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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 11.3.19a

Taylor series and interval of convergence


a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = tan ⁻¹ (x/2), a = 0

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1
Recall that the Taylor series of a function \(f(x)\) centered at \(a\) is given by the formula: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
Since the center is \(a = 0\), this is a Maclaurin series. We need to find the first four nonzero terms of the series for \(f(x) = \tan^{-1}\left(\frac{x}{2}\right)\), so we will compute the derivatives of \(f(x)\) at \(x=0\) up to the order that gives four nonzero terms.
Start by computing the first derivative: \[f'(x) = \frac{d}{dx} \tan^{-1}\left(\frac{x}{2}\right) = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} = \frac{1}{2 \left(1 + \frac{x^2}{4}\right)} = \frac{1}{2 + \frac{x^2}{2}}.\] Evaluate \(f'(0)\) to get the coefficient for the linear term.
Next, find the higher order derivatives \(f''(x)\), \(f^{(3)}(x)\), and \(f^{(4)}(x)\), and evaluate each at \(x=0\). Use these values to write the terms \[\frac{f^{(n)}(0)}{n!} x^n\] for \(n=0,1,2,3\) (or until you have four nonzero terms).
Finally, write out the Taylor series expansion up to the fourth nonzero term by summing these terms: \[f(x) \approx f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \cdots\] This will give you the first four nonzero terms of the Maclaurin series for \(f(x) = \tan^{-1}\left(\frac{x}{2}\right)\).

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Taylor and Maclaurin Series

A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. When centered at zero, it is called a Maclaurin series. Each term involves derivatives evaluated at the center, multiplied by powers of (x - a) and divided by factorials.
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Convergence of Taylor & Maclaurin Series

Derivatives of Inverse Trigonometric Functions

To find the Taylor series of f(x) = arctan(x/2), you need to compute derivatives of arctan and apply the chain rule. The derivatives of arctan(x) follow a known pattern, which helps in finding higher-order terms efficiently.
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Derivatives of Other Inverse Trigonometric Functions

Interval of Convergence

The interval of convergence is the range of x-values for which the Taylor series converges to the function. For arctan(x/2), this depends on the radius of convergence determined by the series' ratio or root test, ensuring the series accurately represents the function within that interval.
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Interval of Convergence
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Taylor series and interval of convergence


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