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# Math 31B Week 7 Notes Math 31B

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LECTURE 15: IMPROPER INTEGRALS (CONTINUED) Friday, February 12 Recall that an integral is improper if either one or more of the limits of integration are inﬁnite, or if the integral is over a ﬁnite interval ra,bs and the integrand has an inﬁnite discontinuity on that interval. We considered the former last time, in this lecture we’ll consider the latter. type 2: discontinuous integrands The following deﬁnite integrals are examples of type 2 improper integrals, » 1 1 » 3 1 ? . 0 x 0 x ▯ 1 Notice that the integrand of the ﬁrst integral has an inﬁnite dis- continuity at an endpoint of the interval r0,1s (at x ▯ 0), while the integrand of the second integral has an inﬁnite discontinuity in the interior of the interval r0,3s (at x ▯ 1). We handle improper integrals of this type in manner similar to the way we handled integrals over inﬁnite integrals. Deﬁnition 0.1. paqIf fpxq is continuous on ra,bq and discontinuous at b, then » » b t fpxqdx ▯ lim▯ fpxqdx a tÑb a provided this limit is a ﬁnite real number. pbq If fpxq is continuous on pa,bs and discontinuous at a, then » » b b fpxqdx ▯ lim fpxqdx a tÑa ▯ t provided this limit is a ﬁnite real number. 1 ‡ The improper integrals b fpxqdx is called convergent if the corre- a sponding limit exists and divergent if the limit does not exist. ‡ pcq If fpxq is discontinuous at a c b and bothcfpxqdx ‡b a and c fpxqdx are convergent, we deﬁne »b » c » b fpxqdx ▯ fpxqdx ▯ fpxqdx. a a c provided this limit is a ﬁnite real number. Let’s consider some examples. »1 1 Example 1. Determine whether the integral ? dx is conver- 0 x gent or divergent. As mentioned before, the integrand in this case has a discontinuity at x ▯ 0. Therefore, we have »1 » 1 1 1 ? x dx ▯ lim▯ ? xdx 0 tÑ0 t ▯ ? ▯1 ▯ lim 2 x ▯ tÑ0▯ ▯t ? ▯ 2 ▯ lim 2 t ▯ 2. tÑ0▯ In this case the integral converges to 2. Let’s consider a another, similar example » 1 1 Example 2. Determine whether the integral x dx is convergent 0 or divergent. The integrand in this case also has a discontinuity atx ▯ 0. There- fore, we have » » 1 1 1 1 dx ▯ lim dx 0 x tÑ0▯ t x 1 ▯ ▯ tÑ0▯ln|x| ▯ t ▯ lim ▯ln|t| ▯ 8. tÑ0▯ In this case the integral diverges. 2 The above examples give the are of the following shaded regions: y ▯ ?1 x 1 1 y ▯ x 1 While the functions ?1 and 1 look similar, the shaded area at the x x top is 2, while the shaded area on the bottom is inﬁnite! As in the previous lecture, this begs the question - for which real numbers p ¡ 0 is the integral » a 1 dx 0 xp convergent? We’ve just shown that this integral is not convergent if p ▯ 1, so let us assume that p ▯ 1. First we know that since p ▯ 1 we have » » a 1 a 1 p dx ▯ li▯ p dx 0 x tÑ0 t x ▯ 1 ▯ ▯ lim x1▯p ▯ tÑ0▯ 1 ▯ p t 1▯p 1▯p a t ▯ ▯ lim ▯ . 