Math Analysis I - Basic Notions

Math Analysis Basic Notions

{XEXIp w - "xiseveu" A -A-The complement is thus obtained by negating the characteristic property:CCA) {XEX(CA) l a p e l }-Reiterating the same process for proper logical connectives, it becomes possible toexpress set operations: intersection, union, difference, symmetric difference,by describing the corresponding characteristic properties.{reXIparthgal}An B-A:{xEX Ipa)} {xeXlpartracx)}A- U B -{REX law}B - 17941}{XEXIA , B - par)A B - {xexlpcxlv.ae)}A-1.

Quantifiers

Given a predicate p(x), with the variable x belonging to a certain set X, one isnaturally lead to ask whether or not p(x) is true for all elements x, or if there existsat least one element x making p(x) true.When posing such questions we are actually considering the following formulas:

Universal q u a n t i t i e sFx "forX, pal holds"E pix)aeese,V x IN,Ee . xxog.Existential quantifierFx X, "thereE pix) leastexists a t a ,o n et h a t h o l d s "such partI x EIN,

5 7Xe. g.11F! X, therepartE e x i s t sx unione a ,aand o n l y,one o n et h a t h o l d s "such partF! I R ,E 3 = 7e . X tg.

By putting a quantifier in front of a predicate it is transformed into a formula,whose truth value may be then determined.

However the effect of negation on a quantified predicate must be handled with attention. Suppose for instance x indicates the generic student of the Polytechnic, and let p(x) = ‘x is an Italian citizen’. The formula ‘∀x, p(x)’ (‘every student of the Polytechnic has Italian citizenship’) is false. Therefore its negation ‘￢(∀x, p(x))’ is true, but beware: the latter does not state that all students are foreign, rather that ‘there is at least one student who is not Italian’.

(tx ⇐pix)) ExE X , -par)X ,E> (Fx ⇐K , (x)) X ,f rE par t- E ap1. Math Analysis Basic Notions 14 of 74

Properties

If a predicate depends upon two or more arguments, each of them may be quantified.

"x y'y)(x, ±P =FyFx I R , y)( x ,E truepIR,Fxfy Y)E pi , FA L S EFxty PA, Y)I R , FALSEEFAY par,'llR ,E T H E1. Math Analysis Basic Notions 15 of 74

SETS NUMBERS

O FIN-nonnegativeSet

This set has the numbers 0, 1, 2,... as elements.

The operations of sum and product are defined on N and enjoy the well-known commutative, associative and distributive properties.

Operations II;

Sum(m,u) → m= t . . . . t e ,t ezta n e ,o o=D Commutativity INtt h Em u tt m m, n2) A s s o c i a t ea)( m t (ut tm,k) INER - m Utt n ,3) N e u t

r u e n t INt m Em t o O f m - m=Prwowctlm.nl#m.n-D Commutativity INt Em . w e n . m,m n2) A s s o c i a t ea)( m . (n. tm,R) INEK - m . U. n ,3) N e u t r o n INF m Em - s - s . mm =4) Distributivity INFm.(mta) K ERk k m u n ,t=1. Math Analysis Basic Notions 16 of 74PropertiesNotation NtWe shall indicate by the set of natural numbers different from 0{o}Nt lIN=A natural number n is usually represented in base 10 by the following expansion: 0¥ Cielo"Z, Cielo" l o " '. . .n = -A t c a , t cotc, lohell INE decimalo f m eC, 5where digits= o n e o f n(Cp 6)C u . ,n " c '= ' (IO)to BASE(Cp 6 )Cn n - i - r - c '= 2 - a (2)BINARY BASERepresentationNatural numbers can also be represented geometrically as points on astraight line.For this it is sufficient to fix a first point O on the line, called origin, andassociate it to the number 0, and then choose another point P different fromO, associated to the number 1. The direction of the line going from O

To P is called positive direction, while the length of the segment OP is taken as unit for measurements. By marking multiples of OP on the line in the positive direction we obtain the points associated to the natural numbers.

to0 1 9 ¥2 3LO1. Math Analysis Basic Notions 17 of 74

Principle Induction

The Principle of Mathematical Induction represents a useful rule for proving properties that hold for any integer number n, or possibly from a certain integer n0 ∈ N onwards. In order to use induction principle, a first heuristic deduction or a very strong conjecture is needed.

