Gauss Elimination & Taylor Series

Cheatsheet Content

Gauss Elimination Method Used to solve systems of linear equations by transforming the augmented matrix into row echelon form. Steps for Gauss Elimination Write Augmented Matrix: Combine the coefficient matrix $A$ and the constant vector $\vec{b}$ into $[A | \vec{b}]$. Forward Elimination: Use row operations to transform the matrix into an upper triangular form (row echelon form). For each column from left to right, make the diagonal element (pivot) 1 (optional, but simplifies calculations). Use row operations to make all elements below the pivot zero. Back Substitution: Solve the resulting system of equations starting from the last equation and working upwards. Example: Solve $Ax = b$ Given system: $x + y + z = 1$ $4x + 3y - z = 6$ $3x + 5y + 3z = 4$ Step 1: Augmented Matrix $$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 4 & 3 & -1 & | & 6 \\ 3 & 5 & 3 & | & 4 \end{bmatrix} $$ Step 2: Make zeros below 1st pivot (R1C1) $R_2 \leftarrow R_2 - 4R_1$ $R_3 \leftarrow R_3 - 3R_1$ $$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & -1 & -5 & | & 2 \\ 0 & 2 & 0 & | & 1 \end{bmatrix} $$ Step 3: Make zero below 2nd pivot (R2C2) $R_3 \leftarrow R_3 + 2R_2$ $$ \begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & -1 & -5 & | & 2 \\ 0 & 0 & -10 & | & 5 \end{bmatrix} $$ Step 4: Back Substitution From $R_3$: $-10z = 5 \implies z = -\frac{1}{2}$ From $R_2$: $-y - 5z = 2 \implies -y - 5(-\frac{1}{2}) = 2 \implies -y + \frac{5}{2} = 2 \implies -y = -\frac{1}{2} \implies y = \frac{1}{2}$ From $R_1$: $x + y + z = 1 \implies x + \frac{1}{2} - \frac{1}{2} = 1 \implies x = 1$ Solution: $x=1, y=\frac{1}{2}, z=-\frac{1}{2}$ Taylor Series for ODEs The Taylor series expansion can be used to find an approximate solution to an initial value problem (IVP) for an ordinary differential equation (ODE). Taylor Series Formula The Taylor series for a function $y(x)$ around $x=a$ is given by: $$ y(x) = y(a) + y'(a)(x-a) + \frac{y''(a)}{2!}(x-a)^2 + \frac{y'''(a)}{3!}(x-a)^3 + \dots $$ For an IVP, we usually expand around the initial point $x=0$, so $a=0$: $$ y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots $$ Steps for Taylor Series Solution Identify $y(0)$: This is given by the initial condition. Calculate $y'(0)$: Use the differential equation and $y(0)$. Calculate higher derivatives $y''(0), y'''(0), \dots$: Differentiate the ODE repeatedly and substitute $x=0, y=y(0), y'=y'(0), \dots$ Substitute into Taylor Series: Plug the calculated derivative values into the Taylor series formula. Example: Solve $\frac{dy}{dx} = x - y^2$, with $y(0) = 1$ Step 1: $y(0)$ Given $y(0) = 1$. Step 2: First derivative $y'(0)$ $y' = x - y^2$ At $x=0, y=1$: $y'(0) = 0 - (1)^2 = -1$. Step 3: Second derivative $y''(0)$ $y'' = \frac{d}{dx}(x - y^2) = 1 - 2y y'$ (using chain rule for $y^2$) At $x=0, y=1, y'=-1$: $y''(0) = 1 - 2(1)(-1) = 1 + 2 = 3$. Step 4: Third derivative $y'''(0)$ $y''' = \frac{d}{dx}(1 - 2yy') = -2(y'y' + y y'') = -2((y')^2 + y y'')$ (using product rule for $yy'$) At $x=0, y=1, y'=-1, y''=3$: $y'''(0) = -2[(-1)^2 + (1)(3)] = -2[1 + 3] = -8$. Step 5: Taylor series up to $x^3$ Substitute values into $y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$ $$ y(x) = 1 + (-1)x + \frac{3}{2!}x^2 + \frac{-8}{3!}x^3 + \dots $$ $$ y(x) = 1 - x + \frac{3}{2}x^2 - \frac{8}{6}x^3 + \dots $$ $$ y(x) = 1 - x + \frac{3}{2}x^2 - \frac{4}{3}x^3 + \dots $$