3 Generalized Multinomial Theorem 3.1 Binomial Theorem Theorem 3.1.1 If
x1 , x2
are real numbers and
n
is a positive integer, then
n
n-r r n x1+ x2 = ΣnＣ r x1 x2
(1.1)
r=0
Binomial Coefficients Binomial Coefficient in (1.1) is a positive number and is described as nＣ r . Here, non-negative integer. nＣ r is the number of ways of picking
r
n
and
unordered outcomes from
r
n
are both
possibilities
and is calculated as follows.
Ｃr =
n
n! (n -r)! r!
Pascal's Triangle About nＣ r , what arranged
n
in the row and
r
in the column is called Pascal's Triangle .
Ｃ0
1
0
Ｃ0
Ｃ1
1
Ｃ0
Ｃ1
2
Ｃ0
Ｃ0
4
Ｃ2
2
Ｃ1
3
Ｃ1
1
2
Ｃ2
3 4
1
1
Ｃ3
3
Ｃ2
4
1
3
Ｃ3
4
Ｃ4
1
4
1 2
3 4
1 3
6
1 4
1
Propertis of the Binomial Coefficient Although a lot of properties of the binomial coefficient are known, fundamental ( understood from Pascal's Triangle immediately ) some are as follows. Among these, ii is used for step-by-step calculation of nＣ r .
ii
Ｃ r = nＣ n-r nＣ r = n -1Ｃ r-1 + n -1Ｃ r
iii
n
iv
n nＣ r = 2 Σ r=0
v
(-1)rnＣ r = 0 Σ r=0
i
n
n -1
Ｃ r = Σ kＣ r-1 k = r-1 n
n
-1-
3.2 Generalized Binomial Theorem 3.2.1 Newton's Generalized Binomial Theorem Theorem 3.2.1 The following formulas hold for a real number .
x r = Σ -r x
Σ r=0
(1+ x)
|x| 1
r
-r
( |x| = 1 is allowed at
>0 )
|x|>1
( 1.1)
( 1.2)
r=0
Proof When n is a natural number, the following expression holds from the binomial theorem. n
(1+ x)n = ΣnＣ r x r r=0
Since nＣ r =
0 for r > n
, this can be written as follows.
(1+ x)n = ΣnＣ r x r r=0
Extending
to real number ,
n
(1+ x) = Σ r=0
r
( +1) xr ( -r +1) r!
xr = Σ r=0
(1.1)
Here, let
( +1) x r ar . ( -r +1) r! Then
(p +1)
r+1
x ar+1 (p -r) (r +1)! = ar (p +1) xr (p -r +1) r!
=
(p -r)x r +1
.
From this,
lim r
ar+1 ar
= lim
r
(p -r)x r +1
= lim
r
(p /r -1)x 1+1/r
According to d'Alembert's ratio test, (1.1) converges absolutely if Next, if
|x|> 1
then x
-1 1+ x = Σ r=0
Multiplying by
x
= |x| =
-x 1
|x|< 1 .
< 1 . Therefore, from (1.1) ,
-1
r
-1 r
x
both sides,
(x +1) = x Σ r=0
The proof in case of
r
|x|= 1
x -r = Σ r=0
x -r -r
is accomplished in the following sub section..
-2-
(1.2)
3.2.2 General Binomial Coefficient The coefficient
r
r
=
in Theorem 3.2.1 is called General Binomial Coefficient and is as follows.
( +1) ( -1)( -2)( -r +1) = r! ( -r +1)(r +1)
(2.0)
The first few are as follows.
0
1
= 1,
=
1!
2
,
( -1)
=
2!
,
3
=
( -1) ( -2) 3!
,
Although properties similar to binomial coefficient also about general binomial coefficient are known, especially an important thing is sum of the general binomial coefficient. We prepare some Lemma, in order to obtain this.
Lemma 3.2.2 When
is not positive integer, binomial series
ar
with Dirichlet series
(-1)r Σ r=1
r
ar Σ r=0
-1
r
converges or diverges simultaneously
.
Source 「岩波数学公式Ⅱ」 p132
Lemma 3.2.3
r
Σ r=0
converges absolutely for non-integer
>0 .
