MATH 4010 HOMEWORK ASSIGNMENT V, 11/10/2015
PROFESSOR CARLOS J MORENO
Problem 20. (Textbook p.642 ) Find several terms in the power series expansion of the
following quotients
!(b)
1
cos(x)
1(c)
1
cos(x) – sin(x)
Problem 21. (Textbook p.632 No. 8) Verify that the inverse hyperbolic sine function
: sinh(x) = e
x-e-x
2
has an inverse (i.e.x = sinh(y)or y = sinh-1
(x)) with a poer series
expansion
sinh-1
(x) = x +
X8
n=1
(-1)n
1 · 3 · · ·(2n – 1)
2 · 4 · · ·(2n)
·
x
2n+1
2n + 1
(Justify your calculation and indicate for which valus of x you are proving the validity of
the expansion, e.g. If the inteval of convergence is a finite interval, what can you say about
the end points?
Problem 22. Obtain the expansion
T =
x
2
+
X8
n=2
(-1)n-1
1 · 3 · · ·(2n – 3)
n!
·
x
n
2
n
for one root of the equation
T
2 + 2T – x = 0,
and show it converges so long as |x| < 1.
Problem 23. Find the radius of convergence for the series
X8
n=1
cnx
n
,
where
cn =
1
v
n2 + 1
+
1
v
n2 + 2
+ · · · +
1
v
n2 + n
.
1
MATH 4010 HOMEWORK ASSIGNMENT V, 11/10/2015
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MATH 4010 HOMEWORK ASSIGNMENT V, 11/10/2015
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MATH 4010 HOMEWORK ASSIGNMENT V, 11/10/2015
PROFESSOR CARLOS J MORENO
Problem 20. (Textbook p.642 ) Find several terms in the power series expansion of the
following quotients
!(b)
1
cos(x)
1(c)
1
cos(x) – sin(x)
Problem 21. (Textbook p.632 No. 8) Verify that the inverse hyperbolic sine function
: sinh(x) = e
x-e-x
2
has an inverse (i.e.x = sinh(y)or y = sinh-1
(x)) with a poer series
expansion
sinh-1
(x) = x +
X8
n=1
(-1)n
1 · 3 · · ·(2n – 1)
2 · 4 · · ·(2n)
·
x
2n+1
2n + 1
(Justify your calculation and indicate for which valus of x you are proving the validity of
the expansion, e.g. If the inteval of convergence is a finite interval, what can you say about
the end points?
Problem 22. Obtain the expansion
T =
x
2
+
X8
n=2
(-1)n-1
1 · 3 · · ·(2n – 3)
n!
·
x
n
2
n
for one root of the equation
T
2 + 2T – x = 0,
and show it converges so long as |x| < 1.
Problem 23. Find the radius of convergence for the series
X8
n=1
cnx
n
,
where
cn =
1
v
n2 + 1
+
1
v
n2 + 2
+ · · · +
1
v
n2 + n
.
1