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If $a_1,a_2,\dots,a_n,b_1,b_2,\dots,b_n$ are complex numbers in $\mathbb C$, and for every $j\in\mathbb N$, we have the power sums $$\sum_{i=1}^n a_i^j=\sum_{i=1}^n b_i^j$$ I want, without applying the Newton's identities to prove that $a_i$ is just $b_k$ with proper subscripts order.

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Why? Newton's identities are a completely transparent proof of this. –  Qiaochu Yuan Sep 23 '11 at 1:03

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We prove this by induction on $n$, the case where $n=1$ being trivial. Without loss of generality, you can assume $b_1$ has the largest magnitude of all the $a_i$'s and $b_i$'s (otherwise just change the names around). Dividing each $a_i$ and each $b_i$ by $b_1$, we may assume $b_1 = 1$.

Adding the equations $\sum_{i = 1}^n a_i^j = \sum_{i = 1}^n b_i^j$ from $j = 1$ to $j = m$ for some $m$ and then dividing by $m$, you obtain $$\sum_{i=1}^n {1 \over m}\sum_{j=1}^{m}a_i^j = \sum_{i=1}^n {1 \over m}\sum_{j=1}^{m}a_i^j$$ Let $k$ denote the number of $a_i$'s equal to $1$, and let $l$ denote the number of $b_i$'s equal to $1$. By the formula for summing geometric series you get $$k + \sum_{\{i: a_i \neq 1\}} {1 \over m} {a_i^{m+1} - a_i \over a_i - 1} = l + \sum_{\{i: b_i \neq 1\}} {1 \over m} {b_i^{m+1} - b_i \over b_i - 1}$$ Now taking the limit of the above as $m$ goes to infinity (recalling each $|a_i|$ and $|b_i|$ are at most $1$) you get that $k = l$. This means that $k$ of the $a_i$'s are equal to $k$ of the $b_i$'s (and are all equal to 1). Thus we can remove them from the sums, and it remains to consider the remaining $n - k$ $a_i$'s and $n -k$ $b_i$'s. The result follows by induction and we're done.

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$\max|a_i|=\max|b_i|$, right? –  gylns Sep 23 '11 at 3:04
    
That follows from that $k = l$ since $l$ is at least 1. And it follows from the overall statement of course too. –  Zarrax Sep 23 '11 at 3:35

Let $A$ and $B$ be the sets $\{a_j\}_{j=1}^n$ and $\{b_j\}_{j=1}^n$ respectively. Note that $\sum_{j=1}^n p(a_j) = \sum_{j=1}^n p(b_j)$ for all polynomials $p$. Now for any $\alpha \in A \cup B$ we have a polynomial $p(z) = \prod \{\frac{z - c}{\alpha - c}: c \in A \cup B \backslash \{\alpha\}\}$ such that $p(\alpha) = 1$ but $p(c) = 0$ for all other $c \in A \cup B$. So $\sum_{j=1}^n p(a_j)$ is the number of times $\alpha$ occurs in the sequence $(a_j)_{j=1}^n$, and this is the same as $\sum_{j=1}^n p(b_j)$ which is the number of times $\alpha$ occurs in $(b_j)_{j=1}^n$.

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