The multi-combiational theory was
first investigated by me in the year 2008. This investigation was later gave
to Nigeria senior lecturer, Dr. Oyetunji, a head of department in University for development
Studies at time for scrutiny. This single investigation was put in an envelop
and it was actually the basis of the mult-combinational theory. The multi-combinational theory has numerous applications in sciences, that are used to compare and
predict various things that are related in sequential order. Below is the basis
of the mult-combinational theory.
THE SEQUENTIAL
COMPARISON RULE:
The sequential
comparison rule which I have developed is intended to treat many of the topics
in mathematics needed by the modern engineer, physicist or applied
mathematician.
Since engineer,
physicist, or applied mathematician are Traditional gatherers for creating
mathematical models for practical consumption, they are being negative
impacted, as they find it difficult to use reliable mathematical rules for
creating models that can solve areas that are not yet explored.
The availability of
the sequential comparison rule will truly help scientist, economist, engineer,
physicist, or applied mathematician to explore areas that are not yet possible.
ADONGO SEQUENTIAL
COMPARISON RULE
Let the sequential
formula pairs
Xo*x1*........*xr-1
= ao
X1* x2*...........xr =
a1
X2* x3*
............* xr+1 = a2
X3 * x4*
............* xr+2 = a3
.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
be given; there exist
unique sequential values xo, x1, x2,
------, which are in comparison with xr, xr+1, xr+2, -------,
respectively and are given as;
Xo =
xr[ ao/a1]
X1 =
xr+1[a1/a2]
X2 =
xr+2[a2/a3]
. . . . . . . . . . .
. . . . . . . . . .
For each i=1,
2, 3, 4, - -, r
EXAMPLE
A eureka can is filled
with water and stones, one, two, three, and four lowered into it each
until each of the stones is fully immersed. The displace water of each stone is
collected and the volume of each stone is measured.
The stones one, two,
three, and four are dried and each of them is weighed’ to find the mass of
each. The density of stones one, two, three, and four, are 100kg/m3,
920kg/m3 and 240kg/m3 110kg/m3respectively.
(a) If the density of stone one is 136kg/m3,
find the density of stone four.
SOLUTION
Let x1 represent
stone one
X2 represent
stone two
X3 represent
stone three
X4 represent
stone four
Applying Adongo
sequential comparison rule, we have
X1 x2 x3
= 100*920*240 = 22,080,000
X2 x3 x4
= 920*240*110 = 24,288,000
This implies that
X1 = x4 [x1 x2 x3/x2 x3 x4]
But if x1 =136kg/m3
Then , 136 = x4 [22,080,000/24,288,000]
136 = x4 [0.909]
X4 = 149.6kg/m3
ADONGO’S RULE OF
INFINITE SOLUTIONS
If xo* x1*x2*x3*-------* xr-1 and
x1* x2 * x3* x4* ------ * xr are
corresponding pairs of line equation and z is a real number (z ≠ 0), and If:
xo* x1* x2* x3*-----------* xr-1 =
ao correspond to x1* x2* x3* x4
------------* xr = a1
x1* x2* x3* -----------
xr-1 = ao/xo correspond to x2*x3* x4*---------------- *xr=
a1/x1
x2*x3*
------------------ *xr = a1/xox1 correspond to x3*x4*
------------------------ *xr = ar/x1x2
……………………………………….......................................................................
Then, there exist
corresponding infinite solution of xo, x1, x2, x3, --------,
xr-1 which are in relative comparison to xr, xr+1, xr+2, xr+3, --------,
xr+k respectively and are given as;
xo=zao or
ao /z or z/ao correspond to xr = za1 or
a1/z or z/a1
x =z(ao/xo) or (ao/xo) /z or z/(ao/xo) correspond to xr+1 =
(z(a1/x1) or (a1/x1) /z or z/(a1/x1)
…………………………………………….......................................................................
