138_Final SOS

138_Final SOS - MATH138: Exam-AID SOS Prepared by Vincent...

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Unformatted text preview: MATH138: Exam-AID SOS Prepared by Vincent Chan v2chan@math.uwaterloo.ca April 8, 2010 I include numbers after each section heading that correspond to the textbook section headings for your convenience. Proofs of theorems are included in your textbook, and will be omitted. Finally, I strongly recommend going through the examples and exercises in your textbook, as they are very helpful in preparing for your exams and understanding the material. 1 Review Example 1.1. Calculate Solution. We have a product of functions, so we try integration by parts: Then The integral is no easier than what we started with, but we can use integration by parts a second time, to get a recurring integral: Then We solve for our original integral using simple algebra now. 1 WATERLOO SOS E XAM -AID: MATH138 F INAL Example 1.2. Evaluate (Winter 2006, Q1) Solution. This is suitable for a partial fraction decomposition. First notice the degree of the numerator is less than that of the denominator, so we do not need to use polynomial division, and the denominator is already factored. Then suppose Clearing the denominator gives comparing denominators or substituting values , we get and . Then Example 1.3. Determine if the following improper integral is convergent or divergent. (Spring 2002, Q2) Solution. Notice and Then by the Comparison Test, is convergent. Example 1.4. Find the volume of the solid obtained by rotating the region bounded by the curves about the line . Solution. First, we calculate the intersection points. We want ). Washer method: 2 , so that , and so and ( or The outer function is and the inner function is and the inner radius is given by . Then Shell method: The top function is Then WATERLOO SOS E XAM -AID: MATH138 F INAL . However, the outer radius is given by and the bottom function is . However, the radius is given by . Example 1.5. A tank contains 120 litres of water in which 200 grams of salt is dissolved. Brine containing 2 grams of salt per litre is then pumped into the tank at a rate of 3 litres per minute; the well-mixed solution is pumped out at the same rate. Find the concentration of salt in the tank at time . Solution. The change in mass is given by the difference of the rate of inflow and the rate of outflow. The rate is given by the product of the flow rate and the concentration, so that rate in and rate out Then we have the differential equation We can solve this in two ways, as a separable or linear differential equation. For the former, we rearrange as then 3 The initial conditions give WATERLOO SOS E XAM -AID: MATH138 F INAL for separable equations, so that and Then concentration is given by Alternatively, we could have written the differential equation as This is the exponential model, which has solution and we get the same answer as before after applying the initial condition. Finally, we can treat the differential equation as a linear differential equation, writing The integrating factor is so that and we get the same answer as before after applying the initial condition. Example 1.6. Determine whether the following sequence is convergent or divergent. If it is convergent, find its limit. (Winter 2006, Q1) Solution. We will prove the more general result that if for constants , then On one hand, 4 WATERLOO SOS E XAM -AID: MATH138 M IDTERM 9 Limits of Sequences (11.1) D EFINITION 9.1. A sequence is an ordered list of objects (numbers in our case), which we may denote as or has the limit D EFINITION 9.2. A sequence , then or if for every there exists an integer such that if In this case we write or and say for every as is convergent. Otherwise, we say that the sequence , there exists an integer such that if , then is divergent. A special case of this is if In this case, we write and say that the sequence diverges to . Limits of sequences are related to limits of functions in a natural way. T HEOREM 9.3. If and , then . For example, this implies if . We also get the following: T HEOREM 9.4. If and the function T IP. Here is a useful sequence to remember: divergent to infinity for . Moreover, is continuous at , then is convergent if if and Limits of sequences behave nicely: T HEOREM 9.5. If and are convergent sequences and is a constant, then (i) (ii) 34 , divergent for if , and . WATERLOO SOS E XAM -AID: MATH138 M IDTERM (iii) (iv) (v) (vi) if (vii) if and The following is a nice result that allows us to find limits without dealing with the limiting behaviour of the sequence directly. T HEOREM 9.6 [S QUEEZE T HEOREM ]. If for and , then . Finally, another theorem that is quite useful. T HEOREM 9.7. E XAMPLE 21. Determine whether the following sequence is convergent or divergent. If it is convergent, find its limit. S OLUTION. Observe that by a theorem, , so that which limits to by a previous