When Voldemort created Horcruxes, he divided the soul into it. Once he died only soul was present.  Then from where exactly his body came from? Peter Pettigrew used Harry's Blood but whose flesh he used to bring back Voldemort?
From a link from the question  to the HP Wikia, I started mulling over something I read about Voldemort and the "regeneration" potion (not a canon potion, but I'll use it for lack of a better way of describing it. . .) and would really like to know:  How did Voldemort drink the regeneration potion to turn his body into the grotesque childlike form we see in Goblet of Fire while he was still in a "mangled spectral state"?  Spectres (i.e. spirits or ghosts) are non-corporeal; they obviously can't consume food or drink liquids. So, how did Voldemort overcome this difficulty and manage to drink the regeneration potion?
Do I need a visa for a connecting flight to Japan? I'm going to the Philippines this October and would like to layover in Japan. I found a reasonably priced ticket with a 15-hour layover to Haneda, however, I have to transfer to Narita. Do I need a visa since I'm going to have to stay at a hotel and ride a bus to travel to Narita airport? Thank you! This question is asked in various forms here but the question boils down to:  I know shrinking a database is risky. In this case, I've removed so much data and I'll never use it again.   How can I shrink my database? What files do I shrink?  What should my considerations be while doing this?  Should I do anything after? What if it is a large database? Can I shrink it in smaller increments? Prove that every Lebesgue measurable function is equal almost everywhere to a Borel measurable function Suppose $(\mathbb{R},\Sigma(m),m)$ is our measure space, where $m$ is Lebesgue measure.  Also, suppose $f : \mathbb{R} \to [-\infty, \infty]$ is a Lebesgue measurable function.  The problem:     Prove that $f$ is equal almost everywhere to a Borel measurable function.   My attempt:  We only have to prove this assertion for a non-negative Lebesgue measurable function $f$ because the result will follow for all Lebesgue measurable functions.  Furthermore, since any Lebesgue measurable function can be approximated by a sequence of non-negative, monotonically increasing, Lebesgue measurable simple functions $s_{n}$, we only need to prove the claim for an arbitrary Lebesgue measurable simple function.  So, let $s: \mathbb{R} \to [-\infty, \infty] $ be a Lebesgue measurable simple function.  We can write $s$ canonically as $$ s(x) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(x)$$ where $\alpha_{i} \in [-\infty, \infty]$, $\bigcup \limits_{i = 1}^{n} A_{i} = \mathbb{R}$, and $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$.  For each $i$, since $A_{i}$ is Lebesgue measurable, we can find a Borel set $B_{i}$ such that $A_{i} \subseteq B_{i}$ and $m(B_{i} \setminus A_{i}) = 0$.  Clearly, this implies that $\bigcup \limits_{i = 1}^{n} B_{i} = \mathbb{R}$.  Now we just need the $B_{i}$'s to be pairwise disjoint, with each $B_{i}$ still retaining $A_{i}$.  To make them pairwise disjoint, I constructed the following sets:  $\tilde{B_{i}} = [B_{i} \setminus (\bigcup \limits_{j \neq i} B_{j})] \cup A_{i}$.  This construction gives us that the $\tilde{B_{i}}$'s are pairwise disjoint (I think....) and $A_{i} \subseteq \tilde{B_{i}}$.  But I don't know that $\tilde{B_{i}}$ is necessarily still a Borel set. :( :(  Am I approaching this problem all wrong? How to find the joint PDF? Question about Angle-Preserving Operators This an exercise out of Spivak's "Calculus on Manifolds".  Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.  Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by  $$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.  A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if $\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.  The exercise as stated:  Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.  Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.  I'm having problems with the $(\Rightarrow)$ direction.  My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption, \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\ &amp; = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\ &amp; = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\ &amp; = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\ \end{align*} may also be calculated as \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; =  \angle(x_{j},x_{k})\\ &amp; = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right). \end{align*}  Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.  Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.  Thanks for any help you can give. Noticing that the "Human Verification" image did not show, I found that the src, , shows that sstatic.net is giving an SSL vertificate for *.stackexchange.com.. Fred and George stumbled upon the Marauder's map in their first year. They passed it on to Harry while in their fifth year. Also note that, while talking to Harry on the train, Ron described Scabbers as being "Percy's old rat".   Hence, we can safely assume that the twins had the Marauder's Map for about five years. How could they not see Peter Pettigrew on the map, with Percy for first two years and with Ron for the next three years?  