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Chapter Ten APPROXIMATING FUNCTIONS In Section ??, for , we found the sum of a geometric series as a function of the common ratio : Viewed from another perspective, this formula gives us a series whose sum is the function "! $#%'& ! ( $# . We now work in the opposite direction, starting with a function and looking for a series of simpler functions whose sum is that function. We first use polynomials, which lead us to Taylor series, and then trigonometric functions, which lead us to Fourier series. Taylor approximations use polynomials, which may be considered the simplest functions. They are easy to use because they can be evaluated by simple arithmetic, unlike transcendental functions, such as )'* and +-, . Fourier approximations use sines and cosines, the simplest periodic functions, instead of polynomials. Taylor approximations are generally good approximations to the function locally (that is, near a specific point), whereas Fourier approximations are generally good approximations over an interval. 446 Chapter Ten APPROXIMATING FUNCTIONS 10.1 TAYLOR POLYNOMIALS In this section, we see how to approximate a function by polynomials. Linear Approximations We already know how to approximate a function using a degree 1 polynomial, namely the tangent line approximation given in Section 4.8 : .0/2143657.0/2893:;.=<>/>893?/@1AB893?C 1DE8 The tangent line and the curve share the same slope at . As Figure 10.1 suggests, the tangent 1 8 line approximation to the function is generally more accurate for values of close to . True value J4K GVL Approximate value of J4K GVL Tangent line W H JNSTK FML K GUQFNL H P GQRF O IH J4K FML J4K FNL I I F G G Figure 10.1: Tangent line approximation of J4K GVL for G near F 8(DYX We first focus on 1RD[X. The tangent line approximation at Taylor approximation at , or as follows: Taylor Polynomial of Degree 1 Approximating 1DZX \! $# is referred to as the first for near 0 .0/@1=35^]`_a/21436D7.0/2XN3:(.=<T/2XN3b1 We/@1=now look at approximations by polynomials of higher degree. Let us compare approxima36D[d?eMfV1 1D[X tions c near by linear and quadratic functions. /21436D^d?eNf91 Example 1 Approximate c Solution The tangent line at 1RD^X , with 1%DXVC XMl in radians, by its tangent line at is just the horizontal line g c If we take 1 , then c /21436D[d?eNf915jhMk for . , as shown in Figure 10.2, so 1 near 0. /2XmC XNln3`DdoeMfp/2XVC XMlM3`DXVC qnqMrsCoCtCuk which is quite close to the approximation c Dih 1D[X doeMf915ih . Similarly, if 1DjAvXmCwh /xAvXmCwhy30D[d?eNfo/zAvXVChp3`D^XmC qMqMl{CtCoC , then 447 10.1 TAYLOR POLYNOMIALS d?eNf91R5jh is close to the approximation . However, if 1RD^XVC | , then c /2XmC |}3DdoeMfp/2XVC |N3`D[XVC qM~VhCoCtCuk d?eMfV15h so the approximation is less accurate. The graph suggests that the farther a point X away from , the worse the approximation is likely to be. is p G P Q 1 G Qs Figure 10.2: Graph of p G and its tangent line at G Quadratic Approximations To get a more accurate approximation, we use a quadratic function instead of a linear function. /@143D^doeMf91 1 X Example 2 Find the quadratic approximation to c Solution To ensure that the quadratic, , is a good approximation to c at , we require doeMf91 that and the quadratic have the same value, the same slope, and the same second derivative at 1RD^X ]'n/>XM3D />XM3 ] < />XM3D ] < < /2XN3D < /2XN3 < < />XM3 c c c . That is, we require , , and . We take the quadratic polynomial ] /@1=36D[:;s_1$:; 1 k for near . /@1=3DdoeMf91 ]'M/2143 and determine , _ , and . Since /@1=36D76{:(s_?1$:( ] ] < /@1=36D7{_6:(~M ] we have < < /@1=36D7~n UD[] /2XN3D s_vD[] < /2XN3D ~n D[] 1DX < < />XM3D 1 and 1 < /@143DAf-ws1 c < < /@143DAd?eNf91k c /2XN3DdoeMfVXDjh c /@1=36DdoeMf91 c UDjh so < /2XM3DjAfzvXD^X c c s_vD^X < < /2XN3DAd?eNfVXDjAhMk DjA _ C Consequently, the quadratic approximation is doeMf91R5^] /@1=36Dih:X?1"A ~ h 1 DihsA 1 ~ k for 1 near 0. doeMf915]M/@1=3 Figure 10.3doeMsuggests that the 1 quadratic approximation is better than the linear f915E] _ /@1=3 approximation for near 0. Let’s compare the accuracy of the two] approximations. ]`_a/@1=3Dh 1 1BDXVC | doeMft/2XVC |N3DXmC qN~9hCtCoC /2XmC |}3vDXVC qM~X Recall that for all . At , we have and , so the quadratic approximation is a significant improvement over the linear approximation. The magnitude of the error is about 0.001 instead of 0.08. 448 Chapter Ten APPROXIMATING FUNCTIONS ¡¥ K G9L Q t G Q O Q G Qs P G Figure 10.3: Graph of p and its linear, ¡¥ K G9L K GVLQ£¢u ¤ ¡ , and quadratic, K GVL ¡ , approximations for G near 0 Generalizing the computations in Example 2, we define the second Taylor approximation at . 1RD^X \! $# Taylor Polynomial of Degree 2 Approximating /@1=36D7.0/2XM3:;. .0/214365[] . < /2XM3x1$: < < /2XN3 ~ for near 0 1 Higher-Degree Polynomials 1D¦X In a small interval around , the quadratic approximation to a function is usually a better approximation than the linear (tangent line) approximation. However, Figure 10.3 shows that the 1 quadratic can still bend away from the original function for large . We can attempt to fix this by .0/2143 using an approximating polynomial of higher degree. Suppose that we approximate a function 1 for near 0 by a polynomial of degree § : .0/21435[]¨/21436D7 :( _ 1$:;o1 :^oty:;¨}© _ 1 ktCoCtC?k nk{_k6 ¨}© _ ¨ :(¨91 ¨ C We need.0to find the values of the constants: . To do this, we] require that ¨ /@143 /@1=3 the function and each of its first at the § derivatives agree with those of the polynomial 1D[X 1RD[X point . In general, the more derivatives there are that agree at , the larger the interval on which the function and the polynomial remain close to each other. Dª To see how to find the constants, let’s take § as an example 1 .0/214365^]«N/@1=36D[:;s_1$:; Substituting 1%D^X Differentiating so substituting gives ]'«N/@143 1RDX yields .0/>XM3D[] « /2XN3D[ ]« < /@143D^s_:;~n shows that . < /2XN3D] :;«o1 C 1$:(ªM«o1 k « < />XM3D7{_aC Differentiating and substituting again, we get ] which gives « < < /@1=36D7~Uhp . < < /2XM3D[] :(ªt~Unhy«?1k « < < />XM36D[~Unhynk « C 449 10.1 TAYLOR POLYNOMIALS so that ] The third derivative, denoted by «< < < . D < < /2XN3 C ~Uh , is ]« < < < /@1=36D^ªp~Unhp«nk so . < < < /2XM3D^] « < < < />XM3D[ªt~Unhy«nk and then . D « < < < /2XN3 C ªt~Unh ]¬M/2143 You can imagine a similar calculation starting with ¬® which would give . />XM3 D ¬ , using the fourth derivative . ¬® , k |tªt~UMh and so on. Using factorial notation,1 we write these expressions as . «D Writing . ¨ ® < < < /2XM3 ªV¯ ¬® . k°¬D /2XN3 C |±¯ . for the §=²2³ derivative of , we have, for any positive integer § . ¨´D ¨ ® 1D[X C ¯ § So we define the §=²>³ Taylor approximation at /2XN3 : Taylor Polynomial of Degree µ Approximating .0/@1=365^] ¨ ¨ for near 0 /@1=3 D[.0/>XM3:;. ] "! $# < />XM3x1$: . < < /2XN3 ~9¯ 1 : . < < < />XM3 ªm¯ 1 « . : ¬® /2XM3 |m¯ /2143 We call1RD^X the Taylor polynomial of degree § centered at . about 1 1D^X ¬ :^oty: . ¨ ® § /2XN3 ¯ .0/@1=36D^f-s1 Construct the Taylor polynomial of degree 7 approximating the function . 1D¶·ª Compare the value of the Taylor approximation with the true value of at . Solution We have f-ws1 < /@143D doeMf91 . . < < < /@143D¹Ad?eMf}1 ¬® /@143D ® .wº /@143D ® .w» .w¼ 1 Recall . < < /@143D¸A%f-1 . . .0/2XM3D giving ® f-ws1 doeMf91 .wº .w» /@143DjAd?eNf}1k .w¼ Âu¾t¿Â , and Èp¾t¿Â . X < < < /2XM3DAh ¬® /@143D¸A%f-1 that ½M¾o¿%½MÀ½ÁÃÂ-ÄMÅxÅxÅÇÆ0Åu . In addition, h < < /2XM3D . . X < /2XM3D . /2XM3D ® X /2XM3D ® ® /2XM3D ¨ , or the Taylor polynomial Example 3 .0/@143D 1 h X /2XM3DiAhnC for 1 near 0. 450 Chapter Ten APPROXIMATING FUNCTIONS Using these values, we see that the Taylor polynomial approximation of degree 7 is f-1%5^] ¼ 1 /21436D[XÉ:hst1$:(X « 1 D^1\A .wÌ ® ªV¯ º 1 : ~9¯ A l9¯ 1 ¼ Ë ¯ « 1 Ahs k ªm¯ 1 for ¬ 1 :(X near 0 14º :^hs |m¯ :(X lV¯ 1±» Ê ¯ Ahs 14¼ Ë ¯ C /2XM3D^X fzs1 Notice that since , the seventh and eighth Taylor approximations to are the same. In Figure 10.4 we show the graphs of the sine function and the approximating polynomial of 1 1 near 0. They are indistinguishable where is close to 0. However, as we look at values degree 7 for 1 of farther away from 0 in either direction, the two graphs move apart. To check the accuracy of this fz/@¶·ªM3D7Í ª}·~D[XVC r ÊnÊ XM~Mla|vCtCoCzC approximation numerically, we see how well it approximates ¡=Ï K GVL G bÐÒÑ Q ¡Ï K GVL Qs G Figure 10.4: Graph of bÐÓÑ G Î bÐÒÑ G and its seventh degree Taylor polynomial, ¶·aªDihnC X| Ë htq Ë Ê Ï K GVL ¡ , for G near CtCoC When we substitute into the polynomial approximation, we obtain XVC r ÊMÊ XN~9htªvCoCoCk which is extremely accurate—to about four parts in a million. /21436D^d?eMfV1 r 1 ] ¼ /2¶·aªN3D Example 4 Graph the Taylor polynomial of degree approximating c Solution We find the coefficients of the Taylor polynomial by the method of the preceding example, giving doeMf91R5[] ] Ì 1 /@1=36DihsA ~V¯ : ¬ 1 1±» |±¯ A Ê ¯ : for 1±Ì rV¯ near 0. C /2143 1 Figure 10.5 shows that Ì ] is/@14close to the ·cosine function for a larger interval of -values than 3DhsAB1 ~ the quadratic approximation in Example 2 on page 447. ¡=Ô K GVL ¡4Ô K GVL t G Q ¡ Figure 10.5: Example 5 ¡ Ô K GVL t G G Qs K GVL approximates p Construct the Taylor polynomial of degree 10 about G ¡ K G9L better than 1RD^X ¡ K GVL for G near for the function .0/@143D[ÕyÖ . 