Suppose that we want to find the square root of a number _N_, and suppose that we have _x_₀ as a guessed square root correct to one figure. For example, _N_ might be 67.2 and _x_₀ 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 _x_₀ into _N_, and obtain _N_/_x_₀. 2. Multiply _x_₀ + _N_/_x_₀ by 0.5 and call the result _x_₁. Now repeat: 1. Divide _x_₁ into _N_ and obtain _N_/_x_₁. 2. Multiply _x_₁ + _N_/_x_₁ by 0.5 and call the result _x_₂. 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 _x_₀ is correct to one figure, then: APPROXIMATION IS CORRECT TO ··· FIGURES _x_₁ 2 _x_₂ 4 _x_₃ 8 _x_₄ 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 _x_₁. Round 2: 8.2 divided into 67.2 gives 8.195122. One half of 8.2 plus 8.195122 is 8.197561. This is _x_₂. Round 3: 8.197561 divided into 67.2 gives 8.197560225. One half of 8.197561 and 8.197560225 is 8.1975606125. This is _x_₃. Checking _x_₃, we find that 8.1975606125 divided into 67.2 gives 8.1975606126 approximately. In this case, then, _x_₃ 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, _y_₇ˌ₂. 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 _y_₇ and _y_₈, 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 _y_₇ = 50, and obtain an estimated _y_₇ˌ₂ = 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.