Not so easy functional equation

Algebra Level 5

For each positive integer n n , we define f ( n ) f(n) such that f ( n ) f(n) is a positive integer, f ( n + 1 ) > f ( n ) f(n+1) > f(n) and ( f ( f ( n ) ) = 3 n (f(f(n)) = 3n . What is the value of f ( 10 ) f(10) ?


The answer is 19.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

It is very easy to see that: f ( f ( 1 ) ) = 3 f ( 1 ) = 2 f(f(1))=3 \rightarrow f(1)=2 Because, if f(1)=1 then f(1)=3 (this is a confusion). f ( f ( 1 ) ) = f ( 2 ) = 3 , f ( f ( 2 ) ) = f ( 3 ) = 6 , f ( f ( 3 ) ) = f ( 6 ) = 9 , f ( 4 ) = 7 , f ( 5 ) = 8 f(f(1))=f(2)=3, f(f(2))=f(3)=6, f(f(3))=f(6)=9, \rightarrow f(4)=7, f(5)=8 f ( f ( 4 ) ) = f ( 7 ) = 12 , f ( f ( 5 ) ) = f ( 8 ) = 15 , f ( f ( 6 ) ) = f ( 9 ) = 18 , f ( f ( 7 ) ) = f ( 12 ) = 21 f(f(4))=f(7)=12, f(f(5))=f(8)=15, f(f(6))=f(9)=18, f(f(7))=f(12)=21 f ( 10 ) = 19 \rightarrow f(10)=\boxed{19}

Can you explain more clearly?

Raven Herd - 5 years, 10 months ago

Log in to reply

The best way to solve this is with a number line (natural nos. 1–21). Simply try to fill up the number line according to the rules of function f f . Let us call the ordering rule "Rule O" ( f ( n + 1 ) > f ( n ) f(n+1)>f(n) ) and the multiple rule "Rule M" ( f ( f ( n ) ) = 3 n f(f(n))=3n ). Your working might progress as follows:

1. Find f ( 1 ) f(1) .

If f ( 1 ) = 1 f(1)=1 , then f ( f ( 1 ) ) = f ( 1 ) = 1 3 f(f(1))=f(1)=1\ne3 . So f ( 1 ) 1 f(1)\ne1 , by Rule M. In fact, f ( 1 ) > 1 f(1)>1 , and so f ( n ) > n f(n)>n for all n n , by Rule O.

If f ( 1 ) 3 f(1)\ge3 , then f ( f ( 1 ) ) = 3 f ( 1 ) f(f(1))=3\le f(1) . So f ( 1 ) < 3 f(1)<3 , as above.

Therefore:

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

Now, applying Rule M, we have the following progression by f f :

1 2 3 6 9 18 54 1\rightarrow2\rightarrow3\rightarrow6\rightarrow9\rightarrow18\rightarrow54\dots

2. Find the f f -trails starting at 4 4 and 5 5 .

By Rule O, we must have that 6 = f ( 3 ) < f ( 4 ) < f ( 5 ) < f ( 6 ) = 9 6=f(3)<f(4)<f(5)<f(6)=9 , so immediately we have the following:

f ( 4 ) = 7 ; f ( 5 ) = 8. f(4)=7;\quad f(5)=8.

Applying Rule M, we have the following progressions by f f :

4 7 12 21 ; 5 8 15 24 4\rightarrow7\rightarrow12\rightarrow21\dots;\quad5\rightarrow8\rightarrow15\rightarrow24\dots

3. Find f ( 10 ) f(10) .

Again applying Rule O, we have that 18 = f ( 9 ) < f ( 10 ) < f ( 11 ) < f ( 12 ) = 21 18=f(9)<f(10)<f(11)<f(12)=21 . So, immediately:

f ( 10 ) = 19 . f(10)=\boxed{19}.

Of course, we could carry on! But at this point, the question is answered.

Joel Toms - 5 years, 10 months ago

Log in to reply

Wow that's an awesome solution @Raven Herd

Devansh Sharma - 4 years ago

Log in to reply

@Devansh Sharma Thanks! I wish I'd worked it through properly before I entered the wrong answer... ;)

Joel Toms - 4 years ago
Sharky Kesa
Aug 3, 2016

We will prove via induction on all non-negative integers n n a claim P n P_n that:

  • For 3 n x 2 3 n 3^n \leq x \leq 2 \cdot 3^n , we have f ( x ) = x + 3 n f(x)=x+3^n
  • For 2 3 n x 3 n + 1 2 \cdot 3^n \leq x \leq 3^{n+1} , we have f ( x ) = 3 x 3 n + 1 f(x)=3x-3^{n+1}

First, we prove that f ( 1 ) = 2 f(1)=2 .

