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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!]-...

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  1. anonymous
    • 5 years ago
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    good question. sorry I can't answer it for you

  2. anonymous
    • 5 years ago
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    Are you sure of the signs of each term?

  3. anonymous
    • 5 years ago
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    Yes. I've figured it out. the answer was 1-sin(3x).

  4. anonymous
    • 5 years ago
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    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)

  5. anonymous
    • 5 years ago
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    yes the question was suppose to be alternating. the answer I posted above was indeed correct according the the answer booklet.

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