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anonymous
 5 years ago
The series [sigma n=0 to infinity] [(1)^(n=2)]*[x^(2*n+1)] is the Taylor series about x=0 for the function A) sin(x) B)cox(x) c)x/(1+x^2) D) 1/(1+x) and E) None of the above. Please help and can you show me the steps please thanks...
anonymous
 5 years ago
The series [sigma n=0 to infinity] [(1)^(n=2)]*[x^(2*n+1)] is the Taylor series about x=0 for the function A) sin(x) B)cox(x) c)x/(1+x^2) D) 1/(1+x) and E) None of the above. Please help and can you show me the steps please thanks...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Taylor_series They have a list of Taylor Series. Personally I don't know enough about them to be able to explain them but the basic ones I can sometimes remember.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A Taylor series is built on the formula f((n))(c)/n! * (xc)^n, with n going from 0 to infinity, where I am using the notation f((n))(c) to mean the nth derivative of f, applied to c. You can use the first few derivatives of sine to build the Taylor series for sine, then compare to the formula you gave above (which has a typo in it, BTW). OR you can just memorize a few basic Taylor series, a half dozen or so, so you recognize them when you see them. The one above looks like sine (minus the typo) but is missing the denominator (2n+1)!.
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