5.3: Fractional changes with general exponents

The fractional-change approximations for changes in \(x^{2}\) (Section 5.2.3) and in \(x^{3}\) (Problem 5.13) are special cases of the approximation for \(x^{n}\)

\[\frac{Δ(x^{n})}{x^{n}} ≈ n \times \frac{Δx}{x}. \label{5.14} \]

This rule offers a method for mental division (Section 5.3.1), for estimating square roots (Section 5.3.2), and for judging a common explanation for the seasons (Section 5.3.3). The rule requires only that the fractional change be small and that the exponent n not be too large (Section 5.3.4).

Rapid mental division

The special case \(n = −1\) provides the method for rapid mental division. As an example, let’s estimate 1/13. Rewrite it as \((x + Δx) − 1\) with \(x = 10\) and \(Δx = 3\). The big part is \(x − 1 = 0.1\). Because \((Δx)/x = 30%\), the fractional correction to \(x − 1\) is roughly −30%. The result is 0.07.

\[frac{1}{13} ≈ \frac{1}{10} − 30% = 0.07, \label{5.15} \]

where the “−30%” notation, meaning “decrease the previous object by 30%,” is a useful shorthand for a factor of \(1 − 0.3\).

Question

How accurate is the estimate, and what is the source of the error?

The estimate is in error by only 9%. The error arises because the linear approximation

\[\frac{Δx^{-1}}{x^{-1}} ≈ −1 \times \frac{Δx}{x} \label{5.16} \]

does not include the square (or higher powers) of the fractional change \((Δx)/x\) (Problem 5.17 asks you to find the squared term).

Question

How can the error in the linear approximation be reduced?

To reduce the error, reduce the fractional change. Because the fractional change is determined by the big part, let’s increase the accuracy of the big part. Accordingly, multiply 1/13 by 8/8, a convenient form of 1, to construct 8/104. Its big part 0.08 approximates 1/13 already to within 4%. To improve it, write 1/104 as \((x + Δx) − 1\) with \(x = 100\) and \(Δx = 4\). The fractional change \((Δx)/x\) is now 0.04 (rather than 0.3); and the fractional correction to \(1/x\) and \(8/x\) is a mere −4%. The corrected estimate is 0.0768:

\[\frac{1}{13} ≈ 0.08 − 4% = 0.08 − 0.0032 = 0.0768. \label{5.17} \]

This estimate can be done mentally in seconds and is accurate to 0.13%!

Multiple problems

Problem 5.16 Next approximation

Multiply 1/13 by a convenient form of 1 to make a denominator near 1000; then estimate 1/13. How accurate is the resulting approximation?

Problem 5.17 Quadratic approximation

Find A, the coefficient of the quadratic term in the improved fractional-change approximation

\[\frac{Δx^{-1}}{x^{-1}} ≈ −1 \times \frac{Δx}{x} + A \times (\frac{Δx}{x})^{2}. \label{5.18} \]

Use the resulting approximation to improve the estimates for 1/13.

Problem 5.18 Fuel efficiency

Fuel efficiency is inversely proportional to energy consumption. If a 55mph speed limit decreases energy consumption by 30%, what is the new fuel efficiency of a car that formerly got 30 miles per US gallon (12.8 kilometers per liter)?

Square roots

The fractional exponent \(n = 1/2\) provides the method for estimating square roots. As an example, let’s estimate \(\sqrt{10}\). Rewrite it as \((x + Δx)^{1/2}\) with \(x = 9\) and \(Δx = 1\). The big part \(x^{1/2}\) is 3. Because \((Δx)/x = 1/9\) and \(n = 1/2\), the fractional correction is 1/18. The corrected estimate is

\[\sqrt{10} ≈ 3 x (1 + \frac{1}{18}) ≈ 3.1667. \label{5.19} \]

The exact value is 3.1622 . . ., so the estimate is accurate to 0.14%.

Multiple problems

Problem 5.19 Overestimate or underestimate?

Does the linear fractional-change approximation overestimate all square roots (as it overestimated 10)? If yes, explain why; if no, give a counterexample.

Problem 5.20 Cosine approximation

Use the small-angle approximation \(\sin θ ≈ θ\) to show that \(\cos θ ≈ 1 − θ^{2}/2\).

Problem 5.21 Reducing the fractional change

To reduce the fractional change when estimating \(\sqrt{10}\), rewrite it as \(\sqrt{360}/6\) and then estimate \(\sqrt{360}\). How accurate is the resulting estimate for \(\sqrt{10}\)?

Problem 5.22 Another method to reduce the fractional change

Because \(\sqrt{2}\) is fractionally distant from the nearest integer square roots \(\sqrt{1}\) and \(\sqrt{4}\), fractional changes do not give a direct and accurate estimate of \(\sqrt{2}\). A similar problem occurred in estimating \(ln2\) (Section 4.3); there, rewriting 2 as (4/3)/(2/3) improved the accuracy. Does that rewriting help estimate \(\sqrt{2}\)?

