Most of my network is business analysts, data analysts/engineers/scientists and software engineers and none of them have jobs requiring security clearance. IMO it coming in many job postings is more likely indicative of which industries you are searching than it being a super common thing
Depending on the target market for the game and how people generally play (online vs in-person), free online simulators can cut pretty significantly into the bottom-line. For an extreme example, imagine someone created a popular Hearthstone simulator where nobody needed to either grind/pay for...
This really sounds like homework you should be doing as a jobseeker, rather than asking the internet. But what does relevant company even mean here? Are you wanting to work in specific industries or do specific kinds of work?
If by relevant company you mean "willing to hire someone with my...
As someone who came from a similar background and currently does engineering / data science work, can vouch that the door for those kinds of gigs should definitely not be closed (e.g. the company I work for almost exclusively hires math and physics undergrads without experience into entry-level...
I actually work in the gaming industry on the engineering and data science sides and regularly interface with professional gamers (defined here as people who play on major eSports teams and get payed for playing on the big stage). Some of the numbers here vary by title and the maturity of the...
Depending on what counts as "doing math" here, this statement is patently untrue. I work in the gaming industry and there are many opportunities for people with undergraduate math degrees:
About 1/3 of our business intelligence team hold only a BA/BS in mathematics or statistics. No programming...
It was not an insult, it was a request. PF is not a place for other people to do the thinking for you. There is an expectation that you think something through before asking.
It would do you good to familiarize yourself with mathspeak, but that point aside I am not sure which part of that...
In the future try thinking about the claim for more than three minutes before asking to have it spelled out for you. In any case, notice that the set An of all real numbers in [0,1] whose decimal expansion has length n is in bijection with the set {0,...,9} x ... x {0,...,9} (this is an n-fold...
This is nonsense. We have perfectly rigorous definitions of real numbers. Axiomatically they are the unique Dedekind-complete ordered field and constructions of the real numbers via Dedekind cuts or Cauchy sequences have been known for over a century. A rigorous definition for π is also easy to...
Since there are only countably many reals with finite decimal representations, and since this observation only accounts for these representations, it will not suffice. Something more sophisticated like the diagonal argument is needed.
Most decent algebra texts will prove these results, but the arguments are actually quite simple so I can sketch them here. For (i) let G be any group and let FG be the free group generated by the elements of G. The universal property of this free group provides a homomorphism FG→G and let K...
Checking journals for your particular series is going to be difficult. It is unlikely something like the series itself would be present in the title, so you would probably need to search specifically for articles about the gamma function, the problem then becoming there are a lot of these...
Oh whoops you are right! Unless I am mistaken it turns out the result is actually false! Take p = n = 2 and consider the map Ω:Z4→Z4 given by Ω(x) = x2. Then Ω(1+1) = Ω(2) = 22 = 0 but Ω(1)+Ω(1) = 12+12 = 2.
Just consider the inverse limit of the sequence ...→Z/p3→Z/p2→Z/p and this produces the p-adic integers. Another way is completing Q with respect to the p-adic norm and then consider all element with p-adic norm less than or equal to one.
Assuming Zp denotes the integers modulo p (and not the p-adic integers), then there is a "natural" projection Zp-1→Z/p obtained simply by looking at the coproduct of the p-1 natural maps Z→Z/p.
Then drop an EC.
The problem sounds like there are not enough hours for ECs and school work combined. So you can either become super efficient and cut down the time needed for school work or you can reduce the number of hours needed for ECs. As I mentioned earlier, the ECs will not make a...
I still stand by my earlier advice, but if sleep deprivation is getting to be a nuisance, then you can always drop some ECs or at least dial your participation down. It will likely have nearly zero impact on your college admissions. Unless the application process has changed significantly in the...
Write n = ∑ak10k and notice that n is divisible by 9 if and only if n ≡ 0 (mod 9). So this means n is divisible by 9 if and only if ∑ak ≡ 0 (mod 9) and the result follows.
One argument that works: Notice there is a fibration G→EG→BG so using the Leray-Serre spectral sequence given information about H*(G;Q) and H*(EG;Q) one can hopefully determine something about H*(BG;Q). Since EG is contractible this gives us one piece of the puzzle and since G has the homotopy...
The first function does not have be an inclusion in the most literal sense per se, but it can be interpreted that way. The second function does not need to be an inclusion in any sense however.
There are lots of reasons and here is one big one: Every group is a quotient of two free groups...
Basically just construct it point-wise. Since forms are sections of tensor bundles evaluating at a point reduces the existence problem to vector spaces. Only a smoothness argument remains and for this local frames come to the rescue.
In this case the winding number is the degree of a map S1→S1. It has connections to complex analysis where there is actually a concrete formula computing this number. Not sure if that is what you thought or not.
Are you stuck using this exact argument? Unless I have misunderstood something the proof should go through easily with a slight modification: Essentially after repeatedly multiplying I∩J by I+J we should end up with something like ∑Ij∩Jk-j and choosing k so large that either Ij or Jk-j vanishes...
Lots of topics below are actually relevant to physics. Unfortunately physics is not really my area of study, so I can only give a fairly shallow indication of results that are immediately of some interest in physics, but hopefully it can get you started until someone more knowledgeable drops by...
Suppose T satisfies (1). Notice that T(nv) = T(v+...+v) = T(v)+...+T(v) = nT(v) and since T(-v) = -T(v) it follows that (2) holds for all integers. Next consider nT(m/n v) = T(m/n v)+...+T(m/n v) = T(m/n v+...+m/n v) = T(mv) = mT(v) and it now follows that (2) holds for all rational numbers...