User:Cmglee/HTML

Common logarithm
The following example uses the bar notation to calculate 0.012 &times; 0.85 = 0.0102:


 * $$\begin{array}{rll}

\text{As found above,}      &\log_{10}0.012\approx\bar{2}.079181                                                    \\ \text{Since}\;\;\log_{10}0.85&=\log_{10}(10^{-1}\times 8.5)=-1+\log_{10}8.5&\approx-1+0.929419=\bar{1}.929419\;,    \\ \log_{10}(0.012\times 0.85) &=\log_{10}0.012+\log_{10}0.85                &\approx\bar{2}.079181+\bar{1}.929419     \\ &=(-2+0.079181)+(-1+0.929419)                &=-(2+1)+(0.079181+0.929419)              \\                             &=-3+1.008600                                 &=-2+0.008600\;^*                         \\                             &\approx\log_{10}(10^{-2})+\log_{10}(1.02)    &=\log_{10}(0.01\times 1.02)              \\ &=\log_{10}(0.0102) \end{array}$$ * This step makes the mantissa between 0 and 1, so that its antilog (10mantissa) can be looked up.

Mode (statistics)
When X has standard deviation ? = 0.25, the distribution of Y is weakly skewed. Using formulas for the log-normal distribution, we find:
 * $$\begin{array}{rlll}

\text{mean}  & = e^{\mu + \sigma^2 / 2} & = e^{0 + 0.25^2 / 2} & \approx 1.032 \\ \text{mode}  & = e^{\mu - \sigma^2}     & = e^{0 - 0.25^2}     & \approx 0.939 \\ \text{median} & = e^\mu                 & = e^0                & = 1 \end{array}$$ Indeed, the median is about one third on the way from mean to mode.

When X has a larger standard deviation, ? = 1, the distribution of Y is strongly skewed. Now
 * $$\begin{array}{rlll}

\text{mean}  & = e^{\mu + \sigma^2 / 2} & = e^{0 + 1^2 / 2} & \approx 1.649 \\ \text{mode}  & = e^{\mu - \sigma^2}     & = e^{0 - 1^2}     & \approx 0.368 \\ \text{median} & = e^\mu                 & = e^0             & = 1 \end{array}$$ Here, Pearson's rule of thumb fails.

Ratio of volumes of a cone, sphere and cylinder of the same radius and height
The above formulas can be used to show that the volumes of a cone, sphere and cylinder of the same radius and height are in the ratio 1 : 2 : 3, as follows.

Let the radius be r and the height be h (which is 2r for the sphere).

$$\begin{array}{llll} \text{Volume of the cone}    & = \tfrac{1}{3} \pi r^2 h & = \tfrac{1}{3} \pi r^2 (2r) & = (\tfrac{2}{3} \pi r^3) \times 1 \\ \text{Volume of the sphere}  & = \tfrac{4}{3} \pi r^3   &                             & = (\tfrac{2}{3} \pi r^3) \times 2 \\ \text{Volume of the cylinder} & = \pi r^2 h             & = \pi r^2 (2r)              & = (\tfrac{2}{3} \pi r^3) \times 3 \end{array}$$

The discovery of the 2 : 3 ratio of the volumes of the sphere and cylinder is credited to Archimedes.

Ratio of surface areas of a sphere and cylinder of the same radius and height
The above formulas can be used to show that the volumes of a sphere and cylinder of the same radius and height are in the ratio 2 : 3, as follows.

Let the radius be r and the height be h (which is 2r for the sphere).

$$\begin{array}{llll} \text{Surface area of the sphere}  & = 4 \pi r^2       &                    & = (2 \pi r^2) \times 2 \\ \text{Surface area of the cylinder} & = 2 \pi r (h + r) & = 2 \pi r (2r + r) & = (2 \pi r^2) \times 3 \end{array}$$

The discovery of this ratio is credited to Archimedes.

I'm My Own Grandpa
Family tree showing how the narrator of the song is his own grandfather.