A new mathematical symbol : the termirial
AA new mathematical symbol : the termirial
Claude-Alexandre Simonetti
1, 2, 3 LPC Caen EAMEA Cherbourg ENSICAEN (Dated: 5 juin 2020)The understanding of probability can be difficult for a few young scientists. Consequently, this newmathematical symbol, related to binomial coefficients and simplicial polytopic numbers, could behelpful to science education. Moreover, one can obtain kinds of remarkable identities and generalizethem to a sort of “Newton’s binomial theorem”. Finally, this symbol could be perhaps useful toother scientific subjects as well, such as computer science. a r X i v : . [ m a t h . G M ] J un I. STUDY OF BINOMIAL COEFFICIENTSA. Reminder
In a set of n elements, the number of combinations of parts of p elements is the following : (cid:18) np (cid:19) = n ! p ! ( n − p )! (1)With (cid:0) np (cid:1) the binomial coefficient, ( n, p ) ∈ N and p ≤ n . And the exclamation point in equation (1) is named“factorial” and is defined as, ∀ n ∈ N ∗ : n ! = n · ( n − · ( n − ... · · n ! = n (cid:89) i =1 i (2)For n = 0, by definition : 0 !=1. B. From factorial to termirial
For its part, the termirial is defined as, ∀ n ∈ N ∗ : n | + = n + ( n −
1) + ... + 3 + 2 + 1 n | + = n (cid:88) i =1 i (3)This is called a triangular number of order n, see reference [1]. The termirial symbol is almost like the factorialsymbol, with just a little difference, however : the dot ( · ) of the interrogation point, which could be a reminder of amultiplication, is replaced by a “plus” (+) sign. Indeed, instead of multiplying factors of a multiplication – factorial– one adds terms of an addition, hence the name of “termirial”. Up to this point, things look quite basic. However,as the termirial is the n th partial sum of an arithmetic sequence ( U n ) n ∈ N ∗ with U = 1 as first term and r = 1 ascommon difference, one can notice a first thing, ∀ n ∈ N ∗ : n | + = n · ( n + 1)2 = (cid:18) n + 12 (cid:19) = (cid:18) n + 1 n − (cid:19) (4)One will get back to it later, with the “generalized” termirial. C. Remarkable identity
It is possible to obtain a kind of remarkable identity. Indeed, ∀ ( n, m ) ∈ N ∗ :( n + m ) | + = ( n + m ) · ( n + m + 1)2( n + m ) | + = n · ( n + 1)2 + 2 · n · m m · ( m + 1)2( n + m ) | + = n | + + n · m + m | + (5) D. Intellectual path
At the beginning, I was studying the black body radiation (Max Planck’s law). In this context, suppose one has n particles to fill up in 2 discrete energy levels determined by quantum mechanics. Suppose that the first energy levelnamed m can contain p particles, and that the second one ( m ) can contain ( n − p ) particles. The number of possiblecombinations is same as equation (1) : (cid:18) np (cid:19) = n ! p ! ( n − p )! (6)Let us start with a simple example : n = 5 and p = 2. The figure 1 tries to explain the intellectual path which hasled to the termirial. One can see the energy level m containing p = 2 particles, with all the ways to fill it up, forparticles particules ranked from 1 to 5. In this case, as a reminder, it is an unordered sampling without replacement,the rank order of the particles has no importance : { } and { } are counted only once. Finally : (cid:0) (cid:1) = 4 | + = 10. Figure
1. This figure represents the intellectual path from the binomial coefficient (cid:0) (cid:1) to the termirial of 4, through a classicaltree view. Here, only the energy level m is represented, which is sufficient, because one only has 2 energy levels, not more. Let us make things a little bit more complicated, with n = 5 and p = 3. In terms of binomial coefficients, one hasthe same result as before : (cid:0) (cid:1) = (cid:0) (cid:1) = 10. But the intellectual path is a bit different, see figure 2.One can see a kind of “termirial of termirials”, which is called a tetrahedral number of order n, see reference [1] : Figure
