- Cuchulainn
**Posts:**58700**Joined:****Location:**Amsterdam-
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Compute the derivative of [$]e^{e^{e^{x}}}[$]. at x = 0.01. It comes from book that says AD is better. Calculus and FD is a nightmare.

- Cuchulainn
**Posts:**58700**Joined:****Location:**Amsterdam-
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Solve the current problem as a least squares Genetic Algorithm optimisation problem using **floating-point genes.**

**[$]min (e^5 - x)^2[$]**

what a puzzle !Compute [$]e^{\pi}[$] to 2 decimal places with pencil and paper. (Gelfond's constant)

A follow-on from Exsan's post andansatz (big conjecture but in the right direction)is whether [$]\pi[$] and [$]e[$] arealgebraically independent? i.e. is there a polynomial relation

[$]a_{n}\pi^{n} + a_{n-1}\pi^{n-1} + ... + a_{0}\pi^{0} = e^5[$]

or

[$]a_{n}e^{n} + a_{n-1}e^{n-1} + ... + a_{0}e^{0} = \pi[$]

where [$]a_{j} , j = 0,..,n[$] are algebraic numbers?

Whichever one takes your fancy.

- Cuchulainn
**Posts:**58700**Joined:****Location:**Amsterdam-
**Contact:**

Have you solved it yet?what a puzzle !Compute [$]e^{\pi}[$] to 2 decimal places with pencil and paper. (Gelfond's constant)

A follow-on from Exsan's post andansatz (big conjecture but in the right direction)is whether [$]\pi[$] and [$]e[$] arealgebraically independent? i.e. is there a polynomial relation

[$]a_{n}\pi^{n} + a_{n-1}\pi^{n-1} + ... + a_{0}\pi^{0} = e^5[$]

or

[$]a_{n}e^{n} + a_{n-1}e^{n-1} + ... + a_{0}e^{0} = \pi[$]

where [$]a_{j} , j = 0,..,n[$] are algebraic numbers?

Whichever one takes your fancy.

I have no idea how to do it. This is very interesting problem to work onHave you solved it yet?what a puzzle !Compute [$]e^{\pi}[$] to 2 decimal places with pencil and paper. (Gelfond's constant)

A follow-on from Exsan's post andansatz (big conjecture but in the right direction)is whether [$]\pi[$] and [$]e[$] arealgebraically independent? i.e. is there a polynomial relation

[$]a_{n}\pi^{n} + a_{n-1}\pi^{n-1} + ... + a_{0}\pi^{0} = e^5[$]

or

[$]a_{n}e^{n} + a_{n-1}e^{n-1} + ... + a_{0}e^{0} = \pi[$]

where [$]a_{j} , j = 0,..,n[$] are algebraic numbers?

Whichever one takes your fancy.

- Cuchulainn
**Posts:**58700**Joined:****Location:**Amsterdam-
**Contact:**

Wed Dec 05, 2018 1:11 pm

Let's take a step backwards.

Gradient descent methods as we know and love them are nothing more than Euler's (ugh) method applied to ODEs (aka*gradient system*):

[$]dx/dt = - grad f(x) = -\nabla f(x)[$] where [$]f[$] is the function to be minimised.

The local minima of [$]f[$] are the*critical points* of this ODE system. (Poincaré-Lyapunov-Bendixson theory).

I have tried it on a number of benchmark unconstrained optimisation problem and solved using C++ boost::odeint without any of the infamous learning rate fudges.

We use more robust ODE solvers than Euler which means that we don't get the well-documented issues with GD methods.

For the evergreen [$]e^5[$] we have ODE [$]dx/dt = -\nabla(x-e^5)^2 = -2(x - e^5)[$]. We get x = 148.413 for any initial value of the ODE.

Let's take a step backwards.

Gradient descent methods as we know and love them are nothing more than Euler's (ugh) method applied to ODEs (aka

[$]dx/dt = - grad f(x) = -\nabla f(x)[$] where [$]f[$] is the function to be minimised.

The local minima of [$]f[$] are the

I have tried it on a number of benchmark unconstrained optimisation problem and solved using C++ boost::odeint without any of the infamous learning rate fudges.

We use more robust ODE solvers than Euler which means that we don't get the well-documented issues with GD methods.

For the evergreen [$]e^5[$] we have ODE [$]dx/dt = -\nabla(x-e^5)^2 = -2(x - e^5)[$]. We get x = 148.413 for any initial value of the ODE.

