Bill Adongo, (25 January, 1983), is a Ghanaian cognitive and innovative thinker. A 21-centrury Ghanaian philosopher, Actuary, Scientist, Pure and Applied Mathematician, is one of the most influential modern thinkers. His ideas have enormous influence nearly all areas of mathematics, sciences, and social sciences. He made much accomplishment in mathematics, sciences and social sciences which were at one time considered as impossible because at that time those ideas were beyond the ability of man to understand or to foresee. Even though he had made numerous achievements in pure and Applied mathematics one of his greatest achievement is his LEAST WHOLE NORMAL FUNCTION and some of his achievements in that field:(that is application of the LEAST WHOLE NORMAL FUNCTION). Below is copy of his works posted.
LEAST WHOLE NORMAL FUNCTION
"As you read this manuscript, you will see that the least whole normal function, l (τ) = α.τ-√τ-β which I have derived from the normal approximation function or equation is one of the most powerful and useful mathematical function at our time.
It provides us with the necessary tools to analysis information, make predictions and arrive at a decision in virtually every area of human endeavor.
The mathematical of the least whole normal function requires two steps:
(1)The construction of mathematical models which approximate the physical situations.
(2)The solutions of the resulting mathematical problems.
To understand God’s purpose, we must study least whole normal model for these are the measure of his purpose. The main purpose of the least whole normal function, like that of any other mathematical function, is to provide a useful model of the real world. All least whole normal models are simplified representations of the reality, ignoring complexities which will have unimportant effects on the final outcome.
The unique feature of least whole normal modeling is that it uses randomness to model those parts of a situation where the details of the process which generate the outcome are unknown, assigning probabilities to possible outcome, rather than predicting definitely which will occur. The least whole normal function constitutes two main categories of models:
a)
Non product model
b)
Product model
(a)NON PRODUCT MODEL
If the normal random variable z which is population available in time or interval or space or quantity τ; and T which is occurrences of each member in the population available in time or interval or space or quantity τ are continuous; then it least whole normal function is given as;
If the normal random variable z which is population available in time or interval or space or quantity τ; and T which is occurrences of each member in the population available in time or interval or space or quantity τ are continuous; then it least whole normal function is given as;
l(τ) = ατ - √τ - β
α=quantile coefficient
β=quantil constant
α=(Θμ)/(Φ-1) (γ%) √(ΘS2+μ2σ2 )
β= T/(Φ-1) (γ%) √( ΘS2+μ2σ2 )
To describe the distribution, we have
Z ∽N (Θτ, σ2τ) ; T∽n (μ, S2)
α=(Θμ)/(Φ-1) (γ%) √(ΘS2+μ2σ2 )
β= T/(Φ-1) (γ%) √( ΘS2+μ2σ2 )
To describe the distribution, we have
Z ∽N (Θτ, σ2τ) ; T∽n (μ, S2)
.Where Θ
and σ2are population mean and variance respectively. μ and S2 are mean of each occurrences in the population
and variance of each occurrences in the population respectively.
LEAST WHOLE TIME OR INTERVAL OR SPACE OR QUANTITY
The parameter τ which represents least whole interval or time or space or quantity is always occur when the function l(τ) is approximately equal to zero. The least whole interval or time or space or quantity is calculated as
LEAST WHOLE TIME OR INTERVAL OR SPACE OR QUANTITY
The parameter τ which represents least whole interval or time or space or quantity is always occur when the function l(τ) is approximately equal to zero. The least whole interval or time or space or quantity is calculated as
√τ=
[1+√(1+4αβ)]/2α
THE TOTAL NUMBER OF EACH OCCURRENCES
At least whole time or interval or space or quantity τ, the total number of each occurrence in the entire population is denoted as T and is Calculated as;
THE TOTAL NUMBER OF EACH OCCURRENCES
At least whole time or interval or space or quantity τ, the total number of each occurrence in the entire population is denoted as T and is Calculated as;
T= Θμτ - Φ-1(γ%)√(ΘS2τ
+μ2σ2τ)
AVERAGE
NUMBER OF INDIVIDUAL OCCURRENCE
Also at least whole time or interval or space or quantity τ, the average number of each occurrences in the entire population is calculated as
Also at least whole time or interval or space or quantity τ, the average number of each occurrences in the entire population is calculated as
μ2
=[ (T2+ Φ-1(γ%)2ΘS2τ)]/[(Θ2τ2-
Φ-1 (γ%)2σ2τ)]
EXAMPLE
(I)
The government of Ghana has assigned a specific duty to roof new build houses in Upper East Region. The size of the new built houses has a Poisson random variable with mean 25. The number of roofing sheets that can complete one house has mean 50 and variance 675. The sizes of the houses and the number of roofing sheets are independent. If the government is taken similar projects in all the regions of Ghana, how many roofing sheets are expected to complete 500 houses at probability greater than 95%.
