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 $N(425268)$ .

#Goldbach'sConjurersGroup #FunctionalAnalysis

Easy Math Editor

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Comments

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 - 7 years, 4 months ago

Nice. A suggestion: the claim you made was obvious, you could have skipped the inductive proof.

A Brilliant Member - 7 years, 4 months ago

It was that or study Sanskrit. :D

Siddhartha Srivastava - 7 years, 4 months ago

the given properties are not satisfied by the given function.In fact the given function is not differentiableat x=1

kshetrimayum saounkisor - 7 years, 4 months ago

The notation isn't implying differentiating the function $n$ times , rather the function is being composed $n$ times.

Jit Ganguly - 7 years, 4 months ago

The notation implies composition, it is written in the question.

A Brilliant Member - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$