When does a polynomial have all pure imaginary roots? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-22T17:42:44Z http://mathoverflow.net/feeds/question/56802 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56802/when-does-a-polynomial-have-all-pure-imaginary-roots When does a polynomial have all pure imaginary roots? Shunyi Liu 2011-02-27T09:16:59Z 2011-02-28T12:18:28Z <p>Let $P(x)=x^{n}+a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}$, where $a_1, a_2, \dots, a_n$ are intergers.</p> <p>Question 1. When does the polynomial $P(x)$ have its zeros all being pure imaginary or zero(here 0 is a root of the given polynomial)?</p> <p>Question 2. Does there exist a characterization that $P(x)$ have its zeros all being pure imaginary or zero?</p> <p>Please point me to some references if this has already been studied. Thanks for your time!</p> http://mathoverflow.net/questions/56802/when-does-a-polynomial-have-all-pure-imaginary-roots/56804#56804 Answer by Abel for When does a polynomial have all pure imaginary roots? Abel 2011-02-27T09:46:24Z 2011-02-27T09:46:24Z <p>This should be a comment but I don't have enough reputation to leave one. First of all, $0$ is a root of $P(x)$ if and only if the independent term $a_n$ is null. Then, you can characterize the coefficients in terms of elementary symmetric polynomials and their degree: $a_i = (-1)^{n-i} \sigma_n$, where $\sigma_n = \sum_{1 \le j_1 &lt; j_2 &lt; \cdots &lt; j_n \le n} r_{j_1}r_{j_2}\dotsm r_{j_n}$, and $r_1, \dotsc , r_n$ are the roots of the polynomial. From this you can see that a necessary condition for $r_1, \dotsc , r_n$ to be imaginary is that $a_k$ is of the form $c_ki$ if $k \equiv n \pmod 4$, $c_k$ if $k \equiv n-1 \pmod 4$, $-c_ki$ if $k \equiv n-2 \pmod 4$, and $-c_k$ if $k \equiv n-3 \pmod 4$, where $c_k$ is a real number for all $k \in \mathbb{N}$.</p> http://mathoverflow.net/questions/56802/when-does-a-polynomial-have-all-pure-imaginary-roots/56812#56812 Answer by Didier Piau for When does a polynomial have all pure imaginary roots? Didier Piau 2011-02-27T12:06:32Z 2011-02-28T12:18:28Z <p>A necessary and sufficient condition is that $P(x)$ is a power of $x$ times a product of terms $x^2+c$ with $c$ real and positive. Hence $P(x)$ is $x^{n-2m}Q(x^2)$ where $m\ge0$ and $Q(t)$ is a unitary polynomial in $t$ of degree $m$ with integer nonnegative coefficients. Finally, a necessary condition is that $a_{k}=0$ for every odd $k$ and one should further require the underlying polynomial $Q$ to be a product of $t+c_i$.</p> <hr> <p><strong>EDIT</strong> It appears that the problem of locating the zeroes of a polynomial resurfaces regularly on MO, see for example <a href="http://mathoverflow.net/questions/9431/when-does-a-real-polynomial-have-a-pair-of-complex-conjugate-roots" rel="nofollow">this question</a> or <a href="http://mathoverflow.net/questions/2837/what-are-conditions-on-real-coefficients-for-zeros-of-a-polynomial-to-be-on-the-u" rel="nofollow">that one</a> or <a href="http://mathoverflow.net/questions/20946/criteria-to-determine-whether-a-real-coefficient-polynomial-has-real-root" rel="nofollow">that one</a>. The answers to these provide the following facts.</p> <p>First, there are Newton's inequalities. For any real numbers $(t_i)_{1\le i\le m}$, their elementary symmetric means $S_k$ are such that, for every $k$, $S_k^2\ge S_{k-1}S_{k+1}$. Here $S_k=\sigma_k/{m\choose k}$ where $\sigma_k$ denotes the $k$th elementary function in the real numbers $(t_i)_{1\le i\le m}$, see <a href="http://en.wikipedia.org/wiki/Newton%27s_inequalities" rel="nofollow">here</a>.