Delightful Posted May 24, 2017 Report Share Posted May 24, 2017 28 minutes ago, StrikerEZ said: When you work out, you burn fat and calories, right? What Allomantic powers do they give? energy! 1 Quote Link to comment Share on other sites More sharing options...
Orlion Blight he/him Posted May 24, 2017 Report Share Posted May 24, 2017 1 hour ago, StrikerEZ said: When you work out, you burn fat and calories, right? What Allomantic powers do they give? In my case, you end up with bloody feet! (Yeah, I'm confused too) 0 Quote Link to comment Share on other sites More sharing options...
The Honor Spren she/her Posted May 24, 2017 Report Share Posted May 24, 2017 (edited) 4 hours ago, Darkness Ascendant said: And I'm pretty sure this is a Synagogue. @Delightfully Smoak comfirmation? Reveal hidden contents I think that's a catholic cathedral/church building. I'm not quite sure what to call it. Edited May 24, 2017 by The Honor Spren 0 Quote Link to comment Share on other sites More sharing options...
The Sovereign Posted May 24, 2017 Report Share Posted May 24, 2017 (edited) 20 minutes ago, The Honor Spren said: I think that's a catholic cathedral/church building. I'm not quite sure what to call it. It is indeed, Roman Catholic be specific. It is Gothic architecture. I believe it is actually a photo of the Reims Kathedrale.@Darkness Ascendant, did you take this in Reims, France? Edited May 24, 2017 by The Sovereign 0 Quote Link to comment Share on other sites More sharing options...
Left he/him Posted May 24, 2017 Report Share Posted May 24, 2017 When you post selfies with windswept hair and a sweaty complexion after a 5k run on the beach, but don't mention the huge amount of food you had for lunch or the third of a pack of oreos devoured 1 Quote Link to comment Share on other sites More sharing options...
Hemalurgic Headshot he/him Posted May 24, 2017 Report Share Posted May 24, 2017 I've had bedhead the whole day and have done nothing to fix it. Living the dream. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 he/him Posted May 24, 2017 Report Share Posted May 24, 2017 9 hours ago, Chaos said: That sounds pretty applied. I dunno, my E&M is really poor and I don't know anything about EE, sorry. Actual math question: I was recently playing around with derivatives, and noticed that while the functions y = ln(x-1) and y = ln(1-x) are different functions with different outputs, they both have the same derivative, which is 1/(x-1). Why is this? 0 Quote Link to comment Share on other sites More sharing options...
Sunbird she/her Posted May 24, 2017 Report Share Posted May 24, 2017 @Darkness Ascendant Love your photos! I wish I could upvote them all individually. 0 Quote Link to comment Share on other sites More sharing options...
Kaymyth she/her Posted May 24, 2017 Report Share Posted May 24, 2017 23 hours ago, Mestiv said: I don't know why you call it purple is blue on all my devices 23 hours ago, The Sovereign said: Gotta go with Mestiv here, it does look blue. Then again, you have way more cones in your eyes than we do, so we should probably just defer to your superior color identification skills. 23 hours ago, bleeder said: I'm pretty sure it's blue, but I don't know how many Breaths you have, so I'll defer to you on this one. I'm just going to go with, "the color balance on my monitor at work is probably off." It really does look purple over here; definitely a blue-heavy purple, almost more of an indigo, but it just doesn't look blue on my screen. Either way, blue is still a superior color to mustard yellow. Not quite as superior as purple, but it is definitely a good color. Maybe somebody will be kind on my birthday later this summer and overlay my Moderator rank with rainbow sparkles for a day. 17 hours ago, Oversleep said: I know it's blue because I tapped into wire magic and looked at the source of the website. Here's what it says: Wait... I can make a meme out of this... EDIT: And I did. I do believe this is the first time I have ever been featured in a meme. I feel honored, and slightly giggly. 0 Quote Link to comment Share on other sites More sharing options...
Mestiv he/him Posted May 24, 2017 Report Share Posted May 24, 2017 14 minutes ago, Kaymyth said: Maybe somebody will be kind on my birthday later this summer and overlay my Moderator rank with rainbow sparkles for a day. Hmm, if only I knew someone who knew how to do that... so when's your birthday exactly? 0 Quote Link to comment Share on other sites More sharing options...
