Solving $2mx^{2m+1}-(2m+1)x^{2m}+1=0$

Question

I set up a problem for myself and came to this equation and I'm wondering if it can be solved algebraically. The variable $m$ is a whole number and we're solving for $x$.

$$2mx^{2m+1}-(2m+1)x^{2m}+1=0$$

One solution is $x=1$ but I believe there should be another root in $(-\infty,0)$.

Answer 1

As a hint$$2mx^{2m+1}-(2m+1)x^{2m}+1=0 \\2mx^{2m+1}-2mx^{2m}-x^{2m}+1=0 \\2m(x^{2m+1}-x^{2m})-(x^{2m}-1)=0\\2mx^{2m}(x-1)-(x^{2m}-1)=0\\$$now factor $(x-1)$
or take $f(x)=2mx^{2m+1}-(2m+1)x^{2m}+1$ $$f'(x)=2m(2m+1)x^{2m}-(2m+1)2mx^{2m-1}=\\2m(2m+1)x^{2m-1}(x-1)=0\\ \to f'=0 \to x=0,1 \\ f(x)=(x-1)^2(somthing)$$ you can draw $f'(x)$ table of sign

and finally with respect to $f'$ sign ,you have $$x=1,1 \ or \ (x-1)^2$$and other negative root

Answer 2

Put $P(x)=2mx^{2m+1}-(2m+1)x^{2m}+1$. Counting each root with its multiplicity you have $2m+1$ roots in $\mathbb C$. $P'(x)=2m(2m+1)x^{2m-1}(x-1)$ so assuming $m>0$ you can see that $P$ is strictly increasing on $]-\infty;0]$ and $P(x)>0$ on $[0;1[\cup]1;+\infty[$ as it reaches its minimum $0$ on $[0;+\infty[$ at $x=1$. $\lim_{x\rightarrow +\infty}P(x)=-\infty$ and $P(0)=1$ so by the intermediate value theorem $P$ has exactly one real root in $]-\infty;0[$ and as $1$ is root of $P'$ with multiplicity $1$ it is root for $P$ with multiplicity $2$. So for any $m>0$ you have two real roots : $1$ and another root $a_m<0$ that depends on $m$, and if $m>1$ you also have roots belonging to $\mathbb C\setminus\mathbb R$.

Answer 3

$$f(x)=2mx^{2m+1}-(2m+1)x^{2m}+1$$

In terms of reals roots, there is a double one at $x=1$ and another which is negative. If $x_{(m)}$ is this root, we have $$-1 \lt x_{(m)} \leq -\frac 1 2$$ To work for this root is not very simple with $f(x)$. What I found more convenient is to let $t=-x$ and search for the zero of function $$g(t)=\log \left(t^{2 m} (1+2 m (t+1))\right)$$

Interesting (but I do not see how to use it) is to notice that the problem is "hust" the inverse of $$2m= \frac{1}{\log (t)}\,W_{-1}\left(\frac{ \log (t)}{t+1}\,t^{\frac{1}{t+1}}\right)-\frac{1}{t+1}$$ where $W(.)$ is Lambert function.

Now, to have an estimate of $t_{(m)}$, we can make one single iteration of a Newton-like method of order $k$ starting with $$t_0=\frac 12+\frac{3 m+1}{6 m (2 m+1)}\log \left(\frac{4^m}{3 m+1}\right)$$