Ex 10.1: (2) a) ๐๐+1 = 1.5๐๐,๐๐ = 1.5 ๐๐0,๐ โฅ 0. b) 4๐๐ = 5๐๐โ1,๐๐ = 1.25 ๐๐0,๐ โฅ 0.
c) 3๐๐+1 = 4๐๐, 3๐1 = 15 = 4๐0,๐0 = 154
,
so ๐๐ = 43
๐๐0 = 4
3
๐ 154
= 5 43
๐โ1,๐ โฅ 0.
d) ๐๐ = 32๐๐โ1,๐๐ = 3
2
๐๐0, 81 = ๐4 = 3
2
4๐0,
so ๐0 = 16 and ๐๐ = 16 32
๐, ๐ โฅ 0.
Ex 10.1: (3) โข ๐๐+1 โ ๐๐๐ = 0,๐ โฅ 0, so ๐๐ = ๐๐๐0. 153
49= ๐3 = ๐3๐0,
13772401
= ๐5 = ๐5๐0 โ๐5๐3
= ๐2 = 949
and ๐ = ยฑ 37.
Ex 10.2: (1.a, 1.b) a) ๐๐ = 5 ๐๐โ1 + 6 ๐๐โ2,๐ โฅ 2,๐0 = 1,๐1 = 3.
Let ๐๐ = ๐๐๐, ๐, ๐ โ 0. Then the characteristic equation is ๐2 โ 5๐ โ 6 = 0 = ๐ โ 6 ๐ + 1 , so ๐ = โ1,6 are the characteristic roots. ๐๐ = ๐ด โ1 ๐ + ๐ต 6 ๐. 1 = ๐0 = ๐ด + ๐ต. 3 = ๐1 = โ๐ด + 6๐ต, so ๐ต = 4
7 and ๐ด = 3
7.
๐๐ = 37
โ1 ๐ + 47
6๐,๐ โฅ 0.
b) ๐๐ = 4 12
๐โ 2 5 ๐,๐ โฅ 0.
Ex 10.2: (1.c) โข ๐๐+2 + ๐๐ = 0,๐ โฅ 0,๐0 = 0,๐1 = 3.
With ๐๐ = ๐๐๐, ๐, ๐ โ 0. The characteristic equation ๐2 + 1 =0 yields the characteristic roots ยฑ๐.
Hence ๐๐ = ๐ด ๐ ๐ + ๐ต โ๐ ๐ = ๐ด cos ๐2
+ ๐๐๐๐ ๐2
๐+
๐ต cos ๐2
+ ๐๐๐๐ โ๐2
๐= ๐ถ๐๐ถ๐ ๐๐
2+ ๐ท๐๐๐ ๐๐
2.
0 = ๐0 = ๐ถ, 3 = ๐1 = ๐ท๐๐ถ๐ ๐2
= ๐ท,
so ๐๐ = 3 sin ๐๐2
,๐ โฅ 0.
Ex 10.2: (1.d) โข ๐๐ โ 6๐๐โ1 + 9๐๐โ2 = 0,๐ โฅ 2,๐0 = 5,๐1 = 12.
Let ๐๐ = ๐๐๐, ๐, ๐ โ 0. Then ๐2 + 6๐ + 9 = 0 = ๐ โ 3 2, so the characteristic roots are 3,3 and ๐ด 3๐ + ๐ต๐ 3๐ . 5 = ๐0 = ๐ด; 12๐1 = 3๐ด + 3๐ต = 15 + 3๐ต,๐ต = โ1. ๐๐ = 5 3๐ โ ๐ 3๐ = 5 โ ๐ 3๐ ,๐ โฅ 0.
Ex 10.2: (1.e) โข ๐๐ + 2๐๐โ1 + 2๐๐โ2 = 0,๐ โฅ 2,๐0 = 1,๐1 = 3. ๐2 + 2๐ + 2 = 0, ๐ = โ1 ยฑ ๐. โ1 + ๐ = 2 cos 3๐
4+ ๐๐๐๐ 3๐
4.
โ1 โ ๐ = 2 cos 3๐4
โ ๐๐๐๐ 3๐4
.
