Multinomial Expansion and “Pascal?s Tetrahedron” The story:
Not long after the Pascal?s Triangle was introduced to me, I realized it was a sequence in 2 dimensions. Then I started to thi
nk about a “Pascal Tetrahedron”, which is a sequence in 3 dimensions. This then lead to another question when I thought about it and discussed it with some fellow students: What?s the formula for multinomial expansion, and how does it relate to the tetrahedron. And thanks to my resourceful peer, Mitchell Ding, we were able to figure out a little something about this topic.
To work out this problem, we need to look back on how we solved the binomial expansion problem. Binomial Expansion: ( + ) The problem of binomial expansion is solved when the coefficient of every term is solved. And it is certain that the coefficient of terms is closely related to the exponent of ?a? and ?b? in each term. Let?s say that the exponent of ?a? is ?i?, the exponent of ?b? is ?k?, and the coefficient of the term is m: mai bk This is the general expression for every term in the expansion. Since k+i =n (n ?a? or ?b? multiplied together, so we get a total exponent of 6) k=n?i So mai bk = mai bn ?i Now we have k out of the way, we will need to figure out the relation between m and i. In order to work out this problem, we need to view it this way: We have n pairs of ?a? and ?b?, we have to choose i ?a?s from it, how many ways are there for us to pick? 1. a b 2. a b 3. a b 4. a b ...... n. a b This leads us directly to combination, for this is basically a question of ?n choose i?, which is expressed in the form of
n i So m = (For there are
n i n i
ways of getting the term ?aibn-i?) So
ai bn ?i (i<n)
And that is the expression of every term in the expansion, it can be used to find the final expansion. And this solves that binomial expansion problem. Trinomial Expansion: ( + + ) To solve the problem of trinomial expansion, we will use the same technique of solving the binomial expansion problem—finding the general expression for every term after the expansion. Suppose the exponent of a to be ?i?, the exponent of b to be ?k?, then the exponent of c is ?n-i-k?; suppose the coefficient of the term is ?m? mai bk c n ?i ?k (i+k<n) And this is the part where it differs from the binomial expansion, we now have two values to determine the coefficient ?m?, because m is no longer a simple choosing problem, it is a variable that is dependent to ?i? and ?k?. The solution to this problem is to think of it this way: every term is formed by multiplying n terms of ?a? or ?b? or ?c? together, we can list the choices out and choose which ones to multiply: 1. a b c 2. a b c 3. a b c 4. a b c ...... n. a b c We are to choose i ?a?s from n ?a b c?s and then choose k ?b?s from the remaining ?a b c?s. To demonstrate, let?s say that we need to expand the trinomial to the exponent of 4, and we need the coefficient of the term a2bc: 1. a b c 2. a b c 3. a b c 4. a b c First, we pick 2 ?a?s from the 4 ?a b c?s, and there are
= 6 ways of doing this
And after choosing 2 ?a?s from the four ?a b c?s, we have 2 ?a b c? left to choose from, and we need 1 ?b? from it, how many ways are there for us to choose?
1. a b c 2. a b c The answer is simple, basically
= 2. Finally, we multiply 6 and 2 together to
get the coefficient of the term ?a2bc?, for there are 6 × 2 = 12 ways of getting ?a2bc? when we expand. And when we looked back at this whole process of working, we found out that the coefficient ?m? is determined by the ways of multiplying up to a certain term. And in trinomial expansion, when the term is determined by the exponents ?a? ?b?, we will have to multiply two combinations together to get the number of ways of multiplying. So ultimately, we get this equation that n n?i m= × (k + i < ) i k n n ? i i k n ?i ?k × ab c i k Now since the process of prove above is really not that convincing and lack some clarity, the prove by induction is presented below:
n ?i k
ai bk c n ?i ?k (i and k are constants and i+k<n)is true for
Step1: Prove Sn is true for n=0.
a0 b 0 c 0 = 1
(a + b + c)0 =1 So Sn is true for n=0 Step2: Suppose Sn is true for all n=j, i.e.
j+1 i j+1?i k
j ?i k
ai bk c j ?i ?k
To show that Sn is true for all n=j+1
L.H.S.= = =
ai bk c j ?i ?k+1 ai bk c j ?i ?k+1
j+1 ! j+1?i !×j! j+1 !
j+1?i ! j+1?i ?k !×k!
j!× j+1 ?i ?k !×k!
a = j j ? i i k j ?i ?k j j ? i i k j ?i ?k j j ? i i k j ?i ?k × ab c +b × ab c +c × ab c i k i k i k
j ?i k
ai+1 bk c j ?i ?k +
j ?i k
ai bk+1 c j ?i ?k +
j ?i k
ai bk c j ?i ?k+1
j i ?1 j i ?1
j ?i+1 k j ?i+1 k
ai bk c j ?i ?k+1 + +
j ?i k ?1
ai bk c j ?i ?k+1 + ai bk c j ?i ?k+1
j! j ?i ! i !
j ?i k
ai bk c j ?i ?k+1
j ?i k ?1
j ?i k
j ?i+1 ! i ?1 !
