Hi folks,

I have the following expansion of $(1 + x - 2x^2)^{-1}$ = $\dfrac{1}{3} [ (1 - x)^{-1} + 2(1 + 2x)^{-1}] $ by partial fractions.

the result is:

$1 - x + 3x^2 - 5x^3 + 11x^4$ ...............(1)

I have to find a suitable value of x to evaluate $(1.0098)^{-1}$ to 7 decimal places.

I calculate that the range of values of x for which the series converges is $x \lt \dfrac{1}{2}$

so $1 + x - 2x^2 = 1.0098$

$x -2x^2 = 0.0098$

$2x^2 - x + 0.0098 = 0$

$x = \dfrac{1 \pm \sqrt{1 - 4.2.*(0.0098)}}{4}$

$x = \dfrac{1 \pm \sqrt{1 - 0.0784}}{4}$

$x = \dfrac{1 \pm \sqrt{0.9216}}{4} $

$x = \dfrac{1 \pm 0.96}{4}$

$x = 0.49$ or $x = 0.01$

given the range of allowable values of x both values are permissible and I was expecting that when I substitute them into the expansion (1) , I would get the same result.

But I do not

$x = 0.01$ gives 0.99029511 which is the correct answer

$x = 0.49$ gives 1.27618311 which I guess is wrong.

I was expecting the range of permissible values of x for a convergent series to eliminate one of my solutions. Is that reasonable?

Thanks