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A comment on Visual proof of Euler's identity (vid).
For those who aren't looking forward to a lame video, the basic idea is this:
e^(pi)(i) = -1.
Why?
Because you can express e^x as a taylor series:
e^x = 1 + x + x^2/2! + x^3/3! + ... thus e^ix = 1 + ix + ix^2/2! + ix^3/3! + .. e^ix = (1 - x^2/2! + x^4/4! + ..) + i(x - x^3/3!+x^5/5!+...) implies e^ix = cosx + isinx
because those parenthetical quantities are the taylor series for cosx and sinx respectively
So, e^i(pi) = cos(pi) + isin(pi). = -1 + i(0) = -1
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For those who aren't looking forward to a lame video, the basic idea is this:
e^(pi)(i) = -1.
Why?
Because you can express e^x as a taylor series:
e^x = 1 + x + x^2/2! + x^3/3! + ... thus e^ix = 1 + ix + ix^2/2! + ix^3/3! + .. e^ix = (1 - x^2/2! + x^4/4! + ..) + i(x - x^3/3!+x^5/5!+...) implies e^ix = cosx + isinx
because those parenthetical quantities are the taylor series for cosx and sinx respectively
So, e^i(pi) = cos(pi) + isin(pi). = -1 + i(0) = -1