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A formula giving a root 2".

2015-06-11 18:16:07 | Mathematics
Theorem 4. yi,ziR,wi=yi+zi√-1 (1≤i≤N) wi≠wk (i≠k) ⇒ (∃R≥0,≠0,R≥r≥0,≠0⇒
∃R(r)≥0,≠0,R(r)≥r"≥0,≠0⇒(|wi+r+r"√-1|-|wk+r+r"√-1|≥0,≠0⇔(|wi|-|wk|≥0,≠0)
∨(|wi|=|wk|,yi-yk≥0,≠0)∨(|wi|=|wk|,yi=yk,zi-zk≥0,≠0))
∧(wi+r+r"√-1≠0)
Proof. ∃R≥0,≠0, R≥r≥0,≠0⇒(|wi|-|wk|≥0,≠0⇒|wi+r|-|wk+r|≥0,≠0)
|wi|=|wk|⇒|wi+r|2-|wk+r|2=2r(yi-yk) So |wi+r|-|wk+r|≥0,≠0 ⇔ yi-yk≥0,≠0 then.
∃R(r)≥0,≠0, R(r)≥r"≥0,≠0⇒(|wi+r|-|wk+r|≥0,≠0⇒|wi+r+r"√-1|-|wk+r+r"√-1|≥0,≠0)
∧(wi+r+r"√-1≠0)
|wi+r|=|wk+r|⇒(|wi|=|wk|)∧(yi=yk) So |wi+r+r"√-1|2-|wk+r+r"√-1|2=2r"(zi-zk)
|wi+r+r"√-1|-|wk+r+r"√-1|≥0,≠0⇔zi-zk≥0,≠0 then.
Remark that |wi+r+r"√-1|≠|wk+r+r"√-1| for i≠k. The direction of the inequalities are
decided by only wi's and don't depend on r or r".♦
Theorem 5. If the equation f(x)=∑0≤i≤Naixi=0 (aN=1) has no multiple root,
x= lim  lim  lim xh,m,n+1/xh,m,n
   h→∞ m→∞ n→∞

(xh,m,n=∫0≤t≤2π1≤s≤n(∑1≤k≤N(-∑N-k≤i≤Nai(-1/h-√-1/m)i-N+kiCN-k)e√-1kt)s-1e√-1(N-n)tdt)
is a root of f(x)=0.
Proof. Put f(x)=(x-w1)…(x-wN). By theorem 4,
∃R≥0,≠0, R≥r≥0,≠0⇒∃R(r)≥0,≠0, R≥r"≥0,≠0⇒|wi+r+r"√-1|≠|wk+r+r"√-1| (i≠k)
Put u=r+r"√-1, vi=wi+u (1≤i≤N),e1=1,ei=0 (2≤i≤N),
ck=-∑N-k≤i≤Nai(-u)i-N+kiCN-k (1≤k≤N).
(x-v1)…(x-vN)=f(x-u)=∑0≤i≤N ai(x-u)i=∑0≤i≤Nai0≤q≤ixq(-u)i-qiCq
=∑0≤q≤Nxqq≤i≤Nai(-u)i-qiCi-q=xN-c1xN-1-…-cN (Put q=N-k.)
You may assume that |v1|-|vi|≥0,≠0 (2≤i≤N). By theorem 3,
xn+2/xn+1→v1 (n→∞, R≥r≥0,≠0, R(r)≥r"≥0,≠0))
v1→w1+r (r"→0, R(r)≥r"≥0,≠0,R≥r≥0,≠0)
w1+r→w1 (r→0, R≥r≥,0≠0)
Putting r=1/h and r"=1/m (h,m∈N),you get the theorem by theorem 2.♦
xh,m,n+1/xh,m,n may not converge when h or m isn't large enough.You are to skip
such h or m.
                                       3

Using theorem 5 and synthetic division,you get the formulas giving the roots of the equation of
degree N for every natural number N. You can express the algebraic functions as explicit functions.
Forget the inconsistency theorem which make the above nonsense.(Refer to the articles "Defect of the
formal algebra 1",2",3"" in this blog.)
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