Fixed Point Method || Undetermined Coefficients

一阶线性递推数列 的 不动点法和待定系数法

two most common algebraic techniques for solving first-order linear non-homogeneous recurrence relations of the form: $a_{n+1}=p{a}_n+q(p\mathrm{\ne }1)$

🔵 1. Fixed Point Method

Core Idea:
Find the equilibrium value (fixed point) $x$ that the sequence would stay at forever if it ever reached it. By shifting the sequence relative to $x$, the constant term $q$ disappears, leaving a pure geometric progression.

Steps:

1. Solve for the fixed point: Set $a_{n+1}=a_n=x$ in the recurrence: $x=px+q\Rightarrow x=\frac{q}{1-p}$​
2. Shift the recurrence: Subtract $x$x from both sides: $a_{n+1}-x=p(a_n-x)$
3. Identify the geometric sequence: Let $b_n=a_n-x$. Then $b_{n+1}=p{b}_n$, so $\{b_n\}$ is geometric with ratio $p$.
4. Write the closed form: $b_n=b_1{p}^{n-1}=(a_1-x)p^{n-1}\Rightarrow \boxed{{\displaystyle a_n=x+(a_1-x)p^{n-1}}}$

Intuition: The fixed point is the “steady state” of the recurrence. Translating the sequence by $x$x centers it at zero, turning the affine map into a linear (scaling) map.

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🔴 2. Method of Undetermined Coefficients

Core Idea:
Assume the recurrence can be algebraically rewritten as a scaled shift: $a_{n+1}+\lambda =p(a_n+\lambda )$. Solve for the unknown constant $\lambda$λ that makes this identity hold.

Steps:

1. Assume the structure: $a_{n+1}+\lambda =p(a_n+\lambda )$
2. Expand and match coefficients: $a_{n+1}=p{a}_n+(p-1)\lambda$ Compare with the original $a_{n+1}=p{a}_n+q$: $(p-1)\lambda =q\Rightarrow \lambda =\frac{q}{p-1}$​
3. Substitute back: $a_{n+1}+\frac{q}{p-1}=p(a_n+\frac{q}{p-1})$
4. Define a new sequence: Let $c_n=a_n+\frac{q}{p-1}$. Then $c_{n+1}=p{c}_n$, so $\{c_n\}$is geometric.
5. Recover an $c_n=c_1{p}^{n-1}=(a_1+\frac{q}{p-1})p^{n-1}$ $\boxed{{\displaystyle a_n=(a_1+\frac{q}{p-1})p^{n-1}-\frac{q}{p-1}}}$​​

Intuition: This is a purely algebraic “guess-and-check” strategy. We postulate a form that eliminates the constant term, then solve for the parameter that makes the postulate valid.

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🔗 3. Relationship & Comparison

Aspect	Fixed Point Method	Undetermined Coefficients
Starting point	Solve $x=px+q$	Assume $a_{n+1}+\lambda =p(a_n+\lambda )$
Key parameter	$x={\displaystyle \frac{q}{1-p}}$​	$\lambda ={\displaystyle \frac{q}{p-1}}=-x$
Nature	Geometric / dynamical systems intuition	Algebraic / structural matching
Result	Identical closed form	Identical closed form
Generalization	Extends naturally to nonlinear recurrences $a_{n+1}=f(a_n)$	Extends to $a_{n+1}=p{a}_n+f(n)$by matching $f(n)$‘s form

✅ Bottom line: Both methods are algebraically equivalent. The fixed point method emphasizes why the shift works; the undetermined coefficients method emphasizes how to systematically find the shift.

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⚠️ 4. Edge Cases & Practical Tips

• p=1 Both methods fail (division by zero). The recurrence becomes $a_{n+1}=a_n+q$, which is an arithmetic sequence: $a_n=a_1+(n-1)q$.
• p=0: $a_{n+1}=q$. The sequence is constant from $n\ge 2$.
• Non-constant qq (e.g., qn=kn,rn,n2): Use the extended undetermined coefficients method. Assume a shift of the same functional form (e.g., $a_{n+1}+\alpha n+\beta =p(a_n+\alpha n+\beta )$for linear $q_n$​), then solve for $\alpha ,\beta$.
• Always verify: Plug $n=1,2$ into your closed form to check against the original recurrence.

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📝 Quick Reference Formula

For $a_{n+1}=p{a}_n+q$a ($p\mathrm{\ne }1$): $\boxed{{\displaystyle a_n=\frac{q}{1-p}+(a_1-\frac{q}{1-p})p^{n-1}}}$This compact form directly combines the steady-state value and the transient geometric decay/growth.

Let me know if you’d like a worked example in English, or an extension to second-order recurrences ($a_{n+2}=p{a}_{n+1}+q{a}_n+r$).

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训练题

1

已知$$a_{n+1}=\frac{a_n+6}{2},\quad a_1=10.$$an+1​=2an​+6​,a1​=10.

求 $a_n$an​ 的通项公式。

2

已知$$a_{n+1}=\frac{3a_n+4}{5},\quad a_1=1.$$an+1​=53an​+4​,a1​=1.

求通项公式，并求 $\lim_{n\to\infty} a_n$limn→∞​an​。

3

已知$$a_{n+1}=2a_n-3,\quad a_1=7.$$an+1​=2an​−3,a1​=7.

求通项公式。

4

已知$$a_{n+1}=\frac{a_n+9}{3},\quad a_1=0.$$an+1​=3an​+9​,a1​=0.

求通项公式。

5

已知$$a_{n+1}=\frac{1}{4}a_n+5,\quad a_1=4.$$an+1​=41​an​+5,a1​=4.

