Determine The First Four Terms Of The Taylor Series For $f(x)=\tan^{-1}(x)$ Centered At $a=1$. What Is The Coefficient Of $ ( X − 1 ) 3 (x-1)^3 ( X − 1 ) 3 [/tex] In That Series?A. $2/3$ B. $1/6$ C.

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Introduction

The Taylor series is a powerful tool in calculus that allows us to approximate a function at a given point using an infinite series of terms. In this article, we will determine the first four terms of the Taylor series for the function $f(x)=\tan^{-1}(x)$ centered at $a=1$. We will also find the coefficient of $(x-1)^3$ in that series.

The Taylor Series Formula

The Taylor series formula for a function $f(x)$ centered at $a$ is given by:

f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+⋯f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

Finding the Derivatives of $f(x)=\tan^{-1}(x)$

To find the Taylor series for $f(x)=\tan^{-1}(x)$, we need to find the derivatives of $f(x)$ at $a=1$. We will start by finding the first four derivatives of $f(x)$.

First Derivative

The first derivative of $f(x)=\tan^{-1}(x)$ is given by:

f′(x)=11+x2f'(x) = \frac{1}{1+x^2}

Evaluating the first derivative at $a=1$, we get:

f′(1)=11+12=12f'(1) = \frac{1}{1+1^2} = \frac{1}{2}

Second Derivative

The second derivative of $f(x)=\tan^{-1}(x)$ is given by:

f′′(x)=−2x(1+x2)2f''(x) = \frac{-2x}{(1+x^2)^2}

Evaluating the second derivative at $a=1$, we get:

f′′(1)=−2(1)(1+12)2=−12f''(1) = \frac{-2(1)}{(1+1^2)^2} = -\frac{1}{2}

Third Derivative

The third derivative of $f(x)=\tan^{-1}(x)$ is given by:

f′′′(x)=2(1+3x2)(1+x2)3f'''(x) = \frac{2(1+3x^2)}{(1+x^2)^3}

Evaluating the third derivative at $a=1$, we get:

f′′′(1)=2(1+3(1)2)(1+12)3=88=1f'''(1) = \frac{2(1+3(1)^2)}{(1+1^2)^3} = \frac{8}{8} = 1

Fourth Derivative

The fourth derivative of $f(x)=\tan^{-1}(x)$ is given by:

f′′′′(x)=−24x(1+x2)+12(1+3x2)2(1+x2)4f''''(x) = \frac{-24x(1+x^2)+12(1+3x^2)^2}{(1+x^2)^4}

Evaluating the fourth derivative at $a=1$, we get:

f′′′′(1)=−24(1)(1+12)+12(1+3(1)2)2(1+12)4=−2416+12(16)16=7216−2416=4816=3f''''(1) = \frac{-24(1)(1+1^2)+12(1+3(1)^2)^2}{(1+1^2)^4} = -\frac{24}{16} + \frac{12(16)}{16} = \frac{72}{16} - \frac{24}{16} = \frac{48}{16} = 3

Finding the Taylor Series

Now that we have found the derivatives of $f(x)=\tan^{-1}(x)$ at $a=1$, we can find the Taylor series for $f(x)$.

The Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$ is given by:

tan⁡−1(x)=tan⁡−1(1)+f′(1)1!(x−1)+f′′(1)2!(x−1)2+f′′′(1)3!(x−1)3+f′′′′(1)4!(x−1)4+⋯\tan^{-1}(x) = \tan^{-1}(1) + \frac{f'(1)}{1!}(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f''''(1)}{4!}(x-1)^4 + \cdots

Substituting the values of the derivatives, we get:

tan⁡−1(x)=π4+1/21!(x−1)−1/22!(x−1)2+13!(x−1)3+34!(x−1)4+⋯\tan^{-1}(x) = \frac{\pi}{4} + \frac{1/2}{1!}(x-1) - \frac{1/2}{2!}(x-1)^2 + \frac{1}{3!}(x-1)^3 + \frac{3}{4!}(x-1)^4 + \cdots

Finding the Coefficient of $(x-1)^3$

To find the coefficient of $(x-1)^3$ in the Taylor series, we need to find the coefficient of the term $(x-1)^3$.

