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Clarified the definition of $g_\eta$ as from the original paper. There was an errant $s$ in one of the fractions.

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edit approved Dec 12, 2022 at 12:47

Fabian Wirth

1.2k
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In an article on the Routh-Hurwitz test, I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$$$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$$\eta=\frac{a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The title of the article is "Elementary proof of the Routh-Hurwitz test" (DOI link), and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

In an article on the Routh-Hurwitz test, I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The title of the article is "Elementary proof of the Routh-Hurwitz test" (DOI link), and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

In an article on the Routh-Hurwitz test, I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The title of the article is "Elementary proof of the Routh-Hurwitz test" (DOI link), and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

Minor grammar improvements and formatting

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edited Dec 11, 2022 at 17:10

Daniele Tampieri

6.4k
7
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In thean article with the proof ofon the Routh-Hurwitz test, I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The nametitle of the article is "Elementary proof of the Routh-Hurwitz test" (DOI link), and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

In the article with the proof of the Routh-Hurwitz test I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The name of the article is "Elementary proof of the Routh-Hurwitz test" DOI link, and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

In an article on the Routh-Hurwitz test, I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The title of the article is "Elementary proof of the Routh-Hurwitz test" (DOI link), and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

formatting, added link, added tag

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edited Dec 11, 2022 at 10:56

YCor

63.9k
5
187
286

Some calculations with polynomials in the proof of the Routh Hurwitz Test-Hurwitz test

In the article with the proof of the Routh Hurwitz-Hurwitz test I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The name of the article is "Elementary proof of the Routh-Hurwitz test" DOI link, and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

Some calculations with polynomials in the proof of the Routh Hurwitz Test

In the article with the proof of the Routh Hurwitz test I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The name of the article is "Elementary proof of the Routh-Hurwitz test", and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!

Some calculations with polynomials in the proof of the Routh-Hurwitz test

In the article with the proof of the Routh-Hurwitz test I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$ Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\frac{s a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.

The name of the article is "Elementary proof of the Routh-Hurwitz test" DOI link, and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!