๐๐ = 2๐๐ด๐๐ถ๐ 3๐
4+ ๐ต๐๐๐ 3๐
4.
1 = ๐0 = ๐ด. 3 = ๐1 = 2 cos 3๐
4+ ๐ต๐๐๐ 3๐
4= 2 โ1
2+ ๐ต 1
2, ๐๐ถ 3 =
โ 1 + ๐ต,๐ต = 4. ๐๐ = 2
๐cos 3๐๐
4+ 4 sin 3๐๐
4,๐ โฅ 0.
Ex 10.2: (3) โข ๐ = 0 : ๐2 + ๐ ๐1 + ๐ ๐0 = 0 = 4 + ๐ 1 + ๐ 0 , so ๐ = โ4. ๐ = 1 : ๐3 โ 4 ๐2 + ๐ ๐1 = 0 = 37 โ 4 4 + ๐, so ๐ = โ21. ๐๐+2 โ 4 ๐๐+1 โ 21 ๐๐ = 0. ๐2 โ 4๐ โ 21 = 0 = ๐ โ 7 ๐ + 3 , ๐ = 7,โ3. ๐๐ = ๐ด 7 ๐ + ๐ต โ3 ๐. 0 = ๐0 = ๐ด + ๐ต โ ๐ต = โ๐ด. 1 = ๐1 = 7๐ด โ 3๐ต = 10๐ด, so ๐ด = 1
10,๐ต = โ 1
10 and ๐๐ = 1
107๐ โ โ3 ๐ ,๐ โฅ 0.
Ex 10.2: (4) โข ๐๐ = ๐๐โ1 + ๐๐โ2,๐ โฅ 2,๐0 = ๐1 = 1. ๐2 โ ๐ โ 1 = 0, ๐ = 1ยฑ๐
2,๐ = 5.
๐0 = ๐ด 1+๐2
๐+ ๐ต 1โ๐
2
๐.
๐0 = ๐1 = 1 โ ๐ด = 1+๐2๐
,๐ต = ๐โ12๐
.
๐๐ = 1๐
1+๐2
๐+1โ 1โ๐
2
๐+1
= 15
1+ 52
๐+1โ 1โ 5
2
๐+1.
Ex 10.2: (20.b) โข Proof: (By Mathematical Induction) For ๐ = 1, we have ๐ผ๐ = ๐ผ1 = ๐ผ = ๐ผ โ 1 + 0 = ๐ผF1 + ๐น0 = ๐ผF๐ + ๐น๐โ1, so the result is true in this case. This establishes the basis step. Now we assume for an arbitrary (but fixed) positive integer ๐ that ๐ผ๐ = ๐ผF๐ + ๐น๐โ1. This is our indutive step. Considering ๐ = ๐ + 1, at this time, we find that ๐ผ๐+1 = ๐ผ ๐ผ๐ = ๐ผ ๐ผF๐ + ๐น๐+1 (Bythe inductive step) = ๐ผ2๐น๐ + ๐ผ๐น๐โ1 = ๐ผ + 1 ๐น๐ + ๐ผF๐โ1[by part (a)] = ๐ผ ๐น๐ + ๐น๐โ1 + ๐น๐ = ๐ผF๐+1 + ๐น๐. Since the given result is ture for ๐ = 1 and the truth for ๐ = ๐ + 1 follows from that for ๐ = ๐, it follows bt the Principle of Mathematical Induction that ๐ผ๐ = ๐ผF๐ + ๐น๐โ1for all ๐ โ ๐+.
Ex 10.2: (31) โข Let ๐๐ = ๐๐2 , ๐0 = 16,๐1 = 169.
This yields the linear relation ๐๐+2 โ 5๐๐+1 + 4๐๐ = 0 with characteristic roots ๐ = 4,1, so ๐๐ = ๐ด 1 ๐ + ๐ต 4 ๐. ๐0 = 16,๐1 = 169 โ ๐ด = โ35,๐ต = 51 and ๐๐ = 51 4 ๐ โ 35. Hence ๐๐ =, 51 4 ๐ โ 35, ๐ โฅ 0.
