I have uninstalled lamp but I want to install it again. But when I run this command  sudo apt-get install lamp-server^   But when I run this command through this error message  Unable to correct problems, you have held broken packages. how to fix that?
After upgrading from 10.04 to 12.04 I am trying to install different packages.  For instance ia32-libs and skype (4.0).  When trying to install these, I am getting the 'Unable to correct problems, you have held broken packages' error message.   Output of commands:  sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   After running this:  sudo dpkg --configure -a foo@foo:~$ sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.
Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$ I am being asked to prove that $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$ I have some progress made, but I am stuck and could use some help. What I did: It holds that $$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re\left(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right) $$ For any $z_{1},z_{2}\in\mathbb{C}$ we have it that if $z_{1}=a+bi,z_{2}=c+di$ then $$\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z2}}{|z_{2}|^{2}}=\frac{(a+bi)(c-di)}{|z_{2}|^{2}}=\frac{ac-bd+i(bc-ad)}{|z_{2}|^{2}}$$ hence $$Re\left(\frac{z_{1}}{z_{2}}\right)=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$ Thus, $$Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}= \frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)} $$ This is the part where I am stuck, I would appriciate any help or hint on how to continue. Edit: Given the corrections by André I get: $$(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+1+\sin(nx+x)\sin(x)$$ so $$\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$$ Edit 2: I found anoter mistake in the above, I will try to correct Edit 3: When multiplying correctly the above it works out :-) I am an Indian citizen studying in Italy on an Italian visa that expires 19 November, 2015. I have permesso di soggiorno. I want to visit the United Kingdom for my vacations for a week during March. I know the procedures already, however I am a little bit confused about the amount that I must show as a proof of support with my bank statements. I will stay for a week with my friends who are also students in Italy and will be in London for 5 days. What would be the amount in Euros that I need to show for my visit to UK? What’s (not) spam? Somebody asked the following :    Clearly this is just a troll post so I (and other users) flagged it as spam. The spam flag was subsequently disputed (which I think simply means that a moderator rejected it):    Huh? Puzzled, I took another look at the list of possible flagging reasons:     I am flagging this question because         it is spam   This question is effectively an advertisement with no disclosure. It is not useful or relevant, but promotional.   it is offensive, abusive, or hate speech   This question contains content that a reasonable person would deem inappropriate for   respectful discourse.   it is a duplicate...   This question has been asked before and already has an answer.   it should be closed for another reason...   This question does not meet this site's standards and should be closed.   it is very low quality   This question has severe formatting or content problems. This question is unlikely to be salvageable through editing, and might need to be removed.   other (needs ♦ moderator attention)   This question needs a moderator's attention. Please describe exactly what's wrong.      Of these, only a few are applicable at all: “spam”, “should be closed for another reason”, “low quality” and “requires moderator attention”.  But:   it shouldn’t be closed (it has no value – it should be deleted), it’s not “low quality” (the description seems to fit actual, albeit bad, questions), and it doesn’t require moderator attention (I certainly didn’t want to bother a mod with this worthless post).   So I still think “spam” is the most fitting reason. Maybe it would help to have a more apt reason “troll post”, or maybe we could explicitly subsume this under one of the other points? As I'm learning about iptables I've made a couple of mistakes and locked myself out.  What method(s) do you use to test rules without locking yourself out?  I'm using ubuntu server 12.04 LTS    All the answers below were helpful.  In the end I used a combination of options.  It also helps to have IPMI access to your remote server just in case!  But ideally test the rules locally on a replicated environment and test that first.  Vagrant helps in this regard to get test setups working quickly. Show that if $(b,c)=1$, then $(a,bc)=(a,b)(a,c)$. How to access raw texture data in DirectX 11? I want to get access to a texture's pixel color data on the CPU. When trying to map the texture, I get E_INVALIDARG in return from ID3D11DeviceContext::Map.  ID3D11ShaderResourceView* resourceView; //represents my texture, already loaded up //...  //Getting the resources of the texture view... ID3D11Resource* res = nullptr; ID3D11Texture2D* tex = nullptr; resourceView-&gt;GetResource(&amp;res); res-&gt;QueryInterface(&amp;tex); D3D11_TEXTURE2D_DESC desc; tex-&gt;GetDesc(&amp;desc); //Correct data gets filled out D3D11_RESOURCE_DIMENSION dim; res-&gt;GetType(&amp;dim); //value gets set as Texture2D which it should  D3D11_MAPPED_SUBRESOURCE mappedResource; HRESULT h; h = context-&gt;Map(tex,0,D3D11_MAP_READ,0,&amp;mappedResource); //This always fails h = context-&gt;Map(res,0,D3D11_MAP_READ,0,&amp;mappedResource); //This also fails   The idea is that I keep only shaderResourceViews for my textures, because that is all I needed until now, and get the relevant resources here.  I tried mapping as WRITE_DISCARD also, but to no success. I am really confused, because I never had any problem with mapping constant buffers and vertex buffers. I seem to be certainly missing something? By the way, I am using Wic Texture loader and DDS Texture loader for the DirectXTex toolkit, and tried this code for png, dds textures, with and without mip-maps. Is it possible to obtain AES-128 key from a known ciphertext-plaintext pair? I have a file, which was encrypted with AES-128 in ECB mode. I know the format of the original file and know that all files in this format have the same headers. So, I have an encrypted block and the original block.  