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Американский Научный Журнал IS IT POSSIBLE TO SOLVE THE ANGLE TRISECTION PROBLEM WITH COMPASSSTRAIGHTEDGE CONSTRUCTION (47-48)
If you pose the question given in the title of this note you will listen a negative answer. In the Google searcher you will receive about 6 million results. It means to find something new in the problem formulated by Greek mathematician is useless. This problem alongside with the circle squaring is considered as undecidable problem. Скачать в формате PDF
American Scientifi c Journal № ( 37 ) / 2020 47
ФИЗИКО -МАТЕМАТИЧЕСКИ Е НАУКИ
IS IT POSSIBLE TO SO LVE THE ANGLE TRISEC TION PROBLEM WITH CO MPASS -
STRAIGHTEDGE CONSTRU CTION?
(short note)
Nigmatullin R.R.
Prof.
Kazan National Research Technical University (KNRTU -KAI)
Radioelectrnics and Informative Measurements technics depa rment
K. Marx str., 10, 420111, Kazan, RF
DOI: 10.31618/asj.2707 -9864.2020.1.37.7
If you pose the question given in the title of this
note you will listen a negati ve answer. In the Google
searcher you will receive about 6 million results. It
means to find something new in the problem
formulate d by Greek mathematician is useless. This
problem alongside with the circle squaring is
considered as undecidable problem.
As stated by Pierre Laurent Wantzel (1837), the
solution of the angle trisection problem corresponds to
an implicit solution of the cubic equation �3−3�−
1= 0, which is algebraically irreducible, and so this
statement is equivalent to the geometric solutio n of the
angle trisection problem. A classic geometric solution
was given in paper with references therein [1].
Nevertheless, in this short note we want to
demonstrate the simplest solution of the angle trisection
problem (ATP) with the help of the compass -
straightedge construction, adding a pencil and sheet of
paper. Two last items are needed only for
demonstration purposes.
The s implest solution will be based on the
following geometric sums
��(+�)= �⋅(1+�+�2+...+��−1)= ( �
1−�)⋅(1−��),
��(−�)= �⋅(1−�+�2+...+(−�)�−1)= (
1+)⋅(1−(−�)�). (1)
In order to receive the desired solution, we put
�= 1/4and present the initial sum in the form
0⋅��(1
4)= 0⋅[1
4+ 1
16 + 1
64 +...+(1
4)
�
]≡ 0
3 (1−�(1
4)),
�(1
4)= (1
4)�. (2)
Here 0 is the initial angle that is needed to be
trisected, N(1/4) – is a small correction. This sum is
converged to the value 0/3 at N >>1. Having a simple
ruler and pencil one can pose the following question:
how many terms it is necessary to keep in (2) for
receiving the trisected angle with the given accuracy?
We su ppose that the pencil -point has 1 mm (or a bit
less) and the trace lef t by the pencil on a sheet of paper
has the same length. Therefore, we have 4 N=1000 that
gives N value located between 4 and 5. It implies that
in the normal conditions having the A4 for mat sheet of
paper, normal sharp pencil or ball pen one can keep
only 4 and 5 terms in (2) for solving the ATP with the
given accuracy 4(1/4) = (1/4) 4100% 0.4%. If we
keep the next term then one can receive 5(1/4) 100%
0.098%. Therefore, supposing that our divider
compass can provide the infinite division of any angl e
on two equal parts ( N >>1) we showed how to solve the
ATP with the given accuracy.
Attentive analysis show that putting x=2 -q
(q=2, 3,…) one can provide more interesting solutions
and solve the angle n-section solution for some odd
numbers as n=5,7,9,…But before it is interesting to
note also that ATP admits another solution. For this aim
we consider the second sum in (1).
0⋅��(−1
2)= 0⋅[1
2−1
4+1
8− 1
16 ...+(−1
2)
�+1
]=
03(1−�(1
2)) =�>>103,�(1
2)= (1
2)�. (3)
48 American Scientific Journal № ( 37 ) / 2020
However, this solution is a poor choice and can be
omitted because of the slow convergence of the sum
SN(-1/2) to the value 0/3.
It is convenient to pre sent other possible angle n -
section solutions in the form of the table when any
chosen angle admits division on two equal parts with
the help of a straightedge ruler and divider compass.
Table showing a possible angle n-section realized
with the help of a ruler and divider.
q-value
x=2 -q
0
2�−1 0
2�+1 Value of the error
(,�)= (2−��)⋅100%
2 0
3 0
5 E(2,4)=3.91 10 -1(%)
E(2,5)=9.77 10 -2(%)
3 0
7 0
9 E(3,3)=1.95 10 -1(%)
E(3,4)=2.44 10 -2(%)
4 0
15 0
17 E(4,3)=2.44 10 -2(%)
E(4,4)=1.53 10 -3(%)
As it follows from this table with increasing of the
n-th angle -sected value of an admissible error value is
decreasing. The author things that this simplest solution
put a final point in soluti on of the ATP attracting the
attention of many mathematicians.
