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anonymous

  • 5 years ago

Which of the following represents the Maclaurin series expansion of the functions y=e^(sin 2x) truncated to the third term?

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  1. anonymous
    • 5 years ago
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    have you taken the derivatives? for the third term you will need to take until the third second derviative

  2. anonymous
    • 5 years ago
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    whoops, i meant for the third terms take until the second derivative

  3. anonymous
    • 5 years ago
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    first term, just plug in 0 into the function e^(sin 2*0), which is e^0 or just 1, so that is the first term

  4. anonymous
    • 5 years ago
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    I meant "at the third term"

  5. anonymous
    • 5 years ago
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    ok, then take the y'' it is nasty, but can be done

  6. anonymous
    • 5 years ago
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    y'=2cos(2x)*e^(sin 2x)

  7. anonymous
    • 5 years ago
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    now use product rule to find y'' y''=-4sin(2x)*e^(sin 2x)+4cos^2(2x)*e^(sin 2x) plug in 0 and you should get 4

  8. anonymous
    • 5 years ago
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    now since the third term will have x^2, you also have to divide by 2!, which means the coefficient will become 4/2=2 so the third term should be 2x^2 if I haven't made an algebra mistake

  9. anonymous
    • 5 years ago
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    Would this answer be correct? 1+2x+2x^(2)

  10. anonymous
    • 5 years ago
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    if you wanted all 3 terms, then yes that is correct

  11. anonymous
    • 5 years ago
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    Thank you !!!!

  12. anonymous
    • 5 years ago
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    hi alma,do you have the selections of answers?

  13. anonymous
    • 5 years ago
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    f(x)=f(0)+f ' (0)x +f ''(0)x^2/(1*2) the 3rd term is =f ''(0)x^2/(1*2) therefore after doing the 2nd derivative set x=0

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