2
Section 1.2
2. Compute each of the following limits and determine the corresponding rate of convergence. (a) (b) (c) (d)
ex −1 x limx→0 sinx x x x−x limx→0 e −cos x2 1+x x2 /2−x4 /24 limx→0 cos x−1+ x6
limx→0
(a) From Taylor’s Theorem, ex = 1 + x x + + 21 x2 eξ for some ξ ξ between between 0 and x. Hence, ex 1 1 = 1 + xeξ . x 2 Because ex 1 1 1 = x eξ < x x 2
−
− −
||
||
for all x all x satisfying x < ln 2, it follows that
| |
lim
ex
− 1 = 1 with rate of convergence O O((x).
x
x→0
3
(b) From Taylor’s Theorem, sin x = x x6 cos ξ ξ for some ξ ξ between 0 and x. Then, sin x x 2 =1 cos ξ x 6 and sin x 1 1 2 1 = x2 cos ξ x . x 6 6
−
−
Finally,
−
|
|≤
sin x = 1 with rate of convergence O O((x2 ). x→0 x lim
(c) From Taylor’s Theorem, we have 1 1 ex = 1 + x + x2 + x3 eξ 2 6
1
and cos x = 1
− 12 x2 + 16 x3 sin ξ 2
for some ξ some ξ 1 and and ξ ξ 2 between 0 and x. Then ex
− cos x − x = 1 + x eξ − sin ξ 2 . 6 x2 For sufficiently small x x,, e ξ < 2 2,, so |eξ − sin ξ 2 | < 2 + 1 = 3. 3. Thus, ex − cos x − x |x| ξ − 1 = e − sin ξ 2 ≤ 1 |x|, 1
1
1
1
x2
6
2
Convergence
3
and lim
ex
− cos x − x = 1
with rate of convergence convergence O O((x).
x2
x→0
(d) From Taylor’s Theorem, we have cos x = 1
1 6 1 − 12 x2 + 241 x4 − 720 x + x8 cos ξ 8!
for some ξ some ξ between between 0 and x x.. Hence, cos x
− 1 + 21 x2 − 241 x4 = − x6
and
1 2 2 6
cos x − 1 + x − x
1 4 24 x
1 1 2 + = x cos ξ 720 8!
It therefore follows that lim
x→0
cos x
1 1 + x2 cos ξ, 720 8!
|
− 1 + 21 x2 − 241 x4 = − x6
| ≤ 8!1 |x2|.
1 720
with rate of convergence O O((x2 ). 3. Numerically determine which which of the following sequences approaches approaches 1 faster, and then confirm the numerical evidence by determining the rate of convergence of each sequence. sin x2 (sin x)2 lim v e r s u s l i m . x→0 x2 x→0 x2 The values in the following table suggest that 2 than (sinx2x) .
x 1.000 1.000 0.100 0.1 00 0.010 0.0 10 0.001 0.0 01
sin x2 x2 0.8414 0.8 414709 709848 848079 0790 0 0.9999 0.9 999833 833334 334166 1667 7 0.9999 0.9 999999 999983 983333 3333 3 0.9999 0.9 999999 999999 999998 9983 3
sin x2 x2
converges towa toward rd 1 more rapidly
(sin x)2 x2 0.7080 0.7 080734 734182 182735 7357 7 0.9966 0.9 966711 711079 079379 3792 2 0.9999 0.9 999666 666671 671111 1111 1 0.9999 0.9 999996 996666 666667 6671 1
To confirm this conclusion, note that by Taylor’s Theorem, sin u = = u u
− 16 u3 cos ξ,
for some ξ some ξ between between 0 and u and u.. Using the substitution u = = x x 2 , we find sin x2 = x2
− 16 x6 cos ξ
4
Section 1.2
for some ξ some ξ between between 0 and x 2 . Consequently Consequently,, 2
sin x x
1 − 1 = 6 x | cos ξ | ≤ 16 x . 4
2
4
Starting from f from f ((x) = (sin x)2 , we find
f ′ (x) = 2sin x cos x = sin2x, sin2x, f ′′ (x) = 2cos 2cos 2x, f ′′′ (x) = and f and f (4) (x) =
−4sin2 4sin2x, x,
−8cos2 8cos2x x. Therefore, (sin x)2 = x 2
− 13 x4 cos2 cos2ξ ξ
for some ξ some ξ between between 0 and x x,, and 2
(sin x) x 2
Finally,
1 1 − 1 = 3 x | cos2 cos2ξ ξ | ≤ x . 3 2
sin x2 = 1 + O(x4 ) 2 x
2
(sin x)2 and = 1 + O(x2 ). 2 x
