What is a difference equation?

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Difference equation is an equation composed of multiple equations, unknown functions and their derivatives.

difference equation

Specific instructions:

meaning

Difference equation is discretization of differential equation. A differential equation may not get an exact solution. If we turn it into a difference equation, we can get an approximate solution.

For example, dy+y*dx=0, y(0)= 1 is a differential equation, and x takes the value [0, 1].

(Note: the solution is y (x) = e (-x));

By discretizing the differential equation, the interval of X can be divided into many cells [0, 1/n], [1/n, 2/n], ... [(n- 1)/n, 1].

In this way, the above differential equation can be discretized as:

y((k+ 1)/n)-y(k/n)+y(k/n)*( 1/n)= 0，k = 0， 1，2，...，n- 1。

Using the condition of y(0)= 1 and the difference equation above, the approximate value of y(k/n) can be calculated.

The basic theory of 1

difference equation

1. Difference

2. An arbitrary sequence {xn} is defined as difference operator δ as follows:

δxn = xn+ 1-xn

For the new series of application of difference operator, there are

δ2xn =δ(δkxn)。

nature

Attribute 1δ k (xn+yn) = δ kxn+δ kyn.

Property 2 δ k (cxn) = cδ kxn

Property 3 δ kxn = ∑ (- 1) jcjkxn+k-j

Infinitely differentiable function with general term of sequence n with property 4 for any k >; = 1, η exists, η kxn = f (k) (η) exists.

difference equation

Definition 8. On the k-order difference equation of 1 equation of sequence;

xn-a 1xn- 1-a2xn-2-……aBxn-k = b(n = k，k+ 1，…)

Where a 1, a2, -AK is a constant, and ak≠0. If b=0, the equation is homogeneous.

On the Algebraic Equation of λ

λk-a 1λk- 1-AK- 1λ-AK = 0

Is the corresponding characteristic equation, and the root is the eigenvalue.

Edit this example

1. Experimental contents and exercises

2. 1 variance

Example 1 Xn={n3}, find the difference sequence of each order:

xn △xn △2xn △3xn △4xn

1 7 12 6 0

8 19 18 6 0

27 37 24 6 0

64 6 1 30 6

125 9 1 36

2 16 127

343

It can be seen that {n3}, the third-order difference sequence is a constant sequence, and the fourth-order difference sequence is 0.

Exercise 1 Find the difference sequence of each order for {1}, {n}, {n2}, {n4} and {n5} respectively.

Exercise 2 {c0n-1} {c1n-1} {c2n-1} to find the difference sequence of each order.

The cubic function whose general term of {Xn} is n,

Xn=a3n3+a2n2+a 1n+a0

It is proved that it is a constant sequence.

It is proved that Xn=a3n3+a2n2+a 1n+a0 can be directly calculated.

Theorem 8. 1 If the general term of the sequence is a polynomial of degree k about n, then the difference sequence of order k is non-zero and the difference sequence of order k+ 1 is 0.

Exercise 3 Prove Theorem 8. 1。

Theorem 8. 2 If the k-order interpolation of {Xn} is divided into non-zero constant series, then {Xn} is a k-order polynomial of n,

Exercise 4 Prove Theorem 8 according to the property of difference. 2

Example 2. Find σi3

Example 4

Solving Sn=∑i3 table

Sn △Sn △2Sn △3Sn △4Sn △5Sn

1 8 19 18 6 0

9 27 37 24 6 0

36 64 6 1 30 6 0

100 125 9 1 36 6 0

225 2 16 127 42

44 1 343 169

784 5 12

1296

Let sn = a4n4+a3n3+a2n2+a1n+A0, S 1 = 1, S2 = 9, S3 = 36, S4 = 100, S5 = 225.

a0=0，a 1=0，a2= 1/4，a3= 1/2，a4= 1/4。

Therefore, sn = (1/4) n4+(1/2) n3+(1/4) N2.

The general term Xn of exercise {Xn} is a polynomial of degree k of n, and it is proved that ∑xi is a polynomial of degree k+/kloc-0 of n; Find σi4.

From Exercise 2 {Crn- 1}.

2.2 difference equation

For a difference equation, if we can find such a general term of the sequence and bring it into the difference equation, the equation will become an identity, and this general term is called the solution of the difference equation.

Example 3 For the difference equation xn-5xn- 1+6xn-2=0, it can be directly proved that xn=c 13n+c22n is the solution of this equation.

The solution in Example 3 contains arbitrary constants, and the number of arbitrary constants is the same as the order of the difference equation. Such a solution is called the general solution of the difference equation.

If the k-order difference equation gives the value of the first k term in the sequence, we can determine any constant of the general solution and get the difference.

The special solution of.

Example 4 The difference equation xn-5xn- 1+6xn-2=0. If x 1 = 1 and x2 = 5 are known, the special solution of the difference equation can be obtained as xn=3n-2n.

Firstly, we study the solutions of homogeneous linear difference equations.

xn=rxn- 1

For the first order difference equation

x 1=a

Obviously xn=arn- 1. Therefore, if the sequence satisfies the first-order difference equation, it is a geometric series.

