Bazerka (NA)
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=sRNB8ALI,comment-id=00150000000000000000,timestamp=2019-08-20T20:21:05.631+0000) > > Alright, how long do I have? I’ll finish them if I get picked :) gotta be done by friday so we can register the team, and you did get picked lol. I updated the thread :)
> [{quoted}](name=Bazerka,realm=NA,application-id=yrc23zHg,discussion-id=sRNB8ALI,comment-id=001500000000000000000000,timestamp=2019-08-20T20:43:23.331+0000) > > gotta be done by friday so we can register the team, and you did get picked lol. I updated the thread :) Eeeey looks like I’m playing ranked tonight. Wanna duo queue? I get off work at 5, and I’ll be available by 7 (PST).
Bazerka (NA)
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=sRNB8ALI,comment-id=001500000000,timestamp=2019-08-20T20:17:20.215+0000) > > Just a quick question: do I have to finish ranked placements to play? yes you do :)
> [{quoted}](name=Bazerka,realm=NA,application-id=yrc23zHg,discussion-id=sRNB8ALI,comment-id=0015000000000000,timestamp=2019-08-20T20:20:19.493+0000) > > yes you do :) Alright, how long do I have? I’ll finish them if I get picked :)
Bazerka (NA)
: > [{quoted}](name=Jopboby377,realm=NA,application-id=yrc23zHg,discussion-id=sRNB8ALI,comment-id=0015,timestamp=2019-08-20T00:51:12.342+0000) > > Bazerka, do you happen to know when we will know who you decide to play with? Just so if anyone doesn't get chosen can find others to team with. Ah good call. I kinda wrote this over lunch on a whim so didn't give it too much thought. Thanks for the call out! editing now. TL;DR i'll rando select 4 ppl tomorrow afternoon and close the thread when I choose :)
> [{quoted}](name=Bazerka,realm=NA,application-id=yrc23zHg,discussion-id=sRNB8ALI,comment-id=00150000,timestamp=2019-08-20T03:03:44.811+0000) > > Ah good call. I kinda wrote this over lunch on a whim so didn't give it too much thought. Thanks for the call out! editing now. TL;DR i'll rando select 4 ppl tomorrow afternoon and close the thread when I choose :) Just a quick question: do I have to finish ranked placements to play?
Bazerka (NA)
: Anyone wanna play in the clash beta?
I’m interested, but is that PST or EST? Pretty sure I’m available either way, but just checking. As far as ranking goes, I was Plat 5 in season 7, Gold 4 in season 8, and I’m unranked this season with a preliminary rank of Silver 3 after one game. I main MID/ADC with no real preference. I can play almost every ADC, with the exceptions being Draven, Kog’Maw, and Kai’Sa. I’ve also been known to take some mages into bot lane ;)
: Then how is the client able to have all of the damage stats and KDAs from the game??? There is, of course, a connection between the two to tell the client to come back up after the game ends also.
I don’t think it’s a given that the two are connected just from that. Since other sites have access to the same data (or at least most of it), it’s more likely that both the client and those sites access Riot’s servers to retrieve the game information, so if mute data isn’t included, there’s nothing to grab (not that it can’t be added). Still, there doesn’t have to be a direct connection between the client and the game for the client to be able to get your KDA. As far as the client restart after game goes, it’s possible that there’s a background process that runs even when it’s “closed”. It could ping Riot’s servers for game status and check to see if the game closes, which could also explain why it takes so long to get out of game. Then again, I haven’t seen their code, so all of this is just speculation, but it’s definitely possible to accomplish those things without ever establishing direct communication between the client and the game.
: > [{quoted}](name=JuiceBoxP,realm=EUNE,application-id=3ErqAdtq,discussion-id=n6E7bW05,comment-id=0000000200000000,timestamp=2019-08-10T14:12:08.502+0000) > > w/e your claims are as valid as you want to believe they are, but that is not the point. > the outplays in the video are a bonus, that is not the topic of the thread. > > the topic of the thread is that pyke gets too much gold from his ult, and that's it. > i would've probably played it differently but it doesn't matter. > > bottom line, he got destroyed this game and he still got BY FAR the most gold in the game. and youre not understand is the reason he got destroyed was his use of said gold. he was literally braindead and youre boasting about it lmfao. get off your high horse. pyke is fine the way he is.
Okay, but, if he was “literally brain dead” and still got 2K more gold than the most fed member on his team, isn’t that kind of a problem in and of itself? His ult being his only source of income doesn’t matter if said income is as massive as it is for what little effort he has to put in to get it.
rujitra (NA)
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=oXeKrYPb,comment-id=000000000000000000000002,timestamp=2019-07-20T14:23:55.105+0000) > > https://boards.na.leagueoflegends.com/en/c/general-discussion/oViZ4Aed-my-account-is-beyond-broken-at-this-point-the-game-is-literally-unplayable > > This guy only has his problem when logged into his main account, so yes, I’d say it’s possible for it to behave differently depending on who logs in. Hell, it already does that naturally by displaying different friends, BE, RP, etc. When it comes to programming, bugs can come from anywhere, and can do some really weird things. That is the air client - i.e. the friends, lobby, etc. client. The issues being referred to are with the *game* client itself.
