f ( f ( n ) ) = g ( n ) f(f(n))=g(n)

Algebra Level 5

f ( x ) f(x) and g ( x ) g(x) are two strictly increasing function that take in Natural numbers and results in Natural number outputs,such that

f ( f ( x ) ) = g ( x ) f(f(x))=g(x)

For g ( 1 ) = k g(1)=k for some natural number k k ,infinitely many values of f ( n ) f(n) ,including n = 1 n=1 is concretely determinable,

find k k


The answer is 3.

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2 solutions

Alex Burgess
Dec 11, 2018

I understand the problem a bit more now, it should say something like: for determined or fixed g ( x ) g(x) with g ( 1 ) = k g(1) = k . i.e if g ( x ) g(x) is fixed from this point onwards you can determine infinitely many values of f ( x ) f(x) from g ( x ) g(x) . But this has two solutions: 3 , 4 3, 4 .

If we restrict g ( x ) g(x) to be linear in x x , and state all values of f ( x ) f(x) are uniquely determinable, this has the unique answer of k = 3 k = 3 .


g ( 1 ) = k f ( 1 ) = a g(1) = k \implies f(1) = a and f ( a ) = k f(a) = k . If k 3 , a = 2 k \geq 3, a = 2 or 3 3 are both valid.

k = 1 , 2 k = 1, 2 lead to contradictions of strictly increasing f ( x ) f(x) .

k = 4 k = 4 fixes f ( 1 ) = 2 , f ( 2 ) = 4 f(1) = 2, f(2) = 4 . This leads to f ( 4 ) = g ( 2 ) f(4) = g(2) . This is a valid answer: eg g ( x ) = 4 x g(x) = 4x , determines f ( 2 n ) = 2 n + 1 f(2^n) = 2^{n+1} . (It uniquely determines f m ( 1 ) m f^m(1) \forall m . (Note that f ( 1 ) = 3 f ( 3 ) = 4 f ( 2 ) f(1) = 3 \implies f(3) = 4 \implies f(2) doesn't exist )

k = 3 k = 3 fixes f ( 1 ) = 2 , f ( 2 ) = 3 f(1) = 2, f(2) = 3 .


Assuming k = 3 k = 3 , and g ( x ) = a x + b g(x) = ax + b (linear):

g ( x ) = a x + ( 3 a ) g(x) = ax + (3 - a) .

f ( 1 ) = 2 f(1) = 2

f ( 2 ) = 3 f(2) = 3

f ( 3 ) = g ( 2 ) = a + 3 f(3) = g(2) = a + 3

We can determine f ( a + 3 ) = g ( 3 ) = 2 a + 3 f(a + 3) = g(3) = 2a + 3 .

Also, f ( 2 a + 3 ) = a 2 + 2 a + 3 , f ( a 2 + 2 a + 3 ) = 2 a 2 + 2 a + 3 f(2a + 3) = a^2 + 2a + 3, f(a^2 + 2a + 3) = 2a^2 + 2a + 3 .

..

Let S n = 2 ( a n + a n 1 + . . . + a + 1 ) + 1 , T n = S n + a n + 1 S_n = 2(a^n + a^{n-1} + ... + a + 1) + 1, T_n = S_n + a^{n+1} .

Further, f ( S n ) = g ( T n 1 ) = T n f(S_n) = g(T_{n-1}) = T_n and f ( T n ) = g ( S n ) = S n + 1 f(T_n) = g(S_n) = S_{n+1} n \forall n .

Noting that T n S n = a n + 1 = S n + 1 T n T_n - S_n = a^{n+1} = S_{n+1} - T_n , and f ( x ) f(x) is strictly increasing, we can populate f ( S n + c ) = T n + c f(S_n + c) = T_n + c for c < a n + 1 c < a^{n+1} .

Now we need to populate the values of f ( x ) f(x) for x x between T n T_n and S n + 1 S_{n+1} :

f ( T n + c ) = S n + 1 + c a f(T_n + c) = S_{n+1} + c*a , for c < a n + 1 c < a^{n+1} .

T n + a n + 1 = S n + 1 T_n + a^{n+1} = S_{n+1} . Hence we have populated \f(x)) for all x x between S n S_n and S n + 1 n S_{n+1} \forall n . Which is all values of x x .

I guessed 0,1,and 3 and got it right lol

Nate Kantorski - 2 years, 5 months ago

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I guessed ,0 1 and 3 but last moment I given 2 😥

Arka Dutta - 2 years, 2 months ago

The question doesn't make sense.

Alapan Das - 2 years, 2 months ago

I accidentally deleted my report and can't rewrite it. The question is flawed, and should probably say f f and g g are strictly increasing linear functions of x x .

