MY OWN WORKS(2007-2010)

1. MATHEMATICIANS FAILED TO MEET CURRENT ECONOMICAL SITUATION AND SCIENTIFIC DEVELOPMENT IN GHANA

As many of you may be aware, I have spent a very broad time in recent years (twenty first century) in order to fine out the facts and connection between math and science research (innovation) and the political, social and economical institutions of Ghana, so that I can bring new inventions to our mathematical society. I hope very well that during this next year I will double those efforts and help to produce new changes in the math, science, engineering and academic society that will enable you to continue to flourish in these challenging periods of times.

I would like to explore these vital areas with you today and I wish that this is just the start of a continuing dialogue with you as individuals and the mathematical and scientific society in Ghana. To start this discussion, I would like to provide some background and content for the situation in which we find ourselves.

The scientific community finds itself in the midst of great challenge and change because of poor mathematical background and our inability to think critical to produce innovation ideas.

The concept of mathematics as part of our scientific and economical development is degraded at an accelerating rate due to lack of competent and reliable theories, rules and models to carry out our scientific and mathematical investigation in areas that are not explored.

With the systematic promotion of mathematics in Ghana; University graduates, Polytechnic graduates, teacher trainees and Senior High School graduates are only thought how to be the best parrots without knowing how to create new mathematical ideas.

Due to lack of inventive competent; the economical, insurance, statistical, scientific, psychological, sociological and engineering industries over the last decade from now demand on mathematical concepts, theories, and models supply have been limited that applied mathematicians, scientists, engineers, economists, statisticians, psychologists, and sociologists still used concepts, theories and models that do not meet our current situations that we are confronting today in Ghana.

THE QUESTIONS

The questions are: what inventions (theories, rules, and models) did I bring into existence so that above problem can be solved? How did my inventions helped to solve the problem?

SOLUTION TO THE QUESTION

For the matter of facts, I have made an attempt to develop new mathematics namely; multi-combinational mathematics, central(gravitational) point theory and least whole normal mathematics. These inventions, mention above can be used to solve problems in everyday life, in science, in industry and in business more especially areas that are not yet

accomplished.

                                                        

PART ONE

MULTI-COMBINATIONAL MATHEMATICS

INVENTED (YEAR: 2008-2009)

BY

WILLIAAM AYINE ADONGO

ACTUARIAL SCIENCE STUDENT

University For Development Studies: GHANA

                                                      

                                                                   CHAPTER ONE

                       MULTI-COMBINATIONAL MATHEMATIICS

1.0THE SEQUENTIAL COMPARISON RULE

The sequential comparison rule which I have developed is intended to treat many of thetopics in mathematic needed by the modern engineer, physicist or applied mathematician. Since engineers, physicists, or applied mathematicians 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. 

THE RULE

Let the sequential formulae pairs

X_o*x_1*........*x_r-1  = a_o

X_1* x_{2*...........}x_r = a₁

X_2* x_{3* ............*} x_r+1  = a₂

X_{3 *} x_{4* ............*} x_r+2 = a₃

_{. . . . . . . . . . . . . . . . . . . . . . . . . . .}

 X_i-1*x_i*...._*x_r+j = a_i-1

X_i*x_i+1*..._*x_r+k = a_i

be given; there exist unique sequential values x_o, x₁, x₂, ------, which are in comparison with x_r, x_r+1, x_r+2, -------, respectively and are given as;

X_o  = x_r [ a_o/a₁]

X₁  =  x_r+1  [a₁/a₂]

X₂  = x_r+2  [a₂/a₃]

. . . . . . . . . . . . . .

X_i-1=x_r+k[a_i-1/a_i]

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/m³ respectively.

(a)    If we change the eureka can and measured the density of stone one to be 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

X₁x₂x₃  = 100_*920_*240 = 22,080,000

X₂x₃x₄  = 920_*240_*110 = 24,288,000

This implies that

X₁ = x₄ [x₁ x₂ x_3/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

1.1 MULTI-COMBINATIONAL ALGEBRA

One of the primary objectives of me developing this algebra is to give the student substantial experience in modeling and solving real world problems. Enough applications are included to convince the most skeptical student that the multi-combination algebra which I have developed is really useful.

