The Central Idea
A recurrence describes how terms are related.
For example,
\[ a_n=a_{n-1}+3 \]says that each term is three greater than the preceding term.
But the rule alone does not tell us where the sequence starts.
If
\[ a_1=2, \]then the sequence is
\[ 2,\ 5,\ 8,\ 11,\ldots \]while if
\[ a_1=10, \]the same recurrence gives
\[ 10,\ 13,\ 16,\ 19,\ldots \]The recurrence controls the step-to-step behavior. The initial condition determines the starting state.
What Is an Initial Condition?
An initial condition is a specified starting value used to begin a recursive process.
In
\[ a_1=4,\qquad a_n=a_{n-1}+5\quad(n\ge2), \]the statement
\[ \boxed{a_1=4} \]is the initial condition.
The recurrence then produces
\[ a_2=9,\qquad a_3=14,\qquad a_4=19. \]Initial Conditions Are Part of the Definition
A recursive sequence is not fully specified unless enough starting information is supplied.
The relation
\[ a_n=2a_{n-1} \]is incomplete because many sequences satisfy it.
For example, both
\[ 1,\ 2,\ 4,\ 8,\ldots \]and
\[ 5,\ 10,\ 20,\ 40,\ldots \]satisfy the same recurrence.
Adding an initial condition such as \(a_1=1\) selects one particular sequence.
One-Step Recurrences Usually Need One Initial Value
A recurrence that depends only on the immediately preceding term is often called a first-order recurrence.
For example,
\[ a_n=3a_{n-1}-2. \]To calculate \(a_2\), we need \(a_1\).
If
\[ a_1=4, \]then
\[ a_2=3(4)-2=10, \] \[ a_3=3(10)-2=28, \] \[ a_4=3(28)-2=82. \]Thus,
\[ \boxed{4,\ 10,\ 28,\ 82,\ldots}. \]Two-Step Recurrences Usually Need Two Initial Values
Suppose
\[ a_n=a_{n-1}+a_{n-2}. \]To calculate \(a_3\), we need both \(a_2\) and \(a_1\).
Therefore, two initial values are required.
For example,
\[ a_1=1,\qquad a_2=1 \]gives
\[ \boxed{1,\ 1,\ 2,\ 3,\ 5,\ 8,\ldots}. \]Why the Number of Initial Conditions Matters
The amount of initial data required depends on how far back the recurrence reaches.
If
\[ a_n=F(a_{n-1}), \]one starting value is typically enough.
If
\[ a_n=F(a_{n-1},a_{n-2}), \]two starting values are typically required.
If
\[ a_n=F(a_{n-1},a_{n-2},a_{n-3}), \]three starting values are typically needed.
Order of a Recurrence
The order of a recurrence describes how far back its dependencies reach.
The recurrence
\[ a_n=4a_{n-1} \]is first order.
The recurrence
\[ a_n=2a_{n-1}+a_{n-2} \]is second order.
The recurrence
\[ a_n=a_{n-1}+a_{n-3} \]is third order because the most distant dependency is \(a_{n-3}\).
Example: A Second-Order Recurrence
Let
\[ a_1=2,\qquad a_2=5, \] \[ a_n=2a_{n-1}-a_{n-2}\qquad(n\ge3). \]Then
\[ a_3=2(5)-2=8, \] \[ a_4=2(8)-5=11, \] \[ a_5=2(11)-8=14. \]Therefore,
\[ \boxed{2,\ 5,\ 8,\ 11,\ 14,\ldots}. \]The Same Recurrence Can Produce Different Sequences
Consider
\[ a_n=a_{n-1}+2. \]If \(a_1=0\), then
\[ 0,\ 2,\ 4,\ 6,\ldots \]results.
If \(a_1=1\), then
\[ 1,\ 3,\ 5,\ 7,\ldots \]results.
The recurrence is identical. Only the initial condition changes.
Initial Conditions Select a Solution
It is useful to think of a recurrence as defining a family of possible sequences.
For
\[ a_n=a_{n-1}+4, \]every choice of \(a_1\) gives a different sequence.
The initial condition chooses one member of that family.
\[ \boxed{\text{recurrence}+\text{initial condition}\longrightarrow\text{specific sequence}}. \]Initial Conditions Can Begin at Index Zero
The first defined term does not have to be \(a_1\).
