My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010. Primary 00A99 ; Secondary 39A60.
Key Words and Phrases. A difference equation, the algebraic equation.
The abstract. An analytic formula giving a root of the algebraic equation exists.

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I show a formula giving a root of the algebraic equation,which is written by the coefficients of the equation
and some analytic symbols.

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Set up that x_n=e_n (1≤n≤N) and x_n=∑_1≤k≤N c_kx_n-k (n≥N+1)
Theorem 1. n≥N+1⇒ x_n=∑_1≤p≤N e_p∑_{k1+…+ks=n-p,1≤ku≤N,ks≥N+1-p} ∏_1≤u≤sc_ku.
Proof. n≥N+1⇒ x_n=∑_1≤n-k1≤Ne_n-k1c_k1+∑_n-k1≥N+1 c_k1x_n-k1.
Put n-k₁-…-k_s=p. If s=1,k_s=n-p≥N+1-p.
By induction, you may assume that
n-k₁≥N+1 ⇒ x_n-k1=∑_1≤p≤N e_p∑_{k2+…+ks=(n-k1)-p,1≤ku≤N,ks≥N+1-p} ∏_1≤u≤s c_ku.
Substituting these, you get theorem 1.♦
Theorem 2. n≥N+1⇒ x_n=∑_1≤p≤Ne_p∑_1≤s≤n-p∫_0≤t≤2π(∑_1≤k≤Nc_ke^ikt)^s-1(∑_{N+1-p≤k≤N}c_ke^ikt)e^it(p-n)dt/2π.
Proof. ∑_{k1+…+ks=n-p, 1≤ku≤N, ks≥N+1-p} ∏_1≤u≤sc_ku is the coefficient of x^n-p in
(∑_1≤k≤Nc_kx^k)^s-1(∑_{N+1-p≤k≤N}c_kx^k) etc..♦
Put d(c₁,…c_N,n)=(∑_1≤s≤n∫_0≤t≤2π(∑_1≤k≤Nc_ke^ikt)^s-1e^i(N-n)tdt)/(∑_1≤s≤n-1∫_0≤t≤2π(∑_1≤k≤Nc_ke^ikt)^s-1e^i(N-n+1)tdt).
Theorem 3. e₁=1, e_u=0 (2≤u≤N), x^N-c₁x^n-1-…-c_N=(x-v₁)…(x-v_N),
|v₁|>|v_u|≠0 (2≤u≤N),v_u≠v_k (u≠k) ⇒ d(c₁,…,c_N,n)→v₁ (n→∞).
Proof. By theory of the difference equation, you can write x_n=b₁v₁^n-1+…+b_Nv_N^n-1.
(1,0,…0)=(b₁,…,b_N)A. A=(v_u^k-1)_{1≤u≤N,1≤k≤N} is a matrix. detA=∏_u>k(v_u-v_k)≠0.
(b₁,…,b_N)=(1,0,…0)A^-1. b₁ is the (1,1)component of A^-1 which is v₂…v_N/∏_u≥2(v_u-v₁)≠0.
(v_u/v₁)^m→0 (m→∞). x_n+2/x_n+1=(b₁v₁+b₂(v₂/v₁)ⁿv₂+…)/(b₁+b₂(v₂/v₁)ⁿ+…)→v₁ (n→∞).
By theorem 2,d(c₁,…,c_N,n)→v₁ (n→∞).♦
Set up that f(x)=x^N+a₁x^N-1+…+a_N=(x-w₁)…(x-w_N), w_u=y_u+z_ui (y_u,z_u∈R. 1≤u≤N.),
f(x-1/r-i/r")=x^N-c₁x^N-1-…-c_N, c_u=-∑_0≤k≤u a_{k N-k}C_N-u(-1/r-i/r")^u-k (a₀=1).
Theorem 4. w_u≠w_k (u≠k),|w₁|≥|w_u|≠0(1≤u≤N)⇒ ∃M((M∈N)∧∀r((M≤r)→∃M(r)((M(r)∈N)∧∀r"((M(r)≤r")
→(|w₁+1/r+1/r"i|>|w_u+1/r+1/r"i|≠0))))).
Proof. ∃M((M∈N)∧∀r((r≥M)→((|w₁|>|w_u|)→(|w₁+1/r|>|w_u|+1/r|≠0)))).
If |w₁+1/r|=|w_u+1/r| (u≠1),|w₁|=|w_u|. (y₁+1/r)²+z₁²=(y_u+1/r)²+z_u², y₁²+z₁²=y_u²+z_u².
y₁=y_u,z₁=-z_u. ∃M(r)((M(r)∈N)∧∀r"((r"≥M(r))→((|w₁+1/r|>|w_u+1/r|)→(|w₁+1/r+i/r"|>|w_u+1/r+i/r"|≠0)))).
If |w₁+1/r+i/r"|=|w_u+1/r+i/r"|(u≠1),|w₁+1/r|=|w_u+1/r|. y₁=y_u,z₁=-z_u,(y₁+1/r)²+(z₁+1/r")²=(y_u+1/r)²+(z_u+1/r")². z₁=z_u=-z_u=0. w₁=w_u. This is contradiction. It is easy to set up |w_u+1/r+i/r"|≠0 by a suitable M(r).♦
Theorem 5. If f(x)=0 (a_N≠0) has no double root,

lim lim lim d(c1,…,cN,n) is a root of it.
 r→∞  r"→∞ n→∞

Proof. By theorem 3 and theorem 4, d(c₁,…,c_N,n)→w₁+1/r+i/r" (n→∞,r≥M,r"≥M(r).)
w₁+1/r+i/r"→w₁+1/r (r"→∞,r"≥M(r),r≥M.) w₁+1/r→w₁ (r→∞,r≥M.)♦
x_n+1/x_n may not converge when r and r" aren't large enough. You are to skip such r and r".

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Does the above become an answer to the 12th problem of Hilbert?