Solutions to Homework Assignment #2

1. This can be solved by writing:
   

2. Golden section algorithm in Forth (what else?)
   
   Minimum of function f(x) = –xe^–x×sinh²(x)/[1+cosh²(x)]:
   

   use( f5 0e0 4e0 20 )tabulate
   .000000 -.000000
   .200000 -.003253
   .400000 -.020859
   .600000 -.055489
   .800000 -.101666
   1.00000 -.150270
   1.20000 -.192479
   1.40000 -.222515
   1.60000 -.238507
   1.80000 -.241696
   2.00000 -.234948
   2.20000 -.221470
   2.40000 -.204064
   2.60000 -.184862
   2.80000 -.165341
   3.00000 -.146443
   3.20000 -.128722
   3.40000 -.112465
   3.60000 -.097781
   3.80000 -.084670  ok
   use( f5 -2e0 0e0 4e0 )golden
   x_min:    1.75506
   f(x_min): -.241960
   number of iterations needed: 33  ok
   

   
   
   Minimum of function f(x) = x⁴+x³+8x+8  :
   

   use( f4 -4e0 4e0 20 )tabulate
   -4.00000 168.000
   -3.60000 100.506
   -3.20000 54.4896
   -2.80000 25.1136
   -2.40000 8.15360
   -2.00000 .000000
   -1.60000 -2.34240
   -1.20000 -1.25440
   -.800000 1.49760
   -.400000 4.76160
   -.000000 8.00000
   .400000 11.2896
   .800000 15.3216
   1.20000 21.4016
   1.60000 31.4496
   2.00000 48.0000
   2.40000 74.2016
   2.80000 113.818
   3.20000 171.226
   3.60000 251.418
   4.00000 360.000  ok
   use( f4 -2e0 0e0 1e0 )golden
   x_min:    -1.56578
   f(x_min): -2.35434
   number of iterations needed: 32  ok
   

3. Quadratic minimization
   
   Minimum of first function:
   

   use( f5 -2e0 4e0 )quadmin
   x_min: 1.75506
   f(x_min): -.241960
   number of iterations needed: 35  ok
   

   Minimum of 2nd function:
   

   use( f4 -2e0 2e0 )quadmin
   x_min: -1.56578
   f(x_min): -2.35434
   number of iterations needed: 126  ok
   

4. 1-dimensional simulated annealing    What simulated annealing does
   
   2-dimensional simulated annealing     Results of 2-dim minimization