I am trying to determine why I am getting a NullPointerException with my program. I need to determine what is null in my NodeLibrary class
What is a NullPointerException, and how do I fix it?
How can I see the history of stories I've read on Apple News? Sometimes I'll be interrupted in the middle of reading a story on Apple News and later when I open the app, it didn't save the story I was previously reading. I want to be able to resume reading the article, how can I find it again? How do we delete subscribers (emails) from Auto suppression list using SSJS. Also, Auto suppression list maintain at SF end or application level. I was watching the film A Game of Shadows starring Robert Downey Junior and Jude Law when this line came up, "...neither you nor I is in control..." (I can’t remember the exact words that ended the phrase; if anyone can supply them, it would be much appreciated.)   It immediately struck me as being odd. When I mentioned this to my boyfriend; an English native speaker, with a PhD in something sciency, he told me it was correct. When I asked why, he couldn't explain but put forward the following sentence as an example. "Neither of us is in control..." However, I disagreed with the wording and said the sentence should have been: "Neither one of us is in control..." The genderless third person, one, is singular, and hence, so too the verb that follows. And we say, "one is" NOT "one are".      "Neither you nor I are in control over ..."    I believe the above sounds better, but it is more logical to say:     "Neither you nor I am in control..."    because the verb agrees with the subject closest to it; however, I'm not sure if I have ever read or heard this solution. So, I tried inverting the subject and auxiliary verb order and the resulting phrase was:    "Neither you nor am I in control..."    which sounds very formal, pompous, and ... odd. Am I imagining things?  The two subjects you and I are two separate individuals who are not joined together, I agree the verb should be in the singular; but why should we use "is"? We don't say, "you is" or "I is". Before anyone rushes off to claim my question is a duplicate of a previous Stackexchange one:  I want to explain why the answer to that question fails to explain why my boyfriend and the film makers consider the  sentence below correct:     Neither you nor I is in control...   As I see it, Michael and Albert are two separate individuals, it therefore stands to reason that the verb should be singular, and "is" agrees with the subject. If I rephrase the neither and nor construction, I would get:          Michael is not correct but neither is Albert   OR          Michael is not correct and Albert isn't either.      Which have the same meaning, (either being the negative equivalent of neither) the sentences are grammatical which leads me to conclude that the verb must be singular and hence         Neither Michael nor Albert is correct      However, when I rephrase the "neither you nor I" sentence following the same procedure I have to say this:        You are not in control but neither am I (‘I am’ is ungrammatical)   OR              You are not in control and I’m not either.      which leads me to surmise that the sentence below must be grammatically correct        Neither you nor I am in control"      So, why does my boyfriend insist that the sentence ought to be:     Neither you nor I is in control   Which one of us is right and why? I'm an android developer and have laptop with 4GB RAM. When I compile my project the RAM is used at 100%.    If I add more RAM will it solve the problem? How to reset evolution settings to defaults? I have not used Evolution for long time now and I would like to reset all the settings and start from scratch. How would I be able to do this? Altium de-highlighting nets after using 'Find Similar Objects...' command Good evening! I'm studing something about finite fields and I didn't find anything that can answer my question.  I have a finite field $\mathbf{F}_q$ and the equation to solve is $x^n=a$. Which are the condition to resolve this equation? For which values of $q, n, a$ there exist a solution?  I will be very happy if anyone can answer this question, possibly with a reference of a book (or an article) that explai this with more details.  Thank in advance to everyone who can answer this one, I really need to understand this. Cardinality of the quotient ring $\mathbb{Z}[x]/(x^2-3,2x+4)$ This problem is from a practice exam I was working on.     What is the cardinality of the quotient $\mathbb{Z}[x]/(x^2-3,2x+4)$ ?   Thoughts. If I find a ring that is easier to handle then this then I can go from there. So I think this is isomorphic to $\mathbb{Z}_2[x]/(x^2-3,x+2)$. And then I am stuck.  Am I correct so far? What should I be looking to do from here?  Thanks for your time and your answers. Why are my villagers attempting to breed (they have the red hearts) but then stop breeding suddenly and gray "angry" particles pop up? I set up a villager breeder and set down 4 beds with 2 villagers in the middle and a carrot farmer supplying them with carrots. The first time both of those villagers bred the breeder worked perfectly fine. After that, however, the gray angry particles popped up every time they would try to breed (they would engage in breeding at a normal pace, just could not actually produce a baby). I even killed the one villager they produced in case it took a bed and even then, the two villagers would not breed anymore. Is there something I have done wrong? This seems odd considering they were only able to breed once and even by eliminating their offspring the villagers could not breed again. Edit: version is 1.16.4 Showing that $f_0 (x_1, \ldots, x_m) \mathrm tr A =\displaystyle{ \sum_{i=1}^n} f_0(x_1, \ldots, Ax_i,\ldots, x_m)$ Question:  Consider $f: (-\epsilon, \epsilon) \to \mathbb R^{m^2}$ a differentiable path of matrices $m \times m$ such that $f(0) = I_m$ and the function $g: I \to \mathbb R$ is defined by $$g(t) = \det f(t)$$  Show that $g'(0) = \mathrm{tr} \,a $ (trace of the matrix $a$), where $a = f'(0)$.   