Euler EquationDate: 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/ |
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