10
$\begingroup$

Consider a polynomial of the form: $$f\left(z\right)=1+z^{p_{0}}+z^{p_{1}}+\ldots+z^{p_{N}}$$ where the $p_{n}$s are distinct positive integers. Are the roots of $f$ (in $\mathbb{C}$) necessarily simple (i.e., must they all have multiplicities of $1$)?

$\endgroup$
  • 6
    $\begingroup$ Compare this function with it's derivative. That's a good start. You know a root is simple if it's not shared between the function and it's derivative. $\endgroup$ – астон вілла олоф мэллбэрг Sep 19 '17 at 23:07
  • $\begingroup$ Obviously. But I am working with polynomials where the exponents are undetermined quantities. I cannot simply compare it that way. I was just wondering if there is some general result which bars polynomials of this form from having repeated roots. $\endgroup$ – MCS Sep 19 '17 at 23:20
  • $\begingroup$ @астонвіллаолофмэллбэрг How can this fact be used? $\endgroup$ – miracle173 Sep 20 '17 at 6:05
  • $\begingroup$ @miracle173 I've explained the usage of this fact in an answer below. $\endgroup$ – астон вілла олоф мэллбэрг Sep 20 '17 at 6:36
  • 1
    $\begingroup$ Of course the polynomial $z^2$ is the smallest example! Ah, upon reading more carefully, I see you require the zeroth-order coefficient to be $1$. $\endgroup$ – Jeppe Stig Nielsen Sep 20 '17 at 7:56
7
$\begingroup$

.I know there's been an answer accepted, but nevertheless since somebody in the comments has asked how this would help, I would like to see if it can fuel a discussion.

So we know that a root of a polynomial is simple if it's not a root of the derivative of that polynomial.

Let $f(z) = 1 + z^{a_0} + ... + z^{a_n}$, where $a_i$ are positive integers, without loss of generality $a_n$ being the largest.

Now, the derivative of $f(z)$ is $a_0z^{a_0 - 1} + a_1z^{a_1-1} + ... + a_{n}z^{a_n - 1}$, as we know it.

We note carefully that $f(-1) = 1 + (-1)^{a_0} + (-1)^{a_1} + ... + (-1)^{a_n}$. This is zero precisely when the number of odd $a_i$ exceeds the number of even $a_i$ by precisely one.

Similarly, $f'(-1) = a_0(-1)^{a_0-1} + a_1(-1)^{a_1-1} + ... + a_n(-1)^{a_n}$. This is zero precisely when this sum is zero, and in that case, we can conclude that $-1$ is a repeated root of $f$.

Observing the sum $f'(-1)$, we note that all odd $a_i$ are being added, while all even $a_i$ are being subtracted. Hence, $f'(-1) = 0$ is the same as saying the sum of all the odd $a_i$ is the same as the sum of all the even $a_i$.

Hence, we want just the following to happen : some number of even $a_i$, just one more number of odd $a_i$, and the sums of these must be the same.

Indeed, in the example given in the other answer, $a_1 = 3,a_2 = 5,a_3 = 8$. We see that the number of odd $a_i$ exceeds the number of even $a_i$ by $1$, and the sum of even and odd $a_i$ are actually equal.

Note that if this condition has to happen, then the number of odd $a_i$ has to be even and the number of even $a_i$ has to be odd (can you see why?)

Finally, we summarize this in a lemma:

Let $n$ be an even number, $\{a_1,...,a_n\}$ be a set of distinct odd numbers , and $\{b_1,...,b_{n-1}\}$ a set of distinct even numbers such that $\sum a_i = \sum b_j$. Then, the polynomial $1 + \sum z^{a_i} + \sum z^{b_j}$ has a repeated root $-1$.

To give a slightly more involved example that the one in the other answer, consider $\{1,3,5,7\}$ and $\{2,4,10\}$. Then, I claim the polynomial $1 + z + z^2 + z^3 + z^4 + z^5 + z^7+ z^{10}$ has the factor $(1+z)^2$. You can check it satisfies all given conditions of the lemma, and indeed: $$ 1 + z + z^2+z^2+z^4+z^5+z^7+z^{10} = (1+z)^2(1+z^2)(1-z+z^2)(1-z^3+z^4) $$

You can use this method to generate all the counter examples you like. If you want a family of counterexamples, you could just do the following : consider $\{3m+1,3m+3,3m+5,3m+7\}$ as a set of odd numbers, and $\{4m+2,4m+4,4m+10\}$ as a set of even numbers for $m$ even, and this again will satisfy all the conditions of the lemma.For example, with $m=2$ you would get the sets $\{7,9,11,13\}$ and $\{10,12,18\}$, which again satisfy the propery.

$\endgroup$
  • $\begingroup$ Beautifully presented. $\endgroup$ – Prune Sep 20 '17 at 15:19
30
$\begingroup$

No. We can construct as follows: The polynomial $(1+z^a)(1+z^b)$ will have two factors of $1+z$ if $a$ and $b$ are both odd. So if we choose $b$ a good bit bigger than $a$, the cross terms will miss each other and all coefficients will be $1$. For instance,

$$f(z) = (1+z^3)(1+z^5) = 1+z^3+z^5+z^8$$

has $-1$ as a double root.

$\endgroup$
  • $\begingroup$ As soon as $b>a>0$, the expansion of $(1+z^a)(1+z^b)$ will have the correct form $1+z^a+z^b+z^{a+b}$. The smallest choice for odd exponents is $a=1,b=3$ which gives $1+z+z^3+z^4$. $\endgroup$ – Jeppe Stig Nielsen Sep 20 '17 at 7:59
  • 1
    $\begingroup$ I think you can do the same for other repeated roots than $-1$. For example pick a square root of $-1$ (such as $i$), then $(1+z^2)(1+z^6)$ has that number as a non-simple root. Similarly for any $n$th root of $-1$, we can find a polynomial of the desired form that has that number as a repeated root, namely $(1+z^n)(1+z^{3n})$. $\endgroup$ – Jeppe Stig Nielsen Sep 20 '17 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.