10.1 TAYLOR POLYNOMIALS Solution .0/2XM3Djh ÕyÖ 451 ÕyÖ ® ® derivatives are equal WeÕyhave . Since the derivative of is . equal to , all the . higher-order DØhnk~VkoCtCoCukohpX Ö Ù /@1=3sDÕyÖ Ù />XM3{DÕ Dh to . Consequently, for any × , and . Therefore, the Taylor polynomial approximation of degree 10 is given by Õ Ö 5]`_b}/@1=36Dih:1$: 1 ~9¯ « 1 : ªm¯ ¬ 1 : :^ooy: |±¯ _b 1 k hpXV¯ for 1 near 0 _ ÕD[Õ C D[~VC Ë htrN~rVhprM~nrsCoCtC To check1; the accuracy]0of_xMthis approximation, we use it ]` to_bapproximate . DZh /zhp3D~9C Ë htrN~rmhtrnXmh Substituting gives . Thus, yields the first seven decimal places Õ 1 ÕyÖ for . For large1E values of , however, the accuracy diminishes because grows faster than any ÚÜÛ .0/2143\DÝÕpÖ polynomial as . Figure 10.6 shows graphs of and the Taylor polynomials of DXVkthnk~9k-ªmkz| degree § . Notice that each successive approximation remains close to the exponential 1 curve for a larger interval of -values. ¡=ß Î á ¢¡=ßâ¡ ¡ o ¡ ¥ ¡ ¡=à ¡=à G á ¢ ¡ QÞ ¥ ¡ Q Þ Qs Î Figure 10.6: For G near , the value of á ¢ is more closely approximated by higher-degree Taylor polynomials Example 6 Construct the Taylor polynomial of degree § approximating Solution Differentiating gives means h .0/2XN3sDØh ¨ 5[] hsAB1 , . < /2XM3sDãh , /21436Dih:1´:1 . < < />XM3sDi~ « :1 . , :^otp:1 for hsAâ1 < < < />XM3vDiªV¯ ¬ :1 h .0/2143D , ¨ ¬® . k 1 near 0. /2XN3sD|m¯ for 1 , and so on. This near 0, Let us compare the Taylor polynomial with the formula obtained on page ?? for the sum of a finite ¨nä _ geometric series: hsAB1 hsAB1 1 1 ¨ä Djh:1$:1 :1 « :1 ¬ ¨ :toa:1 C _ If is close to 0 and is small enough to neglect, the formula for the sum of a finite geometric series gives us the Taylor approximation of degree § : h hsAB1 5h:1$:1 :1 « :1 ¬ :^ooy:1 ¨ C 452 Chapter Ten APPROXIMATING FUNCTIONS Taylor Polynomials Around åæèç .0/@1=3D^éws1 Suppose we want to approximate by a Taylor polynomial. This function has no Taylor 1DX 1;êiX because the function is not defined for1%D8 . However, it turns out that polynomial about we can construct a polynomial centered about some other1point, . D^8 First, let’s look at the equation of the tangent line at : g D[.0/>8}3:;.=<T/2893?/21"A8}3?C This gives the first Taylor approximation .0/@1=365^.0/>8}3:(.=<T/2893?/21\A8}3 . 1 for 84C near < />8}3o/@1%A893 . 1 The 8 term is a correction term which approximates the change in as moves away from . ]¨/2143 17DÝ8 .0/>893 Similarly, the Taylor 1 polynomial centered at is set up as plus correction D8 /@1%A(893 . This is achieved by writing the polynomial in powers of terms which are zero1 for instead of powers of : ¨ .0/@1=365] /@1\A893 /21436D[6s:(s_/@1\A893:( ¨ :^otp:; /@1\A893 ¨ C ]'¨/@1=3 § If we require derivatives of the approximating polynomial and the original function 1D[8 1D[8 to agree at , we get the following result for the § ²2³ Taylor approximations at : "! ´# Taylor Polynomial of Degree µ Approximating .0/214365^] ¨ for near ë /@143 . D[.0/2893:;.=<>/>893?/@1"AB893: < < />8}3 ~V¯ /@1AB893 ¨ ® . :[top: ]'¨/@1=3 /2893 /@1\A893 ¯ § 1Di8 We call the Taylor polynomial of degree § centered at 1D[8 nomial about . ¨ , or the Taylor poly- You can derive the formula for these coefficients in the same way that we did for Problem 28, page 454.) Example 7 Construct the Taylor polynomial of degree 4 approximating the function Solution We have .0/@1=3D és1 < /@1=36D hy·a1 . . . < < < /@1=36D . ¬® /@1=36DjA . « ~N·y1 Ê ·y1 . ¬ k /@1A;hy3 és1R5]¬N/21436D[XÉ:[/@1\Ahp3'A D/@1"A;hy3'A /21"Ahp3 ~ . The Taylor polynomial is therefore ~9¯ : .0/@143D£és1 8RDX for 1 . (See near 1. .0/xhp3D^é/xhy3D^X so < < /@1=36DìAhy·a1 . és1 .0/2143 :;~ < /xhp3D h < < /xhp3D Ah < < < /xhp3D ~ ¬® /@1"A;hy3 /@1"A;hy3 ª /xhp3D « ªV¯ A A C Ê « A Ê /@1\Ahp3 | /@1"A;hy3 ¬ |m¯ ¬ k for 1 near 1 C Graphs of and several of its Taylor polynomials are shown 1 in Figure 10.7. Notice that és1 1 stays reasonably close to for near 1 1,ê; but bends away as gets farther from 1. Also, X éws1 note that the Taylor polynomials are defined for , but is not. ]¬N/2143 453 10.1 TAYLOR POLYNOMIALS Î K G9L ¡ ¡¥ K GVL î Ñ G K GVL ¡ G ¡¥ K GVL í ¡=ß K GVL P ¡ ß K GVL K GVL ¡ Î K GVL ¡ î G Ñ Figure 10.7: Taylor polynomials approximate î Ñ G closely for G near , but not necessarily farther away The examples in this section suggest that the following results are true for common functions: 1âDj8 .0/@1=3 1 8 Taylor polynomials centered at give good approximations to for near . Farther away, they may or may not be good. The higher the degree of the Taylor polynomial, the larger the interval over which it fits the function closely. ï ï Exercises and Problems for Section 10.1 Exercises For Problems 1–10, find the Taylor polynomials of degree ð approximating the given functions for G near 0. (Assume ñ is a constant.) 1. 3. 5. 7. òG , òG ÷oø ú -ù ÷ Ñ `QRG Þ ð , ð G , , ð òG , í ð 9. ,ó ,ô ð 2. ,í ,Þ , QRG p G 4. Þ 6. ,í ,Þ ,í ,Þ 8. 10. Problems ¡ K , ù ÷ î Ñ Ñ G , ð , K òG9L2ý ð K }òûGVL í ,õ , ,Þ ,ó í ð , ð ö , For Problems 11–14, find the Taylor polynomial of degree for G near the given point F . Þ Þ ð õ ,ö ,ü 13. G bÐÒÑ 11. á ¢ , , F 4þ , p G 12. í ð FU , ð Þ 14. ð , ð , 'òG í F , F ð =þ Þ , , ,í ,Þ GVL[FÉòBÿ-Gò G For Problems 15–18, suppose is the J second degree Taylor polynomial for the function about J G . What can you say about the signs of F , ÿ , if has the graph given below? 15. J4K GVL J4K GVL 16. 17. G G J4K GVL 18. G G J4K G9L 19. Suppose the function J4K G9L is approximated near G 454 Chapter Ten APPROXIMATING FUNCTIONS by a sixth degree Taylor polynomial Î K G9LíoGQ"ÞtG ¡ ò G õ Give the value of J4K yL (a) JNSTK pL (b) JNS S SÇK pL (c) J (d) -K yL J (e) -K pL 20. Suppose is a function which has continuous derivaK S K LâQ S S K LB tives, and that õ Lâ í õ , õ , S S S K L£Q6í õ . (a) What is the Taylor polynomial of degree for near í õ ? What is the Taylor polynomial of degree for near 5? (b) Use the two polynomials that you found in part (a) K to approximate Þ ü L . J4K GVLû J4K 23. (a) Based on your observations in Problems 21–22, make a conjecture about Taylor approximations in J the case when is itself a polynomial. (b) Show that your conjecture is true. , for G í G near 0, to explain why î Ð ¢ G bÐÒÑ à G 25. Use the fourth-degree Taylor approximation ß ØQ î Ð ¢ G `Q G ò à p G G for Þ G near G bÐÒÑ . . 26. Use a fourth degree Taylor approximation for á , for near 0, to evaluate the following limits. Would your answer be different if you used a Taylor polynomial of higher degree? (b) 27. If , and , , and , , , calculate the following limits. Explain your reasoning. (a) (b) î (a) î Ð Ð á à K pL{ S K pL p î ¢ Ð Î Q í ¥ "! K ##$ J ¥ J J Î ¥ 32. One of the two sets of functions, , , , or , , Î , is graphed in Figure 10.8; the other set is graphed in each have GØ . Taylor Figure 10.9. Points and polynomials of degree 2 approximating these functions near G are as follows: % ' ( ( J Î K GVL & ( ( ( ( ò%Gòâ ?G `òGQG `òGÉòG ¥ K GVL òGÉò oG K GVL òGÉòG Î K GVL `QRGÉòG (a) Which group of functions, the s or the s, is represented by each figure? (b) What are the coordinates of the points and ? (c) Match the functions with the graphs (I)-(III) in each figure. % & K pLsE J S K tLs J S S K tLí K GVL III III II II ¤ J4K GVL G and GQ á vQQ à Q`Q J4K tLs J J K GVL to explain why 31. The integral à bÐÒÑ þ L is difficult to approximate using, for example, left Riemann sums or the trapezoid K rule because the integrand bÐÒÑ L þ is not defined at . However, this integral converges; its value is Þ Estimate the integral using Taylor polynoü ó ô mials for bÐÓÑ about of (a) Degree 3 (b) Degree 5 J ¥ K G9L t G G (a) How many zeros does each equation have? (b) Which of the zeros of the two equations are approximately equal? Explain. 22. Find the third-degree Taylor polynomial for GVL Î G ò G Q G`ò about G . What do you notice? ö õ GQ 30. Consider the equations bÐÒÑ 21. Find the second-degree Taylor polynomial for ÞpG Q Gò about G . What do you notice? ö Î 24. Show how you can use the Taylor approximation (b) Compare this result to the Taylor polynomial apJ4K proximation of degree 2 for the function GVL(á ¢ G about . What do you notice? (c) Use your observation in part (b) to write out the Taylor polynomial approximation of degree 20 for the function in part (a). (d) What is the Taylor polynomial approximation of de J4K ¢ ? gree 5 for the function GVLá Î ¢ î Ð S S K pL õ S K tLsE NS S K tL ö J4K GVL K GVL 28. Derive the formulas given in the box on page 452 for the coefficients of the Taylor polynomial approximating J a function for G near F . 