Oversleep Posted May 24, 2017 Report Share Posted May 24, 2017 12 hours ago, Delightfully Smoak said: What's the origin of these images? That's my second Expanding Brain meme (the first was about contractions few pages back). I take them from other memes. Those other memes take them from older memes... I'm not sure where the images for the first Expanding Brain memes came from. Some sources suggest they were taken from various articles on transhumanism. 0 Quote Link to comment Share on other sites More sharing options...
Mestiv he/him Posted May 24, 2017 Report Share Posted May 24, 2017 It's my 2000th post ^^ hurray 3 Quote Link to comment Share on other sites More sharing options...
Exalted Posted May 24, 2017 Report Share Posted May 24, 2017 4 hours ago, Silverblade5 said: Actual math question: I was recently playing around with derivatives, and noticed that while the functions y = ln(x-1) and y = ln(1-x) are different functions with different outputs, they both have the same derivative, which is 1/(x-1). Why is this? I don't know how much you know about derivatives, so I'll try my best to explain (and apologies if I take happen to detail something you already knew). When you take the derivative of a function that can be described as f(g(x)) involving the sub-functions f(x) and g(x), you obtain the result using something called the chain rule. The derivative of any such function is f'(g(x))*g'(x). For your first example, if we define f(x) as ln(x) and g(x) as (x-1), then f'(g(x))*g'(x) is (1/(x-1))*(1), or 1/(x-1). For your second example, if we define f(x) as ln(x) and g(x) as (1-x), then f'(g(x))*g'(x) is (1/(x-1))*(-1), where we distribute the negative to the denominator of the fraction in order to swap the terms, leaving us with 1/(x-1), which is the same as the above answer. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 he/him Posted May 24, 2017 Report Share Posted May 24, 2017 1 hour ago, Exalted Dungeon Master said: I don't know how much you know about derivatives, so I'll try my best to explain (and apologies if I take happen to detail something you already knew). When you take the derivative of a function that can be described as f(g(x)) involving the sub-functions f(x) and g(x), you obtain the result using something called the chain rule. The derivative of any such function is f'(g(x))*g'(x). For your first example, if we define f(x) as ln(x) and g(x) as (x-1), then f'(g(x))*g'(x) is (1/(x-1))*(1), or 1/(x-1). For your second example, if we define f(x) as ln(x) and g(x) as (1-x), then f'(g(x))*g'(x) is (1/(x-1))*(-1), where we distribute the negative to the denominator of the fraction in order to swap the terms, leaving us with 1/(x-1), which is the same as the above answer. I understand that completely. What I don't understand is this: I was always told that only one closed form integral existed for a continuous function, excluding the constant of integration. What confuses me is that two functions, which have different graphs from each other, could both be valid answers for S|1/(x-1)|dx. 0 Quote Link to comment Share on other sites More sharing options...
Darkness Ascendant he/him Posted May 24, 2017 Report Share Posted May 24, 2017 Thanks guys! Mm our teacher called it a synagogue and I was confused ha 0 Quote Link to comment Share on other sites More sharing options...
Orlion Blight he/him Posted May 25, 2017 Report Share Posted May 25, 2017 (edited) 36 minutes ago, Darkness Ascendant said: Thanks guys! Mm our teacher called it a synagogue and I was confused ha It's time to teach...*takes out a lead pipe* the teacher! *sets lead pipe aside and replaces it with a polymer pipe. What? I was doing some much needed plumbing renovation whilst commenting. It's called multitasking! Oh cruds, I punctured the main water line!* Edited May 25, 2017 by Orlion On a Cob 0 Quote Link to comment Share on other sites More sharing options...
Exalted Posted May 25, 2017 Report Share Posted May 25, 2017 4 hours ago, Silverblade5 said: I understand that completely. What I don't understand is this: I was always told that only one closed form integral existed for a continuous function, excluding the constant of integration. What confuses me is that two functions, which have different graphs from each other, could both be valid answers for S|1/(x-1)|dx. I don't have too much knowledge of calculus, so the most I can say is that the constant of integration does weird things when you mix it with logarithms (like how the derivative of ln(5x) = (1/x), but the integral of (1/x) = ln(x) + C. Technically, somehow, you can mess with the constant of integration to make the two graphs the same, and the same is probably true for the example you gave.) 0 Quote Link to comment Share on other sites More sharing options...
