1.1 Example with 4 points

1.2 The general formula for a line

\[y = mx + b\]

1.3 Determine the difference between observed and actual

The error for each point of the line is the distance between \(y_i\) and the calculated \(mx_i + b\)

\[ \epsilon_i = y_i - (mx_i+b)\]

1.4 The sum error of a middle line cancels all errors.

\[\sum_{i=1}^n \epsilon_i = \sum \epsilon_i = 0\]

1.5 Sum of all deviations

\[\begin{eqnarray} \epsilon &=& \sum \left( y_i - (mx_i+b) \right)^2\\ &=& \sum \left(y_i^2 - 2mx_i y_i - 2b y_i + m^2 x_i^2 + 2mb x_i + b^2 \right)\\ &=& \sum y_i^2 - 2m \sum x_i y_i - 2b\sum y_i + m^2 \sum x_i^2 +\\ & & \quad \quad 2mb \sum x_i + n b^2\\ \end{eqnarray}\]

1.6 Find the minimum of the partial in respect to \(b\)

\[\begin{eqnarray}\frac{\partial}{\partial b} \sum y_i^2 - 2m \sum x_i y_i - 2b\sum y_i + m^2 \sum x_i^2 +\\ \quad \quad 2mb \sum x_i + n b^2 &=& 0\\ \\ - 2\sum y_i + 2m \sum x_i + 2n b &=&0 \\ - \sum y_i + m \sum x_i + n b &=& 0\\ \end{eqnarray}\]

2.1 Find the minimum of the partial in respect to \(m\)

\[\begin{eqnarray}\frac{\partial}{\partial m} \sum y_i^2 - 2m \sum x_i y_i - 2b\sum y_i + \\ m^2 \sum x_i^2 + 2mb \sum x_i - n b^2 &=& 0\\ - 2 \sum x_i y_i + 2m \sum x_i^2 + 2b\sum x_i &=& 0\\ \sum x_i y_i - m \sum x_i^2 - b \sum x_i &=& 0\\ \sum x_i \left(y_i - m x_i - b \right) &=& 0\\ \end{eqnarray}\]

2.2 Simplify

\[\begin{eqnarray} \sum x_i \left(y_i - m x_i - \left(\bar y - m \bar x\right) \right) &=& 0\\ \sum x_i \left(y_i - \bar y - m\left(x_i - \bar x\right) \right) &=& 0\\ \sum x_i \left(y_i - \bar y\right) - \sum x_i m\left(x_i - \bar x\right) &=& 0\\ \sum x_i \left(y_i - \bar y\right) &=& m \sum x_i \left(x_i - \bar x\right)\\ \frac{\sum x_i \left(y_i - \bar y\right)}{\sum x_i \left(x_i - \bar x\right)} &=& m\\ \end{eqnarray}\]

2.3 Solve for \(m\)

\[\begin{eqnarray} m &=& \frac{\sum x_i \left(y_i - \bar y\right)}{\sum x_i \left(x_i - \bar x\right)} \\ m &=& \frac{\sum \left(x_i -\bar x\right)\left(y_i - \bar y\right)}{\sum \left(x_i -\bar x\right) \left(x_i - \bar x\right)} \\ \\ m &=& \frac{\sum \left(x_i -\bar x\right)\left(y_i - \bar y\right)}{\sum \left(x_i -\bar x\right)^2}\\ \end{eqnarray}\]

2.4 Least means squares

\[\begin{eqnarray} m &=& \frac{\sum \left(x_i -\bar x\right)\left(y_i - \bar y\right)}{\sum \left(x_i -\bar x\right)^2}\\ b &=& \bar y - m \bar x\\ \end{eqnarray}\]

2.5 Using the formula

\[m=\frac{11}{20} = 0.55\] \[b=5.25 - 0.55 \times 5.0 =2.5\]

2.6 Linear Regression Model

 Coefficients:
                Estimate Std. Error t value Pr(>|t|)
    Intercept   2.5000     1.4230   1.757    0.221
    Slope       0.5500     0.2598   2.117    0.168
    
 Residuals:    1    2    3    4 
             -0.6  1.3 -0.8  0.1 
 
 Residual standard error: 1.162 on 2 degrees of freedom
 R-squared: 0.6914 (multiple),  0.5371 (Adjusted) 
 F-statistic: 4.481 on 1 and 2 DF,  p-value: 0.1685

