ABTRACT
Sometimes in life, we are faced
with the challenges of permutation and combination of diverse scenarios. This
led to the evolvement of the Binomial Theorem for solving equations of the form
(X + Y)n. After a critical study of this theorem, it was discovered
that some modifications are required to solve for multiple binomials of the
form (X + V1)(X + V2)(X + V3)…(X + Vn).
In this project, a valid formula for dealing with these types of binomials was
postulated and a step by step methodology of its application was documented. It
is believed that this theorem will find wide applications in resolving some
scientific challenges.
INTRODUCTION
Scientists of old did a lot of
researches to better their lives and that of those around them. These efforts
are responsible for the relative comfort we enjoy today. From the few
researches I have done which are most times triggered by challenges before me,
I discovered that what are yet to be known are more than all presently known.
Who will discover them? Formula to directly expand multiple binomial was one of
them.
OMOROGHOMWAN MULTIPLE BINOMIAL THEOREM
A MULTIPLE BINOMIAL OF THE FORM (X
+ V1)(X + V2)(X + V3)…(X + Vn) CAN
BE EXPANDED DIRECTLY WITHOUT FIRST OPENING BRACKETS BY SYSYEMATICALLY COMBINING
THE CONCERNED VARIABLES IN THEIR NORMAL POWER EQUIVALENT AND ORDERLY ADDED IN
THE NUMBER OF TIMES OF THEIR CORRESPONDING BINOMIAL COEFICIENTS AND OPERATING
THE CONSTANT AS USUAL.
i.e. (X + V1)(X + V2)(X + V3)…(X
+ Vn) = r=0n∑ { Xn-r ∑
(nCr li Vr )} - (1)
nCr: This is the number of times the variables (V)
should be added when combined in r’s which is the normal binomial coefficient
X: The
constant
V: The variable eg. a ,b, c.
i:
First variable(V1
) which is a in our case
l:
Last variable (Vn)
n: no of brackets in a given case
r:
The number of the variable that should be combined as an entity e.g.
when r is 2 then the combination should look like this ab + bc+ cd+…….and when
it is 3, the combination should be like this abc + bcd+ cde +…………
ANALYSIS
SIMPLE BINOMIAL is an expression
in which the sum of two constants are raised to a given power e.g.
(y+q)2, (t-u)3,(s+r)21,
and it has a formula eg.
(a+b)n=
r=0n∑
nCr an-r
br - - - (2)
A MULTIPLE BINOMIAL is the
product of more than one bracket which carries the sum of a constant and
variables e.g.
(x+a)(x-b)(x+c) - - - - (3)
Here x is the constant while
a,b,and c are the variables (V’s)
SYSTEMATIC COMBINATION
DIRECT EXPANSION is a process by
which a simple or multiple binomial expression is expanded in which the
constant to a given degree carries the sum of all its coefficients in one place
e.g.
i.
(x+a)3 =x3 + x2(3q)
+ x1(3q2) + x0(q3) - - - (4)
ii.
(x + a) (x + b) (x + c)= x3 + x2(a
+ b + c) + x1(ab + ac+bc) + x0(abc) - -
(5)
We assumed x to be the constant
in equation 4 above and also assuming q to be variables while in equation 5 we
also assume x is the constant but a, b and c are the variables. . In equation 4
above, x2 is a constant to a given power (degree 2) with the sum of
coefficient (q + q + q =3q) while in equation 5, x2 was a constant to degree 2 with the
sum of its coefficients (a + b + c).
NORMAL POWER EQUIVALENT:
Consider equation 6
(x+q)4 - - - - - - (6)
Recall (a+b)n=
r=0n∑ nCr an-r br
Where n = 4
= x4 +
x3(4q) + x2(6 q2) + x(4q3)
+ q4
the sum of x2 coefficients
is 6q2 (ie q2+
q2 + q2+ q2+ q2+ q2)
also consider equation 7 below
(x + a) (x + b) (x + c) (x + d) - - - - (7)
The sum of x2 coefficients
are (ab + bc + cd + da + ac +bd). Refer to
equations 9 and 17.
This is true because if
q = a = b =c = d
then
ab + bc + cd + da + ac +bd
= a2+ a2+ a2+
a2+ a2+ a2 (6a2)
= b2+ b2+ b2+
b2+ b2+ b2 (6b2)
= c2+ c2+ c2+
c2+ c2+ c2 (6c2)
=d2+ d2+ d2+
d2+ d2+ d2 (6d2)
= q2+ q2+ q2+
q2+ q2+ q2 (6q2)
We can then have a relationship
like this
(x+q)4 = (x+a) (x +
b) (x + c) (x + d)
This time we focus on the
variables. The truth is that degree 2 in a simple binomial will combine in twos
in the corresponding multiple binomial likewise degree 3 will combine in threes
and so on, that is the normal power equivalent.
