55.6 Solving Recursive Relation by Iteration - The method of Iteration
Ex: Arithmetic Sequence 1 As - I
LII#IIfranymtegrk¥# What is at ?
§ A. = t
? Az -12=(1-124242)+2 i
guess An = It 2h ing
To prove it is true . We need Mathematical Induction .
if " Def
'Iff I a constant d Sio .
Apeak, -1 d . Abel"
if and only if "
Ao -- A Ape f. Apu , for all integer k3l . -
Ao -- a
Def : A sequence as . A .
- - - -
.
It follows that
Ana PM do for all integer h 31 .
- Remark 's fete free to Use it for HW . Quit . Exam -
Ex: ( Rich Auntie ) Ao=Sb0i0oo@ Ak -- 4.04) Apr, . for all integer RHw -
Q: how many years to reach $1000 , ooo ?
This is glegnefrxc Sequence . So An = 4.044. As -16437105 n31
Suppose after t years , it reaches $ tooo . ooo ,
then At -= 106
( I. 04T. 105 slob ( 1.044 = to Ln (1.04)t= Into t . In 11,04) = In b
t = Into Into, I 58,7
At year 59 , 2 Gets tooo , ooo .
-
wasguess explicit formula for Mn ? Pause & Think by yourself M ,
2 I
Mz = 2 -Mitt = 2. It I w un
the 2- that 1=2/2.1+1) th' 24241 Mff 2. M31-1=21242*1--231*2161
f w
. . . .
Exo. Q : finer in vertices .
How many edges can you get by Correcting vertices ?
s. ...¥¥¥÷÷÷÷:÷.4 3+326 • IIT
÷ ,a÷. .
.
I 51=0 I Sn ' Saint . 7h32 .
guess explicit for Sn £20 54=53 -1.3=(0+1+2) -13 §=s,tl did SEE' §=9YMIuNTkm 5,22
. Se MAI? r
Wrong ⇐ for a" integer Ksi .
#
. .
-
Cutie Lcr -4kt'D -
= 264k'HRtD = 2K" -12kt Rtl = 2Mt 3kt ) e Gapan mm 2kt 't shek 2K thtPekin : Cr¥ 2kt't 1km ¥
* ReadIt means Pikul is valid only if k .
Better Idea: check first 4 terms . .
Monday Discussion Session 1. Some Exam 2 Problem ?
2- Some confusing Mistakes .
§4.6w
A} No .
→ Show Phi
. Ply . . - Pcbs
Then PlkL
. . .
.
Exams? Problem 3: show An 33 ,
2.3+3.4-1 . -+ ftp.n.MY/Nt2ht3J -
: **÷÷÷÷÷i÷E D8 Given a large * ,
check if it is prime ? my
⇐¥÷÷÷÷*e
V-n.rsEE'
:*::i÷÷÷:÷÷÷÷
proof fibs : ( Def at Composite . Su
¥¥⇒ .si#isenn:::i::::m7*n .
to that factor affairs .EE?::.ces:sm.iFi.Tf# 2' . If FEN -
in . Tap Ff . rin ⇒ pln . PIRI
u - . . Pff
- Def : A set is a collect
⇐ '¥b¥ .
.
- Notation : let us suppose U is a set .
.
.
TEE subset: A is a subset of B
denote that as
.
! IXEB sit . X EA KH
. Z E E IR fitts Not a set
.
ta }=¥ 'll . 23 1 E fi , 23 X 1- E fi , 23 V
= Element proof: Let X. Y be two sets .
To prove X EY c-
% . Shppose X is particular but arbitrarily chosen fromX X E Y
Ex 5 Define Aa f int Z l Ms Grtk for some re-Z) uan
B - f n EZ / ne 35 for some SEZ) improveAIB .
Proof : D . ShowAEB_ Suppose XEA . then
Xi Gy t- K for somey EZ .
Nota that K 69+12=34+41 . . XEZ
.
3=3.1 :EB in
To show 3 TEA .
Then 3a fort 12 for some REZ .
e- e-
Contradiction ! . So 3¥ '
qq.pe,- Set Equality : Def : Two sets AaB are equal Iff They ane subsets of each other.
In other words AB ⇐ "
AIB "
A "
BEA "
.
Then X=6rtk = Gfrtz) ,
YE 65=-66-2/-1/2 . S-26-2 .
Universe "
Def : let A and B subsets of Uw → l ! The union of A andB . Is the set that Contains
all dents that are in A or in B .
AUB -
c-- - - "
.
An B = fxt U / xGA and XEB ) .M 3. Complement of A (w.rt U)-
W is the set of all elements (in U ) but NOT in A .
err
"xI¥¥¥¥. I
4 . The difference of B from A is a set of mum
debuts that is in B but NOT in A .
