Math, you're doing it right
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MAT Ill.
When you absolutely, positively must piss sameone off.
MAT Ill.
When you absolutely, positively must piss sameone off.
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#3
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anon (05/28/2015) [-]
It's not e^(2pie), I'm prette sure it's e^(i pi), where i is an imaginary unit; in calculations on complex numbers you very often encounter such exponents of e. It so happens that e^(i*pi) = -1.
The other part about limit of sum is correct. So it's 0.002 + (-1) + 1, so it's $0.002.
If you want to know more e^(a+ib) = e^a(cos(b)+i sin(b)).In this case a=0 and b=pi, so you get e^(0+i pi) = e^0(cos(pi) + i sin(pi) ) = 1(-1 + i*0) = -1.
Have a nice day!
The other part about limit of sum is correct. So it's 0.002 + (-1) + 1, so it's $0.002.
If you want to know more e^(a+ib) = e^a(cos(b)+i sin(b)).In this case a=0 and b=pi, so you get e^(0+i pi) = e^0(cos(pi) + i sin(pi) ) = 1(-1 + i*0) = -1.
Have a nice day!
