In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.
1. f(x)=1+x+x2f(x)=1+x+x2 at a=1a=1
2. f(x)=1+x+x2f(x)=1+x+x2 at a=−1a=−1
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f(−1)=1;f′(−1)=−1;f''(−1)=2;f(x)=1−(x+1)+(x+1)2f(−1)=1;f′(−1)=−1;f''(−1)=2;f(x)=1−(x+1)+(x+1)2
3. f(x)=cos(2x)f(x)=cos(2x) at a=πa=π
4. f(x)=sin(2x)f(x)=sin(2x) at a=π2a=π2
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f′(x)=2cos(2x);f''(x)=−4sin(2x);p2(x)=−2(x−π2)f′(x)=2cos(2x);f''(x)=−4sin(2x);p2(x)=−2(x−π2)
5. f(x)=√xf(x)=√x at a=4a=4
6. f(x)=lnxf(x)=lnx at a=1a=1
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f′(x)=1x;f''(x)=−1x2;p2(x)=0+(x−1)−12(x−1)2f′(x)=1x;f''(x)=−1x2;p2(x)=0+(x−1)−12(x−1)2
7. f(x)=1xf(x)=1x at a=1a=1
8. f(x)=exf(x)=ex at a=1a=1
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p2(x)=e+e(x−1)+e2(x−1)2p2(x)=e+e(x−1)+e2(x−1)2
In the following exercises, verify that the given choice of n in the remainder estimate |Rn|≤M(n+1)!(x−a)n+1|Rn|≤M(n+1)!(x−a)n+1, where M is the maximum value of |f(n+1)(z)||f(n+1)(z)| on the interval between a and the indicated point, yields |Rn|≤11000|Rn|≤11000. Find the value of the Taylor polynomial pn of ff at the indicated point.
9. [T] √10;a=9,n=3√10;a=9,n=3
10. [T] (28)13;a=27,n=1(28)13;a=27,n=1
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d2dx2x13=−29x53≥−0.00092…d2dx2x13=−29x53≥−0.00092… when x≥28x≥28 so the remainder estimate applies to the linear approximation x13≈p1(27)=3+x−2727x13≈p1(27)=3+x−2727, which gives (28)13≈3+127=3.¯037(28)13≈3+127=3.¯¯¯¯¯¯¯¯037, while (28)13≈3.03658(28)13≈3.03658.
11. [T] sin(6);a=2π,n=5sin(6);a=2π,n=5
12. [T] e2; a=0,n=9a=0,n=9
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Using the estimate 21010!<0.00028321010!<0.000283 we can use the Taylor expansion of order 9 to estimate exex at x=2x=2. as e2≈p9(2)=1+2+222+236+⋯+299!=7.3887…e2≈p9(2)=1+2+222+236+⋯+299!=7.3887… whereas e2≈7.3891e2≈7.3891.
13. [T] cos(π5);a=0,n=4cos(π5);a=0,n=4
14. [T] ln(2);a=1,n=1000ln(2);a=1,n=1000
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Since dndxn(lnx)=(−1)n−1(n−1)!xn,R1000≈11001dndxn(lnx)=(−1)n−1(n−1)!xn,R1000≈11001. One has p1000(1)=1000∑n=1(−1)n−1n≈0.6936p1000(1)=1000∑n=1(−1)n−1n≈0.6936 whereas ln(2)≈0.6931⋯ln(2)≈0.6931⋯.
15. Integrate the approximation sint≈t−t36+t5120−t75040sint≈t−t36+t5120−t75040 evaluated at πt to approximate ∫10sinπtπtdt∫10sinπtπtdt.
16. Integrate the approximation ex≈1+x+x22+⋯+x6720ex≈1+x+x22+⋯+x6720 evaluated at −x2 to approximate ∫10e-x2dx∫10e-x2dx.
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∫10(1−x2+x42−x66+x824−x10120+x12720)dx∫10(1−x2+x42−x66+x824−x10120+x12720)dx
=1−133+1510−1742+199⋅24−111120⋅11+113720⋅13≈0.74683=1−133+1510−1742+199⋅24−111120⋅11+113720⋅13≈0.74683 whereas ∫10e-x2dx≈0.74682∫10e-x2dx≈0.74682.