> [{quoted}](name=rujitra,realm=NA,application-id=yrc23zHg,discussion-id=oXeKrYPb,comment-id=0000000000000000000000020000,timestamp=2019-07-21T00:40:00.061+0000) > > That is the air client - i.e. the friends, lobby, etc. client. The issues being referred to are with the *game* client itself. To quote you: "For the same client to behave differently depending on what authentication token was active? Not really." It seems I miss-understood what you were referring to. However, my point is still relevant: _The main client can exhibit different behavior for different accounts, so you can't dismiss the possibility that the game client can as well. _
rujitra (NA)
: > [{quoted}](name=MessyStuff,realm=NA,application-id=yrc23zHg,discussion-id=oXeKrYPb,comment-id=0000000000000000,timestamp=2019-07-20T01:09:13.175+0000) > > I mean it did happen and there is definitely a way in coding for this to happen. For the same client to behave differently depending on what authentication token was active? Not really.
> [{quoted}](name=rujitra,realm=NA,application-id=yrc23zHg,discussion-id=oXeKrYPb,comment-id=00000000000000000000,timestamp=2019-07-20T01:09:44.276+0000) > > For the same client to behave differently depending on what authentication token was active? Not really. https://boards.na.leagueoflegends.com/en/c/general-discussion/oViZ4Aed-my-account-is-beyond-broken-at-this-point-the-game-is-literally-unplayable This guy only has his problem when logged into his main account, so yes, I’d say it’s possible for it to behave differently depending on who logs in. Hell, it already does that naturally by displaying different friends, BE, RP, etc. When it comes to programming, bugs can come from anywhere, and can do some really weird things.
Ieafboom (NA)
: My account is beyond broken at this point. The game is literally unplayable
I can’t see a way for this to be a problem with your client. My first thought was that there’s a folder in the launcher related to accounts, and the one for your specific account is corrupted. However, I couldn’t find anything after looking through the RiotGames folder, and the problem would’ve disappeared after you reinstalled. It might be possible that the bug is being reinstalled into the folder every time, but that would require the bug to exist on Riot’s servers anyways. Perhaps it’s something to do with how the program handles your specific summoner name. It’s unlikely, but you never know with spaghetti code.
JuiceBoxP (EUNE)
: > [{quoted}](name=Spiced Rum,realm=NA,application-id=3ErqAdtq,discussion-id=hyRvBTPE,comment-id=0005000000000000,timestamp=2019-07-15T13:46:10.904+0000) > > So zed, talon, fizz, Diana, Ekko and kassadin Except every single one of them has incredibly effective ranged poke and ranged wave clear. Katarina has only Q, for ranged waveclear which hits exactly 3 minions and she has to use it either on minions or the enemy laner. Or if the laner is dumb enough to stand ontop of the dying minions then it would hit him too. add that to your list of unfavorable matchups {{champion:84}} {{champion:69}}{{champion:39}} {{champion:38}} {{champion:50}} {{champion:8}} {{champion:157}} {{champion:7}} {{champion:79}} {{champion:74}} They fucking dumpster on her.
I mean, sure, her Q is her only technical form of ranged wave clear, but it's not her only form of wave clear in general since she does AOE damage around her after picking up any dropped daggers. I'd argue that from levels 1-3 or 1-4, she isn't really trying to wave clear and roam because she isn't strong enough to assassinate anyone yet, meaning she's really just trying to last hit and dodge poke under tower. In this case, it doesn't really matter that her only form of ranged wave clear is her Q because, well, eventually she won't mind going melee range with her blink to clear the wave before heading bot lane to try and pick up a double kill. I don't play Katarina myself, so I can't really say how the matchups you listed go (though, judging by the kits, probably not well for Katarina), but it doesn't matter unless they shut her down super hard in the early game. She only needs to get enough gold to hit a power spike so she can start snowballing off of champions she can abuse.
: I can’t spell anymore. That’s how much of a tole it’s taken on me
> [{quoted}](name=Laughing Fish,realm=NA,application-id=Ir7ZrJjF,discussion-id=OeHqieiR,comment-id=00000000,timestamp=2019-04-01T17:08:34.436+0000) > > I can’t spell anymore. > > That’s how much of a tole it’s taken on me Don’t look now, but you also used the wrong form of “its”. ;(
: Laughing Fish’s Birthday Spectacular: Come and get your free skins!
What's this?! My "party hard" reaction gifs folder is finally relevant?! YES! http://i.imgur.com/em2AYOL.gif[\img] This is the only one in there, though...
