- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

What I'm really saying I suppose: I did not come down in the last shower.

I think that you won't get a response unless you challenge them publicly.

I'm not angry, just sad.

I'm not angry, just sad.

I hope that's not an answer you'd give in a live seminarWhat I'm really saying I suppose: I did not come down in the last shower.

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

Never had a reason to. In speakers, honesty is the best policy. Don't dig yourself into a hole.I hope that's not an answer you'd give in a live seminarWhat I'm really saying I suppose: I did not come down in the last shower.

Sure. What does it have to do with the question whether your answer was relevant? The paper was discussing properties of minimas, not optimisation.

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

Weird answer.The paper was discussing properties of minimas, not optimisation.

Not at all; it's about DL-PDE(See the DL-PDE thread). Besides, what is your Venn diagram for the above?

I fear my questions will remain unanswered..

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

Prove that [$]e[$] is the number that maximises [$]\sqrt[x] {x}[$] for [$]x > 0[$].

[$]x^ {1/x}=\exp\left[\frac{\log x}{x}\right][$] so ,maximize [$]\frac{\log x}{x}[$]Prove that [$]e[$] is the number that maximises [$]\sqrt[x] {x}[$] for [$]x > 0[$].

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

And if we continue with [$]log x/x < 1 [$] we conclude [$]\sqrt[x] {x} < e [$] for any [$]x>=0[$].[$]x^ {1/x}=\exp\left[\frac{\log x}{x}\right][$] so ,maximize [$]\frac{\log x}{x}[$]Prove that [$]e[$] is the number that maximises [$]\sqrt[x] {x}[$] for [$]x > 0[$].

Compare with Steiner's proof in Crelle's journal (overkill?)

http://www2.washjeff.edu/users/mwolterm ... rie/89.pdf

BTW [$]\pi(x) = x/log x[$] is the (conjectured) number of primes that do not exceed x. Collector, are you there?

// The Gozilla number cruncher minimises [$]- \sqrt[x] {x}[$] as 2.7182818..

he does seem to have made a mountain out of a molehill.Compare with Steiner's proof in Crelle's journal (overkill?)

http://www2.washjeff.edu/users/mwolterm ... rie/89.pdf

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

[$]e^x[$] is difficult with standard GD (bad answers) but [$]e^{-x}[$] is better. But we see how the learning rate is important.The only solution left is tie solve this problem using (a hand-crafted?) ML algorithm?

Can I just look at the diagram and give the answer?

(using sigmoid here)

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

to seven significant figures

Ref

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

"π⁴ + π⁵ ≈ e⁶

to seven significant figures

Ref

Analysis Fact@AnalysisFact

it's true but not necessarily useful. You want [$]e^{5}[$] and this is a formula for [$]e^{6}[$]"π⁴ + π⁵ ≈ e⁶

to seven significant figures

Ref

Analysis Fact@AnalysisFactDas ist nicht nur nicht richtig; es ist nicht einmal falsch!".

Computing [$]\pi^4[$] and [$]\pi^5[$] may be difficult as well

On my machine

[$]\pi^4+\pi^5= 403.4287761[$]

[$]e^6=403.4287935[$]

- Cuchulainn
**Posts:**58122**Joined:****Location:**Amsterdam-
**Contact:**

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.*

A follow-on from Exsan's post and

or

GZIP: On