My discovery about mathematics

Naoto Meguro. Amateur. MSC 39A06.

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You can express the solution of the difference equation by the coefficients of it.
So there is an analytic formula giving a root of the algebraic equation.
These don't appear on the list of the formulas now.

                                  2

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≤s c_ku.
Proof. n≥N+1⇒ x_n=∑_1≤n-k1≤N e_n-k1c_k1 +∑_n-k1≥N+1 c_k1x_n-k1.
Put n-k₁-…-k_s=p. If s=1,k_s=k₁=n-p≥N+1-p.
By induction, you may assume that
n-k₁≥N+1 ⇒ x_n-k1=∑_1≤p≤Ne_p∑_{k2+…+ks=(n-k1)-p,1≤ku≤N, ks≥N+1-p} ∏_2≤u≤s c_ku.
Substituting these, you get theorem 1.♦
Theorem 2. n≥N+1 ⇒
x_n=∑_1≤p≤N e_p∑_1≤s≤n-p∫_0≤t≤2π(∑_1≤k≤N c_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≤s c_ku is the coefficient of x^n-p in
(∑_1≤k≤N c_kx^k)^s-1(∑_{N+1-p≤k≤N} c_kx^k) etc..♦
(v∈C)∧(w∈C)∧(r∈R-{0})∧(r"∈R-{0})∧(v≠w)∧(|v|=|w|)∧(|v+r|=|w+r|)⇒
(Re(v)=Re(w))∧(Im(v)= - Im(w))∧(|v+r+r"i|≠|w+r+r"i|). ( i²=- 1.)
When f(x)=x^N+a₁x^N-1+…+a_N-1x+a_N=(x-w₁)…(x-w_N) and w_k≠w_h (k≠h) and a_N≠0,
you can set up that |w_k+p^{- 1}+q^{- 1}i|≠|w_h+p^{- 1}+q^{- 1}i|≠0 (k≠h) for p∈N and q∈N
for (|w_k|≠|w_h|)∨(|w_k+p^{- 1}|≠|w_h+p^{- 1}|)⇒(|w_k+p^{- 1}+q^{- 1}i|≠|w_h+p^{- 1}+q^{- 1}i|≠0)
if p and q are large enough.
When f(x-p^{- 1}-q^{- 1}i)=x^N-c₁x^N-1-…-c_N-1x-c_N and e₁=1 and e_d=0 (2≤d≤N),
x_n+1=∑_1≤k≤N d_k(w_k+p^{- 1}+q^{- 1}i)ⁿ⁺¹. (d₁,…,d_N)≠(0,…,0).
x_n+1/x_n→w+p^{- 1}+q^{- 1}i (n→∞). f(w)=0. w+p^{- 1}+q^{- 1}i → w+q^{- 1}i (p→ ∞)
w+q^{- 1}i→w (q→∞).
You get a formula giving a root of the algebraic equation written by the coefficients of it
and ∑, ∫ ,e^t, and lim.

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You can express the algebraic functions as explicit functions.