1 ▯ p tÑ0 1 ▯ p Now, if p 1, then 1 ▯ p ¡ 0, which implies 1▯p t tÑ0m▯ 1 ▯ p ▯ 0. 3 On the other hand, if p ¡ 1, then 1 ▯ p 0, which implies 1▯p lim t ▯ 8. tÑ0▯ 1 ▯ p We’ve just proven the following. Theorem 0.2. Let a ¡ 0. Then $ » ’ a 1▯p a 1 & if p 1 pdx ▯ 1 ▯ p 0 x % diverges if p ¥ 1. Let’s look at one more example before me move on. Notice that in the previous two examples, if one was to blindly integrate without any consideration of the discontinuities of the function, they would have arrived at the same conclusions we came to. Unfortunately, we will not always be so lucky. In the next example we consider the second function mentioned at the beginning of the lecture. This function has an inﬁnite discontinuity occurring in the interior of the interval over which the function is being integrated, and we’ll see that if one does not take this discontinuity into consideration, you will get an incorrect answer. »3 1 Example 3. Determine whether the integral x ▯ 1dx converges 0 or diverges. There is a discontinuity at x ▯ 1, so we have » 3 » 1 »3 1 1 1 x ▯ 1dx ▯ x ▯ 1dx ▯ x ▯ 1dx, 0 0 1 provided both of the above integrals are convergent. Let’s consider the ﬁrst integral to determine if it is in fact convergent. » » 1 1 t 1 dx ▯ lim▯ dx 0 x ▯ 1 tÑ1 0x ▯ 1 ▯ ▯t ▯ lim ln|x ▯ 1|▯ tÑ1▯ ▯0 ▯ ▯ tÑ1▯ ln|t ▯ 1| ▯ 8 ‡1 ‡3 Since 0x▯1 dx diverges, the whole integral0 x▯1dx diverges - we don’t even have to consider the second integral! 4 Note that if we would have just taken the integral with out con- sidering the discontinuity, we would have gotten » 3 ▯3 1 ▯ dx ▯ ln|x ▯ 1| ▯ ▯ lnp2q. 0 x ▯ 1 0 1 a comparison test for improper integrals It may be the case that we cannot determine the value of an improper integral, but we would like to know if it diverges or converges. The following theorem gives us a simple way of determining this in certain cases. Theorem 1.1 (Comparison Theorem) . Suppose that f and g are continuous functions on the interval pa,bq with fpxq ¥ gpxq ¥ 0 for all x P pa,bq. Then we have: » » b b 1. If fpxqdx is convergent, then gpxqdx is convergent. a a »b » b 2. If gpxqdx is divergent, then fpxqdx is divergent. a a This theorem holds if a ▯ ▯8 and/or b ▯ 8. » 8 2 Example 4. Show that e▯x dx is convergent. 