How come Harry didn't see Peter? How was Lupin the only one able to see Peter on the map? Find the supremum of the following set Find the supremum of the following set:  $$A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}$$  So I have as my answer that $\sup(A) = \frac12$ but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound.   For step $1$: if $a$ is a member of $A$ then $a = (n-1)/(2n+3)$ for some natural number $n$ and $\frac{n-1}{2n+3} \leq \frac{n}{2n} = \frac12$ so $\frac12$ is an upper bound for the set  Step 2 is where I am uncertain. I think I need to show that for any $y&lt;\frac12$ there exists a member of the set between $y$ and $\frac12$ but not sure how to do this. Thanks for all the answers so far! How can I import a .c4d static model (Cinema 4d) on Blender? Is there any plugin could help me? I didn't notice anything on Internet. If I push or hit an object in space will it rotate or move along a straight line? I understand that in a lot of big-lab fields it is common for the principal investigator to append their name to a paper even if they did not write the paper, design the experiment, or collect data since they spend energy securing funding, and managing the whole lab. What about for small labs?  What are the requirements for a supervisor to be included as an author on a paper, as opposed to just appearing in the acknowledgements? If you are working on your own projects independently of your supervisor, but using funding provided by your supervisor (how does this change when the funding provides resources versus just your salary), are you suppose to add them as authors or just acknowledge the source of funding? Short Question:  Why can we manipulate a read-only file in Vim using : + w + q + ! even without being an administrator?  Long Question:  I have a text file (myFile.txt) which is read-only for everyone:  navid@navid-ThinkPad-T530:~/ubuntuTest$ ls -l myFile.txt  -r--r--r-- 1 navid navid 26 Aug 22 21:21 myFile.txt   I can open it with Vim without having admin privileges:  navid@navid-ThinkPad-T530:~/ubuntuTest$ vi myFile.txt    I modify it and press: Esc + : + w + q + Enter and I see this error message:  E45: 'readonly' option is set (add ! to override)   So far, everything makes sense.  But when I press: Esc + : + w + q + ! + Enter, Vim saves the changes.  I'm using Ubuntu 16.04 and VIM 7.4. Probability of the union of $3$ events? I need some clarification for why the probability of the union of three events is equal to the right side in the following:  $$P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)$$  What I don't understand is, why is the last term(intersection of all) added back just once, when it was subtracted three times as it appears from a Venn Diagram?  Here on page 3, this is explained but not in enough details that I can understand it: Every $k$ vertices in an $k$ - connected graph are contained in a cycle. Let $G$ be a $k$-connected graph. Meaning, $G$ has no fewer than $k$ vertices, and for every set of $k-1$ or fewer vertices, if we remove them from $G$, the graph stays connected (Of course, $G$ itself is also connected).  I want to prove that for any $k&gt;1$, if $G$ is $k$-connected, then every set of $k$ vertices is contained in a cycle.  I have tried some ways - mainly using induction by removing one of the vertices of the set from the graph, and/or using Menger's theorem to construct the cycle. But I always encounter problems with making sure that the cycle I'm building deosn't have repeating edges etc.  Help would be greatly appreciated :) Thanks! How to pass variable number of arguments to printf/sprintf Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,...,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$. Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,...,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$.  a) Find the base and dimension of fundamental subspaces of matrix A  b) Using the results given by a) find the general solution for $Ax - z_1$.  Any hint would be appreciated. I have recently been reading a lot on this site (@Aniko, @Dikran Marsupial, @Erik) and elsewhere about the problem of overfitting occuring with cross validation - (Smialowski et al 2010 Bioinformatics, Hastie, Elements of statistical learning).  The suggestion is that any supervised feature selection (using correlation with class labels) performed outside of the model performance estimation using cross validation (or other model estimating method such as bootstrapping) may result in overfitting.   This seems unintuitive to me - surely if you select a feature set and then evaluate your model using only the selected features using cross validation, then you are getting an unbiased estimate of generalized model performance on those features (this assumes the sample under study are representive of the populatation)?  With this procedure one cannot of course claim an optimal feature set but can one report the performance of the selected feature set on unseen data as valid?  