Consider two moving objects along X-axis direction with different speeds and different initial starting points. Let the initial distance of the objects be d0 and v1 is greater than v2 i.e v1>v2.   Physics says:  x1 = v1 * t + x01 = v1 * t  x2 = v2 * t + x02 = v2 * t + d0  If we put x1 = x2 we obtain :   t = d0 / (v1 - v2)  and O1 because of its greater speed will eventually reach to O2.  But philosophy says something else:  O1 in t1=(something calculable) will reach to O2's current position but in this time interval O2 will move x1=(something calculable) toward.  Again in the new situation O1 in t2=(something calculable) will reach to O2's current position b but in this time interval O2 will be x2=(something calculable) toward.  By repeating this procedure tn and xn approaches to zero but does not become zero, therefore O1 will never reach O2.  How do you explain this?  Thank you so much.
It seems to me, perhaps naïvely, that Aristotle resolved  well, when he said that,      Time is not composed of indivisible nows any more than any other magnitude is composed of indivisibles,    and that Aquinas clarified the matter for the (relatively) modern reader when he wrote      Instants are not parts of time, for time is not made up of instants any more than a magnitude is made of points, as we have already proved. Hence it does not follow that a thing is not in motion in a given time, just because it is not in motion in any instant of that time.     For me, any further attempts at explanation and resolution, such as Russell's, (bar rather tenuous appeals to connections with quantum physics) simply reiterate, and rather labour Aristotle's initial erudite explanation.  What serious arguments exist that counter Aristotle's objection, if any, and why is Aristotle's objection not considered a resolution to this paradox?
So the complex number z is given as   $z=1-cos(2\theta)-isin(2\theta)$  And the question is to find the argument of z in terms of $\theta$.   So, I used the formula for calculating the argument of a complex number:  $arg(z)=arctan (\frac{sin(2\theta)}{1-cos(2\theta)})$  $=arctan (\frac{2sin\theta cos\theta}{1-(1-2sin^2\theta)})$  $=arctan (\frac{2sin\theta cos\theta}{2sin^2\theta})$  $=arctan (\frac{cos\theta}{sin\theta})$  $=arctan (cot\theta)$  $=arctan (tan(\frac{\pi}{2}-\theta))$  $=\frac{\pi}{2}-\theta$  But apparently the answer is $\theta-\frac{\pi}{2}$.  Where might have I gone wrong?  Edit: the domain given in the question for $\theta$ was $0≤\theta≤\pi$ I am doing some homework exercises and stumbled upon this question. I don't know where to start.      Prove that the union of countably many countable sets is countable.   Just reading it confuses me.   Any hints or help is greatly appreciated! Cheers! Why does a light beam in a moving light clock follows a V shaped path? If I am standing on a platform moving along the X-axis and I throw a basketball vertically upward along the z-axis, the ball also travels along the x-axis. Does the same apply to a photon? According to a special relativity postulate  speed of light is not affected by the velocity of the inertial frame or the velocity of the emitter. If this postulate is true, shouldn't a photon fired vertically as above should travel only along the z-axis but not along the x-axis? Is it possible to create bootable linux on USB drive using windows command line? Diacritics and non-English letters in anglicized loan words: keep 'em, dump 'em, italicize the words, or what? When I take the pth root, the bottom of the p comes very close to the \sqrt symbol.  In the New Century Schoolbook font (which I am using), the p actually intersects the \sqrt symbol, which looks even more terrible.  Is there a clean way to fix this?  \documentclass{article} \usepackage{fouriernc} % use the New Century Schoolbook font \begin{document} When I write $a^{(1/p)}$ as $\sqrt[p]{a}$, the bottom of the $p$ touches the top of the root symbol, which looks ugly. Even when the root symbol is larger, such as with $\sqrt[p]{\frac{1}{n}}$, it still looks bad (since the tail of the p almost hits that line). \end{document} Different font color for question status => closed <= When I stared at  () and I even thought [question refined] as offical status of the post.    &lt;somebody  it now, So, I've edited this question to get same effect>  How about official status [closed] as   different font color or  different font size or put the text [closed] outside the question link   Or   put a special tag as suggested in the    to clarify from those? Firstly, I apologise that this question is specific to two polynomials. I understand that this post will not help a lot of people, and for that I am sorry. I will make it up to you all by anwsering lots more questions in the future! However, as my lecturers enjoy time off on the weekend, I am unable to ask this to anyone else who would know what the heck I'm talking about. Sorry if my notation is not widely used, I can elaborate more if required. Please excuse my poor Latex too.  Let:  $B = Y^2 - X^3 + X$, and   $F = (X^2 + Y^2)^3 - 4X^2Y^2$.  I wish to find the intersection number of $B$ and $F$ at $P = (0,0)$  i.e $I(P,F \cap B). $  However, I want to use the following proposition to do this:  If $P$ is simple on $B$ then $I(P,F \cap B) = ord_P^B(F) $.  I will omit proving that $P$ is simple on $B$ as I hope it is clear to see it is.  I know that $I(P,F \cap B) &gt; M_p(F) * M_p(B) = 1 * 4 $, where $M_P(G)$ is the multiplicity of $P$ on $G$.  Now, consider the Image of $X$ and $Y$:  $K[X,Y] \longrightarrow \frac{K[X,Y]}{Y^2 - X^3 + X} \longrightarrow O_P(Y^2 - X^3 + X)$  $X \longrightarrow x$  $Y \longrightarrow y$  Where lower case denotes the image.  