ICS/midterm/midterm.tex

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\title{Midterm -- Introduction to Computational Science}
\author{Claudio Maggioni}

\begin{document}
\maketitle
\section*{Question 1}
\subsection*{Point a)}
$$7.125_{10} = (1 + 2^{-1} + 2^{-2} + 2^{-5}) * {2^2}_{10} = 0 | 1100 1000 0000 | 110_F$$
$$0.8_{10} = (1 + 2^{-1} + 2^{-4} + 2^{-5}+ 2^{-8} + 2^{-9} + 2^{-12}) * {2^{-1}}_{10} \approx 0 | 100110011001|011_F$$
$$0.046875_{10} = (2^{-2} + 2^{-3}) * {2^{-3}}_{10} = 0 | 0011 0000 0000|000_F$$

For the last conversion, we assume that the denormalized mode of this floating point representation implicitly includes an exponent of $-3$ (this makes the first bit in the mantissa of a denormalized number weigh $2^{-3}$). This behaviour is akin to the denormalized implementation of IEEE 754 floating point numbers.

\subsection*{Point b)}
$$1|011010111000|110_F = -(1+2^{-2} +2^{-3} +2^{-5} +2^{-7} +2^{-8} + 2^{-9}) * {2^{2}}_{10} \approx
-5.6796875$$
$$1|101010101010|010_F = -(1+2^{-1} +2^{-3} +2^{-5} +2^{-7} +2^{-9} + 2^{-11}) * {2^{-2}}_{10} \approx
-0.4166259766$$

\subsection*{Point c)}
$$1|0000 0000 0000|001_F = 2^{-3}_{10} = 0.125_{10}$$

\subsection*{Point d)}
$$1|1111 1111 1111|111_F = $$
$$=(1 + 2^{-1}+ 2^{-2} + 2^{-3} +2^{-4} + 2^{-5} +2^{-6} + 2^{-7} + 2^{-8} + 2^{-9}+ 2^{-10}+ 2^{-11}+ 2^{-12}) * {2^3}_{10} = $$
$$ = 15.998046875_{10}$$

\subsection*{Point e)}
With 12 independent binary choices (bits to flip), there are $2^{12}$ different denormalized numbers in this encoding.

\subsection*{Point f)}
With 12 independent binary choices (bits to flip) and 3 extra bits for the exponent, there are $2^{15}-1$ different denormalized numbers in this encoding. We subtract 1 in order to be consistent with the assumption in point \textit{a)}, since $0.125_{10}$ would be representable both as $1|0000 0000 0000|001_F$ and
as $1|1000 0000 0000|000_F$

\section*{Question 2}

\subsection*{Point a)}
$$ \sqrt[3]{1 + x} - 1 =  (\sqrt[3]{1 + x} - 1) \cdot
 \frac{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1}{ \sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} = \frac{(1 + x) - 1}{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} =$$
$$ \frac{x}{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} $$
\subsection*{Point b)}
$$ \frac{1 - cos(x)}{sin(x)} = \frac{sin^2(x)cos^2(x) - cos(x)}{sin(x)} \cdot \frac{sin(x)}{cos(x)} \cdot \frac{cos(x)}{sin(x)} = (sin^2(x)cos(x) - 1)\cdot\frac{cos(x)}{sin(x)}$$ 
\subsection*{Point c)}
$$ \frac{1}{1-\sqrt{x^2-1}} = \frac{1+\sqrt{x^2-1}}{(1-\sqrt{x^2-1})(1+\sqrt{x^2-1})} = 
\frac{1+\sqrt{x^2-1}}{1 - (x^2-1)} = -\frac{1+\sqrt{x^2-1}}{x^2} $$
\subsection*{Point d)}
$$ x^3\cdot\left(\frac{x}{x^2-1}-\frac{1}{x}\right) = x^3\cdot\left(\frac{x^2-x^2+1}{x^3-x}\right) =
\frac{x^2}{x^2-1}$$
\subsection*{Point e)}
$$ \frac{1}{x} - \frac{1}{x+1} = \frac{x + 1 - x}{x^2 + x} = \frac{1}{x^2 + x}$$

