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December 22, 2014

December 22, 2014

Posted by **CHRIS** on Saturday, November 9, 2013 at 1:53pm.

- Maths -
**Reiny**, Saturday, November 9, 2013 at 3:04pmA

_{1}= 22/47

A_{2}= (22/47 - 1)/(22/47

= -25/22

A_{3}= (-25/22 - 1)/(-25/22)

= 47/25

A_{4}= (47/25 - 1)/(47/25)

= 22/47

A_{5}= -25/22

A_{6}= 47/25

A_{7}= 22/47

...

looks like if the subscript is divisible by 4, the answer is 22/47

since 2012 divides evenly by 4, we get

22/47 as a/b, so a = 22, b = 47

- Maths -
**Steve**, Saturday, November 9, 2013 at 3:12pmf(A) = 1 - 1/A

so,

f(m/n) = (m/n - 1)/(m/n)

= (m-n)/m

f((m-n)/m) = (m-n-m)/(m-n) = -n/(m-n)

f(-n/(m-n)) = (-n+n-m)/(-n) = m/n

so, f(f(f(A))) = A

every 3rd iteration, we are back to A.

2012/3 = 670 with remainder 2.

Thus, f^{2012}(A) = f^{2}(A)

so, take a look:

f(22/47) = 1 - 47/22 = -25/22

f(-25/22) = 1 + 22/25 = 47/25

- Steve is right - Maths -
**Reiny**, Saturday, November 9, 2013 at 3:28pmSteve is right, it is a cycle of 3 terms, not 4 terms like I hastily concluded.

so -25/22

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