Has this been discovered before?

[Anonymous No.8531692]

[Anonymous No.8531701]

>>8531692
Unless you are a retard who actually posts his scientific discoveries on 4chan, it is safe to assume that it has been discovered before

[Anonymous No.8531705]

>>8531692

What the fuck am I looking at.

Why do you need the root2 and the other square root? is this some kind of meme?

[Anonymous No.8531707]

>>8531692
It has been studied and done before by anyone that studied harmonic oscillators, and they are pretty popular in physics

[Anonymous No.8531719]

Baby fresh out of calc II tries to solve a differential equation.

[Anonymous No.8531729]

>>8531705
[math]
\frac{d^2y}{dx^2} = f(y)\\
\frac{d^2y}{dx} = f(y) dx\\
d(\frac{dy}{dx}) = f(y) \frac{dx}{dy} dy\\
\frac{dy}{dx} d(\frac{dy}{dx}) = f(y) dy\\
\frac{(\frac{dy}{dx})^2}{2} = \int f(y)\,dy + C\\
(\frac{dy}{dx})^2 = 2\int f(y)\,dy + C\\
\frac{dy}{dx} = \sqrt{2}\sqrt{\int f(y)\,dy + C}\\[/math]

[Anonymous No.8531733]

>>8531729
what did he mean by this

[Anonymous No.8531735]

>>8531729
this_post_gave_me_cancer.jpeg

[Anonymous No.8531738]

>>8531701
>retard
This will be how fundamental discoveries are shared in the future. It will be up to us to find the gems in the rough.

[Anonymous No.8531743]

>>8531729
Wow that was kinda cool, holding dy/dx as an integratable function

[Anonymous No.8531746]

>>8531729
have you tried differentiating the result to show that it is indeed equal to y''?

[Anonymous No.8531748]

>>8531729
wow op you're really a brainlet aren't you

[Anonymous No.8531751]

>>8531692
shit, finally I can solve my diff. eq. concerning Barnett integrable functions

[Anonymous No.8531756]

>>8531729
Instead of explaining why you're a retard I'd encourage you to actually go solve some differential equations for different values of f(y) and seeing if you are right (hint: you're not)

[Anonymous No.8531765]

>>8531729
Here I rewrote it without the horrendous notation.
[math]\displaystyle y''(x) = f'(y(x))[/math]
[math]\displaystyle y'(x) y''(x) = f'(y(x)) y'(x)[/math]
[math]\displaystyle \frac{y'(x)^2}{2} = f(y(x)) + C[/math]
[math]\displaystyle y'(x) = \sqrt{2} \sqrt{f(y(x)) + C}[/math]

[Anonymous No.8531794]

>>8531746
[math]\frac{dy}{dx}=\sqrt{2}\sqrt{\int f(y)\,dy+C}\\
\frac{d^2y}{dx^2}=\sqrt{2}\frac{d}{dx}\sqrt{\int f(y)\,dy+C}\\
\frac{d^2y}{dx^2}=\sqrt{2}\frac{f(y)\frac{dy}{dx}}{2\sqrt{\int f(y)\,dy+C}}\\
\frac{d^2y}{dx^2}=\frac{f(y)\frac{dy}{dx}}{\sqrt{2}\sqrt{\int f(y)\,dy+C}}\\
\frac{d^2y}{dx^2}=\frac{f(y)\frac{dy}{dx}}{\frac{dy}{dx}}\\
\frac{d^2y}{dx^2}=f(y)\\
[/math]

[Anonymous No.8531809]

>>8531756
[math]\frac{d^2x}{dt^2}=-\omega^2x\\
\frac{dx}{dt}=\sqrt{2}\sqrt{\frac{-\omega^2x^2}{2}+C_0}\\
\frac{dx}{dt}=\sqrt{-\omega^2x^2+C_1}\\
\frac{dx}{dt}=\omega\sqrt{-x^2+C_2}\\
\frac{dx}{dt}=\omega\sqrt{-x^2+A^2}\\
\frac{dx}{dt}=\omega\sqrt{A^2-x^2}\\[/math]

[Anonymous No.8531812]

>>8531756
\frac{d^2x}{dt^2}=-\omega^2x\\
\frac{dx}{dt}=\sqrt{2}\sqrt{\frac{-w^2x^2}{2}+C_0}\\
\frac{dx}{dt}=\sqrt{-w^2x^2+C_1}\\
\frac{dx}{dt}=w\sqrt{-x^2+C_2}\\
\frac{dx}{dt}=w\sqrt{-x^2+A^2}\\
\frac{dx}{dt}=w\sqrt{A^2-x^2}\\