Ex 10.3: (1.a, 1.b) a) ๐๐+1 โ ๐๐ = 2๐ + 3,๐ โฅ 0,๐0 = 1.
๐1 = ๐0 + 0 + 3. ๐2 = ๐1 + 2 + 3 = ๐0 + 2 + 2 3 . ๐3 = ๐2 + 2 2 + 3 = ๐0 + 2 + 2 2 + 3 3 . ๐4 = ๐3 + 2 3 + 3 = ๐0 + 2 + 2 2 + 2 3 + 4 3 . โฆ ๐๐ = ๐0 + 2 1 + 2 + 3 + โฏ+ ๐ โ 1 + ๐ 3 = 1 +2 ๐ ๐โ1
2+ 3๐ = 1 + ๐ ๐ โ 1 + 3๐ = ๐2 + 2๐ + 1 =
๐ + 1 2,๐ โฅ 0. b) ๐๐ = 3 + ๐ ๐ โ 1 2,๐ โฅ 0.
Ex 10.3: (1.c, 1.d) c) ๐๐+1 โ 2๐๐ = 5,๐ โฅ 0,๐0 = 1.
๐1 = 2๐0 + 5 = 2 + 5. ๐2 = 2๐1 + 5 = 22 + 2 โ 5 + 5. ๐3 = 2๐2 + 5 = 23 + 22 + 2 + 1 5. โฆ ๐๐ = 2๐ + 5 1 + 2 + 22 + โฏ+ 2๐โ1 = 2๐ + 5 2๐ โ 1 =6 2๐ โ 2,๐ โฅ 0.
d) ๐๐ = 22 + ๐ 2๐โ1 ,๐ โฅ 0.
Ex 10.3: (2) โข ๐๐ = โ ๐2๐
๐=0 . ๐๐+1 = ๐๐ + ๐ + 1 2,๐ โฅ 0,๐0 = 0. ๐๐+1 โ ๐๐ = ๐ + 1 2 = ๐2 + 2๐ + 1. ๐๐โ = ๐ด,๐๐
๐ = ๐ต๐ + ๐ถ๐2 + ๐ท๐3. ๐ต ๐ + 1 + ๐ถ ๐ + 1 2 + ๐ท ๐ + 1 3 = ๐ต๐ + ๐ถ๐2 + ๐ท๐3 +๐2 + 2๐ + 1 โ ๐ต๐ + ๐ต + ๐ถ๐2 + 2๐ถ๐ + ๐ถ + ๐ท๐3 + 3๐ท๐2 +3๐ท๐ + ๐ท = ๐ต๐ + ๐ถ๐2 + ๐ท๐3 + ๐2 + 2๐ + 1. By comparing coefficients on like powers of ๐, we find that ๐ถ + 3๐ท = ๐ถ + 1, so ๐ท = 1
3. Also ๐ต + 2๐ถ + 3๐ท = ๐ต + 2, so ๐ถ = 1
2. Finally,
๐ต + ๐ถ + ๐ท = 1 โ ๐ต = 16. So ๐๐ = ๐ด + 1
6๐ + 1
2๐2 + 1
3๐3. With
๐0 = 0, it follows that ๐ด = 0 and ๐๐ = 16๐ 1 + 3๐ + 2๐2 =
16๐ ๐ + 1 2๐ + 1 ,๐ โฅ 0.
Ex 10.3: (4) โข Let ๐๐ be the value of the account ๐ months after January 1 of
the year the account is started. ๐0 = 1000. ๐1 = 1000 + .005 1000 + 200 = 1.005 ๐0 + 200. ๐๐+1 = 1.005 ๐๐ + 200, 0 โค ๐ โค 46. ๐48 = 1.005 ๐47. ๐๐+1 โ 1.005 ๐๐ = 200, 0 โค ๐ โค 46. ๐๐โ = ๐ด 1.005 ๐,๐๐
๐ = ๐ถ. ๐ถ โ 1.005๐ถ = 200 โ ๐ถ = โ400000. ๐0 = ๐ด 1.005 0 โ 40000 = 1000, ๐๐ถ ๐ด = 41000. ๐๐ = 41000 1.005 ๐ โ 40000. ๐47 = 41000 1.005 47 โ 40000 = 11830.90. ๐48 = 1.005๐47 = 11890.05.
