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Chapter 8: Problem 8

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.

Short Answer

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

Step by step solution

01

- Find the first derivative

The first derivative of \(\sin(x^2)\) with respect to \(x\) is obtained using the chain rule: \[ f'(x) = \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x = 2x \cos(x^2) \]
02

- Find the second derivative

Apply the product rule and chain rule: \[ f''(x) = \frac{d}{dx}(2x \cos(x^2)) = 2 \cos(x^2) + 2x \cdot \frac{d}{dx}(\cos(x^2)) = 2 \cos(x^2) - 4x^2 \sin(x^2) \]
03

- Find the third derivative

Using the product rule and chain rule again: \[ f'''(x) = \frac{d}{dx}( 2 \cos(x^2) - 4x^2 \sin(x^2)) = -8x \sin(x^2) - 8x \sin(x^2) - 8x^3 \cos(x^2) = -16x \sin(x^2) - 8x^3 \cos(x^2) \]
04

- Find the fourth derivative

Continuing with the chain rule and product rule: \[ f''''(x) = \frac{d}{dx}( -16x \sin(x^2) - 8x^3 \cos(x^2)) = -16 \sin(x^2) - 32x^2 \cos(x^2) - 24x^2 \cos(x^2) + 16x^4 \sin(x^2) = -16 \sin(x^2) - 56x^2 \cos(x^2) + 16x^4 \sin(x^2) \]
05

- 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, \(P_4(x) = x^2 \)
06

- 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: \( x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - ... \)
07

- 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 \( x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

We have shown that if \(\sum(-1)^{k+1} a_{k}\) is a convergent alternating series, then the sum \(S\) of the series lies between any two consecutive partial sums \(S_{n}\). This suggests that the average \(\frac{S_{n}+S_{n+1}}{2}\) is a better approximation to \(S\) than is \(S_{n}\). a. Show that \(\frac{S_{n}+S_{n+1}}{2}=S_{n}+\frac{1}{2}(-1)^{n+2} a_{n+1}\). b. Use this revised approximation in (a) with \(n=20\) to approximate \(\ln (2)\) given that $$ \ln (2)=\sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} . $$ Compare this to the approximation using just \(S_{20} .\) For your convenience, \(S_{20}=\frac{155685007}{232792560}\).

In the Limit Comparison Test we compared the behavior of a series to one whose behavior we know. In that test we use the limit of the ratio of corresponding terms of the series to determine if the comparison is valid. In this exercise we see how we can compare two series directly, term by term, without using a limit of sequence. First we consider an example. a. Consider the series \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k}\) We know that the series \(\sum \frac{1}{k^{2}}\) is a \(p\) -series with \(p=2>1\) and so \(\sum \frac{1}{k^{2}}\) converges. In this part of the exercise we will see how to use information about \(\sum \frac{1}{k^{2}}\) to determine information about \(\sum \frac{1}{k^{2}+k} .\) Let \(a_{k}=\frac{1}{k^{2}}\) and \(b_{k}=\frac{1}{k^{2}+k} .\) i) Let \(S_{n}\) be the \(n\) th partial sum of \(\sum \frac{1}{k^{2}}\) and \(T_{n}\) the \(n\) th partial sum of \(\sum \frac{1}{k^{2}+k}\). Which is larger, \(S_{1}\) or \(T_{1}\) ? Why? ii) Recall that $$ S_{2}=S_{1}+a_{2} \text { and } T_{2}=T_{1}+b_{2} $$ Which is larger, \(a_{2}\) or \(b_{2}\) ? Based on that answer, which is larger, \(S_{2}\) or \(T_{2}\) ? iii) Recall that $$ S_{3}=S_{2}+a_{3} \text { and } T_{3}=T_{2}+b_{3} $$ Which is larger, \(a_{3}\) or \(b_{3}\) ? Based on that answer, which is larger, \(S_{3}\) or \(T_{3}\) ? iv) Which is larger, \(a_{n}\) or \(b_{n}\) ? Explain. Based on that answer, which is larger, \(S_{n}\) or \(T_{n}\) ? v) Based on your response to the previous part of this exercise, what relationship do you expect there to be between \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k} ?\) Do you expect \(\sum \frac{1}{k^{2}+k}\) to converge or diverge? Why? b. The example in the previous part of this exercise illustrates a more general result. Explain why the Direct Comparison Test, stated here, works.

Sequences have many applications in mathematics and the sciences. In a recent paper \(^{3}\) the authors write The incretin hormone glucagon-like peptide- 1 (GLP-1) is capable of ameliorating glucose-dependent insulin secretion in subjects with diabetes. However, its very short half-life (1.5-5 min) in plasma represents a major limitation for its use in the clinical setting. The half-life of GLP-1 is the time it takes for half of the hormone to decay in its medium. For this exercise, assume the half-life of GLP-1 is 5 minutes. So if \(A\) is the amount of GLP-1 in plasma at some time \(t,\) then only \(\frac{A}{2}\) of the hormone will be present after \(t+5\) minutes. Suppose \(A_{0}=100\) grams of the hormone are initially present in plasma. a. Let \(A_{1}\) be the amount of GLP-1 present after 5 minutes. Find the value of \(A_{1}\). b. Let \(A_{2}\) be the amount of GLP-1 present after 10 minutes. Find the value of \(A_{2}\). c. Let \(A_{3}\) be the amount of GLP-1 present after 15 minutes. Find the value of \(A_{3}\). d. Let \(A_{4}\) be the amount of GLP-1 present after 20 minutes. Find the value of \(A_{4}\). e. Let \(A_{n}\) be the amount of GLP-1 present after \(5 n\) minutes. Find a formula for \(A_{n}\). f. Does the sequence \(\left\\{A_{n}\right\\}\) converge or diverge? If the sequence converges, find its limit and explain why this value makes sense in the context of this problem. g. Determine the number of minutes it takes until the amount of GLP-1 in plasma is 1 gram.

Based on the examples we have seen, we might expect that the Taylor series for a function \(f\) always converges to the values \(f(x)\) on its interval of convergence. We explore that idea in more detail in this exercise. Let \(f(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}} & \text { if } x \neq 0, \\ 0 & \text { if } x=0 .\end{array}\right.\) a. Show, using the definition of the derivative, that \(f^{\prime}(0)=0\). b. It can be shown that \(f^{(n)}(0)=0\) for all \(n \geq 2\). Assuming that this is true, find the Taylor series for \(f\) centered at 0 . c. What is the interval of convergence of the Taylor series centered at 0 for \(f ?\) Explain. For which values of \(x\) the interval of convergence of the Taylor series does the Taylor series converge to \(f(x) ?\)

Find the first four terms of the Taylor series for the function \(\cos (x)\) about the point \(a=\) \(-\pi / 4\). (Your answers should include the variable \(\mathrm{x}\) when appropriate.) \(\cos (x)=\) \(+\ldots\)
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