References
[1] C. Rediske, The Trisection of an Arbitrary
Angle: A Classical Geometric Solution, J. of Advances
in Mathematics (2018) pp. 7640 -7669. DOI:
10.24297/jam.v14i2.7402.
К ВОПРОСУ НЕЙТРИННОГО ИЗЛУЧЕНИ Я В РАСШИРЯЮЩЕЙСЯ С ОХЛА ЖДЕНИЕМ
ВСЕЛЕННОЙ
Кошман Валентин Семенович
канд. техн. наук, доцент,
Пермский государственный аграрно -технологический университет,
г. Пермь , Россия
ON THE QUESTION OF N EUTRINO RADIATION IN THE UNIVERSE EXPA NDING WITH
COOLING
Valentin Koshman
Cand. Tech. Sci., Associate Professor,
Perm State Agrarian and Technological University,
Perm , Russia
Аннотация. Рассмотрена однородная модель Вселенной в виде газовой смеси из фотонов, барионов
и нейтрино. По мере а нализа изуч аемой физической системы воспроизведена математическая структура
для описания ее космологической эволюции. Выполнена оценка средней величины энергии реликтового
нейтрино, которая совпала с известной оценкой профессора О. Лахав (2002 г.) по поряд ку величины.
Приведены аргументы в пользу первичного термоядерного взрыва у истока расширения Вселенной на
планковском масштабе времени.
Abstract. A homogeneous model of the Universe in the form of a gas mixture of photons, baryons and
neutrinos is conside red . As the physical system under study is analyzed, the mathematical structure for describing
its cosmological evolution is reproduced. The average value of the relic neutrino energy was estimated, which
coincided with the well -known estimate of Professor O. Lahav (2002) in order of magnitude. Arguments are given
in favor of a primary thermonuclear explosion at the source of the expansion of the Universe on the Planck time
scale.
Ключевые слова: модель расширяющейся Вселенной, реликтовое излучение, планков ски е
величины, закон Стефана – Больцмана, объемная плотность энергии нейтрино, масса реликтового
нейтрино.
Keywords : model of the expanding Universe, relic radiation, Planck quantities, Stefan – Boltzmann law,
volume density of neutrino energy, mass of the relic neutrino.
«Структура математического описания
выявляется
по мере анализа физической системы»
П. Девис [1, с. 260]
Как известно, космология, которая изучает
свойства Вселенной в целом, – одна из немногих
естественных наук, где присутствует эволю ция в
явном виде. Изучение эволюции Вселенной
осложнено тем, что её динамика отнюдь не
представляет собой нечто непосредствен но
наблюдаемое. По этой причине объект
исследования может быть дан как целое лишь
ФИЗИКО -МАТЕМАТИЧЕСКИ Е НАУКИ
IS IT POSSIBLE TO SO LVE THE ANGLE TRISEC TION PROBLEM WITH CO MPASS -
STRAIGHTEDGE CONSTRU CTION?
(short note)
Nigmatullin R.R.
Prof.
Kazan National Research Technical University (KNRTU -KAI)
Radioelectrnics and Informative Measurements technics depa rment
K. Marx str., 10, 420111, Kazan, RF
DOI: 10.31618/asj.2707 -9864.2020.1.37.7
If you pose the question given in the title of this
note you will listen a negati ve answer. In the Google
searcher you will receive about 6 million results. It
means to find something new in the problem
formulate d by Greek mathematician is useless. This
problem alongside with the circle squaring is
considered as undecidable problem.
As stated by Pierre Laurent Wantzel (1837), the
solution of the angle trisection problem corresponds to
an implicit solution of the cubic equation �3−3�−
1= 0, which is algebraically irreducible, and so this
statement is equivalent to the geometric solutio n of the
angle trisection problem. A classic geometric solution
was given in paper with references therein [1].
Nevertheless, in this short note we want to
demonstrate the simplest solution of the angle trisection
problem (ATP) with the help of the compass -
straightedge construction, adding a pencil and sheet of
paper. Two last items are needed only for
demonstration purposes.
The s implest solution will be based on the
following geometric sums
��(+�)= �⋅(1+�+�2+...+��−1)= ( �
1−�)⋅(1−��),
��(−�)= �⋅(1−�+�2+...+(−�)�−1)= (
1+)⋅(1−(−�)�). (1)
In order to receive the desired solution, we put
�= 1/4and present the initial sum in the form
0⋅��(1
4)= 0⋅[1
4+ 1
16 + 1
64 +...+(1
4)
�
]≡ 0
3 (1−�(1
4)),
�(1
4)= (1
4)�. (2)
Here 0 is the initial angle that is needed to be
trisected, N(1/4) – is a small correction. This sum is
converged to the value 0/3 at N >>1. Having a simple
ruler and pencil one can pose the following question:
how many terms it is necessary to keep in (2) for
receiving the trisected angle with the given accuracy?
