Xs7 - Second order homogeneous recurrences | OCR A-Level Further Mathematics B (MEI)

Solve second order homogeneous recurrences using an auxiliary equation. Handle distinct, repeated and complex roots, and use two sequence values to determine the constants.

Two roots, two geometric sequences

The sequence begins $3,0,-6,-18,\ldots$ , with each term equal to three times the previous term minus twice the term before it. Can two geometric sequences be hiding inside that list?

They can. A second order linear recurrence can combine two geometric patterns even when the combined sequence has no fixed ratio.

You already know

First order homogeneous recurrences have solutions $Cr^n$ . In Core Pure, auxiliary equations identified the solution families of second order differential equations. Here the variable is an integer index and the basic solution is $r^n$ , rather than $e^{rx}$ .

Work with the recurrence as given. For $Au_{n+2}+Bu_{n+1}+Cu_n=0$ , substitute $u_n=r^n$ to obtain

r^n(Ar^2+Br+C)=0.

The quadratic factor is the auxiliary equation. This also explains why the coefficients stay in their original order.

Distinct real roots

For a genuine second order recurrence with constant coefficients, if its auxiliary equation has distinct non-zero real roots $r_1,r_2$ , the general solution is

u_n=Ar_1^n+Br_2^n.

The letters $A,B$ here are arbitrary constants. Use two independent sequence conditions to select a particular solution.

Not in the formula booklet: learn it.

A zero auxiliary root needs care at the starting index: the $0^n$ component represents an independently chosen initial term followed by zeros. It is safer to obtain that starting term directly than to write an undefined $0^0$ . The examples below have non-zero roots.

Worked example 1: uncovering two powers

Question. Solve $u_{n+2}-3u_{n+1}+2u_n=0$ , for $n\ge0$ , with $u_0=3$ , $u_1=0$ .

Plan. Find the two auxiliary roots and apply both conditions to their sum.

\begin{aligned} r^2-3r+2&=(r-1)(r-2)=0 && \text{auxiliary equation}\\ u_n&=A+B2^n && \text{general family}\\ A+B&=3,\quad A+2B=0 && \text{conditions at }0,1\\ B&=-3,\quad A=6 && \text{subtract and solve} \end{aligned}

Why this step? Each starting value constrains the complete sum; neither constant is determined by a condition in isolation.

Answer. $u_n=6-3\,2^n$ .

Check. It gives $u_0=3$ , $u_1=0$ , $u_2=-6$ . The two roots make the recurrence residual zero separately, so their sum does too.

Change only the middle coefficient from $-3$ to $-1$ and the final coefficient from $2$ to $-2$ , so the positive root $1$ is replaced by the negative root $-1$ . Keep the same initial values.

Worked example 2: one root is negative

Question. Solve $v_{n+2}-v_{n+1}-2v_n=0$ , $v_0=3$ , $v_1=0$ .

Plan. Preserve the negative root inside the powered base.

\begin{aligned} r^2-r-2&=(r-2)(r+1)=0 && \text{new auxiliary equation}\\ v_n&=A2^n+B(-1)^n && \text{general family}\\ A+B&=3,\quad2A-B=0 && \text{same initial values}\\ 3A&=3\implies A=1,\quad B=2 && \text{add equations} \end{aligned}

Why this step? The term $B(-1)^n$ alternates. Writing $-B$ would remove that root's behaviour.

Answer. $v_n=2^n+2(-1)^n$ .

Check. $v_2=6$ , agreeing with $v_2=v_1+2v_0=6$ . Also $v_1=2-2=0$ .

The deliberate change is the sign of one root. It changes the oscillating component while retaining the geometric component $2^n$ .

Watch out

Wrong for root $-1$ : use a constant term $-B$ . Right: use $B(-1)^n$ . The sign changes with the index.

Retain the negative powered base in this new pair of roots.

A repeated root needs a new factor

Writing $Ar^n+Br^n$ gives only one effective constant. A repeated root needs a second independent shape, $nr^n$ .

Repeated non-zero root

If the auxiliary equation has a repeated non-zero root $r$ , then

u_n=(A+Bn)r^n.

Both $r^n$ and $nr^n$ satisfy the recurrence. Two conditions determine $A$ and $B$ .

Not in the formula booklet: learn it.

For example, substitution into $u_{n+2}-2ru_{n+1}+r^2u_n$ gives the factor

(n+2)-2(n+1)+n=0

for the $nr^n$ component. The second solution is therefore a consequence of the recurrence, not a guess based only on resemblance to differential equations.

Worked example 3: a repeated positive root

Question. Solve $u_{n+2}-4u_{n+1}+4u_n=0$ , $u_0=1$ , $u_1=6$ .

Plan. Identify the squared factor and use the repeated-root family.

\begin{aligned} r^2-4r+4&=(r-2)^2=0 && \text{repeated root}\\ u_n&=(A+Bn)2^n && \text{general solution}\\ A&=1,\quad2(A+B)=6 && \text{initial conditions}\\ B&=2 && \text{solve} \end{aligned}

Why this step? The coefficient at $n=1$ is $A+B$ , but the whole expression is also multiplied by $2$ .

Answer. $u_n=(1+2n)2^n$ .

Check. $u_2=5(4)=20$ , while the recurrence gives $4(6)-4(1)=20$ .

Change only the root's sign, from $2$ to $-2$ , by changing the middle coefficient. Keep both starting values.

Worked example 4: a repeated negative root

Question. Solve $v_{n+2}+4v_{n+1}+4v_n=0$ , $v_0=1$ , $v_1=6$ .

Plan. Use $(A+Bn)(-2)^n$ , including its sign when $n=1$ .

\begin{aligned} r^2+4r+4&=(r+2)^2=0 && \text{repeated root }-2\\ v_n&=(A+Bn)(-2)^n && \text{general solution}\\ A&=1,\quad-2(A+B)=6 && \text{initial conditions}\\ A+B&=-3\implies B=-4 && \text{solve} \end{aligned}

Why this step? The same positive value $v_1=6$ now requires a negative linear factor at $n=1$ , because the powered base contributes $-2$ .

Answer. $v_n=(1-4n)(-2)^n$ .

Check. $v_2=-7(4)=-28$ , agreeing with $v_2=-4(6)-4(1)$ .

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Xs7 - Second order homogeneous recurrences

Two roots, two geometric sequences

A repeated root needs a new factor

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