# Taylor expansion

1. Aug 5, 2005

### Callisto

Hi

How do you expand

(1-exp(-1))^-1

as Taylor series

Callisto

Last edited: Aug 5, 2005
2. Aug 5, 2005

### James R

Try long division to get a Taylor (or McClaurin) series for f(x) = 1/(1-x) for a start.

Then put x=exp(-1) into that series, and there you go!

3. Aug 5, 2005

### Callisto

Taylor series for

1/(1-x) = (1-x)^-1 = 1 + x^2 + x^3 + x^4 +...= SUMi x^i

Then sub in exp(-1) for x
That's it?

4. Aug 5, 2005

### James R

Yes. That's it.

5. Aug 5, 2005

### Callisto

Cheers James R!

6. Aug 5, 2005

### lalbatros

a constant has a trivial Taylor expansion

The expression you submitted is a constant, it doesn't depend on x.
All derivatives of this function with respect to x (the assumed variable) are zero.

Therefore the Taylor expansion contains only one term (and is exact):

f(x) = 1/(1-exp(-1)) + 0 + 0 + 0 + 0 + 0 + 0 + ... = 1/(1-exp(-1))

(unless you did not formulate your problem correctly)

Last edited: Aug 5, 2005
7. Aug 6, 2005

### James R

Technically, you're right of course, lalbatros, but your solution doesn't seem to fit the spirit of the question which was asked. Maybe the question was a little unclear, though...

8. Aug 7, 2005

### lalbatros

You are probably right James.
The question probably was: "how can I get an approximate value for 1/(1-exp(-1)) using Taylor expansions".
Then, there are numerous ways to choose how to expand.
For example:

1/(1-exp(-x)) = 1 /(1-sumi((-x)^i / i!))

or explicitely:

1/(1-exp(-x)) = 1/(x-x^2/2+x^3/6-x^4/24+x^5/120 ...)