ORDERLY ADDED ……Some binomials
(simple and multiple) may contain negative variables. The additional of these
special types have to be done orderly .When negative variables combine in odd
numbers of time, it take – and takes
+ when combining in even numbers of ways or
power as seen in equation 8 and 9
i.
(x-q)4 - - - - - (8)
= x4 -
x3 (4q) + x2(6q2)
-
x(4q3) + q4
ii.
(x-a)
(x-b) (x-c) (x-d) - - - - (9)
= x4 -x3(a + b + c + d) + x2(ab+bc+cd+da+ac+bd) - x (abc
+ bcd + cda + dab) + abcd.
Note the relationship between equations 8 and
9.
IN THE NUMBER OF TIMES OF THEIR
CORRESPONDING SIMPLE BINOMIAL COEFFICIENTS: Look again at equation 8 and 9
above, they have something in common which is x, the constant. Let take x4 in i to be the same x4
in equation 9 likewise x3, x2,x, now the simple
binomial coefficient in equation 8 were
1, 4, 6, 4, 1 respectively alongside with the variable q to various degrees but
the case in equation 9 is more complicated, therefore it has to be
systematically combined and added in the number of times of its equivalent
simple binomial coefficients. This was why the coefficient of x3 in equation 8 changed from 4q (q+ q+ q
+q ) to a + b + c + d in equation 9 correspondingly x2 ‘s changed from 6q (ie
q2+ q2+ q2+ q2+ q2+ q2)
in i to ab + bc + cd + da + ac +bd) in equation 9 just as x’s changed from 4q3 (ie q3+ q3+ q3+
q3+ q3+ q3) in equation 8 to abc + bcd + cda +
dab) in equation 9.
It therefore follows that
equation 8 = equation 9
IF q = a = b = d.
OPERATING THE CONSTANT AS USUAL means
that the constant follows normally whether in simple or multiple binomial
operations. This has already been dealt with as seen above
OMOROGHOWMAN’S RULES FOR MULTIPLE BINOMIAL SYSTEMATIC
COMBINATION
1. In
each bracket, each variable must appear in the same number of times as the
rest. This is equal to the coefficient number directly above the bracket in the
normal binomial form when using the Omoroghomwan’s triangle as shown in fig. 2.
Omoroghomwan’s Triangle is a
modification of Pascal’s Triangle of Fig 1. below.
1
1 1
1 1
1 2 1
1
3 3
1
1 4 6 4 1
1
5 10 10 5
1
1 6 15 20 15
6 1
Fig. 1 Pascal’s Triangle
1
1
1
* 1 2 1
β 1 3 3 1
1 4 6 4
1
1 5
10 10 5
1
1 6 15
20 15 6 1
Fig. 2. Omoroghomwan’s Triangle
See equations 10 and 11 for
explanation.
Eg (x-q)3 - - - - - (10)
= x3 _ x2 (3q) + x(3q3) – q3
Binomial coefficient ( β ) = 1 3 3 1
(x-a) (x-b) (x-c) - - - - (11)
= x3 - x2(a - b - c) +
x (ab + ab + ca) –abc
The number of each variable
appearances
In each bracket (*) = 1 2 1
Where *
stands for the number of times each variable should /must appear in the
bracket as predicted by Omoroghomwan following his triangle .
While β represents the normal binomial
coefficients as derived from Pascal's Triangle.
2. There
must not be more than one identical combination
eg ace =
eca = eac
= aec. Only one must appear.
3. One
variable must not appear more than once in each combination eg adba is wrong
just as apqbq is
4. For
easy and accurate combination note the following steps;
1st , start by
combining the variables alphabetically until the last become the head e.g.
combine abcdef in fours as follows abcd
,bcde, cdef,defa, efab, fabc, but when this has ended then try the next step.
2nd, combine
alphabetically but skipping the next alphabet to the head until the cycle is
completed e.g. acde, bdef,cefa,dfab,
eabc,fbcd, when this try to skip 2, 3, 4, and so on until you discover that
such no longer holds.
3rd, combine by
keeping the first two variables constant before skipping the third alphabet
till the cycle is completed e.g. abcde, bcdef, cdefa.
If there are still more try 2, 3, alphabets and so on until it no longer holds.
Then start again by keeping the first 3, 4 alphabets and so on using the above
manipulation i.e. the skipping of 1, 2, 3 and so on.
4th, combine alphabetically
skipping one after each variable e.g. In combining abcdef in threes, we have
something like abcde and bcdef.
If there are still more then keep the first 2, 3 alphabets constant before
Skipping one, two, three…… n alphabets.
5th, stop the
combination if the numbers of combined entities are equal to the numerical
coefficient in the corresponding normal binomial using the Pascal’s Triangle.
