§8.6 Quantifying the accuracy of approximations

Introduction

Alternating series of real numbers

Error Approximations for Taylor Polynomials

Summary

For a convergent alternating series with decreasing terms, the error after summing $n$ terms is bounded by

Using $\sin(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$, the estimate $\sin(1) \approx 1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}$ is guaranteed accurate to within

If $|f^{(n+1)}(x)| \le M$ on the interval between $a$ and $x$, the Lagrange error bound states $|f(x)-T_n(x)| \le$

Expanding $\frac{4}{1+x^2} = 4(1-x^2+x^4-\cdots)$ and integrating term by term over $[0,1]$ gives a series equal to the integral. The result is

For $S = \sum_{k=0}^{\infty}(-1)^k\left(\frac{4}{5}\right)^k$, the geometric limit is $S = \frac{1}{1+\frac{4}{5}} = \frac{5}{9}$. Report this value, rounded to three decimals.

For $\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}$, the alternating error bound guarantees $|S_{100}-S| \le \frac{1}{101}$. Report this bound, rounded to four decimals.

Compute the alternating estimate $1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}$ for $\sin(1)$, rounded to four decimals.

Using $\int_0^1e^{-x^2}dx = 1-\frac{1}{3}+\frac{1}{2!\cdot5}-\frac{1}{3!\cdot7}+\frac{1}{4!\cdot9}-\frac{1}{5!\cdot11}+\frac{1}{6!\cdot13} = \frac{1614779}{2162160}$, report this estimate, rounded to four decimals.

Using $\cos(x^2) = 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots$, the series estimate of $\int_0^1\cos(x^2)\,dx$ through the $\frac{1}{13\cdot6!}$ term is $\frac{25399}{28080}$. Report this value, rounded to four decimals.

Using the degree $10$ Taylor polynomial of $\sin(x)$ at $a=0$ to approximate $\sin(2)$, the Lagrange bound with $M = 1$ gives error at most $\frac{2^{11}}{11!}$. Report this bound, rounded to seven decimals.

Using the degree $10$ polynomial of $e^x$ at $a=0$ to estimate $e^2$ gives $T_{10}(2) \approx 7.388995$, while $e^2 \approx 7.389056$. Compute the actual error of the estimate, rounded to six decimals.

To estimate $\cos(0.5)$ within $0.000001$ using the Taylor polynomial at $0$, the Lagrange bound with $M = 1$ requires the smallest degree $n$ with $\frac{0.5^{n+1}}{(n+1)!} \lt 0.000001$. Report $n$.

To estimate $\ln(1.5)$ within $0.0001$ using the Taylor polynomial of $\ln(1+x)$ at $0$, the Lagrange bound gives error at most $\frac{0.5^{n+1}}{n+1}$, which first drops below the tolerance at degree $n$. Report $n$.

The integral $\int_0^1\frac{4}{1+x^2}dx$ equals $\pi$. Report this value, rounded to three decimals.

At the endpoint $x = 1$, the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ converges by the Alternating Series Theorem, and its sum is exactly $\ln(2)$.