Currently k = 3 k = 3 fixes f ( 1 ) = 2 , f ( 2 ) = 3 f(1) = 2, f(2) = 3 . But f ( x ) = ( x + a ) f(x) = (x + a) for a 1 , x 3 , g ( x ) = ( x + 2 a ) a \geq 1, x \geq 3, g(x) = (x + 2a) for x 2 x \geq 2 is not unique.

Alex Burgess - 2 years, 2 months ago

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If f f was a linear function, it would be forced to be equal to x + 1 x + 1 . For other values of k 3 k \geq 3 you would have a choice of b x + c bx + c that satisfies b 2 + b c + c = k , b > 0 , c 0 b^2 + bc + c = k, b > 0, c \geq 0 . But this has unique solutions for at least k = 3 , 4 , 5 k = 3, 4, 5 .

Alex Burgess - 2 years, 2 months ago

I'm not completely sure what the question means but i've seen a similar question to this and it's solution didn't have a linear solution to it, but rather a picewise function with some exponents and stuff. the question I just thought it was something like this and put 3

Akshay Singh - 2 years, 2 months ago

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That question is also not well defined. But it does give insight to what the TRUE question is.

Alex Burgess - 2 years, 2 months ago

Don't know about "infinitely many f(n)'s" but i got, If f(1)=1, then only f(1) is determinable and not more. So let f(1)=j, j>1. f(1)=j implies f(j)=k. This implies 1<j<k. For k=1 and 2, no such j exists. For k=3, j=2 exists "concretely" For k>3, there are multiple values of j and just one is indeterminable. So k=3 is a must to have multiple f(n)'s concrete solutions (including f(1) and f(2)) but that just doesn't say anything about infinitely many f(n)'s

Akshay Singh - 2 years, 2 months ago

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Actually k = 4 k = 4 also fixes f ( 1 ) = 2 f(1) = 2 and f ( 2 ) = 4 f(2) = 4 .

Alex Burgess - 2 years, 2 months ago

The question should be for strictly increasing f ( x ) f(x) , with f ( f ( x ) ) = k x f(f(x)) = kx . k = 3 k = 3 uniquely determines all values of f ( x ) f(x) .

Alex Burgess - 2 years, 2 months ago

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f(f(x))=2x+1 also determines all values of f(x).

Akshay Singh - 2 years, 2 months ago

k = 3 k = 3 is the only value that uniquely determines f ( 1 ) f(1) and f ( 2 ) f(2) (as 2 , 3 2, 3 respectively).

One can then work out that this also determines f ( 3 n ) = 2 3 n f(3^n) = 2*3^n and f ( 2 3 n ) = 3 n + 1 f(2*3^n) = 3^{n+1} .

The gap between f ( 3 n ) f(3^n) and f ( 2 3 n ) f(2*3^n) is 3 n 3^n , the same gap as between 3 n 3^n and 2 3 n 2*3^n . Hence, in this range, f ( 3 n + a ) = 2 3 n + a f(3^n + a) = 2*3^n + a .

From this we can determine, for the same values of a a , f ( 2 3 n + a ) = 3 ( 3 n + a ) = 3 n + 1 + 3 a f(2*3^n + a) = 3*(3^n + a) = 3^{n+1} + 3a . Thus determining all values of f ( x ) f(x) between 3 n 3^n and 3 n + 1 3^{n+1} . Doing this for all n n uniquely identifies all values of f ( x ) f(x) .

Alex Burgess - 2 years, 2 months ago

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But can't it be solved for f(f(x))=2x+1,i saw it on a forum once (aops-inmo 2019 group) it also gives same values of f(1) and f(2)?

Akshay Singh - 2 years, 2 months ago

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@Akshay Singh Oh sorry I misread, maybe.

Alex Burgess - 2 years, 2 months ago

@Akshay Singh Yes, I agree, all linear g ( x ) g(x) uniquely determine ALL values of f ( x ) f(x) .

Alex Burgess - 2 years, 2 months ago

I tried putting g(n) as any natural polynomial function which might give g(1) as 3 and i took 4 linear and 2 quadratic and in all cases i could get a lot of f(n)'s determined but i think the question could have been phrased better

Akshay Singh - 2 years, 2 months ago

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Ok fixed my "solution" above, and will edit it in a bit.

For all fixed g ( x ) g(x) , we can determine infinite many f ( x ) f(x) uniquely.

If g ( x ) g(x) is linear we can determine ALL f ( x ) f(x) .

If g ( x ) g(x) is higher order we cannot.

Alex Burgess - 2 years, 2 months ago
Richard Desper
Mar 28, 2019

I don't know what this question is supposed to mean. I don't know what "concretely determinable" is supposed to mean. There appear to be far too few constraints here to determine infinitely many values of f. It is pretty easy to show that k 3 k \geq 3 . It's also easy to show that, with no other information, if k 4 k \geq 4 , then f ( 1 ) f(1) is not uniquely determined. I cannot see what else is supposed to be going on here.

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