Day 1: Advanced Polynomials & Complex Dynamics

1.1 Generalized Vieta’s Formulas
Generalized Vieta’s FormulaThe general sum of k roots taken at a time:r1r2rk=(1)kankan\begin{aligned} &\textbf{\Large Generalized Vieta's Formula} \\ &\rule{220pt}{0.3mm} \\\\ &\text{The general sum of } k \text{ roots taken at a time:} \\\\ &\quad \sum r_1 r_2 \cdots r_k = (-1)^k \cdot \frac{a_{n-k}}{a_n} \end{aligned}
1.2 Symmetric Power Sums (Newton's Sums)
Newton’s SumsThe Polynomial:ax4+bx3+cx2+dx+e=0The Definitions: a,b,c,d,e=The coefficients Sk=The sum of the k-th powers of the roots (r1k+r2k+)The Recurrence System: aS1+b=0 aS2+bS1+2c=0 aS3+bS2+cS1+3d=0 aS4+bS3+cS2+dS1+4e=0\begin{aligned} &\textbf{\Large Newton's Sums} \\ &\rule{220pt}{0.3mm} \\\\ &\textbf{The Polynomial:} \\ &\quad ax^4 + bx^3 + cx^2 + dx + e = 0 \\\\ &\textbf{The Definitions:} \\ &\quad \bullet\ a, b, c, d, e = \text{The coefficients} \\ &\quad \bullet\ S_k = \text{The sum of the } k\text{-th powers of the roots } (r_1^k + r_2^k + \dots) \\\\ &\textbf{The Recurrence System:} \\ &\quad \bullet\ aS_1 + b = 0 \\ &\quad \bullet\ aS_2 + bS_1 + 2c = 0 \\ &\quad \bullet\ aS_3 + bS_2 + cS_1 + 3d = 0 \\ &\quad \bullet\ aS_4 + bS_3 + cS_2 + dS_1 + 4e = 0 \end{aligned}
1.3 Simon’s Favorite Factoring Trick (SFFT)
Simon’s Favorite Factoring Trick (SFFT)Original Formula:xy+bx+ay = CWe want to find a number that’s the product of aband add it to both sides:xy+bx+ay+ab = C+abThe Perfect Factors:(x+a)(y+b) = C+ab\begin{aligned}&\textbf{\Large Simon's Favorite Factoring Trick (SFFT)} \\&\rule{220pt}{0.3mm} \\\\&\textbf{Original Formula:} \\&\quad xy + bx + ay \ = \ C \\\\&\text{We want to find a number that's the product of } \mathbf{a \cdot b} \\&\text{and add it to both sides:} \\\\&\quad xy + bx + ay + \mathbf{ab} \ = \ C + \mathbf{ab} \\\\&\textbf{The Perfect Factors:} \\&\quad \mathbf{(x + a)(y + b) \ = \ C + ab}\end{aligned}
1.4 Sophie Germain Identity
Sophie Germaina4+4b4 = (a22ab+2b2)(a2+2ab+2b2)\begin{aligned}&\textbf{\Large Sophie Germain} \\&\rule{220pt}{0.3mm} \\\\&\quad a^4 + 4b^4 \ = \ (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2)\end{aligned}
1.5 Roots of Unity & De Moivre’s Theorem
Roots of Unity & De Moivre’s TheoremThe Cyclotomic Drops:xn1 = (x1)(xω)(xω2)(xωn1)xn1+xn2++x+1 = (xω)(xω2)(xωn1)De Moivre’s Transformation Masterkey:[r(cosθ+isinθ)]n = rn(cos(nθ)+isin(nθ))The 3-Step Routine to Solve:1) Find your radius r and starting angle θ2) Plug them straight into the baseline equation3) nθ = The target degree you need to hit\begin{aligned}&\textbf{\Large Roots of Unity \& De Moivre's Theorem} \\&\rule{240pt}{0.3mm} \\\\&\textbf{The Cyclotomic Drops:} \\&\quad x^n - 1 \ = \ (x-1)(x-\omega)(x-\omega^2)\cdots(x-\omega^{n-1}) \\\\&\quad x^{n-1} + x^{n-2} + \dots + x + 1 \ = \ (x-\omega)(x-\omega^2)\cdots(x-\omega^{n-1}) \\\\&\textbf{De Moivre's Transformation Masterkey:} \\&\quad \left[r(\cos\theta + i\sin\theta)\right]^n \ = \ r^n(\cos(n\theta) + i\sin(n\theta)) \\\\&\textbf{The 3-Step Routine to Solve:} \\&\quad \mathbf{1)} \ \text{Find your radius } r \text{ and starting angle } \theta \\&\quad \mathbf{2)} \ \text{Plug them straight into the baseline equation} \\&\quad \mathbf{3)} \ n\theta \ = \ \text{The target degree you need to hit}\end{aligned}
1.6 Descartes' Rule of Signs
Descartes’ Rule of SignsThe General Form:P(x)=anxn+an1xn1++a1x+a0What We Track: Sign variations in the coefficient sequence (an,an1,,a0)The Roots Rules: Positive Roots: Sign variations in P(x) give the max count. Negative Roots: Sign variations in P(x) give the max count.Note: The actual count drops from the maximum by even integers due to complex pairs.\begin{aligned}&\textbf{\Large Descartes' Rule of Signs} \\&\rule{220pt}{0.3mm} \\\\&\textbf{The General Form:} \\&\quad P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \\\\&\textbf{What We Track:} \\&\quad \bullet\ \text{Sign variations in the coefficient sequence } (a_n, a_{n-1}, \dots, a_0) \\\\&\textbf{The Roots Rules:} \\&\quad \bullet\ \text{Positive Roots: Sign variations in } P(x) \text{ give the max count.} \\&\quad \bullet\ \text{Negative Roots: Sign variations in } P(-x) \text{ give the max count.} \\\\&\text{Note: The actual count drops from the maximum by even integers due to complex pairs.}\end{aligned}
1.7 The Remainder and Factor Theorems
The Remainder TheoremThe General Form:P(x)=(xc)Q(x)+RWhat We See: P(x)=The polynomial (xc)=The divisor Q(x)=The quotient R=The remainderThe Identity Rules: Evaluating P(c) forces the Q(x) term to drop to zero. This immediately isolates the constant value: P(c)=R\begin{aligned}&\textbf{\Large The Remainder Theorem} \\&\rule{220pt}{0.3mm} \\\\&\textbf{The General Form:} \\&\quad P(x) = (x - c) \cdot Q(x) + R \\\\&\textbf{What We See:} \\&\quad \bullet\ P(x) = \text{The polynomial} \\&\quad \bullet\ (x - c) = \text{The divisor} \\&\quad \bullet\ Q(x) = \text{The quotient} \\&\quad \bullet\ R = \text{The remainder} \\\\&\textbf{The Identity Rules:} \\&\quad \bullet\ \text{Evaluating } P(c) \text{ forces the } Q(x) \text{ term to drop to zero.} \\&\quad \bullet\ \text{This immediately isolates the constant value: } P(c) = R\end{aligned}
1.8 Coefficient Inversion (Flipped Roots)
