updated jump processes

Mathematics/2nd/Numerical_methods/Numerical_methods.tex

@@ -317,6 +317,7 @@
       \centering
       \includestandalone[mode=image|tex,width=0.95\linewidth]{Images/runge}
       \captionof{figure}{Runge's phenomenon. In this case $f(x)=\frac{1}{1+25x^2}$. $p_5(x)$ is the 5th-order Lagrange interpolating polynomial with equally-spaced interpolating points; $p_9(x)$, the 9th-order Lagrange interpolating polynomial with equally-spaced interpolating points, and $p_{13}(x)$, the 13th-order Lagrange interpolating polynomial with equally-spaced interpolating points.}
+      \label{NM:fig_runge}
     \end{minipage}
   \end{center}
   \begin{definition}[Spline]
@@ -438,7 +439,7 @@
     \end{gather*}
     with $\alpha_n=\frac{\langle\phi_n,x\phi_n\rangle}{\langle\phi_n,\phi_n\rangle}$ $\forall n\in\NN\cup\{0\}$ and $\beta_n=\frac{\langle\phi_n,\phi_n\rangle}{\langle\phi_{n-1},\phi_{n-1}\rangle}$ $\forall n\in\NN$.
   \end{theorem}
-  \begin{definition}[Chebyshev polynomials]
+  \begin{definition}[Chebyshev polynomials]\label{NM:chebyshev_poly}
     \emph{Chebyshev polynomials} $T_n$ are the orthogonal polynomials defined on $[-1,1]$ with the weight $\omega(x)=\frac{1}{\sqrt{1-x^2}}$. These can be defined recursively as:
     \begin{gather*}
       T_0(x)=1\quad T_1(x)=x\\T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)
@@ -563,12 +564,12 @@
     Let's start with the $\norm{\cdot}_1$ and $\norm{\cdot}_\infty$ norms. Let $\displaystyle A_1:=\max_{1\leq j\leq n} \sum_{i=1}^n\abs{a_{ij}}$ and $\displaystyle A_\infty:=\max_{1\leq i\leq n} \sum_{j=1}^n\abs{a_{ij}}$ and suppose they are attained at $j=j_0$ and $i=i_0$. Then, for all $\vf{v}=(v_j),\vf{u}=(u_j)\in\RR^n$ such that $\norm{\vf{v}}_1=\norm{\vf{u}}_\infty=1$ we have:
     \begin{align*}
       \begin{split}
-        \norm{\vf{Av}}_1      & =\sum_{i=1}^n\abs{\sum_{j=1}^n a_{ij}v_j}\leq\sum_{j=1}^n\abs{v_j}\sum_{i=1}^n\abs{a_{ij}}\leq\\
-        &\hspace*{4.5cm}\leq \sum_{j=1}^n\abs{v_j}A_1 = A_1
+        \norm{\vf{Av}}_1 & =\sum_{i=1}^n\abs{\sum_{j=1}^n a_{ij}v_j}\leq\sum_{j=1}^n\abs{v_j}\sum_{i=1}^n\abs{a_{ij}}\leq \\
+                         & \hspace*{4.5cm}\leq \sum_{j=1}^n\abs{v_j}A_1 = A_1
       \end{split} \\
       \begin{split}
-        \norm{\vf{Au}}_\infty & =\max_{1\leq i\leq n}\abs{\sum_{j=1}^n a_{ij}u_j}\leq\max_{1\leq i\leq n}\sum_{j=1}^n\abs{a_{ij}}\abs{u_j}\leq\\
-        &\hspace*{4cm}\leq \max_{1\leq i\leq n}\sum_{j=1}^n\abs{a_{ij}}= A_\infty
+        \norm{\vf{Au}}_\infty & =\max_{1\leq i\leq n}\abs{\sum_{j=1}^n a_{ij}u_j}\leq\max_{1\leq i\leq n}\sum_{j=1}^n\abs{a_{ij}}\abs{u_j}\leq \\
+                              & \hspace*{4cm}\leq \max_{1\leq i\leq n}\sum_{j=1}^n\abs{a_{ij}}= A_\infty
       \end{split}
     \end{align*}
     And taking $\vf{v}=\vf{e}_{j_0}$ and $\vf{u}=(\sign{a_{i_01}}, \ldots, \sign{a_{i_0n}})$ we have that $\norm{\vf{Av}}_1=A_1$ and $\norm{\vf{Au}}_\infty=A_\infty$. So $\norm{\vf{A}}_1=A_1$ and $\norm{\vf{A}}_\infty=A_\infty$. Now, let's do the $\norm{\cdot}_2$ norm. Observe that $\transpose{\vf{A}}\vf{A}$ is symmetric, and therefore it diagonalizes in an orthonormal basis of eigenvectors $\vf{v}_1, \ldots, \vf{v}_n$ with eigenvalues $\lambda_1, \ldots, \lambda_n$. Note that for each of these eigenvectors we have:

Mathematics/4th/Harmonic_analysis/Harmonic_analysis.tex

@@ -75,10 +75,10 @@
     Let $f,g\in L^1(\RR)$ and $\alpha,\beta\in\RR$. Then:
     \begin{enumerate}
       \item $\widehat{(\alpha f+\beta g)}(\xi)=\alpha\widehat{f}(\xi)+\beta \widehat{g}(\xi)$
-            \item\label{HA:FTprop2} Let $h\in\RR$. We define $T_hf(x)=f(x+h)$. Then: $$\widehat{T_hf}(\xi)=\exp{2\pi\ii \xi h}\widehat{f}(\xi)$$
-            \item\label{HA:FTprop3} If $g(x)=\exp{2\pi\ii x h}f(x)$, then: $$\widehat{g}(\xi)=\widehat{f}(\xi-h)$$
-            \item\label{HA:FTprop4} If $\lambda\in\RR^*$, then: $$\frac{1}{\lambda}\widehat{f\left(\frac{x}{\lambda}\right)}(\xi)=\widehat{f}(\lambda\xi)$$
-            \item\label{HA:FTprop5} If $g(x)=\overline{f(x)}$, then: $$\widehat{g}(\xi)=\overline{\widehat{f}(-\xi)}$$
+      \item\label{HA:FTprop2} Let $h\in\RR$. We define $T_hf(x)=f(x+h)$. Then: $$\widehat{T_hf}(\xi)=\exp{2\pi\ii \xi h}\widehat{f}(\xi)$$
+      \item\label{HA:FTprop3} If $g(x)=\exp{2\pi\ii x h}f(x)$, then: $$\widehat{g}(\xi)=\widehat{f}(\xi-h)$$
+      \item\label{HA:FTprop4} If $\lambda\in\RR^*$, then: $$\frac{1}{\lambda}\widehat{f\left(\frac{x}{\lambda}\right)}(\xi)=\widehat{f}(\lambda\xi)$$
+      \item\label{HA:FTprop5} If $g(x)=\overline{f(x)}$, then: $$\widehat{g}(\xi)=\overline{\widehat{f}(-\xi)}$$
     \end{enumerate}
   \end{proposition}
   \begin{sproof}
@@ -139,12 +139,12 @@
     $$\lim_{n\to\infty}f^{(k-1)}(a_n)=\lim_{n\to\infty}f^{(k-1)}(b_n)=0$$
     Hence using integration by parts:
     \begin{align*}
-      \widehat{f^{(k)}}(\xi) & =\lim_{n\to\infty}\int_{a_n}^{b_n} f^{(k)}(x)\exp{-2\pi\ii \xi x}\dd{x} \\
+      \widehat{f^{(k)}}(\xi) & =\lim_{n\to\infty}\int_{a_n}^{b_n} f^{(k)}(x)\exp{-2\pi\ii \xi x}\dd{x}   \\
       \begin{split}
-        & =\lim_{n\to\infty}  f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\Big|_{a_n}^{b_n}\dd{x} +\\
-        &\hspace{2cm}+2\pi\ii \xi\lim_{n\to\infty}\int_{a_n}^{b_n}f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\dd{x}
-      \end{split}  \\
-                             & =\left(2\pi\ii \xi\right)\widehat{f^{(k-1)}}(n)                         \\
+         & =\lim_{n\to\infty}  f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\Big|_{a_n}^{b_n}\dd{x} +                   \\
+         & \hspace{2cm}+2\pi\ii \xi\lim_{n\to\infty}\int_{a_n}^{b_n}f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\dd{x}
+      \end{split} \\
+                             & =\left(2\pi\ii \xi\right)\widehat{f^{(k-1)}}(n)                           \\
                              & ={\left(2\pi\ii \xi\right)}^k\widehat{f}(\xi)
     \end{align*}
   \end{proof}
@@ -310,7 +310,7 @@
             \begin{multline*}
               \lim_{R\to \infty}\sup_{\abs{x}\geq \delta}F_R(x)=\lim_{t\to 0}\sup_{\abs{x}\geq \delta}P_t(x)=\\=\lim_{t\to 0}\sup_{\abs{x}\geq \delta}W_t(x)=0
             \end{multline*}
-            \item\label{HA:propsKernelsitem4} For all $\delta>0$, we have:
+      \item\label{HA:propsKernelsitem4} For all $\delta>0$, we have:
             \begin{multline*}
               \lim_{R\to \infty}\int_{\abs{x}\geq \delta}F_R(x)\dd{x}=\lim_{t\to 0}\int_{\abs{x}\geq \delta}P_t(x)\dd{x}=\\=\lim_{t\to 0}\int_{\abs{x}\geq \delta}W_t(x)\dd{x}=0
             \end{multline*}
@@ -371,8 +371,8 @@
                                     & \leq\int_{-\infty}^{\infty}\!\!{\left[\int_{-\infty}^{\infty}\!{\phi_\varepsilon(\vf{y})}^p\abs{f(\vf{x}-\vf{y})-f(\vf{x})}^p\dd{\vf{x}}\right]}^{\!\frac{1}{p}}\!\!\dd{\vf{y}} \\
                                     & = \int_{-\infty}^{\infty}\phi_\varepsilon(\vf{y})\norm{f-T_{-\vf{y}}f}_p\dd{\vf{y}}                                                                                             \\
       \begin{split}
-        & \leq \int_{\abs{\vf{y}}< \delta}\phi_\varepsilon(\vf{y})\norm{f-T_{-\vf{y}}f}_p\dd{\vf{y}}+\\
-        &\hspace{3cm}+2\norm{f}_p\int_{\abs{\vf{y}}\geq\delta}\phi_\varepsilon(\vf{y})\dd{\vf{y}}
+         & \leq \int_{\abs{\vf{y}}< \delta}\phi_\varepsilon(\vf{y})\norm{f-T_{-\vf{y}}f}_p\dd{\vf{y}}+ \\
+         & \hspace{3cm}+2\norm{f}_p\int_{\abs{\vf{y}}\geq\delta}\phi_\varepsilon(\vf{y})\dd{\vf{y}}
       \end{split}
     \end{align*}
     Given $\varepsilon>0$, by \mcref{HA:translated} $\exists\delta>0$ such that the first integral is bounded by $\varepsilon$. Now use this $\delta$ and \mcref{HA:propsKernelsitem4} to conclude that the second integral goes to 0 as $R\to\infty$.
