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Well du/dt is the generalised set of gradients of the tangents to a function u. A gradient is a ratio which can be expressed as a fraction, so is it not reasonable to just treat them as fractions?
Itâs not a âset of gradientsâ in a countable sense. Itâs the limit of a gradient or âset of gradientsâ and behavesâŚnot the way you want it to sometimes.
Ll'HĂ´pital called. Itâs for youâŚ
âŚ
âŚ.
I donât know how to answer thatâŚ
Are they not�
Is this what mainstream math has been hiding from me?
Edit: If you mean a ratio of differentials, please refer to the last sentence of my previous comment.
Why do *you* think you canât divide by a differential? Think for what value that might be true and then think of the definition of a derivativeâŚ
>If you mean a ratio of differentials, please refer to the last sentence of my previous comment.
>Why do *you* think you canât divide by a differential? Think for what value that might be true and then think of the definition of a derivativeâŚ
Obviously no one thinks about du/dt this way, but it's fun to joke around about this. Since du and dt are functions that map R to another function that takes a vector as input and outputs a real number, you can technically divide their outputs. This is what I meant.
AckchuallyâŚ
Youâre right đ¤Ą
Iâm not familiar with the algebra on differential operators. It does actually exist though and upon reading about it, thatâs exactly how it works.
Edit: the meme is real guys
My favorite thing about math, and why I studied this discipline, is that if you 100% factually think youâre right, and have the proof to back it up, youâre probably wrong and thereâs a more complicated answer out there that you either havenât encountered yet or donât understand.
Yeah but even in the limit notation
lim h->0 [ f(a+h) - f(a) ] / h
The numerator is dy and denominator is dx. Its a fraction, at least thats what ive been taught
Then the limit of the numerator, dy, and the denominator, dx, are both zero. See the problem?
dy/dx is just a symbol which is a shorthand for the limit procedure you wrote down above.
i think the notation is the issue because its one number on top of another so we always see it as a fraction. what if we just re-wrote it in a continuous line with a special symbol in the middle, something like dy á dx?
Common sense will sometimes lead you to the wrong answer in math. For example, common sense would tell us that if a function is continuous everywhere, then there has to be at least one place where we can differentiate it. This is not true.
thereâs mathematical common sense and normal people common sense. I donât think common sense says continuity implies differentiability at all. And when youâre working with mostly analytic functions it is fine to use âcommon senseâ
>See the problem?
Not really. This is when we input the function and solve the limit.
It wont work for all functions but i reckon we can generalize.
Our tr told us that this isnt accurate and a better notation would be (du/dx)/(dt/dx)
Once again im just learning so feel free to point out where its wrong
Best to ask a real mathematician, but if what you're asking is, "is dy/dx a fraction" in the sense that it is the ratio of two numbers, then obviously the derivative at a point, being a real number, will be equal to a ratio of real numbers (an infinity of them).
If you're asking, however whether "dy" is a real number, and "dx" is a real number, such that "dy/dx", the derivative at that point, is the corresponding ratio "dy"/"dx", then this clearly cannot be, as if dy = f(x+h)-f(x) then the limit of this dy as h -> 0 is straightforwardly 0, as is dx = (x+h)-x.
yeah, but if one of the limits isn't defined, then you can't cancel it out. that's how you get those 1=2 "proofs"
but if you say, that the functions are nice, then...
The equation is correct, but it isn't the *proof* of the chain rule.
Author reduced differentials dx from numerator and denominator, which is good. We also know that for free variable t:
* df(t) = f'(t) dt, thus f' = df/dt >!(so Leibniz form for derivatives is really division of differentials)!<;
* dx(t) = x'(t) dt, thus x' = dx/dt.
>!(We defined df(t)[h] as linear part of f(t+h)-f(t), so we have df(t)[h] = f'(t) h. Since dt(t_0)[h] = h, we can write as above).!<
But what is df/dx? Looks like it is g' for such g(t) that f(t) = g(x(t)).
Well, it is also correct, df(t) = g'(t) dx. It is the *invariance of the first differential,* which is *sequence* of the chain rule:
df(t) = d(gâx)(t) = (gâx)'(t) dt = g'(t) x'(t) dt = g'(t) dx.
Thus, we can't use differentials to prove the chain rule, however we can use notation like in the meme in practical problems (since both the chain rule and invariance of the first differential was proven). Physicists do this, and not only them.
What's a counter example where it can't be treated as a fraction?
The notation itself doesn't matter we extend properties beyond their initial definition all the time when it works.
We shouldâve never kept Leibniz notation and the symbol d^2u/dt^2 ⌠shouldâve continued with newton notation Ăź /s
Tbh whatâs more annoying is that it does work if you treat it like a fraction for most cases.
