FP2 6.1, 6.2 - First-order recurrence relations | Edexcel A-Level Further Mathematics

Turn a rule for the next term into a formula for any term (FP2 6.1, 6.2). You will combine complementary functions and particular solutions, handle changing coefficients, and interpret a population model.

From a next-term rule to a closed form

A population model keeps $80\%$ of its population each year and adds $40$ newcomers. Starting from $100$ , can it ever reach $200$ ? Generating terms gives clues; a formula for the term after any number of years will settle it.

You already know

An index labels a term, so $u_{n+1}$ means the term after $u_n$ , whereas $u_n+1$ means add one to its value. Powers satisfy $a^{n+1}=a\,a^n$ . We will state the starting index each time.

For instance, $u_{n+1}=2u_n+3,\ u_1=4$ gives $4,11,25,53,\ldots$ . This is a rule for building the sequence, rather than a formula that jumps straight to its hundredth term.

First-order recurrence relation

A first-order recurrence determines a new term using the preceding term. The linear form

u_{n+1}+f(n)u_n=g(n)

has specified functions $f,g$ and one initial value. A closed form expresses $u_n$ directly in terms of $n$ , without unknown earlier terms. For example, $u_n=7\cdot2^{n-1}-3$ is a closed form for the sequence above.

For $u_{n+1}-a u_n=b$ , split the work. A particular solution, $p_n$ , satisfies the full equation. A complementary function, $h_n$ , satisfies the associated homogeneous equation $h_{n+1}-a h_n=0$ . Their sum satisfies the full equation because the homogeneous contribution cancels.

Constant coefficient and constant forcing

For $u_{n+1}=a u_n+b$ , starting at $n=n_0$ , with $a\ne0,1$ , set

L=\frac{b}{1-a},\qquad u_n=L+C a^{n-n_0}

Here $L$ is a constant particular solution and $C=u_{n_0}-L$ . For $a=1$ , use

u_n=u_{n_0}+b(n-n_0)

For $a=0$ , every term after the initial term equals $b$ ; keep the supplied initial term separately.

Not in the formula booklet: learn it.

Subtracting $L=aL+b$ from the recurrence gives $u_{n+1}-L=a(u_n-L)$ . Each step multiplies the displacement from $L$ by $a$ , which explains the power.

Worked example 1: use the given starting index

Question. Determine a closed form for $u_{n+1}-2u_n=3$ , $u_1=4$ , for $n\geq1$ .

Plan. Find a constant particular solution, add the complementary function, then use the value at index $1$ .

\begin{aligned} p-2p&=3 &&\text{constant trial}\\ p&=-3 &&\text{particular solution}\\ h_n&=A2^{n-1} &&\text{homogeneous solution}\\ 4&=A-3 &&\text{use }u_1\\ A&=7 &&\text{determine the constant} \end{aligned}

Why this step? Using $2^{n-1}$ makes the exponential equal to $1$ at the supplied starting index.

Answer. $u_n=7\cdot2^{n-1}-3$ , for $n\geq1$ .

Check. The formula gives $u_1=4$ , $u_2=11$ , and $11-2(4)=3$ .

Change only the initial value; the recurrence and its particular solution stay the same.

Worked example 2: another member of the same family

Question. Determine the closed form when $u_{n+1}-2u_n=3$ , $u_1=6$ .

Plan. Keep the complementary function and particular solution, then apply the new initial value.

\begin{aligned} u_n&=A2^{n-1}-3 &&\text{same general solution}\\ 6&=A-3 &&\text{new initial condition}\\ A&=9 &&\text{solve}\\ u_n&=9\cdot2^{n-1}-3 &&\text{closed form} \end{aligned}

Why this step? Initial conditions select the constant; they do not change the homogeneous multiplier.

Answer. $u_n=9\cdot2^{n-1}-3$ , for $n\geq1$ .

Check. $u_1=6$ and $u_2=15=2(6)+3$ .

The constant $L$ is also called a fixed point: a sequence starting exactly there stays there. A fixed point need not attract nearby sequences; here errors from $-3$ double.

Watch out

Writing $u_n=9\cdot2^n-3$ for the second example gives $u_1=15$ , not $6$ . Either use $2^{n-1}$ , or use $A2^n$ and solve $2A-3=6$ . Keep the index and constant together.

Change just the starting value again.

Choose a particular solution for a polynomial input

The right side need not be constant. For $u_{n+1}-a u_n=P(n)$ , try a polynomial of the same degree when $a\ne1$ , and compare coefficients after replacing every $n$ by $n+1$ .

Polynomial forcing and the unit multiplier

For constant $a\ne1$ , a polynomial right side of degree $d$ admits a polynomial particular solution of degree at most $d$ . Include all lower powers in the trial.

When $a=1$ , the left side is the difference $p_{n+1}-p_n$ : differencing lowers a nonconstant polynomial's degree by one, so a degree- $d$ input needs a degree- $d+1$ trial. Its arbitrary constant is absorbed into the complementary function.

