If we suppose that r is the square root of the given number, then that number can be represented by the area of a square whose side is r = OR OR' as in the figure below.

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If d is the number which we first try, or one of the numbers determined as an approximation by the rule, then the next approximation is obtained by dividing the given number by d and taking the average a of the quotient q and the divisor d. We' shall suppose that d is larger than q. Thus d and q give the lengths of the sides OD and OQ', respectively, of a rectangle

having the same area as the square and such that OD is larger and OQ' smaller than the side OR of the square. The rectangle is shown in the figure placed upon the square so that the two have the right angle at O and the overlapping rectangle OP in common. It follows at once that the area of the rectangle R'P is the same as that of the rectangle PD. Add to each of these latter rectangles the small adjoining rectangle PD', and it will be seen that the areas of rectangles Q'D' and SD are the same. Since R'D' is greater than D'D, it follows that Q'R' is less than RD, and if we lay off on DR the length DB Q'R', the point B will fall within DR. Mark off also on OR the length OQ = OQ' = q. The average a of OQ and OD is given by OA where A is the mid-point of QD and also of RB since QR = BD. Hence the average a = OA is greater than r = OR. It should be observed that this is a proof that the arithmetic mean of any two positive quantities is greater than their geometric mean, for it shows that (q+d)/2 = a > r = √qd.

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The error in taking a as the root is the length RA, which will be denoted by e', while the error in d is RD, denoted by e1, and

2.

the error in q is QR, denoted by e2. The figure gives a good idea of how much smaller e' is than either e, or e2. Only in case of a selection of d obviously too large can e' be greater than e2. We shall now obtain from the figure a very useful relation between e' and e,which enables us to test the accuracy of the approximations at each step. The figure to the right shows the rectangle RD' with the rectangle Q'D' placed upon it in the position BS' so that Q'R' coincides with DB. Since these two rectangles have equal areas the area of the thin rectangle SB is equal to that of the rectangle TD', which is almost square, lacking only ST to be a complete square. Adding then the little rectangle ST to each of the rectangles SB and TD' we have the final result that the rectangle RT and the square SS' have equal areas, i. e., RBXBT SD'2, or, since RB = 2e', BT = d and SD' = e1,

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This shows how rapidly the error decreases from one approximation to the next. For example, if d is correct in the first three decimals, then e1 < .001 and hence e' < .0000005/d so that a will be correct in six decimals, or more if d is much larger than 5. Thus we may state the rough rule that the correct number of decimals in any approximation is in general twice the number in the preceding approximation.

In a similar manner it may be shown from the figure that

and with a little more trouble but in an analogous manner that

This last result may also be obtained by combining the equations in (1) and (2). Since a > r and r > q we deduce from (3) the inequalities

(4)

h2

h2

which are more convenient as tests of accuracy than the others since we always know h. To illustrate the use of these inequalities, let us approximate 3 by taking first for d the value 2; we find mentally that q = 1.5 and a 1.75. Applying the rule again and taking now d = 1.74, since 1.75 is too large, we find

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1.7241379310 and a = 1.7320689655.

that q Computing roughly the value of the expressions in (4) we find that .00001815 <e < .000018199+, and, reducing a by the smaller number, we have 1.7320508 as correct figures of the root. This shows how we may use the expression to the left in (4) as a correction in order to obtain more correct figures, or, in other words, that h2

is a better approximation to r than a. It may be seen from what follows that (5) gives really more correct figures than were indicated above, in fact, we may use ten decimals of the correction .0000181579, thus obtaining as correct figures of the root 1.7320508076.

It is possible to find a second correction, a third, and so on, but it is not so easy to do this graphically, and so we shall proceed algebraically. The error in (5) is

which may be factored by replacing (a-r) by its value in (3) and reduced further by the same substitution, thus

The expressions on the right and left of the inequality sign do not differ greatly and hence we may use the one to the left as a second correction. We obtain in this way the still better approximation

By subtracting r from this expression, and factoring and reducing as before in order to find the error, we shall find a third correction and the approximation,

The reductions become troublesome when we proceed much further in this way, but by the use of the binomial theorem for fractional exponents all of these terms and as many more as may

be desired may be easily obtained. For we have only to write the equation (3) in the form r2 = a2-h2/4 = a2(1-h2/4a2), or r = a(1-h2/4a2), and expand the second factor in order to obtain the above development. This development by the binomial theorem will yield very easily any number of terms, but it depends upon the demonstration of the validity of the binomial theorem for fractional exponents, and this demonstration is quite difficult. But for any ordinary computation only one or two of the correction terms would ever be needed since they diminish. very rapidly when h is small, as in the computation of 3 above, and in this case the first method of derivation has the advantage of using only simple algebraic reductions and of furnishing very convenient upper and lower limits for the error. Thus the original process of successive divisions and taking the average may be supplemented by the above explained method of subtracting from the last average found one or more corrections. However, for general use the unsupplemented method is by far preferable as a rule easily remembered and understood, and sufficiently rapid.

The simpler rule has in addition an important theoretical significance, for it gives a very simple and satisfactory definition of the square root of any positive number. Thus if N is any positive number we may define N in the following way: Divide N by any convenient positive number d and denote the quotient by q and the average of q and d by a. Now divide N by a, and find the corresponding quotient q, and average a2. Proceeding in this way we find two endless sequences of numbers: (A) A1, A2, A3, A4,

(B) 91, 92, 93, 94,

Ai, ai+1,

a1+qi

2

qi, gi+1,

aiqi=n.

such that the sequence A decreases while the sequence B increases, and the difference between a corresponding pair of terms of A and B, ai-qi, is positive and approaches zero as i increases. Also the square of any term in A is greater than N, while the square of any term in B is less than N. The sequences A and B approach therefore the same limit and this limit is N.

These facts may be proved without assuming the existence of VN, as is desirable in such a definition. The proofs are not difficult and are left to the reader as an exercise. A similar definition may be formulated for any root of a positive number.

TWO METHODS OF LOCATING THE GERMAN SUPER GUN.
BY HARRIS F. MAC NEISH,
College of the City of New York.

METHOD BY MEANS OF CIRCLES.

Three observation stations, A, B, and C, are established near the front line, Station A being somewhat in advance of B and C. Station A is connected by wire with Stations B and C, and an instrument is set up at A so that the pushing of a button will start clocks going at B and C. When the discharge of the gun is heard at A the button is pushed, and when the discharge is heard at B and C the time is recorded by the observers.

Suppose the time recorded at B is two seconds, and at C three seconds. The velocity of sound is accurately determined in advance, but for simplicity assume the velocity to be 1,000 feet per second. It is evident then that A is 2,000 feet nearer to the gun than B, and 3,000 feet nearer than C.

A, B, and C are located accurately on a military map (see Figure 1), and two circles are drawn on the map to scale, one with B as center and 2,000 feet as radius, and the other with C as center and 3,000 feet as radius. Two circles are then drawn passing through A and tangent externally to the circles about B and C. Call the centers of these circles G1 and G2. G, and G. are evidently 2,000 feet farther from B than from A, and 3,000