微积分I(标准路径) · free preview

§8.6 Quantifying the accuracy of approximations

对于一个收敛的无穷级数,当我们截断级数的有限项去用有限和近似其值时,会损失多少精度?换句话说,要用有限和得到一个好的近似,需要多少项才够?

1. Introduction

Introduction

2. Alternating series of real numbers

Alternating series of real numbers

3. Error Approximations for Taylor Polynomials

Error Approximations for Taylor Polynomials

4. Summary

Summary

5. Consider the alterna…

Consider the alternating geometric series

S=∑k=0∞(−1)k(45)k=1−45+1625−⋯+(−1)n−1(45)n−1+⋯S = \sum_{k=0}^{\infty} (-1)^k \left( \frac{4}{5} \right)^k = 1 - \frac{4}{5} + \frac{16}{25} - \cdots + (-1)^{n-1} \left( \frac{4}{5} \right)^{n-1} + \cdots

. We want to explore how the partial sums of the series compare to and approximate the exact sum of the series, which is

S=a1−r=11−(−45)=195=59S = \frac{a}{1-r} = \frac{1}{1-\left(\frac{-4}{5}\right)} = \frac{1}{\frac{9}{5}} = \frac{5}{9}

.

Recall that the nn th partial sum, SnS_n, is the sum of the first nn terms of the infinite geometric series SS. This means that

Sn=∑k=0n−1(−1)k(45)k=1−45+1625−⋯+(−1)n−1(45)n−1S_n = \sum_{k=0}^{n-1} (-1)^k \left( \frac{4}{5} \right)^k = 1 - \frac{4}{5} + \frac{16}{25} - \cdots + (-1)^{n-1} \left( \frac{4}{5} \right)^{n-1}

, which we view as being in the form

Sn=a0−a1+a2−⋯+(−1)n−1an−1S_n = a_0 - a_1 + a_2 - \cdots + (-1)^{n-1}a_{n-1}

. Note that the exact fractional values of S1S_1, S2S_2, …\ldots, S6S_6, have been recorded below along with their decimal representations. Your task is to use this information to do some related computations that help us understand the behavior of the series and its partial sums.

First, by computing the differences between SnS_n and SS for several different values of nn (recalling that S=59=0.5‾S = \frac{5}{9} = 0.\overline{5}), fill in the first column of blank spaces provided below. In addition, fill in decimal representations of a3a_3, …\ldots, a6a_6 to compare to the differences in the preceding adjacent column. For decimal representations you enter, be accurate to within 0.000010.00001.

n=1S1=1S1−S=a1=−0.8n=2S2=15=0.2S2−S=a2=0.64n=3S3=2125=0.84S3−S=a3=n=4S4=41125=0.328S4−S=a4=n=5S5=461625=0.7376S5−S=a5=n=6S6=12813125=0.40992S6−S=a6=\begin{aligned} n &= 1 & S_1 &= 1 & S_1 - S &= & a_1 &= -0.8 \\ n &= 2 & S_2 &= \frac{1}{5} = 0.2 & S_2 - S &= & a_2 &= 0.64 \\ n &= 3 & S_3 &= \frac{21}{25} = 0.84 & S_3 - S &= & a_3 &= \\ n &= 4 & S_4 &= \frac{41}{125} = 0.328 & S_4 - S &= & a_4 &= \\ n &= 5 & S_5 &= \frac{461}{625} = 0.7376 & S_5 - S &= & a_5 &= \\ n &= 6 & S_6 &= \frac{1281}{3125} = 0.40992 & S_6 - S &= & a_6 &= \end{aligned}

6. Determine how well t…

Determine how well the 100th100^{\text{th}} partial sum S100S_{100} of

∑k=0∞(−1)kk+1\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+1}

approximates the value of the converging alternating series.