1.1.a ADONGO’S RULE OF INFINITE SOLUTIONS

If x_o* x_1*x_2*x_3*-------* x_r-1 and x_1* x_{2 *} x_3* x_4* ------ _* x_r are corresponding pairs of line equation and z is a real number (z ≠ 0), and if: 

x_o* x_1* x_2* x_{3*-----------*} x_r-1= a   

 X_1* x_2* x_3* x_{4 ------------*} x_r  = a₁   }............. (1)

 X_1* x_2* x_3* ----------- x_r-1 = a_o/x_o     

 X_2*x_3* x_4*---------------- _*x_r= a₁/x₁ }.........(2)

X_2*x_{3* ------------------ *}x_r = a₁/x_o*x₁         

X_3*x_{4* ------------------------ *}x_r = a_r/x_1*x₂   }.............(3)

_{.......................................................................................}

Then there exist corresponding infinite solution of x_o,  x_1,  x_2,  x_3, --------, x_r-1  which are in relative comparison to  x_r,  x_r+1,  x_r+2,  x_r+3, --------,  x_r+k respectively and are given as;

x_o = (za_o or a_o /z or z/a_o)               

x_r = (za₁ or a₁/zor z/a₁)   }................(1)

                      

x₁ = (z(a_o/x_o) or (a_o/x_o) /z or z/(a_o/x_o)   

x_r+1 = (z(a₁/x₁) or (a₁/x₁) /z or z/(a₁/x_o)   }...........(2)            

 ------------------------------------------------------          

For each, i = 1,2,3, ------- , r,  l = 1,2,3, ----, k , and  j = 1, 2, 3, ------,n

EXAMPLE

i Find the possible solution of the multi-combinational equation below: if  x_ox₁x₂x₃x₄x₅ = 5

[Take z = 2]

1/2(x₀x₁x₂x₃x₄x₅)+1/3(x₁x₂x₃x₄x₅x₆)=3/2

SOLUTION

The corresponding pairs of the multiple variables are;

X_ox₁x₂x₃x₄x₅ = 5

X₁x₂x₃x₄x₅x₆ =-3

Applying Adongo’s rule of infinite solutions in equation (1) and (2), we have

X_o = 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/x_o ------- (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  x_o, x_1, x_2, x_3, x_4, x₅ and x₆ are 10,1,1,1,1,1/2, and -6 respectively

1.1.b GENERAL RULE OF MULTI-COMBINATIONAL SYSTEM

Given the system

(a_¡-1)_1*(x_0*x_1*---_*x_r-1) + (a_¡)_1*(x_1*x_2*---_*x_r) + ---+ (a_r+k)1_*(x_¡-1*x_¡*--_*x_r+k) = k₁

(a_¡-1)₂*(x_0*x_1*---_*x_r-1) + (a_¡)_2*(x_1*x_2*---_*x_r) + ---+ (a_r+k)_2*(x_¡-1*x_¡*--_*x_r+k) =k₂

_{......................................................................................................................................................................................}

(a_¡-1)_m*(x_0*x_1*--_*x_r+1) + (a_¡)_{m *}(x_1*x_2*--_*x₁) + ... + (a_r+k)_{m *}(x_¡-1* x_¡* --_*x_r+k) =k_m

Where a_¡-1, a_¡, up to a_r+k are real numbers of (¡-1)^th- term, ¡^th- term, up to (r+k)^th_-term reapectively, has a solution

X₀ = x_r* [x_0*x_1* --_*x_r-1/x_1*x_2*--_*x_r]

X₁ = x_r+1[x_1*x_2*--_*x_r/x_2*x_3*--_*x_r+1]

X₂ = x_r+2* [x_2*x_3*--_*x_r+1/x_2*x_3*--_*x_r+2]

X_r-j = x_r+k*[x_{j-1 *}x_i*--_*x_i+j/x_1* x_i+1*--_*x_r+k]

For each i=1,2,3,---,r, j = -1,0,1,2,---,k, and l = 1,2,3,---, m.

EXAMPLE

If the initial value x₀ of the multi-combinational equations below is x₀=2/3, find the final value x_10.

