Improved syntax.

docs/en/mathematics/multivariable-calculus/differentation.md

@@ -22,7 +22,12 @@ $$
     \partial_{12} f(P) = \partial_{21} f(P),
 $$
 
-*Proof*: will be added later.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 ## Total derivatives
 
@@ -50,6 +55,13 @@ $$
 
 with $\nabla f(\mathbf{a})$ the gradient of $f$.
 
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
 ## Chain rule
 
 *Definition*: let $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) and let $f: D \to \mathbb{R}$, also let $g: \mathbb{R} \to \mathbb{R}$ given by
@@ -78,7 +90,10 @@ The direction of the gradient is the direction of steepest increase of $f$ at $\
 
 *Theorem*: gradients are orthogonal to level lines and level surfaces.
 
-*Proof*: let $\mathbf{r}(t) = \big(x(t),\; y(t) \big)^T$ be a parameterization of the level curve of $f$ such that $\mathbf{r}(0) = \mathbf{a}$. Then for all $t$ near $0$, $f(\mathbf{r}(t)) = f(\mathbf{a})$. Differentiating this equation with respect to $t$ using the chain rule, we obtain
+<details>
+<summary><em>Proof</em>:</summary>
+
+let $\mathbf{r}(t) = \big(x(t),\; y(t) \big)^T$ be a parameterization of the level curve of $f$ such that $\mathbf{r}(0) = \mathbf{a}$. Then for all $t$ near $0$, $f(\mathbf{r}(t)) = f(\mathbf{a})$. Differentiating this equation with respect to $t$ using the chain rule, we obtain
 
 $$
     \partial_1 f(\mathbf{x}) \dot x(t) + \partial_2 f(\mathbf{x}) \dot y(t) = 0,
@@ -91,6 +106,8 @@ $$
 $$
 
 obtaining that $\nabla f$ is orthogonal to $\mathbf{\dot r}$. 
+</details>
+<br>
 
 ## Directional derivatives
 
@@ -138,4 +155,10 @@ We have two interpretations:
 * the composition of linear maps,
 * the matrix multiplication of the Jacobian.
 
-*Proof*: will be added later.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+    

docs/en/mathematics/multivariable-calculus/implicit-equations.md

@@ -26,7 +26,12 @@ $$
     \phi' (x) = - \frac{\partial_1 f\big(x,\phi(x)\big)}{\partial_2 f\big(x,\phi(x)\big)}.
 $$
 
-*Proof*: will be added later.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 ## General case
 
@@ -51,4 +56,9 @@ $$
     D \mathbf{\phi}(\mathbf{x}) = - \Big(D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) \Big)^{-1} D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big).
 $$
 
-*Proof*: will be added later.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>

docs/en/mathematics/multivariable-calculus/taylor-polynomials.md

@@ -24,12 +24,18 @@ $$
     f(\mathbf{x}) = T(\mathbf{x}) + \frac{1}{(n+1)!} \partial_\mathbf{h}^{n+1} f(\mathbf{a} + \theta \mathbf{h}).
 $$
 
-*Proof*: Apply Taylor’s theorem in 1D and the chain rule to the function $\phi : [0, 1] \to \mathbb{R}$ given by
+<details>
+<summary><em>Proof</em>:</summary>
+
+apply Taylor’s theorem in 1D and the chain rule to the function $\phi : [0, 1] \to \mathbb{R}$ given by
 
 $$
     \phi(\theta) := f(\mathbf{a} + \theta \mathbf{h}).
 $$
 
+</details>
+<br>
+
 ## Other methods
 
 Creating multivariable Taylor polynomials by using 1D Taylor polynomials of the different variables and composing them. 

docs/en/mathematics/ordinary-differential-equations/laplace-transform.md

@@ -18,7 +18,12 @@ $$
 
 on the interval where both are defined. 
 
-*Proof*: will be added sometime.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 If $c \in \mathbb{R}$ then $cf$ also has a Laplace transform and, 
 
@@ -40,7 +45,12 @@ $$
 
 on this interval
 
-*Proof*: will be added sometime.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 **More shifting**: let $a>0$, if $f$ has a Laplace transform $F$ on $s_0, \infty$ then the function $g$ given by
 
@@ -56,7 +66,12 @@ $$
 
 on this interval.
 