Suppose
\[ a_0=7, \] \[ a_n=a_{n-1}+3\qquad(n\ge1). \]Then
\[ a_1=10,\qquad a_2=13,\qquad a_3=16. \]Thus,
\[ \boxed{7,\ 10,\ 13,\ 16,\ldots}. \]The starting index and the starting value must both be read carefully.
The Recurrence Must Start after the Initial Data
Suppose
\[ a_0=2,\qquad a_1=3, \] \[ a_n=a_{n-1}+a_{n-2}. \]Because the first two values are supplied, the recurrence can first be applied at \(n=2\).
A complete statement is
\[ \boxed{a_0=2,\qquad a_1=3,\qquad a_n=a_{n-1}+a_{n-2}\quad(n\ge2)}. \]Example: Indexing from Zero
Let
\[ a_0=1,\qquad a_1=4, \] \[ a_n=2a_{n-1}-a_{n-2}\qquad(n\ge2). \]Then
\[ a_2=2(4)-1=7, \] \[ a_3=2(7)-4=10, \] \[ a_4=2(10)-7=13. \]Therefore,
\[ \boxed{1,\ 4,\ 7,\ 10,\ 13,\ldots}. \]Initial Conditions Can Be Zero
Zero is a valid initial value.
For example,
\[ a_1=0,\qquad a_n=a_{n-1}+5 \]gives
\[ \boxed{0,\ 5,\ 10,\ 15,\ldots}. \]An initial condition does not need to be positive or nonzero unless the problem imposes such a restriction.
Initial Conditions Can Be Negative
Let
\[ a_1=-4, \] \[ a_n=a_{n-1}+6. \]Then
\[ a_2=2,\qquad a_3=8,\qquad a_4=14. \]Thus,
\[ \boxed{-4,\ 2,\ 8,\ 14,\ldots}. \]Initial Conditions Can Be Fractions or Decimals
Suppose
\[ a_0=2.5, \] \[ a_n=1.1a_{n-1}\qquad(n\ge1). \]Then
\[ a_1=2.75, \] \[ a_2=3.025, \] \[ a_3=3.3275. \]Recursive definitions work with whatever numerical values are appropriate to the system being modeled.
A Special Initial Value Can Produce a Special Sequence
Consider
\[ a_n=2a_{n-1}. \]If \(a_1=0\), then
\[ a_2=0,\qquad a_3=0,\qquad a_4=0. \]Therefore,
\[ \boxed{0,\ 0,\ 0,\ 0,\ldots}. \]This constant zero sequence is sometimes called a trivial solution of the recurrence.
Fixed Values under a Recurrence
Some recurrences have values that reproduce themselves.
Consider
\[ a_n=\frac12a_{n-1}+3. \]If every term were the same number \(L\), then
\[ L=\frac12L+3. \]Solving,
\[ \frac12L=3, \] \[ L=6. \]Therefore, if \(a_1=6\), then
\[ \boxed{6,\ 6,\ 6,\ 6,\ldots}. \]Later lessons will study repeated state transitions in more detail.
Changing an Initial Value Can Shift a Sequence
Consider two sequences with the same recurrence:
\[ a_n=a_{n-1}+4, \] \[ b_n=b_{n-1}+4. \]Let
\[ a_1=3,\qquad b_1=10. \]Then
\[ a_n: 3,\ 7,\ 11,\ 15,\ldots \]and
\[ b_n: 10,\ 14,\ 18,\ 22,\ldots \]The difference between corresponding terms remains \(7\). The same additive change preserves the initial offset.
Changing an Initial Value Can Change Scale
Consider
\[ a_n=2a_{n-1}. \]With \(a_1=1\), we obtain
\[ 1,\ 2,\ 4,\ 8,\ldots \]while with \(a_1=5\), we obtain
\[ 5,\ 10,\ 20,\ 40,\ldots \]Every term in the second sequence is five times the corresponding term in the first.
Initial Conditions Can Affect Long-Term Behavior
In some recurrences, changing the initial condition changes more than the starting location.
Consider
\[ a_n=(a_{n-1})^2. \]If \(a_1=2\), then
\[ 2,\ 4,\ 16,\ 256,\ldots \]grows rapidly.
If \(a_1=\frac12\), then
\[ \frac12,\ \frac14,\ \frac1{16},\ \frac1{256},\ldots \]moves toward \(0\).
The same recursive rule can therefore have very different behavior under different initial conditions.
