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What is Euler's identity really saying? 
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#1
May412, 06:42 PM

P: 18

So it is true that e^{i∏}+1=0. But what does this mean? Why are all these numbers linked?



#2
May412, 06:50 PM

P: 606

They are linked precisely by that equation, and since the equality [itex]e^{i\theta}:=\cos\theta+i\sin\theta\,\,,\,\,\theta\in\mathbb{R}\,\,[/itex] follows at once say from the definition of the complex exponential function as power series (or as limit of a sequence), the above identity is really trivial. DonAntonio 


#3
May412, 07:30 PM

Math
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Thanks
PF Gold
P: 39,553

Look at the MacLaurin series for those functions:
[tex]e^x= 1+ x+ x^2/2!+ x^3/3!+ \cdot\cdot\cdot+ x^n/n![/tex] [tex]cos(x)= 1 x^2/2!+ x^4/4! x^6/6!+ \cdot\cdot\cdot+ (1)^nx^{2n}/(2n)![/tex] [tex]sin(x)= x x^3/3!+ x^5/5! x^7/7!+ \cdot\cdot\cdot+ (1)^nx^{2n+1}/(2n)![/tex] If you replace x with the imaginary number ix (x is still real) that becomes [tex]e^{ix}= 1+ ix+ (ix)^2/2!+ (ix)^3/3!+ \cdot\cdot\cdot+ (ix)^n/n![/tex] [tex]e^{ix}= 1+ ix+ i^2x^2/2!+ i^3x^3/3!+ \cdot\cdot\cdot+ i^nx^n/n![/tex] But it is easy to see that, since [itex]i^2= 1[/itex], [itex](i)^3= (i)^2(i)= i[/itex], [itex](i)^4= (i^3)(i)= i(i)= (1)= 1[/itex] so then it starts all over: [itex]i^5= (i^5)i= i[/itex], etc. That is, all even powers of i are 1 if the power is 0 mod 4 and 1 if it is 2 mod 4. All odd powers are i if the power is 1 mod 4 and i if it is 3 mod 4. [tex]e^{ix}= 1+ ix x^2/2! ix^3/3!+ \cdot\cdot\cdot[/tex] Separating into real and imaginary parts, [tex]e^{ix}= (1 x^2/2!+ x^4/4! x^6/6!+ \cdot\cdot\cdot)+ i(x x^3/3!+ x^5/5!+ \cdot\cdot\cdot)[/tex] [tex]e^{ix}= cos(x)+ i sin(x)[/tex] Now, take [itex]= \pi[/itex] so that [itex]cos(x)= cos(\pi)= 1[/itex] and [itex]sin(x)= sin(\pi)= 0[/itex] and that becomes [tex]e^{i\pi}= 1[/tex] or [tex]e^{i\pi}+ 1= 0[/tex] I hope that is what you are looking for. Otherwise, what you are asking is uncomfortably close to "number mysticism". 


#4
May512, 04:43 AM

P: 242

What is Euler's identity really saying?



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