observation, and so E XAMPLE 22. Determine whether the following sequence is convergent or divergent. If it is convergent, find its limit. S OLUTION. We will prove the more general result that if for constants , then 35 WATERLOO SOS E XAM -AID: MATH138 M IDTERM First, consider Then the first inequality following since . Since , we have and the second following since the function and so is increasing for We also have so by the Squeeze Theorem, Now, so by limit arithmetic, In particular, for and , we get that E XERCISE 13. Determine whether the following sequence is convergent or divergent. If it is convergent, find its limit. S OLUTION. Notice both the numerator and the denominator approach infinity as pital’s Rule to the related function and obtain Thus by a theorem, so diverges to infinity. 36 . We apply lHos- WATERLOO SOS E XAM -AID: MATH138 F INAL using properties of logarithm. On the other hand, Since and so , we have By continuity of logarithm, Then by the Squeeze Theorem, In particular, for 2 and , we get that Sequences (cont’d) (11.1) Definition 2.1. Suppose is a sequence. (i) is called increasing if for all . (ii) is called decreasing if for all . (iii) is called non-decreasing if for all . (iv) is called non-increasing if for all . It is called monotonic if any of these properties hold. Definition 2.2. A sequence is bounded above if there is a number for all . Similarly, it is bounded below if there is a number for all such that . If both properties hold, it is bounded. such that Theorem 2.3 [M ONOTONE C ONVERGENCE T HEOREM ]. Every bounded, monotonic sequence is convergent. Actually, this theorem is the combination of two statements: (i) If is a non-decreasing sequence that is bounded above, then it is convergent. (ii) If is a non-increasing sequence that is bounded below, then it is convergent. 5 WATERLOO SOS E XAM -AID: MATH138 F INAL Note that this theorem is not true in the converse: certainly there are sequences which are not monotonic that are convergent, for example . However, boundedness is always a necessary condition for the limit to exist. (Indeed, suppose the limit of a sequence is . Then for , there is an such that if , then . That is, for . Then as there are finitely many terms we have neglected, consider , and we will have for every .) Example 2.4. Determine whether the sequence defined as follows converges or diverges. If it converges, find its limit. for (Spring 2002, Q3) and (Spring 2001, Q2 - modified), (Spring 2000, Q6 - modified) Solution. Making some rough estimates, , and . We can see that the sequence seems to be increasing to . In fact, if the limit does exist, say , . That is, , and factoring, this gives then by limit arithmetic, we have that , so that or . Then a limit of seems like a likely candidate, since we know the sequence begins above and tends to increase. However, this is not enough for a solution! We have not yet proved the limit exists, nor have we proved that should a limit exist, it must necessarily be positive. This is important; we can get situations where the “limit” can be calculated much as above, but it does not even exist. To confirm our guesses, we use mathematical induction to show for all . The base case of is already done, since . If we assume this inequality is true for , then In the second and third steps, we have used the fact that addition by a constant and taking the square root does not affect the inequality, as they are increasing functions. The last step came from our recurrence . Thus, we have proven the inequality for , and so the result holds for relation, all by mathematical induction. This shows that the sequence is increasing and bounded (above), so that the Monotone Convergence Theorem guarantees it has a limit, and since all the terms are at least , we also know the limit is positive. So suppose the limit is . The recurrence relation gives by limit arithmetic. Since , we also have . Then solving for , we get , or , and thus or . Since is positive, we have , by uniqueness of the limit. Aside: The reason we had two possible limits was because we took a square when solving for , which instead, the sequence ignores sign. In fact, if we had defined the recurrence relation by would still have a limit, and this time it would be . However we cannot use the Monotone Convergence Theorem to solve it, so we won’t try it. 6 WATERLOO SOS E XAM -AID: MATH138 F INAL Exercise 2.5. Determine whether the sequence defined as follows converges or diverges. If it converges, find its limit. for Solution. If we na¨vely say the limit is given by ı and use it in our recurrence relation by limit . However, the sequence looks like , which does arithmetic, we get and hence not have a limit. In this case, notice the sequence is not monotonic, so we cannot apply the Monotone Convergence Theorem. Exercise 2.6. Determine whether the sequence defined as follows converges or diverges. If it converges, find its limit. for Solution. If we na¨vely say the limit is given by ı and use it in our recurrence relation by limit arithmetic, we get and hence . However, the sequence looks like , which does not have a limit. In this case, notice the sequence is not bounded, so we cannot apply the Monotone Convergence Theorem. 