Null Object Reference Error inspite of Initialising Object
What is a NullPointerException, and how do I fix it?
Proof by Induction: For any integer $n > 0$, $3|(5^{2n} − 1)$. Is there any difference in proving "for any integer $n &gt; 0$, $3|(5^{2n}-1)$" and proving "for any integer $n \ge 0$, $3|(5^{2n} -1)$"? Or are they the same? How to prevent a scrollview from scrolling to a webview after data is loaded? Good Number Theory books to start with? I don't know if this is a well-known fact, but I have observed that every number, no matter how large, that is equally divided by $9$, will equal $9$ if you add all the numbers it is made from until there is $1$ digit. A quick example of what I mean:  $9*99 = 891$ $8+9+1 = 18$ $1+8 = 9$  This works even with really long numbers like $4376331$ Why is that? This doesn't work with any other number. Similarly for $11$ and alternating digits sums. $$Rad(I)=\{a \in R | \exists n \in \mathbb{N} \text{ such that } a^n \in I\}$$   R is a commutative ring, I is an ideal.   To show that $Rad(I)$ is an ideal of $I$, we have to show that for $a,b \in Rad(I)$ it stands that $a-b \in Rad(I)$ and for $r\in R$ it stands that $a\cdot r\in Rad(I)$, right??   When $a,b \in Rad(I)$ we have that $a,b \in R$ and $a^n, b^m \in I$.   $a-b \in R$.   But how can we show that $(a-b)^x \in I, x \in \mathbb{N}$ ?? Can a git submodule be created retroactively? $E$ is a measurable set of finite measure and $1 \le p_{1} &lt; p_{2} &lt; \infty$. Consider the linear space $L^{p_{2}}(E)$ normed by $||.||_{p_{1}}$. Is this space Banach? POSTFIX main.cf permission denied I am trying to set up a postfix server for smtp relay. Centos 5.5  I have followed  but am getting a FAILED error when trying to restart or reload postfix.  When I do:  cat maillog   I get:  fatal: open /etc/postfix/main.cf: Permission denied   Here is what I have tried:  chown postfix /etc/postfix/main.cf chmod u+rwx /etc/postfix/main.cf   After that, I do:  ls -l /etc/postfix/main.cf   Which gives me:  -rwx------. 1 postfix anotheruser 27531 Apf 29 12:19 /etc/postfix/main.cf   No dice. Same error. I even tried temporarily doing chmod 777 but got same error.  Any ideas on the permissions problem? I am assuming it is running the service as the postfix user.  Thanks Can I connect copper wire to aluminum wire? Proving any number of the form $2^n$ can't be written as the sum of k consecutive numbers. I think i have proved that any number of the form $2^n$ can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not.  Let $g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))$, hence g is the sum of k consecutive positive integers.  $g=kx+\sum_{d=1}^{k-1}d$  Hence, $g=kx+\frac{k(k-1)}{2}$  Rearranging for x we find, $x=\frac{g}{k}-\frac{(k-1)}{2}$  So, for certain values of $k$, we can be sure that a positive integer value of $x$ can be found, and hence $g$ can be written as the sum of $k$ consecutive positive integers.There are two cases for $k$ to considor:  Case 1)  $k$ is odd. If $x$ exists, it must be the case $k$ | $g$, as for odd values of $k$, $\frac{(k-1)}{2}$ will alwyas be an integer, and hence $x$ will.  Case 2)  $k$ is even. $k$ does not divide $g$, but $\frac{g}{k}$ must be a multiple of $\frac{1}{2}$, and hence equals $\frac{h}{2}$, for some odd $h$.  Letting $j=2^n$, assume that we can write $j$ as the sum of $k$ consecutive positive integers.  If $k$ is odd, then if a $k$ exists, it must divide $j$, by Case 1. But, $j$ has no odd factors as it is a power of two, hence $k$ could not be odd.  If $k$ is even, by Case 2 it mut be the case $k$ does not divide $j$. Also, $\frac{j}{k}=\frac{h}{2}$, for some odd $h$.  But, this implies $2j$ has an odd factor, $h$. but, as $2j$ is a power of $2$, it can't have odd factors. So, $k$ can't be odd.  Hence, $k$ is neither odd or even, and this is a contradiction to our assumption. Hence no $x$ exists, and hence and number of the form $2^n$ can't be written as the sum of $k$ consecutive positive integers.  Any feedback on this proof would be appreciated. How can I find and replace specific words in a text file using command line? What is wrong with my mob spawner? Why won't my taskbar display on my secondary monitor? On my secondary monitor (currently using Windows 10), my taskbar will barely show up. I've tried adjusting the resolution to lower ones (it recommends 1920x1080) and it doesn't do much. Is there a way to adjust the taskbar so I can see it?  