Math Problem Statement

Question 8 options: Given f = {(a,b),(b,a),(c,b)}, a function from X = {a,b,c} to X: Write f ◦f and f ◦f ◦f as sets of ordered pairs. Define f to the power of n = f ◦f ◦···◦f to be the n-fold composition of f with itself. Write f to the power of 9 and f to the power of 623 as sets of ordered pairs.

Solution

Let's solve this step-by-step.

1. Understanding the function ff

The given function is: f={(a,b),(b,a),(c,b)}f = \{(a,b), (b,a), (c,b)\} This means:

  • f(a)=bf(a) = b
  • f(b)=af(b) = a
  • f(c)=bf(c) = b

2. Computing fff \circ f (i.e., ff composed with itself)

The composition fff \circ f means applying ff twice. In other words, ff(x)=f(f(x))f \circ f(x) = f(f(x)) for each element xXx \in X.

  • ff(a)=f(f(a))=f(b)=af \circ f(a) = f(f(a)) = f(b) = a
  • ff(b)=f(f(b))=f(a)=bf \circ f(b) = f(f(b)) = f(a) = b
  • ff(c)=f(f(c))=f(b)=af \circ f(c) = f(f(c)) = f(b) = a

Thus, fff \circ f is: ff={(a,a),(b,b),(c,a)}f \circ f = \{(a,a), (b,b), (c,a)\}

3. Computing ffff \circ f \circ f

Now, we compose fff \circ f with ff again, meaning we apply ff to the result of fff \circ f.

  • fff(a)=f(ff(a))=f(a)=bf \circ f \circ f(a) = f(f \circ f(a)) = f(a) = b
  • fff(b)=f(ff(b))=f(b)=af \circ f \circ f(b) = f(f \circ f(b)) = f(b) = a
  • fff(c)=f(ff(c))=f(a)=bf \circ f \circ f(c) = f(f \circ f(c)) = f(a) = b

Thus, ffff \circ f \circ f is: fff={(a,b),(b,a),(c,b)}f \circ f \circ f = \{(a,b), (b,a), (c,b)\} Notice that this is exactly the original function ff. This means that fff=ff \circ f \circ f = f.

4. Generalizing fnf^n (composition of ff with itself n times)

Since fff=ff \circ f \circ f = f, we can observe that:

  • f1=f={(a,b),(b,a),(c,b)}f^1 = f = \{(a,b), (b,a), (c,b)\}
  • f2=ff={(a,a),(b,b),(c,a)}f^2 = f \circ f = \{(a,a), (b,b), (c,a)\}
  • f3=fff=f={(a,b),(b,a),(c,b)}f^3 = f \circ f \circ f = f = \{(a,b), (b,a), (c,b)\}

From this, we can conclude that the composition alternates between ff and fff \circ f.

5. Finding f9f^9 and f623f^{623}

Since the composition alternates between ff and fff \circ f, the general pattern is:

  • f1=ff^1 = f
  • f2=fff^2 = f \circ f
  • f3=ff^3 = f
  • f4=fff^4 = f \circ f
  • And so on...

Thus:

  • f9=ff^9 = f because 9 is odd.
  • f623=ff^{623} = f because 623 is odd.

Final Answers:

  • ff={(a,a),(b,b),(c,a)}f \circ f = \{(a,a), (b,b), (c,a)\}
  • fff={(a,b),(b,a),(c,b)}f \circ f \circ f = \{(a,b), (b,a), (c,b)\}
  • f9={(a,b),(b,a),(c,b)}f^9 = \{(a,b), (b,a), (c,b)\}
  • f623={(a,b),(b,a),(c,b)}f^{623} = \{(a,b), (b,a), (c,b)\}

Would you like a more detailed breakdown of any part of this solution?


Related Questions:

  1. How can I find the powers of more complex functions?
  2. How does function composition work when the functions have different domains and ranges?
  3. What is the difference between fff \circ f and f2f^2?
  4. What happens when we compose a function with itself an even number of times versus an odd number of times?
  5. How can I visualize function composition on a graph?

Tip:

When working with function compositions, always track the pattern for small values of nn first—this can often reveal periodicity or repetition that simplifies calculations for larger nn.

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Math Problem Analysis

Mathematical Concepts

Function Composition
Set Theory
Exponents in Functions

Formulas

f ◦ f(x) = f(f(x))
f^n = f ◦ f ◦ ... ◦ f (n times)

Theorems

Composition of functions
Properties of function powers

Suitable Grade Level

Grades 9-12