Jamaree (NA)
: > [{quoted}](name=Tuition Fee,realm=NA,application-id=3ErqAdtq,discussion-id=EGAHMEbn,comment-id=00000000,timestamp=2019-03-27T00:25:34.230+0000) > > I don't play JG... I’m just fucking tired of jungle mains complaining about how hard shit is and acting like it isn’t hard for everywhere else. Why is it the laners are expected to go out of their way to help them when they are getting invaded, putting themselves behind to do so most of the time, yet junglers aren’t expected to them gank afterwards? Why are they allowed to have an entire area they can farm without interaction at all and if they are, are then expected to also get help? Why does no one else remember the cancer of junglers getting to be level 3 full life from getting both their major buffs while laners were still at level two. I’m just getting tired of this constant whining about junglers having it so fucking hard and acting like they are these second class citizens when they have so much shit slanted towards them. Is it fair they get blamed if the laners are shit? God no, but it isn’t fair that they think hey should just be able to mindlessly carry anyway if they do nothing when the other jungler is ganking, nor is it fair that they should expect to just be able to carry every game just because they are the jungle, downvote me for it but fuck it, I’m tired of the constant bitch about their perceived weakness
> [{quoted}](name=Jamaree,realm=NA,application-id=3ErqAdtq,discussion-id=EGAHMEbn,comment-id=000000000000,timestamp=2019-03-27T00:37:06.148+0000) > Why is it the laners are expected to go out of their way to help them when they are getting invaded, putting themselves behind to do so most of the time, yet junglers aren’t expected to them gank afterwards? Watafool already outlined most of the points I wanted to make, but here are a few questions I found myself asking after reading this: - Gank whom, the people that came to help? Which one should he gank, all of them? I'm pretty sure he was going to do that in the first place considering he's, you know, the jungler. - If he is going to gank all of them "because he's expected to", which one should he gank first? What if one of those lanes doesn't need help and is already ahead, but a different lane that didn't come to help is starting to fall behind and could use some assistance? What if the only lane that came to help is top lane, and we know that the enemy jungler is bottom side ready to make a play for dragon? If we expect the jungler to just immediately gank top lane because they came to help him on the invade, doesn't that mean we give up dragon (and probably one or two kills in bot lane)? - When should he gank? If he ganks twenty minutes later, does that count as "returning the favor" or just playing his role? At the end of the day, the jungler's job isn't just to gank lanes. It's to create pressure on the map by being in the right place at the right time, securing vision of the enemy jungler, and potentially stealing resources from the enemy team (or securing some for your team). A good jungler can identify when to go for what, and doesn't need to gank to apply pressure on the map. TL;DR: It's not always in your team's best interest for the jungler to gank a certain lane, but it is always in your team's best interest to collapse if your jungler is invaded (assuming the enemy team can't also collapse easily).
: Well, Do you accept the unknown game or do you decline it :x? The PEOPLE WANT TO KNOW
> [{quoted}](name=TheGreatNinx,realm=NA,application-id=LqLKtMpN,discussion-id=qPsmEXnm,comment-id=00000000,timestamp=2019-03-27T16:11:54.319+0000) > > Well, Do you accept the unknown game or do you decline it :x? The PEOPLE WANT TO KNOW I ACCEPT THE CHALLENGE
: I have a, um, slight problem
Slight update: https://gyazo.com/ffa9b6d9b48ad744b3eca22863f59f90
Rioter Comments
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=v3Lcf4wR,comment-id=0000000000010000,timestamp=2019-03-24T18:48:34.044+0000) > > Alright, do you have anything specific you want cleared up from there? I'm not quite sure what your class has covered. Ill have to take a look at my exam. I had issues with solving for current and voltage of a double battery circuit.
> [{quoted}](name=How Do You Meta,realm=NA,application-id=yrc23zHg,discussion-id=v3Lcf4wR,comment-id=00000000000100000000,timestamp=2019-03-24T18:49:21.818+0000) > > Ill have to take a look at my exam. I had issues with solving for current and voltage of a double battery circuit. I had a feeling it would be Kirchoff's Laws :) I don't quite remember them that well, but essentially all you need to know is that you can split the circuit into two loops. Then, (and this is the part I'm kind of fuzzy), the sum of the potential differences across each section of the circuit should add to zero; or, if you recognize that the EMF of the battery is negative relative to the resistors, the sum of the potentials around the loop should be the EMF of the battery on that loop, exactly like it is for a single circuit. There's a similar rule for current, and I think resistors. This results in a system of linear equations. You could solve it by solving for each variable individually, but you can also just make a matrix equation, which is typically more useful for really complicated systems. Edit: There is no rule for resistors, but you should be able to obtain a third equation using the first two. At the end of the day, if you have n unknowns, you need n equations to solve it.
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=v3Lcf4wR,comment-id=00000000,timestamp=2019-03-24T18:29:16.637+0000) > > OOF! That's rough, buddy. If it makes you feel better, the class average on my Statistical Mechanics final was a 69. I might have some tips and tricks, but do you have anything specific in mind? We are all struggling with electromagnetism
> [{quoted}](name=How Do You Meta,realm=NA,application-id=yrc23zHg,discussion-id=v3Lcf4wR,comment-id=000000000001,timestamp=2019-03-24T18:45:55.317+0000) > > We are all struggling with electromagnetism Alright, do you have anything specific you want cleared up from there? I'm not quite sure what your class has covered.
: > 69 Nice
> [{quoted}](name=Timethief49,realm=EUW,application-id=yrc23zHg,discussion-id=v3Lcf4wR,comment-id=000000000000,timestamp=2019-03-24T18:36:27.247+0000) > > Nice https://gyazo.com/29b045321201bff62dcb09911225dd6b Closer to 70, but we'll use a floor function ;)
: Do you have physics II tips and tricks? Our class average is a 55 :/ BTW I LOVE CALCULUS! It's a fun course because it's all about the language.
> [{quoted}](name=How Do You Meta,realm=NA,application-id=yrc23zHg,discussion-id=v3Lcf4wR,comment-id=0000,timestamp=2019-03-24T18:27:30.851+0000) > > Do you have physics II tips and tricks? Our class average is a 55 :/ > BTW I LOVE CALCULUS! It's a fun course because it's all about the language. OOF! That's rough, buddy. If it makes you feel better, the class average on my Statistical Mechanics final was a 69. I might have some tips and tricks, but do you have anything specific in mind?
Rioter Comments
: Riot, In case you still don't know, after several years...
Now that I think about it, I might be able to explain this. What’s most likely happening here is that the percentage shows what percentage of files have been downloaded, not the actual status of the update. Basically, once it hits 100% of files downloaded, it then takes forever to actually apply them. Either that, or once it hits 100%, it then has to go through a bunch of shutdown checks or something. Either way it’s poorly designed, though...
