Chapter 350

On the other side, Huaguo.

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

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

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

Since it is not feasible to impose the two lemma in the proof process of Bertrand's hypothesis, Cheng Nuo wonders whether he can draw several inferences from these two lemma and then apply them to Bertrand's hypothesis.

In this case, although it turns a corner, it seems to be a lot more troublesome than Chebyshev's method. But until the real results are out, no one dares to say it 100%.

Cheng Nuo felt that he should still give it a try.

The tools were already ready, and he groaned for a while, and began to experiment with various things on the scratch paper.

Whether he is God or not does not know with certainty which of the inferences derived from the lemma is useful and which is not. The safest way to do this is to try them one by one.

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

Boom~~

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

[Let m be the maximum natural number satisfying pm ≤ 2n, then it is obvious that i > m, floor(2n/pi) - 2floor(n/pi) = 0 - 0 = 0, the sum ends at i = m, and the total m term is m. Since floor(2x)- 2floor(x) ≤ 1, each of these m terms is either 0 or 1......]

From the above, it follows that 1: [Let n be a natural number and p be a prime number, then it is divisible by (2n)!/(n!n!) The highest power of p is: s =Σi≥1 [floor(2n/pi)- 2floor(n/pi)]. 】

[Because n ≥ 3 and 2n/3 2n, the sum is only i = 1 term, i.e.: s = floor(2n/p)- 2floor(n/p). Due to 2n/3 < p ≤ n also indicates 1 ≤ n/p < 3/2, so s = floor(2n/p) - 2floor(n/p) = 2 - 2 = 0. 】

From this, it follows that 2 follows: [Let n ≥ 3 be a natural number, p be a prime number, s be divisible by (2n)!/(n!n!) the highest power of p, then: (a) ps ≤ 2n; √2n, then s ≤ 1;(c) if 2n/3 < p ≤ n, then s = 0. 】

One row at a time, one column at a time.

In addition to classes, Cheng Nuo spent the whole day in the library.

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

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

It was the result of a day of his labor.

Tomorrow's work is to find out from these dozen inferences that are useful for Bertrand's hypothesis proof work.

............

Nothing was said all night.

The next day, it was another sunny day and spring flowers blooming.

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

Cheng Nuo has enough time to wave...... Oh no, to perfect his dissertation.

The progress of the dissertation proceeded according to the plan planned by Cheng Nuo, and on this day, he found five inferences from the dozen inferences deduced that proved the important role of Bertrand's hypothesis.

At the end of this busy day, the next day, Cheng Nuo began to formally prove the Bertrand hypothesis.

It's not an easy job.

Cheng Nuo was not very sure that he could do it in one day.

But as the old saying goes, one blow, then decline, and three exhaustion. Now the momentum is strong, and it is better to win it in one day.

At this time, Cheng Nuo had to prepare to start the Dafa of cultivating immortals again.

And the immortal cultivation artifact, "Kidney Treasure", Cheng Nuo has already been prepared.

Liver, boy!

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

When Chershev proved Bertrand's hypothesis, he took the scheme of directly deriving the known theorem by hard deduction, without any skill at all.

Of course, Cheng Nuo can't do that.

For the Bertrand hypothesis, he is prepared to use the method of counterproof.

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

Especially...... When it is proved that a conjecture is not true!

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

Chershev has already proved the validity of this hypothesis, and the use of counter-evidence is nothing more than a simplification of the proof step.

Cheng Nuo is full of confidence.

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

Step 2: (2n)!/(n!n!) decomposition of (2n)!/(n!n!) =Π ps(p)(s(p) is the power of the quality factor p.

The third step is to know p < by inference 5; 2n, p ≤ n is assumed by the counterargument, and p ≤ 2n/3 is known by inference 3, so (2n)!/(n!n!) =Πp≤2n/3 ps(p)。

..................

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

The train of thought was smooth, Cheng Nuo wrote it 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'm already so powerful now, unconsciously!!

Cheng Nuo crossed his waist for a while.

Then, he bowed his head and continued to list the proof formula.

In the eighth step, since the number of multiplied factors in the first group of the product is the number of primes within √2n, i.e. no more than √2n/2 - 1 (because even numbers and 1 are not primes)...... This result: (2n)!/(n!n!) &lt; (2n)√2n/2-1 · 42n/3。

Step 9, (2n)!/(n!n!) is the largest of the (1+1)2n expansions, which have a total of 2n terms (we combine the first and last two terms, 1 into 2), so (2n)!/(n!n!) ≥ 22n / 2n = 4n / 2n。 Take the logarithm of both ends and further simplify them to obtain: √2n ln4 < 3 ln(2n)。

Now, that's the final step.

Since the power function √2n grows much faster with n than the logarithmic function ln(2n), the above equation is obviously not true for a sufficiently large n.

At this point, it can be shown that Bertrand's hypothesis is true.

The draft part of the paper is officially completed.

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

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

!!

Bang Bang Bang ~~

Cheng Nuo's fingers tapped on the keyboard, and 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 draft of the defense, Cheng Nuo did not prepare this thing.

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

If at my brother's level, I can't even pass a graduation defense, then I might as well 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 PDF format in English and submitted it to an academic journal in Deguguo: "Mathematical Communication Symbols".

One of the SCI journals, ranked in Zone 1.

The impact factor is 5.21, which is at the upper average level even among many well-known academic journals in District 1.

........................

PS: "Love Apartment", hey~~