(j ?i+1)! j ?i+1?k ! k !
j ?i ! i !
j ?i ?k+1 ! k ?1 !
(j ?i)! j ?i ?k ! k !
ai bk c j ?i?k+1
=j! j ?i+1 !×i+j! j ?i+1 !×k+j! j ?i+1 !×k+j! j ?i+1 !(j ?i ?k+1) j ?i+1 !×i!× j ?i+1?k !×k!
ai bk c j ?i ?k+1
j!× j ?i+1 !×(j+1) j ?i+1 !×i!× j ?i+1?k !×k! (j+1)!
ai bk c j ?i ?k+1
i!× j ?i+1?k !×k!
ai bk c j ?i ?k+1 =L.H.S.
If Sn is true for n=j Then Sn is true for n=j+1
Step3: ∵If Sn is true for n=j
Also Then Sn is true for n=j+1 ∵Sn is true for n=0 ∴Sn is true for all n∈N
Multinomial Expansion: And on to the final part of the expansion package of this essay, multinomial expansion ( + + + ? + ) And let?s just jump to the part of defining terms of the final expansion. Suppose ?mi? to be the exponent of ?ai?, and ?y? to be the coefficient of each term
y(a1 )m 1 (a2 )m 2 (a3 )m 3 … (ai )n ?m 1 ?m 2 ?m 3 ????m i ?1 y= n n ? m1 n ? m1 ? m2 × × ×… m1 m2 m3 × n ? m1 ? m2 ? ? ? mi ?2 mi ?1
And that is the general expression of terms in a multinomial expansion problem; this can be used to work out any multinomial expansion problem.
Same as the multinomial expansion, to solve the problem of ?Pascal Tetrahedron?, we need to look back into ?Pascal?s Triangle?. Pascal?s Triangle: After the exploration, we should all know its relationship between binomial expansions. 1 11 121 1331 14641 … Or if we combine binomial expansion: 1 × (a+b) 1a 1b × (a+b) 2 2 1a 2ab 1b × (a+b) 3 2 2 3 1a 3a b 3ab 1b × (a+b) 4 3 2 2 3 4 1a 4a b 6a b 4ab 1b × (a+b) … And the general expression of
ak bn ?k
And the triangle is related to this expression because of the special property of k k k+1 + = r?1 r r Also known as the ?Pascal Theorem?, this equation states how each number in the triangle is derived and why it relates to binomial expansion, can be proved through the following steps: k! k! k! × r + k! × (k ? r + 1) + = k?r+1 !× r?1 ! k?r !× r ! k?r+1 !× r ! k+1 ! k+1 = = k?r+1 !× r ! r This equation is the definition equation of Pascal?s Triangle, because it explains everything in the triangle. It?s the general formula of a 2-Dimension sequence called Pascal?s Triangle. ?Pascal Tetrahedron? If sequences can appear in 2-Dimensions, it can also appear in 3-Dimensions. And this finally leads to what I and my peer, Mitchell, called ?Pascal Tetrahedron?. Notice it?s not ?Pascal?s Tetrahedron? but “Pascal Tetrahedron?, because this thing belongs to me, it belongs to Mitchell, not Pascal.
A typical tetrahedron A ?Pascal Tetrahedron? is basically a ?Pascal?s Triangle? in 3-Dimensions. Every number in the tetrahedron is obtained by adding the 3 numbers above it together. To clarify, the first 4 levels of the triangle will be listed below: 1 Lvl.1 1 11 Lvl.2 1 2 2 1 2 1 Lvl.3 1 3 3 3 6 3 1 3 3 1 Lvl.4 For further clarification: 1 4 4 6 12 6 4 1212 4 1 4 6 4 1 Lvl.5
The Pascal Tetrahedron in 3-D And as you can see here, the levels are stack on top of each other, forming the tetrahedron which obeys the given rule: each number in the tetrahedron is obtained by adding the three numbers above it.