7. Use the fact that $\…$

Use the fact that sin⁡(x)=x−13!x3+15!x5−⋯\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots to estimate sin⁡(1)\sin(1) to within 0.00010.0001. Do so without entering “sin⁡(1)\sin(1)” on a computational device. After you find your estimate, enter “sin⁡(1)\sin(1)” on a computational device and compare the results.

8. Recall our recent wo…

Recall our recent work with ∫01e−x2 dx\int_0^1 e^{-x^2} \, dx which can be expressed as the series

∫01e−x2 dx=1−13+12!⋅5−13!⋅7+⋯+(−1)nn!⋅(2n+1)+⋯\int_0^1 e^{-x^2} \, dx = 1 - \frac{1}{3} + \frac{1}{2! \cdot 5} - \frac{1}{3! \cdot 7} + \cdots + \frac{(-1)^n}{n! \cdot (2n+1)} + \cdots

. Use this series representation to estimate ∫01e−x2 dx\int_0^1 e^{-x^2} \, dx to within 0.00010.0001. Then, compare what a computational device reports when you use it to estimate the definite integral.

9. Find the Taylor seri…

Find the Taylor series for cos⁡(x2)\cos(x^2) and then use the Taylor series and to estimate the value of ∫01cos⁡(x2) dx\int_0^1 \cos(x^2) \, dx to within 0.00010.0001. Compare your result to what a computational device reports when you use it to estimate the definite integral.

10. Recall we know that …

Recall we know that if ∣x∣<1|x| \lt 1, then

ln⁡(1+x)=x−12x2+13x3−⋯+(−1)n−11nxn+⋯\ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots + (-1)^{n-1} \frac{1}{n} x^n + \cdots

. What happens if x=1x = 1?

Explain why the series 1−12⋅12+13⋅13−⋯+(−1)n−11n⋅1n+⋯1 - \frac{1}{2} \cdot 1^2 + \frac{1}{3} \cdot 1^3 - \cdots + (-1)^{n-1} \frac{1}{n} \cdot 1^n + \cdots must converge and estimate its sum to within 0.010.01. What is the exact sum of this series?

11. Determine the maximu…

Determine the maximum error possible when using the degree 1010 Taylor polynomial centered at a=0a = 0 for sin⁡(x)\sin(x) to approximate the value of sin⁡(2)\sin(2).

12. Use the degree $10$ …

Use the degree 1010 Taylor polynomial (centered at a=0a = 0) of f(x)=exf(x) = e^x to estimate the value of e2e^2. What is the maximum error of your estimate, according to the Lagrange Error Bound? How does this compare to the actual error between e2e^2 and T10(2)T_{10}(2) as reported by a computer algebra system?

13. Use a degree $n$ Tay…

Use a degree nn Taylor polynomial (centered at a=0a = 0) of f(x)=cos⁡(x)f(x) = \cos(x) to estimate the value of cos⁡(1)\cos(1) within an accuracy of 0.000000010.00000001. According to the Lagrange Error Bound, what value of nn is needed to achieve this accuracy? What is the resulting approximate value of cos⁡(1)\cos(1)?

14. Recall that for $f(x…$

Recall that for f(x)=ln⁡(1+x)f(x) = \ln(1+x), its Taylor series centered at a=0a = 0 is given by

ln⁡(1+x)=x−12x2+13x3−⋯+(−1)n−11nxn+⋯\ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots + (-1)^{n-1}\frac{1}{n}x^n + \cdots

and that the nthn^{\text{th}} derivative of f(x)=ln⁡(1+x)f(x) = \ln(1+x) is given by

f(n)(x)=(−1)n−1(n−1)!(1+x)nf^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n}

. If we want to estimate f(0.5)=ln⁡(1.5)f(0.5) = \ln(1.5) to within an accuracy of 0.00010.0001, what value of nn is needed to achieve this accuracy from computing Tn(0.5)T_n(0.5), according to the Lagrange Error Bound?

15. In this exercise we …

In this exercise we consider the definite integral

∫0141+x2 dx\int_0^1 \frac{4}{1+x^2} \, dx

from two different perspectives.

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