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Let \(f(x) = 0\) if \(x \leq 1\), and \(f(x) = \log_2{x}\) if \(x>1\),

and let \(f^{(n+1)}(x) = f(f^{(n)}(x)) \) for \(n\geq 1\) with \(f^{(1)}(x)=f(x)\). Let \(N(x) = \min \{n \geq 1 : f^{(n)}(x) = 0\} \). Compute \(N(425268)\).

Note by Paramjit Singh
3 years, 5 months ago

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Let \( d(x) ( x > 1) \) denote the smallest\( n \) for which \( x \leq 2^{2^{2\ldots}} \) where n is the number of 2s in the exponent number. (Eg. d(4) = d(2^2) = 1, d(5) = 2 )

Claim :- \( N(x) = d(x) + 2 \)

Base case :- For d(x) = 0, Therefore \( 1 < x \leq 2 \)

\( log_2 1 < log_2(x) \leq log_2 \)

\( 0 <log_2(x) \leq 1 \)

Therefore, \(f(log_2x) = 0 \)

\( f^2(x) = 0 \)

\( N(x) = 2 = d(x) + 2 \)

Inductive Case :- If true for d(x) = n, then,for d(x) = n+1,

\( 2^{2^{2\ldots }}(n twos) < x \leq 2^{2^{2\ldots}}((n+1) twos) \)

\( log_2 2^{2^{2\ldots}}(n twos) < log_2(x) \leq 2^{2^{2\ldots}}((n+1) 2s) \)

\( 2^{2^{2\ldots}}((n-1) twos) < y \leq 2^{2^{2\ldots}}(n 2s)\) where \(( log_2x = y)\)

Therefore \(d(y) = n, f^{n+2}(y) = 0 \)

\( f^{n+2}(log_2x) = 0 \)

\( f^{n+3}(x) = 0 \)

\(N(x) = n+3 = d(x) + 2 \)

Now, \( 2 = 2, 2^2 = 4, 2^{2^2} = 16, 2^{2^{2^2}} = 35536, 2^{2^{2^{2^2}}} = 2^{35536} >> 425268 \)

Therefore \( d(x) = 4, N(x) = 6 \) Siddhartha Srivastava · 3 years, 5 months ago

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@Siddhartha Srivastava Nice. A suggestion: the claim you made was obvious, you could have skipped the inductive proof. Paramjit Singh · 3 years, 5 months ago

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@Paramjit Singh It was that or study Sanskrit. :D Siddhartha Srivastava · 3 years, 5 months ago

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the given properties are not satisfied by the given function.In fact the given function is not differentiableat x=1 Kshetrimayum Saounkisor · 3 years, 5 months ago

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@Kshetrimayum Saounkisor The notation implies composition, it is written in the question. Paramjit Singh · 3 years, 5 months ago

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@Kshetrimayum Saounkisor The notation isn't implying differentiating the function \(n\) times , rather the function is being composed \(n\) times. Jit Ganguly · 3 years, 5 months ago

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