6 Commits

Author SHA1 Message Date
42f85c1eb7 Wrote a rough draft of problem 5 2023-09-02 15:50:13 +02:00
0430b0fcf0 Merge pull request #13 from FYS3150-G2-2023/1-solve-problem-1
1 solve problem 1
2023-09-01 15:27:18 +02:00
2f9a39021b Add Janita's last name 2023-09-01 14:52:00 +02:00
c86c1dc73b Finish problem 1 2023-09-01 14:50:31 +02:00
cadee04b9f Add latex files 2023-09-01 14:49:44 +02:00
ab06e99c0d Merge pull request #12 from FYS3150-G2-2023/11-set-up-structure-in-latex
Set up the basic structure of latex file
2023-09-01 14:19:30 +02:00
3 changed files with 52 additions and 8 deletions

6
.gitignore vendored
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@@ -30,3 +30,9 @@
*.exe
*.out
*.app
# Latex
*.aux
*.log
*.out
*.bib

BIN
latex/assignment_1.pdf Normal file

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@@ -74,8 +74,8 @@
\begin{document}
\title{Title of the document} % self-explanatory
\author{Cory Balaton \& Janita} % self-explanatory
\title{Project 1} % self-explanatory
\author{Cory Balaton \& Janita Willumsen} % self-explanatory
\date{\today} % self-explanatory
\noaffiliation % ignore this, but keep it.
@@ -87,6 +87,38 @@
\section*{Problem 1}
% Do the double integral
\begin{align*}
u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
&= \int \int -100 e^{-10x} dx^2 \\
&= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
&= \int 10 e^{-10x} + c_1 dx \\
&= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
&= -e^{-10x} + c_1 x + c_2
\end{align*}
Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
\begin{align*}
u(0) &= 0 \\
-e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
-1 + c_2 &= 0 \\
c_2 &= 1
\end{align*}
\begin{align*}
u(1) &= 0 \\
-e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
-e^{-10} + c_1 + c_2 &= 0 \\
c_1 &= e^{-10} - c_2\\
c_1 &= e^{-10} - 1\\
\end{align*}
Using the values that we found for $c_1$ and $c_2$, we get
\begin{align*}
u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
&= 1 - (1 - e^{-10}) - e^{-10x}
\end{align*}
\section*{Problem 2}
@@ -104,11 +136,19 @@
\subsection*{a)}
% Phrase it better
$n = m - 2$, since $\textbf{A}$ is used to solve for all of the points in
between the end-points $0$ and $1$. For the complete solution, we need to add
$u(0)$ and $u(1)$.
\subsection*{b)}
\section{Problem 6}
When solving for $\vec{v}$, we find the approximate solutions for $u(x)$
that are in between the end-points, but not the end-points themselves.
\subsection{a)}
\section*{Problem 6}
\subsection*{a)}
% Use Gaussian elimination, and then use backwards substitution to solve the equation
@@ -116,9 +156,7 @@
% Figure it out
\section*{Problem 7}
% Link to relevant files on gh and possibly add some comments
% Linelevant files on gh and possibly add some comments
\section*{Problem 8}
@@ -126,7 +164,7 @@
\section*{Problem 9}
% Show the algorithm, then calculate FLOPs, then link to relevant files
% Shon*{Proalgorithm, then calculate FLOPs, then link to relevant files
\section*{Problem 10}