Pythagorean Means

A short comparison of Pythagorean Means: the arithmetic mean (AM), the geometric mean (GM), and the harmonic mean (HM).
Arithmetic Mean (AM)
Geometric Mean (GM)
Harmonic Mean (HM)
Initial Value
Growth Factor (e.g., 180%)
Growth Rate (e.g., 80%)
Time Elapsed (e.g., Years)
$$\def\AM{\tip{am}{\mathrm{AM}}} \def\GM{\tip{gm}{\mathrm{GM}}} \def\HM{\tip{hm}{\mathrm{HM}}} \def\DistanceUnit{{\ \mathrm{(km)}}} \def\TimeUnit{{\ \mathrm{(hr)}}} \def\SpeedUnit{{\ \mathrm{(km/hr)}}} \def\MassUnit{{\ \mathrm{(g)}}} \def\VolumeUnit{{\ \mathrm{(m^3)}}} \def\DensityUnit{{\ \mathrm{(g/m^3)}}} \def\VoltageUnit{{\ \mathrm{(V)}}} \def\CurrentUnit{{\ \mathrm{(I)}}} \def\ResistanceUnit{{\ \mathrm{(\Omega)}}}$$

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

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Comparison

Use cases

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 variables

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

General form of $$n$$ variables

Type General form Note
Arithmetic mean $$\displaystyle\frac{1}{n}\sum_{i=1}^nx_i$$ $$=\displaystyle\frac{x_1+x_2+\cdots+x_n}{n}$$
Geometric mean $$\displaystyle(\prod_{i=1}^nx_i)^{\frac{1}{n}}$$ $$=\displaystyle\sqrt[n]{x_1x_2\cdots x_n}$$
Harmonic mean $$\displaystyle\left(\frac{\displaystyle\sum_{i=1}^nx_i^{-1}}{n}\right)^{-1}$$ $$=\displaystyle\frac{n}{\displaystyle\sum_{i=1}^nx_i^{-1}}=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}$$

Assuming all values are positive, the following relationship holds:

$\min\le \HM\le \GM\le \AM\le\max$

Examples

Arithmetic Mean

Statistics mean:

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

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

Geometric Mean

Proportional growth:

An orange tree yields $$\tip{initial}{100}$$, $$180$$, $$210$$, $$300$$ in each year within a 4 year time frame.

The growth factors are: $$\frac{180}{\tip{initial}{100}}=180\%$$, $$\frac{210}{180}\approx116.67\%$$, $$\frac{300}{210}\approx142.86\%$$.

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

Simulate an orange tree that grows $$\tip{am}{146.51\%}$$ for 3 years:

 Initial $$\tip{initial}{100}$$ $$1^{st}$$ Year $$\tip{initial}{100}\cdot\tip{am}{146.51\%}\approx147$$ $$2^{nd}$$ Year $$\tip{initial}{100}\cdot(\tip{am}{146.51\%})^2\approx215$$ $$3^{rd}$$ Year $$\tip{initial}{100}\cdot(\tip{am}{146.51\%})^3\approx314$$

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

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

Simulate an orange tree that grows $$\tip{gm}{144.22\%}$$ for 3 years:

 Initial $$\tip{initial}{100}$$ $$1^{st}$$ Year $$\tip{initial}{100}\cdot\tip{gm}{144.22\%}\approx144$$ $$2^{nd}$$ Year $$\tip{initial}{100}\cdot(\tip{gm}{144.22\%})^2\approx208$$ $$3^{rd}$$ Year $$\tip{initial}{100}\cdot(\tip{gm}{144.22\%})^3\approx300$$

The result $$300$$ accurately describes the final yield.

The intuition behind this is that the growth factors are multiplied together: $$180\%\cdot116.67\%\cdot142.86\%\approx300\%$$, and the final yield is calculated as exponential growth $$f(\tip{time_elapsed}{t})=\tip{initial}{a}(\tip{growth_factor}{1+\tip{growth_rate}{r}})^\tip{time_elapsed}{t}$$.

Your goal is to find $$x$$ in $$\tip{initial}{100}\cdot(x)^3=\tip{initial}{100}\cdot(180\%\cdot116.67\%\cdot142.86\%)$$, therefore $$x=\GM$$ is the natural fit.

Harmonic Mean

Speed:

$\mathrm{Speed}\SpeedUnit=\frac{\mathrm{Distance}\DistanceUnit}{\mathrm{Time}\TimeUnit}$

Starting at home, you travel with $$60\SpeedUnit$$ to a location and return with $$20\SpeedUnit$$. We denote the travel distance as $$d\DistanceUnit$$.

Assume $$d=120\DistanceUnit$$, the total time spent is $$\frac{120\DistanceUnit}{60\SpeedUnit}+\frac{120\DistanceUnit}{20\SpeedUnit}=2\TimeUnit+6\TimeUnit=8\TimeUnit$$.

The arithmetic mean is $$\frac{60+20}{2}=\tip{am}{40\SpeedUnit}$$. However, this also overestimates the mean.

The overestimation of speed causes underestimation of time $$\frac{120\cdot2\DistanceUnit}{\tip{am}{40\SpeedUnit}}=6\TimeUnit$$.

The harmonic mean is $$(\frac{60^{-1}+20^{-1}}{2})^{-1}=\frac{2}{\frac{1}{60\SpeedUnit}+\frac{1}{20\SpeedUnit}}=\tip{hm}{30\SpeedUnit}$$.

The estimation $$\frac{120\cdot2\DistanceUnit}{\tip{hm}{30\SpeedUnit}}=8\TimeUnit$$ accurately describes the total time spent.

The intuition behind this is that the speeds are ratios, and the total time is calculated as $$t\TimeUnit=\frac{d\DistanceUnit}{s_1\SpeedUnit}+\frac{d\DistanceUnit}{s_2\SpeedUnit}$$. The multiplicative inverse of speed can be interpreted as “slowness”.

Your goal is to find $$x$$ in $$\frac{2d\DistanceUnit}{x\SpeedUnit}=\frac{d\DistanceUnit}{60\SpeedUnit}+\frac{d\DistanceUnit}{20\SpeedUnit}$$, 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:

$\mathrm{Density}\DensityUnit=\frac{\mathrm{Mass}\MassUnit}{\mathrm{Volume}\VolumeUnit}$

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 $$\frac{2m}{x}=\frac{m}{d_1}+\frac{m}{d_2}$$.

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:

• $$(V)=(\mathrm{kg\cdot m^2\cdot s^{-3}\cdot A^{-1}})$$.
• $$(I)=(\mathrm{A})$$.
• $$(\Omega)=(\mathrm{kg\cdot m^2\cdot s^{-3}\cdot A^{-2}})$$.
$\mathrm{Resistance}\ResistanceUnit=\frac{\mathrm{Voltage}\VoltageUnit}{\mathrm{Current}\CurrentUnit}$

To calculate the equivalent resistance of two resistors connect in parallel (Same voltage difference), we can use $$x=\HM$$ to find the equivalent resistance $$\frac{2V}{x}=\frac{V}{x}+\frac{V}{x}=\frac{V}{R_1}+\frac{V}{R_2}$$. The equivalent resistance $$2x$$ 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.

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