Calculus Self-Assessment Practice Worksheet 11 - Answer Key

Math 121 Self-Assessment Practice Worksheet #11 Answer Key Find the Sum for each of the following series. 1. 1−2+ 8 16 32 22 23 24 25 4 − + − + ... = 1 − 2 + − + − + ... 2! 3! 4! 5! 2! 3! 4! 5! = 1 + (−2) + (−2)2 (−2)3 (−2)4 (−2)5 + + + + ... 2! 3! 4! 5! = e−2 Recall: ex = 1 + x + x2 x3 x4 + + + ... 2! 3! 4! 2. 1 1 1 1 1 1 1 1 − + − + ... = − + − + ... 2 3 4 2 3 3π 18π 81π 324π 3π 2 · 9π 3 · (27)π 4 · (81)π 4 = 1 1 1 1 − + − + ... 3π 2(3π)2 3(3π)3 4(3π)4   1 = ln 1 + 3π Recall: ln(1 + x) = x − x2 x3 x4 + − + ... 2 3 4 3. ∞ X (−1)n (ln 9)n 2n+1 n! n=0  n   ln 9 ln 9 −   ∞ ∞ 1 X (− ln 9)n 1X 1 − 2 2 = = = e 2 n=0 2n n! 2 n=0 n! 2  1 9− 2 1 ln = e
 2 x Recall: e = ∞ X xn n=0 n! 1  −1  1 = 9 2 = 2 2  1 √ 9  1 = 2   1 1 = 3 6 and that the series for ex is not alternating. Absorb the minus. 4. 1 − 1 1 1 1 π + − + − . . . = arctan(1) = 3 5 7 9 4 Recall: arctan x = x − x3 x5 x7 + − + ... 3 5 7   π2 π4 π6 π8 −1 :   cos π − 1 = −1 − 1 = −2 − + − ... =  5. − + 2! 4! 6! 8! Recall: cos x = 1 − x2 x4 x6 + − + ... 2! 4! 6! Note: this is the formula for cos(π) but we’re missing the first term of 1 so we subtract the π2 π4 π6 π8 1 from the full sum cos 1 = 1 − + − + − ... 2! 4! 6! 8! 6. ∞ X n=0 n 2n (−1) π = 9n (2n + 1)! ∞ X n=0 3 = π Recall: sin x = n 2n (−1) π = 32n (2n + 1)! ∞ X (−1)n n=0 ∞ X  π 2n  π  3 3 · π (2n + 1)! 3 (−1)n n=0  π 2n+1 3 (2n + 1)! π  3 3 = sin = π 3 π ∞ X (−1)n x2n+1 n=0 (2n + 1)! ∞ ∞ ∞ X X X (−1)n π 2n (−1)n π 2n (−1)n π 2n 7. = = 24n (2n)! (22 )2n (2n)! n=0 42n (2n)! n=0 n=0 = ∞ (−1)n X n=0 Recall: cos x = ∞ X (−1)n x2n n=0 (2n)!  π 2n 4 (2n)! = cos π  4 √ 2 = 2 √ ! √ 3 3 3 = 2 2π


    1 2 1 3 1 4 1 1 1 7 1 1 1 6 6 6 8. − + − + . . . = ln 1 + = ln + − +... = − 2 3 4 6 2·6 3·6 4·6 6 2 3 4 6 6 Recall: ln(1 + x) = x − 9. ∞ X n=0 1 1 = 3!π n 3! Recall: ∞ X xn = n=0 ∞  X n=0 x2 x3 x4 + − + ... 2 3 4  1 π n 1 =  6    1  1 1  1 = = 1 6 π−1 6 1− π π  π π−1  = 1 1−x ∞ ∞ X (−1)n+1 π 2n+1 π X (−1)n π 2n π π π 10. =− = − cos π = − (−1) = 9 (2n)! 9 n=0 (2n)! 9 9 9 n=0 Recall: cos x = ∞ X (−1)n x2n n=0 (2n)!   π4 π6 π8 π3 π5 π7 :0 11. π − + − + ... = π π − + − + . . . = π sinπ = 0 3! 5! 7! 3! 5! 7! 2 x 3 x5 x7 Recall: sin x = x − + − + ... 3! 5! 7! 