Find this closed form?
$$I=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos{(2kx)}x^{m-1}e^{-ax}dx,m\ge 1,a>0$$
use
$$\cos{(2kx)}e^{-ax}=e^{-ax}\cdot\dfrac{e^{2kix}+e^{-2kix}}{2}=\dfrac{1}{2}(e^{(2ki-a)x}+e^{-(a+2ki)x})$$
so
$$I=\dfrac{1}{2}\sum_{k=1}^{n}\int_{0}^{\infty}(e^{(2ki-a)x}+e^{-(a+2ki)x})x^{m-1}dx...
@math110 That one is a simple application of the polygamma function. The higher order of $m$ are obtained by considering higher orders of polygamma function. It should flow naturally by considering the derivatives with respect to $a$ ....
For $m=1$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^1}+\frac{1}{(a+2 i k)^1}\right)=\frac{1}{2} i \left(\psi ^{(0)}\left(1+\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1-\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1+\frac{i a}{2}\right)+\psi ^{(0)}\left(1-\frac{i a}{2}\right)\right)$$
For $m=2$
$$\sum _{k=...
@Chris'ssis Have you seen this integral before $$\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$
Hallo @DanielFischer!!! :D I want to describe an algorithm that given an unsorted array $B$ that stores $m$ integers, and any integer number $y$, determines if there are two elements of the array of which the quotient is equal to $y$. The time complexity of the algorithm should be $O(m \log m)$.
Couldn't we sort tha array with the algorithm Heapify?
How could we check if there are two elements of the array of which the quotient is equal to y?
@evinda Sorting is good. Whether you use heap sort or any other algorithm with a guaranteed $O(m\log m)$ bound is your choice. Now think about how you would check whether a sorted array contains two elements with quotient $y$ in $O(m)$ time.
@Mathy I see Mr. Gruber helped you, that method certainly is better than my tedious solution considering $4,5,6$ edges distinctly (really tedious. More tedious than that.)
@evinda That depends on how you interpret "check as pairs from the beginning till the end". You need not look at all pairs (of course, that would be $\Theta(m^2)$), and if you find a pair, you can stop.
@evinda Yes. You want the quotient to be $y$. (By quotient, is the integer quotient $\left\lfloor \frac{a}{b}\right\rfloor$ meant or the rational quotient $\frac{a}{b}$? That decides whether you look for $yb \leqslant a < (y+1)b$ or $a = yb$.)
So I would appreciate some hint. Anyhow, I have a continuous unto function $u:X\to Y$. Given that for every topological space $Z$ and every function $w:Y\to Z$, if $wu$ is continuous then $w$ is continuous, I want to show that $X/u$ is homeomorphic to $Y$. I've been thinking of setting $Z$ to be $X$, then I have that $f^{-1}$ is continuous, or $Z$ to be discrete space, but I have no clue how that helps me (either)
@BalarkaSen @r9m I'm just wondering till this day, how did he got those answers? He can answer every problems appears on timeline within minutes, probably less than 3 minutes.
@Balarka I really have no clue how to build it up. I've been trying to think what do I need to know about $u$ that will allow me said isomorphism - but it's so general I don't think it's gonna help. I mean, I need the distinct $u(x)$'s to fit open groups in $Y$ according to sources in $X$ and vice-versa, but it's not helpful as I said..
@evinda That doesn't have an important impact, it just means we write the test as a < y*b in the loop and check for a == y*b, otherwise we'd test a/b < y and check a/b == y.
It seems some mods undeleted an answer of mine I'd rather see deleted. It's about this question. Actually, it has two answers of mine, and I am quite sure to have deleted both. The other deleted one has undelete votes too, but as the question has changed, undeleting would be bad.
Originally, the question was as interesting as indicated by its title. I gave an answer, then found that it did not exactly answer the question (which seemed to not allow a nontrivial multiplier system), so I deleted it and gave an updated reference in the comments. The OP than included the criteria given in my early answer and changed the question into asking why one would need integrality of the exponents in there.
@evinda Well, due to the array being sorted, we can rule out a huge number of pairs without looking at them. Just in case: If the array can contain negative as well as positive numbers, we need to take care of that and treat the part with negative numbers and the one with positive numbers separately.
I gave an answer to the updated question too. That's the currently undeleted one. The answer is really un-enlightened and involves guessing; and the question it answers stems from a change that actually moved it out of my field of interest. I felt that it did not look good, particularly not in my answers list, and so I deleted it. Now I find it undeleted. It might be stupid to enter another delete-undelete cycle. Suggestions?
@ccorn Since it was undeleted by ordinary 10k users, you can delete it yourself. You should then edit in a note "I'd rather this be deleted because ...". Another option is to flag for moderator attention and ask for the answer to be disassociated from your account [hmm, or does one need community team involvement for that and hence use the "contact us" link?].
@robjohn It seems it's a joke because I found this interesting fact $$ \int_{-1}^{1}\frac{\mathrm dx}{\left ( x^2+1\right )\left ( e^x+1\right )}=\frac{\pi}{4} $$
@evinda To get the idea, suppose all entries are positive. You have a $y > 1$ (the case $y = 1$ is clear), and look at two array elements A[k] and A[r] with r > k. Where do you look next if A[r] < y*A[k]?
@MikeMiller If $G$ is a nonabelian group with every subgroup normal, isn't it true that there exists a nontrivial subgroup $H$ of $G$ such that every nontrivial subgroup of $G$ contains $H$?
@DanielFischer I love The Doors!!!! My dad got me into 60's and 70's music in middle school and I still haven't grown out of it! Although, my musical taste is currently more diverse than it was back then...
@DanielFischer So do we have to find the mid of the array and if A[mid] < yA[mid+1], we will check if $A[p] == yA[p+1]$ for $p \in [mid+1,m-1]$ and elsewhise we check if $A[p] == y*A[p+1]$ for $p \in [0, mid-1 ]$ ?
@evinda No. If we have negative and positive entries, we need to find $0$ or the position of the first positive entry. If we have only positive entries, we begin at the start, and if only negative entries, we begin at the end.
@evinda It gives you a split of the array into two parts. [Aside, I haven't mentioned it yet, if the desired quotient $y$ is negative, you do things a little differently; if $y = 0$, it basically comes out to checking whether $0$ is in the array but not all elements are $0$.] For $y > 0$, you look at both parts separately.
While trying to find several references to answer Pranav's problem, I encounter the following multiple integrals
$$I=\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$
Question :
Does the above integral have a closed form...
@DanielFischer Ok.. But when we consider the two parts of the array, don't we have to look at the first element with the second, the second with the third and so on? So, won't the time complexity be again O(n^2)? Or am I wrong?
@DanielF Can you guess what the OP is having trouble with here? I tried to be a little bit more explicit than the author of their book, but I doubt my answer will help much without knowing where they're actually stuck.