2. This figure represents the intellectual path from the binomial coefficient (cid:0) (cid:1) to a sum of termirials, a kind of “termirialof termirials” or a 2 nd termirial. From now, one can generalize to the 3 rd termirial, the 4 th termirial, etc. n (2) | + = n | + + ( n − | + + ... + 3 | + + 2 | + + 1 | + n (2) | + = n (cid:88) k =1 k | + = n (cid:88) k =1 k (cid:88) i =1 i = n (cid:88) k =1 k ( k + 1)2! (7) n (2) | + = (cid:18) n + 23 (cid:19) = (cid:18) n + 2 n − (cid:19) = n ( n + 1)( n + 2)3! (8)The transition from (7) to (8) can be done with mathematical induction. Indeed, if the following proposition P ( n ) istrue : n (2) | + = n (cid:88) k =1 k ( k + 1)2! = n ( n + 1)( n + 2)3! (9)And, for P (1) : 1 (2) | + = (cid:88) k =1 k ( k + 1)2 = 1 · (1 + 1)2 = 1 (cid:18) (cid:19) = 1(1 + 1)(1 + 2)3! = 1Finally, with P ( n + 1) : ( n + 1) (2) | + = n (cid:88) k =1 k ( k + 1)2 + ( n + 1)( n + 2)2= n ( n + 1)( n + 2)3! + ( n + 1)( n + 2)2= n ( n + 1)( n + 2)6 + 3 ( n + 1)( n + 2)6( n + 1) (2) | + = ( n + 1)( n + 2)( n + 3)3! (10) QED . E. Remarkable identity of the nd termirial Like the 1 st termirial, it is possible to obtain a kind of remarkable identity. Indeed, ∀ ( n, m ) ∈ N ∗ :( n + m ) (2) | + = ( n + m + 2)3 · ( n + m ) | + = ( n + m + 2)3 · ( n | + + n · m + m | + )( n + m ) (2) | + = n (2) | + + n · m | + + m · n | + + m (2) | + (11)From now on, one can generalized to the p th termirial. II. GENERALIZATION OF THE TERMIRIALA. The p th termirial Obviously, the the p th termirial, which is called a (p+1)-simplicial polytopic number, see reference [1], will bedefined as, ∀ ( n, p ) ∈ N ∗ × N ∗ : n ( p ) | + = n (cid:88) m =1 m (cid:88) l =1 ... k (cid:88) j =1 j (cid:88) i =1 i (12) n ( p ) | + = n (cid:88) k =1 k ( p − | + = n (cid:88) k =1 p ! p − (cid:89) i =0 ( k + i ) (13) n ( p ) | + = (cid:18) n + pp + 1 (cid:19) = (cid:18) n + pn − (cid:19) (14) n ( p ) | + = 1( p + 1)! p (cid:89) i =0 ( n + i ) (15)With, for the equation (12), a number of sigmas ( (cid:80) ) symbols equal to p .For p = 0, one can define the 0 th termirial as : n (0) | + = n = (cid:18) n (cid:19) For p = −
1, one can define the − th termirial as : n ( − | + = 1 = (cid:18) n − (cid:19) The proof to go from (13) to (14) (or from (13) to (15)) can be done with mathematical induction as well. Indeed,admitting that the following P ( n ) proposition is true : n (cid:88) k =1 p ! p (cid:89) i =0 ( k + i ) = 1( p + 1)! p (cid:89) i =0 ( n + i ) (16)One have, for P (1) : 1 ( p ) | + = 1 p ! p − (cid:89) i =0 (1 + i ) = p ! p ! = 11( p + 1)! p (cid:89) i =0 (1 + i ) = ( p + 1)!( p + 1)! = 1 (17)And, for P ( n + 1) : ( n + 1) ( p ) | + = n (cid:88) k =1 p ! p − (cid:89) i =0 ( k + i ) + 1 p ! p − (cid:89) i =0 ( n + 1 + i )= 1( p + 1)! p (cid:89) i =0 ( n + i ) + ( p + 1)( p + 1)! p (cid:89) i =1 ( n + i )= ( n + p + 1)( p + 1)! p (cid:89) i =1 ( n + i )( n + 1) ( p ) | + = 1( p + 1)! p +1 (cid:89) i =1 ( n + i ) = (cid:18) n + p + 1 p + 1 (cid:19) (18) QED . B. Pascal’s rule
With Pascal’s rule as a reminder : (cid:18) np (cid:19) + (cid:18) np + 1 (cid:19) = (cid:18) n + 1 p + 1 (cid:19) It is possible to adapt it to termirials : ( n + 1) ( p ) | + + ( n ) ( p +1) | + = ( n + 1) ( p +1) | + C. Newton’s binomial theorem
