- Cuchulainn
**Posts:**60757**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.

http://www.datasimfinancial.com

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

- Cuchulainn
**Posts:**60757**Joined:****Location:**Amsterdam-
**Contact:**

Solve the current problem as a least squares Genetic Algorithm optimisation problem using **floating-point genes.**

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

http://www.datasimfinancial.com

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

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:**60757**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.

http://www.datasimfinancial.com

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

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:**60757**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.

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

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:**60757**Joined:****Location:**Amsterdam-
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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?)

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

- FaridMoussaoui
**Posts:**473**Joined:****Location:**Genève, Genf, Ginevra, Geneva

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,

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

- katastrofa
**Posts:**8625**Joined:****Location:**Alpha Centauri

- Cuchulainn
**Posts:**60757**Joined:****Location:**Amsterdam-
**Contact:**

How fast can it compute [$]e^5[$]?

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

- katastrofa
**Posts:**8625**Joined:****Location:**Alpha Centauri

I cannot test it now, but it looks charming, doesn't it?

The article is old, so we were wondering with a couple of guys here if such methods haven't been implemented in modern compilers (e.g. --ffast-math flat in gcc). If no, the concept is worth soft-max bros' attention.

The article is old, so we were wondering with a couple of guys here if such methods haven't been implemented in modern compilers (e.g. --ffast-math flat in gcc). If no, the concept is worth soft-max bros' attention.

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