- #1

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Hi

How do you expand

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

as Taylor series

Callisto

How do you expand

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

as Taylor series

Callisto

Last edited:

- Thread starter Callisto
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- #1

- 41

- 0

Hi

How do you expand

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

as Taylor series

Callisto

How do you expand

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

as Taylor series

Callisto

Last edited:

- #2

James R

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

- #3

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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

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Cheers James R!

- #6

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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:

- #7

James R

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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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