Proof
r
=
-1
-r
r
Then
Σ r=0 Let
ar =
r
=Σ r=0
-r
. Then -r -1
ar Σ r=0
r
(-1)r Σ r=1
ar r
=Σ r=0
-1
r
r
= Σ(-1)r r=1
(s1)
r ( -r)
(s2)
Here, let
(-1)r
br r ( -r)
Then
-3-
(r +1) ( -r -1) r ( -r) = (r +1)( -r -1) r (-1) r ( -r)
(-1)r+1
br+1 = br
From this
= lim
br+1 lim r br
= lim r
r
r ( -r) (r +1)( -r -1)
r
+1
+1
-1
r r +1 (r +1) (r +1) -1 (r +1)+1
r -r +1
= lim (r +1) -(r +1) +1
r
= lim r
1 1 1+ r
+1
-1
r
-1
r +1
=1
Since the judgment is impossible, we try Raabe's test for convergence.
(r +1)( -r -1) -1 r ( -r)
r(r +1) r(r +1) = lim r r ( -r) - r
br -1 = lim r r br+1
lim r r
r
- r 1 1 = lim 1+ r + 1- /r - r r
= lim r
r 1+
1 r
- 1+
1 r
r -r
r
1- /r > 0
Since
for sufficiently large
r
,
lim r r
1 = lim r 1+ r
br -1 = lim r br+1
1 1+ r
+ lim 1+ r
1
-r
1- /r
-1
r
1 r + 1+ r
1 r
1
1- /r
Here,
1 r
1+
=Σ s=0
s r
1 s
=1+
1 1! r
+
1
( -1) 1 2!
r
2
+
( -1)( -2) 1 3!
r
Then,
r
1 1+ r
-1
= rΣ s=1
s r
1 s
=Σ s=1
s r
1 s-1
= +Σ s=2
Therfore,
lim r r
1+
1 r
- 1 = + lim Σ r s=2
s r
Moreover,
1 1+ lim r r
1 1- /r
=1
After all,
-4-
1 s-1
=
s r
1 s-1
3
+
br -1 = +1 br+1
lim r r Thus, if
>0 ,
(s2) converges absolutely. Then, (s1) also converges absolutely according to Lemma 3.2.2 .
Theorem 3.2.4 The following expressions hold for arbitrary real number
>0 .
r = 2 Σ(-1) r = 0
Σ r=0
(2.1)
r
(2.2)
r=0
Proof
Σ r=0
According to Lemma 3.2.3 ,
r
converges absolutely for non-integer
>0 .
Therefore, from Theorem 3.2.1 (1.1) ,
r Σ r (-1) = (1-1)
Σ r=0
1r = (1+1) = 2
r
(2.1)
=0
(2.2)
r=0
Note In fact, it is known that (2.1) holds if For example, when
= -0.9 ,
> -1 .
the right side is
( Where, it is conditional convergence. )
2-0.9 = 0.53588673
and the left side seems to
converge to this value. However, the confirmation is difficult as the convergence is very slow. So, we apply Knopp Transformation to this and accelerate the convergence. It is as follows.
We can see that (2.1) converges to
2-0.9
.
3.2.3 Generalized Binomial Theorem Theorem 3.2.1 can be further generalized.
Theorem 3.2.5 When
is a real number, the following expression holds for
x1+ x2 = Σ r=0
r
x1 -rx2r
x1 , x2
s.t. x 1 x 2 .
( x 1 = x 2 is allowed at
>0 )
Proof If x 1 x 2 then
1>
x2 x1
=
x2 x1
. Therefore, using Theorem 3.2.1 (1.1) ,
-5-
(3.1)
= x x x x + =x 1 + + 1 x 2 x 3 x + = x + x x + x x + 0 1 2 3 x x + =Σ r x x
x1+ x2 =
x
1
x2 1+ x1
x2 1+ x1
1
2
3
2
2
2
1
1
1
1
-1 1
-2 2
-3 3
1
1
1
1
r=0
2
2
-r r
1
2
Note As is clear from the process of the proof, if x 1 x 2 then (3.1) converges absolutely. Where, x 1 x2 is allowed at
>0 .
This becomes important in Generalized Multinomial Theorem.