4. Suppose that 0 < 0 < a < b. b. (a) Show that if α α n = α + O(1 (1/n /nb ), then α then α n = α + O(1 (1/n /na ). (b) Show that if f ( f (x) = L + O(xb ), then f then f ((x) = L + O(xa ). (a) Suppose αn = α + + O O(1 (1/n /nb ). The Then, n, there exists exists a con consta stant nt λ such that for 1 sufficiently large n n,, αn α λ n . Beca Because use a a < b, it follows that n a < nb and n1 > n1 for all n all n > 1 1.. Thus,
| − | ≤
a
b
b
|αn − α| ≤ λ n1b < λ n1a , and α and α n = = α α + O(1 (1/n /na ). (b) Suppose f Suppose f ((x) = L + O(xb ). Then, there exists a constant K K such such that for all sufficiently small x small x,, f f ((x) L K x b . Because a Because a < b, b , it follows that for all x 1, x b x a . Thus, for sufficiently small x x,,
| | ≤ | | ≤ | |
|
− |≤ | |
|f f ((x) − L| ≤ K |x|b ≤ K |x|a, and f and f ((x) = L + O (xa ). 5. Suppose that f that f 1 (x) = L 1 + O(xa ) and f and f 2 (x) = L2 + O(xb ). Show that c1 f 1 (x) + c2 f 2 (x) = c 1 L1 + c2 L2 + O(xc ),
Convergence
5
where c where c = = min(a, min(a, b). Suppose f 1 (x) = L1 + O(xa ) and Suppose f and f f 2 (x) = L = L 2 + O(xb ). Then, there there exist constants constants a K 1 and and K K 2 such that for all sufficiently small x x,, f 1 (x) L1 K 1 x and f 2 (x) b L2 K 2 x . Let c Let c 1 and and c c 2 be any real numbers. Applying the triangle inequality, we find
|
|≤ | |
− |≤ | |
|
−
|c1f 1(x) + c2f 2(x) − (c1L1 + c2L2)| ≤ |c1||f 1(x) − L1| + |c2||f 2(x) − L2| ≤ |c1|K 1|xa| + |c2|K 2|xb|. Now, let c let c = = min(a, min( a, b). Then, for |x| < 1 1,, |c1|K 1|xa| + |c2|K 2|xb| < |c1|K 1|xc| + |c2|K 2|xc| = ( |c1|K 1 + |c2|K 2)|xc|. Consequently,
c1 f 1 (x) + c2 f 2 (x) = c 1 L1 + c2 L2 + O(xc ).
6. The table below lists the errors of successive iterates for three different methods for approximating 5. Estimate the order of convergence of each method, and explain how you arrived at your conclusions.
√ 3
Method 1 4.0 10−2 9.1 10−4 4.8 10−7
× × ×
Method 2 3.7 10−4 1.2 10−15 1.5 10−60
× × ×
Method 3 4.3 10−3 1.8 10−8 1.4 10−24
× × ×
If a sequence converges of order α, then the error in each term of the sequence is roughly the error in the previous term raised to the power α. Fr From om the data data for for “Method 1,” we see that each error is roughly the previous error squared; therefore, we estimate the order of convergence to be α = 2. Fr From om the data data for “Method “Method 2,” we see that each error is roughly the previous error raised to the fourth power; therefore, we estimate the order of convergence to be α α = = 4. Finally, from the data for “Method 3,” we see that each error is roughly the previous error raised to the third power; therefore, we estimate the order of convergence to be α = 3. 7. Let pn be a sequence which converges to the limit p.
{ }
(a) If
lim
n→∞
| p pn+1 − p| = 0, | p pn − p|α 0 ,
what can be said about the order of convergence of pn to p? (b) If lim
n→∞
| p pn+1 − p| → ∞, | p pn − p|α
{ }
what can be said about the order of convergence of pn to p?
