The problem:
$\displaystyle \[f(x) = 1 / (1+9x^2)\]$
My attempt:
$\displaystyle f(x) = 1 / (1-(-9x^2)) = \sum (-9x^2)^n = \sum (-1)^n(9x^2)^n$
With |-9x^2| < 1, 9x < 1, x < 1/9
R = 1/9, I = (-1/9, 1/9)
Not sure if this is correct. Thanks
The problem:
$\displaystyle \[f(x) = 1 / (1+9x^2)\]$
My attempt:
$\displaystyle f(x) = 1 / (1-(-9x^2)) = \sum (-9x^2)^n = \sum (-1)^n(9x^2)^n$
With |-9x^2| < 1, 9x < 1, x < 1/9
R = 1/9, I = (-1/9, 1/9)
Not sure if this is correct. Thanks
That's a good way to start: you recognized that $\displaystyle \frac{1}{1- r}$ is the sum of the geometric series $\displaystyle \sum r^n$ and here r= -9x^2[/tex]. However, $\displaystyle 9x^2< 1$ does NOT give 9x< 1, it gives $\displaystyle 9x^2< 1$ so that $\displaystyle -\frac{1}{3}< x< \frac{1}{3}$