摘要：Inthe highest full-form fraction, how many anti-high-definitionfgsxorder reduction factors does it have at most, and how manycutti

"Five Chapters"

Section 4 3.

In the highest full form fraction , the penalty effect of anti-high shrinkage fgsx order

We already know that the anti-high shrinkage fgsx order causes impairment of the order fraction , including two aspects:

1. The cutting penalty caused by anti-high shrinkage fgsx cutting point G number,

2. The reduction penalty caused by the number of anti-high reduction fgsx order reduction factors.

Now that we have proven that the highest total fraction of an even-order expression ,

Evaluation formula :

{2ny+[2nysx(ny-1)+2(ny)^2 (2ny-1)}/sx(sx+2ny)}* qxz紀* fgsx紀

So,

In the highest full-form fraction , how many anti-high-definition fgsx order reduction factors does it have at most, and how many cutting points are there for the anti-high-definition fgsx ?

In actual calculations, it was found that in each order field on the even positive and negative bidirectional number axis segment ,

It has the following performance:

1. The length of the highest full-digit field (containing the number of odd digits ), on the forward number axis segment, at the same position and between vertical columns, contains the number of endpoints of the inverse high-shrunk fgsx order field, which is the fgsx cut it generates Number of points G ;

2. The size of the full-order degree field of the end-point order is proportional to its division by the even midpoint X. It determines the number of inverse high-shrinking fgsx orders in the highest total fraction, with or without cutting .

3. Since the highest full-form fraction number field contains cutting inverse high-shrinking fgsx orders , they are all inverse high-shrinking fgsx orders higher than the end-point stage . In the same position between vertical columns , the more the number of anti-high-shrinking fgsx orders contained in the end segment , the fewer the cutting anti-high-shrinking fgsx G- points in its highest full- digit segment ; the reduction of the anti-high-shrinking fgsx order The more factors there are . This is because as the order of the anti-high shrinkage fgsx order field increases, its anti-high shrinkage fgsx order field becomes larger and larger than the number field of the highest full-form fraction .

Therefore, the penalty effect of the anti-high shrinkage fgsx order in the highest full-form fraction must be calculated from these three aspects .

Since the even midpoint​​​​​​​​​The cutting position and the amount of loss are also different.

However, the solution to the even number [ guess ] contains the general rule of minimum number of occurrences, because the order level of X does not change at the midpoint of the even number. It just causes them to compare the even numbers in the same order field . The even numbers in different positions, which contain the minimum number of [guessed] solutions, are different in the same order field.

It follows that

" Part 2", five chapters, conclusion 2,

It has an order fraction cut by inverse high-shrinking fgsx . The inverse high-shrinking fgsx it contains only has a penalty effect on the special position (middle position) of the number field of some order fractions . Most of them do not have a penalty effect. .

Among the total fractions of each order of the order formula, the number of anti-high shrinkage fgsx cutting points G in the highest full-style fraction, and its anti-high shrinkage fgsx order reduction factor, have the least content, and some may not. .

Section 5,

The highest perfect form is a fraction of a sister order, and its anti-high contraction fgsx order has the maximum number of penalties

Section 5 1.

Without considering the anti-high-definition fgsx cutting and only the anti-high-definition fgsx order reduction effect, the highest total fraction of an even number contains the minimum number of [guessed] solutions.

We have deduced that,

For any even number, the highest full-form fraction value of order ≥11 , without the action of inverse high-shrinking fgsx order , their fraction value is ≥2 , and there are at least 2 sets of [guess] solutions.

So, the highest total fraction of the sister order is the one with the smallest fraction value . After the reduction and cutting of the anti-high shrinkage fgsx order , do they still have a [ guessed ] solution ?

Let's first look at the fractional values ​​of the highest even-numbered full-form fractions when there is only inverse high-shrinking fgsx order reduction and no inverse high-shrinking fgsx cutting .

Under the action of no anti-high-shrinking fgsx cutting and only anti-high-shrinking fgsx order reduction factors ,

In the formula for evaluating the highest total fraction of an even number ,

=2ny *qxz tired *fgsx tired +[2nysx(ny-1)+2(ny)^2 (2ny-1)]/sx(sx+2ny)*qxz tired *fgsx tired ,

set up:

Take the minimum value of ny , ny =1 ,

Substituting into the formula we get:

2ny *qxz tired *fgsx tired +[2nysx(ny-1)+2(ny)^2 (2ny-1)]/sx(sx+2ny)*qxz tired * fgsx tired

=2*1*qxz tired *fgsx tired +[2*1* sx(1-1)+2*1^2(2*1-1)]/sx(sx+2*1)* qxz Tired * fgsx Tired

=2 *qxz tired *fgsx tired +2/sx(sx +2)*qxz tired *fgsx tired

In its second addition , 2/sx(sx+2)*qxz tired *fgsx tired , in,

Because there is,

sx≥1

2/sx(sx+2)>0

If there is more, l* qxz tired l＞l* fgsx tired l

It can be concluded that

2/sx(sx+2)*qxz tired *fgsx tired >0

That is :

[2nysx(ny-1)+2(ny)^2(2ny -1)]/sx(sx+2ny)*qxz紀*fgsx紀>0

From this, we can get even-numbered order formulas. The highest total fractions of order ≥11 are available.

{2ny+[2nysx(ny-1)+2(ny)^2(2ny-1)]/sx(sx+2ny)}* qxz紀* fgsx紀

=2ny *qxz tired *fgsx tired +[2nysx(ny-1)+2(ny)^2(2ny-1)]/sx(sx+2ny)*qxz tired *fgsx tired

=2ny *qxzti *fgsxti

≥2ny

Because, ny =1, is its minimum value,

So it follows that,

"Part Two", five chapters,

Conclusion 3.

In the absence of anti-high shrinkage fgsx cutting function,

Any highest total fraction value of order ≥11 is: ≥2ny

In this way, in the range of the highest full-form fraction value of an even-order expression : ≥2ny *qxz cumulative *fgsx cumulative , in the inverse high-reduction fgsx order reduction cumulative, and the inverse high-reduction fgsx cutting the number of impairments, it becomes Calculates an even-order equation, the only unknown factor of the highest total fraction value .

Then, the highest total fractions of various even-order expressions ,

1. What is the maximum number of anti-high shrinkage fgsx order reduction factors that they each have?

2. They all exist at most. How many anti-high-shrink fgsx cutting points G are there?

3. What is the maximum amount of damage caused by the anti-high-shrinking fgsx cutting and shrinking they produce ?

4. What is the increase and decrease ratio of qxz and fgsx in the highest total fraction ?

5. Can the highest total fraction of each sister order with the smallest fraction value still have a [guessed] solution that is ≥1 after the inverse high-shrinking fgsx order ?

With these questions in mind we continue to calculate and deduce.

The full text of paragraph 58 is completed.

Thanks for reading!

Welcome to discuss!

See you next week!

来源：科学大雷雷