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What is Euler's identity really saying?

  1. May 4, 2012 #1
    So it is true that ei∏+1=0. But what does this mean? Why are all these numbers linked?
  2. jcsd
  3. May 4, 2012 #2

    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.

  4. May 4, 2012 #3


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    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]
    [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".
    Last edited by a moderator: May 4, 2012
  5. May 5, 2012 #4
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