Difference between revisions of "2018 Putnam B Problems/Problem 2"
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==Solution== | ==Solution== | ||
| + | |||
| + | Given | ||
| + | <cmath> | ||
| + | f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}, | ||
| + | </cmath> | ||
| + | we want to show \( f_n(z) \neq 0 \) for all \( |z| \leq 1 \). | ||
| + | |||
| + | Rewrite \( f_n(z) \) as | ||
| + | <cmath> | ||
| + | f_n(z) = \sum_{k=0}^{n-1} (n-k)z^k. | ||
| + | </cmath> | ||
| + | Multiply by \( 1-z \): | ||
| + | <cmath> | ||
| + | (1-z)f_n(z) = \sum_{k=0}^{n-1} (n-k)z^k - \sum_{k=0}^{n-1} (n-k)z^{k+1}. | ||
| + | </cmath> | ||
| + | Shift index in the second sum: | ||
| + | <cmath> | ||
| + | = n + \sum_{k=1}^{n-1} (n-k)z^k - \sum_{k=1}^{n} (n-(k-1))z^k. | ||
| + | </cmath> | ||
| + | Write out terms: | ||
| + | <cmath> | ||
| + | = n + \sum_{k=1}^{n-1} (n-k)z^k - \sum_{k=1}^{n} (n-k+1)z^k. | ||
| + | </cmath> | ||
| + | Combine sums for \( k=1 \) to \( n-1 \): | ||
| + | <cmath> | ||
| + | = n + \sum_{k=1}^{n-1} [(n-k) - (n-k+1)]z^k - (n-n+1)z^n, | ||
| + | </cmath> | ||
| + | which simplifies to | ||
| + | <cmath> | ||
| + | = n + \sum_{k=1}^{n-1} (-1)z^k - z^n = n - \sum_{k=1}^{n} z^k. | ||
| + | </cmath> | ||
| + | So | ||
| + | <cmath> | ||
| + | (1-z)f_n(z) = n - \frac{z(1-z^n)}{1-z} \quad \text{if } z \neq 1. | ||
| + | </cmath> | ||
| + | Multiply both sides by \( 1-z \): | ||
| + | <cmath> | ||
| + | (1-z)^2 f_n(z) = n(1-z) - z(1-z^n). | ||
| + | </cmath> | ||
| + | Suppose \( f_n(z) = 0 \) with \( |z| \leq 1 \) and \( z \neq 1 \). Then | ||
| + | <cmath> | ||
| + | n(1-z) = z(1-z^n). | ||
| + | </cmath> | ||
| + | Rearranged: | ||
| + | <cmath> | ||
| + | n = \frac{z(1-z^n)}{1-z}. | ||
| + | </cmath> | ||
| + | Consider \( |z| \leq 1 \). If \( |z| = 1 \), then \( |1-z^n| \leq 2 \) and denominator \( |1-z| \) is at most 2, but the right side stays bounded by something less than \( n \) (except at \( z=1 \), excluded). This can't equal \( n \) exactly. | ||
| + | |||
| + | If \( |z| < 1 \), the geometric sum \( \frac{1-z^n}{1-z} \) has magnitude less than \( \frac{1}{1-|z|} \), which is finite but smaller than \( n \). So the equation can't hold for any \( z \neq 1 \). | ||
| + | |||
| + | At \( z=1 \), \( f_n(1) = n + (n-1) + \dots + 1 = \frac{n(n+1)}{2} \neq 0 \). | ||
| + | |||
| + | Thus, \( f_n(z) \neq 0 \) for \( |z| \leq 1 \). :-) | ||
| + | |||
| + | ~Pinotation | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:06, 18 August 2025
Problem
Let \( n \) be a positive integer, and let \( f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1} \). Prove that \( f_n \) has no roots in the closed unit disk \( \{z \in \mathbb{C}: |z| \leq 1 \} \).