In the following exercises, find the smallest value of n such that the remainder estimate |Rn|≤M(n+1)!(x−a)n+1|Rn|≤M(n+1)!(x−a)n+1, where M is the maximum value of |f(n+1)(z)||f(n+1)(z)| on the interval between a and the indicated point, yields |Rn|≤11000|Rn|≤11000 on the indicated interval.
17. f(x)=sinxf(x)=sinx on [-π,π],a=0[-π,π],a=0
18. f(x)=cosxf(x)=cosx on [−π2,π2],a=0[−π2,π2],a=0
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Since f(n+1)(z)f(n+1)(z) is sinzsinz or coszcosz, we have M=1M=1. Since |x−0|≤π2|x−0|≤π2, we seek the smallest n such that πn+12n+1(n+1)!≤0.001πn+12n+1(n+1)!≤0.001. The smallest such value is n=7n=7. The remainder estimate is R7≤0.00092R7≤0.00092.
19. f(x)=e−2xf(x)=e−2x on [−1,1],a=0[−1,1],a=0
20. f(x)=e-xf(x)=e-x on [−3,3],a=0[−3,3],a=0
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Since f(n+1)(z)=±e-zf(n+1)(z)=±e-z one has M=e3M=e3. Since |x−0|≤3|x−0|≤3, one seeks the smallest n such that 3n+1e3(n+1)!≤0.0013n+1e3(n+1)!≤0.001. The smallest such value is n=14n=14. The remainder estimate is R14≤0.000220R14≤0.000220.
In the following exercises, the maximum of the right-hand side of the remainder estimate |R1|≤max|f''(z)|2R2|R1|≤max|f''(z)|2R2 on [a−R,a+R][a−R,a+R] occurs at a or a±Ra±R. Estimate the maximum value of R such that max|f''(z)|2R2≤0.1max|f''(z)|2R2≤0.1 on [a−R,a+R][a−R,a+R] by plotting this maximum as a function of R.
21. [T] ex approximated by 1+x,a=01+x,a=0
22. [T] sinxsinx approximated by x, a=0a=0
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Since sinxsinx is increasing for small x and since sin''x=-sinxsin''x=-sinx, the estimate applies whenever R2sin(R)≤0.2R2sin(R)≤0.2, which applies up to R=0.596R=0.596.
23. [T] lnxlnx approximated by x−1,a=1x−1,a=1
24. [T] cosxcosx approximated by 1,a=01,a=0
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Since the second derivative of cosxcosx is -cosx-cosx and since cosxcosx is decreasing away from x=0x=0, the estimate applies when R2cosR≤0.2R2cosR≤0.2 or R≤0.447R≤0.447.
In the following exercises, find the Taylor series of the given function centered at the indicated point.
26. 1+x+x2+x31+x+x2+x3 at a=−1a=−1
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(x+1)3−2(x+1)2+2(x+1)(x+1)3−2(x+1)2+2(x+1)
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Values of derivatives are the same as for x=0x=0 so cosx=∞∑n=0(−1)n(x−2π)2n(2n)!cosx=∞∑n=0(−1)n(x−2π)2n(2n)!
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cos(π2)=0,-sin(π2)=−1cos(π2)=0,-sin(π2)=−1 so cosx=∞∑n=0(−1)n+1(x−π2)2n+1(2n+1)!cosx=∞∑n=0(−1)n+1(x−π2)2n+1(2n+1)!, which is also -cos(x−π2)-cos(x−π2).
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The derivatives are f(n)(1)=ef(n)(1)=e so ex=e∞∑n=0(x−1)nn!ex=e∞∑n=0(x−1)nn!.
33. 1(x−1)21(x−1)2 at a=0a=0 (Hint: Differentiate 11−x.11−x.)
34. 1(x−1)31(x−1)3 at a=0a=0
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1(x−1)3=-(12)d2dx211−x=-∞∑n=0((n+2)(n+1)xn2)1(x−1)3=-(12)d2dx211−x=-∞∑n=0((n+2)(n+1)xn2)
35. F(x)=∫x0cos(√t)dt;f(t)=∞∑n=0(−1)ntn(2n)!F(x)=∫x0cos(√t)dt;f(t)=∞∑n=0(−1)ntn(2n)! at a=0a=0 (Note: ff is the Taylor series of cos(√t).cos(√t).)