\section*{Question 3}
\subsection*{Point a)}

Consider the Taylor expansion with $a=x$ of $f(x+h)$ and $f(x-h)$:

$$f(x+h) \geq f(x) + \frac{f'(x)}{1}\cdot h + \frac{f''(x)}{2}\cdot h^2 + 
\frac{f'''(x)}{6}\cdot h^3$$

  $$f(x-h) \geq f(x) - \frac{f'(x)}{1}\cdot h + \frac{f''(x)}{2}\cdot h^2 -
\frac{f'''(x)}{6}\cdot h^3$$ 
 
Then, we can derive that:

$$\frac{f(x + h) - f(x - h)}{2h} \geq 
\frac{1}{2h} \cdot 
\left(2hf'(x) + \frac{2h^3f'''(x)}{6} \right) =
f'(x) + \frac{h^2f'''(x)}{6}
$$

So:

$$ \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| \leq \left|f'(x) - \left(f'(x) + \frac{h^2f'''(x)}{6}\right)
\right| = \frac{h^2|f'''(x)|}{6}
 \Rightarrow
C = \frac{f'''(x)}{6}$$

\section*{Question 4}
\subsection*{Point a)}
\[
A_1 = 
 \begin{bmatrix}
   1 & 1  & 1  & 1  & 1 \\
   2 & 4  & 4  & 4  & 4 \\
   3 & 7  & 10 & 10 & 10 \\
   4 & 10 & 16 & 20 & 20 \\
   5 & 13 & 22 & 30 & 35 \\
 \end{bmatrix},
 b = 
 \begin{bmatrix}
 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
 \end{bmatrix}
\]
\[
l_1 = 
  \begin{bmatrix}
 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
 \end{bmatrix},
u_1 = 
  \begin{bmatrix}
 1 & 1 & 1 & 1 & 1 \\
 \end{bmatrix},
 A_2
 \begin{bmatrix}
   0 & 0  & 0  & 0  & 0 \\
   0 & 2  & 2  & 2  & 2 \\
   0 & 4  & 7  & 7  & 7 \\
   0 & 6  & 12 & 16 & 16 \\
   0 & 8  & 17 & 25 & 30 \\
 \end{bmatrix}
\]
\[
l_2 = 
  \begin{bmatrix}
 0 \\ 1 \\ 2 \\ 3 \\ 4 \\
 \end{bmatrix},
u_2 = 
  \begin{bmatrix}
 0 & 2 & 2 & 2 & 2 \\
 \end{bmatrix},
 A_3
 \begin{bmatrix}
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 3  & 3  & 3 \\
   0 & 0  & 6 & 10 & 10 \\
   0 & 0  & 9 & 17 & 22 \\
 \end{bmatrix}
\]
\[
l_3 = 
  \begin{bmatrix}
 0 \\ 0 \\ 1 \\ 2 \\ 3 \\
 \end{bmatrix},
u_3 = 
  \begin{bmatrix}
 0 & 0 & 3 & 3 & 3 \\
 \end{bmatrix},
 A_4
 \begin{bmatrix}
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0 & 4 & 4 \\
   0 & 0  & 0 & 8 & 13 \\
 \end{bmatrix}
\]
\[
l_4 = 
  \begin{bmatrix}
 0 \\ 0 \\ 0 \\ 1 \\ 2 \\
 \end{bmatrix},
u_4 = 
  \begin{bmatrix}
 0 & 0 & 0 & 4 & 4 \\
 \end{bmatrix},
 A_5
 \begin{bmatrix}
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0  & 0  & 0 \\
   0 & 0  & 0 & 0 & 0 \\
   0 & 0  & 0 & 0 & 5 \\
 \end{bmatrix}
\]
\[
l_5 = 
  \begin{bmatrix}
 0 \\ 0 \\ 0 \\ 0 \\ 1 \\
 \end{bmatrix},
u_5 = 
  \begin{bmatrix}
 0 & 0 & 0 & 0 & 5 \\
 \end{bmatrix},
 L = 
 \begin{bmatrix}
   1 & 0 & 0 & 0 & 0 \\
   2 & 1 & 0 & 0 & 0 \\
   3 & 2 & 1 & 0 & 0 \\
   4 & 3 & 2 & 1 & 0 \\
   5 & 4 & 3 & 2 & 1 \\
 \end{bmatrix},
  U = 
 \begin{bmatrix}
   1 & 1 & 1 & 1 & 1 \\
   0 & 2 & 2 & 2 & 2 \\
   0 & 0 & 3 & 3 & 3 \\
   0 & 0 & 0 & 4 & 4 \\
   0 & 0 & 0 & 0 & 5 \\
 \end{bmatrix} 
 \]
\subsection*{Point b)}
\[
l_1 = 
