how can my arraylist of object get updated on the onCreateview of a fragment
Is Java "pass-by-reference" or "pass-by-value"?
Recently I'm working on a project aiming to convert the vibration of sound waves into electricity to charge batteries using a piezoelectric material (MFC type). I was trying to use the human voice as an input, but theoretically it was not enough, so do you have any ideas about some places that I can apply my project on it (that must be noisy or can give me vibrations)?  Moreover, has the MFC material been used before in this application? If I used many pieces of it in parallel would that help? What Happens When A Gravitational Wave Interacts With Another One? If two gravitational waves came in contact with each other what would happen? In another question entirely, what happens when a higher gravitational field interacts with a weaker one. Why is my iPhone screen stuck on the 'connect to iTunes' screen? I need help with my iPhone. It's stuck on the 'connect to iTunes' screen.   iTunes does recognize the iPhone, but when I select 'restore', it goes through a few progress bars, then errors again. I've already tried to hard reset the phone, but that doesn't work either. Integrating Factor Method. Why Can We Set the Constant of Integration to Zero? My textbook gives a demonstration of the use of the integrating factor method for solving first order ordinary differential equations:  $\dfrac{dx}{dt} + p(t)x = r(t)$  $\implies g(t)\dfrac{dx}{dt} + g(t)p(t)x = g(t)r(t)$  $\dfrac{d}{dt}[g(t)x]$  We use the product rule and compare the expression to the left-hand side of the above equation:  $g(t)\dfrac{dx}{dt} + \dfrac{dg}{dt}x = g(t)\dfrac{dx}{dt}+g(t)p(t)x$  $\therefore \dfrac{dg}{dt} = g(t)p(t)$  The above equation is separable.  $\therefore \int\dfrac{dg}{g(t)} = \int p(t) dt$  $\implies ln[g(t)] = \int p(t)dt$  My textbook then says that, "We can set the constant of integration to zero -- any function that satisfies $\dfrac{dg}{dt} = g(t)p(t)$ will do for our purposes". However, it gives no explanation as to why this is the case. Why can we set the constant of integration to zero? Why is any function that satisfied $\dfrac{dg}{dt} = g(t)p(t)$ appropriate for our purposes?  I would greatly appreciate it if people could please take the time to elaborate on this. Why does `ls -d *[!e]*` display all files instead of omitting all files that contain an e? ls -d *e* prints out all files containing an "e" somewhere in it. But the opposite, ls -d *[!e]* does not seem to work. Why exactly is that?  is it because ls takes * as its own argument and expands all the files?  I understand I can do: ls -d !(*e*) (in Bash with extglob set, or in ksh). Can $\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}$ have two values? Show that $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot 2^{\aleph_1}$ So basically I want to show:  $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot2^{\aleph_1}$ , Where $\omega\leq\alpha&lt;\omega_1$  There is a proof that I found;    But I don't understand his proof for the limit case of $\alpha$.  To be more specific, I dont understand why $ \left( \sup_{\omega \le \beta &lt; \alpha} \aleph_{\beta}\right)^{\aleph_1} \le \left( \prod_{\omega \le \beta &lt; \alpha} \aleph_{\beta}\right)^{\aleph_1}$  Any insights or help is deeply appreciated.  Cheers Evaluate the limit      $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$   My Attempt:  $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$  How can I solve the problem from this point? Explicit solution for linear regression with two predictors I have some samples of data of the form $x,y$ and $z=f(x,y)$. I wish to fit a plane $$z = Ax + By + C$$ to the data with the smallest mean square errors.  I have found an "answer" in section 3 of  document, but it is left in the form of some equations left to be solved. I just about have the ability to solve these equations, but the process gets so messy that the likelihood of me making a mistake is quite high. Surely someone has written out the full solution longhand somewhere (it might be called "closed form") anyway, in the form $$A=\ldots,$$ $$B=\ldots,$$ $$C=\ldots.$$  EDIT: Maybe "closed form" is the wrong phrase. So let me be clear. I want an explicit solution for $A$, $B$, and $C$, and not a solution that ends with "So now if you can solve these equations then you can discover the values of $A$, $B$ and $C$". $$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor. This is trigonometry from double angles and compound angles. Thus far I've simplified it up to this point. $$2 \sin^2θ – 3\sinθ +1 = 0$$  I'm trying to factorize this to solve for General Solution. I tried taking out a common but that +1 is giving me a problem. I used the quadratic equation to get 1 and 0.5 as an answer, but i don't know how to write it in terms of 2sin(squared)θ and -3sinθ.  Any help will be appreciated. $E[e_te_s\Delta B_t\Delta B_s]$ for $\Delta B_t$ Brownian motion increments and $e_t(\omega)$ a measurable function. I have the following question:    $\textbf{Question}:$ Let $\mathcal{F}_1$ and $\mathcal{F}_2$ be two algebras. Is $\mathcal{F}_1 \cap \mathcal{F}_2$ a $\sigma$-algebra?  I believe the answer is no. I know the definition of an algebra and the definition of $\sigma$-algebra.    I'm assuming I need to show that the intersection of the finite unions of each algebra is finite, but then I get stuck trying to work out how to show it is not a $\sigma$-algebra.  All help is appreciated.  Many thanks, John Problem in notation: $(\vec{v}\cdot\vec{\nabla})\vec{v}$ versus $L_v(\vec{v})$ Is it implied that Porpetina and Queenie Goldstein are Jewish? If you have an $m \times n$ matrix and the rows are linearly independent, are the columns linearly independent too?  I know that it's the case for $n \times n$ matrices. I've yet tried to think about it, and I think the answer is yes, for example the matrix: $\begin{bmatrix} 1 &amp; 2 &amp; 3 \\ a &amp; 2a &amp; 3c \end{bmatrix}$ has linearly dependent rows and columns, but I'm not sure that it works for every $m \times n$ matrix. How do I get a Realtek RTL8723BE wireless card to work? I would like to clarify the differences between by value and by reference.  I drew a picture     So, for passing by value,  a copy of an identical object is created with a different reference, and the local variable is assigned the new reference, so to point to the new copy  How to understand the words: " If the function modifies that value, the modifications appear also within the scope of the calling function for both passing by value and by reference "  Thanks! I would like to solve the beginner's standard exercise which claims that a point of $\mathrm{Spec} \ R$ is closed iff it is a maximal ideal.  The reverse implication is easy. The direct one seems much subtler. If $P = V(I)$ then, since there exist a maximal ideal $M$ containing $P$, it would follow that $\{P, M\} \subseteq V(I) = P$, whence $P=M$. In order to show the existence of $M$ I applied Zorn's lemma to the set of ideals containing $P$.  The question is the following: is there any other proof of the above implication that does not (indirectly) use the axiom of choice? My concern is that I may be using a cannon to shoot a fly. Also, it would be the first time that I see closedness of points to require the axiom of choice. If the axiom is indeed needed in general, are there "nice" classes of rings for which we could get away without it? jQuery Galleria 'fatal error: no theme found.' in IE