I need a word that describes the love of learning, discovering, or it could even be a word that describes a curious person.
Is there a term that means "a person who enjoys learning"? This term might be used to describe someone who:   Is a self-motivated learner. Is curious, wants to understand many things.   I understand the term "philosopher" might be a good fit, in terms of its root words, however, the general population has an inconsistent understanding of this term, so I am looking for a more precise term.
If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$ I'm trying to prove the following:     If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n&lt;\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2&lt;\infty$, then $\sum_{n=1}^\infty a_n^2 &lt;\infty$.   The context here is functional analysis homework, in the subject of Hilbert spaces.  Here's what I've thought:  Let $f=(a_n)&gt;0$. Then the problem reads: if $\int f\overline{g}&lt;\infty$ for all $g&gt;0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.  I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle &lt;\infty$ for all $g&gt;0,g\in \ell^2$ then $f\in \ell^2$.  I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P  From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: $$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$  but it's not useful here. htdigest file format Cancelling upvote on comment? As mentioned on :     For something that can be done so   easily by accident, there should be   the ability to undo mistakes.   I know that I have clicked several times on a comment up-vote without actually wanting to.  I would suggest being able to cancel the up-vote (maybe only for a couple of minutes or even less than a minute, no problem, as long as I can undo a miss-click). The IPA symbol for the labiodental flap (ⱱ) UTF-8 encoding a servlet form submission with Tomcat How can I maximize the "blurry background, sharp subject" (bokeh) effect? Switching back to last *used* tab on Chrome How do I save a variable or function definition to a file? Can I bring a pack of razor blades on a plane? I have to take a flight soon and I want to know if I can take a pack of razors with me. They are in a pack that is designed to not let any out unlesss a person really wants to open it. I'm going to the USA so, can I?  I mean plain old razor blades: When I upload an image to Stack Overflow, I cannot see my JPEG file in my post. It's all blank. I've tried it with different files, but gotten the same result. Proving that $A_n$ is the only proper nontrivial normal subgroup of $S_n$, $n\geq 5$ There is a famous Theorem telling that:  For $n≥5$, $A_n$ is the only proper nontrivial normal subgroup of $S_n$.  For the proof, we firstly start with assuming a subgroup of $S_n$ which $1≠N⊲S_n$. We proceed until at the last part of proof's body, we assume $N∩A_n=\{1\}$. This assumption should be meet a contradiction with normality of $N$ in $S_n$. There; we get $N=\{1,\pi $} in which $\pi$ is an odd permutation of order $2$. Now for meeting desire inconsistency, I have two approaches:  (a) Since every normal subgroup, having two elements, lies in the center of $G$ so, our $N⊆ Z(S_n)=\{1\}$ for $n≥5$ and then $N=\{1\}$.  (b) Clearly, $1≠N$ acting on set $\Omega=\{1,2,...,n\}$ is intransitive wherein $|\Omega|≥5$ and according to the following Proposition $S_n$ would be imprimitive.  Proposition 7.1: If the transitive group $G$ contains an intransitive normal subgroup different from $1$, then $G$ is imprimitive (Finite Permutation Groups by H.Wielandt).  May I ask if the second approach is valid? I am fond of knowing new approach if exists. Thanks. Factor 2 in Heisenberg Uncertainty Principle: Which formula is correct? Some websites and textbooks refer to $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ as the correct formula for the uncertainty principle whereas other sources use the formula $$\Delta x \Delta p \geq \hbar.$$ Question: Which one is correct and why?  The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.  This is the mentioned derivation:     $\sin \theta = \frac{\lambda}{\Delta y}$      assuming $\theta$ is small:      $\tan \theta = \frac{\lambda}{\Delta y}$      de Broglie equation:      $\lambda = \frac{h}{p_x}$      $\Rightarrow \tan \theta \approx \frac{h}{p_x \cdot \Delta y}$      but also:      $\tan \theta = \frac{\Delta p_y}{p_x}$      equalize $\tan \theta$:      $\frac{h}{p_x \cdot \Delta y} \approx \frac{\Delta p_y}{p_x}$      $\Rightarrow \frac{h}{\Delta y} \approx \Delta p_y \Rightarrow \Delta p_y \Delta y \approx h$      $\Rightarrow \Delta p_y \Delta y \geq \frac{h}{2 \pi}$      $\Rightarrow \Delta p \Delta x \geq \frac{h}{2 \pi}$   So: Which one is correct and why? What could cause java.lang.reflect.InvocationTargetException? How to choose the number of hidden layers and nodes in a feedforward neural network? Is there a standard and accepted method for selecting the number of layers, and the number of nodes in each layer, in a feed-forward neural network? I'm interested in automated ways of building neural networks. Using "the" instead of "a" for something that isn't a specific object I've been curious about this for a while.  Generally, one uses "the" if they mean a specific object ("The copy of Lord of the Rings I bought yesterday") instead of any object that fits a description ("I have a copy of Lord of the Rings").  However I've seen "the" used in some cases where the phrase doesn't refer to a specific object. I've noticed it mostly being about technology, such as "This game requires the Classic Controller" (seen on some Wii Virtual Console games). While saying "This game requires a Classic Controller" seems to be correct as well, using "the" doesn't feel wrong either.  The controller's page on Wikipedia also uses "the":   What is the rule that allows the use of "the" in this context?  I presume it has something to do with "Classic Controller" being a proper noun. Can't see a glass object through another glass object in Eevee I was just testing Eevee and found that you cannot view a glass object through another glass. Is there any workaround? It works fine only If I rotate the glass window or put my camera to the back side of the scene How do I get console output in C++ with a Windows program? Cardioid in coffee mug? Recently I had to change the folder permissions of /usr folder. Now the whole sudo is not working. If I try to update my system using this command:  sudo apt-get update   I get the following message:  sudo: /usr/lib/sudo/sudoers.so must be only be writable by owner sudo: fatal error, unable to load plugins   So I tried to change the folder permissions by opening Nautilus as root by using this command  gksudo nautilus   But nothing happened. Nautilus didn't open. This is an emergency as the whole of sudo is not working. when I tried searching the solution from Google many suggested re-installation. I don't prefer re-installation. so is there any other solution? What is the Python equivalent of static variables inside a function? What is the idiomatic Python equivalent of this C/C++ code?  void foo() {     static int counter = 0;     counter++;     printf("counter is %d\n", counter); }   specifically, how does one implement the static member at the function level, as opposed to the class level? And does placing the function into a class change anything?