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Chapter 8: Problem 3

We can use power series to approximate definite integrals to which known techniques of integration do not apply. We will illustrate this in this exercise with the definite integral \(\int_{0}^{1} \sin \left(x^{2}\right) d s\) a. Use the Taylor series for \(\sin (x)\) to find the Taylor series for \(\sin \left(x^{2}\right) .\) What is the interval of convergence for the Taylor series for \(\sin \left(x^{2}\right) ?\) Explain. b. Integrate the Taylor series for \(\sin \left(x^{2}\right)\) term by term to obtain a power series expansion for \(\int \sin \left(x^{2}\right) d x\) c. Use the result from part (b) to explain how to evaluate \(\int_{0}^{1} \sin \left(x^{2}\right) d x\). Determine the number of terms you will need to approximate \(\int_{0}^{1} \sin \left(x^{2}\right) d x\) to 3 decimal places.

Short Answer

Expert verified
\[ \int_{0}^{1} \sin(x^2) dx \approx \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} \] Summing sufficient terms to get an approximation to 3 decimal places.

Step by step solution

01

Taylor Series for \(\sin(x)\)

The Taylor series for \(\sin(x)\) about \(x = 0\) is given by: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \]
02

Taylor Series for \(\sin(x^2)\)

Substitute \(x^2\) for \(x\) in the Taylor series of \(\sin(x)\): \[ \sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} \] Since the Taylor series for \(\sin(x)\) converges for all \(x\), the Taylor series for \(\sin(x^2)\) also converges for all \(x\), including on the interval \([0, 1]\).
03

Integrate the Taylor Series Term-by-Term

Integrate the series term-by-term to find the power series for the integral: \[ \int \sin(x^2) dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int x^{4n+2} dx \] Integrating each term, we get: \[ \int x^{4n+2} dx = \frac{x^{4n+3}}{4n+3} \] Thus, \[ \int \sin(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \]
04

Evaluate the Definite Integral \(\int_{0}^{1} \sin(x^2) dx\)

Evaluate the power series at the bounds from 0 to 1: \[ \left[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \right]_{0}^{1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} (1^{4n+3} - 0^{4n+3}) \] Since any power of 0 is 0, this simplifies to: \[ \int_{0}^{1} \sin(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)! (4n+3)} \]
05

Approximate to 3 Decimal Places

Calculate the sum to determine the number of terms needed for an approximation to 3 decimal places. This typically requires summing the series until the terms become sufficiently small. Summed terms are iteratively computed until the change between successive sums is less than 0.001.

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Key Concepts

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

Taylor series
The Taylor series is a way to represent functions as infinite sums of polynomial terms. For our exercise, we use the Taylor series for \(\text{sin}(x)\), expanded around zero \((x = 0)\). The series is given by: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \]
This mathematical representation lets you approximate \(\text{sin}(x)\) with high accuracy by using enough terms of the series. By substituting \((x^2)\) into our \(\text{sin}(x)\) Taylor series, we find the Taylor series for \(\text{sin}(x^2)\): \[ \sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} \]
The interval of convergence for this series is all real numbers, including our target interval [0, 1]. Instead of integrating \(\text{sin}(x^2)\) directly, we use its Taylor series expansion.
Power series integration
A power series is a series of the form: \[ \sum_{n=0}^{\infty} a_n x^n \]
Each term is a power of \((x)\) with a coefficient. Integrating a power series term-by-term simplifies complex integrals. First, we take our Taylor series for \(\text{sin}(x^2)\): \[ \sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} \]
Then, we integrate each term separately: \[ \int \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int x^{4n+2} dx \] After integrating each term, we have: \[ \int x^{4n+2} dx = \frac{x^{4n+3}}{4n+3} \] Substituting this back, the integrated series is: \[ \int \sin(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \]
This new series represents the indefinite integral of \(\text{sin}(x^2)\).
Definite integrals
Definite integrals calculate the net area between the function and the x-axis over a given interval. In our case, the interval is \([0, 1]\). To find the definite integral \(\text{\textbackslash int_{0}^{1} \text{sin}(x^2)dx}\), we use the result of our power series integration: \[ \int_{0}^{1} \sin(x^2) dx = \left[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \right]_{0}^{1} \]
Evaluating the series at the bounds: \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} (1^{4n+3} - 0^{4n+3}) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} \]
To approximate this integral to 3 decimal places, we sum enough terms until the change in successive sums is less than 0.001. This usually means computing several terms until the value stabilizes. By summing sufficiently many terms, you achieve a very close approximation of the definite integral.

