▲ 764 ▼ UwU brat mathematician behavior (mander.xyz) submitted 1 year ago by fossilesque@mander.xyz to c/science_memes@mander.xyz 125 comments fedilink hide all child comments
[–] Zagorath@aussie.zone 274 points 1 year ago (27 children) Fake and gay. No way the engineer corrects the mathematician for using j instead of i. permalink fedilink source hideshow 27 child comments replies: [–] LeFrog@discuss.tchncs.de 53 points 1 year ago (9 children) As an engineer I fully agree. Engineers¹ aren't even able to do basic arithmetics. I even cannot count to 10. ¹ Except maybe Electrical engineers. They seem to be quite smart. permalink fedilink source parent hideshow 9 child comments replies: [–] boonhet@sopuli.xyz 44 points 1 year ago (1 child) Engineer here, I can definitely count to 10 tho 0 1 10 permalink fedilink source parent hideshow 1 child comment replies: [–] gandalf_der_12te@discuss.tchncs.de 8 points 1 year ago 0 1 everything that comes after is simply summarizes as "many" permalink fedilink source parent [–] gnutrino@programming.dev 31 points 1 year ago (1 child) Electrical engineers are the ones that use j though (because i is used for current) permalink fedilink source parent hideshow 1 child comment replies: [–] Klear@sh.itjust.works 2 points 1 year ago I am used for current permalink fedilink source parent [–] thomasloven@lemmy.world 11 points 1 year ago 10? That’s the name some put to 1e1, right? permalink fedilink source parent [–] exasperation@lemmy.dbzer0.com 10 points 1 year ago* (1 child) Except maybe Electrical engineers. Yup, I can count just fine to the 10th number in a zero-indexed counting system: black, brown, red, orange, yellow, green, blue, violet, gray, white. permalink fedilink source parent hideshow 1 child comment replies: [–] alt_xa_23@lemmy.world 4 points 1 year ago https://xkcd.com/227/ permalink fedilink source parent [–] Baguette@lemmy.blahaj.zone 6 points 1 year ago The inner machinations of an electrical engineer is too complicated for me to understand, I think they might be thinking on a higher order to understand these circuits Thats why I barely passed my electrical engineering class lol permalink fedilink source parent [–] Fedizen@lemmy.world 5 points 1 year ago Having worked with electrical engineers, some of them are quite smart, the rest have lead poisoning. permalink fedilink source parent [–] TheSlad@sh.itjust.works 19 points 1 year ago (3 children) Right? They got that shit backwards. Op is a fraud. i is used in pure math, j is used in engineering. permalink fedilink source parent hideshow 3 child comments replies: [–] Chakravanti@monero.town 2 points 1 year ago (2 children) That's hilarious. You're not seeing what's going on backwards just like that (as I point at the point going nowhere shitty) in an equation that is finding as many clAEver ways to say something you actually not caring about talking about. That's like, "How many time van express the only thing that van't be done until the 'verse itself tries to do what can't be done and sever your... ...Oh, I see...you don't have ([of course, because you can't have to give {is}) nothing)] to give. Unable to sea time doesn't mean we can't see(k)ER the mAETh.ac(k).cc(k).08 The only thin(g):(k) that doesn't ever be never, is not at alla hack(g)in(g).G your lackthereof to divi... permalink fedilink source parent hideshow 2 child comments replies: [–] TheSlad@sh.itjust.works 3 points 1 year ago (1 child) Is this a copypasta or are you having a stroke? permalink fedilink source parent hideshow 1 child comment replies: [–] Chakravanti@monero.town 1 point 1 year ago Wah, wah, wah... Not my problem. Keep trying. /s permalink fedilink source parent [–] Hoimo@ani.social 37 points 1 year ago (3 children) How do we know it's gay though? OP could be a girl (male) permalink fedilink source parent hideshow 3 child comments replies: [–] SippyCup@feddit.nl 57 points 1 year ago Because it's 4chan. And there are no women on the Internet on 4chan permalink fedilink source parent [–] floquant@lemmy.dbzer0.com 17 points 1 year ago Sure OP is a girl. Guy In Real Life permalink fedilink source parent [–] ByteJunk@lemmy.world 10 points 1 year ago Newfag. (sorry! seemed like the appropriate 4chan reply) permalink fedilink source parent [–] kogasa@programming.dev 18 points 1 year ago (7 children) The mathematician also used "operative" instead of, uh, something else, and "associative" instead