DHS Recurrence Relations (9649) (Revision Solutions)

Recurrence Relations 1 H2 Double Math (TOPICAL REVISION – SUGGESTED SOLUTION) RECURRENCE RELATIONS Question 1 [Solution] 1 (i) Solving 18 20 , 17 18p q p q= + = + => 1 ,82pq== . (ii) (Method 1) For 1p  , rewrite the RR 1nnx px q+ =+ as 1 ( ),nnx k p x k+ − = − where k R . 1 ( ) 1 nn qx p x k k q k kp k p + = − +  = − = −Then and . Let .nnu x k=− Then 1n n u pu + = is a constant and the sequence {}nu is a geometric sequence with first term 00=u x k − and common ratio p. For the sequence {}nu to be convergent, 11 p−   . (Method 2) Alternatively, limits may be used to explain the condition using the similar approach of rewriting the RR 1nnx px q+ =+ as 1 ( ),nnx k p x k+ − = − where k R . 23 1 2 3 0( ) ( ) ( ) ( ) n n n n nx k p x k p x k p x k p x k− − −− = − = − = − = = − 00lim( ) = lim ( ) = ( ) limnn nn n n x k p x k x k p → → → − − − For the sequence to be convergent, 11 p−   so that 00lim ( ) ( ) lim 0nn nn p x k x k p → → − = − = Then lim =nn xk → (iii) Since 161 qk p==− , lim = 16nn x k = → (iv) ( ) ( ) ( ) 1 2 22 21 00 1 11 ... 1 116 4 , 02 rr rr r r r r r r x px q p px q q p x q p pp x q p p p p x q p xr − −− − =+ = + + = + +  −= + + + + + = +  − = +  

Recurrence Relations 2 Question 2 [Solution] 2 “KillPest”: 10.35 500nnkk −=+ , in the long run, lim 769nn k →  “PestKill”: 10.15 650nnpp −=+ , in the long run, lim 765nn p →  Pestkill will be more effective in the long run since there will be fewer pests on the trees. Question 3 [Solution] 3 ( ) ( ) 11 1 2 1 1 1 7 15 0 7 7 8 0 7 8 0 n n n n n n n n n n n n n x x y y y yy x y x x x x ++ + + + + + + + + = + + + + =     − + + =          21 21 6 8 0 6 8 0 nn nn n n n yy y y y yy ++ ++ − + =  − + = Characteristic equation: 2 6 8 0 2 or 4 m m m m − + = == 11 ) (4 ) ) (4 ) 7)( (2 ( ( 4) 2 2 nn n nn n nn A A B Bx A y B ++ + += + = − Question 4 [Solution] 4 1 1 1 1 1 3 2 8 3 ( 1) 5 35 nn nn nn nn nn yy nyy yy nnyy uu − + − + − = − +  − = − − +  =+ Thus (3 )n nu A B=+ , where 535 2B B B= +  =− Now 1 1 2 3111 22 yu y= − = − = 1 1 5 1 5(3 ) 3 12 2 2u A A A= −  = −  = 53 2 n nu = −

Recurrence Relations 3 1 53 2 nn n y ny + = + − Question 5 [Solution] 5 114 3 (1)n n nW W I −−=+ 11 2 (2)n n nI I W −−=−  1 2 24 3( 2 ) (3)n n n nW W I W− − −= + − From (1), 2 1 234n n nI W W− − −=− Put into (3): 1 1 2 24 ( 4 ) 6n n n n nW W W W W− − − −= + − − 12 12 5 10 5 10 0 (shown) n n n n n n W W W W W W −− −− =−  − + = Auxiliary Equation: 2 5 10 0− + =  5 15 5 15 i2 2 2 −= =  1 12 1510; tan 0.65906 5r   −= = = = ( ) ( )( )210 cos 0.65906 sin 0.65906 n nW A n B n=+ 0 10 10WA=  = 1 0 0 14 3 55W W I W= +  = ( ) ( )( )55 10 10cos 0.65906 sin 0.65906 15.4919 B B =+ = ( ) ( )( )210 10cos 0.659 15.5sin 0.659 n nW n n=+ Question