Pythagorean Means
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:
- \((V)=(\mathrm{kg\cdot m^2\cdot s^{-3}\cdot A^{-1}})\).
- \((I)=(\mathrm{A})\).
- \((\Omega)=(\mathrm{kg\cdot m^2\cdot s^{-3}\cdot A^{-2}})\).
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.