For each, i = 1,2,3,
------- , r, l = 1,2,3, ----, k ,
EXAMPLE
Find the possible solution of the
multi-combinational equation below: if xo x1x2 x3x4 x5 =
5
i)½(x0x1x2x3x4x5)
+ 1/3(x1x2x3x4x5x6)
=3/2
[Take z = 2]
SOLUTION
The corresponding pairs of the multiple
variables are;
Xox1x2x3x4x5 =
5
X1x2x3x4x5x6 =-3
Applying Adongo’s rule of infinite solutions
in equation (1) and (2), we have
Xo = 2*5 = 10
X6 = 2*-3 = -6
Making x1 x2 x3 x4 x5 and
x2 x3 x4 x5 x6 the
subject, we have
X1x2x3x4x5 = 5/xo -------
(3)
X2x3x4x5x6 =
-3/x1 ------ (4)
Applying Adongo’s rule of infinite solution in
equation (3) and (4), we
have
x1 = 1
Making x2x3x4x5 and
x3x4x5x6 in
equation (3) and (4),the subject we
have
x2x3x4x5 =
1/2x1 -----(5)
x3x4x5x6 = 3/ x2 ------(6)
Applying Adongo’s rule of infinite solution in
equation (5) and (6), we
have
x2 = 1
Making x3x4x5 and
x4x5x6 in
equation (5) and (6), we have
X3x4x5 = 1/2x2 ----
(7)
X4x5x6 = 3/x3 ------(8)
Applying Adongo’s rule of infinite solution in
equation (7) and (8), we have
X3 = 1
Making x4x5 and x5x6 in
equation (7) and (8), we have
X4x5 = 1/2x3
X5x6 = 3/x4
Applying Adongo’s rule of infinite solution,
we have,
X4 = 1
X5 = ½
Hence, the possible solutions of xo,
x1, x2, x3, x4, x5 and
x6 are 10,1,1,1,1,1/2, and -6 respectively
ADONGO’S RULE OF DETERMINANT
TWO SYSTEM EQUATIONS:
If A is an invertible 2*r matrix,
the solution to the system
a1(xo*x1* -----* xr-1)
+ b1 (x1*x2* -----*xr)
= k1
a2 (xo* x1*........*xr-1)
+ b2 (x1* x2* ------*xr)
= k2
Of 2 equations in the variable xo which
is in relative comparison to the variable xrand are given
as;
Xo = xr* [det A1/det A2]
Where;
det A1 = k1b2 -
k2b1
det A2 = k2a1 – k1a2
For each j = 1,2,3,
------, r
EXAMPLE
If the initial value xo of
multi-combinational equations below is xo =2/3; find the final
value
X10.
2(xox1x2x3x4x5x6x7x8x9)
+ 3(x1x2x3x4x5x6x7x8x8x9x10)
= 1 ------(1)
3( xox1x2x3x4x5x6x7x8x9)
+ 3(x1x2x3x4x5x6x7x8x9x10)
= 6 --------(2)
SOLUTION
det A1 =
-15
det A2 = 9
But xo = x10* [det
A1/detA2]
This implies that
½ = x10* [-15/9]
X10 = -2/5
Hence, x10 is -2/5 if xo is
2/3.
THREE SYSTEM EQUATIONS
If A is an invertible
3*r matrix, the solution to the system
A2(xo*x1* ------*xr-1)
+ b1(x1*x2* ----*xr)
+ c1(x2*x3* ----*xr+1)
= k1
A2(xox1 * -----*xr-1)
+ b2 (x1*x2* -----*xr)
+ c2(x2*x3*----*xr+1 )
= k2
A3(xo*x1* -----*xr-1)+
b2 (x1*x2* -----*xr)
+ c3(x2*x3* ----*xr+1)
= k
Of 3 equation in the variables xo,
x1, which are in relative comparison to the variables
Xr, xr+1 respectively
and are given as
Xo = xr* [detA1/ detA2]
X1=xr+1[detA2/detA3]
Where;
det A1 = k1[c3b2 –
c2b3] – b1[c3k2 –c2k3]+c1[b3k2-b2k3]
det A2 = a1[c3k2-c2k3]
– k1[c3a2-c2a3]+c1[k3a2 –
k2a3]
det A3 = a1[k3b2
– k2b3] – b1[k3a2 –
k2a3] + k1[b3a2 – b2a3]
For each i = 1,2,3, ------, r
EXAMPLE
If the initial values xo, x1, of
the multi-combinational equations below are 1/2, 1 respectively,
Find the final values x6, x7.
