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.

So, as soon as the day's class was over, Cheng Nuo hurried to the library, randomly picked a place where no one was there, took out a pen and paper, and tested his thoughts.

Since it is not feasible to impose two lemmas into the proof process of the Bertrand hypothesis, what Cheng Nuo thought was whether to draw some inferences based on these two lemmas, and then apply them to the Bertrand hypothesis.

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

Cheng Nuo felt that he should give it a try.

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

Whether he is God or not, he cannot clearly know which inferences drawn through lemmas are useful and which ones are not. The safest way is to try them one by one.

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

Swish Swish Swish

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

[Assuming that m is the largest natural number that satisfies pm≤2n, it is obvious that for iamp;gt;m, floor(2n/pi)-2floor(n/pi)=0-0=0, the summation ends at i=m, the total item m. Since floor(2x)-2floor(x)≤1, each of these m items is either 0 or 1...]

From the above, inference 1: [Assuming 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, the summation only has i=1 item, namely: s=floor(2n/p)-2floor(n/p). Since 2n/3amp;lt;p≤n also shows that 1≤n/pamp;lt;3/2, so s=floor(2n/p)-2floor(n/p)=2-2=0. 】

thus,

Inference 2: [Assume that 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.

Apart from attending classes, Cheng Nuo spent the whole day in the library.

When the library closed at ten o'clock in the evening, Cheng Nuo reluctantly left with his schoolbag on his back.

And on the draft paper he held in his hand, there were already a dozen inferences densely listed.

This is the fruit of his day's labor.

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

…………

Nothing to say all night.

The next day was another sunny day with flowers blooming in spring.

The date was early March, and there were still more than ten days left in the one-month vacation that Professor Fang gave Cheng Nuo.

Cheng Nuo had enough time to waste... Oh, no, to perfect his graduation thesis.

The progress of the thesis was carried out according to the plan planned by Cheng Nuo. On this day, he found five inferences that proved the important role of Bertrand's hypothesis from the dozen inferences he derived.

After finishing this busy day, the next day, Cheng Nuo began to formally prove Bertrand's hypothesis without stopping.

It's not an easy job.

Cheng Nuo didn't have much confidence that he could finish it in a day.

But there is an old saying that is good, one vigorous effort, then decline, and three exhaustion. It is gaining momentum now, and it is best to win it one day.

At this time, Cheng Nuo had no choice but to prepare to start cultivating immortals again.

Cheng Nuo had already prepared the artifact for cultivating immortals, the "Kidney Treasure".

Liver, boy!

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

When Chershev proved the Bertrand hypothesis, the solution he adopted was to directly carry out known theorems for hard derivation, without any tricks at all.

Of course Cheng Nuo couldn't do that.

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

This is the most commonly used proof method besides the direct derivation 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 a counterexample to prove that Bertrand's hypothesis was not true.

Chershev has already proved the establishment of this hypothesis, and using the method of counter-evidence is nothing more than simplifying the proof steps.

Cheng Nuo was full of confidence.

The first step is to use the method of proof by contradiction, 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!) into (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, to know p≤n from the hypothesis of counter-evidence, and to know p≤2n/3 from Corollary 3, so (2n)!/(n!n!)=Πp≤2n/ 3ps(p).

………………

The seventh step, using Corollary 8, we can get: (2n)!/(n!n!)≤Πp≤√2nps(p) Π√2namp;lt;p≤2n/3p≤Πp≤√2nps(p) Πp≤2n/3p!

With a clear mind, Cheng Nuo wrote down 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 now, unknowingly, already so powerful! ! !

Cheng Nuo sat proudly for a while.

After that, he bowed his head and continued to list the proof formulas.

The 8th step, because the multiplied factor number of the first group in the product is the prime number number within √2n, promptly is no more than √2n/2-1 (because even number and 1 are not prime numbers)... thus obtain: (2n )!/(n!n!)amp;lt;(2n)√2n/2-1·42n/3.

The ninth step, (2n)!/(n!n!) is the largest item in the (1+1)2n expansion, and the expansion has 2n items (we combine the first and last two 1 into 2), So (2n)!/(n!n!)≥22n/2n=4n/2n. Take the logarithm of both ends and further simplify: √2nln4amp;lt;3ln(2n).

Next, is the last step.

Since the growth rate of the power function √2n with n is much faster than that of the logarithmic function ln(2n), the above formula is obviously impossible for a sufficiently large n.

So far, it can be explained that Bertrand's hypothesis is established.

The draft part of the thesis is officially completed.

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

In this case, the document version of the graduation thesis can be produced while it is hot.

Do it! Do it! Do it!

clap clap

Cheng Nuo tapped the keyboard with his fingers. After more than four hours, the graduation thesis was officially completed.

Cheng Nuo made another PPT, which he will use in his graduation defense.

As for the defense draft, Cheng Nuo didn't prepare it.

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

If Yige's level can't even pass a graduation defense, then it's better to just find a piece of tofu and kill him.

Oh yes, one more thing.

Cheng Nuo patted his head, as if he remembered something.

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

The impact factor is 5.21, even among many famous academic journals in the first district, it belongs to the upper-middle level.

……………………

PS: "Love Apartment", hey