Complete integers

Complete Integers Davide Peressoni version 0.1.1 January 2021 ©May 2015 Davide Peressoni [section 1.5, section 1.6] ©May 2017 Davide Peressoni [section 1.3, chapter 3, section F.4] ©April 2018 Davide Peressoni [chapter 1] ©March 2019 Davide Peressoni [rev. chapter 3] ©April 2019 Davide Peressoni [chapter 2] ©November 2020 Davide Peressoni [rev. chapter 2] ©January 2021 Davide Peressoni [rev. chapter 1] i Abstract In this document we will see the complete integers (ZC): a superset of integer numbers (Z), which also contains the dual of Zalong parity (dis-integers, ZD). Then we will see some properties of these numbers and of the even unit (l∈ZD). Moreover we will define perenomials and hyper-perenomials, an extension of polynomials, which allow us to write functions defined by cases with a single formula. Resumo En ˆci tiu dokumento ni vidos la kompletajn entjerojn (ZC): superaro de entjeraj nombroj (Z), kiu anka˘u enhavas la dualon de Z la˘u pareco (neentjerojn, ZD). Tiam ni vidos iujn ecojn de ˆci tiuj nombroj kaj de la para unuo (l∈ZD). Cetere ni difinos la perenomojn kaj la hiperperenomojn, etendon de polinomoj, kiujn ni uzos por skribi funkciojn per kazoj kun ununura formulo. Somari In chest document i definirìn i intîrs complets (ZC) che i numars intîrs (Z) a son un lôr sotinsiemi, e che a tegnin dentri ancje il duâl di Zcun rispiet a la paritât (disintîrs, ZD). Po daspò i viodarìn diviersis proprietâts di chescj numars e de unitât pâr (l∈ZD). Cun di plui i definirìn i perenomis e i iper-perenomis, une estension dai polinomis, che o doprarìn par scrivi funzions par câs dome cuntune formule. Sommario In questo documento vedremo gli interi completi (ZC): un sovrainsieme dei numeri interi (Z), il quale contiene anche il duale di Zrispetto alla parità (disinteri, ZD). In seguito vedremo alcune proprietà di questi numeri e dell’unità pari (l∈ZD). Inoltre definiremo perenomi, ed iper-perenomi: un’estensione dei polinomi che ci permetterà di scrivere funzioni per casi usando un’unica formula. Contents Contents ii Nomenclature v 1 Complete integer numbers 1 1.1 Valueandparity ........................ 3 1.2 Relatedsets........................... 7 1.3 Ratio between even numbers . . . . . . . . . . . . . . . . . . 12 1.4 Cartesian plot of half-complete rationals . . . . . . . . . . . 13 1.5 Evenunit ............................ 14 1.6 Exponential of complete integers . . . . . . . . . . . . . . . . 17 2 Perenomials 21 2.1 Parity of perenomials . . . . . . . . . . . . . . . . . . . . . . 23 2.2 Perenomial matrix form . . . . . . . . . . . . . . . . . . . . 29 2.3 Binomial theorem . . . . . . . . . . . . . . . . . . . . . . . . 35 2.4 Interpolation .......................... 36 2.5 Hyper-perenomials . . . . . . . . . . . . . . . . . . . . . . . 39 3 Selector function 45 3.1 Signfunction .......................... 45 3.2 Heaviside step function . . . . . . . . . . . . . . . . . . . . . 45 3.3 Rectfunction .......................... 47 3.4 Selectorfunction ........................ 48 3.5 Indicator function . . . . . . . . . . . . . . . . . . . . . . . . 50 3.6 Functions defined by cases with selector function . . . . . . . 52 Future work 57 F.1 Applications........................... 57 F.2 Parityextension......................... 57 F.3 Perenomial extension . . . . . . . . . . . . . . . . . . . . . . 57 CONTENTS iii F.4 Complete integers and complex numbers . . . . . . . . . . . 58 References 59 Nomenclature Costants oOdd zero, see Remark 1.7, page 15. lEven unit, see Definition 1.11, page 14. Variables n(Complete) natural number. xReal number. z(Complete) integer number. zComplex number. pParity, see Notation. vValue, see Notation. v Vector. MMatrix. Functions par(z)Parity function, see Definition 1.4, page 5. val(z)Value function, see Definition 1.3, page 3. sgn(x)Sign function, see section 3.1, page 45. 