SOLUTION
Θ = 25
σ 2 =25
μ =50
S2=675
τ =500 houses
The government of Ghana has assigned a specific duty to roof new build houses in Upper East Region. The size of the new built houses has a Poisson random variable with mean 25. The number of roofing sheets that can complete one house has mean 50 and variance 675. The sizes of the houses and the number of roofing sheets are independent. If the government is taken similar projects in all the regions of Ghana, how many roofing sheets are expected to complete 500 houses at probability greater than 95%.
SOLUTION
Θ = 25
σ 2 =25
μ =50
S2=675
τ =500 houses
Applying
equation above, we have
T= Θμτ - Φ-1(γ%)√(ΘS2τ +μ2.σ2τ)
T= 25*50*500 - Φ-1 (95%)√(25*675*500+502*25*500)
T= 625000-10363.1776
T=614636.8264≈ 614637.00
Hence, the number of roofing sheets that can complete 500 houses is 614627.
T= Θμτ - Φ-1(γ%)√(ΘS2τ +μ2.σ2τ)
T= 25*50*500 - Φ-1 (95%)√(25*675*500+502*25*500)
T= 625000-10363.1776
T=614636.8264≈ 614637.00
Hence, the number of roofing sheets that can complete 500 houses is 614627.
EXAMPLE
(2)
The number of health insurance claims reported per month at certain health insurance company has mean 110 and variance 750.Individual losses have mean 1101 and variance 4900. The number of claims and amount of individual losses are independent. Estimate the expected losses to be incurred a year at a probability greater than 95%
The number of health insurance claims reported per month at certain health insurance company has mean 110 and variance 750.Individual losses have mean 1101 and variance 4900. The number of claims and amount of individual losses are independent. Estimate the expected losses to be incurred a year at a probability greater than 95%
SOLUTION
Θ = 110
σ2=750
μ=1101
S2 = 4900
τ = 12months
Θ = 110
σ2=750
μ=1101
S2 = 4900
τ = 12months
Applying equation
above, we have;
T= Θμτ - Φ-1(γ%)√(ΘS2τ +μ2σ2τ)
T= 110*1101*12- Φ-1 (95%)√(110*4900*12+11012*750*12)
T= 1453320-171871.2264
T=1,251,448.774≈ 1,251, 449.00
Hence, the losses that are expected to be incurred a year is 1,251, 449.00
T= Θμτ - Φ-1(γ%)√(ΘS2τ +μ2σ2τ)
T= 110*1101*12- Φ-1 (95%)√(110*4900*12+11012*750*12)
T= 1453320-171871.2264
T=1,251,448.774≈ 1,251, 449.00
Hence, the losses that are expected to be incurred a year is 1,251, 449.00
EXAMPLE
(3)
The genetic of certain fungal (fungal infections) has average number of chiastmata to be 5.5 per micro interval and variance 14 per micro interval. Average frequency of each chiasmata occurring per micro interval has mean 0.08 and variance 0.23. Estimate the least whole number of micro intervals the chiasmata frequency 0.4 will occur at a probability greater than 99%.
The genetic of certain fungal (fungal infections) has average number of chiastmata to be 5.5 per micro interval and variance 14 per micro interval. Average frequency of each chiasmata occurring per micro interval has mean 0.08 and variance 0.23. Estimate the least whole number of micro intervals the chiasmata frequency 0.4 will occur at a probability greater than 99%.
(b)PRODUCT
MODEL
For many centuries the problem of physics and its causes had puzzled scientists – it was not until the time Galileo, Newton, Einstein etc. that a real progress in its explanation was made.
