Assume that a continuous function $f:]1, 2[ \rightarrow \mathbb{R}$ satisfies the conditions      $$x\to1⁺ = -\infty \text{ and } x\to2^- = -\infty$$      Show that there exists and $c \in ]1, 2[$, for which every $x \in ]1, 2[$ holds $f(x) \leq f(c)$   This is essentially a special case of the Weierstrass theorem (Extreme Value Theorem) where they don't give us the actual function. How do we find the closed interval so that we can show the existense of $f(c)$?
Suppose that f is continuous on $(a,b)$ and $\lim_{x\to a^+}  f(x) = \lim_{x\to b^-} f(x) = \infty$. Prove that $f$ has a minimum on all of $(a,b)$.  I was thinking of applying the extreme value theorem, but I realized that it applies only to closed intervals. So I was thinking of defining a closed interval that is in $(a,b)$, and using the definition of one sided limits to go about the problem, but I don't know how to proceed. Help please?
How can I disable the purple splash screen at boot? How can I disable the purple "Ubuntu" splash screen that displays at boot before the login screen has loaded? I'd prefer to just see plain text scroll by. Get text from pressed button Deleted downvotes should appear in Achievements dialog "Good morning.  Here is something awesome that you got last night!", said the Top Bar:    "21?  That must be two answer upvotes and, um, an answer that I downvoted that was deleted?", I thought.  Ok, let's look at the details and find out:    The +20 on my egrep answer is new and I've seen the other Achievements on the list, so where did the extra +1 come from?  In order to avoid confusion, the recent reputation changes in the dropdown should add up to the number in the notification.      I suspect this particular case is a bug.  Assuming the +1 is a downvote reversed by an answer deletion, I could see the argument for excluding it from the Achievements dropdown.  After all, it's not exactly something you accomplished.  Except it kinda is.  Part of the reason for downvoting is to  low-quality content.  The ultimate sign that such content was deemed beneath community standards is when it gets deleted.  I'm not sure that we should , but putting a note in your achievements in such cases seems like it would reinforce positive (albeit negative) behaviour.  This question is a feature request that rep gains from deleted downvotes be listed in the Achievements dialog, without getting into broader issues of displaying other activity on deleted posts such as those discussed in .  Thus, this is not intended to completely address the issue of the Achievements sum not adding up, but is intended to help encourage good community moderation by appropriate downvoting of low-quality content. Can an irrational number raised to an irrational power be rational? Can an irrational number raised to an irrational power be rational?  If it can be rational, how can one prove it? I tried to export my Geogebra graphs into tikz-code. In General it works fine, but plot x^1/3 doesn't work:  \documentclass[10pt]{article} \usepackage{pgf,tikz} \usetikzlibrary{arrows} \pagestyle{empty} \begin{document} \definecolor{ccqqqq}{rgb}{0.8,0,0} \definecolor{qqttcc}{rgb}{0,0.2,0.8} \definecolor{cqcqcq}{rgb}{0.75,0.75,0.75} \begin{tikzpicture}[line cap=round,line join=round,&gt;=triangle 45,x=1.0cm,y=1.0cm] \draw [color=cqcqcq,dash pattern=on 1pt off 1pt, xstep=1.0cm,ystep=1.0cm] (-4,-4) grid (4,4); \draw[-&gt;,color=black] (-4,0) -- (4,0); \foreach \x in {-4,-3,-2,-1,1,2,3} \draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$}; \draw[-&gt;,color=black] (0,-4) -- (0,4); \foreach \y in {-4,-3,-2,-1,1,2,3} \draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$}; \draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$}; \clip(-4,-4) rectangle (4,4); \draw[color=qqttcc, smooth,samples=100,domain=-4.0:4.0] plot(\x,{(\x)*(\x)*(\x)}); \draw[color=ccqqqq, smooth,samples=100,domain=-4.0:4.0] plot(\x,{((\x))^(1/(3))}); \draw [color=qqttcc](2.06,4.18) node[anchor=north west] {$f(x)=x^3$}; \draw [color=ccqqqq](2.06,1.18) node[anchor=north west] {$f(x)=\sqrt[3]{x}$}; \begin{scriptsize} \draw[color=qqttcc] (-1.74,-7.16) node {$f$}; \end{scriptsize} \end{tikzpicture} \end{document}      Error line 21: !Missing number, treated as zero.   So plot has problems with the negative x I suppose, but I don't know. Any help would be great! I googled it for nearly 1 hour - but no real answers.  