1(x)Heaviside step function, see section 3.2, page 45. χA(x)Indicator function, see section 3.5, page 50. Π(x)Rect function, see section 3.3, page 47. 4CHAPTER 1. COMPLETE INTEGER NUMBERS 4. Completely multiplicative. val(1) = 1 val(x·y) = val(x)·val(y)(1.9) Proof. 1. x∈Z⇒ −x∈Z⇒val(−x)=−x=−val(x) x="v p#∈ZC⇒val(−x) = val −"v p#!= val "−v p#!=−v=−val(x) 2. a) x, y ∈Z val(x)=x; val(y)=y val(x+y) = x+y= val(x) + val(y) b) x="vx px#, y ="vy py#∈ZC val(x) + val(y)=vx+vy val(x+y) = val "vx px#+"vy py#!= val "vx+vy px⊕py#!=vx+vy And then val(nx) = val(x+···+x | {z } n ) = val(x)+···+ val(x) | {z } n =nval(x) for n∈N. Finally val(zx) =    zval(x)z∈N −val(−zx)=zval(x)z∈Z\N 3. val(x)∈Z⇒val(val(x)) = val(x) 4. a) x, y ∈Z val(x)=x; val(y)=y val(x·y) = x·y= val(x)·val(y) b) x="vx px#, y ="vy py#∈ZC val(x)·val(y) = vx·vy val(x·y) = val "vx px#·"vy py#!= val "vx·vy px·py#!=vx·vy 1.1. VALUE AND PARITY 5 Definition 1.4 (Parity function). par: Z∪ZC→F2 Ze= 2Z={2z:z∈Z} Zo= 2Z+ 1 = ={2z+ 1: z∈Z} par(z):=   0z∈Ze 1z∈Zo∀z∈Z par "v p#!:=p (1.10) Remark 1.1.Observe that par |Z(z) = χZo(z). Theorem 1.2 (Properties of parity).Given x, y ∈Z∨x, y ∈ZCand z∈Z 1. Parity is even. par(−x) = par(x)(1.11) 2. Linearity. Since par(x)∈F2the sum operator must be replaced by exclusive or (⊕) and so the homogeneity property is valid modulus 2. par(x+y) = par(x)⊕par(y)(1.12) par(z·x)≡2zpar(x)(1.13) 3. Idempotence of the parity. par ◦par = par (1.14) 4. Completely multiplicative. par(1) = 1 par(x·y) = par(x)·par(y)(1.15) Proof. 1. a) x∈Z par(−x) =    0−x∈Ze 1−x∈Zo =   0x∈Ze 1x∈Zo = par(x) b) x="v p#∈ZC par(−x) = par −"v p#!= par "−v p#!=p= par(x) 6CHAPTER 1. COMPLETE INTEGER NUMBERS 2. a) x, y ∈Z x y par(x) par(y) par(x)⊕par(y)x+ypar(x+y) ZeZe0 0 0 Ze0 ZeZo0 1 1 Zo1 ZoZe1 0 1 Zo1 ZoZo1 1 0 Ze0 b) x="vx px#, y ="vy py#∈ZC par(x+y) = par "vx px#+"vy py#!= par "vx+vy px⊕py#!=px⊕py par(x)⊕par(py)=px⊕py And then for n∈N par(nx) = par(x+···+x | {z } n ) = par(x)⊕···⊕par(x) | {z } n≡2npar(x) Finally par(zx)  ≡2zpar(x)z∈N = par(−zx)≡2−zpar(x)≡2zpar(x)z∈Z\N 3. par(x)∈F2⊂Z⇒par(par(x)) =    0 par(x)∈Ze 1 par(x)∈Zo = =   0 par(x)=0 1 par(x)=1= par(x) 4. a) x, y ∈Z x y par(x) par(y) par(x)·par(y)x·ypar(x·y) ZeZe0 0 0 Ze0 ZeZo0 1 0 Ze0 ZoZe1 0 0 Ze0 ZoZo1 1 1 Zo1 b) x="vx px#, y ="vy py#∈ZC par(x)·par(y)=px·py par(x·y) = par "vx px#·"vy py#!= par "vx·vy px·py#!=px·py 1.2. RELATED SETS 7 Lemma 1.2 (Parity of powers). par(zn) = par(z)∀n∈N+ Proof. From 1.15 and since 0n= 0 and 1n= 1 par(zn) = [par(z)]n= par(z) 1.2 Related sets Now we will see some subsets of ZCwhich will be useful later. Definition 1.5 (Integers prime).Let’s define the set of the integers prime as Z′:=("v p#∈ZC:p= par(v))(1.16) Definition 1.6 (Dis-ntegers).Let’s define the set of the dis-integers as ZD:=("v p#∈ZC:p= par(v))(1.17) Remark 1.2.{Z′,ZD}is a partition of ZC. A⊔B=C⇔A∪B=C A∩B=∅ Z′⊔ZD=ZC Theorem 1.3 (Zand Z’ are isomorphic).The function fZ:Z→Z′defined as fZ(z) = "z par(z)#(1.18) is a isomorphism. Proof. Let’s check if fZis a homomorphism: a, b ∈Z •For addition: fZ(a+b) = "a+b par(a+b)#="a+b par(a)⊕par(b)# fZ(a)+fZ(b) = "a par(a)#+"b par(b)#="a+b par(a)⊕par(b)# 8CHAPTER 1. COMPLETE INTEGER NUMBERS •For multiplication: fZ(a·b) = "a·b par(a·b)#="a·b par(a)·par(b)# fZ(a)fZ(b) = "a par(a)#" b par(b)#="a·b par(a)·par(b)# •For multiplicative identity: fZ(1) = "1 1#= 1 Now let’s check if fZis an isomorphism, that is if ∃f−1 Z:Z′→Zhomomorphism. Picking z∈Z′we can write z="v par(v)#with v= val(z)∈Z, so let’s define g:Z′→Z g " v par(v)#!:=v Now let’s prove that g(z)=f−1 Z(z), in other words fZ◦g= id: fZ g " z par(z)#!!=f(z) = "z par(z)# and g−1(z)=fZ(z), or g◦fZ= id: g(fZ(z))=g " z par(z)#!