Currently my Nexus 5 is on Android M preview. I know that developer preview won't get the OTA update, that's why I am trying to manually install OTA of Android M public release, but I get these errors on 50%:  failed to verify whole-file signature  signature verification failed Installation aborted   adding more details:  no root no custom recovery just bootloader unlocked no changes in system files   Does manually installing Android M final release OTA work on device running on Android M preview? Or do I need to downgrade it to Lollipop first?
If I install the Dev Preview 3 on my Nexus 5 right now, will I get an OTA to the stable build or will I have to flash it as well?
My PS3 controllers have died and are not charging How to create vector graphics In Photoshop? I know people prefer using CorelDraw or Adobe Illustrator, but I want to know if there is a technique I can follow to use Adobe Photoshop to make vector Images. Why would parametric statistics ever be preferred over nonparametric? Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists? Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ?   I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance Create WiFi hotspot on ubuntu "Stay home" or "stay at home" – which is correct and why? Which one of these two statements is correct?     Our staff do ...      Our staff does ...   And is staffs ever correct? Let $n\in\mathbb N$ and let $a_1,\ldots,a_n$ be natural numbers smaller than $10^n$. Write each $a_k$ in base $10$ and add $0$'s to the left of each decimal expansion, if needed, so that each $a_k$ is written with $n$ digits. Consider the matrix such that the entries of the $k$th row are the digits of $a_k$ (written in the same order). Let $d\in\mathbb N$ be such that $d$ divides each $a_k$. Prove that $d\mid\det A$.   For instance, suppose that $n=4$ and that your numbers are $3876$, $2784$, $684$, and $8388$, each of which is a multiple of $12$. Then$$A=\begin{bmatrix}3 &amp; 8 &amp; 7 &amp; 6 \\  2 &amp; 7 &amp; 8 &amp; 4 \\ 0 &amp; 6 &amp; 8 &amp; 4 \\  8 &amp; 3 &amp; 8 &amp; 8\end{bmatrix}$$and $\det A=-360$, which is, in fact, a multiple of $12$.  I learned about this problem yesterday. I found it quite cute and I decided to share it with all of you.  Note: I know how to prove it. I'm developing a GIS application with java Spring Framework and GeoTools library. I have some vector layer. I want to get this layers with wms and wfs protocols. I've downloaded geoserver code. It includes module as bellow:   Main Module Core Platform Module Web Map Service Module Web Feature Service Module and etc.   Can I use Web Map Service Module and Web Feature Service Module of Geoserver in my project? If yes, How? I want a sample code to use these modules apart from geoserver. Help me please. How do i enter my numeric password in iPhone 4 in VoiceOver mode? My iPhone 4 is in VoiceOver mode. It`s currently locked and not accepting my passcode which is absolutely correct. And now it's saying iPhone is disabled.  Any solution? How do I make my program watch for file modification in C++? I'm working with Beamer and I would like to create boxes with round corners. I use \ovalbox, but I would like to control the thickness of the line. How can I do this? I also tried \thinlines, but I still not understand the sintax since I didn't find a minimal example. From the  on the dual of a norm:  $X$ and $Y$ are normed spaces, and we associate with each $f\in L(X,Y)$ (the space of bounded linear operators from $X$ to $Y$) the number $$\|f\| = \sup\{|f(x)|:x\in X, \|x\| \leq 1\}.$$  At some point in the proof that $L(X,Y)$ is bounded, it seems like they use the inequality $$\|fx\| \leq \|f\|\cdot\|x\|,$$ where $\|x\|\leq 1$.  I can't see why this should holds.   $\|f|\|$ is the supremum of what $|f(x)|$ can be, given that $\|x\|\leq 1$. So, given that $\|x\| \leq 1$, $|fx|$ should be bounded by $\|f\|$. But why should it be bounded by $\|f\|\cdot\|x\|$? Understanding the definition of the order of an entire function in Ahlfors's Complex Analysis Let $f: \mathbb C \to \mathbb C$ be an entire function. The order of $f$ is defined by $$\lambda=\limsup_{r \to \infty} \frac{\log \log M(r)}{\log r}, $$ where $$M(r)=\max_{|z|=r} |f(z)| .$$  Ahlfors in his Complex Analysis claims that     "According to this definition $\lambda$ is the smallest number such that       $$M(r)\leq e^{r^{\lambda+\varepsilon}} $$ for any given $\varepsilon &gt; 0$ as soon as $r$ is sufficiently large."   Why is this true?  My attempt:  We know that $$\lambda=\lim_{\rho \to \infty} \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}. $$  From the definition of the limit we have that for any $\varepsilon&gt;0$, there exists some $\rho_0&gt;0$, such that $$\left\lvert \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}-\lambda \right\rvert \leq \varepsilon ,$$ for every $\rho \geq \rho_0$. In other words $$\frac{\log \log M(r)}{\log r} \leq \lambda+\varepsilon $$ for every $r \geq \rho_0$. From here it is easy to see that $$M(r)\leq e^{r^{\lambda+\varepsilon}}, $$ for all $r \geq \rho_0$. I cannot see why $\lambda$ is the smallest number with this property.  Thanks in advance. Is Poly1305 an information-theoretically secure MAC? I know  that uses $x={\sqrt{2}}^{\sqrt{2}}$ to prove that there must exist an irrational to an irrational power that evaluates to a rational. But I don't know if $x$ itself is known to be irrational or rational.  Could someone enlighten me? Thanks! Probability of a 2 pair in poker Remove windows XP after installing Ubuntu I have a question about the following text:     Last week Alex Knapp at Forbes published an article criticizing my Nikola Tesla comic. I don't normally respond to these kinds of thing, but since it's' Forbes I figured a proper response was in order.    I would rather say "this kind of things" rather than the version in the text above. Are orbiting planets an  example of perpetual motion?