Then $ord_P^{Y^2 - X^3 + X}((X^2 + Y^2)^3 - 4X^2Y^2)$  Can be found by substituting $y^2 = x^3 - x$ into $(x^2 + y^2)^3 - 4x^2y^2$  Which gives:  $ord_P^{Y^2 - X^3 + X}(x^9+3x^8 -5x^6 -4x^5 +3x^4 +3x^3)=3$  i.e $I(P,F \cap B) = 3$, a contradiction to  $I(P,F \cap B) &gt; 4$ as above.  The correct answer I know is 6, so where have I gone wrong? Why is it necessary to sample from the posterior distribution if we already KNOW the posterior distribution? My understanding is that when using a Bayesian approach to estimate parameter values:   The posterior distribution is the combination of the prior distribution and the likelihood distribution. We simulate this by generating a sample from the posterior distribution (e.g., using a Metropolis-Hasting algorithm to generate values, and accept them if they are above a certain threshold of probability to belong to the posterior distribution). Once we have generated this sample, we use it to approximate the posterior distribution, and things like its mean.   But, I feel like I must be misunderstanding something. It sounds like we have a posterior distribution and then sample from it, and then use that sample as an approximation of the posterior distribution. But if we have the posterior distribution to begin with why do we need to sample from it to approximate it? Project Euler - 34 / Find a mathematical approach for upper bound 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find   the sum of all numbers which are equal to the sum of the factorial of   their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not   included.   So I tried a random enough large upper bound (100000) in my program and it worked.  I deleted this part to not spoil the answer  My question is : How to find the closest upper bound using a mathematical approach ? Why does "Object reference not set to an instance of an object" not refer the variable name or column? Proving that $G$ is a group if $a*x=b$ and $y*a=b$ have solutions. (Reference : Fraleigh, A first course in abstract algebra)     Prove that a nonempty set $G$, together with an associative binary operation * on $G$ such that $a*x=b$ and $y*a=b$ have solutions in G, $\forall a,b\in G$, is a group.   Because * is an associatie operation, we have to prove that: there exists $e\in G$ such that $e*a=a=a*e$ for any $a\in G$, and that for any $a\in G$ there exists $\alpha \in G$ such that $a*\alpha=e=\alpha *a$.  This is very confusing, first I tried to prove the identity exists. If we consider a=b, then $a*x=a=y*a$, and because of the hypothesis $x,y\in G$ always exists, now we have to show that $x=y$, but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck.  And I tried to get unstuck, so I thought particular cases, first if $G$ had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated.  What can I do? Would it be wrong to copy-paste questions from Quora? Is is possible to restore the MFT of a NTFS partition after a format? I've searched through several questions but didn't quite find anyone treating mine.  So, I have a 1TB external hard drive I use for backup purposes with only 1 NTFS partition. Today using DiskPart I formated that partition by mistake (who would have thought the disc count started in 0 and not in 1). The complete command I used was:  format recommended override quick label=Windows   So I performed a quick format, and from which I've read after a quick format it's possible to recover the data stored there. Anyway, as soon as I realized it wasn't the partition I wanted to format the one I specified I removed the access to my external drive to minimize possible damages.  Now, the MFT is the metadata part where the file name, creation date, access permissions and contents are registered; and it's mirrored too (so there's a copy of it somewhere in the disk). What I don't know is if a quick format erases that MFT mirror too, I hope not.  I'm running a program to recover files on it but I'll have to save the files to another location (once it's finished scanning, ~12 hours from now) and I don't quite have over 900GB of free storage where to put the data.  I would like to know if there's a program or procedure by which I could restore the old MFT in-place and have all the data that used to be in there accessible once again.  Any help or light you could shed on how to proceed or what programs to use is really appreciated. Exposing database IDs - security risk? How can I get this 'Version: 0.1.0 - 94" value from the  tag in selenium webdriver? How can I get this Version: 0.1.0 - 94 value from the &lt;div&gt; tag in selenium webdriver? and here is my code:  &lt;div _ngcontent-qbk-c3="" style="text-align: center;"&gt;Version: 0.1.0 - 94&lt;/div&gt; Why does jQuery or a DOM method such as getElementById not find the element? Why is the upper Riemann integral the infimum of all upper sums? I was reading the theory of Riemann integration when I cam across the following ,   If $f$ is bounded on $[a,b]$, and $P = \{x_0,x_1,x_2.......x_n\}$ is a partition of $[a,b]$, let $$M_j = \sup_{x_{j-1}\leq x\leq x_j}f(x)$$ The upper sum of f over P is $$S(P) = \sum_{j=1}^{n} M_j(x_j-x_{j-1})$$ and the upper integral of $f$ over $[a,b]$, denoted by $$\int_{a}^{b^-} f(x)dx$$ is the infimum of all upper sums.  The theorem similarly goes on to state the result for lower sums.  My doubt is : I do not understand how is $$\int_{a}^{b^-} f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partition P, then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P (but still being the lower limit does not prove that it is the infimum) and what about those partitions for which P itself is the refinement of?  How do I know that it will be a lower limit for those, let alone a infimum? Get rows with most recent date for each different item Prove the map sending a matrix to it's inverse is continuous.