Math 2300 Section 005 – Calculus II – Fall 2010
Quiz – Friday, November 12, 2010
We will be having a review in class every day until the next midterm exam on Wednesday of next week. For this quiz, you are given 10 minutes to decide which problems you think you can solve and which ones you want to see solved during our review session today.

If , , , , and , calculate
Solution: Using L'Hopital's rule we have

Find the first four nonzero terms of the Taylor series about 0 for the function
Solution: Notice that
Since this is a binomial series with we will expand this series and then integrate it term by term to obtain the series that we need. We have
Now integrate term by term and find
Since we're only looking for the first four terms, we're done.

Find the Taylor polynomials of degree approximating for near 0.
Solution: This is actually one of the Taylor series that you should remember. In any case, we first find the derivatives using the chain rule. We'll denote and then
But, notice that all of these derivatives evaluated at zero are . Also, since , the constant term of is zero. (Thus, we write the Taylor polynomials as a sum starting at since the first degree term is the first that will appear in the expansion.) So, we form the th degree Taylor polynomial as

Use a fourth degree Taylor approximation for for near 0 to evaluate the limit
Solution: We know that the fourth degree Taylor polynomial for is
Using this in the limit we find

Find the first four nonzero terms of the Taylor series about 0 for the function .
Solution: Just take the Taylor series for and replace the by .

Find the thirddegree Taylor polynomial for about .
Solution: Remember that the first four terms make up . We discussed the Taylor polynomials of polynomials a long time ago. The result really shouldn't come as any surprise. We have
Then
That is, the th degree Taylor polynomial (and all subsequent higher degree Taylor polynomials) of an degree polynomial is just the polynomial itself.

Find the first four terms of the Taylor series for the given function about .
Solution: Notice that this is a binomial series
So, the first four terms are given as

By recognizing the series as a Taylor series evaluated at a particular value of , find the sum of the following convergent series.
Solution: Notice that this is just the Taylor series for centered at zero with .

Use the Taylor series expansion of to approximate .
Solution: We know that
Thus, if we can find the Taylor series for this function, we can integrate it and get back to . We can perform a substitution and find
Then, we integrate and find
So, we now have a Taylor series for around . Approxmating is then a simple trick. We recall that
Using our Taylor series we then have

Find the first four terms of the Taylor series for about .
Solution: This can be done directly. The derivatives are
We evaluate each derivative at and find (some reducing happens in the first couple of terms)

Find the radius and interval of convergence for the Taylor series of around .
Solution: Whenever you want to find the radius and interval of convergence of a Taylor series, you're really just using the fact that a Taylor series is a power series. So, we use the ratio test. We know that the Taylor series for around is given as
That is, a general term of the series is given as
Thus, by the ratio test, the series will converge whenever
That final step may seem a bit confusing. Why in the world would we rewrite as ? The reason is, as I've mentioned before, that provides a convenient way to say “the distance between and .” So represents the distance between zero and . Saying just says “the distance between and is less than one.” Now, by the ratio test, we have that the given Taylor series converges when , and so the radius of convergence is 1.
We now find the interval of convergence. It will be one of the following intervals:
We need to plug in and to the series and determine convergence individually at these points. At we have
does not converge. So, we do not include in the interval of convergence. At we have
converges by the alternating series test. Thus, the interval of convergence is .