Notes from Friday, October 9

2245 days ago by amelia7

We know the polynomial of odd degree must have a real root. 

Because of the the intermediate value theorem:

Proof: < 0 for some x

         > 0 for some x

   + IVT

2 things to point  out: 

i)not necessarily true for even degree polynomials (for example, $x^2 +1 = 0$ has no real root)

ii)not necessarily true for rational numbers. $x^3 +2=0$ has no rational root.

Suppose $f(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4+ ... + a_nz^n$

Is there a $z_0$ element of C such that $f(z_0)=0$?

That is, does every complex polynomial have a root in C?

Question: What does Newton's method do for $x^2+1$ starting at $x_0=1?$

Theorem: every f(z) above has n roots of C. 

Let's assume a_n=1 (divide out by it otherwise)

1. Prove that each has a single root

2. Pull it out, by deduction, it will have n-1 single roots.

so we can write $f(z)=(z-r_1)(z-r_2)...(z-r_n)$         $r_j$element of C

if $f(r_1)=0$, then we can write $f(z)=(z-r_1)q(z)+c$

and then $f(r_1)=0 => (r_1-r_1)q(r_1) +c = 0$

                         c = 0

so since q(z) has degree n-1, we can use induction hypothesis to get $q(z) = (z-r_2)...(2-r_n)$

 

so it is enough to show that $f(z)=0$ has at least one root!

 $f(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4+ ... + a_nz^n$

if a_0 = 0, then z=0 is a root. And we would be done.

So we may assume that a_0 does not equal 0. 

 
       

$f(z) = z^n ( 1 + a_{n-1}/z + a_{n-2}/z^2 + ...)$

$f(z) = z^n$

$z=Re^{i*theta}$

where R is huge

g(z) = z^n     which is n copies of a circle of radius R^n

if we change our radius continuously from E (not containing 0) to R (containing 0) at some  point we must change from not containing to containing  the origin. Thus there must be a value r_0 and a rho =theta_0 so 

$f(r_0e^{i*theta_0})=0$ so $z=r_0e^{i*theta_0}$ is a root.

solve(x^3+3*x+1,x) 
       
[x == -1/2*(1/2*sqrt(5) - 1/2)^(1/3)*(I*sqrt(3) + 1) + 1/2*(-I*sqrt(3) +
1)/(1/2*sqrt(5) - 1/2)^(1/3), x == -1/2*(1/2*sqrt(5) -
1/2)^(1/3)*(-I*sqrt(3) + 1) + 1/2*(I*sqrt(3) + 1)/(1/2*sqrt(5) -
1/2)^(1/3), x == (1/2*sqrt(5) - 1/2)^(1/3) - 1/(1/2*sqrt(5) -
1/2)^(1/3)]
[x == -1/2*(1/2*sqrt(5) - 1/2)^(1/3)*(I*sqrt(3) + 1) + 1/2*(-I*sqrt(3) + 1)/(1/2*sqrt(5) - 1/2)^(1/3), x == -1/2*(1/2*sqrt(5) - 1/2)^(1/3)*(-I*sqrt(3) + 1) + 1/2*(I*sqrt(3) + 1)/(1/2*sqrt(5) - 1/2)^(1/3), x == (1/2*sqrt(5) - 1/2)^(1/3) - 1/(1/2*sqrt(5) - 1/2)^(1/3)]