- The first few terms of the Maclaurin series for are
- Plug in and subtract :
- Multiply by , neglect all but the dominant term, and conclude
Various exercises relating to the application of Taylor Series to other problems of interest.
- First combine logarithms to simplify a bit:
- Write out the first few terms of the Maclaurin series for and substitute in the appropriate expression involving :
- Multiply by , neglect all but the dominant term (as to conclude
- Compute the first few terms of the Taylor series of numerator and denominator centered at :
- Neglect all but the dominant terms in numerator and denominator. Note that dominant here means as . Then take the limit :
- The remainder formula says that for some point which depends on and is somewhere in the interval . The upper bound for the magnitude of on this interval is .
- For which values of do we have (Check manually for small values of ; use the fact that is between and .)
- The Maclaurin series of the function converges conditionally at to . 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 to an error strictly less than ? Answer: Taylor polynomial used should have degree or greater.
- A similar but distinct strategy would be to compute instead because we know . Evaluating the Maclaurin series at and doing a little simplification, we see from the above series that (Once again, reindex if your answer does not already start at .)
- The presence of a factor exponential in suggests that this second series converges to much slowerslightly slowerslightly fastermuch faster than the first series. Of the two series, then, the firstsecond series presents a more efficient way to compute numerically.
- Guided by the example above, we will use the Maclaurin series First, compute the derivatives of . Find a pattern which holds for all : When is between and , the largest value of the -st derivative is what?
- Using this upper bound for the -st derivative, we know that (use the upper bound and the remainder formula.)
- For which values of is the expression you just found less than ? Check by hand for smallish values of to find the smallest one which works.
Use Taylor series to estimate the value of to within an error of at most . (Hints will not be revealed until you choose a response.)
Use Taylor series to estimate the value of to within an error of at most .