Ex 10.3: (5) a) ๐๐+2 + 3๐๐+1 + 2๐๐ = 3๐,๐ โฅ 0,๐0 = 0,๐1 = 1.
With ๐๐ = ๐๐๐, ๐, ๐ โ 0, the characteristic equation ๐2 + 3๐ + 2 =0 = ๐ + 2 ๐ + 1 yields the characteristic roots ๐ = โ1,โ2. Hence ๐๐
โ = ๐ด โ1 ๐ + ๐ต โ2 ๐, while ๐๐๐ = ๐ถ 3 ๐.
๐ถ 3 ๐+2 + 3๐ถ 3 ๐+1 + 2๐ถ 3 ๐ = 3๐ โ 9๐ถ + 9๐ถ + 2๐ถ = 1 โ๐ถ = 1
20.
๐๐ = ๐ด โ1 ๐ + ๐ต โ2 ๐ + 120
3๐. 0 = ๐0 = ๐ด + ๐ต + 1
20.
1 = ๐1 = โ๐ด โ 2๐ต + 320
. Hence 1 = ๐0 + ๐1 = โ๐ต + 4
20 and ๐ต = โ4
5. Then ๐ด = โ๐ต โ 1
20=
34
.๐๐ = 34โ1 ๐ + โ4
5โ2 ๐ + 1
203 ๐,๐ โฅ 0.
b) ๐๐ = 29โ2 ๐ โ 5
6๐ โ2 ๐ + 7
9,๐ โฅ 0.
Ex 10.3: (11.a) โข Let ๐๐2 = ๐๐,๐ โฅ 0. ๐๐+2 โ 5 ๐๐+1 + 6 ๐๐ = 7๐. ๐๐โ = ๐ด 3๐ + ๐ต 2๐ , ๐๐
๐ = ๐ถ๐ + ๐ท. ๐ถ ๐ + 2 + ๐ท โ 5 ๐ถ ๐ + 1 + ๐ท + 6 ๐ถ๐ + ๐ท = 7๐ โ ๐ถ = 7
2,๐ท = 21
4.
๐๐ = ๐ด 3๐ + ๐ต 2๐ + 7๐2
+ 214
. ๐0 = ๐02 = 1, ๐1 = ๐12 = 1. 1 = ๐0 = ๐ด + ๐ต + 21
4.
1 = ๐1 = 3๐ด + 2๐ต + 72
+ 214
. 3๐ด + 2๐ต = โ31
3.
2๐ด + 2๐ต = โ344
. ๐ด = 3
4,๐ต = โ5.
๐๐ = 34
3๐ โ 5 โ 2๐ + 7๐2
+ 214
12 ,๐ โฅ 0.
Ex 10.3: (11.b) โข ๐๐2 โ 2๐๐โ1 = 0,๐ โฅ 1,๐0 = 2. ๐๐2 = 2๐๐โ1. log2 ๐๐2 = log2(2๐๐โ1) = log2 2 + log2 ๐๐โ1. 2 log2 ๐๐ = 1 + log2 ๐๐โ1. Let ๐๐ = log2 ๐๐. The solution of the recurrence relation 2๐๐ = 1 + ๐๐โ1is ๐ = ๐ด 1
2
๐+ 1.๐0 = log2 ๐0 = log22 = 1, so 1 = ๐0 = ๐ด + 1
and ๐ด = 0. Consequently, ๐๐ = 1,๐ โฅ 0, and ๐๐ = 2,๐ โฅ 0.
Ex 10.4: (1.a, 1.b) a) ๐๐+1 โ ๐๐ = 3๐,๐ โฅ 0,๐0 = 1.
Let ๐ ๐ฅ = โ ๐๐๐ฅ๐โ๐=0 .
โ ๐๐+1๐ฅ๐+1โ๐=0 โ โ ๐๐๐ฅ๐+1โ
๐=0 = โ 3๐๐ฅ๐+1โ๐=0 .