求通项公式，并判断是否收敛。

6

已知$$a_{n+1}=1-a_n,\quad a_1=2026.$$an+1​=1−an​,a1​=2026.

求 $a_n$an​ 的通项公式。

7

已知$$a_{n+1}=\frac{2a_n-1}{3},\quad a_1=5.$$an+1​=32an​−1​,a1​=5.

求通项公式。

8

已知$$a_{n+1}=\frac{5a_n+12}{7},\quad a_1=2.$$an+1​=75an​+12​,a1​=2.

求通项公式，并求极限。

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提升题

9

已知$$a_{n+1}=\frac{a_n+2}{2},\quad a_1=x.$$an+1​=2an​+2​,a1​=x.

用 $x$x 表示 $a_n$an​，并讨论当 $n\to\infty$n→∞ 时的极限。

10

已知$$a_{n+1}=ra_n+c,\quad r\neq 1.$$an+1​=ran​+c,r=1.

试写出通项公式，并说明为什么一定要先找“固定点”。

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这类题的统一做法

先令不动点 $L$L 满足$$L=rL+c,$$L=rL+c,

然后设$$b_n=a_n-L.$$bn​=an​−L.

递推会变成$$b_{n+1}=rb_n,$$bn+1​=rbn​,

于是$$b_n=b_1r^{n-1},$$bn​=b1​rn−1,

最后还原：$$a_n=L+(a_1-L)r^{n-1}.$$an​=L+(a1​−L)rn−1.

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标准答案

1

1. 求不动点（平衡值）
   令 $x={\displaystyle \frac{x+6}{2}}$x=2x+6​，解得 $x=6$x=6。
2. 构造等比数列
   将递推式改写为： $a_{n+1}-6={\displaystyle \frac{a_n+6}{2}}-6={\displaystyle \frac{a_n-6}{2}}$an+1​−6=2an​+6​−6=2an​−6​ 即 $\{a_n-6\}${an​−6} 是公比为 ${\displaystyle \frac{1}{2}}$21​ 的等比数列。
3. 求首项与通项
   首项：$a_1-6=10-6=4$a1​−6=10−6=4
   故：$a_n-6=4\times {\left({\displaystyle \frac{1}{2}}\right)}^{n-1}=2^2\times 2^{-(n-1)}=2^{3-n}$an​−6=4×(21​)n−1=22×2−(n−1)=23−n
4. 最终通项公式
   $\boxed{{\displaystyle a_n=6+2^{3-n}}}\text{或}{a}_n=6+\frac{4}{2^{n-1}}$an​=6+23−n​或an​=6+2n−14​

2

$$L=\frac{4}{1-3/5}=10.$$L=1−3/54​=10.

所以$$a_n=10+(1-10)\left(\frac35\right)^{n-1} =10-9\left(\frac35\right)^{n-1}.$$an​=10+(1−10)(53​)n−1=10−9(53​)n−1.

极限是$$\lim_{n\to\infty}a_n=10.$$n→∞lim​an​=10.

3

$$L=2L-3 \Rightarrow L=3.$$L=2L−3⇒L=3. $$a_n=3+(7-3)\cdot 2^{n-1}=3+4\cdot 2^{n-1}.$$an​=3+(7−3)⋅2n−1=3+4⋅2n−1.

4

$$L=\frac{9}{2}.$$L=29​. $$a_n=\frac92+\left(0-\frac92\right)\left(\frac13\right)^{n-1} =\frac92-\frac92\left(\frac13\right)^{n-1}.$$an​=29​+(0−29​)(31​)n−1=29​−29​(31​)n−1.

5

$$L=\frac{5}{1-1/4}=\frac{20}{3}.$$L=1−1/45​=320​. $$a_n=\frac{20}{3}+\left(4-\frac{20}{3}\right)\left(\frac14\right)^{n-1} =\frac{20}{3}-\frac{8}{3}\left(\frac14\right)^{n-1}.$$an​=320​+(4−320​)(41​)n−1=320​−38​(41​)n−1.

收敛，极限为 $\frac{20}{3}$320​。

6

$$L=1-L \Rightarrow L=\frac12.$$L=1−L⇒L=21​. $$a_n=\frac12+\left(2026-\frac12\right)(-1)^{n-1}.$$an​=21​+(2026−21​)(−1)n−1.

它不收敛，因为在两个值之间来回跳。

7

$$L=\frac{2L-1}{3}\Rightarrow 3L=2L-1\Rightarrow L=-1.$$L=32L−1​⇒3L=2L−1⇒L=−1. $$a_n=-1+(5+1)\left(\frac23\right)^{n-1} =-1+6\left(\frac23\right)^{n-1}.$$an​=−1+(5+1)(32​)n−1=−1+6(32​)n−1.

8

$$L=\frac{5L+12}{7}\Rightarrow 7L=5L+12\Rightarrow L=6.$$L=75L+12​⇒7L=5L+12⇒L=6. $$a_n=6+(2-6)\left(\frac57\right)^{n-1} =6-4\left(\frac57\right)^{n-1}.$$an​=6+(2−6)(75​)n−1=6−4(75​)n−1.

极限是 $6$6。

9

这里$$L=\frac{2}{1-1/2}=4.$$L=1−1/22​=4.

所以$$a_n=4+(x-4)\left(\frac12\right)^{n-1},$$an​=4+(x−4)(21​)n−1,

极限是 $4$4。

10

固定点$$L=rL+c \Rightarrow L=\frac{c}{1-r}.$$L=rL+c⇒L=1−rc​.

设 $b_n=a_n-L$bn​=an​−L，则$$b_{n+1}=r b_n,$$bn+1​=rbn​,

所以$$a_n=L+(a_1-L)r^{n-1}.$$an​=L+(a1​−L)rn−1.