of "commutative" permalink fedilink source parent hideshow 7 child comments replies: [–] Zagorath@aussie.zone 3 points 1 year ago (6 children) "operative" instead of, uh, something else I think they meant "operand". As in, in the way dy/dx can sometimes be treated as a fraction and dx treated as a value. permalink fedilink source parent hideshow 6 child comments replies: [–] Chrobin@discuss.tchncs.de 1 point 1 year ago (5 children) I think you mean operator. The operand is the target of an operator. permalink fedilink source parent hideshow 5 child comments replies: [–] Zagorath@aussie.zone 2 points 1 year ago (4 children) The operand is the target of an operator Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy). permalink fedilink source parent hideshow 4 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (3 children) I'd say the $\int dx$ is the operator and the integrand is the operand. permalink fedilink source parent hideshow 3 child comments replies: [–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent [–] NeatNit@discuss.tchncs.de 12 points 1 year ago My thoughts exactly lol permalink fedilink source parent
[–] LeFrog@discuss.tchncs.de 53 points 1 year ago (9 children) As an engineer I fully agree. Engineers¹ aren't even able to do basic arithmetics. I even cannot count to 10. ¹ Except maybe Electrical engineers. They seem to be quite smart. permalink fedilink source parent hideshow 9 child comments replies: [–] boonhet@sopuli.xyz 44 points 1 year ago (1 child) Engineer here, I can definitely count to 10 tho 0 1 10 permalink fedilink source parent hideshow 1 child comment replies: [–] gandalf_der_12te@discuss.tchncs.de 8 points 1 year ago 0 1 everything that comes after is simply summarizes as "many" permalink fedilink source parent [–] gnutrino@programming.dev 31 points 1 year ago (1 child) Electrical engineers are the ones that use j though (because i is used for current) permalink fedilink source parent hideshow 1 child comment replies: [–] Klear@sh.itjust.works 2 points 1 year ago I am used for current permalink fedilink source parent [–] thomasloven@lemmy.world 11 points 1 year ago 10? That’s the name some put to 1e1, right? permalink fedilink source parent [–] exasperation@lemmy.dbzer0.com 10 points 1 year ago* (1 child) Except maybe Electrical engineers. Yup, I can count just fine to the 10th number in a zero-indexed counting system: black, brown, red, orange, yellow, green, blue, violet, gray, white. permalink fedilink source parent hideshow 1 child comment replies: [–] alt_xa_23@lemmy.world 4 points 1 year ago https://xkcd.com/227/ permalink fedilink source parent [–] Baguette@lemmy.blahaj.zone 6 points 1 year ago The inner machinations of an electrical engineer is too complicated for me to understand, I think they might be thinking on a higher order to understand these circuits Thats why I barely passed my electrical engineering class lol permalink fedilink source parent [–] Fedizen@lemmy.world 5 points 1 year ago Having worked with electrical engineers, some of them are quite smart, the rest have lead poisoning. permalink fedilink source parent
[–] boonhet@sopuli.xyz 44 points 1 year ago (1 child) Engineer here, I can definitely count to 10 tho 0 1 10 permalink fedilink source parent hideshow 1 child comment replies: [–] gandalf_der_12te@discuss.tchncs.de 8 points 1 year ago 0 1 everything that comes after is simply summarizes as "many" permalink fedilink source parent
[–] gandalf_der_12te@discuss.tchncs.de 8 points 1 year ago 0 1 everything that comes after is simply summarizes as "many" permalink fedilink source parent
[–] gnutrino@programming.dev 31 points 1 year ago (1 child) Electrical engineers are the ones that use j though (because i is used for current) permalink fedilink source parent hideshow 1 child comment replies: [–] Klear@sh.itjust.works 2 points 1 year ago I am used for current permalink fedilink source parent
[–] thomasloven@lemmy.world 11 points 1 year ago 10? That’s the name some put to 1e1, right? permalink fedilink source parent
[–] exasperation@lemmy.dbzer0.com 10 points 1 year ago* (1 child) Except maybe Electrical engineers. Yup, I can count just fine to the 10th number in a zero-indexed counting system: black, brown, red, orange, yellow, green, blue, violet, gray, white. permalink fedilink source parent hideshow 1 child comment replies: [–] alt_xa_23@lemmy.world 4 points 1 year ago https://xkcd.com/227/ permalink fedilink source parent
