250 Followers problem!

Algebra Level 2

$\begin{aligned} f_1 (x)&=&x\\ f_2 (x)&=&1-x\\ f_3 (x)&=&\frac{1}{x}\\ f_4 (x)&=&\frac{1}{1-x}\\ f_5 (x)&=&\frac{x}{x-1}\\ f_6 (x)&=&\frac{x-1}{x}\\ \end{aligned}$

If we know that $f_6 (f_m(x))=f_4(x)$ and $f_n (f_4(x))=f_3(x)$ , find the minimum value of $m+n$ .

The answer is 11.

2 solutions

Curtis Clement
Apr 19, 2015

The best way to approach this is to use group theory to prove that the functions form a complete group, (closed under the composite function operation):

1) Identity element is $\ f_{1}$

2) Inverses: $\ f_{1}$ , $\ f_{2}$ , $\ f_{3}$ and $\ f_{5}$ are all their own inverses and.. $f^{-1}_{4} (x) = f_{6} (x) \ .... \ f^{-1}_{6} (x) = f_{4} (x)$ 3) The function are all associative

4) A Cayley table can be used to show that the group is closed under operation such that: $(f_{6}(x))^2 = f_{4} (x) \ ....... \ f_{5} (f_{4} (x) ) = f_{3} (x)$ $\Rightarrow\ m = 6 \ , \ n = 5 \therefore \ m+n = 11$

I just realized that this question is similar to the question in the MEI FP3 paper for June 2010

Curtis Clement - 6 years, 1 month ago

I love this group. If you make the assignment, f1(x) = x = cos^2 a, for an arbitrary angle a, then the other 5 functions generate the other 5 trig functions in this order, sin^2 a, sec^2 a, csc^2 a, -cot^2 a, and -tan^2 a. The composite of any pair, hence any number, yields one of the other functions in the group. For any selected argument and output function, there is only 1 other function that will form the correct composite. In that sense, there is no minimum value of m+n. There is only 1 choice for each.

Tom Capizzi - 4 years, 11 months ago

Info Web
Aug 30, 2020

If you DON'T read the question carefully, you will get it very simply without Group Theory :

First f 6( f m(x)) = f 4(x) therefore, ( f m(x)-1) / f m(x) = 1/(1-x) from which we get f m(x) = (x-1)/x, which is the function f 6.
m = 6

now, just compose the function f 4(x) in every given function, we find that f 5( f 4(x)) = f 3(x) . so n = 5

250 Followers problem!

The answer is 11.

2 solutions

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