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/sci/ - Science & Math


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8922141 No.8922141 [Reply] [Original]

Why can dy/dx be treated as a fraction only in certain contexts?

>> No.8922143

>>8922141
nonstandard analysis

>> No.8922151
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8922151

there are people on /sci/ that will defend this

>> No.8922154

Pure coincidence. The notation was invented just because things like that work out. It's just notation; don't get too caught up in thinking about math as mystical runes obeying mysterious rules.

>> No.8922155

>>8922143
This construction soert of allows you to treat it as a fraction, but it's abstract and complicated if you never had done real anal or some sort of calculus with proofs. The thing behind everything is what does it mean for something to get arbitrarily close to some other thing? You can get plenty of intuition by looking at the graph of a function and saying look it goes to there, but that is not sufficiently rigorous. Read up the most common definition of limit (delta-epsilon definition by Weirstrass) and then look at the definition of a derivative and you will clearly see that it makes no sense to use it as a fraction because dy/dx is literally just notation and nothing else.

>> No.8922160

>>8922141
Its only mathematicians autistic screeching, engineers and scientists do it all the time and bridges dont fall and reactors dont explode.

>> No.8922187

>>8922141
coincidence. it's literally just a way of letting engineers do "maths". in every case there's a way to avoid it and work it out properly

>> No.8922225

>>8922151
what's invalid about it?

>> No.8922236

>>8922225
lines 4 & 6

>> No.8922242

I'll just provide a counterexample.
>y=(pi)x
>dy/dx=pi
>pi cannot be written as a fraction

>> No.8922246

>>8922242
>being this much of a brainlet

>> No.8922253

>>8922236
how would you do it without dx=du part?

>> No.8922268

>>8922151
stupid person here, why is he assuming dx/du = 1? Am i missing something here?

>> No.8922270

>>8922242
kek

>> No.8922272

>>8922268
Do you know how to differentiate a linear function?

>> No.8922281

>>8922272
sort of, i know that you need to break down the equations into the basic forms of integrals to solve them. Haven't messed around with calculus for a while now (economics major lel)

>> No.8922310

>>8922253
[eqn]\mathcal{I}(x)=\int f(\varphi(x))\varphi'(x)\,\mathrm{d}x[/eqn] where [math]f(x)=(x-1)/\sqrt{x}[/math] and [math]\varphi(x)=x+1[/math]. By the chain rule
[eqn]\mathcal{I}(x)=\int \frac{u-1}{\sqrt{u}}\,\mathrm{d}u=\cdots[/eqn]

>> No.8922313

>>8922268
Derivate u - 1. You'll get du. Then just divide both sides by du and you'll get dx/du = 1

>> No.8922315

>>8922253
By doing the exact same thing and just not writing down that step. He's being a pedantic ass, and will get over it after a couple years in undergrad.

>> No.8922320

>>8922310
your first line is completely generic and would be written over and over again with different functions f and phi. It's not really necessary to write.

>> No.8922327

>>8922141
it just works. For example: take
[eqn]\frac{dy}{dx}=2[/eqn]
now cancel the "d"
[eqn]\frac{y}{x}=2[/eqn]
and solve for y:
[eqn]y=2x[/eqn]
so it's true. taking the derivative of y=2x gives you 2.

>> No.8922356
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8922356

>>8922327

>> No.8922464

you may have done something

>> No.8922504

This is the most retarded thread i have seen here in a long time. So by the logic I see here d is just some variable. Pro-tip its not!

>> No.8922511

These silly physics students. I'm always dying a little inside if i see people calculating with differentials as with numbers, even through it works fine in some context if everything is nice enough.

>> No.8922514

>>8922327
lmfao

>> No.8922539

>>8922236
What's wrong with saying [math]\frac {\mathrm{d}x} {\mathrm{d}u} = 1 \implies \mathrm{d}x=\mathrm{d}u[/math]? Can you give a case where this isn't true?

>> No.8922547

>>8922327
Top kek

>> No.8922621

>>8922313
thanks breh

>> No.8922682

>>8922242
>pi cannot be written as a fraction
are you retarded?
pi/1? 2pi/2?

>> No.8922695

>>8922539
dx and du are just notation without meaning (unless you're working with differential forms, which you aren't)

>> No.8922699

>>8922682
pi cannot be written as a fraction of integers

>> No.8922705

>>8922151
This is definitely cancer

>> No.8922710

>>8922160
this

>> No.8922711

>>8922699
22/7
Did you pass 7th grade maths?

>> No.8922718

>>8922711
[eqn]\frac{22}{7} \gt \pi[/eqn]

>> No.8922727

>>8922711

sounds like you never went further than 7th grade

>> No.8923293
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8923293

>>8922695
>just notation without meaning

this is an oxymoron, you oxymoron. dx means a tiny change in the x direction, du means a tiny change in the u direction. dx = du means that a tiny change in x leads to a proportional tiny change in u.

>> No.8923697

>>8922268

Is this place full of highschoolers or what?

>> No.8923730

>>8922695
You are a retard, it's the same as x'(u)

>> No.8923745

>>8922327

>[math]\frac{dy}{dx} = \frac{y}{x}[/math]

If only this was true, calculus would be a lot simpler.

>> No.8923750

>>8922160
/thread

>> No.8923784

>>8922699
And who said dy or dx are integers?

>> No.8923915

>>8922160
Actual mathematicians don't care, and they use the same kind of shortcuts when they know they'll work. It's just first year undergrads who are overly obsessed with rigor.

>> No.8924001

>>8923293
>t. physist

>> No.8924087

Good thread.

>> No.8924099

>>8923915

>implying mathematicians ever do integrals

Lol

>> No.8924107

it works bc of the chain rule

>> No.8924172

>>8924099
I mean, they do come up. Even in shit like number theory.

>> No.8924179

it literally is a fraction though

>> No.8924180

>>8922151
LMAFO

I literally write the exact same way as left side

with the let statement and therefores

I even do the dx/du=1 shit

>> No.8924204

shit ive realized i actually never understood this. How come its sometimes valid to separate it and other times not? What makes it rigorous?

>> No.8924263

>>8924204
suppose you have some integral [eqn] I=\int f(g(x))g'(x)\,dx\,. [/eqn] then by the chain rule
[eqn] I=\int f(u)\,du\quad (u=g(x))\,, [/eqn]
which is integration by substitution.
by the other way you have something like: let [math] u=g(x) [/math]. then [math] \frac{du}{dx}=g'(x) [/math] so [math] dx=\frac{du}{g'(x)} [/math]. therefore [eqn] I=\int f(u)g'(x)\,\frac{du}{g'(x)}=\int f(u)\,du\,, [/eqn]
which is the same result.

idk about differential equations as it varies in each case, but the "proof" will be similar.

>> No.8924266

>>8924263
i should add that that doesnt make the method rigorous, but it shows why it works