Chaos he/him Posted May 25, 2017 Report Share Posted May 25, 2017 11 hours ago, Silverblade5 said: I understand that completely. What I don't understand is this: I was always told that only one closed form integral existed for a continuous function, excluding the constant of integration. What confuses me is that two functions, which have different graphs from each other, could both be valid answers for S|1/(x-1)|dx. At first this had me turn my head, but the answer to your question is simple: |x-1| and |1-x| are the same function. You see, when you integrate 1/x, the antiderivative is not ln(x), but actually ln|x|. That absolute value is important because now those two functions have the same domains. That is essential here because when you integrate 1/(x-1) you get ln|x-1|. But really, by looking ln(x-1) vs. ln(1-x) they are the same function except for some domain issues. The absolute values fix that. ln|x-1| contains those smaller domained functions. Basically those two functions are the two halves of ln|x-1|. Of course, if you use Wolfram Alpha plotting you'll see that their graph of their real part (if you allow complex solutions) are exactly the same: http://www.wolframalpha.com/input/?i=y+%3D+ln(x-1) http://www.wolframalpha.com/input/?i=y+%3D+ln(1-x) 3 Quote Link to comment Share on other sites More sharing options...
The Honor Spren she/her Posted May 25, 2017 Report Share Posted May 25, 2017 My favorite youtube artist uploaded a song in response to the manchester bombing: Spoiler 0 Quote Link to comment Share on other sites More sharing options...
Exalted Posted May 25, 2017 Report Share Posted May 25, 2017 14 hours ago, Chaos said: At first this had me turn my head, but the answer to your question is simple: |x-1| and |1-x| are the same function. You see, when you integrate 1/x, the antiderivative is not ln(x), but actually ln|x|. That absolute value is important because now those two functions have the same domains. That is essential here because when you integrate 1/(x-1) you get ln|x-1|. But really, by looking ln(x-1) vs. ln(1-x) they are the same function except for some domain issues. The absolute values fix that. ln|x-1| contains those smaller domained functions. Basically those two functions are the two halves of ln|x-1|. Of course, if you use Wolfram Alpha plotting you'll see that their graph of their real part (if you allow complex solutions) are exactly the same: http://www.wolframalpha.com/input/?i=y+%3D+ln(x-1) http://www.wolframalpha.com/input/?i=y+%3D+ln(1-x) Ah, yes, the absolute value signs. I can never remember that those exist, as evidenced by the fact that I keep leaving them off on tests and losing points. 0 Quote Link to comment Share on other sites More sharing options...
Chaos he/him Posted May 25, 2017 Report Share Posted May 25, 2017 Just now, Exalted Dungeon Master said: Ah, yes, the absolute value signs. I can never remember that those exist, as evidenced by the fact that I keep leaving them off on tests and losing points. I would take one point out of ten for that Also remember your differentials!! Don't make me cry. 0 Quote Link to comment Share on other sites More sharing options...
Delightful Posted May 26, 2017 Report Share Posted May 26, 2017 On 25 May 2017 at 2:52 AM, Darkness Ascendant said: Thanks guys! Mm our teacher called it a synagogue and I was confused ha There is a synagogue roundabout that area, but thats not it I'm pretty sure. 0 Quote Link to comment Share on other sites More sharing options...
Sunbird she/her Posted May 26, 2017 Report Share Posted May 26, 2017 I acquired 13 mosquito bites last night... and I'm not even certain that I've found all of them yet. But on the plus side, I enjoyed roasting hot dogs and marshmallows, making smores, and seeing an American Kestrel going to her nest box repeatedly. 0 Quote Link to comment Share on other sites More sharing options...
Darkness Ascendant he/him Posted May 26, 2017 Report Share Posted May 26, 2017 26 minutes ago, Sunbirb said: I acquired 13 mosquito bites last night... and I'm not even certain that I've found all of them yet. But on the plus side, I enjoyed roasting hot dogs and marshmallows, making smores, and seeing an American Kestrel going to her nest box repeatedly. *winces mm you'd hate mosquitoes in aus XD And that sounds like a tonne of fun! Did you get many pics of the birbs? 1 Quote Link to comment Share on other sites More sharing options...
A Budgie she/her Posted May 26, 2017 Report Share Posted May 26, 2017 3 hours ago, Sunbirb said: I acquired 13 mosquito bites last night... and I'm not even certain that I've found all of them yet. But on the plus side, I enjoyed roasting hot dogs and marshmallows, making smores, and seeing an American Kestrel going to her nest box repeatedly. Sounds fun. Smores are amazing, and birbs going to nesting boxes sound cool. 0 Quote Link to comment Share on other sites More sharing options...
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