๐ ๐ฅ โ ๐0 โ ๐ฅ๐ ๐ฅ = ๐ฅ โ 3๐ฅ ๐โ๐=0 = ๐ฅ
1โ3๐ฅ.
๐ ๐ฅ โ 1 โ ๐ฅ๐ ๐ฅ = ๐ฅ1โ3๐ฅ
.
๐ ๐ฅ = 11โ๐ฅ
+ ๐ฅ1โ๐ฅ 1โ3๐ฅ
=1
1โ๐ฅ+ โ1
21
1โ๐ฅ+ 1
21
1โ3๐ฅ= 1
21
1โ๐ฅ+ 1
2( 11โ3๐ฅ
),
and ๐๐ = 12
1 + 3๐ ,๐ โฅ 0.
b) ๐๐ = 1 + ๐ ๐โ1 2๐โ16
,๐ โฅ 0.
Ex 10.4: (1.c) โข ๐๐+2 โ 3 ๐๐+1 + 2 ๐๐ = 0,๐ โฅ 0,๐0 = 1,๐1 = 6. โ ๐๐+2๐ฅ๐+2โ๐=0 โ 3โ ๐๐+1๐ฅ๐+2โ
๐=0 + 2โ ๐๐๐ฅ๐+2โ๐=0 = 0.
โ ๐๐+2๐ฅ๐+2โ๐=0 โ 3๐ฅโ ๐๐+1๐ฅ๐+1โ
๐=0 + 2๐ฅ2 โ ๐๐๐ฅ๐โ๐=0 = 0.
Let ๐ ๐ฅ = โ ๐๐๐ฅ๐โ๐=0 . Then
๐ ๐ฅ โ 1 โ 6๐ฅ โ 3๐ฅ ๐ ๐ฅ โ 1 + 2๐ฅ2๐ ๐ฅ = 0, and ๐ ๐ฅ 1 + 3๐ฅ + 2๐ฅ2 = 1 + 6๐ฅ โ 3๐ฅ = 1 + 3๐ฅ, Consequently, ๐ ๐ฅ = 1+3๐ฅ
1โ2๐ฅ 1โ๐ฅ= 5
1โ2๐ฅ+ โ4
1โ๐ฅ= 5โ 2๐ฅ ๐โ
๐=0 โ 4โ ๐ฅ๐โ๐=0 ,
and ๐๐ = 5 2๐ โ 4,๐ โฅ 0.
Ex 10.4: (1.d) โข ๐๐+2 โ 2 ๐๐+1 + ๐๐ = 2๐,๐ โฅ 0,๐0 = 1,๐1 = 2. โ ๐๐+2๐ฅ๐+2โ๐=0 โ 2โ ๐๐+1๐ฅ๐+2โ
๐=0 + โ ๐๐๐ฅ๐+2โ๐=0 =
โ 2๐๐ฅ๐+2โ๐=0 .
Let ๐ ๐ฅ = โ ๐๐๐ฅ๐โ๐=0 . Then
๐ ๐ฅ โ ๐0 โ ๐1๐ฅ โ 2๐ฅ ๐ ๐ฅ โ ๐0 + ๐ฅ2๐ ๐ฅ = ๐ฅ2 โ 2๐ฅ ๐โ๐=0 .
๐ ๐ฅ โ 1 โ 2๐ฅ โ 2๐ฅ๐ ๐ฅ + 2๐ฅ + ๐ฅ2๐ ๐ฅ = ๐ฅ2
1โ2๐ฅ.
๐ฅ2 โ 2๐ฅ + 1 ๐ ๐ฅ = 1 + ๐ฅ2
1โ2๐ฅโ ๐ ๐ฅ =
11โ๐ฅ 2 + ๐ฅ2
1โ2๐ฅ 1โ๐ฅ 2 = 1โ2๐ฅ+๐ฅ2
1โ๐ฅ 2 1โ2๐ฅ= 1
1โ2๐ฅ= 1 + 2๐ฅ +
2๐ฅ 2 + โฏ, so ๐๐ = 2๐,๐ โฅ 0.