For each nonnegative integer \(n\), define \[ f_n(z)=\sum_{k=0}^n\frac{z^k}{k!}. \] Prove that \(f_n\) has no roots in the open unit disk. (Hint: check \(n=1\) and \(n=2\) directly.)
Solution
1. The case \(n=0\) has no roots.
Here \[ f_0(z)=1. \]
2. For every integer \(n\geq1\),\[ (1-z)f_n(z) = 1-H_n(z), \]where\[ H_n(z) = \sum_{k=2}^{n} \frac{k-1}{k!}z^k + \frac{z^{n+1}}{n!}. \]
Starting from \[ f_n(z)=\sum_{k=0}^n\frac{z^k}{k!}, \] one has \[ \begin{aligned} (1-z)f_n(z) &= \sum_{k=0}^n\frac{z^k}{k!} -\sum_{k=0}^n\frac{z^{k+1}}{k!}\\ &= 1 +\sum_{k=1}^{n} \left( \frac1{k!}-\frac1{(k-1)!} \right)z^k -\frac{z^{n+1}}{n!}. \end{aligned} \] The coefficient of \(z\) is zero, and for \(k\geq2\), \[ \frac1{k!}-\frac1{(k-1)!} = -\frac{k-1}{k!}. \] This is the stated identity.
3. The coefficients occurring in \(H_n\) satisfy\[ \sum_{k=2}^{n}\frac{k-1}{k!} +\frac1{n!} = 1. \]
For \(k\geq2\), \[ \frac{k-1}{k!} = \frac1{(k-1)!}-\frac1{k!}. \] Hence the sum telescopes: \[ \begin{aligned} \sum_{k=2}^{n}\frac{k-1}{k!} +\frac1{n!} &= \sum_{k=2}^{n} \left( \frac1{(k-1)!}-\frac1{k!} \right) +\frac1{n!}\\ &= 1. \end{aligned} \] For \(n=1\), the sum from \(k=2\) to \(1\) is empty and the same identity reads \(1/1!=1\).
4. Fix \(0<r<1\). On the circle \(\abs{z}=r\),\[ \abs{H_n(z)}<1. \]
By step 3 and the positivity of the coefficients, \[ \begin{aligned} \abs{H_n(z)} &\leq \sum_{k=2}^{n} \frac{k-1}{k!}r^k + \frac{r^{n+1}}{n!}\\ &< \sum_{k=2}^{n} \frac{k-1}{k!} + \frac1{n!}\\ &= 1. \end{aligned} \] The inequality is strict because \(0<r<1\).
5. For every \(0<r<1\), the polynomial \((1-z)f_n(z)\) has no zeros in\(\abs{z}<r\).
By step 2, \[ (1-z)f_n(z)=1-H_n(z). \] On \(\abs{z}=r\), step 4 gives \[ \abs{-H_n(z)}<\abs{1}. \] Rouché’s theorem therefore implies that \(1-H_n(z)\) and the constant function \(1\) have the same number of zeros in \(\abs{z}<r\), namely zero.
6. For every \(n\geq1\), the polynomial \(f_n\) has no roots in the openunit disk.
Suppose \(f_n(z_0)=0\) with \(\abs{z_0}<1\). Choose \(r\) such that \[ \abs{z_0}<r<1. \] Then \[ (1-z_0)f_n(z_0)=0, \] contradicting step 5.
7. For every nonnegative integer \(n\), the polynomial \(f_n\) has no rootsin the open unit disk.
Step 1 handles \(n=0\), and step 6 handles every \(n\geq1\).
8. Q.E.D.
Step 7 is the required conclusion.