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### Re: exp(5) = [$]e^5[$]

Posted: September 24th, 2018, 8:10 am
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.

### Re: exp(5) = [$]e^5[$]

Posted: October 3rd, 2018, 8:31 pm
Solve the current problem as a least squares Genetic Algorithm optimisation  problem using floating-point genes.

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

### Re: exp(5) = [$]e^5[$]

Posted: October 25th, 2018, 10:40 pm
Compute [$]e^{\pi}[$] to 2 decimal places with pencil and paper. (Gelfond's constant)
A follow-on from Exsan's post and ansatz (big conjecture but in the right direction) is whether [$]\pi[$] and [$]e[$] are algebraically 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?
what a puzzle !

### Re: exp(5) = [$]e^5[$]

Posted: December 4th, 2018, 11:31 pm
Compute [$]e^{\pi}[$] to 2 decimal places with pencil and paper. (Gelfond's constant)
A follow-on from Exsan's post and ansatz (big conjecture but in the right direction) is whether [$]\pi[$] and [$]e[$] are algebraically 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?
what a puzzle !
Have you solved it yet?

### Re: exp(5) = [$]e^5[$]

Posted: December 5th, 2018, 10:56 am
Compute [$]e^{\pi}[$] to 2 decimal places with pencil and paper. (Gelfond's constant)
A follow-on from Exsan's post and ansatz (big conjecture but in the right direction) is whether [$]\pi[$] and [$]e[$] are algebraically 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?
what a puzzle !
Have you solved it yet?
I have no idea how to do it. This is very interesting problem to work on

### Re: exp(5) = [$]e^5[$]

Posted: December 5th, 2018, 1:26 pm
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.

### Re: exp(5) = [$]e^5[$]

Posted: December 5th, 2018, 2:15 pm
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 [$]\mathbb {Q}[$].

### Re: exp(5) = [$]e^5[$]

Posted: December 15th, 2018, 12:29 pm
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.
Maestro !

### Re: exp(5) = [$]e^5[$]

Posted: February 6th, 2019, 2:24 pm
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?)

### Re: exp(5) = [$]e^5[$]

Posted: February 6th, 2019, 2:37 pm
Mathematica said [$] u(t) = \sqrt{2} ~ InverseErf (\sqrt{\frac{2}{\pi}} ( t + C)) [$] where InverseErf is the inverse error function.

### Re: exp(5) = [$]e^5[$]

Posted: February 6th, 2019, 4:16 pm
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?)
bow not bough? ship's bow

it's first order separable
[$]du/dt = e^{u^2/2}[$]

[$]e^{-u^2/2}du=dt[$]
integrate both sides
[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)=t+C[$]

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

### Re: exp(5) = [$]e^5[$]

Posted: February 9th, 2019, 10:01 pm
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?)
bow not bough? ship's bow

it's first order separable
[$]du/dt = e^{u^2/2}[$]

[$]e^{-u^2/2}du=dt[$]
integrate both sides
[$]\sqrt{\pi/2}\,\textrm{erf}\left(u/\sqrt{2}\right)=t+C[$]

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!!!

### Re: exp(5) = [$]e^5[$]

Posted: July 30th, 2019, 1:26 pm

### Re: exp(5) = [$]e^5[$]

Posted: August 1st, 2019, 2:10 pm
How fast can it compute [$]e^5[$]?

### Re: exp(5) = [$]e^5[$]

Posted: August 2nd, 2019, 12:24 am
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.