Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 24"
 
(One intermediate revision by the same user not shown)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
Notice that for each <math>f_{n}(3)</math> where <math>n</math> is odd, the value is <math>1/5</math> and for each value of <math>f_{n}(3)</math> where <math>n</math> is even, the value is <math>3</math>. It follows that the answer is <math>1/5</math>.
+
<math>f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}</math>.  Then <math>f_1(3) = f(-\frac12) = \frac{1 - \frac12}{1 + 3\cdot\frac12} = \frac15</math>, <math>\displaystyle f_2(3) = f(\frac15) = \frac{1 + \frac15}{1 - 3\cdot\frac15} = 3</math> and <math>f_3(3) = f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}</math>.
 +
 
 +
It follows immediately that the function cycles and <math>f_n(3) = -\frac12</math> if <math>n = 3k</math>, <math>f_n(3) = \frac15</math> if <math>n = 3k + 1</math> and <math>f_n(3) = 3</math> if <math>n = 3k + 2</math>.  Since <math>1993 = 3\cdot 664 + 1</math>, <math>f_{1993}(3) = \frac 15 \Longrightarrow \mathrm{(D)}</math>.
 +
 
 +
 
 +
Alternatively, <math>f_1(x) = f(f(x)) = \frac{1 + f(x)}{1 - 3f(x)} = \frac{1 + \frac{1 + x}{1 - 3x}}{1 - 3\frac{1 + x}{1 - 3x}} = \frac{(1 - 3x) + (1 + x)}{(1 - 3x) - 3(1 + x)} = \frac{2 - 2x}{-2 - 6x} = \frac{x - 1}{3x + 1}</math>.  Thus <math>f_2(x) = f(f_1(x)) = \frac{1 + \frac{x - 1}{3x + 1}}{1 - 3\frac{x - 1}{3x + 1}} = \frac{(3x + 1) + (x - 1)}{(3x + 1) - 3(x - 1)} = \frac{4x}{4} = x</math>, so <math>f_{3k + 2}(x)  = f(f(f(f_{3k - 1})))(x) = f_{3k - 1}(x) = \ldots = f_2(x) = x</math>. Thus <math>f_{1993}(x) = f(f(f_{1991}(x))) = f(f(x))</math> so <math>f_{1993}(3) = f(f(3)) = \frac15</math>.
  
 
----
 
----

Latest revision as of 14:27, 19 August 2006

Problem

If $f(x) = \frac{1 + x}{1 - 3x}, f_1(x) = f(f(x)), f_2(x) = f(f_1(x)),$ and in general $f_n(x) = f(f_{n-1}(x)),$ then $f_{1993}(3)=$

$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }1993 \qquad \mathrm{(C) \ }\frac 12 \qquad \mathrm{(D) \ }\frac 15 \qquad \mathrm{(E) \ } -2^{-1993}$

Solution

$f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}$. Then $f_1(3) = f(-\frac12) = \frac{1 - \frac12}{1 + 3\cdot\frac12} = \frac15$, $\displaystyle f_2(3) = f(\frac15) = \frac{1 + \frac15}{1 - 3\cdot\frac15} = 3$ and $f_3(3) = f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}$.

It follows immediately that the function cycles and $f_n(3) = -\frac12$ if $n = 3k$, $f_n(3) = \frac15$ if $n = 3k + 1$ and $f_n(3) = 3$ if $n = 3k + 2$. Since $1993 = 3\cdot 664 + 1$, $f_{1993}(3) = \frac 15 \Longrightarrow \mathrm{(D)}$.


Alternatively, $f_1(x) = f(f(x)) = \frac{1 + f(x)}{1 - 3f(x)} = \frac{1 + \frac{1 + x}{1 - 3x}}{1 - 3\frac{1 + x}{1 - 3x}} = \frac{(1 - 3x) + (1 + x)}{(1 - 3x) - 3(1 + x)} = \frac{2 - 2x}{-2 - 6x} = \frac{x - 1}{3x + 1}$. Thus $f_2(x) = f(f_1(x)) = \frac{1 + \frac{x - 1}{3x + 1}}{1 - 3\frac{x - 1}{3x + 1}} = \frac{(3x + 1) + (x - 1)}{(3x + 1) - 3(x - 1)} = \frac{4x}{4} = x$, so $f_{3k + 2}(x) = f(f(f(f_{3k - 1})))(x) = f_{3k - 1}(x) = \ldots = f_2(x) = x$. Thus $f_{1993}(x) = f(f(f_{1991}(x))) = f(f(x))$ so $f_{1993}(3) = f(f(3)) = \frac15$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_24&oldid=9716"