: his argment is blatently wrong, and i honestly feel bad that you spent the time to type all this out for that sad soul
> [{quoted}](name=HiredGûn,realm=NA,application-id=3ErqAdtq,discussion-id=EsXj1vG3,comment-id=00010003000000010001,timestamp=2019-03-09T00:49:17.274+0000) > > his argment is blatently wrong, and i honestly feel bad that you spent the time to type all this out for that sad soul I wouldn’t say his argument is blatantly wrong, though at the end of the day he is making a huge mistake (and no, I’m not talking about how many times he made “you are” possessive). He only discussed pyke’s engage in the context of his E, and ignored his W. Pyke is (semi-) punishable after his E (though you have to actually get away from the stun, which a good Pyke can use to disengage if he misses), but if his W is up and he gets enough space he’s gone. Any damage you did will be mostly healed up, and he’ll come back in 15 seconds. Herein lies the problem: both Pyke’s E and W can function as engage and disengage, and as long as they’re both up when he goes in, he’s not really punishable without hard CC (and even then it can be tricky).
: The funny thing is there was that whole thing with riot august saying the boards are useless since there is so much bandwagons and look at this someone makes a post about a ok champion being boken with 0 proof and I make a post saying he is wrong then people say im wrong while giving reason that make no sense. He was right the people on the board will just follow hate even if it makes no sense.
> [{quoted}](name=boricCentaur1,realm=NA,application-id=3ErqAdtq,discussion-id=EsXj1vG3,comment-id=000100030000,timestamp=2019-03-08T14:40:01.405+0000) > > The funny thing is there was that whole thing with riot august saying the boards are useless since there is so much bandwagons and look at this someone makes a post about a ok champion being boken with 0 proof and I make a post saying he is wrong then people say im wrong while giving reason that make no sense. He was right the people on the board will just follow hate even if it makes no sense. Hello, professional calculus here, I thought I might shed some light on why people downvoted you. 1) Formatting: Your original post has quite a few grammar mistakes. Most people can look past something like usage (wrong “to”, “there”, or “your”), but when you post what could easily be 3-4 sentences with two periods and no commas, it’s going to sound like you’re rambling (even if you’re not). Now, this shouldn’t be a factor in your downvotes since this would normally lead to people just ignoring your comment (or being sassy and asking you to speak English), but points (2), (3), and (4), coupled with the fact that your post is readable (just not pleasantly readable), it actually plays a factor. 2) Attitude: Your attitude in many of your comments comes off as... less than pleasant. I’m on a phone, so I’m only going to pick apart your first comment since it’s the one I remember best and garnered the most downvotes. You start off with “Ummm, no” which is already enough to piss some people off, but then your argument basically boils down to “lol, what’s wrong with you, just learn how to play against him.” At this point, you’ve already done enough damage to make people want to stop reading and downvote you, even if anything afterwards is completely factual. Even if this wasn’t enough to make people downvote you on its own, pairing it with point (1) makes for a deadly combo. 3) Blind Downvoting: Yes, this can happen. People agree with something so much, they try to downplay anyone who disagrees with them. The only way to combat this is by writing a well founded, cohesive argument... which has been thrown out the window by points (1) and (2). 4) Aggressive Diction and Punctuation: Phrases like “you’re stupid” offer nothing to an argument and are just begging for downvotes. Not to mention some of your comments have phrases in all caps with several exclamation marks following them. This just makes it seem like you’re yelling insults at them, and when was the last time you actually listened to someone who was only shouting insults? With all of these combined, it’s not hard to see how you garnered so many downvotes, regardless of how correct your argument may have been. Edit: Damn autocorrect kept changing “downvotes” to “downvoted”. That, or my thumbs are too fat. Either way, let me know if you see any errors.
: Story behind your league names?
I like Calculus just a little too much.
: Let me rephrase. Do you think it's possible to find a formula to predict the Nth prime number?
> [{quoted}](name=PrismalDawn,realm=NA,application-id=yrc23zHg,discussion-id=AETPnRux,comment-id=0002,timestamp=2019-02-05T06:53:52.205+0000) > > Let me rephrase. > > Do you think it's possible to find a formula to predict the Nth prime number? I... genuinely don't know. I'm a physics graduate student, not a math graduate student. I just like vector spaces and differential algebra/geometry, sorry :(
Hotarµ (NA)
: how do I save my parent's marriage
> [{quoted}](name=Hotarµ,realm=NA,application-id=yrc23zHg,discussion-id=AETPnRux,comment-id=0000,timestamp=2019-02-05T05:20:10.374+0000) > > how do I save my parent's marriage Invest in a physics PhD, discover time travel, and wing it. {{sticker:slayer-jinx-wink}}
Rioter Comments
Ligseo (EUW)
: If you still want some: here is mine: You have 9 pennies Each are visually identical One is a fake The fake one is either lighter or heavier You have a Balance scale to check them Can you identify the fake one with only 3 uses of the balance?
> [{quoted}](name=Ligseo,realm=EUW,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0009,timestamp=2019-01-02T01:19:35.145+0000) > > If you still want some: here is mine: > > You have 9 pennies > Each are visually identical > One is a fake > The fake one is either lighter or heavier > > You have a Balance scale to check them > > Can you identify the fake one with only 3 uses of the balance? UUUUUGH I just spent an hour and a half trying to find the answer. I remember solving this once, but I forgot what the solution was.