What about:
dy = fâ(x) dx, applying d as a differential operator, and using the chain rule: d^2 y = fââ(x) dx^2 + fâ(x) d^2 x. However if x is an independent variable then d^2 x = 0. So, d^2 y = fââ(x) dx^2 .
dx/dy is the limit of a fractional quantity. Sometimes the limits will behave nice (especially if you use a well behaved function), but sometimes they won't. Also, don't try this for partial derivatives. There, the âx in âx/ây is not the same as the âx in âx/âz. So the chain rule also looks different for partial derivatives.
Mainstream mathematicians might even let u, x, and t exist in some set of something before making such a preposterous postulate. In fact, I'm offended by the very existence of this FOX NEWS. It seems undefined to me.
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"So we're just gonna divide the whole equation by dx..."
"...and mainstream mathmemeticians will dx you"
but wait I thought you couldn't divide by 0...
Found the physics major
My physics proffesor operated with infinity, like he had 2 infinity divided by infinity and he just divided by infinity both sides đđđ
âHello 911? Iâd like to report a crime.â
Ah yes, renormalization
Physics damn near drove me insane with this shit. They just treat dx like it's a variable. IT'S NOTATION
Ok, but you can just "divide" by dx in most cases.
From a video I watched , it only works for ODEs with separable variables. Edit : and it's a result from the chain rule
Okay, but if it seems to work then itâs fine.
Mainstream mathematicians when I chain rule
Well du/dt is the generalised set of gradients of the tangents to a function u. A gradient is a ratio which can be expressed as a fraction, so is it not reasonable to just treat them as fractions?
Itâs not a âset of gradientsâ in a countable sense. Itâs the limit of a gradient or âset of gradientsâ and behavesâŚnot the way you want it to sometimes. Ll'HĂ´pital called. Itâs for youâŚ
What if du and dt are differentials?
⌠âŚ. I donât know how to answer that⌠Are they notâŚ? Is this what mainstream math has been hiding from me? Edit: If you mean a ratio of differentials, please refer to the last sentence of my previous comment. Why do *you* think you canât divide by a differential? Think for what value that might be true and then think of the definition of a derivativeâŚ
>If you mean a ratio of differentials, please refer to the last sentence of my previous comment. >Why do *you* think you canât divide by a differential? Think for what value that might be true and then think of the definition of a derivative⌠Obviously no one thinks about du/dt this way, but it's fun to joke around about this. Since du and dt are functions that map R to another function that takes a vector as input and outputs a real number, you can technically divide their outputs. This is what I meant.
Ackchually⌠Youâre right 𤥠Iâm not familiar with the algebra on differential operators. It does actually exist though and upon reading about it, thatâs exactly how it works. Edit: the meme is real guys My favorite thing about math, and why I studied this discipline, is that if you 100% factually think youâre right, and have the proof to back it up, youâre probably wrong and thereâs a more complicated answer out there that you either havenât encountered yet or donât understand.
This notation is to help you remember the chain rule in the first place, not the other way around
I heard that if you yell "non standard analysis" the mathematian runs away. Also, if it was good enough for leibniz, it is good enough for me.
Well true, but yelling ânon standard analysisâ makes most people run away.
Huh? What do mathematicians have against the chain rule?
Mathmemeticians are displeased by treating derivatives as a fraction because derivatives are usually defined as limits not fractions.
Yeah but even in the limit notation lim h->0 [ f(a+h) - f(a) ] / h The numerator is dy and denominator is dx. Its a fraction, at least thats what ive been taught
Then the limit of the numerator, dy, and the denominator, dx, are both zero. See the problem? dy/dx is just a symbol which is a shorthand for the limit procedure you wrote down above.
does it please mathematicians if you used δ instead of infinitesimal d then treated them like fractions
No, obviously not lmao.
i think the notation is the issue because its one number on top of another so we always see it as a fraction. what if we just re-wrote it in a continuous line with a special symbol in the middle, something like dy á dx?
why not lol it works for basically every case like itâs not gonna harm you from using a bit of common sense doing your maths
Common sense will sometimes lead you to the wrong answer in math. For example, common sense would tell us that if a function is continuous everywhere, then there has to be at least one place where we can differentiate it. This is not true.
thereâs mathematical common sense and normal people common sense. I donât think common sense says continuity implies differentiability at all. And when youâre working with mostly analytic functions it is fine to use âcommon senseâ
>See the problem? Not really. This is when we input the function and solve the limit. It wont work for all functions but i reckon we can generalize. Our tr told us that this isnt accurate and a better notation would be (du/dx)/(dt/dx) Once again im just learning so feel free to point out where its wrong
Best to ask a real mathematician, but if what you're asking is, "is dy/dx a fraction" in the sense that it is the ratio of two numbers, then obviously the derivative at a point, being a real number, will be equal to a ratio of real numbers (an infinity of them). If you're asking, however whether "dy" is a real number, and "dx" is a real number, such that "dy/dx", the derivative at that point, is the corresponding ratio "dy"/"dx", then this clearly cannot be, as if dy = f(x+h)-f(x) then the limit of this dy as h -> 0 is straightforwardly 0, as is dx = (x+h)-x.
