Download - Solution Week 1
Ex 10.1: (2) a) ππ+1 = 1.5ππ,ππ = 1.5 ππ0,π β₯ 0. b) 4ππ = 5ππβ1,ππ = 1.25 ππ0,π β₯ 0.
c) 3ππ+1 = 4ππ, 3π1 = 15 = 4π0,π0 = 154
,
so ππ = 43
ππ0 = 4
3
π 154
= 5 43
πβ1,π β₯ 0.
d) ππ = 32ππβ1,ππ = 3
2
ππ0, 81 = π4 = 3
2
4π0,
so π0 = 16 and ππ = 16 32
π, π β₯ 0.
Ex 10.1: (3) β’ ππ+1 β πππ = 0,π β₯ 0, so ππ = πππ0. 153
49= π3 = π3π0,
13772401
= π5 = π5π0 βπ5π3
= π2 = 949
and π = Β± 37.
Ex 10.2: (1.a, 1.b) a) ππ = 5 ππβ1 + 6 ππβ2,π β₯ 2,π0 = 1,π1 = 3.
Let ππ = πππ, π, π β 0. Then the characteristic equation is π2 β 5π β 6 = 0 = π β 6 π + 1 , so π = β1,6 are the characteristic roots. ππ = π΄ β1 π + π΅ 6 π. 1 = π0 = π΄ + π΅. 3 = π1 = βπ΄ + 6π΅, so π΅ = 4
7 and π΄ = 3
7.
ππ = 37
β1 π + 47
6π,π β₯ 0.
b) ππ = 4 12
πβ 2 5 π,π β₯ 0.
Ex 10.2: (1.c) β’ ππ+2 + ππ = 0,π β₯ 0,π0 = 0,π1 = 3.
With ππ = πππ, π, π β 0. The characteristic equation π2 + 1 =0 yields the characteristic roots Β±π.
Hence ππ = π΄ π π + π΅ βπ π = π΄ cos π2
+ ππππ π2
π+
π΅ cos π2
+ ππππ βπ2
π= πΆππΆπ ππ
2+ π·πππ ππ
2.
0 = π0 = πΆ, 3 = π1 = π·ππΆπ π2
= π·,
so ππ = 3 sin ππ2
,π β₯ 0.
Ex 10.2: (1.d) β’ ππ β 6ππβ1 + 9ππβ2 = 0,π β₯ 2,π0 = 5,π1 = 12.
Let ππ = πππ, π, π β 0. Then π2 + 6π + 9 = 0 = π β 3 2, so the characteristic roots are 3,3 and π΄ 3π + π΅π 3π . 5 = π0 = π΄; 12π1 = 3π΄ + 3π΅ = 15 + 3π΅,π΅ = β1. ππ = 5 3π β π 3π = 5 β π 3π ,π β₯ 0.
Ex 10.2: (1.e) β’ ππ + 2ππβ1 + 2ππβ2 = 0,π β₯ 2,π0 = 1,π1 = 3. π2 + 2π + 2 = 0, π = β1 Β± π. β1 + π = 2 cos 3π
4+ ππππ 3π
4.
β1 β π = 2 cos 3π4
β ππππ 3π4
.
ππ = 2ππ΄ππΆπ 3π
4+ π΅πππ 3π
4.
1 = π0 = π΄. 3 = π1 = 2 cos 3π
4+ π΅πππ 3π
4= 2 β1
2+ π΅ 1
2, ππΆ 3 =
β 1 + π΅,π΅ = 4. ππ = 2
πcos 3ππ
4+ 4 sin 3ππ
4,π β₯ 0.
