Fixed the explanation for renaming f
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@ -2,7 +2,7 @@
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% Show that each iteration of the discretized version naturally creates a matrix equation.
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% Show that each iteration of the discretized version naturally creates a matrix equation.
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The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation we can approximate the value of $f(x_{i}) = f_{i}$. This will result in a set of equations
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The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation we solve for $\vec{v}$. This will result in a set of equations
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\begin{align*}
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\begin{align*}
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- v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\
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- v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\
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- v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
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- v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
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@ -10,7 +10,7 @@ The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation
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- v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\
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- v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\
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\end{align*}
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\end{align*}
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Rearranging the first and last equation, moving terms of known boundary values to the RHS
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where $v_{i} = v(x_{i})$ and $f_{i} = f(x_{i})$. Rearranging the first and last equation, moving terms of known boundary values to the RHS
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\begin{align*}
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\begin{align*}
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2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\
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2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\
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- v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
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- v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
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@ -40,5 +40,5 @@ We now have a number of linear eqations, corresponding to the number of unknown
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\right.
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\right.
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\right]
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\right]
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\end{align*}
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\end{align*}
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where $g_{i} = h^{2} f_{i}$. An augmented matrix can be represented as $\boldsymbol{A} \vec{x} = \vec{b}$. In this case $\boldsymbol{A}$ is the coefficient matrix with a tridiagonal signature $(-1, 2, -1)$ and dimension $n \cross n$, where $n=m-2$.
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Since the boundary values are equal to $0$ the RHS can be renamed $g_{i} = h^{2} f_{i}$ for all $i$. An augmented matrix can be represented as $\boldsymbol{A} \vec{x} = \vec{b}$. In this case $\boldsymbol{A}$ is the coefficient matrix with a tridiagonal signature $(-1, 2, -1)$ and dimension $n \cross n$, where $n=m-2$.
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