2(x₀x₁x₂x₃x₄x₅x₆x₇x₈x₉) + 3(x₁x₂x₃x₄x₅x₆x₇x₈x₉x₁₀) =1

3(x₀x₁x₂x₃x₄x₅x₆x₇x₈x₉) + 3(x₁x₂x₃x₄x₅x₆x₇x₈x₉x₁₀) = 6

Solution

 X₀ = x₁₀ [x₀x₁x₂x₃x₄x₅x₆x₇x₈x₉/x₁x₂x₃x₄x₅x₆x₇x₈x₉x₁₀]

But x₀ =213, x₀x₁x₂x₃x₄x₅x₆x₇x₈x₉ = 5, and x₁x₂x₃x₄x₅x₆x₇x₈x₉x₁₀ = -3

Therefore, 213 = x_10* [-5/3]

213 = -5/_3* x₁₀

X₁₀ = -2/5.

1.1.c ADONGO’S RULE OF DETERMINANT

 1.1.c1 TWO SYSTEM EQUATIONS:

If A is an invertible 2_*r matrix, the solution to the system

a₁(x_o*x_1* -----_* x_r-1) + b₁ (x_1*x_2* -----_*x_r) = k₁

a₂ (x_o* x_1*........_*x_r-1) + b₂ (x_1* x_2* ------_*x_r) = k₂              

Of 2 equations in the variable x_o which is in relative comparison to the variable x_r and are given as;

X_o = x_r* [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 x_o of multi-combinational equations below is x_o =2/3; find the final value X_10.

2(x_ox₁x₂x₃x₄x₅x₆x₇x₈x₉) + 3(x₁x₂x₃x₄x₅x₆x₇x₈x₈x₉x₁₀) = 1 ------(1)

3( x_ox₁x₂x₃x₄x₅x₆x₇x₈x₉) + 3(x₁x₂x₃x₄x₅x₆x₇x₈x₉x₁₀) = 6 --------(2)

SOLUTION

det A₁ = -15                       

det A₂ = 9

But x_o = x_10* [det A₁/detA₂]

This implies that

½ = x_10* [-15/9]

X₁₀ = -2/5

Hence, x₁₀ is -2/5 if x_o is 2/3.

1.1.c2 THREE SYSTEM EQUATIONS:

If A is an invertible 3_*r matrix, the solution to the system

a₂(x_o*x_1* ------_*x_r-1) + b₁(x_1*x_2* ----_*x_r) + c₁(x_2*x_3* ----_*x_r+1) = k

a₂(x_ox_{1 *} -----_*x_r-1) + b₂ (x_1*x_2* -----_*x_r) + c₂(x_2*x_3*----_*x_r+1 ) = k₂

a₃(x_o*x_1* -----_*x_r-1)+ b₂ (x_1*x_2* -----_*x_r) + c₃(x_2*x_3* ----_*x_r+1) = k

Of 3 equation in the variables x_o, x_1, which are in relative comparison to the variables

 X_r, x_r+1 respectively and are given as

X_o = x_r* [detA₁/ detA₂]

X₁=x_r+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[k₃b_{2 –} k₂b₃] – b₁[k₃a₂ – k₂a₃] + k₁[b₃a₂ – b₂a₃]

For each i = 1,2,3, ------, r

EXAMPLE

If the initial values x_o, x_1, of the multi-combinational equations below are 1/2, 1 respectively,

Find the final values x_6,  x₇.

5x_ox₁x₂x₃x₄x₅  +  x₁x₂x₃x₄x₅x₆   -  x₂x₃x₄x₅x₆x₇   = 4 ---(1)

9x_ox₁x₂x₃x₄x₅ + x₁x₂x₃x₄x₅x_{6 –} x₂x₃x₄x₅x₆x₇ = 1----(2)

X_ox₁x₂x₃x₄x_{5 –} x₁x₂x₃x₄x₅x₆ + 5x₂x₃x₄x₅x₆x₇ = 2 ----(3)

SOLUTION

detA₂  = -166                                                                                                                                                           

Det A₁ = 12

                                                      

detA₃  = -42                

But 1/2 = x_2* [12/-166]

X₆ = 83/12

And

1 = x_7*[-166/-42]

X₇ = 21/83

Hence, x_{6 ,} x₇ are 83/12, 21/83 respectively if x_o, x_, are 1/2 , 1 respectively.

                                                     

1.2 MILTI–COMBINATIONAL TRIGONDMETRY

Multi-Combinational Trigonometry, is a branch of mathematics that deals with the relationships between the sides and angles of triangles which are related in multiple sequential order from triangle to triangle and used mainly for comparison and predictive purpose. It also provides method of comparing and predicting these sides and angles.