Can I get the encryption key, using this information? Is a droid a ‘he/she’ or an ‘it’? Inspired by  question.  Are droids in the Star Wars universe referred to with gendered pronouns (i.e. he or she), or are they referred to as objects (i.e. it)? What exactly are the Python scoping rules?  If I have some code:  code1 class Foo:    code2    def spam.....       code3       for code4..:        code5        x()   Where is x found?  Some possible choices include the list below:   In the enclosing source file In the class namespace In the function definition In the for loop index variable Inside the for loop   Also there is the context during execution, when the function spam is passed somewhere else. And maybe  pass a bit differently?  There must be a simple reference or algorithm somewhere.  It's a confusing world for intermediate Python programmers. Because I'm still living like it's 2008, I run a website with a .in TLD. The website has nothing to do we India and does not target Indian users.  Unfortunately, when I go to Google Webmaster Tools I see that Google have locked my geotargeting status to India:    On the , Google state that:     Sites with country-coded top-level domains (such as .ie) are already associated with a geographic region, in this case Ireland. In this case, you won't be able to specify a geographic location.   This seems implausibly inflexible, so I want to know if there are any sneaky ways of removing auto-geotargeting, or changing it to a more appropriate region? MPI Get Processor with Minimum value "Above"/"below" before/after a noun How do I use the Natural Earth SQLite DB with QGIS? I just downloaded the Natural Earth data in SQLite format from . I had assumed that this would be a SpatiaLite database, but it appears not to be! QGIS is failing to recognize it as a spatial database. OGR supposedly supports reading geometries even when stored in a plain SQLite database, but maybe QGIS doesn't use OGR for SpatiaLite?  The SQLite database does have geometry_columns and spatial_ref_sys tables. Is there some way to convert it to a full SpatiaLite database? Combining values in one column (or part) when values in another column (or part) match Why are mirrorless cameras much slower than DSLRs? This is a follow-up to a .  I've been comparing cycle times (or shot to shot times in single shot mode) of mirrorless cameras with interchangeable lenses with traditional DSLRs and all of them, even the highest end ones (like the Olympus OM-D EM-5 or Panasonic GH3) are slower than even entry level DSLRs.  How's that possible? The electronics should be the same and they don't have the burden of a mirror going up and down between each shot. I think that there must be a technical reason for that (maybe some cpu time is going to feed the live-view before the image is encoded?), because from a commercial point of view it does not make sense.  I've seen the cycle times under the performance tab in the reviews pages of .  The average CSC has cycle times around 0.7~1.2s, the flagship ones around 0.5s. The average DSLR has cycle times around 0.4s and the flagships around 0.25s. jQuery call function from a string How to fix task manager memory column if it shows lies UK Visa in an expired Nigerian passport - what to use? How to see if certain representation of $SU(2)$ is symmetric/antisymmetric/mixed? I am looking at the tensor product of $n$ spin halfs (fundamental of $SU(2)$): \begin{equation} \left(\frac{1}{2}\right)^n = \frac{1}{2} \times \frac{1}{2} \times ... \times \frac{1}{2} = \frac{n}{2} + ... \end{equation} For example, for $n = 3$ \begin{equation} \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \times \left(1 + 0\right) = \left(\frac{3}{2}\right)_1 + \left(\frac{1}{2}\right)_1 + \left( \frac{1}{2} \right)_0 \end{equation} where lower indexes on the right hand side specify from which tensor product representation came from. My question is how do I read off if a certain term on the right is a symmetric, antisymmetric or a mixed symmetry representation of $SU(2)$? I came up with two ways of approaching the problem.  First is Young Tableaux. As far as I understand rows of Young Tableaux correspond to symmetric indexes while columns to antisymmetric ones. So, for example, a diagram with just one row (like one that corresponds to the highest spin $\frac{3}{2}$ from the example - three boxes in a row) describes a totally symmetric representation. A diagram that has one column (maximum 2 boxes since we are dealing with $SU(2)$) describes totally antisymmetric representation. Everything that has both rows and columns is a mixed symmetry representation. Following this logic it looks like only the highest representation in the product can be totally symmetric, singlet is antisymmetric and everything in between has mixed symmetry. So, for the example \begin{equation} \left(\frac{3}{2}\right)_1: \text{symmetric rep}\\ \left(\frac{1}{2}\right)_{0,1}: \text{mixed symmetry reps}\\ \end{equation} Is this is a correct logic?  Another way to think about this is the following. Let's build representation the way we do in QM starting from the highest one, when all the spins are up. Then all components of the highest multiplet can be generated by applying lowering operators to the highest state. So, all the relative signs in the expressions are &quot;+&quot; and thus the highest representation is always symmetric as 1. also suggests. Then all the other representations must be constructed by writing down expressions orthogonal to components of the highest multiplet, so they will necessary get some &quot;-&quot; signs and as a result must be either antisymmetric or have mixed symmetry, but can not be totally symmetric.   Is this correct logic? Is it true that only the highest representation in the product can be symmetric and all the others have mixed symmetry or antisymmetric? How is using a transformer for isolation safer than directly connecting to the power grid? How is using a 1:1 transformer safer than using the mains straight off? Is it because you can limit the current coming from the transformer whereas straight from the mains its not current limited? I fail to see how its "safer" when playing with electricity is dangerous. Could someone please explain why it is considered safer to be isolated by a transformer.