Algebraic independence of known constants

Although both [$]\pi[$] and [$]e[$] are known to be transcendental, it is not known whether the set of both of them is algebraically independent over [$]\mathbb {Q}[$]. In fact, it is not even known if [$]\pi +e[$] is irrational. Nesterenko proved in 1996 that:

the numbers [$]π[$], [$]e^{\pi}[$], and [$]\Gamma(1/4)[$] are algebraically independent over [$]\mathbb {Q}[$].

the numbers [$]π[$], [$]e^{\pi\sqrt{3}}[$], and [$]\Gamma(1/3)[$] are algebraically independent over [$]\mathbb {Q}[$].

for all positive integers [$]n[$], the numbers [$]π[$], [$]e^{\pi\sqrt{n}}[$] are algebraically independent over [$]\mathbb {Q}[$].

Maestro !Wed Dec 05, 2018 1:11 pm

Let's take a step backwards.

Gradient descent methods as we know and love them are nothing more than Euler's (ugh) method applied to ODEs (akagradient system):

[$]dx/dt = - grad f(x) = -\nabla f(x)[$] where [$]f[$] is the function to be minimised.

The local minima of [$]f[$] are thecritical pointsof this ODE system. (Poincaré-Lyapunov-Bendixson theory).

I have tried it on a number of benchmark unconstrained optimisation problem and solved using C++ boost::odeint without any of the infamous learning rate fudges.

We use more robust ODE solvers than Euler which means that we don't get the well-documented issues with GD methods.

For the evergreen [$]e^5[$] we have ODE [$]dx/dt = -\nabla(x-e^5)^2 = -2(x - e^5)[$]. We get x = 148.413 for any initial value of the ODE.

- Cuchulainn
**Posts:**58700**Joined:****Location:**Amsterdam-
**Contact:**

Does the ODE

[$]du/dt = e^{u^2/2}[$] have a soluion in which interval?

(e.g. as a first shot across boughs, maybe a symbolic solution in Maple?)

[$]du/dt = e^{u^2/2}[$] have a soluion in which interval?

(e.g. as a first shot across boughs, maybe a symbolic solution in Maple?)

- FaridMoussaoui
**Posts:**364**Joined:**

Mathematica said [$] u(t) = \sqrt{2} ~ InverseErf (\sqrt{\frac{2}{\pi}} ( t + C)) [$] where InverseErf is the inverse error function.

bow not bough? ship's bowDoes the ODE

[$]du/dt = e^{u^2/2}[$] have a soluion in which interval?

(e.g. as a first shot across boughs, maybe a symbolic solution in Maple?)

it's first order separable

[$]du/dt = e^{u^2/2}[$]

[$]e^{-u^2/2}du=dt[$]

integrate both sides

[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)=t+C[$]

edit: added

as to the domain, you know [$]-1\le \textrm{erf}(x)\le 1[$]

so if you have an initial condition [$]u_{0}[$] at [$]t_{0}[$],

[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)=t-t_{0}[$]

[$]t=t_{0}+\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]

so the solution is valid for [$]t[$] in the range

[$]t_{0}-\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)\le t

\le t_{0}+\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]

or in terms of [$]\textrm{erfc}(x)=1-\textrm{erf}(x)[$] which is between 0 and 2

[$]t_{0}-\sqrt{\pi/2}\,\textrm{erfc}\left(-u_{0}/\sqrt{2}\right)\le t

\le t_{0}+\sqrt{\pi/2}\,\textrm{erfc}\left(u_{0}/\sqrt{2}\right)[$]

a very rough limit would be to replace erfc by 2,

[$]t_{0}-\sqrt{2\pi}\le t\le t_{0}+\sqrt{2\pi}[$]

good!!!bow not bough? ship's bowDoes the ODE

[$]du/dt = e^{u^2/2}[$] have a soluion in which interval?

(e.g. as a first shot across boughs, maybe a symbolic solution in Maple?)

it's first order separable

[$]du/dt = e^{u^2/2}[$]

[$]e^{-u^2/2}du=dt[$]

integrate both sides

[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)=t+C[$]

edit: added

as to the domain, you know [$]-1\le \textrm{erf}(x)\le 1[$]

so if you have an initial condition [$]u_{0}[$] at [$]t_{0}[$],

[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)=t-t_{0}[$]

[$]t=t_{0}+\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]

so the solution is valid for [$]t[$] in the range

[$]t_{0}-\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)\le t

\le t_{0}+\sqrt{\pi/2}-\sqrt{\pi/2}\,\textrm{erf}\left(u_{0}/\sqrt{2}\right)[$]

or in terms of [$]\textrm{erfc}(x)=1-\textrm{erf}(x)[$] which is between 0 and 2

[$]t_{0}-\sqrt{\pi/2}\,\textrm{erfc}\left(-u_{0}/\sqrt{2}\right)\le t

\le t_{0}+\sqrt{\pi/2}\,\textrm{erfc}\left(u_{0}/\sqrt{2}\right)[$]

a very rough limit would be to replace erfc by 2,

[$]t_{0}-\sqrt{2\pi}\le t\le t_{0}+\sqrt{2\pi}[$]

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