proof of equivalence of Fermat’s Last Theorem to its analytic form
Consider the Taylor expansion of the cosine function. We have
For the sequence is decreasing as the denominator grows faster than the numerator. Hence for the sequence is increasing as and . So if for some , we have . Conversely if no such exists then for , so its limit, , is also less than or equal to . However as this expression cannot be negative we would have .
Similarly for the sequence is decreasing, and for the sequence is increasing. So if for some we have . Conversely if no such exists then . However as this expression cannot be negative we would have .
Note that precisely when . Also precisely when .
So the form of the theorem may be read:
If for positive reals we have for some odd integer , then either or not in or not in .
Clearly this only fails if for positive integers and some odd , we have
Dividing through by we see that .
Conversely suppose we have non-zero integers satisfying for some . If we have , contradicting example of Fermat’s last theorem. Hence if is even we may replace with and with , which will be odd and greater than 1 (and hence greater than 2 as it is odd). So without loss of generality we may assume odd.
Finally replace with their absolute values and if reorder to obtain a positive integer solution. This would be a counterexample to the form of the theorem as stated above.
|Title||proof of equivalence of Fermat’s Last Theorem to its analytic form|
|Date of creation||2013-03-22 16:19:04|
|Last modified on||2013-03-22 16:19:04|
|Last modified by||whm22 (2009)|