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In Genius Season 1 Episode 1 at 9:00, Young Albert Einstein defies his teacher by solving the equation on board and states

Natural log of constant multiplied by x equals natural log of one plus v squared. And since v equals y over x. That gives us the final function: x squared plus y squared minus c x cubed equals zero.

What is he talking about?

P.S.

The question was asked first on Movies/TV stackexchange but it was suggested from the members that the question would fare well on maths stackexchange. However if someone considers this question unsuitable for the stackexchange let me know.

Edit:

I am perfectly aware what does the context translates into. What I am asking is whether the equation is known for something. Like "$c^2=a^2+b^2$" instantly make us recall of "Pythagoras Theorem". In the same sense, does the provided context above hold any significant meaning?

mathnoob123
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    $\log(cx) = \log(1+v^2)$ and $v = \frac{y}{x}$ implies $x^2+y^2 - cx^3 = 0$. I think the PD of that episode doesn't know any genius in real life, there is nothing special about this. – achille hui Jun 03 '17 at 21:47
  • Please see the edit. – mathnoob123 Jun 03 '17 at 21:49
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    I doubt there is any special meaning in that sentence. Most PD/script writer in TV series/movies doesn't really know math/science enough to generate any sensible math/science quotes. Up to what I heard, the people from "big bang theory" may be an exception. If you want to look for something meaningful, try that instead. – achille hui Jun 03 '17 at 21:57
  • Tv and movies are making a point to be more accurate. And they are getting better. But they don't seem to get what is significant or not. This equation and solution is perfectly true and makes perfect sense but it isn't particularly hard or interesting. Hard to understand why this would be "defying" the teacher. There's no indication in what context the equation arose. – fleablood Jun 04 '17 at 00:11

2 Answers2

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What's described is simple algebra.

Natural log of constant multiplied by x equals natural log of one plus v squared.

So$$\ln(cx)=\ln(1+v^2).$$

And since v equals y over x.

So$$\ln(cx)=\ln\left(1+\left(\frac{y}{x}\right)^2\right).\qquad(1)$$

That gives us the final function: x squared plus y squared minus c x cubed equals zero.

So$$x^2+y^2-cx^3=0,\qquad(2)$$which comes from exponentiating both sides of (1):$$cx=1+\left(\frac{y}{x}\right)^2,$$multiplying both sides by $x^2$:$$cx^3=x^2+y^2,$$and subtracting $cx^3$ from both sides, which gives (2).

Equation (2) looks like (but is not) the spacetime interval invariant:

$$s^2 = −c^2t^2 + x^2 + y^2 + z^2.$$

Geremia
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The equations related to the text seem to be:

$$\log(cx)=\log(1+v^2)\tag{1}$$

$$v=\frac{y}{x}\tag{2}$$

$$cx=1+\left(\frac{y}{x}\right)^2\tag{3}$$

$$cx^3=x^2+y^2\tag{4}$$

$$x^2+y^2-cx^3=0\tag{5}$$

Equations $(3)$ and $(4)$ are possible intermediate steps not mentioned.

Jaume Oliver Lafont
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