If $d_{1}(x,y)$ and $d_{2}(x,y)$ are distance metrics, is $\sqrt{d_{1}(x,y)^{2} + d_{2}(x,y)^{2}}$ also a metric? How do you prove this? Thank you.
$d_1(x_1,y_1)$ and  $d_2(x_2,y_2)$ are metric on $X$ and $d(x,y)$ is defined as: $$d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$ I am trying to show this is a metric. Can you give me some clue about proving the triangle inequality for $d$ ? Thank you for your help.
Allow bounty to be set with reputation from another site? Is it consistent with ZF for there to exist a set $S$ such that the power set $P(S)$ is countable? If so, what is the weakest form of the axiom of choice needed to prove that no such set exists? Different batteries connected in parallel If we have 2 batteries one of emf x and the other is of emf y and we connect them in series we get an effective emf of x+y.  But what if we connect them in parallel, how to calculate the emf now? Mouse cursor flickering and disappearing Elmah: How to get JSON HTTP request body from error report In an integral domain every non-zero prime ideal is invertible implies every non-zero proper ideal is invertible. How can I reset my Ability Points? I'm also curious about being able to reset my points, should I wish to change my ability configuration later.  So far (10h-ish) I've only seen one way, which is a potion sold by Kiera Metz for 1000g. However she's no longer available as a store, due to a story choice, so I'm not sure where else I could buy this potion.  Where might I buy this potion, and is it the only way to reset your ability points? Removal of lines with no more or fewer than 'N' fields? I'm trying to re-learn my undergrad math, and I'm using Stephen Abbot's Understanding Analysis. In section 2.7, he has the following exercise:  Exercise 2.7.5 (a) Show that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ also converges absolutely. Does this proposition hold without absolute convergence?  I'm posting about this here because my answer to that last question about whether the proposition holds without absolute convergence is "Yes", but my suspicions are raised by him merely asking the question. Usually questions like this are asked to point out that certain conditions are necessary in the statement of propositions, theorems, etc. I just want to see if I'm missing something here.  Anyway, here's how I prove the absolute convergence of $\sum{a_n^2}$, and note that I never use the fact that $\sum{a_n}$ is absolutely convergent:  Proof: Let $s_n = \sum_{i=1}^n{a_n^2}$. I want to show that $(s_n)$ is a Cauchy sequence. So, let $\epsilon &gt; 0$ and $n &gt; m$, and consider, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |(a_1^2 + a_2^2 + \cdots a_n^2) - (a_1^2 + a_2^2 + \cdots a_m^2)|\\                 &amp;= |(a_{m+1})^2 + (a_{m+2})^2 + \cdots + a_n^2|\\                 &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\   \end{aligned} \end{equation} Now, since $\sum{a_m}$ converges, the sequence $(a_m)$ has limit 0. Therefore I can choose an $N$ such that $|a_m| &lt; \sqrt{\epsilon/n}$ for all $m &gt; N$. So for all $n &gt; m&gt; N$,  we have, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n}}\right)^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2\\                 &amp;= \epsilon   \end{aligned} \end{equation} Therefore $(s_n)$ is Cauchy and $\sum{a_n^2}$ converges. And since $\sum{a_n^2} = \sum{|a_n^2|}$, the series $\sum{a_n^2}$ is absolutely convergent. &#x220E;  Am I doing something wrong here? Am I right in thinking that $\sum{a_n}$ need not be absolutely convergent for $\sum{a_n^2}$ to be absolutely convergent? A well-known application of the MVT is to prove that any $f: \mathbb{R} \to \mathbb{R}$ with $f'= 0$ is constant. But of course, the MVT relies fundamentally on the properties of $\mathbb{R}$, and in particular does not hold in $\mathbb{Q}$. A standard counter-example is the function $f :\mathbb{Q} \to \mathbb{Q}$ defined by $ f(x)=0 $ if $ x &lt;\sqrt{2}$, and $f(x)=1$ if $ x &gt; \sqrt{2}$, for this function is locally constant. There are several ways of thinking about the difference between $\mathbb{R}$ and $\mathbb{Q}$ : $\mathbb{R}$ is connected while $\mathbb{Q}$ is totally disconnected, closed bounded intervals are compact in $\mathbb{R}$, but not in $\mathbb{Q}$, $\mathbb{R}$ satisfies the least upper bound axiom, $\mathbb{Q}$ does not.   Now, I wanted to see how far one could push this counterexample - is there a function $f :\mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing? Clearly, any slight modification of the usual counterexample where $f$ is locally constant will not do. Also, one can show that if $f$ is uniformly differentiable with derivative zero, it must be constant, even in $\mathbb{Q}$, so I knew I was looking for non-uniformly differentiable but still differentiable $f$.   