  \begin{bmatrix}
 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
 \end{bmatrix},
e^{T}_1 = 
  \begin{bmatrix}
 1 & 0 & 0 & 0 & 0 \\
 \end{bmatrix},
 L_1 = 
 \begin{bmatrix}
   0 & 0 & 0 & 0 & 0 \\
   -2 & 1 & 0 & 0 & 0 \\
   -3 & 0 & 1 & 0 & 0 \\
   -4 & 0 & 0 & 1 & 0 \\
   -5 & 0 & 0 & 0 & 1 \\
 \end{bmatrix}
 \]
 \[
l_2 = 
  \begin{bmatrix}
 0 \\ 1 \\ 2 \\ 3 \\ 4 \\
 \end{bmatrix},
e^{T}_2 = 
  \begin{bmatrix}
 0 & 1 & 0 & 0 & 0 \\
 \end{bmatrix},
 L_2 = 
 \begin{bmatrix}
   1 & 0 & 0 & 0 & 0 \\
   0 & 0 & 0 & 0 & 0 \\
   0 & -2 & 1 & 0 & 0 \\
   0 & -3 & 0 & 1 & 0 \\
   0 & -4 & 0 & 0 & 1 \\
 \end{bmatrix}
 \]
  \[
l_3 = 
  \begin{bmatrix}
 0 \\ 0 \\ 1 \\ 2 \\ 3 \\
 \end{bmatrix},
e^{T}_3 = 
  \begin{bmatrix}
 0 & 0 & 1 & 0 & 0 \\
 \end{bmatrix},
 L_3 = 
 \begin{bmatrix}
   1 & 0 & 0 & 0 & 0 \\
   0 & 1 & 0 & 0 & 0 \\
   0 & 0 & 0 & 0 & 0 \\
   0 & 0 & -2 & 1 & 0 \\
   0 & 0 & -3 & 0 & 1 \\
 \end{bmatrix}
 \]
   \[
l_4 = 
  \begin{bmatrix}
 0 \\ 0 \\ 0 \\ 1 \\ 2 \\
 \end{bmatrix},
e^{T}_4 = 
  \begin{bmatrix}
 0 & 0 & 0 & 1 & 0 \\
 \end{bmatrix},
 L_4 = 
 \begin{bmatrix}
   1 & 0 & 0 & 0 & 0 \\
   0 & 1 & 0 & 0 & 0 \\
   0 & 0 & 1 & 0 & 0 \\
   0 & 0 & 0 & 0 & 0 \\
   0 & 0 & 0 & -2 & 1 \\
 \end{bmatrix}
\] 
\subsection*{Point c)}
$$Ly = b$$
\[
 L = 
 \begin{bmatrix}
   1 & 0 & 0 & 0 & 0 \\
   2 & 1 & 0 & 0 & 0 \\
   3 & 2 & 1 & 0 & 0 \\
   4 & 3 & 2 & 1 & 0 \\
   5 & 4 & 3 & 2 & 1 \\
 \end{bmatrix},
b = 
 \begin{bmatrix}
 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
 \end{bmatrix}
 \]
 \[
 L = 
 \begin{bmatrix}
   1 & 0 & 0 & 0 & 0 \\
   2 & 1 & 0 & 0 & 0 \\
   3 & 2 & 1 & 0 & 0 \\
   4 & 3 & 2 & 1 & 0 \\
   5 & 4 & 3 & 2 & 1 \\
 \end{bmatrix},
b = 
 \begin{bmatrix}
 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
 \end{bmatrix}
 \]
 $$y_1 = \frac{1}{1} = 1$$
 $$y_2 = \frac{2 - 2 \cdot 1}{1} = 0$$
 $$y_3 = \frac{3 - 3 \cdot 1 - 2 \cdot 0}{1} = 0$$
 $$y_4 = \frac{4 - 4 \cdot 1 - 3 \cdot 0 - 2 \cdot 0}{1} = 0$$
 $$y_5 = \frac{5 - 5 \cdot 1 - 4 \cdot 0 - 3 \cdot 0 - 2 \cdot 0}{1} = 0$$
 \[
y = 
 \begin{bmatrix}
 1 \\ 0 \\ 0 \\ 0 \\ 0 \\
 \end{bmatrix} 
 \]
 \subsection*{Point d)}
$$Ux = y$$
\[
  U = 
 \begin{bmatrix}
   1 & 1 & 1 & 1 & 1 \\
   0 & 2 & 2 & 2 & 2 \\
   0 & 0 & 3 & 3 & 3 \\
   0 & 0 & 0 & 4 & 4 \\
   0 & 0 & 0 & 0 & 5 \\
 \end{bmatrix} 
y = 
 \begin{bmatrix}
 1 \\ 0 \\ 0 \\ 0 \\ 0 \\
 \end{bmatrix}
 \]
 $$x_5 = \frac{0}{5} = 0$$
 $$x_4 = \frac{0 - 4 \cdot 0}{4} = 0$$
 $$x_3 = \frac{0 - 3 \cdot 0 - 3 \cdot 0}{3} = 0$$
 $$x_2 = \frac{0 - 2 \cdot 0 - 2 \cdot 0 - 2 \cdot 0}{2} = 0$$
 $$x_1 = \frac{0 - 1 \cdot 0 - 1 \cdot 0 - 1 \cdot 0 - 1 \cdot }{1} = 0$$
 \[
x = 
 \begin{bmatrix}
 0 \\ 0 \\ 0 \\ 0 \\ 0 \\
 \end{bmatrix} 
 \]
 