5xox1x2x3x4x5 + x1x2x3x4x5x6 - x2x3x4x5x6x7 =
4 ---(1)
9xox1x2x3x4x5 +
x1x2x3x4x5x6
– x2x3x4x5x6x7 =
1----(2)
Xox1x2x3x4x5
– x1x2x3x4x5x6 +
5x2x3x4x5x6x7 =
2 ----(3)
SOLUTION
Det A1 = 12
detA2 = -166
detA3 =
-42
But 1/2 = x2* [12/-166]
X6 = 83/12
And
1 = x7*[-166/-42]
X7 = 21/83
Hence, x6 ,x7 are
83/12, 21/83 respectively if xo, x, are 1/2 , 1
respectively.
APPLICATIONS:
THE
MULTI-COMBINATIONAL THEORY AS A PREDICTIVE MODEL IN CHEMISTRY
My multi-combinational theory
(mathematics) has useful effects on chemistry. One of the areas
to be considered now is the Atomic structure. The multiple combinations of
copper, chlorine and oxygen is the multiple combinations of their masses,
relative atomic masses and percentage abundances. The relative percentage
abundance or decimal fractions of the combine elements should sum up to 100(r+1)
where r+1 is denoted as number of combined elements. The
combined element x1x2*…*xi*….*xr has
the combined isotopes A and B. The combined relative atomic masses ofx1x2*….*xi*…*xr is
x1x2*….*xi*…*xr of Ar=[A*%(1) +B*%(2)]100(R+1), where
A= combined masses of isotope A
B=combined masses of isotope B
1=combined relative abundances of A
2=combined relative abundances of B
Given the respective
combined percentage abundances be x1x2*….*xi*…*xr and x2x3*..*xi*…*xr.
This implies that;
x1x2*….*xi*…*xr+
x2x3*..*xi*…*xr=100(r+1)
EXAMPLE
Copper, chlorine, and
oxygen have the four isotopes 63cu and 65cu for
copper; 35cland 37cl for
chlorine. The relative atomic masses of the naturally occurring copper and
chlorine are 63.55 and 35.0 respectively.
What will be the
relative percentage abundance of 37cl if the
relative percentage abundance of 63cu is 72.5%
SOLUTION
Let, 63cu65cu35cl=x1x2x3 and 65cu35cl37cl=
x2x3x4
x1x2x3 + x2x3x4=100(4)
combined-Ar = [(63cu65cu35cl*
x1x2x3)+( 65cu35cl37cl*
x2x3x4)]/400
63.5*35 = [(63*65*35 x1x2x3)+(65*35*37*
x2x3x4)]/400
2224.22 = [143325x1x2x3+
84175(400- x1x2x3)]/400
x1x2x3 =
554.1896
x2x3x4 =
400-554.1896=154.1896
Applying the
sequential comparison rule, we have;
x4=x1[x2x3x4/x1x2x3]
x4=72.5[154.189/554.1896]
x4=20.17%.
Hence, the
isotope 37cl is 20.17% if
the percentage abundance of isotope 63cu is
72,5%.
NOTE: All negations are nullifie
INTRODUCTION TO MULTI-COMBINATIONAL CALCULUS
NOTE: All negations are nullifie
INTRODUCTION TO MULTI-COMBINATIONAL CALCULUS
One of
the most useful concepts which I am developing is called relative comparison
derivative. The concept is to help physicist, economist, scientist and engineer
to compare different rate of changes of quantitative information as a fitted
models for predictive purpose.
POWER
RULE OF RELATIVE DERIVATIVE
For the
multi-combinational derivative functions f′
(xo * x1 *---- * xr-1) and
f′ (x1*x2*-----* xr), there exist unique initial derivative f′ (xo) ,which is in relative comparison of the final derivative f ′(xr) and is given as;
f′ (x1*x2*-----* xr), there exist unique initial derivative f′ (xo) ,which is in relative comparison of the final derivative f ′(xr) and is given as;
f′(xo) = f′ (xr) [f′ (xo*x1---*xr-1)/
f′(x1*x2*------*xr)]
EXAMPLE
The gross
domestic product (GDP) of Ghana, Nigeria, USA, and Indian was
N′ (to t1 t2)
= to2 t12 t22 +
106 billion
dollars after 2007
N′(t1 t2 t3)
= t21 t22 t23
+ 106 billion
dollars after 2007
(a) At what rate was the GDP
changing with respect to the multiple time to t1 t2
and t1 t2 t3.