29. (a) Find the Taylor polynomial approximation of degree J4K 4 about G for the function GVLá ¢ ¤ . I % Figure 10.8 I & Figure 10.9 10.2 TAYLOR SERIES 10.2 455 TAYLOR SERIES In the previous section we saw how to approximate a function near a point by Taylor polynomials. Now we define a Taylor series, which is a power series that can be thought of as a Taylor polynomial that goes on forever. Taylor Series for cos )+* sin )+*-, 1%DX We have the following Taylor polynomials centered at for d?eNf91 : doeMf91R5[]M/@1=36Dih doeMf91R5[] ~V¯ 1 ¬ /@1=36DihsA doeMf91R5[] doeMf91R5[] 1 /@1=36DihsA doeMf91R5[] /@1=36DihsA » : Ê ¯ 1±» A |±¯ /2143 ] 1±» A |±¯ ¬ 1 : ~V¯ ]M/@1=3 |±¯ ¬ 1 ~V¯ 1 /@1=36DihsA Ì ~V¯ 1 ¬ 1 : Ê ¯ ]¬N/@1=3 1±Ì : ] C rV¯ /2143 ] /@143 Here we have a sequencedoeMoff91 polynomials, , , , » , Ì , ..., each of which is a 1 better approximation to than the last, for near 0. When we go to a higher-degree polynomial ] ] 1±Ìp·rV¯ (say from » to Ì ), we add more terms ( , for example), but the terms of lower degree don’t change. Thus, each polynomial includes the information from all the previous ones. We represent doeMf91 the whole sequence of Taylor polynomials by writing the Taylor series for : h{A 1 ~V¯ 1 : ¬ 1±» A |m¯ 1±Ì : Ê ¯ A;tooC rV¯ ]'¨/@1=3 Notice that the partial sums of this series are the Taylor polynomials, . fz{1 ÕyÖ We define the Taylor series for and 1 similarly. It turns out that, for these functions, the Taylor series converges to the function for all , so we can write the following: 1 f-w{1RD^1\A d?eNf91RDihsA Õ Ö 1 « ªV¯ ~V¯ Dih:1´: 1±º : 1 : 1 ~V¯ 1±¼ l9¯ ¬ A Ë ¯ 1±» A |m¯ : 1 « ªm¯ /. 1 : qm¯ 1±Ì : Ê ¯ : 1 ¬ |m¯ rV¯ A;to A;to :^ot f-ws1 d?eMfV1 ÕpÖ 1D[X These series are also called Taylor expansions of the functions , , and about .¨ The 1 · ¯ in the series. For example, § general term of a Taylor series is a formula which gives any term ÕyÖ /zAhp3-Ùp1 ÙM·9/T~ 3?¯ × is the general d?term in the Taylor expansion for , and is the general term in the eNf91 expansion for . We call § or × the index. Taylor Series in General . Any. function , all of whose derivatives.0/@exist at 0, has a Taylor series. However, the Taylor series 143 1 1 for does not necessarily converge to for all values of . For the values of for which the .0/2143 series does converge to , we have the following formula: 456 Chapter Ten APPROXIMATING FUNCTIONS Taylor Series for "! $# .0/2143D7.0/2XM3:;. è about . < /2XM3x1$: < < /2XN3 0 ~9¯ 1 . : < < < /2XM3 ªV¯ 1 « :^oty: . ¨ ® /2XN3 § ¨ 1 ¯ :to X In addition, just as we have 1 Taylor polynomials centered at points . other than1D , we can also D 8 8 (provided all the derivatives of exist at ). For the have a Taylor series centered at 1 .0/@1=3 values of for which the series converges to , we have the following formula: Taylor Series for J4K GVL J4K FML±ò "! $# è about J S S K FNL J S K FML K GQFNLVò ë J S S S K FML K GQFML ò í 0#010 Î K GQ"FML ò 32 J uò ð K FML K GQFML 2 40$0#0 ò The Taylor series is a power series whose partial sums are the Taylor polynomials. As we saw in 1[DZ8 Section ??, power series generally converge1 on an interval centered at . The Taylor series for such a function can be interpreted when is replaced by a complex number. This extends the domain of the function. See Problem 41.1 . For a given function and a given , even if the Taylor series converges, it might not converge .0/@143 to , though it does for most commonly1encountered functions. Functions for which this is DZ8 true for the Taylor series about every point in their domain are called analytic functions. ÕyÖ d?eMfV1 f-s1 1 , and converge to the original function for all , so Fortunately, the Taylor series for , they are analytic. See Section 10.4. Intervals of Convergence of Taylor Series és1 Let us look again at the Taylor polynomial for about page ??. A similar calculation gives the Taylor Series éws1D/@1\Ahp3'A 65 75 85 /@1"A;hy3 ~ : /@1\Ahp3 ª « A /21"Ahp3 | 1D h that we derived in Example 7 on ¬ :top:[/xAhy3 ¨}© _ /@1"A;hy3 ¨ :totC § Example 2 onX page1449 and Example 5 on page 451 show that this power series has interval of êè~ convergence . However, although we know that the series converges in this interval, éws1 we do notX yet know that its sum is . The fact that in Figure 10.10éws the polynomials fit the curve 71 ~ 1 X 71(êj~ well for suggests that the Taylor series does converge to for . For such 1 -values, a higher-degree polynomial gives, in general, a better approximation. 1 ~9k However, when the polynomials move away from the curve and the approximations get worse as the degree oféwsthe polynomial increases. Thus, the Taylor polynomials are effective only 1 1 as approximations to for values of between 0~ and 2; outside that interval, they should not X be used. Inside the interval, but near the ends, or , the polynomials converge very slowly. This és1 means we might have to take a polynomial of very high degree to get an accurate value for . 9: 95 457 10.2 TAYLOR SERIES P O Interval of convergence / ¡ ¡ ¡ K V G L K GVL Ï K V G L Ô K GVL ¡ K GVL î G Ñ G ¡ Ï K GVL ¡ í Þ O O O î Ñ O ¡4Ô K GVL G ; ; K GVL ¡ Figure 10.10: Taylor polynomials ; K GVL ¡ ¡=Ï K GVL ; ## converge to Ô K G9L ¡ K GVL ¡ î Ñ =< ?> G for âG B and diverge outside that interval To compute the interval of convergence exactly, we first compute1the radius of convergence 1D[X D[~ using the method on page ??. Convergence at the endpoints,és1 and , has to be determined separately. However, proving that the series converges to on its interval of convergence, as Figure 10.10 suggests, requires the error term introduced in Section 10.4. Example 1 Find the Taylor series for Solution Taking derivatives of éw/zh:143 é/zh:1=3 about 1RD^X , and calculate its interval of convergence. 1D[X and substituting 1 é/xh{:1=3D1\A : ~ « 1 leads to the Taylor series ª ¬ 1 A Notice that this is the same series that we get by substituting és1RDj/21"Ahp3'A /@1\Ahp3 : ~ éws1 /@1\Ahp3 « X /xh:143 1 for ¬ in the series for :^ot | @5 A5 1âDh /21\A;hy3 A ª :^ootC | ?5 X for (1ê~ és1 : . [1ê7~ about 1RDconverges for , the interval of convergence for the Since the series for éw/zh:1=3 X Ah ;1êEh Taylor series for about is . Thus we write é/zh:1=36D£1"A 1 ~ : « 1 ª A | :^ot Notice that the series could not possibly converge to defined there. Interval of convergence P A5 ¬ 1 for é/zhs:;1=3 B ¡ O Ah for K GVL (1êEh 1(êAh ¡Ï K GVL ¡ . since é/xhv:;143 K G9L K òGVL î Ñ G Qs ¡ ¡ K GVL K GVL ¡Ï K V G L ¡ Ô K GVL O O O ¡ O P P î Ñ B K GVL K 'òG9L ¡=Ô K GVL Figure 10.11: Interval of convergence for the Taylor series for ¡ î Ñ K GVL K 0ò%GVL is C< -> Qs âG is not 458 Chapter Ten APPROXIMATING FUNCTIONS The Binomial Series Expansion ED , with F 1RDX .0/2143D/xh:143 We now find the Taylor series about for the function not necessarily a positive integer. Taking derivatives: .0/@143Di/zh:1=3 . . F < /2143D . / © / Ahp3?/ ED F . ©4 ED © A~M3?/zh:1=3 « k D ] « /@1=3uk-] F 5[]'«M/21436Djh: F F / Ahp3 1 ~9¯ / F F < < < />XM3D near 0 is 1$: F F < < />XM3D . 1 F < />XM3D . Thus, the third-degree Taylor polynomial for /zh:1=3 .0/>XM3Dih so _ Ahp3?/zh:1=3 F F < < < /2143D ED /xh:143 F F < < /2143D D a constant, but / F F / : Ahp3o/ A;hy3?/ ªm¯ F ?5 G5 ¬ /@143?koCtCoC Ahp3 F F A~n3?C A~M3 « 1 C Graphing .0/@1=3 for various specific values of suggests that the Taylor polynomials Ah E1 h converge to for . (See Problems 24–23, page 459.) This can be confirmed using .0/@1=3Di/zh:1=3 1RD[X the radius of convergence test. The Taylor series for about is as follows: ED The Binomial Series D /xh:143 Djh: F F F / 1û: Ahp3 ~9¯ 1 F F / : F Ahp3o/ ªV¯ A~n3 « 1 :^ot for =5 H5 Ah 1 ID 7h . F /xh:R143 In fact the binomial series gives the same result as multiplying out when is a positive integer. (Newton discovered that the binomial series can be used for noninteger exponents.) Example 2 Use the binomial series with Solution The series is « /xh:143 1 F D[ª to expand Dh:(ª1$: ªUp~ ~9¯ 1 /zh:1=3 « . ªt~Unh : 1 ªV¯ « ªp~UhspX : |m¯ 1 ¬ :tooC ¬ The term and all terms beyond it turn out to be zero, because each coefficient contains a factor « « of 0. Simplifying gives /xh:143 Djh:(ª1$:(ª1 :1 k which is the usual expansion obtained by multiplying out Example 3 Find the Taylor series about Solution Since h Di/zh:1=3 h:1 h h:1 © 1D[X for _ F /xAhy3?/zAÉ~n3 Dih:^/zAhp3b1û: DihsAâ1$:1 A5 95 . , use the binomial series with © . h h:1 _ D/xh:143 « /xh:143 ~9¯ Aâ1 « :to DAh 1 : for . Then /xAhp3o/xAÉ~M3?