1，q:( 1/ 1)3+( 1/2)3+( 1/3)3+( 1/4)3+。

More generally: when k is an odd number,

q:( 1/ 1)k+( 1/2)k+( 1/3)k+( 1/4)k+( 1/5)。

Euler worked out:

( 1/ 1)^2+( 1/2)^2+( 1/3)^2+( 1/4)^2+( 1/5)^2+…+( 1/n)^2=(π^2)/6

The expression when k is an even number is given.

So, so he raised the above questions.

2. Transcendence of e+π:

Background: This topic is a special case of Hilbert problem 7.

The transcendence of e π has been proved, but so far no one has proved the transcendence of e+π.

3. Prime number problem (also called Riemann conjecture).

Prove:

ζ(s)= 1+( 1/2)s+( 1/3)s+( 1/4)s+( 1/5)s+ ...

The zeros of the definition function zeta (s) all have the real part of 1/2 except negative integer real numbers.

Background: This is Hilbert's eighth question.

It has been proved that the first 3 million zeros of ζ(s) function really conform to the conjecture.

The extended question is: the expression formula of prime number? What is the essence of prime numbers?

4. Is there an odd perfect number?

Background:

A perfect number is equal to the sum of its factors.

The first three perfect numbers are:

6= 1+2+3

28= 1+2+4+7+ 14

496= 1+2+4+8+ 16+3 1+62+ 124+248

The known 32 perfect numbers are all even numbers.

1973 concludes that if n is an odd perfect number, then:

n & gt 10^50

5. Except 8 = 2 3 and 9 = 3 2, are there no two consecutive integers that can be expressed as powers of other positive integers?

Background:

This is Catalonia's guess (1842).

1962, China mathematician Ke Zhao independently proved that there is no continuous integer that can be expressed as the power of other positive integers.

1976, Dutch mathematicians proved that any two positive integer powers greater than a certain number are discontinuous. So just check whether any positive integer power less than this number is continuous.

But because this number is too large, there are more than 500 bits, which is beyond the calculation range of the computer.

So this conjecture is almost correct, but no one can confirm it so far.

6. Given a positive integer n, if n is even, change it to n/2, and if it becomes odd after division, multiply it by 3 plus 1 (that is, 3n+ 1). Repeat this operation, and after a limited number of steps, will you definitely get 1?

Background:

This old conjecture (1930).

People have never found a counterexample through a lot of checking computations, but no one can prove it.

Three unsolved problems in Hilbert's 23 problem.

1, problem 1 continuum hypothesis.

There is no other cardinality between the cardinality of all positive integers (called countable set) and the cardinality of real number set (called continuous set).

BACKGROUND: 1938, Austrian mathematician Godel proved that this hypothesis is not falsifiable in the axiomatic system of set theory, that is, the axiomatic system of Zeromolo-Fo Runkle.

1963, American mathematician Cohen proved that this assumption cannot be proved to be correct in this axiomatic system.

So far, no one knows whether this assumption is right or wrong.

2. Question 2 Compatibility of arithmetic axioms.

Background: Godel proved the incompleteness of the arithmetic system, which dashed Hilbert's idea of proving that the arithmetic axiom system is not contradictory with meta-mathematics.

3. Question 7: Unreasonability and transcendence of some figures.

See 2 above.

5, problem 8 prime number problem.

See article 2 ter above.

6. The coefficient of the problem 1 1 is the quadratic form of any algebraic number.

Background: German and French mathematicians made great progress in 1960s.

7. The generalization of Kroneck's theorem on the problem 12 Abelian field on arbitrary algebraic rational field.

Background: This problem has only some scattered results and is far from being completely solved.

8. Question 13: It is impossible to solve the general algebraic equation of degree 7 only with binary functions.

Background: 1957 Soviet mathematicians solved the case of continuous function. If you need to parse the function, this problem has not been completely solved.

9. Question 15 The strict foundation of Schubert counting calculus.

Background: The number of intersections in algebra. Related to algebraic geometry.

10, Problem 16 Topology of algebraic curves and surfaces.

Algebraic curves need to contain the maximum number of closed branch curves. And the maximum number and relative position of limit cycles of differential equations.

1 1, problem 18 constructing space with congruent polyhedron.

The problem of the closest arrangement of infinite polyhedron in a given form has not been solved yet.

12, the 20th general boundary value problem.

Boundary value problems of partial differential equations are developing vigorously.

13, further development of the variational method in question 23.

Four thousand and seven difficult problems

In 2000, the American Clay Association for the Advancement of Mathematics proposed. In memory of 23 questions raised by Hilbert a hundred years ago. The reward for each question is millions of dollars.

1, Riemann conjecture.

See page 3 of 2.

Through this conjecture, mathematicians believe that the mystery of prime number distribution can be solved.

This problem is one of Hilbert's 23 unsolved problems. By studying Riemann conjecture number

Scientists believe that in addition to solving the mystery of prime number distribution, it is of great significance to analytic number theory, function theory,

Elliptic function theory, group theory and prime number test will all have substantial influence.

2. Young-Mills Theory and Mass Gap

Hypothetically)

In 1954, Yang-Mills gauge theory was put forward by and Mills.

At the beginning of mathematics, a normative theoretical framework was put forward, and later it gradually developed into quantum.

The important theory of physics also makes him an important figure in the foundation of modern physics.

In the theory put forward by Yang Zhenning and Mills, particles that transfer force will be produced. They

The difficulty is the mass of this particle. The result they got through mathematical deduction.

Yes, this particle has a charge, but no mass. However, the difficulty is that if this is charged.

Our particles have no mass, so why is there no experimental evidence? If we assume

If the particle has mass, the gauge symmetry will be destroyed. General physicists believe in quality.

