Believe me guys. First person was right. It is 0.002 + e^(2*pi) + infinite_sum_starting_in_one_of_(1/2^n). So it's not e^(2*i), nor the infinite_sum adds to zero; just take 1/2 + 1/4 and you'll have 0.75; whole infinite_sum adds to 1. Total check sum = $536.49.
6 comments:
$536.49
not true, it's $0.002 or 0.2 cents.
e^(i*pi) = -1
sum of 1/2^n from 1 to infinity is 1
You are wrong, e^(i*pi) is a complex number with real part=0.
1/2^n from 1 to infinity is 0.
It's 0.2 cents.
no the second comment is really right in the reasoning, third is just wrong
Believe me guys. First person was right. It is 0.002 + e^(2*pi) + infinite_sum_starting_in_one_of_(1/2^n). So it's not e^(2*i), nor the infinite_sum adds to zero; just take 1/2 + 1/4 and you'll have 0.75; whole infinite_sum adds to 1. Total check sum = $536.49.
Last commentor: Read the check again, the exponent of e is "i*Pi", not "2*Pi".
My 0.2 cents.
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