{ }
6
Section 1.2
(a) If lim
n→∞
| p pn+1 − p| = 0,0 , | p pn − p|α
then the numerator numerator app approach roaches es zero fast faster er than the denominato denominator. r. In order to achieve a nonzero limit, we must increase the power in the denominator. Therefore Theref ore,, the order of convergence convergence must be grea greater ter than α α.. (b) If pn+1 p p lim , n→∞ p pn p α then the denominator denominator approaches approaches zero faster than the numer numerato ator. r. In order to achieve a nonzero limit, we must decrease the power in the denominator. Therefore, the order of convergence must be less than α α..
| − | →∞ | − |
8. Suppose theory indicates that the sequence pn converges to p of order 1.5. Explain how you would numerically verify this order of convergence.
{ }
To numerically verify the order of convergence, calculate the ratio
| p pn+1 − p| | p pn − p|1.5
for several successive values of n. If the order order of converge convergence nce is α = 1.5, these ratios should approach a constant, specifically the asymptotic error constant.
√
9. Theory indicates that the following sequence should converge to 3 of order 1.618. 1.61 8. Does the seque sequence nce actually actually achieve achieve an order of convergen convergence ce of 1.618 1.618?? If not, what is the actual order? n
0 1 2 3 4 5
pn 2.0000 2.0 000000 000000 000000 00000 00 1.6666 1.6 666666 666666 666666 66667 67 1.7272 1.7 272727 727272 272727 72727 27 1.7321 1.7 321428 428571 571428 42857 57 1.7320 1.7 320506 506804 804317 31722 22 1.7320 1.7 320508 508075 075654 65499 99
Because the values in the third column of the following table appear to be approaching a constant, the evidence suggests that the sequence does, in fact, converge toward 3 with order of convergence α α = = 1.618 618..
√
n 1 2 3 4 5 6
pn 2.0000 2.0 000000 000000 000000 00000 00 1.666 1.6 6666 6666 6666 6666 6666 6667 67 1.727 1.7 2727 2727 2727 2727 2727 2727 27 1.732 1.7 3214 1428 2857 5714 1428 2857 57 1.732 1.7 3205 0506 0680 8043 4317 1722 22 1.732 1.7 3205 0508 0807 0756 5654 5499 99
| p pn − √ 3|/| p pn 1 − √ 3|1.618 −
0.5506 0.55 0660 6002 0295 9531 3142 42 0.39 0. 3942 4292 9299 9985 8515 1516 16 0.52 0. 5235 3588 8803 0316 1628 2884 84 0.43 0. 4310 1007 0791 9144 4414 1420 20 0.48 0. 4852 5255 5581 8157 5793 9327 27
Convergence
7
10. Theory indicates that the following sequence should converge to 4 /3 of order 1.618.. Does the seque 1.618 sequence nce actually actually achieve achieve an order of convergen convergence ce of 1.61 1.618? 8? If not, what is the actual order? pn 1.4986640 1.4986 640985 985800 80016 16 1.4973 1.4 973539 539977 977922 92205 05 1.4288 1.4 288019 019773 773353 35339 39 1.4010 1.4 010929 929153 153895 89552 52 1.3764 1.3 764936 936760 760514 51456 56 1.3613 1.3 613457 457455 455731 73130 30 1.3510 1.3 510344 344825 825008 00881 81 1.3444 1.3 444798 798506 506950 95066 66
n
0 1 2 3 4 5 6 7
Because the values in the third column of the following table are increasing with n, the evidence suggests that the sequence does not have order of convergence α = 1.618 618,, but rathe ratherr that the order of convergence convergence is less than 1.618. Becau Because se the values in the fourth column appear to be approaching a constant, these values suggest that the sequence is converging to 4 4/ /3 with order of convergence α α = = 1. n 1 2 3 4 5 6 7 8
pn 1.4986 1.4 986640 640985 985800 80022 1.49735399779221 1.42880197733534 1.40109291538955 1.37649367605146 1.36134574557313 1.35103448250088 1.34447985069507
| p pn − 4/3|/| p pn 1 − 4/3|1.618 | p pn − 4/3|/| p pn 1 − 4/3| −
−
3.01718763541581 1.77891367138598 3.03079120639280 3.36181849329742 4.52671513900300 5.75689539760301 7.61855893491390
0.99207588021590 0.58205253781266 0.70975745769255 0.63696294174768 0.64903127444432 0.63190377951100 0.62970586012393
11. Show that the convergence of the sequence generated by the formula xn+1 =
x3n + 3x 3 xn a 2 3xn + a
√ a is third-order. What is the asymptotic error constant?
toward Note
xn+1
3 3 xn a √ − √ a = xn + 3x − a
3x2n + a
= =
Thus, lim
n→∞
|xn+1 −√ √ a| = lilimm |xn − a|3 n
− 3x2n√ a + 3x 3xn a − a3/2 3x2n + a √ (x − a)3 x3n
n
3x2n + a
.
