Ok, I’ve not quite given up yet. I’ll keep doing this blog a little bit more and see if something comes out of it. Since I don’t have anything special to say at this point, I’ll post about something quite simple and well known. It is a proof of the fundamental theorem of algebra that uses only freshman calculus. Actually I learned about it during my second year of prep school, but it is really quite simple, an doesn’t use any complex analysis. Let me recall the result.
Fondamental theorem of algebra
Let be a polynomial with a non negative integer, complex numbers.
Then there exist such that .
The proof goes like this. You consider the function on the complex plane. As the fact that is a field does not play any role at this point, you can identify to . You prove that this fuction admits a global minimum. You then prove that for any complex number such that , there exists a neighbouring number such that . But this means that the global minimum that we found must be a zero of , and that is the expected result.
The fact that has a global minimum is a consequence of the following lemma
Let be a continuous function such that . Then has a global minimum.
Proof: There is an such that for all , . Now has a minimun on the compact set (the closed ball of radius ). Let be a point where the minimal value is reached. For any point , either , in which case , or , and then . Either way, , and therefore is a point of global minimum.
Now to apply this lemma, we need only to check that , and this is made obvious by the reversed triangle inequality :
This part is the closest in spirit to complex analysis, but since we deal with polynomials we won’t need the whole theory. We will prove a version of the minimum principle.
Let be a polynomial that satisfies the hypothesis of the fondamental theorem (with complex coefficients and not a constant), and a complex number such that . Then there exists a complex number such that .
Proof: If we divide by , we only need to study the case where . By the usual factorisation argument, there is an integer and a polynomial with such that for any ,
We will now use the polar representation to find such that . By continuity of , for any given , there is an such that for any , . We now set , and we choose to get
(which of course can be done). Then by the triangle inequality,
If, at the beginning of this argument , we choose to be , we get
This ends the proof of the second lemma and of the theorem.