Second-order recurrence relationsEdexcel A-Level Further Maths: Revision notes

A second-order linear recurrence relation links each term to the two before it: $p\,u_{n+2}+q\,u_{n+1}+r\,u_n=f(n),$ where $p$, $q$ and $r$ are constants. It is homogeneous if the right-hand side is $0$ and non-homogeneous if it is a non-zero constant. Two initial conditions, such as $u_1$ and $u_2$, are needed to fix the sequence, because the solution has two arbitrary constants.

For the homogeneous relation try $u_n=m^n$. Dividing by $m^n$ gives the auxiliary equation $p\,m^2+q\,m+r=0$.

• Two distinct real roots $m_1,m_2$: complementary function $A\,m_1^n+B\,m_2^n$.
• One repeated root $m$: complementary function $(A+Bn)\,m^n$. Example: $2u_{n+2}+7u_{n+1}-15u_n=0$ gives $2m^2+7m-15=0$, so $(2m-3)(m+5)=0$ and $m=\frac32$ or $m=-5$. The complementary function is $A\left(\frac32\right)^n+B(-5)^n$. Example (repeated): $u_{n+2}-6u_{n+1}+9u_n=0$ gives $(m-3)^2=0$ and $(A+Bn)3^n$.

For a constant right-hand side, the general solution is the complementary function plus a particular solution. Try $u_n=\lambda$, a constant, in every term: $u_{n+2}$, $u_{n+1}$ and $u_n$ all become $\lambda$. Example: $2u_{n+2}+7u_{n+1}-15u_n=6$. Substituting, $2\lambda+7\lambda-15\lambda=6$, so $-6\lambda=6$ and $\lambda=-1$. The general solution is $u_n=A\left(\tfrac32\right)^n+B(-5)^n-1.$

Substitute $n=1$ and $n=2$ (or whichever values are given) into the general solution to form two simultaneous equations for $A$ and $B$. Continuing: $u_1=10$ gives $\frac32A-5B-1=10$, and $u_2=-17$ gives $\frac94A+25B-1=-17$. Solving, $A=4$ and $B=-1$, so $u_n=4\left(\tfrac32\right)^n-(-5)^n-1.$ Check with the relation: $2u_3+7(-17)-15(10)=6$ gives $u_3=137.5$, and the formula gives $4\times\frac{27}{8}+125-1=137.5$.

As $n\to\infty$, a term $m^n$ with $|m|<1$ tends to $0$, so the solution tends to the particular solution $\lambda$. If any root has $|m|>1$ the solution grows without bound, dominated by the root of largest magnitude. Method: (1) write the auxiliary equation and solve it; (2) state the CF; (3) find the PS; (4) form the general solution; (5) use both initial conditions for $A$ and $B$; (6) check with $u_3$ and interpret in context. Use the words 'auxiliary equation', 'complementary function' and 'particular solution'.