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Collector
Posts: 4104
Joined: August 21st, 2001, 12:37 pm

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

One glaring omission is that no one has mentioned how $e$ came to be. Everyone blames Euler (some blame APL) but its roots are probably due to a 17th century banker or trader who stumbled on compound interest:

$\displaystyle\lim_{n\to\infty} (1 + r/n)^{nt} \rightarrow e^{rt}$

Then the special case gives us

$\displaystyle\lim_{n\to\infty} (1 + 1/n)^{n} == e$
This means (from atomist quantum) point of view that e is time (or alternatively space) dependent $2\ge e <= \lim_{n\to\infty} (1 + 1/n)^{n}$.
e for one second is $(1 + \frac{1}{\frac{c}{l_p}})^{\frac{c}{l_p}} =(1 + \frac{l_p}{c})^{\frac{c}{l_p}} == \mbox{True } e \approx (1 +5.39106\times 10^{-44} )^{1.85492\times10^{43}} \approx \lim_{n\to\infty} (1 + r/n)^{n}\approx 2.718281828$
e for one Planck second must be $(1 + 1/n)^{n}= (1 + 1)^{1} == 2$ (me-2)

True $e^5==2^5==32$ for one Planck second.

(PS The limit of n is 1 inside one Planck second and inside one Planck mass particle, the turning point of light.

everything else is opinions, bad smelling onions and approximations

Dont worry your approximation methods will work very well until we are closing in on the Planck scale.

"How to compute $e^5$ with two decimal places?"  ==32.00 ($or kr your choice) or are u guys working on the elementary charge? sorry I just discovered this thread, is it new? Skål! Last edited by Collector on November 5th, 2017, 9:49 pm, edited 6 times in total. Cuchulainn Posts: 58126 Joined: July 16th, 2004, 7:38 am Location: Amsterdam Contact: Re: exp(5) = $e^5$ In Orson Scott Card’s Ender novels, one of the main characters grows very rich and lives for thousands of years by spending most of his time in interstellar travel at relativistic speeds. It seemed to me like it was so easy to earn fantastic sums through compound interest that more people would adopt that tactic and interest rates would plummet so that it no longer worked. The cost of course is that of travelling into the future and thus being left behind by social change, having your friends and family age and die, etc. Collector Posts: 4104 Joined: August 21st, 2001, 12:37 pm Re: exp(5) = $e^5$ Mathematica high precision 1 second-max resolution e = Overflow occurred in computation Mathematica high precision 1 second electron-second-resolution e = Overflow occurred in computation Mathematica exact precision 1 Planck-second e == 2.00 each space-time window has its own e The very depth of reality is simple, binary and even e is 2 there! How can we talk about a limit in $n$ without also a limit in time (a observational time window limit), totally absurd and only possible with math-magic in Mathmagic Land! Cuchulainn Posts: 58126 Joined: July 16th, 2004, 7:38 am Location: Amsterdam Contact: Re: exp(5) = $e^5$ Put$1 in a bank account yielding exactly 2% continuously compounded and then check the account balance at noon in 252 years and 179 days.
Very good answer. I had not read this until this morning. But funnily the same idea crossed my mind yesterday.

So, the question was: How long does it take to earn $148.413 by investing$1 at a fixed rate of 2%?
Yep!

I also pondered if other exponential processes might be adapted to answer this question.  But base-2 doubling requires an inconvenient 5/ln(2) = 7.213475 doublings to get the answer.  And looking for the right seed for a Fibonacci process seemed to yield an answer of starting with 7457 and 12067 which after 10 cycles reaches 10000*e^5 with requisite accuracy but it feel like a hack!
It is conjectured that e is normal, i.e. when expressed in e.g. base 2 the digits are uniformly distributed, Homework.
On a related solution, how about computing e by Montmort's derangements (hat check problem).

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

* Another solution; $e$ is the global minimum of $x^x$ for $x \geq 0$

So, yer standard optimisers are a load of junk, we use differential evolution witth a population of 1000 and 1000 generations to give 0.3678944 80131139.

Cuchulainn
Posts: 58126
Joined: July 16th, 2004, 7:38 am
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Re: exp(5) = $e^5$

each space-time window has its own e

Big $e^5$ndians and little $e^5$ndians, what?

Collector
Posts: 4104
Joined: August 21st, 2001, 12:37 pm

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

each space-time window has its own e

Big $e^5$ndians and little $e^5$ndians, what?
right on!

Cuchulainn
Posts: 58126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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Re: exp(5) = $e^5$

* An opportunity for interval arithmetists

$(1 + 1/x)^{x} < e < (1 + 1/x)^{x+1}, x > 0.$

1, Show it is unique
2. Find an algorithm to bracket e to 0.01 accuracy.
Last edited by Cuchulainn on November 6th, 2017, 11:10 am, edited 1 time in total.

Cuchulainn
Posts: 58126
Joined: July 16th, 2004, 7:38 am
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Re: exp(5) = $e^5$

Another way is Bernouili trials. e.g. use formula 7.3 page 281 (take n big) in @Collector's fine book (tusen takk) and it will approximate (put k = 0).

$1/e == \displaystyle\lim_{n\to\infty} (1 - 1/n)^{n}$.

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

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

* An opportunity for interval arithmetists

$(1 + 1/x)^{x} < e < (1 + 1/x)^{x+1}, x > 0.$

1, Show it is unique
2. Find an algorithm to bracket e to 0.01 accuracy.
Or one can go straight for a solution to the OP's question:
$(1 + 1/x)^{5*x} < e^{5} < (1 + 1/x)^{5*x+5}, x > 0.$

Cuchulainn
Posts: 58126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

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

* An opportunity for interval arithmetists

$(1 + 1/x)^{x} < e < (1 + 1/x)^{x+1}, x > 0.$

1, Show it is unique
2. Find an algorithm to bracket e to 0.01 accuracy.
Or one can go straight for a solution to the OP's question:
$(1 + 1/x)^{5*x} < e^{5} < (1 + 1/x)^{5*x+5}, x > 0.$
That would be the corollary, indeed. The function is monotone, so we are in luck.
And tight bounds is the 148 million dollar question. How to compute, Will Smith.

Collector
Posts: 4104
Joined: August 21st, 2001, 12:37 pm

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

$\pi^5$ more interesting ?

"good" old speculative numerology:

Cuchulainn
Posts: 58126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

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

...n/a
Last edited by Cuchulainn on November 7th, 2017, 9:59 pm, edited 1 time in total.

Cuchulainn
Posts: 58126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

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

$\pi^5$ more interesting ?

"good" old speculative numerology:

http://www.espenhaug.com/BlackHoleHedgeFund.html

Collector
Posts: 4104
Joined: August 21st, 2001, 12:37 pm

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

$e^w=e^5=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$

This corresponds to a rapidity of w=5, that again correspond to a quite high relativistic doppler shift and a velocity ratio of
$v/c=\tanh(5)\approx$ c x 0.999909204/c

someone would possibly call this a Pseudorapidity , but that is something different Pseudorapidity

in my max velocity theory: for rest-mass particles with this as max rapidity correspond to a hypothetical particle with reduced Compton wavelength of $\bar{\lambda}=\frac{1}{2}e^5l_p$

anyway I leave $e^5$ to people doing sinful usury and mathematicians "that only are dealing with the structure of the reasoning...that don't even need to know what they are talking about...or what they they say is true.."

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Last edited by Collector on November 7th, 2017, 11:55 pm, edited 9 times in total.

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