0 This is a famous integral because it is related to the Normal distribution (the most common probability distribution), and it is difﬁcult to integrate. Now, in order to show that this integral converges we ﬁrst note that » 8 2 » 1 2 »8 2 e▯x dx ▯ e ▯x dx ▯ e ▯x dx. 0 0 1 We’ll see why one would want to break the integral up like this in a moment. Note that the ﬁrst integral is the deﬁnite integral of 5 a continuous function over the ﬁnite interval r0,1s, and as such this integral is just some ﬁnite real number. Therefore, to show ‡8 2 the integral 0 e▯x dx converges, it sufﬁces to show the integral ‡8 ▯x2 1 e dx converges. The reason we broke the integral up the way we did is because we will use the fact that ▯x ▯x2 e ¥ e ¥ 0 ‡or al2 x P r1,8q. By the comparison ‡heorem, the integral 8 e▯x dx must converge if the integral8 e▯x dx converges. 1 1 We have » » 8 ▯x t ▯x e dx ▯ lim e dx 1 tÑ8 1 ▯t ▯x▯ ▯ ltÑ8▯e ▯ ▯ 1 ▯t ▯1 ▯1 ▯ lim ▯e ▯ e ▯ e . tÑ8 » 8 1 ▯ ex Example 5. Show that dx is divergent. 1 x Note that 1 ▯ ex 1 ¥ ¥ 0 x x ‡or all x P r1,8q. By the comparison th‡orem, the integral 8 1▯e▯x dx must converge if the integral8 1dx diverges. 1 x 1 x We have » » 8 1 t 1 dx ▯ lim dx 1 x tÑ8 1 x ▯t ▯ ▯ ltÑ8ln|x| ▯ 1 ▯ lim ln|t| ▯ 8. tÑ8 6 LECTURE 14: IMPROPER INTEGRALS Wednesday, February 10 For this lecture we’re returning to integration. Up to this point, every deﬁnite integral that we’ve considered has been over a ﬁnite interval with an integrand that is ﬁnite on this interval. In this lecture, we’ll deﬁne deﬁnite integrals over inﬁnite intervals or deﬁnite integrals over ﬁnite intervals in which the integrand tends to 8. Integrals of this type are called improper integrals. type 1: infinite intervals We begin by considering integrals over an inﬁnite interval. Let’s look at a speciﬁc example. Suppose we want to know the area of 1 under the curve y ▯ x2 that lies above the x-axis, and to the left of the vertical line x ▯ 1. That is, the area of the shaded region below: y ▯ 2 x 1 Let us denote this area by » 8 1 x2 dx. 1 Keep in mind this ‡s just notaiton at this point, we still need to deﬁne what we mean by 8 1 dx. To do so, we will apply a common theme 1 x2 of calculus - approximate what you want to know, then take the limit. For any real number R ¡ 1, the deﬁnite integral »R 1 dx 1 x2 1 gives the are of the shaded region below: 1 y ▯ x2 1 R This deﬁnite integral gives an approximation of the shaded area over the inﬁnite intervalr1,8q. In fact, we see that the larger thatR is, the better the approximation: y ▯ 2 x 1 R 1 y ▯ x2 1 R So taking the limit of these approximations as R Ñ 8 will give us ‡8 1 the shaded area denoted by 1 x2 dx, that is, we should deﬁne » » 8 1 R 1 dx ▯ lim dx, 1 x2 RÑ8 1 x2 assuming that this limit exists. In this case we see that ▯ ▯▯ ▯ » 8 1 » R 1 1▯ 1 dx ▯ lim dx ▯ lim ▯ ▯ ▯ lim 1 ▯ ▯ 1. 1 x2 RÑ8 1 x2 RÑ8 x 1 RÑ8 R We follow this intuition in deﬁning integrals over inﬁnite intervals. 