Why does my camera lens only work with a Polaroid .43x lens?
Why do I get blurry pictures when I take a lens attachment off?
diophantine equation: $x^2 +y^2 = z^n$ Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :) Can Java's NullPointerException be changed to report on which variable was null? Using contracted forms ("don't", "let's") in a formal text How compelled should I feel to use non-contracted forms (do not rather than don't and so on) when writing in a rather formal text, say an academic paper? In one case I am afraid to seem too stilted, in the other, too casual. Are there good guidelines?  And are there differences in this regard between British and American English?  (In this  there was not much more than "Contractions are more frequent in informal than formal contexts".) I installed ubuntu 16.04 LTS, but there is no ubuntu one software, not in ubuntu software center also..Though it is showing selected in unity tweak tool.How can I get it. Please reply. What do the different colors mean in ls? I am encountering two definitions of the special orthogonal lie algebra, and I would like to know if they are equivalent, and if there are advantages to working with one over the other.  If we begin with an $n$-dimensional vector space $V$ over a field $k$ and a chosen basis, we can define a bilinear form on $V$ by a matrix $S\in M_n(k)$, ie, let $\langle v,w\rangle=v^tSw$ for all $v,w\in V$.  Now $g\in GL_n(k)$ preserves the form ($\langle g(v),g(w)\rangle=\langle v,w\rangle$) if and only if $g^tSg=S$, so all such $g$ form a linear algebraic group $G$.  The tangent space at the identity of $G$ will be contained in that of $GL_n(k)$, so $T_eG\subset T_eGL_n(k)=M_n(k)$, and in fact, $T_eG=\{B\in M_n(k)\mid B^tS+SB=0\}$.  $T_eG$ becomes a lie algebra, $Lie(G)$, if we define the bracket to be the commutator of two matrices.  Now, if $S=I_n$, it follows that $G=O_n(k)$ is the orthogonal group of matrices satisfying $g^tg=I_n$, and $Lie(G)=\mathfrak{so}_n$ is the lie algebra of antisymmetric matrices.  In Humphrey's Introduction to Lie Algebras and Representation Theory, he defines $\mathfrak{so}_n$ to be all matrices $B$ satisyfing $B^tS+SB=0$, where $$ S=\begin{pmatrix} 1&amp;0&amp;0\\ 0&amp;O&amp;I_l\\ 0&amp;I_l&amp;O \end{pmatrix} \hspace{.5in}\text{or}\hspace{.5in} S=\begin{pmatrix} O&amp;I_l\\ I_l&amp;O \end{pmatrix} $$ depending on the parity of $n$.  The matrices obtained in this way are not antisymmetric, nor is the group $G$ preserving the form defined by $S$ the orthogonal group $O_n(k)$.  Are the two groups obtained from considering different $S$ isomorphic?  Are the two lie algebras isomorphic?  If so, why would one prefer one form to the other? Can somebody help me with the execute commands in Minecraft? Recently, Minecraft had removed the /testfor command, which broke these sets of commands that I was trying to execute today.  /testfor @p[r=10] {SelectedItem:{id:"minecraft:feather",tag:{display:{Name:"Roc's Feather",Lore:[Jump higher using this feather.]}}}}`   Now normally, from Minecraft 1.13 onwards, the new /execute command would've been the replacement for the /testfor command, but as I soon found out, the command failed to detect my Roc's Feather and thus could not execute an /effect command that gave me a Jump Boost, for 1 second, at level 3.  I tried remaking my feather from scratch with this command here:  /give @p minecraft:feather{display:{Name:"{\"text\":\"Roc's Feather\",\"color\":\"light_purple\",\"bold\":true,\"italic\":true,\"underlined\":false,\"strikethrough\":false,\"obfuscated\":false}",Lore:["{\"text\":\"Jump higher using this feather.\",\"bold\":false,\"italic\":true,\"underlined\":false,\"strikethrough\":false,\"obfuscated\":false}"]}} 1   But the execute command I placed did not detect this type of uniqueness.  /execute if entity @p[distance=10,nbt={SelectedItem:{id:"minecraft:feather",Count:1b,tag:{display:{Name:"{\"text\":\"Roc's Feather\",\"color\":\"light_purple\",\"bold\":true,\"italic\":true,\"underlined\":false,\"strikethrough\":false,\"obfuscated\":false}",Lore:["{\"text\":\"Jump higher using this feather.\",\"bold\":false,\"italic\":true,\"underlined\":false,\"strikethrough\":false,\"obfuscated\":false}"]}}}}] run effect give @s minecraft:jump_boost 1 3 true   Can somebody tell me what I'm doing wrong here because I cannot even tell what's the issue. Proof Check: Every Cauchy Sequence is Bounded Sorry if I keep asking for proof checks. I'll try to keep it to a minimum after this. I know this has a well-known proof. I understand that proof as well but I thought I'd do a proof that made sense to me and seemed, in some ways, simpler. Trouble is I'm not sure if it's totally correct. It's quite short though. I was just hoping someone could look it over and see if it is a valid proof. Thank you!  Lemma: Every Cauchy sequence is bounded.  Proof: Let $(a_{n})$ be Cauchy. We choose $ 0&lt;\epsilon_{0}$. So $ \forall \; n&gt;m\geq N_{0}$ we have that $\vert a_{n}-a_{m} \vert &lt; \epsilon_{0}$. Therefore $(a_{n})$ is bounded for all $ m \geq N_{0} $ by $ \epsilon_{0} $. Since $ \mathbb{N}_{N_{0}}$ is finite, it is bounded. So, for all $ m&lt;N_{0} $, $ (a_{n})$ is bounded. Therefore $(a_{n})$ is bounded.  