My question is this what will happen if I will flash a beam of light just say in an dark area where nothing is present ( vaccum ) there to reflect light. Just say I am standing and someone flashed a light in front of me would I be able to see that light.( As there is no particle there to reflect light. )
This is a question thats been bothering me a while. I don't even know if it makes sense or not (like if it is a physics question or becoming a philosophical one). But here it goes. The crux of my question basically is that we all know that we can't see light (like in its photon or electromagnetic wave form) directly when it is traveling past us. However, we also know that the way we see objects is by light reflecting off them. This then means that we are "seeing" the light reflecting from the object which then sends the signal to our brain saying that we are seeing a particular object. We know that both light traveling past us and light reflected from objects are made of photons (so they are the same kind)? So then my question is that what is happening to the photon of a light after it is reflected from the objects, that causes us to see it or the object, but on the other hand we can't see light as it is directly traveling past us.
Windows 10 thinks it's up to date, but it's not mixed declarations and codes A few weeks back I was using std::ifstream to read in some files and it was failing immediately on open because the file was larger than 4GB. At the time I couldnt find a decent answer as to why it was limited to 32 bit files sizes, so I wrote my own using native OS API.  So, my question then: Is there a way to handle files greater than 4GB in size using std::ifstream/std::ostream (IE: standard c++)  EDIT: Using the STL implementation from the VC 9 compiler (Visual Studio 2008).  EDIT2: Surely there has to be standard way to support file sizes larger than 4GB. Prove the existence of an analytic function Let $\Delta=\{z\in \mathbb C: |z| &lt; 1\}$. Let $f: \Delta \rightarrow \mathbb C$ be a one-to-one analytic function fixing the origin. Prove that there is a one-to-one analytic function $g: \Delta \rightarrow \mathbb C$ such that $[g(z)]^2=f(z^2)$. Further, show that such a function is odd.  The only idea which comes to my mind is the following (not sure whether it is the right direction though). Let $h: \Delta \rightarrow \Delta \rightarrow \mathbb C,\ z\mapsto z^2\mapsto f(z^2)$. Note that $h'(0)=0,\ h''(z)=2f'(z^2)+4z^3f''(z^2)$; since $f$ is one-to-one and analytic, its derivative never vanishes, so $h''(0)\ne 0$. So $h$ is an analytic function of $\Delta$ with a zero of order two at the origin. Then there is an analytic function $g$ defined in an open disc about the origin with a simple zero at the origin such that $[g(z)]^2=f(z^2)$.   This is the only thing I came up with, but it seems to be a little bit irrelevant since I don't see any way of extending this $g$ to the whole $\Delta$, and also the injectivity and oddness of such a $g$ are unclear. A lot of people recommend to use the maximum node for glass setup, but I can't find that node. Nothing comes up when I search for it.  Where do I find it? Find duplicate records in MySQL 3/4 circle with plus in it? From this paper:     Shouldn't $du$ be $dt$? And also how do you get from that line to the final result if $du$ is not $dt$? Now that I have Firefox 49 or higher, how do I watch Netflix? Just got the Firefox 49 update (yay!)  But Netflix still wants to install Silverlight (boo!)  How can I watch Netflix in Firefox, as ? Double dual mappings How do I uninstall Pandoc in Ubuntu 19.10 (and install most recent version) I have an old version of Pandoc installed, but I'd like to install the latest version of Pandoc.   To begin I've run the following to get rid of the old version:  sudo apt-get autoremove --purge pandoc  But when I check that everything is uninstalled by running: pandoc --version Then pandoc 2.2.3.2 pops up.  I can't seem to get rid of this old version of pandoc so that I can install a new one.  Could you advise me on how I can get rid of the old version and install the new one please. Bridged adapter in Virtual Box - cannot obtain IP I have virtual box with RHEL installed. My virtual box is running on windows 7 64 -bit. I configured bridged connection, but when I am restarting network interfaces, I receive following error:  Determining IP information for eth1... failed   On Windows XP it works without problems. I disabled Windows Firewall and installed latest Virtual Box, still without success.   NAT is working without any problems. What did I missed?  UPDATE  Ouput of ifconfig -a is:  eth0      Link encap:Ethernet  HWaddr 08:00:27:B6:FC:69           inet addr:10.0.2.15  Bcast:10.0.2.255  Mask:255.255.255.0           inet6 addr: fe80::a00:27ff:feb6:fc69/64 Scope:Link           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:430 errors:0 dropped:0 overruns:0 frame:0           TX packets:346 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000           RX bytes:36520 (35.6 KiB)  TX bytes:31455 (30.7 KiB)  eth1      Link encap:Ethernet  HWaddr 08:00:27:C3:84:4A           inet6 addr: fe80::a00:27ff:fec3:844a/64 Scope:Link           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:1464 errors:0 dropped:0 overruns:0 frame:0           TX packets:17 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000           RX bytes:135419 (132.2 KiB)  TX bytes:4230 (4.1 KiB)  lo        Link encap:Local Loopback           inet addr:127.0.0.1  Mask:255.0.0.0           inet6 addr: ::1/128 Scope:Host           UP