/xAvªN3 ªm¯ =5 H5 Ah 1 7h 1 « :to . This series is A both a special case of the binomial series and an example of a geometric series. It h ;1 Eh converges for . 459 10.2 TAYLOR SERIES Exercises and Problems for Section 10.2 Exercises For Problems 1–4, find the first four terms of the Taylor series for the given function about . QRG 13. 'òG 1. 2. `QG òG 3. 4. òG ú `Q K QGVLQGQ î Ñ K 'òGVLGQ For Problems 5–11, find the first four terms of the Taylor series for the function about the point F . 8. 11. G bÐÒÑ 5. , 4þ Þ ù ÷ F U Ñ þ G Qs G Fû , 9. 4þ Þ , 6. KJ , t F Fè , bÐÒÑ 7. 4þ Þ þ G LJ , 15. FUQ þ G 10. , =þ Þ F 16. 17. FY M Find an expression for the general term of the series in Problems 12–19 and give the starting value of the index (ð or , for example). ÷oø á ¢ ¤ 18. 19. -ù òG G ÷ G ò í í ò Þ G ò Ô G ò ß G Q ò ö G ò G Ï G Q õ ß G ò ö Î G p G Q õ Ï G Q õ ò G ò Þ P0#0#0 P0#0#0 P0#0#0 P0#010 G ò í ß G Q í Q Þ Î G ò O0#010 H010#0 ß G Q í GGQ Ñ N0#0$0 Q Î G Q ò ß òG G Î G GGQ bÐÒÑ QG G ß òG Î `QGÉòG î Ñ 14. òG 40#0$0 Î òGÉòG 12. Þ Ô G Q ò ó Problems 20. (a) Find the terms up to degree ó of the Taylor series for J4K GVL K G L bÐÒÑ about G by taking derivatives. (b) Compare your result in part (a) to the series for bÐÒÑ G . How could you have obtained your answer to part (a) from the series for bÐÒÑ G ? J4K K î 21. (a) Find the Taylor series for G9L' Ñ 'òB oG9L about G by taking derivatives. (b) Compare your result in part (a) to the series for î Ñ K òG9L . How could you have obtained your anî K swer to part (a) from the series for Ñ 0ò%GVL ? (c) What do you expect the interval of convergence for î K the series for Ñ 'ò oGVL to be? By recognizing each series in Problems 27–35 as a Taylor series evaluated at a particular value of G , find the sum of each of the following convergent series. " P0$010 27. 'ò 29. J4K GVLU 31. and several 32. of its Taylor polynomials, estimate the interval of convergence of the series you found in Problem 3. 33. 23. By graphing the function 24. By graphing the function J4K GVL 'òG and several of 34. its Taylor polynomials, estimate the interval of convergence of the series you found in Problem 1. Compute the radius of convergence analytically. 35. `QG Þ ð ò Q ò Î ò ò Þ ¥ Î K aL 40$0#0 ¥ ß K L Q í ?ò Q ò Þ 40$0#0 " P0#0#0 P0$010 " 40#0$0 Q% 0ò ü 'òíò Q% 2 010#0 " 28. ?ò Þ ¥ K nL Q Q ò í ò J4K 22. By graphing the function GVL 0ò%G and several of its Taylor polynomials, estimate the interval of convergence of the series you found in Problem 2. ô ò í ò õ Q Þ 40#0$0 ò oò K ð ò Þ ò ó Q tn í ò 26. (a) Write the general term of the binomial series for K òGVL2ý about G . (b) Find the radius of convergence of this series. 36. 'òGÉòG G ò í 40#0$0 P010$0 Î òG ò G Î ò a õ ò ?ò 2 0 U2 T P0#0#0 ¥ K nL K Qs?L ò ô ò Q Þ 0ò ö í ò Þ In Problems 36–37 solve exactly for the variable. GQ ö uptp òò Î ò oò p 25. Find the radius of convergence of the Taylor series î K around zero for Ñ `QGVL . 37. Q 30. RQ 'S RQ 'S P0#0#0 RQ "S 2 40#0$0 40$010 'ò Þ ò ò?L ¥ ò 2 0#010 ; 2 0 2 P0 ; K Qs?L K ð ò ò?L K Qs?L K ð tu L ò 460 Chapter Ten APPROXIMATING FUNCTIONS J ¥ J J Î ¥ 38. One of the two sets of functions, , , , or , , Î is graphed in Figure 10.12; the other set is graphed in Figure 10.13. Taylor series for the functions about a point corresponding to either or are as follows: % & J ¥ K GVLíò K GQ?LQ K GQ?L J K GVLíQ K GQ?L±ò K GQ?L J Î K GVLísQ% K GQ?L±ò K GQ?L 0#010 0#010 0#010 39. Suppose that you are told that the Taylor series of G á ¢ ¤ about G is ¥ K GVL õ Q õ Î K GVL õ & ò J tion is J S K pL G GÉò Find I , ¢ Î ß G ò ò Þ G uò J à about G ò ð J S S S K yL J ¥ à K yL á in the . Use this definition2 to explain Eu- á ¢ [\ á KJ Z ]J ò p III Figure 10.12 á ¢ ¤ oG aG à í II III ò Þ P0#0#0 2 40$0#0 , , and . We define [\ by substituting Z J ò J S S K yL Z & í ¥ à G ò G 41. Let Qs Taylor series for ler’s formula II 10.3 Ô G ò exist at , and that Taylor series for I % G ¢ (a) Which group of functions is represented in each figure? (b) What are the coordinates of the points and ? (c) Match the functions with the graphs (I)-(III) in each figure. % 0#0#0 P0#0#0 0#0#0 ! Q ! Q # 0 # 0 0 Find ! SWVV X and ! SYVV X VV VV 40. Suppose you know that all the derivatives of some func ß K G QÞaL G òG ò K GQÞaLmò K GQÞaL K GQÞaLmò K G QÞaL tG á ¢ ¤ aG K GQÞaL=Q Q K GVL J4K G9L bÐÒÑ Figure 10.13 FINDING AND USING TAYLOR SERIES Finding a Taylor series for a function means finding the coefficients. Assuming the function has all its derivatives defined, finding the coefficients can always be done, in theory at least, by differenÕyÖ tiation. That is how we derived the four most important Taylor series, those for the functions , f-ws1 doeMf91 /zh:£1=3 , , and . For many functions, however, computing Taylor series coefficients by differentiation can be a very laborious business. We now introduce easier ways of finding Taylor series, if the series we want is closely related to a series that we already know. ED New Series by Substitution © Õ Ö 17DèX ¤ Suppose we want to find the Taylor© series for about . We could find the coefficients © Ö ¤ Õ Ö ¤ AÉ~1±Õ by differentiation. Differentiating by the chain rule gives , and differentiating again © © Ö ¤ :£|M1 Ö ¤ AÉ~nÕ Õ gives . Each time we differentiate we use the product rule, and the number of © © Õ Ö ¤ Õ Ö ¤ Finding the tenth or twentieth derivative of _x grows. terms , and thus the series for up to 1 1 the or terms, by this method is tiresome (at least without a computer or calculator that can differentiate). Fortunately, there’s a quicker way. Recall that ^ Õ Dh: DjAs1 Substituting g Õ 2 Complex © Ö ¤ g : g « ~V¯ : g ¬ ªV¯ : g |±¯ :^ot for all g C tells us that Dh:[/xAs1 3: /xAs1 ~9¯ numbers are discussed in Appendix B. 3 : /zAs1 ªV¯ 3 « : /xAs1 |m¯ 3 ¬ :to for all 1C 461 10.3 FINDING AND USING TAYLOR SERIES Simplifying shows that the Taylor series for © Õ Ö ¤ DhsAB1 Ö ¤ ¬ 1 : © Õ A ~V¯ is » 1 ªm¯ Ì 1 : Example 1 Find the Taylor series about Solution The binomial series tells us that h Di/zh: h: g D1 Substituting g g © 3 _ for h:1 : g or 1 for all . 1 terms. h .0/@1=36D DjhsA _b 1 Using this method, it is easy to find the series up to the 1D[X :^ot |m¯ « A g . ¬ : g :to g A5 5 Ah for EhC g gives h DihsAB1 h:1 ¬ :1 » AB1 Ì :1 :to 5 H5 Ah for (1 7h , h which is the Taylor series for h:1 . These examples demonstrate that we can get new series from old ones by substitution. More advanced texts show that series obtained by this method are indeed correct. Dã1 In Example 1, we made the substitution g . We can also substitute an entire series into another one, as in the next example. _ Example 2 Find the Taylor series about Solution For all g and , we know that D[X _ ^ Õ Õ `ba cd 4e _ _ Dhn: A « ªm¯ : _ _ A£ot : lV¯ g g D « : ~9¯ A g ¬ ªm¯ _ « ªV¯ : : º g :^oo |m¯ A;totC l9¯ for g : f º : _ _ _ f- f-w 3D^Õ Dih: and Let’s substitute the series for U`ba c'd . _ / for c h ~V¯ e_ _ A _ « ªm¯ : º Aot lV¯ f : h ªV¯ e_ _ A « ªm¯ h : _ º lV¯ Aot f « :ÃotoC To simplify, we multiply out and collect terms. The only constant term is the , and there’s only one ·~9¯ « term. The only term is the first termA we« ·get by multiplying out the square, namely . There ªV¯ « are two contributors to the term: the from within the first parentheses, and the first term ·ªV¯ we get from multiplying out the cube, which is . Thus the series starts _ _ Õ _ _ `ba c/d Djh: Djh: _ _ : _ _ : _ ~9¯ : e _ ~9¯ « A :X _ « ªV¯ _ f « : ªV¯ :^oo :^oo for all _ C _ 462 Chapter Ten APPROXIMATING FUNCTIONS New Series by Differentiation and Integration Just as we can get new series by substitution, we can also get new series by differentiation and integration. Here again, proof that this method gives the correct series and that the new series has the same interval of convergence as the original series, can be found in more advanced texts. Example 3 Solution Find the Taylor Series about We know that g e for /xhsAB143 /zhsAâ1=3 h from the series for C hsAB1 , so we start with the geometric series Djh:1$:1 hsAB1 h h [D h{AB1 1 g f h h 1D[X « :1 ¬ :1 :to =5 95 Ah for 1 EhnC Differentiation term by term gives the binomial series h /zhsAâ1=3 g e D 1 g Example 4 Find the Taylor series about Solution We know that 1j g "h i h g hg i j h g / d {143 / d 1=3 1D[X for h"i 1j h d « :|n1 :to h {1 from the series for hi j h where lk h 1RD h:1 ¬ :1 g d The series for {1D^1\A hi j h d 1 1 « ª : 1±º l « 1 1RD[:1\A » Aâ1 A 14º : ª is the constant of integration. Since h"i 1j h Ì :1 Aoo l hi j h d 1±¼ Ë A : 14¼ Ë {XD[X . 1 q . 