Tessellate S.T.E.M.S - Computer Science - School - Set 3 - Subjective Problem

Consider the function $f$ described below.

$f : \mathbb{N} \to \mathbb{N} \\ f(1) = 1 \\ f(2n) = 2 f(n) \\ f(2n + 1) = 4 f(n)$

Define $g(n) = f(n) - f(n-1)$ .

Is $g(n)$ bounded? If yes, what is the maximum value that $g$ attains? If not, give a proof of the claim.

This problem is a part of Tessellate S.T.E.M.S.

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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`	$\sum_{i=1}^3$
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Comments

It is not bounded since if we set $n=2^x+1$ then $f(n)=f(2^x+1)=4f(2^{x-1})=2^{x+1}$ and $f(n-1)=f(2^x)=2^x$ . Thus $f(n)-f(n-1)=2^x$ which is obviously not bounded since we can increase the value of x

Haran Mouli - 3 years, 5 months ago

Nice argument. We can work out the details by proving smaller claims with induction.

Agnishom Chattopadhyay - 3 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}`	$\boxed{123}$