Problem 5.23 Cube root

Estimate \(2^{1/3}\) to within 10%.

A reason for the seasons?

Summers are warmer than winters, it is often alleged, because the earth is closer to the sun in the summer than in the winter. This common explanation is bogus for two reasons. First, summers in the southern hemisphere happen alongside winters in the northern hemisphere, despite almost no difference in the respective distances to the sun. Second, as we will now estimate, the varying earth–sun distance produces too small a temperature difference. The causal chain—that the distance determines the intensity of solar radiation and that the intensity determines the surface temperature is most easily analyzed using fractional changes.

Intensity of solar radiation: The intensity is the solar power divided by the area over which it spreads. The solar power hardly changes over a year (the sun has existed for several billion years); however, at a distance r from the sun, the energy has spread over a giant sphere with surface \(\text{area} ∼ r^{2}\). The intensity I therefore varies according to \(I ∝ r^{-2}\). The fractional changes in radius and intensity are related by

\[\frac{ΔI}{I} ≈ −2 \times \frac{Δr}{r}. \label{5.20} \]

Surface temperature: The incoming solar energy cannot accumulate and returns to space as blackbody radiation. Its outgoing intensity depends on the earth’s surface temperature T according to the Stefan–Boltzmann law \(I = σT^{4}\) (Problem 1.12), where σ is the Stefan–Boltzmann constant. Therefore \(T ∝ I^{1/4}\). Using fractional changes,

\[\frac{ΔT}{T} ≈ \frac{1}{4} x \frac{ΔI}{I}. \label{5.21} \]

This relation connects intensity and temperature. The temperature and distance are connected by \((ΔI)/I = −2 \times (Δr)/r\). When joined, the two relations connect distance and temperature as follows:

The next step in the computation is to estimate the input \((Δr)/r\) namely, the fractional change in the earth sun distance. The earth orbits the sun in an ellipse; its orbital distance is

\[r = \frac{1}{1 + \mathcal{E}\cosθ}, \label{5.22} \]

where \(\mathcal{E}\) is the eccentricity of the orbit, θ is the polar angle, and l is the semilatus rectum. Thus r varies from \(r_{min} = l/(1 + \mathcal{E})\) (when \(θ = 0^{◦}\)) to \(r_{max} = l/(1 − \mathcal{E})\) (when \(θ = 180^{◦}\)). The increase from \(r_{min}\) to l contributes a fractional change of roughly \(\mathcal{E}\). The increase from l to \(r_{max}\) contributes another fractional change of roughly \(\mathcal{E}\). Thus, r varies by roughly \(2\mathcal{E}\). For the earth’s orbit, \(\mathcal{E} = 0.016\), so the earth–sun distance varies by 0.032 or 3.2% (making the intensity vary by 6.4%).

Multiple problems

Problem 5.24 Where is the sun?

The preceding diagram of the earth’s orbit placed the sun away from the center of the ellipse. The diagram to the right shows the sun at an alternative and perhaps more natural location: at the center of the ellipse. What physical laws, if any, prevent the sun from sitting at the center of the ellipse?

Problem 5.25 Check the fractional change

Look up the minimum and maximum earth–sun distances and check that the distance does vary by 3.2% from minimum to maximum./

A 3.2% increase in distance causes a slight drop in temperature:

\[\frac{ΔT}{T} ≈ -\frac{1}{2} x \frac{Δr}{r} = -1.6%. \label{5.23} \]

However, man does not live by fractional changes alone and experiences the absolute temperature change \(ΔT\).

\[ΔT = −1.6% × T. \lbael{5.24} \nonumber \]

If our calculation predicts that \(ΔT ≈ 0^{◦}C\), it must be wrong. An even less plausible conclusion results from measuring \(T\) in Fahrenheit degrees, which makes \(T\) often negative in parts of the northern hemisphere. Yet \(ΔT\) cannot flip its sign just because \(T\) is measured in Fahrenheit degrees!

Fortunately, the temperature scale is constrained by the Stefan–Boltzmann law. For blackbody flux to be proportional to \(T^{4}\), temperature must be measured relative to a state with zero thermal energy: absolute zero. Neither the Celsius nor the Fahrenheit scale satisfies this requirement.

In contrast, the Kelvin scale does measure temperature relative to absolute zero. On the Kelvin scale, the average surface temperature is \(T ≈ 300 K\); thus, a 1.6% change in T makes \(ΔT ≈ 5K\). A \(5K\) change is also a \(5^{◦}C\) change Kelvin and Celsius degrees are the same size, although the scales have different zero points. (See also Problem 5.26.) A typical temperature change between summer and winter in temperate latitudes is \(20^{◦}C\) much larger than the predicted \(5^{◦}C\) change, even after allowing for errors in the estimate. A varying earth–sun distance is a dubious explanation of the reason for the seasons.

