Register to reply 
Why does e^(i*pi)+1=0?by jsmith
Tags: eipi 
Share this thread: 
#1
Jan2114, 01:28 PM

P: 4

I will first say that I fully understand how to prove this equation from the use of power series, what I am interested in though is why e and pi should be linked like they are. As far as I know pi comes from geometry (although it does have an equivalent analytical definition), and e comes from calculus. I cannot see any reason why they should be linked and the proof doesn't really give any insights as to why the equation works. Is there some nice way of explaining this?



#2
Jan2114, 01:39 PM

Mentor
P: 15,067

It's just a special case of ##\exp(ix) = \cos x + i\sin x## at x=pi.
The Taylor expansions of the exponential function and the trigonometric functions are very similar. The Taylor series of sin(x) is very similar to odd terms of the Taylor series of exp(x). The only difference is that the sine series alternates in sign. The same is true for the Taylor series of cos(x) and the even terms of the Taylor series of exp(x). There's obviously a connection between the exponential function and those trigonometric functions. Those sign alternations vanish when one looks at exp(ix) as opposed to exp(x). The real and imaginary parts of exp(ix) are the trigonometric functions. 


#3
Jan2114, 02:51 PM

Sci Advisor
P: 6,040

If you look at the complex plane, then exp(ix) represents points on the unit circle. When x = π, the point on the circle is 1.



#4
Jan2114, 03:00 PM

P: 2,179

Why does e^(i*pi)+1=0?



#5
Jan2114, 03:15 PM

P: 181

A geometric way to think of the identity [itex] e^{i\pi}=1 [/itex] is that magnitudes multiply and arguments add when multiplying complex numbers.
[tex] e^{i\pi}= \lim_{n \to \infty} (1+\frac{i\pi}{n})^n= \lim_{n \to \infty} (\sqrt{1+(\frac{\pi^2}{n^2})}z)^n [/tex] where [itex] z [/itex] is a complex number with unit magnitude and argument [itex] \frac{\pi}{n} [/itex] (equal to ([itex] cos\frac{\pi}{n})+isin(\frac{\pi}{n})[/itex]). So, [tex] \lim_{n \to \infty} (\sqrt{1+\frac{\pi^2}{n^2}}z)^n= \lim_{n \to \infty} (1+(\frac{\pi^2}{n^2}))^\frac{n}{2}z^n=\lim_{n \to \infty} z^n [/tex]. Since [itex] z [/itex] has magnitude 1 and argument [itex] \frac{\pi}{n} [/itex], the limit has magnitude 1 and argument [itex] \pi [/itex]. Therefore [itex] e^{i\pi}=1 [/itex]. Note: I adapted this argument from The Princeton Companion to Mathematics. I thought it was neat and that I should share it. Choosing a general angle instead of [itex] \pi [/itex] gives the more general formula [itex] e^{i\theta}=cos\theta + isin\theta [/itex]. 


#6
Jan2214, 10:34 AM

P: 1,291

The fact that we don't see any reason why the two (or three!) concepts should be linked is what makes the equality so famous.
There is a deep, underlying unity in a lot of math, that may be hard to see as long as we look at these things from a viewpoint of what problems they were invented to solve. 


#7
Jan2214, 01:00 PM

P: 4

Thanks for the replies. I do understand fully why this identity holds. I guess my question is why should a completely geometric consept (pi) be linked so nicely with a completely analytical constant (e). Thinking about it I am wondering if this is anything to do with the fact that e is defined by e^x is its own derivative, and using the geometric interpretation of differentiation? Does this somehow give a link between e^x and the trig functions?



#8
Jan2314, 10:54 AM

P: 82

I thought about this a lot too once, though the best I could come up with is like this. Multiplying by i turns a number 90 degrees around the complex plane. Exponentiation with real multiples of i is essentially continuous multiplication such that you are "smoothly" turning around the complex plane in a spiral shape. As mathman said:



#9
Jan2614, 02:12 AM

P: 5




Register to reply 