exercise 26, problem 1

solutions/Exercises 26_sol.ipynb

 %% Cell type:markdown id: tags:
 
 # Exercises 26
 
-%% Cell type:code id: tags:
+# problem 1
 
-``` haskell
+(This is problem 4 of the SIAM 100-digit challenge.)
+
+Consider the function
+
+$$f(x, y) = e^{\sin(50x)} + \sin(60 e^y) + \sin(70\sin(x)) + \sin(\sin(80y)) - \sin(10(x+y)) + \frac{x^2 + y^2}{4}.$$
+
+The function, its gradient and is Hessian are are implemented by the following Haskell functions:
+
+```haskell
+f :: Double -> Double -> Double
+f x y = exp (sin (50*x))
+      + sin (60 * exp y)
+      + sin (70 * sin x)
+      + sin (sin (80*y))
+      - sin (10 * (x+y))
+      + (x^2 + y^2)/4
+
+gradf :: Double -> Double -> [Double]
+gradf x y = [dfx, dfy]
+    where dfx = 50 * cos (50*x) * exp (sin (50*x))
+              + 70 * cos x * cos (70 * sin x)
+              - 10 * cos (10 * (x+y))
+              + x/2
+          dfy = 60 * exp y * cos (60 * exp y)
+              + 80 * cos (80*y) * cos (sin (80*y))
+              - 10 * cos (10 * (x+y))
+              + y/2
+
+hessf :: Double -> Double -> [[Double]]
+hessf x y = [[hfxx, hfxy], [hfxy, hfyy]]
+    where hfxx = 2500 * cos (50*x) ^ 2 * exp (sin (50*x))
+               - 2500 * sin (50*x) * exp (sin (50*x))
+               - 70 * sin x * cos (70 * sin x)
+               - 4900 * cos x ^ 2 * sin (70 * sin x)
+               + 100 * sin (10 * (x+y))
+               + 1/2
+          hfyy = 60 * exp y * cos (60 * exp y)
+               - 3600 * exp y ^ 2 * sin (60 * exp y)
+               - 6400 * sin (80*y) * cos (sin (80*y))
+               - 6400 * cos (80*y) ^ 2 * sin (sin (80*y))
+               + 100 * sin(10 * (x+y))
+               + 1/2
+          hfxy = 100 * sin (10 * (x+y))
 ```
+
+Try to find the global minimum of the function by solving the equation
+
+$$\nabla f(x, y) = 0$$
+
+using e.g. Newton's method. Since the function oscillates very quickly and therefore has multiple local extrema, you should solve the equation for a sufficiently large number of initial values.
+
+It may also be useful to try to refine the set of initial points until some heuristic criterion is fulfilled. Try to formulate such a criterion.
+
+For reference, the global minimum is -3.306868647.