-6-
2
3.3 Multinomial Theorem Theorem 3.3.0 For real numbers
x1 , x2 , , xm n
x1+ x2++ xm = Σ where
n
where
, the followings hold.
n! r r r x11 x22 xmm r1! r2! rm!
denotes the sum of all combinations of
x1+ x2++ xm = Σ
n , r1 , r2 , , rm
and non negative integers
(0.1)
r1 , r2 , , rm s.t. r1+ r2++ rm= n .
n! n-r -r r r x1 1 m-1 x21 xmm-1 n -r1 -rm-1! r1! rm-1!
denotes the sum of all possible combinations of
(0.2)
n , r1 , r2 , , rm-1 .
Since (0.1) is well known, the proof is omitted. In addition, it is also clear (0.2) and (0.1) are synonymous. These are near a definition rather than a theorem.
How to generate multinomial coefficients Theorem 3.3.0 is not difficult in theory. Difficulty is its proviso. This is to actually generate combinations ( m choose n ) with repetition. But this is not easy when it becomes more than 3 terms. Since I found out the formulas which generates these without leak, I present it here as a theorem. (1.2) realizes the provis by an iterated series (multiple series) and (1.1) realizes it by a diagonal series (half-multiple series).
Theorem 3.3.1 For real numbers
x1 , x2 , , xm n
and a natural number
n
r1=0 r2=0
r r r n
rm -2
r1
x1+ x2++ xm = Σ Σ Σ
rm -1=0
n , the following expressions hold.
r1
1
rm-2
rm-2-rm-1 rm-1 n -r1 r1-r2 x2 xm-1 xm
x1
m-1
2
(1.1) n
n
n
= Σ Σ Σ
r1=0 r2=0 rm -1=0
r + r + + r n
1
r1+ r2+ + rm-1 r2+ + rm-1
m-1
2
n -r1- - rm-1
x1
rm-2 + rm-1 rm-1
r
x21 x32 xmm-1 r
r
(1.2)
Proof According to Theorem 3.1.1 , the following expressions hold. n
n
n-r1
x1+ x2+ x3+ x4++ xm = Σ nＣ r1 x1 r1=0
r
1 x2+ x3+ x4++ xm
r1
(2)
r2=0
r
r
= Σ r2Ｃ r3 x32-r3 x4++xm 3
x3+ x4++ xm
(1)
r
r
= Σ r1Ｃ r2 x21-r2 x3+x4++xm 2 r2
r2
r1
x2+ x3+ x4++ xm
(3)
r3=0
r
m-3 xm-2 + xm-1 + xm
rm -3
=
Σ r Ｃr r =0 m-3
m-2
m -2
-7-
r
-rm-2
m-3 xm-2
rm-2
xm-1 + xm
(m -2)
rm -2
r
m-2 xm-1 + xm
=
r
Σ r Ｃr r =0
m-1
m-2
r
-rm-1
m-2 xm-1
xmm-1
(m -1)
m -1
Substituting (2), (3), ,(m -2) ,(m -1) for (1) one by one, we obtain (1.1) . Next, according to Theorem 3.4.1 ( later ) , when
x1+ x2++ xm = Σ Σ Σ
r1=0 r2=0 rm -1=0
x1 x2+ x3++ xm , r1+ r2+ + rm-1
r + r + + r 1
r2+ + rm-1
m-1
2
rm-2 + rm-1 rm-1
-r1-- rm -1 r1 r2
x1 Replacing the real number
n
with non-negative integer
n
r1=0 r2=0 rm -1=0
r + r + + r 1
r1+ r2+ + rm-1 r2+ + rm-1
m-1
2
n -r1-- rm -1
x1 Since
r
x2 x3 xmm-1
,
x1+ x2++ xm = Σ Σ Σ
rm-2 + rm-1 rm-1
r
x21 x32 xmm-1 r
r
r = 0 for r > n, r =1, 2, 3, , this is a definite multiple series. Therefore, the condition n
x1 x 2+ x3++ xm is unnecessary. Although this is not bad as it is, replacing on the with
n
, we obtain (1.2) .
cf. (1.2) results in (0.2) . Because,
r + r + + r n
1
2
r1+ r2+ + rm-1 r2+ + rm-1
m-1
Example 1: The expasion of x1+ x2+ x3
rm-2 + rm-1 rm-1
=
n! n -r1 - - rm-1! r1!+ + rm-1!
