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Difference between revisions of "1997 AHSME Problems/Problem 27"

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==Problem==
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Consider those functions <math>f</math> that satisfy <math> f(x+4)+f(x-4) = f(x) </math> for all real <math>x</math>. Any such function is periodic, and there is a least common positive period <math>p</math> for all of them. Find <math>p</math>.
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<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32 </math>
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==Solution==
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Recall that <math>p</math> is the fundamental period of function <math>f</math> iff <math>p</math> is the smallest positive <math>p</math> such that <math>f(x) = f(x + p)</math> for all <math>x</math>.
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In this case, we know that <math>f(x+ 4) + f(x - 4) = f(x)</math>.  Plugging in <math>x+4</math> in for <math>x</math> to get the next equation in the recursion, we also get <math>f(x + 8) + f(x) = f(x + 4)</math>.  Adding those two equations gives <math>f(x + 8) +  f(x - 4) = 0</math> after cancelling out common terms.
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Again plugging in <math>x + 4</math> in for <math>x</math> in that last equation (in order to get <math>f(x)</math>), we find that <math>f(x) = -f(x + 12)</math>.  Now, plugging in <math>x+12</math> for <math>x</math>, we get <math>f(x + 12) = -f(x + 24)</math>.  This proves that <math>f(x)  = f(x + 24)</math>, so there is a period of <math>24</math>, which gives answer <math>\boxed{D}</math>.  We now eliminate answers <math>A</math> through <math>C</math>.
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Let <math>f(x) = x</math> for <math>x \in \{1, 2, 3, 4, 5, 6, 7, 8\}</math>.  Plugging in <math>x=5</math> into the initial equation gives <math>f(9) + f(1) = f(5)</math>, which implies that <math>f(9) = 4</math>.  Since <math>f(1) \neq f(1 + 8)</math>, the function does not have period <math>8</math>.
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Continuing, <math>f(10) = f(6) - f(2) = 4</math>, <math>f(11) = f(7) - f(3) = 4</math>, and <math>f(12) = f(8) - f(4) = 4</math>.  Once we hit <math>f(13)</math>, we have <math>f(13) = f(9) - f(5) = 4 - 5 = -1</math>.  Since <math>f(1) \neq f(1 + 12)</math>, the function does not have period <math>12</math>.
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Finally, <math>f(14) = f(10) - f(6) = 4 - 6 = -2</math>, <math>f(15) = f(11) - f(7) = 4 - 7 = -3</math>, <math>f(16) = f(12) - f(8) = 4 - 8 = -4</math>, and <math>f(17) = f(13) - f(9) = -1 - 4 = -5</math>.  Since <math>f(1) \neq f(1 + 16)</math>, the function does not have period <math>16</math>.
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To confirm that our original period works, we may see that <math>f(18) = -6</math>, <math>f(19) = -7</math>, <math>f(20) = -8</math>, <math>f(21) = -4</math>, <math>f(22) = -4</math>, <math>f(23) = -4</math>, and <math>f(24) = -4</math>.  Finally, <math>f(25) = 1</math>, which is indeed the same as <math>f(1)</math>.
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== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=26|num-a=28}}
 
{{AHSME box|year=1997|num-b=26|num-a=28}}

Revision as of 15:43, 21 August 2011

Problem

Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$. Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$.

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$

Solution

Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$.


In this case, we know that $f(x+ 4) + f(x - 4) = f(x)$. Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$. Adding those two equations gives $f(x + 8) + f(x - 4) = 0$ after cancelling out common terms.


Again plugging in $x + 4$ in for $x$ in that last equation (in order to get $f(x)$), we find that $f(x) = -f(x + 12)$. Now, plugging in $x+12$ for $x$, we get $f(x + 12) = -f(x + 24)$. This proves that $f(x) = f(x + 24)$, so there is a period of $24$, which gives answer $\boxed{D}$. We now eliminate answers $A$ through $C$.


Let $f(x) = x$ for $x \in \{1, 2, 3, 4, 5, 6, 7, 8\}$. Plugging in $x=5$ into the initial equation gives $f(9) + f(1) = f(5)$, which implies that $f(9) = 4$. Since $f(1) \neq f(1 + 8)$, the function does not have period $8$.

Continuing, $f(10) = f(6) - f(2) = 4$, $f(11) = f(7) - f(3) = 4$, and $f(12) = f(8) - f(4) = 4$. Once we hit $f(13)$, we have $f(13) = f(9) - f(5) = 4 - 5 = -1$. Since $f(1) \neq f(1 + 12)$, the function does not have period $12$.

Finally, $f(14) = f(10) - f(6) = 4 - 6 = -2$, $f(15) = f(11) - f(7) = 4 - 7 = -3$, $f(16) = f(12) - f(8) = 4 - 8 = -4$, and $f(17) = f(13) - f(9) = -1 - 4 = -5$. Since $f(1) \neq f(1 + 16)$, the function does not have period $16$.

To confirm that our original period works, we may see that $f(18) = -6$, $f(19) = -7$, $f(20) = -8$, $f(21) = -4$, $f(22) = -4$, $f(23) = -4$, and $f(24) = -4$. Finally, $f(25) = 1$, which is indeed the same as $f(1)$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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