diff --git a/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf b/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf index ce63095..c7fce0d 100644 Binary files a/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf and b/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf differ diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/00_intro.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/00_intro.tex index 7dd78c4..ffaf0d3 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/00_intro.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/00_intro.tex @@ -5,3 +5,5 @@ $f' - a = 0$ has only solution $f(x) = \int_{x_0}^{x} a(t) \dx t$ \setcounter{all}{6} \shorttheorem Let $F: \R^2 \rightarrow \R$ be a differential function of two variables. Let $x_0 \in \R$ and $y_0 \in \R^2$. The Ordinary Differential Equation (ODE) $y' = F(x, y)$ has a unique solution $f$ defined on a ``largest'' interval $I$ that contains $x_0$ such that $y_0 = f(x_0)$ + +A diffeq is ordinary if it has only one variable and is evaluated at the same point. diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/00_intro.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/00_intro.tex index 1cc50a4..bc43a83 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/00_intro.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/00_intro.tex @@ -1,6 +1,6 @@ \newsectionNoPB \subsection{Linear Differential Equations} -An ODE is considered linear if and only if the $y$s are only scaled and not part of powers.\\ +An ODE is considered \bi{linear} if and only if the $y$s are only scaled and not part of powers.\\ \compactdef{Linear differential equation of order $k$} (order = highest derivative) $y^{(k)} + a_{k - 1}y^{(k - 1)} + \ldots + a_1 y' + a_0 y = b$, with $a_i$ and $b$ functions in $x$. If $b(x) = 0 \smallhspace \forall x$, \bi{homogeneous}, else \bi{inhomogeneous}\\ diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/02_constant-coefficient.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/02_constant-coefficient.tex index 649acf9..1a8aac6 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/02_constant-coefficient.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/diffeq/linear-ode/02_constant-coefficient.tex @@ -29,5 +29,5 @@ The homogeneous equation will then be all the elements of the set summed up.\\ \item \bi{Separation of variables} For equations of form $y' = a(y) \cdot b(x)$ (Note: Not linear), we transform into $\frac{y'}{a(y)} = b(x)$ and then integrate by substituting $y'(x) dx = dy$, changing the variable of integration. Solution: $A(y) = B(x) + c$, with $A = \int \frac{1}{a}$ and $B(x) = \int b(x)$. - To get final solution, solve for the above equation for $y$. + To get final solution, solve the above equation for $y$. \end{itemize} diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/02_differential.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/02_differential.tex index fae6e08..ca057f6 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/02_differential.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/02_differential.tex @@ -49,12 +49,16 @@ Finally, evaluate $\frac{\partial g}{\partial \phi}$ at the required points and % ──────────────────────────────────────────────────────────────────── \setLabelNumber{all}{11} \compactdef{Tangent space} The graph of the affine linear approximation $g(x) = f(x_0) + u(x - x_0)$, or the set -\vspace{-0.75pc} +\rmvspace \begin{align*} \{ (x, y) \in \R^n \times \R^m : y = f(x_0) + u(x - x_0) \} \end{align*} -\dnrmvspace +\drmvspace +\shade{gray}{Computing the tangent space} Also called the \bi{Tangent plane} in 3D. +We only need to compute $g(x) = f(x_0) + J_f(x_0) \cdot (x - x_0)$, where both $x$ and $x_0$ are vectors (and $x_0$ is the point at which we compute the tangent space). +All there is left to do is state the space: $\{ (x, y, \ldots) \in \R^n | z = g() \}$ + % ──────────────────────────────────────────────────────────────────── \stepLabelNumber{all} \compactdef{Directional derivative} $f$ has a directional derivative $w \in \R^m$ in the direction of $v \in \R^n$, @@ -71,7 +75,7 @@ Suppose we know the dir. der. $w_1$ and $w_2$ in directions $v_1$ and $v_2$, the \shade{gray}{Computing a directional derivative} Always normalize the vector! We can compute a directional derivative using the differential $\limit{h}{0} \frac{f(x_0 + hv) - f(x_0)}{h}$ or using a $1$-dimensional helper function $g: h \mapsto f(x_0 + hv)$, calculating the derivative of it and evaluating $g'(0)$. That corresponds to the directional derivative. E.g. for function $f: x, y \mapsto x^2 + y^2$, we have $g: h \mapsto (x_0 + h)^2 + (y_0 + h)^2$. -A final option is to compute it using a matrix-vector product: $D_v f(x_0) = J_f(x_0) v$ +An \bi{easy option} is to use this property: $D_v f(x_0) = J_f(x_0) \cdot v = \nabla f \cdot v$ \rmvspace \begin{center} diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/04_change_of_variable.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/04_change_of_variable.tex index b0d8916..3418892 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/04_change_of_variable.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/04_change_of_variable.tex @@ -1,4 +1,4 @@ -\newsectionNoPB +\newsection \subsection{Change of variable} The idea is to substitute variables for others that make the equation easier to solve. A common example is to switch to polar coordinates from cartesian coordinates, as already demonstrated with continuity checks diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/05_taylor_polynomials.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/05_taylor_polynomials.tex index 58fdcc1..723ea83 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/05_taylor_polynomials.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/05_taylor_polynomials.tex @@ -1,4 +1,4 @@ -\newsection +\newsectionNoPB \subsection{Taylor polynomials} \compactdef{Taylor polynomials} Let $f : X \rightarrow \R$ with $f \in C^k(X, \R)$ and $y \in X$. The Taylor-Polynomial of order $k$ of $f$ at $y$ is: diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex index 27f5f4e..9b9263b 100644 --- a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex +++ b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex @@ -27,7 +27,7 @@ $\gamma_i$ as above, then Thus, the sum of all line integrals is the same thing as the Riemann-Integral of the curl. We can use Green's Formula to compute integrals. For that we need the set of curves that define the set. -For the unit circle, that is just one curve, +For the \bi{unit circle}, that is just one curve, being $\gamma(t) = \begin{pmatrix} R \cdot \cos(t) \\ R \cdot \sin(t) @@ -53,9 +53,8 @@ That set is derived from the image that is given for the line. Be cognizant of what direction the integral goes, if the set is on the right hand side of the curve, the final result has to be negated to change the direction of the integral. If the curve doesn't fully enclose the set, then we can simply compute the line integrals of the missing sections and subtract them from the final result. -We can also use known formulas to compute the area of discs, etc (like $r^2 * \pi$ for a circle). -To calculate the area enclosed by a curve using Green's formua, we can use the vector field -% TODO: Finish +We can also use known formulas to compute the area of discs, etc (like $r^2 \cdot \pi$ for a circle). +To calculate the area enclosed by a curve using Green's formua, if not given a vector field, we can use the vector field $F(x, y) = (0, x)$. \shade{gray}{Center of mass} The center of mass of an object $\cU$ is given by $\displaystyle \overline{x}_i = \frac{1}{\text{Vol}(\cU)} \int_{\cU} x_i \dx x$.