This is what I have done so far. Although the problem in this code is that in 6th line, after 9 it prints 10,11 and so on instead of starting again at 0,1... etc.
How do I generate a random int value in a specific range? I have tried the following, but those do not work: Attempt 1: randomNum = minimum + (int)(Math.random() * maximum);  Bug: randomNum can be bigger than maximum. Attempt 2: Random rn = new Random(); int n = maximum - minimum + 1; int i = rn.nextInt() % n; randomNum =  minimum + i;  Bug: randomNum can be smaller than minimum.
Right place to ask a "discussion" question (I'm setting up a blog, what should....) Is there any module or some configuration that I can define the meta tags for the views page display in drupal 7 using  3.0? How to enable default file manager(nautilus) to handle desktop icons How to arrive at $\int_{0}^{1} f^3(x) \, dx < (\int_{0}^{1} f(x) \, dx)^2$? Define $f : [0,1] \to \mathbb{R} $ differentiable satisfying $f(0)=0$ and $0&lt;f'(x)&lt;1$. Prove:  $\displaystyle\int_{0}^{1} f^3(x) \, dx &lt; \left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2$ I found the following answer,  Define $F(x)=\displaystyle\int_{0}^{x} f^3(t) \, dt - \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)^2$ We have $$F'(x)=f^3(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)f(x)=f(x)\left[f^2(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)\right]$$ and $$\left[f^2(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)\right]'=2f(x)\left[f'(x)-1\right]$$ from which we can arrive at the conclusion.  But I felt a little confused. In the past I usually proved the inequality by classical inequalities and something like Taylor expansion to use the condition about derivative. But this answer is little out of my expectation. Do we have other proofs or ideas to arrive at $\displaystyle\int_{0}^{1} f^3(x) \, dx &lt; \left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2$? And other thoughts about solving these types of issues? I think when we talk about $\left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2 &gt;\ldots \,$ it's a little difficult to deal with. I don't quite know what tools to use. Can less be used without any comparison? Can I use &quot;less&quot; in sentences like this:  1 Why do we have so less number of students for this class? 2 My song collection is very less.  &quot;Small&quot; sounds better in both examples, but I would like to know if using &quot;less&quot; in these sentences is correct? How to upgrade Git on Ubuntu Hardy? What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to? This problem comes from another equation on another question (. I tried to split it in half but I found out that $$\sum_{k=0}^\infty \frac{k}{2^k}$$ can't be divided. Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ I wrote that $$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$ But that's not what I wanted. Could anyone help me? Runtime's exec() method is not redirecting the output Word or idiom for someone holding (seemingly) contradictory views I think some good words would be conflicted, or confused. As an example of context, think of someone who believes in freedom of speech, but, at the same time, they would support banning comedians who make jokes about certain issues that person cares about. In the review page, it shows stats of recent reviewers. But it is showing wrong stats. Because today (on Monday) it shows week count 53 (on Meta Stack Overflow) which is impossible as there is a limit of 20 reviews per day.    PS: This bug   and . But the bug has occurred again. Let $C$ be the class of all real-valued continuously differentiable functions $f$ on the interval $[0,1]$ with $f(0)=0$ and $f(1)=1$. How to show that $$\inf_{f\in C}\int_0^1|f'(x)-f(x)|dx=\frac{1}{e}?