I have a string:  java.lang.IllegalArgumentException: local part cannot be "null" when creating a Qname   which I'm pulling from a log file on a unix server.   and I have this same exact string which I'm pulling from a windows .txt file to compare against the string from the unix server:  java.lang.IllegalArgumentException: local part cannot be "null" when creating a Qname  I'm trying to determine if they are equal, but the .equals() method in java is letting me know that they are not equal.   How can I go about determining that they are indeed equal?
I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?
I have a set of values $x$ and $y$ which are theoretically related exponentially:  $y = ax^b$  One way to obtain the coefficients is by applying natural logarithms in both sides and fitting a linear model:  &gt; fit &lt;- lm(log(y)~log(x)) &gt; a &lt;- exp(fit$coefficients[1]) &gt; b &lt;- fit$coefficients[2]   Another way to obtain this is using a nonlinear regression, given a theoretical set of start values:  &gt; fit &lt;- nls(y~a*x^b, start=c(a=50, b=1.3))   My tests show better and more theory-related results if I apply the second algorithm. However, I would like to know the statistical meaning and implications of each method.  Which of them is better? Can we use CAPITAL LETTERS to indicate actions? While designing for the web or the mobile applications, a natural free flowing text can be used for the general content, but when it comes to actions... can we use CAPITAL LETTERS. Simple & easy way to jail users I need a simple and easy way to jail users in their home directories in Oneiric. Do you have a simple configuration for jailing users, with full help, or some good web links?  I would be offering an online free public server with 10 to 20 GB free space. I don't know how many users. I want to give them SSH and SFTP so that they can connect through FileZilla. How can I find the number of partitions of $n$ into exactly $k$ distinct parts, where each part is at most $M$?  The number of partitions $p_k(\leq M,n)$ of $n$ into at most $k$ parts, each of size at most $M$, is given by the generating function: $$  \binom{M+k}{k}_{x} = \prod_{j=1}^{k}\frac{1-x^{M+k-j+1}}{1-x^j}= \sum_{n=0}^{kM} p_{k}(\leq M,n) x^n $$  For the number of the partitions $p_k(\mathcal{D},n)$ of $n$ into at most $k$ parts there is the recurrence relationship: $$   p_{k}(\mathcal{D},n) = p_{k}(\mathcal{D},n-k) + p_{k-1}(\mathcal{D},n) $$  But what, if I want to count only the partitions with distinct parts and restricted number of parts and restricted part size?  Update: Now I know the generating function for the number of distinct restricted partitions $p_k(\leq M, \mathcal{D},n)$ of $n$ into exactly $k$ distinct parts, all at most $M$ is $$    \prod_{j=1}^{M} (1+xq^{j}) = \sum_{k,n=0}^{\infty}p_k(\leq M, \mathcal{D},n)x^{k}q^{n} $$ and there is also a recurrence relation $$    p_k(\leq M, \mathcal{D},n) =         p_{k-1}(\leq M-1, \mathcal{D},n-k) +         p_k(\leq M-1, \mathcal{D},n-k) $$ How can I prove this? Could you recommend a book, where I could read about this? Can a multiclass character use either of its spellcasting modifiers for spells? Real Analysis Riemann Integration - Strict Monotonicity for Integrals If $f,g$ are Riemann integrable on $[a,b]$, and $f(x) &lt; g(x)$ for all $x \in [a,b]$, prove that $$ \int_a^b f(x) \,dx &lt; \int_a^b g(x) \,dx$$  This is a strict inequality. I know how to prove the monotonicity of integrals with the non strict inequality, but I am not sure how to do this. Should the eigenkets be weighted in $|P\rangle = \sum\limits_{r}|\xi^r\rangle$? Page 37 of Dirac's book The Principles of Quantum Mechanics, states  The condition for the eigenstates of $\xi$ to form a complete set must thus be formulated, that any ket $|P\rangle$ can be expressed as an integral plus a sum of eigenkets of $\xi$, i.e. $$|P\rangle = \int |\xi'c\rangle\; d\xi' + \sum\limits_{r}|\xi^rd\rangle$$ where the $|\xi'c\rangle$, $|\xi^rd\rangle$ are all eigenkets of $\xi$, the labels c and d being inserted to distinguish them when the eigenvalues $\xi'$ and $\xi^r$ are equal, and where the integral is taken over the whole range of eigenvalues and the sum is taken over any selection of them.  I find it inconsistent that the eigenkets under the integral are weighted by what appears to be a differential eigenvalue $d\xi'$, yet all the eigenkets under the summation are weighted by the value 1. Is this correct? existence of sequence of polynomial Is there a sequence of polynomials $P_n$ such that   $\displaystyle\lim_{n \to \infty} P_n(z) = \begin{cases} 1, &amp; \text{Im } z &gt; 0\\ 0, &amp; z \text{ is real,} \\ -1, &amp; \text{Im } z &lt; 0 \end{cases}$  I have no clue where to start from. Please provide some hints. Thanks in advance. How do I make my program watch for file modification in C++? How to help a toddler overcome the fear of water on head? Does existence of $\gcd$ implies PID? We know that if $A$ is a PID, then we can guarantee the existence of a $\gcd$ between two elements. My question is, does the converse hold? If I know that every two elements have a $\gcd$, do I know that the ring is a PID? My intuition says no, in that case I would like a counter-example.  