3 Series (11.2) Definition 3.1. A series is the sum of the terms of an infinite sequence , It is denoted by or We define the nth partial sum to be It should be noted that although these definitions start with an initial term of , we may sometimes begin with if the notation is cleaner, for example, the geometric series which will come shortly. Definition 3.2. If the sequence is convergent, say is said to be convergent and we write The number is called the sum of the series. Notice that Otherwise, the series is said to be divergent. 7 for a real number , then the series WATERLOO SOS E XAM -AID: MATH138 F INAL Tip. Here is a useful series to remember: the geometric series is convergent if If and its sum is , then the geometric series is divergent. Tip. Here is another useful series to remember: the harmonic series is divergent. Some theorems about convergence of series: Theorem 3.3 [D IVERGENCE T EST ]. If the series is convergent, then . Notice the converse to this theorem is false, and the harmonic series is a counterexample. This theorem gives a nice test for divergence however, since it implies that if does not exist or is not , then the series is divergent, hence the name of the theorem. Series, like limits of sequences, have the expected behavior when performing arithmetic on them: Theorem 3.4. If and . Moreover, are convergent series, then so are the series (where is a constant) and (i) (ii) 4 . Convergence Tests and Error Estimates (11.3 - 11.6) Integral Test: Theorem 4.1 [I NTEGRAL T EST ]. Suppose is a continuous, positive, decreasing function on and let . Then the series is convergent if and only if the improper integral is convergent. Note that the sum of the series is not the same as the value of the integral. Tip. Recall the improper integral Test, the p-series is convergent if is convergent if and divergent if harmonic series diverges. 8 and divergent if . Using the Integral . In particular, notice this implies the WATERLOO SOS E XAM -AID: MATH138 F INAL By the method of proof of the Integral Test, we also get an error estimate for certain series. Theorem 4.2 [I NTEGRAL T EST R EMAINDER E STIMATE ]. Suppose , where is a continuous, positive, decreasing function for and is convergent. If the th remainder term is given by , then If we wish for an estimate the sum itself, we may write Comparison Test: Theorem 4.3 [C OMPARISON T EST ]. Suppose that (i) If for all and is convergent, then (ii) If for all and and is divergent, then are series with positive terms. We have: is also convergent. is also divergent. Tip. Our two important series will be very useful when used in conjunction with the Comparison Test (recall the geometric series and the -series). We can use the Integral Test combined with the Comparison Test in order to get an error estimate for other series. In particular, suppose is a series which works well with the integral test, and for every . The th remainder term is given by . Since for every , we have , so getting a bound on using something like the Integral Test will give the same bound on . Limit Comparison Test: Theorem 4.4 [L IMIT C OMPARISON T EST ]. Suppose that where is a finite number and and are series with positive terms. If , then either both series converge or both series diverge. Tip. This theorem will be extremely useful in testing series whose terms are algebraic functions of , so that they are similar to -series. Alternating Series Test: Theorem 4.5 [A LTERNATING S ERIES T EST ]. If the alternating series satisfies 9 WATERLOO SOS E XAM -AID: MATH138 F INAL 1. for all 2. then the series converges. Tip. Here is another useful series to remember: the alternating harmonic series is convergent. We also get an estimate for the remainder of alternating series. Theorem 4.6 [A LTERNATING S ERIES R EMAINDER E STIMATE ]. If series that satisfies is the sum of an alternating (i) (ii) then Not a Test, But Needed (Other Kinds of Convergence): Before we begin the other tests of convergence, we should introduce a new form of convergence. Definition 4.7. A series is absolutely convergent if the series of absolute values is conditionally convergent if the series is convergent, but not absolutely convergent. Theorem 4.8. If a series is convergent. It is absolutely convergent, then it is convergent. Notice the converse is not true, that is, it is possible to be conditionally convergent (and so our definition is not wasted). Indeed, the Alternating Series Test shows that the alternating harmonic series is convergent, while the series of its absolute values is the harmonic series, which is not convergent. Ratio Test: Theorem 4.9 [R ATIO