Humans (homo sapiens) have been on the earth for thousands of generations, and we have kept evolving throughout that time. Why don't we just keep evolving so that, let's say, we live for an average of 300 years? Or why don't we evolve so that we don't need to sleep?  This could also apply to many other species.
Why does evolution not make life longer for humans or any other species?  Wouldn't evolution favour a long life?
I encountered a problem today to prove that $(X_n)$ with $X_n = \cos(n!)$ does not have a limit (when $n$ approaches infinity). I have no idea how to do it formally. Could someone help? The simpler the proof (by that I mean less complex theorems are used) the better. Thanks Building a matrix by merging the same row vector multiple times I am trying to find out how to approach a relief type project. I have a background image and a flat plane mesh to sculpt on, but I actually don´t see any changes looking straight on top. When I tilt the perspective a bit the image disappears.  Is there a way to make the image appear at all perspectives? seem to be well-regarded for their versatility and accuracy of metering.  What puzzles me about them is how can they account for the different light transmittance of different lenses? It's easy for an in-camera metering system that meters  so any peculiarity in transmittance is already applied to light when it reaches the metering sensor. But a standalone meter does not have the privilege of looking at the world through the lens it meters for. I tried looking it up in  and , but this issue was not touched upon at all.  This leaves several variants how it could be handled -   light meter uses  for aperture display/input, and photographer is expected to perform the adjustments to and from  (which are used by cameras); light meter assumes some median transmittance of lenses of its era, and automatically accounts for that; photographers simply ignore the matter - even a 75% transmitting lens will be off only by about half a stop, no big deal; each light meter used is expected to know that you have to dial in some exposure compensation to account for the light loss; ...   So how does it work in real world? No module named numpy I am trying to run Jupyter notebook from my Ubuntu subsystem in Windows 10. I've installed numpy, scipy, and pandas using the following commands:  pip install scipy numpy pip install pandas   When I load python in Ubuntu, there is no issue importing numpy in Ubuntu cmd.   But, when I run my Jupyter Notebook from Ubuntu terminal, and try to import numpy as np, or import pandas as pd, they report  ModuleNotFoundError                       Traceback (most recent call last) &lt;ipython-input-1-a9340201ed9f&gt; in &lt;module&gt;       5 import dash_html_components as html       6 import plotly.graph_objs as go ----&gt; 7 import numpy as np       8 from dash.dependencies import Input, Output       9   ModuleNotFoundError: No module named 'numpy'   Is there anyway to solve this issue? Thanks a lot. I'm stuck underground.   I can't mine any further to reach lava. There is no water nearby, and zombies can't reach me.  I can't see to find any blocks to suffocate my character.  NBTEdit and similar programs don't work on Mac, so I can't edit my save file to escape. I have hardly any blocks so cannot jump and lay my way up. I can't see, and do not have any torches, so can't make a staircase back up. I don't want to make new save as I have an awesome house filled with stuff that took me ages to find.    How can I get back to the surface? How do I change the angle of the axis without changing the angle of the object? Hey guys I am currently working on a scene with multiple objects.  There is one object (a wheel) that I need to rotate the origin for without rotating the wheel itself.  I have the exact angle that I need so a manual solution would be the best.  Any help would be much appreciated. I am used to parametric modeling and Blender is a little out of my depth. How assign shortcut to custom function? Gradient steepest direction and normal to surface? From this , I now understand the gradient to be the directional derivative that returns the steepest slope at a point. However, reading my textbooks, they all say that the gradient is normal to the tangent plane. In the above SE question, the gradient sounds like it should be parallel to a direction in the $xy$ plane, not actually up on the surface.   My question is therefore do I have a misunderstanding of the tangent plane, gradient or directional derivative?   In addition, I question how the gradient always passes through the origin or the $xy$ plane? If for instance in the graphic, the gradient had been another direction (say horizontal), then it wouldn't pass through the origin! In addition, the gradient of other points in the graphic would satisfy this example. Many thanks! Cyclic Group Generators of Order $n$ How many generators does a cyclic group of order $n$ have? I know that a cyclic group can be generated by just one element while using the operation of the group.  I am having trouble coming up with the generators of a group of order $n$.  Any help would be great!  Thanks! given a polynomial $p(x,y)$ from $\mathbb Z[x,y]$. I want to substitute $x=1$ leaving $y$ as it is. The command Value(p,[1,y]) does not work. Can you give me a hint?  After that it gets a little more complicated. Let  $$p(x,y)=x^6+y^6$$ After substituting $x=1$ we have $p(1,y)=1+y^6$. Given a natural number $j$. I want to find all expressions $y^j$ in $p$ and substitute these $y^j$ by $1$ and the others by $0$. In this case (and for $j=6$) we would get $1+1=2$. If $$p(x,y)=x^4y^2+x^2y^6$$ we should get $0+1=1$. Is there any way to do this?  