Not in the formula booklet: learn it.

For example, $p_n=qn+r$ gives $p_{n+1}-p_n=q$ . This is why a constant input with multiplier $1$ needs a linear particular solution.

Worked example 3: a linear input

Question. Solve $u_{n+1}-2u_n=n$ , $u_0=0$ , for $n\geq0$ .

Plan. Try $p_n=an+b$ , compare coefficients and then add $C2^n$ .

\begin{aligned} p_{n+1}-2p_n &=a(n+1)+b-2(an+b) &&\text{shift the entire trial}\\ &=-an+(a-b) &&\text{collect powers of }n\\ -a&=1,\qquad a-b=0 &&\text{compare coefficients}\\ a&=-1,\qquad b=-1 &&\text{particular constants}\\ u_n&=C2^n-n-1 &&\text{add the complementary function}\\ 0&=C-1 &&\text{use }u_0 \end{aligned}

Why this step? A trial $an$ alone cannot cancel the constant produced when $n$ becomes $n+1$ .

Answer. $u_n=2^n-n-1$ , for $n\geq0$ .

Check. The first values are $0,0,1,4$ ; the next-term rule gives $4=2(1)+2$ .

Double only the right side, keeping the coefficient and initial value unchanged.

Worked example 4: scale the input

Question. Solve $u_{n+1}-2u_n=2n$ , $u_0=0$ .

Plan. Use a complete linear trial again and apply the same initial index.

\begin{aligned} -an+(a-b)&=2n &&\text{substitute the linear trial}\\ a&=-2,\qquad b=-2 &&\text{compare coefficients}\\ u_n&=C2^n-2n-2 &&\text{CF plus particular solution}\\ 0&=C-2 &&\text{initial value}\\ C&=2 &&\text{select the solution} \end{aligned}

Why this step? The particular solution doubles, and the initial condition then changes the complementary constant too.

Answer. $u_n=2^{n+1}-2n-2$ .

Check. $u_2=2$ , $u_3=8$ , and $8-2(2)=4=2(2)$ .

For a quadratic input, start with $an^2+bn+c$ ; solve its coefficients before applying the initial value.

Watch out

The wrong line $p_{n+1}=an+b+1$ shifts the value rather than the index. The correct line is $p_{n+1}=a(n+1)+b$ . Substitute into every occurrence of the index before expanding.

Scale the input once more.

When an exponential trial meets the complementary function

For an exponential input $K b^n$ , a multiple of $b^n$ is the natural trial. Check first whether it already solves the homogeneous equation.

Exponential input

For $u_{n+1}-a u_n=K b^n$ , with $a,b\ne0$ , try

p_n=Db^n

If $b\ne a$ , substitution gives $D(b-a)=K$ .

If $b=a$ , the trial cancels identically. Instead use

p_n=Dn a^n,\qquad Da=K

The extra factor $n$ removes the overlap with the complementary function. This overlap is often called resonance.

Not in the formula booklet: learn it.

The rule comes from substituting into the recurrence, rather than from dividing by a coefficient that might be zero.

Worked example 5: distinct input and homogeneous multipliers

Question. Solve $u_{n+1}-2u_n=3^n$ , $u_0=2$ .

Plan. Try $D3^n$ , then add the complementary function and use the initial condition.

\begin{aligned} D3^{n+1}-2D3^n&=3^n &&\text{particular trial}\\ D(3-2)&=1 &&\text{divide by }3^n\\ D&=1 &&\text{particular coefficient}\\ u_n&=C2^n+3^n &&\text{general solution}\\ 2&=C+1 &&\text{use }u_0\\ C&=1 &&\text{solve} \end{aligned}

Why this step? $3^n$ is independent of the homogeneous term $2^n$ , so the simple exponential trial can work.

Answer. $u_n=2^n+3^n$ .

Check. $u_1=5$ , and $5-2(2)=1=3^0$ .

Change only the input base from $3$ to $2$ . It now matches the homogeneous multiplier.

Worked example 6: repair the overlapping trial

Question. Solve $u_{n+1}-2u_n=2^n$ , $u_0=2$ .

Plan. Replace the cancelled trial $D2^n$ by $Dn2^n$ .

\begin{aligned} D(n+1)2^{n+1}-2Dn2^n&=2^n &&\text{modified trial}\\ 2D\,2^n&=2^n &&\text{cancel the terms in }n\\ D&=\frac12 &&\text{match coefficients}\\ u_n&=C2^n+\frac n2\,2^n &&\text{add the CF}\\ C&=2 &&\text{use }n=0 \end{aligned}

Why this step? Multiplication by $n$ creates a nonzero difference where a plain multiple of $2^n$ would give zero.

Answer. $u_n=(2+n/2)2^n$ .

Check. $u_1=5$ , $u_2=12$ , and $12-2(5)=2=2^1$ .

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FP2 6.1, 6.2 - First-order recurrence relations

From a next-term rule to a closed form

Choose a particular solution for a polynomial input

When an exponential trial meets the complementary function

Continue this lesson.

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