1 : q Use the series for h"i 1j h d 1 7hMC C A;to for Aot , we have for A5 H5 Ah ¶ to estimate the numerical value of . ;1 ED[X , so ;1 . A5 95 5 95 Ah 7h A;h for was discovered by James Gregory (1638–1675). Applications of Taylor Series Example 5 h:1 1 , so we use the series from Example 1 on page 461: DhsAB1 h:1 Antidifferentiating term by term gives d A5 95 Ah for h h6:1 h D 1 7Dh:;~a1´:ªn1 hsAB1 D 1 f h EhnC 1 7hMk Solution j Since hi h 10.3 FINDING AND USING TAYLOR SERIES 463 1j from Example 4. We assume—as is the , we use the series for h"i h case—that the series does converge to j at , the endpoint of its interval of convergence. Substituting into the series for hi h gives We f hi j h d h;D ¶·y| d ¶·a| 1£DÝh d 1%Djh ¶D£| d 1 hÉD^| Table 10.1 Approximating m2 {1 h =hsA ¶ h : ª h A l h : Ë using the series for Aot C q hi j h d 1 4 5 25 100 500 1000 10,000 2.895 3.340 3.182 3.132 3.140 3.141 3.141 ð n ¨ Table 10.1 shows the value of¨ the § ²2³ partial sum, ,¶ obtained by summing the nonzero terms from D^ªmCwht|mhCtCoCC 1 through § . The values of do seem to converge to However, this series converges very slowly, meaning that we have to take a large number of terms to get an accurate estimate for ¶ ¶ . So this way of calculating is not particularly practical. (A better one is given in Problem 1, ¶ page 475.) However, the expression for given by this series is surprising and elegant. n A basic question we can ask about two functions is which one gives larger values. Taylor series can often be used to answer this question over a small interval. If the constant terms of the series for two functions are the same, compare the linear terms; if the linear terms are the same, compare the quadratic terms, and so on. Example 6 _ smallest, for near 0. Solution d By looking at their Taylor series, decide which of the following functions is largest, and which is h _ D[X The Taylor expansion about _ D[X d Õ _ The Taylor expansion about Í h: So, substituting _ D/xh: _ 3 D[X po © _ _ d Õ _ _ º _ _ hy· Í _ h _ ~ º _ : « 3 ª r ¼ Ë ¯ :^otoC ¬ :^ootC |m¯ _ l A _ h Ê _ ~V¯ : _ A « ªm¯ :totC Ë ¯ lV¯ _ 3 /xA o : ~ ¼ _ hsA~ Í is /zA h DihsA _ : _ h: _ : ªm¯ ~V¯ _ A « A : (c) l9¯ is Dih: for : ªV¯ _ _ Dih: for _ « A DihsA AÉ~ is _ _ _ h:f- The Taylor expansion about _ fz D So Õ (b) for f- h _ h:f-w (a) « /zA : _ 3 /xA o « 3?/xA ªm¯ :tooC Í for : h hsA~ _ h DihsA Dih: /xAÉ~ _ ~ : ª ~ _ ª _ 3: /xAÉ~ : r l ~ _ « _ 3 A :^ot?C l h Ê /xAÉ~ _ 3 « :^oo º 3 _ « :^oo 464 Chapter Ten APPROXIMATING FUNCTIONS _ For near 0, we can neglect terms beyond the second degree. We are left with the approximations: _ h:(f-w Õ d Since _ 5 h: _ Eh: _ _ : ~ ª ~ _ ~ _ 5 h : : 5ih: _ hsA~ _ _ 5ih: h Í _ 5ih: C _ [h: ª : _ ~ k _ and since the approximations are valid for near 0, we conclude that, for near 0, _ 5 d 5 h:(fz Example 7 h Õ C _ hsA~ Í Two electrical charges of equal magnitude and opposite signs located near one another are called A and are a distance apart. (See Figure 10.14.) The electric an electrical dipole. The charges ] field, , at the point is given by q s s t r q tq D A t q r / : C 3 t « Use series to investigate the behavior of the electric field far away from the dipole. Show that when hy· is large in comparison to , the electric field is approximately proportional to . r u ¡ P w O w v P v ¡ w w Q O Figure 10.14: Approximating the electric field at due to a a distance apart dipole consisting of charges and Q Solution t r t In order to use a series approximation, we need a variable whose value is small. Although we know · know that do not itself is small. The quantity is, however, that is much smaller than ,hawe ·9/ : 3 · very small. Hence we expand in powers of so that we can safely use only the first few terms of the Taylor series. First we rewrite using algebra: r t r h t / : r h t D 3 r Now we use the binomial expansion for t Q h h: tr S t e ©4 D h t e h D :(ª · 3 Aâ| tr s tq A / t q r : 3 D D s ~ A h hsA~ Aª :| DjAÉ~ : ªm¯ Q tr S C , we have q xt t e y t q e tr tr h C /zAÉ~n3o/xAvªM3o/xAs|}3 : :to « · « So, substituting the series into the expression for D 1D ~9¯ ©= h: /xAÉ~M3?/zAvªM3 : t Q tr S r t and F Q tr S f h D /zh:1=3 Q tr S tr tr hsA~ r t ED with /xh: h:^/zAÉ~n3 r t tr tr :ª « « tr AB| Aot f C tr « « :^ot f{z « :^ot f 465 10.3 FINDING AND USING TAYLOR SERIES t Since r r t converges. We are is smaller than 1, the binomial expansion for in t t interested the electric field far away from the dipole. The quantity r is small there, and r and higher · /zh9: · ©4 3 · / · 3 powers are smaller still. Thus, we approximate by disregarding all terms except the first, giving tq e t r f s ~ Since q 5 k r and are constants, this means that s r t In the previous example, we say that · in the expansion is . s s so ~ 5 t qr « C is approximately proportional to is expanded in terms of r t · hy· t « C , meaning that the variable Exercises and Problems for Section 10.3 Exercises Find the first four nonzero terms of the Taylor series about 0 for the functions in Problems 1–12. 1. 5. 9. `Q% oG î Ñ | á K yQ ?NL ¤ 2. 6. 10. p ÷?ø K J bÐÒÑ K ò 3. L á ¢ 4. G 7. 11. L bÐÒÑ }| `Q 8. á p 12. 14. Î p 13. ò ~ lems 13–15. Give the general term. ò K ~ 15. Î K 'òG9L K L4Q nbÐÒÑ Î Q L LJ bÐÒÑ For Problems 16–17, expand the quantity about 0 in terms of the variable given. Give four nonzero terms. 16. G in terms of òG F 17. F ò%G in terms of where G F F â Find the Taylor series around 0 for the functions in Prob- Problems 18. For F , ÿ positive constants, the upper half of an ellipse has equation U U J4K G9Lÿ Q G F J J J small positive . (a) 'ò bÐÒÑ (b) t (c) `Q J4K (a) Find the second degree Taylor polynomial of GVL about . (b) Explain how you could have predicted the coefficient of G in the Taylor polynomial from a graph of J . (c) For G near , the ellipse is approximated by a parabola. What is the equation of the parabola? What are its G -intercepts? (d) Taking FUí and ÿ , estimate the maximum difJ4K ference between GVL and the second-degree Taylor Q6 G B polynomial for . {> ?> 19. By looking at the Taylor series, decide which of the following functions is largest, and which is smallest, for J 20. For values of near 0, put the following functions in increasing order, using their Taylor expansions. K î K (a) Ñ tòv L (b) bÐÒÑ L (c) aQ p 21. Figure 10.15 shows the graphs of the four functions below for values of G near 0. Use Taylor series to match graphs and formulas. (a) (d) Q"G QRG (b) K uòGVL ¥ ß (c) 'ò G 466 Chapter Ten APPROXIMATING FUNCTIONS M (IV) | v Expand term. Figure 10.15 22. Consider the functions Uá ¢ ¤ and U þ K 0òG L (a) Write the Taylor expansions for the two functions about G[ . What is similar about the two series? What is different? (b) Looking at the series, which function do you predict K will be greater over the interval Qs uL ? Graph both and see. (c) Are these functions even or odd? How might you see this by looking at the series expansions? (d) By looking at the coefficients, explain why it is rea¢ ¤ converges for sonable that the series for á G þ K {òG L all values of , but the series for K converges only on Qs ?L . U 23. The hyperbolic sine and cosine are differentiable and satisfy the conditions p v and bÐÒÑ v , and ! K aG GVL p bÐÒÑ ! ! K G bÐÒÑ aG GVL p K K u w w 25. An electric dipole on the G -axis consists of a charge at G and a charge Q at GQs . The electric field, G , at the point on the G -axis is given (for ) by K u Mw QâuL ; u Show that, for large Q K u Mw ò?L u F QG as far as the second nonzero u K u òF u Q L , u F ~ 28. One of Einstein’s most amazing predictions was that light traveling from distant stars would bend around the sun on the way to earth. His calculations involved solving for in the equation ~ ~ òÿ K ò p ò p ~ L where ÿ is a very small positive constant. ~ (a) Explain why the equation could have a solution for which is near 0. (b) Expand the left-hand side of the equation in Taylor m series about E , disregarding terms of order and higher. Solve for . (Your answer will involve ÿ .) J4K G Q `òâFGVL þ 0òâÿG9L (a) Let G9L6 , where F and ÿ are constants. Write J4down the first three terms of the K Taylor series for G9L about G . (b) By equating the first three terms of the Taylor seJ4K ries about G for GVL and for á ¢ , find F and ÿ J4K G9L so that approximates á ¢ as closely as possible G near . òG as a series in bÐÒÑ 24. Padé approximants are rational functions used to approximate more complicated functions. In this problem, you will derive the Padé approximant to the exponential function. is given by 27. The electric potential, , at a distance along the axis perpendicular to the center of a charged disc with radius F and constant charge density , is given by G (a) Using only this information, find the Taylor approxJ4K imation of degree ð ô about G for G9Ls G p . (b) Estimate the value of p . (c) Use the result from part (a) to find a Taylor polynomial approximation of degree ð ö about Gâ K for G9L= bÐÒÑ G . | F | 26. Assume F is a positive constant. Suppose the expression (II) ! u where is a positive constant whose value depends on the units. Expand