Question

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)`?

Answer 1

`1/(1-x) = -1/2 sum_(n=0)^oo ((3-x)/2)^n`

Explanation:

Using the power rule we can determine the derivatives of `f(x)` of all orders:

`f(x) = (1-x)^(-1)`

`(df)/dx = (-1)(-1)(1-x)^(-2)= (1-x)^(-2)`

`(d^2f)/(dx^2) = (-1)(-2)(1-x)^(-3)= 2(1-x)^(-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:

`1/(1-x) = sum_(n=0)^oo (-1)^(n+1)(x-3)^n/2^(n+1)`

or:

`1/(1-x) = -1/2 sum_(n=0)^oo ((3-x)/2)^n`

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

`1/(1-x) = 1/(1-t-3) = -1/(t+2) = -1/2 1/(1+t/2)`

Now, `1/(1+t/2)` is the sum of a geometric series of ratio `(-t/2)`, so:

`1/(1-x) = -1/2 sum_(n=0)^oo (-t/2)^n`

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

`1/(1-x) = -1/2 sum_(n=0)^oo ((3-x)/2)^n`