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anonymous
 5 years ago
Find the function with the following Maclaurin series starting with the Maclaurin series for sinx: 1[((5^3)(x^3))/3!][((5^5)(x^5))/5!]+[((5^7)(x^7))/7!]...
anonymous
 5 years ago
Find the function with the following Maclaurin series starting with the Maclaurin series for sinx: 1[((5^3)(x^3))/3!][((5^5)(x^5))/5!]+[((5^7)(x^7))/7!]...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good question. sorry I can't answer it for you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure of the signs of each term?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. I've figured it out. the answer was 1sin(3x).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01[((5^3)(x^3))/3!][((5^5)(x^5))/5!]+[((5^7)(x^7))/7!]... The signs are not correct...and sorry, it cannot be 1sin(3x). The series expansion for sin x is sinx = x  (x^3)/3! + (x^5)/5!  (x^7)/7! + ... If i assume the signs are alternating, the answer might be 1  3x + sin(3x) or 1 + 3x  sin(3x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes the question was suppose to be alternating. the answer I posted above was indeed correct according the the answer booklet.
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