Proving $frac 1 {max{h_1,h_2}}(n-frac{jh_1} {p-1})leq α+frac {σ(k_1)+…+σ(k_m)-(p-1)} {p-1}+ frac...
$begingroup$
Let $F_1,F_2in mathbb{Z}[x_1,...,x_n]$ where $d_i=deg(F_i)$ and $d_1+d_2<n$.
Write $F_1(x)=a_1x^{e_1}+...+a_mx^{e_m}$ and $F_2(x)=b_1x^{ε_1}+...+b_μx^{ε_μ}$, where each $e_i,ε_i$ are vectors in $mathbb{Z}^n$. Write $e_1=begin{bmatrix} e_{11} \ ... \ e_{1n} end{bmatrix}$ and so on.
Let $p$ be prime. Let $0leq jleq p-1$ and let $k_1,...,k_min mathbb{Z}_{geq 0}$ such that $k_1+...+k_m=jp^n$.
Let $0leq l_1,...l_μleq p-1$.
Let $h_1=max{σ(e_{i1})+...+σ(e_{in}):1leq i leq m}$ and $h_2=max{σ(ε_{i1})+...+σ(ε_{in}):1leq i leq μ}$, where $σ$ returns the sum of digits in the base-p expansion.
Let $α=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}=0}$.
Let $β=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}$ isn't zero but is divisible by $p-1}$.
Assume further that $α+β=n$.
I want to show that $frac 1 {max{h_1,h_2}}(n-frac{jh_1} {p-1})leq α+frac {σ(k_1)+...+σ(k_m)-(p-1)} {p-1}+ frac {l_1+...+l_μ} {p-1}$
I see that the sum of coordinates in the vector $σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_1+...+l_με_μ$ is bounded below by $β(p-1)$ since each coordinate is divisible by $p-1$, and bounded above by $[σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$.
Hence we have the inequality $β(p-1)leq [σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$, which I'm hoping to be able to use to prove the highlighted inequality.
number-theory inequality polynomials prime-numbers vectors
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
$endgroup$
add a comment |
$begingroup$
Let $F_1,F_2in mathbb{Z}[x_1,...,x_n]$ where $d_i=deg(F_i)$ and $d_1+d_2<n$.
Write $F_1(x)=a_1x^{e_1}+...+a_mx^{e_m}$ and $F_2(x)=b_1x^{ε_1}+...+b_μx^{ε_μ}$, where each $e_i,ε_i$ are vectors in $mathbb{Z}^n$. Write $e_1=begin{bmatrix} e_{11} \ ... \ e_{1n} end{bmatrix}$ and so on.
Let $p$ be prime. Let $0leq jleq p-1$ and let $k_1,...,k_min mathbb{Z}_{geq 0}$ such that $k_1+...+k_m=jp^n$.
Let $0leq l_1,...l_μleq p-1$.
Let $h_1=max{σ(e_{i1})+...+σ(e_{in}):1leq i leq m}$ and $h_2=max{σ(ε_{i1})+...+σ(ε_{in}):1leq i leq μ}$, where $σ$ returns the sum of digits in the base-p expansion.
Let $α=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}=0}$.
Let $β=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}$ isn't zero but is divisible by $p-1}$.
Assume further that $α+β=n$.
I want to show that $frac 1 {max{h_1,h_2}}(n-frac{jh_1} {p-1})leq α+frac {σ(k_1)+...+σ(k_m)-(p-1)} {p-1}+ frac {l_1+...+l_μ} {p-1}$
I see that the sum of coordinates in the vector $σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_1+...+l_με_μ$ is bounded below by $β(p-1)$ since each coordinate is divisible by $p-1$, and bounded above by $[σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$.
Hence we have the inequality $β(p-1)leq [σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$, which I'm hoping to be able to use to prove the highlighted inequality.
number-theory inequality polynomials prime-numbers vectors
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
$endgroup$
add a comment |
$begingroup$
Let $F_1,F_2in mathbb{Z}[x_1,...,x_n]$ where $d_i=deg(F_i)$ and $d_1+d_2<n$.