With Newton’s binomial theorem as a reminder, ∀ ( a, b ) ∈ R × R and ∀ n ∈ N :( a + b ) n = n (cid:88) p =0 (cid:18) np (cid:19) a n − p · b p It is possible to adapt it as well to termirials, ∀ ( n, m ) ∈ N ∗ × N ∗ and ∀ p ∈ {{− } ∪ N } :( n + m ) ( p ) | + = p (cid:88) i = − n ( i ) | + · m ( p − i − | + (19)The proof can be done with mathematical induction. Indeed, admitting that the following P ( p ) proposition is true :( n + m ) ( p ) | + = p (cid:88) i = − n ( i ) | + · m ( p − i − | + First of all, equations (5) and (11) respectively hold P (1) and P (2), even if only P (1) – or only P (2) – is enough forthis mathematical induction. Considering P ( p + 1) :( n + m ) ( p +1) | + = n + m + p + 1 p + 2 ( n + m ) ( p ) | + = n + m + p + 1 p + 2 p (cid:88) i = − n ( i ) | + · m ( p − i − | + = n + m + p + 1 p + 2 p (cid:88) i = − n ( i ) | + · m ( p − i − | + = 1 p + 2 p (cid:88) i = − { [ n + ( i + 1)] + [ m + p + 1 − ( i + 1)] } · n ( i ) | + · m ( p − i − | + = 1 p + 2 p (cid:88) i = − [ n + ( i + 1)] · n ( i ) | + · m ( p − i − | + + 1 p + 2 p (cid:88) i = − [ m + p + 1 − ( i + 1)] · n ( i ) | + · m ( p − i − | + = 1 p + 2 p (cid:88) i = − ( i + 2) · n ( i +1) | + · m ( p − i − | + + 1 p + 2 p (cid:88) i = − [ p + 2 − ( i + 1)] · n ( i ) | + · m ( p +1 − i − | + = 1 p + 2 p +1 (cid:88) i =0 ( i + 1) n ( i ) | + m ( p +1 − i − | + + p (cid:88) i = − n ( i ) | + m ( p +1 − i − | + − p + 2 p (cid:88) i = − ( i + 1) n ( i ) | + m ( p +1 − i − | + = n ( p +1) | + · m ( − | + + p (cid:88) i = − n ( i ) | + m ( p +1 − i − | + ( n + m ) ( p +1) | + = p +1 (cid:88) i = − n ( i ) | + m ( p +1 − i − | + QED . III. PRATICAL APPLICATIONSA. Computational complexity
In computer science, the termirial can be used to calculate the complexity of computer programs with intricated“for” loops as follows : for example, with 4 “for” loops, with :- for i from 1 to n=100 ;- for j from 1 to i ;- for k from 1 to j ;- for l from 1 to k.The complexity will be : 100 (4 − | + = (cid:18)
100 + 33 + 1 (cid:19) = (cid:18) (cid:19) = (cid:18) (cid:19) = 4 421 275So, approximatey 4,4 millions of operations.More generally, the complexity of this kind of program will be, according to equation (15) and ∀ ( n, p ) ∈ N ∗ × N ∗ : n ( p − | + = Θ( n p ). B. Pseudo-fractal aspect
Figure
3. This figure tries to convert p th termirial of 4 into fractal figures, for p=0, p=1 and p=2. Only the grey squares arecounted. The figure 3 tries to convert p th termirial of 4 into fractal figures, for p=0, p=1 and p=2. Only the grey squaresare counted. For example, for 4 (2) | + , 20 squares are in grey, which corresponds to 4 (2) | + = 20. For each incrementation ofp, the side of each square is divided by 2, and the same basic pattern is repeated :- 4 | + when the 4 previous squares are in grey ;- 3 | + when the 3 previous squares are in grey ;- 2 | + when the 2 previous squares are in grey ;- 1 | + when only 1 previous square is in grey.But the Hausdorff dimension (D) of this presumed fractal is not constant, and the limit of D as p approachesinfinity is 2. Indeed, let a the side length of each square related to the p th termirial. The total surface S p of thesquares in grey will be : S p = a · n ( p ) | + = a · p + 1)! p (cid:89) i =0 ( n + i )And for the S p − surface, related to the ( p − th termirial, for a side length of each square doubled in size : S p − = 4 a · n ( p − | + = 4 a · p − (cid:89) i =0 ( n + i )Hence, for p > S p − S p = 4 · p + np + 1 = 4 · n/p /p Finally, this surface ratio, as p approaches infinity, is 4 :lim p →∞ S p − S p = 4Consequently, the Hausdorff dimension (D) is 2, as 2 D = 4, when p approaches infinity. To conclude, as D is notconstant, even if figure 3 looks interesting, termirials do not seem to be related to fractals. IV. CONCLUSION
The termirial is a symbol which could be helpful to scientific students, for the understanding of probability. As well,it could be perhaps helpful in other subjects than science education, that are out of this pr´e-publication. [1] L. E. Dickson,
History of the Theory of Numbers, Volume 2, Chapter 1 (Carnegie Institution of Whashington, 1919).[2] B. Pascal,
Trait´e du triangle arithm´etique (Arvensa Editions, 2019).[3]