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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0have you taken the derivatives? for the third term you will need to take until the third second derviative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoops, i meant for the third terms take until the second derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I meant "at the third term"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, then take the y'' it is nasty, but can be done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y'=2cos(2x)*e^(sin 2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would this answer be correct? 1+2x+2x^(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you wanted all 3 terms, then yes that is correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hi alma,do you have the selections of answers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(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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