Finding Derivatives Using Taylor/Maclaurin Polynomials

In summary, to find the 6th derivative of f(x) = arctan((x^2)/4) at x = 0, we can use the Maclaurin series for f(x) and substitute x^2/4 for x. The general term of the series is ((-1)^n)*x^{4n+2}/(4^{2n+1})(2n+1), and we need to find the value of n for which the exponent of x is 6. By setting 4n+2 equal to 6, we get n = 1. The coefficient of x^6 is then -(1/192).
  • #1
HypeBeast23
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Homework Statement



Compute the 6th derivative of f(x) = arctan((x^2)/4) at x = 0. Hint: Use the Maclaurin series for f(x).

Homework Equations



The maclaurin series of arctanx which is ((-1)^n)*x^(2n+1)/2n+1

The Attempt at a Solution



I subbed in x^2/4 for x into the maclaurin series and got *summation* ((-1)^n)*x^(4n+2)/(4^2n+1)(2n+1). After that, I'm not sure what to do. I know that the 6th derivative in a general maclaurin series would be represented by *the 6th derivative of f at 0*/n! and x is to the power of 6. My prof said to find the value where the exponent of x in the series is 6.

Any help would be greatly appreciated. Thanks!
 
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  • #2
Okay, so you know that the general term is [itex]((-1)^n)*x^{4n+2}/(4^{2n+1})(2n+1)[/itex].
And you know that the coefficient is the nth derivative over n!. Okay. For what n is
4n+2= 6? What is the coefficient of [itex]x^6[/itex]?
 
  • #3
Yes, I did as such and found n = 1 to get x^6. Thus, the coefficient of x would be -(1/192), no? Apparently, the answer was wrong.
 

FAQ: Finding Derivatives Using Taylor/Maclaurin Polynomials

What is the purpose of using Taylor/Maclaurin polynomials to find derivatives?

Taylor and Maclaurin polynomials are used to approximate a function by calculating its derivatives at a specific point. This allows us to find the value of the function at that point and to estimate its behavior around that point.

How do you find the Taylor/Maclaurin polynomial for a given function?

To find the Taylor/Maclaurin polynomial for a given function, you need to first determine the derivatives of the function at a specific point. Then, plug in those values into the formula for the polynomial, which is a sum of the derivatives multiplied by the corresponding powers of x. Finally, simplify the polynomial to get the final result.

Can the Taylor/Maclaurin polynomial be used to find derivatives at any point?

Yes, the Taylor/Maclaurin polynomial can be used to find derivatives at any point. However, the accuracy of the approximation will depend on the distance between the point and the point at which the derivatives were calculated.

What is the difference between a Taylor polynomial and a Maclaurin polynomial?

The main difference between a Taylor polynomial and a Maclaurin polynomial is the point at which the derivatives are calculated. A Taylor polynomial is centered around a specific point, while a Maclaurin polynomial is centered around the origin (x=0).

How can Taylor/Maclaurin polynomials be used in real-world applications?

Taylor/Maclaurin polynomials are used in many real-world applications, particularly in engineering, physics, and economics. They can be used to approximate complicated functions, which allows for easier calculations and predictions. They are also used in numerical analysis to improve the accuracy of numerical methods.

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