Solution
Given
![\[f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1},\]](http://latex.artofproblemsolving.com/d/1/f/d1f62f8a20858f9ac218b3abe08a16c36abc6897.png)
we want to show \( f_n(z) \neq 0 \) for all \( |z| \leq 1 \).
Rewrite \( f_n(z) \) as
![\[f_n(z) = \sum_{k=0}^{n-1} (n-k)z^k.\]](http://latex.artofproblemsolving.com/b/c/a/bca1fd0ecdfafa40d9d9e02a555e05e895318f5b.png)
Multiply by \( 1-z \):
![\[(1-z)f_n(z) = \sum_{k=0}^{n-1} (n-k)z^k - \sum_{k=0}^{n-1} (n-k)z^{k+1}.\]](http://latex.artofproblemsolving.com/6/4/6/6468023dddb1764fe7fa90f7b4d5bcd930b4d61e.png)
Shift index in the second sum:
![\[= n + \sum_{k=1}^{n-1} (n-k)z^k - \sum_{k=1}^{n} (n-(k-1))z^k.\]](http://latex.artofproblemsolving.com/3/5/9/359311e3061df013ecf042ba734fe0a0f81d4cd5.png)
Write out terms:
![\[= n + \sum_{k=1}^{n-1} (n-k)z^k - \sum_{k=1}^{n} (n-k+1)z^k.\]](http://latex.artofproblemsolving.com/7/1/1/711e9b70aaa0c2d152458fc8e091e18372e34ace.png)
Combine sums for \( k=1 \) to \( n-1 \):
![\[= n + \sum_{k=1}^{n-1} [(n-k) - (n-k+1)]z^k - (n-n+1)z^n,\]](http://latex.artofproblemsolving.com/e/b/7/eb7d59c84a4c5d9c8d4485a38f5f97c0e52c65df.png)
which simplifies to
![\[= n + \sum_{k=1}^{n-1} (-1)z^k - z^n = n - \sum_{k=1}^{n} z^k.\]](http://latex.artofproblemsolving.com/3/9/3/393683270ff33c354b2e8d9e63e25a1ec31e485a.png)
So
![\[(1-z)f_n(z) = n - \frac{z(1-z^n)}{1-z} \quad \text{if } z \neq 1.\]](http://latex.artofproblemsolving.com/b/5/6/b56203fcb1ea8633a1fd28550fdcecaa583f935e.png)
Multiply both sides by \( 1-z \):
![\[(1-z)^2 f_n(z) = n(1-z) - z(1-z^n).\]](http://latex.artofproblemsolving.com/b/c/7/bc7c5d7dfbe494bc0a6af03a46aebb70c9e487ed.png)
Suppose \( f_n(z) = 0 \) with \( |z| \leq 1 \) and \( z \neq 1 \). Then
![\[n(1-z) = z(1-z^n).\]](http://latex.artofproblemsolving.com/2/7/9/279bc651b50a754d64da21b109cf9a481a267adf.png)
Rearranged:
![\[n = \frac{z(1-z^n)}{1-z}.\]](http://latex.artofproblemsolving.com/d/0/8/d08545c50e8678f419a81cbf990415484e47aa57.png)
Consider \( |z| \leq 1 \). If \( |z| = 1 \), then \( |1-z^n| \leq 2 \) and denominator \( |1-z| \) is at most 2, but the right side stays bounded by something less than \( n \) (except at \( z=1 \), excluded). This can't equal \( n \) exactly.
If \( |z| < 1 \), the geometric sum \( \frac{1-z^n}{1-z} \) has magnitude less than \( \frac{1}{1-|z|} \), which is finite but smaller than \( n \). So the equation can't hold for any \( z \neq 1 \).
At \( z=1 \), \( f_n(1) = n + (n-1) + \dots + 1 = \frac{n(n+1)}{2} \neq 0 \).
Thus, \( f_n(z) \neq 0 \) for \( |z| \leq 1 \). :-)
~Pinotation
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