T EST ]. (i) If , then the series is absolutely convergent (and hence convergent). (ii) If (iii) If or , then the series is divergent. , then the Ratio Test is inconclusive, that is, the series 10 can be convergent or divergent. WATERLOO SOS E XAM -AID: MATH138 F INAL The canonical example of (iii) is comparing the -series when and . In both cases, the limit of the ratio of terms is , but the case is divergent and the case is convergent. It should be noted that the case when the limit does not exist will not be covered, but for interest’s sake, there is a version of the Ratio Test that uses rather than , which always exists (and is a term you do not need to know). Tip. If clusive. are rational or algebraic functions of , then , and so the Ratio Test will be incon- Root Test: Theorem 4.10 [R OOT T EST ]. (i) If , then the series is absolutely convergent (and hence convergent). or (ii) If (iii) If , then the series is divergent. , then the Root Test is inconclusive, that is, the series can be convergent or divergent. Tip. It turns out that if in the Ratio Test, then in the Root Test as well. Conversely, if in the Root Test, then in the Ratio Test as well. Thus, if one of these tests gives an inconclusive result (i.e. ), then don’t attempt the other test. As in the Ratio Test, it should be noted that the case when the limit does not exist will not be covered, but for interest’s sake, there is a version of the Root Test that uses rather than , which always exists (and is a term you do not need to know). 5 Strategy for Testing Series (11.7) As in integration, it can be difficult to determine when a series converges, but we have many tools. Also as in integration, there is no set algorithm for solving these kinds of problems, but here is a guide you can use to focus your thoughts. 1. Check if converge. . If not, the Divergence Test immediately tells us that the series cannot 2. Check using known series. If the series has a form similar to a geometric series or a -series, we may be able to manipulate the sum to get precisely these series. If not, we may be able to use one of the Comparison Tests. For example, if the terms are rational functions or algebraic functions (involving roots of polynomials) of , you can try comparing with a -series. 3. If the series is in the form , then the Alternating Series Test may be applicable. 4. Series that involve products, such as factorials and raising to the th power, may be approachable with Ratio Test. A notable exception: if are rational or algebraic functions of , then , and so the Ratio Test will be inconclusive. 5. If is of the form , then the Root Test may yield some useful information. 11 6. If where is easily evaluated and then the Integral Test will be of use. WATERLOO SOS E XAM -AID: MATH138 F INAL is a continuous, positive, decreasing function, Example 5.1. Determine whether the following series converges or diverges. (Winter 2006, Q4) Solution. Notice , so the Divergence Test tells us that the series diverges. Example 5.2. Determine whether the following series converges or diverges. (Winter 2006, Q4) Solution. Notice , so the Divergence Test tells us nothing. However, the terms are algebraic functions of , so we try the Comparison Test with a -series. Notice all of these terms are positive, and as , we have , so by the Comparison Test, converges. To Now, converges by the -series with be precise, we only have converges, whereas our sum starts at . However, when dealing with series, the first finitely many terms never matter. Alternatively, we can try the Limit Comparison Test. We have Now, converges by the -series with , so by the Limit Comparison Test, Example 5.3. Determine whether the following series converges or diverges. (Winter 2006, Q4 - modified) 12 converges. WATERLOO SOS E XAM -AID: MATH138 F INAL Solution. As in the previous example, terms are algebraic functions of , so we try the Comparison Test with a -series. However, the presence of the negative sign will not allow us to compare each term with for . Then as before. Instead, notice Now, converges by the -series with , so by the Comparison Test, converges. To be precise, we only have converges, whereas our sum starts at . However, when dealing with series, the first finitely many terms never matter. Alternatively, we can try the Limit Comparison Test. We have Now, converges by the -series with , so by the Limit Comparison Test, converges. Example 5.4. Determine whether the following series converges absolutely, converges conditionally, or diverges. (Winter 2006, Q6) Solution. It is clear that the terms limit to , so the Divergence Test will tell us nothing. The function is positive, decreasing, and continuous, so the Integral Test may be applied. Using the substitution or by inspection, so this series does not converge absolutely. However, it may still be conditionally convergent; the presence of the factor indicates we should try the Alternating