Ex 10.2: (3) β’ π = 0 : π2 + π π1 + π π0 = 0 = 4 + π 1 + π 0 , so π = β4. π = 1 : π3 β 4 π2 + π π1 = 0 = 37 β 4 4 + π, so π = β21. ππ+2 β 4 ππ+1 β 21 ππ = 0. π2 β 4π β 21 = 0 = π β 7 π + 3 , π = 7,β3. ππ = π΄ 7 π + π΅ β3 π. 0 = π0 = π΄ + π΅ β π΅ = βπ΄. 1 = π1 = 7π΄ β 3π΅ = 10π΄, so π΄ = 1
10,π΅ = β 1
10 and ππ = 1
107π β β3 π ,π β₯ 0.
Ex 10.2: (4) β’ ππ = ππβ1 + ππβ2,π β₯ 2,π0 = π1 = 1. π2 β π β 1 = 0, π = 1Β±π
2,π = 5.
π0 = π΄ 1+π2
π+ π΅ 1βπ
2
π.
π0 = π1 = 1 β π΄ = 1+π2π
,π΅ = πβ12π
.
ππ = 1π
1+π2
π+1β 1βπ
2
π+1
= 15
1+ 52
π+1β 1β 5
2
π+1.
Ex 10.2: (20.b) β’ Proof: (By Mathematical Induction) For π = 1, we have πΌπ = πΌ1 = πΌ = πΌ β 1 + 0 = πΌF1 + πΉ0 = πΌFπ + πΉπβ1, so the result is true in this case. This establishes the basis step. Now we assume for an arbitrary (but fixed) positive integer π that πΌπ = πΌFπ + πΉπβ1. This is our indutive step. Considering π = π + 1, at this time, we find that πΌπ+1 = πΌ πΌπ = πΌ πΌFπ + πΉπ+1 (Bythe inductive step) = πΌ2πΉπ + πΌπΉπβ1 = πΌ + 1 πΉπ + πΌFπβ1[by part (a)] = πΌ πΉπ + πΉπβ1 + πΉπ = πΌFπ+1 + πΉπ. Since the given result is ture for π = 1 and the truth for π = π + 1 follows from that for π = π, it follows bt the Principle of Mathematical Induction that πΌπ = πΌFπ + πΉπβ1for all π β π+.
Ex 10.2: (31) β’ Let ππ = ππ2 , π0 = 16,π1 = 169.
This yields the linear relation ππ+2 β 5ππ+1 + 4ππ = 0 with characteristic roots π = 4,1, so ππ = π΄ 1 π + π΅ 4 π. π0 = 16,π1 = 169 β π΄ = β35,π΅ = 51 and ππ = 51 4 π β 35. Hence ππ =, 51 4 π β 35, π β₯ 0.
Ex 10.3: (1.a, 1.b) a) ππ+1 β ππ = 2π + 3,π β₯ 0,π0 = 1.
π1 = π0 + 0 + 3. π2 = π1 + 2 + 3 = π0 + 2 + 2 3 . π3 = π2 + 2 2 + 3 = π0 + 2 + 2 2 + 3 3 . π4 = π3 + 2 3 + 3 = π0 + 2 + 2 2 + 2 3 + 4 3 . β¦ ππ = π0 + 2 1 + 2 + 3 + β―+ π β 1 + π 3 = 1 +2 π πβ1
2+ 3π = 1 + π π β 1 + 3π = π2 + 2π + 1 =
π + 1 2,π β₯ 0. b) ππ = 3 + π π β 1 2,π β₯ 0.
Ex 10.3: (1.c, 1.d) c) ππ+1 β 2ππ = 5,π β₯ 0,π0 = 1.
π1 = 2π0 + 5 = 2 + 5. π2 = 2π1 + 5 = 22 + 2 β 5 + 5. π3 = 2π2 + 5 = 23 + 22 + 2 + 1 5. β¦ ππ = 2π + 5 1 + 2 + 22 + β―+ 2πβ1 = 2π + 5 2π β 1 =6 2π β 2,π β₯ 0.
d) ππ = 22 + π 2πβ1 ,π β₯ 0.