The coefficient of the term $(x-1)^3$ is given by:

f′′′(1)3!=13!=16\frac{f'''(1)}{3!} = \frac{1}{3!} = \frac{1}{6}

Therefore, the coefficient of $(x-1)^3$ in the Taylor series is $\frac{1}{6}$.

Conclusion

In this article, we have determined the first four terms of the Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$. We have also found the coefficient of $(x-1)^3$ in that series, which is $\frac{1}{6}$.

The Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$ is given by:

tan⁡−1(x)=π4+1/21!(x−1)−1/22!(x−1)2+13!(x−1)3+34!(x−1)4+⋯\tan^{-1}(x) = \frac{\pi}{4} + \frac{1/2}{1!}(x-1) - \frac{1/2}{2!}(x-1)^2 + \frac{1}{3!}(x-1)^3 + \frac{3}{4!}(x-1)^4 + \cdots

The coefficient of $(x-1)^3$ in the Taylor series is $\frac{1}{6}$.

References

  • [1] "Taylor Series" by Math Is Fun. Retrieved 2023-02-20.
  • [2] "Derivatives of Trigonometric Functions" by Purplemath. Retrieved 2023-02-20.
  • [3] "Taylor Series Expansion" by Wolfram MathWorld. Retrieved 2023-02-20.

Introduction

In our previous article, we determined the first four terms of the Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$ and found the coefficient of $(x-1)^3$ in that series. In this article, we will answer some frequently asked questions about the Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$.

Q1: What is the Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$?

A1: The Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$ is given by:

tan⁡−1(x)=π4+1/21!(x−1)−1/22!(x−1)2+13!(x−1)3+34!(x−1)4+⋯\tan^{-1}(x) = \frac{\pi}{4} + \frac{1/2}{1!}(x-1) - \frac{1/2}{2!}(x-1)^2 + \frac{1}{3!}(x-1)^3 + \frac{3}{4!}(x-1)^4 + \cdots

Q2: What is the coefficient of $(x-1)^3$ in the Taylor series?

A2: The coefficient of $(x-1)^3$ in the Taylor series is $\frac{1}{6}$.

Q3: How do I find the Taylor series for a function $f(x)$ centered at $a$?

A3: To find the Taylor series for a function $f(x)$ centered at $a$, you need to find the derivatives of $f(x)$ at $a$ and substitute them into the Taylor series formula.

Q4: What is the Taylor series formula?

A4: The Taylor series formula for a function $f(x)$ centered at $a$ is given by:

f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+⋯f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

Q5: How do I find the derivatives of a function $f(x)$?

A5: To find the derivatives of a function $f(x)$, you need to apply the power rule and the product rule of differentiation.

Q6: What is the power rule of differentiation?

A6: The power rule of differentiation states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

Q7: What is the product rule of differentiation?

A7: The product rule of differentiation states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Q8: How do I evaluate the derivatives of a function $f(x)$ at a point $a$?

A8: To evaluate the derivatives of a function $f(x)$ at a point $a$, you need to substitute $a$ into the derivative.

Q9: What is the significance of the Taylor series in mathematics?

A9: The Taylor series is a powerful tool in mathematics that allows us to approximate a function at a given point using an infinite series of terms.

Q10: How do I use the Taylor series to approximate a function?

A10: To use the Taylor series to approximate a function, you need to find the Taylor series for the function centered at a point and then substitute the point into the series.

Conclusion

In this article, we have answered some frequently asked questions about the Taylor series for $f(x)=\tan^{-1}(x)$ centered at $a=1$. We have also provided some additional information about the Taylor series and how to use it to approximate a function.

References

  • [1] "Taylor Series" by Math Is Fun. Retrieved 2023-02-20.
  • [2] "Derivatives of Trigonometric Functions" by Purplemath. Retrieved 2023-02-20.
  • [3] "Taylor Series Expansion" by Wolfram MathWorld. Retrieved 2023-02-20.