1 2 2 1 2 1 Lvl.3 1 3 3 3 6 3 1 3 3 1 Lvl.4
This shows how the number 6 is obtained
1 3 3 3 6 3 1 3 3 1 Lvl.4 1 4 4 6 12 6 41212 4 1 4 6 4 1 Lvl.5 This shows how the number 12 is obtained Now we have that out of the way, consider this question: how does the Pascal Tetrahedron relate to the trinomial expansion. To make define this sequence, we will need a reference scale, a grid running through the tetrahedron to express the terms of this amazing 4-D sequence. k | axis 01 0 i-axis Lvl.1 k | axis 11
0 11 01 i-axis Lvl.2 k | axis 21 12 2 01 2 1 0 1 2 i-axis Lvl.3 k | axis 31 23 3 13 6 3 01 3 3 1 0 1 2 3 i-axis Lvl.4 We have now 3 axis: the i-axis, the k-axis, and the n-axis. Those 3 axis runs through the tetrahedron in 3 different ways: the n-axis goes vertically up and down, it measures which level the number is in; the i-axis and the k-axis both goes horizontally, but they go perpendicular to each other, creating a coordinate on each level. To express the number in the tetrahedron, we will need three values. For instance, the highlighted 1 in the 4th level, is expressed as (3,0,0).(the n-axis also starts with 0) Simple. Now, since ?Pascal?s Triangle? is related to binomial expansion in a particular way, we will assume that the ?Pascal Tetrahedron? is related to trinomial expansion in the same way, except entangled in a 3-D way. So each number: (l, i, k) is the coefficient of each term in trinomial expansion: (n,i,k)=
n ?i k
To prove this is true, we will have to prove that each term can be obtained by adding the three terms above it together, which gives us: n n?i+1 n n?i n n?i n+1 n+1?i × + × + × = × i?1 k i k?1 i k i k Oh wait a second… does the equation above look familiar to you? Yes! It appeared in the induction that was used to prove the formula of trinomial expansion! So there is no need to prove it true, because it was already proved. And this is not a coincidence. Now the 3-D sequence is well defined by its formula: (n,i,k)= ?Pascal tri????
n ?i k
Isn?t it fascinating that the 3-D version of Pascal?s Triangle is related to trinomial expansion, the tri version of binomial expansion? But that?s not the end. The final frontier of the human mind is cut clearly on the edge of 3-D and 4-D, anything with a higher dimension than 3-D, we cannot understand. The Pascal theorem will have its 4-D version, its 5-D version, its i-D version, and they will all align perfectly with the equation of multinomial expansion, only we will not be able to imagine. And the Pascal theorem in i-D is shown below:
n n ? m1 n ? m1 ? m2 × × ×… m1 m2 m3 × = n?1 × m1 ? 1 n ? m1 ? m2 ? ? ? mi ?2 mi ?1 n ? 1 ? (m1 ? 1) ? m2 m3
n ? 1 ? (m1 ? 1) × m2
× …× + n?1 × m1 × + n?1 × m1 × + n?1 × m1 × + n?1 × m1 ×
n ? 1 ? m1 ? 1 ? m2 ? m3 ? ? ? mi ?2 mi ?1 n ? 1 ? m1 ? (m2 ? 1) ×… m3
n ? 1 ? m1 × m2 ? 1
n ? 1 ? m1 ? (m2 ? 1) ? m3 ? ? ? mi ?2 mi ?1 n ? 1 ? m1 × m2 n ? 1 ? m1 ? m2 ×… m3 ? 1 +?
n ? 1 ? m1 ? m2 ? (m3 ? 1) ? ? ? mi ?2 mi ?1 n ? 1 ? m1 × m2
n ? 1 ? m1 ? m2 ) ×… m3
n ? 1 ? m1 ? m2 ? m3 ? ? ? (mi ?2 ? 1) mi ?1 n ? 1 ? m1 × m2 n ? 1 ? m1 ? m2 ×… m3
n ? 1 ? m1 ? m2 ? m3 ? ? ? mi ?2 mi ?1 ? 1
This also suggest that all the mathematical theories that is true in 2-D will have its
3-D, its i-D version, that our world goes on and on, vastly beyond the frontier of any possible imagination. That math itself is so beautiful.
Special regards to Mitchell Ding, who worked and discussed and figured the entire thing out. No citation required, purely generated by our mind.