A few months ago I asked where the details of a custom toolbar were located with a view to restoring a backup. With the assistance of Psycogeek I got to the bottom of it.  I offer you the batch scripts which solved the problem.
A few years ago Umber Ferrule wrote that the toolbars in the taskbar are defined in   HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\Streams.   This looks very plausible but cannot be the entire story.  Deleting this key will in no way affect the taskbar and on rebooting the system the key magically reappears.  Why do I want to know?  Simple.  I would like to export the key to a reg. file as a convenient backup.
Name of component on output side cable of a DC power adapter? Is this an equivalent formulation of "surjective" resp. "epimorphism"? Cardinality of $\mathbb{R}^\mathbb{R}$ by diagonal argument I want to prove that the set of all real functions $\mathbb{R}^\mathbb{R}$ has a higher cardinality than the real numbers $\mathbb R$, by Cantor's diagonal argument.  I'm having difficulties with approaching this problem.  What I'm looking for is a hint in the right direction.  I've seen an example where it is shown that the power set $2^S$ of a countable set $S$ is uncountable, by the diagonal argument. There's an election going on. What's happening and how does it work? I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks          Show that the polynomial $h(x) = (x-1)(x-2)\cdots(x-n) + 1$ is irreducible over $\mathbb Z$ for all $n\ge1$ and $ n\ne4$.      Suppose $h(x) = f(x) g(x)$, then we must have $f(i)g(i) = 1$ for all $i = 1,2,...n$. So both $f(i)$ and $g(i)$ are $1$ or $-1$. In either case, $m(x) = f(x) - g(x)$ has degree smaller than $n$ and have $n$ different roots ($1,2,...,n$). So we must have $m(x) = 0$. Then $h(x) = f(x)^{2}$. So $n$ must be even. Let $n = 2k$. Because $f(x)$ has degree $k$, there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.  This is where I got stuck. Hope some one can help me solve this. Thanks. $\sum \cos$ when angles are in arithmetic progression Possible Duplicate:          Prove $$\cos(\alpha) + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \dots + \cos[\alpha + (n-1)\beta] = \frac{\cos(\alpha + \frac{n-1}{2}\beta) \cdot \sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}} $$ I am running an analysis on the probability of loan default using logistic regression and random forests.    When I use logistic regression, the prediction is always all '1' (which means good loan).  I have never seen this before, and do not know where to start in terms of trying to sort out the issue.  There are 22 columns with 600K rows.  When I decrease the # of columns I get the same result with logistic regression.   Why could the logistic regression be so wrong?  **Actual from the data**  0 :   41932  1 :   573426  **Logistic regression output**   prediction for 1 when actually 0: 41932 prediction for 1 when actually 1:573426  A**s you can see, it always predicts a 1**   **Random forests does better:**  actual 0, pred 0 : 38800  actual 1, pred 0 : 27  actual 0, pred 1 : 3132 actual 1, pred 1 : 573399 Conditional expectation of $\mathbb{E}(X - Y | (X, Y)\in\mathcal{A})$ Given two independent random variables $X \sim \mathcal{U}[-1,5]$ and $Y \sim \mathcal{U}[-5,5]$, what is $$\mathbb{E}\{Y - X | X \le 1, Y &gt; X, Y \in [-1,1] \}\,?$$  I managed to compute the respective probability   $$ \mathbb{P}\{ X \le 1, Y &gt; X, Y \in [-1,1] \} = \mathbb{P}(X \le 1) \mathbb{P}(Y &gt; X) \mathbb{P}(Y \in [-1,1]) = 1/3 * 1/2 * 1/5 = 1/30$$  How do I continue from there? Why are doctors addressed as Mr. in the UK? In the US most physicians, surgeons and dentists are addressed as "doctor". Very few other professionals receive the same title.  In the UK, however, surgeons and dentists seem to prefer to be referred to as Mr. If that is true, why?  And who is addressed as Dr. in the UK?  