In the following exercises, compute the Taylor series of each function around x=1x=1.
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2−x=1−(x−1)2−x=1−(x−1)
38. f(x)=(x−2)2f(x)=(x−2)2
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((x−1)−1)2=(x−1)2−2(x−1)+1((x−1)−1)2=(x−1)2−2(x−1)+1
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11−(1−x)=∞∑n=0(−1)n(x−1)n11−(1−x)=∞∑n=0(−1)n(x−1)n
41. f(x)=12x−x2f(x)=12x−x2
42. f(x)=x4x−2x2−1f(x)=x4x−2x2−1
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x∞∑n=02n(1−x)2n=∞∑n=02n(x−1)2n+1+∞∑n=02n(x−1)2nx∞∑n=02n(1−x)2n=∞∑n=02n(x−1)2n+1+∞∑n=02n(x−1)2n
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e2x=e2(x−1)+2=e2∞∑n=02n(x−1)nn!e2x=e2(x−1)+2=e2∞∑n=02n(x−1)nn!
[T] In the following exercises, identify the value of x such that the given series ∞∑n=0an∞∑n=0an is the value of the Maclaurin series of f(x)f(x) at xx. Approximate the value of f(x)f(x) using S10=10∑n=0anS10=10∑n=0an.
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x=e2;S10=34,9134725≈7.3889947x=e2;S10=34,9134725≈7.3889947
47. ∞∑n=0(−1)n(2π)2n(2n)!∞∑n=0(−1)n(2π)2n(2n)!
48. ∞∑n=0(−1)n(2π)2n+1(2n+1)!∞∑n=0(−1)n(2π)2n+1(2n+1)!
Show Solution
sin(2π)=0;S10=8.27×10−5sin(2π)=0;S10=8.27×10−5
The following exercises make use of the functions S5(x)=x−x36+x5120S5(x)=x−x36+x5120 and C4(x)=1−x22+x424C4(x)=1−x22+x424 on [-π,π][-π,π].
49. [T] Plot sin2x−(S5(x))2sin2x−(S5(x))2 on [-π,π][-π,π]. Compare the maximum difference with the square of the Taylor remainder estimate for sinxsinx.
50. [T] Plot cos2x−(C4(x))2cos2x−(C4(x))2 on [-π,π][-π,π]. Compare the maximum difference with the square of the Taylor remainder estimate for cosxcosx.
Show Solution
The difference is small on the interior of the interval but approaches 11 near the endpoints. The remainder estimate is |R4|=π5120≈2.552|R4|=π5120≈2.552.
51. [T] Plot |2S5(x)C4(x)−sin(2x)| on [-π,π].
52. [T] Compare S5(x)C4(x) on [−1,1] to tanx. Compare this with the Taylor remainder estimate for the approximation of tanx by x+x33+2x515.
Show Solution
The difference is on the order of 10−4 on [−1,1] while the Taylor approximation error is around 0.1 near ±1. The top curve is a plot of tan2x−(S5(x)C4(x))2 and the lower dashed plot shows t2−(S5C4)2.
53. [T] Plot ex−e4(x) where e4(x)=1+x+x22+x36+x424 on [0,2]. Compare the maximum error with the Taylor remainder estimate.
54. (Taylor approximations and root finding.) Recall that Newton’s method xn+1=xn−f(xn)f′(xn) approximates solutions of f(x)=0 near the input x0.
- If f and g are inverse functions, explain why a solution of g(x)=a is the value f(a)off.
- Let pN(x) be the Nth degree Maclaurin polynomial of ex. Use Newton’s method to approximate solutions of pN(x)−2=0 for N=4,5,6.
- Explain why the approximate roots of pN(x)−2=0 are approximate values of ln(2).
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a. Answers will vary. b. The following are the xn values after 10 iterations of Newton’s method to approximation a root of pN(x)−2=0: for N=4,x=0.6939...; for N=5,x=0.6932...; for N=6,x=0.69315...;. (Note: ln(2)=0.69314...) c. Answers will vary.
In the following exercises, use the fact that if q(x)=∞∑n=1an(x−c)n converges in an interval containing c, then limx→cq(x)=a0 to evaluate each limit using Taylor series.
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ln(1−x2)x2→-1
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cos(√x)−12x≈(1−x2+x24!−⋯)−12x→−14
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