Currently, I have realized that more development is needed to solve the scientific situations that we are confronting today.
I have no turning point than to restructure the existing concepts of physics to a new branch of concept called “Least Whole Normal Concept.” The development of the least whole normal concept in physics is mainly depends on a basic theory called “theory of quantitative products” which is discovered by me, whose previous name was Adongo Ayine William and now known as Bill Adongo in the year 2009. The detail of the least whole normal concept in physics will be explained later.
For many centuries the problem of physics and its causes had puzzled scientists – it was not until the time Galileo, Newton, Einstein etc. that a real progress in its explanation was made.
Currently, I have realized that more development is needed to solve the scientific situations that we are confronting today.
I have no turning point than to restructure the existing concepts of physics to a new branch of concept called “Least Whole Normal Concept.” The development of the least whole normal concept in physics is mainly depends on a basic theory called “theory of quantitative products” which is discovered by me, whose previous name was Adongo Ayine William and now known as Bill Adongo in the year 2009. The detail of the least whole normal concept in physics will be explained later.
We have
seen how the least whole normal function used to solve a long standing technical
problems. One of its useful application is the product modeling ( i.e used to
fit two or more quantitative products)
Two Quantitative Product
|
Three Quantitative Product
|
Power=force x velocity
P=fv
E(P)=μf *τv
|
Electric energy=volt*current*time
E=Vit
E(E)= Θv*μI*
τt
|
Mass=density*volume
M=d*v
E(M)= μd *τv
|
Heat energy=mass*specific
heat*temperature
E=mct
E(E)= Θm*μc*
τt
|
· 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.
TWO QUANTITATIVE
PRODUCTS
Let us consider E ( Q) = (1* τi )*µi
where 1 is denoted as a unit
non-subjective quantity and τ the
radical and µi is the
subjective quantity. The non-subjective unit quantity is 1. The quantitative
equation E ( Q ) = (Ii * τ )*µi has a distribution.
Tj ̴ N( τ , 0) ; Tj ̴ n (µi , Si2 )
If the
normal random quantity T1 is total non-subjective unit quantity and Ti is total subjective quantity, then the least
whole normal function for the above distribution is
ℓ (τ) = α * τ - √ (τ ) – β
Where the
constant α and β are denoted as quantile
coefficient of τ
and quantile constant respectively are calculated as;
α = µi / {ϕ -1(γ%) (√Si2)}
β = Ti / {ϕ -1(γ%) (√Si2)}
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
THE TOTAL SUBJECTIVE
QUANTITY
The quantity Ti which
represents total subjective quantity is calculated as;
Ti = µi
* τ0 – ϕ -1 (γ %) √ (Si2
* τ0 )
MEAN SUBJECTIVE
QUANTITY
The quantity µi which represents the mean subjective
quantity is calculated as;
µi2
= [ (1 / τ2)] [Ti2 + ϕ -1( γ%) S2 * τ]
APPLYING TWO
QUANTITATIVE PRODUCT
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) = µf * τ
Applying the theory of two
quantitative products, we have the distribution
T1 ~ N ( τ, 0); Tf ~ n (µf , Sf 2)
Where
µf denoted as mean force and Sf 2 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 Tf which is denoted as total number of force,
then it least whole normal function is calculated as;
ℓ (τ0) = α * τ0 – ( √τ0 ) – β
where α and β are calculated as;
α
= µf -1 / {ϕ-1 (γ%) (√Si2)}
β = Tf / {ϕ -1(γ%) (√Si2)}
THE LEAST WHOLE
DISTANCE (RADICAL QUANTITY)
The quantity Tf which represents
