# Review Quiz 1

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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.

1. If , , , , and , calculate

Solution: Using L'Hopital's rule we have

2. 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.

3. 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

4. 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

5. 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 .

6. Find the third-degree 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.

7. 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

8. 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 .

9. 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

10. 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)

11. 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 .