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

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.
 
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Cuchulainn
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Re: exp(5) = [$]e^5[$]

October 3rd, 2018, 8:31 pm

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

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

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ExSan
Posts: 4521
Joined: April 12th, 2003, 10:40 am

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

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?
Whichever one takes your fancy.
what a puzzle !  
 
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Cuchulainn
Posts: 57288
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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Re: exp(5) = [$]e^5[$]

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?
Whichever one takes your fancy.
what a puzzle !  
Have you solved it yet?
 
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ExSan
Posts: 4521
Joined: April 12th, 2003, 10:40 am

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

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?
Whichever one takes your fancy.
what a puzzle !  
Have you solved it yet?
I have no idea how to do it. This is very interesting problem to work on 
 
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Cuchulainn
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Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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Re: exp(5) = [$]e^5[$]

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.
 
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ppauper
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Joined: November 15th, 2001, 1:29 pm

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

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 [$]\mathbb {Q}[$].
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