-*Proof*: will be added sometime.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 **Scaling**: let $a > 0$. If $f$ has a Laplace transform $F$ on $(s_0, \infty)$ then the function $g$ given by 
 
@@ -72,7 +87,12 @@ $$
 
 on this interval.
 
-*Proof*: will be added sometime.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 **Derivatives**: if $f$ has a derivative $g$ having a Laplace transform $G$ on the interval $(s_0,\infty)$ then $f$ has a Laplace transform on the same interval, and
 
@@ -86,7 +106,10 @@ $$
 \mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0)
 $$
 
-*Proof*: for large enough $s$, the case $n=1$ follows by integration by parts
+<details>
+<summary><em>Proof</em>:</summary>
+
+for large enough $s$, the case $n=1$ follows by integration by parts
 
 $$
 \begin{align*} \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ &= sF(s) - f(0) \end{align*},
@@ -98,6 +121,9 @@ $$
 \begin{align*} \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt , \\ &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0) \end{align*}.
 $$
 
+</details>
+<br>
+
 ## Examples
 
 **Solving a second order linear ODE**: with $y: \mathbb{K} \to \mathbb{R}$ given by 

docs/en/mathematics/ordinary-differential-equations/second-order-ode.md

@@ -52,7 +52,12 @@ $$
 y(t) = (c_1 + c_2t) e^{\lambda_1 t}.
 $$
 
-*Proof*: will at some point be added.
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
 
 #### Example
 
@@ -75,19 +80,25 @@ $$
 *Theorem*: let $y_p$ be a particular solution to $(*)$. Then the general solution to $(*)$ is given by
 
 $$
-y = y_H + y_p,
+y = y_h + y_p,
 $$
 
-with $y_H$ the solution to the homegeneous case.
+with $y_h$ the solution to the homegeneous case.
 
-*Proof*: let $y$ be a solution to $(*)$, then $L[y - y_p] = L[y] - L[y_p] = f - f = 0$. Therefore $y = (y - y_p) + y_p = y_H + y_p$.
+<details>
+<summary><em>Proof</em>:</summary>
+
+let $y$ be a solution to $(*)$, then $L[y - y_p] = L[y] - L[y_p] = f - f = 0$. Therefore $y = (y - y_p) + y_p = y_h + y_p$.
+
+</details>
+<br>
 
 #### Method of variation of parameters
 
 We need the general solution to the homogeneous case 
 
 $$
-y_H(t) = c_1 y_1(t) + c_2 y_2(t), \qquad c_1,c_2 \in \mathbb{C}. 
+y_h(t) = c_1 y_1(t) + c_2 y_2(t), \qquad c_1,c_2 \in \mathbb{C}. 
 $$
 
 Ansatz: let $y_p(t) = c_1(t) y_2(t) + c_2(t) y_2(t)$, then taking the derivative of $y_p(t)$

docs/en/mathematics/ordinary-differential-equations/systems-of-linear-ode.md

@@ -44,13 +44,19 @@ $$
 \mathbf{\dot y}(t) = A \mathbf{y}(t) + \mathbf{f}(t), \qquad t \in I. \qquad (*)
 $$
 
-*Theorem*: let $\mathbf{y}_p: I \to \mathbb{R}^n$ a particular solution for $(*)$ and $\mathbf{y}_H$ the general solution to the homegeneous system. Then the general solutions of the inhomogeneous system $(*)$ is given by
+*Theorem*: let $\mathbf{y}_p: I \to \mathbb{R}^n$ a particular solution for $(*)$ and $\mathbf{y}_h$ the general solution to the homegeneous system. Then the general solutions of the inhomogeneous system $(*)$ is given by
 
 $$
-\mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_H(t), \qquad t \in I
+\mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_h(t), \qquad t \in I
 $$
 
-*Proof*: similar to 1d case, will possibly be added later.
+<details>
+<summary><em>Proof</em>:</summary>
+
+similar to 1d case, will be added later.
+
+</details>
+<br>
 
 ### Method of variation of parameters