n次多项式所有根都是实数，求证实根上界的充要条件

abababa

设$f(x)$是首一实系数$n$次多项式，且$n$个根都是实根，求证实数$b$是$f(x)$的实根上界的充要条件是：$f(b),f'(b),\cdots,f^{(n)}(b)>0$。

hbghlyj

$\Longleftarrow$
若$f(b),f'(b),\dots,f^{(n)}(b)>0$，则$\forall x>b:$
\[x-b>0\implies f(x)=f(b)+\sum_{k=1}^n\frac{f^{(k)}(b)}{k!}(x-b)^k\ge f(b)>0\]所以$b$是$f(x)$的实根上界。

$\Longrightarrow$
设$f$的根为$x_1\le x_2\le\dots\le x_n$，由Rolle中值定理，$f'$在每个区间$[x_1,x_2],\dots,[x_{n-1},x_n]$有根，但$f'$至多有$n-1$个实根，故$f'$在每个区间$[x_1,x_2],\dots,[x_{n-1},x_n]$恰有一个根。以此類推$\forall k=1,\dots,n$，$f^{(k)}$有$n-k$个实根，都属于$[x_1,x_n]$，
若$b>x_n$，$f^{(k)}$的根都$<b$，且$f^{(k)}$首一，所以 $f^{(k)}(b)>0$.

abababa

hbghlyj 发表于 2024-4-4 03:41
$\Longleftarrow$
若$f(b),f'(b),\dots,f^{(n)}(b)>0$，则$\forall x>b:$
\[x-b>0\implies f(x)=f(b)+\sum_ ...

原来如此，多项式用泰勒展开到$\deg(f)$后余项是零。谢谢。

hbghlyj

Lagrange's and Cauchy's bounds
Lagrange and Cauchy were the first to provide upper bounds on all complex roots. Lagrange's bound is
$$ \max \left\{1,\sum _{i=0}^{n-1}\left|{\frac {a_{i}}{a_{n}}}\right|\right\}, $$
and Cauchy's bound is
$$ 1+\max \left\{\left|{\frac {a_{n-1}}{a_{n}}}\right|,\left|{\frac {a_{n-2}}{a_{n}}}\right|,\ldots ,\left|{\frac {a_{0}}{a_{n}}}\right|\right\} $$
Proof of Lagrange's and Cauchy's bounds

If z is a root of the polynomial, and |z| ≥ 1 one has
$$ |a_{n}||z^{n}|=\left|\sum _{i=0}^{n-1}a_{i}z^{i}\right|\leq \sum _{i=0}^{n-1}|a_{i}z^{i}|\leq \sum _{i=0}^{n-1}|a_{i}||z|^{n-1}. $$
Dividing by $ |a_{n}||z|^{n-1}, $ one gets
$$ |z|\leq \sum _{i=0}^{n-1}{\frac {|a_{i}|}{|a_{n}|}}, $$
which is Lagrange's bound when there is at least one root of absolute value larger than 1. Otherwise, 1 is a bound on the roots, and is not larger than Lagrange's bound.

Similarly, for Cauchy's bound, one has, if |z| ≥ 1,
$$ |a_{n}||z^{n}|=\left|\sum _{i=0}^{n-1}a_{i}z^{i}\right|\leq \sum _{i=0}^{n-1}|a_{i}z^{i}|\leq \max |a_{i}|\sum _{i=0}^{n-1}|z|^{i}={\frac {|z|^{n}-1}{|z|-1}}\max |a_{i}|\leq {\frac {|z|^{n}}{|z|-1}}\max |a_{i}|. $$
Thus
$$ |a_{n}|(|z|-1)\leq \max |a_{i}|. $$
Solving in |z|, one gets Cauchy's bound if there is a root of absolute value larger than 1. Otherwise the bound is also correct, as Cauchy's bound is larger than 1.

hbghlyj

abababa 发表于 2024-4-4 04:00
原来如此，多项式用泰勒展开到$\deg(f)$后余项是零。谢谢。

$f^{(k)}$的根都属于$[x_1,x_n]$，
这个是Gauss–Lucas theorem的特殊情况。

Special cases
In addition, if a polynomial of degree n of real coefficients has n distinct real zeros $ x_{1}<x_{2}<\cdots <x_{n}, $ we see, using Rolle's theorem, that the zeros of the derivative polynomial are in the interval $ [x_{1},x_{n}] $ which is the convex hull of the set of roots.

hbghlyj

hbghlyj 发表于 2024-4-4 19:25
if a polynomial of degree n of real coefficients has n distinct real zeros $ x_{1}<x_{2}<\cdots <x_{n}, $

这里说，这n个实根需要不同，1#没有这个条件呀？

使用 Rolle's theorem 需要这n个实根不同

那么上面的证明正确吗

hbghlyj

abababa 发表于 2024-4-4 04:00
原来如此，多项式用泰勒展开到$\deg(f)$后余项是零。

没有用到“多项式”这个条件，有没有一般的实解析函数，满足$f^{(k)}\ge0,\forall k\ge0$？

hbghlyj

hbghlyj 发表于 2024-4-4 20:12
没有用到“多项式”这个条件，有没有一般的实解析函数，满足$f^{(k)}\ge0,\forall k\ge0$？ ...

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hbghlyj

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hbghlyj

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