|
|
Archived: 07 Aug 2014, 09:52
|
Author |
Message |
Azrael84
Joined: 22 Sep 2011, 10:17 Posts: 9
|
 Exercise [5.10]e
This is my attempt at a solution essentially it boils down to:
If we want (e^a)^b=e^{ab} (1) to hold then after taking Log(e)=log(e)=1 on the RHS we are forced into Log(e^a)=alog(e)=a on the LHS as mentioned by Penrose in [5.15].
So in order for (1) to hold we are assuming the branch Log(e^(1+i2k.pi))=(1+i2k.pi). However the original assertion of the paradox was that e=e^{(1+i2k.pi)} and taking logarithms each side: Log(e)=Log(e^{1+i2k.pi}) . Now because we took Log(e)=log(e)=1 earlier we should do again for consistency forcing us to conclude Log(e^{1+i2k.pi}) =1 contrary to our earlier conclusion that Log(e^(1+i2k.pi))=(1+i2k.pi) that we had to assume for the validity of (1).
Attachments:
paradox.pdf [76.36 KiB]
Downloaded 146 times
|
23 Sep 2011, 16:08 |
|
 |
Azrael84
Joined: 22 Sep 2011, 10:17 Posts: 9
|
 Re: Exercise [5.10]e
Another solution based on a different method:
|
23 Sep 2011, 17:24 |
|
 |
|
|