ASRJC_9649_2025_Prelim_P2_Solutions

1 Consider 21nnnUUU−− += 1 22 1 nn nn UU UU − −− += 11 2 12 1 n nn n nn U UU U UU −− − −− +=   ( )111 n nnR RR−−+= If sequence converges, 1nRL− → , nRL→ when n→∞ where L is the limit of the sequence 21 LL∴+ = 2 10LL−−= ( )( )1 1 41 1 15 22L ±− − ±= = 15 2 += since all the Fibonacci numbers are positive, where a = 1 and b = 5 2 (a) [ ] 2 3 f () f () 0 f ()0 o r f ()0 2f( ) nn nn n xx xx x ′′ ′′= ⇒= = ′ This implies that the function f is a linear function, i.e. it is of the form f( )x ax b= + . (b) 1f () l n,f() l n 1 ,f()x xx x x x x ′ ′′= = += n nx (Newton-Raphson) nx (Chebyshev) 1 5 5 2 1.916121 1.551663 3 1.161072 1.027829 4 1.010204 1.000013 5 1.000051 1.000000 At each iteration, Chebyshev’s results are markedly closer to the actual root, demonstrating a better convergence rate compared to Newton-Raphson method. (c) Using Chebyshev’s method, 2 2.5277868 0x = −< , which is outside the domain of f. We cannot continue to compute 3x as 2ln x is undefined. 3 (a) ( ) ( ) 22f 4 11 2 7x xx y x y= +− + + − ( ) ( ) 22f 2 11 4 7y x y yx y= +− + + − ( ) ( ) ( ) ( ) ( )( ) 22 22 41 2 1 1 21 2 7 36 9f 1, 2 432 82 1 2 11 4 2 1 2 7  +− + + − −  ∇= == −  − +− + + −   

For steepest descent, the unit vector is in the opposite direction of the gradient vector, hence unit vector is 91 8145   . ( )f 1,2 4 145uD =− (b) ( ) 22f 4 11 8 2xx xy x= +− + + ( ) 22f 28 4 7yy y xy= + + +− f 44xy xy= + At ( )( )3 ,0 ,f 3 ,0 , ( )f 3, 0 20= , f 32x =− , f4y =− , f 66xx = , f 14yy =− , f 12xy = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 21, f 3, 0 3 f 3, 0 f 3, 0 3 f 3, 0 3 f 3, 0 2 x y xx xyQxy x y x yx≈ +− + + − + − ( ) 21 f 3, 02 yyy+ ( ) ( ) ( ) ( ) 2 2, 20 32 3 4 33 3 12 3 7Q x y xy x y xy≈− −− + − + −− For saddle point, ( ) ( ) ( ), 0 32 66 3 12 0 33 6 115 1xQ xy x y x y= ⇒− + − + = ⇒ + = −−−−− ( ) ( ) ( ), 0 4 12 3 14 0 6 7 20 2yQ xy x y x y= ⇒− + − − = ⇒ − = −−−−− Solving gives 925 103.46, 0.112267 89xy= ≈= ≈ 266 0, 14, 12 1068 0xx yy xy xx yy xyQ Q Q QQ Q= > = − = ⇒ − = − <⇒ saddle point, shown Coordinates are ( )3.46,0,112,13.7 . 4 (a) ( ) ( ) ( )det 1 7 12 2 0 4 3 0 1 0= − − −+ −=E Since det 0 =E , the word vectors are linearly dependent. (b) Set containing any 2 columns in the matrix E. (c) 13 13 123 6 123 6 1236 014 9 014 9 0149 1 3 7 18 0 1 4 12 0 0 0 3 RR RR−+ −+         → →          The system is inconsistent (last row is not a whole row of zeros). The vector v cannot be represented in the model’s current embedding space. The model lacks the capacity to understand or process this particular word/concept, as it lies outside the span of the stored word vectors. (d) 12310 1 0 1 40140 21 151 5137 − −−   = =       AE . Note that the two rows ar