5/7 ANSWERS TO THE EXERCISES 



5/7. 1 : A possible set of transformations is: 



5/9. 1: fl = (100,100); D turns it to (110,110)— i.e. Si = 10, §2 = 10; T is 

 given; it is not stable. 2: a and D are as before, but Tis changed, and 

 the system is stable. 3: Usually the limit will be some state other than 

 a; it is not stable to such Z)'s. 4: Yes; the deviations tend to zero, 

 which is a state of equiUbrium. 5: No; the deviations increase to a 

 degree limited only by extraneous factors such as the shape of the 

 couplings. 6: To make any deviation wane rather than wax. 7: It 

 is self-aggravating — a perpetual headache to route managers. 8: Any 

 displacement from the state of equilibrium would increase until some 

 other limiting factor came in. 9: Yes; for all displacements D; thus 

 if D displaces the state to (8i, 82), then at's successive values are Si, 

 282, 4S1, iS2, . . . which obviously converges to 0; similarly for >'. 



5/13. 1: No; for y would have to be in equilibrium at under some value of 

 jS; thus ^ would have to satisfy = 2^0 + 3, which is impossible. 



6/3. 1: See 5.6/5. 



6/5. 1: ^j->f. 2: j-^f-^h (the protocol gives no evidence about 



f/ 

 transitions from g with input at /3). 3: No, the transition from C is 

 not single-valued. 4: Yes, so far as the evidence goes. 

 5:{x,y) j 00 01 02 10 II 12 20 21 22 



(x'y) 01 00 11 11 00 21 11 20 11 

 6: For each input value, n transitions have to be observed, taking at 

 least n steps; so the whole set of transformations cannot be observed in 

 fewer than inn steps. 7: Select any two values for x and a, and find 

 what value of x' ensues. Thus "a = 1, x = 4, and x' = 4 " shows the 

 Box to be /. An even simpler test is to set a = and see whether x 

 increases or decreases in value. 



6/7. 1 : y dominates x. 



6/9. 1: j « * ^ ^ ^ 2: Six. 

 ^ t p r q s 



3: Two variables are necessary, the dial reading (v) and its rate of change 

 (v); dvldt = V, dv/dt = k(u — v) — /v, where k represents the strength 

 of the spring and moment of inertia of the mass, and /is the coefficient 

 of friction; (ii) dy/dt = y, dy/dt = -Ry/L - y/CL + x. To be 

 isomorphic in the strict sense defined above, they must have / = R/L 

 and k = XjCL. If this is so they can be shown isomorphic by the one- 

 one transformation 



I 



y y x 

 V V ku 

 u V w 



4: i 



' z X y 



6/10. 1 : They are identical : /J :j± ^ —> r. 2: ii and iv may be changed; i.iii, and 

 V unchanged. 3: All are unchanged. 



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