4
Using (1.1) ,
r s 4 0 = x 0 Σ s x 4 1 + x 1 Σ s x 4 3 + x Σ 3 s x 4
r
4
r
4 x1+ x2+ x3 = ΣΣ r=0 s=0
4 1
3 1 1 1
=
r-s
x14-rx2 x3s
0
s=0 1
s=0 3
s=0
0-s s 2 x3 1-s s 2 x3 3-s s 2 x3
2 s x 4 4 + x Σ 4 s x 4
+
2
x12Σ s=0
0 1
4
s=0
x14 + 4x13 x2 + x3 + 6x12 x22 + 2x2 x3 + x32 + 4x1x23 + 3x22 x3 + 3x2 x32 + x33 + x24 + 4x23 x3 + 6x22 x32 + 4x2 x33 + x34
-8-
2
2-s s 2 x3 4-s s 2 x3
Using (1.2) ,
r + s s 4 0+ s = Σ 0+ s s x 4 1+ s +Σ 1+ s s x 4 3+ s +Σ 3+ s s x 4
4
4 x1+ x2+ x3 = ΣΣ
r+s
4
r=0 s=0 4
4-s 0 1 x2
s=0 4
x3s
3-s 1 s 1 x2 x3
s=0 4
1-s 3 s 1 x2 x3
s=0
=
x14-r-s x2r x3s
x14
+
4x13 x3
2+ s s x 4 4+ s +Σ 4+ s s x
6x12 x32
+
4
+Σ
4
2+ s
s=0 4
0-s 4 s 1 x2 x3
s=0
+ 4x1x33 + x34
+ 4x13 x2 + 12x12 x2 x3 + 12x1 x2 x32 + 4x2x33
+ 0
+ 6x12 x22 + 12x1 x22 x3 + 6x22 x32
+
0
+ 0
+ 4x1 x23 + 4x23 x3 +
x24
+
0
2-s 2 s 1 x2 x3
+
0
+
0
+ 0
+
0
+
0
+ 0
we can see that what totaled this along the diagonal line is equal to the above.
Example 2: The expasion of x1+ x2+ x3+ x4
3
Now, the formulas of the theorem are expanded using mathematical software. (1.1) and (1.2) are expanded and are verified respectively.
-9-
Sum of multinomial coefficients n
r m -2
r1
Σ Σ Σ nＣ r1 r1Ｃ r2 rm-2Ｃ rm-1 = m n r1=0 r 2=0
(1.1")
rm -1=0
Proof n
n
n-r r
nＣ r =ΣnＣ r1 Σ r=0 r=0 n
r
n
r
1 = (1+1)n = 2n
n
r
n
nＣ r Σ rＣ s nＣ r1 Σ ΣnＣ r rＣ s =Σ s=0 = Σ r=0 s=0 r=0 r=0 s
n
r
s
n-r r
2 = (1+2)n = 3n
n
=ΣnＣ r1n-r3r = (1+3)n = 4n nＣ r rＣ s sＣ t =ΣnＣ r ΣΣrＣ s sＣ t Σ Σ Σ r=0 s=0 t=0 r=0 s=0 t=0 r=0 Hereafter, by induction we obtain the desired expression.
Example: Sum of multinomial coefficients of x1+x2+x3
4
Let's calculate sum of multinomial coefficients in Example 1 . Then it is as follows.
1+(4+4)+(6+12+6)+(4+12+12+4)+(1+4+6+4+1) = 81 = 34
- 10 -
3.4 Generalized Multinomial Theorem Although I do not know whether the theorem like generalized multinomial theorem exists or not , since this is essential for Higher Calculus of Function Product, I present this here.
