Math note (2)-sums (1)

This section mainly describes the basic methods of summation, including the conversion of summation and recursion, multiple and calculation.

I. Notation

This section describes some symbols in the summation form. For example, general form:

(2.1)

Or delimited form:

(2.2)

Or generalized Sigma-notation.

(2.3)

We usually use 2.2 to present a problem or result, while we prefer 2.3 in the Process of calculation.

We usually use 2.4 to represent all sums, where k satisfies the property P (k ).

(2.4)

Using iversion's Convention (for example, :), we can re-write 2.4 as 2.5:

(2.5)

Ii. sums and recurrences

This section mainly discusses the relationship between Recursion and summation.

1. sum is equivalent to 2.6.

 
(2.6)

If a constant is a multiple of K, 2.6 can be expressed in the following form:

 
(2.7)

We can write the following form:

(2.8)

We can use the Repertoire method in the previous chapter to set them separately.

2. Likewise, recursive expressions can be expressed as summation, taking the following form into account:

(2.9)

We multiply on both sides, which satisfies, that is. So, expand the recursive formula and we can get:

 

The final solution of the 2.9 formula is:

 
(2.10)

By expanding the formula, we can get 2.11 or any constant to multiply it.

(2.11)

3. The number of times of Quick Sort and comparison in the following application analysis is shown in 2.12:

 
(2.12)

Both sides multiply N:

.

We can use n-1 to replace N to obtain:

Two subtraction methods can be used to obtain:

. It also works when n = 1. To sum up:

We can know from the comparison formula 2.9, then we can get the value based on 2.11:

The final result can be obtained based on the formula 2.10:

We introduce a harmonic number ):

 
(2.13)

So

Then we can write the number of comparisons for quick sorting as follows :.

3. manipulation of sums

This part focuses on the basic nature of sum.

1. If K is any finite Integer Set, it satisfies three properties: Distributive Law (2.15), Associative Law (2.16), and commutative law (2.17)

(2.15)

 
(2.16)

(2.17)

2. For 2.17, for each integer N, there is only one integer k so that P (K) = n.

When two sets are different, they have the following properties:

 
(2.20)

We split the sum:

(2.23)

This is the basis of the so-called perturbation method.

Through the above split and simple calculation, we can reach the Ric progression summation formula:

(2.25)

3. the following uses perturbation technology to solve a slightly difficult problem:

We add:

The first item on the right is, and the second item is the geometric series above, which is equal to. Finally, we can get the result:

.

When we replace 2 with X, we can get

(2.26)

We can get the same result using the differential method. Consider the following equation:

We can find the following information for the two sides:

4. Multiple Sums

This section mainly explains how to deal with multiple and.

1. First, let's look at the basic features of heavy weights:

• Interchanging the order of summation:
  (2.27)
• Distributive Law:
  (2.28)

Type 2.27 has some deformation:

• Vanilla version:
  (2.29)
• Rocky-road formula:
  (2.30 ).

The Set meets the following requirements :. Based on this nature, we can get a useful formula:
(2.31) So:
(2.32)

2. The following is an example. For the square matrix (symmetric), calculate the sum of the diagonal lines and the above elements :.

Yi Zhi, so
(2.33)

3. The following application is a complicated problem .. When J and K are exchanged, they have symmetry :. We add the two and follow the relationship between the subscripts:

We can see that, obviously, the next item is 0. The preceding item is split and calculated by simple means :. The following interesting formula is returned:

 
(2.34)

Based on 2.34, we can obtain the Chebyshev's monotonic inequalities ):

4. The subscript transformations of Multiple Sums are considered below.

Consider the 2.17 commutative law and set F to any function: We use F (j) to replace K, then we can get the following formula:

 
(2.35)

The number of elements that meet the right formula :.

The following is a specific application :. Through simple calculation, we can obtain the formula:

• Add J first:
• Accumulate K first:
• Use K + J to replace K: before conversion :. (The range of K following the Third Equal sign is that it is easier to calculate it)

Based on the above calculation, we get an additional formula:
(2.36)

5. General methods

This part is mainly used to consolidate the sum of squares of the First n numbers in different ways:
(2.37)

1. You cocould look it up

You can easily find the answer to this question:
(2.38)

2. Guess the answer, prove it by industry

First, we start to look for a law from a small scale to propose a conjecture, and then use mathematical induction to prove it. This is also relatively simple, but sometimes requires inspiration.

3. perturb the sum

We seek for some relationships through the last separate item: that is, we can get:

.

Then we can get the following information based on the Cube:

According to the above-mentioned result.

4. Build a repertoire

That is, the so-called undetermined coefficient method. The form of the result is :.
(2.41)

It is similar to 2.8 at the time. We know the values of a (n), B (n), and C (n) from 2.8. We can get D (n) in the command ).

5. Replace sums by Integrals

Look for the relationship between points and sum: by drawing the area meaning, we can find that it is very close to it. So we can find the recursive formula:

In addition, we can use the integral method to calculate the expression:

Then you can calculate it.

6. expand and contract

We extend single coincidence to multiple weights. We know, so we can split the square:

Multiple operations and processing may be difficult, but sometimes they are relatively simple. You can't scale the highest mountain peaks by climbing only uphill.

7. Use finite calculus 8 and use Generating Functions: these are analyzed in subsequent chapters.