Omnipotent Data

Chapter 350

On the other side, China.

After a night of thinking, confused Cheng Nuo finally had a new idea for his graduation thesis.

Regarding the application of the two lemmas, Cheng Nuo has his own unique insights.

Therefore, as soon as the daytime class was over that day, Cheng Nuo hurried to the library, randomly picked a no-one position, took out paper and pen to verify his ideas.

Since imposing two lemmas into the proof process of Bertrand's hypothesis does not work in this direction, what Cheng Nuo wonders is whether he can draw several inferences based on these two lemmas and then apply them to Bertrand's hypothesis.

In this case, although it turned a corner, it seemed more troublesome than Chebyshev's method. But before the real results come out, no one dare to say 100% like this.

Cheng Nuo thinks he should try it.

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

Whether he is a God or not, it is not clear which inferences drawn by lemmas are useful and which are not. The safest way is to try one by one.

Since time is enough, Cheng Nuo is not in a hurry.

Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

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

[Let m be the largest natural number satisfying pm≤2n, then obviously for iamp;gt;m, floor(2n/pi)-2floor(n/pi)=0-0=0, and the sum ends at i=m, total m items. Since floor(2x)-2floor(x)≤1, each of these m items is either 0 or 1...]

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

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

Thus, inference 2: [Suppose n≥3 is a natural number, p is a prime number, and s is the highest power of p that can divide (2n)!/(n!n!), then: (a)ps≤ 2n; (b) If pamp;gt;√2n, then s≤1; if 2n/3amp;lt;p≤n, then s=0. 】

Row by row, column by column.

Except for classes, Cheng Nuo spends all day in the library.

When the library closed at ten in the evening, Cheng Nuo reluctantly left with his schoolbag.

On the draft paper in his hand, a dozen inferences are already densely lined up.

This is the result of his day of labor.

Cheng Nuo's job tomorrow is to find out the useful inferences for Bertrand's hypothesis from these dozen inferences.

…………

No words for a night.

The next day was another sunny day with warm spring flowers.

The date is early March, and there are more than ten days left in the one-month holiday given to Cheng Nuo by Professor Fang.

Cheng Nuo has enough time to spend...oh no, to perfect his graduation thesis.

The progress of the thesis proceeded in accordance with the plan planned by Cheng Nuo. On this day, he found five inferences that proved the important role of Bertrand hypothesis from a dozen inferences derived.

At the end of this busy day, the next day, Cheng Nuo started to prove Bertrand's hypothesis without stopping.

This is not an easy job.

Cheng Nuo didn't have much confidence to get it done in one day.

But an old saying goes well, it's a rush, it fades again, and it is exhausted. Now that the momentum is in full swing, it's best to take it one day.

At this time, Cheng Nuo had to prepare again to start the cultivation method.

And Cheng Nuo had already prepared for the god-cultivation artifact, the "Kidney Treasure".

Liver, boy!

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

When Chershev proved Bertrand's hypothesis, he adopted the scheme to directly carry out the known theorems for rigid derivation, without any skill at all.

Of course Cheng Nuo could not do this.

For Bertrand hypothesis, he is going to use the method of proof by contradiction.

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

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

But Cheng Nuo was not looking for counterexamples to prove Bertrand's hypothesis was not true.

Chershev has already proved the validity of this hypothesis, and the use of proof by contradiction is nothing more than simplifying the proof steps.

Cheng Nuo is confident.

The first step is to use the contradiction 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.

In the second step, the decomposition of (2n)!/(n!n!) (2n)!/(n!n!)=Πps(p) (s(p) is the power of prime factor p.

The third step is to know pamp;lt;2n from Corollary 5, assume that p≤n by the contradiction method, and then know that p≤2n/3 from Corollary 3, so (2n)!/(n!n!)=Πp≤2n/ 3ps(p).

………………

In the seventh step, use Corollary 8 to obtain: (2n)!/(n!n!)≤Πp≤√2nps(p)·Π√2namp;lt;p≤2n/3p≤Πp≤√2nps(p)· Πp≤2n/3p!

With smooth thoughts, Cheng Nuo wrote them all the way, without 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, unconsciously! ! !

Cheng Nuo was proud of his arms for a while.

Afterwards, he lowered his head and continued to forcefully list the proof formula.

The eighth step, because the number of multiplied factors in the first group of the product 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)... From this we get: (2n )!/(n!n!)amp;lt;(2n)√2n/2-1·42n/3.

In the ninth step, (2n)!/(n!n!) is the largest term in the (1+1)2n expansion, and the expansion has 2n terms (we merge the first and last two terms 1 into 2), Therefore (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 final step.

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

So far, it can be explained that Bertrand hypothesis holds.

The draft part of the paper is officially completed.

And the completion time was a full half time earlier than Cheng Nuo expected.

In this case, you can get the document version of the graduation thesis while it's hot.

Do it! Do it! Do it!

Pop pop

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

Cheng Nuo made another copy~www.readwn.com~ which will be used in graduation defense.

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

Anyway, when the time comes, soldiers will come to block, and water will come to cover.

If you can't even pass a graduation defense at your brother's level, you might as well just find a piece of tofu and kill him.

Oh, yes, one more thing.

Cheng Nuo slapped his head, as if remembering something.

After searching the Internet for a while, Cheng Nuo converted the paper into English PDF format, packaged it and submitted it to an academic journal in Germany: "Mathematics Communication Symbols". One of the CI journals, ranked first.

The impact factor of 5.21, even among the many well-known academic journals in the first district, belongs to the upper middle level.

……………………

PS: "Love Apartment", hey