How to install Ubuntu alongside Windows 8.1 with multiple hard drives
How to use manual partitioning during installation?
Why is inductance (L) proportional with turns-square (N²)? We start from the   $$ \mathbf{\nabla} \times \mathbf{B} = \mu \mathbf{J} + \overbrace{\mu \epsilon \dfrac{\partial \mathbf{E}}{\partial t}}^0. $$  We take surface integration of both sides, for the surface (\$s\$) inside the mean path (\$c\$) of the core.  $$ \int_s \left( \mathbf{\nabla} \times \mathbf{B} \right) \cdot d\mathbf{s} = \mu \int_s \mathbf{J} \cdot d\mathbf{s} $$  We use the  to rewrite the left hand side; where \$c\$ is in the same direction with the magnetic flux \$\Phi\$.  $$ \oint_c \mathbf{B} \cdot d \mathbf{\ell} = \mu N I $$  (The integral on the left hand side results \$NI\$, because there are \$N\$ different wires on the winding.)  The magnetic field density inside these kind of cores are considered to be uniform. So, we can write  $$ B \ell_c \overset\sim= \mu NI \implies B = \dfrac{\mu NI}{\ell_c}; $$  where \$\ell_c\$ is the mean path lenght of the core.  We can find magnetic flux from the magnetic flux density we have found by using the cross sectional area of the core \$A_c\$.  $$ \Phi = BA_c = \dfrac{\mu NIA_c}{\ell_c} $$  By definition, inductance is the amount of magnetic flux generated per applied current, that is  $$ L \overset\triangle= \dfrac{\Phi}{I}. $$  So, we find inductance of the system as  $$ \boxed{ L = \dfrac{\Phi}{I} = \dfrac{\dfrac{\mu NIA_c}{\ell_c}}{I} = \dfrac{\mu NA_c}{\ell_c} }. $$  But, all other sources () give inductance of an inductor like this as  $$ L = \dfrac{\mu N^2A_c}{\ell_c}. $$  What is the mistake I did in my derivation? Please explain in detail. On Ubuntu 12.04 LTS on a Dell Vostro 2420 laptop, each time I start Ubuntu, the screen brightness is set to the maximum level, which gives a high glare. (I have recently installed Ubuntu on the laptop, but did not change any default settings related to brightness, IIRC.) I am currently lowering the brightness manually each time after starting, using the System Settings icon on the left of the screen. Looking for a permanent way to set it to a lower value. I did the following proof and I was wondering if its valid. It feels wrong because I didn't actually test the case when purportedly $n$ is not prime, but please feel free to correct me.  Assume there exists $n$ such that $\phi(n) = 14$. Assume $n$ is prime. Then $\phi(n) = n-1$. Then here, n-1 = 14, so n = 15.  We know since Euler's totient function is multiplicative that $\phi(xy)$ = $\phi(x)\phi(y)$, so $\phi(15) = \phi(3)\phi(5)$, but alas $\phi(3)\phi(5) = 2\cdot4 = 8 \ne 14$. If $n$ is not prime a similar argument follows since we know then that $n$ must be composed of prime numbers by the prime factorization theorem. How use spaces instead of underscores in file names? I'm referring to this "feature"   It was not happening before, but it is now, and it is messing up the structure of some folders I previously created. Is there a way to disable it? $$\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}= 5$$ Substitute for onions and garlic I love the taste of onions and garlic and it seems lots of other people do too.  But they upset my stomach so much that I can't really cook with them.  What can I use in their place to give my food a similar flavour? No more than 10,000 results are visible in search If you make a search for something that returns more than 10,000 items in result. For example, search for  return 161,338 results on MSE. But if you select a page where 10,000+ result should be shown:        In the edge case  shows zero results:        But this is actual for any page where per_page * page_number is more than 10,000.  Though this does not happen for Stack Overflow in English. It looks like all other sites are affected.  Related bug report on ruSO.meta: My 12.04 x64 system won't boot into GUI, but I can still use virtual console with my own account, so apparently it's the Windows Management System that's causing it, how can I reinstall it from tty, what's the apt-get command? Is there Ubuntu's analogue of snapshot.debian.org? I wonder  whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic?   for every two real numbers $\alpha,\beta,\alpha\lt\beta$,  $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$   I can't  say such a function does not exist, neither can I construct a example  Thanks a lot! Fastest Square Root Algorithm What is the fastest algorithm for finding the square root of a number?  