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Stephen Wirkus, Richard Rand, and Andy Ruina

Pumping a swing from a standing position amounts to increasing the amplitude of a pendulum by modifying its length in a systematic way. Pumping from a seated position, however, involves a sudden rotation of the rider's body when the swing momentarily comes to a stop, which lifts the rider slightly above the previously highest level. Both of these standard pumping strategies are modeled with second-order autonomous differential equations, and qualitative analysis reveals that seated pumping adds a fixed quantity to the amplitude per swing cycle, while standing pumping multiplies the amplitude by a fixed factor greater than 1. Thus, for small amplitudes seated pumping gives a greater increase in amplitude per cycle but at larger amplitudes standing pumping is more efficient. The authors propose a new strategy for pumping from a seated position that promises a geometric increase in amplitude like standing pumping.

**The Fundamental Theorem of Algebra**

Michael D. Hirschhorn

Given a polynomialp(z), consider the surfaceSabove the complex plane given in parametric form by(x,y,t)= (Re{p(Re^{iq})}, Im{p(Re^{iq})},R),0 £R, 0 £ q £ 2p. In general,Sis a cone-like surface rising from the pointp(0) = (re{p(0)}, Im{P(0)}), whose cross-section at heightRis the image of the circle |z| =Runder the mappingzÃ?p(z). Thus these cross-sections are closed curves that up high wind around the vertical axisntimes, wheren= deg(p(z)), and down low they remain nearp(0) and hence do not reach the vertical axis. Since the surface is continuous, there must be some pointsz=Re^{i}^{q}at which the surface intersects the vertical axis, and therep(z) = 0. The proof is illustrated with the plot of the coneSfor a sample polynomial.

**How Much Money Do You (or Your Parents) Need for Retirement?**

James W. Daniel

This introduction to actuarial science considers the fundamental question: For a given yearly rate of return, how much do retirees need to have invested now in order to provide $1 at the start of each year from their date of retirement as long as they live? Employing mortality tables, a satisfactory is found for a large group of retirees of a given age and gender. However, for an individual to have a given confidence level of not running out of money during his or her lifetime, a much larger initial investment is needed because many will live longer than the average survival time for the group. A statistical argument shows how by pooling their risks individuals can greatly reduce the investment needed to achieve a given level of confidence that they will not run out of money during their retirement years.

**Making Squares from Pythagorean Triangles**

Charles Jepsen and Roc Yang

In this report on an undergraduate research project, the authors show a square exists that can be dissected intomPythagorean triangles (right triangles with integer sides) if and only ifm£ 5. This is shown by displaying a square that can be dissected into five Pythagorean triangles, and observing that a dissection intomtriangles implies a dissection of a (usually larger) square intom+ 1 squares. It then remains to be shown that no square can be dissected into 2, 3 or 4 Pythagorean triangles. The existence of such a dissection in the casem= 4 would imply the existence of a rational point (x,y) withx> 4 on a certain elliptic curve, and algebraic geometers have shown that the curve has no such rational points.

**On Factoring with the b-algorithm**

Vincent Lucarelli

An algorithm suggested by Fermat can be used to factor integersnthat are the product of two nearly equal factors. Here a more efficent algorithm is proposed for factoring such special numbers, and it is shown to be more efficient than the Fermat method. The key step in the algorithm is to search for integersbof a certain parity that make an expressionassume an integer value, where the parity of band the values of the integer parametersp, q, are easily computed directly fromn. When such a value ofnis found, the factorization ofnfollows at once. This special algorithm implies that the modulus chosen for implementing the RSA encryption algorithm should not be a product of two nearly equal factors.

**Egyptian Fractions and the Inheritance Problem**

Premchand Anne

The inheritance problem (W. Frederick and J. Hersberger, The mathematical judge: A fable,CMJ26:5 (1995) 377-381) is closely related to the ancient problem of completing a given sum of "Egyptian fractions" (fractions with unit denominator), with other such fractions to make the sum equal 1. A special case of particular interest is when the final Egyptian fraction is required to be the product of all the others:Ã?(1/n) + 1/(P_{i}n) =1. The solutions with 7 or fewer_{i}n_{i}are known, but few general results are known for solutions of length greater than 7. This note surveys the known results and the current status of related conjectures.

Given a one-to-one function

**Norton Starr, Nothing Counts for Something**

Students often overlook the empty set of the vacuous state in counting problems. A good example is the problem of counting the number of possible lighting levels that can be produced by a floor lamp with a central bulb that can shine with say 50, 100 or 150 watts, and three 60 watt bulbs controlled by a second switch so that any number of them can be on simultaneously. The most common student response is 9 possible illumination levels, but once their mistake is pointed out they recognize that the off position for each switch is just as important a state to consider as any of the others.

**Yukio Kobayashi, A Geometric View of a Vector Identity**

The vector identity

**Bob Palais, An Example Demonstrating the Fundamental Theorem of Calculus**

By appealing to the area formulas for triangles and sectors of a circle, one can show directly (using implicit differentiation) that if *A*(*t*) is the area under the graph of the function *f*(*x*) = Ã·1 - *x*^{2} for -1 £ *x* £ *t* , then *A'* (*t*) = *f*(*t*) . Such a calculation in a specific case sets the stage for the general proof of the fundamental theorem of calculus.

**Sidney H. Kung, More Coconuts**

Another solution is given to the linear Diophantine problem of dividing a pile of coconuts, discussed in a recent article: S. Singh and D. Bhattacharya, On Dividing Coconuts, *CMJ* 28:3 (1997) 203-204.

To enliven the discussion of arc length, calculus students are challenged to find three examples of continuous nonnegative functions on the unit interval whose values at both endpoints are zero, each with a unit area bounded by the

**Van Bain, An Algorithm for Drawing the n-cube**

The