The Means

Consider two similar problems:

Problem 1

A fellow travels from city $A\;$ to city $B.\;$ For the first hour, he drove at the constant speed of $20\;$ miles per hour. Then he (instantaneously) increased his speed and, for the next hour, kept it at $30\;$ miles per hour. Find the average speed of the motion.

Problem 2

A fellow travels from city $A\;$ to city $B.\;$ The first half of the way, he drove at the constant speed of $20\;$ miles per hour. Then he (instantaneously) increased his speed and traveled the remaining distance at $30\;$ miles per hour. Find the average speed of the motion.

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

$\displaystyle\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}\;$ or $\displaystyle S = \frac{D}{T}.$

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 $\displaystyle S=\frac{50}{2}=25\;$ m/h.

The second problem is only a little more complex. There are three ways to write the formula above: $\displaystyle S=\frac{D}{T},\;$ $D=ST,\;$ $\displaystyle T=\frac{D}{S}.\;$ I apply the latter one. Let $d\;$ be half the distance between the two cities. The first leg of the journey took $\displaystyle\frac{d}{20}\;$ hours, the second $\displaystyle\frac{d}{30}.\;$ Therefore, on the whole, the fellow was on the road $\displaystyle T=\frac{d}{20}+\frac{d}{30}\;$ hours. During that time he covered $D=2d\;$ miles. It follows that his average speed is given by $\displaystyle S=\frac{2d}{\displaystyle\frac{d}{20}+\frac{d}{30}}\;$ or, after cancelling out the common factor $d,\;$ and a few arithmetic operations $S=24\;$ m/h. If the distance between the cities were, as in the first problem, $50\;$ miles, then the journey would take $T=\displaystyle\frac{50}{24}\;$ hours or $2\;$ hours and $5\;$ minutes, of which $1\;$ hour and $15\;$ minutes $(=\displaystyle\frac{25}{20}\;$ hours) were spent on the first $25\;$ miles and $50\;$ minutes $(=\displaystyle\frac{25}{30}\;$ 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 $\left(S = \displaystyle\frac{D}{T}\right)\;$ 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 $\displaystyle\frac{20+30}{2}\;$ is known as their arithmetic mean while $\displaystyle\frac{2}{\displaystyle\frac{1}{20}+\frac{1}{30}}\;$ 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 $a_{1},a_{2}, \ldots, a_{n}\;$ in a vector form $\mathbf{a}=(a_{1},a_{2},\ldots, a_{n}).\;$ Then I am going to use the following shorthands:

$\displaystyle A(\mathbf{a}) = \frac{1}{n}\sum\mathbf{a} = \frac{1}{n}\sum_{i=1}^{n}a_{i}\;$ and $\displaystyle H(\mathbf{a}) = \frac{n}{\sum\displaystyle\frac{1}{\mathbf{a}}} = \frac{n}{\displaystyle\sum_{i=1}^{n}\frac{1}{a_{i}}},$

Observe that $\displaystyle H(\mathbf{a})=\frac{1}{A(\frac{1}{\mathbf{a}}}).\;$ (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

$M_{r}(\mathbf{a}) = (\frac{1}{n}\sum_{i=1}^{n}\mathbf{a}^{r})^{1/r} = (\frac{1}{n}\sum_{i=1}^{n}a_{i}^{r})^{1/r}$

where I assume that all $a_{i}$'s are non-negative and $r\;$ is a real number different from $0.\;$ For example, $A(\mathbf{a}) = M_{1}(\mathbf{a})\;$ whereas $H(\mathbf{a}) = M_{-1}(\mathbf{a}).\;$

In general, $M_{r}(\mathbf{a})\;$ is the mean value of numbers $a_{1}, a_{2}, \ldots, a_{n}\;$ with the exponent $r.\;$ $M_{2}(\mathbf{a})\;$ 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 $\mathbf{a}\;$ 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 $M_{r}(\mathbf{a})\;$ 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 $M_{0}(\mathbf{a}).\;$ 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, $\displaystyle\lim_{r\rightarrow 0}M_{r}(\mathbf{a})\;$ exists and is equal to the geometric mean of numbers $a_{1}, a_{2}, \ldots, a_{n}.\;$ Thus the following complements the definition of $M_{r}(\mathbf{a}):$

$M_{0}(\mathbf{a}) = (a_{1}a_{2} \ldots a_{n})^{1/n}=\sqrt[n]{a_{1}a_{2} \ldots a_{n}}.$

Lemma

For $r > 0,\;$ $M_{r}(\mathbf{a}) < M_{2r}(\mathbf{a}),\;$ provided not all $a_{i}$'s are equal. If they are then $M_{r}(\mathbf{a}) = M_{2r}(\mathbf{a}).\;$

Proof

Inequality $M_{r}(\mathbf{a}) < M_{2r}(\mathbf{a})\;$ can be rewritten as $M_{1}(\mathbf{a}^{r}) < M_{2}(\mathbf{a}^{r}),\;$ where, by convention, $\mathbf{a}^{r}= (a_{1}^{r},a_{2}^{r}, \ldots, a_{n}^{r}).\;$ The latter follows from a beautiful identity