=z Since Z′is isomorphic to Z, and so the two cannot be distinguished [Art91], we won’t write Z′anymore and we will use the notation hv par(v)ito denote elements in Ztoo. More precisely we will write, with an abuse of notation, Z′=Zand hv par(v)i= vmeaning respectively Z′=fZ(Z)and hv par(v)i=fZ(v). Remark 1.3.Note that, by definition, val(z) = val(fZ(z)); par(z) = par(fZ(z)) 1.2. RELATED SETS 9 Remark 1.4 (ZDis not a ring).It’s easy to see that ZDis not closed under addition: "1 0#+"1 0#="2 0# Definition 1.7. We can define the restrictions of ZDand ZCto natural numbers: ND:=("v p#∈ZD:v∈N)(1.19) NC:=("v p#∈ZC:v∈N)(1.20) We can also extend the sets of even and odd numbers: ZDe:=("v 0#∈ZD)={z∈ZD: par(z)=0}(1.21) ZDo:=("v 1#∈ZD)={z∈ZD: par(z)=1}(1.22) ZCe:=("v 0#∈ZC)={z∈ZC: par(z)=0}=Ze⊔ZDe(1.23) ZCo:=("v 1#∈ZC)={z∈ZC: par(z)=1}=Zo⊔ZDo(1.24) Definition 1.8 (Half-complete rational numbers).We now define the set of half-complete rational numbers as QH:=Q×F2=("v p#:v∈Q, p ∈F2)(1.25) And we extend val and par in such a way val hv pi=v∀hv pi∈QH(1.26) par hv pi=p∀hv pi∈QH(1.27) Remark 1.5. Z⊔ZD=ZC⊂QH Definition 1.9 (Ring QH).Let’s define QHas a commutative ring with unit which extends ZC: 10 CHAPTER 1. COMPLETE INTEGER NUMBERS Given "a b#,"c d#∈QH "a b#+"c d#:="a+c b⊕d#(1.28) "a b#·"c d#:="a·c b·d#(1.29) Theorem 1.4 (Division of complete integers).The division between a complete integer hv1 pi∈ZCand a complete odd hv2 1i∈ZCo\h0 1iis an halfcomplete rational: hv1 pi hv2 1i="v1/v2 p#∈QH Besides, QH=x y:x∈ZC, y ∈ZCo\h0 1i. Proof. Let’s start checking hv1 pi hv2 1i="v1/v2 p#: "v1 p#=hv1 pi hv2 1i"v2 1#="v1/v2 p#"v2 1#="v1 v2v2 p#="v1 p# Now let’s check that "v1/v2 p#∈QH: v1 v2∈Qsince v1, v2∈Zand p∈F2 Finally we are going to prove that all QHcan be generated in such way: "v p#∈QH=⇒v∈Q=⇒v=a b:a, b ∈Z=⇒"v p#=ha pi hb 1i Corollary 1.4.1. The division between an half-complete rational hv1 pi∈QH and a complete odd hv2 1i∈ZCo\h0 1iis an half-complete rational: hv1 pi hv2 1i="v1/v2 p#∈QH The proof is similar to the one above. 1.2. RELATED SETS 11 Definition 1.10 (Complete rational numbers).Defining QH:=     q h1 0i:q∈QH    (1.30) we can define the set of complete rational numbers as QC:=QH⊔QH(1.31) Theorem 1.5 (Complete division of complete integers).The division between two complete integers x, y ∈ZC,val(y)= 0 is a complete rational and QC=nx y:x, y ∈ZC,val(y)= 0o. Proof. If y∈ZCothen for Theorem 1.4 we know x y∈QH⊂QCand that QH=x y:x∈ZC, y ∈ZCo\h0 1i. Otherwise, if y∈ZCethen ∃z∈ZCosuch that y=zh1 0i(z=hval(y) 1i). Now for Theorem 1.4 q=x z∈QHand so x y=x zh1 0i=q h1 0i∈QH⊂QC. Then QH=     q h1 0i:q∈QH    =     x zh1 0i:x∈ZC, z ∈ZCo\h0 1i    =(x y:x∈ZC, y ∈ZCe\{0}) And finally QC=QH∪QH=(x y:x, y ∈ZC,val(y)= 0) Lemma 1.3. We can now see that Qis a subset of QC: Q⊂QC Proof. Each element of Qis a ratio between two integers: ∀q∈Q∃x, y ∈Z, y = 0: q=x y but then q=x y∈QCbecause x, y ∈Z⊂ZCand val(y)= 0. 12 CHAPTER 1. COMPLETE INTEGER NUMBERS 1.3 Ratio between even numbers Lemma 1.4 (Parity of the ratio between even numbers).The ratio between two even numbers could be either odd or even. Proof. If x y=zwe know that x=yz, but since yis even xwill be always even, independently of the parity of z. Another way to see this is: z=x y=hval(x) 0i hval(y) 0i="val(x) val(y) 0/0# where val(z) = val(x) val(y)and par(z) = 0 0which could be either 0 or 1. Now that we know the ratio between even numbers could be either even or odd, we expect that such ratio has two results. The two results will have the same value but one will be even and the other odd; such change of parity could be done adding an odd zero h0 1i: Theorem 1.6 (Ratio between even numbers). a b= val a b∨val a b+"0 1#∀a, b ∈ZCe Example 1.1. val 2 2= 1 2 2= 1 ∨1 + "0 1#= 1 ∨"1 0# In facts 2·1=2and 2·h1 0i= 2. Example 1.2. val 4 6=2 3=h2 0i h3 1i="2/3 0# 4 6=2 3∨2 3+"0 1#=2 3∨"2/3 1# In facts 6·2 3= 4 and 6·h2/3 1i= 4. 