I can't get my program to test: "java.lang.NullPointerException"
What is a NullPointerException, and how do I fix it?
Commandline shortcut for current directory similar to ~ for home directory? I'm somewhat new to using logistic regression, and a bit confused by a discrepancy between  my interpretations of the following values which I thought would be the same:    exponentiated beta values  predicted probability of the outcome using beta values.    Here is a simplified version of the model I am using, where undernutrition and insurance are both binary, and wealth is continuous:  Under.Nutrition ~ insurance + wealth   My (actual) model returns an exponentiated beta value of .8 for insurance, which I would interpret as:     "The probability of being undernourished for an insured individual is .8 times the probability of being undernourished for an uninsured individual."   However, when I calculate the difference in probabilities for individuals by putting in values of 0 and 1 into the insurance variable and the mean value for wealth, the difference in undernutrition is only .04. That is calculated as follows:   Probability Undernourished = exp(β0 + β1*Insurance + β2*Wealth) /                              (1+exp(β0 + β1*Insurance + β2*wealth))   I would really appreciate it if someone could explain why these values are different, and what a better interpretation (particularly for the second value) might be.     Further Clarification Edits As I understand it, the probability of being under-nourished for an uninsured person (where B1 corresponds to insurance) is:  Prob(Unins) = exp(β0 + β1*0 + β2*Wealth) /               (1+exp(β0 + β1*0+ β2*wealth))   While the Probability of being under-nourished for an insured person is:  Prob(Ins)= exp(β0 + β1*1 + β2*Wealth) /            (1+exp(β0 + β1*1+ β2*wealth))   The odds of being undernourished for an uninsured person compared to an insured person is:  exp(B1)   Is there a way to translate between these values (mathematically)? I'm still a bit confused by this equation (where I should probably be a different value on the RHS):  Prob(Ins) - Prob(Unins) != exp(B)   In layman's terms, the question is why doesn't insuring an individual change their probability of being under-nourished as much as the odds ratio indicates it does?  In my data, Prob(Ins) - Prob(Unins) = .04, where the exponentiated beta value is .8 (so why is the difference not .2?) I want to shade the area of overlap for adjacent normal curves, shown in the lower graph. I can get shading between the first two curves but not for subsequent curves. I need shading at overlapping areas indicated by the arrows. I applied the "logic" of the first fill for the subsequent curves but that does not work.   The arrows and letters in the example image are not part of the MWE.   \documentclass{article}  \usepackage{pgfplots} \pgfplotsset{ticks=none} \pgfplotsset{compat=1.7}  \usepgfplotslibrary{fillbetween}  \pgfmathdeclarefunction{gauss}{2}{%normal distribution where #1 = mean and #2 = sd}     \pgfmathparse{exp(-((x-#1)^2)/(2*#2^2))}% }  \pgfplotsset{baseplot/.style={%     no markers,      domain=1:4.5,      samples=100,     smooth,     axis lines*=left,     height=5cm, width=12cm,     enlargelimits=upper, clip=false, axis on top,     xlabel = near ticks,     xlabel={Resource} }}  \begin{document}  \begin{tikzpicture}   \begin{axis}[baseplot]   \addplot+[name path=A]{gauss(2.7,0.4)};   \addplot+[name path=B]{gauss(3.2,0.4)};   \fill[gray!20, intersection segments ={of= B and A}]; \end{axis} \end{tikzpicture}  \vspace*{3\baselineskip}  \begin{tikzpicture}   \begin{axis}[baseplot]   \addplot+[name path=A]{gauss(1.7,0.15)};   \addplot+[name path=B]{gauss(2.2, 0.15)};   \addplot+[name path=C]{gauss(2.7, 0.15)};   \addplot+[name path=D]{gauss(3.2, 0.15)};   \addplot+[name path=E]{gauss(3.7, 0.15)};   \fill[gray!20, intersection segments ={of= B and A}];   \fill[gray!20, intersection segments ={of= C and B}];   \fill[gray!20, intersection segments ={of= D and C}];   \fill[gray!20, intersection segments ={of= E and D}]; \end{axis} \end{tikzpicture}  \end{document} If $f(\frac{1}{n})=0$ all $ n\to \mathbb{Z^{+}}$ then find $ f'(0)$? Let $f: \mathbb{R}\to \mathbb{R}$ be a twice differntiable function and $f(\frac{1}{n})=0$ all $n\in \mathbb{Z^{+}}$? then which of the statement (s) are correct from below.  (a)$f(0)=0$  (b)$f^{'}(0)=0$  (c)$f^{''}(0)=0$  (d)$f$ is a non zero polynomial  I tried of using Tyler series but I wasn't able to do this any help would be appreciated thank Will tcpdump see packets that are being dropped by iptables? After I factory reset my Nokia 6, this icon appeared. I've never seen it before. What is this icon in the highlighted part? Lualatex and bm package I've updated my latex packages yesterday and have now issues with lualatex. One problem arising for me now is, that lualatex does not print bold characters using the bm package.  