Question About the Eth Root of E
Michael Ejercito
e rational?
Michael
Tim Little
> If e is the base of the natural logarithm, then is the eth root of e
> rational?
I don't know, and no simple arguments eitehr way have occurred to me.
I do know that it is not known whether some possibly "simpler"
expressions are rational, such as 2^e.
- Tim
A N Niel
<eb44ea05-e3b3-4ec2...@w40g2000yqd.googlegroups.com>,
Michael Ejercito <meje...@hotmail.com> wrote:
Of course it is irrational. Fortunately for me, you did not ask for a
proof!
Michael Ejercito
> In article
> <eb44ea05-e3b3-4ec2-94e3-90c4fe1f5...@w40g2000yqd.googlegroups.com>,
>
> Michael Ejercito<mejer...@hotmail.com> wrote:
> > If e is the base of the natural logarithm, then is the eth root of
> > e rational?
>
> > Michael
>
> Of course it is irrational. Fortunately for me, you did not ask for a
> proof!
Michael
pubkeybreaker
It would follow immediately from Shanuel's Conjecture.
If e^1/e is rational, then
e = 1/(log a - log b) for some integers a,b.
The LHS has transcendence degree 1, while the rhs has transcendence
degree 2
by Shanuel.
Otherwise:
I suspect that this is provable using the same technique
that Hermite used to prove e is transcendental. Treat
1/(log(x) - log(k)) as a infinite series for some fixed k
and show that its partial sums do not approximate e sufficiently
well for all k.
Michael Ejercito
> On Apr 10, 11:26 am,Michael Ejercito<mejer...@hotmail.com> wrote:
>
> > On Apr 10, 4:49 am, A N Niel <ann...@nym.alias.net.invalid> wrote:> In article
> > > <eb44ea05-e3b3-4ec2-94e3-90c4fe1f5...@w40g2000yqd.googlegroups.com>,
>
> > >Michael Ejercito<mejer...@hotmail.com> wrote:
> > > > If e is the base of the natural logarithm, then is the eth root of
> > > > e rational?
>
> > > > Michael
>
> > > Of course it is irrational. Fortunately for me, you did not ask for a
> > > proof!
>
> > What is the proof?
>
> It would follow immediately from Shanuel's Conjecture.
>
> If e^1/e is rational, then
> e = 1/(log a - log b) for some integers a,b.
both sides to the e power to get e=a^e/b^e, and then taking the common
logarithm of both sides yields log (e)=e log(a) + e log (b)
(approximately 0.434294481).
>
> The LHS has transcendence degree 1, while the rhs has transcendence
> degree 2
> by Shanuel.
>
> Otherwise:
>
> I suspect that this is provable using the same technique
> that Hermite used to prove e is transcendental. Treat
> 1/(log(x) - log(k)) as a infinite series for some fixed k
> and show that its partial sums do not approximate e sufficiently
> well for all k.
Michael
G. A. Edgar
<64389956-39e8-4ffa...@d38g2000prn.googlegroups.com>,
Michael Ejercito <meje...@hotmail.com> wrote:
> On Apr 10, 9:32 am, pubkeybreaker <pubkeybrea...@aol.com> wrote:
> > On Apr 10, 11:26 am,Michael Ejercito<mejer...@hotmail.com> wrote:
> >
> > > On Apr 10, 4:49 am, A N Niel <ann...@nym.alias.net.invalid> wrote:> In
> > > article
> > > > <eb44ea05-e3b3-4ec2-94e3-90c4fe1f5...@w40g2000yqd.googlegroups.com>,
> >
> > > >Michael Ejercito<mejer...@hotmail.com> wrote:
> > > > > If e is the base of the natural logarithm, then is the eth root of
> > > > > e rational?
> >
> > > > > Michael
> >
> > > > Of course it is irrational. Fortunately for me, you did not ask for a
> > > > proof!
> >
> > > What is the proof?
> >
> > It would follow immediately from Shanuel's Conjecture.
> >
> > If e^1/e is rational, then
> > e = 1/(log a - log b) for some integers a,b.
> How do you derive that? I put in the equation e^(1/e)=a/b, raised
> both sides to the e power to get e=a^e/b^e, and then taking the common
> logarithm of both sides
"log" is used for the natural logarithm by mathematicians
> yields log (e)=e log(a) + e log (b)
> (approximately 0.434294481).
log(e) = 1, solve for e
> >
> > The LHS has transcendence degree 1, while the rhs has transcendence
> > degree 2 by Shanuel.
> >
> > Otherwise:
> >
> > I suspect that this is provable using the same technique
> > that Hermite used to prove e is transcendental. Treat
> > 1/(log(x) - log(k)) as a infinite series for some fixed k
> > and show that its partial sums do not approximate e sufficiently
> > well for all k.
> What is the proof for Shaunel's Conjecture?
Unknown. That is why it is called a conjecture. But the "log" in it
is the natural logarithm.
>
>
> Michael
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Toni...@yahoo.com
> On Apr 10, 9:32 am, pubkeybreaker <pubkeybrea...@aol.com> wrote:
>
>
>
> > On Apr 10, 11:26 am,Michael Ejercito<mejer...@hotmail.com> wrote:
>
> > > On Apr 10, 4:49 am, A N Niel <ann...@nym.alias.net.invalid> wrote:> In article
> > > > <eb44ea05-e3b3-4ec2-94e3-90c4fe1f5...@w40g2000yqd.googlegroups.com>,
>
> > > >Michael Ejercito<mejer...@hotmail.com> wrote:
> > > > > If e is the base of the natural logarithm, then is the eth root of
> > > > > e rational?