1.4. CARTESIAN PLOT OF HALF-COMPLETE RATIONALS 13 val(x) par(x) −1 0 1 2 3 4 5 6 7 1 0 Figure 1.1: 1 = h1 1i;h1 0i;h2 1i;5 = h5 1i;h7 0i=h2 1i+h5 1i val(x) par(x) −1 0 1 2 3 4 5 6 7 1 0 1 Figure 1.2: x= 1;x=h1 0i;x= 3 ·1;x= 3 ·h1 0i 1.4 Cartesian plot of half-complete rationals We can see the sum between half-compelte rationals as a vector sum on a bi-dimensional cartesian plane with the value on x-axis and parity on y-axis (Figure 1.1). Obviously QHis not a vector space because Qand F2are different fields. In fact on x-axis we could have only rational numbers and on y-axis we have only 0 and 1 cyclically repeated. The previous representation is valid also for "scalar product", which is the product by an integer (Figure 1.2). Remark 1.6.Repeating parity on the negative side of the y-axis we notice that the "vectors" which goes up by one or goes down by one alongside y-axis are indeed the same (Figure 1.3).So we don’t need to repeat parity (Figure 1.4). 20 CHAPTER 1. COMPLETE INTEGER NUMBERS Antiderivative F(x) = Zxa+bldx=x(a+1)+bl (a+ 1) + bl +c=xa+1+bl a+b+ 1 +cwith c∈R Like for derivative. Lemma 1.10. Zxzdx=xz+1 val(z)+1+cwith c∈R;∀z∈ZC Proof. d dx xz+1 val(z)+1+c= val(z+ 1) xz val(z)+1 =  val(z)+1 xz   val(z)+1 Chapter 2 Perenomials In this chapter we will extend polynomials in order to have also dis-integers exponents. Definition 2.1 (Polynomial ring).Let’s call Pthe set of polynomials over field K[Hal72]: P:=K[x]={p(x) = n X 0ipixi:n∈N, p ∈Kn+1} Definition 2.2 (Dis-polynomials).Now we will define the dis-polynomils as: PD:={q(x) = m X 0iqixi+o:m∈N, q ∈Km+1} Lemma 2.1 (Relationship beetwen polynomials and dis-polynomials). q(x)∈PD=⇒ ∃ s(x)∈P:q(x) = s(x)xo(2.1) p(x)∈P=⇒ ∃ t(x)∈PD:p(x)=t(x)xo(2.2) Proof. 1. Given q(x) = m X 0iqixi+o we can define s(x):= m X 0iqixi∈P and so s(x)xo= m X 0iqixixo= m X 0iqixi+o=q(x) 22 CHAPTER 2. PERENOMIALS 2. We can define t(x):=p(x)xo∈PD and so t(x)xo=p(x)x2o=p(x)  x0=p(x) Definition 2.3 (Degree of a dis-polynomial).given PD∋q(x) = s(x)xo deg(q):= deg(s) Lemma 2.2 (Polynomials and dis-polynomials). P∩PD={0} Proof. p∈P∩PD⇔p∈P∧p∈PD p∈P⇔p(x) = n X 0ipixi q∈PD⇔q(x) = m X 0iqixi+o ∃(p, q),(n, m): p=q? n X 0ipixi= m X 0iqixi+o⇔ n X 0ipixi− m X 0iβixi+o= 0 ∀x assuming N= max{n, m}and pi=qj= 0 ∀i=n+ 1, . . . , N;j=m+ 1, . . . , N N X 0ipixi−qixi+o= N X 0ipixi−qixixo= = N X 0i(pi−qixo)xi= 0 ∀x⇔p =q = 0 Definition 2.4 (Perenomials).We define the set of complete polynomials (or perenomials) as: PC:={r(x) = p(x) + q(x): p∈P, q ∈PD} Theorem 2.1 (Polynomials, dis-polynomials and perenomials). 2.1. PARITY OF PERENOMIALS 23 1. P⊂PC 2. PD⊂PC 3. P∪PD⊊PC Proof. From Lemma 2.2 we know that 0∈P∩PD 1. p(x) = p(x)+0∧0∈PD=⇒p∈PC∀p∈P 2. q(x) = 0 + q(x)∧0∈P=⇒q∈PC∀q∈PD 3. From previous points we see that P∪PD⊂PC, moreover ∃t(x) = 1 + xo∈PC;t∈ P∪PD 2.1 Parity of perenomials Theorem 2.2 (Even–odd decomposition of a perenomial).Given r(x) = p(x)+s(x)xo, calling fe(x) = f(x)+f(−x) 2the even part of fand fo(x) = f(x)−f(−x) 2the odd part of f[IEE02]: 1. re(x)=pe(x)+so(x)xo 2. ro(x) = po(x)+se(x)xo Proof. 1. re(x) = r(x) + r(−x) 2=p(x)+s(x)xo+p(−x) + s(−x)(−x)o 2= =[p(x)+p(−x)] + s(x)xo−s(−x)xo 2= =[p(x)+p(−x)] + [s(x)−s(−x)]xo 2= =p(x)+p(−x) 2+s(x)−s(−x) 2xo=pe(x)+so(x)xo 24 CHAPTER 2. PERENOMIALS 2. ro(x) = r(x)−r(−x) 2=p(x)+s(x)xo−p(−x)−s(−x)(−x)o 2= =[p(x)−p(−x)] + s(x)xo+s(−x)xo 2= =[p(x)−p(−x)] + [s(x) + s(−x)]xo 2= =p(x)−p(−x) 2+s(x)+s(−x) 2xo=po(x)+se(x)xo Definition 2.5 (Even-nomials).Let’s define the set of even-nomials as: Pe:={v(x): v∈PC;veven} vis even if v(x) = ve(x)or, in an equivalent way, v(x)=v(−x). Definition 2.6 (Odd-nomials).Let’s define the set of odd-nomials as: Po:={w(x): w∈PC;wodd} wis odd if w(x) = wo(x)or, in an equivalent way, w(x)=−w(−x). Lemma 2.3 (Evenand odd-nomials). Pe∩Po={0} Proof. t(x)∈Pe∩Po⇔t(−x) = −t(−x)⇔t(−x) = 0 Theorem 2.3 (Even-, oddand perenomials). 1. Pe⊂PC 2. Po⊂PC 3. Pe∪Po⊊PC Proof. 1. By definition. 2.1. PARITY OF PERENOMIALS 25 2. By definition. 3. From previous points we see that Pe∪Po⊂PC, moreover ∃t(x) = 1 + xo∈PC te(x) = 1; to(x)=xo=⇒t=te∧t=to=⇒t∈ Pe∪Po Lemma 2.4 (Perenomial decomposition by parity). r∈PC⇔r(x) = v(x) + w(x); v∈Pe, w ∈Po Proof. =⇒r(x)=re(x)+ro(x) ⇐=by definition. Definition 2.7 (Perenomial units).let’s define: •x0= 1 as perenomial even unit. •xoas perenomial odd unit. Remark 2.1 (Unit).