2 Deﬁnition 0.1. Fix a real number a. ‡t paq If afpxqdx exists for every t ¥ a, then »8 » t fpxqdx ▯ limtÑ8 fpxqdx a a provided this limit is a ﬁnite real number. ‡ pbq If afpxqdx exists for every t ⁄ a, then t »a » a fpxqdx ▯ tÑ▯8 fpxqdx ▯8 t provided this limit is a ﬁnite real number. ‡ ‡ The improper integrals8 fpxqdx and a fpxqdx are called conver- a ▯8 gent if the corresponding limit exists and divergent if the limit does not exist. ‡8 ‡a pcq If botha fpxqdx and ▯8 fpxqdx are convergent, we deﬁne » 8 »8 »a fpxqdx ▯ fpxqdx ▯ fpxqdx. ▯8 a ▯8 provided this limit is a ﬁnite real number. For pcq, any real number a can be used. Let’s look at some examples: » 8 Example 1. Determine whether the integral 1 dx is conver- 1 x gent or divergent. Note that the deﬁnite integral » t ▯t 1 dx ▯ ln|t| ▯ ▯ ln|t|, 1 x ▯ 1 is deﬁned for all t ¥ 1. Therefore, by deﬁnition, we have » 8 1 » t 1 dx ▯ lim dx ▯ lim ln|t| ▯ 8. 1 x tÑ8 1 x tÑ8 In this case the integral is divergent. 3 Comparing the above example to the example we looked at in the beginning of the lecture is interesting. Consider the following shaded regions: 1 y ▯ x 1 1 y ▯ x 1 While the functions 1{x and 1{x look similar, and the area under their graphs above the x-axis and to the left of 1 looks similar, the shaded area at the top is inﬁnite, while the shaded area on the bottom is just 1! This begs the question, for which real numbers p ¡ 0 is the integral » 8 1 x pdx 1 convergent? We’ve just shown that this integral is not convergent if p ▯ 1, so let us assume that p ▯ 1. First we know that since p ▯ 1 we have » ▯ t 1 1 ▯ t1▯p 1 1 dx ▯ x1▯p ▯ ▯ ▯ ▯ pt1▯p ▯ 1q, 1 xp 1 ▯ p 1 1 ▯ p 1 ▯ p 1 ▯ p ‡t 1 so we know that the integral 1 x dx exists for all t ¥ 1. Therefore, by deﬁnition we have » 8 »t 1 1 1 1▯p p dx ▯ lim p dx ▯ lim pt ▯ 1q. 1 x tÑ8 1 x tÑ8 1 ▯ p 4 Now, if p 1, then 1 ▯ p ¡ 0, which implies lim 1 pt▯p ▯ 1q ▯ 8. tÑ8 1 ▯ p On the other hand, if p ¡ 1, then 1 ▯ p 0, which implies lim 1 pt1▯p▯ 1q ▯ 1 . tÑ8 1 ▯ p 1 ▯ p We record a slight generalization of what we’ve just shown in the following theorem. Theorem 0.2. Let a ¡ 0. Then $ » ’ a1▯p 8 1 & if p ¡ 1 p dx ▯ 1 ▯ p a x % diverges if p ⁄ 1. » 0 x Example 2. Determine whether the integral xe dx converges ▯8 or diverges. Using integration by parts, with u ▯ x du ▯ dx x x dv ▯ e dx v ▯ e we see that for all t ⁄ 0 » ▯ » ▯ ▯ 0 ▯0 0 ▯0 ▯0 xe dx ▯ xe x▯ ▯ e dx ▯ xe x▯ ▯ ex▯ ▯ ▯ te ▯ 1 ▯ e .t t ▯t t ▯t ▯t ‡0 x Hence the integral txe dx exists, and we have »0 »0 x x t t xe dx ▯ tÑ▯8 xe dx ▯ tÑ▯8 p▯te ▯ 1 ▯ e q ▯8 t ▯ ▯ lim te ▯ lim e ▯ lim 1. tÑ▯8 tÑ▯8 tÑ▯8 Now, the latter to limits are easy to determine, but the ﬁrst is indeterminate of type 0 ▯ 8. So we employ l’Hospital’s rule, t t 1 tÑ▯8 te ▯ tÑ▯8 e▯t ▯ tÑ▯8 ▯e ▯t ▯ 0. Hence, lim p▯te ▯ e ▯ 1q ▯ ▯ 1. tÑ▯8 5 »8 Example 3. Evaluate 1 dx. ▯8 x ▯ 1 Recall that » 8 » a »8 1 dx ▯ 1 dx ▯ 1 dx x ▯ 1 x ▯ 