Math 1320-6 Lab 5
Name:
T.A.: Kyle Steffen
25 February 2016
uNID:
Instructions and due date:
• Due: 3 March 2016 at the start of class.
• For full credit: Show all of your work, and simplify your final answers.
• Work together! However, your work should be your own (not copied
from a group member).
1. Give an example of a power series with the following radii of convergence: (a) R = 0; (b) R = 3;
(c) R = ∞. Justify your answer.
(a) R = 0
Solution: Consider
∞
X
n!xn .
n=1
an+1 (n + 1)!xn+1 =
= (n + 1)|x| < 1.
Ratio test =⇒ an n!xn
1
.
Solving for |x| =⇒ |x| < n+1
Taking the limit as n → ∞ =⇒ |x| < 0 =⇒ radius of convergence = 0.
(b) R = 3
Solution: Consider ln(3 + x) = ln (3 (1 + x/3)) = ln 3 + ln (1 + x/3).
∞
∞
X
X
(−1)n−1 (x/3)n
(−1)n−1 xn
Then ln(3 + x) = ln 3 +
= ln 3 +
.
n
n3n
n=1
n=1
an+1 xn+1 /((n + 1)3n+1 ) = |x| · n < 1.
Ratio test: for n > 0, we have =
3(n + 1)
n
n
an
x /(n3 )
Solving for |x| =⇒ |x| < 3(n + 1)/n.
Taking the limit as n → ∞ =⇒ |x| < 3 =⇒ radius of convergence = 3.
(c) R = ∞
x
Solution: Consider e =
∞
X
xn
n!
.
n=0n+1
x
/(n
an+1 +
1)!
|x|
=
Ratio test =⇒ =
< 1.
n
an
x /n!
(n + 1)
Solving for |x| =⇒ |x| < (n + 1).
Taking the limit as n → ∞ =⇒ |x| < ∞ =⇒ radius of convergence = ∞.
Math 1320-6
Lab 5 - Page 2 of 6
2. Consider the function f (t) = (arctan t)/t.
Rx
(a) Compute
a power series expansion of I(x) = 0 f (t)dt. Write your answer in the form
P∞
2k−1 for an appropriate choice of a . Compute the radius of convergence of the
k
k=1 ak x
resulting power series expansion.
Solution: Recall that arctan(x) = x −
Then
x3 x5 x7 x9
+
−
+
− ···.
3
5
7
9
x
t2 t4 t7 t8
1 − + − + − · · · dt
I(x) =
3
5
7
9
0
x
3
5
7
9
t
t
t
t
= t − 2 + 2 − 2 + 2 − · · · 3
5
7
9
t=0
Z
x3 x5 x7 x9
+ 2 − 2 + 2 − ···
32
5
7
9
∞
X
x2k−1
=
(2k − 1)2
=x−
(1)
(2)
(3)
(4)
k=1
2
ak+1 x2k+1 /(2k + 1)2 =
< 1, or |x2 | · (2k − 1) < 1.
Ratio test =⇒ 2k−1
2
ak
(2k + 1)2
x
/(2k − 1)
Solving for |x| =⇒ |x| < (2k + 1)/(2k − 1).
Taking the limit as k → ∞ =⇒ |x| < 1 =⇒ radius of convergence = 1.
P
(b) Let In (x) = nk=1 ak x2k−1 be the n-th partial sum of the series in part (a) (i.e. it consists
of the first n terms of the power series). Define En to be the error of approximating I(1)
by In (1): En = |I(1) − In (1)|.
What is the minimum number of terms N required so that EN < 0.01? What if we require
EN < 0.001? EN < 0.000001 = 10−6 ? (Hint: Use the Alternating Series Estimation
Theorem. Also: All three of your answers should be positive integers.)
Solution:
Alternating Series Estimation
Theorem =⇒ En ≤ the (n + 1)-th term of the series,
i.e., x2n+1 /(2n + 1)2 x=1 = 1/(2n + 1)2 .
Therefore, we want En ≤ 1/(2n + 1)2 = ( = .01, .001, or .000001).
Solve for n as a function of : n = (−1/2 − 1)/2.
Plug in = .01 =⇒ n = 4.5. Round up =⇒ min number of terms required for
En < .01 is N = 5.
Similarly:
• = .001 =⇒ n ≈ 15.3 =⇒ min number of terms required is N = 16, and
• = .000001 =⇒ n = 499.5 =⇒ min number of terms required is N = 500.
Math 1320-6
Lab 5 - Page 3 of 6
3. Consider the following graph (in blue) of a function f (x). Answer the following questions, and
justify your answer with a short sentence or two.
3
2
y
1
0
−1
−2
−3
−5 −4 −3 −2 −1
(a) Is
0
x
1
2
3
4
5
3
1
1
5
− x + x2 + x3 − x4 + · · · the Maclaurin series of f (x)?
4
2
2
8
Solution: Recall that the Maclaurin series of f (x) is
∞
f (x) = f (0) + f 0 (0)x +
X f (k) (0)
f 00 (0) 2 f 000 (0) 3
x +
x + ··· =
xk .
