Exercises: Taylor Series Applications

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Various exercises relating to the application of Taylor Series to other problems of interest.

Compute the limit $\lim _{n \rightarrow \infty } n^3 \left [ \arctan \frac {1}{n} - \frac {1}{n} \right ].$

The first few terms of the Maclaurin series for $\arctan x$ are $\cos x = \answer {x} - \answer {\frac {x^3}{3}} + \answer {\frac {x^5}{5}} + \cdots$
Plug in $x = \frac {1}{n}$ and subtract $1/n$ : $\arctan \frac {1}{n} - \frac {1}{n} = - \frac {1}{3n^3} + \answer {\frac {1}{5n^5}} + \cdots$
Multiply by $n^3$ , neglect all but the dominant term, and conclude $\lim _{n \rightarrow \infty } n^3 \left [ \arctan \frac {1}{n} - \frac {1}{n} \right ] = \answer {-\frac {1}{3}}$

Compute the limit $\lim _{n \rightarrow \infty } \frac {\ln (n+3) - \ln n}{\ln (n+1) - \ln n}.$

First combine logarithms to simplify a bit: $\ln (n+3) - \ln n = \ln \answer {1 + \frac {3}{n}} \ \ \text { and } \ \ \ln (n+1) - \ln n = \ln \answer {1 + \frac {1}{n}}.$
Write out the first few terms of the Maclaurin series for $\ln (1+x)$ and substitute in the appropriate expression involving $n$ : $\ln (1 + x) = \answer {x} - \answer {\frac {x^2}{2}} + \answer {\frac {x^3}{3}} - \cdots$ $\ln (n+3) - \ln n = \answer { \frac {3}{n}} - \answer {\frac {9}{2n^2}} + \answer {\frac {9}{n^3}} - \cdots$ $\ln (n+1) - \ln n = \answer {\frac {1}{n}} - \answer {\frac {1}{2n^2}} + \answer {\frac {1}{3n^3}} - \cdots$
Multiply by $n$ , neglect all but the dominant term (as $n \rightarrow \infty )$ to conclude $\lim _{n \rightarrow \infty } \frac {\ln (n+3) - \ln n}{\ln (n+1) - \ln n} = \answer {3}.$

Compute the limit $\lim _{x \rightarrow 1} \frac {\ln x - x + 1}{\sin \pi x}.$

Compute the first few terms of the Taylor series of numerator and denominator centered at $x = 1$ : $\ln x - x + 1 = \answer {0} + \answer {0} (x-1) + \answer {-\frac {1}{2}} (x-1)^2 + \answer {\frac {1}{3}} (x-1)^3 + \cdots$ $\sin \pi x = \answer {0} + \answer {-\pi }(x-1) + \answer {0} (x-1)^2 + \answer {\frac {\pi ^3}{6}}(x-1)^3 + \cdots$
Neglect all but the dominant terms in numerator and denominator. Note that dominant here means as $x \rightarrow 1$ . Then take the limit $x \rightarrow 1$ : $\lim _{x \rightarrow 1} \frac {\ln x - x + 1}{\sin \pi x} = \answer {0}.$

Use the remainder formula for Taylor series to determine which partial sums of the series $\sum _{n=0}^\infty \frac {1}{n!}$ differ from $e$ by at most $1/100$ ?

The remainder formula says that $e^1 - \sum _{n=0}^{N} \frac {1}{n!} = \frac {e^{\xi }}{(N+1)!} (1-0)^{N+1}$ for some point $\xi$ which depends on $N$ and is somewhere in the interval $(\answer {0},\answer {1})$ . The upper bound for the magnitude of $e^\xi$ on this interval is $M = \answer {e}$ .
For which values of $N$ do we have $\frac {M}{(N+1)!} \leq \frac {1}{50}?$ (Check manually for small values of $N$ ; use the fact that $e$ is between $2$ and $3$ .) $N \geq \answer {4}.$

In this exercise, we will investigate two different ways of numerically approximating the value of $\ln 2$ using infinite series.

The Maclaurin series of the function $\ln 1 + x$ converges conditionally at $x = 1$ to $\ln 2$ . Compute the Maclaurin series: $\ln (1 + x) = \sum _{n=0}^\infty \answer {\frac {(-1)^n}{n+1}} x^{n+1}$ (Reindex your answer to match the template above if your answer doesn’t work as-is.)
What degree Taylor polynomial would you need to use to approximate the value of $\ln 2$ to an error strictly less than $10^{-3}$ ? Answer: Taylor polynomial used should have degree $N = \answer {1000}$ or greater.
A similar but distinct strategy would be to compute $\ln (1/2)$ instead because we know $- \ln (1/2) = \ln 2$ . Evaluating the Maclaurin series at $x=-1/2$ and doing a little simplification, we see from the above series that $\ln 2 = \sum _{n=1}^\infty \answer {\frac {1}{n 2^n}}.$ (Once again, reindex if your answer does not already start at $n=1$ .)
The presence of a factor exponential in $n$ suggests that this second series converges to $\ln 2$ than the first series. Of the two series, then, the series presents a more efficient way to compute $\ln 2$ numerically.

For the first estimation, use the series estimation features of the Alternating Series Test.