12. ∞ X n=0 ∞ X ∞ X (−1)n+1 π 2n (−1)n π 2n = − =− 4n (2n + 1)! 22n (2n + 1)! n=0 n=0 2 =− π Recall: sin x = ∞ X n=0 ∞ X (−1)n x2n+1 n=0 (2n + 1)! (−1) n  π 2n+1 2 (2n + 1)!  π 2n  π  (−1) 2 2 · π (2n + 1)! 2 n π  *1 2 2   = − sin = − π 2 π π 6(π − 1) 13. ∞ X (−1)n 3n+1 (ln 4)n n! n=0 =3 ∞ X (−1)n 3n (ln 4)n n=0 n! =3 ∞ X (−3 ln 4)n n=0 n! 3 3 ln(4−3 ) = 3 (4−3 ) = 3 = = 3e−3 ln 4 = 3e
 4 64 x Recall: e = ∞ X xn n=0 14. 1 − e + n! and that the series for ex is not alternating. Absorb the minus. (−e)2 (−e)3 (−e)4 e2 e3 e4 − + − . . . = 1 + (−e) + + + + . . . = e−e 2! 3! 4! 2! 3! 4! Recall: ex = 1 + x + x2 x3 x4 + + + ... 2! 3! 4!   π3 π5 π7 π3 π5 π7 :0 sinπ = 0 15. −π + − + − ... = − π − + − + . . . = − 3! 5! 7! 3! 5! 7! Recall: sin x = ∞ X (−1)n x2n+1 n=0 (2n + 1)!   1 1 1 1 1 1 1 1 16. −1 + − + − + . . . = − 1 − + − + − . . . = − ln(1 + 1) = − ln 2 2 3 4 5 2 3 4 5 Recall: ln(1 + x) = x − x2 x3 x4 + − + ... 2 3 4   2 2 2 2 2 1 1 1 1 1 17. − + − + − + . . . = 2 − + − + − + . . . 2 3 4 5 6 2 3 4 5 6 or = 2 [ln(1 + 1) − 1] = 2 [ln 2 − 1] = 2 ln 2 − 2 Recall: ln(1 + x) = x − Recall: ln(2) = 1 − x2 x3 x4 + − + ... 2 3 4 1 1 1 + − + . . . so we have the full sum minus the first typical term of 1 2 3 4 18. 1 + 1 − π2 π4 π6 π8 : −1   + − + − . . . = 1 + cosπ =1−1= 0 2! 4! 6! 8! Recall: cos x = 1 − 19. x2 x4 x6 + − + ... 2! 4! 6! 1 1 1 1 (−1)2 (−1)3 (−1)4 (−1)5 − + − + ... = + + + + . . . = e−1 2! 3! 4! 5! 2! 3! 4! 5! Recall: ex = 1 + x + x2 x3 x4 + + + ... 2! 3! 4! 1 1 :0 1  Note: e−1 =  1 + (−1) + − + − . . . and by chance the first two terms 1 − 1 end up 2! 3! 4! cancelling giving the original sum. 20. ∞ X n=0 √ ∞ ∞ X X (−1)n (−1)n (−1)n 3 √ √ = = ·√ n (2n + 1) 3 (2n + 1) ( 3)2n (2n + 1) ( 3)2n 3 n=0 n=0 = √ 3 ∞ X n=0 = Recall: arctan x = √ ∞ X  √1 3 2n+1 (−1) √ (−1)n √ = 3 2n + 1 (2n + 1) ( 3)2n+1 n=0  3 arctan 1 √ 3  √ √ π  3π = 3 = 6 6 ∞ X (−1)n x2n+1 n=0 n 2n + 1 21. Find the MacLaurin Series for ln (1 + 6x). P.S. Z > 6 Hint: ln (1 + 6x) = dx. Yes, solve for +C. 1 + 6x  First derive it using substitution and integration. Z Z 6 6 dx = dx ln(1 + 6x) = 1 + 6x 1 − (−6x) =6 Z X ∞ Z X ∞ (−1)n 6n xn dx (−6x) dx = 6 n n=0 n=0 = Z X ∞ (−1)n 6n+1 xn dx n=0 = ∞ X (−1)n 6n+1 xn+1 n+1 n=0 ∞ 0 X (−1)n 6n+1 xn+1 n+1 n=0 = +C To solve for +C, first expand this equation ln(1 + 6x) = 6x − 62 x2 63 x3 64 x4 65 x5 + − + − ... + C 2 3 4 5 Test x = 0 into both sides of the equation above. Note that x = 0 is in the Interval of Convergence for this series because it is the Center point of this power series. * 0 ln 1=  0− 0 0 0 0 + − + − ... + C 2 3 4 5 That is, 0 = 0 − 0 + 0 − 0 + 0 − . . . + C ⇒ C = 0, Substitute above. Finally, ln(1 + 6x) = ∞ X (−1)n 6n+1 xn+1 n=0 n+1 22. Find the MacLaurin Series for ln (9 + x2 ). P.S. Z > 2x 2 Hint: ln (9 + x ) = dx. Yes, solve for +C. 9 + x2  First derive it using substitution and integration.   ln(9 + x2 ) = Z 2x dx = 9 + x2  2x 1 9 + x2  Z dx = 2x   9 1 2 1+ x 9   dx  Z 2x   9  Z 2x X (−1)n x2n dx = 2 9 n=0 9n = =  Z  1  2  dx = x  1− − 9 ∞ Z n ∞  2x X x2 − dx 9 n=0 9 Z X ∞ (−1)n x2n+1 dx 9n+1 n=0 ∞ ∞ ln 9 X X (−1)n x2n+2 (−1)n x2n+2  = 2 + C + ln 9 =2 n+1 (2n + 2) n+1 (2n + 2) 9 9 n=0 n=0 To solve for +C, first expand this equation  2  4 6 8 x x x x + − + ... + C ln(9 + x2 ) = 2 − 9 · 2 92 · 4 93 · 6 94 · 8 Test x = 0 into both sides of the equation above. Note that x = 0 is in the Interval of Convergence for this series because it is the Center point of this power series. 0 0 0 0 − 2 + 3 − 4 +C 9·2 9 ·4 9 ·6 9 ·8 :  ln 9  ln(9 +0) =   That is, ln 9 = 0 − 0 + 0 − 0 + 0 − . . . + C ⇒ C = ln 9, Substitute above. Finally, ln(9 + x2 ) = ∞ X (−1)n x2n+2 + ln 9 n+1 (2n + 2) 9 n=0 ∞
X x2n+1 1 x e − e−x = 23. Show that the MacLaurin Series for 2 (2n + 1)! n=0 ! ∞ ∞
1 X xn X (−x)n 1 x e − e−x = − 2 2 n=0 n! n=0 n! 1 = 2     x2 x3 x4 x5 (−x)2 (−x)3 (−x)4 (−x)5 1+x+ + + + + . . . − 1 + (−x) + + + + + ... 2! 3! 4! 5! 2! 3! 4! 5! 1 = 2     x2 x3 x4 x5 x2 x3 x4 x5 1+x+ + + + + ... − 1 − x + − + − + ... 2! 3! 4! 5! 2! 3! 4! 5!    x2
x3 x4
x5 x2
x3 x4
x5  1
+ x +
+ +
+ + ...  −1 + x −
+ −
+ + ... 3!
4! 5! 3!
4! 5!
2!
2!   1 x3 x5 x 3 x5 = x+ + + ... + x + + + ... 2 3! 5! 3! 5! 1 = 2   ∞ X 1 2
x3 2
x5 x 3 x5 x2n+1 = 2
x + + + ... = x + + + ... = 2
3! 5! 3! 5! (2n + 1)! n=0 ∞
X x2n 1 x −x e +e = 24. Show that the MacLaurin Series for 2 (2n)! n=0 ! ∞ ∞
1 X 1 x xn X (−x)n −x e +e = + 2 2 n=0 n! n=0 n! 1 = 2     (−x)2 (−x)3 (−x)4 (−x)5 x2 x3 x4 x5 + + + + . . . + 1 + (−x) + + + + + ... 1+x+ 2! 3! 4! 5! 2! 3! 4! 5!     x2 x3
x4 x5
x2 x3
x4 x5
1 + x+ +
+ +
+ . . . + 1 − x+ −
+ −
+ ... 2!
3! 4!
5! 2!
3! 4!
