Chapter 350

On the other side, China.

After a night of thinking, Cheng Nuo was puzzled and finally had a new idea about his graduation thesis.

Cheng Nuo has his own unique insights on the use of two lemmas.

So, as soon as the daytime class ended, Cheng Nuo hurried to the library, picked a place where no one was, took out paper and pen to verify his ideas.

Since this direction does not work during the proof of the two lems into the bertrand hypothesis, Cheng Nuo thought about whether he could draw several inferences based on these two lems and then apply them to the bertrand hypothesis.

In this case, even though it takes a turn, it seems to be much more troublesome than Chebischev's method. But before the real result comes out, no one dares to say this 100%.

Cheng Nuo thinks he should try it.

The tools were already ready, and he pondered for a while and began to make various attempts on the draft paper.

He is not God, and he cannot clearly know which inferences derived from lemma is useful and which one is useless. The safest way is to try them one by one.

Anyway, there is enough time, so Cheng Nuo is not in a hurry.

Swish swish swish

With his head down, he listed the next line of formulas.

[Suppose m is the maximum natural number that satisfies pm≤2n, then obviously for iamp;gt;m, floor(2n/pi)-2floor(n/pi)=0-0=0, the sum is stopped at i=m, and the total number of m terms is m. Since floor(2x)-2floor(x)≤1, each item in these m terms is either 0 or 1...]

From the above, we can infer that 1: [Suppose n is a natural number and p is a prime number, then the highest power of p that can be divisible (2n)!/(n!n!) is: s=Σi≥1[floor(2n/pi)-2floor(n/pi)].]

[Because n≥3 and 2n/3amp;lt;p≤n indicates p2amp;gt;2n, there is only one i=1 to sum, that is: s=floor(2n/p)-2floor(n/p). Since 2n/3amp;lt;p≤n also indicates 1≤n/pamp;lt;3/2, s=floor(2n/p)-2floor(n/p)=2-2=0.]

From this, we can infer that 2: [Suppose n≥3 is a natural number, p is a prime number, and s is the highest power of p that can be divisible (2n)!/(n!n!), then: (a) ps≤2n; (b) If pamp;gt;√2n, then s≤1; if 2n/3amp;lt;p≤n, then s=0.]

Rows, rows, columns.

Except for classes, Cheng Nuo spent the whole day in the library.

It was not until the store closed at 10 o'clock in the evening that Cheng Nuo left reluctantly with his schoolbag on his back.

And on the draft paper he was holding, there were already a dozen inferences.

This is the result of his day of labor.

Cheng Nuo's work tomorrow is to find out the inferences that are useful for the work of proving the bertrand hypothesis from these dozen inferences.

…………

No words all night.

The next day, it was another day when the sun was shining and the flowers were blooming in spring.

The date is early March, and Professor Fang has more than ten days left for Cheng Nuo for one month vacation.

Cheng Nuo has enough time to go to bed... Oh, no, it is to perfect his graduation thesis.

The progress of the paper was carried out according to the plan planned by Cheng Nuo. On this day, he found five inferences that proved that the bertrand hypothesis had an important role.

After the busy day, Cheng Nuo started to prove the formal bertrand hypothesis the next day.

This is not an easy job.

Cheng Nuo is not sure that he can handle it in one day.

But the old saying goes well, one move will decline again, and then it will be exhausted. Now that the momentum is in full swing, it is best to win it one day.

At this time, Cheng Nuo had to prepare to activate the magic method of cultivating immortals again.

Cheng Nuo has already prepared the magical artifact of cultivating immortals, "Kidney Treasure".

Liver, boy!

Cheng Nuo's carbon pen with his right hand and kidney treasure with his left hand began to overcome the last difficulty.

When Chelshev proved the bertrand hypothesis, he adopted a solution to directly conduct known theorems for hard derivation, without any skill at all.

Of course, Cheng Nuo cannot do this.

For the bertrand hypothesis, he was ready to use the counter-proof method.

This is the most commonly used proof method besides direct derivation of proof method, and it is very important when facing many conjectures.

Especially... when proving that a certain conjecture is not true!

But Cheng Nuo was not looking for counterexamples at the time to prove that the bertrand hypothesis was not valid.

Chershev has proved that this hypothesis is true, and using the counter-proof method is nothing more than simplifying the proof steps.

Cheng Nuo is full of confidence.

The first step is to use the reverse proof method, assuming that the proposition is not true, that is, there is a certain n≥2, and there is no prime number between n and 2n.

The second step is to decompose (2n)!/(n!n!)(2n)!/(n!n!)=Πps(p)(s(p) is the power of the mass factor p.

The third step is to know pamp;lt;2n from the inference method, and to assume that p≤n is known from the inference method, and then from the inference 3, p≤2n/3, so (2n)!/(n!n!)=Πp≤2n/3ps(p).

……………………

Step 7, using Inference 8, we can get: (2n)!/(n!n!)≤Πp≤√2nps(p)·Π√2namp;lt;p≤2n/3p≤Πp≤√2nps(p)·Πp≤2n/3p!

The idea was smooth, Cheng Nuo wrote down all the way, without seeing any resistance, and completed more than half of the proof steps in about an hour.

Even Cheng Nuo himself was surprised for a while.

It turns out that I am already so powerful now!!!

Cheng Nuo put his hips on his hips and was proud for a while.

Then, he bowed his head and continued to list the proof formula in a bitter way.

In the eighth step, since the number of multiplication factors in the first group of multiplication is the number of prime numbers within √2n, that is, no more than √2n/2-1 (because even numbers and 1 are not prime numbers)... This results in: (2n)!/(n!n!)amp;lt; (2n)√2n/2-1·42n/3.

In the ninth step, (2n)!/(n!n!) is the largest item in the (1+1)2n expansion formula, and the expansion formula has a total of 2n terms (we combine the first and last terms 1 into 2), so (2n)!/(n!n!)≥22n/2n=4n/2n. Take the logarithm at both ends and further simplify it to get: √2nln4amp;lt;3ln(2n).

Next is the last step.

Since the power function √2n grows much faster with n than the logarithmic function ln(2n), the above formula is obviously impossible for a large enough n.

At this point, it can be explained that the bertrand hypothesis is true.

The draft part of the paper is officially completed.

Moreover, the completion time was half earlier than Cheng Nuo expected.

In this way, you can also create the document version of your graduation thesis while it is hot.

Do it! Do it! Do it! Do it!

Slap

Cheng Nuo tapped his finger on the keyboard, and after more than four hours, the graduation thesis was officially completed.

Cheng Nuo made another ppt, which he would use during graduation defense.

As for the defense draft, Cheng Nuo did not prepare this thing.

Anyway, when the soldiers come to block the enemy, the water comes to cover the earth.

If I can’t even pass a graduation defense with my brother’s level, then it’s better to just find a piece of tofu to kill him.

Oh, by the way, there is one more thing.

Cheng Nuo slapped his head, as if he had remembered something.

After searching online for a while, Cheng Nuo converted his paper into English PDF format and submitted it to an academic journal located in Degu Country: one of the journals "Mathematical Communication Symbols". It is ranked first in the first district.

Influencing factor 5.21, even among many famous academic journals in District 1, is of a medium or upper level.

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ps: "Love Apartment", Alas
Chapter completed!