For an alternating series, the error of approximating the series by a partial sum is never greater than the magnitude of the first term omitted.

Use the remainder formula for Taylor series to determine which partial sums of the series $\sum _{n=1}^\infty \frac {1}{n 3^n}$ differ from $\ln (3/2)$ by at most $10^{-2}$ ?

Guided by the example above, we will use the Maclaurin series $- \ln (1-x) = \sum _{n=1}^\infty \frac {x^n}{n}.$ First, compute the derivatives of $- \ln (1-x)$ . Find a pattern which holds for all $N$ : $\frac {d^{N+1}}{dx^{N+1}} \left ( - \ln (1 - x) \right ) = \answer { \frac {N!}{(1-x)^{N+1}}}.$ When $x$ is between $0$ and $1/3$ , the largest value of the $(N+1)$ -st derivative is what? $\left | \frac {d^{N+1}}{dx^{N+1}} \left ( - \ln (1 - x) \right ) \right | \leq \answer {N! 3^{N+1}/2^{N+1}}.$
Using this upper bound for the $(N+1)$ -st derivative, we know that $\left | \ln \frac {3}{2} - \sum _{n=0}^N \frac {1}{n 3^n} \right | \leq \answer { \frac {1}{(N+1) 2^{N+1}}}$ (use the upper bound and the remainder formula.)
For which values of $N$ is the expression you just found less than $1/100$ ? Check by hand for smallish values of $N$ to find the smallest one which works. $N \geq \answer {4}.$

Sample Quiz Questions

Use Taylor series to estimate the value of $\sqrt [3]{\frac {11}{10}}$ to within an error of at most $1/900$ . (Hints will not be revealed until you choose a response.)

$\displaystyle \frac {31}{30}$ $\displaystyle \frac {47}{45}$ $\displaystyle \frac {19}{18}$ $\displaystyle \frac {16}{15}$ $\displaystyle \frac {97}{90}$ $\displaystyle \frac {49}{45}$

We may use the remainder formula for Taylor series to approach this problem. Suppose $p_n(x)$ is the degree $n$ Taylor polynomial of the function $f(x) = \sqrt [3]{1+x}$ with center $a=0$ . Then the error $E_n(x)$ , i.e., the difference between the polynomial and the function, does not exceed $\frac {f^{(n+1)}(\xi )}{(n+1)!}x^{n+1}$ , where $\xi$ is some unknown point in the range $0 \leq \xi \leq x$ .

In this case one should take $x = 1/10$ and determine how many derivatives are required to make this error estimate less than the given threshold.

This means checking by hand for small numbers of derivatives. For the specific problem at hand, if we approximate $f(x)$ by the Taylor polynomial of degree $n = 1$ , we have $\left | \frac {x^{n+1} }{(n+1)!} f^{(n+1)}(\xi ) \right | = \left | \frac {x^{n+1}}{(n+1)!} \left ( -\frac {2}{9} (\xi +1)^{-5/3} \right ) \right | \leq \left | \frac {x^{n+1}}{(n+1)!} \left ( \frac {2}{9} \right ) \right | = \frac {1}{900}$ when $x = 1/10$ .

We conclude that the correct Taylor approximation is $p_n \left ( \frac {1}{10}\right ) = \left (\frac {1}{10}\right )^{0} + \frac {1}{3}\left (\frac {1}{10}\right )^{1} = 1 + \frac {1}{30} = \frac {31}{30}.$

Use Taylor series to estimate the value of $e^{-\frac {1}{3}}$ to within an error of at most $1/162$ .

$\displaystyle \frac {5}{9}$ $\displaystyle \frac {13}{18}$ $\displaystyle \frac {8}{9}$ $\displaystyle \frac {19}{18}$ $\displaystyle \frac {11}{9}$ $\displaystyle \frac {25}{18}$

We may use the remainder formula for Taylor series to approach this problem. Suppose $p_n(x)$ is the degree $n$ Taylor polynomial of the function $f(x) = e^{-x}$ with center $a=0$ . Then the error $E_n(x)$ , i.e., the difference between the polynomial and the function, does not exceed $\frac {f^{(n+1)}(\xi )}{(n+1)!}x^{n+1}$ , where $\xi$ is some unknown point in the range $0 \leq \xi \leq x$ .

In this case one should take $x = 1/3$ and determine how many derivatives are required to make this error estimate less than the given threshold.

This means checking by hand for small numbers of derivatives. For the specific problem at hand, if we approximate $f(x)$ by the Taylor polynomial of degree $n = 2$ , we have $\left | \frac {x^{n+1} }{(n+1)!} f^{(n+1)}(\xi ) \right | = \left | \frac {x^{n+1}}{(n+1)!} \left ( e^{-\xi } \right ) \right | \leq \left | \frac {x^{n+1}}{(n+1)!} \left ( 1 \right ) \right | = \frac {1}{162}$ when $x = 1/3$ .

We conclude that the correct Taylor approximation is $p_n \left ( \frac {1}{3}\right ) = \left (\frac {1}{3}\right )^{0} -1\left (\frac {1}{3}\right )^{1} + \frac {1}{2}\left (\frac {1}{3}\right )^{2} = 1 -\frac {1}{3} + \frac {1}{18} = \frac {13}{18}.$