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Cuchulainn
Posts: 62635
Joined: July 16th, 2004, 7:38 am
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### Re: exp(5) = $e^5$

What I'm really saying I suppose: I did not come down in the last shower.
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ISayMoo
Posts: 2340
Joined: September 30th, 2015, 8:30 pm

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

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

I'm not angry, just sad.
What I'm really saying I suppose: I did not come down in the last shower.
I hope that's not an answer you'd give in a live seminar

Cuchulainn
Posts: 62635
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### Re: exp(5) = $e^5$

What I'm really saying I suppose: I did not come down in the last shower.
I hope that's not an answer you'd give in a live seminar
Never had a reason to. In speakers, honesty is the best policy. Don't dig yourself into a hole.
Step over the gap, not into it. Watch the space between platform and train.
http://www.datasimfinancial.com
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ISayMoo
Posts: 2340
Joined: September 30th, 2015, 8:30 pm

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

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: 62635
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### Re: exp(5) = $e^5$

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..
Step over the gap, not into it. Watch the space between platform and train.
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Cuchulainn
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### Re: exp(5) = $e^5$

Prove that $e$ is the number that maximises $\sqrt[x] {x}$ for $x > 0$.
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ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

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

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 $\frac{\log x}{x}$

Cuchulainn
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### Re: exp(5) = $e^5$

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 $\frac{\log x}{x}$
And if we continue with $log x/x < 1$ we conclude $\sqrt[x] {x} < e$ for any $x>=0$.

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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ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

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

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

http://www2.washjeff.edu/users/mwolterm ... rie/89.pdf
he does seem to have made a mountain out of a molehill.

Cuchulainn
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### Re: exp(5) = $e^5$

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?

$e^x$ is difficult with standard GD (bad answers) but $e^{-x}$ is better. But we see how the learning rate is important.
(using sigmoid here)

Step over the gap, not into it. Watch the space between platform and train.
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Cuchulainn
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### Re: exp(5) = $e^5$

cntd
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ExSan
Posts: 4574
Joined: April 12th, 2003, 10:40 am

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

π⁴ + π⁵ ≈ e⁶
to seven significant figures

Ref
Analysis Fact  @AnalysisFact

Cuchulainn
Posts: 62635
Joined: July 16th, 2004, 7:38 am
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### Re: exp(5) = $e^5$

π⁴ + π⁵ ≈ e⁶
to seven significant figures

Ref
Analysis Fact  @AnalysisFact
"Das ist nicht nur nicht richtig; es ist nicht einmal falsch!".
Step over the gap, not into it. Watch the space between platform and train.
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ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

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

π⁴ + π⁵ ≈ e⁶
to seven significant figures

Ref
Analysis Fact  @AnalysisFact
"Das ist nicht nur nicht richtig; es ist nicht einmal falsch!".
it's true but not necessarily useful. You want $e^{5}$ and this is a formula for $e^{6}$
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: 62635
Joined: July 16th, 2004, 7:38 am
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### Re: exp(5) = $e^5$

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.
Step over the gap, not into it. Watch the space between platform and train.
http://www.datasimfinancial.com
http://www.datasim.nl

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