[Anonymous No.8531816]

>>8531756
[math]\frac{d^2x}{dt^2}=-w^2x\\
\frac{dx}{dt}=\sqrt{2}\sqrt{\frac{-w^2x^2}{2}+C_0}\\
\frac{dx}{dt}=\sqrt{-w^2x^2+C_1}\\
\frac{dx}{dt}=w\sqrt{-x^2+C_2}\\
\frac{dx}{dt}=w\sqrt{-x^2+A^2}\\
\frac{dx}{dt}=w\sqrt{A^2-x^2}\\[/math]

[Anonymous No.8531822]

>>8531756
[math]\frac{d^2x}{dt^2}=-w^2x\\
\frac{dx}{dt}=\sqrt{2}\sqrt{\frac{-w^2x^2}{2}+C}\\
\frac{dx}{dt}=\sqrt{-w^2x^2+C}\\
\frac{dx}{dt}=w\sqrt{-x^2+C}\\
\frac{dx}{dt}=w\sqrt{-x^2+A^2}\\
\frac{dx}{dt}=w\sqrt{A^2-x^2}\\[/math]

[Anonymous No.8531825]

>>8531822
[math]\frac{d^2x}{dt^2}=-w^2x\\
\frac{dx}{dt}=\sqrt{2}\sqrt{\frac{-w^2x^2}{2}+C}\\
\frac{dx}{dt}=\sqrt{-w^2x^2+C}\\
\frac{dx}{dt}=w\sqrt{-x^2+C}\\
\frac{dx}{dt}=w\sqrt{-x^2+A^2}\\
\frac{dx}{dt}=w\sqrt{A^2-x^2}[/math]

[Anonymous No.8531832]

>>8531756

[Anonymous No.8531881]

It's usually done this way.

[Anonymous No.8531887]

>>8531832
That's my boy harmonic oscillator

[Anonymous No.8531894]

>>8531822
>>8531825
>>8531832
You don't actually know how to solve a differential equation, do you?

[Anonymous No.8532219]

>>8531894
The math in the failed posts are exactly the same.

Plug it into http://quicklatex.com/ and get the same fucking thing

[Anonymous No.8532229]

>>8532219
sci has a built in latex preview if you click on the upper left of the comment box

[Anonymous No.8532250]

>>8532219
>>8532229
[math]\int \left(\sum _{123123=123123}^{1231231} 123123\right) \, d123123123[/math]

[/A/nonymous !cPUZU5OGFs No.8532595]

>>8532229
[math]\frac{d^2x}{dt^2}=-w^2x\\
\frac{dx}{dt}=\sqrt{2}\sqrt{\frac{-w^2x^2}{2}+C}\\
\frac{dx}{dt}=\sqrt{-w^2x^2+C}\\
\frac{dx}{dt}=w\sqrt{-x^2+C}\\
\frac{dx}{dt}=w\sqrt{-x^2+A^2}\\
\frac{dx}{dt}=w\sqrt{A^2-x^2}[/math]

I put >>8531825 into the viewer.

[/A/nonymous !cPUZU5OGFs No.8532599]

>>8532595
What's this? It works in the viewer but not the post. I think /sci/ has broke.

[Anonymous No.8532647]

>>8532599
Add more spaces.

[Anonymous No.8532658]

>>8531735
>>8531748
>>8531756
>somebody posts scientific discovery on /sci/
>FUUUUUUUUUGCCCKCKCK YOU FUCKING AGGGGGGGGGGH RETARDED!!!!!!!!! FUUUUUUUUUUUUUUU 4CAHAAAA MEEEEEEMES SMUG ANIME GIRL
this board is retarded

[Anonymous No.8532670]

>>8532658
this isn't a scientific discovery, he's trolling you retarded fuck

[Anonymous No.8532683]

this is now a big number thread

post some big ones

[Anonymous No.8532697]

>>8531881
what happens in the last step?

[Anonymous No.8532728]

>>8531729
lol

[Anonymous No.8532764]

>>8531729
you see if you did what you're trying now 400 years ago, you might make some menial contributions in formulating concepts of calculus/diff eq. Except you're 400 years too late, and it's time to stand on shoulder's of giants. If you keep messing around with trivial equations such as this one and refuses to learn already developed + super refined concepts (which btw are very elementary), you wouldn't even have enough time to get yourself a math degree. pretty unfortunate huh.

