SOME OPERATIONS OF ALGEBRA
One of the operations of algebra that is important for a mechanical brain is approximation, the problem of getting close to the right value of a number. Take, for example, finding square root (see the end of this supplement). The ordinary process taught in school is rather troublesome. We can set down another process, however, using a desk calculator to do division, which gives us square root with great speed.
Suppose that we want to find the square root of a number N, and suppose that we have x₀ as a guessed square root correct to one figure. For example, N might be 67.2 and x₀ might be 8, chosen because 8 × 8 is 64, and 9 × 9 is 81, and it seems as if 8 should be near the square root of 67.2. Here is the process:
- 1. Divide x₀ into N, and obtain N / x₀.
- 2. Multiply x₀ + N / x₀ by 0.5 and call the result x₁.
Now repeat:
- 1. Divide x₁ into N and obtain N / x₁.
- 2. Multiply x₁ + N / x₁ by 0.5 and call the result x₂.
Every time this process is repeated, the new x comes a great deal closer to the correct square root. In fact it can be shown that, if x₀ is correct to one figure, then:
| Approximation | Is Correct To ··· Figures |
|---|---|
| x₁ | 2 |
| x₂ | 4 |
| x₃ | 8 |
| x₄ | 16 |
Let us see how this actually works out with 67.2 and a 10-column desk calculator.
Round 1: 8 divided into 67.2 gives 8.4. One half of 8 plus 8.4 is 8.2. This is x₁.
Round 2: 8.2 divided into 67.2 gives 8.195122. One half of 8.2 plus 8.195122 is 8.197561. This is x₂.
Round 3: 8.197561 divided into 67.2 gives 8.197560225. One half of 8.197561 and 8.197560225 is 8.1975606125. This is x₃.
Checking x₃, we find that 8.1975606125 divided into 67.2 gives 8.1975606126 approximately.
In this case, then, x₃ is correct to more than 10 figures. In other words, with a reasonable guess and two or three divisions we can obtain all the accuracy we can ordinarily use. This process is called rapid approximation, or rapidly convergent approximation, since it converges (points or comes together) very quickly to the number we are seeking.
Another important operation of algebra is interpolation, the problem of putting values smoothly in between other values. For example, suppose that we have the table:
| x | y |
|---|---|
| 5 | 26 |
| 6 | 37 |
| 7 | 50 |
| 8 | 65 |
| 9 | 82 |
Suppose that we want to find the value that y (or yₓ) ought to have when x has the value of 7.2. This is the problem of interpolating y so as to find y at the value of 7.2, y₇ˌ₂.
One way of doing this is to discover the formula that expresses y and then to put x into that formula. This is not always easy. Another way is to take the difference between y₇ and y₈, 15, and share the difference appropriately over the distance 7 to 7.2 and 7.2 to 8. We can, for example, take ²/₁₀ of 15 = 3, add that to y₇ = 50, and obtain an estimated y₇ˌ₂ = 53. This is called linear interpolation, since the difference 0.2 in the value of x is used only to the first power. It is a good practical way to carry out most interpolation quickly and approximately.
Actually here y = x² + 1, and so the true value of y₇ˌ₂ is (7.2 × 7.2) + 1, or 52.84, which is rather close to 53. Types of interpolation procedures more accurate than linear interpolation will come much nearer still to the true value.