Series Test. The terms are certainly positive, , and the terms are decreasing. Then by the Alternating Series Test, converges. Therefore, this series converges conditionally. Example 5.5. Determine whether the following series converges absolutely, converges conditionally, or diverges. (Fall 2000, Q4 - modified) Solution. The presence of the for odd, . That is, is simply there to trick you. Notice that for even, and , and we have a usual alternating series. It is not trivial to 13 WATERLOO SOS E XAM -AID: MATH138 F INAL see why the terms decrease to , but the fact that we are dealing with products (in the form of a factorial) implies we might be able to try the Ratio Test. so the series converges absolutely. Notice in this case, we do not need to check conditional convergence. (Had we started with using Alternating Series Test to check for convergence instead of absolute convergence, we would have wasted time since absolute convergence implies convergence.) Example 5.6. How many terms of the following series must you sum so that the partial sum differs from the actual sum by no more than ? Solution. The previous example shows that this series satisfies the Alternating Series Test, so by the Alternating Series Remainder Estimate Theorem, the error is given by Then to get an error of at most , we simply need to find such that expression on the left simplifies to , which is approximately . Then as expect around to work, and simple calculation gives works (since . Notice the , we would ). Example 5.7. Determine whether the following series converges or diverges. (Spring 2000, Q1) Solution. We will illustrate several ways of solving this. First, notice is positive, decreasing, and continuous, so the integral test applies. We can compute the integral in two ways: directly by parts, or using and the known integral . For the former method, we try the “1” trick: 14 WATERLOO SOS E XAM -AID: MATH138 F INAL Then and this second term does not converge (this is a little sloppy here, since we should really take a limit to solve an improper integral). Alternatively, we could manipulate the integrand prior to integrating. To illustrate, the indefinite integral can be computed using: which yields the same result as before (the gets absorbed into the constant anyways). The second method involves noticing that if we exponentiate, the logarithms will disappear and the sum becomes a product. To make this precise, consider the partial sums . Since is a continuous function, will converge to the same value as . Calculating, so . The easiest method is to use the logarithm property outlined in the second Integral Test method, but directly on the series. That is, the partial sums give: Example 5.8. Determine whether the following series converges or diverges. Solution. It is not trivial to see why the terms decrease to , but the fact that we are dealing with products (in the form of a factorial) implies we might be able to try the Ratio Test. 15 To handle the WATERLOO SOS E XAM -AID: MATH138 F INAL term, notice binomial expansion gives Then so the series converges absolutely. Notice in this case, we do not need to check conditional convergence. (Had we started with using Alternating Series Test to check for convergence instead of absolute convergence, we would have wasted time since absolute convergence implies convergence.) Exercise 5.9. Determine whether the following series converges absolutely, converges conditionally, or diverges. (Spring 2002, Q3 - modified) ˆ , l’Hospital’s Rule gives Solution. Let us check if the terms go to . Since so by a theorem about sequences, , and thus by another theorem. Then the Divergence Theorem will be inconclusive, and we should begin checking for both convergence and absolute convergence. Of course, if we are lucky enough for it to be absolutely convergent, it is automatically convergent by a theorem. For , we have and diverges (this is the harmonic series starting from the second term). Then by the Comparison Test, diverges, and hence so does . Thus, this series does not converge absolutely. However, it may still be conditionally convergent; the presence of the factor indicates we should try are certainly positive and as proved before, so the Alternating Series Test. The terms we need only check that the terms are decreasing, so consider again the function . Recall a function is decreasing if its derivative is negative, so we calculate This derivative is negative when Alternating Series Test, converges conditionally. , that is, when . Thus, converges, and hence so does is decreasing for . By the . Therefore, this series Exercise 5.10. Determine whether the following series converges absolutely, converges conditionally, or diverges. 