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000000050000,timestamp=2019-01-02T20:10:12.735+0000) > > Hoooo boy. Well, there's a LOT of interesting things that involve i. Quantum mechanics is a pretty big one, since the time evolution operator in the Schrodinger picture looks like e^{-iEt/hbar}, where E is the energy of an energy eigenstate and hbar is Planck's constant divided by 2\*pi. > > Another cool thing to note is that a + b\*i is a vector in the complex plane, meaning the "number" i can be used to define a vector space. However, I'm going to choose my personal favorite application: > > Claim: e^{i\*pi} = -1 > > Proof: First, recall the taylor expansion of e^x as SUM{n=0,infty} (x^n)/(n!). Now recall the taylor expansions of sine and cosine: > > sin(x) = SUM{n=0,infty} (-1)^(n)\*x^(2n+1)/(2n+1)! > cos(x) = SUM{n=0,infty} (-1)^(n)\*x^(2n)/(2n)! > > *If you don't know what a Taylor Series is, it's basically an infinite sum that exactly equals the function in question, though it can be truncated to estimate the function to arbitrary accuracy* > > Note that, if we add the Taylor expansion of cos(x) and sin(x) together, we ALMOST get the Taylor Series for e^x. However, we're off by some minus signs. Using the "number" i, it's pretty easy to check that: > > e^(ix) = cos(x) + i\*sin(x) (sub in ix for x in the Taylor expansion of e^x and see what happens). > > Therefore, e^(i\*pi) = cos(pi) + i\*sin(pi) = -1 + 0 = -1. > > Corollary: ln(-1) = i\*pi I don't get any of this. I'm 15 and I'm learning Algebra 2.
> [{quoted}](name=Monstermushmush,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000000500000000,timestamp=2019-01-02T22:58:58.607+0000) > > I don't get any of this. I'm 15 and I'm learning Algebra 2. Well, the only thing we need to understand here is a Taylor Series. Consider this series: 1 + 2 + 3 + 4 + 5 + ... This is an infinite series that doesn't converge to any value (it has no solution). It can be expressed as: S = SUM{n=1,infty} n In other words, start at n = 1 and add n + (n+1) + (n+2) + ... A TAYLOR series uses calculus to find a polynomial that represents a function. The exponential function can be represented by: e^x = 1 + x + (x^2)/2 + (x^3)/6 + ... Here's a plot to compare them: https://gyazo.com/828205b36a0ba249a8680aefd36dbe29 And so on. If you're still with me, try writing out a few terms of the sine and cosine series representations I gave you. Then, multiply the sine series by i and add it to the cosine series. Next, take the terms of the e^x series above, sub in (ix) for x, and compare :).
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00110000,timestamp=2019-01-02T21:46:33.120+0000) > > Assume both cubes have the same density (mass per volume). The initial mass of the system is: > > rho*(X^3 + Y^3) > > Where rho is the density mentioned above. Suppose the new side length of the second cube is Y + M. We seek M: > > rho\*(Y+M)^3 + rho\*(X-N)^3 = rho\*(X^3 + Y^3) (total mass doesn't change) > > => M = [X^3 + Y^3 - (X-N)^3]^{1/3} - Y > > This formula assumes that the volume of the solid in question can be exactly written as the cube of one of its sides (in other words, it assumes the solid is a cube). A rectangular prism's volume cannot be written in this way, so THIS formula definitely does not apply. > > Question: Are the ratios of the sides of the rectangular prism ALWAYS the same, even after adding the mass? While we were forced to add the same length to each side of the cube to keep it a cube, we're able to change the sides of a rectangular prism any way we want (length wise, at least) and it would still be a rectangular prism. A similar technique should be applicable in the case of the prism, but we would need to know that the dimensions of the prisms are kept proportional after adding mass. For what I was thinking of, the proportions would be roughly humanoid - so about 7x tall, 3x wide, 1x deep. If I downscaled one prism so that it was X inches shorter, how much taller would I be able to make the other prism. I would want to keep the proportions the same, so if one prism started out 140cm tall, 60cm wide, 20cm deep and they became 70cm shorter I'd expect their dimensions to be 70cm, 30cm, 10cm. What I don't know is if I was to take that missing mass and add it to the other prism what it's new dimensions would be.
> [{quoted}](name=DrCyanide,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=001100000000,timestamp=2019-01-02T22:39:48.922+0000) > > For what I was thinking of, the proportions would be roughly humanoid - so about 7x tall, 3x wide, 1x deep. If I downscaled one prism so that it was X inches shorter, how much taller would I be able to make the other prism. > > I would want to keep the proportions the same, so if one prism started out 140cm tall, 60cm wide, 20cm deep and they became 70cm shorter I'd expect their dimensions to be 70cm, 30cm, 10cm. What I don't know is if I was to take that missing mass and add it to the other prism what it's new dimensions would be. If all you did was make the prism 70cm shorter, how are you changing the dimensions of the width and depth? Are you saying to halve every dimensions? I was think you could have dimensions A x B x C and remove N, M, and P, respectively. If the ratios of the prisms are kept the same (in other words, the ratio of their heights is X:Y), and the ratio is kept the same after adding the mass to the other one? Actually, I don't think that would be possible if we did it for all three sides. Actually, if we can only add mass to the second, then the ratios can't be kept constant for any side. At best, we could flip the ratio to be Y:X. The biggest questions that need to be addressed here are: 1) HOW are we taking mass off the first prism? 2) WHERE are we putting this mass on the second prism? In the cube question, the answers were easy: 1) Same amount taken off each side. 2) Same amount put on each side. The question doesn't exactly give us enough information to work with, here.
5 V (NA)
: Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.
