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Solve $\frac{x+\dots+x^K}{K} = \frac{1}{2}$ for large values of $K$

I am interested in the unique solution $x$ for the equation :

$$
p_K(x)=\frac{x+\dots+x^K}{K}=\frac{1}{2},
$$

for large values of $K$. When $K$ is small ($K=1$ and $K=2$) we can solve this equation explicitly and find :
$$
x=\frac{1}{2}, \frac{\sqrt{5}-1}{2}.
$$

For $K=3$ we still get an explicit solution which is more complicated and from $K=4$ on I do not find an explicit solution. As $K$ tends to infinity, we find that the unique solution $x_K$ of $p_K(x_K)=0$ tends to one, i.e. $\lim_{K \rightarrow \infty} x_K = 1$. I would like to find an asymptotic approximation of $p_K(x)$ denoted by $\tilde p_K(x)$ for which the solutions $\tilde x_K$ satisfy:
$$
\lim_{K\rightarrow \infty} K \cdot \log(x_K)
=
\lim_{K\rightarrow \infty} K \cdot \log(\tilde x_K).
$$

My idea was to use the approximation :
$$
\frac{x+\dots+x^K}{K} \approx x^{\frac{\sum_{j=1}^K j }{K}} = x^{\frac{K+1}{2}}.
$$

Using this approximation, we find $\tilde x_K = \left( \frac{1}{2} \right) ^{\frac{2}{K+1}}$ and we find for the limit :
$$
\lim_{K\rightarrow \infty} K \cdot \log(\tilde x_K) = -2 \log(2) \approx -1.38
$$

but by numerical approximation, find :
$$
\lim_{K\rightarrow \infty} K \cdot \log(x_K) \approx – 1.592
$$

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