Is there any way to recover this?  Can I change it without knowing my current password?
My girlfriend forgot the password for her MacBook (don't ask), so I reset it using the Utilities.  But as you may be aware, this doesn't reset the Keychain password. Her MacBook is now virtually unusable due to Keychain demands continuously provoking a request for the Keychain password.  There are lots of other people asking this around the web and they're usually pointed towards this , which doesn't apply in our situation because the support advice involves entering the old password.  We don't know the old password.  Is it possible to reset Keychain without knowing the old password? We're not bothered about what's in there and what will be lost, we just want the continuous password prompts to go away.
I found this problem on a well-known problem solving website. It is apparently from Ramanujan.     With   $$\LARGE{a = \frac{1}{1 + \frac{e^{-2\pi}}{1 + \frac{e^{-4\pi}}{1 + \ddots}}}},$$   what is $a$?   The basic methods I know for questions of this type were not helpful to me.  I do not require a full solution, but feel free to post one if you would like. Hints or lecture notes on how to solve problems of this nature would be appreciated.   Keep in mind that I am at an undergraduate level in mathematics. I get a message:     Sorry, you are only allowed to ask 50 questions in a 30 day period   Why is there a question limit? 50/month is just not enough :) When I receive a message on an SE post, I get a notification like this:    If I double tap this notification, the SE application opens, but my feed is empty approximately half the time. It looks like this:    If I refresh the feed (slide from the top), the feed does refresh and shows the new notification.   The notification, above, leads to  comment.  I haven't figured out the pattern yet, but this occurs roughly half the time I open a notification. It's occurred for chat messages, comments and answers - but not consistently. I have received notifications for the same post (for example, the two comments after the one linked above) that do not show an empty feed. But, a chat message barely 10 minutes later will show the empty feed again.  I am utilizing the beta app - version 1.0.72 on Android 5.0.1. Here's a screen capture from a site where I can view /site-analytics:    The orange "all posts" curve should be strictly larger than the blue "questions" curve, but they are mistakenly displaying the same data. ls command operating differently depending on recipient Keyboard backlight not working on an Asus N56V See .  The FN keys are not working  Using answer in  I installed the asus-wmi-dkms_0.2_all.deb.gz package  Now, using xev I see that the keys emit keycodes (for backlight keys xev says for fn-f3 and fn-f4 keycode 237)  But no keyboard backlight, also volume keys have no effect  ================ original question ========================================  I cannot enable the keyboard backlight Seems that the leds class is missing I've not found the classes used in   Description of my framework:  ===========================================================  Machine: Asus N56V laptop  ===========================================================  Software   Kubuntu 12.04 amd64  kernel (3.5.0.5-generic) and drivers from xorg-edgers fresh X crack ppa   ===========================================================  Hardware and problems   after installation do not boot - SOLVED using boot-repair every couple of hours the system freezes - SOLVED using 3.5 kernel keybord FN keys - NOT WORKING Keyboard backlight - NOT WORKING Nvidia GEFORCE GT - NOT USED glxgears says 60 fps Ethernet Atheros card - SOLVED install alx driver external subwoofer - SOLVED adding options snd-hda-intel model=asus-mode4   ==========================================================  Solutions  see the answers to the  and  question.  ========================================================== $ \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} dx dy $ Question : The integral   $$ \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} \mathop{dx}\mathop{dy} $$  is   (a) infinite   (b) finite but can not be evaluated in closed form   (c) 1  (d) 2 .  I tried substituting $u=x+y$ and $v=y$ that led me no where . I'm not even sure about convergence of integral .Any help would be greatly appreciated . How do you get the caret position in a &lt;textarea&gt; using JavaScript?  For example: This is| a text  This should return 7.  How would you get it to return the strings surrounding the cursor / selection?  E.g.: 'This is', '', ' a text'.  If the word “is” is highlighted, then it would return 'This ', 'is', ' a text'. Similar to this question  and an fellow up on this one    What was the official reason by Canonical to axed the project when it was love by some many of their users and innovative? Can I safely charge my laptop with a non-standard, third-party charger? I have a Toshiba Satellite laptop. My charger has stopped working.  I have access to a Lenovo charger. Can I use this charger on my laptop? Always wanting more , of anything, never has enough,never contented with what they have Is it possible to do vector calculus in Mathematica? Lisp Executable I've just started learning Lisp and I can't figure out how to compile and link lisp code to an executable.  I'm using clisp and clisp -c produces two files:   .fas .lib   What do I do next to get an executable? I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.  