Can any one explain e^i^pi=-1?

Status
Not open for further replies.
Re: maths

with \[i=\sqrt{-1}\], \[e^{-i \pi}= \cos(\pi) + i \sin(\pi)=-1+0=-1\]. this is called Euler's Identity. You can derive it by looking at the Taylor series expansion of \[e^x\] and substitute \[x=i \pi\]. It is explained very well here

Euler's formula - Wikipedia, the free encyclopedia
(Make sure you enter the correct URL as the bit after the ' is not highlighted)
(look down the page where it says "Using Taylor Series")

I hope this helps you out.

Best regards,
v_c
 
Last edited by a moderator:

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…