1 x ▯ 1 ▯8 ▯8 a for any real number a. Furthermore, »a » t 1 dx and 1 dx s x ▯ 1 a x ▯ 1 exists for all s ⁄ a and t ¥ a. To make things easier, let us take a ▯ 0. Then we have » 8 »0 »8 1 dx ▯ 1 dx ▯ 1 dx ▯8 x ▯ 1 ▯8 x ▯ 1 0 x ▯ 1 » » 0 1 t 1 ▯ lim 2 dx ▯ lim 2 dx sÑ▯8 ▯s x ▯ 1 ▯ tÑ8 ▯ x ▯ 1 ▯ ▯0 ▯ ▯ lim tan▯1pxq ▯ lim tan▯1pxq▯ sÑ▯8 ▯ tÑ8 ▯ s 0 ▯1 ▯1 ▯ ▯ lisÑ▯8n psq ▯tÑ8m tan ptq ▯ ▯ ▯▯ ▯ ▯ ▯ ▯ ▯ ▯ 2 2 Note that this integral can be interpreted as the area between the 2 curve y ▯ 1{px ▯ 1q and the x-axis, that is, the shaded area below: y ▯ 21 x ▯1 That is, the area of the shaded above, extending inﬁnitely in both directions, is exactly ▯! 6 TAYLOR POLYNOMIALS: ESTIMATING ERROR Recall Taylor’s theorem, which gives us a way to determine the accuracy of approximating a functionf by the n th Taylor polynomial. Theorem 0.1 (Taylor’s Theorem). Let f be deﬁned on the interval pa,bq and let c be a point in this interval phere we allow a ▯ ▯8 or b ▯ 8q. st pn▯1q Suppose the pn ▯ 1q derivative f exists on pa,bq and let T n denote the n th Taylor polynomial of f centered at c. Then for each x ▯ c in pa,bq there is some ▯ between c and x such that fpn▯1q p▯q fpxq ▯ Tnpxq ▯ px ▯ cq n▯1 . pn ▯ 1q! Let’s look at several examples using this theorem. th Example 1. Let T benthe n Taylor polynomial of fpxq ▯ cospxq centered at c ▯ 0. Find a value of n for which the following inequality is satisﬁed: |cosp0.1q ▯ T p0.1q| ⁄ 10 ▯7 . n Solution: By Taylor’s theorem, we know that for each n ¥ 0 we have ▯ ▯ ▯fpn▯1qp▯ q ▯ |cosp0.1q ▯ T n0.1q| ▯ ▯ n p0.1qn▯1 ▯ ▯ pn ▯ 1q! ▯ where ▯ in some real number between0 and 0.1. Here we denote this real number by ▯ ranher than ▯ to emphasize the fact that it is not necessarily the same real number for each n. Now, we don’t know what ▯ is, onny that it exists. Therefore, we shouldn’t hope to have an exact value for the error, but rather a bound. Speciﬁcally, ▯ ▯ ▯fpn▯1qp▯ q ▯ p0.1qn▯1 p0.1qn▯1 ▯ n p0.1qn▯1 ▯ ▯ ▯ |f▯1p▯ n| ⁄ K (1) ▯ pn ▯ 1q! ▯ pn ▯ 1q! pn ▯ 1q! 1 where K is an upper bound on|f pn▯1qpxq| over the intervalp0,0.1q, preferably the least upper bound. The reason this inequality holds is because we know that ▯ P p0,0.1q. Therefore, if f pn▯1qpxq ⁄ K n pn▯1q for all x P p0,0.1q, we know that f p▯nq ⁄ K. Let’s determine the bound (1) for each n. When n ▯ 0 we have fp0▯1pxq ▯ ▯ sinpxq. Over the interval p0,0.1q we know that p1q |f pxq| ▯ |sinpxq| ⁄ sinp0.1q. So the bound (1) becomes p0.1q▯1 0.1 ▯7 K ▯ sinp0.1q ▯ 0.00998 ¡ 10 . pn ▯ 1q! 1! Since the possible error between the value of the |cosp0.1q ▯ ▯7 T0p0.1q| is greater than 10 , we see that n ▯ 0 will not work. So we try n ▯ 1. When n ▯ 1 we have fp1▯1pxq ▯ ▯ cospxq Over the interval p0,0.1q we know that p2q |f pxq| ▯ |cospxq| ⁄ cosp0q ▯ 1. So the bound (1) becomes n▯1 2 p0.1q p0.1q ▯7 K pn ▯ 1q! ▯ 2! ▯ 0.005 ¡ 10 . Since the possible error between the value of the |cosp0.1q ▯ T p0.1q| is greater than 10 ▯7, we see that n ▯ 1 will not work. 2 So we try n ▯ 2. 