small updates

Mathematics/4th/Dynamical_systems/Dynamical_systems.tex

@@ -159,7 +159,7 @@
     \begin{itemize}
       \item {stable node} if $\lambda_1,\lambda_2\in\RR$ and $\abs{\lambda_1},\abs{\lambda_2}<1$.
       \item {unstable node} if $\lambda_1,\lambda_2\in\RR$ and $\abs{\lambda_1},\abs{\lambda_2}>1$.
-      \item {saddle point} if $\lambda_1,\lambda_2\in\RR$ and $\abs{\lambda_1}<1$ and $\abs{\lambda_2}>1$ (or viceversa).
+      \item {saddle point} if $\lambda_1,\lambda_2\in\RR$ and $\abs{\lambda_1}<1$ and $\abs{\lambda_2}>1$ (or vice versa).
       \item {stable focus} if $\lambda_1,\lambda_2\in\CC$ and $\abs{\lambda_1}<1$.
       \item {unstable focus} if $\lambda_1,\lambda_2\in\CC$ and $\abs{\lambda_1}>1$.
     \end{itemize}

Mathematics/4th/Stochastic_processes/Stochastic_processes.tex

@@ -234,7 +234,7 @@
     with $q_0=1$ and $q_a=0$, whose solution is straightforward.
   \end{sproof}
   \begin{proposition}
-    Consider the Gambler's ruin problem. Suppose that we play against another player (and so when we lose, he wins and viceversa). Let $p_z^*$, $q_z^*$ be the respective probabilities for the other player. Then:
+    Consider the Gambler's ruin problem. Suppose that we play against another player (and so when we lose, he wins and vice versa). Let $p_z^*$, $q_z^*$ be the respective probabilities for the other player. Then:
     $$q_z+q_z^*=1$$
     Hence, $D_z\almoste{<}\infty$.
   \end{proposition}
@@ -370,9 +370,9 @@
       p_{ij}^{(n+1)} & =\Prob(X_{n+1}=j\mid X_0=i)                     \\
                      & =\sum_{k\in I}\Prob(X_{n+1}=j, X_n=k\mid X_0=i) \\
       \begin{split}
-         & =\sum_{k\in I}\Prob(X_{n}=k\mid X_0=i)\cdot          \\
-         & \hspace{2.5cm}\cdot\Prob(X_{n+1}=j\mid X_n=k, X_0=i)
-      \end{split}          \\
+        & =\sum_{k\in I}\Prob(X_{n}=k\mid X_0=i)\cdot          \\
+        & \hspace{2.5cm}\cdot\Prob(X_{n+1}=j\mid X_n=k, X_0=i)
+      \end{split}           \\
                      & =\sum_{k\in I}p_{ik}^{(n)}p_{kj}^{(1)}
     \end{align*}
     where the penultimate equality follows from \mcref{SP:lema1Markov} and the last equality follows from \mcref{SP:lema2Markov} and the Markov property because if $D=\{X_n=k, X_0=i\}$ we have that:
@@ -590,9 +590,9 @@
     \begin{align*}
       \begin{split}
         p_\ell & =\Prob_i(X_{m_1+\cdots+m_\ell}=i,X_{m_1+\cdots+m_\ell-1}\ne i,\ldots, \\
-               & \hspace{4cm}X_{m_1+\cdots+m_{\ell-1}+1}\ne i\mid A)
+        & \hspace{4cm}X_{m_1+\cdots+m_{\ell-1}+1}\ne i\mid A)
       \end{split} \\
-       & =\Prob_i(X_{m_\ell}=i,X_{m_\ell-1}\ne i,\ldots,X_{1}\mid X_{0}=i)                                     \\
+       & =\Prob_i(X_{m_\ell}=i,X_{m_\ell-1}\ne i,\ldots,X_{1}\mid X_{0}=i)           \\
        & =\Prob(\tau_i=m_\ell)
     \end{align*}
   \end{proof}
@@ -652,14 +652,14 @@
   \begin{proof}