Theorem 3.4.1 The following expressions hold for real numbers and .
r r r
r m -2
r1
x1+ x2++xm = Σ Σ Σ r1=0 r 2=0
x1 , x2 , , xm s.t.x 1 x2+ x3++ x m .
rm -1=0
r1
1
rm-2
-r1 r1-r2
x1
x2
r
-rm -1 rm -1 xm
m -2 xm-1
m-1
2
(1.1)
= Σ Σ Σ
r 1=0 r 2=0 rm -1=0
r + r + + r 1
r1+ r2+ + rm-1 r2+ + rm-1
m-1
2
rm-2 + rm-1 rm-1
-r1- - rm-1 r1 r2
x1
r
x2 x3 xmm-1 (1.2)
>0 .
Where, x 1 x 2+ x 3++ x m is allowed at
Proof From Theorem 3.2.5 , when x 1 x 2+ x 3++ x m , the following expression holds.
-r1
x1 x1+ x2+ x3+ x4++ xm = Σ r1=0 r1
x2+ x3+ x4++xm
r1
Here, the right side converges absolutely. On the other hand, from Theorem 3.3.1 (1.1) , the following expression holds. r1
r1
r2
r m -1=0
r2
r3
rm -2
r1
x1+ x2++ xm = Σ Σ Σ r2=0 r 3=0
r1
r m -2
r2
rm-2 rm-1
r -r2
x21
r -r3
x32
r
-rm -1 rm -1 xm
r
-rm -1 rm -1 xm
m -2 xm-1
Substituting the latter for the former ,
r1
x1+ x2++ xm = Σ Σ Σ r 1=0 r 2=0
r m -1=0
r r r x rm-2
-r1 r1-r2
1
1
x2
m -2 xm-1
m-1
2
(1.1) Naturally, the right side also converges absolutely. Next, let us describe a multiple series and its iterated series as follows respectively.
Σ ar ,r ,,r r , r , ,r =0 1
1
2
m
,
m
2
Σ Σ Σ ar ,r ,, r r =0 r =0 r =0 1
1
m -1 ,rm
2
m
2
In order to convert the iterated series to its diagonal series, we should just perform the following operations. ( See " 02 Multiple Series & Exponential Function " ). Replace
rm-1 with rm-1- rm ,
Replace
rm-2 with rm-2- rm-1 ,
and replace the 1st
and replace the 2nd
with
rm-1
with
from the right.
rm-2
from the right.
Replace
r1 with r1- r2 ,
and replace the (m-1)th
with
r1
from the right.
If so, in order to return the diagonal series to the original iterated series, we should just perform this opposite operation. That is, Replace
r1 with r1+ r2 ,
and replace
r1
on the 2nd
with from the left.
Replace
r2 with r2+ r3 ,
and replace
r2
on the 3rd
with from the left.
- 11 -
Replace
rm-1 with rm-1+ rm ,
on the (m -1)t h
and replace
rm-1
r1
with from the left.
For example,
r r x x x r +r r r r + r ,ΣΣ; = Σ Σ Σ r +r r r x x x r +r +r r +r = r r + r ,Σ Σ; Σ Σ Σ r +r +r r +r r x
r1
r2
x1+ x2+ x3+ x4 = Σ Σ Σ r r1=0 r 2=0 r 3=0 1
1
1
r1
2
2
3
4
3
1
2
2
2
r2
2
-r1 r1-r2 r2-r3 r3
x1
r2
-r1-r2 r1 r2-r3 r3
1
r1=0 r2=0 r 3=0
2
r2
1
2
2
2
3
x4
3
1
2
3
2
-r1-r2-r3 r1 r2 r3
3
3
x2 x3 x4
1
r1=0 r 2=0 r3=0
1
2
3
2
3
3
Thus, performing this operation to (1.1) , we obtain the following.
x1+ x2++ xm = Σ Σ Σ
r 1=0 r 2=0 rm -1=0
r + r + + r 1
r1+ r2+ + rm-1
m-1
2
r2+ + rm-1
rm-2 + rm-1 rm-1
- r + + rm-1 r1 r2 r x1 1 x2 x3 xmm-1
(1.2)
Since (1.1) converges absolutely, this rearrangement is allowed.
