1 is in the modified version I provided:  e^(t*i) - 1 = 0

Unless you were reading that as e^(t*i) +  (-1) = 0

Also, if the more "important numbers" that can be included, the more
beautiful you find the equation, we can also throw in 2, arguably the next
most important number: e^(2*t*i) - 1 = 0, but I don't think trying to
include as many important numbers into one equation as possible is what
makes for an elegant equation.  What makes for an elegant equation is
showing an important connection between two concepts.  e^(t*i) = e^(0) = 1,
but t*i != 0.  This is much more surprising than if you try the same with
Pi, as you will find ln(e^(Pi*i)) = Pi*i, but ln(e^(t*i)) = 0.

Jason



On Thu, Jul 11, 2013 at 1:46 PM, John Clark <johnkcl...@gmail.com> wrote:

> On Thu, Jul 11, 2013 at 1:44 PM, Jason Resch <jasonre...@gmail.com> wrote:
>
>  > If you want to see all the constants at once there is an easy
>>>> correction:  e^(t*i) - 1 = 0
>>>>
>>>
>>>  Then it has the additive identity but not the multiplicative identity
>>> and I still prefer Euler's original.
>>>
>>>
>>
>> What is the mutliplicative identity in the original that is missing from
>> this one?
>>
>
> 1.
>
>    John K Clark
>
>
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