anonymous
  • anonymous
Which of the following represents the Maclaurin series expansion of the functions y=e^(sin 2x) truncated to the third term?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
have you taken the derivatives? for the third term you will need to take until the third second derviative
anonymous
  • anonymous
whoops, i meant for the third terms take until the second derivative
anonymous
  • anonymous
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

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anonymous
  • anonymous
I meant "at the third term"
anonymous
  • anonymous
ok, then take the y'' it is nasty, but can be done
anonymous
  • anonymous
y'=2cos(2x)*e^(sin 2x)
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
Would this answer be correct? 1+2x+2x^(2)
anonymous
  • anonymous
if you wanted all 3 terms, then yes that is correct
anonymous
  • anonymous
Thank you !!!!
anonymous
  • anonymous
hi alma,do you have the selections of answers?
anonymous
  • anonymous
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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