as a series in þ , giving the first two nonzero terms. (III) (I) ~ ~ ~ 29. The Michelson-Morley experiment, which contributed to the formulation of the Theory of Relativity, involved the ¥ difference between the two times and that light took to travel between two points. If is the velocity of light; ¥ ¥ , , and are constants; and , then the times and are given by ¥ < 6 K Q þ L Q 6Q ¥ þ ¥ `Q Q (a) Find an expression for , and give its Taylor expansion in terms of þ up to the second nonzero term. (b) For small , to what power of is proportional? What is the constant of proportionality? þ Q K Q ¥ þ L 467 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS 32. When a body is near the surface of the earth, we usually assume that the force due to gravity on it is a constant , where is the mass of the body and is the acceleration due to gravity at sea level. For a body at a distance above the surface of the earth, a more accurate expression for the force is 30. A hydrogen atom consists of an electron, of mass , orbiting a proton, of mass , where is much smaller than . The reduced mass, , of the hydrogen atom is defined by ò (a) Show that . (b) To get a more accurate approximation for , express as times a series in . (c) The approximation is obtained by disregard- U 4 e and think of Assume H< F Q F the other quantities constant. K F ò L u u ¥ à á ¢ ¤ aG sums for both left-hand and right-hand sums with ð õ subdivisions. J4K (b) Approximate the function GVLá ¢u¤ with a Taylor polynomial of degree 6. (c) Estimate the integral in part (a) by integrating the Taylor polynomial from part (b). (d) Indicate briefly how you could improve the results in each case. as a function of F , with (a) Expand as a series in F þ . Give the first two nonzero terms. (b) Show that the approximation for obtained by using only the first nonzero term in the series is independent of the radius, F . (c) If F$ y , by what percentage does the approximation in part (a) differ from the approximation in part (b)? 10.4 ! using Riemann (a) Estimate the value of 33. ¥ L f ò (a) Show that . multiplied by a series in þ . (b) Express as (c) The first-order correction to the approximation is obtained by taking the linear term in the series but no higher terms. How far above the surface of the earth can you go before the first-order correction changes the estimate by more than u ? (Assume ó Þpp km.) K u u u 31. A thin disk of radius F and mass lies horizontally; a particle of mass is at a height directly above the center of the disk. The gravitational force, , exerted by the disk on the mass is given by where is the radius of the earth. We will consider the situation in which the body is close to the surface of the earth so that is much smaller than . ing all but the constant term in the series. The firstorder correction is obtained by including the linear í þ ô ó , by what term but no higher terms. If percentage does including the first-order correction ? change the estimate u þ 34. Use Taylor series to explain how the following patterns arise: (a) üpô p tpíttÞp õ ó y otÞy ö #$ ô ó íy ó Þ #1 (b) Q S ütü THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS In order to use an approximation intelligently, we need to be able to estimate the size of the error, which is the difference between the exact answer (which we do not know) and the approximate value. ]¨/2143 .0/@143 When we use , the §=²2³ degree Taylor polynomial, to approximate , the error is the difference ¨/@143D^.0/@1=30A]¨/2143?C s s ¨ s ¨ If is positive, the approximation is smaller than the true value. If is negative, the approxima¨ tion is too large. Often we are only interested in the magnitude of the error, . ] ¨ /@143 Recall that we constructed so that its first § derivatives equal¨ ® the corresponding deriva¨ ¨ /2XN3D .0/@1=3 X X ot />XM3DãX ] ¨ /@1=3 <¨ /2XN3D ¨ < < /2XN3DãX tives of . Therefore, , , , ä _z® , is ¨ ¨ä _z® . Since ¨ / :hy3 /@143ÃDZ. /2143 ¨ä _z® its . In addition, an §=²2³ degree.polynomial, § ² derivative is 0, so /@1=3 1 X supposeX´that is bounded by a positive constant , for all positive values of near , ê(1ê say for , so that ¨nä _x® s VV This means that g VV s` A A ê ê£. s s /2143{ê ¨ ¨ä _z® /@1=3{ê for for s s X´ê1ê X$ê;1ê s C g g C 468 Chapter Ten APPROXIMATING FUNCTIONS X 1 Writing for the variable, we integrate this inequality from to , giving Ok k s g s Ö Ö A so A ¨ä ¨ ê / 3 ê ¨ ® /@143sê X 1 X´ê(1ê 1 Hk l g k s Ö Ö ¨ ê so h A ~ 1 s ê ¨}© ¨ g k l g h / :^hp3?¯ § ¨ä 1 _ s ê g Ö ¨ ® / 3 ê for _z® h /@1=3sê ~ 1 X´ê(1ê X´ê(1ê for h ¨/@1=3{ê / :hy3u¯ § ¨ä 1 _ for g k C g By repeated integration, we obtain the following estimate: A k C g We integrate this inequality again from to , giving A X´ê1ê for ¨ 1ê g k g for Ö _x® X´ê1ê g k which means that s ¨ /@143 D ¨ .0/21430A] 1 A h /@1=3 ê / :^hp3u¯ § ¨nä 1 _ for Xûê1ê êj1£êiX g g C When is to the left of 0, so , and when the Taylor series is centered at similar calculations lead to the following result: Theorem 10.1: Bounding the Error in ¡ If ]¨/@143 is the §=²2³ Taylor approximation to s where D max ¨/@1=3 ¨nä . D .0/@1=30AB]'¨/@1=3 DØX , ! $# .0/2143 8 8 about , then ê / _x® 8 on the interval between § :^hp3?¯ 1\A8 ¨ä _ k 1 and . Using the Error Bound for Taylor Polynomials Example 1 Give a bound on the error, AvXVC lûê;1ê£XVC l . Solution Let .0/2143D[ÕyÖ s ¬ ÕyÖ , when ¢ is approximated by its fourth-degree Taylor polynomial for .wº . Then the fifth derivative is . º ® /2143 êÕ Thus º D s ¬ This means, for example, that on _ A5 /2XmC lM3xº D Í 5 /@1=36DÕyÖ Õ ~ .0/@1=30AB]¬}/@143 5h:1$: XVC XnXnX Ê C ÕyÖ . Since is increasing, AvXVC lûê(1ê£XVC l}C for AvXVC lûê;1ê£XVC l Õ Ö has an error of at most ® ~ ê 1 l9¯ º C , the approximation 1 ~V¯ : 1 « ªV¯ : 1 ¬ |m¯ 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS 469 The error formula for Taylor polynomials can be used to bound the error in a particular numer1 ical approximation, or to see how the accuracy of the approximation depends on the value of or / :^hp3 ² § . Observe that the error for a Taylor polynomial of degree § depends on the § the value of 1 ¨ä _ § power of 1 . That means, for example, with a Taylor polynomial of~ degree centered at 0, if we decrease by a factor of 2, the error bound decreases by a factor of . ` Example 2 Compare the errors in the approximations ¢ Õ Solution _ h 5jh:XmCwh: ~9¯ /2XVChp3 Õ and ¢ º h 5jh:[/2XmC XNln3: ~V¯ ÕyÖ />XVC XMlM3 C 1jD¦XmCwh We are approximating by its second-degree Taylor polynomial, first at , and then at 1RD^XVC XMl 1 « . Since we have decreased by a factor of 2, the error bound decreases by a factor of about ~ D[r . To see what actually happens to the errors, we compute them: ¢ £e ¢ e Õ Õ Since º _ A h A h:XmCwh6: h:(XVC XMls: ~9¯ h />XVC XMlM3 ~9¯ /2XmC XMXnXmh Ë hy3-·V/2XVC XnXMXnXN~9hp3D[rVCh /2XmCwhy3 f [DihnChtXNl9h Ë h6AhnChtXNlXnXMXUD^XVC XnXnXmh Ë h f DihnC XMlVhp~ Ë h6AhnC XMlVhp~nlnXUD^XVC XnXnXMXM~Vh , the error has also decreased by a factor of about 8. Convergence of Taylor Series for cos ) 1%D[X doeMf91 We have already seen that the Taylor polynomials centered at 1 for are good approxima1 near 0. (See Figure 10.16.) In fact, for any value of , if we take a Taylor polynomial tions for 1 D[X centered at of high enough degree, its graph is nearly indistinguishable from the graph of the cosine near that point. ¡ Ô K GVL ¡ p G Q ¡ ¡ G ] ] » Ì K GVL and two Taylor polynomials for G near 1D7¶·~ hsA/2¶·~M3 ]¬}/@¶·n~n3DihsA/2¶·~n3 G Qs K GVL Let’s see what happens numerically. Let d?eNfo/@¶·n~n3D[X 1D[X mations to about are ]N/@¶·n~n3D p G Figure 10.16: Graph of p Ô K GVL . The successive Taylor polynomial approxi- ·n~9¯ ·n~9¯t:[/@¶·~M3 DjAvXmC ~nªnª Ë XsCoCtC ¬ ·y|±¯MD XVC XVhpqnq Ë CoCtC /@¶·n~n3D oo DjAvXmC XMXnXnrMqsCoCtC /@¶·n~n3D oo D XVC XnXMXnXN~{CoCtCC 470 Chapter Ten APPROXIMATING FUNCTIONS d?eNfo/@¶·n~n3D[X It appears that the approximations converge to the true value,d?eNf9¶DjAh , very rapidly. Now take 1 X 1RD^¶ a value of somewhat farther away from , say , then and /2¶36DihsA/2¶3 ] ·~9¯}D AvªmC qMª|NrnXsCtCoC ]¬M/2¶36D ot D XmCwhy~ªMqVhCtCoC ] /2¶36D ot D AhMC ~VhnhpªMl{CtCoC /2¶36D ot D AvXmC q Ë Ê XM~{CtCoC ]0_xM/2¶36D ot D AhMC XMXVhprnªsCtCoC ] _ n/2¶36D ot D AvXmC qMqnqMqnXsCtCoC ] _x¬ /2¶36D ot DjAhMC XMXnXnXMX|vCtCoC » ] Ì ho| We seedoeMthat the rate of convergence is somewhat slower; it takes a ²2³ d?degree polynomial to approxf9¶ r eNfo/@¶·n~n3 1 ²>³ degree polynomial approximates imate as accurately as an . If were taken still X farther away from , then we would need still more terms to obtain as accurate an approximation of d?eNf91 . d?eMfV1 1 Using the ratio test, we can show the Taylor series for converges for all values of . In doeMf91 addition, we will prove that it converges to using Theorem 10.1. Thus, we are justified in writing the equality: ¬ 1 doeMf91DihsA 1 : ~V¯ 1±» A |m¯ 1±Ì : Ê ¯ 1 A;to rV¯ for all . Showing the Taylor Series for cos å Converges to cos å The error bound in Theorem 10.1 allows us to see if the Taylor series for a function converges to d?eMfV1 that function. In the series for , the odd powers are missing, so we assume §¨ is even and write s ¨ ¨ /@143D^doeMf91\A] giving 1 hsA ~9¯ Thus, for the Taylor series to converge to doeMf91 o ¨ :^otp:^/zAhp3 ¨ 1 / , we must have ©ª¨¬« ` § : 3?