Multi-combinational trigonometry has applications in such theoretical sciences as physics and astronomy, and in such practical fields as surveying and navigation.

                                                    

1.2.a RATIO DEFINITION

Let ө_1* ө_2*.......* ө_i be a multiple angles of i’s-right-angles triangles in standard positions with any multiple point P[x_0*x_1*---*x_r-1,   x_1*x_2*--*x_r] on the multiple terminal side, a multiple distance x_2*x_3*---x_r+1 from the origin x_2*x_3*---*x_r+1≠ 0.

The relations of the multi-combinational functions are defined as followed.

X_r = x_0* [sin (ө_1* ө_2*.......* ө_i) / cos(ө_1* ө_2*------ ө_i)]

X_r = x_0* [sec(ө_1* ө_2*...._* ө_i) / csc(ө_1* ө_2*......ө_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 = x_0, BC = x₁ and CA = x₂ for right angle triangle ABC: BC = x₁ CD = x_2, DB = x₃ for right-angle triangle BCD: CD = x_2, 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

    ө_1* ө_2* ө₃ = 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 negative sign and write 8.66m.

                         

                           1.3 MULTI-COMBINATIONAL CALCULUS

One of the most useful concept 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.

1.3.a POWER RULE OF RELATIVE DERIVATIVE

For the multi-combinational derivative functions f′ (x_{o *} x_{1 *----*} x_i-1) and f′ (x_1*x_2*-----* x_i), there exist unique initial derivative f′ (x_o),which is in relative comparison of the final derivative f ′(x_r) and is given as;

f′(x_o) = f′ (x_r) [f′ (x_o*x_1---*x_i-1)/ f′(x_1*x_2*------*x_i)]

EXAMPLE

The gross domestic product (GDP) of Ghana, Nigeria, USA, and Indian was

N′ (t_ot₁t₂) = t²₁t²₂t²₃ + 106billion dollars after 2007

N′(t₁t₂t₃) = t²₁t²₂t²₃ + 106billion dollars after 2007

(a)    At what rate was the GDP changing with respect to the multiple time t_o t₁ t₂ and t₁ t₂ t_3.

[take t_ot₁t₂ = 8 and t₁t₂t₃ = 6]

(b)   If the gross domestic product of Ghana is N (t_o) = t²_o - 5 t_o + 10 and change respectively to time to and if the gross domestic product occurs at time t_o, t_1, t₂ and t₃ for Ghana, Nigeria, USA and Indian respectively; Find the rate of change of GDP in  Indian.

(Take t_o = 3)

SOLUTION

(a)   N′ ( t_ot₁t₂ ) = 8 ( t_ot₁t₂ )

(b)   N′ (t₁t₂t₃) = 8 (t₁t₂t₃)

(c)    Applying power rule of relative comparison, we have

N′(t_o) = N′(t₃) [N′ (t_ot₁t₂ )/N′ (t₁t₂t₃ )]

But N (t_o) = 2 t_o -5

      N′(t_o) = 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

1.3.b RELATIVE COMPARISON INTEGRATION

Knowing the rate of inflation we may wish to estimate the relative future prices of different kind of commodity.

This process of comparing the sequential functions of the commodities from their derivatives is called relative comparison integration.

RULE OF RELATIVE INTEGRATION

When f΄(x_i-1), f΄(x_i),...,f(x_i-j), are a relative integral of f(x_r), f(x_r+1), ---, f(x_r+k), the result is the integral of f, if

f(x_i-1) = f(x_r) [∫f(x_0*x_1* -- _*x_r-1)d(x_0*x_1* ---_*x_r-1)]/[∫f(x_1*x_{2* ---*}x_r)d(x_1*x_{2* --- *}x_r)]

f(x_i) = f(x_r+1) [∫f(x_1*x_{2* ---*}x_r) d(x_1*x_{2* ---*}x_r)]/[∫f(x_2*x_{3* ---*}x_r+1)d(x_2*x_3*...._*x_r+1)]

­­­­-----------------------------------------------------------------------------------------------

F(x_i-;) f(x_r+k) [∫f(x_i-1* x_{i* --- *} x_i+j)d(x_i-1* x_{i* ---*}x_i+j)]/[∫f(x_i* x_{i+1* ---*}x_r+k)dx_1*x_{i+1* ---} x_r+k)]                 

for each i = 1,2,3, ---, r,  j = -1, 0, 1, 2, ---, k.