Initial Conditions in Modeling
In applications, an initial condition usually represents the observed or chosen state of a system at the beginning of a model.
- an initial population;
- an initial account balance;
- an initial inventory level;
- an initial temperature;
- an initial position or velocity;
- an initial probability distribution or state.
The recurrence describes how the system evolves from that starting state.
Example: Savings Balance
Suppose an account begins with \(\$500\), and \(\$50\) is added at the end of each month.
Let \(B_n\) be the balance after \(n\) monthly deposits.
Then
\[ B_0=500 \]is the initial condition, and
\[ \boxed{B_n=B_{n-1}+50\qquad(n\ge1)} \]is the recurrence.
The first balances are
\[ 500,\ 550,\ 600,\ 650,\ldots \]Example: Repeated Percentage Growth
Suppose a population begins at \(1200\) and grows by \(5\%\) each year.
A \(5\%\) increase means multiplying by \(1.05\).
If \(P_n\) is the population after \(n\) years, then
\[ P_0=1200 \]and
\[ \boxed{P_n=1.05P_{n-1}\qquad(n\ge1)}. \]The initial condition records the population before any yearly growth step occurs.
Example: Inventory Process
A warehouse begins with \(300\) units. Each day, \(40\) units are shipped and \(25\) new units arrive.
The net daily change is
\[ -40+25=-15. \]Let \(I_n\) be inventory after \(n\) days.
Then
\[ I_0=300 \]and
\[ \boxed{I_n=I_{n-1}-15\qquad(n\ge1)}. \]Initial Conditions and Closed Forms
When a recurrence can be converted to an explicit formula, the initial conditions determine the constants appearing in that formula.
For example,
\[ a_n=a_{n-1}+3 \]has constant-step form
\[ a_n=C+3(n-1), \]where \(C\) depends on the starting value.
If \(a_1=7\), then \(C=7\), so
\[ \boxed{a_n=7+3(n-1)}. \]Later lessons will develop closed forms and recurrence-solving methods more systematically.
Initial Conditions Determine Constants
Consider the family
\[ a_n=C\cdot2^{n-1}. \]At \(n=1\),
\[ a_1=C. \]Therefore, the initial condition determines \(C\) directly.
If \(a_1=6\), then
\[ \boxed{a_n=6\cdot2^{n-1}}. \]Initial Conditions Must Be Consistent with Other Descriptions
Sometimes a problem gives both an explicit formula and additional starting information. They must agree.
Suppose
\[ a_n=2n+1 \]and someone also claims
\[ a_1=4. \]But the explicit formula gives
\[ a_1=2(1)+1=3. \]Therefore, the stated initial condition is inconsistent with the formula.
Initial Data Can Be Given Later in the Index
A recurrence does not always need to be anchored at index \(0\) or \(1\).
Suppose
\[ a_5=12, \] \[ a_n=a_{n-1}+2\qquad(n\ge6). \]Then
\[ a_6=14,\qquad a_7=16,\qquad a_8=18. \]Here \(a_5=12\) serves as the starting condition for forward recursion.
Forward Recursion
Most introductory recurrence problems proceed forward. We begin with known initial values and repeatedly apply the recurrence to obtain later terms.
For example,
\[ a_1=3,\qquad a_n=2a_{n-1}+1 \]gives
\[ a_2=7, \] \[ a_3=15, \] \[ a_4=31. \]Each new value depends on a term already known.
Backward Recovery May Be Possible
Some recurrences can be rearranged to recover earlier terms.
Suppose
\[ a_n=a_{n-1}+4. \]Then
\[ a_{n-1}=a_n-4. \]If \(a_5=20\), then
\[ a_4=16,\qquad a_3=12,\qquad a_2=8,\qquad a_1=4. \]Not every recurrence is equally easy to reverse, but known conditions can sometimes determine terms in more than one direction.
Initial Conditions versus Boundary Conditions
An initial condition is usually specified at the beginning of an indexed process.
In more advanced mathematics, conditions imposed at other locations are often called boundary conditions.
The shared idea is that additional conditions select a particular solution from a family allowed by a rule.
Initial Conditions as State Information
A recurrence can be viewed as a rule for moving from one system state to the next.
If the recurrence is first order, the current state may be represented by
\[ a_n. \]If the recurrence is second order, the current state may require a pair:
\[ (a_{n-1},a_n). \]The initial conditions supply the first complete state from which the system can evolve.
