5.1: Multiplication using one and few

In almost every quantitative problem, the analysis simplifies when you follow the proverbial advice of doing first things first. First approximate and understand the most important effect the big part then refine your analysis and understanding. This procedure of successive approximation or “taking out the big part” generates meaningful, memorable, and usable expressions. The following examples introduce the related idea of low- entropy expressions (Section 5.2) and analyze mental multiplication (Section 5.1), exponentiation (Section 5.3), quadratic equations (Section 5.4), and a difficult trigonometric integral (Section 5.5).

Multiplication using one and few

The first illustration is a method of mental multiplication suited to rough, back-of-the-envelope estimates. The particular calculation is the storage capacity of a data CD-ROM. A data CD-ROM has the same format and storage capacity as a music CD, whose capacity can be estimated as the product of three factors:

\[\underbrace{1 \mathrm{hr} \times \frac{3600 \mathrm{~s}}{1 \mathrm{hr}}}_{\text {playing time }} \times \underbrace{\frac{4.4 \times 10^{4} \text { samples }}{1 \mathrm{~s}}}_{\text {sample rate }} \times \underbrace{2 \text { channels } \times \frac{16 \text { bits }}{1 \text { sample }}}_{\text {sample size }} \label{5.1} \]

(In the sample-size factor, the two channels are for stereophonic sound.)

Multiple problems

Problem 5.1 Sample rate

Look up the Shannon–Nyquist sampling theorem [22], and explain why the sample rate (the rate at which the sound pressure is measured) is roughly 40 kHz.

Problem 5.2 Bits per sample

Because \(2^{16} ∼ 10^{5}\), a 16-bit sample as chosen for the CD format requires electronics accurate to roughly 0.001%. Why didn’t the designers of the CD format choose a much larger sample size, say 32 bits (per channel)?

Problem 5.3 Checking units

Check that all the units in the estimate divide out except for the desired units of bits.

Back-of-the-envelope calculations use rough estimates such as the playing time and neglect important factors such as the bits devoted to error detection and correction. In this and many other estimates, multiplication with 3 decimal places of accuracy would be overkill. An approximate analysis needs an approximate method of calculation.

Question

What is the data capacity to within a factor of 2?

The units (the biggest part!) are bits (Problem 5.3), and the three numerical factors contribute \(3600 \times 4.4 \times 10^{4} \times 32\). To estimate the product, split it into a big part and a correction.

The big part: The most important factor in a back-of-the-envelope product usually comes from the powers of 10, so evaluate this big part first:

3600 contributes three powers of 10, \(4.4 \times 10^{4}\) contributes four, and 32 contributes one. The eight powers of 10 produce a factor of 10 .

The correction: After taking out the big part, the remaining part is a correction factor of \(3.6 \times 4.4 \times 3.2\). This product too is simplified by taking out its big part. Round each factor to the closest number among three choices: 1, few, or 10. The invented number few lies midway between 1 and 10: It is the geometric mean of 1 and 10, so \((few)^{2} = 10\) and few ≈ 3. In the product \(3.6 \times 4.4 \times 3.2\), each factor rounds to few, so \(3.6 \times 4.4 \times 3.2 ≈ (few)^{3}\) or roughly 30.

The units, the powers of 10, and the correction factor combine to give

\[\text{capacity} ∼ 10^{8} \times 30 bits = 3 \times 10^{9} bits. \label{5.2} \]

This estimate is within a factor of 2 of the exact product (Problem 5.4), which is itself close to the actual capacity of \(5.6 \times 10^{9}\) bits.

Multiple problems

Problem 5.4 Underestimate or overestimate?

Does \(3 \times 10^{9}\) overestimate or underestimate \(3600 \times 4.4 \times 10^{4} \times 32\)? Check your reasoning by computing the exact product.

Problem 5.5 More practice

Use the one-or-few method of multiplication to perform the following calculations mentally; then compare the approximate and actual products.

a. \(161 \times 294 \times 280 \times 438\). The actual product is roughly \(5.8 \times 10^{9}\).

b. Earth’s surface area \(A = 4\pi R^{2}\), where the radius is \(R ∼ 6 \times 10^{6}\)m. The actual surface area is roughly \(5.1 \times 10^{14}m^{2}\).