2016 Update

Task/Euler-method/00DESCRIPTION

@@ -1,50 +1,61 @@
-Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in [[wp:Euler method|the wikipedia page]].
+Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value.   It is an explicit method for solving initial value problems (IVPs), as described in [[wp:Euler method|the wikipedia page]].
+
 The ODE has to be provided in the following form:
 
-:<math>\frac{dy(t)}{dt} = f(t,y(t))</math>
+::: <big><math>\frac{dy(t)}{dt} = f(t,y(t))</math></big>
 
 with an initial value
 
-:<math>y(t_0) = y_0</math>
+::: <big><math>y(t_0) = y_0</math></big>
 
-To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
+To get a numeric solution, we replace the derivative on the &nbsp; LHS &nbsp; with a finite difference approximation:
 
-:<math>\frac{dy(t)}{dt}  \approx \frac{y(t+h)-y(t)}{h}</math>
+::: <big><math>\frac{dy(t)}{dt}  \approx \frac{y(t+h)-y(t)}{h}</math></big>
 
 then solve for <math>y(t+h)</math>:
 
-:<math>y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}</math>
+::: <big><math>y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}</math></big>
 
 which is the same as
 
-:<math>y(t+h) \approx y(t) + h \, f(t,y(t))</math>
+::: <big><math>y(t+h) \approx y(t) + h \, f(t,y(t))</math></big>
 
 The iterative solution rule is then:
 
-:<math>y_{n+1} = y_n + h \, f(t_n, y_n)</math>
+::: <big><math>y_{n+1} = y_n + h \, f(t_n, y_n)</math></big>
+
+where &nbsp; <big><math>h</math></big> &nbsp; is the step size, the most relevant parameter for accuracy of the solution. &nbsp; A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
 
-<math>h</math> is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
 
 '''Example: Newton's Cooling Law'''
 
-Newton's cooling law describes how an object of initial temperature <math>T(t_0) = T_0</math> cools down in an environment of temperature <math>T_R </math>:
-
-:<math>\frac{dT(t)}{dt} = -k \, \Delta T</math>
+Newton's cooling law describes how an object of initial temperature &nbsp; <big><math>T(t_0) = T_0</math></big> &nbsp; cools down in an environment of temperature &nbsp; <big><math>T_R</math></big>:
 
+::: <big><math>\frac{dT(t)}{dt} = -k \, \Delta T</math></big>
 or
+::: <big><math>\frac{dT(t)}{dt} = -k \, (T(t) - T_R)</math></big>
 
-:<math>\frac{dT(t)}{dt} = -k \, (T(t) - T_R)</math>
-
-It says that the cooling rate <math>\frac{dT(t)}{dt}</math> of the object is proportional to the current temperature difference <math>\Delta T = (T(t) - T_R)</math> to the surrounding environment.
+<br>
+It says that the cooling rate &nbsp; <big><math>\frac{dT(t)}{dt}</math></big> &nbsp; of the object is proportional to the current temperature difference &nbsp; <big><math>\Delta T = (T(t) - T_R)</math></big> &nbsp; to the surrounding environment.
 
 The analytical solution, which we will compare to the numerical approximation, is
+::: <big><math>T(t) = T_R + (T_0 - T_R) \; e^{-k t}</math></big>
 
-:<math>T(t) = T_R + (T_0 - T_R) \; e^{-k t}</math>
 
-'''Task'''
+;Task:
+Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
+:::* &nbsp; 2 s
+:::* &nbsp; 5 s &nbsp; &nbsp; &nbsp; and
+:::* &nbsp; 10 s
+and to compare with the analytical solution.
 
-The task is to implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of 2 s, 5 s and 10 s and to compare with the analytical solution.
-The initial temperature <math>T_0</math> shall be 100 °C, the room temperature <math>T_R</math> 20 °C, and the cooling constant <math>k</math> 0.07. The time interval to calculate shall be from 0 s to 100 s.
 
-A reference solution ([[#Common Lisp|Common Lisp]]) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
+;Initial values:
+:::* &nbsp; initial temperature &nbsp; <big><math>T_0</math></big> &nbsp; shall be &nbsp; 100 °C
+:::* &nbsp; room temperature &nbsp; <big><math>T_R</math></big> &nbsp; shall be &nbsp; 20 °C
+:::* &nbsp; cooling constant &nbsp; &nbsp; <big><math>k</math></big> &nbsp; &nbsp; shall be &nbsp; 0.07
+:::* &nbsp; time interval to calculate shall be from &nbsp; 0 s &nbsp; ──► &nbsp; 100 s
+
+<br>
+A reference solution ([[#Common Lisp|Common Lisp]]) can be seen below. &nbsp; We see that bigger step sizes lead to reduced approximation accuracy.
 [[Image:Euler_Method_Newton_Cooling.png|center|750px]]