Plot two functions with different ranges into one plot
1 Plot, 2 Scale/Axis
$f \in L^1$, but $f \not\in L^p$ for all $p > 1$ "Find an $f \in [0,1]$ such that $f \in L^1$ but $f \not\in L^p$ for any $p &gt; 1$."  I've thought about doing something like $$f(x) = \frac{1}{x}$$ where $|f|^p = \frac{1}{x^p}$ doesn't converge when $p &gt; 1$.  But this function isn't itself in $L^1$.  Could someone please give me a hint for how to solve this problem?  I wish there were a situation where you had convergence on the closed half disc $[0,1]$ and divergence on $(1, \infty)$, rather than my current predicament where I have convergence on the open half-disc $[0, 1)$ and divergence on $[1, \infty)$. Word for person who looks for flattery I am building a regression model and I need to calculate the below to check for correlations   Correlation between 2 Multi level categorical variables Correlation between a Multi level categorical variable and continuous variable  VIF(variance inflation factor) for a Multi level categorical variables   I believe its wrong to use Pearson correlation coefficient for the above scenarios because Pearson only works for 2 continuous variables.   Please answer the below questions   Which correlation coefficient works best for the above cases ?  VIF calculation only works for continuous data so what is the alternative?  What are the assumptions I need to check before I use the correlation coefficient you suggest?  How to implement them in SAS &amp; R? Virtual vs Real image For a while now I've noticed that if I take a cup of hot tea and pour it into two cups and leave it then both cups will cool faster than the single cup.  How is it that nature cools two cups of tea in parallel more quickly? It's still the same amount of tea but it gets cooled quicker if I have two cups. Let $A \in Mat_{n \times n}(F)$ a matrix which satisfies $A^n=0$ and $A^{n-1} \ne 0$ for some positive integer $n$. Let $v \in F^n$ be such that $A^{n-1}v \ne 0$.  Prove that $\{v, Av, A^2v,...,A^{n-1}v\}$ is a basis for $F^n$.     So, from definition of $A$, I know that $A^iv \ne 0$ for all $0\le i \le n-1$. I know I need to prove the set is linearly independent and span $F^n$ to prove it's a basis, but I just don't know how. Can virtual particle have mass? Do all virtual particle travel at light speed in a vacuum? else wouldn't that imply they should have rest mass however tiny? When they pop back out of existence do their mass disappear instantly? BTW what is the heaviest virtual particle ever found? The only good way of creating circles in ArcMap is using true arc, (compound curves) but I need to convert those to vertex (densifying) because of a compatibility problem. I need info for geodatabase and SDO_geometry (arcsde)  and yes there is a way. I use:  UPDATE layer1 a SET arctype = 'compound' WHERE has_compound_curves(a.shape) &lt;&gt; 0;   and then using ET Geotools to densify  layer by layer.  But there should be easier way of doing this straightforward in Oracle or ArcMap, and I just want to densify the features that have true arc. I don't want any unnecessary vertices on straight lines and so on.  How can I identify true arcs in ArcMap, and how can I densify them as fast and easy as possible?  Same for Oracle SQL. I used to say "sharp sign" to refer to the # sign.   Today a friend told me that the correct term is number sign or hash sign or even just hash.  What is the difference between these options and what's the correct usage of the names for this sign? PHP replacing special characters like à->a, è->e In Peter Jackson's film The Fellowship of the Ring, when the Council of Elrond takes place it shows a total of 22 members:    We all know the names of the ones who were part of the Fellowship, but I want to know who are the rest of the people in the council. Who are the elves that came with Legolas? The men next to the dwarves? The three old men sitting next to Boromir? Or the two elves next to Aragorn?  Starting from left to right:   Frodo Gandalf Elf Elf Legolas Elf Old Man Old Man Dwalin Gloin Gimli Nori Old Man Boromir Old Man Old Man Elf Elf Aragorn Elladan Elrond Elrohir How can I assign the output of a command to a shell variable? How to make nautilus forget saved password? I am browsing Shared folders on a Windows network through Samba.  When I first opened a system on the network I was asked for a login username, password and I checked the save-password setting. Now the shared folder opens directly without asking for authentication.  I want to browse the folder using a different user which has more access rights. How should I do that? Random Forest mtry Spit-to-Split Variable Sampling Procedure Strictly Upper Triangular $n\times n$ Matrices form a Nilpotent Lie Algebra over $\mathbb{C}$ for $n\geq 2$ I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints.  