∀r∈PC 1. |r(x)x0|=|r(x)| 2. |r(x)xo|=|r(x)| Proof. 1. |r(x)x0|=|r(x)·1|=|r(x)| 2. |r(x)xo|=|r(x)||xo|=|r(x)| Remark 2.2 (Parity of the units). 1. x0∈Pe 2. xo∈Po 26 CHAPTER 2. PERENOMIALS Lemma 2.5 (Relationship between evenand odd-nomials). w(x)∈Po=⇒ ∃ t(x)∈Pe:w(x)=t(x)xo(2.3) v(x)∈Pe=⇒ ∃ t(x)∈Po:v(x)=t(x)xo(2.4) Proof. 1. Given w(x) = p(x) + s(x)xo w∈Po=⇒w(x) = wo(x)=po(x)+se(x)xo defining t(x):=se(x)+po(x)xo t(−x) = se(−x)+po(−x)(−x)o=se(x)−po(x)(−xo) = t(x) =⇒t∈Pe and so t(x)xo=se(x)xo+po(x) x2o=w(x) 2. Defining t(x):=v(x)xo∈Po so t(x)xo=v(x) x2o=v(x) Theorem 2.4 (Expression of even-nomials). v(x)∈Pe⇔v(x) = n X 0ivixil;n∈N,v ∈Kn+1 Proof. =⇒v(x) = ve(x) = pe(x)+so(x)xo= = n X 0ipi[xi+ (−x)i] + m X 0isi[xi−(−x)i]xo 2 Since xi+ (−x)i=   2xii∈ZCe 0x∈ZCo 2.1. PARITY OF PERENOMIALS 27 we can say n X 0ipi[xi+ (−x)i] = 2 n X 0i∈ZCe pixi= 2 n X 0i∈ZCe pixil Since xi−(−x)i=   0x∈ZCe 2xii∈ZCo we can write m X 0isi[xi−(−x)i]xo= 2 m X 0i∈ZCo sixixo= 2 m X 0i∈ZCo sixi+o and from par(i+o) = par(i)⊕1=0 so i+o= (i+o)l=il +0=il and so 2 m X 0i∈ZCo sixi+o= 2 m X 0i∈ZCo sixil Defining vi:=       pii∈ZCe, i ⩽n sii∈ZCo, i ⩽m 0 N:= max{n, m} we can write 2 n X 0i∈ZCe pixil = 2 N X 0i∈ZCe vixil; 2 m X 0i∈ZCo sixil = 2 N X 0i∈ZCo vixil Finally v(x) = n X 0ipi[xi+ (−x)i] + m X 0isi[xi−(−x)i]xl−1 2=  2 N X 0i∈ZCe vixil + 2 N X 0i∈ZCo vixil  2= N X 0ivixil 28 CHAPTER 2. PERENOMIALS ⇐= v(−x) = n X 0ivi(−x)il = n X 0ivi(−x)li= n X 0ivixli= n X 0ivixil =v(x) Theorem 2.5 (Expression of odd-nomials). w(x)∈Po⇔w(x) = n X 0iwixil+o;n∈N, w ∈Kn+1 Proof. ∃t(x)∈Pe:w(x)=t(x)xo t(x)∈Pe⇔t(x) = n X 0iwixil w(x) = n X 0iwixilxo= n X 0iwixil+o Lemma 2.6 (Sum of even-nomials). v1, v2∈Pe=⇒v1+v2∈Pe Proof. (v1+v2)(−x) = v1(−x)+v2(−x) = v1(x)+v2(x)=(v1+v2)(x) Lemma 2.7 (Sum of odd-nomials). w1, w2∈Po=⇒w1+w2∈Po Proof. (w1+w2)(−x) = w1(−x)+w2(−x)=(−w1(x))+(−w2(x)) = −(w1+w2)(x) Lemma 2.8 (Product of evenand odd-nomials). 1. v1, v2∈Pe=⇒v1v2∈Pe 2.2. PERENOMIAL MATRIX FORM 29 2. w1, w2∈Po=⇒w1w2∈Pe 3. v∈Pe, w ∈Po=⇒vw ∈Po Proof. 1. v1v2(−x)=v1(−x)v2(−x) = v1(x)v2(x)=v1v2(x) 2. w1w2(−x)=w1(−x)w2(−x)=(−w1(x))(−w2(x)) = w1w2(x) 3. vw(−x)=v(−x)w(−x) = v(x)(−w(x)) = −vw(x) Definition 2.8 (Degree of evenand odd-nomials). t(x) = n X 0itixil ∈Pe∨ n X 0itaixil+o∈Po So, analogously with polynomial, we will define: deg(t):= min{i:tj= 0 ∀j > i} Remark 2.3 (Degree of dis-polynomials).This definition can be used also for dis-polynomials, inasmuch it is coherent with Definition 2.3: q∈PD=⇒deg(q) = deg(s): s∈P∧q(x) = s(x)xo q(x) = s(x)xo= n X 0isixixo= n X 0isixi+o deg(q) = deg(s) = min{i:sj= 0 ∀j > i} 2.2 Perenomial matrix form Lemma 2.9 (Perenomial coefficients matrix). r∈PC⇔ ∃ R:r(x) = N X 0i 1 X 0jri,jxhi ji Proof. For Lemma 2.4 PC∋r(x) = v(x) + w(x); v∈Pe, w ∈Po v(x) = n X 0ivixhi 0i;w(x) = m X 0iwixhi 1i 36 CHAPTER 2. PERENOMIALS Proof. Every (x−zi)oin the product introduces a change of sign in zi, as it is in f(x). Another way to see this is that n Y 1i(x−zi)is a polynomial which zeros are {z1, . . . , zn}, all with multiplicity one. So the polynomial changes sign in the same points of f(x). Then n Y 1i(x−zi)!o = n Y 1i(x−zi)o. Finally we have to adjust the initial sign, and since limx→∞(x−zi)o= 1 then limx→∞ s n Y 1i(x−zi)o=s. To have sgn (f(x)) = s n Y 1i(x−zi)owe now must put s= limx→∞ sgn (f(x)). Lemma 2.13 (Perenomial transofrmations). Horizontal translation this is not possible: r(x−k)=p(x−k)+s(x−k)(x−k)o but for Theorem 2.10 this is not a perenomial. For more informatons see Lemma 2.15. Vertical translation r(x)+kis a perenomial: r(x) + k= (p(x)+k)+q(x) Reflection r(−x)is a perenomial: r(−x)=p(−x)+s(−x)(−x)o=p(−x)−s(−x)(x)o Horizontal stretch or shrink r(kx)is a perenomial: r(kx) = p(kx)+s(kx)(kx)o=p(kx) + (sgn(k)s(kx)) (x)o Vertical stretch or shrink k·r(x)is a perenomial: k·r(x)=k·p(x)+k·s(x)(x)o 2.4 Interpolation Lemma 2.14 (Lagrange polynomial).Given n+1 distinct points (xi, yi)we can always build a polynomial of minimal