i have programmed a library in java i want to protect from reverse engineering. I read about obfuscation and understood that it is only making it a little more difficult to read instead of making it completely impossible to decompile.  Iam not very expierenced in it-security/ software protection.  I found  a free tool for java/ android obfuscation. I have tried different ways to obfsucate the jar file. My problem and also question is, can i actually import a jar file in to a different project after obfuscation??  My library was not working anymore as i imported it in a project.  Do i have to obfusacate the whole java application at the end? or can i obfuscate my library i want to protect and use it in different projects?
How do I lock compiled Java classes to prevent decompilation?  I know this must be very well discussed topic on the Internet, but I could not come to any conclusion after referring them.  Many people do suggest obfuscator, but they just do renaming of classes, methods, and fields with tough-to-remember character sequences but what about sensitive constant values?  For example, you have developed the encryption and decryption component based on a password based encryption technique. Now in this case, any average Java person can use  to decompile the class file and easily retrieve the password value (defined as constant) as well as  and in turn can decrypt the data by writing small independent program!  Or should such sensitive components be built in native code (for example, VC++) and call them via ?
Mathematical symbol for "and" I have found some pretty complete lists (I think) of mathematical symbols  and , but I don't see a symbol for the word "and" on either list.  A person could easily just write the word "and" or use an ampersand, but I was wondering if there was an actual mathematical symbol for the word "and".  Also, if anyone knows any lists that are more complete than the ones I have linked to please provide a link. What *is* an Artificial Neural Network? Consider a block cipher with a key of size 128 bits but a small block size, say 32 bits.  Is this kind of block cipher secure?  I would like to know to what extent I can use a small block cipher, like a 32 bit block cipher. What are the constraints so that the use is acceptable in a real life application.  For instance, if we use a one-time key, that is, to encrypt only one message. Besides, the message could be small.  Suppose that when we want so send one file, we perform a key agreement so that we use the key for one message. This scenario is not efficient, but the CPA find-then-guess game is not adequate in this case. Convergence of $\int\limits_1^{\infty}x^\alpha\sin(x^\beta)dx$ I need to show that $$ \int\limits_1^{\infty}x^\alpha\sin(x^\beta)dx $$ converges if and only if $\alpha + 1 &lt; \beta$. I used substitution and integration to get $$ \lim\limits_{t\to\infty}\int\limits_1^tx^\alpha\sin(x^\beta)dx \overset{u = x^\beta}{\underset{du = \beta x^{\beta-1}}{=} }\lim\limits_{t\to\infty}\left(\frac{1}{\beta}\int\limits_1^{t^\beta}u^{\frac{\alpha + 1 - \beta}{\beta}}\sin{u}du\right) $$ If $\alpha + 1 -\beta &lt; 0$, then we have a product if a function with bounded anti-derivative and a differentiable decreasing function which tend to 0, and from Dirichlet's criterion, the integral converges.   For the converse, I used integration by parts $$  = \frac{1}{\beta}\lim\limits_{t\to\infty}\left(\left.-u^{\frac{\alpha + 1-\beta}{\beta}}\cos{u}\right|_1^{t^\beta} + \frac{\alpha+1-\beta}{\beta}\int\limits_1^{t^\beta}u^{\frac{\alpha + 1-2\beta}{\beta}}\cos{u}du\right) $$ For $\alpha + 1 -\beta\geq 0$, I get that the first term indeed diverges, but this is not enough to show total divergence. How can one show it? How many authors should attend a conference? I wrote a paper with  3 other students regarding my final year project in mechanical engineering. It was recently selected for a conference in NIT Delhi. Minimum one author is required to present and the price is pretty steep for the participation certificate.Should all of us attend the conference or just one of us? Does the certificate hold any value? I absolutely hate it when I watch a video on YouTube and it automatically loads another video page after it's done playing.  This is especially annoying when I'm in the middle of typing a comment as it will cause me to lose what I have typed.  I'd like the video to simply stop and for no further action to take place once it is finished playing.  Is there any way to accomplish this? Considering this code, can I be absolutely sure that the finally block always executes, no matter what something() is?  try {       something();       return success;   }   catch (Exception e) {        return failure;   }   finally {       System.out.println("I don't know if this will get printed out"); } Using a naked black hole as a mirror? The YouTube video is a good example how the gravity of this merging binary black holes bend light around themselves. Shell script to delete directories older than n days What does "very approximate" mean? An approximate answer is one which is close to the correct answer.  