> [{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000) > > Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot. Boop
: A bit of an odd ball problem to throw at you, but I figure you might enjoy it. As you probably know, going from a cube with sides of X to a cube with sides of 2\*X, the mass doesn't increase 2 times, but rather 8 times. Similarly, going to sides of 0.5\*X makes the mass 1/8th. Lets say you had 2 cubes, one with sides of length X, the other length Y. If you decrease the side of the first cube by N and added that mass to the other cube, what formula would describe how much longer the other cube's sides are now (M)? Can this formula be adapted to two rectangular prisms, assuming the prisms are proportional in their dimensions?
> [{quoted}](name=DrCyanide,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0011,timestamp=2019-01-02T20:50:59.042+0000) > > A bit of an odd ball problem to throw at you, but I figure you might enjoy it. > > As you probably know, going from a cube with sides of X to a cube with sides of 2\*X, the mass doesn't increase 2 times, but rather 8 times. Similarly, going to sides of 0.5\*X makes the mass 1/8th. > > Lets say you had 2 cubes, one with sides of length X, the other length Y. If you decrease the side of the first cube by N and added that mass to the other cube, what formula would describe how much longer the other cube's sides are now (M)? > Can this formula be adapted to two rectangular prisms, assuming the prisms are proportional in their dimensions? Assume both cubes have the same density (mass per volume). The initial mass of the system is: rho*(X^3 + Y^3) Where rho is the density mentioned above. Suppose the new side length of the second cube is Y + M. We seek M: rho\*(Y+M)^3 + rho\*(X-N)^3 = rho\*(X^3 + Y^3) (total mass doesn't change) => M = [X^3 + Y^3 - (X-N)^3]^{1/3} - Y This formula assumes that the volume of the solid in question can be exactly written as the cube of one of its sides (in other words, it assumes the solid is a cube). A rectangular prism's volume cannot be written in this way, so THIS formula definitely does not apply. Question: Are the ratios of the sides of the rectangular prism ALWAYS the same, even after adding the mass? While we were forced to add the same length to each side of the cube to keep it a cube, we're able to change the sides of a rectangular prism any way we want (length wise, at least) and it would still be a rectangular prism. A similar technique should be applicable in the case of the prism, but we would need to know that the dimensions of the prisms are kept proportional after adding mass.
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0004000000010000,timestamp=2019-01-02T20:11:29.466+0000) > > DAMMIT, I FORGOT HOW TO DO FLOATING POINT CALCULATIONS! AAAAAAGH! > > Also, I don't really have a GitHub link, if that's what you're asking for here. I took a 1 hour "programming for physicists" course at my university, where we were taught basic python. In the class, we had assignments to turn in, so I would look up the packages I needed to understand and look at examples they had. I basically copied enough examples to learn the language, and then started winging everything else xD. Just tell me whether it's 0.3 or not. :^) I meant some sample of programs you've written, that sort of thing. Do you only write specific, one-off scripts for whatever physics problem you're solving at the moment? Not that there's anything wrong with that, of course. I mean, my dad programs all day for work, and yet has no GitHub or other outside programming presence; it's not a hobby for him. He actually just learned git a few weeks ago so he can do some basic version control on his local machine and hopefully never again have to debug obfuscated source code after his boss accidentally saves over a release version of his code. :)
> [{quoted}](name=KFCeytron,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00040000000100000000,timestamp=2019-01-02T20:44:03.882+0000) > > Just tell me whether it's 0.3 or not. :^) > > I meant some sample of programs you've written, that sort of thing. Do you only write specific, one-off scripts for whatever physics problem you're solving at the moment? Not that there's anything wrong with that, of course. I mean, my dad programs all day for work, and yet has no GitHub or other outside programming presence; it's not a hobby for him. He actually just learned git a few weeks ago so he can do some basic version control on his local machine and hopefully never again have to debug obfuscated source code after his boss accidentally saves over a release version of his code. :) Yeah, it's 0.3. If you PM me your e-mail, I can send you some sample codes I've written. Be warned: none of them are commented (except maybe a bit). EDIT: I found something to remind me of how to convert to binary: 0.1\*2 = 0.2: 0 0.2\*2 = 0.4: 0 0.4\*2 = 0.8: 0 0.8\*2 = 1.6: 1 0.6\*2 = 1.2: 1 0.2\*2 = 0.4: 0 0.4\*2 = 0.8: 0 0.8\*2 = 1.6: 1 (we've reached 8 bits, but this could continue infinitely, suggesting that 0.1 can't be exactly represented in binary). 0.2\*2 = 0.4: 0 0.4\*2 = 0.8: 0 0.8\*2 = 1.6: 1 0.6\*2 = 1.2: 1 0.2\*2 = 0.4: 0 0.4\*2 = 0.8: 0 0.8\*2 = 1.6: 1 0.6\*2 = 1.2: 1 Therefore, in binary, 0.1 + 0.2 is given by 0.00011001 + 0.00110011 = 0.01001100 = 2^{-2} + 2^{-5} + 2^{-6} = 0.296875, which is almost 0.3, but that's because 0.1 and 0.2 have infinite representations in binary.
Kythers (NA)
: how do i get a gf
> [{quoted}](name=Kythers,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0010,timestamp=2019-01-02T16:57:37.622+0000) > > how do i get a gf With the power of math!
5 V (NA)
: Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.
> [{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000) > > Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot. SURPRISE! My phone couldn't access the second page yesterday, so here's more responses.