Write $F_1(x)=a_1x^{e_1}+...+a_mx^{e_m}$ and $F_2(x)=b_1x^{ε_1}+...+b_μx^{ε_μ}$, where each $e_i,ε_i$ are vectors in $mathbb{Z}^n$. Write $e_1=begin{bmatrix} e_{11} \ ... \ e_{1n} end{bmatrix}$ and so on.
Let $p$ be prime. Let $0leq jleq p-1$ and let $k_1,...,k_min mathbb{Z}_{geq 0}$ such that $k_1+...+k_m=jp^n$.
Let $0leq l_1,...l_μleq p-1$.
Let $h_1=max{σ(e_{i1})+...+σ(e_{in}):1leq i leq m}$ and $h_2=max{σ(ε_{i1})+...+σ(ε_{in}):1leq i leq μ}$, where $σ$ returns the sum of digits in the base-p expansion.
Let $α=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}=0}$.
Let $β=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}$ isn't zero but is divisible by $p-1}$.
Assume further that $α+β=n$.
I want to show that $frac 1 {max{h_1,h_2}}(n-frac{jh_1} {p-1})leq α+frac {σ(k_1)+...+σ(k_m)-(p-1)} {p-1}+ frac {l_1+...+l_μ} {p-1}$
I see that the sum of coordinates in the vector $σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_1+...+l_με_μ$ is bounded below by $β(p-1)$ since each coordinate is divisible by $p-1$, and bounded above by $[σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$.
Hence we have the inequality $β(p-1)leq [σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$, which I'm hoping to be able to use to prove the highlighted inequality.
number-theory inequality polynomials prime-numbers vectors
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
$endgroup$
Let $F_1,F_2in mathbb{Z}[x_1,...,x_n]$ where $d_i=deg(F_i)$ and $d_1+d_2<n$.
Write $F_1(x)=a_1x^{e_1}+...+a_mx^{e_m}$ and $F_2(x)=b_1x^{ε_1}+...+b_μx^{ε_μ}$, where each $e_i,ε_i$ are vectors in $mathbb{Z}^n$. Write $e_1=begin{bmatrix} e_{11} \ ... \ e_{1n} end{bmatrix}$ and so on.
Let $p$ be prime. Let $0leq jleq p-1$ and let $k_1,...,k_min mathbb{Z}_{geq 0}$ such that $k_1+...+k_m=jp^n$.
Let $0leq l_1,...l_μleq p-1$.
Let $h_1=max{σ(e_{i1})+...+σ(e_{in}):1leq i leq m}$ and $h_2=max{σ(ε_{i1})+...+σ(ε_{in}):1leq i leq μ}$, where $σ$ returns the sum of digits in the base-p expansion.
Let $α=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}=0}$.
Let $β=card{i:σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_{1i}+...+l_με_{μi}$ isn't zero but is divisible by $p-1}$.
Assume further that $α+β=n$.
I want to show that $frac 1 {max{h_1,h_2}}(n-frac{jh_1} {p-1})leq α+frac {σ(k_1)+...+σ(k_m)-(p-1)} {p-1}+ frac {l_1+...+l_μ} {p-1}$
I see that the sum of coordinates in the vector $σ(k_1)e_{11}+...+σ(k_m)e_{mi}+l_1ε_1+...+l_με_μ$ is bounded below by $β(p-1)$ since each coordinate is divisible by $p-1$, and bounded above by $[σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$.
Hence we have the inequality $β(p-1)leq [σ(k_1)+...+σ(k_m)]d_1+[l_1+...l_μ]d_2$, which I'm hoping to be able to use to prove the highlighted inequality.
number-theory inequality polynomials prime-numbers vectors
number-theory inequality polynomials prime-numbers vectors
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
asked Dec 20 '18 at 18:07
Pascal's WagerPascal's Wager
364315
364315
add a comment |
add a comment |
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