Difference between revisions of "Aufgaben:Problem 8"
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As for the alternative version: yes, it was a good idea, so everybody can choose the one she/he prefers. | As for the alternative version: yes, it was a good idea, so everybody can choose the one she/he prefers. | ||
− | For \( L^1 \): it is, since \( e^{-x} \) is in Schwartz space, but I've added it to clarify | + | For \( L^1 \): it is, since \( e^{-x} \) is in Schwartz space (over positive real numbers), but I've added it to clarify. I didn't have this requirement in the formulation of the exercise class, but I think you're right. |
Cheers, '[[User:Nick|Nick]] ([[User talk:Nick|talk]]) 17:24, 12 January 2015 (CET)' | Cheers, '[[User:Nick|Nick]] ([[User talk:Nick|talk]]) 17:24, 12 January 2015 (CET)' |
Revision as of 16:27, 12 January 2015
Concerning part a)
First thing: 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:
$$ \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$
Second thing: Nevermind, confused myself...
Greetings A.
I didn't write that: I meant that the limit of the different quotient is the derivative. I was a bit inconsistent with the variable x (now it's fixed, I've used a y). If you prefer Bernoulli it's the same for me. Nick
Okay, I see what you meant now. I personally find it a bit confusing. But I guess it's a matter of taste.
A.
To Part b)
Question:
Since we had an whole exercise sheet on Lebesgue-convergence and i would assume they would ask of us to find an upper bound function g to prove that we're allowed to use that theorem. Why don't you do that? (It might be obvious but those just learning it by heart won't notice.)
Answer:
The upper bound function is for dominated convergence, not for monotone convergence. But I agree it is not so clearly stated, I'll fix it, thanks
Again to part b) As I remember (and according to Wikipedia http://de.wikipedia.org/wiki/Satz_von_der_monotonen_Konvergenz) the function to which \(f_k\) converges, has to be \(\in L^1 \). That should maybe shown too?
Trubo-Warrior (talk) 12:20, 5 January 2015 (CET)
Since we all like alternative versions here on the wiki, and I've encountered some confusion among my friends about the limit-thing we discussed above, I felt so free to put the alternative version in. Best, A.
As for the alternative version: yes, it was a good idea, so everybody can choose the one she/he prefers. For \( L^1 \): it is, since \( e^{-x} \) is in Schwartz space (over positive real numbers), but I've added it to clarify. I didn't have this requirement in the formulation of the exercise class, but I think you're right. Cheers, 'Nick (talk) 17:24, 12 January 2015 (CET)'