                                         

                                                           

                                                               PART TWO

LEAST WHOLE NORMAL MATHEMATICS

FORMULATED (YEAR: 2009-2010)

BY

WILLIAM AYINE ADONGO

ACTUARIAL SCIENCE STUDENT

University For Development Studies: GHANA

                                                      

                                                             CHAPTER TWO

                       ADONGO’S LEAST WHOLE NORMAL MATHEMATICS

                                          1.0 HISTORICAL BACKGROUND

The normal curve was devised around 1720 by the mathematician Abraham De Moivre in other to solve problems connected with games of chance. Around 1870 the Belgian mathematician Adolph Quetelled had the idea of using the curve as an ideal histogram, to which histogram for data could be compared.

Around 2009, I had created a mathematical function from the normal approximation equation which could be used in modeling problems in medical science, insurance, economics, material science and biological science.

I also realized that more development is needed to solve the scientific situations that I was confronting in my time. In view of this I was able to restructure the formalism of the existing concepts of physics to a new branch of formalism called Least Whole Normal Formalism which we are to discuss next.

1.1 ADONGO’S LWN FORMALISM

Definition:

·         Radical quantity: the radical quantity is any quantity that has main influence in the quantitative product. Let us consider the quantitative equation for velocity which is given as

V = S * (1/t): the radical quantity is the time t.

·         Subjective quantity: the subjective quantity is any quantity that is determined or has main relationship with the radical quantity. Let us also consider the quantitative equation for velocity which is given as V = S * (1/t):  the subjective quantity is the distance S.

·         Non-subjective quantity: the non-subjective quantity is a chosen quantity in the quantitative product which cannot be related by the radical quantity. The quantitative equation for force which is given as

      F = m*v*t ^–1:  the non-subjective quantity could be v if m is chosen as subjective quantity or it could be m if v is chosen as subjective quantity.

1.2 TWO QUANTITATIVE PRODUCTS

Let us consider Q ( i , j ) = (I _j * τ )*µ_i  where I _j  is denoted as a unit non-subjective quantity and τ the radical and µ_i  is the subjective quantity. The non-subjective unit quantity  I _j  is always equal to one (i.e. I _j = 1)

The quantitative equation Q ( i , j ) = (I _j * τ )*µ_i    has a distribution 

T_j  ̴  N( τ , 0)    

T_j  ̴  n (µ_i , S_i² )  }...........................(1)  for direct relation and

 T_j  ̴  N( τ ^-1 , 0)    

 T_j  ̴  n (µ_i , S_i² )    }    .................. ( 2)  for inverse relation.

Where j is one ( j = 1) and i  represents any quantity.

If the normal random quantity T _j   is total non-subjective unit quantity and T _i  is total subjective quantity, then the least whole normal function for the above distribution is

ℓ (τ) = α * τ - √ (τ) – β (i.e. when directly related)

 ℓ (τ ^-1) = α *τ ^-1 - √ (τ ^-1) – β  (i.e. when inversely related)

Where the constant α  and β are denoted as quantile coefficient of  τ  or  τ ^-1 and quantile constant respectively are calculated as;

α = µ_i  / {ϕ ^-1(γ%)  (√S_i²)}

β = T_i  /  {ϕ ^-1(γ%)  (√S_i²)}

1.3 THE LEAST WHOLE NORMAL QUANTITY

The quantity which represents least whole radical quantity is always approximately equal to zero. The least whole radical quantity is calculated as;

√ ( τ₀) = [1+√ (1 + 4αβ)] / 2α  for direct relation and

√ ( τ₀^-1 ) = [1+√ (1 + 4αβ)] / 2α  for inverse relation

1.4 THE TOTAL SUBJECTIVE QUANTITY

The quantity T_i   which represents total subjective quantity is calculated as;

T_i = µ_i * τ₀ – ϕ ^-1 (γ %) √ (S_i² * τ₀ )  for direct relation and

T_i = µ_i * τ₀^-1 – ϕ ^-1 (γ %) √ (S_i² * τ₀)for inverse relation.

1.5 MEAN SUBJECTIVE QUANTITY

The quantity µ_i which represents the mean subjective quantity is calculated as;

µ_i² = [ (1 / τ₀²)] [T_i² + ϕ ^-1( γ%) S²  * τ₀]  for direct relation and

µ_i² = [ (1 / τ₀²)] [T_i² + ϕ ^-1( γ%) S²  * τ₀^-1]  for inverse relation.

