Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Use your knowledge of the binomial series to find the \(n\) th degree Taylor polynomial for \(f(x)\) about \(x=0 .\) Give the radius of convergence of the corresponding Maclaurin series. One of these "series" converges for all \(x\). $$ f(x)=(1-x)^{\frac{2}{3}}, \quad n=3 $$

Step-by-Step Solution

Answer

The third degree Taylor polynomial for \(f(x)=(1-x)^{\frac{2}{3}}\) around \(x=0\) is \(P_3(x) = 1 - \frac{2}{3}x + \frac{1}{9}x^2 - \frac{4}{27}x^3\). The radius of convergence of the corresponding Maclaurin series is 1. The function that converges for all x is the original \(f(x)=(1-x)^{\frac{2}{3}}\).

Step 1: Find the first few derivatives at x=0

Firstly, the Maclaurin series requires calculating the derivative of the function multiple times at \(x=0\). Because the function is \((1-x)^{\frac{2}{3}}\), the first four derivatives at \(x=0\) will be needed. Using the general formula for the derivative of the binomial term, the results are as follows:\n\nFirst derivative: \(f'(x)=\frac{-2}{3}(1-x)^{-\frac{1}{3}}\), thus \(f'(0)=-\frac{2}{3}\)\n\nSecond derivative: \(f''(x)=\frac{2}{9}(1-x)^{-\frac{4}{3}}\), thus \(f''(0)=\frac{2}{9}\)\n\nThird derivative: \(f'''(x)=-\frac{8}{27}(1-x)^{-\frac{7}{3}}\), thus \(f'''(0)=-\frac{8}{27}\)\n\nFourth derivative: \(f^{(4)}(x)=\frac{160}{243}(1-x)^{-\frac{10}{3}}\), even though you don't need it for n=3, it's useful to calculate in our analysis phase for the radius of convergence.

Step 2: Calculate the third-degree Taylor Polynomial

The Taylor polynomial around \(x=0\) (the Maclaurin series) can be calculated using the formula: \(P_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!}\), which simplifies to: \n\n\(P_3(x) = 1 - \frac{2}{3}x + \frac{2}{18}x^2 - \frac{8}{54}*\frac{x^3}{6} = 1 - \frac{2}{3}x + \frac{1}{9}x^2 - \frac{4}{27}x^3\)

Step 3: Define the radius of convergence

The convergence of a series is controlled by its ratio of consecutive terms. Convergence occurs when this limit is less than one; for a binomial series, \(|(1-x)/1|<1\) holds true, which simplifies to \(-1

Key Concepts

Maclaurin Series

The Maclaurin Series is a special case of the Taylor series and is centered at zero. Imagine it as a powerful tool for expressing functions in terms of polynomials. Essentially, it allows complex functions to be represented by the sum of infinitely many polynomials. This approach is particularly beneficial since polynomials are much easier to work with. - The Maclaurin series of a function is defined using derivatives of the function evaluated at zero.- The expansion is of the form: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots \]- In our specific exercise, we use this formula to express \( f(x) = (1-x)^{\frac{2}{3}} \) with given derivatives up to the third degree.

Radius of Convergence

When dealing with series, especially in calculus, the concept of convergence is key. The Radius of Convergence helps determine the range over which a series will reliably give the correct value of the function it represents. Consider it as the boundary within which you can trust the series to approximate the function.- For the given function, \(f(x) = (1-x)^{\frac{2}{3}}\),- The series converges when \(-1 < x < 1\), giving a Radius of Convergence of \(r = 1\).- Series convergence relies on the ratio of consecutive terms and ensuring this limit stays below one.- Within this radius, the Maclaurin series and other similar expansions will hold true.