2 When n ▯ 2 we have fp2▯1qpxq ▯ sinpxq Over the interval p0,0.1q we know that |f3qpxq| ▯ |sinpxq| ⁄ sinp0.1q. So the bound (1) becomes p0.1qn▯1 p0.1q ▯7 K ▯ sinp0.1q ▯ 0.0000166 ¡ 10 . pn ▯ 1q! 3! Since the possible error between the value of the |cosp0.1q ▯ T2p0.1q| is greater than 10 ▯7, we see that n ▯ 2 will not work. So we try n ▯ 3. When n ▯ 3 we have fp3▯1qpxq ▯ cospxq Over the interval p0,0.1q we know that |fp4pxq| ▯ |cospxq ⁄ cosp0q ▯ 1. So the bound (1) becomes p0.1qn▯1 p0.1q K ▯ ▯ 0.0000041 ¡ 10 ▯7. pn ▯ 1q! 4! Since the possible error between the value of the |cosp0.1q ▯ T3p0.1q| is greater than 10 ▯7, we see that n ▯ 3 will not work. So we try n ▯ 4. When n ▯ 4 we have p4▯1q f pxq ▯ ▯ sinpxq Over the interval p0,0.1q we know that p4q |f pxq| ▯ |sinpxq| ⁄ sinp0.1q. So the bound (1) becomes n▯1 5 K p0.1q ▯ sinp0.1q p0.1q ▯ 0.00000000831 10 ▯7 . pn ▯ 1q! 5! We’ve ﬁnally found an n that works! What this means is that the possible error between the value of ▯7 the |cosp0.1q ▯ T4p0.1q| is less than 0.00000000831 10 . So the answer to the question is 4. 3 Example 2. Let T be the n th Taylor polynomial of fpxq ▯ lnpxq n centered at c ▯ 1. Find a value of n for which the following inequality is satisﬁed: ▯4 |lnp1.3q ▯ n p1.3q| ⁄ 10 . Solution: We’ll repeat our analysis as in the previous example. When n ▯ 0 we have fp0▯1pxq ▯ 1 . x Over the interval p1,1.3q we know that |f1qpxq| ▯ 1 ⁄ 1 ▯ 1. x 1 By Taylor’s theorem we have 0▯1 p0.3q ▯4 |lnp1.3q ▯ 0 p1.3q| ⁄ p0 ▯ 1q! ▯ 0.3 ¡ 10 . Since the possible error between the value of the|lnp1.3q▯T p0.3q| is greater than 10▯4 , we see that n ▯ 0 will not work. So we try n ▯ 1. When n ▯ 1 we have p1▯1q 1 f pxq ▯ ▯ 2 x Over the interval p1,1.3q we know that p2q 1 1 |f pxq| ▯ 2 ⁄ 2 ▯ 1. x 1 By Taylor’s theorem we have p0.3q1▯1 |lnp1.3q ▯ 1 p1.3q| ⁄ ▯ 0.045 ¡ 10 ▯4 . p1 ▯ 1q! Since the possible error between the value of the|lnp1.3q▯T p1.3q| ▯4 is greater than 10 , we see that n ▯ 1 will not work. So we try n ▯ 2. 4 When n ▯ 2 we have 2 f2▯1qpxq ▯ x3 Over the interval p1,1.3q we know that p3q 2 2 |f pxq| ▯ 3 ⁄ 3 ▯ 2. x 1 By Taylor’s theorem we have 2▯1 p0.3q ▯4 |lnp1.3q ▯ T2p1.3q| ⁄ 2 ▯p2 ▯ 1q! ▯ 0.009 ¡ 10 . Since the possible error between the value of the|lnp1.3q▯T 21.3q| is greater than 10▯4, we see that n ▯ 2 will not work. So we try n ▯ 3. When n ▯ 3 we have fp3▯1qpxq ▯ ▯ 6 x4 Over the interval p1,1.3q we know that 6 6 |f4pxq| ▯ ⁄ ▯ 6. x4 14 By Taylor’s theorem we have p0.3q3▯1 ▯4 |lnp1.3q ▯ 3 p1.3q| ⁄ 6 ▯ ▯ 