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

A recurrence relation defines each term from the previous one, for example $u_{n+1}=1.1u_n+30$ with a starting value such as $u_1=400$. It is first order because $u_{n+1}$ depends only on $u_n$, and linear because $u_n$ appears only to the first power. They model repeated change: a population that grows by a percentage and has a fixed number added or removed each year; a loan with interest and repayments; a drug that decays between doses. A percentage increase of $r\%$ multiplies by $1+\frac{r}{100}$; a fixed addition or removal is $+b$ or $-b$. Be careful with the order of events: 'grows by 10% and then 30 are added' gives $1.1u_n+30$, whereas adding first gives $1.1(u_n+30)$.

A first-order relation $u_{n+1}=a\,u_n$ (written $u_{n+1}-a\,u_n=0$) is homogeneous. Try $u_n=m^n$: it gives the auxiliary equation $m-a=0$, so $m=a$ and the solution is $u_n=A\,a^n$, where $A$ is a constant found from the initial condition. This is also called the complementary function (CF). Example: $u_{n+1}=3u_n$ with $u_1=2$: $u_n=A\times3^n$ and $2=3A$, so $A=\frac23$ and $u_n=\frac23\times3^n=2\times3^{n-1}$.

For $u_{n+1}-a\,u_n=b$ (a non-zero constant $b$) the general solution is $u_n=\text{complementary function}+\text{particular solution}.$ To find a particular solution (PS) try a constant, $u_n=\lambda$: then $\lambda-a\lambda=b$, so $\lambda=\dfrac{b}{1-a}$ (for $a\ne1$). Example: $u_{n+1}-5u_n=8$, $u_1=1$. CF: $A\times5^n$. PS: $\lambda-5\lambda=8$, so $\lambda=-2$. General solution $u_n=A\times5^n-2$. Using $u_1=1$: $5A-2=1$, so $A=\frac35$ and $u_n=\frac35\times5^n-2=3\times5^{n-1}-2$. Check: $u_2=5(1)+8=13$ and $3(5)-2=13$.

Substitute the starting term into the general solution to find $A$, then check by computing $u_2$ directly from the relation and from your formula. If $|a|<1$ then $a^n\to0$, so $u_n$ tends to the particular solution $\lambda=\frac{b}{1-a}$. For example $u_{n+1}=0.75u_n+200$ gives $\lambda=800$ and the amount tends to $800$. If $|a|>1$ the CF grows without bound, so $u_n$ diverges (up or down, depending on the sign of $A$). Be careful with the index: if the question gives $u_0$ then $n=0$ goes into the formula, and if it gives $u_1$ then $n=1$ does. Sometimes you must solve an inequality using logarithms; reverse the inequality when dividing by a negative logarithm.

1. Rearrange to $u_{n+1}-a\,u_n=b$.
2. Write the auxiliary equation $m-a=0$ and the CF $A\,a^n$.
3. Try $u_n=\lambda$ for the PS and solve for $\lambda$.
4. Write the general solution, use the initial condition to find $A$, and state $u_n$ in full.
5. Check against $u_2$, and interpret in context (units, long-term value, first term exceeding a value).