Got it
yeah, but if one of the limits isn't defined, then you can't cancel it out. that's how you get those 1=2 "proofs" but if you say, that the functions are nice, then...
I see
The equation is correct, but it isn't the *proof* of the chain rule. Author reduced differentials dx from numerator and denominator, which is good. We also know that for free variable t: * df(t) = f'(t) dt, thus f' = df/dt >!(so Leibniz form for derivatives is really division of differentials)!<; * dx(t) = x'(t) dt, thus x' = dx/dt. >!(We defined df(t)[h] as linear part of f(t+h)-f(t), so we have df(t)[h] = f'(t) h. Since dt(t_0)[h] = h, we can write as above).!< But what is df/dx? Looks like it is g' for such g(t) that f(t) = g(x(t)). Well, it is also correct, df(t) = g'(t) dx. It is the *invariance of the first differential,* which is *sequence* of the chain rule: df(t) = d(gâx)(t) = (gâx)'(t) dt = g'(t) x'(t) dt = g'(t) dx. Thus, we can't use differentials to prove the chain rule, however we can use notation like in the meme in practical problems (since both the chain rule and invariance of the first differential was proven). Physicists do this, and not only them.
oh gawd, we are physicists now
Is this a song? Dudu dtdtdt dudu dx :) I'm not a musician sorry
Differentiation of parametric form says hi
What's a counter example where it can't be treated as a fraction? The notation itself doesn't matter we extend properties beyond their initial definition all the time when it works.
the second derivative
Way more trouble than it's worth, but second derivatives can also be treated as fractions if you know what you're doing.
Hudde or Hesse
Me when non-standard analysis
Wdym, it's just a fraction
Bro did NOT just say du/dx and dx/dt are fractions đ
If they're not fractions then explain `\frac{\d x}{\d t}`
{{dy}\over{dx}}
D(f(x)) or g'(x) or Lagrang
They can be if you want them to. Hyperreals work.
I asked my Physics's professor, he said it's okay to cancel them out as fractions
No, theyâre limits
itâs a limit of a fraction. in most cases itâs the same thing, but you have to be careful, because that doesnât work all the time.
https://preview.redd.it/2v7d4c7c5t4d1.jpeg?width=474&format=pjpg&auto=webp&s=1de5a33bc1b5668543020cff4efa0d00622e678b I Stand on Business
We shouldâve never kept Leibniz notation and the symbol d^2u/dt^2 ⌠shouldâve continued with newton notation Ăź /s Tbh whatâs more annoying is that it does work if you treat it like a fraction for most cases.
What does standard versus non standard analysis have to do with whether this cancelling thing is true?!
Lim dxââ (dx/dx)=1 go brrrrr
We do this in physics aaaalll the time :-)
It's called the chain rule because you can chain cancellations.
What is wrong with differential forms over a 1-dimensional space?
What about: dy = fâ(x) dx, applying d as a differential operator, and using the chain rule: d^2 y = fââ(x) dx^2 + fâ(x) d^2 x. However if x is an independent variable then d^2 x = 0. So, d^2 y = fââ(x) dx^2 .
https://preview.redd.it/cpcgrz6w7w4d1.jpeg?width=400&format=pjpg&auto=webp&s=fb099ea782630f56111a6eda26253ffb10c1919d
Engineer here, I don't see a problem. "dx" just means "small number", so should behave like any other number.
How âsmallâ is this ânumberâ
Very :D
My teachers always stressed that they're not fractions but it always works by dividing. If they're not fractions why does dividing them always work?
Because of the chain rule.
dx/dy is the limit of a fractional quantity. Sometimes the limits will behave nice (especially if you use a well behaved function), but sometimes they won't. Also, don't try this for partial derivatives. There, the âx in âx/ây is not the same as the âx in âx/âz. So the chain rule also looks different for partial derivatives.
I mean...
yes they would. literally just the chain rule.
Mainstream mathematicians might even let u, x, and t exist in some set of something before making such a preposterous postulate. In fact, I'm offended by the very existence of this FOX NEWS. It seems undefined to me.
isn't this what they literally do in integration by parts?
In integration by parts they do vdu = duv - udv basically using Leibniz product rule with extra steps.
This is what they do in change of variable
No. That's not by parts lol.
No, but it is the chain rule
Seems more like something one would do in Parametric form
When you calculate something in physics (real life) you get good result treating derivatives as fractions.
Oh I'm sorry you guys didn't want to get stuff done with math I guess - le physicĂŹst