degree not greater than n, which passes trough all points, with the following formula [Dav75]: p(x) = n X 0iyi n Y 0j=i x−xj xi−xj (2.5) 2.4. INTERPOLATION 37 Theorem 2.12 (Lagrange dis-polynomial).Given n+ 1 distinct points (xi, yi)we can always build a dis-polynomial of minimal degree not greater than n, which passes trough all points, with the following formula: q(x) = n X 0iyixo i n Y 0j=i x−xj xi−xj xo(2.6) Proof. If q(x)is a dis-polynomial then p(x)=q(x)xois a polynomial and so we can apply Lemma 2.14 to find pand then obtain q(x) = p(x)xo First of all obtain the points of pfrom those of q:(xi, yi)7→ (xi, yixo i) = (xi, y′ i) Then find pwith Equation 2.5: p(x) = n X 0iy′ i n Y 0j=i x−xj xi−xj = n X 0iyixo i n Y 0j=i x−xj xi−xj Finally obtain q(x) = p(x)xo: q(x) = p(x)xo= n X 0iyixo i n Y 0j=i x−xj xi−xj xo Theorem 2.13 (Perenomial interpolation).Given distinct (xi, yi)with |{i:xi⩽0}| = |{i:xi⩾0}| =n+ 11we can always build a perenomial of minimal degree not greater than n, which passes trough all points, with the following algorithm: 1. Interpolate the polynomial a(x)with (xi, yi): xi⩾0 2. Interpolate the polynomial c(x)with (xi, yi): xi⩽0 3. Obtain r(x) = p(x) + q(x)xowith p=a+c 2and q=a−c 2 Proof. Begin proving the theorem for x= 0: r(x)=p(x)+q(x)xo=p(x)+q(x) =:a(x)∀x>0 r(x)=p(x)+q(x)xo=p(x)−q(x) =:c(x)∀x<0 So aand care two polynomials of degree nwhich can be interpolated whit n+ 1 points (Lemma 2.14). 1Namely n+ 1 points with negative xand n+ 1 points with positive xor npoints with negative x, one point with x= 0 and npoints with positive x. 38 CHAPTER 2. PERENOMIALS Then we can obtain pand qwith: p=a+c 2=(p+q) + (p−q) 2= 2p  2=p q=a−c 2=(p+q)−(p−q) 2= 2q  2=q Finally r(0) = p0+q00ois defined if and only if q0= 0 and so q(0) = 0 and thus a(0) = c(0). For this a point (0, r(0)) could be used to interpolate both aand c. Example 2.1. Begin interpolating a polynomial: r(x) = x2. It is of grade two so we need 3 points with non-negative and 3 points with non-positive x: i xiyi 0 -5 25 1 -2 4 2 0 0 3 3 9 4 5 25 Now let’s interpolate a(x): a(x) = 4 X 2iyi 4 Y 2j=i x−xj xi−xj = 0 +  3 9x(x−5)  3(3 −5) + >5 25 x(x−3)  5(5 −3) = =x−3x+  3·5 + 5x−  5·3 5−3=x25−3 5−3=x2 Now let’s interpolate c(x): c(x) = 2 X 0iyi 2 Y 0j=i x−xj xi−xj = >5 25 (x+ 2)x (−5 + 2)(− 1 5) + 2 4(x+ 5)x (−2 + 5)(− 1 2) + 0 = =x5x+  5·2−2x−  2·5 5−2=x25−2 5−2=x2 Then p(x) = a(x)+c(x) 2= 2x2  2=x2and q(x) = a(x)−c(x) 2= 0, so r(x) = p(x)+q(x)xo=p(x) = x2. Example 2.2. Now let’s find a perenomial which interpolates the following points: 2.5. HYPER-PERENOMIALS 39 i xiyi 0 -5 5 1 -2 2 2 2 0 3 3 1 Begin interpolating a(x): a(x) = 3 X 2iyi 3 Y 2j=i x−xj xi−xj = 0 + 1x−2   3−2=x−2 Then interpolate c(x): c(x) = 1 X 03iyi 1 Y 0j=i x−xj xi−xj = 5 x+ 2 −5+2+ 2 x+ 5 −2+5 = =−5x−  5·2 + 2x+  2·5 5−2=−3x 3=−x Now p(x) = a(x)+c(x) 2=−1and q(x) = a(x)−c(x) 2=−2x−2 2=x−1, finally r(x)=p(x)+q(x)xo=−1−xo+xlwhich degree is one. We can find a polynomial of degree 3 which interpolates the same points (see ˜pin Figure 2.1), but it’s clear that we cannot find a polynomial of degree 1 which interpolates them (the best linear approximation ˜p′can be seen in the same figure). 2.5 Hyper-perenomials Definition 2.10 (Hyper-perenomial).An hyper-perenomial h(x)has the form: h(x) = p0(x) + n X 1ipi(x)(x−zi)o where each pi(x)is a polynomial and each zia constant number. Remark 2.8.We can write h(x) = n X 0ipi(x)(x−zi)owith z0→ −∞. Remark 2.9.We can see that an hyper-perenomial is the sum of a perenomial and shifted (horizontal translated) dis-polynomials: h(x) = p0(x) + n X 1ipi(x)(x−zi)o=p0(x)+pj(x)xo+ n X 1i=jpi(x)(x−zi)o= =r(x) + n X 1i=jqi(x−zi) 40 CHAPTER 2. PERENOMIALS x y Figure 2.1: r(x)=xl−1−xo;˜p(x) = 1 35x3+3 14x2−43 70x+1 7;˜p′(x)=−22 41x+71 41 because pi(x)=si(x−zi)and so si(x−zi)(x−zi)o=qi(x−zi). Also a sum of shifted perenomials is an hyper-perenomial: n X 1iri(x−zi) = n X 1isi(x−zi)+ti(z−zi)(x−zi)o= = n X 1isi(x−zi) + n X 