Has this been discovered before?

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

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

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

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)

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]

[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]

[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]

[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]

...

It's usually done this way.

That's my boy harmonic oscillator

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

The math in the failed posts are exactly the same.

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

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

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

[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 into the viewer.

What's this? It works in the viewer but not the post. I think Veeky Forums has broke.

Add more spaces.

>somebody posts scientific discovery on Veeky Forums
>FUUUUUUUUUGCCCKCKCK YOU FUCKING AGGGGGGGGGGH RETARDED!!!!!!!!! FUUUUUUUUUUUUUUU 4CAHAAAA MEEEEEEMES SMUG ANIME GIRL
this board is retarded

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

this is now a big number thread

post some big ones

what happens in the last step?

lol

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.

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.

I just discovered a GUT. AMA

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

foryou

>all these people taking this seriously

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

G^G^G^G....^G (G times)

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....

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

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

N-NANI!?

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

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

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]

youtube.com/watch?v=SrU9YDoXE88

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.

Counter-example?

Method?

Graduate student in mathematics here.

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

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:

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

And:

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?

I meant PDEs. Hahahaha

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

ITT: Math majors discover kinetic and potential energy.

>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

>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]

>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

-1/12

>... 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 $(dy)^{2}$?

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

zozzle

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

>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

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=)(=

What happens from line 4 to line 5?

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

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.

can we make this a new forced meme?