## Re: [MATH for FUN] i

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• Oh yeah.... Ø = pi/2 + k.2.pi....!!!! Carlos, the king of typos.... ... Ø = pi/2 + k.2.pi ? ... for ... http://docs.yahoo.com/info/terms/ ... To unsubscribe
Message 1 of 7 , May 5, 2003
Oh yeah....
Ø = pi/2 + k.2.pi....!!!!
Carlos, the king of typos....

----- Original Message -----

--- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
wrote:
> Analitically:
>
> By Euler's Formula
>
> e^(i.Ø)=cos Ø + i.sin Ø
>
> If we make i = e^(i.Ø) then
> i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi

Ø = pi/2 + k.2.pi ?

>
> Taking just the first solution:
>
> i = e^(i.pi/2)
>
>
> So i^i = [e^(i.pi/2)]^i
> i^i = e^(i².pi/2)
> i î = e^(-pi/2) = 0.207879576351......
>
> Carlos
> PS: Please somebody explain the other solutions for k<>0
>
> ----- Original Message -----
> > According to my trusted calculator: 0.207879576351
> > Carlos
> >
> > ----- Original Message -----
> > > anyone knows the value of i raised to the i th power? thanks
for
> >
> >
> >
> > To unsubscribe from this group, send an email to:
> > mathforfun-unsubscribe@yahoogroups.com
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >
> >

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• King of typos? You should see some of mine! ... http://docs.yahoo.com/info/terms/
Message 2 of 7 , May 6, 2003
King of typos? You should see some of mine!

--- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
wrote:
> Oh yeah....
> Ø = pi/2 + k.2.pi....!!!!
> Carlos, the king of typos....
>
> ----- Original Message -----
>
> --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
> wrote:
> > Analitically:
> >
> > By Euler's Formula
> >
> > e^(i.Ø)=cos Ø + i.sin Ø
> >
> > If we make i = e^(i.Ø) then
> > i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi
>
> Ø = pi/2 + k.2.pi ?
>
> >
> > Taking just the first solution:
> >
> > i = e^(i.pi/2)
> >
> >
> > So i^i = [e^(i.pi/2)]^i
> > i^i = e^(i².pi/2)
> > i E= e^(-pi/2) = 0.207879576351......
> >
> > Carlos
> > PS: Please somebody explain the other solutions for k<>0
> >
> > ----- Original Message -----
> > > According to my trusted calculator: 0.207879576351
> > > Carlos
> > >
> > > ----- Original Message -----
> > > From: "grouptheory2002" <no_reply@yahoogroups.com>
> > > > anyone knows the value of i raised to the i th power? thanks
> for
> > > > your help.
> > >
> > >
> > >
> > > To unsubscribe from this group, send an email to:
> > > mathforfun-unsubscribe@yahoogroups.com
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
> > >
> > >
> > >
> > >
>
>
>
> To unsubscribe from this group, send an email to:
> mathforfun-unsubscribe@yahoogroups.com
>
>
>
> Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
• hehehe....not easy to have 5 thumbs when at the keyboard....hehehe.... Carlos ... From: clooneman King of typos? You should see
Message 3 of 7 , May 6, 2003
hehehe....not easy to have 5 thumbs when at the keyboard....hehehe....
Carlos

----- Original Message -----
King of typos? You should see some of mine!

--- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
wrote:
> Oh yeah....
> Ø = pi/2 + k.2.pi....!!!!
> Carlos, the king of typos....
>
> ----- Original Message -----
>
> --- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
> wrote:
> > Analitically:
> >
> > By Euler's Formula
> >
> > e^(i.Ø)=cos Ø + i.sin Ø
> >
> > If we make i = e^(i.Ø) then
> > i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi
>
> Ø = pi/2 + k.2.pi ?
>
> >
> > Taking just the first solution:
> >
> > i = e^(i.pi/2)
> >
> >
> > So i^i = [e^(i.pi/2)]^i
> > i^i = e^(i².pi/2)
> > i E= e^(-pi/2) = 0.207879576351......
> >
> > Carlos
> > PS: Please somebody explain the other solutions for k<>0
> >
> > ----- Original Message -----
> > > According to my trusted calculator: 0.207879576351
> > > Carlos
> > >
> > > ----- Original Message -----
> > > From: "grouptheory2002" <no_reply@yahoogroups.com>
> > > > anyone knows the value of i raised to the i th power? thanks
> for
> > > > your help.
> > >
> > >
> > >
> > > To unsubscribe from this group, send an email to:
> > > mathforfun-unsubscribe@yahoogroups.com
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
> > >
> > >
> > >
> > >
>
>
>
> To unsubscribe from this group, send an email to:
> mathforfun-unsubscribe@yahoogroups.com
>
>
>
> Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/

To unsubscribe from this group, send an email to:
mathforfun-unsubscribe@yahoogroups.com

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• After all the nudists/Mother s Day/SARS/Iragi tees postings, back to math (BTW I am not new to the list) Multiple values of i^i are explained by multivalued
Message 4 of 7 , May 6, 2003
After all the nudists/Mother's Day/SARS/Iragi tees postings, back to
math (BTW I am not new to the list)

Multiple values of i^i are explained by multivalued complex
functions. e^z is periodic with period 2*pi*i, so e^(z+2*pi*i)=e^z.
When you define a^z as e^(z*log(a)) then the logarithm (the inverse
of e^z) is multivalued and so is a^z. Nothing much else to explain,
it is like the plus or minus for square roots.

--- In mathforfun@yahoogroups.com, "Carlos Assale" <c.assale@t...>
wrote:
> Analitically:
>
> By Euler's Formula
>
> e^(i.Ø)=cos Ø + i.sin Ø
>
> If we make i = e^(i.Ø) then
> i = cos Ø + i.sin Ø and Ø = pi/2 + k.pi
>
> Taking just the first solution:
>
> i = e^(i.pi/2)
>
>
> So i^i = [e^(i.pi/2)]^i
> i^i = e^(i².pi/2)
> i î = e^(-pi/2) = 0.207879576351......
>
> Carlos
> PS: Please somebody explain the other solutions for k<>0
>
> ----- Original Message -----
> > According to my trusted calculator: 0.207879576351
> > Carlos
> >
> > ----- Original Message -----
> > > anyone knows the value of i raised to the i th power? thanks
for
> >
> >
> >
> > To unsubscribe from this group, send an email to:
> > mathforfun-unsubscribe@yahoogroups.com
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >
> >
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