Math Problem Statement

Find f(1), f(2), f(3), and, f(4) if f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, . . . 1

ICS 141: Discrete Mathematics I (Fall 2014) a) f(n+1)=f(n)+2 f(1) = f(0 + 1) = f(0) + 2 = 1 + 2 = 3 f(2) = f(1 + 1) = f(1) + 2 = 3 + 2 = 5 f(3) = f(2 + 1) = f(2) + 2 = 5 + 2 = 7 f(4) = f(3 + 1) = f(3) + 2 = 7 + 2 = 9 b) f(n+1)=3f(n) f(1) = f(0 + 1) = 3f(0) = 3(1) = 3 f(2) = f(1 + 1) = 3f(1) = 3(3) = 9 f(3) = f(2 + 1) = 3f(2) = 3(9) = 27 f(4) = f(3 + 1) = 3f(3) = 3(27) = 81

Solution

Given the recursive function definitions for both cases (a) and (b), we can now compute f(1)f(1), f(2)f(2), f(3)f(3), and f(4)f(4).

Case (a) f(n+1)=f(n)+2f(n+1) = f(n) + 2 and f(0)=1f(0) = 1:

We can compute the values step by step as follows:

  • f(0)=1f(0) = 1
  • f(1)=f(0)+2=1+2=3f(1) = f(0) + 2 = 1 + 2 = 3
  • f(2)=f(1)+2=3+2=5f(2) = f(1) + 2 = 3 + 2 = 5
  • f(3)=f(2)+2=5+2=7f(3) = f(2) + 2 = 5 + 2 = 7
  • f(4)=f(3)+2=7+2=9f(4) = f(3) + 2 = 7 + 2 = 9

Thus, for case (a), we have:

  • f(1)=3f(1) = 3
  • f(2)=5f(2) = 5
  • f(3)=7f(3) = 7
  • f(4)=9f(4) = 9

Case (b) f(n+1)=3f(n)f(n+1) = 3f(n) and f(0)=1f(0) = 1:

Similarly, compute the values step by step:

  • f(0)=1f(0) = 1
  • f(1)=3f(0)=31=3f(1) = 3 \cdot f(0) = 3 \cdot 1 = 3
  • f(2)=3f(1)=33=9f(2) = 3 \cdot f(1) = 3 \cdot 3 = 9
  • f(3)=3f(2)=39=27f(3) = 3 \cdot f(2) = 3 \cdot 9 = 27
  • f(4)=3f(3)=327=81f(4) = 3 \cdot f(3) = 3 \cdot 27 = 81

Thus, for case (b), we have:

  • f(1)=3f(1) = 3
  • f(2)=9f(2) = 9
  • f(3)=27f(3) = 27
  • f(4)=81f(4) = 81

Would you like any more details or explanations?

Related questions:

  1. What is the general solution for f(n)f(n) in case (a)?
  2. How would you solve a recursive function where f(n+1)=f(n)+nf(n+1) = f(n) + n?
  3. Can you express case (b) in terms of a closed-form equation?
  4. How would the solution change if f(0)1f(0) \neq 1 in either case?
  5. How do you approach finding recursive sequences in general?

Tip: When dealing with recursive functions, it's helpful to compute the first few terms to observe any patterns, which may help in formulating a closed form.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Recursion
Arithmetic Progressions
Geometric Progressions

Formulas

f(n+1) = f(n) + 2 (Arithmetic Progression)
f(n+1) = 3f(n) (Geometric Progression)

Theorems

Recursion Formula

Suitable Grade Level

Grades 9-12