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

Solve second-order recurrence relations using auxiliary roots, complementary functions and particular solutions (FP2 6.1, 6.2). You will handle all three root types, repeated-root overlap and a population model with a one-year delay.

Two previous terms give two arbitrary constants

Can a sequence have a closed form full of square roots and trigonometric functions, yet every term be an integer? The auxiliary equation will explain how. First, two simpler roots show why a second-order rule needs two initial values.

You already know

A complementary function solves the homogeneous equation; adding a particular solution solves the full equation. Auxiliary roots also organised second-order differential equations, but recurrence solutions use powers $r^n$ , rather than $\mathrm{e}^{rx}$ . Initial values must use their stated indices.

A second-order linear recurrence has the form

u_{n+2}+f(n)u_{n+1}+g(n)u_n=h(n)

and uses two preceding terms. For example, $u_{n+2}=3u_{n+1}-2u_n$ , with $u_0=1,u_1=4$ , gives $u_2=10$ . One starting value alone would leave $u_1$ , and hence later terms, undetermined.

For constant coefficients, try $u_n=r^n$ in the homogeneous equation. With a nonzero root,

r^{n+2}+pr^{n+1}+qr^n =r^n(r^2+pr+q)

so the bracket gives the auxiliary equation.

Distinct auxiliary roots

For

u_{n+2}+pu_{n+1}+qu_n=0

with constant real $p,q$ , $q\ne0$ , solve

r^2+pr+q=0

Distinct roots $r_1,r_2$ give

u_n=Ar_1^n+Br_2^n

Two initial values determine $A,B$ .

Not in the formula booklet: learn it.

If the coefficient of $u_{n+2}$ is not $1$ , divide through first or retain it in the auxiliary quadratic. These constant-coefficient forms do not automatically solve arbitrary changing coefficients.

Worked example 1: distinct real roots

Question. Solve $u_{n+2}-3u_{n+1}+2u_n=0$ , $u_0=1,u_1=4$ .

Plan. Factor the auxiliary quadratic, write both power terms and apply both initial conditions.

\begin{aligned} r^2-3r+2&=(r-1)(r-2)=0 &&\text{auxiliary equation}\\ u_n&=A+B2^n &&\text{one term per root}\\ A+B&=1,\qquad A+2B=4 &&\text{indices }0,1\\ B&=3,\qquad A=-2 &&\text{solve simultaneously} \end{aligned}

Why this step? The first equation alone fixes only a sum; the next initial term separates the two constants.

Answer. $u_n=3\cdot2^n-2$ , for $n\geq0$ .

Check. It gives $u_0=1,u_1=4,u_2=10$ , and $10-3(4)+2(1)=0$ .

Change only $u_1$ , leaving the roots and $u_0$ unchanged.

Worked example 2: use both initial values again

Question. Solve the same recurrence with $u_0=1,u_1=5$ .

Plan. Use the same two-constant family and the changed second equation.

\begin{aligned} u_n&=A+B2^n &&\text{same complementary function}\\ A+B&=1,\qquad A+2B=5 &&\text{new second initial value}\\ B&=4,\qquad A=-3 &&\text{subtract and solve}\\ u_n&=4\cdot2^n-3 &&\text{closed form} \end{aligned}

Why this step? Changing an initial value changes the member of the solution family, not the auxiliary equation.

Answer. $u_n=4\cdot2^n-3$ .

Check. $u_2=13=3(5)-2(1)$ , and both supplied values agree.

A zero root needs care at the starting index. For $u_{n+2}=2u_{n+1}$ , $u_0=5,u_1=3$ , simply use $u_n=3\cdot2^{n-1}$ for $n\geq1$ , with $u_0=5$ separately. This avoids an ambiguous $0^0$ ; the first value no longer influences later terms.

Watch out

The line $u_n=A\mathrm{e}^n+B\mathrm{e}^{2n}$ uses the differential-equation form. For recurrence roots $1,2$ , the correct line is $u_n=A1^n+B2^n=A+B2^n$ .

Change only the second initial term once more.

A repeated root needs an extra factor of n

Two identical power terms would merge: $Ar^n+Br^n=(A+B)r^n$ . A repeated root must still provide two independent sequences.

For $r\ne0$ , substitute $u_n=r^nv_n$ into $u_{n+2}-2r u_{n+1}+r^2u_n=0$ . Cancelling $r^{n+2}$ gives

v_{n+2}-2v_{n+1}+v_n=0

so consecutive differences of $v_n$ are constant. Thus $v_n=A+Bn$ .

Repeated auxiliary root

A repeated nonzero root $r$ gives

u_n=(A+Bn)r^n

The factor $n$ keeps the second solution independent.

Not in the formula booklet: learn it.

A negative repeated root is allowed; retain $(-a)^n$ with brackets. If the auxiliary equation is $r^2=0$ , the recurrence is $u_{n+2}=0$ : retain the two initial values separately, then all later terms vanish.

Worked example 3: two independent repeated-root terms

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

Plan. Use the repeated-root form before applying the conditions.

\begin{aligned} r^2-4r+4&=(r-2)^2=0 &&\text{repeated root}\\ u_n&=(A+Bn)2^n &&\text{independent terms}\\ A&=1,\qquad 2(A+B)=6 &&\text{initial values}\\ B&=2 &&\text{solve} \end{aligned}

Why this step? Repeating $2^n$ twice cannot fit two independently chosen initial values; $n2^n$ supplies the missing freedom.