Example: State Pair for a Second-Order Recurrence
Consider
\[ a_n=a_{n-1}+a_{n-2}. \]If
\[ a_1=2,\qquad a_2=5, \]then the initial state can be regarded as
\[ (2,5). \]The next state is
\[ (5,7), \]then
\[ (7,12), \]and so on.
This viewpoint connects recurrences to later ideas about iteration and state transitions.
Common Error: Giving Too Few Initial Values
Consider
\[ a_n=a_{n-1}+a_{n-2}. \]A student gives only
\[ a_1=3. \]This is not enough to calculate \(a_3\), because \(a_2\) is still unknown. A second initial value is required.
Common Error: Applying the Rule before Enough Terms Exist
Suppose
\[ a_1=2,\qquad a_2=4, \] \[ a_n=a_{n-1}+a_{n-2}\qquad(n\ge3). \]The recurrence first applies at \(n=3\). Trying to apply it at \(n=2\) would require \(a_0\), which has not been defined.
Common Error: Treating the Initial Value as a Repeated Rule
In
\[ a_1=7,\qquad a_n=a_{n-1}+2, \]the statement \(a_1=7\) applies only to the first term. It does not mean every term equals \(7\). The recurrence determines the later terms.
Common Error: Changing the Initial Condition without Noticing
Suppose
\[ a_1=5,\qquad a_n=2a_{n-1}. \]The correct sequence begins
\[ 5,\ 10,\ 20,\ 40,\ldots \]Writing \(1,2,4,8,\ldots\) follows the same recurrence but uses the wrong initial condition.
Common Error: Ignoring the Starting Index in a Model
Suppose \(P_0=1000\) is the population before any annual growth step.
Then \(P_1\) represents the population after one step.
If the indexing is shifted accidentally, every later interpretation can be off by one period. Always define what index \(0\) or index \(1\) means in context.
A Reliable Procedure for Using Initial Conditions
- Identify the recurrence relation.
- Determine how far back the recurrence depends on earlier terms.
- Identify the required initial values.
- Check the starting index of each supplied value.
- Check the index at which the recurrence becomes valid.
- Generate later terms in order.
- Verify each term against the recurrence.
A Reliable Procedure for Determining How Much Initial Data Is Needed
- Find the most distant earlier term appearing in the recurrence.
- Determine which earlier positions must be known before the rule can first be applied.
- Make sure those values are supplied or otherwise determined.
For example,
\[ a_n=2a_{n-1}-a_{n-3} \]reaches back to \(a_{n-3}\), so three consecutive starting values are normally needed for forward recursion.
Example: Determine the Required Initial Data
Consider
\[ a_n=3a_{n-1}+2a_{n-2}-a_{n-3}. \]To calculate \(a_4\), we need
\[ a_3,\qquad a_2,\qquad a_1. \]Therefore, a natural complete setup is
\[ \boxed{ a_1=\text{given},\quad a_2=\text{given},\quad a_3=\text{given},\quad a_n=3a_{n-1}+2a_{n-2}-a_{n-3}\ (n\ge4) }. \]Example: Compare Different Initial Conditions
Consider
\[ a_n=a_{n-1}+3. \]With \(a_1=1\),
\[ 1,\ 4,\ 7,\ 10,\ldots \]results.
With \(a_1=6\),
\[ 6,\ 9,\ 12,\ 15,\ldots \]results.
Every corresponding term differs by \(5\). The initial difference persists because both sequences undergo the same additive change at every step.
Example: Same Recurrence, Different Qualitative Behavior
Consider
\[ a_n=(a_{n-1})^2. \]If \(a_1=1\), then
\[ \boxed{1,\ 1,\ 1,\ 1,\ldots}. \]If \(a_1=-1\), then
\[ -1,\ 1,\ 1,\ 1,\ldots \]If \(a_1=2\), the terms grow rapidly.
If \(a_1=\frac12\), the terms shrink toward \(0\).
Initial conditions can affect not only values, but also the qualitative behavior of a recursive system.
Practice 1: One Initial Condition
Let
\[ a_1=4,\qquad a_n=a_{n-1}+6\quad(n\ge2). \]Find the first five terms.
Practice 2: Different Starting Values
Use the recurrence
\[ a_n=2a_{n-1}. \]Generate the first four terms when:
- \(a_1=1\);
- \(a_1=3\).