total subjective distance is always approximately equal to zero. The least
whole distance is calculated as;
√τ0 = [1 + √ (1 + 4αβ) ] / 2α
TOTAL FORCES
(SUBJECTIVE QUANTITY)
The quantity Tf which represents total subjective quantity or
total forces is calculated as;
Tf = µf * τ – ϕ -1
[γ%] [√(Si2*τ) ]
MEAN FORCE (MEAN
SUBJECTIVE QUANTITY)
The quantity µf which represent mean force is calculated as;
µf
2 = [ (1 / τ02)]
[Tf 2 + ϕ -1( γ%) S 2 * τ0-1 ]
.............(*)
EXAMPLE(1)
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 above, 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 = 136 * 19500 − ϕ -1 [95%] [√(729 ×19500)]
Tf =
2,652,000 – 1.645 × √(14,215,500)
Tf = 2,645,797.783N
Hence, the total force is 2,645,797.783N
EXAMPLE(2)
Given the formula p=fv, where p is power, f is force and v
is velocity: an engineer finds that the mean force of different bodies and
standard deviation of a moving bodies are 1400 and 320 respectively per
velocity. Calculate
1)
The
total force at velocity 150m/s
2)
The
average force at velocity 150m/s
3)
Average
power at velocity 150m/s
4)
Total
power at velocity 150m/s
(Take γ =95)
SOLUTION
(i)Total force at velocity 150m/s is:
Tf=
µ f * τv − ϕ -1 [γ%][√(Sf 2 *
τv) ]
Tf=1400*150-1.645*√(3202*150)
Tf=203,552.945N
(ii) µi2 = [ (1 / τ2)] [Ti2
+ ϕ -1( γ%) S2 * τ]
µi2
=1/150[203,552.9432+1.645*3202*150]
EXAMPLE(3)
Given M=d*v, where
M is mass, d is density and V is volume. Scientist finds that mean
density of different bodies is 800kg/m3 and standard
deviation 150kg/m3 per
volume. Calculate
i)the total density of the bodies at volume 500m3
ii)the average density of the bodies at volume 500m3
iii)average mass of the bodies at volume 500m3
iv)total masses of the bodies at volume 500m3
i)TD=394482.5023kg/m3
ii) µD2=789.0119kg/m3
THREE QUANTITATIVE PRODUCTS
Let us
consider E (Q) = (ÄI*τ ) * µi where Äj is denoted as non-subjective quantity µi is denoted as subjective quantity and τ is the radical quantity.
If the
non-subjective quantity Äj is not equal to one, but represents any
quantitative value. Then we have the distribution of the quantitative equation E(Q) which is given as;
Tj ~ N (Äj* τ , σ 2* τ ) ; Tj ~ n ( µi , Si2 )
If the
normal random quantity Tj is total non-subjective quantities and Ti is total subjective quantity, then the least
whole normal function is given as
ℓ (τ) = α * τ – ( √τ ) – β
)
the total
subject quantity is calculated as:
T= Θμτ - Φ-1(γ%)√(ΘS2τ
+μ2σ2τ)
Averge subjective
quantity is:
μ2
= (T2+ Φ-1(γ%)2ΘS2τ)/(Θ2τ2-
Φ-1 (γ%)2σ2τ)
The
radical quantity is calculated as:
√τ=
[1+√(1+4αβ)]/2α
EXAMPLE(1)
Energies=volts*currents*time:
scientist finds that 5.4v with standard deviation 2.10v per second and current 42A
with standard deviation 18A per second. Calculate the seconds required for
total current 110000A.
(Take γ =95)
SOLUTION
α=5.4*42/[1.645√(5.4*182+422*2.102)]
= 1.412398
β =110000/[1.645√(5.4*182+422*2.102)]
= 685.02554
√τ=
[1+√(1+4αβ)]/2α
√τ=22.3797
√τ=22.3797
τ=500.85"
EXAMPLE(2)
NOTE:THESE WORKS ARE QUOTED FROM BILL ADONGO
EXAMPLE(2)
Heat Energies =masses*specific capacities*change
in temperature: If
the average specific heat capacity is 12.8 and standard deviation is 2.4 per
change in temperature and average mass is 2.08 and standard 0.06 per change in
temperature.
1) Calculate
temperature change if the specific heat capacity is 80
SOLUTION
α=2.08*12.4/[1.645√(2.08*2.42+12.82*0.062)]
α=4.42222
β=80/[1.645√(2.08*2.42+12.82*0.062)
β=13.7165745
√τ=[1+√(1+4*4.42222*13.7165745)]/2*(4.42222)
τ=3.526 0C
NOTE:THESE WORKS ARE QUOTED FROM BILL ADONGO
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