Likewise, we can talk of an approximate model, or approximate methods in mathematics.  The etymology is from the Latin ad, "to" and proximus, "closest", see e.g. .  I don't think this can be disputed.  Occasionally proximate is used as a synonym, though that word has a different meaning in legal English.  So if an approximate answer is close, a very approximate answer should be very close.  In practice, "very approximate" or "very approximately" is overwhelmingly used with the opposite meaning of "roughly", indicating that an answer or estimate is not necessarily accurate.  For examples,   Is there any justification for this use? Is there any justification for my suggested use?  Should we avoid the phrase altogether? Should the results of a calculation happen automatically, or require a button push? This is a problem I've ran into several times, and I imagine that others have ran across it too.    Basically, I have a complicated equation, and I want the user to type in the values needed to calculate the results.  (For what it's worth, there are good default values for the variables in the equation.)  I've constructed a GUI that shows the user the names of the variables/fields that need to be provided, along with a control (like a numeric up/down, slider, etc.) to enter the values of the field.  Below that, I have an area that shows the output of the equation.  So my question is this.  Should my results area automatically change the calculated results whenever the user makes any changes to any of the input fields, or should they push a "Calculate" button to generate the results?  One other thing that probably applies is that the user is not trying to "get the right answer" out of the results.  They shouldn't be using the output to then go back and modify the input to get a better answer.  Are there any benefits to requiring the button press?  Is the extra click to see the results worth those benefits? Alternative Method to Acquire El Capitan Upgrade Can you be sued for using publicly used images on a blog without copyright permission? How to convert hex string to float in Java? How to create a scope? I'd like to create a custom scope that searches and returns results from an online source for personal use.  Is this possible? And if so, how? By considering:  $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4$$  Determine if this is true: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac 1 {{m+1}}$$  If it is, prove it.  If it is not, evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{{m+1}}$. Disable OS X startup sound How can I permanently disable the startup sound (or chime) in Yosemite?  (I am using a late-2014 13" macbook pro)  I already tried several options found here and there :    installed StartNinja.app => did not work installed Auto-Mute.app  => did not work tried to install the old StartupSound prefPane => did not work tried the command "sudo nvram SystemAudioVolume=%80" => volume does not change (I also tried other values than 80, such as 00, 01, or nothing)   NOTE: I know that it is possible to mute the next startup sound by turning the volume down, but this is not a permanent solution. Ralink MT7601U (148f:7601) Wi-Fi adapter installation I am trying to install a Ralink wifi adapter in Ubuntu Desktop 14.04.  The system is not detecting the wifi.   lsusb lists the device as:  Bus 002 Device 004: ID 148f:7601 Ralink Technology, Corp.  I have the rt2800 USB driver listed in "/etc/modules"; Following some instructions on the web, I tried this command:  echo 148F 7601 | sudo tee /sys/bus/usb/drivers/rt2800usb/new_i  lsmod | grep rt2800 returns the following:  rt2800usb              26581  0  rt2x00usb              20041  1 rt2800usb rt2800lib              83150  1 rt2800usb rt2x00lib              48886  3 rt2x00usb,rt2800lib,rt2800usb mac80211              545990  3 rt2x00lib,rt2x00usb,rt2800lib crc_ccitt              12627  1 rt2800lib Differentiability implies continuity - A question about the proof I have a question, to aid my understanding, about the proof that differentiability implies continutity.$\mathstrut$    Differentiability Definition  When we say a function is differentiable at $x_0$, we mean that the limit:  $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  Continuity Definition  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.  Proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.  My Question  The proof seems to execute the following steps:   Assume the function is continuous at $x_0$ Show that, with little algebra, we can change this into an equivalent question about differentiability at $x_0$. With this little bit of algebra, we can show that if a function is differentiable at $x_0$ it is also continuous.   What I am slightly unsure about is the apparent circularity. In my mind it seems to say, if a function is continuous, we can show that if it is also differentiable, then it is continuous. Rather than what I was expecting, namely, if a function is differentiable, we can show it must be continuous.  Hopefully my confusion is clear. Any help will be greatly appreciated. How should I understand "$A$ unless $B$"?