1.6 APPLYING THE FORMALISM

DOING WORK:

Doing works are way of transferring energies using forces. The amount of energies transferred. The amount of works done equal to sizes of the forces times the distance move.

Expected work= mean force *mean distance

i.e E (w) = µ_f * τ

Applying the theory of two quantitative products, we have the distribution

T₁~ N ( τ,0)

T_f  ~ n (µ_f , S_f ²)    }    .................. ( 1)

Where µ_f  denoted as mean force and S_f²   is the variance of the forces.

If the normal random quantity T₁  which is total number of units quantity I which mean is always equal to one and T_f which is denoted as total number of force, then it least whole normal function is calculated as;

ℓ (τ₀) = α * τ₀  – ( √τ₀ ) – β

 (i.e when directly related)  where α andβ are calculated as;

α = µ_f  ^-1 / {ϕ^-1  (γ%)  (√S_i²)}

β = T_f  /  {ϕ ^-1(γ%)  (√S_i²)}

1.7 THE LEAST WHOLE DISTANCE (RADICAL QUANTITY)

The quantity T_f which represents total subjective distance is always approximately equal to zero. The least whole distance is calculated as;

√τ₀ = [1 + √ (1 + 4αβ) ] / 2α .........................( 0 )

TOTAL FORCES (SUBJECTIVE QUANTITY)

The quantity T_f  which represents total subjective quantity or total forces is calculated as;

T_f= µ_f  *τ – ϕ ^-1 [γ%] [√(S_i²*τ) ].........................(□)

1.8 MEAN FORCE (MEAN SUBJECTIVE QUANTITY)

The quantity µ_f  which represent mean force is calculated as;

µ_f ²  = [ (1 / τ₀²)] [T_f ² + ϕ ^-1( γ%) S ²  * τ₀^-1] .............(*)

EXAMPLE

Horses pull cars with constant horizontal forces of mean 136N and standard deviation 27N per distance. Calculate the

         i.            Total forces at distance 19500m

       ii.            Average forces at distance 19500m

(Take γ = 95% )

SOLUTION

i)                     Applying equation (□), we have

T_f = µ_f * τ₀ − ϕ ^-1 [γ%][√(S _f ² * τ₀) ]

T_f  = 136 * 19500 − ϕ ^-1 [95%]  [√(729 ×19500)]

T_f= 2,652,000 – 1.645 × √(14,215,500)

T_f  = 2,645,797.783N

Hence, the total force is 2,645,797.783N

1.9 THREE QUANTITATIVE PRODUCTS

Let us consider Q (i , j) = (Ä_j *τ ) * µ_i   where Ä_j  is denoted as non-subjective quantity µ_i is denoted as subjective quantity and τ is the radical quantity.

If the non-subjective quantity Ä_j ,  is not equal to one but represents any quantitative values. Then we have the distribution of the quantitative equation Q(i, j)  which is given as;

T_j ~ N (Ä_j* τ , σ ² * τ )

T_j ~ n ( µ_i , S_i² )                  }    .................. ( 1)   for direct relation and

T_j ~ N (Ä_j* τ^-1  , σ ² * τ^-1 )

T_j ~ n ( µ_i , S_i² )                       }    .................. ( 2)  for inverse relation

If the normal random quantity T_j  is total non-subjective quantities and T_i  is total subjective quantity, then the least whole normal function is given as

ℓ (τ) = α * τ  – ( √τ ) – β  (i.e when directly related)

ℓ (τ^-1) = α * τ^-1  – ( √τ ^-1 ) – β  (i.e when inversely related)

         

1.10 THE RADICAL QUANTITY

The parameter τ or τ^-1 which represents least whole radical quantity is always occur when the function τ or τ^-1  is near to zero.The least whole radical quantity τ or τ^-1  is calculated as;

τ^0.5 = [1+ (1+4αβ)^0.5]/2α

For direct relation

For inverse relation and

τ^-0.5= [1 +(1+4αβ)^0.5]/2α

1.11 TOTAL SUBJECTIVE QUANTITY

At least whole radical quantity τ or τ^-1 the total subjective quantity T_i is calculated as

T_i = Θ *μ* τ - Φ^-1(γ%)*(Θ*S²* τ + μ²* σ^-2* τ)^0.5

For direct relation and

T_i = Θ* μ* τ^-1 – Φ^-1 (γ%)*(Θ *S²* τ^-1 + μ²*σ²* τ^--1)^0.5

For inverse relation.