$\displaystyle\sum_{i=1}^{n}a_{i}^{2}\sum_{i=1}^{n}b_{i}^{2} - (\sum_{i=1}^{n}a_{i}b_{i})^{2} = \frac{1}{2}\sum_{i=1}^{n}(a_{i}b_{j} - a_{j}b_{i})^{2}$

whose right-hand side is non-negative and is equal to zero only when vectors $\mathbf{a}\;$ and $\mathbf{b}\;$ are proportional. The lemma follows from the latter with $b_{i} = 1\;$ for all $i = 1, 2, \ldots, n.\;$

Applying Lemma repeatedly we get

$\displaystyle M_{r}(\mathbf{a}) > M_{\frac{r}{2}}(\mathbf{a}) > M_{\frac{r}{2^{2}}}(\mathbf{a}) > M_{\frac{r}{2^{3}}}(\mathbf{a}) \ldots\;\rightarrow\;M_{0}(\mathbf{a}),$

as above (remember the L'Hôpital Rule!). From here it follows that for any $r \gt 0,\;$ $M_{r}(\mathbf{a}) \gt M_{0}(\mathbf{a})\;$ (except, of course, for the case where all $a_{i}$'s are equal). This is a generalization of the Arithmetic mean - Geometric mean inequality $M_{1}(\mathbf{a}) \gt M_{0}(\mathbf{a}).\;$

On the other hand, if $r \lt 0\;$ and $s = -r,\;$ we can write

$\displaystyle M_{r}(\mathbf{a}) = M_{-s}(\mathbf{a}) = \frac{1}{M_{s}(\frac{1}{\mathbf{a}})} \lt \frac{1}{M_{0}(\frac{1}{\mathbf{a}})} = M_{0}(\mathbf{a}).$

Therefore, for any two real $s\;$ and $r,\;$ $s \lt r,\;$ we have a further generalization: $M_{s}(\mathbf{a}) \lt M_{r}(\mathbf{a}).\;$

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.


If you are reading this, your browser is not set to run Java applets. Try IE11 or Safari and declare the site https:///www.cut-the-knot.org as trusted in the Java setup.

various means


What if applet does not run?

It's also easily seen that, unless all $a_{i}$'s are the same,

$\min\{a_{i}\} \lt M_{r}(\mathbf{a}) \lt \max\{a_{i}\}$

Furthermore, it can be shown that $M_{r}(\mathbf{a})\;$ approaches $\max\{a_{i}\}\;$ as $r\;$ grows without limit. Also, $M_{r}(\mathbf{a})\;$ approaches $\min\{a_{i}\}\;$ as $r\;$ tends to $-\infty.\;$ For this reason, we also define

$M_{-\infty}(\mathbf{a}) = \min\{a_{i}\}\;$ and $M_{\infty}(\mathbf{a}) = \max\{a_{i}\}.\;$

Finally we can claim that, for any two real $s\;$ and $r,\;$ $s < r,\;$ and not all $a_{i}$'s equal,

$M_{-\infty}(\mathbf{a}) \lt M_{s}(\mathbf{a}) \lt M_{r}(\mathbf{a}) \lt M_{\infty}(\mathbf{a}).$

Here's a proof for $M_{s}(\mathbf{a}) \lt M_{r}(\mathbf{a}):$

Hölder's inequality tells us that: $\sum \mathbf{u}^{\alpha}\mathbf{v}^{1-\alpha}\le \left(\sum \mathbf{u}\right)^{\alpha}\cdot\left(\sum \mathbf{v}\right)^{1-\alpha},\,$ for $\alpha\in (0,1).$

Let $s=r\cdot\alpha,\,$ $\mathbf{u}=\displaystyle \frac{1}{n}\mathbf{a}^r,\,$ $\mathbf{v}=\displaystyle \frac{\mathbf{1}}{n}:$

$\mathbf{u}^{\alpha}=n^{-\alpha}\mathbf{a}^s,\,$ $\sum \mathbf{u}^{\alpha}\mathbf{v}^{1-\alpha}=\sum n^{-\alpha}\mathbf{a}^s\cdot n^{\alpha-1}=\displaystyle \frac{\sum \mathbf{a}^s}{n}.$

$\displaystyle\begin{align}\left(\sum \mathbf{u}\right)^{\alpha}\cdot\left(\sum \mathbf{v}\right)^{1-\alpha}&=\left(\sum \mathbf{a}^r\right)^{\frac{s}{r}}\cdot\left(\sum\frac{1}{n}\right)^{1-\frac{s}{r}}\\ &=\left(\frac{\sum \mathbf{a}^r}{n}\right)^{\frac{s}{r}}\cdot\left(\sum\frac{1}{n}\right)\\ &=\left(\frac{\sum \mathbf{a}^r}{n}\right)^{\frac{s}{r}}. \end{align}$

The combination yieds $\displaystyle \frac{\sum \mathbf{a}^s}{n}\le\left(\frac{\sum \mathbf{a}^r}{n}\right)^{\frac{s}{r}},\,$ and the conclusion follows.

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 $\displaystyle f(x) = \frac{x^{2} - 1}{x^{5} + 1}.\;$ For a fixed $\mathbf{a},$ $f(x) = M_{x}(\mathbf{a})\;$ 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 $O.\;$ $CT\perp AB,\;$ $CF\perp OT,\;$ $OR\perp AB.\;$ Let $AC=a,\;$ $BC=b,\;$ $OR=\displaystyle\frac{a-b}{2}.\;$ 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, $AR \ge AO=OT\ge CT\ge TF.$ It is worth remembering that, unless all $a_{i}$'s are equal

QM greater than AM greater than GM greater than HM

When all $a_{i}$'s are equal, so are the means.

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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