If f ( 1 ) = 1 f(1)=1 , we have f ( f ( 1 ) ) = 1 = 3 f(f(1))=1=3 , a contradiction, hence f ( 1 ) > 1 f(1)>1 and with the condition f ( n + 1 ) > f ( n ) f(n+1)>f(n) , we have f ( n ) > n f(n)>n for all positive integers n n .

If f ( 1 ) 3 f(1) \geq 3 , we'd have f ( f ( 1 ) ) f ( 3 ) > 3 f(f(1)) \geq f(3) >3 , also a contradiction. Hence we must have f ( 1 ) = 2 f(1)=2 . Also, we have f ( 2 ) = f ( f ( 1 ) ) = 3 f(2)=f(f(1))=3 and f ( 3 ) = f ( f ( 2 ) ) = 6 f(3)=f(f(2))=6 .

We note that we have our claim P n P_n be true for n = 0 n=0 , given our values of f ( 1 ) , f ( 2 ) , f ( 3 ) f(1),f(2),f(3) . Now assume for some non-negative integer k k , we have P k P_k be true, i.e. that:

  • For 3 k x 2 3 k 3^k \leq x \leq 2 \cdot 3^k , we have f ( x ) = x + 3 k f(x)=x+3^k
  • For 2 3 k x 3 k + 1 2 \cdot 3^k \leq x \leq 3^{k+1} , we have f ( x ) = 3 x 3 k + 1 f(x)=3x-3^{k+1}

We then consider the claim P k + 1 P_{k+1} :

For 3 k + 1 x 2 3 k + 1 3^{k+1} \leq x \leq 2 \cdot 3^{k+1} , consider the x x where x = 3 y x=3y for some integer 3 k y 2 3 k 3^k \leq y \leq 2 \cdot 3^k . We then have x = f ( f ( y ) ) = f ( y + 3 k ) x=f(f(y))=f(y+3^k) , and consequently f ( x ) = f ( f ( y + 3 k ) ) ) = 3 ( y + 3 k ) = 3 y + 3 k + 1 = x + 3 k + 1 f(x)=f(f(y+3^k)))=3(y+3^k)=3y+3^{k+1}=x+3^{k+1} .

For 3 k + 1 x 2 3 k + 1 3^{k+1} \leq x \leq 2 \cdot 3^{k+1} , there also exist x , x + 1 x,x+1 where 3 y < x < x + 1 < 3 ( y + 1 ) 3y<x<x+1<3(y+1) for some ( 3 k y 2 3 k 1 (3^k \leq y \leq 2 \cdot 3^k-1 . We can infer that x = 3 y + 1 , x + 1 = 3 y + 2 x=3y+1,x+1=3y+2 . Then using our results that f ( 3 y ) = 3 y + 3 k + 1 , f ( 3 y + 3 ) = 3 y + 3 + 3 k + 1 f(3y)=3y+3^{k+1},f(3y+3)=3y+3+3^{k+1} from just now, we have:

3 y + 3 k + 1 = f ( 3 y ) < f ( x ) < f ( x + 1 ) < f ( 3 y + 3 ) = 3 y + 3 + 3 k + 1 3y+3^{k+1}=f(3y)<f(x)<f(x+1)<f(3y+3)=3y+3+3^{k+1}

We thus have f ( x ) = 3 y + 1 + 3 k + 1 = x + 3 k + 1 , f ( x + 1 ) = 3 y + 2 + 3 k + 1 = x + 1 + 3 k + 1 f(x)=3y+1+3^{k+1}=x+3^{k+1},f(x+1)=3y+2+3^{k+1}=x+1+3^{k+1} , hence we have it that for all 3 k + 1 x 2 3 k + 1 3^{k+1} \leq x \leq 2 \cdot 3^{k+1} , we have f ( x ) = x + 3 k + 1 f(x)=x+3^{k+1}

For 2 3 k + 1 x 3 k + 2 2 \cdot 3^{k+1} \leq x \leq 3^{k+2} , we note that 3 k + 1 x 3 k + 1 2 3 k + 1 3^{k+1} \leq x-3^{k+1} \leq 2 \cdot 3^{k+1} and from our previous result we have f ( x 3 k + 1 ) = x 3 k + 1 + 3 k + 1 = x f(x-3^{k+1})=x-3^{k+1}+3^{k+1}=x . Consequently, we have f ( x ) = f ( f ( x 3 k + 1 ) ) = 3 x 3 k + 2 f(x)=f(f(x-3^{k+1}))=3x-3^{k+2} .

Thus we have proven that with P 0 P_0 being true and P k P_k being true \Rightarrow P k + 1 P_{k+1} being true, we have proven our claim that P n P_n is true for all non-negative integers n.

Using the above formula, we have f ( 10 ) = 19 f(10)= 19 .

Typo on your 7th line: you say 'f(1)=3', but it equals 2.

Miles Koumouris - 3 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...