1.12 MEAN SUBJECTIVE QUANTITY

Also, at least whole radical quantity τ or τ^-1, the mean subjective quantity μ²_i is calculated as;

μ²_i = (T²_i + Φ^-1 (γ%)²*Θ*s²*τ)/(Θ²*τ² – Φ^-1 (γ%)²*σ²*τ)

For direct relation and

μ²_i = (T²_i + Φ^-1 (γ%)^.Θ*s²* τ^-1 )/(Θ²*τ^-2_ Φ^-1 (γ%)²*σ².τ^-1)

For inverse relation.

                                                            

                                                                  PART THREE

 

CENTRAL (GRAVITATIONAL) POINT THEORY

DISCOVERED (YEAR:2007-2008)

BY

WILLIAM AYINE ADONGO

ACTTUARIAL SCIENCESTUDENT

University For Development Studies: GHANA

                                                     

                                                    CHAPTER THREE

                        GRAVITATIONAL POINT OF A LINE IN A PLANE

                                                 1.0 ARGUMENT

Most mathematical scientists argued that, since there are infinite many points in a line equation we cannot locate or predict a unique point of a straight line equation as our central (or point of gravity), because of that we cannot locate or determine the ends of a straigtht line in a plane; unless a range of points are given that the mid-point of a line can be determined.

In fact, I have disagreed with that philosophy. A line equation has a unique point which I strongly believed as the central of gravity point of a line in plane, using a concept called equal pairing concept, which was discovered by me between the years 2007-2008.

1.1 MATHEMATICAL CHARACTERISTICS OF THE CENTRAL(GRAVITATIONAL) 

                                                             POINT

1)      It is a unique point at which the axial term of x is equal to the axial term of y in a line equation  Ax + By = C.

2)      At gravitational point of a line equation Ax + By = C in a plane, the value of the axial coordinates x and y is calculated as;

X_G= C / 2A

Y_G= C / 2B

3)      At gravitational point of a line equation Ax + By = C in a plane; the slope of a line equation is given as;

S_G = - ( Y_G / X_G )

1.2 ADONGO’S GRAVITATIONAL POINT VALUES INTERVAL THEOREM.

 If  x_GÎ X_G  and  y_GÎ Y_G, where  X_G  contains the possible value of  x_G  and  Y_G  contains the possible value of  y_G. If  X_G  is a corresponding pair of  Y_G  and  x_G  is a corresponding pair of  y_G, then;

( X_G ; Y_G ) = [ ( x_G , x_G – x_G , x_G – x_G – x_G ,- - - - - -) ;( y_G ,y_G + y_G ,y_G + y_G + y_G , - - - - - ) ]

Or[ ( x_G , x_G + x_G , x_G + x_G + x_G ,- - - - - -) ;( y_G , y_G – y_G , y_G – y_G – y_G , - - - - - ) ]

1.3 INTERCEPT OF A LINE

In fact, with the help of the gravitational point vales theorem the intercept of the line is calculated as;

( x_G; 0 ) Î  ( X_G , 0 ) = (2x_G, 0 )

( 0 ; y_G  ) Î  ( 0 , Y_G) =  (0,2y_G  )

EXAMPLE

If  3x + y = 6 ,  find

         i.            The gravitational pairs, x_Gand y_Gof the equation.

       ii.            The slope of its graph using Adongo’s gravitations slope formula approach.

      iii.            The intercept of the line.

     iv.            The possible values of  x_G  and  y_G  to fourth interval.

