Update notes

Author: cpraveen

leastsquare.md

@@ -710,7 +710,7 @@ Since $F(\theta)$ is even, the sine terms vanish, and this is also the full Four
 
 ---
 
-Now if $f(x)$ is piecewise continuous, so is $F(\theta)$, and we know that
+Now if $f(x)$ is piecewise continuous, so is $F(\theta)$, and we know from [](#thm:fdecay) that
 
 $$
 |a_j| = \order{\frac{1}{j}}, \qquad j \to \infty
@@ -720,7 +720,7 @@ In this case, we have $\norm{f - C_n}_2 \to 0$.
 
 ---
 
-If $f \in \cts^{\nu-1}[-1,1]$ for some $\nu \ge 1$ and $f^{(\nu)}$ is piecewise continuous, then $F \in \cts^{\nu-1}_p[-\pi,\pi]$ and $F^{(\nu)}$ is piecewise continuous. This means that
+If $f \in \cts^{\nu-1}[-1,1]$ for some $\nu \ge 1$ and $f^{(\nu)}$ is piecewise continuous, then $F \in \cts^{\nu-1}_p[-\pi,\pi]$ and $F^{(\nu)}$ is piecewise continuous. From [](#thm:fdecay), this means that
 
 $$
 |a_j| = \order{\frac{1}{j^{\nu+1}}}, \qquad j \to \infty

myst.yml

@@ -60,10 +60,9 @@ project:
       - file: aquad.md
     - file: ode.md
       children:
-      - file: ode/euler0.md
-      - file: ode/euler1.md
+      - file: ode/euler.md
       - file: ode/trapezoid1.md
-      - file: ode/multistep_instability.md
+      - file: ode/multistep.md
       - file: ode/fe_be_stability.md
       - file: ode/euler2.md
       - file: ode/ode_stability.md

newton_cotes.md

@@ -232,5 +232,5 @@ print("sum(w)   = ", sum(w))
 print("sum(|w|) = ", sum(abs(w)))
 ```
 
-Because the weights are large in size, there is also more roundoff error when we add them, the sum is not exactly one.
+Because the weights are large in size and take both signs, there is also more roundoff error when we add them, and the sum is not exactly one. This error increases as we use more points.
 :::

ode/euler1.md ode/euler.mdode/euler1.md renamed to ode/euler.md

@@ -1,21 +1,98 @@
 ---
-jupytext:
-  text_representation:
-    extension: .md
-    format_name: myst
-    format_version: 0.13
-    jupytext_version: 1.16.7
+exports:
+  - format: pdf
 kernelspec:
-  display_name: Python 3
+  display_name: Python 3 (ipykernel)
   language: python
   name: python3
+numbering:
+  code: false
+  equation: true
+  title: true
+  headings: true
 ---
 
-# ODE using forward Euler
+# Forward Euler
+
+```{include} ../math.md
+```
+
+```{code-cell}
+import numpy as np
+from matplotlib import pyplot as plt
+```
+
+:::{prf:example}
+Consider the ODE problem
+
+\begin{gather}
+y' = y, \qquad t \ge 0 \\
+y(0) = 1
+\end{gather}
+
+whose exact solution is
+
+$$
+y(t) = \exp(t)
+$$
+
+Right hand side function
+
+```{code-cell}
+def f(t,y):
+    return y
+```
+
+Exact solution
+
+```{code-cell}
+def yexact(t):
+    return np.exp(t)
+```
+
+This implements Euler method
+
+$$
+y_n = y_{n-1} + h f(t_{n-1},y_{n-1})
+$$
+
+```{code-cell}
+def euler(t0,T,y0,h):
+    N = int((T-t0)/h)
+    y = np.zeros(N)
+    t = np.zeros(N)
+    y[0] = y0
+    t[0] = t0
+    for n in range(1,N):
+        y[n] = y[n-1] + h*f(t[n-1],y[n-1])
+        t[n] = t[n-1] + h
+    return t, y
+```
+
+```{code-cell}
+t0,y0,T = 0.0,1.0,5.0
+
+te = np.linspace(t0,T,100)
+ye = yexact(te)
+plt.plot(te,ye,'--',label='Exact')
+
+H = [0.2,0.1,0.05]
+
+for h in H:
+    t,y = euler(t0,T,y0,h)
+    plt.plot(t,y,label="h="+str(h))
+
+plt.legend(loc=2)
+plt.xlabel('t')
+plt.ylabel('y')
+plt.grid(True);
+```
+:::
 
 +++
 
-Conside the ODE
+:::{prf:example}
+Consider the ODE
 
 $$
 y' = -y + 2 \exp(-t) \cos(2t)
@@ -33,21 +110,16 @@ $$
 y(t) = \exp(-t) \sin(2t)
 $$
 
-```{code-cell} ipython3
-import numpy as np
-from matplotlib import pyplot as plt
-```
-
 Right hand side function
 
-```{code-cell} ipython3
+```{code-cell} 
 def f(t,y):
     return -y + 2.0*np.exp(-t)*np.cos(2.0*t)
 ```
 
 Exact solution
 
-```{code-cell} ipython3
+```{code-cell}
 def yexact(t):
     return np.exp(-t)*np.sin(2.0*t)
 ```
@@ -57,7 +129,7 @@ $$
 y_n = y_{n-1} + h f(t_{n-1},y_{n-1})
 $$
 
-```{code-cell} ipython3
+```{code-cell}
 def euler(t0,T,y0,h):
     N = int((T-t0)/h)
     y = np.zeros(N)
@@ -70,9 +142,9 @@ def euler(t0,T,y0,h):
     return t, y
 ```
 
-## Solve for a given h
+**Solve for a given h**
 
-```{code-cell} ipython3
+```{code-cell}
 t0,y0 = 0.0,0.0
 T  = 10.0
 h  = 1.0/20.0
@@ -87,13 +159,11 @@ plt.grid(True)
 plt.title('Step size = ' + str(h));
 ```
 
-## Solve for several h
-
-+++
+**Solve for several h**
 
 Study the effect of decreasing step size. The error is plotted in log scale.
 
-```{code-cell} ipython3
+```{code-cell}
 hh = 0.1/2.0**np.arange(5)
 err= np.zeros(len(hh))
 for (i,h) in enumerate(hh):
@@ -115,3 +185,4 @@ plt.loglog(hh,err,'o-')
 plt.xlabel('h')
 plt.ylabel('Error norm');
 ```
+:::