Question

Determine the Taylor Series about x=2 for 1/(1-x) and the radius of convergence

Answer 1

So I did the derivatives of f(x) and arrive that the general derivative pattern is n!/(1-x)^(n+1)
So what's next?

Answer 2

use the taylor series formula
$$ \sum_{n=0}^{\infty } \frac{f^{(n)}(a)(x-a)^n}{n!} $$

Answer 3

f^(n)(a)=(-1)^(n+a)n!
will continue..

Answer 4

by the way I got a different taylor expansion

Answer 5

In regards to my first comment or third?
My third are the coefficients, what is that called?

Answer 6

$$ \frac{1}{1-x }= -1+(x-2)- \left( x-2 \right) ^{2}+ \left( x-2 \right) ^{3}- \left( x-2
\right) ^{4}+ \left( x-2 \right) ^{5}+...$$

Answer 7

So, what's the purpose of finding the general equation for the derivatives if we have to plug in 2 for each one of them?
1. find multiple derivatives
2. plug in point.
3. find general equation of that.
4. divide by n! and multiply by (x-x0)^n
Is that a general case?
Wouldn't that just result in -1^(n+1)*(x-2)^n ?

Answer 8

I agree with your general derivative $$ f(x) = \frac{1}{\left( 1-x \right) }
\\ f '(x) = \frac{1}{\left( 1-x \right) ^{2}}
\\ f ''(x) = \frac{2}{\left( 1-x \right) ^{3}}
\\ f '''(x) = \frac{6}{\left( 1-x \right) ^{4}}
\\ ...
\\
\\ f^{(n)}(x) = \frac{n!}{\left( 1-x \right) ^{n+1}}
$$

Answer 9

Why does one even bother solving for the general derivative?

Answer 10

How were you able to ascertain your third comment without fully solving the problem?

Answer 11

I am using a computer algebra system, like wolfram or Maple

Answer 12

Oh.

Answer 13

but it can be done 'by hand', it is instructive to do a few of these problems by hand

Answer 14

yeah, I have do this by hand..

Answer 15

ok so we have the general derivative, now we need to plug in x=2 and find a pattern

Answer 16

$$f(x) = \frac{1}{\left( 1-x \right) } \implies f(2) = -1
\\ f '(x) = \frac{1}{\left( 1-x \right) ^{2}} \implies f '(2) = 1
\\ f ''(x) = \frac{2}{\left( 1-x \right) ^{3}} \implies f ''(2) = -2
\\ f '''(x) = \frac{6}{\left( 1-x \right) ^{4}} \implies f '''(2) = 6
\\ ...
\\
\\ f^{(n)}(x) = \frac{n!}{\left( 1-x \right) ^{n+1}} \implies f^{(n)}(2) = (-1)^{n+1}~ n!
$$

Answer 17

I assumed \( n = 0,1,2,3... \)

Answer 18

Yep, I think I have this above, and solved it above.
Thanks for your help

Answer 19

Mistakenly wrote a in place of 1