16 WATERLOO SOS E XAM -AID: MATH138 F INAL (Spring 2002, Q3) . Indeed, squaring gives Solution. Notice this series converges absolutely by the Geometric Series with . Thus, . Exercise 5.11. Determine whether the following series converges absolutely, converges conditionally, or diverges. (Winter 2002, Q3) Solution. The presence of the is simply there to trick you. Notice that for even, and for odd, . That is, , and we have a usual alternating series. The series does not are positive, converge absolutely, since this is a -series with , but we do have that the terms decreasing, and . Then by the Alternating Series Test, this series converges conditionally. Exercise 5.12. Determine whether the following series converges or diverges. (Spring 2000, Q4) Solution. This looks almost like a -series, except the exponent of is not a constant. Instead of directly for using a -series, we might try using the Comparison Test, except that all we can say is that , or for , which does not help. We could try using the Limit Comparison Test: This limit might not be familiar to you, so we will show this (this question was also on a past Math 138 exam). First, continuity of the exponential gives ˆ This limit can be evaluated using l’Hospital’s Rule, to get Thus, , and so the corresponding sequence result holds as well. Now, (this is the harmonic series), so by the Limit Comparison Test, diverges. 17 diverges WATERLOO SOS E XAM -AID: MATH138 F INAL 6 Power Series (11.8) Definition 6.1. A power series is a series of the form where is a variable and are constants called the coefficients of the series. For each fixed , this power series is a series of constants which we can test for convergence. More generally, a series of the form is called a power series centered at or a power series about . As a side note, we use the definition , even when for are , and so the power series always converges when else the series converges. Theorem 6.2. For a power series . Notice that in this case, all the terms . The natural question is to ask where , there are only three possibilities: (i) The series converges only when . (ii) The series converges for all . (iii) There is a positive number such that the series converges if and diverges if . Definition 6.3. The number in (iii) is called the radius of convergence. By convention, in (i) and in (ii). The interval of convergence of a power series is the interval consisting of all points for which the series converges. In (i), this interval is the singleton . In (ii), the interval is . Case (iii) is more complicated, as convergence may or may not hold at the endpoints . You must check the endpoints separately, and your final interval of convergence will be one of four possibilities: To determine radii and intervals of convergence, you should appeal to the Ratio or Root Tests to get a radius of convergence, then test the endpoints (if necessary) using other methods. Example 6.4. Find the radius and interval of convergence of the following series. Solution. Comparing consecutive terms in the series, By the Ratio Test, the series converges if , so that the radius of convergence is , and diverges if . That is, we have convergence for all and the interval of convergence is . 18 WATERLOO SOS E XAM -AID: MATH138 F INAL Example 6.5. Find the radius of convergence of the following series. (Winter 2001, Q3) Solution. Comparing consecutive terms in the series, By the Ratio Test, the series converges if , and diverges if . That is, we have convergence for and divergence for , so that the radius of convergence is . Example 6.6. Find the radius and interval of convergence of the following series. (Spring 2006, Q7) Solution. Comparing consecutive terms in the series, , and diverges if . That is, we have By the Ratio Test, the series converges if convergence for and divergence for , so that the radius of convergence is . To find the interval of convergence, we must check what happens at the endpoints and . At , the series becomes which diverges (the harmonic series). At , the series becomes which converges (the alternating harmonic series). Thus, the interval of convergence is . 19 WATERLOO SOS E XAM -AID: MATH138 F INAL Exercise 6.7. Find the radius and interval of convergence of the following series. (Spring 2002, Q4) Solution. Comparing consecutive terms in the series, By the Ratio Test, the series converges if , and diverges if . That is, we have convergence for and divergence for , so that the radius of convergence is . To find the interval of convergence, we must check what happens at the endpoints and . At , the series becomes which converges upon using the Limit Comparison Test with the -series where series becomes . At , the which converges since the previous case shows this series converges absolutely. Thus, the interval of convergence is . Exercise 6.8. Find the radius and interval of convergence of the following series. (Winter 2006, Q5) Solution. Comparing consecutive terms in the series, By the Ratio Test, the series converges if , and diverges if . That is, we have convergence for and divergence for , so that the radius of convergence is . To find the interval