Question:
Investigate the value of $ \Psi(x, \frac{T}{2}) $.
Solution:
Just like homework 3 bonus question, we have the exponential term reverse the sign of all the odd terms and leave the even terms alone.
The odd terms are odd functions, so $ \psi_{2n+1}(-x) = -\psi_{2n+1}(x) $.
The even terms are even functions, os $ \psi_{2n}(x) = \psi_2n(-x) $.
So we see by simply substituting -x, we get what we want!
The last thing to notice is that we need to compensate for the extracted out phase term (which I forget), therefore the end result is $ \Psi(x, \frac{T}{2}) = -\Psi(-x, 0) $.
Investigate the value of $ \Psi(x, \frac{T}{2}) $.
Solution:
Just like homework 3 bonus question, we have the exponential term reverse the sign of all the odd terms and leave the even terms alone.
The odd terms are odd functions, so $ \psi_{2n+1}(-x) = -\psi_{2n+1}(x) $.
The even terms are even functions, os $ \psi_{2n}(x) = \psi_2n(-x) $.
So we see by simply substituting -x, we get what we want!
The last thing to notice is that we need to compensate for the extracted out phase term (which I forget), therefore the end result is $ \Psi(x, \frac{T}{2}) = -\Psi(-x, 0) $.
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