Taylor expansion

  • Thread starter Callisto
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  • #1
Callisto
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Hi

How do you expand

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

as Taylor series

Callisto
 
Last edited:

Answers and Replies

  • #2
James R
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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
Callisto
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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
James R
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Yes. That's it.
 
  • #5
Callisto
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Cheers James R!
 
  • #6
lalbatros
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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)
 
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  • #7
James R
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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
lalbatros
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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 ...)
 

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