We su ppose that the pencil -point has 1 mm (or a bit
less) and the trace lef t by the pencil on a sheet of paper
has the same length. Therefore, we have 4 N=1000 that
gives N value located between 4 and 5. It implies that
in the normal conditions having the A4 for mat sheet of
paper, normal sharp pencil or ball pen one can keep
only 4 and 5 terms in (2) for solving the ATP with the
given accuracy 4(1/4) = (1/4) 4100% 0.4%. If we
keep the next term then one can receive 5(1/4) 100%
0.098%. Therefore, supposing that our divider
compass can provide the infinite division of any angl e
on two equal parts ( N >>1) we showed how to solve the
ATP with the given accuracy.
Attentive analysis show that putting x=2 -q
(q=2, 3,…) one can provide more interesting solutions
and solve the angle n-section solution for some odd
numbers as n=5,7,9,…But before it is interesting to
note also that ATP admits another solution. For this aim
we consider the second sum in (1).
0⋅��(−1
2)= 0⋅[1
2−1
4+1
8− 1
16 ...+(−1
2)
�+1
]=
03(1−�(1
2)) =�>>103,�(1
2)= (1
2)�. (3)
48 American Scientific Journal № ( 37 ) / 2020
However, this solution is a poor choice and can be
omitted because of the slow convergence of the sum
SN(-1/2) to the value 0/3.
It is convenient to pre sent other possible angle n -
section solutions in the form of the table when any
chosen angle admits division on two equal parts with
the help of a straightedge ruler and divider compass.
Table showing a possible angle n-section realized
with the help of a ruler and divider.
q-value
x=2 -q
0
2�−1 0
2�+1 Value of the error
(,�)= (2−��)⋅100%
2 0
3 0
5 E(2,4)=3.91 10 -1(%)
E(2,5)=9.77 10 -2(%)
3 0
7 0
9 E(3,3)=1.95 10 -1(%)
E(3,4)=2.44 10 -2(%)
4 0
15 0
17 E(4,3)=2.44 10 -2(%)
E(4,4)=1.53 10 -3(%)
As it follows from this table with increasing of the
n-th angle -sected value of an admissible error value is
decreasing. The author things that this simplest solution
put a final point in soluti on of the ATP attracting the
attention of many mathematicians.
References
[1] C. Rediske, The Trisection of an Arbitrary
Angle: A Classical Geometric Solution, J. of Advances
in Mathematics (2018) pp. 7640 -7669. DOI:
10.24297/jam.v14i2.7402.
К ВОПРОСУ НЕЙТРИННОГО ИЗЛУЧЕНИ Я В РАСШИРЯЮЩЕЙСЯ С ОХЛА ЖДЕНИЕМ
ВСЕЛЕННОЙ
Кошман Валентин Семенович
канд. техн. наук, доцент,
Пермский государственный аграрно -технологический университет,
г. Пермь , Россия
ON THE QUESTION OF N EUTRINO RADIATION IN THE UNIVERSE EXPA NDING WITH
COOLING
Valentin Koshman
Cand. Tech. Sci., Associate Professor,
Perm State Agrarian and Technological University,
Perm , Russia
Аннотация. Рассмотрена однородная модель Вселенной в виде газовой смеси из фотонов, барионов
и нейтрино. По мере а нализа изуч аемой физической системы воспроизведена математическая структура
для описания ее космологической эволюции. Выполнена оценка средней величины энергии реликтового
нейтрино, которая совпала с известной оценкой профессора О. Лахав (2002 г.) по поряд ку величины.
Приведены аргументы в пользу первичного термоядерного взрыва у истока расширения Вселенной на
планковском масштабе времени.
Abstract. A homogeneous model of the Universe in the form of a gas mixture of photons, baryons and
neutrinos is conside red . As the physical system under study is analyzed, the mathematical structure for describing
its cosmological evolution is reproduced. The average value of the relic neutrino energy was estimated, which
coincided with the well -known estimate of Professor O. Lahav (2002) in order of magnitude. Arguments are given
in favor of a primary thermonuclear explosion at the source of the expansion of the Universe on the Planck time
scale.
Ключевые слова: модель расширяющейся Вселенной, реликтовое излучение, планков ски е
величины, закон Стефана – Больцмана, объемная плотность энергии нейтрино, масса реликтового
нейтрино.
Keywords : model of the expanding Universe, relic radiation, Planck quantities, Stefan – Boltzmann law,
volume density of neutrino energy, mass of the relic neutrino.
«Структура математического описания
выявляется
по мере анализа физической системы»
П. Девис [1, с. 260]
Как известно, космология, которая изучает
свойства Вселенной в целом, – одна из немногих
естественных наук, где присутствует эволю ция в
явном виде. Изучение эволюции Вселенной
осложнено тем, что её динамика отнюдь не
представляет собой нечто непосредствен но
наблюдаемое. По этой причине объект
исследования может быть дан как целое лишь