0.002025 ¡ 10 . p3 ▯ 1q! Since the possible error between the value of the|lnp1.3q▯T 31.3q| is greater than 10▯4, we see that n ▯ 3 will not work. So we try n ▯ 4. When n ▯ 4 we have p4▯1q 24 f pxq ▯ 5 x Over the interval p1,1.3q we know that |f5qpxq| ▯ 24 ⁄ 24 ▯ 24. x5 15 5 By Taylor’s theorem we have 4▯1 p0.3q ▯4 |lnp1.3q ▯ T4p1.3q| ⁄ 24 ▯ p4 ▯ 1q! ▯ 0.000486 ¡ 10 . Since the possible error between the value of the|lnp1.3q▯T p1.4q| is greater than 10 ▯4 , we see that n ▯ 4 will not work. So we try n ▯ 5. When n ▯ 5 we have p5▯1q 120 f pxq ▯ ▯ 6 x Over the interval p1,1.3q we know that p6q 120 120 |f pxq| ▯ 6 ⁄ 6 ▯ 120. x 1 By Taylor’s theorem we have p0.3q▯1 ▯4 |lnp1.3q ▯ T5p1.3q| ⁄ 120 ▯ ▯ 0.0001215 ¡ 10 . p5 ▯ 1q! Since the possible error between the value of the|lnp1.3q▯T p1.5q| is greater than 10 ▯4 , we see that n ▯ 5 will not work. So we try n ▯ 6. When n ▯ 6 we have 720 fp6▯1qpxq ▯ ▯ x7 Over the interval p1,1.3q we know that |f p7qpxq| ▯ 727 is decreas- x ing. Therefore, ▯ ▯ ▯720 ▯ 720 ▯ 7 ▯ ⁄ 7 ▯ 720. x 1 By Taylor’s theorem we have p0.3q6▯1 ▯4 |lnp1.3q ▯ T6p1.3q| ⁄ 720 ▯ ▯ 0.00003124285 10 . p6 ▯ 1q! This works! So the answer is n ▯ 6. 6 ? Example 3. Let T n be the n thTaylor polynomial of fpxq ▯ x centered at c ▯ 1. Find a value of n for which the following inequality is satisﬁed: ? ▯6 | 1.3 ▯ T n1.3q| ⁄ 10 . Solution: When n ▯ 0 have p0▯1q 1 f pxq ▯ 2 x . Over the interval p1,1.3q we know that ▯ ▯ ▯ 1 ▯ 1 ▯ ? ▯ ⁄ . 2 x 2 Therefore, by Taylor’s theorem we have ? 0▯1 | 1.3 ▯ T p1.3q| ⁄ 1 ▯0.3q ▯ 0.15 ¡ 10 ▯6 . 0 2 p0 ▯ 1q! So n ▯ 0 will not work, let’s try n ▯ 1. When n ▯ 1 we have p1▯1q 1 f pxq ▯ ▯ 3{2 4x Over the interval p1,1.3q we know that ▯ ▯ ▯ 1 ▯ 1 ▯ 3{2▯ ⁄ . 4x 4 By Taylor’s theorem we have ? 1▯1 | 1.3 ▯ T p1.3q| ⁄ 1▯p0.3q ▯ 0.01125 ¡ 10 ▯6. 1 4 p1 ▯ 1q! So n ▯ 1 will not work, let’s try n ▯ 2. When n ▯ 2 we have p2▯1q 3 f pxq ▯ 8x5{2. 7 Over the interval p1,1.3q we know that ▯ ▯ ▯ 3 ▯ 3 ▯ ▯ ⁄ . 8x5{2 8 By Taylor’s theorem we have ? 2▯1 | 1.3 ▯ T p1.3q| ⁄ 3 ▯0.3q ▯ 0.0016875 ¡ 10 ▯6. 2 8 p2 ▯ 1q! Since n ▯ 2 does not work, let’s try n ▯ 3. When n ▯ 3 we have fp3▯1qpxq ▯ ▯ 15 . 16x 7{2 Over the interval p1,1.3q we know that ▯ ▯ ▯ 15 ▯ ⁄ 15 . ▯ 16x 7{2▯ 16 By Taylor’s theorem we have ? 15 p0.3q 3▯1 ▯6 | 1.3 ▯ T3p1.3q| ⁄ ▯ ▯ 0.000316 ¡ 10 . 16 p3 ▯ 1q! Since n ▯ 3 does not work, let’s try n ▯ 4. When n ▯ 4 we have p4▯1q 105 f pxq ▯ 9{2. 32x Over the interval p1,1.3q we know that ▯ ▯ ▯ 105 ▯ 105 ▯ 9{2▯ ⁄ . 32x 32 By Taylor’s theorem we have ? 105 p0.3q 4▯1 | 1.3 ▯ T 41.3q| ⁄ ▯ ▯ 0.000066 ¡ 10 ▯6 . 32 p4 ▯ 1q! Since n ▯ 4 does not work, let’s try n ▯ 5. 8 When n ▯ 5 we have fp5▯1qpxq ▯ ▯ 945 . 64x 11{2 Over the interval p1,1.3q we know that ▯ ▯ ▯ 945 ▯ 945 ▯▯ 64x 11{2▯ ⁄ 64 . By Taylor’s theorem we have ? 