Taylor series to find value of nth derivative

In summary, the conversation is about finding the 40th and 41st derivatives of the function f(x) = x^5*cos(x^6) at x=0 using the Taylor series. The value of the derivatives is not simply the coefficient, but rather the 40th derivative of the x^40 term in the series. This can be found by equating the original Taylor series formula to the one for this function and solving for the derivative.
  • #1
Panphobia
435
13

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?
 
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  • #2
You are evaluating the 40th and 41st derivatives at 0. I am not seeing how this is related to the Taylor series.
 
  • #3
RUber said:
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
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
Panphobia said:

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
Panphobia said:

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
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.
 

Related to Taylor series to find value of nth derivative

What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, with each term corresponding to a derivative of the function evaluated at a single point.

Why would one use a Taylor series to find the value of an nth derivative?

A Taylor series can be used to approximate the value of a function at a certain point, even if the function is not easily calculable at that point. By taking higher order derivatives, the approximation becomes more accurate.

How is a Taylor series calculated?

A Taylor series is calculated by taking the derivatives of a function at a specific point, plugging those derivatives into the Taylor series formula, and summing the terms to get the approximate value of the function at that point.

What is the formula for a Taylor series?

The formula for a Taylor series is: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... + f^(n)(a)(x-a)^n/n! where f^(n) represents the nth derivative of the function f at the point a.

What is the significance of the nth derivative in a Taylor series?

The nth derivative plays a crucial role in the accuracy of the Taylor series approximation. The higher the value of n, the more accurate the approximation becomes. However, calculating higher order derivatives can be time-consuming and may not always be necessary for a desired level of accuracy.

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