why is e^(iπ)+1=0?
Thank you guys~
\[e^{i \pi} =\cos(\pi)+i \sin(\pi)=-1+i (0)=-1+0=-1\]
\[-1+1=0\]
e iπ =cos(π)+isin(π) but why?Satisfy my curiosity~
I can't visit the wikipedia I'm in China
use proxy
I tried but the network in my house in veeeery slow
\[e ^{i x}=cosx+isinx\]
that's the short answer above you can prove that\[e^{ix}=\cos x+i\sin x\]by using\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\]\[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\]\[\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...\] try plugging in \[e^{ix}\]into that expansion for e^x and you will see that it will amount to \[\cos x+i\sin x\]
Wikepedia is banned in china? !!!
everything is banned in China
is wikileaks banned in US?
no it's not
Is any news network other than Fox banned in the US? :-P
haha :D
Look at that ----so many people Hi!~
That awkward moment when you say "Hi!" and not a single other comment is made... :-P Sup chum.
it's a cool question, people like to answer it.
Or just read the cool answers provided.
Thanks again
here is the wikipedia article in pdf
you won't believe that the net-speed in my house is less than1M 200kb/s
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