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

Compute the limit
• The first few terms of the Maclaurin series for $\arctan x$ are
• Plug in $x = \frac {1}{n}$ and subtract $1/n$:
• Multiply by $n^3$, neglect all but the dominant term, and conclude
Compute the limit
• First combine logarithms to simplify a bit:
• Write out the first few terms of the Maclaurin series for $\ln (1+x)$ and substitute in the appropriate expression involving $n$:
• Multiply by $n$, neglect all but the dominant term (as $n \rightarrow \infty )$ to conclude
Compute the limit
• Compute the first few terms of the Taylor series of numerator and denominator centered at $x = 1$:
• Neglect all but the dominant terms in numerator and denominator. Note that dominant here means as $x \rightarrow 1$. Then take the limit $x \rightarrow 1$:
Use the remainder formula for Taylor series to determine which partial sums of the series differ from $e$ by at most $1/100$?
• The remainder formula says that 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 (Check manually for small values of $N$; use the fact that $e$ is between $2$ and $3$.)
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: (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 (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 $\ln 2$ much slowerslightly slowerslightly fastermuch faster than the first series. Of the two series, then, the firstsecond 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 differ from $\ln (3/2)$ by at most $10^{-2}$?
• Guided by the example above, we will use the Maclaurin series First, compute the derivatives of $- \ln (1-x)$. Find a pattern which holds for all $N$: When $x$ is between $0$ and $1/3$, the largest value of the $(N+1)$-st derivative is what?
• Using this upper bound for the $(N+1)$-st derivative, we know that (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.

### Sample Quiz Questions

Use Taylor series to estimate the value of to within an error of at most $1/900$. (Hints will not be revealed until you choose a response.)

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

Use Taylor series to estimate the value of to within an error of at most $1/162$.

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