@@ -839,8 +839,8 @@
     \begin{align*}
       \abs{\phi_\varepsilon(\varphi)-\delta_{\vf{0}}(\varphi)} & \leq \int_\Omega \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}} \\
       \begin{split}
-        & =\int_{\norm{\vf{x}}<\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}+\\
-        &\hspace{1cm}+\int_{\norm{\vf{x}}\geq\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}
+         & =\int_{\norm{\vf{x}}<\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}+               \\
+         & \hspace{1cm}+\int_{\norm{\vf{x}}\geq\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}
       \end{split}
     \end{align*}
     Now use the properties of approximation of identity to see that each interval goes to zero as $\varepsilon\to 0$.
@@ -1063,15 +1063,15 @@
   \begin{proof}
     Clearly $T*\psi$ is linear. Let $\varphi_n\overset{\mathcal{S}}{\longrightarrow}0$. Then, it suffices to see that $\tilde\psi*\varphi_n\overset{\mathcal{S}}{\longrightarrow}0$. For the sake of simplicity we only do the case $d=1$. For all $\alpha,\beta\in{\NN\cup\{0\}}$ we have:
     \begin{align*}
-      \abs{\vf{x}^\alpha \partial^\beta(\tilde\psi*\varphi_n)(\vf{x})} & =\abs{\vf{x}^\alpha (\partial^\beta\tilde\psi*\varphi_n)(\vf{x})}                                                                 \\
-                                                                       & \leq \int_{\RR^d} \abs{\vf{x}^\alpha \partial^\beta\psi(\vf{y})\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}                               \\
+      \abs{\vf{x}^\alpha \partial^\beta(\tilde\psi*\varphi_n)(\vf{x})} & =\abs{\vf{x}^\alpha (\partial^\beta\tilde\psi*\varphi_n)(\vf{x})}                                                                  \\
+                                                                       & \leq \int_{\RR^d} \abs{\vf{x}^\alpha \partial^\beta\psi(\vf{y})\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}                                \\
       \begin{split}
-        & \leq 2^m\!\!\int_{\RR^d}\!\abs{\vf{x}-\vf{y}}^\alpha \abs{\partial^\beta\psi(\vf{y})}\abs{\varphi_n(\vf{x}-\vf{y})}\!\dd{\vf{y}} \\
-        &\quad\;+2^m\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})} \abs{\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}} \\
+         & \leq 2^m\!\!\int_{\RR^d}\!\abs{\vf{x}-\vf{y}}^\alpha \abs{\partial^\beta\psi(\vf{y})}\abs{\varphi_n(\vf{x}-\vf{y})}\!\dd{\vf{y}} \\
+         & \quad\;+2^m\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})} \abs{\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}            \\
       \end{split} \\
       \begin{split}
-        & \leq 2^m\sup_{\vf{x}\in\RR^d}\abs{\vf{x}}^\alpha\abs{\varphi_n(\vf{y})}\int_{\RR^d}  \abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}} \\
-        &\;\;\;+2^m\sup_{\vf{x}\in\RR^d}\abs{\varphi_n(\vf{y})}\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}} \\
+         & \leq 2^m\sup_{\vf{x}\in\RR^d}\abs{\vf{x}}^\alpha\abs{\varphi_n(\vf{y})}\int_{\RR^d}  \abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}}  \\
+         & \;\;\;+2^m\sup_{\vf{x}\in\RR^d}\abs{\varphi_n(\vf{y})}\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}} \\
       \end{split}
     \end{align*}
     where in the second inequality we have used \mcref{HA:lemma_aMbM} with $m=\abs{\alpha}+1$. Note that this latter terms tend to zero as $n\to\infty$ because of the properties of the Schwartz space.
@@ -1434,10 +1434,10 @@
     Averaging those terms we have:
     \begin{align*}
       \begin{split}
-        \widehat{f^2}+2\F{(\H(f\H f))} & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)[1+m(\xi)\cdot\\
-          &\hspace{2cm}\cdot(m(\xi-\eta)+m(\eta))]\dd{\eta}
+        \widehat{f^2}+2\F{(\H(f\H f))} & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)[1+m(\xi)\cdot \\
+                                       & \hspace{2cm}\cdot(m(\xi-\eta)+m(\eta))]\dd{\eta}
       \end{split} \\
-       & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)m(\xi-\eta)m(\eta)\dd{\eta}                 \\
+       & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)m(\xi-\eta)m(\eta)\dd{\eta}                              \\
        & =\widehat{\H f}*\widehat{\H f}=\widehat{(\H f)^2}
     \end{align*}
     where the second equality follows for all $\xi,\eta\in\RR^2\setminus\{(0,0)\}$