2.7 Statistical interpretation

2.8 Experimental results

3.1 Total number of cases

\[\begin{eqnarray} C &=&\int axe^{(-bx)} dx\\ &=& - \frac{a(bx + 1)e^{(-bx)}}{b^2}\\ &=& 326 \left(\frac{100000000 - 1049600000 e^{-(1187/125)}}{1408969}\right)\\ \\ &=& 23041\\ \end{eqnarray}\]

3.2 Peak day of peak infection

\[\begin{eqnarray} 0 &=& f'(x) = \frac{d\left( axe^{(-bx)}\right)}{dx}\\ 0 &=& ae^{(-bx)} - abxe^{(-bx)} = −a\left(bx-1\right)e^{-bx}\\ 0 &=& bx -1\\ bx &=& 1\\ \\ x &=& \frac{1}{b} = \frac{1}{0.1187} =8.42\\ \end{eqnarray} \]

3.3 Point of inflection

\[\begin{eqnarray} 0 &=& f''(x) = \frac{d(ae^{-bx} - abxe^{-bx})}{dx}\\ 0 &=& -abe^{(-bx)} + −ab\left(bx-1\right)e^{-bx}\\ 0 &=&ab\left(bx-1\right)e^{-bx}-abe^{-bx}\\ 0 &=&ab\left(bx-2)\right) e^{-bx}\\ 0 &=& bx - 2\\ \\ x &=& \frac{2}{b} = \frac{2}{0.1187}= 16.85 days\\ \end{eqnarray}\]

3.4 Duration of the infection

86 days

3.5 Statistical analysis

3.6 Covid 19 Chiang Mai

Formula: cases ~ a1 * cdays * exp(-b1 * cdays)

Parameters:
    Estimate Std. Error t value Pr(>|t|)    
a1 99.181963   5.538972   17.91 6.44e-13 ***
b1  0.265549   0.008625   30.79  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 9.996 on 18 degrees of freedom

Number of iterations to convergence: 7 
Achieved convergence tolerance: 1.959e-07

3.7 Modelled data

3.8 Early warning

8.1

10.1 United States population

x=c(1790,1800,1810,1820,1830,1840,1850,1860, 1870,1880,1890,1900,1910, 1920,1930,1940,1950,1960,1970,1980,1990, 2000)

y = c(3.9,5.3,7.2,9.6,12.9,17.1,23.1,31.4, 38.6,50.2,62.9,76.0,92.0, 105.7,122.8,131.7,150.7,179.3,203.3,248.7,281.4) )

11.1 Elasticity of American demand of farm products

11.2 Revenue as a function of price

\[\eqalign{\frac{dR}{dp} &= \frac{d}{dp} (pq)\cr &= p\frac{dq}{dp}+\frac{dp}{dp}q\cr &= p\frac{dq}{dp}+ q\cr} \] \[p\frac{dq}{dp} + q = 0\] \[p\frac{dq}{dp}= - q\] \[1= \frac{p}{q} \frac{dq}{dp} = E\]

11.3 US Cable subscribers

x= c(1977:2003) y = c(16.6,17.9,19.4,22.6,28.3,35.0,40.5, 43.7,46.2,48.1,50.5,53.8,57.1,59.0, 60.6,61.5,62.6,63.4,65.7,66.7,67.3, 67.4,68.0,67.8,69.2,68.9,68.0)

11.4 Recorded vehicle speed in secs

18.1 Separation of Variables: Ex 2

\[\eqalign{\frac{dy}{dt} & = - \frac{x}{y} \cr y dy &= - x dx\cr \int y dy &= - \int x dx\cr \frac{y^2}{2} &= -\frac{x^2}{2}\cr x^2 + y^2 &= C\cr}\]

18.2 Separation of Variables: Ex 3

\[\eqalign{\frac{dy}{dx} &= k(y-A)\cr \frac{1}{y-A} &= k dx\cr \int \frac{1}{y-A} &= \int k dx\cr \ln(|y - A|) &= kt + D\cr |y - A| &= e^{kt} e^{D}\cr y &= A + C e^{kt}\cr}\]

18.3 Separation of Variables: Ex 4

\[\eqalign{\frac{dP}{dt} &= 2P - 2Pt\cr \frac{1}{P}} &= (2-2t) dt\cr \int \frac{1}{P}} &= \int (2-2t) dt\cr \ln(|P|) &= 2t - t^2 + C\cr |P| &= e^{2t - t^2 + C}\cr P &= B e^{2t-t^2}\cr }\]