CHAPTER 1
INTRODUCTORY CONCEPTS AND CALCULUS REVIEW

1.1 BASIC TOOLS OF CALCULUS

1. 2. What is the third-order Taylor polynomial for , about ?

   Solution: We have and

   so that , , . Therefore

2. 3. What is the sixth-order Taylor polynomial for , using ? Hint: Consider the previous problem.
3. 4. Given that

   for , where is between and 0, find an upper bound for , valid for all , that is independent of and .

4. 5. Repeat the above, but this time require that the upper bound be valid only for all .

   Solution: The only significant difference is the introduction of a factor of in the denominator:

5. 6. Given that

   for , where is between and 0, find an upper bound for , valid for all , that is independent of and .

6. 7. Use a Taylor polynomial to find an approximate value for that is accurate to within .

   Solution: There are two ways to do this. We can approximate and use , or we can approximate and use . In addition, we can be conventional and take , or we can take in order to speed convergence.

   The most straightforward approach (in my opinion) is to use a Taylor polynomial for about . The remainder after terms is

   We quickly have that

   and a little playing with a calculator shows that

   but

   So we would use

   To fourteen digits, , and the error is , much smaller than required.

7. 8. What is the fourth-order Taylor polynomial for , about ?

   Solution: We have and

   so that , , . Thus,

8. 9. What is the fourth-order Taylor polynomial for , about ?
9. 10. Find the Taylor polynomial of third-order for , using:
   1. .
      Solution: We have

      so

   2. ;
   3. .
10. 11. For each function below construct the third-order Taylor polynomial approximation, using , and then estimate the error by computing an upper bound on the remainder, over the given interval.
    1. , ;
    2. , ;
    3. , ;
    4. , ;
    5. , .

    Solution:

    1. The polynomial is with remainder This can be bounded above, for all , by
    2. The polynomial is with remainder We can't bound this for all , because of the potential division by zero.
    3. The polynomial is with remainder This can be bounded above, for all , by
    4. The polynomial is the same as in (b), of course, with remainder For all this can be bounded by
    5. The polynomial is with remainder This can be bounded above, for all , by Obviously, this is not an especially good approximation.
11. 12. Construct a Taylor polynomial approximation that is accurate to within , over the indicated interval, for each of the following functions, using .
    1. , ;
    2. , ;
    3. , ;
    4. , ;
    5. , .

    Solution:

    1. The remainder here is for . Therefore, we have Simple manipulations with a calculator then show that but Therefore the desired Taylor polynomial is
    2. The remainder here is for . Therefore, we have Simple manipulations with a calculator then show that but Therefore the desired Taylor polynomial is
    3. , .
    4. Solution: The remainder is now and makes the error small enough.
    5. , .
12. 13. Repeat the above, this time with a desired accuracy of .
13. 14. Since

    we can estimate by estimating . How many terms are needed in the Gregory series for the arctangent to approximate to 100 decimal places? 1,000? Hint: Use the error term in the Gregory series to predict when the error gets sufficiently small.

    Solution: The remainder in the Gregory series approximation is

    so to get 100 decimal places of accuracy for , we require

    thus, we have to take terms. For 1,000 places of accuracy we therefore need terms.

    Obviously, this is not the best procedure for computing many digits of !

14. 15. Elementary trigonometry can be used to show that

    This formula was developed in 1706 by the English astronomer John Machin. Use this to develop a more efficient algorithm for computing . How many terms are needed to get 100 digits of accuracy with this form? How many terms are needed to get 1,000 digits? Historical note: Until 1961, this was the basis for the most commonly used method for computing to high accuracy.

    Solution: We now have two Gregory series, thus complicating the problem a bit. We have

    Define as the approximation generated by using an term Gregory series to approximate and an term Gregory series for . Then we have

    where is the remainder in the Gregory series. Therefore,

    To finish the problem we have to apportion the error between the two series, which introduces some arbitrariness into the problem. If we require that they be equally accurate, then we have that

    and

    Using properties of logarithms, these become

    and

    For , these are satisfied for , . For , we get , . Changing the apportionment of the error doesn't change the results by much at all.

15. 16. In 1896, a variation on Machin's formula was found:

    and this began to be used in 1961 to compute to high accuracy. How many terms are needed when using this expansion to get 100 digits of ? 1,000 digits?

    Solution: We now have three series to work with, which complicates matters only slightly more compared to the previous problem. If we define based on

    taking terms in the series for , terms in the series for , and terms in the series for , then we are led to the inequalities

    and

    For , we get , , and ; for we get , , and .

    Note: In both of these problems a slightly more involved treatment of the error might lead to fewer terms being required.

16. 17. What is the Taylor polynomial of order 3 for , using ?

    Solution: This is very direct:

    so that

17. 18. What is the Taylor polynomial of order 4 for , using ? Simplify as much as possible.
18. 19. What is the Taylor polynomial of order 2 for , using ?
19. 20. What is the Taylor polynomial of order 3 for , using ? Simplify as much as possible.

    Solution: We note that , so we have (using the solution from the previous problem)

    The polynomial is its own Taylor polynomial.

20. 21. Let be an arbitrary polynomial of degree less than or equal to . What is its Taylor polynomial of degree , about an arbitrary ?
21. 22. The Fresnel integrals are defined as

    and

    Use Taylor expansions to find approximations to and that are accurate for all with . Hint: Substitute into the Taylor expansions for the cosine and sine.

    Solution:

    We will show the work for the case of , only. We have

    Looking more carefully at the remainder term, we see that it is given by

    Therefore,

    A little effort with a calculator shows that this is less than for ; therefore the polynomial is

22. 23. Use the Integral Mean Value Theorem to show that the "pointwise" form (1.3) of the Taylor remainder (usually called the Lagrange form) follows from the "integral" form (1.2) (usually called the Cauchy form).
23. 24. For each function in Problem 11, use the Mean Value Theorem to find a value such that

    is valid for all , in the interval used in Problem 11.

    Solution: This amounts to finding an upper bound on over the interval given. The answers are as given below.

    1. , ; .
    2. , ; is unbounded, since and is possible.
    3. , ; .
    4. , ; .
    5. , . .
24. 25. A function is called monotone on an interval if its derivative is strictly positive or strictly negative on the interval. Suppose is continuous and monotone on the interval , and ; prove that there is exactly one value such that = 0.

    Solution: Since is continuous on the interval and , the Intermediate Value Theorem guarantees that there is a point where , i.e., there is at least one root. Suppose now that there exists a second root, . Then . By the Mean Value Theorem, then, there is a point between and such that

    But this violates the hypothesis that is monotone, since a monotone function must have a derivative that is strictly positive or strictly negative. Thus we have a contradiction, thus there cannot exist the second root.

    A very acceptable argument can be made by appealing to a graph of the function.

25. 26. Finish the proof of the Integral Mean Value Theorem (Theorem 1.5) by writing up the argument in the case that is negative.

    Solution: All that is required is...