Let n0 ≥0 be an integer and denote by P(n) a predicate defined for every integer n≥n0. (n)fu% %no no, po

Suppose the following conditions hold:

1. (no) true
2. p INDUCTION
3. is BASIS I N D U CT I O N
4. ii) Fu 1)pcu)% par t truen o , - A is steptu INtheorempen)p(u) true-a i nis is% no, e

Proof (byington) Cn-l) t u(n-2)u! EIN(n) - 2 . su==p . . .a) o ! b y definition= Liidcutil! (htt).. s - n ! (htt)Cu-z)....n ( n - 1 )-

Theorem

Gausso fSumeelwumbersfnomitoi53 42II, to tototot o 99 9698 97100 L O L L O LL O L L O L L O L 50lol.S o m e =5050=1. Math Analysis Basic Notions 18 of 74Generalisation GaussofSumofalluumbersjwnstont1) w i s e 53 42I| to totototor n - I n - 3n-2u u - u I(htt)h t th t t sum--h t th t th t t1)n i add computatiousftotnenti)thet oo 3 420 I| to totototor n - I n - 3n-2u h - hsum-u.tn#Pnooj/hhhhhbyiuow-tiou)u 412¥[pansPln) e n o -06¥-o-Egan,e) pro) - h a t hIopa,ii)p(a) = 2"I§pCn)=ChH){nt#=(utµ)inJpluti) -Induction:¥§pCn) =h(nz#turtle-I£p(a) t u t ,=n(nt¥)=(htY{u¥ f u EIN, pal1. Math Analysis Basic Notions 19 of 74StatisticsSuppose we have n ≥ 2 balls of different colours in a box. In how many ways canwe extract the balls from the box?When taking the first ball we are making a choice among the n balls in the box;the second ball will be chosen among the n−1 balls left, the third one among n−2and so on.Altogether we have

n(n-1) • • • • 2 • 1 = n! different ways to extract the balls: n! represents the number of arrangements of n distinct objects in a sequence, called permutations of n ordered objects.

If we stop after k extractions, 0 < k < n, we end up with n(n-1) • • • (n-k + 1):

K select boxes to suppose out of u:

1st the first box, there are u possibilities to choose from.

2nd the second box, there are (u-1) possibilities to choose from.

3rd the third box, there are (u-2) possibilities to choose from.

• • •

Kth the Kth box, there are (u-K+1) possibilities to choose from.

After K extractions: n(n-1) • • • (n-K+1) = (u-K)!

That is the number of possible permutations of n distinct objects in a sequence of K objects.

If we allow repeated colours, for instance by reintroducing in the box a ball of the same colour as the one just extracted, each time we choose among n.h "h "After k > 0 choices there are then possible sequences of colours: is the number of permutations of n objects in

sequences of k, with repetitions (i.e., allowing an object to be chosen more than once).

1. Math Analysis Basic Notions 20 of 74

Therefore the number of permutations of j object is equal to j factorial.

l e t objects boxes

J b e i n 11st object choices

I2' object choices

I - 1r n•Tth choice

o b j e c t A( 1 1 - g !f r- I l y - l ) . . .

Given n balls of different colours, let us fix k with 0 ≤ k ≤ n. How many different sets of k balls can we form?

Extracting one ball at a time for k times, we already know that there are (n−1). . .(n−k + 1) outcomes. On the other hand the same k balls, extracted in a different order, will yield the same set. In simple permutations every choice has been counted many times, correspondingly to the possible ordering of k boxes, that is, the possible exchanges of permutations of k objects.

Since the possible orderings of k elements are k!, we see that the number of distinct sets of k balls chosen from n is:

(n-Kti)(n-l)... n!h- - Cu-K)!K! K !

The coefficient represents the number of combinations of n objects taken k at a time. Equivalently, it represents the number of subsets of k elements of a set of cardinality n.

n! N E W TO N ' s(%)=M R . BINOMIAL1. Math Analysis Basic Notions 21 of 74)(Newton's Theorem Given two natural numbers n and k such that 0 ≤ k ≤ n, one calls binomial coefficient the number: -hln-dji.la#=-"nffifey" (f) =~hII. - I -Ch-d!J ! I!(u-d!cytu-II.lu#= U == ( a n t s ) ! (n-y)!cnn.ffi.cn#-- (La) ( I )= -181-1¥-1=121.11-e . g..a.tn#--1!iiII=u--nH(n - I - N n - i ) C n # n (all•ii. ← # ÷(u-z)!2!⇐ (n-1)(u-2)l n - 3 1 . l u - 3 !n= # =3!(4-3)/ -"la-1) (n-z,=1. Math Analysis Basic Notions 22 of 74Remarks: f u z e ta1 t h a tsuch o r g e t44=1511+1%4The expression provides a convenient means for comp