Question

Let h(x) 1/(5-x) The taylor series = sum from 0 to infinity of an(x-3)^n

What is an?

Answer 1

h(x) = 1/(5-x) *

Answer 2

The goal here is to write the function as one with a geometric Taylor series.

Recall that \(\dfrac{1}{1-x}=\sum x^n\) for \(|x|<1\).

Your desired series is centered at \(x=3\), so a slight change needs to be made to the given function:
\[\frac{1}{5-x}=\frac{1}{2-(x-3)}=\frac{1}{2}\cdot\frac{1}{1-\dfrac{x-3}{2}}=\frac{1}{2}\sum_{n=1}^\infty \left(\frac{x-3}{2}\right)^n\]
Simplify what you can to find \(a_n\).

Answer 3

So would it be (1/2)^(n+1)?

Answer 4

Yes

Answer 5

Alright, one more small question

Answer 6

Shoot

Answer 7

The series converges for |x-3| < what?

Answer 8

It's apparently not 1, so I'm a bit confused

Answer 9

Applying the ratio test, we see that
\[\lim_{n\to\infty}\left|\frac{(x-3)^{n+1}}{2^{n+2}}\cdot\frac{2^{n+1}}{(x-3)^n}\right|=|x-3|\lim_{n\to\infty}\left|\frac{2^{n+1}}{2^{n+2}}\right|=\frac{1}{2}|x-3|\]
which converge if it's less than 1, which in turn means convergence when \(|x-3|<2\).

Answer 10

Ah. I see. Thank you so much.

Answer 11

yw