Find the undetermined coefficient method to derive the series an

When doing this question during the exam, you should pay attention to speed. So I offer a very prescriptive way to solve the problem:

1. According to fx, it can be judged that it is symmetrical about x=3 4, and it is an odd function, so it is also symmetrical about x=-3 4. It can be analogized to a sinusoidal function that achieves a peak at x=3 4, so the period is 3

Second, the substitution method is used during the examination to calculate a5=-31 and a6=-63 in turn

So the original equation = f(2) + f(0) = -3

sn=2an+n

s(n-1)=2a(n-1)+n-1

Subtract the two formulas to get.

an=2an-2a(n-1)+1

an=2a(n-1)-1

an-1=2(a(n-1)-1)

So it's a proportional series, and the common ratio is 2

a1-1=-2

So an-1=-2 n

an=1-2^n

a5=-31,a6=-63

f(3/2+x)=f(-x)=-f(x)

So f(3+x)=-f(3 2+x)=f(x), i.e. f(x) is a periodic function, and the period is 3

f(-63)=f(0)=0

f(-31)=f(-1)=f(2)=-f(-2)=3, so f(-31)+f(-63)=3

SequencesPending coefficient methodYesEqual difference seriesFind the formula for the general term. As long as the tolerance and the first term are set first, and two systems of equations are listed according to the general term formula of the equal difference series, the tolerance and the first term can be solved, and then the general term formula comes out, and the sum of the first n terms also comes out.

The general usage is to let a polynomial be.

All or part of the coefficients are unknown, utilizing two polynomial identities.

The principle of equality of coefficients of the same kind or other known conditions determines these coefficients and thus obtains the value to be sought.

For example, factoring a known polynomial.

The coefficients of some factors can be set to be unknowns, and the unslag modulus numbers can be obtained by using the condition of identity, such as answering slowly. Finding the equation for a conic curve passing through certain points can also be done using the undetermined coefficient method. In a broader sense, the undetermined coefficient method is a method of treating some constants of an analytical formula as unknowns, using known conditions to determine these unknowns, so that the problem can be solved.

To find the expression of the function, put a rational fraction.

Decompose it into the sum of several simple fractions to find the differential equation.

can be used in this way.

Conditions for the use of the pending coefficient method:

The undetermined coefficient method should design the form of the function, obtain the system of equations, find the value of the undetermined coefficient, and finally write the analytical method of the function.

The method of finding unknowns for undetermined coefficients is to express a polynomial in the form of another undetermined coefficient to obtain an identity.

In the steps of the undetermined coefficient method, the form of the function should be set and the equation should be substituted into the analytic formula. Find the value of the coefficient to be determined. Write out the analytic formula of the function.

the use of the method of the coefficient to be determined; These coefficients are determined using the principle of equality of the identities of two polynomials. Get a value to be evaluated.

The undetermined coefficient method is to express a polynomial into another new form with undetermined coefficients, so that an identity is obtained. Then, according to the properties of the identity, the equation or system of equations that the coefficients should satisfy is obtained, and then the undetermined coefficients can be found by solving the equations or systems of equations, or the relations satisfied by some coefficients can be found, and this method of solving the problem is called the undetermined coefficient method.

Introduction

The undetermined coefficient method is an important method for junior high school mathematics. To use the undetermined coefficient method to decompose the cause-to-heart formula is to first assume the original formula as the product of several factors according to known conditions, and the coefficients in these factors can be expressed by letters first, and their values are to be determined.

Since the product of these factors is the same as the original formula, then according to the principle of identity, the system of equations of the undetermined coefficients is established, and finally the value of the undetermined coefficients can be obtained by solving the system of equations. It is often seen in junior high school competitions.

a(n+1)+k=2(an+k)

The original form of comparison evokes has k = 1, so.

an+1 is the proportional chain year, a1+1=2, and the band is an+1=2 nan=2 n-1

Answer: In the sequence AN, a1=1 and a2=2

Condition satisfied: a(n+2)=(2 3)a(n+1)+(1 3)an.........1）

Let a(n+2)-sa(n+1)=t*[a(n+1)-san].

1) and (2) were compared:

t+s=2/3

ts=1/3

Solution: t=1, s=-1 3

t=-1/3，s=1

Take the second group:

a(n+2)-a(n+1)=-(1/3)*[a(n+1)-an ]

So: a(n+1)-an is a proportional series, and the common ratio q=-1 3

The first term a2-a1=2-1=1

So: a(n+1)-an=(-1 3) (n-1).

a2-a1=1

a3-a2=(-1/3)^1

a4-a3=(-1/3)^2

All of the above are added up:

a(n+1)-a1=[1-(-1/3)^n ] / [1-(-1/3)]

a(n+1)=(3/4)*[1-(-1/3)^n]+1=7/4 -(3/4)*(1/3)^n

So: an=(7 4)-(3 4)*(1 3) (n-1).

The meaning of the undetermined coefficient is that we set a polynomial (a sequence is an expression of a sequence, whether it is recursive or the first n term sum or relation).One or more of the middle coefficients are unknown.

Next, let's use the following property to show that polynomials are equal.

The sufficient and necessary conditions for the equality of two polynomials are the coefficients of each term and the number of an unknown number, which correspond to the same one-to-one!

Let's take a concrete example.

a(n+1)=7an + 5 --1)

This equation is much like the deformation of a proportional series, so let's set the pending coefficient, a(n+1)+x]=7[ an + x].

a(n+1) +x =7an +7x

a(n+1)=7an +6x---2)

1) and (2) are the same formula, then according to the correspondence of each term, there is 6x=5

x=5 6So[a(n+1)+ 5 6]=7[ an + 5 6].

Now we caught the proportional number series! It is a proportional series with a common ratio of 7 (of course, it is necessary to verify that a1+5 and 6 are not equal to 0).

All in all, it doesn't matter if it's a series of numbers, or somewhere else (a split or something.) The meaning of the coefficients to be determined is the same: the sufficient and necessary conditions for the equality of polynomials!