945 p0.3q5▯1 | 1.3 ▯ T5p1.3q| ⁄ ▯ ▯ 0.000015 ¡ 10 ▯6 . 64 p5 ▯ 1q! Since n ▯ 5 does not work, let’s try n ▯ 6. When n ▯ 6 we have 10395 fp6▯1qpxq ▯ 13{2. 128x Over the interval p1,1.3q we know that ▯ ▯ ▯ 10395 ▯ 10395 ▯ ▯ ⁄ . 128x 13{2 128 By Taylor’s theorem we have 6▯1 ? 10395 p0.3q ▯6 | 1.3 ▯ T 61.3q| ⁄ 128 ▯ p6 ▯ 1q! ▯ 0.0000035 ¡ 10 . Since n ▯ 6 does not work, let’s try n ▯ 7. When n ▯ 7 we have p7▯1q 135135 f pxq ▯ ▯ 256x 15{2. Over the interval p1,1.3q we know that ▯ ▯ ▯ 135135 ▯ 135135 ▯ 256x 15{2▯ ⁄ 256 . By Taylor’s theorem we have ? 135135 p0.3q 7▯1 | 1.3 ▯ T7p1.3q| ⁄ ▯ ▯ 0.00000085 10 ▯6 . 256 p7 ▯ 1q! So n ▯ 7 works! 9 th x Example 4. Let T n be the n Taylor polynomial of fpxq ▯ e centered at c ▯ 0. Find a value of n for which the following inequality is satisﬁed: ▯0.1 ▯6 |e ▯ Tnp▯0.1q| ⁄ 10 . Solution: When n ▯ 0 have fp0▯1qpxq ▯ e . x Over the interval p▯0.1,0q we know that x e | ⁄ 1. Therefore, by Taylor’s theorem we have p0.1q0▯1 |e▯0.1▯ T 0▯0.1q| ⁄ ▯ 0.1 ¡ 10 ▯6. p0 ▯ 1q! So n ▯ 0 will not work, let’s try n ▯ 1. When n ▯ 1 have p1▯1q x f pxq ▯ e . Over the interval p▯0.1,0q we know that ex | ⁄ 1. Therefore, by Taylor’s theorem we have 1▯1 ▯0.1 p0.1q ▯6 |e ▯ T1p▯0.1q| ⁄ p1 ▯ 1q! ▯ 0.005 ¡ 10 . So n ▯ 1 will not work, let’s try n ▯ 2. When n ▯ 2 have fp2▯1qpxq ▯ e . x Over the interval p▯0.1,0q we know that ex | ⁄ 1. Therefore, by Taylor’s theorem we have ▯0.1 p0.1q2▯1 ▯6 |e ▯ T 2▯0.1q| ⁄ ▯ 0.000167 ¡ 10 . p2 ▯ 1q! So n ▯ 2 will not work, let’s try n ▯ 3. 10 When n ▯ 3 have fp3▯1qpxq ▯ e .x Over the interval p▯0.1,0q we know that |ex| ⁄ 1. Therefore, by Taylor’s theorem we have 3▯1 |e▯0.1▯ T p▯0.1q| ⁄ p0.1q ▯ 0.000004167 ¡ 10 ▯6. 3 p3 ▯ 1q! So n ▯ 3 will not work, let’s try n ▯ 4. When n ▯ 4 have p4▯1q x f pxq ▯ e . Over the interval p▯0.1,0q we know that x |e | ⁄ 1. Therefore, by Taylor’s theorem we have ▯0.1 p0.1q4▯1 ▯6 |e ▯ T 4▯0.1q| ⁄ ▯ 0.0000000834 10 . p4 ▯ 1q! So n ▯ 4 works! rd Example 5. Determine the accuracy of the 3 Taylor polynomial T3of fpxq ▯ lnp1 ▯ 2xq centered at c ▯ 1 over the intervalp0.5,1.5q. Solution: Since we want to understand the accuracy of the 3 rd th Taylor polynomial, we need to determine the 4 derivative of f: p1q 2 ▯1 f pxq ▯ ▯ 2p1 ▯ 2xq 1 ▯ 2x fp2pxq ▯ ▯ 4p1 ▯ 2xq ▯2 fp3pxq ▯ 8p1 ▯ 2xq ▯3 p4q ▯4 24 f pxq ▯ ▯ 24p1 ▯ 2xq ▯ ▯ 4 p1 ▯ 2xq 11 Since ▯ ▯ ▯p4q ▯ 24 3 ▯ pxq▯ ▯ p1 ▯ 2xq 4 ⁄ 2 , for all x P p0.5,1.5q, Taylor’s theorem tells us that ▯ ▯ |lnp1 ▯ 2xq ▯ T pxq| ▯ ▯fp4qp▯ q▯▯ 1 ▯ |x| 3 3 4! 3 1 ⁄ ▯ ▯ p1.5q 2 24 ▯ 0.31640625 for all x P p0.5,1.5q. That is, the function f and its 3 rd Taylor polynomial T wil3 differ by at most 0.31640625 over the interval p0.5,1.5q. 12

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