Example 5 Find the general term {Fn} of Fibonacci sequence, where f 1 = 1, F2 = 1, fn = fn- 1+fn-2.

The first term of Fibonacci sequence is: 1, 1, 2, 3, 5, 8, 13, 2 1, 34, 55, 89, ... This series has a very wide range of applications.

The difference equation satisfied by Fibonacci sequence is Fn-Fn- 1-Fn-2=0,

Its characteristic equation is λ2-λ- 1=0.

Its roots are λ 1=, λ2=. Using λ 1λ2, the difference equation can be written as

Fn-(λ 1+λ2)Fn- 1+λ 1λ2Fn-2 = 0，

Namely fn-λ1fn-1= λ 2 (fn-1-λ1fn-2).

The sequence {Fn-λ 1Fn- 1} satisfies the first-order difference equation. Obviously ()

The same reason ()

A general term that can be solved by the above two formulas.

Exercise 9 proves that if the sequence {} satisfies the second-order difference equation and its characteristic equation consists of two unequal roots, it is two special solutions of the difference equation. So its general solution is.

It can be seen from Exercise 9 that if the characteristic equation of the second-order difference equation has two unequal roots, the general formula of its general solution can be written. Then we can work out the coefficient from the value of and write the special solution of the difference equation.

Exercise 10 specifically find the general term of Fibonacci sequence and prove it. Then, if the second-order linear homogeneous difference equation has two equal roots, how to find its solution?

Let the characteristic equation of the second-order linear homogeneous difference equation have two equal roots, then the difference equation can be written as. When both sides of the difference equation are divided at the same time, there are. Set, and then (n & gt=3). Because this formula holds when n> equation =3, we rewrite it as (n >；; = 1)。 (8.2)

Equation (8.2) has a second-order difference on the left, so it is a linear function of n. If it is set, it is. The upper part is the general solution of the difference equation.

Exercise 1 1 proves that if the characteristic equation of the third-order difference equation satisfied by the sequence {0} consists of three equal roots, the general solution of the difference equation is.

Generally speaking, let … be all the different solutions of the characteristic equation of the difference equation, and their multiplicity is

For the general k-order homogeneous linear difference equation, we can get k special solutions of the above form through its characteristic equation, and then get the general solution of the difference equation.

Exercise 12 If the series {} satisfies the difference equation.

And find the general term of {}.

Example 6 If the root of the difference equation with real coefficients is an imaginary number, its solution is also represented by an imaginary number, which brings inconvenience to the discussion. difference equation

xn-2xn- 1+4xn-2=0

The eigenvalue of is i. If x 1 = 1 and x2 = 3, the special solution can be easily obtained by the following program:

xn =()( 1+I)n+(-)( 1-I)n

Clear[x 1, x2, c 1, c2, l 1, l2, solution];

x 1 = 1； x2 = 3；

solution=solve[ 1^2-2l+4==0, 1]；

l 1=l/。 Solution [[ 1,1]];

l2=l/。 Solution [[2,1]];

c=solve[{c 1*l 1+c2*l2==x 1,c 1*l 1^2+c2*l2^2==x2},{c 1,c2}]；

c 1 = Simplify[Re[c 1]]+Simplify]* I；

C2 = Simplify[Re[C2]]+Simplify]* I；

print["xn=(",c 1,")(",l 1,")^n+(",c2,")(",l2,")^n "]

The form of the solution is quite complicated. Can they be represented by real numbers?

If =rei, then =re, we can rewrite the expression in (8.4) as follows.

xn=re (2e )n+re (2e )\n

=2r Cos()

=(2rCos)

As you can see, this general term can be written as. So, is sum a special solution of difference equation?

Exercise 13 Verify that sum is a special solution of difference equation (8.3).

For the difference equation (8.3), we find two special solutions of its real form, so that the general solution can be expressed in the form of real numbers. This method is also true for general equations.

The two eigenvalues set in exercise 14 are. It is proved that the general solution of difference equation can be expressed as.

Exercise 15 express the special solution of the difference equation with real numbers.

Last time we discussed the solution of the second order linear difference equation. So, can the non-homogeneous linear difference equation be transformed into homogeneous linear difference equation?

Exercise 16 If the nonhomogeneous linear difference equation is known,

(8.5)

A special solution of is verification: if it is, it satisfies the homogeneous difference equation.

From the exercise 16, we can know that if a special solution of the non-homogeneous linear difference equation (8.5) is known, it can be transformed into a homogeneous linear difference equation.

Obviously, the simplest form of equation (8.5) is (where p is a constant), which is obtained by substituting (8.5).

If ... yes.

Let p = be the equilibrium value of non-homogeneous linear difference equation (8.5). In (8.5), the order has

By, by

In this way, the original non-homogeneous linear difference equation can be transformed into homogeneous linear difference equation.

If the equilibrium value of equation (8.5) does not exist, all n in equation (8.5) can be replaced by n+ 1, and we can get

(8.6)

Equations (8.6) and (8.5) are subtracted.

Therefore, the original non-homogeneous linear difference equation can be transformed into a higher-order homogeneous linear difference equation.

