Aufgaben:Problem 8

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Concerning part a)

It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \). Wouldn't it be much easier to just:

$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$

and then use Bernoulli-L'Hopital to get:

$$ \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$

Also I think you might mean this one:

$$ ((1-\dfrac{x}{k})^{k}\cdot ... \cdot (1 - x) \cdot 1)^{\frac{1}{k+1}} \leq^{AM-GM} \dfrac{1}{k+1}((1-\dfrac{x}{k})^k + ... + (1 - x) +1) \leq \dfrac{1}{k+1}((1-\dfrac{x}{k}) + ... + (1 - x) +1) $$

where we took away the exponents since \( n \in \mathbb{N}, x \in \left( 0, n \right) : (1 - \frac{x}{n}) < 1 \Rightarrow (1 - \frac{x}{n})^n \leq (1 - \frac{x}{n}) \). (Here I actually see a problem, since this would generally hold for \( x \in \left( 0, 1 \right) \) and not on [0, n]) So:

$$ ((1-\dfrac{x}{k})^{k}\cdot ... \cdot (1 - x) \cdot 1)^{\frac{1}{k+1}} \leq \frac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = (1 - \dfrac{x}{k+1}) $$

Otherwise wouldn't you have to use:

$$ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{2}} \leq \dfrac{1}{2}(k(1-\dfrac{x}{k})^k+1) $$

to use the GM-AM?

Greetings A.