5!   1 x 2 x4 x2 x4 = 1+ + + ... + 1 + + + ... 2 2! 4! 2! 4! 1 = 2   ∞ X 1 2
x2 2
x4 x2 x4 x2n = + + ... = 1 + + + ... = 2
+ 2
2! 4! 2! 4! (2n)! n=0 1 Z 25. Use Series to Estimate 0 ∞ X
1 x3 sin x2 dx with error less than . Justify. 100 n 2n+1 (−1) x (2n + 1)! n=0 Z 1 X Z 1 Z 1 X ∞ ∞ 2n+1
(−1)n (x2 ) (−1)n x4n+2 3 3 2 3 x x sin x dx = dx = x dx (2n + 1)! (2n + 1)! 0 0 0 n=0 n=0 Recall: sin x = Z = 0 = ∞ ∞ 1X X (−1)n x4n+6 (−1)n x4n+5 dx = (2n + 1)! (2n + 1)!(4n + 6) n=0 n=0 x6 x10 x14 − + − ... 6 3!(10) 5!(14) 1 0 1 0 = 1 1 1 − + + . . . − (0 − 0 + 0 − . . .) 6 60 1680 ≈ 1 10 1 9 1 3 − = − = = ← estimate 6 60 60 60 60 20 Using the Alternating Series Estimation Theorem, if we approximate the actual sum with only the first two terms, the error from the actual sum will be at most the absolute value of 1 1 1 the next (first neglected) term, . Here < as desired. 1680 1680 100 Z 26. Use Series to Estimate 0 Recall ex = ∞ X n=0 xn n! 1 3 x4 e−x dx with error less than 1 . Justify. 10 Z 1 4 −x3 xe Z dx = 0 x 0 Z = 0 = 1 4 ∞ X (−x3 )n n=0 ∞ 1X n=0 n! Z 1 x dx = 4 0 ∞ X (−1)n x3n n=0 n! ∞ X (−1)n x3n+5 (−1)n x3n+4 dx = n! n!(3n + 5) n=0 x5 x8 x11 − + − ... 0!(5) 1!(8) 2!(11) dx 1 0 1 0 = 1 1 1 − + − . . . − (0 − 0 + 0 − 0 + . . .) 5 8 22 ≈ 1 1 8 5 3 − = − = 5 8 40 40 40 ← estimate Note this is an alternating series. Use the Alternating Series Estimation Theorem (ASET). If we approximate the actual sum with only the first two terms, the error from the actual 1 1 1 < sum will be at most the absolute value of the next (first neglected) term, . Here 22 22 10 as desired. 1 1 27. Estimate √ with error less than . Justify. 100 e ∞ X xn x2 x3 x 4 =1+x+ + + + ... Recall ex = n! 2! 3! 4! n=0


1 2 1 3 1 4 − − − 1 1 1 2 2 2 √ = e− 2 = 1 − + + + + ... 2 2! 3! 4! e =1− 1 1 1 1 + 4 − 8 + 16 + . . . 2 2! 3! 4! =1− 1 1 1 1 + − + + ... 2 8 48 384 ≈1− 1 1 29 1 48 24 6 1 ← estimate + − = − + − = 2 8 48 48 48 48 48 48 Using ASET we can estimate the full sum using only the first four terms with error at most 1 1 (the first neglected term in absolute value) < as desired. 384 100 28. Estimate sin(1) with error less than 1 . Justify. Hint: 7! = 5040 1000 x 3 x5 x7 + − + ... 3! 5! 7! 1 1 1 sin 1 = 1 − + − + . . . 3! 5! 7! Recall: sin x = x − =1− 1 1 1 + − + ... 6 120 5040 ≈1− 1 1 120 20 1 101 + = − + = ← estimate 6 120 120 120 120 120 Using ASET we can estimate the full sum using only the first four terms with error at most 1 1 < as desired. (the first neglected term in absolute value) 5040 1000   1 1 29. Estimate arctan . Justify. with error less than 2 100 x3 x5 x7 Recall: arctan x = x − + − + ... 3 5 7  3  5  7 1 1 1   1 1 2 2 2 arctan = − + − + ... 2 2 3 5 7 1 1 1 8 32 − ... = − + 2 3 5 = 1 1 1 − + − ... 2 24 160 ≈ 1 1 12 1 11 − = − = ← estimate 2 24 24 24 24 Using ASET we can estimate the full sum using only the first four terms with error at most 1 1 (the first neglected term in absolute value) < as desired. 160 100