Ex 10.3: (2) β’ ππ = β π2π
π=0 . ππ+1 = ππ + π + 1 2,π β₯ 0,π0 = 0. ππ+1 β ππ = π + 1 2 = π2 + 2π + 1. ππβ = π΄,ππ
π = π΅π + πΆπ2 + π·π3. π΅ π + 1 + πΆ π + 1 2 + π· π + 1 3 = π΅π + πΆπ2 + π·π3 +π2 + 2π + 1 β π΅π + π΅ + πΆπ2 + 2πΆπ + πΆ + π·π3 + 3π·π2 +3π·π + π· = π΅π + πΆπ2 + π·π3 + π2 + 2π + 1. By comparing coefficients on like powers of π, we find that πΆ + 3π· = πΆ + 1, so π· = 1
3. Also π΅ + 2πΆ + 3π· = π΅ + 2, so πΆ = 1
2. Finally,
π΅ + πΆ + π· = 1 β π΅ = 16. So ππ = π΄ + 1
6π + 1
2π2 + 1
3π3. With
π0 = 0, it follows that π΄ = 0 and ππ = 16π 1 + 3π + 2π2 =
16π π + 1 2π + 1 ,π β₯ 0.
Ex 10.3: (4) β’ Let ππ be the value of the account π months after January 1 of
the year the account is started. π0 = 1000. π1 = 1000 + .005 1000 + 200 = 1.005 π0 + 200. ππ+1 = 1.005 ππ + 200, 0 β€ π β€ 46. π48 = 1.005 π47. ππ+1 β 1.005 ππ = 200, 0 β€ π β€ 46. ππβ = π΄ 1.005 π,ππ
π = πΆ. πΆ β 1.005πΆ = 200 β πΆ = β400000. π0 = π΄ 1.005 0 β 40000 = 1000, ππΆ π΄ = 41000. ππ = 41000 1.005 π β 40000. π47 = 41000 1.005 47 β 40000 = 11830.90. π48 = 1.005π47 = 11890.05.
Ex 10.3: (5) a) ππ+2 + 3ππ+1 + 2ππ = 3π,π β₯ 0,π0 = 0,π1 = 1.
With ππ = πππ, π, π β 0, the characteristic equation π2 + 3π + 2 =0 = π + 2 π + 1 yields the characteristic roots π = β1,β2. Hence ππ
β = π΄ β1 π + π΅ β2 π, while πππ = πΆ 3 π.
πΆ 3 π+2 + 3πΆ 3 π+1 + 2πΆ 3 π = 3π β 9πΆ + 9πΆ + 2πΆ = 1 βπΆ = 1
20.
ππ = π΄ β1 π + π΅ β2 π + 120
3π. 0 = π0 = π΄ + π΅ + 1
20.
1 = π1 = βπ΄ β 2π΅ + 320
. Hence 1 = π0 + π1 = βπ΅ + 4
20 and π΅ = β4
5. Then π΄ = βπ΅ β 1
20=
34
.ππ = 34β1 π + β4
5β2 π + 1
203 π,π β₯ 0.
b) ππ = 29β2 π β 5
6π β2 π + 7
9,π β₯ 0.
Ex 10.3: (11.a) β’ Let ππ2 = ππ,π β₯ 0. ππ+2 β 5 ππ+1 + 6 ππ = 7π. ππβ = π΄ 3π + π΅ 2π , ππ
π = πΆπ + π·. πΆ π + 2 + π· β 5 πΆ π + 1 + π· + 6 πΆπ + π· = 7π β πΆ = 7
2,π· = 21
4.
ππ = π΄ 3π + π΅ 2π + 7π2
+ 214
. π0 = π02 = 1, π1 = π12 = 1. 1 = π0 = π΄ + π΅ + 21
4.
1 = π1 = 3π΄ + 2π΅ + 72
+ 214
. 3π΄ + 2π΅ = β31
3.
2π΄ + 2π΅ = β344
. π΄ = 3
4,π΅ = β5.
