Very Awesome Nonlinear ODE

[TylerH]

y^2=y' \Rightarrow y=\frac{y'}{y} \Rightarrow \int y dx = ln \left( y \right) \Rightarrow y=e^{\int y dx}=e^{\int e^{\int y dx} dx}=e^{\int e^{\int e^{\int y dx} dx} dx}=\cdots

Is that correct?

[hunt_mat]

Why not just say:

\frac{y'}{y^{2}}=1\Rightarrow -\frac{1}{y}=x+C

[HallsofIvy]

Which is the same as y= -1/(x+C) so that \int ydx= -ln(x+C)+ C_2 so that
e^{\int ydx}=C_2 \frac{1}{x+ C}= \frac{-1}{x+ C}= y
so, yes, your chain of exponentials is correct- but a very complicated way of writing a very simple function.

[pmsrw3]

This reminds me of a problem in a math contest I once saw. Solve:

x^{x^{x^{x^\ldots}}} = 2

The answer is \sqrt{2}.

[TylerH]

Why not just say:

\frac{y'}{y^{2}}=1\Rightarrow -\frac{1}{y}=x+C


Yeah, I found the real solution. But, it's the fact that something so contrived is equal to something so simple that makes it awesome.

[TylerH]

This reminds me of a problem in a math contest I once saw. Solve:

x^{x^{x^{x^\ldots}}} = 2

The answer is \sqrt{2}.

That was the first thing that came to mind. Btw, you would could solve that "infinite power tower" using (4) here: http://mathworld.wolfram.com/LambertW-Function.html.

[HallsofIvy]

This reminds me of a problem in a math contest I once saw. Solve:

x^{x^{x^{x^\ldots}}} = 2

The answer is \sqrt{2}.
If
x^{x^{x^{x^\ldots}}}= 2
then
x^{\left(x^{x^{x^\ldots}}\right)}= x^2=2

More generally, if
x^{x^{x^{x^\ldots}}}= a> 0
then x=\sqrt{a}.

[mheslep]

If
x^{x^{x^{x^\ldots}}}= 2
then
x^{\left(x^{x^{x^\ldots}}\right)}= x^2=2

More generally, if
x^{x^{x^{x^\ldots}}}= a> 0
then x=\sqrt{a}.
almost:
x=\sqrt[a]{a}

[pmsrw3]

If
x^{x^{x^{x^\ldots}}}= 2
then
x^{\left(x^{x^{x^\ldots}}\right)}= x^2=2

More generally, if
x^{x^{x^{x^\ldots}}}= a> 0
then x=\sqrt{a}.
Actually, it should be a^{1/a}, no?

But this is true only if there is a solution. When I substitute \sqrt{2} for x and iterate it, it does indeed converge to 2. But when I try it for cube root of 3, it doesn't converge to 3. It converges, but to 2.47805. What's more, when I ask Mathematica for -\frac{\text{ProductLog}[-\text{Log}[z]]}{\text{Log}[z]}, which is supposed to give the infinite power tower of z (see http://mathworld.wolfram.com/PowerTower.html), it does in fact come to 2.47805. MathWorld says it converges only up to e^{1/e}. The interesting thing is, 2.47805^{1/2.47805} equals the cube root of 3.