(r=1/3\). Other rational powers can be handled in exactly the same way. Choose a base where \(a^r\) and \(r a^{r-1}\) are manageable. A nearby perfect power often provides such a base.

Why the Tangent Line Gives a Local Estimate

The reason linearization works is built into differentiability. At a point \(a\), the derivative describes the limiting rate of change as the input change approaches zero. For a small change \(h\), the difference quotient is close to \(f'(a)\):

Multiplying by \(h\) and adding \(f(a)\) gives \(f(a+h)\approx f(a)+f'(a)h\), the linearization formula. More precisely, differentiability means the error remaining after this linear estimate, divided by \(h\), approaches zero as \(h\) approaches zero. This is why the estimate is local: the guarantee concerns inputs near \(a\), not inputs arbitrarily far away.

For rational powers, the derivative calculation makes this reasoning explicit. If \(f(x)=x^r\), then \(f(a)=a^r\) and \(f'(a)=r a^{r-1}\). Substitution into the definition of linearization gives \(L(x)=a^r+r a^{r-1}(x-a)\). For square roots, setting \(r=1/2\) gives \(a^{1/2}+\frac12a^{-1/2}(x-a)\), which is the square-root formula above.

Curvature and the Direction of the Estimate

For a twice-differentiable function, the sign of its second derivative determines its concavity on an interval. A concave-down function lies below each of its tangent lines on an interval where it is concave down; a concave-up function lies above its tangent lines there. This provides a way to predict the direction of a linear estimate, provided the function remains in the relevant interval.

For \(f(x)=x^r\), the second derivative is \(f''(x)=r(r-1)x^{r-2}\). On \(x>0\), square roots and cube roots have exponents \(0<r<1\), so \(r(r-1)<0\): these functions are concave down. Their tangent-line estimates are therefore above the function values. For an exponent \(r>1\), the sign is positive on \(x>0\), so the power function is concave up and its tangent lines lie below the function.

Choosing a Base and Avoiding a Common Error

A useful base is close to the target input and makes both the function value and derivative convenient. For a square root, a nearby perfect square often works. For an \(n\)th root, a nearby perfect \(n\)th power may be effective. For a general rational power, look for a positive base whose root or power is easy to evaluate. A very convenient base that is far from the target can give a poor local estimate, so convenience alone is not enough.

A common mistake is to compute the derivative correctly but then omit the factor \(x-a\). For example, at \(a=49\), the slope of \(\sqrt{x}\) is \(1/14\). It would be wrong to claim \(\sqrt{50}\approx 1/14\): this uses the slope as though it were the function value. The line must start at \(f(a)=7\) and change by slope times input change:

not just \(1/14\). The base value anchors the tangent line; the slope determines how the line changes away from that base.

Another safeguard is to keep the computation exact until the end. Substitute the target input into the linearization, simplify using fractions, and round only the final estimate. Then check that the line has value \(f(a)\) and slope \(f'(a)\) at the base point. When the sign of the second derivative is known, compare the estimate's direction with the concavity as an additional reasonableness check.