Sundusk (NA)
: i dunno if youre still answering but like math problem an airplane travels west at 180 km/h and returns east with the jet stream at 300 km/h. what was the average speed in km/h for the whole trip? math homework over break woO
> [{quoted}](name=Sundusk,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000b,timestamp=2019-01-02T06:22:18.200+0000) > > i dunno if youre still answering > but like > math problem > an airplane travels west at 180 km/h and returns east with the jet stream at 300 km/h. what was the average speed in km/h for the whole trip? > math homework over break woO There's a nice and easy formula for doing this, but let's derive it! A quick google search will result in this image: https://www.onlinemath4all.com/images/averagetspeed1.png Let's prove it: Proof: First note that average speed is defined as total distance divided by total time. Suppose the aircraft travels a distance d in both directions (which I assume we are able to do since it, if it doesn't start and end in the same place, this problem becomes much more complicated, and I'm not even sure if you can solve it). Anyways, we have: S_{avg} = 2d/T Where T = t_1 + t_2, t_1 is the amount of time spent flying at speed x, and t_2 the amount of time spent flying at speed y. Therefore, t_1 = d/x, and t_2 = d/y. So: S_{avg} = 2d/(d/x + d/y) = 2/(1/x + 1/y) = 2/((x+y)/xy) = 2xy/(x+y).
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000000000000,timestamp=2019-01-01T21:43:15.060+0000) > > (it’s how I learned to program in python) GitHub/etc. link? > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00040000,timestamp=2019-01-01T23:03:08.772+0000) > > We can make this problem more complicated by converting to base 2. In base 2, 9 = 2^0 + 2^3, written in binary as 00001001. Ten is given by 00001010. Added together, this is 00010011. This is 2^0 + 2^1 + 2^4 = 16 + 2 + 1 = 19. This is how your computer would have calculated it. How about 0.1 + 0.2? :^)
> [{quoted}](name=KFCeytron,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000400000001,timestamp=2019-01-02T06:04:56.699+0000) > > GitHub/etc. link? > > How about 0.1 + 0.2? :^) DAMMIT, I FORGOT HOW TO DO FLOATING POINT CALCULATIONS! AAAAAAGH! Also, I don't really have a GitHub link, if that's what you're asking for here. I took a 1 hour "programming for physicists" course at my university, where we were taught basic python. In the class, we had assignments to turn in, so I would look up the packages I needed to understand and look at examples they had. I basically copied enough examples to learn the language, and then started winging everything else xD.
: > [{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000) > Idk how to quote. Anyways, what is the point of the imaginary number (i)? It doesn't exist in real life.
> [{quoted}](name=Monstermushmush,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00000005,timestamp=2019-01-02T05:44:58.473+0000) > > Idk how to quote. Anyways, what is the point of the imaginary number (i)? It doesn't exist in real life. Hoooo boy. Well, there's a LOT of interesting things that involve i. Quantum mechanics is a pretty big one, since the time evolution operator in the Schrodinger picture looks like e^{-iEt/hbar}, where E is the energy of an energy eigenstate and hbar is Planck's constant divided by 2\*pi. Another cool thing to note is that a + b\*i is a vector in the complex plane, meaning the "number" i can be used to define a vector space. However, I'm going to choose my personal favorite application: Claim: e^{i\*pi} = -1 Proof: First, recall the taylor expansion of e^x as SUM{n=0,infty} (x^n)/(n!). Now recall the taylor expansions of sine and cosine: sin(x) = SUM{n=0,infty} (-1)^(n)\*x^(2n+1)/(2n+1)! cos(x) = SUM{n=0,infty} (-1)^(n)\*x^(2n)/(2n)! *If you don't know what a Taylor Series is, it's basically an infinite sum that exactly equals the function in question, though it can be truncated to estimate the function to arbitrary accuracy* Note that, if we add the Taylor expansion of cos(x) and sin(x) together, we ALMOST get the Taylor Series for e^x. However, we're off by some minus signs. Using the "number" i, it's pretty easy to check that: e^(ix) = cos(x) + i\*sin(x) (sub in ix for x in the Taylor expansion of e^x and see what happens). Therefore, e^(i\*pi) = cos(pi) + i\*sin(pi) = -1 + 0 = -1. Corollary: ln(-1) = i\*pi
: Not really a question but a math video I found really interesting: https://youtu.be/DGpwkWhnWAI Thought I'd share it.
> [{quoted}](name=Triistana,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000a,timestamp=2019-01-02T04:21:02.010+0000) > > Not really a question but a math video I found really interesting: > > https://youtu.be/DGpwkWhnWAI > > Thought I'd share it. I, too, have a math video I love. Thanks for sharing: https://www.youtube.com/watch?v=_s5RFgd59ao
: > [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00040000,timestamp=2019-01-01T23:03:08.772+0000) > > We can make this problem more complicated by converting to base 2. In base 2, 9 = 2^0 + 2^3, written in binary as 00001001. Ten is given by 00001010. Added together, this is 00010011. This is 2^0 + 2^1 + 2^4 = 16 + 2 + 1 = 19. This is how your computer would have calculated it. fucking right, count. ######im a count too, how's your day, brother?
> [{quoted}](name=Ćount Kledula,realm=EUW,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000400000000,timestamp=2019-01-02T00:27:27.393+0000) > > fucking right, count. > > ######im a count too, how's your day, brother? It was pretty good. I actually couldn't access this page of responses on my phone for some reason, so I'm responding to them now.
: If 3x−y=12, what is the value of 8x2y? A) 212 B) 44 C) 82 D) The value cannot be determined from the information given.