2!
3!
k!
(5)
k=0
The first term of the Maclaurin series is 34 , suggesting that f (0) = 34 . Compared to the
graph of f (x), however, this is wrong: the first term should be negative, since f (0) < 0.
5 5
5
5
1
(b) Is − + (x + 1) + (x + 1)2 + (x + 1)3 − (x + 1)4 + · · · the Taylor series of f (x) at
7 9
4
12
7
the point x = −1?
Solution: Recall that the Taylor series of f (x) centered at x = −1 is
∞
f (x) = f (−1)+f 0 (−1)(x+1)+
X f (k) (−1)
f 00 (−1)
f 000 (−1)
(x+1)2 +
(x+1)3 +· · · =
(x+1)k .
2!
3!
k!
k=0
(6)
• The first term of the Taylor series is − 57 , suggesting that f (−1) = − 75 . Compared
to the graph of f (x), this is plausible, since −1 < f (−1) < 0.
• The second term of the Taylor series is 59 (x + 1), suggesting that f 0 (−1) = 59 .
Compared to the graph of f (x), this is false, since f 0 (−1) ≈ 0.
Math 1320-6
Lab 5 - Page 4 of 6
8 8
3
15
3
(c) Is + (x − 3) + (x − 3)2 + (x − 3)3 − (x − 3)4 + · · · the Taylor series of f (x) at
5 9
18
31
11
the point x = 3?
Solution:
• The first term of the Taylor series is 85 , suggesting that f (3) = 85 . Compared to
the graph of f (x), this is plausible, since 1 < f (3) < 2.
• The second term of the Taylor series is 89 (x − 3), suggesting that f 0 (3) = 89 .
Compared to the graph of f (x), this is plausible, since f 0 (3) ≈ 1 (and 89 ≈ 1).
3
3
(x − 3)2 , suggesting that f 00 (3) = .
• The third term of the Taylor series is
18
9
Compared to the graph of f (x), this is wrong: f is concave down at x = 3, so we
should have f 00 (3) < 0.
Math 1320-6
Lab 5 - Page 5 of 6
4. The following limits represent some derivative of f (x). Use Taylor series to determine which
derivative it is.
1
2
h→0 h
(a) lim
[f (x + h) + f (x − h) − 2f (x)]
Solution: Compute the Taylor series of f (x + h), centered at the point “x”:
[(x + h) − x]2 00
[(x + h) − x]3 000
f (x) +
f (x) + · · ·
2!
3!
3
2
h
h
=⇒ f (x + h) = f (x) + hf 0 (x) + f 00 (x) + f 000 (x) + O(h4 ).
(7)
2!
3!
f (x + h) = f (x) + [(x + h) − x]f 0 (x) +
Do a similar calculation for f (x − h):
f (x − h) = f (x) − hf 0 (x) +
h3
h2 00
f (x) − f 000 (x) + O(h4 )
2!
3!
(8)
Combine Eqns. 7, 8 with the original limit:
1
[f (x + h) + f (x − h) − 2f (x)]
h2
1
h2 00
h3 000
0
4
= lim 2
f (x) + hf (x) + f (x) + f (x) + O(h ) + · · ·
h→0 h
2!
3!
2
h3 000
h 00
4
0
(9)
· · · + f (x) − hf (x) + f (x) − f (x) + O(h ) − 2f (x)
2!
3!
2
1
h 00
h2
= lim 2 (f (x) + f (x) − 2f (x)) + (hf 0 (x) − hf 0 (x)) +
f (x) + f 00 (x) · · ·
h→0 h
2!
2!
3
3
h
h 000
f (x) − f 000 (x) + O(h4 )
(10)
··· +
3!
3!
1 = lim 2 h2 f 00 (x) + O(h4 )
(11)
h→0 h
= lim f 00 (x) + O(h2 )
(12)
lim
h→0
h→0
= f 00 (x)
(13)
Math 1320-6
(b) limh→0
Lab 5 - Page 6 of 6
1
2h
[f (x + h) − f (x − h)]
Solution: Combine Eqns. 7, 8 with the original limit:
1
[f (x + h) − f (x − h)]
h→0 2h
1
h2 00
h3 000
0
4
= lim
f (x) + hf (x) + f (x) + f (x) + O(h ) · · ·
h→0 2h
2!
3!
2
h 00
h3 000
0
4
· · · − f (x) − hf (x) + f (x) − f (x) + O(h )
2!
3!
2
h 00
1
h2 00
0
0
(f (x) − f (x)) + (hf (x) + hf (x)) +
= lim
f (x) − f (x) · · ·
h→0 2h
2!
2!
3
3
h 000
h
··· +
f (x) + f 000 (x) + O(h4 )
3!
3!
1
= lim
2hf 0 (x) + O(h3 )
h→0 2h
= lim f 0 (x) + O(h2 )
(17)
= f (x)
(18)
lim
h→0
0
(14)
(15)
(16)