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anonymous
 4 years ago
Simplify problem
anonymous
 4 years ago
Simplify problem

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1, n=odd}^{\infty} (i/(\pi*n))((i) ^{n}+(i) ^{n})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the final answer contains something with (1)^n in the numerator

phi
 4 years ago
Best ResponseYou've already chosen the best response.0We could first simplify the expression in paretheses \[  i^n + i^{n} \] find a common denominator, and add: \[\frac{1}{i^n}\frac{i^{2n}}{i^n}= \frac{1 (i^2)^n}{i^n}= \frac{1 (1)^n}{i^n}\] n is odd so we know 1^(odd_power) is always going to be 1. Therefore the expression becomes \[ \frac{2}{i^n} \]

phi
 4 years ago
Best ResponseYou've already chosen the best response.0using the above in your summation\[\sum_{n=1, n=odd}^{\infty} (i/(\pi*n))((i) ^{n}+(i) ^{n})= \frac{2}{\pi}\sum_{n=1, odd}^{\infty}\frac{1}{n i^{n1}}\] I do not like summation indices that go up by 2, so replace the summation index n with m=0,1,2,... n= 2m+1 rewrite the summation to get \[\frac{2}{\pi}\sum_{m=0}^{\infty}\frac{1}{(2m+1) i^{2m}}=\frac{2}{\pi}\sum_{m=0}^{\infty}\frac{1}{(2m+1) (i^{2})^{m}}\] of course, i^2 is 1, so this becomes a series with alternating signs: \[\frac{2}{\pi}(1\frac{1}{3}+\frac{1}{5}\frac{1}{7}+\frac{1}{9}...)\] This is a wellknown series http://mathworld.wolfram.com/LeibnizSeries.html You can easily find the final answer =\( \frac{1}{2} \)
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