Consider two similar problems:

Problem 1

Problem 2

The two problems are often confused and the difference between them may not be immediately obvious. The second one, as a mathematical conundrum, has been included in many math puzzles books.

By definition, the average speed S of the motion that lasted time T over the distance D is

The definition applies directly to the first problem. The fellow was on the road for the total of T = 2 hours. Going at 20 mi/h for the first hour the fellow covered the distance of 20 miles. Similarly, in the second hour he covered the distance of 30 miles. Therefore, in 2 hours he traveled the total of D = 20 + 30 = 50 miles. The average speed then is found to be S = 50/2 = 25 m/h.

The second problem is only a little more complex. There are three ways to write the formula above: S = D/T, D = ST, T = D/S. I apply the latter one. Let d be half the distance between the two cities. The first leg of the journey took d/20 hours, the second d/30. Therefore, on the whole, the fellow was on the road T = d/20 + d/30 hours. During that time he covered D = 2d miles. It follows that his average speed is given by S = 2d/(d/20 + d/30) or, after cancelling out the common factor d, S = 2/(1/20 + 1/30) = 24 m/h. If the distance between the cities were, as in the first problem, 50 miles, then the journey would take T = 50/24 hours or 2 hours and 5 minutes, from which 1 hour and 15 minutes (= 25/20 hours) were spent on the first 25 miles and 50 minutes (= 25/30 hours) on the second 25 mile stretch.

What we found is that, depending on the circumstances, to determine the average speed of motion, computations based on the same basic formula (S = D/T) may have to follow different routes.

Now there are three means in music: first the arithmetic, secondly the geometric, and thirdly the subcontrary, the so-called harmonic. Archytas, cited by Porphyry in his Commentary on Ptolemy's Harmonics I. Thomas, Greek Mathematical Works, v1, Harvard University Press, 2006, p 113

For the given two quantities, 20 and 30, the number (20 + 30)/2 is known as their arithmetic mean while 2/(1/20 + 1/30) is the harmonic mean of the two numbers. Of course the are more general definitions. For convenience, it is customary to group a finite sequence of numbers a₁, a₂, ..., a_n in a vector form a = (a₁, a₂, ..., a_n). Then I am going to use the following shorthands:

where the summation is being carried from i = 1 through i = n. Observe that H(a) = 1/A(1/a). (For those who are not very comfortable or familiar with the summation formulas, it might be useful to verify this statement for small values of n, say 2, 3, and 4.) This leads to a more general definition

where I assume that all a_i's are positive and r is a real number different from 0. For example, A(a) = M₁(a) whereas H(a) = M_-1(a).

In general, M_r(a) is the mean value of numbers a₁, a₂, ..., a_n with the exponent r. M₂(a) is known as the quadratic average. It differs by a factor of n^1/2 from the Euclidean distance from the end of the vector a to the origin. Do other means have special names? Not that I am aware of. Are they of any use? I must say that I dislike this question. I am appalled by the current tendency to place an emphasis (while teaching and learning) on the so-called useful things. (Should, for example, a math teacher be concerned with how and whether factoring is used in real world? I do not know whether superellipses were of any use until Piet Hein discovered their esthetic properties.)

In the case of the means, considering M_r(a) for nonzero real r almost immediately proves to be a nice idea. It appears that once the means were defined for nonzero exponents, it becomes also possible to define M₀(a). A 2-step proof is very simple but needs an application of what's known as the L'Hôpital Rule, a subject covered in any Calculus I course. By the L'Hôpital Rule, it follows that the limit, limM_r(a), as r approaches 0, exists and is equal to the geometric mean of numbers a₁, a₂, ..., a_n. Thus the following complements the definition of M_r(a):

Lemma

For r > 0, M_r(a) < M_2r(a) provided not all a_i's are equal. If they are then M_r(a) = M_2r(a).

Proof

Inequality M_r(a) < M_2r(a) can be rewritten as M₁(a^r) < M₂(a^r), where, by convention, a^r= (a₁^r, a₂^r, ..., a_n^r). The latter follows from a beautiful identity

whose right-hand side is non-negative and is equal to zero only when vectors a and b are proportional. The lemma follows from the latter with b_i = 1 for all i, i = 1, 2, ..., n.

Applying Lemma repeatedly we get

as above (remember the L'Hôpital Rule!). From here it follows that for any r > 0, M_r(a) > M₀(a) (except, of course, for the case where all a_i's are equal). This is a generalization of the Arithmetic mean - Geometric mean inequality M₁(a) > M₀(a).

On the other hand, if r < 0 and s = -r, we can write

Therefore, for any two real s and r, s < r, we have a further generalization: M_s(a) < M_r(a).

The applet below illustrates this property of the means. The number bars are draggable up and down, while the exponent r can be changed by dragging the horizontal slider.

It's also easily seen that, unless all a_i's are the same,

Furthermore, it can be shown that M_r(a) approaches max{a_i} as r grows without limit. Also, M_r(a) approaches min{a_i} as r tends to -∞. For this reason, we also define

Finally we can claim that, for any two real s and r, s < r, and not all a_i's equal,

Remark

A question is often raised whether or not it is useful to study continuous functions that are not given by a single formula like f(x) = (x² - 1)/(x⁵ + 1). For a fixed a, f(x) = M_x(a) serves an example of a continuous function that can't be represented in such a manner. As we saw above, f(0) requires an essentially different formula than other values of x.

A Nice Exercise

Given a trapezoid with bases of lengths a and b. Consider 4 line segments parallel to the bases. Prove that

1. the length of the segment that divides the trapezoid into two of equal area is the quadratic mean of a and b,
2. the length of the segment half way between the bases is the arithmetic mean of a and b
3. the length of the segment through the point of intersection of diagonals is the harmonic mean of a and b
4. the length of the segment that divides the trapezoid into two similar ones is the geometric mean of a and b.

Another Variant

In the diagram on the right, AB is a diameter of the semicircle ATB with center O. CT ⊥ AB, CF ⊥ OT, OR ⊥ AB. Let AC = a, BC = b, OR = (a-b)/2. Then

1. CT is the geometric mean of a and b,
2. OT is the arithmetic mean of a and b,
3. TF is the harmonic mean of a and b,
4. AR is the quadratic mean of a and b.

Obviously, AR ≥ AO = OT ≥ CT ≥ TF.

References

1. E. F. Beckenbach, R. Bellman, Introduction to Inequalities, Random House, 1975
2. S. L. Greitzer, Arbelos, v6, MAA, 1988
3. G. H. Hardy, J.E.Littlewood, G.Pólya, Inequalities, Cambridge Univ Pr., 1998
4. O. A. Ivanov, Easy as p, Springer, 1998

Related material Read more...
	The Means
	Averages, Arithmetic and Harmonic Means
	Expectation
	The Size of a Class: Two Viewpoints
	Averages of divisors of a given integer
	Family Statistics: an Interactive Gadget
	Averages in a sequence
	Arithmetic and Geometric Means
	Geometric Meaning of the Geometric Mean
	A Mathematical Rabbit out of an Algebraic Hat
	AM-GM Inequality
	The Mean Property of the Mean
	Harmonic Mean in Geometry

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