LOOPBACK RUNNING  MTU:16436  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0           RX bytes:0 (0.0 b)  TX bytes:0 (0.0 b)   Output of route -n  Kernel IP routing table Destination     Gateway         Genmask         Flags Metric Ref    Use Iface 10.0.2.0        0.0.0.0         255.255.255.0   U     0      0        0 eth0 169.254.0.0     0.0.0.0         255.255.0.0     U     1002   0        0 eth0 0.0.0.0         10.0.2.2        0.0.0.0         UG    0      0        0 eth0   Output of dhclient -v eth1 is:  Listening on LPF/eth1/08:00:27:c3:84:4a Sending on   LPF/eth1/08:00:27:c3:84:4a Sending on   Socket/fallback DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 6 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 9 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 15 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 18 (xid=0x6afb6c4c) DHCPDISCOVER on eth1 to 255.255.255.255 port 67 interval 13 (xid=0x6afb6c4c) No DHCPOFFERS received. No working leases in persistent database - sleeping.   Moreover the same machine with the same configuration works well on Windows XP. The problem is only on Windows 7. Backup of clients' directory to Windows Server 2008 What are the current travel restrictions on individuals entering the United States from regions affected by Covid-19? Check if this proof about real numbers with an irrational product is correct. Can anyone confirm if my proof is correct, please?  Claim:- “If $x$ and $y$ are real numbers and their product is irrational, then either $x$ or $y$ must be irrational.”  Proof:-   Assume that both $x$ and $y$ are rational. Now, let $x = \dfrac pq$ and $y = \dfrac mn$ since both of them are rational. $xy =\dfrac pq * \dfrac mn = \dfrac{pm}{qn}$ Thus, if the product $xy$ can be written as a fraction, it's not a irrational number. Therefore if one of $x$ and $y$ is not irrational, then the product is not irrational.  By the principle of proof by contraposition, If $x$ and $y$ are real numbers and their product is irrational, then either $x$ or $y$ must be irrational. Set Theory: Tree Property Why does the tree property hold for regular cardinals but not singular cardinals? (I.e. There exists a tree of height $\kappa$ with countable levels and no cofinal branch for $\kappa$ a singular cardinal) Does $G\cong G/H$ imply that $H$ is trivial? Testing for coefficients significance in Lasso logistic regression [A similar question was asked  with no answers] I have fit a logistic regression model with L1 regularization (Lasso logistic regression) and I would like to test the fitted coefficients for significance and get their p-values. I know Wald's tests (for instance) are an option to test the significance of individual coefficients in full regression without regularization, but with Lasso I think further problems arise which do not allow to apply the usual Wald formulas. For instance, the variance estimates neded for the test do not follow the usual expressions. The original Lasso paper:  suggests a bootstrap-based procedure to estimate the coefficients variance, which (again, I think) may be needed for the tests (section 2.5, last paragraph of page 272 and beginning of 273):  One approach is via the bootstrap: either $t$ can be fixed or we may optimize over $t$ for each bootstrap sample. Fixing $t$ is analogous to selecting the best subset (of features) and then using the least squares standard error for that subset  What I understand is: fit a Lasso regression repeatedly to the whole dataset until we find the optimal value for the regularization parameter (this is not part of the bootstrap), and then use only the features selected by the Lasso to fit OLS regressions to subsamples of the data and apply the usual formulas to compute the variances from each of those regressions. (And then what should I do with all those variances of each coefficient to get the final variance estimate of each coefficient?) Furthermore, is it correct to use the usual significance tests (for instance Wald's test which makes use of the estimated betas and variances) with the Lasso estimates of the coefficients and the bootstrap-estimated variances? I am fairly sure it is not, but any help (use a different test, use a more straightforward approach, whaterever...) is more than welcome. According to the answers  I suspect inference and p-values just cannot be obtained. In my case, p-values are an external requirement (although the use of L1 regularization was my choice). Thanks a lot EDIT What if I fit an OLS logistic regression using only the variables selected by a previous run of the Lasso logistic regression? Apparently (see ),  There's no need to run the model again after doing cross-validation (you just get the coefficients from the output of cv.glmnet), and in fact if you fit the new logistic regression model without penalisation then you're defeating the purpose of using lasso  But what if I do this with the sole purpose of being able to compute p-values while keeping the number of variables low? Is it a very dirty approach? :-) What happens if you use a Clone Effect on a Double Faced Card that's Transformed? Say someone has , which is a 0/3 Defender, and after some mechanics, transforms into a 13/13 Lizard Horror with Trample.  Say this someone managed to transform said card into its 13/13 Trample form, and then cast  on it (or any other clone type card if it would matter between different ones). Does the Metamorph/Clone/etc. come into play transformed, or as the 0/3 Defender? My keyboard has no "media" keys; can I control media without them?