When I compile the following MWE with pdflatex everything looks okay, but for lualatex it doesn't:  \documentclass{article} \usepackage{bm} \begin{document} % Bold $\bm{E}$ \\ Normal $E$ % \end{document}   Output using pdflatex    using lualatex    Also trying to fall back to an earlier version of lualatex by using \RequirePackage{luatex85} does not work for me.  When having a look at the log file, there are the suspicious messages   LaTeX Font Info:    Redeclaring math alphabet \mathbf on input line 142. LaTeX Info: Redefining \bm on input line 208. [...] Missing character: There is no х (U+0445) in font cmmi10!   Can anyone tell me, what is going on here and what I need to do to solve this problem, since I am quite sure lualatex and bm worked before? Increasing notebook magnification pushes output onto a new line Does an Animagus's animal form change along with their Patronus? In the two examples shown of an Animagus capable of casting a corporeal Patronus (Minerva McGonagall and James Potter) both had the same animal as their Patronus and as their Animagus animal.  Now, it is also shown that extreme circumstances can change one's Patronus. I think the only example of this was Tonks, which turned from a jackrabbit into a wolf in response for her love of Lupin.  If Tonks (or any other wizard who had their Patronus change) had been an animagus, would her animagus have changed as well? Can it happen that the image of a functor is not a category? Suppose I have a node in the shape of a circle. How can I make its border thicker?   \node[circle, draw=blue!80, inner sep=0pt, minimum size=12pt] (1) at (0,0) {1}; Ubuntu Software (Software Center) Crash - Ubuntu 16.04 How do trees lift water higher than 10 meters? The atmosphere pressure is 10 meters of water (approx). This means that it is impossible to lift water higher than 10 meters with vacuum or сapillary action (on Earth, under normal conditions).  There are trees higher than 10 meters.  How do they lift water to their tops?  UPDATE  In other words: how cohesion-tension theory can be true if it apparently contradicts the laws of physics?   UPDATE 2  Atmospheric pressure helps to rise the water, not resists rising. What is resisting is water weight. When water column is 10 meters high, then atmospheric pressure can't help anymore.   Any adhesion/cohesion mechanism can't help here too, because it acts only in thin molecular layer. To transfer action force further the pressure is required, which is insufficient at 10 meters.  UPDATE 3  If we had capillary small enough to rise water to 10 meters and then we will build smaller capillary which we expect will rise water higher, we will fail. Water column will break and does not climb higher than 10 meters.     Menisci acts like small piston and can't help rising water higher than 10 meters.  UPDATE 4  Common pressure distribution in capillary is follows:    $P_0$ is atmospheric pressure. As you see, right under menisci, the pressure is lowered by $2 \sigma / R$ where $R$ is the radius of menisci and $\sigma$ is surface tension. The entire term is called "Laplace pressure". As you see, it can't supersede atmospheric pressure, because water continuity will be broken in the case.  I.e. no any menisci can rise water higher than 10 meters.  The existence of higher trees PROVES that there are some other significant mechanisms, not adhesion/cohesion, not capillary.  UPDATE 5  Current version, as I understood it, is based on a declaration, that a water, if put into thin capillary, can behave like a solid body. Particularly, it can resist tension up to minus 15 atmospheres.   This is a tensile strength of concrete, so I don't believe that without additional proofs.  I think it is just not hard to make thin tube, put water into it and check, how high it can climb.  Was it done ever? Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$:   $$f(f(x)f(y))+f(x+y)=f(xy).$$   It was six hours ago in IMO 2017 (problem 2).  I tried  the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.   Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?   Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and  $1-(1-x)(1-y)+1-x-y=1-xy$. Why doesn't Steam start? When I try to start Steam nothing happens. Reinstall doesn't help.  $ steam Running Steam on ubuntu 13.04 64-bit STEAM_RUNTIME is enabled automatically Installing breakpad exception handler for appid(steam)/version(1378500910_client) Installing breakpad exception handler for appid(steam)/version(1378500910_client) Installing breakpad exception handler for appid(steam)/version(1378500910_client) unlinked 0 orphaned pipes removing stale semaphore last operated on by process 7278 with name 0eBlobRegistryMutex_DC7EBB4AEA41D3C6C74E5EFFD6757CEC removing stale semaphore last operated on by process 7278 with name 0eBlobRegistrySignal_DC7EBB4AEA41D3C6C74E5EFFD6757CEC removing stale semaphore last operated on by process 7278 with name 0emSteamEngineInstance removing stale semaphore last operated on by process 7278 with name 0eSteamEngineLock Gtk-Message: Failed to load module "overlay-scrollbar" Installing breakpad exception handler for appid(steam)/version(1378500910_client) [0913/182910:WARNING:proxy_service.cc(958)] PAC support disabled because there is no system implementation   How to fix it? If I want to solve the equation $4x = e^x$, is there a way to solve for $x$ without graphing and looking for intersection? pyPdf for IndirectObject extraction Following this example, I can list all elements into a pdf file  import pyPdf pdf = pyPdf.PdfFileReader(open("pdffile.pdf")) list(pdf.pages) # Process all the objects. print pdf.resolvedObjects   now, I need to extract a non-standard object from the pdf file.   