[Anonymous No.8532769]

>>8531794
Your problem is you assume the primitive of f(y) is a function of y. It isn't, in fact the primitive is just a notational shortcut. In reality you should write the integral of f from some arbitary lower bound to a an upper bound that is the independent variable. The y featured in the function and as part of "dy" is just a dummy variable. So two different things can be done: One is you can say the upper bound variable is x, in which case your final result is y'' = f(x)/(root(2)*dy/dx) or you can say the upper bound variable is y, in which case you will get y'' = f(x). In either case its not what you originally are trying to assert.

[Anonymous No.8532943]

I just discovered a GUT. AMA

[Anonymous No.8532971]

>>8532683
It doesn't say how many times to iterate that process.

[Anonymous No.8533272]

>>8532683
foryou

[Anonymous No.8533329]

>>8531729
>all these people taking this seriously

[Anonymous No.8533338]

>>8532683
[math] 10^{10^{200}}+1[/math]

[Anonymous No.8533356]

>>8532683
G^G^G^G....^G (G times)

[Fields Medalist No.8533376]

>>8531729
wtf you can't just assume dx and dy to be equivalent to fractions. dy/dx is a function not one function over the other....

[Anonymous No.8533706]

>>8531881
You mean that's how you do it without the hand wavy brainlet manipulation

[Anonymous No.8533717]

>>8532683
[math]\aleph_{\aleph_{\aleph_{._{._.}}}}[/math]

[Anonymous No.8533735]

>>8533338
N-NANI!?

[Anonymous No.8533772]

>>8533376
Not sure what everyone's on about, looks right to me.
Y'all know it's f(y) and not f(x) right?

[Anonymous No.8533774]

>>8533376
Also you can do that but you should understand what you're actually doing so as to not make mistakes and provide reassurance.

[Anonymous No.8534221]

>>8531692
You didn't even finish solving it.
[math]\displaystyle y''(x) = f(y(x)) \iff y(x) = F^{-1}(x),F(x) = C_1 + \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\int f(x) dx + C_2} }[/math]

[/A/nonymous !cPUZU5OGFs No.8534230]

>>8532683
https://www.youtube.com/watch?v=SrU9YDoXE88

[Anonymous No.8534273]

>>8531729
You should avoid manipulating dx like fractions in any serious mathematical context.

Treating Leibniz's notation like a fraction in fine in engineering/physics, but should be avoided.

[Anonymous No.8534378]

>>8533376
Counter-example?

[Anonymous No.8534382]

>>8534221
Method?

[Anonymous No.8534617]

Graduate student in mathematics here.

1. PDFs are not my area of study, and I'm honestly a bit confused about the conclusion here:

>>8531756
Seems to be confident that OP is wrong. And honestly, I am not exactly comfortable with the way he treats Liebniz notation as a fraction either.

However:

>>8531707
Seems to be confirming that this is a known fact in physics.

And:

>>8531881

Gives an argument that I can certainly follow for why it is so (apart from the square root symbol limiting this to R+ in OPs statement). That z'z thing was a cool trick btw, never seen it before.

So, PDF fags: Is this an accurate statement (ignoring the square rooting), or not?

2. Can any of you provide an example of where treating Liebniz notation as a fraction fails?

I am often asked about this from my students at lower levels. Someone here said it's a function, I think of it as a limit of symbols, but afaik the inventors of calculus thought of it as a ratio at an infitesimal level.

Does thinking of it as a ratio - a fraction - ever fail outside of completely anal constructions in analysis?

[Anonymous No.8534636]

>>8534617
I meant PDEs. Hahahaha

(And "infinitesimals" while I'm at it)

[Anonymous No.8534703]

ITT: Math majors discover kinetic and potential energy.

[Anonymous No.8534936]

>>8534617
> I am not exactly comfortable with the way he treats Liebniz notation as a fraction either.

That's because OP did it in a moronic handwavy style, however it's possible to show OPs identity is true:
[math]\frac{d^2 y}{dx^2}=f(y)[\math]
A simples substation of [math]\frac{dx}{dy}=\dot{y}] will lead to
[math]\frac{d \dot{y}}{dx}=f(y)[/math]
[math]\frac{d \dot{y}}{dy} \frac{dy}{dx} =f(y)[/math]

Which as you can see will lead to OPs result. As I said, this a VERY famous did equation in physics, in fact they're equations of motion for an harmonic oscillator if you define [math]f(y)=-ky[/math], where k is a positive real number.

