Problem 16 - Essential Calculus Early Transcendentals

Question

Find the Taylor series for \(f(x)\) centered at the given value of \(a\) . [Assume that \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=\sin x, \quad a=\pi / 2$$

Step-by-Step Solution

Answer

The Taylor series for \(\sin x\) centered at \(a = \pi/2\) is \(1 - \frac{1}{2}(x-\frac{\pi}{2})^2 + \frac{1}{24}(x-\frac{\pi}{2})^4 + \cdots\).

Step 1: Recall the Definition of the Taylor Series

The Taylor series of a function \(f(x)\) centered at \(a\) is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f(x)\) evaluated at \(x = a\).

Step 2: Calculate the Derivatives of \(f(x) = \sin x\)

We will calculate the derivatives up to the fourth derivative, as higher derivatives will follow a cyclic pattern:- \(f(x) = \sin x\)- \(f'(x) = \cos x\)- \(f''(x) = -\sin x\)- \(f'''(x) = -\cos x\)- \(f^{(4)}(x) = \sin x\) and the cycle repeats.

Step 3: Evaluate the Derivatives at \(x = \frac{\pi}{2}\)

Now substitute \(x = \frac{\pi}{2}\) into the derivatives:- \(f(\frac{\pi}{2}) = \sin \frac{\pi}{2} = 1\)- \(f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} = 0\)- \(f''(\frac{\pi}{2}) = -\sin \frac{\pi}{2} = -1\)- \(f'''(\frac{\pi}{2}) = -\cos \frac{\pi}{2} = 0\)- \(f^{(4)}(\frac{\pi}{2}) = \sin \frac{\pi}{2} = 1\)

Step 4: Calculate the Taylor Series Coefficients

Using the evaluated derivatives, compute each coefficient for the Taylor series:- \(\frac{f(\frac{\pi}{2})}{0!} = \frac{1}{1} = 1\)- \(\frac{f'(\frac{\pi}{2})}{1!} = \frac{0}{1} = 0\)- \(\frac{f''(\frac{\pi}{2})}{2!} = \frac{-1}{2} = -\frac{1}{2}\)- \(\frac{f'''(\frac{\pi}{2})}{3!} = \frac{0}{6} = 0\)- \(\frac{f^{(4)}(\frac{\pi}{2})}{4!} = \frac{1}{24} = \frac{1}{24}\)

Step 5: Formulate the Taylor Series

Substitute the coefficients back into the Taylor series formula:\[ \sin(x) = 1 + 0 \times (x-\frac{\pi}{2}) - \frac{1}{2}(x-\frac{\pi}{2})^2 + 0 \times (x-\frac{\pi}{2})^3 + \frac{1}{24}(x-\frac{\pi}{2})^4 + \cdots \]Simplified, it becomes:\[ \sin(x) = 1 - \frac{1}{2}(x-\frac{\pi}{2})^2 + \frac{1}{24}(x-\frac{\pi}{2})^4 + \cdots \]

Key Concepts

Maclaurin Series

A Maclaurin series is a special case of the Taylor series where the function is expanded about zero, that is, where the point \( a = 0 \). This means all the terms in the Taylor series formula are evaluated at zero. This particular form is especially useful because it often simplifies the evaluation of derivatives, particularly for functions like exponential, sine, and cosine.When you see a function like \( \sin(x) \) in its Maclaurin series form, it gives you an approximation around \( x = 0 \). Implementing this form involves the same steps as a Taylor series, which includes calculating derivatives, but instead of evaluating at a generic \( a \), you simply use the value zero for all terms. For instance, the Maclaurin series for \( \sin(x) \) is:

• \( \sin(0) = 0 \)
• \( \cos(0) = 1 \)

This results in only the odd powers of \( x \) appearing in the series since even derivatives of \( \sin \) at zero are zero.Although in our original problem we worked with a Taylor series, understanding the Maclaurin series provides a concrete example of how powerful these series expansions can be for approximations.

Derivatives

The derivative of a function is a measure of how the function value changes as its input changes. In the context of Taylor series, derivatives play a crucial role because each term in the series involves a derivative of a function. Specifically, the \( n \)-th term of a Taylor series for a function \( f(x) \) at \( a \) is determined by the \( n \)-th derivative evaluated at \( a \).For the function \( \sin(x) \), which we focused on, the derivatives exhibit a predictable cycle:

• First derivative: \( \cos x \)
• Second derivative: \( -\sin x \)
• Third derivative: \( -\cos x \)
• Fourth derivative: \( \sin x \)

This cyclic pattern means higher-order derivatives are essentially repetitions, simplifying the computation of derivatives for the series. Evaluating these at \( a = \pi/2 \) further simplifies the coefficients in the expanded series by filtering terms (only those derivatives not equal to zero contribute significant coefficients).Understanding derivatives and their behavior helps in constructing the Taylor series effectively, aiding in numerous approximations and applications in scientific calculations.

Power Series Expansion

Power series expansion is a method of expressing a function as an infinite sum of terms powered by a variable, such as \( x-a \) in a Taylor series. Every function that can be represented through such a power series can be expressed in terms of its derivatives at a point, which allows for broad utility when approximating complex functions with simpler polynomial expressions.This expansion is especially vital when dealing with functions that are too difficult to work with directly. Instead of grappling with the entire function, the series expansion allows for a usable approximation with a finite number of terms.In practice, the power series expansion transforms difficult integrals, solves differential equations, and even finds application in physics and engineering scenarios. Knowing how to transform a function into its power series form enables these computations to be performed with greater ease. Through mastery of the Taylor series, you get a strong grasp of approximation methods essential not only in mathematics but across quantitative sciences. In our example, the function \( \sin(x) \) was converted into a power series expanded around \( x = \pi/2 \), simplifying it for particular applications by only considering the significant contributing terms from the expanded series.