$$  I have been able to show that $1/e$ is a lower bound. Indeed, $$\begin{align*} \int_0^1|f'(x)-f(x)|dx &amp;= \int_0^1|f'(x)e^{-x}-f(x)e^{-x}|e^xdx \\ &amp;\geq \int_0^1\left(f'(x)e^{-x}-f(x)e^{-x}\right) dx \\ &amp;= \int_0^1 \frac{d\left(f(x)e^{-x}\right)}{dx}dx  \\ &amp;= f(1)e^{-1}-f(0)e^{0}\\ &amp;=\frac{1}{e}. \end{align*}$$ But how to show this is the infimum? Is there a function $f\in C$ such that we get $\int_0^1|f'(x)-f(x)|dx=1/e$? Partition around bad drive sectors? I've seen a lot of hard drives with only a few bad (unrecoverable) sectors (as indicated by SMART). Of course the proper thing to do with a bad drive is replace it, as there's a good chance of the drive crashing completely in the near future; I'm aware of that. However, in the interim, or for those who don't want to spend $70 to save data they don't value highly anyway (sigh, non-technical people), or just to get some use out of old hardware instead of throwing it away, I'm wondering if there's a way to identify where the unrecoverable sectors are ("where" meaning numerically, from 0 to the number of sectors on the drive), so I could stick a blank partition in that general area. Yes, this would involve low-level-formatting the drive and reinstalling the OS, of course.  So:  Would partitioning around unrecoverable sectors even be a workable way of reducing the disk of losing data?  If so, is there a way to find the location of these unrecoverable sectors? How can a distribution have infinite mean and variance? It would be appreciated if the following examples could be given:   A distribution with infinite mean and infinite variance. A distribution with infinite mean and finite variance. A distribution with finite mean and infinite variance.  A distribution with finite mean and finite variance.   It comes from me seeing these unfamiliar terms (infinite mean, infinite variance) used in an article I am reading, googling and reading a thread on the , and not finding it a sufficiently clear explanation. I also haven't found any explanations in any of my own textbooks. If $\int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then every line $y=mx$ Intersect $ y^2+\int_{0}^{x}f(t)dt=2$ If $f$ is a continuous function with $\displaystyle \int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then show that every line $y=mx$ Intersect the curve $\displaystyle y^2+\int_{0}^{x}f(t)dt=2$  $\bf{My\; Try::}$ Put $y=mx$ in $$\displaystyle y^2+\int_{0}^{x}f(t)dt=2\;,$$ We get $$\displaystyle m^2x^2+y^2+\int_{0}^{x}f(t)dt=2.$$ We can write it as $$\displaystyle 2\int_{0}^{x}2m^2tdt+\int_{0}^{x}f(t)dt=2$$ So $$\displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2 =0$$ Now Let $$g(x) = \displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2.\;,$$ When $x\rightarrow \infty\;,$ Then $g(x)\rightarrow \infty.$ (Where $g(x)$ is an Continuous function.) Now I did not Understand How can i solve that, Help me Thanks How to validate an email address in PHP What word means what many people think 'ironic' means? 'Ironic' is often used to mean an unusual coincidence rather than its true meaning which is closer to sarcastic. That being said, is there a word that would be a good replacement for what many people seem to think it means? How do you calculate this limit $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{n^2+k^2}$? How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$   is unable to calculate it.  This is a question from a questions booklet, and the options for the answer are--  $\begin{align} &amp;A) \dfrac{\pi}{2} \\ &amp;B) \log 2 \\ &amp;C) \dfrac{\pi}{4} \\ &amp;D) \dfrac{1}{2} \log 2 \end{align}$ "At least one" - singular or plural subject? Smart phone Smartphone Smart-phone   I am really looking for some help regarding hyphen (-) issue in text found in research article. Is there any standard method how to use it? Movie featuring  space crew, murders, a forbidden door and ending with a revelation I searched wide and far and couldn't come up with a title for a movie I watched somewhere around the mid-90ies up to early 00's as a rerun in late-night television (in Germany, specifically on Vox; but I'm sure it was an American movie to begin with).  Basically, what I remember from the story revolves around a crew in a space station or maybe space ship of some sort. Some crew members murder other crew members and there is some kind of door they are not supposed/unable to go through. Late in the movie they do go through the door and I'm pretty sure one of the first views is a skeleton of a dead security guard. It turns out the whole thing was some kind of simulation gone wrong and they find themselves to be in a (nuclear?) wasteland. And that's pretty much the revelation/wrap-up of the movie, as I recall it.  I would be delighted if somebody had any idea what movie this could be :) Thanks!  EDIT: The whole movie was in color and I think it was feature-length, not just 45 mins.