ππ = 34
3π β 5 β 2π + 7π2
+ 214
12 ,π β₯ 0.
Ex 10.3: (11.b) β’ ππ2 β 2ππβ1 = 0,π β₯ 1,π0 = 2. ππ2 = 2ππβ1. log2 ππ2 = log2(2ππβ1) = log2 2 + log2 ππβ1. 2 log2 ππ = 1 + log2 ππβ1. Let ππ = log2 ππ. The solution of the recurrence relation 2ππ = 1 + ππβ1is π = π΄ 1
2
π+ 1.π0 = log2 π0 = log22 = 1, so 1 = π0 = π΄ + 1
and π΄ = 0. Consequently, ππ = 1,π β₯ 0, and ππ = 2,π β₯ 0.
Ex 10.4: (1.a, 1.b) a) ππ+1 β ππ = 3π,π β₯ 0,π0 = 1.
Let π π₯ = β πππ₯πβπ=0 .
β ππ+1π₯π+1βπ=0 β β πππ₯π+1β
π=0 = β 3ππ₯π+1βπ=0 .
π π₯ β π0 β π₯π π₯ = π₯ β 3π₯ πβπ=0 = π₯
1β3π₯.
π π₯ β 1 β π₯π π₯ = π₯1β3π₯
.
π π₯ = 11βπ₯
+ π₯1βπ₯ 1β3π₯
=1
1βπ₯+ β1
21
1βπ₯+ 1
21
1β3π₯= 1
21
1βπ₯+ 1
2( 11β3π₯
),
and ππ = 12
1 + 3π ,π β₯ 0.
b) ππ = 1 + π πβ1 2πβ16
,π β₯ 0.
Ex 10.4: (1.c) β’ ππ+2 β 3 ππ+1 + 2 ππ = 0,π β₯ 0,π0 = 1,π1 = 6. β ππ+2π₯π+2βπ=0 β 3β ππ+1π₯π+2β
π=0 + 2β πππ₯π+2βπ=0 = 0.
β ππ+2π₯π+2βπ=0 β 3π₯β ππ+1π₯π+1β
π=0 + 2π₯2 β πππ₯πβπ=0 = 0.
Let π π₯ = β πππ₯πβπ=0 . Then
π π₯ β 1 β 6π₯ β 3π₯ π π₯ β 1 + 2π₯2π π₯ = 0, and π π₯ 1 + 3π₯ + 2π₯2 = 1 + 6π₯ β 3π₯ = 1 + 3π₯, Consequently, π π₯ = 1+3π₯
1β2π₯ 1βπ₯= 5
1β2π₯+ β4
1βπ₯= 5β 2π₯ πβ
π=0 β 4β π₯πβπ=0 ,
and ππ = 5 2π β 4,π β₯ 0.
Ex 10.4: (1.d) β’ ππ+2 β 2 ππ+1 + ππ = 2π,π β₯ 0,π0 = 1,π1 = 2. β ππ+2π₯π+2βπ=0 β 2β ππ+1π₯π+2β
π=0 + β πππ₯π+2βπ=0 =
β 2ππ₯π+2βπ=0 .
Let π π₯ = β πππ₯πβπ=0 . Then
π π₯ β π0 β π1π₯ β 2π₯ π π₯ β π0 + π₯2π π₯ = π₯2 β 2π₯ πβπ=0 .
π π₯ β 1 β 2π₯ β 2π₯π π₯ + 2π₯ + π₯2π π₯ = π₯2
1β2π₯.
π₯2 β 2π₯ + 1 π π₯ = 1 + π₯2
1β2π₯β π π₯ =
11βπ₯ 2 + π₯2
1β2π₯ 1βπ₯ 2 = 1β2π₯+π₯2
1βπ₯ 2 1β2π₯= 1
1β2π₯= 1 + 2π₯ +
2π₯ 2 + β―, so ππ = 2π,π β₯ 0.