JC1 H2 Mathematics: Recurrence Relations

Find constants, then work backwards

A tank is monitored once an hour. Its readings satisfy $U_{t+1}=p+qU_t$, with $U_3=82,U_4=66,U_5=58$. Find $p,q$ and the initial reading $U_0$.

Use each reading as the input to the next:

Subtract (2) from (1), then substitute:

Make the earlier reading the subject:

Thus $p=25,q=1/2,U_0=306$. Forward substitution gives $306,178,114,82,66,58$, checking both the rule and the index.

Find a formula and the first reading below a threshold

For the same tank, find $U_t$, describe the long-term behaviour and find the first integer hour at which the reading is less than 51.

First find the fixed value $L$:

Subtract 50 from both sides of the recurrence:

So $V_t=U_t-50$ is a GP starting at $V_0=306-50=256$, with common ratio $1/2$.

The positive excess above 50 halves each hour. Hence $U_t$ decreases and converges to 50, while remaining above 50.

The first integer hour is 9. At hour 8 the reading is exactly 51, which fails the strict inequality; at hour 9 it is 50.5.

Compare starting values in an unstable model

A sequence satisfies $u_1=a$ and $u_{n+1}=3u_n-8$. Describe its behaviour for $a=5,4,3$, and find $u_3+u_6+\cdots+u_{30}$.

For $a=5$, the terms increase and diverge to $+\infty$. For $a=4$, every term is 4. For $a=3$, the terms decrease and diverge to $-\infty$. Solving $L=3L-8$ alone gives the possible limit 4; it does not prove convergence from an arbitrary starting value.

There are 10 selected terms. Their exponential parts have first term $3^2$ and common ratio $3^3=27$:

Check $a=4$: the sum becomes $10(4)=40$.

Count complete cycles and leftover terms

Let $u_1=3$ and $u_{n+1}=1-1/u_n$. Find $u_{2026}$ and $\sum_{r=1}^{50}u_r$.

Since the same deterministic rule is applied again, the three non-zero terms repeat. The period is 3.

A remainder of zero means the last term in a cycle, not a nonexistent $u_0$. A periodic sequence with unequal repeating values does not converge to one limit.

A forbidden starting value and a constant sequence

Let $u_1=p$ and $u_{n+1}=10-16/u_n$. Explain why $p=40/21$ does not define an infinite sequence, and find all constant sequences.

Then $u_4=10-16/0$ is undefined. Hence this starting value cannot generate an infinite sequence. Checking only $u_2$ would miss the later failure.

For a constant sequence all terms equal the same non-zero $p$:

Both values are valid and remain unchanged under the rule.

A local sign conclusion is not a global claim

A sequence satisfies $x_{n+1}=x_n^2-2$. Assuming it converges to $L$, find the possible limits. Show that if $-1<x_n<2$, then $x_{n+1}<x_n$.

On $-1<x_n<2$, the first factor is positive and the second negative. Thus $x_{n+1}-x_n<0$, as required.

This result is conditional on the current term being in that interval. It does not prove every later term remains there: for example, $x_n=0$ gives $x_{n+1}=-2$. To claim a whole sequence is decreasing, first establish the required bounds for all its terms.