How do you expand
as Taylor series
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!
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
Yes. That's it.
Cheers James R!
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)
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...
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.
1/(1-exp(-x)) = 1 /(1-sumi((-x)^i / i!))
1/(1-exp(-x)) = 1/(x-x^2/2+x^3/6-x^4/24+x^5/120 ...)
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