exercise 26

exercises/Exercises 26.ipynb

+%% Cell type:markdown id: tags:
+
+# Exercises 26
+
+# problem 1
+
+Implement Newton's method for functions from $\mathbb{R}^n$ to $\mathbb{R}^n$ given as a function of the form
+
+```haskell
+newton :: (Vector Double -> Vector Double) -> (Vector Double -> Matrix Double) -> Vector Double -> Vector Double
+newton f df initial = [...]
+```
+
+Here, `f` is the function whose root should be found, and `df` is its derivative (i.e. Jacobian). `initial` is an initial guess, and should return the first iterate for which the norm of the function value is smaller than some fixed bound.
+
+%% Cell type:markdown id: tags:
+
+# problem 2
+
+(This is problem 4 of the SIAM 100-digit challenge.)
+
+Consider the function
+
+$$f(x, y) = e^{\sin(50x)} + \sin(60 e^y) + \sin(70\sin(x)) + \sin(\sin(80y)) - \sin(10(x+y)) + \frac{x^2 + y^2}{4}.$$
+
+1. Plot the function using the `vis.py` script from the source repository.
+
+2. Try to find the global minimum of the function by solving the equation
+
+    $$\nabla f(x, y) = 0$$
+
+    using Newton's method from problem 1.
+
+    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. Use a grid of $10 \times 10$ points in the square $(-0.5,\, 0.5)^2 \subset \mathbb{R}^2$ and run Newton's method on each of these points.
+
+3. 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.
+
+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))
+```

solutions/Exercises 26_sol.ipynb

 %% Cell type:markdown id: tags:
 
 # Exercises 26
 
 # problem 1
 
+Implement Newton's method for functions from $\mathbb{R}^n$ to $\mathbb{R}^n$ given as a function of the form
+
+```haskell
+newton :: (Vector Double -> Vector Double) -> (Vector Double -> Matrix Double) -> Vector Double -> Vector Double
+newton f df initial = [...]
+```
+
+Here, `f` is the function whose root should be found, and `df` is its derivative (i.e. Jacobian). `initial` is an initial guess, and should return the first iterate for which the norm of the function value is smaller than some fixed bound.
+
+%% Cell type:markdown id: tags:
+
+# problem 2
+
 (This is problem 4 of the SIAM 100-digit challenge.)
 
 Consider the function
 
 $$f(x, y) = e^{\sin(50x)} + \sin(60 e^y) + \sin(70\sin(x)) + \sin(\sin(80y)) - \sin(10(x+y)) + \frac{x^2 + y^2}{4}.$$
 
+1. Plot the function using the `vis.py` script from the source repository.
+
+2. Try to find the global minimum of the function by solving the equation
+
+    $$\nabla f(x, y) = 0$$
+
+    using Newton's method from problem 1.
+
+    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. Use a grid of $10 \times 10$ points in the square $(-0.5,\, 0.5)^2 \subset \mathbb{R}^2$ and run Newton's method on each of these points.
+
+3. 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.
+
 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.