HOMEWORK ASSIGNMENT 4
ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071]
Each problem will be marked out of 4 points.
Exercise 1 ([1, Exercise 8.4]). Observe that (r + 1)3 − r3 = 3r2 + 3r + 1. Show that
n n
n
X
X
X
(n + 1)3 − 1 =
(r + 1)3 − r3 = 3
r2 + 3
r + n,
r=1
and use it to show that
n
X
r=1
r=1
r2 =
r=1
n(n + 1)(2n + 1)
6
for every integer n > 0. Use the same methods to find formulae for
Solution. Put
fr (n) =
n
X
Pn
r=1 r
3
and
Pn
r=1 r
4.
ir
i=1
for every positive integers r and n. We have f0 (n) = n. And
n(n + 1)
2
3
3
2
by [1, Example 8.6]. Since (r + 1) − r = 3r + r + 1, we see that the equalities
n n n
n
X
X
X
X
r2 + 3
r + n = 3f2 (n) + 3f1 (n) + n
3r2 + r + 1 = 3
(r + 1)3 − r3 =
f1 (n) =
r=1
r=1
r=1
r=1
hold. On the other hand, we have
n X
(r+1)3 −r3 = (n+1)3 −n3 + n3 −(n−1)3 +· · ·+ 33 −23 + 23 −13 = (n+1)3 −1,
r=1
which implies that (n + 1)3 − 1 = 3f2 (n) + 3f1 (n) + n. Then
−n
(n + 1)3 − 1 − 3 n(n+1)
(n + 1)3 − 1 − 3f1 (n) − n
n3 n2 n
n(n + 1)(2n + 1)
2
=
=
+ + =
,
3
3
3
2 6
6
which is the same formula obtained in [1, Example 8.7].
To complete the solution, we should find formulae for f3 (n) and f4 (n).
Since (r + 1)4 − r4 = 4r3 + 6r2 + 4r + 1, we see that the equalities
n n X
X
(r + 1)4 − r4 =
4r3 + 6r2 + 4r + 1 = 4f3 (n) + 6f2 (n) + 4f1 (n) + n
f2 (n) =
r=1
r=1
hold. On the other hand, we have
n X
(r+1)4 −r4 = (n+1)4 −n4 + n4 −(n−1)4 +· · ·+ 34 −24 + 24 −14 = (n+1)4 −1,
r=1
This assignment is due on Thursday 22nd October 2015.
1
which implies that (n + 1)4 − 1 = 4f3 (n) + 6f2 (n) + 4f1 (n) + n. Then
(n + 1)4 − 1 − 6f2 (n) − 4f1 (n) − n
n4 n3 n2
n2 (n + 1)2
=
+
+
=
4
4
2
4
4
for every positive n (we plugged in there formulae for f2 (n) and f1 (n)).
Since (r + 1)5 − r5 = 5r4 + 10r3 + 10r2 + 5r + 1, we see that
n n X
X
(n+1)4 −1 =
(r+1)5 −r5 =
5r4 +10r3 +10r2 +5r+1 = 5f4 (n)+10f3 (n)+10f2 (n)+5f1 (n)+n,
f3 (n) =
r=1
r=1
which implies that (n +
1)5
− 1 = 5f4 (n) + 10f3 (n) + 10f2 (n) + 5f1 (n) + n. Then
(n + 1)5 − 1 − 10f3 (n) − 10f2 (n) − 5f1 (n) − n
,
5
where we already know formulae for f3 (n), f2 (n), f1 (n). Then
f4 (n) =
n5 n4 n3
n
n(2n + 1)(n + 1)(3n2 + 3n − 1)
+
+
−
=
,
5
2
3
30
30
which completes the solution.
f4 (n) =
Exercise 2 ([1, Exercise 8.10]). (a) Prove than
1
1
7
1
+
+ ··· +
>
n+1 n+2
2n
12
for every positive integer n > 2.
(b) Prove than
√
1
1
1
1 + √ + √ + ··· + √ 6 2 n
n
2
3
for every positive integer n > 1.
Solution. Let us prove the first inequality by induction. If n = 2, then
1 1
1
1
1
7
7
= + =
+
+ ··· +
> ,
12
3 4
n+1 n+2
2n
12
so the inequality holds in this case. Suppose it holds for some n > 2. We must show that
1
1
1
7
+
+ ··· +
>
n+2 n+3
2(n + 1)
12
to show that the inequality holds for n + 1. Notice that the equality
!
1
1
1
1
1
1
1
1
1
+
+···+
=
+
+···+
−
+
+
n+2 n+3
2(n + 1)
n+1 n+2
2n
n + 1 2n + 1 2n + 2
holds. Now using the induction assumption we see that the inequality
1
1
1
7
1
1
1
+
+ ··· +
>
−
+
+
n+2 n+3
2(n + 1)
12 n + 1 2n + 1 2n + 2
holds. One the other hand, the equality
1
1
1
1
1
1
−
+
+
=
−
=
n + 1 2n + 1 2n + 2
2n + 1 2n + 2
(2n + 1)(2n + 2)
holds. Thus, we see that
1
1
1
7
1
7
+
+ ··· +
>
+
> ,
n+2 n+3
2(n + 1)
12 (2n + 1)(2n + 2)
12
which implies that the required inequality holds for n + 1.
2
Now let us prove that
√
1
1
1
1 + √ + √ + ··· + √ 6 2 n
n
2
3
for every positive integer n > 1. This inequality holds for n = 1. In fact, we have
√
1
1 + √ 6 2 2,
2
√
as can be easily seen by multiplying both sides by 2. This means that the inequality
holds for n = 2 as well (but we do not need to know this to get the solution).
Suppose the inequality we want to prove holds for some n > 1. We must prove that
√
1
1
1
1
62 n+1
1 + √ + √ + ··· + √ + √
n
n+1
2
3
to show that the inequality holds for n + 1. By induction, we have
√
1
1
1
1
1
62 n+ √
,
1 + √ + √ + ··· + √ + √
n
n+1
n+1
2
3
√
√
1
and to complete the solution it is enough to show that 2 n + √n+1
6 2 n + 1. But
√
√
√
√
√
1
1
1
6 2 n + 1 ⇐⇒ 2 n 6 2 n + 1− √
⇐⇒ 4n 6 2 n + 1− √
2 n+ √
n+1
n+1
n+1
√
√
√
because 2 n and 2 n + 1 − 1/ n + 1 are positive real numbers. Since
!2
√
1
1
1
2 n+1− √
= 4(n + 1) − 4 +
= 4n +
> 4n,
n+1
n+1
n+1
we see that all these inequalities hold, which completes the solution.
Exercise 3 ([1, Exercise 9.2]). (a) Prove that for a convex polyhedron with V vertices,
E edges and F faces, the following inequalities are true:
2E > 3F and 2E > 3V.
(b) Deduce using Euler’s formula that
2V > F + 4, 3V > E + 6, 2F > V + 4 and 3F > E + 6.
(c) Give an example of a convex polyhedron for which all these inequalities are equalities:
2E = 3V = 3F, 2V = F + 4, 3V = E + 6, 2F = V + 4, 3F = E + 6.
Solution. It should be pointed out that both inequalities 2E > 3F and 2E > 3V easily
follow from the proof of [1, Theorem 9.3] if you drop the regularity assumption there.
We will give a detailed proof anyway. Let introduce new notations as follows:
• let F1 , F2 , . . . , FF be faces of the convex polyhedra,
• let ni be the number of edges in the face Fi for every i ∈ {1, 2, . . . , F },
• let E11 , E21 , . . . , En1 1 be the edges in the face F1 ,
• let E12 , E22 , . . . , En2 2 be the edges in the face F2 ,
• . . .,
• let E1F , E2F , . . . , EnFF be the edges in the face FF .
How many symbols Eji we introduced? Exactly n1 + n2 + · · · + nF . But
F
X
ni >
i=1
F
X
i=1
because ni > 3 for every i ∈ {1, . . . , F }.
3
3 = 3F,
!2
,
Since each edge is contained in exactly two faces, the set
o
n
nF
n2
n1
1
2
1
2
1
2
E1 , E1 , . . . , E1 , E2 , E2 , . . . , E2 , . . . , EF , E F , . . . , E F
has exactly twice less elements as number of symbols Eji we introduced. But the latter
set is just the set of all edges of our polyhedra. Then
F
E=
1X
3
ni > F,
2
2
i=1
since ni > 3 for every i ∈ {1, . . . , F }. This gives us 2E > 3F .
Now let us prove that 2E > 3V . Let introduce new notations as follows:
• let V1 , V2 , . . . , VV be vertices of the convex polyhedra,
• let mi be the number of edges attached to the vertex Vi for every i ∈ {1, 2, . . . , V },
1
• let E11 , E21 , . . . , Em
1 be the edges attached to the vertex V1 ,
m
• let E12 , E22 , . . . , E2 2 be the edges attached to the vertex V2 ,
• . . .,
V
• let E1V , E2V , . . . , Em
be the edges attached to the vertex VV .
V
How many symbols Eji we introduced this time? Exactly m1 + m2 + · + mV . But
V
X
mi >
i=1
V
X
3 = 3V,
i=1
because mi > 3 for every i ∈ {1, . . . , V }.
Since each edge joins exactly two vertices, the set
n
o
mV
m2
1
2
1
2
1
E11 , E21 , . . . , Em
,
E
,
E
,
.
.
.
,
E
,
.
.
.
,
E
,
E
,
.
.
.
,
E
2
2
V
V
1
2
V
has exactly twice less elements as number of symbols Eji we introduced. But the latter
set is just the set of all edges of our polyhedra. Then
V
E=
3
1X
mi > V,
2
2
i=1
since mi > 3 for every i ∈ {1, . . . , V }. This gives us 2E > 3V .
Now let us use 2E > 3F , 2E > 3V and Euler’s formula to show that
2V > F + 4, 3V > E + 6, 2F > V + 4 and 3F > E + 6.
Since V − E + F = 2 by Euler’s formula and 2E > 3F , we have
2V + 2F = 4 + 2E > 4 + 3F,
which gives 2V > F + 4. Since V − E + F = 2 by Euler’s formula and 2E > 3V , we have
2V + 2F = 4 + 2E > 4 + 3V,
which gives 2V > V + 4. Since V − E + F = 2 by Euler’s formula and 2E > 3F , we have
6 = 3V − 3E + 3F 6 3V − 3E + 2E = 3V − E,
which gives 3V > E + 6. Since V − E + F = 2 by Euler’s formula and 2E > 3V , we have
6 = 3V − 3E + 3F 6 2E − 3E + 3F = −E + 3F,
which implies that 3F > E + 6.
For a tetrahedron, we have V = 4, E = 6, and F = 4, which implies that
2E = 3V = 3F,
4
and 2V = F + 4, 3V = E + 6, 2F = V + 4, 3F = E + 6. Vice versa, if
2V = F + 4, 3V = E + 6, 2F = V + 4, 3F = E + 6,
then Linear Algebra tells us that V = 4, E = 6, and F = 4, which implies that the convex
polyhedra must look like an old Soviet style milk container on this picture
.
which is a tetrahedron of course.
Exercise 4 ([1, Exercise 9.6]). Draw all connected plane graphs with 4 edges, and all
the connected plane graphs with 4 vertices.
Solution. Let us first draw all connected plane graphs with 4 edges:
(a) a graph that looks like a chain:
F
F
F
F
F
(b) a graph that looks like a snail (bit curly and with long tail)
F
F
F
F
F
(c) a graph that looks like triangle with a tail or kite
F
F
F
F
5
(d) a graph that forms a convex polygon with four edges
F
F
F
F
(e) a Sputnik style shaped graph
F
F
F
F
F
One can easily show that these graphs are “all” possible plane graphs with 4 edges up
to a natural “equivalence” of course (say two chain look graphs with the same number of
edges are considered to be the same).
Now let us draw all connected plane graphs with 4 vertices:
(a) a graph that looks like a chain:
F
F
F
F
(b) a graph that looks like rotated letter Y (shorter that the one above)
F
F
F
F
(c) a graph that looks like triangle with a tail
F
F
F
F
(d) a graph that forms a polygon with four edges
F
F
F
F
6
(e) a graph that that looks like two triangles with one common edge
F
F
F
F
(f) a graph that can be obtained by projecting tetrahedron to the plane
F
F
F
F
One can easily show that these graphs are “all” possible plane graphs with 4 vertices.
Exercise 5 ([1, Exercise 9.8]). Prove that every connected plane graph has a vertex that
is joined to at most five other vertices1.
Solution. Suppose that there exists a connected plane graph Γ such that all its vertices
are joined to at least six other vertices. Let us show that this assumptions leads to a
contradiction.
Let v be the number of vertices of Γ, and let e be the number of edges of Γ. Then
v > 6 > 3. Thus, it follows from [1, Exercise 9.3] that
e 6 3v − 6,
which also can be proved in a similar way to the inequality 3F > E + 6 in Exercise 3.
On the other hand, arguing just as in the proof of the inequality 2E > 3V in Exercise 3,
we see that e > 3v, which is impossible, since e 6 3v − 6.
To be pedantic, let us prove that e > 3v. Let introduce new notations as follows:
• let v1 , v2 , . . . , vv be vertices of the graph Γ,
• let mi be the number of edges in Γ attached to the vertex vi for every i ∈
{1, 2, . . . , v},
1
• let e11 , e21 , . . . , em
1 be the edges in Γ attached to the vertex v1 ,
m2
1
2
• let e2 , e2 , . . . , e2 be the edges attached to the vertex v2 ,
• . . .,
v
• let e1v , e2v , . . . , em
v be the edges attached to the vertex vv .
How many symbols eji we introduced this time? Exactly
v
X
i=1
mi >
v
X
6 = 6v,
i=1
since mi > 6 for every i ∈ {1, . . . , v} by assumption (we assume that all vertices in Γ are
joined to at least six other vertices in Γ).
1Hint: for a plane graph with v vertices and e edges, if v > 3, then e 6 3v − 6 by [1, Exercise 9.3].
7
Since each edge in Γ joins exactly two vertices, the set
o
n
m2
1
2
mv
1 2
1
,
.
.
.
,
e
,
e
,
.
.
.
,
e
,
e
,
e
,
.
.
.
,
e
e11 , e21 , . . . , em
V
V
v
2 2
2
1
has exactly twice less elements as number of symbols eji we introduced. But the latter set
is just the set of all edges of our polyhedra. Then
v
1X
6
e=
mi > v = 3v,
2
2
i=1
since mi > 6 for every i ∈ {1, . . . , v}. This gives us e > 3v.
References
[1] M. Liebeck, A concise introduction to pure mathematics
Third edition (2010), CRC Press
8