Binomial Series

The Binomial Series is particularly interesting when you're dealing with functions like \((1-x)^p\).- This expansion is a form of Maclaurin series for binomial expressions where the exponent \((p)\) might not be a positive integer.- The function \((1-x)^{\frac{2}{3}}\) in our exercise is a perfect candidate for a binomial series.- The Binomial Series formula is: \[ (1-x)^p = 1 + px + \frac{p(p-1)x^2}{2!} + \frac{p(p-1)(p-2)x^3}{3!} + \ldots \]- It's a great way to expand functions into polynomial forms even when dealing with fractional exponents.

Derivatives

Derivatives are a way of determining how a function changes as its input changes. In calculus, derivatives are crucial for constructing Taylor and Maclaurin series. They tell you the slope of a function at any given point.- For a Maclaurin series, you need the derivatives of your function evaluated at zero. - First derivative: Tells you how your function changes. - Second derivative: Gives information about the curvature. - Third derivative and so on: Provide deeper insights into the function's behavior.- For the function \((1-x)^{\frac{2}{3}}\),- Those derivatives are \[ f'(x) = \frac{-2}{3}(1-x)^{-\frac{1}{3}}, f''(x) = \frac{2}{9}(1-x)^{-\frac{4}{3}}, f'''(x) = -\frac{8}{27}(1-x)^{-\frac{7}{3}} \]- Calculating these derivatives is essential as they form the coefficients in the Taylor polynomial.

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Other exercises in this chapter

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Use a second degree Taylor polynomial centered appropriately to approximate the expression given. $$ \sqrt[3]{8.3} $$

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Problem 16

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Problem 17

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Key Concepts

Maclaurin Series

The Maclaurin Series is a special case of the Taylor series and is centered at zero. Imagine it as a powerful tool for expressing functions in terms of polynomials. Essentially, it allows complex functions to be represented by the sum of infinitely many polynomials. This approach is particularly beneficial since polynomials are much easier to work with. - The Maclaurin series of a function is defined using derivatives of the function evaluated at zero.- The expansion is of the form: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots \]- In our specific exercise, we use this formula to express \( f(x) = (1-x)^{\frac{2}{3}} \) with given derivatives up to the third degree.

Radius of Convergence

When dealing with series, especially in calculus, the concept of convergence is key. The Radius of Convergence helps determine the range over which a series will reliably give the correct value of the function it represents. Consider it as the boundary within which you can trust the series to approximate the function.- For the given function, \(f(x) = (1-x)^{\frac{2}{3}}\),- The series converges when \(-1 < x < 1\), giving a Radius of Convergence of \(r = 1\).- Series convergence relies on the ratio of consecutive terms and ensuring this limit stays below one.- Within this radius, the Maclaurin series and other similar expansions will hold true.

Binomial Series

The Binomial Series is particularly interesting when you're dealing with functions like \((1-x)^p\).- This expansion is a form of Maclaurin series for binomial expressions where the exponent \((p)\) might not be a positive integer.- The function \((1-x)^{\frac{2}{3}}\) in our exercise is a perfect candidate for a binomial series.- The Binomial Series formula is: \[ (1-x)^p = 1 + px + \frac{p(p-1)x^2}{2!} + \frac{p(p-1)(p-2)x^3}{3!} + \ldots \]- It's a great way to expand functions into polynomial forms even when dealing with fractional exponents.

Derivatives

Derivatives are a way of determining how a function changes as its input changes. In calculus, derivatives are crucial for constructing Taylor and Maclaurin series. They tell you the slope of a function at any given point.- For a Maclaurin series, you need the derivatives of your function evaluated at zero. - First derivative: Tells you how your function changes. - Second derivative: Gives information about the curvature. - Third derivative and so on: Provide deeper insights into the function's behavior.- For the function \((1-x)^{\frac{2}{3}}\),- Those derivatives are \[ f'(x) = \frac{-2}{3}(1-x)^{-\frac{1}{3}}, f''(x) = \frac{2}{9}(1-x)^{-\frac{4}{3}}, f'''(x) = -\frac{8}{27}(1-x)^{-\frac{7}{3}} \]- Calculating these derivatives is essential as they form the coefficients in the Taylor polynomial.