I've been asked to prove the following from Spivak's Calculus $$\lim_{n\to\infty}\sqrt[n]{a^n+b^n}=\max(a,b); a,b &gt; 0$$  I understand that this is a proof by cases, and that our cases are $a=b$, $a&gt;b$, and $b&gt;a$. I have done the $a=b$ case, but I am stuck on the $a&gt;b$ and $b&gt;a$ cases.  Some hints would be appreciated-  Thanks!
If a>0 and b>0,  $ \lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ? $  What I was trying to do: Suppose a>b. Then, for sufficiently large values of x, $ a^x &gt;&gt; b^x $; so $\sqrt[x]{a^x+b^x} \rightarrow \sqrt[x]{a^x} \rightarrow a $ when $x \rightarrow +\infty$.   Is that idea correct? How can I formalize it?
Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[  \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$  I have a question about the following solution:  We may write it in the form:  $$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$  Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 &lt; \frac{1}{n} &lt; \ldots &lt; \frac{n}{n}$.  Can you explain to me to reach this conclusion? I installed Ubuntu 12.10 and entered a password in SeaHorse. I would like to remove the password of "Default" but I get this error :  GDBus.Error:org.freedesktop.Secret.Error.NoSuchObject: The collection does not exist   Is there a fix or a way to bypass modify the password ? File Save dialog unreadable I just updated my iMac to Mojave, and am still trying to setup my preferences.  I tried dark mode, but did not like it (half of me apps seem to ignore it anyway).  I switched to light mode, but this too has problems. The File Save dialog shows a list of Directories and existing files, but the latter are very light grey, almost indistinguishable from the background.    I have searched the Preferences menu, but can't seem to find any setting to set normal contrast levels like in High Sierra. Let $A$ be the $\mathbb{C}$-subalgebra of $M_4(\mathbb C)$ consisting of matrices of the form   \begin{pmatrix} * &amp; * &amp; * &amp; *\\ * &amp; * &amp; * &amp;*\\ 0 &amp; 0 &amp; *&amp;0 \\ 0 &amp; 0&amp;* &amp;* \end{pmatrix}   Then how would I find the Jacobson radical $J(A)$ of $A$?    I don't really understand finding the Jacobson radical of $\mathbb{C}$-subalgebras.  Any help on this would be greatly appreciated, thank you! Sufficient Identification for Intra Schengen flights If one takes a direct flight from one Schengen member state to another, what's the minimum sufficient ID that needs to be provided to board the flight? mathpazo + siunitx: π turns into ß When I use mathpazo and siunitx (version 2.3h) together, if I use \pi inside the numerical argument to \SI, it will show up in the output as "ß".  \documentclass{article} \usepackage{mathpazo, siunitx} \begin{document}     $\pi + \SI{\pi}{\ohm}$ \end{document}   Produces:    But when I don't use mathpazo:    Why does this happen and how can I fix it? I'm having some trouble proving the following statement using mathematical induction:   $$\frac{1}{2}n^{\frac{3}{2}} \leq \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + ... + \sqrt{n}      ,\text{ (for all sufficiently large n)}$$   I'm sort of confused because $n^{\frac{3}{2}}$ = $n  n^{\frac{1}{2}}$ which means that there are n of the largest possible root values on the left hand side while the right hand side has values like $\sqrt{1}$, $\sqrt{2}$, etc. As n grows sufficiently large wouldn't the left hand side outgrow the right hand side no matter what the constant multiple is? If I'm thinking about this correctly just let me know, if not, any help would be appreciated. Where can I find the location of folders for installed programs? A closed ball in a metric space is a closed set Is multiplication not just repeated addition? If multiplication is not repeated addition, what is it? How do you define multiplication? All of us had to memorize the multiplication table in elementary school, but how did they come up with it if so many people claim that it is wrong to think of it as repeated addition? Voodoo magic? I'm studying machine learning and every book I open I bump into chi-squared distribution, gamma-function, t-distribution, Gaussian, etc.   Every book I have opened so far only defines what the distributions are: they don't explain or give the intuition on where the specific formulas for the functions come from.   For example, why is chi-squared distribution the way it is? What is the t-distribution? What is the intuition behind the distribution? Proofs? etc.   I would like to have a clear and fundamental understanding of the most commonly used distributions so that every time later on when I see them, I truly understand what is a t-distribution, what is a Gaussian distribution and most importantly why are they the way they are.   It would be nice if the books / tutorials  can explain the concepts to a layman so that in order to understand them you don't already need to understand them x) Many books are like this, they don't fit for beginners :( How to add some non-standard font to a website? Is there a way to add some custom font on a website without using images,  or some other graphics?  For example, I was working on a wedding website, and I found a lot of nice fonts for that subject. But I can't find the right way to add that font on the server. And how do I include that font with CSS into the HTML? Is this possible to do without graphics? The  spell says time stops for other creatures.  What all is affected by the spell? Is gravity affected? Air movement?  I know that the spell only does what it says, and nothing else, but just wondering.   In addition, is it saying that the other creatures are not affected by environmental effects and the like while time is stopped?  For instance, if a wind blew really hard at one of the creatures you're fighting, would it affect that foe while time stop is active? If you were to place a bear trap behind the person, and a hard wind blew enough to tip them, would they then fall over onto the bear trap and be damaged without the spell ending?  This is a very curious spell... The  (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:  $$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$  The basic approach to solving this problem outlined by Mordell [Ref1] is described below  By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that  a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$  In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$  b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus  $$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$  will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.  Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$  The argument continues...  Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$  Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$  Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n&lt;1009$.  This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.   Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.  The question is:  Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?  [Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290. I would like to prove that the equation $ 3^x+4^x=5^x $ has only one real solution ($x=2$)  I tried to study the function $ f(x)=5^x-4^x-3^x $ (in order to use the intermediate value theorem) but I am not able to find the sign of $ f&#39;(x)= \ln(5)\times5^x-\ln(4)\times4^x-\ln(3)\times3^x $ and I can't see any other method to solve this exercise... This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$. Minimum bounty on a question is now 100 rep? Cannot replace a packed image in the uv image editor I was trying to replace an image in my UV image editor, but the image defaults to a different image. how do I fix this? How to conduct an Accent Sensitive search in MySql I have a MySQL table with utf8 general ci collation. In the table, I can see two entries:  abad abád  I am using a query that looks like this:  SELECT *  FROM `words` WHERE `word` = 'abád'   The query result gives both words:  abad abád  Is there a way to indicate that I only want MySQL to find the accented word? I want the query to only return    abád    I have also tried this query:  SELECT *  FROM `words` WHERE BINARY `word` = 'abád'   It gives me no results. Thank you for the help. linux: touch date precision