Euler Equation

Date: 09/13/97 at 17:19:49
From: Oliver Dale
Subject: Euler function and radians

I have seen the famous Euler equation e^i*pi-1 = 0 and know something 
abut its derivation. Specifically, it follows from the infinite series 
that define sin, cos and e^x. My question is: does the equation still 
work if we decide to work in degrees (i.e. is e^i*90-1=0 true)?  

I ask because the mysterious appearance of e, i, and pi in one 
equation isn't so strange if it is partly definitional. I believe the 
question boils down to this: do the infinite series definitions of sin 
and cos 

   (cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ..., and 
    sin(x) = x - x^3/3!  + x^5/5!  - x^7/7! + ...) 

work if x is expressed in degrees, or are they only valid when x is a 
radian expression?

Date: 09/13/97 at 19:04:56
From: Doctor Anthony
Subject: Re: Euler function and radians

The Euler equation is  e^(i.pi)+1 = 0

The series you quote for cos(x) and sin(x) require x to be in radians.  
Radian measure is non-dimensional, being the ratio of two lengths, 
arc length divided by radius, just as sin and cos are non-dimensional, 
again being ratios of lengths.  It is necessary, of course, that the 
two sides of the equation have the same dimensions - in this case both 
are non-dimensional. 

Degrees are not mathematical in any real sense.  They are arbitrary 
conventions and, historically, mathematicians could have agreed to 
have 500 'degrees' in a circle or any number you care to think of.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 09/18/97 at 21:39:35
From: Oliver Dale
Subject: Re: Euler function and radians

Thank you very much for your reply. It was quite helpful. I am still
a little confused.  I have heard in a number of places that the Euler
function is in some way arbitrary. From the information in your email,
it seems to me that one could fairly easily prove the truth of the
equation. To me that indicates that the Euler function tells us
something "real" and not constructed. Am I missing something?

Thanks again for the help you have already given.

Oliver Dale

Date: 09/19/97 at 07:41:35
From: Doctor Anthony
Subject: Re: Euler function and radians

The Euler equation is indeed something real. The derivation is not too 
difficult if you are familiar with the basics of complex numbers and 
exponential functions.

Start with        z = cos(x) + i.sin(x)    ......(1)

              dz/dx = -sin(x) + i.cos(x)

                    = i(cos(x) + i.sin(x))    (since i^2 = -1)

                    = i.z

          So   dz/z = i.dx          now integrate both sides

              ln(z) = i.x + const.  from (1) when x= 0, z= 1  
                                  so const. = 0
              ln(z) = i.x

                  z = e^(i.x)    but z = cos(x) + i.sin(x) so

  cos(x) + i.sin(x) = e^(i.x) .........(2)

Put x = pi in (2) and we get:

             -1 + 0 = e^(i.pi)

and so e^(i.pi) + 1 = 0

As you can see there is nothing contrived about this equation.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/