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

Check. $u_2=20$ , and $20-4(6)+4(1)=0$ .

Change only the second initial value.

Worked example 4: the same repeated root

Question. Solve the same recurrence with $u_0=1,u_1=8$ .

Plan. Keep the repeated-root form and replace the second condition.

\begin{aligned} u_n&=(A+Bn)2^n &&\text{same root structure}\\ A&=1 &&\text{use }u_0\\ 2(1+B)&=8 &&\text{use }u_1\\ B&=3 &&\text{solve} \end{aligned}

Why this step? The value at $n=0$ eliminates the $Bn$ term, but the value at $n=1$ includes both terms and the multiplier $2$ .

Answer. $u_n=(1+3n)2^n$ .

Check. $u_2=28=4(8)-4(1)$ .

The same root can produce different growth rates within its family: the extra linear factor matters, even though the exponential base stays fixed.

Watch out

Writing $u_1=A+B$ omits the factor $2^1$ . The correct second condition is $u_1=2(A+B)$ .

Change just the second initial value again.

Complex roots can describe an integer sequence

For conjugate roots, use their modulus and argument. The real part of a recurrence root is not the base of an exponential envelope.

Conjugate auxiliary roots

If the roots are $\rho(\cos\theta\pm\mathrm{i}\sin\theta)$ , with $\rho>0$ , their real complementary function is

u_n=\rho^n\bigl(A\cos n\theta+B\sin n\theta\bigr)

Here $A,B$ are real for a real-valued sequence. This follows by combining the conjugate powers using De Moivre's theorem, which is supplied; this recurrence solution form is not supplied.

Not in the formula booklet: learn it.

For roots $1\pm\mathrm{i}$ , the modulus is $\sqrt2$ and the angles are $\pm\pi/4$ . Their powers combine into a real expression.

Worked example 5: real values from complex roots

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

Plan. Write the roots in polar form, then use the two initial values in the real trigonometric form.

\begin{aligned} (r-1)^2&=-1 &&\text{auxiliary equation}\\ r&=1\pm\mathrm{i}=\sqrt2\,(\cos(\pi/4)\pm\mathrm{i}\sin(\pi/4)) &&\text{polar roots}\\ u_n&=(\sqrt2)^n\bigl(A\cos(n\pi/4)+B\sin(n\pi/4)\bigr) &&\text{real CF}\\ A&=1,\qquad A+B=0 &&\text{use }u_0,u_1\\ B&=-1 &&\text{solve} \end{aligned}

Why this step? At $n=1$ , $\sqrt2\cos(\pi/4)=\sqrt2\sin(\pi/4)=1$ , so the second condition is especially simple.

Answer. $u_n=(\sqrt2)^n\bigl(\cos(n\pi/4)-\sin(n\pi/4)\bigr)$ .

Check. It gives $u_2=-2$ , agreeing with $2(0)-2(1)$ .

Change only $u_1$ , keeping the roots and $u_0$ unchanged.

Worked example 6: another real combination

Question. Solve the same recurrence with $u_0=1,u_1=2$ .

Plan. Keep the polar-root form and apply the new second initial value.

\begin{aligned} u_n&=(\sqrt2)^n\bigl(A\cos(n\pi/4)+B\sin(n\pi/4)\bigr) &&\text{same CF}\\ A&=1,\qquad A+B=2 &&\text{initial values}\\ B&=1 &&\text{solve} \end{aligned}

Why this step? The two conditions select a different sine coefficient, while the modulus and angular step stay fixed.

Answer. $u_n=(\sqrt2)^n\bigl(\cos(n\pi/4)+\sin(n\pi/4)\bigr)$ .

Check. $u_2=2$ , and $2-2(2)+2(1)=0$ .

The first sequence begins $1,0,-2,-4,-4,0,8,16,\ldots$ . Every term is an integer: its initial values and recurrence coefficients are integers. The square roots and trigonometric values in its closed form cancel exactly.

[DIAGRAM: asset_name: Integer terms from conjugate auxiliary roots; asset_slug: conjugate-root-integer-sequence; caption: A trigonometric closed form can produce integer terms at every integer index.; description: Discrete unjoined blue points for the first conjugate-root example from n0 to12 show the values1,0,minus2,minus4,minus4,0,8,16,16,0,minus32,minus64,minus64. Pale sage dashed upper and lower envelopes have heights plus and minus root2 to the power n+1. The expression and roots1 plus or minus i are labelled in ink; axes have integer indices.]
A trigonometric closed form can produce integer terms at every integer index.

The envelope in the diagram has modulus base $\sqrt2$ , rather than the root's real part $1$ .

Watch out

For roots $1\pm\mathrm{i}$ , $u_n=\mathrm{e}^n(A\cos n+B\sin n)$ is the wrong solution type. Use the root modulus $\sqrt2$ and argument $\pi/4$ in the recurrence form.

Change only the second initial term once more.

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FP2 6.1, 6.2 - Second-order recurrence relations

Two previous terms give two arbitrary constants

A repeated root needs an extra factor of n

Complex roots can describe an integer sequence

Continue this lesson.

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