1iti(z−zi)(x−zi)o= =p0(x) + n X 1ipi(x)(x−zi)o with p0(x) = n X 1isi(x−zi)and pi(x) = ti(x−zi)polynomials. Lemma 2.15 (Hyper-perenomial transofrmations). Horizontal translation h(x−k)is an hyper-perenomial: h(x−k) = p0(x−k) + n X 1ipi(x−k)(x−zi−k)o Vertical translation h(x)+kis an hyper-perenomial: h(x)+k= (p0(x)+k) + n X 1ipi(x)(x−zi)o 2.5. HYPER-PERENOMIALS 41 Reflection h(−x)is an hyper-perenomial: h(−x) = p0(−x)+ n X 1ipi(−x)(−x−zi)o=p0(−x)− n X 1ipi(−x)(x+zi)o Horizontal stretch or shrink h(kx)is an hyper-perenomial: h(kx)=p0(kx) + n X 1ipi(kx)(kx −zi)o and (kx −zi)o= (k(x−zi/k))o= sgn(k)(x−zi/k)o Vertical stretch or shrink k·h(x)is an hyper-perenomial: k·h(x) = k·p0(x) + n X 1ik·pi(x)(x−zi)o Theorem 2.14 (Sum of hyper-perenomials).Given h1(x)=p1,0(x) + n X 1ip1,i(x)(x−z1,i)oand h2(x)=p2,0(x)+ m X 1jp2,j(x)(x−z2,j)o, then h1(x)+ h2(x)is an hyper-perenomial. Proof. h(x) = h1(x)+h2(x) = =p1,0(x) + n X 1ip1,i(x)(x−z1,i)o+p2,0(x) + m X 1ip2,j(x)(x−z2,j)o= = (p1,i(x)+p2,0(x)) + n X 1ip1,i(x)(x−z1,i)o+ m X 1jp2,j(x)(x−z2,j)o= =p0(x) + c X 1kpk(x)(x−zk)o with p0(x) = p1,i(x)+p2,0(x)and 1. ∀z1,i =z2,j ∃zk=z1,i =z2,j :pk=p1,i +p2,j 2. ∀z1,i :∄z2,j =z1,i ∃zk=z1,i :pk=p1,i 3. ∀z2,j :∄z1,i =zj∃zk=z2,j :pk=p2,j Thus c=|{zk∀k}| =|{z1,i, z2,j ∀i, j}| ⩽n+m. 42 CHAPTER 2. PERENOMIALS Lemma 2.16. (x−a)o(x−b)o= 1 −(x−x1)o+ (x−x2)o with x1,2=a, b and x1< x2. The result is an hyper-perenomial. Proof. (x−a)o(x−b)o= ((x−a)(x−b))o=x2−(a+b)x+abo which is the sign of a U-shaped parabola, so equal to 1 if it does not intersect the x axis (or intersects it only in the vertex); otherwise it is always 1, except between the two points of intersection with the x axis where the sign value is -1. x y So with ∆=(a+b)2−4ab = (a−b)2the points of intersection are x1,2=a+b±√∆ 2=a, b. We can then see that the hyper-perenomial 1−(x−x1)o+ (x−x2)o with x1< x2is the result we wanted, when the parabola intersects the x axis. This is valid also in the case of ∆ = 0, being a=band so x1=x2. Since the delta can never be negative, we do not need to cover the case where the parabola does not intersect the x axis. Lemma 2.17 (Product between polynomial and hyper-perenomial).The product between a polynomial and an hyper-perenomial is an hyper-perenomial. Proof. p(x)h(x) = p(x) n X 0ipi(x)(x−zi)o!= n X 0ip(x)pi(x)(x−zi)o 2.5. HYPER-PERENOMIALS 43 Theorem 2.15 (Product between hyper-perenomials).The product of two (or more) hyper-perenomials is an hyper-perenomial. Proof. p1(x) = n X 0ip1,i(x)(x−z1,i)o;p2(x) = m X 0jp2,j(x)(x−z2,j)o p(x)=p1(x)p2(x) = n X 0ip1,i(x)(x−z1,i)o m X 0jp2,j(x)(x−z2,j)o= = n X 0i m X 0jp1,i(x)(x−z1,i)op2,j(x)(x−z2,j)o= = n X 0i m X 0jp1,i(x)p2,j(x)(x−z1,i)o(x−z2,j)o= = n X 0i m X 0jpi,j(x)hi,j(x) where pi,j =p1,ip2,j is a polynomial and hi,j(x)=(x−z1,i)o(x−z2,j)o an hyper-perenomial, for Lemma 2.16. Then for Lemma 2.17 pi,j(x)hi,j(x) is an hyper-perenomial and for Theorem 2.14 the sum is itself an hyperperenomial. Chapter 3 Selector function 3.1 Sign function sgn(x):=xo x y −1 1 y= sgn(x) As we can see, this function (which is the odd unit of perenomials, as already seen in chapter 2) returns the sign of its argument. Remark 3.1.Sign is an odd function, in fact all the exponents (o) of its perenomial form are odd, and so it is an odd-nomial. 3.2 Heaviside step function 1(x):=(0 : x<0 1 : x>0 52 CHAPTER 3. SELECTOR FUNCTION Proof. せb a=1(x−a)1(b−x)=1(x−a)[1 −1(−(b−x))] = 1(x−a)[1 −1(x−b)] Lemma 3.10. If a < b then: 1(x−a)1(x−b) = 1(x−b) Proof. 1(x−a)1(x−b) = せ∞ a(x)せ∞ b(x)=せ∞ b(x) = 1(x−b) because [a, ∞)∩[b, ∞) = [b, ∞) Lemma 3.11. Se a < b allora: せb a=1(x−a)−1(x−b) Proof. せb a=1(x−a)[1−1(x−b)] = 1(x−a)−1(x−a)1(x−b)=1(x−a)−1(x−b) Theorem 3.2 (Hyper-perenomial form of the selector function). せb a(x) = 1 2(x−a)o−1 2(x−b)o Proof. せb a=1(x−a)−1(x−b) = " 1 2+1 2(x−a)o#−" 1 2+1 2(x−b)o# 3.6 Functions defined by cases with selector function Theorem 3.3 (Functions defined by cases as a single formula).Thanks to the selector function we can write each function defined by cases (undefined in the extremities of the cases for which the limit does not exist) as a single formula. 