3.3 Functions, Sequences & Custom Operations

3.3 Functions, Sequences & Custom Operations

Functions, sequences, and defined symbol operations test your ability to apply strict mathematical rules, work with input-output mappings, and recognize patterns. GRE questions often introduce unfamiliar notation; mastering the underlying principles allows you to execute these problems methodically.

---

1. Function Notation, Domain, and Range

A function $f$ is a rule that assigns to each input element $x$ in a set called the domain exactly one output element $f(x)$ in a set called the range.

Evaluating Functions

Evaluating $f(a)$ means substituting $a$ for every occurrence of $x$ in the function rule:

• Numerical input: If $f(x) = x^2 - 3x$, then $f(4) = 4^2 - 3(4) = 16 - 12 = 4$.
• Algebraic input: If $f(x) = x^2 - 3x$, then $f(a + 1) = (a + 1)^2 - 3(a + 1) = a^2 + 2a + 1 - 3a - 3 = a^2 - a - 2$.

Domain and Range Restrictions

On the GRE, unless otherwise specified, functions deal exclusively with real numbers.

---

2. Composite Functions

A composite function applies one function to the result of another. The notation $f(g(x))$ (read "$f$ of $g$ of $x$") means:

1. First, evaluate the inner function $g(x)$.
2. Second, substitute the output $g(x)$ into the outer function $f$.

Note: In general, function composition is not commutative: $f(g(x)) \neq g(f(x))$.

Worked Example 3.3.1: Composite Function Evaluation

Let $f(x) = 3x - 5$ and $g(x) = x^2 + 2$. Evaluate $f(g(4))$ and $g(f(4))$.

Solution:

1. Evaluate $f(g(4))$:
   • Inner evaluation: $g(4) = 4^2 + 2 = 16 + 2 = 18$.
   • Outer evaluation: $f(18) = 3(18) - 5 = 54 - 5 = 49$.
2. Evaluate $g(f(4))$:
   • Inner evaluation: $f(4) = 3(4) - 5 = 12 - 5 = 7$.
   • Outer evaluation: $g(7) = 7^2 + 2 = 49 + 2 = 51$. Notice that $f(g(4)) = 49$, whereas $g(f(4)) = 51$.

---

3. Arithmetic Sequences & Series

An arithmetic sequence is a ordered list of numbers where the difference between consecutive terms is constant. This constant is called the common difference ($d$):

$$d = a_{k+1} - a_k$$

Formulas for Arithmetic Sequences

• nth Term Formula: $$a_n = a_1 + (n - 1)d$$
• Sum of the First $n$ Terms ($S_n$): $$S_n = \frac{n}{2}(a_1 + a_n) = n \times \text{Average of first and last terms}$$

Worked Example 3.3.2: Arithmetic Sequence Sum

An arithmetic sequence has first term $a_1 = 7$ and common difference $d = 4$. Calculate the 20th term $a_{20}$ and the sum of the first 20 terms $S_{20}$.

Solution:

1. Compute 20th term $a_{20}$: $$a_{20} = a_1 + (20 - 1)d = 7 + 19(4) = 7 + 76 = 83$$
2. Compute sum $S_{20}$: $$S_{20} = \frac{20}{2}(a_1 + a_{20}) = 10 \times (7 + 83) = 10 \times 90 = 900$$

---

4. Geometric Sequences

A geometric sequence is a list of numbers where each term after the first is found by multiplying the previous term by a non-zero constant called the common ratio ($r$):

$$r = \frac{a_{k+1}}{a_k}$$

Formulas for Geometric Sequences

• nth Term Formula: $$a_n = a_1 \cdot r^{n-1}$$

Worked Example 3.3.3: Finding a Term in a Geometric Sequence

A geometric sequence has first term $a_1 = 5$ and common ratio $r = 3$. Which term $n$ of the sequence is equal to $405$?

Solution:

1. Set up equation using the nth term formula: $a_n = a_1 \cdot r^{n-1} \implies 405 = 5 \cdot 3^{n-1}$.
2. Divide both sides by $5$: $\frac{405}{5} = 3^{n-1} \implies 81 = 3^{n-1}$.
3. Express $81$ as a power of $3$: $81 = 3^4$.
4. Equate exponents: $n - 1 = 4 \implies n = 5$. Therefore, $405$ is the 5th term of the sequence.

---

5. Defined Symbol / Custom Operations

GRE questions frequently define custom operations using non-standard symbols such as $\star, \diamond, \otimes, \Delta$. These are self-contained rule definitions.

How to Solve Custom Operator Problems

1. Identify the Rule: Treat the symbol purely as an instruction template. For example, if $x \star y = 2x - y^2$, the left operand is doubled, and the square of the right operand is subtracted.
2. Follow Standard Parentheses Order: Work from the innermost parentheses outward.
3. Do NOT Invent Properties: Never assume commutative ($a \star b = b \star a$) or associative properties unless explicitly proven for that symbol.

Worked Example 3.3.4: Nested Custom Operator Evaluation

For all real numbers $a$ and $b$, let $a \star b = 2a - b^2$. Evaluate $(4 \star 3) \star 2$.

Solution:

1. Evaluate inner parentheses $(4 \star 3)$: Here $a = 4$ and $b = 3$. $$4 \star 3 = 2(4) - (3)^2 = 8 - 9 = -1$$
2. Substitute result into outer operation $(-1) \star 2$: Here $a = -1$ and $b = 2$. $$(-1) \star 2 = 2(-1) - (2)^2 = -2 - 4 = -6$$ The final value is $-6$.

---

6. GRE Function Tactics & Pitfalls

• Undefined Inputs Trap: When quantitative comparison options involve functions with denominators or square roots, check if variable choices create division by zero or negative square roots (making Quantity A or B undefined).
• Sequence Indexing Trap: Pay attention to whether a sequence starts at $n = 1$ ($a_1$) or $n = 0$ ($a_0$). On the GRE, standard indexing starts at $a_1$.