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!]-...

Question

anonymous · Answer

good question. sorry I can't answer it for you

anonymous · Answer

Are you sure of the signs of each term?

anonymous · Answer

Yes. I've figured it out. the answer was 1-sin(3x).

anonymous · Answer

1-[((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 1-sin(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 · Answer

yes the question was suppose to be alternating. the answer I posted above was indeed correct according the the answer booklet.