I realize I haven't said what the bounds are but I think that's sort of irrelevant. So long as we know it's bounded. Any help is much appreciated! Should I compile release builds with debug info as "full" or "pdb-only"? How does Python evaluate the expression 1+++2?  How many ever + I put in between, it is printing 3 as the answer. Please can anyone explain this behavior  And for 1--2 it is printing 3 and for 1---2 it is printing -1 On page 19 of the textbook  (by James, Witten, Hastie and Tibshirani--it is freely downloadable on the web, and very good), the following is stated:  Consider a given estimate $$\hat{Y} = \hat{f}(x)$$ Assume for a moment that both $$\hat{f}, X$$ are fixed. Then, it is easy to show that: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ $$ = [f(X) - \hat{f}(X)]^2 + \mathrm{Var}(\epsilon)$$  It is further explained that the first term represents the reducible error, and the second term represents the irreducible error. I am not fully understanding how the authors arrive at this answer. I worked through the calculations as follows: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ This simplifies to $[f(X) - \hat{f}(X) + \mathrm{E}[\epsilon]]^2 = [f(X) - \hat{f}(X)]^2$ assuming that $\mathrm{E}[\epsilon] = 0$. Where is the $\mathrm{Var}(x)$ indicated in the text coming from? Any suggestions would be greatly appreciated. let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$   show that   $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$   This problem is from Secrets In Inequalities volume 1  page 30,example 1.24. the comment.the  author this case is a bit more diffcult,But this author can't post solution.Thank you For aer.bst, citations with 3 or more authors are listed using et al. I need to modify the bibliography so that citations with 4 or more authors are listed using et al., but that citations with 3 authors have all the names written out. For example, I want "See Adams, Brown, and Clark (2003)" as opposed to "See Adams et al. (2003)." I'm using natbib with LaTex: \usepackage{natbib,har2nat}  In order to do that, I need to modify the function format.lab.names.abbr.  Might someone know how to do this?  FUNCTION {format.lab.names.abbr} { 'name.list :=   name.list num.names$ 'numnames :=   numnames #1 &gt;     { numnames #2 &gt;         { name.list #1 "{vv~}{ll}" format.name$ " et al." * }         { name.list #2 "{ff }{vv }{ll}{ jj}" format.name$ "others" =             { name.list #1 "{vv~}{ll}" format.name$ " et al." * }             { name.list #1 "{vv~}{ll}" format.name$ " and " *               name.list #2 "{vv~}{ll}" format.name$ *             }           if$         }       if$       field.used editor.field = {", eds" *} {} if$     }     {         name.list #1 "{vv~}{ll}" format.name$         field.used editor.field = {", ed" *} {} if$     }   if$ }   FUNCTION {format.lab.names.full} { 'name.list :=   #1 'nameptr :=   name.list num.names$ 'numnames :=   numnames 'namesleft :=     { namesleft #0 &gt; }     { name.list nameptr "{vv~}{ll}" format.name$ 'temp :=       nameptr #1 &gt;         { namesleft #1 &gt;             { ", " * temp * }             { temp "others" =                 { " et~al." * }                 { " and " * temp * }               if$             }           if$         }         'temp       if$       nameptr #1 + 'nameptr :=       namesleft #1 - 'namesleft :=     }   while$   numnames #1 &gt; field.used editor.field = and {", eds" *} {} if$   numnames #1 = field.used editor.field = and {", ed" *} {} if$ } A dollar amount $\${-}67.9{-}$ is divisible by 72. What is this number? A client buys 72 turkeys. On the receipt, 2 digits are missing, the first and the last. So, the total price is $\${-}67.9{-}$. Find a way to make the price of each turkey round at the second digit without trial-and-error method.  So, the answer is obviously $\$5.11$/turkey, so $\$367.92$ total. I found that with with 2-3 trial-and-error.  I bypassed his problem by using python and made a script that tries every combination and assures it is divisible by 0.01.   Please help me find a mathematical way of solving this problem. It's a standard result that given $X_1,\cdots ,X_n $ random sample from $N(\mu,\sigma^2)$, the random variable $$\frac{(n-1)S^2}{\sigma^2}$$ has a chi-square distribution with $(n-1)$ degrees of freedom, where $$S^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2.$$ I would like help in proving the above result. Thanks. Firmware vs. Software For x < 5 what is the greatest value of x It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be? What is the name of the problem where lines appear in a photo where there should just be a smooth gradient? What is the correct position of the word 'also'?     I would like to also talk about ...      I would also like to talk about ... Preventing robots from crawling specific part of a page