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# How would you do this at hand?

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, 7 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 $$d(x) = 4, N(x) = 6$$ · 3 years, 7 months ago

Nice. A suggestion: the claim you made was obvious, you could have skipped the inductive proof. · 3 years, 7 months ago

It was that or study Sanskrit. :D · 3 years, 7 months ago

the given properties are not satisfied by the given function.In fact the given function is not differentiableat x=1 · 3 years, 7 months ago

The notation implies composition, it is written in the question. · 3 years, 7 months ago

The notation isn't implying differentiating the function $$n$$ times , rather the function is being composed $$n$$ times. · 3 years, 7 months ago

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