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Most popular questions from this chapter

Conditionally convergent series exhibit interesting and unexpected behavior. In this exercise we examine the conditionally convergent alternating harmonic series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) and discover that addition is not commutative for conditionally convergent series. We will also encounter Riemann's Theorem concerning rearrangements of conditionally convergent series. Before we begin, we remind ourselves that $$ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=\ln (2) $$ a fact which will be verified in a later section.a. First we make a quick analysis of the positive and negative terms of the alternating harmonic series. i. Show that the series \(\sum_{k=1}^{\infty} \frac{1}{2 k}\) diverges. ii. Show that the series \(\sum_{k=1}^{\infty} \frac{1}{2 k+1}\) diverges. iii. Based on the results of the previous parts of this exercise, what can we say about the sums \(\sum_{k=C}^{\infty} \frac{1}{2 k}\) and \(\sum_{k=C}^{\infty} \frac{1}{2 k+1}\) for any positive integer \(C ?\) Be specific in your explanation. b. Recall addition of real numbers is commutative; that is $$ a+b=b+a $$ for any real numbers \(a\) and \(b\). This property is valid for any sum of finitely many terms, but does this property extend when we add infinitely many terms together? The answer is no, and something even more odd happens. Riemann's Theorem (after the nineteenth-century mathematician Georg Friedrich Bernhard Riemann) states that a conditionally convergent series can be rearranged to converge to any prescribed sum. More specifically, this means that if we choose any real number \(S\), we can rearrange the terms of the alternating harmonic series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) so that the sum is \(S\). To understand how Riemann's Theorem works, let's assume for the moment that the number \(S\) we want our rearrangement to converge to is positive. Our job is to find a way to order the sum of terms of the alternating harmonic series to converge to \(S\). i. Explain how we know that, regardless of the value of \(S\), we can find a partial sum \(P_{1}\) $$ P_{1}=\sum_{k=1}^{n_{1}} \frac{1}{2 k+1}=1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n_{1}+1} $$ of the positive terms of the alternating harmonic series that equals or exceeds \(S\). Let $$ S_{1}=P_{1} $$ii. Explain how we know that, regardless of the value of \(S_{1}\), we can find a partial sum \(N_{1}\) $$ N_{1}=-\sum_{k=1}^{m_{1}} \frac{1}{2 k}=-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\cdots-\frac{1}{2 m_{1}} $$ so that $$ S_{2}=S_{1}+N_{1} \leq S $$ iii. Explain how we know that, regardless of the value of \(S_{2}\), we can find a partial sum \(P_{2}\) $$ P_{2}=\sum_{k=n_{1}+1}^{n_{2}} \frac{1}{2 k+1}=\frac{1}{2\left(n_{1}+1\right)+1}+\frac{1}{2\left(n_{1}+2\right)+1}+\cdots+\frac{1}{2 n_{2}+1} $$ of the remaining positive terms of the alternating harmonic series so that $$ S_{3}=S_{2}+P_{2} \geq S $$iv. Explain how we know that, regardless of the value of \(S_{3}\), we can find a partial sum $$ N_{2}=-\sum_{k=m_{1}+1}^{m_{2}} \frac{1}{2 k}=-\frac{1}{2\left(m_{1}+1\right)}-\frac{1}{2\left(m_{1}+2\right)}-\cdots-\frac{1}{2 m_{2}} $$ of the remaining negative terms of the alternating harmonic series so that $$ S_{4}=S_{3}+N_{2} \leq S $$ v. Explain why we can continue this process indefinitely and find a sequence \(\left\\{S_{n}\right\\}\) whose terms are partial sums of a rearrangement of the terms in the alternating harmonic series so that \(\lim _{n \rightarrow \infty} S_{n}=S\).