Coefficient Inversion (Flipped Roots)The Original Polynomial:P(x)=ax3+bx2+cx+d=0with roots: (r1,r2,r3)What We See: a,b,c,d=The default sequence of coefficientsThe "Fliparonni" Rule:To create a new polynomial with inverted roots (1r1,1r2,1r3),completely reverse the order of the coefficients:Q(x)=dx3+cx2+bx+a=0\begin{aligned}&\textbf{\Large Coefficient Inversion (Flipped Roots)} \\&\rule{220pt}{0.3mm} \\\\&\textbf{The Original Polynomial:} \\&\quad P(x) = a x^3 + b x^2 + c x + d = 0 \quad \text{with roots: } (r_1, r_2, r_3) \\\\&\textbf{What We See:} \\&\quad \bullet\ a, b, c, d = \text{The default sequence of coefficients} \\\\&\textbf{The "Fliparonni" Rule:} \\&\text{To create a new polynomial with inverted roots } \left(\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}\right), \\&\text{completely reverse the order of the coefficients:} \\\\&\quad Q(x) = d x^3 + c x^2 + b x + a = 0\end{aligned}
1.9 Lagrange Interpolation Formula
Lagrange InterpolationThe Grid Points:(2,3), (3,5), (4,4)The On/Off Switches: To turn on the first point (x=2):(x3)(x4)(23)(24)  Equals 1 at x=2. Drops to 0 at x=3,4. To turn on the second point (x=3):(x2)(x4)(32)(34)  Equals 1 at x=3. Drops to 0 at x=2,4.The Final Formula Puzzle:P(x)=3[Switch 1]+5[Switch 2]+4[Switch 3]\begin{aligned}&\textbf{\Large Lagrange Interpolation} \\&\rule{220pt}{0.3mm} \\\\&\textbf{The Grid Points:} \\&\quad (2,3), \ (3,5), \ (4,4) \\\\&\textbf{The On/Off Switches:} \\&\quad \bullet\ \text{To turn on the first point } (\mathbf{x=2}): \\&\quad\quad \frac{(x-3)(x-4)}{(2-3)(2-4)} \ \longrightarrow \ \text{Equals } 1 \text{ at } x=2. \text{ Drops to } 0 \text{ at } x=3, 4. \\\\&\quad \bullet\ \text{To turn on the second point } (\mathbf{x=3}): \\&\quad\quad \frac{(x-2)(x-4)}{(3-2)(3-4)} \ \longrightarrow \ \text{Equals } 1 \text{ at } x=3. \text{ Drops to } 0 \text{ at } x=2, 4. \\\\&\textbf{The Final Formula Puzzle:} \\&\quad P(x) = \mathbf{3} \cdot \left[ \text{Switch 1} \right] + \mathbf{5} \cdot \left[ \text{Switch 2} \right] + \mathbf{4} \cdot \left[ \text{Switch 3} \right]\end{aligned}
1.10 Polynomial Sequence Difference Tables
Polynomial Difference TablesThe Value Grid:x12345Y25101726The Subtraction Ladder:[Row 1:]52=3,105=5,1710=7[Row 2:]53=2,75=2  Constant RowThe Rule: Count the ladder rows taken to hit a constant value. The number of rows = highest power of the polynomial. Since it took 2 steps here, the equation is a quadratic (x2).\begin{aligned} &\textbf{\Large Polynomial Difference Tables} \\ &\rule{220pt}{0.3mm} \\\\ &\textbf{The Value Grid:} \\ &\quad \begin{array}{c|ccccc} x & 1 & 2 & 3 & 4 & 5 \\ \hline Y & 2 & 5 & 10 & 17 & 26 \end{array} \\\\ &\textbf{The Subtraction Ladder:} \\ &\quad \mathbf{[Row\ 1:]} \quad 5-2=\mathbf{3}, \quad 10-5=\mathbf{5}, \quad 17-10=\mathbf{7} \\ &\quad \mathbf{[Row\ 2:]} \quad 5-3=\mathbf{2}, \quad 7-5=\mathbf{2} \ \longrightarrow \ \text{Constant Row} \\\\ &\textbf{The Rule:} \\ &\quad \bullet\ \text{Count the ladder rows taken to hit a constant value.} \\ &\quad \bullet\ \text{The number of rows } = \text{ highest power of the polynomial.} \\ &\quad \bullet\ \text{Since it took } \mathbf{2} \text{ steps here, the equation is a quadratic } (x^2). \end{aligned}