[–] Baguette@lemmy.blahaj.zone 6 points 1 year ago The inner machinations of an electrical engineer is too complicated for me to understand, I think they might be thinking on a higher order to understand these circuits Thats why I barely passed my electrical engineering class lol permalink fedilink source parent
[–] Fedizen@lemmy.world 5 points 1 year ago Having worked with electrical engineers, some of them are quite smart, the rest have lead poisoning. permalink fedilink source parent
[–] TheSlad@sh.itjust.works 19 points 1 year ago (3 children) Right? They got that shit backwards. Op is a fraud. i is used in pure math, j is used in engineering. permalink fedilink source parent hideshow 3 child comments replies: [–] Chakravanti@monero.town 2 points 1 year ago (2 children) That's hilarious. You're not seeing what's going on backwards just like that (as I point at the point going nowhere shitty) in an equation that is finding as many clAEver ways to say something you actually not caring about talking about. That's like, "How many time van express the only thing that van't be done until the 'verse itself tries to do what can't be done and sever your... ...Oh, I see...you don't have ([of course, because you can't have to give {is}) nothing)] to give. Unable to sea time doesn't mean we can't see(k)ER the mAETh.ac(k).cc(k).08 The only thin(g):(k) that doesn't ever be never, is not at alla hack(g)in(g).G your lackthereof to divi... permalink fedilink source parent hideshow 2 child comments replies: [–] TheSlad@sh.itjust.works 3 points 1 year ago (1 child) Is this a copypasta or are you having a stroke? permalink fedilink source parent hideshow 1 child comment replies: [–] Chakravanti@monero.town 1 point 1 year ago Wah, wah, wah... Not my problem. Keep trying. /s permalink fedilink source parent
[–] Chakravanti@monero.town 2 points 1 year ago (2 children) That's hilarious. You're not seeing what's going on backwards just like that (as I point at the point going nowhere shitty) in an equation that is finding as many clAEver ways to say something you actually not caring about talking about. That's like, "How many time van express the only thing that van't be done until the 'verse itself tries to do what can't be done and sever your... ...Oh, I see...you don't have ([of course, because you can't have to give {is}) nothing)] to give. Unable to sea time doesn't mean we can't see(k)ER the mAETh.ac(k).cc(k).08 The only thin(g):(k) that doesn't ever be never, is not at alla hack(g)in(g).G your lackthereof to divi... permalink fedilink source parent hideshow 2 child comments replies: [–] TheSlad@sh.itjust.works 3 points 1 year ago (1 child) Is this a copypasta or are you having a stroke? permalink fedilink source parent hideshow 1 child comment replies: [–] Chakravanti@monero.town 1 point 1 year ago Wah, wah, wah... Not my problem. Keep trying. /s permalink fedilink source parent
[–] TheSlad@sh.itjust.works 3 points 1 year ago (1 child) Is this a copypasta or are you having a stroke? permalink fedilink source parent hideshow 1 child comment replies: [–] Chakravanti@monero.town 1 point 1 year ago Wah, wah, wah... Not my problem. Keep trying. /s permalink fedilink source parent
[–] Chakravanti@monero.town 1 point 1 year ago Wah, wah, wah... Not my problem. Keep trying. /s permalink fedilink source parent
[–] Hoimo@ani.social 37 points 1 year ago (3 children) How do we know it's gay though? OP could be a girl (male) permalink fedilink source parent hideshow 3 child comments replies: [–] SippyCup@feddit.nl 57 points 1 year ago Because it's 4chan. And there are no women on the Internet on 4chan permalink fedilink source parent [–] floquant@lemmy.dbzer0.com 17 points 1 year ago Sure OP is a girl. Guy In Real Life permalink fedilink source parent [–] ByteJunk@lemmy.world 10 points 1 year ago Newfag. (sorry! seemed like the appropriate 4chan reply) permalink fedilink source parent
[–] SippyCup@feddit.nl 57 points 1 year ago Because it's 4chan. And there are no women on the Internet on 4chan permalink fedilink source parent
[–] floquant@lemmy.dbzer0.com 17 points 1 year ago Sure OP is a girl. Guy In Real Life permalink fedilink source parent
[–] ByteJunk@lemmy.world 10 points 1 year ago Newfag. (sorry! seemed like the appropriate 4chan reply) permalink fedilink source parent