Practice 3: Two Initial Conditions
Let
\[ a_1=2,\qquad a_2=4, \] \[ a_n=a_{n-1}+a_{n-2}\qquad(n\ge3). \]Find \(a_3,a_4,a_5,a_6\).
Practice 4: Determine Required Initial Values
How many consecutive initial values are normally needed for
\[ a_n=2a_{n-1}-a_{n-3}? \]Practice 5: Index from Zero
Let
\[ a_0=5,\qquad a_n=a_{n-1}+4\quad(n\ge1). \]Find \(a_1,a_2,a_3,a_4\).
Practice 6: Missing Information
A student writes
\[ a_n=a_{n-1}+a_{n-2} \]together with only
\[ a_1=7. \]Explain why the definition is incomplete.
Practice 7: Model an Initial State
A machine begins with \(80\) units of material and uses \(6\) units per cycle. Let \(M_n\) be the amount remaining after \(n\) cycles.
Write an initial condition and recursive rule.
Practice 8: Special Initial Value
Consider
\[ a_n=3a_{n-1}. \]What sequence results if \(a_1=0\)?
Practice 9: Check Consistency
The explicit rule is
\[ a_n=4n-2. \]Is the initial condition \(a_1=2\) consistent with the formula?
Practice 10: Second-Order Recurrence
Let
\[ a_0=1,\qquad a_1=3, \] \[ a_n=2a_{n-1}-a_{n-2}\qquad(n\ge2). \]Find \(a_2,a_3,a_4,a_5\).
Practice Answers
Practice 1
\[ \boxed{4,\ 10,\ 16,\ 22,\ 28}. \]Practice 2
If \(a_1=1\),
\[ \boxed{1,\ 2,\ 4,\ 8}. \]If \(a_1=3\),
\[ \boxed{3,\ 6,\ 12,\ 24}. \]Practice 3
\[ a_3=4+2=6, \] \[ a_4=6+4=10, \] \[ a_5=10+6=16, \] \[ a_6=16+10=26. \]Therefore,
\[ \boxed{a_3=6,\quad a_4=10,\quad a_5=16,\quad a_6=26}. \]Practice 4
The recurrence reaches back to \(a_{n-3}\), so three consecutive initial values are normally needed.
\[ \boxed{3\text{ initial values}}. \]Practice 5
\[ a_1=9,\qquad a_2=13,\qquad a_3=17,\qquad a_4=21. \]Practice 6
The recurrence uses two previous terms. To calculate \(a_3\), both \(a_1\) and \(a_2\) must be known. Since only \(a_1\) is supplied, one more initial value is required.
Practice 7
Before any cycles occur,
\[ M_0=80. \]Each cycle removes \(6\) units:
\[ \boxed{M_0=80,\qquad M_n=M_{n-1}-6\quad(n\ge1)}. \]Practice 8
Starting from zero, every later term remains zero:
\[ \boxed{0,\ 0,\ 0,\ 0,\ldots}. \]Practice 9
Evaluate the explicit formula at \(n=1\):
\[ a_1=4(1)-2=2. \]Therefore,
\[ \boxed{\text{Yes, the condition is consistent.}} \]Practice 10
\[ a_2=2(3)-1=5, \] \[ a_3=2(5)-3=7, \] \[ a_4=2(7)-5=9, \] \[ a_5=2(9)-7=11. \]Therefore,
\[ \boxed{a_2=5,\quad a_3=7,\quad a_4=9,\quad a_5=11}. \]What to Remember
Initial conditions provide the starting information required by a recursive definition.
- A recurrence alone may describe many possible sequences.
- Initial conditions select the particular sequence being studied.
- First-order recurrences typically need one initial value.
- Second-order recurrences typically need two initial values.
- The required initial data depends on how far back the recurrence reaches.
- The starting index is part of the definition.
- Changing initial conditions can change offsets, scale, or long-term behavior.
- In applications, initial conditions represent the starting state of a system.
The core principle is
\[ \boxed{\text{recurrence}+\text{enough initial data}\longrightarrow\text{determined sequence}}. \]Where We Go Next
Initial conditions tell us where a recursive process begins. The next step is to study two of the most common recurrence patterns: repeated addition and repeated multiplication.
The next lesson, Arithmetic and Geometric Recurrences, develops recursive forms for constant differences and constant ratios, connects them with familiar explicit sequence formulas, and shows how these patterns arise in repeated-change models.