In this exercise we will begin with a strange power series and then find its sum. The Fibonacci sequence \(\left\\{f_{n}\right\\}\) is a famous sequence whose first few terms are$$ f_{0}=0, f_{1}=1, f_{2}=1, f_{3}=2, f_{4}=3, f_{5}=5, f_{6}=8, f_{7}=13, \cdots, $$ where each term in the sequence after the first two is the sum of the preceding two terms. That is, \(f_{0}=0, f_{1}=1\) and for \(n \geq 2\) we have $$ f_{n}=f_{n-1}+f_{n-2} $$ Now consider the power series $$ F(x)=\sum_{k=0}^{\infty} f_{k} x^{k} $$ We will determine the sum of this power series in this exercise. a. Explain why each of the following is true. $$ \text { i. } x F(x)=\sum_{k=1}^{\infty} f_{k-1} x^{k} $$ ii. \(x^{2} F(x)=\sum_{k=2}^{\infty} f_{k-2} x^{k}\) b. Show that $$ F(x)-x F(x)-x^{2} F(x)=x $$ c. Now use the equation $$ F(x)-x F(x)-x^{2} F(x)=x $$ to find a simple form for \(F(x)\) that doesn't involve a sum. d. Use a computer algebra system or some other method to calculate the first 8 derivatives of \(\frac{x}{1-x-x^{2}}\) evaluated at \(0 .\) Why shouldn't the results surprise you?

In this exercise we investigation the Taylor series of polynomial functions. a. Find the 3 rd order Taylor polynomial centered at \(a=0\) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Does your answer surprise you? Explain. b. Without doing any additional computation, find the 4 th, 12 th, and 100 th order Taylor polynomials (centered at \(a=0\) ) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Why should you expect this? c. Now suppose \(f(x)\) is a degree \(m\) polynomial. Completely describe the \(n\) th order Taylor polynomial (centered at \(a=0\) ) for each \(n\).

We return to the example begun in Preview Activity 8.1 .1 to see how to derive the formula for the amount of money in an account at a given time. We do this in a general setting. Suppose you invest \(P\) dollars (called the principal) in an account paying \(r \%\) interest compounded monthly. In the first month you will receive \(\frac{r}{12}\) (here \(r\) is in decimal form; e.g., if we have \(8 \%\) interest, we write \(\frac{0.08}{12}\) ) of the principal \(P\) in interest, so you earn $$ P\left(\frac{r}{12}\right) $$ dollars in interest. Assume that you reinvest all interest. Then at the end of the first month your account will contain the original principal \(P\) plus the interest, or a total of $$ P_{1}=P+P\left(\frac{r}{12}\right)=P\left(1+\frac{r}{12}\right) $$ dollars. a. Given that your principal is now \(P_{1}\) dollars, how much interest will you earn in the second month? If \(P_{2}\) is the total amount of money in your account at the end of the second month, explain why $$ P_{2}=P_{1}\left(1+\frac{r}{12}\right)=P\left(1+\frac{r}{12}\right)^{2} $$ b. Find a formula for \(P_{3}\), the total amount of money in the account at the end of the third month in terms of the original investment \(P\). c. There is a pattern to these calculations. Let \(P_{n}\) the total amount of money in the account at the end of the third month in terms of the original investment \(P\). Find a formula for \(P_{n}\)

The examples we have considered in this section have all been for Taylor polynomials and series centered at 0 , but Taylor polynomials and series can be centered at any value of \(a\). We look at examples of such Taylor polynomials in this exercise. a. Let \(f(x)=\sin (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=\frac{\pi}{2}\). Then find the Taylor series for \(f(x)\) centered at \(x=\frac{\pi}{2}\). Why should you have expected the result? b. Let \(f(x)=\ln (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=1\). Then find the Taylor series for \(f(x)\) centered at \(x=1\).
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