Taylor series to find value of nth derivative

  • Thread starter Panphobia
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  • #1
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Homework Statement


If f(x) = x^5*cos(x^6) find f40(0) and f41(0)

The Attempt at a Solution


So we are supposed to get the Taylor series and use that to get the value of the derivatives I just manipulated the Taylor series for cosx to get the one for this. Would the value be the coefficient?
 

Answers and Replies

  • #2
RUber
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You are evaluating the 40th and 41st derivatives at 0. I am not seeing how this is related to the Taylor series.
 
  • #3
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You are evaluating the 40th and 41st derivatives at 0. I am not seeing how this is related to the Taylor series.
Well this unit is all about Taylor series, and in class he told us to use the Taylor series to get the values of the 40th and 41st derivatives at 0.
 
  • #4
RUber
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Once you have. The full Taylor series for f(x), you should be able to tell what 40 derivatives would do. Evaluating at 0 will leave only one term.
 
  • #5
vela
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Homework Statement


If f(x) = x^5*cos(x^6) find f40(0) and f41(0)

The Attempt at a Solution


So we are supposed to get the Taylor series and use that to get the value of the derivatives I just manipulated the Taylor series for cosx to get the one for this. Would the value be the coefficient?
Not quite. What's the general formula for the Taylor series?
 
  • #6
Dick
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Homework Statement


If f(x) = x^5*cos(x^6) find f40(0) and f41(0)

The Attempt at a Solution


So we are supposed to get the Taylor series and use that to get the value of the derivatives I just manipulated the Taylor series for cosx to get the one for this. Would the value be the coefficient?

The value of the derivative f40(0) is the 40th derivative of the x^40 term in the Taylor series. Similar for f41(0). So, no, it's not just the coefficient. Taking the 40 derivatives will give you an extra factorial.
 
  • #7
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Yea I figured it out, it is 41!/6!, you have to equate the original taylor series formula to the one for this function, and then solve for the derivative.
 

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