>
> > > > > Michael
>
> > > > Of course it is irrational. Fortunately for me, you did not ask for a
> > > > proof!
>
> > > What is the proof?
>
> > It would follow immediately from Shanuel's Conjecture.
>
> > If e^1/e is rational, then
> > e = 1/(log a - log b) for some integers a,b.
>
> How do you derive that? I put in the equation e^(1/e)=a/b, raised
> both sides to the e power to get e=a^e/b^e, and then taking the common
> logarithm of both sides yields log (e)=e log(a) + e log (b)
> (approximately 0.434294481).
>
You got it wrong: log(A/B) = log A - log B, so:
e = a^e/b^e ==> 1 = log e = e[log a - log b], and from here the
expression for e.
Regards
Tonio
> > The LHS has transcendence degree 1, while the rhs has transcendence
> > degree 2
> > by Shanuel.
>
> > Otherwise:
>
> > I suspect that this is provable using the same technique
> > that Hermite used to prove e is transcendental. Treat
> > 1/(log(x) - log(k)) as a infinite series for some fixed k
> > and show that its partial sums do not approximate e sufficiently
> > well for all k.
>
> What is the proof for Shaunel's Conjecture?
>
> Michael- Hide quoted text -
>
> - Show quoted text -
Noone
<meje...@hotmail.com> wrote:
>On Apr 10, 9:32 am, pubkeybreaker <pubkeybrea...@aol.com> wrote:
>> On Apr 10, 11:26 am,Michael Ejercito<mejer...@hotmail.com> wrote:
>>
>> > On Apr 10, 4:49 am, A N Niel <ann...@nym.alias.net.invalid> wrote:> In article
>> > > <eb44ea05-e3b3-4ec2-94e3-90c4fe1f5...@w40g2000yqd.googlegroups.com>,
>>
>> > >Michael Ejercito<mejer...@hotmail.com> wrote:
>> > > > If e is the base of the natural logarithm, then is the eth root of
>> > > > e rational?
>>
>> > > > Michael
>>
>> > > Of course it is irrational. Fortunately for me, you did not ask for a
>> > > proof!
>>
>> > What is the proof?
>>
>> It would follow immediately from Shanuel's Conjecture.
>>
>> If e^1/e is rational, then
>> e = 1/(log a - log b) for some integers a,b.
> How do you derive that? I put in the equation e^(1/e)=a/b, raised
>both sides to the e power to get e=a^e/b^e, and then taking the common
>logarithm of both sides yields log (e)=e log(a) + e log (b)
>(approximately 0.434294481).
Why raise things to a power? Simply take the natural log of both
sides and realize that log(e) = 1.
e^(1/e) = a/b
1/e log(e) = log(a/b)
Solve for e.
>>
>> The LHS has transcendence degree 1, while the rhs has transcendence
>> degree 2
>> by Shanuel.
>>
>> Otherwise:
>>
>> I suspect that this is provable using the same technique
>> that Hermite used to prove e is transcendental. Treat
>> 1/(log(x) - log(k)) as a infinite series for some fixed k
>> and show that its partial sums do not approximate e sufficiently
>> well for all k.
> What is the proof for Shaunel's Conjecture?
A proof it not known, which is why it remains a conjecture.
>
>
> Michael
Michael Ejercito
> On Sat, 11 Apr 2009 09:48:19 -0700 (PDT), Michael Ejercito
>
>
>
> >On Apr 10, 9:32 am, pubkeybreaker <pubkeybrea...@aol.com> wrote:
> >> On Apr 10, 11:26 am,Michael Ejercito<mejer...@hotmail.com> wrote:
>
> >> > On Apr 10, 4:49 am, A N Niel <ann...@nym.alias.net.invalid> wrote:> In article
> >> > > <eb44ea05-e3b3-4ec2-94e3-90c4fe1f5...@w40g2000yqd.googlegroups.com>,
>
> >> > >Michael Ejercito<mejer...@hotmail.com> wrote:
> >> > > > If e is the base of the natural logarithm, then is the eth root of
> >> > > > e rational?
>
> >> > > > Michael
>
> >> > > Of course it is irrational. Fortunately for me, you did not ask for a
> >> > > proof!
>
> >> > What is the proof?
>
> >> It would follow immediately from Shanuel's Conjecture.
>
> >> If e^1/e is rational, then
> >> e = 1/(log a - log b) for some integers a,b.
> > How do you derive that? I put in the equation e^(1/e)=a/b, raised
> >both sides to the e power to get e=a^e/b^e, and then taking the common
> >logarithm of both sides yields log (e)=e log(a) + e log (b)
> >(approximately 0.434294481).
>
> Why raise things to a power? Simply take the natural log of both
> sides and realize that log(e) = 1.
>
when I enter 2.718281828 into my computer's calculator and click
the log function key, I get 0.434294481.
Michael
A N Niel
<d6337140-0c4d-4b2b...@d7g2000prl.googlegroups.com>,
Michael Ejercito <meje...@hotmail.com> wrote:
> when I enter 2.718281828 into my computer's calculator and click
> the log function key, I get 0.434294481.
Your calculator seems to use the common log. Mathematicians use the
natural log. See if your calculator has a "ln" key for the natural
log.