SOLUTION

         
 i.            The gravitation pairs  x_G  and  y_G  is calculated as;

x_G= C / 2A

x_G= 6 / [ 2 (3) ]

x_G= 1

y_G= C / 2B

y_G= 6 / [ 2 ( 1)]

y_G= 3

 ii.            Slope of the line equation or graph is

S_G= - ( Y_G / X_G )

S_G= - ( 3 / 1 )

S_G= - 3

iii.            The intercept of the line is calculated as

( x_G; 0 ) Î  ( X_G , 0 ) = (2x_G, 0 )

                                          = [ 2 (1) , 0 ]  

                                         = ( 2 , 0 )

And

( 0 ; y_G  ) Î  ( 0 , Y_G ) =  ( 0 , 2y_G  )

                               = (0 , 2 * 3 )

                               = (0 , 6 )

    
 iv.            Applying the gravitational point values interval theorem (GPVIT), we have,

( x_G; y_G) Î [( X_G ; Y_G ) ] = [ ( 1 , 0 , -1 , 2 , -3 ) ; ( 3 , 6 , 9 , 12 , 15 ) ]

                                  [ ( 1 , 2 , 3 , 4 , 5 ) ; ( 3 , 0 , -3 , -6 , -y )]

                                               

                                           


   1.4 ADONGO CENTRAL PREDICTION 

Applying The gravitational point equation can also used to predict mean value (central values) of two data pair (x_g,y_g).

This equation is developed mainly to enhance the best accuracy for predicting the mean value (Central values) of two data pair.  For example, we can predict the value of y_g when the value x_g is known.

The equation constitutes the variables y_g, x_g, c and k which I named as predictive value, suggestive value, constant and coefficient of a suggestive value respectively. The developed formula for the equation is given as;

Y_g= Kx_g +C

Where;

K = Σy/2Σx

C = Σy/2n

1.5 PROOF AND ANALYSIS OF THE FORMULARS

 For an apparent relationship between x and y values from a sample or population data, the total sum of x is apparently related to the total sum of y (i.e Σy is relative to Σx).

This implies that

Σy = kΣx + nc

But the relationship between k and C are given as;

Nc = Σy - kΣx ---(1)

Or

kΣx = Σy – nc --- (2)

Adding equation (1) and (2) together, we have

nc + kΣx = (Σy - kΣx) + (Σy – nc)

Arranging equal pairs together, we have:

nc – (Σy - kΣx) = (Σy – nc) - kΣx

By principle of equal pairs, we have:

Nc = Σy - nc

2nc = Σy

C = Σy/2n ---- (3)

And:

(Σy - kΣx) = kΣx

Σy - kΣx = kΣx

K2Σx = Σy

K = Σy/2Σx --- (4)

EXAMPLE

The table below contained the apparent relationshipbetween students high school average (HAS) and grade point  average (GPA) after the freshman year of college.

HAS(X)                               GPA(Y)

80                                        2.4

85                                        2.8

88                                         3.3

90                                         3.1

95                                         3.7

92                                         3.0

82                                         2.5

75                                          2.3

78                                          2.8

85                                         3.1

(a)    If the students have average HAS of 85, find the best estimate of their average GPA

(b)   If they students have average HAS of 92, find the best estimate their average GPA

SOLUTION

   
HAS(x)                 GPA(y)

   80                          2,4

   85                          2.8

   88                           3.3

   90                           3.1

   95                           3.7

   92                            3.0

   82                            2.5

  75                            2.3

  78                            2.8

    85                          3.1

ΣX= 850                  ΣY= 29

K = Σy/2Σx

K = 29/2(850)

K = 0.0171

And:

C = Σy/2n

C = 29/2(10)

C = 1. 45

Hence, the predictive equation is given as:

y_p = 0.0171x_p + 1.45

(a)   y_p = 0.0171 (85) + 1.45

y_p = 2.90

(b)   y_p = 0.0171 (92) + 1.45

             
         y_p = 3.0

REFERENCE
*Adongo Ayine William(Me), Transcript, Posted(EMS-Bolga Branch) to Mathematical Association of Ghana in the Year 2008.(Tittle: New Mathemacal Concept....)
*Adongo Ayine William(Me), Diary(2008), Posted(EMS-Bolga Branch) to Ghana Academy of Arts and Sciences in the year 2010.
*Adongo Ayine William(Me), Diary(2009), Posted(EMS=-Bolga Branch) to Ghana Academy of Arts and Sciences in the Year 2010.
*D T Whiteside(ed). "The Mathematical Papers of Isaac Newton(Volume1). Cambridge University Press, 1967.
*Galton, Francis.(1886). 'Regression Towards Mediocrity in  Hereditary Stature'. Volume 15.
*Rene Descartes.(1637). "The Geometry".
*Cramer, Gabriel(1750). "Introduction Analysis of Linear Algebra"

BY WILLIAM AYINE ADONGO

ayinewilliam@yahoo.com