universal data

Chapter 347 Finishing the Graduation Thesis
350 chapter

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 Bertrand's hypothesis, what Cheng Nuo thought was whether to draw some inferences based on these two lemmas, and then apply them to Bertrand's hypothesis.

In this case, even though it has turned a corner, it seems to be a lot more troublesome than Chebyshev's method.But before the real results come out, no one dares to say so 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 has God or not, he cannot clearly know which of the inferences drawn through the lemma is useful and which is useless.The safest way is to try them one by one.

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

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

[Assuming m is the largest natural number satisfying pm ≤ 2n, obviously for i > m, floor(2n/pi)- 2floor(n/pi)= 0 - 0 = 0, the summation ends at i = m, A total of m items.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/3 < p ≤ n show that p2 > 2n, the summation only has i = 1, namely: s = floor(2n/p)- 2floor(n/p) .Since 2n/3 < p ≤ n also implies that 1 ≤ n/p < 3/2, s = floor(2n/p) - 2floor(n/p) = 2 - 2 = 0. 】

From this, we can get Corollary 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 p >√2n, then s ≤ 1; (c) If 2n/3 < 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 out from these dozen inferences useful inferences for proving Bertrand's hypothesis.

…………

Silent 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 or so 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!
Cheng Nuo started to overcome the last difficulty with a carbon pen in his right hand and a kidney treasure in his left hand.

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

Of course Cheng Nuo couldn't do that.

For Bertrand's hypothesis, he was prepared to use proof by contradiction.

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

Especially... when a certain conjecture is proved to be false!

But Cheng Nuo was not looking for a counterexample to prove that Bertrand's hypothesis was not valid.

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 false, that is, there exists a certain n ≥ 2, and there is no prime number between n and 2n.

In the second step, decompose (2n)!/(n!n!) into (2n)!/(n!n!)=Π ps(p)(s(p) is the power of prime factor p.

The third step is to know p & lt; 5n from Corollary 2, to know p ≤ n from the assumption of counter-evidence, and to know p ≤ 3n/2 from Corollary 3, so (2n)!/(n!n!)=Πp≤2n /3 ps(p).

………………

The seventh step, using Corollary 8, we can get: (2n)!/(n!n!)≤Πp≤√2n ps(p) Π√2n<p≤2n/3 p ≤Πp≤√2n ps( p)·Πp≤2n/3 p!

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.

Afterwards, he lowered his head and continued to painstakingly list out the proof formulas.

The eighth step, because the number of multiplicated factors of the first group in 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)... thus obtained: (2n )!/(n!n!)<(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.Taking the logarithm of both ends and further simplifying can get: √2n ln4 < 3 ln(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 cannot be established 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 SCI 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~~
(End of this chapter)