Sifting polynomials
$begingroup$
Let's say that polynomial $P(x)$ sifts another polynomial $Q(x)$ if three conditions are satisfied:
- $deg P(x) + 1 = deg Q(x) = n$
$Q(x)$ has $n$ distinct real roots, $P(x)$ has $n - 1$ distinct real roots- Roots of $Q(x)$ and $P(x)$ alternate, so if $alpha_1, alpha_2, dots, alpha_n$ are roots of $Q(x)$, $beta_1, beta_2, dots, beta_{n - 1}$ are roots of $P(x)$, then $alpha_1 < beta_1 < alpha_2 < beta_2 < dots < beta_{n - 1} < alpha_n$.
There's a statement that if $P(x)$ sifts $Q(x)$ and $R(x)$ is the remainder of $Q(x)$ divided by $P(x)$, then $R(x)$ sifts $P(x)$, but I failed to come up with an idea to prove it. I tried to build something used factorised forms of polynomials, but it seems to lead nowhere.
real-analysis polynomials
asked Jan 15 at 17:09
NikromNikrom
1376
$endgroup$
add a comment |
$begingroup$
Let's say that polynomial $P(x)$ sifts another polynomial $Q(x)$ if three conditions are satisfied:
- $deg P(x) + 1 = deg Q(x) = n$
$Q(x)$ has $n$ distinct real roots, $P(x)$ has $n - 1$ distinct real roots- Roots of $Q(x)$ and $P(x)$ alternate, so if $alpha_1, alpha_2, dots, alpha_n$ are roots of $Q(x)$, $beta_1, beta_2, dots, beta_{n - 1}$ are roots of $P(x)$, then $alpha_1 < beta_1 < alpha_2 < beta_2 < dots < beta_{n - 1} < alpha_n$.
There's a statement that if $P(x)$ sifts $Q(x)$ and $R(x)$ is the remainder of $Q(x)$ divided by $P(x)$, then $R(x)$ sifts $P(x)$, but I failed to come up with an idea to prove it. I tried to build something used factorised forms of polynomials, but it seems to lead nowhere.
real-analysis polynomials
asked Jan 15 at 17:09
NikromNikrom
1376
$endgroup$
add a comment |
$begingroup$
Let's say that polynomial $P(x)$ sifts another polynomial $Q(x)$ if three conditions are satisfied:
- $deg P(x) + 1 = deg Q(x) = n$
$Q(x)$ has $n$ distinct real roots, $P(x)$ has $n - 1$ distinct real roots- Roots of $Q(x)$ and $P(x)$ alternate, so if $alpha_1, alpha_2, dots, alpha_n$ are roots of $Q(x)$, $beta_1, beta_2, dots, beta_{n - 1}$ are roots of $P(x)$, then $alpha_1 < beta_1 < alpha_2 < beta_2 < dots < beta_{n - 1} < alpha_n$.
There's a statement that if $P(x)$ sifts $Q(x)$ and $R(x)$ is the remainder of $Q(x)$ divided by $P(x)$, then $R(x)$ sifts $P(x)$, but I failed to come up with an idea to prove it. I tried to build something used factorised forms of polynomials, but it seems to lead nowhere.
real-analysis polynomials
asked Jan 15 at 17:09
NikromNikrom
1376
$endgroup$
Let's say that polynomial $P(x)$ sifts another polynomial $Q(x)$ if three conditions are satisfied:
- $deg P(x) + 1 = deg Q(x) = n$
$Q(x)$ has $n$ distinct real roots, $P(x)$ has $n - 1$ distinct real roots- Roots of $Q(x)$ and $P(x)$ alternate, so if $alpha_1, alpha_2, dots, alpha_n$ are roots of $Q(x)$, $beta_1, beta_2, dots, beta_{n - 1}$ are roots of $P(x)$, then $alpha_1 < beta_1 < alpha_2 < beta_2 < dots < beta_{n - 1} < alpha_n$.
There's a statement that if $P(x)$ sifts $Q(x)$ and $R(x)$ is the remainder of $Q(x)$ divided by $P(x)$, then $R(x)$ sifts $P(x)$, but I failed to come up with an idea to prove it. I tried to build something used factorised forms of polynomials, but it seems to lead nowhere.
real-analysis polynomials
real-analysis polynomials
asked Jan 15 at 17:09
NikromNikrom
1376
asked Jan 15 at 17:09
NikromNikrom
1376
edited Jan 15 at 21:52
Nikrom
asked Jan 15 at 17:09
NikromNikrom
1376
asked Jan 15 at 17:09
NikromNikrom
1376
asked Jan 15 at 17:09
NikromNikrom
1376
1376
add a comment |
add a comment |
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