I created one that can find the square root of "$987654321$" to $16$ decimal places in just $20$ iterations   I've now tried Newton's method as well as my own method (Newtons code as seen below)  What is the fastest known algorithm for taking the second root of a number?  My code for Newton's Method (*Edit: there was an error in my code, it is fixed in the comments below):        a=2   //2nd root     b=97654321   //base     n=1   //initial guess     c=0   //current iteration (this is a changing variable)     r=500000   //total number of iterations to run     while (c&lt;r)      {         m = n-(((n^a)-b)/(a*b))  //Newton's algorithm         n=m         c++;         trace(m + "  &lt;--guess   ...   iteration--&gt;  " + c)     } Why do we use ./filename to execute a file in linux?  Why not just enter it like other commands gcc, ls etc... Prove a differentiable function between banach spaces is continuous I really don't know where to start with this problem. It intuitively makes sense from basic calculus knowledge but I have absolutely no clue how to show it. , When I spoke with a friend of mine, I had noticed she used sentences similar to "I need to go buy food."   Is it correct to say "to go buy," or should I say "to go to buy"?    I know that I could say "I need to buy food," but I think that in "I need to go buy food," who is speaking is putting in evidence s/he needs to go and that s/he cannot stay any longer, or that her/his time to stay is limited. What is the easiest way to swap occurrences of two strings in Vim? For a given inner product space $V$, I would like to prove the inequality $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).$$  We see that $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,$$  which by linearity $$=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.$$  I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement? Let $f(z)=\sum\limits_{n=0}^{\infty}a_n z^n$ be an analytic function in $|z|\leq R$ such that $|f(z)|\leq M$. Prove an inequality about $f$'s zero Has there been any progress made on granting a developer account for SF and HubExchange Devs? It feels conceptually wrong to me a SF product doesn't allow developers to perfect and develop a wider solution offering without paying +10K license fee per year. Even without CRM the HubExchange market could be a richer portal for consumers and developers if we were allowed a means to test our concept.  I think we'd all be perfectly happy with an account that was blackholed or limited in such a way where we could only send a finite qty of emails to an email address associated to the user.  If there's a forum or avenue where we can assist in raising the value proposition please let us know as I know there are many people looking. Normal Subgroup Counterexample Im having trouble with the second part of this question,  Let $H$ be a normal subgroup of $G$ with $|G:H| = n$,  i) Prove $g^n \in H$ $\forall g \in G$ (which i have done)  ii) Give an example to show that this all false when $H$ is not normal in $G$.(which I am having trouble with showing)  Any suggestions? Proving bessel identity $[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$ and $|J_0(x)|\le 1$ The generating function for a Bessel equation is:      $$g(x,t) = e^{(x/2)(t-1/t))}$$      Using the product $g(x,t)\cdot g(x,-t)$ show that:      a) $$[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$$      and consequently:      b)      $$|J_0(x)|\le 1, \forall x$$      c) $$|J_n(x)| \le \frac{1}{\sqrt{2}}; n=1,2,3,\cdots$$   For a) I tried the product:  $$e^{(x/2)(t-1/t))}\cdot e^{(x/2)(-t+1/t))} = 1$$  I at least arrived at the right side of the equation. Since this generates the bessel functions, I should arrive at something related to $J_n$ in the left side. I know that  $$e^{(x/2)(t-1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)t^n$$  $$e^{(x/2)(-t+1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)(-t)^n$$  but it's not just a matter of multiplying coefficients from the two infinite series, right?  For $b$, I tried to use  but it's not as obvious  I have no idea how to deal with $c$, could somebody help me?