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Given the function $$ \phi(x) = (x - x_1) (x - x_2) \cdots (x - x_m) $$ where $m$ is odd, and the points $x_1, x_2, \cdots, x_m$ are symmetric wrt the midpoint of its domain, show that the function is odd wrt the midpoint of its domain. We define $$ a \le x_1 < x_2 < \cdots < x_m \le b $$ so that the midpoint of its domain $[a, b]$ is $\frac{a+b}{2}$.

What we can do is define a transform $$ \tau = x - \frac{a + b}{2}$$ that recentres the domain onto $$ \left[\frac{a - b}{2}, \frac{b - a}{2}\right] $$ which has midpoint the origin.

It remains to show that $$ \phi(-\tau) = -\phi(\tau). $$

I understand the function should be odd, but I can't show this algebraically. I've tried using induction on $m$, but if $m=1$ then $$ -\phi(\tau) = -(\tau - x_1),$$ $$ \phi(-\tau) = (-\tau - x_1),$$ and I can't even prove the base case. I've also tried just substituting in $\tau$ and $-\tau$ for arbitrary odd $m$ and doing some algebra but I only get $$ -\phi(\tau) = - (\tau - x_1) (\tau - x_2) \cdots (\tau - x_m),$$ $$ \phi(-\tau) = - (\tau + x_1) (\tau + x_2) \cdots (\tau + x_m).$$

jamesh625
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1 Answers1

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Hint: Let $m = 2k + 1$ for some $k \in \mathbb Z$. Then if $x_1, x_2, \ldots, x_{2k+1}$ are symmetric wrt the origin, it follows that we can relabel these $2k + 1$ roots as: $$ -r_k,~ -r_{k-1},~ \ldots,~ -r_1,~ 0,~ r_1,~ \ldots,~ r_{k-1},~ r_k $$

Adriano
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