Attempt:  First we use that $\det f(t) = \det (f_1(t), \ldots, f_m(t)) $, where $f_i$ is the $i$-th row of the matrix $f(t)$ $m \times m$. Considering the operator $': \mathbb R^m \to \mathbb R^m$, $f(t) \mapsto f'(t)$  With this in mind $$\begin{align}g'(t) = \det'f(t) &amp;= \sum_{i=1}^{m} \det (f_1(t),\ldots,f'_i(t),\ldots, f_m(t)) \\&amp;\underbrace{=}_{(*)}\det(f_1(t), \ldots, f_m(t))\, \mathrm tr f'(t) \end{align}$$  For $t = 0$ we have   $$\begin{align}g'(0) &amp;=\det(e_1, \ldots, e_m) \,\mathrm tr f'(0) \\&amp;= 1 \dot\, \mathrm tr \,a \\&amp; = \mathrm tr \, a\end{align}$$  I'm trying to come up with a proof using the fact that $\det$ is a $m$-linear form.   In order to prove $(*)$ I need the following equality, which according to this  holds.   If $A: V \to V$ is a linear operator and $f_0$ is a alternated $m$-linear form, such that $f_0 (e_1, \ldots, e_m) = 1$ then   $$f_0 (x_1, \ldots, x_m) \mathrm tr A = \sum_{i=1}^m f_0(x_1, \ldots, Ax_i,\ldots, x_m) $$  Could someone give a proof to this? I have an old PowerMac G4 that I installed Lubuntu 16.04 on a while ago. I've been getting notifications in the terminal that 18.04 is available to upgrade, but the upgrader fails to run, saying that the package ubuntu-minimal is not available. I haven't seen any ISOs for PowerPC available, and ports.ubuntu.com doesn't have any bionic distributions available for architecture  powerpc. But while searching I found  that suggests that there is/was a version of Ubuntu 18.04 for PowerPC. Is this just a pre-generated page, or is there somewhere I can get Ubuntu Bionic for PPC? How to overlay images I want to overlay one image with another using CSS. An example of this is the first image (the background if you like) will be a thumbnail link of a product, with the link opening a lightbox / popup showing a larger version of the image.   On top of this linked image I would like an image of a magnifying glass, to show people that the image can be clicked to enlarge it (apparently this isn't obvious without the magnifying glass). For random variable $X$ that follows some distribution, $f(x)$ is the probability density function of that distribution if and only if $$\mathbb{E}[\phi(X)] = \int_{-\infty}^\infty \phi(x) f(x)dx$$ for all functions $\phi$.  Context: My professor used this in lecture to demonstrate a way to find the distribution of $cX$ given random variable $X$ that follows a specific distribution.   What is the name of this theorem? Have I missed any qualifications/caveats? I am in particular curious about whether "for all functions" is correct. My professor mentioned "positive, bounded, deterministic functions," but I am not sure what he meant by that. Where can I find a proof of this theorem? The computer gave me this output in a window:  E: Encountered a section with no Package: header E: Problem with MergeList /var/lib/apt/lists/archive.ubuntu.com_ubuntu_dists_natty_main_binary-i386_Packages E: The package lists or status file could not be parsed or opened.   How can I fix this? How to reconnect a logically disconnected USB device? It is possible for a USB storage device to become logically disconnected: the device is still plugged in, but is invisible from the operating system (e.g. it's not listed under /proc/bus/usb). Maybe this can happen to other kinds of USB or other removable devices as well. This can happen for example:   ; after VirtualBox has claimed a USB storage device (the device disappears from the Linux host when it is attached in the guest, and does not automatically reappear if it is detached from the guest).   What is going on in the driver or in the hardware, and most importantly, (how) can the device be logically reconnected (without physically plugging it out and back in)?  Note that this question is not about mounting. In the “logically disconnected” state, the kernel believes there is nothing on the USB bus (and so of course there is no entry under /dev). What does it mean for a QFT to not be well-defined? It is usually said that QED, for instance, is not a well-defined QFT. It has to be embedded or completed in order to make it consistent. Most of these arguments amount to using the renormalization group to extrapolate the coupling up to a huge scale, where it formally becomes infinite.  First of all, it seems bizarre to me to use the renormalization group to increase resolution (go to higher energies). The smearing in the RG procedure is irreversible, and the RG doesn't know about the non-universal parts of the theory which should be important at higher energies.   I have read in a note by Weinberg that Kallen was able to show that QED was sick using the spectral representation of the propagator. Essentially, he evaluated a small part of the spectral density and showed that it violated the inequalities for the field strength $Z$ that are imposed by unitarity + Lorentz invariance, but I do not have the reference.  So what I would like to know is :   What does it take to actually establish that a QFT is not well-defined?   In other words, if I write down some arbitrary Lagrangian, what test should I do to determine if the theory actually exists. Are asymptotically free theories the only well-defined ones?   My initial thought was: We would like to obtain QED (or some other ill-defined theory) using the RG on a deformation of some UV CFT. Perhaps there is no CFT with the correct operator content so that we obtain QED from the RG flow. However, I don't really know how to make that precise.   Any help is appreciated. How to copy all files of a certain type into a folder from command prompt? I want to copy all files of a certain type (like .docx, .pdf, etc) on my PC into a folder. How can I do that? How to find out who called a method? Disable or prevent multitouch in Activity FileSystemWatcher Changed event is raised twice