Thanks. Does kinetic friction increase as speed increases? Possible Duplicate:          Layman alert...I last did physics in high school, and am just trying to understand something my son is working on.   To keep a box moving across the floor at a constant velocity I need to apply a force equal and opposite to that of the  force working on it.  If I start applying a force greater than the kinetic friction force, the box will speed up. Will it keep speeding up indefinitely, or will the kinetic friction force I need to overcome increase as the box gets faster? I am starting to learn using blender, and at some point I managed to do something to the 3D view camera.  Usually the camera orbits around an object, but now the rotation seems to happen (atleast very close to) around the "camera's origin". Neither the input setting in "User Preferences", the pivot point selection nor Ctrl-Alt-Shift-C (as suggested on some forum) do anything to the issue. So, we know that there is no "natural" isomorphism between a finite dimensional vector space $X$ and its dual $X^{\vee}$. However, given a basis for $X$ $(e_1, \dots, e_n)$ we can always construct a (dual) basis $(e^1, \dots, e^n)$ for $X^{\vee}$ that satisfies $e^i(e_j) = \delta^i_j$. It follows that $V \approx V^{\vee}$ simply because they have the same finite dimension $n$. Now, it seems to me that we can construct an isomorphism by assigning $$ e_i \mapsto e^i $$ and extending by linearity. That is, for an arbitrary vector $v \in X$ given by $v = \sum v^ie_i$ we can specify a linear map   $$\phi:V \rightarrow V^{\vee}$$  by  $$ \phi(\sum v^ie_i) := \sum v^ie^i $$ It is a quick exercise to check that $\phi$ is an isomorphism. Now, although this construction seems correct, there is something that doesn't quite sit right with me about it. For one thing, there is no way to really make sense of the summation convention as the right side of the function will always contain two superscripts thus requiring an explicit summation sign. The other thing that feels rather off is actually taking a lower index and moving it to an upper index, i.e., $e_i \mapsto e^i$ (which, of course, is the reason that the summation convention doesn't work).  So, with this background my questions are   Is there any sense in which the above isomorphism is "favored" or "canonical"? Is there another way to construct an isomorphism between a vector space and its dual, that would "conserve indexes", for lack of a better way to state it. My guess here is that the answer is no but becomes possible if one assumes $V$ has an inner product. USB-C to MagSafe 2 Adapter I have a 2015 MacBook Pro with a MagSafe 2 charging cable. My wife has a new MacBook Pro from 2016 with USB-C. I was wondering whether I could technically (in the matter of electronic feasability) use her charging cable on my MacBook and vice versa with some sort of adapter / dongle, and, if so, where I would get one? If we had an airtight container that was large enough, say extended out to the Karman line, would there still be a pressure gradient? If we assume that the pressure at the bottom was $1\ \mathrm{atm}$, would it have the same exact pressure gradient at 6 miles up on the outside? This is assuming that the gravity is still $9.8\ \mathrm{m/s^2}$. Contextually grey out panel element in python 2.8 I have a panel with split column.  In the 2nd row is 3 boolean variables. I want to grey out the 2nd and 3rd option by default. I want the 2nd to be changeable if 1st is true, and 3rd to be changeable if both 1 and two are true. I've read something about "poll()" but not how to actually use it, or set conditions. Physics should only be enablable if singleMesh, and destructable only if physics enabled.      # Create two columns, by using a split layout.     split = layout.split()      # First column     col = split.column()     col.label(text="Building Size:")     col.prop(bpy.context.scene.user_vars, 'buildingHeight')     col.prop(bpy.context.scene.user_vars, 'buildingWidth')     col.prop(bpy.context.scene.user_vars, 'buildingDepth')      # Second column, aligned     col = split.column()     col.label(text="Dynamics:")     col.prop(bpy.context.scene.user_vars, 'singleMesh')     col.prop(bpy.context.scene.user_vars, 'enablePhysics')     col.prop(bpy.context.scene.user_vars, 'destructable_Building')  #EDIT This is what the specific solution to my example looks like. col = split.column() col.label(text="Dynamics:") col.prop(vars, 'singleMesh') sub = col.row() sub.enabled = vars.singleMesh sub.prop(vars, 'enablePhysics') sub = col.row() sub.enabled = vars.enablePhysics sub.prop(vars, 'destructable_Building') "Can’t help but" vs. "can help but" Is "can’t help but" considered to be a confused mix of the expressions "can but" and "can’t help"? If not, what is the difference between "can help but" and "can’t help but"? Get DOS path instead of Windows path resolution is undefined in XeTeX so the following code will produce different values when compiled with pdfTeX and XeTeX.  \documentclass{article} \usepackage{graphicx}  \newsavebox\IBox \savebox\IBox{\includegraphics[scale=1]{koala.jpg}}  \begin{document} \the\wd\IBox \end{document}   How to force pdfTeX and XeTeX to produce the same measurement for the imported image?