Mathematics/4th/Linear_models/Images/confidence_band.tex

@@ -34,16 +34,16 @@
     \def\beta{-\alpha+\n}
     \def\betb{-\alphb+\n}
     \addplot[domain=0:1,name path=A,color_green3,thick]{exp(x-\alpha)+\beta};
-    \addplot[domain=0:1,name path=B,\sta,thick]{exp(x-\alphb)+\betb};
+    \addplot[domain=0:1,name path=B,\pro,thick]{exp(x-\alphb)+\betb};
 
     \addplot[domain=0:1,name path=C,color_green3,thick]{log10(1+x-0.65)/log10(e)+0.85};
-    \addplot[domain=0:1,name path=D,\sta,thick]{log10(1+x-0.75)/log10(e)+0.75};
-    % \addplot[domain=0:1,name path=B,\sta]{exp(x-\alphb)+\betb};
+    \addplot[domain=0:1,name path=D,\pro,thick]{log10(1+x-0.75)/log10(e)+0.75};
+    % \addplot[domain=0:1,name path=B,\pro]{exp(x-\alphb)+\betb};
     \addplot[color_green3!30,opacity=0.5] fill between[
         of=A and C,
         soft clip={domain=0:1},
       ];
-    \addplot[\sta!30,opacity=0.5] fill between[
+    \addplot[\pro!30,opacity=0.5] fill between[
         of=B and D,
         soft clip={domain=0:1},
       ];

Mathematics/4th/Partial_differential_equations/Images/waves_char_solve.tex

@@ -65,7 +65,7 @@
         (4,4-\eps)
         (4,6-\eps)
       };
-    \addplot[thin,mark=*,only marks, mark size=0.075cm,\sta] coordinates {
+    \addplot[thin,mark=*,only marks, mark size=0.075cm,\pro] coordinates {
         (3,0.5-\eps)
         (3,3.5-\eps)
       };