So, yes OPs identity is true, and I have to recognize his skills on problem solving (although he manipulates the equation in a WRONG hand wavy way). But whatever Fourier also lacked formalism in his work. In the end OP is a moron for thinking he discovered something new and also an idiot for lack of formalism.

[Anonymous No.8534953]

>>8534617
>I am not exactly comfortable with the way he treats Liebniz notation as a fraction either.

That's because OP did it in a moronic handwavy style, however it's possible to show OPs identity is true:
[math]\frac{d^2 y}{dx^2} = f(y)[/math]
By a simple substitution [math]\frac{dy}{dx}=\dot{y}[/math], will lead to
[math]\frac{d \dot{y}}{dx} = f(y)[/math]
[math]\frac{d \dot{y}}{dy} \frac{dy}{dx} = f(y)[/math]
[math]\frac{d \dot{y}}{dy} \dot{y} = f(y)[/math]

Which as you can see will lead to OPs result. As I said, this a VERY famous did equation in physics, in fact they're equations of motion for an harmonic oscillator if you define f(y)=-ky, where k is a positive real number.

So, yes OPs identity is true, and I have to recognize his skills on problem solving (although he manipulates the equation in a WRONG hand wavy way). But whatever Fourier also lacked formalism in his work. In the end OP is a moron for thinking he discovered something new and also an idiot that achieved the right answer by wrong means

[Anonymous No.8534963]

>>8534617
>Does thinking of it as a ratio - a fraction - ever fail outside of completely anal constructions in analysis?
This doesn't really make sense: [math]2^{dy/dx} = \sqrt[dx]{2^{dy}}[/math]

[Anonymous No.8535021]

>>8534617
>Does thinking of it as a ratio - a fraction - ever fail outside of completely anal constructions in analysis?

This is actually done a lot in physics, prof used to joke that it was only allowed when there are no mathematicians in the room. However I've never seen it done with second derivatives the way op did, that probably exceeded the amount of hand wavy math it is allowed to be done by a physicist, so I wouldn't trust that

[Anonymous No.8536080]

>>8532683
-1/12

[Anonymous No.8536085]

>>8534617
>... Liebniz notation as a fraction fails?
The main problem is saying precisely what you mean by the quantities $dy,dx$. If you say they're numbers that are non-zero, you immediately fall into contradiction. If you say they /are/ zero, then what does this notation even mean?

As for an example of where it fails, you probably want to look at second derivatives (I don't think there really are any algebraic problems with viewing $dy/dx$ as a fraction). For instance, what does $$\frac{d^{2}y}{dx^{2}}$$
mean as a ratio? What is $d^{2}y$? How does it relate to $(dy)^{2}$?

[Anonymous No.8536087]

>>8536085
Oh, fuck. How does latex work on here again?

[Anonymous No.8536187]

>>8531729
zozzle

[Anonymous No.8536203]

>>8531729
I think you should explain how to pass from line 2 to line 3.

[Anonymous No.8536205]

>>8531701
>scientific discovery
>reassembling meaningless letters to represent some other meaningless thing¨
HAHAHAHA
tell me, how does this "information" help us build iphones? that's right, it doesn.t

[Anonymous No.8536207]

>>8531729
nice, these symbols mean nothing. it's funny how delusional you math retards are.

that is equivalent to me saying
skk99( = 77a77))
882== S/%& 9
7a890sd78 (( 0a0
a7sd798 == (
My equation proves that 8120// is /)(/)( than 8=)(=

[Anonymous No.8536234]

>>8531729
What happens from line 4 to line 5?

[Anonymous No.8536278]

I don't get it. Isn't the mistake here essentially that he assumed the integral of y'' = y'^2/2, which is only true when y'' = kx

[Anonymous No.8536283]

>>8531692
Yes this is a well known "trick" used to solve differential equations.
y''=f(y)
dy'/dx=f(y)
(dy'/dy)(dy/dx)=f(y) (chain rule)
(dy'/dy)*y'=f(y) (solve this by separation of variables)
y'dy'=f(y)dy
.5y'^2=integral (f(y)dy)
Everyone calling this retarded is either autistic about the notation used or just a complete dumbass.

[Anonymous No.8536327]

>>8531729
can we make this a new forced meme?