1 1 = . 2 →∞ 3x + a 4a n
8
Section 1.2
Consequently, xn constant λ constant λ = 41a .
→ √ a with order of convergence α = 3 and asymptotic error
12. Let a be a non non-ze -zero ro real number. number. For any x0 satisfying 0 < x0 < 2/a /a,, the recursive sequence defined by xn+1 = xn (2
− axn)
converges to 1/a 1 /a.. What are the order of con conver vergenc gencee and the asymptotic asymptotic error constant? Note xn+1
− 1a = xn(2 − axn) − a1
1 −ax2n + 2x 2 xn − a
= =
Thus, lim
n→∞
Consequently, xn constant λ constant λ = = a a..
→
1 a with
−a
x2n
−
2 1 xn + 2 = a a
−a
xn
−
1 a
2
.
|xn+1 − a1 | = lim a = a. |xn − a1 |2 n = a. →∞
order ord er of convergence convergence α = 2 and asymp asymptoti toticc erro errorr
13. Suppose that the sequence pn conv converges erges linearly to the limit p with asymptotic error constant λ constant λ.. Further suppose that p n+1 p p,, p n p p and and p n−1 p p are are all of the same sign. Show that pn+1 pn λ. pn pn−1
{ }
−
−
−
− ≈ −
Suppose the sequen sequence ce pn converges linearly to p with asymptotic error constant λ. Then pn+1 p p lim = λ, n→∞ p pn p
{ }
|
so, for sufficiently large n n,,
− | | − |
| p pn+1 − p| ≈ λ| p pn − p|. Moreover,
| p pn − p| ≈ λ| p pn 1 − p| or | p pn 1 − p| ≈ λ1 | p pn − p|. Because we are given that p n+1 − p p,, p n − p p and and p n 1 − p p are are all of the same sign, −
−
−
we may drop the absolute values values from the above exp expressio ressions. ns. Now Now,, pn+1 pn pn pn−1
−
−
=
pn+1 p ( pn pn p ( pn−1
− − − −
− p p)) − p p))
Convergence
9 λ( pn p p)) ( pn p p)) 1 pn p λ ( pn p) p) λ 1 = λ. 1 λ1
− − − − − − − −
≈ =
14. A sequence pn converges superlinearly to to p provided
{ }
| p pn+1 − p| = 0.0 . n | p pn − p| Show that if p of order α order α for α > 1, then { pn } converges superlinearly to p n → p p of lim
→∞
p.
Suppose the sequence pn converges p of order α > 1 with asymp asymptoti toticc erro errorr constant λ constant λ.. Then pn+1 p p lim = λ. n→∞ p pn p α
{ }
| − | | − |
Then lim
n→∞
| p pn+1 − p| | p pn − p|
| p pn+1 − p| · | p pn − p|α 1 | p pn − p|α | p pn+1 − p| · lim | p pn − p|α | p pn − p|α n −
=
lim
n→∞
=
lim
n→∞
1
−
→∞
= λ 0 = 0.
·
Therefore, pn converges superlinearly to p to p..
{ } 15. Suppose that { pn } converges superlinearly to p (see Exercise 14). Show that | p pn+1 − pn| = 1.1 . lim n | p pn − p| →∞
Note that
pn+1 pn pn+1 p = pn p pn
− −
− − ( pn − p p)) = pn+1 − p − 1. − p pn − p
Because pn converges superlinearly to p to p,, it then follows that
{ }
lim
n→∞
| p pn+1 − pn| = lim pn+1 − p − 1 = |0 − 1| = 1.1 . | p pn − p| n pn − p →∞
16.. (a 16 (a)) Determine the third-degree Taylor polynomial and associated remainder term for the function f function f ((x) = ln(1 x). Use x Use x 0 = 0.
