Find the Taylor Expansion for the function f(x) = 2/(1-x) centered at x = 3.
Give the interval of convergence for this series.
So if I remember correctly, we first take the first four or so derivatives.
$f(x) = 2/(1-x)$
$f'(x) = 2/(1-x)^2$
$f''(x) = 4/(1-x)^3$
$f'''(x) = 12/(1-x)^4$
$f''''(x) = 48/(1-x)^5$
Now at x = 3 for those derivatives are:
$f(3) = -1$
$f'(3) = .5$
$f''(3) = -.5$
$f'''(3) = (3/4)$
$f''''(3) = -(3/2)$
Is this the Taylor Expansion?
$f(x) = -1 + (1/2)(x-3) – (1/2)( (x-3)^{2} / 2! ) + (3/4)( (x-3)^{3} / 3! ) …$
I need to find a general solution for this, but I can't really nail down a pattern. To do the interval of convergence portion I need to figure out that pattern so this is where I am stuck.
Best Answer
The idea is to take $y=x-3$ so $x=y+3$ and then $$f(x)=\frac{2}{1-x}=\frac{-2}{2+y}=\frac{-1}{1+(y/2)}=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{y}{2}\right)^n=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{x-3}{2}\right)^n$$
The ratio test gives $R=2$ so the interval of convergence centred at $3$ is $(1,5)$.