Comic: e = 1

superseniorsmath.png

Inspired by a conversation a bunch of us were having a while back. Can you figure out where the flaw in the logic is?

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3 Responses to Comic: e = 1

  1. Abhinav says:

    The last step is flawed. 1^ 1/(2i*pi) is not exclusively 1 (i.e. there are other roots).

  2. vlad says:

    third step is false

  3. Tim says:

    e^(i*pi)=-1 so e^(2*i*pi)=1 and therefore one of the solutions to the equation x^(2*i*pi)= 1 is x=e. Another solution is x=1. So you contruct the equation using one solution (e) and then present another solution (1). The flaw is to assume that these two solutions must be identical.

    (Explicitly, let z=e^iv. Then, z^(2ipi)=e^(-2piv). For v=0 you get z=1 and for v=-1/(2pi) you get z=e.)

    In fact, if w=e^n for any whole-numbered n, then w solves the equation x^(2*i*pi)= 1.

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