Exercise 17 finds the general solutions of the sum of difference equations respectively.

2.3 Algebraic Equation for Roots

From the property of Fibonacci sequence, we can use it to approximate, and we can use this property to calculate the approximation of. Generally speaking, a>0 can be obtained by constructing the difference equation.

For a given positive number A, let λ 1= and λ2=, then λ 1, λ2 is the root of the equation λ2-2λ+( 1+a)=0. This equation is the characteristic equation of difference equation. Therefore, the selection can construct the sequence {} by using the difference equation.

Exercise 18 proves that if a> 1, for any >; 0,>0, if ≦, then the sequence {} constructed by the above method satisfies.

In this way, we get a calculation method:

1. Given (as error control), take any initial value to make n =1;

2. If

Then the calculation is terminated and the result is output; Otherwise, let n: = n+ 1, and go to step 3;

3. Order, go to step 2.

Exercise 19 for A = 1.5, 10, 12345, use the above method.

The convergence speed of the above method is not fast enough, so we can improve it.

Let the integer u satisfy, so, then it is the two roots of the equation.

Exercise 20 According to the component sequence {x} of the difference equation above, make

Exercise 2 1 A in exercise 19 is calculated by the above method, and the convergence rates of the two methods are compared.

algebraic equation

(8.7)

Is the characteristic equation of the difference equation (8. 1). Can this difference equation be used to solve equation (8.7)?

Let equation (8.7) have k different roots.

, (8.8)

Then the general solution form of the corresponding difference equation is

Exercise 22 Let the root of equation (8.7) satisfy the condition (8.8), and use the difference equation (8. 1) (take b=0) to construct the sequence {}. If the coefficient in the general solution is ≠0, it is proved that:

Using the conclusion obtained in Exercise 22, we can find the root with the largest absolute value of polynomial equation.

Exercise 23 Find the root with the largest absolute value of the equation.

In fact, if the mutually different roots of equation (8.7) satisfy

≥ ≥…≥

The conclusion in Exercise 22 is still valid.

2.4 National income 4 National income stability

A country's national income can be used for consumption, investment in reproduction and so on. Generally speaking, neither consumption nor reproduction investment should be unlimited. Reasonable control of local investment can make the national economy virtuous circle. How to match the investment proportion of each part in order to keep the national economy stable? This is the problem to be discussed in this section.

We first give some assumptions:

1. National income is used for consumption, reproduction investment and public facilities construction.

2. Record the national income level and consumption level in the k-th cycle respectively. The value is in direct proportion to the national income of the previous cycle. That is =A, (8.9) where a is a constant (0 3. Used to indicate the level of investment for reproduction in the k-th cycle, which depends on the change of consumption level, that is, (8. 10).

4.G represents the government's expenditure on public facilities, and let g be a constant. Assume that 1 has. (8. 1 1), the above formula is a difference equation. When a given value is given, the national income level (k=2, 3, ...) can be directly calculated to see if it is stable.

If Example 7 holds, the data in Table 8.3 can be obtained by calculation.

Table 8.3 Changes in values

k 2 3 4 5 6 7 8 9 10 1 1

1 1.0 24.5 35.8 39. 1 32.9 20.3 7.48 0.95 3.93 15.0

k 12 13 14 15 16 17 18 19 20 2 1

28.5 37.8 38.2 29.5 16.0 4.58 0.82 6.65 19.2 32. 1

We can draw a scatter diagram to observe its changes. The calculation and drawing procedures are as follows:

y0 = 2； y 1 = 2； a = 0.5b = 2； g = 10；

y={y0，y 1 }；

For [k= 1, k < = 20, k++,

y2 = a( 1+b)* y 1-b * a * y0+g；

Y=Append[y，y2]；

Y0=y 1，y 1=y2]

YListPlot[y，PlotJoined True，

PlotStyle thickness [0.0 12]]

Figure 8. 1 Changes in national income

From Figure 8. 1, it is found that the data of Example 7 shows signs of periodic changes.

Exercise 24 assumes that for parameters A and B in Table 8.4, (k=2, 3, …) is calculated and the observed changes are plotted.

Table 8.4 Values of Parameters A and B

a 1/2 1/2 1/2 8/9 9/ 10 3/4 4/5

B 1 2 3 1/2 1/2 3 3

It can be seen that the stability of national income level (k=2, 3, …) presents different states with different parameter values.

Then, when the parameters meet what conditions, the national income level is in a stable development state?

The difference equation (8. 1 1) is a non-homogeneous linear difference equation with constant coefficients. It is easy to find its equilibrium value from A< 1

Make it available

Its characteristic value is

Jose

Where is the amplitude.

So the solution of the difference equation is

Where is a constant.

If it is easy to see that {} is a periodic function with a value of-,then {} is in a state of periodic change. As shown in example 7.

Exercise 25 discusses the changing trend of {} in the case of and. Will national income develop steadily?

Exercise 26 If so, under what conditions will the national income develop steadily?

Mathematica software statement description involved in this experiment

1.solution=solve[ 1^2-2l+4==0, 1]；

l 1= 1/。 Solution [[ 1,1]];