Thanks for reading! edit: Well, it turns out I used the function Value wrong. The right code for my problem would be Value(p,[x,y],[1,y]). So my first question is solved. How do I get a file extension in PHP? I have recently registered some domains on the basically brand new .directory gTLD. Because of operating costs to maintain a gTLD, it is assumed that many of the hundreds being rolled out this year and next will fail. I will admit I am not very knowledgeable on how the TLD process works.   What exactly does a new TLD failing mean? I realize that they lose ICANN accreditation, but does that just mean that they will be no longer be sold? If the .directory TLD were to go under, would my .directory domains be affected? I bought it from domain registrar Gandi if that helps. I think it would be good to have a way for the user to view the target URL of a link before following it. I'm pretty sure that I'm not the only one who dislikes blind-clicking -tapping links.   When I use SE in the mobile browser, I routinely long-press links to find out where they go before following them. I wish I could do the same thing in the app, with the same or a similar method.  (It's important that this won't conflict with text selection, since you can't long-press on links to select them now anyway.) A bit of background, I am doing research in atmospheric physics and detection systems  Recently, I received a paper submission back with revisions suggested by a couple of reviewers.  For the most part, the revisions were definitely constructive, justified and most importantly, helpful.    That is, except on one particular crucial aspect of the paper where both reviewers made suggestions, that they labelled as critical for the paper's success.  However, the suggestions were directly opposite.  i.e. Both suggestions are mutually exclusive, but equally feasible.    What is an effective means to address this conflict?  I have spoken to my academic supervisor and he is not sure how to proceed either and we have double checked and both suggested paths are feasible. Any difference between "Good morning, Brian" and "Good morning to you, Brian" I often hear it on TV when they are talking to a guest.  Normally, if we want to greet Brian in morning, we would say "Good Morning Brian". But many times TV presenters say "Good morning to you Brian".  Why is that? Why do they add "to you". There seems to be no need to do it.  Regards, SaveAsActivity is not allowed with whatId that is not supported as a task whatId How can I get back to Blender's default theme? I am trying to incorporate a wms feed in the British national grid projection (EPSG:27700) into a openlayers 3 project. I initially had used this wms layer in an openlayers 2 application using the following syntax:  new OpenLayers.Layer.WMS(         "ukimage",         "http://www.linktowms",         {             "LAYERS": 'ukimage',             "STYLES": '',             format: state.image_format         }, {     buffer: 0,     displayOutsideMaxExtent: true,     isBaseLayer: true } );   But I am having trouble converting this over to openlayers 3. I first tried setting the map projection to British national grid (view: new ol.View({ projection: 'EPSG:27700'}) as I had in the openlayers 2 app, but this produced the error  TypeError: h.B is not a function    and a white map.   When I try and use proj4 to convert the projection to EPSG:3857 as described here:  I get the same error:   TypeError: h.B is not a function    and again a white map.   my wms request is generally a varient of:  new ol.layer.Image({       source: new ol.source.TileWMS({         url: 'http://www.linktowms',         params: {'LAYERS': 'ukimage'},       })     }),   (I tried adding projection but that didn't help)  I am trying to work out if the image is even getting loaded from the WMS feed, but the syntax seems to have changed significantly from openlayers 2, instead of the properties and functions that have rather descriptive names, there are a now lots of short functions with names like Ua and Ig, which leaves me a bit in the dark as to what's actually going on when I try and inspect the layer with firebug.  I don't mind what projection the application is in, as I can re-project my other layers to match the basemap, so basically any way of using this EPSG 27700 WMS layer in openlayers 3 would be great. Proof of linear independence of $e^{at}$ Given $\left\{ a_{i}\right\} _{i=0}^{n}\subset\mathbb{R}$ which are distinct, show that $\left\{ e^{a_{i}t}\right\} \subset C^{0}\left(\mathbb{R},\mathbb{R}\right)$,  form a linearly independent set of functions.   Any tips on how to go about this proof. I tried a working from the definition of an exponential and combining sums but that didn't seem to get me anywhere. I saw a tip on the internet that said write it in the form  $\mu_{1}e^{a_{1}t}+\dots+\mu_{n}e^{a_{n}t}=0$ to try to show $\mu_{1}=\dots=\mu_{n}=0$ considering each term of the left hand side must be positive, but I can't get my head around that because while I understand $e^{x}&gt;0\forall x\in\mathbb{R}$ I cannot see why $\mu_{i}$ must be positive in any case. I have thought about differentiating but that doesn't seem to help. The question did originally ask for a "rigourous" proof but I'll take any hints right now and the provided the solution of 'is obvious' is most unhelpful to me.  Any input would be fantastic. Thank you.