7 8. What is the fourth‐order Taylor polynomial for , about ?Solution: We have andso that , , . Thus,
8 9. What is the fourth‐order Taylor polynomial for , about ?
9 10. Find the Taylor polynomial of third‐order for , using:.Solution: We haveso;.
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., ;, ;, ;, ;, .Solution:The polynomial iswith remainderThis can be bounded above, for all , byThe polynomial iswith remainderWe can't bound this for all , because of the potential division by zero.The polynomial iswith remainderThis can be bounded above, for all , byThe polynomial is the same as in (b), of course,with remainderFor all this can be bounded byThe polynomial iswith remainderThis can be bounded above, for all , byObviously, 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 ., ;, ;, ;, ;, .Solution:The remainder here isfor . Therefore, we haveSimple manipulations with a calculator then show thatbutTherefore the desired Taylor polynomial isThe remainder here isfor . Therefore, we haveSimple manipulations with a calculator then show thatbutTherefore the desired Taylor polynomial is, .Solution: The remainder is nowand makes the error small enough., .
12 13. Repeat the above, this time with a desired accuracy of .
13 14. Sincewe 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 isso to get 100 decimal places of accuracy for , we requirethus, 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 thatThis 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 haveDefine as the approximation generated by using an term Gregory series to approximate and an term Gregory series for . Then we havewhere 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 thatandUsing properties of logarithms, these becomeandFor , 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 ontaking terms in the series for , terms in the series for , and terms in the series for , then we are led to the inequalitiesandFor , 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 asandUse 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 haveLooking more carefully at the remainder term, we see that it is given byTherefore,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 thatis 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., ; ., ; is unbounded, since and is possible., ; ., ; ., . .
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 thatBut 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 to observe that if is negative, then we haveandThe proof is completed as in the text.
26 27. Prove Theorem 1.6, providing all details.
27 28. Let , be given, , and let , . Then, use the Discrete Average Value Theorem to prove that, for any function ,for some .Solution:We can't apply the Discrete Average Value Theorem to the problem as it is posed originally, so we have to manipulate a bit. Definethenand now we can apply the Discrete Average Value Theorem to finish the problem.
28 29. Discuss, in your own words, whether or not the following statement is true: “The Taylor polynomial of degree is the best polynomial approximation of degree to the given function near the point .”
1.2 ERROR, APPROXIMATE EQUALITY, AND ASYMPTOTIC ORDER NOTATION
Exercises:
1. Use Taylor's Theorem to show that
1 2. Use Taylor's Theorem to show that for sufficiently small.Solution: We can expand the cosine in a Taylor series asIf we substitute this into and simplify, we getso that we havewhere . Therefore, .
2 3. Use Taylor's Theorem to show thatfor sufficiently small.Solution:We have, from Taylor's Theorem, with ,for some between 0 and . Sincefor all sufficiently small, the result follows. For example, we havefor all .
3 4. Use Taylor's Theorem to show thatfor sufficiently small.Solution:This time, Taylor's Theorem gives us thatfor some between 0 and . Thus, for all such that ,where .
4 5. Show that
5 6. Recall the summation formulaUse this to prove thatHint: What is the definition of the notation?
6 7. Use the above result to show that 10 terms () are all that is needed to computeto within absolute accuracy.Solution:The