Quantity, so how to fill this loophole is a very challenging mathematical problem.

3.P versus NP problem.

With the increase of calculation scale, the kind of problem that the calculation time will increase in a polynomial way is called "P problem".

P of the p problem is the first letter of polynomial time. already

Given that the size is n, if it can be determined that the calculation time is below cnd (c and D are positive real numbers) time.

When we are sure, we call it "polynomial time determination method". You can use this.

The problem solved by the algorithm is the P problem. On the other hand, if other factors are involved, such as the sixth sense.

The algorithm is called "non-deterministic algorithm", and this kind of problem is "NP problem", NP is

Abbreviation for uncertain polynomial time.

By definition, P problem is a part of NP problem. But is there a NP problem?

What about things that don't belong to the P-level? Or will the NP problem eventually become a P problem? this

Is the quite famous PNP problem.

4. Naville-Stokes equation.

Because Euler's equation is too simplified, it is produced in the process of seeking correction.

New results. French engineer Naville and British mathematician Stoke experienced rigorous mathematics.

Considering the viscous term, the Cashierville-Stoke equation is derived.

Starting from 1943, French mathematician Le Rey proved Naville Stowe.

After the global weak solution of Dirk equation, people always want to know.

Is this solution unique? The result is that if we presuppose the Naville-Stoke party,

If the solution of a process is a strong solution, then this solution is unique. So the question becomes: how big is the gap between the weak solution and the strong solution, and is it possible for the weak solution to be equal to the strong solution? In other words, can we get the full-time smooth solution of Naville-Stoke equation? Then there is the certificate.

The solution will blow up in a limited time.

Solving this problem is not only helpful to mathematics, but also to physics and aerospace engineering, especially chaos.

Flow (turbulence) will have a decisive influence, as well as Naville-Stoke equation and Austria.

The great physicist Pozmann's Pozmann equation is also closely related to Naville's research.

Er-Stoke equation and Boltzmann equation.

The knowledge of the relationship between people is called the limit of fluid mechanics, which is acceptable.

Weil-Stoke equation itself has very rich connotations.

5. Poincare conjecture (Poincare conjecture)

Poincare conjecture is a difficult problem in topology. In mathematical terms: simple connection

Three-dimensional closed manifold and three-dimensional spherical homeomorphism.

In the mathematical sense, this is a seemingly simple problem, but it is not.

This frequently encountered problem was put forward by Poincare in 1904.

Later, it attracted many excellent mathematicians to devote themselves to this research topic.

Soon after Poincare's conjecture (Figure 4) was put forward, mathematicians would naturally

We extend it to high-dimensional space (n4) and call it generalized Poincare conjecture: simply connected.

≥

N(n4)-dimensional closed manifold, if N is used.

A sphere with dimension ≥ has the same basic group, so it must be homeomorphic to an N-dimensional sphere.

After nearly 60 years, in 196 1 year, the American mathematician Smale took

Clever method, he ignored the difficulties of three-dimensional and four-dimensional, and directly proved that five-dimensional (n5) is above.

≥

Generalized Poincare conjecture, so he won the Fields Prize in 1966. After 20 years.

Later, another American mathematician Friedman proved the four-dimensional Poincare hypothesis.

1986, he won the Fields Medal for this achievement. But for us, really.

The three-dimensional space where people live (n3) was still an unsolved mystery at that time.

=

Until April 2003, Russian mathematician perelman was at

MIT made three speeches, in which he answered many mathematicians' questions.

There are many indications that Feldman may have cracked the Poincare conjecture. A few days later, the new york Times appeared for the first time.

The second time, the news was disclosed to the public with the title "Russians solved famous mathematical problems". the same

The headline article published by MathWorld, an influential mathematics website, is Poincare conjecture.

Facts have proved that this time it is true! 」[ 14]。

The examination for mathematicians will not be completed until 2005, and so far, it has not been found.

Feldman couldn't collect the million-dollar loophole in the Clay Institute of Mathematics.

6. White and Swinerton-Dale conjecture (Burch and Swinerton-Dale)

Guess)

General elliptic curve equation y 2 = x 3+ax+b, when calculating the arc length of an ellipse.

You will encounter this curve. Since 1950s, mathematicians have discovered elliptic curves and number theory,

Geometry and cryptography are closely related. For example, wiles proved Fermat.

In the last theorem, the key step is to use the relationship between elliptic curve and module-that is, the Taniyama-Zhicun conjecture, which is related to the white matter-swinton-Dell conjecture.

Elliptic curve.

In the 1960 s, Bai Zhi and swinton Dale of the University of Cambridge, England, calculated some by computer.

Understanding of polynomial equation. There are usually infinite solutions, but how to calculate infinity?

And then what? Its solution is to classify first, and the typical mathematical method is the concept of congruence.

From this, we can get the congruence class, that is, the remainder after dividing by a number, which is infinite.

Multiple numbers cannot be all. Mathematicians naturally choose prime numbers, so this problem has nothing to do with

Zeta function of riemann conjecture. After a long period of calculation and data collection, he

Scientists have observed some laws and patterns, so they put forward this conjecture. They worked out the knot by computer.

Decisively speaking, an elliptic curve will have infinite rational points if and only if it is attached to the curve.

When zeta function zeta (s) =, the value is 0, that is zeta (1).

; When s 1= 0

7. Hodge conjecture

Any harmonic differential form on a nonsingular projective algebraic curve is an algebraic circle.

Rational combination of cohomology classes. 」

The last question, although not the most difficult of the seven major problems in the Millennium, may be

Energy is the most difficult thing for ordinary people to understand. Because there are too many profound, professional and abstract.

References:

100 basic problems of mathematics, mathematics and culture, a review of Hilbert's 23 mathematical problems

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