As suggestet it rotated and mirrored the graph - works great!:  \documentclass[10pt]{article} \usepackage[utf8]{inputenc} \usepackage{pgf,tikz} \usetikzlibrary{arrows} \pagestyle{empty} \begin{document} \definecolor{ccqqqq}{rgb}{0.8,0,0} \definecolor{qqttcc}{rgb}{0,0.2,0.8} \definecolor{cqcqcq}{rgb}{0.75,0.75,0.75} \begin{tikzpicture}[line cap=round,line join=round,&gt;=triangle 45,x=1.0cm,y=1.0cm] \draw [color=cqcqcq,dash pattern=on 1pt off 1pt, xstep=1.0cm,ystep=1.0cm] (-4,-4) grid (4,4); \draw[-&gt;,color=black] (-4,0) -- (4,0); \foreach \x in {-4,-3,-2,-1,1,2,3} \draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize \x}; \draw[-&gt;,color=black] (0,-4) -- (0,4); \foreach \y in {-4,-3,-2,-1,1,2,3} \draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize \y}; \draw[color=black] (0pt,-10pt) node[right] {\footnotesize 0}; \clip(-4,-4) rectangle (4,4); \draw[color=qqttcc, smooth,samples=100,domain=-4.0:4.0] plot(\x,{(\x)*(\x)*(\x)}); \begin{scope}[yscale=-1,xscale=1] \draw[rotate=90, color=ccqqqq, smooth,samples=100,domain=-4.0:4.0] plot(\x,{(\x)*(\x)*(\x)}); \end{scope} \draw [color=qqttcc](2.06,4.00) node[anchor=north west] {$f(x)=x^3$}; \draw [color=ccqqqq](2.06,1.00) node[anchor=north west] {$f(x)=\sqrt[3]{x}$}; \begin{scriptsize} \draw[color=qqttcc] (-1.74,-7.16) node {f}; \end{scriptsize} \end{tikzpicture} \end{document} I'm an Indian. I'm planning to book a journey from Mumbai to Johannesburg, South Africa and back (conference visa). I looked at refundable tickets, while they are priced higher, they give you half your money back, for instance if the to and fro flight costs INR 40,000 they give you INR 20,000 . But this is if you are cancelling for some reason.   What if you have to cancel because your visa application or the transit visa application got rejected. As far as I understand the procedure is :-  a. Buy airline tickets   b. Apply for Visa   c. Wait between 1-6 weeks to get update about your visa application, for both visa and transit visa (although transit visa should be shorter I guess.)  d. If visa is granted then move on to planning other things about the trip.   Now in case, if the visa application fails (for whatever reason ) what recourse does the traveler have ?  Are there any airlines who are better understanding than others about this situation ?  In non-refundable tickets, I guess that all is lost or is there some way to save money there too.   Lastly, I read travel insurance can help in this case. How does travel insurance help in the above scenario ?  Edit :- One of the pre-conditions of getting the visa is   Proof of a Valid Return or onward ticket or purchase of ticket. This is from   and How to make the simple three square box and connect them? I want to make the three square box and simply connect them. In how many ways $5$ different rings can be worn on $4$ fingers ?   Although these is similar question here   But I want to answer it in different way which is like take $4$ fingers like $a,b,c,d$. Now $a$ can be filled in $5$ ways and so are others. So total ways are $5^4$ but answer is $4^5$. What is wrong in my reasoning? Which cases have I left out? FireFox 66 Quantum disabled all my extensions at midnight! Literal &#39; appears in flag panel instead of being escaped Can we model gravitation as a repulsive force? This question is actually related to my earlier question ("").  The fact that objects move a lot in the universe and that the universe is expanding, can imply that gravity is a repulsive force that increases with distance.. so the farthest objects repel us more.  This can still explain several existing observations, e.g., why does the apple fall?   Motion is the result of such repulsion. Two objects unlucky enough not to be moving relative to each other get squished due to the repulsion of the rest of the universe  around them. The earth repels the apple less than the stars so it is pushed towards the earth.   Furthermore, it can explain the expanding universe without the need for dark energy.  This could be demonstrated in a thought experiment. If we take a lot of same-charge particles (with small mass) such as electrons and lock them in a large box at a low enough temperature. The mutual repulsion of the particles may cause similar motion as if due to gravitational attraction.   Another experiment would be to measure the slight changes in our weight during day and night when the sun and earth align (if their masses are large enough to detect the feeble change in repulsion).   [EDIT: the question in the original form may not have been clear. It is "can we model".. with a yes/no answer and why (not). If downvoting, please justify. Without appealing to the axiom of choice, it can be shown that (Proposition:) if $X$ is well-orderable, then $2^X$ is totally-orderable.  Question: can we show the stronger result that if $X$ is well-orderable, then so too is $2^X$?  Proof of Proposition. Pick any well-ordering of $X$. Then the lexicographic order totally orders $2^X$.  More explicitly: for any two $f,g \in 2^X$, define $f &lt; g$ iff   there exists $x \in X$ such that $f(x) \neq g(x)$, and if $x \in X$ is minimal such that $f(x) \neq g(x)$, then $f(x)=0$ and $g(x)=1$.   It can be shown that $&lt;$ totally-orders $2^X$. I wonder if this is a bug, or if I missed something... When a light hits tangentially convex surfaces with "smooth shading", the shadow on this surface seems to "partially ignore" the smoothness of the surface. Here is an example with a simple cylinder :    On the left, a cylinder with "flat shading". Everything's ok. On the right, the same with "smooth". We can see an edge, bad shadowed.  