That is about what I see. I could make it bigger via Snipping Tool and see my entire desktop on the picture but not normally. Smart Phone's camera socket Filter the score of the questions on the homepage before loading them At the moment questions on the homepage are loaded by the top N most recent and then ones with a score of -4 or less are filtered out. This doesn't make much difference for a user who is logged in but consider anonymous users; their home page is only 15 questions in length. If someone was to edit several posts that bump to the top with a score of -4 or less then the home page looks tiny. At the moment on SFF, for example, the homepage for anonymous users only shows 9 questions:  At one point it only showed 3 questions (a few weeks back):  Can the filtering logic be changed to before the questions are loaded to appear on the homepage? Crossroads / Second Chance / Choices I am trying to find a short story involving an alien race of 'Intellectuals.' They possess no opposable thumbs or other limbs to control their world.  I think the basis of the story is their discovery of a way to influence another unintelligent species that can manipulate objects.  This leads to a new start for them.  They may have been at risk of breeding beyond available resources.  The story ends with explorers returning from sea. They may have discovered a new land into which they can expand. The last few scenes have a darker flavor where one character has been presented with a rock that glows in the dark. He considers it for a period before being called away to greet the explorers. The author then takes over and makes some comments about most species passing one crossroads only to meet another.  The story could have been written at any time from the Golden Age though to the 1970's. Is it possible to call a non-exported function that resides in an exe? Proving that $\sum\limits_{n = 0}^{2013} a_n z^n \neq 0$ if $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$ I'm going to teach a preparation course for the complex analysis qualifying exam from my university (which basically consists of me doing some problems from past exams) and I'm trying to solve some questions from previous exams. One of the questions is the following:           Prove that $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ if the coefficients  satisfy $a_0 &gt; a_1 &gt; \dots &gt; a_{2013} &gt; 0$ and $|z| \leq 1$.      I tried to approach the problem by using the reverse triangle inequality to try to isolate the leading term, which I thought should be the constant coefficient, as follows:  \begin{array} . \left|  \sum_{n = 0}^{2013} a_n z^n \right| &amp;= \left| a_0 +  \sum_{n = 1}^{2013} a_n z^n \right| \\ &amp;\geq |a_0| - \left|  \sum_{n = 1}^{2013} a_n z^n \right|\\ &amp;\geq |a_0| -  \sum_{n = 1}^{2013} |a_n z^n|\\ &amp;\geq a_0 -  \sum_{n = 1}^{2013} a_n |z^n|\\ &amp;\geq a_0 -  \sum_{n = 1}^{2013} a_0 |z^n|\\ &amp;= a_0\left( 1  -  \sum_{n = 1}^{2013} |z^n| \right) \end{array}  Then for $|z| &lt; 1$, we have that this last expression is equal to  $$ a_0\left( 1  -  |z| \frac{|z|^{2013} - 1}{|z| - 1} \right) $$  and my hope was to prove that this was always positive, nevertheless, after failing to prove it, I plotted the expression inside the parenthesis as a function of $|z|$, and found out that it changes sign in the interval $(0, 1)$.  Question  Can my approach somehow be made to work? And if not, how can I prove that the sum $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ under the given conditions?  Thank you very much for any help. The X button stopped working on one of my PS3 controllers. It's hardly through abuse, the controller isn't a year old and I don't play a lot.  The button doesn't stick, it presses like the others, only the system isn't aware that I'm using it. I tried resetting it with the tiny reset button in the back, to no avail.  What can I do? How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to , but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.  In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$