September 2017 Update

Task/Euler-method/Jq/euler-method-1.jq

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+# euler_method takes a filter (df), initial condition
+# (x1,y1), ending x (x2), and step size as parameters;
+# it emits the y values at each iteration.
+# df must take [x,y] as its input.
+def euler_method(df; x1; y1; x2; h):
+  h as $h
+  | [x1, y1]
+  | recurse( if ((.[0] < x2 and x1 < x2) or
+                 (.[0] > x2 and x1 > x2)) then
+  		[ (.[0] + $h), (.[1] + $h*df) ]
+             else empty
+             end )
+  | .[1] ;
+
+
+# We could now solve the task by writing for each step-size, $h
+# euler_method(-0.07 * (.[1]-20); 0; 100; 100; $h)
+# but for clarity, we shall define a function named "cooling":
+
+# [x,y] is input
+def cooling: -0.07 * (.[1]-20);
+
+# The following solves the task:
+# (2,5,10) | [., [ euler_method(cooling; 0; 100; 100; .) ] ]

Task/Euler-method/Jq/euler-method-2.jq

@@ -0,0 +1,10 @@
+def euler_solution(df; x1; y1; x2; h):
+  def recursion(exp): reduce recurse(exp) as $x (.; $x);
+  h as $h
+  | [x1, y1]
+  | recursion( if ((.[0] < x2 and x1 < x2) or
+                   (.[0] > x2 and x1 > x2)) then
+  		[ (.[0] + $h), (.[1] + $h*df) ]
+             else empty
+             end )
+  | .[1] ;

Task/Euler-method/Jq/euler-method-3.jq

@@ -0,0 +1 @@
+(1,2,5,10,20) | [., [ euler_solution(cooling; 0; 100; 100; .) ] ]

Task/Euler-method/Jq/euler-method-4.jq

@@ -0,0 +1,6 @@
+$ jq -M -n -c -f Euler_method.jq
+[1,[20.05641373347389]]
+[2,[20.0424631833732]]
+[5,[20.01449963666907]]
+[10,[20.000472392]]
+[20,[19.180799999999998]]