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

Question

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 : $$ \lim_{K\rightarrow \infty} K \cdot \log(x_K) \approx - 1.592 $$

Answer 1

If the main objective is to solve the equation, I would suggest to plug into the equation the approximation for $\tilde{x}_K$ you are looking for. In particular, you appear to be looking after an approximate solution of the equation $p_K(\tilde{x}_K) \approx 1/2$ in the form $$ \tilde{x}_K = e^{C/K} $$ where $C \neq 0$ should be determined later.

Substituting in $p_K$ we have $$ p_K(\tilde{x}_K) = \frac{1}{K} \sum_{i=1}^{K} e^{Ci/K} = \frac{1}{K} \frac{e^C-e^{C/K}}{e^{C/K}-1} \approx \frac{1}{C} \left( e^C-1 \right) $$ for sufficiently large $K$. Using this into $p_K(\tilde{x}_K) \approx 1/2$, we deduce that the constant $C$ should solve the following equation $$ e^C -\frac{C}{2} -1 = 0. $$ Numerical solution finally gives $C \approx -1.592$ as claimed.

Answer 2

Without any hypothesis about a possible functional form of $x_k$, what we know is that the solution is closer and closer to $1$.

So, considering the function $$f(k)=\sum_{i=1}^k x^i -\frac 12 k$$ let us expand it as a series around $x=1$ to get $$f(k)=\frac 12 k+\sum_{n=1}^\infty \Big[\binom{k}{n}+\binom{k}{n+1}\Big](x-1)^n=0$$ and now, using series reversion (I did not the summation up to $\infty$ but simply to $n=15$). What is obtained after truncation is $$x=1+\sum_{m=1}^{p}{a_m}k^{-m}+O\left(\frac{1}{k^{p+1}}\right)$$ and, continuing with Taylor series, then $$k\log(x)=a_1+\frac{2a_2-a_1^2}{2k}+O\left(\frac{1}{k^{2}}\right)$$ that is to say $$k\log(x)=-\frac{14023057827349961}{8799642593280000}+\frac{430560793178719455374528289816022 7}{2013276454006156651030118400000000 }\frac 1k+O\left(\frac{1}{k^{2}}\right)$$ where the constant coefficient is $-1.59359$.

Continuing from @Stefano' answer, we have to solve $$\frac{e^c-1}{c}=\frac 12 \implies c=-2-W\left(-\frac{2}{e^2}\right)\approx -1.59362$$

Edit

In the above, was chosen arbitrarily the value $n=15$ for the expansions. In the table below are reproduced the values of coefficient $a_1^{(n)}$ $$\left( \begin{array}{ccc} n & a_1^{(n)}= & a_1^{(n)}\approx \\ 1 & -1 & -1.000000000 \\ 2 & -\frac{4}{3} & -1.333333333 \\ 3 & -\frac{53}{36} & -1.472222222 \\ 4 & -\frac{829}{540} & -1.535185185 \\ 5 & -\frac{10141}{6480} & -1.564969136 \\ 6 & -\frac{35821}{22680} & -1.579409171 \\ 7 & -\frac{4317869}{2721600} & -1.586518592 \\ 8 & -\frac{519299}{326592} & -1.590054257 \\ 9 & -\frac{66841349}{41990400} & -1.591824536 \\ 10 & -\frac{19311235721}{12124728000} & -1.592714964 \\ 11 & -\frac{6490405418051}{4073908608000} & -1.593164217 \\ 12 & -\frac{36165985229449}{22697490816000} & -1.593391337 \\ 13 & -\frac{607632388456487}{381317845708800} & -1.593506297 \\ 14 & -\frac{414309948666703}{259989440256000} & -1.593564524 \\ 15 & -\frac{14023057827349961}{8799642593280000} & -1.593594021 \end{array} \right)$$ Based on the last values, Aitken acceleration would lead to $$-\frac{5991457555368746604987211351}{3759642426462509750504140800}\approx -1.593624307$$ which almost coincides with the solution based on Lambert function $(-1.593624260)$.