¯ s ¨ s /21436Ú s /2143 D ¨ d?eMfV1\AB] /2143 X / as § f-1 for every § :^hp3?¯ § 1 / 1 ê o § 3u¯ f k ¨/@1=3uC xå Showing 0 as ¨ä _z® doeMf}1 .0/@143vDid?eNf91 / :[hy3 . /@143 ¨nä § _x® Proof Since , the is or1 ² derivative . /2143 êEh X So for all § , we have on the interval between and . By the error bound formula in Theorem 10.1, we have ¨ä _ ¨ ¨ :^otp:[/xAhp3 ~9¯ 1 d?eMfV1Djh:1$: ¤¥C¦ L§=¨ e /@143D^doeMf91\A Ú Û . , no matter what § is. C 1 To show that the errors go to zero, we must show that for a fixed , ¨ä _ 1 / § Ú :hy3u¯ X as Ú § ÛC 1 To see why this is true, think about what happens when § is much larger than . Suppose, for 1RDh Ë C ª example, that . Let’sD^ look at the value of the sequence for § more than twice as big as 17.3, Dª Ê Dª Ë ªMr say § , or § , or § : For § For § For § Dª Ê Dª Ë Dªnr h : ª Ë ¯ : ªnrm¯ : ªnqm¯ h h « ¼ /xh Ë C ªM3 « /xh Ë C ªM3 h Ë Cª Ì /xh Ë C ªM3 « . D h D ªMr h Ë Cª ªMq ª Ë ¯ h Ë C ª ªnr /xh Ë C ªN3 h ª Ë ¯ « ¼ k /xh Ë C ªM3 « ¼ k CtCoC 471 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS h Ë C ªN·aª Ê _ _ we increase « § by 1, the term is multiplied by a number less Since_ is less than , each time /zh Ë C ªN3 ¼ « ¨ ä _ ¼ than . No matter what_ the/zvalue of is, if we keep on dividing it by two, the result gets h Ë C ªM3 ¨nä _x® closer to zero. Thus goes to 0 as § goes to infinity. h Ë C ª 1 ~ 1 We can generalize this by replacing by an arbitrary . For § , the following _ sequence converges to 0 because each term is obtained from its predecessor by multiplying by a number less than : ¨nä _ ¨nä ¨ä « 1® ® 1 / Therefore, the Taylor series : :hy3u¯ § hsAB1 1 k / ¬ ·~V¯p:1 § 1 k :(~M3u¯ / ·a|m¯Aot § :ªN3u¯ koCtCoCoC does converge to d?eNf91 . fzs1 Problems 16 and 151 ask you to show that the Taylor series for and original function for all . In each case, you again need the following limit: ±³°¯ ² ¨ converge to the ¨ 1 é ÕpÖ D^XmC ¯ § Exercises and Problems for Section 10.4 Exercises Use the methods of this section to show how you can estimate the magnitude of the error in approximating the quantities in Problems 1–4 using a third-degree Taylor polynomial about G . Problems J4K ´ 1µ Lá 5. Suppose you approximate by a Taylor polyno on the interval z õ . mial of degree about (a) Is the approximation an overestimate or an underestimate? (b) Estimate the magnitude of the largest possible error. Check your answer graphically on a computer or calculator. 1. ¥ ´ µ J K õ L 3. þ í 4. ¡ ù ÷ Ñ ¥ à K ¶ ¶> LJ J LJ J bÐÒÑ î Ñ ¡ à K GVL ¡ à K GVL bÐÒÑ (a) Where is the approximation an overestimate, and where is it an underestimate? (b) Estimate the magnitude of the largest possible error. Check your answer graphically on a computer or calculator. Î 8. Repeat Problem 7 for the approximation þ í . 2. L (a) The error in approximating p õ by and by õ ¡ à K L õ . (b) The maximum error in approximating t G by ¡ à K GVL for G ( . ¡¥ à K G9L (c) If we want to approximate p G by to an accuracy of within 0.2, what is the largest interval of G -values on which we can work? Give your answer to the nearest integer. 6. Repeat ProblemK 5 using the second-degree Taylor ap¡ L , to á . proximation, 7. Consider the error in using the approximation on the interval Qs u . Î õ Q 9. Use the graphs of â t G and its Taylor polynomi¡ ¥ à K GVL ¡ à K GVL als, and , in Figure 10.17 to estimate the following quantities. í G Qsu Q ó Q6í í ? ó Q6í ¡ ¥ à K G9L ¡ Figure 10.17 ¥ à K GVL 472 Chapter Ten APPROXIMATING FUNCTIONS ¸·º¹ ´ µ ¶ ¶ > , # # , , (b) Add to your table the values of the error , õ , $ # , Q6 Þ Q6 , , Þ for these G -values. (c) Using a calculator or computer, draw a graph of the ¥ bÐÒÑ GQG showing that quantity ¶ ¶ < QR K GVLá ¢ Q ¡ á ¢ Q ¶ ¶ > G â K tòvG±òsG Q6 þ pL BG ? Use the graph to {> ?> and have the (c) Explain why the graphs of Î G for QR G â ¥ ¶ ¶> GQG for shapes they do. ¥ bÐÓÑ G What shape is the graph of confirm that . õ ? Use the graph to 12. (a) Using a calculator, make a table of the values to four decimal places of bÐÒÑ G for ¥ {> ?> (b) Let for ¼> Y. Choose > . a suitable range and graph ¥ 11. What degree Taylor polynomial about G do you need to calculate t to four decimal places? To six decimal places? Justify your answer using the results of Problem 10. GQ6 What shape is the graph of confirm that 10. Give a bound for the maximum possible error for the ð degree Taylor polynomial about GY approximating t G on the interval u . What is the bound for bÐÓÑ G ? G 14. For > -> â , graph the error à p GQ ¡=à K G9L p GQ ¶ ¶ ¶ ¶ > 13. In this problem, you will investigate the error in the ·º¹ 15. Show that the Taylor series about 0 for converges to '½ degree Taylor approximation to for various values of for every . Do this by showing that the error 2 y ½ ¾ . as . (a) Let for ». Using > > a cal- . 16. toShow thatfortheeveryTaylor. series about 0 for converges culator or computer, graph ¥ â pí Q% for õ G B Explain the shape of the graph, using the Taylor expanà sion of t G . Find a bound for for G â . õ á ¢ ð á ¢ á ¢ K GVL G ð Y ð ¥ á ¢ ¥ K GVLá ¢ ¡ Q K ò"GVL Q ¥ Q6 G bÐÒÑ âG â G bÐÓÑ G REVIEW PROBLEMS FOR CHAPTER TEN Exercises For Problems 1–4, find the second-degree Taylor polynomial about the given point. á ¢ 1. 4. ù G 2. î Ñ / G G; / 3. G bÐÒÑ Q LJ , J J KJ ¿ p bÐÒÑ 7. 8. 9. ÞsQRG | QÞ G 4þ Þ For Problems 10–11, expand the quantity in a Taylor series around the origin in terms of the variable given. Give the first four nonzero terms. Ñ ÷ þ Þ 4 6. u v J4K GVL 5. Find the third-degree Taylor polynomial for Î G ò G Q Gò at G . ö õ F 10. ÿ in terms of Fsòÿ F Q 11. u in terms of v In Problems 6–9, find the first four nonzero terms of the Taylor series about the origin of the given functions. Problems 12. Find an exact value for each of the following sums. ö K (b) K (a) ö ò Î p pL ö tL y ö K ò K ö 010#0 ò ?L p pL ò ö K p pL6ò ö ?ò ¥ àbà K p pL ß K ?L K ?L ö ö ò ò í . ö ò K " 40$0#0 ö y tL 18. ò 13. 010#0 mQ í ò Q ü ò ö ô Q 14. ô òÞ=ò´ mò"}ò 0#010 ?ò ò Þ ò ¥ à ò 15. Find the exact value of the sums of the series in Problems 13– yQ yò " A0$0#0 Þ Q ô í ò ó Þ ò 16. mQ ô í ò íp õ Q ? ö -0;010 ô ò íò"íò 17. í õ " í P0$010 í ò í ò í ò Þ K 18. ?L K ò Q uL K Q J4K (b) The force on the particle at the point G is given by S2K GVL Q . For small G , show that the force on the particle is proportional to its distance from the origin. What is the sign of the proportionality constant? Describe the direction in which the force points. ò Ô ò JNS STK ípL íyL 19. A function has ípLûY , õ and J4K Qsu . Find the best estimate you can for í uL . 20. Suppose G is positive but very small. Arrange the following expressions in increasing order: / G / ; K ?òGVL î Ñ G bÐÓÑ / pQ U ¦á ¢ Qs t G ÷?ø ù / G Ñ ÷ 21. By plotting several of its Taylor polynomials and the J4K K function GVLR þ ò£GVL , estimate graphically the interval of convergence of the series expansions for this function about G . Compute the radius of convergence analytically. 22. Use Taylor series to evaluate 23. (a) Find \ bÐÒÑ î Ð à J J K à ¢ L K òGòG G bÐÓÑ î Ñ î Ð L=QG J . Explain your reasoning. J4K J J bÐÓÑ K L L$ (b) Use series to explain why looks ^ like a parabola near . What is the equation of the parabola? J J4K 28. The theory of relativity predicts that when an object moves at speeds close to the speed of light, the object appears heavier. The apparent, or relativistic, mass, , of the object when it is moving at speed is given by the formula à G `QG k ! ¢ à (c) Using your answer to part (b), show that ò í " ò Þ K L " ò K í õ L 40$010 ò ó K Þ L ò a 25. (a) Find the first two nonzero terms of the Taylor series J4K for G9L Þ{QG about G (b) Use your answer to part (a) and the Fundamental Theorem of ¥ Calculus to calculate an approximate value for k Þ{QG à ! aG . (c) Use the substitution G\^ bÐÒÑ to calculate the exact value of the integral in part (b). 