Problem 5.26 Converting to Fahrenheit

The conversion between Fahrenheit and Celsius temperatures is

\[F = 1.8C + 32, \label{5.25} \]

so a change of \(5^{◦}C\) should be a change of \(41^{◦}F\) sufficiently large to explain the seasons! What is wrong with this reasoning?

Problem 5.27 Alternative explanation

If a varying distance to the sun cannot explain the seasons, what can? Your proposal should, in passing, explain why the northern and southern hemispheres have summer 6 months apart.

Limits of validity

The linear fractional-change approximation

\[\frac{Δ (x^{n})}{x^{n}} ≈ n x \frac{Δx}{x} \label{5.26} \]

has been useful. But when is it valid? To investigate without drowning in notation, write \(z\) for \(Δx\); then choose \(x = 1\) to make \(z\) the absolute and the fractional change. The right side becomes \(nz\), and the linear fractional-change approximation is equivalent to

\[(1+z)^{n} ≈ 1 + nz. \label{5.27} \]

The approximation becomes inaccurate when \(z\) is too large: for example, when evaluating \(1 + z\) with \(z = 1\) (Problem 5.22). Is the exponent \(n\) also restricted? The preceding examples illustrated only moderate-sized exponents: \(n = 2\) for energy consumption (Section 5.2.3), −2 for fuel efficiency (Problem 5.18), −1 for reciprocals (Section 5.3.1), 1/2 for square roots (Section 5.3.2), and −2 and 1/4 for the seasons (Section 5.3.3). We need further data.

Question

What happens in the extreme case of large exponents?

With a large exponent such as \(n = 100\) and, say, \(z = 0.001\), the approximation predicts that 1.001100 ≈ 1.1 close to the true value of 1.105...However, choosing the same \(n\) alongside \(z = 0.1\) (larger than 0.001 but still small) produces the terrible prediction

\[\underbrace{1.1^{100}}_{(1+z)^{n}} = 1 + \underbrace{100 x 0.1}_{nz} = 11; \label{5.28} \]

1.1100 is roughly 14,000, more than 1000 times larger than the prediction.

Both predictions used large n and small z, yet only one prediction was accurate; thus, the problem cannot lie in \(n\) or \(z\) alone. Perhaps the culprit is the dimensionless product \(nz\). To test that idea, hold \(nz\) constant while trying large values of \(n\). For \(nz\), a sensible constant is 1 the simplest dimensionless number. Here are several examples.

\[1.1^{10} ≈ 2.59374, \nonumber \]

\[1.01^{100} ≈ 2.70481, \label{5.29} \]

\[1.001^{100} ≈ 2.71692. \nonumber \]

In each example, the approximation incorrectly predicts that \((1 + z)^{n} = 2\).

Question

What is the cause of the error?

To find the cause, continue the sequence beyond 1.0011000 and hope that a pattern will emerge: The values seem to approach \(e = 2.718281828...\), the base of the natural logarithms. Therefore, take the logarithm of the whole approximation.

\[ln(1 + z)n = n ln(1 + z). \label{5.30} \]

Pictorial reasoning showed that \(ln(1 + z) ≈ z\) when \(z ≪ 1\) (Section 4.3). Thus, \(n ln(1 + z) ≈ nz\), making \((1 + z) ≈ e\). This improved approximation explains why the approximation \((1 + z)n ≈ 1 + nz\) failed with large \(nz\): Only when \(nz ≪ 1\) is \(enz\) approximately \(1 + nz\). Therefore, when \(z ≪ 1\) the two simplest approximation are

\[(1+z)^{n} \approx\left\{\begin{array}{ll}
1+n z & (z \ll 1 \text { and } n z \ll 1) \\
e^{n z} & (z \ll 1 \text { and } n z \text { unrestricted) }
\end{array}\right.\label{5.31} \]

The diagram shows, across the whole \(n – z\) plane, the simplest approximation in each region. The axes are logarithmic and n and z are assumed positive: The right half plane shows \(z ≫ 1\), and the upper half plane shows \(n ≫ 1\). On the lower right, the boundary curve is \(n ln z = 1\). Explaining the boundaries and extending the approximations is an instructive exercise (Problem 5.28).

Multiple problems

Problem 5.28 Explaining the approximation plane

In the right half plane, explain the \(n/z = 1\) and \(n ln z = 1\) boundaries. For the whole plane, relax the assumption of positive \(n\) and \(z\) as far as possible.

Problem 5.29 Binomial-theorem derivation

Try the following alternative derivation of \((1+z)^{n} ≈ e^{nz}\) (where \(n ≫ 1\)). Expand \((1 + z)^{n}\) using the binomial theorem, simplify the products in the binomial coefficients by approximating \(n − k\) as \(n\), and compare the resulting expansion to the Taylor series for \(e^{nz}\).