−
10
Section 1.2
(b) Using the results of part (a), approximate ln(0 .25) and compute the theoretical oreti cal error bound associated with this approximat approximation. ion. Comp Compare are the theoretic theor etical al error bound with the actua actuall error error.. (c) Compute the following limit and determine the corresponding rate of convergence: ln(1 x) + x + 21 x2 lim . x→0 x3
−
(a) Let Let f f ((x) = ln(1
− x). Then 1 1 f (x) = − , f (x) = − , f 1−x (1 − x)2 ′
′′
′′′
(x) =
Moreover,
f (0) f (0) = ln ln 1 = 0, 0 , f ′ (0) =
′′
′′′
−1, f (0) = −1, f
− (1 −2 x)3 , and f (4)(x) = − (1 −6 x)4 .
(0) =
Finally, ln(1
− x)
−2, and f (4)(ξ ) = − (1 −6 ξ )4 .
= P 3 (x) + R3 (x) x 2 x 3 = x 2 3
4
− − − − 4(1 x− ξ )4 ,
for some ξ some ξ between between 0 and x and x.. (b) Using the result of part (a),
0..752 0 2
ln(0..25) ln(0
0..753 0 = 3
≈ P 3(0 − −1.171875 (0..75) = −0.75 − 171875.. Because 0 Because 0 < ξ < 0 0..75 75,, (1 − ξ ) 4 ≤ 44 and 0.754 81 |error| = |R3(0 ≤ (0..75)| = = 20. 20 .25 25.. 4 4(1 − ξ ) 4 The actual error is | ln(0 ln(0..25) − P 3 (0 (0..75)| ≈ 0.214419 214419,, which is significantly −
less than the theoretical theoretical error bound. (c) Once again using the result from part (a), we find ln(1
− x) + x + 21 x2 = − 1 − x3
3
Moreover,
ln(1 − x) + x + x
1 2 2x
3
ln(1
x→0
with rate of convergence O O((x).
4(1
− ξ )4 .
1 x + = 3 4 1 ξ 4
for all sufficiently small x x.. Therefore, lim
x
| | ≤ |x|, | − |
− x) + x + 21 x2 = − 1 , x3
3
Convergence
11
17.. (a 17 (a)) Determine the third-degree Taylor polynomial and associated remainder term for the function f function f ((x) = 1 + x. Use x Use x 0 = 0.
√
√
(b) Using the results of part (a), approximate approximate 1.5 and compute the theoretical error bound associated with this approximation. Compare the theoretical error bound with the actual error. (c) Compute the following limit and determine the corresponding rate of convergence: 1 + x 1 21 x lim . x→0 x2
√
(a) Let Let f f ((x) = f ′ (x) =
− −
√ 1 + x. Then
1 (1 + x)−1/2 , f ′′ (x) = 2
and f and f (4) (x) =
− 1516 (1 + x)
7/2
−
− 1516 (1 + ξ )
√ 1 + x
7/2
−
3/2
−
, f ′′′(x) =
− 38 (1 + x)
5/2
−
,
. Moreover,
f (0) f (0) = 1, 1, f ′ (0) = and f and f (4) (ξ ) =
− 14 (1 + x)
1 ′′ , f (0) = 2
− 14 , f
′′′
(0) =
3 , 8
. Finally Finally,,
= P 3 (x) + R3 (x) 1 1 2 1 = 1+ x x + x3 2 8 16
−
5 4 − 128 x (1 + ξ )
7/2
−
,
for some ξ some ξ between between 0 and x x.. (b) Using the result of part (a),
√
1 1 1 ≈ P 3(0 (0..5) = 1 + (0 (0..5) − (0 (0..5)2 + (0 (0..5)3 = 1. 1 .2265625 2265625.. 2 8 16 Because 0 Because 0 < < ξ < 0 0..5, (1 + ξ ) 7/2 ≤ 1 and 5 |error| = |R3(0 (0..5)| ≤ (0..5)4 ≈ 2.44 × 10 3 . (0 128 √ The actual error is | 1.5 − P 3 (0 (0..5)| ≈ 1.82 × 10 3 , which is less than the 1.5
−
−
−
theoretical error bound.
