Confirm that this function is bounded

This is a workout from Problems from guide by Andreescu and also Dospinescu. When it was posted on AoPS a year ago I invested numerous hrs attempting to address it, yet fruitless, so I am wishing a person below can inform me.

Trouble: Confirm that the function $f : [0, 1) \to \mathbb{R}$ specified by

$\displaystyle f(x) = \log_2 (1 - x) + x + x^2 + x^4 + x^8 + ...$

is bounded.

An initial monitoring is that $f$ pleases $f(x^2) = f(x) + \log_2 (1 + x) - x$. I experimented with utilizing this useful formula for some time, yet could not fairly make it function.

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2019-05-05 04:55:17
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Answers: 2

Starting from (the all-natural logarithm of) $(1-x)^{-1} = (1+x)(1+x^2)(1+x^4) \dots$, it comes to be more clear where the $\log(2)$ variable originates from.

One needs to show that $\Sigma (x^{2^k} - C\log(1 + x^{2^k}))$ is bounded amount of favorable terms. The amount of the first $n$ terms comes close to $n - Cn\log(2)$ as $x \to 1-$, so we require $C = 1/\log(2)$ if there is to be boundedness.

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2019-05-08 09:18:34
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OK, a 2nd method is required (yet it in fact ends up the trouble). It behaves and also straightforward sufficient that it's possibly what the writers planned by a "Book" remedy.

Allow $f(x) = x \log(2) - \log(1+x)$. We intend to show that $S(x) = f(x) + f(x^2) + f(x^4) + \dots$ is bounded. Due to the fact that $f(0)=f(1)=0$ and also $f$ is differentiable, we can locate a constant $A$ such that $|f(x)| \leq Ax(1-x) = Ax - Ax^2$. The amount of this bound over the powers $x^{2^k}$ is telescopic.

Notification that the duty of $\log(2)$ was to make certain that $f(1)=0$.

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2019-05-08 08:33:18
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