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

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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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
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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.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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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
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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..

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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

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

- Cuchulainn
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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..

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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
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[$]e^x[$] 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)

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- Cuchulainn
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cntd

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to seven significant figures

Ref

- Cuchulainn
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"π⁴ + π⁵ ≈ e⁶

to seven significant figures

Ref

Analysis Fact@AnalysisFact

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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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
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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

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