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− | In question 2e <math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(x-7k)^2} \ </math> | + | In question 2e |
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+ | <math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(x-7k)^2} \ </math> | ||
<br> | <br> | ||
− | should it be<math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(t-7k)^2} \ </math> | + | should it be like this? |
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+ | <math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(t-7k)^2} \ </math> | ||
<br> | <br> | ||
− | and I was trying to find out what the peak value is for this question but turns out to be very hard to calculate | + | and I was trying to find out what the peak value is for this question but turns out to be very hard to calculate the sum |
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+ | <math> \sum_{t=-\infty}^\infty \frac{1}{1+t^2} \ </math> and wolfram said answer is '''π * coth(π)'''. is there any easier way to do that? | ||
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+ | Yimin. Jan 20 | ||
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Revision as of 04:36, 20 January 2011
In question 2e
$ x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(x-7k)^2} \ $
should it be like this?
$ x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(t-7k)^2} \ $
and I was trying to find out what the peak value is for this question but turns out to be very hard to calculate the sum
$ \sum_{t=-\infty}^\infty \frac{1}{1+t^2} \ $ and wolfram said answer is π * coth(π). is there any easier way to do that?
Yimin. Jan 20