26. (a) Find the Taylor series expansion of ÷oø (b) Use Taylor series to find G ÷oø ÷oø ù ÷ bÐÓÑ Ñ G as ½ G bÐÓÑ G . the limit (a) Write the Taylor polynomial of degree 2 for about G\£ . What can you say about the signs of the coefficients of each of the terms of the Taylor polynomial? þ à is the mass of the v 29. The potential energy, , of two gas molecules separated by a distance is given by à v v f e v S Q v S ¼ Q à Q and v à v à v ¥ à Q are positive constants. v v à (a) Show that if , then takes on its minimum à . value, Q K à L Q (b) Write as a series in up through the quadratic term. à (c) For near , show that the difference between and its minimum value is approximately proK à L Q portional to . In other words, show that à L à Q K Q ò is approximately proportional K à L to Q . (d) The force, , between the molecules is given by à ¦Q þ . What is when ? For à near , show that is approximately proportional K à L to Q . v v v v ¿! v v v v v !v v v v 30. The gravitational field at a point in space is the gravitational force that would be exerted on a unit mass placed there. We will assume that the gravitational field strength at a distance away from a mass is ! G 27. A particle moving along the -axis has potential energy K GVL at the point G given by . The potential energy has a minimum at G . `Q (a) Use the formula for to decide what values of are possible. (b) Sketch a rough graph of against , labeling intercepts and asymptotes. (c) Write the first three nonzero terms of the Taylor series for in terms of . (d) For what values of do you expect the series to converge? where á where is the speed of light and object when it is at rest. L á 24. (a) Find the Taylor series for about . (b) Using your answer to part (a), find a Taylor series expansion about G for 473 REVIEW PROBLEMS FOR CHAPTER TEN ß 0$0#0 uL ö JNS>K ?L í õ J " K ! where is constant. In this problem you will investigate the gravitational field strength, , exerted by a system consisting of a large mass and a small mass , with a distance between them. (See Figure 10.18.) v 474 Chapter Ten APPROXIMATING FUNCTIONS u ¡ P O v P O Figure 10.18 (a) Write an expression for the gravitational field ¡ strength, , at the point . v u v u v u J4K Gvò L K Gsò ! aG K2J4K GVL 32. Use Taylor expansions for firm the chain rule: ! ! K2J4K aG J S K GVL K GVLbL4 K GVL9ò M J4K ò K GVLbLbL J S K 33. Suppose all the derivatives of has a critical point at G . L J4K GVL and /0 K GVLbL S K GVL K Gò L ÁÀ Q < ?> =< ?> G ½y¾ â G to explain why the following sum must approach =þ ð : `Q ò í Q õ ö 0#010 ò K Qs?L ?ò 2UT ¥ ð ¼> ?< ò 36. Find a Fourier polynomial of degree three for á ¢ , for G ( . as Þ J4K GVL$ J4K to con- (a) Write the ð Taylor polynomial for at G . (b) What does the Second Derivative test for local maxima and minima say? (c) Use the Taylor polynomial to explain why the Second Derivative test works.  J 1 (a) If the Fourier coefficients of are F and ÿ , show J S are that the Fourier coefficients of its derivative ÿ Q F and . J (b) How are the amplitudes of the harmonics of and JNS related? J J S (c) How are the energy spectra of and related? M # exist at Gû and that ·º¹ Qs 37. Suppose that G9L is a differentiable periodic function J of period . Assume the Fourier series of is differentiable term by term. S K GVL J4K GVL L and in Taylor series and take 31. Expand a limit to confirm the product rule: ! 35. Use the Fourier polynomials for the square wave (b) Assuming is small in comparison to , expand in a series in þ . K (c) By discarding terms in þ L and higher powers, explain why you can view the field as resulting from ò , plus a correction a single particle of mass term. What is the position of the particle of mass ò ? Explain the sign of the correction term. 34. (Continuation of Problem 33) You may remember that the Second Derivative test tells us nothing when the second derivative is zero at the critical point. In this problem you will investigate that special case. Assume has the same properties as in Problem 33, S S K yLÉj . What does the Taylor and that, in addition, polynomial tell you about whether has a local maximum or minimum at G ? M   1 1 ! % & % »& %  & # % 1 & ! 39. Suppose that is a periodic function of period Pand. that is a horizontal shift of , say Show that and have the same energy. J 38. If the Fourier coefficients of are F and ÿ , and the and , and if and Fourier coefficients of are J ò are real, show that the Fourier coefficients of F `ò ÿ `ò are and . J J K GVLv J4K Gûò uL J PROJECTS Exercises 1. Machin’s Formula and the Value of (a) Use the tangent addition formula j h Äà GÅ / : with à D $3D h"i 1j h d ¶ j h »Ã Pj h »Å Nj h CÃ9j h »Å hsA : /xhy~X}·}hnhpqM3 to show that hi j h h"i 1j h 9e f hi j h 9e f h"i 1j h à Šh"i 1j h (b) Use to show that hi j h e f h"i 1j h e f A d /zhy·n~ªnqN3 hp~nX hnhpq ED and d Å D ~ A D d d d h ~nªnq [D d /xha·ln3 h l D d l hp~ C hnC 475 PROJECTS Use a similar method to show that h"i 1j h 7e f | d [D h"i 1j h }e h"i 1j h formula: /xha·~ªMqM3 h l h"i 1j h d (c) Obtain Machin’s d | d /xhy·nln3'A C hnhtq ¶·y| D . (d) Use the Taylor polynomial approximation of degree 5 to the arctangent function to ap¶ proximate the value of . (Note: In 1873 William Shanks used this approach to calculate ¶ to 707 decimal places. Unfortunately, inlM1946 ~r it was found that he made an error in the ²>³ place.) (e) Why do the two series for arctangent converge so rapidly here while the series used in Example 5 on page 462 converges so slowly? 3 2. Approximating the Derivative In applications, .0/@1=3 frequently known the values of a function 1 1 are 1 ~ CoCtC?C only at discrete values , , , Suppose we are interested in approximating the . < /@1 3 derivative . The definition ÇÆ É È± ¯ .=<>/21ma36Déw suggests that for small as follows: . Æ Æ Æ Æ . 3'A.0/21ma3 Æ < /@1=3 Æ 36D^.0/21ma3:;. < /@1±y3 into the approximation for .0/21m: Æ 3`A.0/21mp3 Æ D7. . Æ < /21 : 3 < /@1ma3: . < < /21ma3 Æ ~ :to , we find . < < /21ma3 Æ ~ :^ot This suggests (and it can be proved) that the error in the approximation is bounded as follows: .0/@1 VV Æ : VV 30A.0/@1 3 Æ A.=<>/21 where . < < /2143 ê 3 From Ë i e Mark Kunka Table 10.2 " K2J4K G à u u 3 VV ÊÆ VV 1\AB1m ê ~ ß ÆÌ ê C õ t ö tp õ y õ tpt ö õ õ õ "u y õ õ ÎÍ =Í AÍ AÍ Error Î ß (a) Using Taylor series, suggest an error bound for each of the following finite-difference approximations. . < /21 .0/21mp3`A.0/21mvA 35 .0/21 .=<>/21ma35 (ii) . : Æ Æ 3'A.0/@1 < /21ma35 AÉ.0/21m:;~ ~ Æ Æ Æ 3 A Æ 3 Æ 3:rN.0/@1m{: 3'ABrN.0/@1±vA hp~ Æ (iv) Use each of the formulas in part (a) to ÕyÖ © _ derivative ©= © of © at « ¬ approximate the first 1èD¹X DihtX kohpX kohpX kohtX for . As is decreased by a factor of 10, how does the error decrease? Does this agree with the error bounds found in part (a)? Which is the most accurate formula? Æ Æ .0/@1=3^D ha·y1 © (v) Repeat part (b) using and 1mBD htX º . Why are these formulas not good approximations anymore? Continue to decrease by factors of 10. How small does have to be before formula (iii) is the best approximation? At these smaller values of , what changed to make the formulas accurate again? Æ Æ Æ ]`_xN/2143 3. (a) Use a computer algebra system to find _bM/@143 polynomials and , the Taylor of degree 1RD[X fz 1 doeMf 1 10 about for and . (b) What similarities do you observe between the two poynomials? Explain your observation in terms of properties of sine and cosine. q k Î J4K G à LbL þ L=Q u u ò ¥ (iii) C Æ ÊÆ (i) Such finite-difference approximations are used frequently in programming a computer to solve differential equations. Taylor series can be used to analyze the error in this approximation. Substituting .0/@1m{: Æ 30A.0/@1±p3 we can approximate .0/21m: < /21ma35 Æ .0/@1±: Æ X Notice that as , the error also goes to zero, provided is bounded. .0/@143DÕ Ö 1 D As an example, we take and X . < /21 3´D¦h , so . The error for various values of are given in Table 10.2. We see that decreasing by a factor of 10 decreases the error by a factor of ·~ about 10, as predicted by the error bound . f hp~X Ú Æ 3:.0/@1± 476 Chapter Ten APPROXIMATING FUNCTIONS 4. (a) Use your computer algebra system to find /@1=3 ] /2143 ¼ and ¼ 1, D¦ theX Taylor polynomials of .0/@1=3%D fzs1 degree 7 about for and /@1=3D^f-s1Éd?eNf91 « c . 1 (b) Find the ratio between the coefficient of in the two polynomials. Do the same for the coef14º 1±¼ ficients of and . (c) Describe the pattern in the ratios that you computed in part (b). Explain it using the identity f-/>~143D7~`fz{1Éd?eNf91 . Ö© _ : Ö .0/2143D . 5. Let . Although the formula for 1D7X . is not defined at , we can make . continuous .0/2XN3D h by setting . If we do this, has Taylor 1D[X series around . ] _b /@1=3 (a) Use a computer algebra system to find , q ÏpÐ 1âD the Taylor polynomial of degree 10 about X . for . (b) What do you notice about the degrees of the . terms in the polynomial? What property of does this suggest? . (c) Prove that has the property suggested by part (b). n /@143D Ö f-w/ g 3 6. Let . ]`_-_n/@1=3 (a) Use a computer algebra system to find , 1âD the Taylor polynomial of degree 11 about X /@1=3 , for . (b) What is the percentage error in the approxima/xhy3 ]0__a/zhp3 tion of by />~M3 ? What about the ap] _-_ />~M3 proximation of by ? n n n