Eventually, I came up with a construction, albeit not a very explicit one: the idea is that, since $\mathbb{Q}$ is countable, one can enumerate it $q_{1}, q_{2}, q_{3}, \cdots$, and then define $f: \mathbb{Q} \to \mathbb{Q}$ inductively and recursively. If one could define $f$ in such a way that $ \forall \ q_{n} \in \mathbb{Q} \ \exists \ c_{n} \in \mathbb{Q} : \forall \ x \in \mathbb{Q} \ |f(x)-f(q_{n})| \leq c_{n}|x-q_{n}|^2$, and $f$ is strictly increasing, we would be done. The recursive definition essentially defines $f(q_{n+1})$, after having defined $f(q_{1}),f(q_{2}),f(q_{3}), \cdots , f(q_{n})$, in such a way that these inequalities hold and $f$ is strictly increasing, and at each step the new $c_{n}$ is chosen large enough so that these inequalities do not become incompatible at subsequent steps ( I am willing to provide more details of this construction if asked).  Note that, as $n \to \infty$, we must have $c_{n} \to \infty$, so that $f$ is not uniformly differentiable. Interestingly, the only key fact about $\mathbb{Q}$ used in the construction is that it is countable, so in a sense the construction provides a proof that $\mathbb{R}$ is uncountable, given the MVT.  While the construction works, it is not very simple nor explicit, which is perhaps to be expected, and the function must be evaluated point by point in the order given by the enumeration of $\mathbb{Q}$. So I was wondering if anyone else knows or can come up with a simpler construction of a function $ f : \mathbb{Q} \to \mathbb{Q}$ such that $f'=0$ and $f$ is strictly increasing. Any thoughts and ideas are welcome.  Edit: I've asked the same question on MO: I received a journal paper for a review. The paper suggests a solution to an important problem, but the solution is very complicated and has mediocre performance. While reading the paper, I thought of a different solution, which is much simpler and has better performance (to give some scale: their solution is 1 page to write and 5 pages to prove; my solution is 5 lines to write and half a page to prove).  What should I do now?  A. Write my solution in the review, hoping the authors will remember to "thank the anonymous reviewer"?  B. Recommend that the paper published as is, then write a paper of my own with my improved solution?  C. Something else?  EDIT: Before I found the superior solution, I thought of recommending "resubmit with major revision", with the reasons that I mentioned above: your solution is too complicated and its performance is too weak.  However, now I am not sure it is the good course of action. A major revision may take a long time. The authors might even decide to delay or stop working on the paper without telling the editor (or me). Meanwhile, I will have to postpone the publishing of the superior solution for an indefinite amount of time. What do you think? If $\alpha$ is a real root of the equation   $$x^5-x^3+x-2=0$$   find the value of $\lfloor {\alpha}^6 \rfloor$.      This one totally stumped me. We are asked to calculate $\lfloor {\alpha}^6 \rfloor$ without actually calculating the root or using wolfram alpha or any other calculator. I found that the above equation has only one real root by sketching its graph using calculus. I was also able to use the intermediate value theorem to conclude that $1&lt;\alpha&lt;2$ , but this is of little use while calculating $\lfloor {\alpha}^6 \rfloor$. Please help! Primary Decomposition of $\mathbb{R^n}$ under T I have a question related to the Primary Decomposition Theorem for a vector space under a linear transformation.  The following question is a direct corollary of this theorem.  However, I wish to prove it without calling upon theorem.  I am currently studying for a comprehensive exam and ran across this question.  I never really felt comfortable with the proof I saw of the Primary Decomposition Theorem so I hoping with your help with this problem I will understand this particular problem and better understand that proof as I believe their proofs will be essentially the same techniques.   Question:  Suppose we have a linear transformation $T: \mathbb{R^n}\to \mathbb{R^n}$ with minimal polynomial $m(x)=f(x)g(x)$ and $f(x),g(x)$ are relatively prime.  Show that we can write $\mathbb{R^n}=M\oplus N$ where $M$ and $N$ are $T$-invariant subspaces s.t $f(x)$ is the minimal polynomial for $T|_M$ and $g(x)$ is the minimal polynomial for $T|_N$.   So I know that I want to use $M=ker(f(T))$ and $N=ker(g(T))$ but after that I really don't have much.  