Let $\mathfrak{u}(n,\mathbb{C})$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $\mathbb{C}$.  I am trying to show that $\mathfrak{u}(n,\mathbb{C})$ is a nilpotent Lie algebra for $n\geq 2$.  This is what I have looked at so far.  I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for $n=2,3,4$ and have come up with some ideas.  Let $L=\mathfrak{u}(n, \mathbb{C})$.  If $n=2$,  $L^1 = [L, L] = &lt;0&gt;_{\mathbb{C}}$  So $L$ is nilpotent.  If $n=3$,  $L^1 = [L, L] = &lt;e_{13}&gt;_{\mathbb{C}}$  $L^2 = [L, L^1] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  If $n=3$  $L^1 = [L,L] = &lt;e_{13},e_{14},e_{24}&gt;_\mathbb{C}$  $L^2 = [L,L^1] = &lt;e_{14}&gt;_{\mathbb{C}}$  $L^3 = [L, L^2] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  So my hypothesis so far is that for an arbitrary $m\in\mathbb{N}$, $L^{m-1}=\{0\}$  My instinct is to do this by induction. My base case, $n=2$ has already been shown. Now I assume that $\mathfrak{u}(n,\mathbb{C})$ is nilpotent for $n=k$ ($k\in\mathbb{N}$) and consider the case for n=k+1.  However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory - J.E. Humphreys that $L^k$ should be spanned by basis vectors $e_{ij}$ where $j-i=k+1$ - however in my calculations for $L^1$, I obtain an $e_{14}$ which does not satisfy 4-1=2.  Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!)  Thanks,  Andy. Darth Vader threw Emperor Palpatine to a pit. Well, , but unlike other times, Palpatine wasn't injured or unconscious when he was thrown.  Why didn't he simply prevent his fall using his lightsaber (I have seen such action in Star Wars: The Clone Wars TV series lots of time) and make his way out? Blending colors in photoshop and illustrator I am working on a print for my wall and I am trying to create something like this:      I have already created the letters in Illustrator and now I am adding the colors in Illustrator and then exporting everything to Photoshop and then using the liquify tool to blend the colors to create a similar effect. I also used the healing patch tool to blend in the hard edges.  Is that the right way of doing it? Or is there another way? Is the harmonic oscillator potential unique in having equally spaced discrete energy levels? I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and quadratic in others, but that's only trivially different. I am not referring to equally spaced as a limiting behavior either, I mean truly integer spaced.  Any ideas? If not, is there a proof for its uniqueness?  If there are other potentials with equally spaced eigenvalues, can one use them as starting points for a free-field QFT? It would be interesting to know if there is a deeper mathematical relation between all of these potentials and whether they could be used to study interacting systems. How do I prove this function is bijective? $$ v(s,p)=2^{p-1}(2s-1). $$ The domain is natural numbers and the codomain is also the natural numbers.  And this one: $$ f(s)=2s-1. $$ The domain is the natural numbers, and the codomain is the odd numbers in the natural numbers.  With this one I would do this to show it's injective: $$ \begin{align} v(s)&amp;=v(s_1)\\ \implies 2s-1&amp;=2s_1-1\\ \implies (2s_1)/2&amp;=(2s_2)/2\\ \implies s=s_1 \end{align} $$ So it's injective since if $v(s)=v(s_1)$ then $s=s_1$.  And to show it's surjective $f(s)=y$: $$ \begin{align} y&amp;=2s-1\\ \implies s&amp;=(y+1)/2 \end{align} $$ Then the function must be surjective since every $y$ is the same as the codomain for $f$.  Am I correct? Why does "for good" mean "forever"? A very recent and similar  was closed asks what "for good" means.  While general reference can answer the question, I became curious as to the etymology of the idiom.   around got me nowhere, which is unusual.  The closest I found was a thread in :     OED   for good (and all): as a valid conclusion; hence, as a final act, finally.    15.. Parl. Byrdes Aij, Than desyred al the Byrdes great and smal to mewe the hauke for good and all. 1603 in Crt. &amp; Times Jas. I (1849) I. 25 D’Auval.. is gone for good and all. 1687 CONGREVE Old Bach. I. i, Ay, you may take him for good-and-all if you will. 1711 SWIFT Jrnl. to Stella 4 July, This day I left Chelsea for good, (that’s a genteel phrase). a 1732 T. BOSTON Crook in Lot (1805) 37 He was obliged for good and all to leave his country. 1850 J. H. NEWMAN Diffic. Anglic. 324 Throw off, for good and all, the illusions of your intellect. 1882 W. E. FORSTER Let. To Gladstone 10 Apr. in T. W. Reid Life (1888) II. viii. 421 This morning we released Parnell—not for good, but on parole.   However, this only traces some historical usages, and suggests that for good and all used to be a more popular version of the current idiom.  Is there anything more substantial?  Why did "for good" come to mean "forever"?