Find the local minima and maxima of $f(x) = 1 + \sum_{i=1}^{n} x^i$

Question

The question: Let $f(x) = 1 + \sum_{i=1}^{n} x^i $. Find all the local minima and local maxima of $f(x)$. For each local minimum or maximum $(c, f(c))$, find the integer $k$ which satisfies $k \le c < k+1$.

Here's my partial solution:

We shall use the property that $f'(c) = 0$.

Consider $f'(x) = 1 + \sum_{i=1}^{n-1} (i+1)x^{i} $. Notice that $f'(x) > 0$ if $x \ge 0$. Since we only wish to find the values of $x$ such that $f'(x) = 0$, we shall assume $x < 0$. \begin{align} \implies f(x) & = \frac{x^{n+1} - 1}{x - 1} \\~\\ \implies f'(x) & = \frac{nx^{n+1} - nx^n - x^n + 1}{(x-1)^2} \\~\\ \text{Let $y = -x$.} \\~\\ \implies f'(x) & = \frac{n(-y)^{n+1} - n(-y)^n - (-y)^n + 1}{(1 + y)^2} \end{align}

If $2\nmid n$, then $ f'(x) = \dfrac{ny^{n+1} + ny^n + y^n + 1}{(1+y)^2} $. This implies that $f'(x) > 0$ if $2\nmid n$. So we shall assume that $2 | n$.

\begin{align} \implies f'(x) = \dfrac{-ny^{n+1} - ny^n - y^n + 1}{(1 + y)^2} \end{align}

So, $f'(x) = 0$ iff $ny^{n+1} + ny^n + y^n - 1 = 0$.

I have no idea how to proceed from here. Is it even possible to find a neat expression for $x$ or $y$ in terms of $n$?

But, it's easy to see that $ 0 < y \le 1$, so $k=-1$ for all local minima and maxima (which exist only when $2 | n$). Also, I'm not sure if this is useful, but I noticed that $f(x) = 1 + nx^{n + 1}$ if $f'(x) = 0$.

Answer 1

Repeating partly what I wrote here $$f_n(x)=\sum_{k=0}^{2n}{x^k}=\frac{x^{2 n+1}-1}{x-1}$$ So, as you wrote it, we need to find the zero of $$g_n(x)=2nx^{2n+1}-(2n+1)x^{2n}+1$$ which is such that $$-1 < x_{(n)} \leq -\frac 12$$ and which quickly tends to $-1$. For example, $x_{(10)}=-0.834053$.

Assuming that the solution is "close" to $-1$, expand $g_n(x)$ as a infinite series around this point $$g_n(x)=-4n+\sum_{p=1}^\infty (-1)^{p+1}\,\Bigg[(2 n+1) \binom{2 n}{p}+2 n \binom{2 n+1}{p} \Bigg]\,(x+1)^p$$

Truncate to some order and use series reversion. For example, for $n=10$ and truncation to $O\left((x+1)^{11}\right)$ and then to $O(t^9)$, we have $$x_{(10)}=-1+\sum_{k=0}^6 a_k\,t^k \qquad \text{where} \qquad t=\frac{g(x)+40}{840}$$ the first coefficients being $$\left\{1,\frac{39}{4},\frac{3119}{24},\frac{125747}{64},\frac{61084573}{1920},\frac{4131 349789}{7680},\frac{3022601082407}{322560},\frac{286996560592451}{1720320}\right\}$$ and we want $g_n(x)=0$ (so $t=\frac{1}{21}$).

Using only these terms, we have $$x_{(10)}=-\frac{57419453859786701}{65067421415915520}=-0.882461$$ Not fantatisc, could you say. Let me repeat the expansion to $O\left((x+1)^{50}\right)$ and then to $O(t^{50})$ : the result is now $x_{(10)}=-0.839714$ which a bit better.

We could also do it using the first iterate of an high order iterative method starting with $x_0=−1$ as an initial guess and obtain what I wrote here.

This was the idea.

But, you must notice that what @Aaron Hendrickson wrote is incredibly better. I am sure that, sooner or later, you will enjoy the paper.