f ( x ) and 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 )
For g ( 1 ) = k for some natural number k ,infinitely many values of f ( n ) ,including n = 1 is concretely determinable,
find k
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I guessed 0,1,and 3 and got it right lol
The question doesn't make sense.
I accidentally deleted my report and can't rewrite it. The question is flawed, and should probably say f and g are strictly increasing linear functions of x .
Currently k = 3 fixes f ( 1 ) = 2 , f ( 2 ) = 3 . But f ( x ) = ( x + a ) for a ≥ 1 , x ≥ 3 , g ( x ) = ( x + 2 a ) for x ≥ 2 is not unique.
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If f was a linear function, it would be forced to be equal to x + 1 . For other values of k ≥ 3 you would have a choice of b x + c that satisfies b 2 + b c + c = k , b > 0 , c ≥ 0 . But this has unique solutions for at least k = 3 , 4 , 5 .
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
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That question is also not well defined. But it does give insight to what the TRUE question is.
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
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Actually k = 4 also fixes f ( 1 ) = 2 and f ( 2 ) = 4 .
The question should be for strictly increasing f ( x ) , with f ( f ( x ) ) = k x . k = 3 uniquely determines all values of f ( x ) .
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f(f(x))=2x+1 also determines all values of f(x).
k = 3 is the only value that uniquely determines f ( 1 ) and f ( 2 ) (as 2 , 3 respectively).
One can then work out that this also determines f ( 3 n ) = 2 ∗ 3 n and f ( 2 ∗ 3 n ) = 3 n + 1 .
The gap between f ( 3 n ) and f ( 2 ∗ 3 n ) is 3 n , the same gap as between 3 n and 2 ∗ 3 n . Hence, in this range, f ( 3 n + a ) = 2 ∗ 3 n + a .
From this we can determine, for the same values of a , f ( 2 ∗ 3 n + a ) = 3 ∗ ( 3 n + a ) = 3 n + 1 + 3 a . Thus determining all values of f ( x ) between 3 n and 3 n + 1 . Doing this for all n uniquely identifies all values of f ( x ) .
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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)?
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@Akshay Singh – Oh sorry I misread, maybe.
@Akshay Singh – Yes, I agree, all linear g ( x ) uniquely determine ALL values of f ( x ) .
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
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Ok fixed my "solution" above, and will edit it in a bit.
For all fixed g ( x ) , we can determine infinite many f ( x ) uniquely.
If g ( x ) is linear we can determine ALL f ( x ) .
If g ( x ) is higher order we cannot.
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 . It's also easy to show that, with no other information, if k ≥ 4 , then f ( 1 ) is not uniquely determined. I cannot see what else is supposed to be going on here.
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I understand the problem a bit more now, it should say something like: for determined or fixed g ( x ) with g ( 1 ) = k . i.e if g ( x ) is fixed from this point onwards you can determine infinitely many values of f ( x ) from g ( x ) . But this has two solutions: 3 , 4 .
If we restrict g ( x ) to be linear in x , and state all values of f ( x ) are uniquely determinable, this has the unique answer of k = 3 .
g ( 1 ) = k ⟹ f ( 1 ) = a and f ( a ) = k . If k ≥ 3 , a = 2 or 3 are both valid.
k = 1 , 2 lead to contradictions of strictly increasing f ( x ) .
k = 4 fixes f ( 1 ) = 2 , f ( 2 ) = 4 . This leads to f ( 4 ) = g ( 2 ) . This is a valid answer: eg g ( x ) = 4 x , determines f ( 2 n ) = 2 n + 1 . (It uniquely determines f m ( 1 ) ∀ m . (Note that f ( 1 ) = 3 ⟹ f ( 3 ) = 4 ⟹ f ( 2 ) doesn't exist )
k = 3 fixes f ( 1 ) = 2 , f ( 2 ) = 3 .
Assuming k = 3 , and g ( x ) = a x + b (linear):
g ( x ) = a x + ( 3 − a ) .
f ( 1 ) = 2
f ( 2 ) = 3
f ( 3 ) = g ( 2 ) = a + 3
We can determine f ( a + 3 ) = g ( 3 ) = 2 a + 3 .
Also, f ( 2 a + 3 ) = a 2 + 2 a + 3 , f ( a 2 + 2 a + 3 ) = 2 a 2 + 2 a + 3 .
..
Let 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 and f ( T n ) = g ( S n ) = S n + 1 ∀ n .
Noting that T n − S n = a n + 1 = S n + 1 − T n , and f ( x ) is strictly increasing, we can populate f ( S n + c ) = T n + c for c < a n + 1 .
Now we need to populate the values of f ( x ) for x between T n and S n + 1 :
f ( T n + c ) = S n + 1 + c ∗ a , for c < a n + 1 .
T n + a n + 1 = S n + 1 . Hence we have populated \f(x)) for all x between S n and S n + 1 ∀ n . Which is all values of x .