# 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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