I accept that selecting features based on the entire data set may resuts in some data leakage between test and train sets. But if the feature set is static after initial selection, and no other tuning is being done, surely it is valid to report the cross-validated performance metrics?   In my case I have 56 features and 259 cases and so #cases > #features. The features are derived from sensor data.  Apologies if my question seems derivative but this seems an important point to clarify.  Edit: On implementing feature selection within cross validation on the data set detailed above (thanks to the answers below), I can confirm that selecting features prior to cross-validation in this data set introduced a significant bias. This bias/overfitting was greatest when doing so for a 3-class formulation, compared to as 2-class formulation.  I think the fact that I used stepwise regression for feature selection increased this overfitting; for comparison purposes, on a different but related data set I compared a sequential forward feature selection routine performed prior to cross-validation against results I had previously obtained with feature selection within CV. The results between both methods did not differ dramatically. This may mean that stepwise regression is more prone to overfitting than sequential FS or may be a quirk of this data set. How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga?  I'm thinking we need permutations? I already tried finding each letter and still don't get it... any help?   So far I've broken it off to:-  $M=1$  $I=2$  $S=4$  $A=2$  $U=1$  $G=1$  However, I don't really know what to do next. Driver for Validity Sensors Fingerprint scanner prakhar@pr3d4tOr:~$ lsusb -s 001:003 -v  Bus 001 Device 003: ID 138a:0018 Validity Sensors, Inc. Fingerprint scanner Couldn't open device, some information will be missing Device Descriptor:   bLength                18   bDescriptorType         1   bcdUSB               1.10   bDeviceClass          255 Vendor Specific Class   bDeviceSubClass        17    bDeviceProtocol       255    bMaxPacketSize0         8   idVendor           0x138a Validity Sensors, Inc.   idProduct          0x0018 Fingerprint scanner   bcdDevice            0.78   iManufacturer           0    iProduct                0    iSerial                 1    bNumConfigurations      1   Configuration Descriptor:     bLength                 9     bDescriptorType         2     wTotalLength           46     bNumInterfaces          1     bConfigurationValue     1     iConfiguration          0      bmAttributes         0xa0       (Bus Powered)       Remote Wakeup     MaxPower              100mA     Interface Descriptor:       bLength                 9       bDescriptorType         4       bInterfaceNumber        0       bAlternateSetting       0       bNumEndpoints           4       bInterfaceClass       255 Vendor Specific Class       bInterfaceSubClass      0        bInterfaceProtocol      0        iInterface              0        Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x01  EP 1 OUT         bmAttributes            2           Transfer Type            Bulk           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0040  1x 64 bytes         bInterval               0       Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x81  EP 1 IN         bmAttributes            2           Transfer Type            Bulk           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0040  1x 64 bytes         bInterval               0       Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x82  EP 2 IN         bmAttributes            2           Transfer Type            Bulk           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0040  1x 64 bytes         bInterval               0       Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x83  EP 3 IN         bmAttributes            3           Transfer Type            Interrupt           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0008  1x 8 bytes         bInterval               4   I have searched high and low but I cannot get this to work. All questions related to 138a:0018 are either unanswered or old. Any help guys? I'm working through Rotman (1994). It's the 2nd chapter where my curiosity lead me to a question that is not from the book. Here's the problem that inspired the question:      2.8. Imbed $S_n$ as a subgroup of $A_{n+2}$, but show, for $n\ge2$, that $S_n$ cannot be imbedded in $A_{n+1}.$    Since the alternate subgroup has the property of even parity, I thought that a little investigation of parity for the second and third symmetric groups was in order. So I constructed a table for each of them, grouped by even and odd number of disjoint products.   I'll forego writing the actual table and simply provide the answers. In symmetric groups with an even idegree, two even permutations come together under composition to produce an even permutation. Two odds also make an even. Two different parities produce odd elements.   When the symmetric group has an odd degree, e.g. $S_3$, two odds make an odd, two evens make an odd, and two different parities make an even element. Exactly the opposite results from the symmetric groups with even index.   Two questions: is this enough to prove the $A_{n+1}$ case? And can this be generalized modulo 2 e.g. $A_{2n+1}$?