[take to t1 t2 =
8 and t1 t2 t3
= 6]
(b) If the gross domestic product of
Ghana is N (to) = t2o - 5 to + 10 and change
respectively to time to and if the gross domestic product occurs at time to,
t1, t2 and t3 for Ghana, Nigeria, USA and
Indian respectively; Find the rate of change of GDP in Indian.
(Take to
= 3)
SOLUTION
(a) N′ ( to t1 t2
) = 8 ( to t1 t2 )
N′ (t1 t2 t3)
= 8 (t1 t2 t3)
(b) Applying power
rule of relative comparison we have
N′(to)
= N′(t3) [N′ (to t1 t2 )/N′ (t1
t2 t3 )]
But N (to)
= 2 to -5
N′(to) = 2(3) – 5
= 1
1 = N′ (t3)
[(8(8))/8(6)]
1 = N′ (t3)
(1.33)
N′(t3)
= 1/1.33 = 0.75 per month after 2007
RELATIVE COMPARISON INTEGRATION
The multi-combinational mathematics (or theory) has great influenced in almost all areas of sciences. One of its influence area is calculus ( named as “multi-combinational calculus”). Themulti-combinational calculus is divided into two main parts. That is relative comparison derivation and relative comparison integration.
The relative comparison integration is a technique I developed for of comparing the initial sequential functions of different variables from their final sequential functions.
APPLYING THE CONCEPT
The speed v0m/s, v1m/s, v2m/s, v3m/s, and v4m/s of bodies 0, 1, 2, 3, and 4 in straight line from a point 0 in the times: t0seconds, t1seconds, t2seconds, t3seconds, and t4seconds respectively is given by the fitted multi-combinational velocity equations,
v0v1v2=24t0t1t2-3t0t1t2 ……(1)
v1v2v3=6t1t2t3+4t1t2t3 ……..(2)
v2v3v4=10t2t3t4+8t2t3t4 ......(3).
Calculate the distance s3 and s4 from 0 at the end of time t0=3s, t1=2s, t2=4s, t3=1s and t4=6s of the bodies 0, 1, 2, 3, and 4 if the initial distance of body 0 and body 1 is s0=108 and s1=144 respective
RELATIVE COMPARISON INTEGRATION
The multi-combinational mathematics (or theory) has great influenced in almost all areas of sciences. One of its influence area is calculus ( named as “multi-combinational calculus”). Themulti-combinational calculus is divided into two main parts. That is relative comparison derivation and relative comparison integration.
The relative comparison integration is a technique I developed for of comparing the initial sequential functions of different variables from their final sequential functions.
APPLYING THE CONCEPT
The speed v0m/s, v1m/s, v2m/s, v3m/s, and v4m/s of bodies 0, 1, 2, 3, and 4 in straight line from a point 0 in the times: t0seconds, t1seconds, t2seconds, t3seconds, and t4seconds respectively is given by the fitted multi-combinational velocity equations,
v0v1v2=24t0t1t2-3t0t1t2 ……(1)
v1v2v3=6t1t2t3+4t1t2t3 ……..(2)
v2v3v4=10t2t3t4+8t2t3t4 ......(3).
Calculate the distance s3 and s4 from 0 at the end of time t0=3s, t1=2s, t2=4s, t3=1s and t4=6s of the bodies 0, 1, 2, 3, and 4 if the initial distance of body 0 and body 1 is s0=108 and s1=144 respective
SOLUTION
s0s1s2=∫(v0v1v2)d(t0t1t2)=3(t0t1t2)2-0.375(t0t1t2)2+k1...............(4)
s1s2s3= ∫(v1v2v3)d(t1t2t3)=0.75(t1t2t3)2+0.5(t1t2t3)2+k2….......(5)
s2s3s4= ∫(v2v3v4)d(t2t3t4)=1.25(t2t3t4)2+(t2t3t4)2+k3…………(6)
At the point 0 I.e the starting point t0=0, t1=0, t2=0, t3=0, t4=0 and s0=0, s1=0, s2=0, s3=0, s4=0.Therefore substituting expression t0=0, t1=0, t2=0, t3=0, t4=0 and s0=0, s1=0, s2=0, s3=0, s4=0into equation(4), (5),and (6) we have k1=0, k2=0, k3=0.