WORKED EXAMPLES
1. i. (x +q)2 - - - - - (12)
i.e (a+b)n = r=0n∑ nCr an-r br
Where n = 2
=
x2+ (2q) + q2
β
= 1
2 1
ii. (x+a) (x+b) - - - (13)
i.e. (X + V1)(X
+ V2) = r=0n∑ { X2-r ∑ (2Cr
ba Vr )}
=
x2 + x(a + b) + ab
* = 1
1
2. i.
(x + q)3 - - - - (14)
i.e (a+b)n=
r=0n∑ nCr an-r br
Where n = 3
=
x3 + x2(3q) +(3q2) + q3
β
= 1 3 3 1
ii. (x + a) (x + b) (x + c) - - - (15)
i.e. (X + V1)(X
+ V2)(X + V3)= r=0n∑ {X3-r ∑ (3Cr
ca Vr)}
= x3 + x2 (a + b+c) + x (ab + bc + ca)+ abc
* = 1 2 1
3. i.
(x-q)4 - - - - (16)
i.e (a+b)n=
r=0n∑
nCr an-r
br
Where n = 4
=
x4 - x3(4q) +
x2(6 q2) + x(4q3) + q4
β = 1 4 6 4 1
ii. (x-a) (x-b) (x-c) (x-d) - - - (17)
i.e. (X + V1)(X
+ V2)(X + V3)(X + V4) = r=0n∑
{X4-r ∑ (4Cr
da Vr )}
= x4 - x3 (a +b+c+d) + x2 (ab+bc+ca +da+ ac+bd) –x( abc
+ bcd + cda + dab) +abcd
* = 1 3 3 1
4. i. (x-q)5
- - - (18)
i.e (a+b)n=
r=0n∑ nCr an-r br
Where n = 5
= x5 - x4(5q) + x3(10q2) +
x2(10q3) + x(5q4) - q4
ii. (x-a)
(x-b) (x-c) (x-d) (x-e) - - - - (19)
i.e. (X + V1)(X
+ V2)(X + V3)(X + V4)(X + V5) = r=0n∑
{X5-r ∑ (5Cr
ea Vr )}
= x6 - x4 (a + b + c
+ d + e) + x3 (ab + bc + cd + de + ea + ac+ bd + ce + da + eb)
-x2(
abc + bcd + cde + dea + eab+ acd + bde + ced + dab + ebc)
+ x(abcd +bcde +cdea + deab + eabc ) – abcde
5.
i. (x-q)6 - - - (20)
i.e (a+b)n=
r=0n∑ nCr an-r br
Where n = 6
= x6 - x5(6q)
+ x4(15q2) +
x3(20q3) + x2(15q4) - x(6q5)
+ q6
ii. (x-a)
(x-b) (x-c) (x-d) (x-e) (x-f) - - (21)
i.e. (X + V1)(X
+ V2)(X + V3)(X + V4)(X + V5)(X + V6)
= r=0n∑ { X6-r ∑ (6Cr
fa Vr )}
= x6 - x5 (a
+ b + c + d + e + f) + x4 (ab + bc + cd + de + ef + fa + ac + bd +
ce
+ df + ea+ fb + ad + be + cf) -x3(
abc + bcd + cde + def + efa + fab + acd
+ bde + cef + dfa + eab + fbc +
ade + bef + cfa + dab + ebc + fcd + ace +bdf )
+ x2 (abcd +bcde
+cdef + defa + efab + fabc + acde + bdef + cefa + dfab
+ eabc + fbcd + abde +bcef +
cdfa ) –x( abcde + bcdef +cdefa + defab
+ efabc +fabcd ) +abcdef
Recall that;
* The number of times each variable
should /must appear in the bracket as predicted by Omoroghomwan following his triangle.
β The normal binomial
coefficient.
QUESTIONS
i.
a, b, c d & e combine ten times in 3’s but
a,b,c ,d,e, & f combine twenty times
in 3’s.
Why?
Answer: Because of the
introduction of another variable (f)
ii.
What does this implies.
Answer: It implies that the new
variable (f) must combine ten times with the rest to make them twenty entities
in that bracket.
Also note that ten is
Omoroghomwan’s Triangle parameter while twenty is Pascal’s.
iii.
Assume x=10, a=1, b=2, c=3, d=4
Evaluate
(x-a)(x-b)(x-c)(x-d) - - - - (22)
Solution:
Substituting values, equation 22
becomes
(10-1)(10-2)(10-3)(10-4)
=9x8x7x6
=3024
Also from worked example 3,
equation 22 becomes,
X4-x3(a+b+c+d)+x2(ab+bc+cd+da+ac+bd)-x(abc+bcd+cda+dab)-abcd -
- (23)
Substituting values into
equation 23, we have.
104-103(1+2+3+4)+102(1x2+2x3+3x4+4x1+1x3+2x4)-10(1x2x3+2x3x4+3x4+1+4x1x2)+(1x2x3x4)
=104-103x10+102x35-10x50+24
=104-104+3500-500+24
=3024
Therefore, equation 22 =
equation 23
CONCLUSION
REFERENCE
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