SOME APPLICATIONS OF BOLZANO’S THEOREM - FUNCTIONS AND LIMITS - What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

CHAPTER VI. FUNCTIONS AND LIMITS

§6. SOME APPLICATIONS OF BOLZANO’S THEOREM

1. Geometrical Applications

Bolzano’s simple yet general theorem may be used to prove many facts which are not at all obvious at first sight. We begin by proving: If A and B are any two areas in the plane, then there exists a straight line in the plane which bisects A and B simultaneously. By an “area” we mean any portion of the plane included within a simple closed curve.

Let us begin by choosing some fixed point P in the plane, and drawing from P a directed ray PR from which to measure angles. If we take any ray PS which makes an angle x with PR, there will exist a directed straight line in the plane bisecting the area A, and with the same direction as the ray PS. For if we take a directed line l1 with the direction of PS and lying wholly on one side of A and move this line parallel to itself until it is in position l2 (see Fig. 173), wholly on the other side of A, then the function whose value is defined to be the area of A to the right of the line (the east direction if the arrow on the line points north) minus the area of A to the left of the line will be positive for position l1 and negative for position l2. Since this function is continuous, by Bolzano’s theorem it must be zero for some intermediate position lx, which therefore bisects A. For each value of x from x = 0° to x = 360°, the line lx which bisects A is uniquely defined.

image

Fig. 173. Simultaneous bisection of two areas.

Now let the function y = f(x) be defined as the area of B to the right of lx minus the area of B to the left of lx. Suppose that the line l0 which bisects A and has the direction of PR has more of B to the right than to the left; then for x = 0°, y is positive. Let x increase to 180°, then the line l180with direction RP which bisects A is the same as l0 but oppositely directed, with right and left interchanged; hence the value of y for x = 180° is the same numerically as for x = 0°, but with opposite sign, and therefore negative. Since y is a continuous function of x as lx turns around, there exists some value α of x between 0° and 180° for which y is zero. It follows that the line lα bisects A and B simultaneously. This completes the proof.

Note that although we have proved the existence of a line with the desired property, we have given no definite procedure for constructing it; this exhibits again the distinguishing feature of mathematical existence proofs as compared with constructions.

A similar problem is the following: Given a single area in the plane, it is desired to cut it into four equal pieces by two perpendicular lines. In order to prove that this is always possible, we return to our previous problem at the stage where we had defined lx for any angle x, but we forget about the area B. Instead, we take the line lx+90 which is perpendicular to lx and which also bisects A. If we number the four pieces of A as shown in Figure 174, then we have

A1 + A2 = A3 + A4

and

A2 + A3 = A1 + A4,

image

Fig. 174.

from which it follows, on subtracting the second equation from the first, that

A1A3 = A3A1

i.e.

A1 = A3,

and hence

A2 =A4.

Thus if we can show the existence of an angle α such that for lα

A1(α) = A2(α),

then our theorem will be proved, since for such an angle all four areas will be equal. To do this, we define a function y = f(x) by drawing Ix, and setting

f(x) = A1(x) – A2(x).

For x = 0°, f(0) = A1(0) − A2(0) may be positive. In that case, for x = 90°, A1(90) − A2(90) = A2(0) − A3(0) = − A2(0) − A1(0) will be negative. Therefore, since f(x) varies continuously as x increases from 0° to 90°, there will be some value a between 0° and 90° for which f(a) = A1(α) — A2(α)= 0. The lines lα and lα+90 then divide the area into four equal pieces.

It is interesting to observe that these problems may be generalized to three and higher dimensions. In three dimensions the first problem becomes: Given three volumes in space, to find a plane which bisects all three simultaneously. The proof that this is always possible again depends on Bolzano’s theorem. In more than three dimensions the theorem is still true but the proof requires more advanced methods.

*2. Application to a Problem of Mechanics

We shall conclude this section by discussing an apparently difficult problem in mechanics that is easily answered by an argument based on continuity concepts. (This problem was suggested by H. Whitney.)

Suppose a train travels from station A to station B along a straight section of track. The journey need not be of uniform speed or acceleration. The train may act in any manner, speeding up, slowing down, coming to a halt, or even backing up for a while, before reaching B. But the exact motion of the train is supposed to be known in advance; that is, the function s = f(t) is given, where s is the distance of the train from station A, and t is the time, measured from the instant of departure. On the floor of one of the cars a rod is pivoted so that it may move without friction either forward or backward until it touches the floor. If it does touch the floor, we assume that it remains on the floor henceforth; this will be the case if the rod does not bounce. Is it possible to place the rod in such a position that, if it is released at the instant when the train starts and allowed to move solely under the influence of gravity and the motion of the train, it will not fall to the floor during the entire journey from A to B?

image

Fig. 175.

It might seem quite unlikely that for any given schedule of motion the interplay of gravity and reaction forces will always permit such a maintenance of balance under the single condition that the initial position of the rod is suitably chosen. Yet we state that such a position always exists.

Paradoxical as this assertion might seem at first sight, it can be proved easily once one concentrates on its essentially topological character. No detailed knowledge of the laws of dynamics is needed; only the following simple assumption of a physical nature need be granted: The motion of the rod depends continuously on its initial position. Let us characterize the initial position of the rod by the initial angle x which it makes with the floor, and by y the angle which the rod makes with the floor at the end of the journey, when the train reaches the point B. If the rod has fallen to the floor we have either y = 0 or y = π. For a given initial position x the end position y is, according to our assumption, uniquely determined as a function y = g(x) which is continuous and has the values y = 0 for x = 0 and y = π for x = π (the latter assertion simply expressing that the rod will remain flat on the floor if it starts in this position). Now we recall that g(x), as a continuous function in the interval 0 ≤ x ≤ π, assumes all the values between g(0) = 0 and g (π ) = π; consequently, for any such values y, e.g. for the value image, there exists a specific value of x such that g(x) = y; in particular, there exists an initial position for which the end position of the rod at B is perpendicular to the floor. (Note: In this argument it should not be forgotten that the motion of the train is fixed once for all.)

Of course, the reasoning is entirely theoretical. If the journey is of long duration or if the train schedule, expressed by s = f(t), is very erratic, then the range of initial positions x for which the end position g(x) differs from 0 or π will be exceedingly small, as is known to anyone who has tried to balance a needle upright on a plate for an appreciable time. Still, our reasoning should be of value even to a practical mind inasmuch as it shows how qualitative results in dynamics may be obtained by simple arguments without technical manipulation.

Exercises: 1. Using the theorem of page 315, show that the reasoning above may be generalized to the case where the journey is of infinite duration.

2. Generalize to the case where the motion of the train is along any curve in the plane and the rod may fall in any direction. (Hint: It is not possible to map a circular disk continuously onto its circumference alone by a mapping which eaves every point of the circumference fixed (see p.255)).

3. Show that the time required for the rod to fall to the floor, if the car is stationary and the rod is released at an angle ε from the vertical position, tends to infinity as ε tends to zero.