> [{quoted}](name=the shy top,realm=OCE,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000c,timestamp=2019-01-02T10:45:03.347+0000) > > If 3x−y=12, what is the value of 8x2y? > > A) 212 > B) 44 > C) 82 > D) The value cannot be determined from the information given. D. You have one equation and two unknowns. 3x - y = 12 gives you information about y in terms of x, but to solve 8 \* 2y (I'm assuming your x means multiplication and is not the variable x), you would need to have a second equation that can be used to find a value for x (either as a number, or in terms of y). Then, you would be able to solve for a numerical value of y.
: I sat down and tried to model the equation algebraically, but I didn't get very far before I got stumped. One thing I find interesting is that with the nature of odd numbers, the list you give will (if I'm not mistaken) give an odd number if every number in the sequence is added up. What I'm still a bit stumped about is how you can choose any number and make the ABSOLUTE VALUE of the difference, making it always be positive. I'm going to model a few more and see what other properties I can find. Certainly an interesting problem. Regarding my problem, 3blue 1brown has a great video on the problem, and explains it better than I ever could, and gives how one would find the solution. I would recommend watching it :D https://www.youtube.com/watch?v=OkmNXy7er84
> [{quoted}](name=Pile Of Pillows,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000300010000,timestamp=2019-01-01T23:38:48.187+0000) > > I sat down and tried to model the equation algebraically, but I didn't get very far before I got stumped. One thing I find interesting is that with the nature of odd numbers, the list you give will (if I'm not mistaken) give an odd number if every number in the sequence is added up. What I'm still a bit stumped about is how you can choose any number and make the ABSOLUTE VALUE of the difference, making it always be positive. I'm going to model a few more and see what other properties I can find. Certainly an interesting problem. > Regarding my problem, 3blue 1brown has a great video on the problem, and explains it better than I ever could, and gives how one would find the solution. I would recommend watching it :D > https://www.youtube.com/watch?v=OkmNXy7er84 I'm not sure what you mean by "What I'm still a bit stumped about is how you can choose any number and make the ABSOLUTE VALUE of the difference, making it always be positive", but if I'm understanding your question correctly: we take the absolute value of the difference to guarantee we add a positive number to the list. I'm actually not sure if that's even necessary, based on the solution I found, but essentially, what we do is: 1) Take two numbers 2) Subtract one from the other 3) If the resulting value is negative, multiply by negative 1. If it's not, leave it alone That's how we take the absolute value. Also, HINT: Consider the number of odd numbers in the list :)
: I’m waiting on my flight at the airport, and I love math.
Alright guys, the plane is about to board. Thanks for the questions!
5 V (NA)
: Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.
> [{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000) > > Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot. Okay, I think I quoted you this time.
: I’m waiting on my flight at the airport, and I love math.
> [{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=,timestamp=2019-01-01T21:22:26.676+0000) > > Ask me any (math) problem! Not the millenium problems, though, if I had the solutions to those I wouldn’t be waiting at an airport. Edit: this is not the person I meant to quote xD
: Is mayonnaise an instrument?
> [{quoted}](name=Bob the Toastr,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0005,timestamp=2019-01-01T23:00:29.624+0000) > > Is mayonnaise an instrument? No, Patrick, mayonnaise is not an instrument. ... Horseradish isn’t an instrument either.
: what's nine plus ten?
> [{quoted}](name=Ćount Kledula,realm=EUW,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0004,timestamp=2019-01-01T22:53:05.907+0000) > > what's nine plus ten? We can make this problem more complicated by converting to base 2. In base 2, 9 = 2^0 + 2^3, written in binary as 00001001. Ten is given by 00001010. Added together, this is 00010011. This is 2^0 + 2^1 + 2^4 = 16 + 2 + 1 = 19. This is how your computer would have calculated it.
: You might have seen this one, but it is one of my favorites :D Suppose you have a sphere of any volume but 0, and create 4 randomly placed points on the surface of the sphere. If you draw lines from the points to the inside of the sphere creating a tetrahedron, what are the exact chances that the tetrahedron made would encapsulate the center of the sphere? From the Putnam, took me about 3 weeks to solve for myself, but I'm not some savant lol. If you can get a piece of paper, drawing this out really helps. Good luck!
> [{quoted}](name=Pile Of Pillows,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0003,timestamp=2019-01-01T22:47:22.129+0000) > > You might have seen this one, but it is one of my favorites :D > Suppose you have a sphere of any volume but 0, and create 4 randomly placed points on the surface of the sphere. If you draw lines from the points to the inside of the sphere creating a tetrahedron, what are the exact chances that the tetrahedron made would encapsulate the center of the sphere? > From the Putnam, took me about 3 weeks to solve for myself, but I'm not some savant lol. If you can get a piece of paper, drawing this out really helps. Good luck! I don’t have paper and I don’t think I’ll be able to answer your question, but I do have a fun Putnam question of my own! Choose any odd number and double it. Make a list of 1 through that number (ex: choose 3, take list 1, 2, 3, 4, 5, 6). Choose any two numbers from the list and take the absolute value of their difference. Remove the two numbers you chose from the list and replace them with the difference you calculated. Repeat this until you only have one number remaining. Claim: the number will always be odd. Provide the proof :) Demonstration: 1, 2, 3, 4, 5, 6 Choose 1,2: 2-1 = 1, new list is 1, 3, 4, 5, 6 Choose 5,6: 6-5 = 1, new list is 1, 1, 3, 4 Choose 1,1: 1-1 = 0, new list is 0, 3, 4 Choose 0,3: 3-0 = 3, new list is 3, 4 Choose 3,4: 4-3 = 1, remaining number is 1.
5 V (NA)
: Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.
> [{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000) > > Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot. Someone asked a math question, so here’s your ping.
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Count Calculus

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