Those who hold doctoral degrees only? Custom attributes - Yea or nay? Can I change some but not all Apple keyboard function keys (dim vs F1, bright vs F2)? I want to override F1 and F2 behavior, since I have external display and Mac can't control display's brightness. The rest of the keys should retain their non Fx  I know about "Use all F1-F12 keys as standard function keys.", but I don't want to change all of the keys to be F when pressed, just some of them. (For me single press music and volume control is necessary.)  Is there a way to pick some of the function keys to default to numbered function instead of the "Apple" definitions system wide? I have an Asus Sabertooth Z170 motherboard, with a Samsung 950 Pro m.2 NVMe drive on it. I've heard in the past that installing Windows 7 on an NVMe drive was a significant amount of work. Its now August 2016, is it still difficult? I've seen updates from Microsoft saying they've added support for NVMe in Windows 7.  Can I just download Windows 7 from Microsoft () and put it on a USB drive and install?  Is it harder than that? Find $x,y$ given $\gcd(x,y)$ and ${\rm lcm}(x,y)$ Setting controller variables not reliable from apex:commandButton I'm trying to set a variable inside my controller via an actionFunction, which is called by some Javascript, which is initially called by either by an actionPanel (with styleClass='btn') or a commandButton.  If the caller is the actionPanel then the variable updates reliably. However with the commandButton I can see the updated value flash briefly before reverting to its previous value. Maybe one out of every six clicks of the button actually causes the new value to stick. Sometimes it can take a dozen or more tries, regardless of click frequency.  Can someone please explain what's going on here?  PAGE  &lt;apex:page controller="MyController"&gt;     &lt;apex:form &gt;         &lt;apex:outputPanel id="panel"&gt;             &lt;apex:outputText value="MyString = {!MyString}"/&gt;             &lt;apex:actionFunction name="SetString" rerender="panel"&gt;                 &lt;apex:param name="param1" value="" assignTo="{!MyString}"/&gt;             &lt;/apex:actionFunction&gt;             &lt;script&gt;                 function JSetString() {                     SetString('{!MyString}' + '.');                 }             &lt;/script&gt;             &lt;apex:outputPanel onclick="JSetString()" styleClass="btn"&gt;                 Grow {!MyString} (works)             &lt;/apex:outputPanel&gt;             &lt;apex:commandButton onclick="JSetString()" value="Grow {!MyString} (unreliable)" rerender="panel"/&gt;         &lt;/apex:outputPanel&gt;     &lt;/apex:form&gt; &lt;/apex:page&gt;   CONTROLLER  global class MyController {      public MyController() { MyString = 'hello'; }      public String MyString {get;set;} } ‘ld: warning: directory not found for option’ Mathematical operators, such as function names, should be set in roman type, not italics. Latex already has commands for some operators, including \max, \min, and \log.  How can I define additional such commands? Determiners: anyone versus someone I'm trying to get through Spivak's Calculus on my own and even though I kinda understand induction I'm not so sure that's the case when it comes to complete induction. So I tried to do a starred problem which involves using it. But I'm not sure that my proof is valid, can someone check this for me?  So the problem is that I have to prove that     For every natural $n&gt;1$, if $n$ is not prime we can write it as a product of primes.   I'm starting with a base case. For $n=2$, $n$ is prime, so the assumption holds.  Let's assume that for some $n &gt; 1$ and also all numbers $p &lt;= n$ the assumption holds.  Then $n+1 = cd$. If $n+1$ is not prime, then $c &lt; n+1$ and $d &lt; n+1$, but as they are natural numbers, we can also write that $c &lt;= n$ and $d &lt;= n$. But we assumed that if $p &lt;= n$, and $p$ is not prime, we can write it as a product of primes. So if $c$ or $d$ are not primes, we can write them as a product of primes. That means that $n+1$ can be written as a product of primes if it's not prime, which completes the proof. How do you run update-grub? Can I travel to the UK unaccompanied on an accompanied child visa now that I am over 18?