Since this isn't a homework problem for a class I am essentially looking for someone to present me a proof of this simpler question and walk me through their logic so that I can hopefully understand the general theorem and its proof.  Thanks in advance for the help! How can I do traffic shaping in Linux by IP? We have a transparent proxy setup. I tried looking for traffic shaping in Linux, and all I could find online was to limit traffic by interface (eth0/eth1...).  I need to limit the bandwidth (never exceeding a specific limit) by IP address or IP ranges and I can't find a way to do that.  Is there any way to do that? What if the hard disk is partitioned into more than 26 sections in Windows? As we know, in Windows the partition location is prefixed with [a-z]:. What happens if there are more than 26 partitions? $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes I am using a Collection (a HashMap used indirectly by the JPA, it so happens), but apparently randomly the code throws a ConcurrentModificationException. What is causing it and how do I fix this problem? By using some synchronization, perhaps?  Here is the full stack-trace:  Exception in thread "pool-1-thread-1" java.util.ConcurrentModificationException         at java.util.HashMap$HashIterator.nextEntry(Unknown Source)         at java.util.HashMap$ValueIterator.next(Unknown Source)         at org.hibernate.collection.AbstractPersistentCollection$IteratorProxy.next(AbstractPersistentCollection.java:555)         at org.hibernate.engine.Cascade.cascadeCollectionElements(Cascade.java:296)         at org.hibernate.engine.Cascade.cascadeCollection(Cascade.java:242)         at org.hibernate.engine.Cascade.cascadeAssociation(Cascade.java:219)         at org.hibernate.engine.Cascade.cascadeProperty(Cascade.java:169)         at org.hibernate.engine.Cascade.cascade(Cascade.java:130) I have a shapefile with 3800 polygons representing the smallest administrative areas in Ireland. Many of these, in rural areas, have tiny populations, under 100 people. In urban areas, all would have adequate populations.  For each area I have a population figure, and various other attributes, all of which can be intelligently summed to give valid results for areas produced by combining the smaller areas.  I need to merge the areas to get say, 500 or 600 areas of reasonable population. I need to leave the larger (in population) areas alone, where possible. I would greatly prefer to do this automatically. This is necessary for model fitting purposes, primarily.  I've looked at a number of the dissolve questions here:   'Batch Dissolve Polygons based on Attributes in QGIS or FWTools' I think would require me to know which areas to merge, when this is what I need to calculate. 'Practical way of managing polygons and unions of them' and 'How do dissolve polygons from shapefile using open source tools?' have the same issue.   Any suggestions? The following is exercise 8.2 in 3rd edition (and exercise 8.19 in 2nd edition) of Goldstein's Classical Mechanics.   Adding the total time derivative of a function of $q_i$ and t to the Lagrangian will not change the the Euler-Lagrangian equation. So if we make the following change to Lagrangian,   $$L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q_1,q_2,...,q_n,t)}{dt}$$ we can get  $$ \frac{d}{dt}\frac{\partial{L'}}{\partial{\dot{q_i}}} - \frac{\partial{L'}}{\partial{q_i}} = 0 $$ from  $$ \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_i}}} - \frac{\partial{L}}{\partial{q_i}} = 0 $$  How can we get the corresponding Hamiltonian equation part? This is to prove  $$ \dot{p'_i} = \frac{\partial{H'}}{\partial q_i} $$ $$ -\dot{q_i} = \frac{\partial{H'}}{\partial p'_i}  $$  from   $$ \dot{p_i} = \frac{\partial{H}}{\partial q_i} $$ $$ -\dot{q_i} = \frac{\partial{H}}{\partial p_i}  $$ where $p'_i = \frac{\partial L'}{\partial \dot q_i}$.  Edit  The corresponding $H'$ is $$ H' = \sum_k{p'_k \dot{q_k}} - L' $$ where $p'_k = \frac{\partial L'}{\partial \dot q_k}$. Ctrl -C, Ctrl -V not working in different keyboard layout When I am in English, Ctrl+C, Ctrl+V works fine when I need to copy paste. But when I turn to Greek, Ctrl+C and Ctrl+V and basically no shortcut almost works in Greek.  So each time I have to alt shift and go back to English in order to user Ctrl+C to copy something. How can I apply this to the Greek keyboard profile?  EDIT  I also realized that when using NetBeans IDE 7.4,  I can copy and paste in Greek but I can't use the tonic system of the language over the words. In Aptana studio I can't use Copy and paste Ctrl+C, Ctrl+V When using gedit everything works fine. Different software use different keyboard layout settings?