Weierstrass product inequality
For any , and any fixed values of the for , is a polynomial of the first degree in . Consequently is minimal either at or . That brings us down to two cases: all the are zero, or at least one of them is . But in both cases it is clear that , QED.
|Title||Weierstrass product inequality|
|Date of creation||2013-03-22 13:58:23|
|Last modified on||2013-03-22 13:58:23|
|Last modified by||Daume (40)|