(c) Once again using the result from part (a), we find
√ 1 + x − 1 − 1 x 2
x2
=
− 18 − 16x (1 + ξ )
5/2
−
.
Moreover,
√ 1 + x − 1 − x 2
1 2x
1 x + = 1 8 16
| | | − ξ |
5/2
−
≤ 161 |x|,
12
Section 1.2 for all sufficiently small x x.. Therefore, lim
x→0
√ 1 + x − 1 − 1 x 2
x2
=
− 18 ,
with rate of convergence O O((x).
In Exercises 18 - 21, verify that Taylor’s Taylor’s theorem produces the indicated formula, formula, where ξ where ξ is is between 0 and x . 18. ex = 1 + x +
x 2 + 2
n+1
n
· · · + xn! + (nx+ 1)! eξ
Let f ((x) = ex . Then f Let f Then f (n) (x) = e x and and f f (n) (0) = 1 for 1 for all n all n.. Therefore, by Taylor’s Theorem with x with x 0 = 0, n
x
e
=
(k)
f
k=0
(0) k f (n+1) (ξ ) n+1 x + x k! (n + 1)!
= 1+x+
x 2 + 2
n+1
n
· · · + xn! + (nx+ 1)! eξ ,
for some ξ some ξ between between 0 and x x.. 19. 3
sin x = = x x
5
n 2n+1
1) x − x3! + x5! − + · · · + ( −(2 (2n n + 1)!
+
( 1)n+1 x2n+3 cos ξ (2n (2 n + 3)!
−
Let f Let f ((x) = sin x. Then f ′ (x) = cos x, f ′′ (x) =
′′′
− sin x, and and f f
(x) =
− cos x.
Moreover, 0 , f ′ (0) = 1, 1, f ′′ (0) = 0, 0, and f (0) = 0, f (0) and f f ′′′(0) =
−1.
As higher-order derivatives are calculated, this cycle of four values repeats indefinitely. In particular, we find (2n n) (2n n+1) f (2 (0) = 0 and f (2 (0) = ( 1)n .
−
Therefore, by Taylor’s Theorem with x 0 = 0, 3
5
n 2n+1
x x ( −1) x sin x = = x x − + − + · · · + 3! 5! (2n (2 n + 1)! for some ξ some ξ between between 0 and x x..
( 1)n+1 x2n+3 + cos ξ, (2n (2 n + 3)!
−
Convergence
13
20. 2
cos x = 1
4
n 2n
1) x − x2! + x4! − + · · · + ( −(2 (2n n)!
+
( 1)n+1 x2n+2 cos ξ (2n (2 n + 2)!
−
Let f Let f ((x) = cos x. Then f ′ (x) =
′′
′′′
− sin x, f (x) = − cos x, and and f f
(x) = sin x.
Moreover, f (0) f (0) = 1, 1, f ′ (0) = 0, 0, f ′′ (0) =
′′′
−1, and and f f
(0) = 0. 0.
As higher-order derivatives are calculated, this cycle of four values repeats indefinitely.. In particular, we find nitely (2n (2n n) n+1) f (2 (0) = ( 1)n and f (2 (0) = 0. 0.
−
Therefore, by Taylor’s Theorem with x 0 = 0, 2
cos x = 1
4
n 2n
1) x − x2! + x4! − + · · · + ( −(2 (2n n)!
+
( 1)n+1 x2n+2 cos ξ, (2n (2 n + 2)!
−
for some ξ some ξ between between 0 and x and x.. 21. 1 =1 1+x Let f Let f ((x) = f ′ (x) =
1 1+x = 1+x
n+1 n+1
− x + x2 − + · · · + (−1)nxn + ( −(11)+ ξ )nx+2
(1 + x)−1 . Then,
−1 · (1 + x)
2
−
, f ′′ (x) = 1 2 (1 + x)−3 , f ′′′(x) =
· · and, in general, f general, f (n) (x) = (−1)n · n! · (1 + x)
n−1
−
n+1 n+1
− x + x2 − + · · · + (−1)nxn + ( −(11)+ ξ )nx+2
for some ξ some ξ between between 0 and x and x..
,
. Therefore, by Taylor’s Theorem
with x with x 0 = 0,
1 =1 1+x
4
−
−1 · 2 · 3 · (1 + x)
,