[–] kogasa@programming.dev 18 points 1 year ago (7 children) The mathematician also used "operative" instead of, uh, something else, and "associative" instead of "commutative" permalink fedilink source parent hideshow 7 child comments replies: [–] Zagorath@aussie.zone 3 points 1 year ago (6 children) "operative" instead of, uh, something else I think they meant "operand". As in, in the way dy/dx can sometimes be treated as a fraction and dx treated as a value. permalink fedilink source parent hideshow 6 child comments replies: [–] Chrobin@discuss.tchncs.de 1 point 1 year ago (5 children) I think you mean operator. The operand is the target of an operator. permalink fedilink source parent hideshow 5 child comments replies: [–] Zagorath@aussie.zone 2 points 1 year ago (4 children) The operand is the target of an operator Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy). permalink fedilink source parent hideshow 4 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (3 children) I'd say the $\int dx$ is the operator and the integrand is the operand. permalink fedilink source parent hideshow 3 child comments replies: [–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Zagorath@aussie.zone 3 points 1 year ago (6 children) "operative" instead of, uh, something else I think they meant "operand". As in, in the way dy/dx can sometimes be treated as a fraction and dx treated as a value. permalink fedilink source parent hideshow 6 child comments replies: [–] Chrobin@discuss.tchncs.de 1 point 1 year ago (5 children) I think you mean operator. The operand is the target of an operator. permalink fedilink source parent hideshow 5 child comments replies: [–] Zagorath@aussie.zone 2 points 1 year ago (4 children) The operand is the target of an operator Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy). permalink fedilink source parent hideshow 4 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (3 children) I'd say the $\int dx$ is the operator and the integrand is the operand. permalink fedilink source parent hideshow 3 child comments replies: [–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Chrobin@discuss.tchncs.de 1 point 1 year ago (5 children) I think you mean operator. The operand is the target of an operator. permalink fedilink source parent hideshow 5 child comments replies: [–] Zagorath@aussie.zone 2 points 1 year ago (4 children) The operand is the target of an operator Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy). permalink fedilink source parent hideshow 4 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (3 children) I'd say the $\int dx$ is the operator and the integrand is the operand. permalink fedilink source parent hideshow 3 child comments replies: [–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Zagorath@aussie.zone 2 points 1 year ago (4 children) The operand is the target of an operator Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy). permalink fedilink source parent hideshow 4 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (3 children) I'd say the $\int dx$ is the operator and the integrand is the operand. permalink fedilink source parent hideshow 3 child comments replies: [–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Chrobin@discuss.tchncs.de 2 points 1 year ago (3 children) I'd say the $\int dx$ is the operator and the integrand is the operand. permalink fedilink source parent hideshow 3 child comments replies: [–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Zagorath@aussie.zone 1 point 1 year ago (2 children) You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances. ∫(dy/dx)dx = ∫dy = y Or the chain rule: (dz/dy)(dy/dx) = dz/dx In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand. permalink fedilink source parent hideshow 2 child comments replies: [–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Chrobin@discuss.tchncs.de 2 points 1 year ago (1 child) I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with: ∫ f(x) dx = ∫ dx f(x) In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined. permalink fedilink source parent hideshow 1 child comment replies: [–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] Zagorath@aussie.zone 0 points 1 year ago Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around. But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea. permalink fedilink source parent
[–] NeatNit@discuss.tchncs.de 12 points 1 year ago My thoughts exactly lol permalink fedilink source parent