Problem 8 - Active Calculus

Question

We can use known Taylor series to obtain other Taylor series, and we explore that idea in this exercise, as a preview of work in the following section. a. Calculate the first four derivatives of \(\sin \left(x^{2}\right)\) and hence find the fourth order Taylor polynomial for \(\sin \left(x^{2}\right)\) centered at \(a=0\) b. Part (a) demonstrates the brute force approach to computing Taylor polynomials and series. Now we find an easier method that utilizes a known Taylor series. Recall that the Taylor series centered at 0 for \(f(x)=\sin (x)\) is $$ \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k+1}}{(2 k+1) !} $$ i. Substitute \(x^{2}\) for \(x\) in the Taylor series \((8.5 .7) .\) Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should give the Taylor series for \(\sin \left(x^{2}\right)\) centered at \(0 .\) ii. What should we expect the interval of convergence of the series for \(\sin \left(x^{2}\right)\) to be? Explain in detail.

Step-by-Step Solution

Answer

The fourth order Taylor polynomial is \(P_4(x) = x^2 \). The interval of convergence is all real numbers.

Step 1 - Find the first derivative

The first derivative of \(\sin(x^2)\) with respect to \(x\) is obtained using the chain rule: \[ f'(x) = \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x = 2x \cos(x^2) \]

Step 2 - Find the second derivative

Apply the product rule and chain rule: \[ f''(x) = \frac{d}{dx}(2x \cos(x^2)) = 2 \cos(x^2) + 2x \cdot \frac{d}{dx}(\cos(x^2)) = 2 \cos(x^2) - 4x^2 \sin(x^2) \]

Step 3 - Find the third derivative

Using the product rule and chain rule again: \[ f'''(x) = \frac{d}{dx}( 2 \cos(x^2) - 4x^2 \sin(x^2)) = -8x \sin(x^2) - 8x \sin(x^2) - 8x^3 \cos(x^2) = -16x \sin(x^2) - 8x^3 \cos(x^2) \]

Step 4 - Find the fourth derivative

Continuing with the chain rule and product rule: \[ f''''(x) = \frac{d}{dx}( -16x \sin(x^2) - 8x^3 \cos(x^2)) = -16 \sin(x^2) - 32x^2 \cos(x^2) - 24x^2 \cos(x^2) + 16x^4 \sin(x^2) = -16 \sin(x^2) - 56x^2 \cos(x^2) + 16x^4 \sin(x^2) \]

Step 5 - Write Taylor polynomial

Using the derivatives, the fourth-order Taylor polynomial centered at \(a = 0\) can be written: \[ P_4(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} \] Since all derivatives evaluated at zero are 0 except the original function, \(P_4(x) = x^2 \)

Step 6 - Substitute \(x^2\) into Taylor series of \(\sin(x)\)

Substitute \(x^2\) for \(x\) in the Taylor series for \(\sin(x)\): \[ \sum_{k=0}^{\infty}(-1)^k \frac{x^{4k+2}}{(2k+1)!} \] The first few terms are: \( x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - ... \)

Step 7 - Interval of convergence

In similar fashion to \(\sin(x)\)'s Taylor series, the interval of convergence will be \( |x^2| < \infty \) implying convergence for all real \( x \).

Key Concepts

derivatives

Derivatives represent the rate of change of a function with respect to a variable. To calculate the first four derivatives of \( \sin(x^2) \), we use two critical rules: the chain rule and the product rule. The chain rule helps us differentiate composite functions. For \( \sin(x^2) \), applying the chain rule gives us: \[ f'(x) = \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x = 2x \cos(x^2) \], which is the first derivative. The product rule is necessary when dealing with products of functions. It states that \( (uv)' = u'v + uv' \). By applying these rules iteratively, we can compute the higher-order derivatives. Each step of differentiation introduces more complexity, but the rules help us manage that complexity.

Taylor polynomial

A Taylor polynomial approximates a function as a sum of terms calculated from the values of the function’s derivatives at a single point. In this case, at \( a = 0 \), the fourth-order Taylor polynomial for \( \sin(x^2) \) is found by including terms up to the \(x^4 \) level: \[ P_4(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} \]. Evaluating the derivatives of \( \sin(x^2) \) at zero, we find that they are zero, except for the \(x^2 \) term. Thus, \[ P_4(x) = x^2 \]. This polynomial provides a good approximate near the point of expansion.

interval of convergence

The interval of convergence denotes the range of input values for which a Taylor series converges to the original function. For the series of \( \sin(x^2) \), we substitute \( x^2 \) into the Taylor series of \( \sin(x) \), resulting in: \[ \sum_{k=0}^{\infty}(-1)^k \frac{x^{4k+2}}{(2k+1)!} \]. This series converges as long as \( |x^2| < \infty \), implying convergence for all real \( x \). This wide interval makes Taylor series valuable for approximating functions over a broad domain.

chain rule

The chain rule is an essential tool for differentiating composite functions. It states: if \( y = f(u) \) and \( u = g(x) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). For \( \sin(x^2) \), we set \( u = x^2 \) and \( y = \sin(u) \), then calculate \[ \frac{dy}{dx} = \cos(x^2) \cdot 2x \]. This means we differentiate \( \sin(u) \) with respect to \( u \), and multiply the result by the derivative of \( u \) with respect to \( x \). Applying the chain rule simplifies the process of deriving complex functions, making it indispensable.

product rule

The product rule is used when differentiating the product of two functions. If \( u(x) \) and \( v(x) \) are functions of \( x \), the rule states: \[ (uv)' = u'v + uv' \]. For example, in finding the second derivative of \( 2x \cos(x^2) \), we differentiate \( 2x \) and \( \cos(x^2) \) separately and apply the product rule: \[ f''(x) = 2 \cos(x^2) - 4x^2 \sin(x^2) \]. The rule generalizes to higher derivatives and complex products, making it a vital technique in calculus.