My object is the one named MYOBJECT and it is a string.  The piece printed by the python script that concernes me is:  {'/MYOBJECT': IndirectObject(584, 0)}   The pdf file is this:  558 0 obj &lt;&lt;/Contents 583 0 R/CropBox[0 0 595.22 842]/MediaBox[0 0 595.22 842]/Parent 29 0 R/Resources   &lt;&lt;/ColorSpace &lt;&lt;/CS0 563 0 R&gt;&gt;     /ExtGState &lt;&lt;/GS0 568 0 R&gt;&gt;     /Font&lt;&lt;/TT0 559 0 R/TT1 560 0 R/TT2 561 0 R/TT3 562 0 R&gt;&gt;     /ProcSet[/PDF/Text/ImageC]     /Properties&lt;&lt;/MC0&lt;&lt;/MYOBJECT 584 0 R&gt;&gt;/MC1&lt;&lt;/SubKey 582 0 R&gt;&gt; &gt;&gt;     /XObject&lt;&lt;/Im0 578 0 R&gt;&gt;&gt;&gt;   /Rotate 0/StructParents 0/Type/Page&gt;&gt; endobj ... ... ... 584 0 obj &lt;&lt;/Length 8&gt;&gt;stream  1_22_4_1     ---&gt;&gt;&gt;&gt;  this is the string I need to extract from the object  endstream endobj   How can I follow the 584 value in order to refer to my string (under pyPdf of course)?? Assume $\bar A\cap\bar B=\emptyset$. Is $\partial (A \cup B)=\partial A\cup\partial B$, where $\partial A$ and $\bar A$ mean the boundary set and closure of set $A$?   I can prove that $\partial (A \cup B)\subset \partial A\cup\partial B$ but for proving $\partial A\cup\partial B\subset \partial (A \cup B)$ it seems not trivial. I tried to show that for $x\in \partial A\cup\partial B$ WLOG, $x\in \partial A$ so $B(x)\cap A$ and $B(x)\cap A^c$ not equal to $\emptyset$ but it seems not enough to show the result. Changing "index.php" Within The Address I understand we cannot state with arbitrary precision the position and momentum of a micro-particle as we superpose infinite waves to create a wave packet at the exact position of the particle and hence cannot say which wave is corresponding to its momentum. This is theoretically said. Also some people say that measurement of one disturbs the other and some say it's reality.  Firstly, the theoretical explanation: why should I consider that a wave out of many superposed will give us the momentum or position? We might use another single wave which is not sine in nature and might get the results. Or why won't a single sine wave suffice? This all considers the micro-particle as wave because of de Broglie.  Secondly, the practical thing seems legit.  Thirdly, the electron must have some kinetic energy and position. It is who we are measuring and can't get it exact so why blame the electron as in theoretical concept. How we quantify the uncertainty, it must depend on the measuring device and various other factors.  I might need to edit this post based on your answer so that I can explain more clearly my thinking.  Edit: From here is what gets the measurement problem:        Text: Imagine that you're blind and over time you've developed a technique for determining how far away an object is by throwing a medicine ball at it. If you throw your medicine ball at a nearby stool, the ball will return quickly, and you'll know that it's close. If you throw the ball at something across the street from you, it'll take longer to return, and you'll know that the object is far away.   The problem i­s that when you throw a ball -- especially a heavy one like a medicine ball -- at something like a stool, the ball will knock the stool across the room and may even have enough momentum to bounce back. You can say where the stool was, but not where it is now. What's more, you could calculate the velocity of the stool after you hit it with the ball, but you have no idea what its velocity was before you hit it.   This is the problem revealed by Heisenberg's Uncertainty Principle. To know the velocity of a quark we must measure it, and to measure it, we are forced to affect it. The same goes for observing an object's position. Uncertainty about an object's position and velocity makes it difficult for a physicist to determine much about the object.   Edit: After seeing this answer one of my doubts is very well cleared, the other answers are also written and helpful:        Text: There is a definine velocity and momentum, we just don't know it.   Nope. There is no definite velocity--this was the older interpretation. The particle has all (possible) velocities at once;it is in a wavefunction, a superposition of all of these states. This can actually be verified by stuff like the double-slit experiment with one photon--we cannot explain single-photon-fringes unless we accept the fact that the photon is in "both slits at once".So, it's not a knowledge limit. The particle really has no definite position/whatever.