     Note that $\{X_n=j\}\subseteq \{\tau_j\leq n\}=\bigsqcup_{m=1}^n\{\tau_j=m\}$. Hence, $\{X_n=j\}=\bigsqcup_{m=1}^n[\{X_n=j\}\cap \{\tau_j=m\}]$. Thus:
     \begin{align*}
-      p_{ij}^{(n)} & =\Prob_i(X_n=j)                                               \\
-                   & =\sum_{m=1}^n \Prob_i(X_n=j,\tau_j=m)                         \\
-                   & =\sum_{m=1}^n \Prob_i(X_n=j\mid\tau_j=m)\Prob_i(\tau_j=m)     \\
+      p_{ij}^{(n)} & =\Prob_i(X_n=j)                                              \\
+                   & =\sum_{m=1}^n \Prob_i(X_n=j,\tau_j=m)                        \\
+                   & =\sum_{m=1}^n \Prob_i(X_n=j\mid\tau_j=m)\Prob_i(\tau_j=m)    \\
       \begin{split}
-         & =\sum_{m=1}^n \Prob_i(X_n=j\mid X_m=j,X_{m-1}\ne j,\ldots, X_1\ne j)\cdot \\
-         & \hspace{7cm}\cdot f_{ij}^{(m)}
+        & =\sum_{m=1}^n \Prob_i(X_n=j\mid X_m=j,X_{m-1}\ne j,\ldots, X_1\ne j)\cdot \\
+        & \hspace{7cm}\cdot f_{ij}^{(m)}
       \end{split} \\
-                   & =\sum_{m=1}^n \Prob_j(X_{n-m}=j)f_{ij}^{(m)}                  \\
+                   & =\sum_{m=1}^n \Prob_j(X_{n-m}=j)f_{ij}^{(m)}                 \\
                    & =\sum_{m=1}^nf_{ij}^{(m)}p_{jj}^{(n-m)}
     \end{align*}
   \end{proof}
@@ -926,9 +926,9 @@
     \begin{align*}
       p_{ij}(t+s) & =\sum_{k\in I} \Prob(X_{t+s}=j,X_s=k\mid X_0=i) \\
       \begin{split}
-         & =\sum_{k\in I} \Prob(X_{t+s}=j\mid X_s=k,X_0=i)\cdot \\
-         & \hspace{3.5cm}\cdot\Prob(X_s=k\mid X_0=i)            \\
-      \end{split}       \\
+        & =\sum_{k\in I} \Prob(X_{t+s}=j\mid X_s=k,X_0=i)\cdot \\
+        & \hspace{3.5cm}\cdot\Prob(X_s=k\mid X_0=i)            \\
+      \end{split}        \\
                   & =\sum_{k\in I} p_{ik}(t)p_{kj}(s)
     \end{align*}
   \end{proof}
@@ -1357,7 +1357,7 @@
     \end{enumerate}
   \end{proposition}
   \begin{definition}
-    Let ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ be two stochastic processes. We say that ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ are \emph{stochastically equivalent} if $\forall t\in T$ we have: $$\Prob(X_t=Y_t)=1$$ In that case we also say that ${(X_t)}_{t\in T}$ is a \emph{version} of ${(Y_t)}_{t\in T}$ (or viceversa). We say that ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ are \emph{indistinguishable} if: $$
+    Let ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ be two stochastic processes. We say that ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ are \emph{stochastically equivalent} if $\forall t\in T$ we have: $$\Prob(X_t=Y_t)=1$$ In that case we also say that ${(X_t)}_{t\in T}$ is a \emph{version} of ${(Y_t)}_{t\in T}$ (or vice versa). We say that ${(X_t)}_{t\in T}$ and ${(Y_t)}_{t\in T}$ are \emph{indistinguishable} if: $$
       \Prob(X_t=Y_t\ \forall t\in T)=1
     $$
   \end{definition}

Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex

@@ -276,10 +276,10 @@
     $$
     By density, it is enough to prove the result for $u\in \mathcal{C}^1(\RR_{\geq 0})$. An easy check shows that $\bar{u}\in \mathcal{C}^1(\RR)$. Moreover, we have:
     \begin{align*}
-      {\norm{\bar{u}}_{W^{1,p}(\RR)}}^p & =\int_{\RR}{\abs{\bar{u}}^p}+{\abs{\bar{u}'}^p}                                                    \\
+      {\norm{\bar{u}}_{W^{1,p}(\RR)}}^p & =\int_{\RR}{\abs{\bar{u}}^p}+{\abs{\bar{u}'}^p}                                                   \\
       \begin{split}
-         & =\!\int_{\RR_{\geq 0}}\!{\abs{u}^p}\!+\!\!{\abs{u'}^p}\!+\!\int_{\RR_{\leq 0}}\![{\abs{-3u(-x)+4u(-x/2)}^p}+ \\
-         & \hspace{2.75cm}+{\abs{3u'(-x)-2u'(-x/2)}^p}]
+        & =\!\int_{\RR_{\geq 0}}\!{\abs{u}^p}\!+\!\!{\abs{u'}^p}\!+\!\int_{\RR_{\leq 0}}\![{\abs{-3u(-x)+4u(-x/2)}^p}+ \\
+        & \hspace{2.75cm}+{\abs{3u'(-x)-2u'(-x/2)}^p}]
       \end{split} \\
                                         & \leq C{\norm{u}_{W^{1,p}(\RR_{\geq 0})}}^p
     \end{align*}

Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex

@@ -24,7 +24,7 @@
     where we have assumed that the boundary of $\Omega$ is smooth enough to define the normal vector $\vf{n}$. The condition is called \emph{homogeneous} if $g=0$. Note that if $\vf{A}=\vf{I}_d$, then the Neumann boundary condition is just $\partial_{\vf{n}} u=g$.
   \end{definition}
   \begin{remark}
-    If the coefficients $a_{ij}\in\mathcal{C}^1$, then we convert the equation from non-divergence form to divergence form and viceversa.
+    If the coefficients $a_{ij}\in\mathcal{C}^1$, then we are able to convert the equation from non-divergence form to divergence form and vice versa.
   \end{remark}
   \begin{definition}
     Let $a_{ij},b_j,c$ be known functions on $\Omega\subseteq \RR^d$. We say that the operator
@@ -161,7 +161,7 @@
     Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then, $\dim \ker(\id-K)<\infty$.
   \end{lemma}
   \begin{proof}
-    We first prove that $\dim\ker(\id-K)<\infty$. If $\dim\ker(\id-K)=\infty$, then $\exists (u_n)\in \ker(\id-K)$ orthonormal, and thus bounded. In particular, $u_n=Ku_n$ and since $K$ is compact, we have that $(Ku_n)$ has a convergent subsequence. But:
+    If $\dim\ker(\id-K)=\infty$, then $\exists (u_n)\in \ker(\id-K)$ orthonormal, and thus bounded. In particular, $u_n=Ku_n$ and since $K$ is compact, we have that $(Ku_n)$ has a convergent subsequence. But:
     \begin{align*}
       0 & =\lim_{k\to\infty}\norm{Ku_{n_k}-Ku_{n_{k+1}}}^2        \\
         & =\lim_{k\to\infty}\norm{u_{n_k}-u_{n_{k+1}}}^2          \\
@@ -180,7 +180,7 @@
     Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then, $\im(\id-K)$ is closed.
   \end{lemma}
   \begin{proof}
-    Let $(v_n)\in \im(\id-K)$ be such that $v_n\to v\in H$. Then, $\exists (u_n)\in H$ such that $v_n=(\id-K)u_n$. Thanks to \mnameref{RFA:projection}, we can write $u_n=u_n^{\text{ker}}+ u_n^{\text{ker}^\perp}$, where $u_n^{\text{ker}}\in \ker(\id-K)$ and $u_n^{\text{ker}^\perp}\in {\ker(\id-K)}^{\perp}$. Thus, $v_n=(\id-K)u_n^{\text{ker}^\perp}$ and by \mcref{INEPDE:lemma2_fredholm}, we have:
+    Let $(v_n)\in \im(\id-K)$ be such that $v_n\to v\in H$. Then, $\exists (u_n)\in H$ such that $v_n=(\id-K)u_n$. By \mnameref{RFA:projection}, we can write $u_n=u_n^{\text{ker}}+ u_n^{\text{ker}^\perp}$, where $u_n^{\text{ker}}\in \ker(\id-K)$ and $u_n^{\text{ker}^\perp}\in {\ker(\id-K)}^{\perp}$. Thus, $v_n=(\id-K)u_n^{\text{ker}^\perp}$ and by \mcref{INEPDE:lemma2_fredholm}, we have:
     $$
       \norm{v_n-v_m}\geq c\norm{u_n^{\text{ker}^\perp}-u_m^{\text{ker}^\perp}}
     $$