\section*{Question 5}
\subsection*{Point a)}
$$f(x) = x \hspace{2cm} K_{abs} = |f'(x)| = 1 \hspace{2cm} K_{rel} = \left|\frac{1 \cdot x}{x}\right| = 1$$

\subsection*{Point b)}
$$f(x) = \sqrt[3]{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{3\sqrt[3]{x^2}} \hspace{2cm}
 K_{rel} = \left|\frac{1}{3\sqrt[3]{x^2}} \cdot \frac{x}{\sqrt[3]{x}}\right| = \frac{1}{3}$$

 \subsection*{Point c)}
$$f(x) = \frac{1}{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{x^2} \hspace{2cm}
 K_{rel} = \left|\frac{-x}{x^2} \cdot \frac{1}{\frac{1}{x}}\right| = 1$$
 
  \subsection*{Point d)}
$$f(x) = e^x \hspace{2cm} K_{abs} = |f'(x)| = e^x \hspace{2cm}
 K_{rel} = \left|\frac{xe^x}{e^x}\right| = |x|$$
 
 \subsection*{Point e)}
 Cases \textit{a)},\textit{b)} and \textit{c)} are well-conditioned for any $x$ since their $K_rel$
 is not defined by x. Case \textit{d)} is well-conditioned only for $x$s whose absolute value is in the order of magnitude of $1$ or less, since $K_rel$ in this case is exactly $|x|$.
\end{document}