U B-A - fxtulxtb but # A 's
. /¥) Ex: ( Intervals ) . -
(a , b) = txt IRI acxcb } Ea
, b) = fx GIRI a * ab } C- bib? = HEIR / x Eb] A- f-hot
. Be [ O , D A- B = tho)
¥#¥→ - I 0 I
rfAi* .Ex ? { [ i. itIf I -- IRB) Is Chad
= { [ 1127 , [2137
( !%Ai¥fxEU/xtAi for allie't) .
Ex : If I = f ahh - " in} Then ( like the Sigma Notating)
¥⇒ Ai) . Hi Ai UADOA.)u¥.. I-- 251 Then
.
Ai -- tf Ai Same goes for !?±Ai - Empty set ( Null set )
Def : A set that Contains No element f } Denoted as Of
Ex; 1h33 n k 's = 0
{ XHRI He -11=9
.
C 'disjoint !
.
• Def . ( Mutually disjoint to
{ Ai Ii #I } are mutually disport ift for any i.j GI , itj then
Ain Aj -0
• (partition ) .
A collection of sets f Ai lieI } is a partition of a set A if A = U Ai
ifI
: f Ai lieI ] is mutually disjoint . Ex I Define
,
• Power Set
Def : Gilan a set A , the power set of A devoted PCA ) , I a set ofall subsets of A .
One dents Exo. -Way) - { f. KITH ,
1h23 } p m
• Cartesian Products . • Def of Ordered N-tuple let next
. Xi ,xziixn be numbers ( "II!;g:)The undead h-tuple cons its of Xi .- in
in the order five X , then Xuthen by , -w '
denoted as (Xi , Xz , - - -
Ex : ( 1,231 ! Ch 3,2) ( I
, Echo
The C.P. is denoted as
A , x Az x Az x - - -X An = { (ai , an -- .am/AoEAo , Khai- in}
Exo. A , a 1h23 . An label .
I xA Kha) . libs .cl .g. caesium . Has } Ai XA .
= ? c-
.
. Because BEC - ←
So ¥ . QED .
"
. Commutative Law :
HEB)EGAUCBUC) (AIB) IC = AMBAD
Distributive Law AU CBND= (AUB) ! AND AI Buc) = AnB) U CAN
. AUD = A .
'
. Unmindful:u=u
hmmm
VAN HIEI , then a)
af proof of.
,
w . show BE AAB
.
Suppose X E B .
NTS X EANB em
.
(AUB) ' -
- AcnBc
proofs show (AVBYEACNBC Suppose X C- (AUB)? NTS Xt Ach Bc m in
Because xe CAUSE . i.e . x ¢ AUB on
£EAoRB is true By de Horgan of Logic ~ (XEA OR x EB) = XxGADAndfk So XEIA and KGB
S XEAC and XEB '
- Empty set : Thang
. If E is empty . then EEA for any set A .
Empty set is unique . ( called • To show a set Ae 0 .
Suppose It has an element , & find Contradiction .
um
if AEB and BE C '
then Iiic =D ,
. XEA , so XEB .
Then XEC . xGE . So XG CNE =
QEP .Contradiction !
.
AUC ! Bi ) = n (AUB: ) 921
Proof : Suppose A. Bi - - ' Bn are any sets , show AOL.niB.ie#lEnEhiHn
@Suppose XE Abitibi) , Then XEA OR Xf ti Bi
l ? YEA : Then XE AUB ; for THR , - - in
so XE in , CAN Bil .
I ? XE IIBi : so XE Bi , forin←
Go XEAUl for it ,4- - in So
xe II #Bi) a Show in CARNE AUC .EBiD Suppose Xf in ( AUB ; )
ht
mm
Hasek . xGA.lt/IiiILAkBiI It means XE AUB,- for i4R.- in
.
9 , = 1
.
g. = a =@L.QE- g.FI --Ee ¥ E- g÷=⇐÷i÷¥3A-- fi .¥i .- TAE GE gf.IT = 7×215
s
" 3¥
-
2 .
.
QB what is an ?
*
2.Sn=nHamHa±⇒can
hwan -r - Tao'EdHa±±TEi=zaIn.de/2Sn=IDtOE2AotAdI÷÷¥÷÷:asn=±k¥tn÷÷a -
3 .
sn-ebsf-kn-n-3.LT#Sn=3t32t3.Ix....t3.zn2E=s¥iaEaE¥E F-
Sabiaact from : - O : *
Sn = 3.213.243K¥.*32 - (Q.IE#5aIii*xxH&nn=3.aM-
Trick :D Multiply Ign at the Common
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