Pythagorean Means

3 minute read

A short comparison of Pythagorean Means: the arithmetic mean ( $AM$ ), the geometric mean ( $GM$ ), and the harmonic mean ( $HM$ ).

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ComparisonPermalink

Use casesPermalink

Type	Value Property	Application	Example
Arithmetic mean	linear	average	statistics mean
Geometric mean	multiplicative, exponential	growth	proportional growth
Harmonic mean	reciprocal	rate, ratio	speed, density, resistance

Special case of two variablesPermalink

Type	First Value	Second Value	Mean	Note
Arithmetic mean	$x$	$y$	$\frac{x + y}{2}$
Geometric mean	$e^{x}$	$e^{y}$	$(e^{x} e^{y})^{\frac{1}{2}}$	$= \sqrt{e^{x} e^{y}} = e^{\frac{x + y}{2}}$
Harmonic mean	$x^{- 1}$	$y^{- 1}$	$(\frac{x^{- 1} + y^{- 1}}{2})^{- 1}$	$= \frac{2}{\frac{1}{x} + \frac{1}{y}} = \frac{2 x y}{x + y}$

General form of $n$ variablesPermalink

Type	General form	Note
Arithmetic mean	$\frac{1}{n} \sum_{i = 1}^{n} x_{i}$	$= \frac{x_{1} + x_{2} + \dots + x_{n}}{n}$
Geometric mean	$(\prod_{i = 1}^{n} x_{i})^{\frac{1}{n}}$	$= \sqrt[n]{x_{1} x_{2} \dots x_{n}}$
Harmonic mean	${(\frac{\sum_{i = 1}^{n} x_{i}^{- 1}}{n})}^{- 1}$	$= \frac{n}{\sum_{i = 1}^{n} x_{i}^{- 1}} = \frac{n}{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{n}}}$

Assuming all values are positive, the following relationship holds:

min \leq HM \leq GM \leq AM \leq max

See Proof on Wikipedia

ExamplesPermalink

Arithmetic MeanPermalink

Statistics mean:

Three people with monthly income: $1000$ , $2000$ , $3000$

The arithmetic mean is $\frac{1000 + 2000 + 3000}{3} = 2000$

Geometric MeanPermalink

Proportional growth:

An orange tree yields $100$ , $180$ , $210$ , $300$ in each year within a 4 year time frame.

The growth factors are: $\frac{180}{100} = 180 %$ , $\frac{210}{180} \approx 116.67 %$ , $\frac{300}{210} \approx 142.86 %$ .

The arithmetic mean is $\frac{180 % + 116.67 % + 142.86 %}{3} \approx 146.51 %$

Simulate an orange tree that grows $146.51 %$ for 3 years:

Initial	$100$
$1^{s t}$ Year	$100 \cdot 146.51 % \approx 147$
$2^{n d}$ Year	$100 \cdot (146.51 %)^{2} \approx 215$
$3^{r d}$ Year	$100 \cdot (146.51 %)^{3} \approx 314$

The result $314$ has $| \frac{314 - 300}{300} | = 4.67 %$ overestimation.

The geometric mean is $\sqrt[3]{180 % \cdot 116.66 % \cdot 142.86 %} \approx 144.22 %$

Simulate an orange tree that grows $144.22 %$ for 3 years:

Initial	$100$
$1^{s t}$ Year	$100 \cdot 144.22 % \approx 144$
$2^{n d}$ Year	$100 \cdot (144.22 %)^{2} \approx 208$
$3^{r d}$ Year	$100 \cdot (144.22 %)^{3} \approx 300$

The result $300$ accurately describes the final yield.

The intuition behind this is that the growth factors are multiplied together: $180 % \cdot 116.67 % \cdot 142.86 % \approx 300 %$ , and the final yield is calculated as exponential growth $f (t) = a (1 + r)^{t}$ .

Your goal is to find $x$ in $100 \cdot (x)^{3} = 100 \cdot (180 % \cdot 116.67 % \cdot 142.86 %)$ , therefore $x = GM$ is the natural fit.

Harmonic MeanPermalink

Speed:

Speed (km / hr) = \frac{Distance (km)}{Time (hr)}

Starting at home, you travel with $60 (km / hr)$ to a location and return with $20 (km / hr)$ . We denote the travel distance as $d (km)$ .

Assume $d = 120 (km)$ , the total time spent is $\frac{120 (km)}{60 (km / hr)} + \frac{120 (km)}{20 (km / hr)} = 2 (hr) + 6 (hr) = 8 (hr)$ .

The arithmetic mean is $\frac{60 + 20}{2} = 40 (km / hr)$ . However, this also overestimates the mean.

The overestimation of speed causes underestimation of time $\frac{120 \cdot 2 (km)}{40 (km / hr)} = 6 (hr)$ .

The harmonic mean is $(\frac{60^{- 1} + 20^{- 1}}{2})^{- 1} = \frac{2}{\frac{1}{60 (km / hr)} + \frac{1}{20 (km / hr)}} = 30 (km / hr)$ .

The estimation $\frac{120 \cdot 2 (km)}{30 (km / hr)} = 8 (hr)$ accurately describes the total time spent.

The intuition behind this is that the speeds are ratios, and the total time is calculated as $t (hr) = \frac{d (km)}{s_{1} (km / hr)} + \frac{d (km)}{s_{2} (km / hr)}$ . The multiplicative inverse of speed can be interpreted as “slowness”.

Your goal is to find $x$ in $\frac{2 d (km)}{x (km / hr)} = \frac{d (km)}{60 (km / hr)} + \frac{d (km)}{20 (km / hr)}$ , therefore $x = HM$ is the natural fit.

Side Note: If the problem is modified to traveling with two speeds given the same elapsed time. Then arithmetic mean is the correct method to use.

Density:

Density (g / m^{3}) = \frac{Mass (g)}{Volume (m^{3})}

Assume we combine 2 objects with the same mass, and the volume will add up (does not hold in most cases). We can use $x = HM$ to find the combined density $x$ in $\frac{2 m}{x} = \frac{m}{d_{1}} + \frac{m}{d_{2}}$ .

Side Note: If the problem is modified to combining two objects with the same volume. Then arithmetic mean is the correct method to use.

Resistance:

$(V) = (kg \cdot m^{2} \cdot s^{- 3} \cdot A^{- 1})$ .
$(I) = (A)$ .
$(Ω) = (kg \cdot m^{2} \cdot s^{- 3} \cdot A^{- 2})$ .

Resistance (Ω) = \frac{Voltage (V)}{Current (I)}

To calculate the equivalent resistance of two resistors connect in parallel (Same voltage difference), we can use $x = HM$ to find the equivalent resistance $\frac{2 V}{x} = \frac{V}{x} + \frac{V}{x} = \frac{V}{R_{1}} + \frac{V}{R_{2}}$ . The equivalent resistance $2 x$ indicates that if we replace the resistors with 2 new resistors with resistance $x$ , the resulting resistance will be equivalent.

Side Note: If the problem is modified to connect two resistors in serial. Then arithmetic mean is the correct method to use.

ReferencesPermalink

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