$$\large \begin{align*} x^n & = -b^n \\ & \Updownarrow \\ x &= e^{i\cdot\left(\pi \cfrac{1+k}{n} \right)} \qquad \quad k=0..n \end{align*}$$ ? Correct, perhaps? If it is, I am still clueless how to explain it...
If I only had time =) I guess I will learn this stuff propperly in my winter vacation. Exam is tommorow. Having 5 exams and reading 2500 pages in a month is tough.
@Matt No, I didn't mean to discourage you. In any case, just add a link to the old question and try to highlight how this is different. You should be fine, I think.
@N3buchadnezzar You want me to be real quick and give a solution?
@N3buchadnezzar BTW, I slightly lost my cool somewhere in between. I guess that's because I was talking to 3 people almost simultaneously. Sorry about that... =)
@Srivatsan One last thing:Can you please see whether my answer is correct or has a flaw? I think it is correct, but then it is best to get one's answer verified by someone competent.
Let's summarize what is given and what others have answered.
$X_1, X_2, \cdots X_n$ are $i.i.d.$ exponential variables with mean $1$.
As told by David Mitra, $S_k$, i.e. sum of k independent exponential variables with mean $1/λ$ has a Gamma distribution with parameters $α=k$ and $λ$ .
Given $Y_...
@NikhilBellarykar Which statement? That $S_n$'s are independent? That's clearly false. Asymptotically, they could be independent -- whatever that means. I don't quite know what that means anyway.
I have been a computer nerd for more than half of my life. If you want to trick me, try harder and when I am not on the bus from my iPhone. Preferably, though, never.
So you'd migrate it there? To me it looks like "what algorithm did X use to produce Y". Answer: Maybe algorithm A or maybe algorithm B... maybe C? To produce a feasible letter set for example to play in English you'd take the English letter distribution and then select the letters according to that at random. Then you'd have to check if that's easy enough to play and then you'd probably find that you need to tune it a bit to make it easier.
Well, I just want to answer the question for the simple reason that I can, upto a level. As I am a newbie here, I thought to give this question a try, thats all :)
See, it's not about whether you are a new user or not. You say you can answer the question upto some level. I just happen to think it's a hard question.
Again it's not that I am discouraging you. You might have an idea that eludes the rest of us.
yes. We have iid exponential variables. We thus know the pdf and cdf of their partial sums i.e. gamma distribution. The cdf of absolute value of difference between a partial sum and its mean is also given in a comment. What is required is the cdf of maximum of the set of absolute value of difference between a partial sum and its mean and that can be calculated from the order statistic approach.
The absolute differences translate to partial sums which are dependent. So that involves some messy calculations. Otherwise, in my opinions, that's all about the problem.
@NikhilBellarykar I see. This looks sound, whatever you wrote till now. I can even imagine writing it down as a giant summation, integrals and whatnot. But it's perhaps not clear that this can be simplified much, is it?
And, btw, I should really get going. Sorry I couldn't stay longer. See you.
@AsafKaragila: logarithms reduced multiplication to addition. Laplace transform reduced differentiation and integration to multiplication and division. Does there exist a method that reduces something more complicated to say, taking transforms of functions?
That startled me, I thought the room was "empty" and expected to hear back in a few hours or so. So the question is related to [this](http://math.stackexchange.com/questions/67370/smooth-functions-with-compact-support-are-dense-in-l1)
Does anyone know if there exist general methods to improve the speed of convergence of a series? I.e., can we find a different series that converges faster to the same number?
I was going to ask if I got the big picture and then while typing I found an inconsistency. Now I'm going to ask the related question and maybe come back with this one. So in that question they claim $J_\varepsilon$ is a mollifier. I looked up the definition and one of the properties that it should have is that $\lim_{\varepsilon \to 0} J_\varepsilon (x) = \delta(x)$, so I tried to verify this. Somehow I get $\infty$ for all $x$ in $\mathbb{R}$.
@tb This morning I noticed that you had edited this question about a month after I'd asked it. Do you re-read old stuff at random to fix typos? Or did you go through my old stuff to see what I hadn't been doing properly?
@JonasTeuwen Yes I was thinking about this while trying to get the big picture of the question. Just going to go over it again now.
@Matt As that is a distributional limit it actually states the same as the convolution with the $\delta$ distribution in $x$ picks up the value of the original function in the point $x$. Roughly speaking.
A quite well-known Dutch (for Dutch standards) algebraist said about analysis that he never has an "Aha Erlebnis" when seeing analysis proofs. He says it is all about computing integrals by splitting them into pieces and calculate them separately... :'D.
@Matt The answer is plain and simple: I was eradicating the typo "Lebesque" from this site, so I searched for it and edited all posts in which I found it. It was around the same time as Srivatsan and I were taking care of the various spellings of continuous...
@JonasTeuwen I don't know either how I wandered off the right path. When I was in first year I was looking forward to not doing analysis anymore and only do discrete maths like e.g. algebra and graph theory. I wonder how I ended up the way I ended up. : P
But I think you can take a not necessarily continuous $f$ in $L^p$ and then mollify it to get a smooth function $f_\varepsilon$ by convoluting it with a mollifier.
@JonasTeuwen Click the up arrow next to your name.
So if you convolve an integrable function with a continuous function the result is continuous. If you convolve an integrable function with a $C^k$ function the result is $C^k$, and so on. The point is essentially that $(f \ast g)' = f' \ast g = f \ast g'$, so you can push the derivatives on the function which is differentiable.
@JonasTeuwen I was badgered into it. I have to have what's called "complementary" courses. I didn't fancy the one's offered so I applied to get this set theory course credited as complementary course.
Q : There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probability //I am unable to understand what does author means by tripled death probability?
@Matt Well, if you know that for every $L^p$ function $f$ and $\varepsilon \gt 0$ you find a continuous function $g$ with compact support such that $\|f-g\|_p \lt \varepsilon$. Then the thing that remains to show is that you can find a smooth function with compact support which is arbitrarily close to $g$.
Yes. Exactly. What you gain by replacing $f$ by $g$ is that now you have a function $g$ which is continuous and of compact support, hence it is uniformly continuous. Now if you take $\delta$ so small that $|x-y| \lt \delta$ implies $|f(x) - f(y)| \lt \varepsilon$ then $J_{\delta}(x) \ast g$ can't be too far away from $g$.
@FreakEnum I think there is some information that was lost in translation.
It has to be 1/3 of dieing plus living which is given by 2/3 * p^3 as p is tripled so raised to 3 probability of die is 1/3 or its get tripled the probability of tripled is 2/3*p*p*p that's 2/3*p^3
here is another problem : If the area was hit by a virus and so the decrease in the population because of death was x/3 and the migration from other places increased a population by 2x then annually it had so many ppl. find our the population in the starting.
@QED If you look at the beginning of that article it says "Logicians spend lots of time carefully defining open variables, closed variables, and so forth..."