Applying the methods of relative comparison, we have,
s0=[∫(v0v1v2)d(t0t1t2) /∫(v1v2v3)d(t1t2t3)]
s1=s4[∫(v1v2v3)d(t1t2t3)/∫(v2v3v4)d(t2t3t4)]
This implies that;
108=s3[1512/80]
S3=5.71m.
and;
144=[80/1296]
S4=2332.8m.
Hence the distances of the bodies 3 and 4 are 5.71 and 2332.8 respectively.
MILTI–COMBINAATIONAL
TRIGONDMETRY
The goal of the
research is to examine the effect of multi-combinational theory in
trigonometry.
The study found that
new rules, formulae, theorems and methods can be created with the influence of
the multi-combinational theory (I.e. sequential comparison rule) and prove very
effectively.
Looking at the
selected topic discussed below, this research of the multi-combinational
paradigm of trigonometry is examined:
(1) Can new mathematical rules, formulae and
methods created in trigonometry in the name of the multi-combinational theory?
(2) Is the availability of the
multi-combinational theory consumption affected positively in trigonometry?
DEFINITION OF THE
MULTI-COMBINATIONAL THEORY
Multi-Combinational
theory is the study of different kinds of mathematical information which are
related in multiples order of sequences and used mainly for comparison and
predictive purposes.
The background of this
theory is mainly depends on the “sequential comparison rule” which has main
influences in all areas of trigonometry.
RATIO DEFINITION
Let Ó©1* Ó©2*.......* Ó©i be
a multiple angles of i’s-right-angles triangles in standard positions with any
multiple point P[x0*x1*---*xr-1, x1*x2*--*xr]
on the multiple terminal side, a multiple distance x2*x3*---xr+1from
the origin x2*x3*---*xr+1≠ 0.
The relations of the
multi-combinational functions are defined as followed.
Xr = x0* [sin
(Ó©1* Ó©2*.......* Ó©i) / cos(Ó©1* Ó©2*------ Ó©i)]
Xr = x0* [sec(Ó©1* Ó©2*.... * Ó©i)
/ csc(Ó©1* Ó©2*......Ó©i)]
For each i=
1,2,3,----, r
EXAMPLE
In the three
right-angle triangles ABC, BCD, and CDE which are in linear mapping (Reference
Tittle: “Maker of modern mathematics”) have sides AB = x0, BC =
x1 and CA = x2 for right angle triangle ABC: BC
= x1 CD = x2, DB = x3 for
right-angle triangle BCD: CD = x2, DE = x3, EC = x4 for
right angle triangle CDE.
If angles ABC, BCD,
and CDE are 900 each and angle CAB =600; angle DBC
=400, and angle ECD = 500.
(i) Find the length EC of right angle triangle CDE if
the length AB of right angle triangle ABC is 5m.
SOLUTION
Ó©1* Ó©2* Ó©3 =
600* 400* 500 =
120,0000
X0 =
5m
Therefore
X4 = 5* [sin(120,0000)
/ cos(120,0000)]
X4 =
-8.66
But since distance
don’t measure in negative sign, we neglect the
MULTI- COMBINATIONAL
SYSTEM ALGEBRAL AS A FITTED MODEL IN ECONOMY:
The economy system of
one society consists of several industries in which each industry depends on
its own and others to produce its own products. Empirically, the interdependent
relationship between one industry and others varies from time to time and the
time period each industry depends on others industries to produce its own
products varies from industry to industry. Because of this, it is difficult for
an economist to determine the gross production of each industry: unless the
gross production of one industry is given that the gross production of others
industries can be determined at a specified future time. This can be done by
applying the concept of the multi- combinational system algebra to
construct an input- output economical models based on past period production
and interdependent relationship to forecast future gross production of each of
the industry.