</p> <p>So, write $Q$ as $Q(t)=t^m+b_1t^{m-1}+\cdots+b_m$ (and recall that $b_k=a_{2k}$). Any $Q$ under consideration is such that $b_k$ must be nonnegative for every $k$ and Newton's inequalities indicate that supplementary necessary condition are $S_k^2\ge S_{k-1}S_{k+1}$ for every $k$, where the $S_k$ are based on the $\sigma_k=b_k$. (At first sight, one should choose $\sigma_k=(-1)^kb_k$ but the $(-1)^k$ disappear.) Hence, a necessary condition is that, for every $k$, $$k(m-k)b_k^2\ge (k+1)(m-k+1)b_{k-1}b_{k+1}.$$ Second, a <em>complete characterization</em> of the polynomials $Q$ with only real negative roots is based on the notion of Hermite forms, see <a href="http://mathoverflow.net/questions/20946/criteria-to-determine-whether-a-real-coefficient-polynomial-has-real-root/30533#30533" rel="nofollow">this answer</a>. </p> <p>Recall that the Hermite form of a polynomial $Q$ of degree $m$ is a symmetric matrix, usually denoted by $H_1(Q)$, of size $m\times m$ with entries $(h_{ij}(Q))$, defined by $$h_{ij}(Q)=s_{i+j-2}(Q),$$ where, for every $k$, $s_k(Q)$ is the sum of the $k$th powers of the roots of $Q$ (and $s_0(Q)=m$), see <a href="http://www.math.ucsd.edu/~njw/Teaching/Math271C/Lecture_11.pdf" rel="nofollow">these lecture notes</a>. Recall that the $s_k(Q)$ are well known functions of the elementary symmetric functions $\sigma_k=(-1)^kb_k$, see <a href="http://en.wikipedia.org/wiki/Symmetric_polynomial" rel="nofollow">here</a>. Recall also that the signature of a symmetric matrix is equal to the number of its positive eigenvalues minus the number of its negative eigenvalues. Then the signature of $H_1(Q)$ is equal to the number of real roots of $Q$.</p> <p>In the context of this question, one already knows that $Q$ has no positive real root because $b_k$ is nonnegative for every $k$ and one wants $Q$ to have $m$ real roots. Hence the signature of $H_1(Q)$ must be $m$, that is, $H_1(Q)$ must be positive definite.</p> <p>Finally, necessary and sufficient conditions are that the odd numbered $a_k$ are zero and that the even numbered $a_k$ are nonnegative and define a polynomial $Q$ such that the symmetric matrix $H_1(Q)$ is positive definite.</p> http://mathoverflow.net/questions/56802/when-does-a-polynomial-have-all-pure-imaginary-roots/56830#56830 Answer by Kevin O'Bryant for When does a polynomial have all pure imaginary roots? Kevin O'Bryant 2011-02-27T15:53:53Z 2011-02-27T15:53:53Z <p>The general study of connections between the coefficients of a polynomial, the locations of its roots, the roots of its derivative, et cetera, is called the <em>Geometry of Zeros</em>. There are books, I believe, with this title. Google turned up <a href="http://www.ams.org/journals/bull/2011-48-01/S0273-0979-2010-01321-5/home.html" rel="nofollow">this survey</a>.</p> <p>As with most things, your exact question is likely not in print anywhere (I like Didier's answer), but there is an active and beautiful field with powerful techniques that are worth study.</p> http://mathoverflow.net/questions/56802/when-does-a-polynomial-have-all-pure-imaginary-roots/56835#56835 Answer by Michael Renardy for When does a polynomial have all pure imaginary roots? Michael Renardy 2011-02-27T16:38:36Z 2011-02-27T16:38:36Z <p>There is a method due to Sturm which allows the determination of the number of roots of a polynomial in any real interval, in particular you may apply this to find the number of negative roots. The following Wikipedia article may serve as an initial reference: <a href="http://en.wikipedia.org/wiki/Sturm%27s_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Sturm%27s_theorem</a></p>