A new very short proof of the fundamental theorem of algebra
I've always been intrigued by the fundamental theorem of algebra (every nonconstant polynomial with complex coefficients has a root), not least because I don't know any proof which uses algebra only. Earlier, I posted an easy proof in this blog, one that uses Cauchy's theorem.There is a recent proof (Oswaldo Rio Branco de Oliveira, Mathem. Intellig., March 2011), which is almost trivial. It goes as follows (and this is a chance for me to see if the embedded LaTeX script works...):
Let be a polynomial of degree . Since is a nonnegative continuous function, tending to as tends to , it has a minimum at some point :
, for all .
By division of by , write
where . Since is nonconstant, the integer is .
Therefore
, for all ,
and, expanding the square,
, for all .
Let , divide by , and let tend to . We obtain
, for all real .
It is an easy exercise in algebra that, if is a complex number such that for all , then . Hence . Since , we obtain .
Nuestras horas son minutos cuando esperamos saber y siglos cuando sabemos lo que se puede aprender. (Our hours are minutes when we wait to learn and centuries when we know what is to be learnt.) --António Machado Αγεωμέτρητος μηδείς εισίτω. (Those who do not know geometry may not enter.) --Plato
Sapere Aude! Habe Muth, dich deines eigenen Verstandes zu bedienen! (Dare to know! Have courage to use your own reason!) --Kant
This happened at Stockholm airport, that is, the airport of the capital of Sweden, not the airport of Ouarzazate in the country of the kind Berber people. I'm not the only one to have problems with SAS and the Stockholm airport. Here's another.
No comments:
Post a Comment