MULTI-COMBINATIONAL
INPUT- OUTPUT MATRIX
Consider a
simple open economy as being based on agricultural products, manufactured goods
and fuels is given as:
OUTPUTS
|
|||
INPUTS
|
Agricultural products
|
Manufactured goods
|
Fuels
|
Agric
|
a11
|
a12
|
a13
|
Manufactured
|
a21
|
a22
|
a23
|
Fuels
|
a31
|
a32
|
a33
|
If we know the
surpluses or final demands of agricultural products to beD1,
manufactured goods to be D2 and fuels to be D3: then
the multi-combinational input- output matrix of above open economy is given as:
A
|
M
|
F
|
Surpluses
|
|
A
M
F
|
a11
a21
a31
|
a12
a22
a32
|
a13
a23
a33
|
D1
D2
D3
|
AM
MF
|
a11a12a21a22
a21a22a31a32
|
a12a13a22a23
a22a23a32a33
|
D1D2
D2D3
|
|
AM
|
MF
|
Where the gross
production matrix is
X =
|
(X1X2, X2X3)
|
All is represented
as; X- AX= D or (I – A)X =D, where I is
identity matrix.
OPEN ECONOMY
The economy of
Ghana is based on agricultural industry, mining industry and oil industry. An
output of 1ton of agricultural products requires an input of 0.1ton
of agricultural products, 0.02 ton of minerals, and 0.05 ton
of oil. An output of 1 ton minerals requires an input of 0.01
ton of agricultural products, 0.13 ton of minerals and 0.18 ton
oil. An output of 1 ton of oil requires an input of 0.01 ton
of agricultural products, 0.2 ton of minerals and 0.05 ton
of oil. The multi-combinational input-output matrix of the above open
economy
which have
surpluses of 40 units of agric, 35 units of
mineral and 25 units of oil.
Agricultural(A)
|
Manufactured(M)
|
Oil(O)
|
Surpluses(D)
|
|
A
M
O
|
0.1
0.02
0.05
|
0.01
0.13
0.18
|
0.01
0.20
0.05
|
40
35
25
|
AM
MO
|
0.1*0.01*0.02*0.13
0.02*0.13*0.05*0.18
|
0.01*0.01*0.13*0.2
0.13*0.2*0.18*0.05
|
40*35
35*25
|
|
AM
|
MO
|
Find the gross
production of oil industry if the gross production of agricultural product is
80
|
X = (I-A) -1 D,
X1x2=
1399.97
X2x3=
875.03
X3=
80[875.03/1399.97]= 50.
Hence, the gross
production of x3 is 50, if the gross production of
x1 is 80.
NOTE: The value of x2 is
non-dependent variable and has no effect in the computation.
CLOSED ECONOMY
In this model no
external demands are needed: inputs and outputs are used within the
system, then such a model is called closed economical models. In such a model
labor is needed so, there is no surplus and D=0
ADONGO’S-RYBCZYNSKI’S
THEOREM
The relative change of
the industries x0, x1,x2,……,xj-1,…,xr-1 depend
the effect on outputs of change in the lands H1,H2,….,
H0,Hj-1,…,Hr and labors, L0,L1,L2,….,Lj-1,….,Lr
-1 endowments.
This theorem states
that if the endowments of some resources increase, the industries use the
resources most intensively will relatively increase their outputs while the
others industries will relatively decrease their outputs. The relative factor
intensity of each industry is measured by the ratio of factors use in each
industry in comparison to others industries 0,1,2,...., and 1,2,3,...,
are higher are relatively initial-orders labors-lands-intensively and
final-orders labors-lands-intensively respectively.
Algebraically,
the Adongo’s-Rybczynski’s theorem is derived from the simple
solution of multi-combinational system equations. In
the example above, the solution of the model are composed by the lands and
labors constraints in their relative order of sequences.
(a0*a1*….aj-1*…ar-1)1*(x0*x1*…*Xj-1*….*xr-1)+(a1*a2*…aj*…ar)1*(x1*x2*….xj*….*xr)=H ...1
(a0*a1*….aj-1*…ar-1)2*(x0*x1*…*Xj-1*….xr-1)+
(a1*a2*…*aj*…ar)2(x1*x2*….xj*….xr)=H.....2
Applying Adongo’s
rule of determinants, we have
X0=xr[detA1/detA2]
Where;
H=H1*H2*…*Hj*….*Hr+H0*H1*….*Hj-1*….Hr-1
L=L1*L2*…*Lj*….*Lr+L0*L1*…*Lj-1*….*Lr-1
DetA1=H(a1*a2*..aj*…ar)2-L(a1*a2*…*aj*…*ar)1
DetA2=L(a0*a1*….*aj-1*….*ar-1)1-H(a0*a1*…*aj-1*…ar-1