3.6. FUNCTIONS DEFINED BY CASES WITH SELECTOR FUNCTION 53 Proof. f(x) =                      f0(x)x < a0 . . . fi(x)ai⩽x⩽bi . . . fn(x)x>bn =f0(x)1(a0−x)+ n−1 X 1ifi(x)せbi ai(x)+fn(x)1(x−bn) Theorem 3.4 (Functions defined by cases as hyper-perenomials).If the functions fi(x)of Theorem 3.3 are hyper-perenomials, then the resulting formula is an hyper-perenomial. Proof. fi(x)are hyper-perenomials by definition. There exists an hyperperenomial expression both for 1(Theorem 3.1) and せ(Theorem 3.2), so for Theorem 2.15 the result is still an hyper-perenomial. Example 3.1. g(x) = x+ 4; h(x) = 3; i(x)=x f(x) =        g(x)x < −1 h(x)−1⩽x⩽3 i(x)x > 3 x y 54 CHAPTER 3. SELECTOR FUNCTION y=f(x) y=せ−1 −∞(x)=1(−x−1) y=せ3 −1(x)y=せ∞ 3(x)=1(x−3) So the function f(x)can be expresses as: f(x) = 1(−x−1) ·g(x)+せ3 −1(x)·h(x) + 1(x−3) ·i(x) = = (x+ 4)[1 −1(x+ 1)] + 3[1(x+1)−1(x−3)] + x1(x−3) = =x+4+(3−x−4)1(x+1)+(x−3)1(x−3) = x+ 4 −(1+x)1(x+ 1) + (x−3)1(x−3) = =x+4−(1+x)"1 2+(x+ 1)o 2#+ (x−3) "1 2+(x−3)o 2#= =x+4−1 2−x 2−(x+ 1)l 2+x 2−3 2+(x−3)l 2= =2+x−|x+ 1| 2+|x−3| 2 Remark 3.8.As we can see the fact the selector function is undefined on the extremes doesn’t matter: all the generated discontinuities are removable, and so in result we have a continue function as we wanted. Example 3.2. g(x) = ex;h(x)=x2;i(x)=x−3 f(x) =        g(x)x < −1 h(x)−1<x<2 i(x)x > 2 3.6. FUNCTIONS DEFINED BY CASES WITH SELECTOR FUNCTION 55 x y y=f(x) y=せ−1 −∞(x) = 1(−x−1) y=せ2 −1(x)y=せ∞ 2(x)=1(x−2) So the function f(x)can be expresses as: f(x) = 1(−x−1) ·g(x)+せ2 −1(x)·h(x) + 1(x−2) ·i(x) = =1 2{[1 −(x+ 1)o]ex+ [(x+ 1)o−(x−2)o]x2+[1+(x−2)o](x−3)}= =1 2[ex+ (x+ 1)o(x2−ex)−(x−2)o(x2−x+ 3) + x−3] Future work In this chapter we will see some ideas for future works on this topic. F.1 Applications Obviously an important topic to explore are the possible real applications of the complete integers. In this document we saw them only for theoretical pleasure, but without any real application they are useless. F.2 Parity extension It could be useful to extend the parity concept to cover all QCand not only QH. Obviously two values will be not enough, so one idea could be to use the parity measure which states how much an integer is even, and then extend it to rationals: g par(z)=0 ∀z∈Zo(F.1) g par(2) = 1 (F.2) g par (a·b) = g par(a) + g par(b)(F.3) g par a b=g par(a)−g par(b)(F.4) This works in Q, but not in QC(e.g. g par(l) = g par 2 2=g par(2)−g par(2) = 0 as for odd numbers). So another path must be followed. F.3 Perenomial extension We saw that perenomials extend polynomials but loosing the binomial expansion closure property (Theorem 2.10). It would be useful to extend perenomials in something closed under binomial expansion. 58 CHAPTER 3. SELECTOR FUNCTION F.4 Complete integers and complex numbers ln(−1) = ln(eπi)=πi ⇒i=ln(−1) π(F.5) so (eai)l=eail =eial =eila =eln(−1) πla = (−1)la π=|−1|a π= 1a π= 1 (F.6) but (eai)l=eail =eali =elai =|e|ai =eai = cos(a)+isin(a)(F.7) In Equation F.6 we commuted lonly with real numbers (al =la and 1 πla = la π) and we get an expected result since |eai|= 1. Besides int Equation F.7 we tried to commute lwith ibut we get another result, so we can guess l and icannot commute. Another future work could be investigating how complete integers and complex numbers work together. References [Art91] Michael Artin. Algebra. Prentice Hall, 1991. isbn: 9780130047632,0130047635. [Dav75] Philip J. Davis. Interpolation and approximation. Dover Publications, 1975. isbn: 9780486624952,0486624951. [Dri84] Lou van den Dries. “Exponential rings, exponential polynomials and exponential functions.” In: Pacific J. Math. 113.1 (1984), pp. 51–66. url:https://projecteuclid.org:443/euclid. pjm/1102709376. 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