of convergence, we must check what happens at the endpoints and . At , the series becomes 20 WATERLOO SOS E XAM -AID: MATH138 F INAL which converges upon using the Alternating Series Test. At , the series becomes which diverges by the -test with Thus, the interval of convergence is 7 . . Functions as Power Series (11.9 - 11.11) Recall the geometric series gives as a power series, at least on a particular interval for . This gives a representation of the function of convergence. Power series serves as a useful representation as it allows for extremely simple differentiation and integration (which plays a role in approximations and differential equations). In particular, we have the following useful theorem, which allows term-by-term differentiation and integration. Theorem 7.1. If the power series has radius of convergence is differentiable and integrable on the interval , then the function defined by and (i) (ii) The radii of convergence of both of these power series . This can be written as (i) (ii) hence the name “term-by-term differentiation and integration”. Using this theorem, we can get various power series representations of functions in the form of deriva, where may be a “nice” function of . Are there other kinds of functions which tives and integrals of have power series representations? How can we find the coefficients in the power series? It turns out that for nice enough functions, they will have power series representations, but we will leave this question for now. For the second question, we have the following theorem. Theorem 7.2. If has a power series representation at , say 21 WATERLOO SOS E XAM -AID: MATH138 F INAL for , then its coefficients are given by Then this series is called the Taylor series of f at a (or about a or centered at a). For the special case Maclaurin series. , the series is known as a To go back to the first question as to whether a function has a power series representation, we use the following two theorems, which relies on the following definition. Definition 7.3. We define the nth-degree Taylor polynomial of f at a to be The remainder of the Taylor series is given by Theorem 7.4. If for , then where . is the th-degree Taylor polynomial of is equal to the sum of its Taylor series for Theorem 7.5 [TAYLOR ’ S I NEQUALITY ]. If Taylor series satisfies the inequality for for . Using these, we can develop some useful Maclaurin series. 22 at and . , then the remainder of the Example 7.6. Find the Taylor series for around WATERLOO SOS E XAM -AID: MATH138 F INAL and determine its radius of convergence. Solution. Notice that Now, we know the power series expansion of , for . Thus, if , we have Then integrating term-by-term, we have To determine , we have the condition that at so . Thus, the Taylor series about , is given by , and the series disappears at , (This may not look like a usual power series due to the constant floating outside of the sum, but think of it as a part of the term. That is, .) To find its radius of convergence, we could go through the argument of using the Ratio Test again, or remark that the power series representation of that we applied had a radius of convergence of , and as we are integrating, the radius of convergence does not change. That is, we have . Thus, the radius of convergence is . , so that Example 7.7. If Solution. Recall that since , find the values of and has a power series about , say . , we have that Then By the previous example, Note that so that corresponds to the term . Thus, , not the term . However, the coefficient of the Likewise, to get we are looking for the coefficient of the term coefficient is given by , and so 23 , so that or term is , . Then the WATERLOO SOS E XAM -AID: MATH138 F INAL Example 7.8. Find the Maclaurin series for Q11) and determine its radius of convergence. (Spring 2006, Solution. The power series representation of is given by with convergence everywhere, so that Then for a constant , that is, for . Integrating does not change the radius of convergence, so this series converges for . Example 7.9. Write the number as the sum of a convergent series. (Spring 2006, Q11) Solution. From the previous example, we have a series representation given by Then the constant cancels when evaluating as a definite integral, giving Example 7.10. Let be the 9th-degree Taylor polynomial of involved in approximating by . (Spring 2006, Q11) at . Estimate the bound Solution. We may be tempted to use Taylor’s inequality, but the derivatives of are not wellbounded. Instead, notice that the previous example showed that this number has a power series which satisfies the alternating series test, and so by the remainder estimate, the error is bounded by: ( which is about , but on the exam the above representation should suffice.) 