In the proof of the theorem, they wrote:     $$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$  ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$  [...]   I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?  Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so? The use of imaginary and complex values comes up in many physics and engineering derivations. I have a question about that: Is the use of complex numbers simply to make the process of derivation easier, or is it an essential ingredient, without which it would be impossible to derive some results?   I can identify two different settings where the answer may be different:   It doesn't look like it is mandatory for results in Newtonian mechanics, General Relativity and classical electrodynamics. Can we say the same thing about quantum mechanics either way for sure?   Could this be a difference in quantum mechanics over the classical picture? What is the proof that the universal constants ($G$, $\hbar$, $\ldots$) are really constant in time and space? Why isn’t singular ‘they’ used with 3Sg verb forms? There are many (duplicate) questions about the acceptance, popularity and history of singular they (and their, them and themself) around here, it even got a tag of its own. If I didn’t miss anything, however, the proper verb form hasn’t been questioned yet.  As we all know, English third person singular pronouns (it, she, he and one, +body), names (Alice, Bob, …) and nouns (student, teacher, …) demand the +(e)s suffix be added to the finite verb form in simple present, where some auxiliary verbs have a “special” form (is &lt; *bes / *ares and has &lt; *haves). All other subjects don’t, including plural third person pronoun they.  When the plural you replaced thou (with thee, thy / thine), the other marked verb form that had remained in English – i.e. suffix +(e)st or +t – vanished, too. The first and second persons only ever appear as pronouns (I, we; you), not names or nouns, so there was no strong inclination to keep the verbal inflection.  The second person precedent would suggest that singular they be used with uninflected verb forms which is how it’s usually encountered in the wild. Assuming that they someday replaces he and she (and maybe it) it would lead to disagreement with the words the pronoun stands in for:        Alice         goes to her   place  by herself.   Bob           goes to his   place  by himself.   Alice and Bob go   to their place  by themselves. – (common)   Alice and Bob go   to their places by themselves. – (separate)                  She  goes to her   place  by herself.   He   goes to his   place  by himself.   They go   to their place  by themselves.   They go   to their places by themselves.                  They ?goes to their place  by themself.   ditto   They  go   to their place  by themselves.   They  go   to their places by themselves.                  They ?go to their place  by ?themselves.   ditto   They  go to their place  by  themselves.   They  go to their places by  themselves.      So why doesn’t singular they afford s on present-tense verbs like all other third person singular subjects do?   I learned about singular they only long after I had been taught English as a second language in school. That’s why it’s still a conscious decision to use it and hence I could easily adapt to use s forms with it, but would that sound and look funny / strange / wrong to native speakers? How to change background and maintain realistic effect? I have like 30 different photos to process like this, but I want to know key things to maintain realistic photo by changing it's background, or maybe just modify background so it would be more pleasing to eye.(If you know tutorial related to my problem, please do provide it to me)  Here is the photo:   Here is my step 2:   Here is background:   Here is the result:   As you can see, it isn't any close to realistic feel. I fooled around with levels, but all I got was this and still it doesn't feels right.  First try:   Second try:   Update  Different background: Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}$\ $(-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$  My attempt :   Assume $f$ is strictly positive  (if not n=0 works). Let $Z(f^{(n)})$ the number of zeros of $f^{(n)}$ on $(-1,1)$. Then we have: $$ \forall n\in\mathbb{N}:\;f^{(n)}(-1)=f^{(n)}(1)=0 $$ so that iterating the Rolle theorem to the successive derivatives of f we have  $$  \;Z(f^{(k+1)}) \geq Z(f^{(k)}) +1 $$ thus, by induction : $$ \forall n\in\mathbb{N}: \;Z(f^{(n+1)}) \geq Z(f^{(n)}) +1 $$ Therefore, $$ Z(f^{(n)}) \geq n $$ How can I find an another zero ?   I'm still looking for an another proof which does not use the lemma ..,  Thank you in advance. Expected number of hamiltonian paths in a tournament The following theorem is from Alon&amp;Spencer's The probabilistic method, in the beginning of chapter 2:  Theorem 2.1.1: There is a tournament $T$ with $n$ vertices and at least $\frac{n!}{2^{n-1}}$ Hamiltonian paths.  Briefly the theorem is proved by looking at $X_\sigma$ the indicator random variable for a permutation $\sigma$ giving a Hamiltonian path, that is, satisfying $(\sigma(i),\sigma(i+1))\in T$ for $1\leq i \lt n$. Then   $\sum \mathbb{E}\left[X_\sigma\right] = \frac{n!}{2^{n-1}}$.  I don't understand where did this value ($\frac{n!}{2^{n-1}}$) came from! could someone explain it please?