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Consider the power series $$ \sqrt{1+z} = \sum_{k=0}^\infty \left( \begin{array}{c} \frac{1}{2} \\ k \end{array} \right) z^k. $$

I am interested in characterizing the roots of the partial sum $$ s_n(z) = \sum_{k=0}^n \left( \begin{array}{c} \frac{1}{2} \\ k \end{array} \right) z^k. $$

The $n$ roots all seem to lie outside the unit circle and asymptotically approach the unit circle (and the roots seem to have equally spaced arguments). For instance, for $n=50$, the roots (red dots) look as follows in the complex plane (solid black circle denotes the unit circle): enter image description here

Is there any way of getting a handle on these roots, either for a fixed (large) $n$ or asymptotically $n \to \infty$? More generally, I am struggling to find useful references for the more general question of understanding the roots of partial sums. Perhaps there is not that much one can say about them in general. (One nice general result I found in the literature constrains where the roots can appear as a function of the radius of convergence; however, I could not find much more.)

EDIT: by calculating derivatives, it is straightforward to prove that the roots of $s_n(z)$ can be written as $z_i = - \frac{1}{\lambda_i}$ with $$ \sum_{i=1}^n \lambda_i^k = \frac{1}{2} \textrm{ for all } k =1,2,\cdots,n. $$ In other words, characterizing the roots is equivalent to solving $$ \left\{ \begin{array}{ccc} \lambda_1 + \lambda_2 + \cdots + \lambda_n & = & \frac{1}{2} \\ \lambda_1^2 + \lambda_2^2 + \cdots + \lambda_n^2 & = & \frac{1}{2} \\ & \vdots & \\ \lambda_1^n + \lambda_2^n + \cdots + \lambda_n^n & = & \frac{1}{2}. \end{array} \right. $$

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    $\begingroup$ Not sure if this helps, but it looks like they are forming some sort of cardioid. $\endgroup$
    – wormram
    Commented Dec 31, 2020 at 1:53
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    $\begingroup$ A similar question has been asked here. $\endgroup$
    – Jean Marie
    Commented Dec 31, 2020 at 2:19
  • $\begingroup$ Certainly caused by the asymptotic expansion of $\frac{{1/2\choose k}}{{1/2\choose n}}$ (for $k\le n$) in powers of $1-k/n$ $\endgroup$
    – reuns
    Commented Jan 1, 2021 at 15:18
  • $\begingroup$ Sorry for entering late in the game, but I guess you are considering the Szegö curve. I am no expertise on this topic, but this keyword would reveal a lot more about zeros of Taylor polynomials. $\endgroup$ Commented Jul 14, 2022 at 12:37

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