# How do you express the nth term of Tn(x) for f(x) centered at a = 3 in terms of n given f(x) = 1/(1-x)?

Feb 28, 2017

$\frac{1}{1 - x} = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(\frac{3 - x}{2}\right)}^{n}$

#### Explanation:

Using the power rule we can determine the derivatives of $f \left(x\right)$ of all orders:

$f \left(x\right) = {\left(1 - x\right)}^{- 1}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(- 1\right) \left(- 1\right) {\left(1 - x\right)}^{- 2} = {\left(1 - x\right)}^{- 2}$

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \left(- 1\right) \left(- 2\right) {\left(1 - x\right)}^{- 3} = 2 {\left(1 - x\right)}^{- 3}$

and we can easily see that:

(d^nf)/(dx^n) =(n!) /(1-x)^(n+1)

For $x = 3$ then the Taylor coefficients are:

f^((n))(3)/(n!) = (n!)/( (1-3)^(n+1)(n!)) = (-1)^(n+1)/2^(n+1)

And the Taylor series is:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n + 1} {\left(x - 3\right)}^{n} / {2}^{n + 1}$

or:

$\frac{1}{1 - x} = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(\frac{3 - x}{2}\right)}^{n}$

We can obtain the same result directly by substituting $t = x - 3$ in the original expression:

$\frac{1}{1 - x} = \frac{1}{1 - t - 3} = - \frac{1}{t + 2} = - \frac{1}{2} \frac{1}{1 + \frac{t}{2}}$

Now, $\frac{1}{1 + \frac{t}{2}}$ is the sum of a geometric series of ratio $\left(- \frac{t}{2}\right)$, so:

$\frac{1}{1 - x} = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- \frac{t}{2}\right)}^{n}$

and substituting back $- t = 3 - x$:

$\frac{1}{1 - x} = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(\frac{3 - x}{2}\right)}^{n}$