I'm working on a series convergence problem and am stuck on this part:
The series converges when $|1-2x| < |1+x|$.
How can I proceed from here to pick the values of $x$ that satisfy this inequality? I tried using the triangle inequalities, but it wasn't helpful and gave silly upper bounds.
On
Edit: Corrected a mistake (turned a $4$ part solution into a $2$ part one). I apologize for the error.
The standard way would be to just consider all cases. That is, $$1-2x<1+x$$ $$2x-1<1+x$$ This will give a $2$-part solution to the inequality.
$0<x<2$
On
Suppose, $x\leq 0$. Then $1-2x>0$, $|1-2x|=1-2x$.
$1-2x < |1+x|$ means that either $$1+x>1-2x$$ or $$1+x<2x-1.$$ And it could be simplified to $$3x>0$$ or $$x>2,$$ which is not true for $x\leq 0$.
Suppose $x>0$. Then $1+x>0$, $|1+x|=1+x$.
$|1-2x| < 1+x$ means that $$-1-x<1-2x<1+x,$$ which is equal to $$-2<-x<2x.$$ $-x<2x$ is obviously true for all $x>0$, but $-2<-x$ means $x<2$.
So, $0<x<2$.
On
I find drawing the graphs to be the quickest.
If I have had a few glasses of wine, or the computation has some life threatening possibilities, I do the following:
If you look at the quantities in the absolute values it will be clear that the two points of 'interest' are $-1, {1 \over 2}$.
If $x \in(-\infty,-1]$, we have $1-2x < -1-x $, or $2 < x$, so no solutions here.
If $x \in (-1,{1 \over 2}]$, we have $1-2x < 1+x$, or $0 < x$, so the solutions here are $(0, {1 \over 2}]$.
Otherwise, we have $2x-1 < 1+x$, or $x<2$, so the solutions here are $({1 \over 2}, 2)$.
Combining gives $(0,2)$.
Note that $$|1-2x|\lt |1+x| \quad\text{if and only if}\quad (1-2x)^2\lt (1+x)^2.$$ If we expand, the inequality on the right turns out to be equivalent to $$3x^2-6x\lt 0.$$ This inequality holds precisely if $x$ is (strictly) between the two roots of $3x^2-6x=0$. So the inequality holds whenever $0\lt x\lt 2$.