Example 1: The expasion of x1+ x2+ x3
3.9
Using (1.1) ,
r s x x x 3.9 0 3.9 1 = + x x x x 0 Σs 1 Σs x x 3.9 2 3.9 3 + x Σ x x + x Σ 2 s 3 s x x 3.9 4 4.1 5 1 1 + + x x 4 x Σs 5 x Σs x
r
3.9 x1+ x2+ x3 = ΣΣ
r
3.9
r=0 s=0
3.9 1
3.9-r r-s s 1 2 3
0
s=0
1.9 1
2
s=0
4
0.1 s=0 1 x13.9
=
1
0-s s 2 3
2.9 1
2-s s 2 3
0.9 1
1-s s 2 3
s=0 3
3-s s 2 3
s=0
5
4-s s 2 3
1.1 s=0 1
5-s s 2 x3
+
3.9 x12.9 x2 + x3
+
5.655 x11.9 x22 + 2x2 x3 + x32
+
+ 3.5815 x10.9 x23 + 3x22 x3 + 3x2 x32 + x33 0.805838 4 3 2 2 3 4 + x2 + 4x2 x3 + 6x2 x3 + 4x2 x3 + x3 0.1 x1 0.0161168 5 4 3 2 3 2 4 5 x2 +5x2 x3 +10x2 x3 +10x2 x3 +5x2 x3 + x3 1.1 x1 Using (1.2) ,
x1+ x2+ x3
3.9
= ΣΣ r=0 s=0
r+s s x 3.9
r+s
3.9-r-s r x2 1
- 12 -
x3s
Σ s=0 0+ s 3.9
s 3.9 2+ s +Σ 2+ s s x 3.9 4+ s +Σ 4+ s s x
=
0+ s
1+ s s x 3.9 3+ s +Σ 3+ s s x 3.9
x13.9-s x20 x3s +Σ
0.9-s 3 s x2 x3 1
s=0
+
-0.1-s 4 s x2 x3 1
s=0
2.9-s 1 s x2 x3 1
s=0
1.9-s 2 s x2 x3 1
s=0
1+ s
4
=
x13.9
2.9 3.9 x1 x3
+
2.9 3.9 x1 x 2
+
1.9 2 5.655 x1 x2
+
0.9 3 3.5815 x1 x2
+
+
1.9 2 5.655 x1 x3
+
0.9 3.5815 x1
x 33
+
0.805838 x3 x10.1 3
+
1.9 11.31 x1 x2 x3
+
0.9 2 10.7445 x1 x2 x3
+
3.22335 x 2x3 x10.1
2 2
0.9 2 10.7445 x1 x2 x3
+
+
3
+
+
x10.1
x10.1
-
-
0.0805838 x2 x 3 x11.1
-
x10.1 3
x11.1
0.161168 x2 x3 x11.1
2
3
0.161168 x2 x3
4
+ 3
0.0590948 x2 x3
+
x11.1 +
0.0805838
2 3
4.83503 x2 x3
4
4
0.805838 x2
3.22335 x 2 x3
-
+
+
x12.1 2
0.0443211 x2 x3 x12.1
4
-
3
0.0310247 x2 x3 x13.1
+
we can see that what totaled this along the diagonal line is equal to the above.
Example 2: The expasion of
Although
a = b +c+d
(a + b + c + d)2.9
in this numerical example, (1.1) and (1.2) are consistent.
Sum of General Multinomial Coefficients
r1
rm -2
ΣΣ Σ r 1=0 r2=0
rm -1=0
r1
1
2
r r r
rm-2
m
m-1
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(1.1")
As expected, the following expression does not hold.
rm -2
r1
ΣΣ Σ r 1=0 r2=0
It is because
rm -1=0
r1
1
2
r r r = m
x 1= x2== xm = 1
rm-2
m-1
does not satisfy the condition x 1 x 2+ x 3++ x m .
Let its partial sum be n
rm -2
r1
Sn = Σ Σ Σ r 1=0 r2=0
Then, when
r m -1=0
n , Sn
r1
1
2
r r r
rm-2 m-1
oscillates and diverges. And
m
is the median of this oscillating divergent series. In fact, applying Knopp Transformation to S n , we can obtain the approximate value of m with high precision. However, it does not become a series but becomes an asymptotic expansion. 2007.07.06 2016.02.13 updated 2016.02.22 renewed Kano. Kono
Alien's Mathematics
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