24 WATERLOO SOS E XAM -AID: MATH138 F INAL Example 7.11. Find the Taylor series for and state its radius of convergence. Solution. We have not specified the center of the series here, so we will consider two possibilities. Since , we can use directly to get Notice this is a series centered at , convergent everywhere. Alternatively, we may be interested in a Maclaurin series, that is, centered at . Then we can write which also converges everywhere. Exercise 7.12. Find the Maclaurin series for Solution. Recall the Maclaurin series for and state its radius of convergence. (Spring 2002, Q5) is given by which converges for all . Then and the radius of convergence is Exercise 7.13. If for . and is the 11th-degree Taylor polynomial, find a constant such that . (Spring 2002, Q5) Solution. By the previous example, This is then multiplied by a power series (easily verified). Notice that gives , which satisfies the Alternating Series Test for this series, by degree argument. Then by the 25 WATERLOO SOS E XAM -AID: MATH138 F INAL remainder estimate for the Alternating Series Test, we have so works. Exercise 7.14. Approximate the integral with error less than . (Spring 2002, Q5) Solution. By the previous exercise, Integrating term by term gives Notice that this is an alternating series and satisfies the hypothesis of the remainder estimate (clearly the terms are positive and decrease to ). Then the error when using the th-degree Taylor polynomial is given by If we want the error to be less than , we need , and we notice that gives precisely. Then we require the 2nd-degree Taylor polynomial to approximate the integral with error less than , that is, 8 Parametric Equations (10.1 - 10.2) Definition 8.1. In , suppose the coordinates are functions of another variable , called the parameter by the equations , called parametric equations. Each determines a point , and varying makes this point trace out a curve called a parametric curve. Example 8.2. A particle travels on the path direction of motion, and points . (Winter 2006, Q2) for 26 . Sketch this curve, indicating the WATERLOO SOS E XAM -AID: MATH138 F INAL Solution. Notice , so this is an ellipse with semi-major axis (along -axis) and semi-minor axis (along -axis). The curve begins at and moves in a clockwise direction since begins to incorrespond to the points and respectively. crease and begins to decrease. The points In the last example, we were fortunate enough to have a familiar identity, but this will not always be the case. To aid in curve sketching, we consider tangent lines. The slope of a tangent to a curve in is given by if , and otherwise the slope is undefined (because the tangent is vertical). In particular, the curve is horizontal when and vertical when . Combining information about the slopes of the tangents with other information such as intercepts makes sketching curves much easier. Example 8.3. Consider the curve given by the parametric equations , for . Find the intercepts of the curve and the points where the tangent to the curve is horizontal or vertical, and include them on a sketch of the curve. (Winter 2002, Q7) Solution. To find the -intercept(s), we are solving , so that or . Solving for , we have and , so the -intercepts are and . To find the -intercept(s), we are solving , so that , and we have already considered this point. We have and . The curve is horizontal when , that is, when and hence . We saw that this was one of the -intercepts, at . The curve is vertical when , that is, when and hence . At both of these points, and , so the curve is vertical at . We should check the slope(s) at the other -intercept (which is the -intercept, remember) to get more at information about the curve. Then we are considering the slope . We have so 27 WATERLOO SOS E XAM -AID: MATH138 F INAL This tells us that there are two tangents at , so the curve intersects itself, with one tangent having negative the slope of the other. Finally, we should check the endpoints of the curve given by the bounds on . At , we have and . Putting this all together, we can sketch the curve. Recall the arc-length formula: if the curve is in the form of then the length of the curve is given by , for (and is continuous), Similarly, one can use the substitution rule to obtain the arc-length formula for parametric curves. Theorem 8.4. Suppose a curve and are continuous on given by is described by the parametric equations and is traversed exactly once as increases from Notice this equation reduces to our previous equation if Example 8.5. A particle travels on the path (Winter 2006, Q2) Solution. We have and , so that for . Then 28 with , where to . Then the length of is is a function of . . Find the length of the curve. WATERLOO SOS E XAM -AID: MATH138 F INAL Exercise 8.6. Consider the curve given by the parametric equations , for . Find the length of the curve. (Winter 2002, Q7) Solution. We have and . Then 29 Proofs Absolute Convergence => Convergence Comparison Test Geometric Series Monotone Sequence Theorem Ratio Test ...
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