CoCalc -- lab 3 report.tex

Project: isaac shaeffer - Math 5_Linear Algebra_Spring_2018
Path: Lab 3: Least Squares Regression/lab 3 report.tex
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\documentclass{article}
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\usepackage{hyperref}
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\usepackage{amsmath}
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\usepackage{verbatim}
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\title{Lab 3 Report}
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\author{Isaac Shaeffer and Satori Kubo}
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% Enable SageTeX to run SageMath code right inside this LaTeX file.
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% http://mirrors.ctan.org/macros/latex/contrib/sagetex/sagetexpackage.pdf
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% \usepackage{sagetex}
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\begin{document}
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\maketitle
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\begin{enumerate}
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% Part 1
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\item 
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% Part 1.a
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\begin{enumerate}
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\item Solving the normal equations, we obtain
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\[
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    A = \left(\begin{array}{rr}
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        1 & 1 \\
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        1 & 2 \\
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        1 & 4 \\
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        1 & 5
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    \end{array}\right),
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    \hat{\beta} = \left(\begin{array}{r}
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        - \frac{3}{5}\\
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        \frac{7}{10}
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        \end{array}\right) \text{ and }
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    \textbf{b} = 
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        \left(\begin{array}{r}
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        0 \\
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        1 \\
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        2 \\
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        3
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    \end{array}\right)
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\]
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% Part 1.b
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\item The line of best fit is $y = \frac{7}{10}x - \frac{3}{5}$.
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    \begin{figure}[h]
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        \includegraphics[width=10cm]{plot1lab3_part1}
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        \centering
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        \captionof{figure}{ $y = \frac{7}{10}x - \frac{3}{5}$ }
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    \end{figure}
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\end{enumerate}
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\clearpage
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% Part 2
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\item Changing the the first two data points, we obtain a line with a negative slope that doesn't fit the data well. \par
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    \begin{minipage}{\linewidth}
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        \includegraphics[width=10cm]{plot1lab3_part2}
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        \centering
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        \captionof{figure}{$y = -\frac{3}{10} x + \frac{73}{20} $}
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    \end{minipage}
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% Part 3
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\item \par
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    \begin{minipage}{\linewidth}
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        \includegraphics[width=10cm]{plot1lab3_part3}
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        \centering
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        \captionof{figure}{$y = \frac{513}{2800} \, x + \frac{261}{280} $}
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    \end{minipage}
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% Part 4
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\item \par
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    \begin{minipage}{\linewidth}
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        \includegraphics[width=10cm]{plot1lab3_part4}
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        \centering
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        \captionof{figure}{Curve of best fit: $y = 2.342\cos{x} + 7.448\sin{x}$}
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    \end{minipage}
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% Part 5
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\item \par
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    \begin{minipage}{\linewidth}
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        \includegraphics[width=10cm]{plot1lab3_part5}
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        \centering
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         \captionof{figure}{Curve of best fit: $y = 19.94e^{-0.02t} + 10.10e^{-0.07t}$}
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    \end{minipage}
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% Part 6
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\clearpage
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\item The systolic blood pressure of a child weighing 100 lbs is 
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\[ p(100)  = 38.77 \, \log\left(10\right) + 17.92 = 107. \] \par
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    \begin{minipage}{\linewidth}
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        \includegraphics[width=10cm]{plot1lab3_part6}
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        \centering
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        \captionof{figure}{Curve of best fit: $p = 19.38\log\left(w\right) + 17.92$}
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    \end{minipage}
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% Part 7
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\item At time $t = 4.5$ seconds, the plane's velocity is 52.7 feet per second. \par
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    \begin{minipage}{\linewidth}
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        \includegraphics[width=10cm]{plot1lab3_part7}
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        \centering
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        \captionof{figure}{Curve of best fit: $y = - 0.00652t^{3} + 5.17t^{2} + 6.60t - 2.52$}
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    \end{minipage}
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\clearpage
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% Part 8
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\item The main idea we explored was how to associate a data set with curve using a matrix. We then translated the matrix equation $A^T A\hat{\beta} = A^T\textbf{b}$ into Python code:
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\begin{verbatim}
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    A = matrix(RDF, [[x-values and constants]])
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    b = matrix([[y-values]])
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    A1 = A.transpose()*A # the left side of the equation
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    b1 = A.transpose()*b # the right side of the equation
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    A1.augment(b1).rref() # the entries of the last 
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                          # column are the solutions 
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                          # to the equation
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\end{verbatim}
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We also learned about several new utility functions for this lab. First, we learned that defining a symbolic function is just like defining a normal python function. We explored plotted  functions and data points with the `plot()` function. We also learned about Pythons anonymous lambda functions which we used to generate the large Matrix in part 7 with the command
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\begin{verbatim}
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   A = matrix(QQ, 13, 4, lambda i, j: i^j)
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\end{verbatim}
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\end{enumerate}
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\end{document}
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