Table of Contents



Linear System of Equations

  1. : \

  2. Linear System of Equations:
    formula

  3. Linear Operations:
    1. Equation \(E_i\) can be multiplied by any nonzero constant \(\lambda\) with the resulting equation used in place of \(E_i\). This operation is denoted \((\lambda E_i) \rightarrow (E_i)\).

    2. Equation \(E_j\) can be multiplied by any constant \(\lambda\) and added to equation \(E_i\) with the resulting equation used in place of \(E_i\). This operation is denoted (\(E_i + \lambda E_j) \rightarrow (E_i)\).

    3. Equations \(E_i\) and \(E_j\) can be transposed in order. This operation is denoted \((E_i) \leftrightarrow (E_j)\).

  4. Asynchronous: \

Matrices and Vectors

  1. Matrix:
    formula

  2. Gaussian Elemenation: \

  3. Algorithm: \


Operation Counts

  1. Asynchronous: \

  2. Multiplications/divisions [for each i]:
    \[(n − i) + (n − i)(n − i + 1) = (n − i)(n − i + 2)\]
  3. Additions/subtractions [for each i]: \
    \[(n − i)(n − i + 1)\]
  4. Summing the operations in Steps 5 and 6:
    formula

  5. Multiplications/divisions [Gauss-Elem]: \
    \[(n − i) + (n − i)(n − i + 1) = (n − i)(n − i + 2)\]
  6. Additions/subtractions [Gauss-Elem]: \

    \(y' = -y + ty^{1/2},\ 2 \leq t \leq 3,\ y(2) = 2,\) with \(h = 0.25\) \(\dfrac{d}{dy}f(t,y) =\)\(\dfrac{d}{dy} (-y +\)\(ty^{1/2})\) \(= \dfrac{t}{2 * \sqrt{y}} - 1\) \(\implies \vert \dfrac{d}{dy} f(t,y) \vert = \vert \dfrac{t}{2 \cdot \sqrt(y)} - 1 \vert,l\)
    Now, since \(t \in [2,3]\), we know that this is maximized at \(t = 3 \\\) \(\implies Max_{f'} = \vert \dfrac{t}{2\dot \sqrt(y)} - 1 \vert \\\) However, since \(y \in [-\infty, \infty]\), at \(y=0\) we get, \
    \(\vert \dfrac{t}{2\dot \sqrt(y)} - 1 \vert = \vert \dfrac{t}{2\dot \sqrt(0)} - 1 \vert = \vert \dfrac{t}{0} - 1 \vert = \infty \\\) Thus, this problem is ill posed and doesn’t satisfy lipschitz condition.

  7. Asynchronous [Gauss-Elem]: \

  8. Asynchronous [Gauss-Elem]: \

  9. … For a total of \(\approx \dfrac{n^3}{3}\) operations, \(\implies \in \mathcal{O}(n^3)\).