4
$\begingroup$

I have a polynomial of the form:

$$\sum_{m=0}^k\frac{(-1)^{m+1}(4k-2m)!x^{2k-2m}}{m!(2k-m)!(2k-2m+1)!}$$ or identically:

$$\sum_{m=0}^k\frac{(-1)^{m+1}(4k-2m)!(x^{2})^{k-m}}{m!(2k-m)!(2k-2m+1)!}$$

where we can see that the sum of the coefficients is always zero.

Now , my question is that :

what can be concluded from this (i.e sum of the coefficients being zero) about zeros (roots) of the polynomial?

edit:

except the fact that 1 is a root.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

The sum of the coefficients is zero if and only if $1$ is a root of the given polynomial.

$\endgroup$
2
  • $\begingroup$ right, but I mean more than that. (Indeed , $x$ here is $cos(\alpha)$ and other solutions( other than 1 and 0) are of interest to me. $\endgroup$
    – Mostafa
    Commented Mar 10, 2013 at 18:16
  • $\begingroup$ from the fact that sum of coefficient is 0 you may conclude only that because of "if and only if" $\endgroup$
    – RiaD
    Commented Nov 16, 2017 at 13:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .