Math 25: Advanced Calculus
UC Davis, Spring 2011
Solutions to homework assignment #5
Reading material.
Read sections 2.4–2.7 in the textbook.
Problems on previous material
1. Answer the following problems in the textbook: A.7.1, A.7.3, A.8.2,
A.8.5, 1.9.2, 1.9.6, 1.10.3, 1.10.5
Solution to A.7.1.
Let
n
= 3. For this
n
, the equation 4
x
2
+
x

n
=
0 becomes 4
x
2
+
x

3 = 0. This equation has a rational root
x
=

1.
Solution to A.7.3.
The converse: “Every continuous function is
diﬀerentiable” (this is false — e.g., the function
f
(
x
) =

x

is continu
ous but not diﬀerentiable). The contrapositive: “Every function that
is not continuous is also not diﬀerentiable”. This is true.
Solution to A.8.2.
I claim that 1 + 3 + 5 +
...
+ (2
n

1) =
n
2
.
The proof is by induction on
n
. For
n
= 1, the lefthand side is 1, and
n
2
= 1, so the statement is true. Assume that it is true for
n

1. This
means that
1 + 3 + 5 +
...
+
(
2(
n

1)

1
)
= 1 + 3 + 5 +
...
+ (2
n

3) = (
n

1)
2
.
It follows, using the induction hypothesis, that
1 + 3 + 5+
...
+ (2
n

1) = (1 + 3 + 5 +
...
+ (2
n

3)) + (2
n

1)
= (
n

1)
2
+ (2
n

1) =
n
2

2
n
+ 1 + (2
n

1) =
n
2
.
Solution to 1.10.5.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '11
 Wiley
 Differential Equations, Calculus, Equations

Click to edit the document details