It cold o r
( Pln ) ) μ-;
Melhemetuid halation is a
o (x ” ‘t ” .
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dimensional
vector degree n
contains has an
↳ hole Hot
to be no htt
dat number
twirly independent Vedas
polynomial
arbitrary m .
often difficile to directly prove
Induction gives u s a crutch . Instead, we prove
the following equivalent proposition : PLO) n the (Pln) → Platt))FG
technique for proving propositions
Plm) for en
-? Suppose they
( P h i l l . Iet
equi val ent
effect is you every
a n infinite
knock over the
“”” PCH: the k domino vill fall .
th (Plats Plath)
by nodus toners,
11 . PCH ) .
is 🙁 1) prove PCO) arbitrary
clarino will daring O
PG) then you ,,
knock over
→ Plz !Plus )
PCH → pity → p
k t prove Plklsplktll usually
w e n t [ base Cinderella step)
if it tells
every danby will fell
• Proof For
at last 19 6″
pint Hf:p, Rft’:O → .
:lR→lR is O ,
derivative
polynomial
” if PCI) r Anza (Plat→ Platt) ) .
This follows from MVT .
‘ lxko f ,
(Bese case)PCH: if f- ( Inductive steel Plklsplket) :
constant . proof. Assam
Plk) , so tht –
function with k ” derivtia O is a polynomial of degree E K 1 .
prove cry with YI
take fRsk ‘ O ,
So any i. f-,= .
Want fpolyhowtouse PCH?
function w/ Ckttlstckiwtineo is poly deg Ek ,
coefficients
x” t C is
Chd f is such
golgofoeg.sk , 2
autiderirofive, art’s
setting Sto the Sam, at1
to give atart
r s t – a r t . n . t o . i t ✓#
” Pxeognice the saltern ”
intuition like n o .
We ‘ll prove r e t , by
fn (Platt wore ,
i n d i c t i o n . a
Let’s give Phil
Plml → Plinth. a Wort
← th to use Pcm) ?
k ‘s . move
+a.r”‘=,a. I
n is inductive )
1 land 31, n=2 #§oIo±01tzG=ITd3
PII (non – n
PII ( inductive )
( Basel ( Inductive)
‘t 3k ‘t 3kt 1
k I k = 3 m
k ‘ -13k ‘
a s s u mp t i o n
So k?3mth, which is divisible by
lest 3h’ -12k : Xm +
available from you choose a
class 417 to
End af &4/10 Dey ,
Last day of melon’d Chet will appear on midterm,
book holes
not open interactive Internet
end review /dis → send questions, rearrests or, Discord clad of thief
Next week will
Tithe structured review
time to Tuesday
to HW problems
Best thing to do to get reedy is to be comfortable with enemy
delayed until Tuesday .
Today’s quiz
30 minutes
Iinishirgclgkrhl Last time, Prove the PIN ) by proving
ideation : -inductive
Plan new ( Mals Plath)
instance , Plath ) .
( Pln)) by PII ) ,.,
also versions with
hckiple base
cases . For (Phil →
you c a n p c o H
me : ” a ” l o s e
Theninparkour, PlasPCH,putsPG),r..
Plot, Might
in a n inductive
21k- Htt and
PHI , step
as well as
(Plat ) . Uk- I)
the broadest version
proof when
c a r e to available I So it’s
Pln, you typically
Pla- H already
all shatter dominos assure
V.ken (Plk)
the inductive Plh)→ Pln))
them (Mull
I aP l t .
pcosnpct, → pcz, pp
(thePCH)→M2) i:c –
> they be ”
product of primes i n .
let ‘s pase
NEMis prime if kln→
Pln ) prove
p . . – proving
htt, whichis
> 1 induction ?
( Pla ) Assuming
pm p; prime he get k: p- – ‘
k-I useless for
Peel . induction
Kellett’re to
: m2 step : pick
prove Plan )
nontrivial
divisor m’ofm
If then ‘ on
we find one
both be written as a
by assumption m and
Ifwe can suchan m,
divisors ere 13in so ,
primes I Thus
Base cases
is to induction as constructions for algorithms)
( 1.51 (2)
is an algorithm thot Vectors n :
there is none
factored . algorithm ?
recursive steer
Using this
(g) Multiply
the kind of
defined functions
toughief: N→ X, first define fish,
then define f-In) in terms of fuel for Ken .
III. Let X – N ,
nk fln- Itt flu- 2)
fat =L , fol- flatfoot – :3:*
let fool:O let ,
fly = full- fly =L :i ¥i÷÷¥±÷
something.
Let X– IR and define fol: I, flat- nofin- 1)formo. ,
fit’ ‘ flat’ h!
Bose age –
define IT recursively
can lleytto) :
” k o ! ‘- 1
Re-enie step (length) :
.. ITH ,xm )
t ) 2- :L fat
fol’- fat – ,,
recall that IRA- { (x ,xz ;xm ) IneHl ,
is the set of strings of red numbers ; it’s {11411341241121
Let’s define intuitively ,
product fanatics
x.IT/l–x.1–x
… . ,xm) = x – xz.
More precisely,
numbers ITCH,- . .,km- e).
#usually Recursively
TIK,y) = x. Ily) = x – y
. . . . . . xml sets The set
xn ) I empty, o r is
For any The
recursively
string is string .
empty string
antimony : Cl,(O)Ll),10,01,(1,01,10,11(1,11,10,411.
Is, a ) for ifA-{0,13, get
He bitsfrigs ! Note he could
compactness .
delete C) and,
words , S :
containing
definition
implicitly
that He A-→ A’
a n element concatenation .
Notethotindose- I ,,
Cs.at. Basically A’ is thesadist
contesting
and a function AKA → A ,
and a surjective
h t t for the set
is basically the set
recursively !
nothing else .
( so injective )
recursively
be the inductive
recursively
function, :
Is, a ) definition ,
string of length O : m , then there are
Oy induction I
tells.cl/z1 > 0 ,
km strings in
definition . so there is only one
empty, o r SEAS and aft,
C) Otherwise , if ( .
, k possibilities for a and
At wich lengthm.
For eng PII By
for s function
Shortland for “assentthe ore
concatenation
AA is defied (2.3.41 – lit, e )
possibilities
– strings of length u flea
s.t.CA, lls
!12,3,4, A) a Ks’s 19’T’ e )’
definition A . = llsal)): lls) – llsttllcl)
structural
induction over
11130 { G. at } ‘
Bese : t —
Inductive step: f-(t’ a) assure
lls.tl lls tht )
, alt ‘ .at )
) t let 1+1 = llsltlltl
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