With or without the "Edge split" modifier, no matters, the edge is still there. The shader is a simple white diffuse shader with no texture nor displace.  I have to add a subsurf modifier with at least 2 levels, which will insanely increase the vertex count... And I can't do that for my project.  I hope someone already saw it, and have a little tip to share. Or if someone already know it's a bug, or a limit of the current Cycles renderer ? I was trying to suggest a retag of  on Seasoned Advice, but the system won't let me; I get the familiar message     Edits must be at least 6 characters; is there something else to improve in this post?   In general, that's not true for retagging; if you don't change the question body at all, it should work. I suspect the logic checking the edit is thrown off guard because of the title warning:     The question you're asking appears subjective and is likely to be closed       I found , but I cannot find any evidence (in SEDE) for invisible characters being present in the question's source. Also, that's supposed to be . Prove that if b is coprime to 6 then $b^2 \equiv 1 $ (mod 24) Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24).  Now I have that as $\gcd(b,6) = 1$, we know that  $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.)  So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd.  Then I'm not sure where to take it from here? Maybe I've gone down the wrong path to start with, and should be looking more along the lines of the Chinese Remainder Theorem? I'm not sure. What is the difference between "speaking" and "talking"? It seems more politically correct to say that someone is speaking rather than talking.  What is the definitive difference between these terms? FoxTrot Bill Amend Problems So I found this on the Wolfram website today:     So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:    $ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $ $ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $  $ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] $  $ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$     Ideas   The degree of the numerator is $\frac{3}{2}$. The degree of the denominator is $\frac{3}{2}$. The expression is of an indeterminate form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule.  No ideas, really.  Derivatives are always pretty easy, although this one is a bit bashy. Basically bash Product and Chain Rules, etc.  Integration by Parts bash? The $\mathrm{d}\theta$ part is very trivial. I mean for the $\rho$ and $\phi$. Algebra of LDA. Fisher discrimination power of a variable and Linear Discriminant Analysis Apparently,     the Fisher analysis aims at simultaneously maximising the   between-class separation, while minimising the within-class   dispersion. A useful measure of the discrimination power of a variable   is hence given by the diagonal quantity: $B_{ii}/W_{ii}$.     I understand that the size (p x p) of the Between (B) and Within-Class (W) matrices are given by the number of input variables, p. Given this, how can $B_{ii}/W_{ii}$ be a "useful measure of the discrimination power" of a single variable? At least two variables are required to construct the matrices B and W, so the respective traces would represent more than one variable.  Update: Am I right in thinking that $B_{ii}/W_{ii}$ is not a trace over a trace, where the sum is implied, but the matrix element $B_{ii}$ divided by $W_{ii}$? Currently that is the only way I can reconcile the expression with the concept. Continuum between addition, multiplication and exponentiation? I noticed this  which attempts to find the shades of grey between a linear and log scale where results are between zero and one.  However, I was looking for the more general case where we find the continuum between the operators themselves - add, multiply, and exponentiation or even higher.  The function could be a hyperoperation$(n,x,y)$, where $x$ and $y$ are the numbers to operate on, and $n$ is allowed to be a non-integer number for the type of operator. So $n=1$ would be 'addition', $n=2$ would be multiplication, and $n=3$ would be exponentiation. But one would also have $n=2.5$ to be the shade of grey BETWEEN multiplication and exponentiation, or $n=3.5$ or $4$ for beyond exponentiation and tetration).  I want the formula to work with real numbers, not just integers. Any thoughts?  These related posts may also be of interest: I find that the minimum in-call volume on my HTC Desire is still far to loud on some calls.  Since I got the phone, I have flashed several different custom ROMs, radio updates, etc, and have had the issue throughout.  Do you know of any way (setting, hack, app, whatever) to reduce the minimum volume?  My device is rooted and I am currently running one of the CyanogenMod 7 (Gingerbread) nightly builds.