Problem Set 2
Due 8 July 2008
MATH 126B
Homework 3: 5, [6], 7
Homework 4: 1, 2, 3, 4, 5, 6, 7, 8
Homework 5: 1, 2, 3, 4, 5, 6, [7], 8, 9
12.1: [7], 8, 9, [11], 13, 16, [19], 25, 30, 33, 34, 40
0.D, [0.E], 0.F, 0.G, 0.H
Homework 3-5 listed above are posted on the Math 126 Materials Web site,
http://www.math.washington.edu/∼m126/ .
0.D Because ex=P∞
k=0
1
k!xkfor all x, evaluating at x= 1 gives that e=e1=1+1+ 1
2! +1
3! +· · · . Use a
Taylor polynomial Tn(for appropriate n) for f(x) = exbased at 0 and Taylor’s Inequality to compute
eto seven decimal places (that is, to within 10−7).
0.E Compute the fourth Taylor polynomial of erf (x) (see Problem 0.C) based at b= 0. (Hint: Instead of
using the method of Problem 0.C, derive the Taylor series of the function based at b= 0.)
0.F Let g(x) = ex8. Compute g2008(0). (Hint: Consider the Taylor series of gbased at b= 0.)
0.G Recall that the hyperbolic tangent is defined as
tanh x=ex−e−x
ex+e−x.(1)
(a) The function tanh is invertible; find its inverse. The inverse, called the inverse hyperbolic tangent,
is denoted artanh (or tanh−1).
(b) Use known Taylor series to show that the Taylor series for artanh based at b= 0 is
∞
X
k=0
x2k+1
2k+ 1 =x+x3
3+x5
5+· · · .
Remark: Note that, besides the factor of (−1)k, this is the same as the Taylor series for arctanx.
0.H (Problems Plus, Chapter 11, Problem 7)
(a) Show for xand ysatisfying xy 6=−1 that
arctan x−arctan y= arctan x−y
1 + xy .
if the left-hand side lies between −π
2and π
2. (Hint: Use the angle sum identity for tangent.)
(b) Use part (a) to show that
arctan 120
119 −arctan 1
239 =π
4.
(c) Use (b) to show that
4 arctan 1
5−arctan 1
239 =π
4.
(Hint: Show that tan(4 arctan 1
5) = 120
119 by applying the double-angle identity for tangent twice.)
(d) Use the Taylor series for arctan based at b= 0 to compute both arctan 1
5and arctan 1
239 to nine
decimal places. (Use the method of Problem 0.D.)
(e) Use parts (c) and (d) to compute πto seven decimal places.