Solved problems on functions and sequences

This text contains solved problems on functions and sequences including bijective proofs, arithmetic and geometric progressions, and more.

Question 1
We check whether the domains and co-domains of f and g asbelow;
f : { 0 ,1 } → {0 ,1 } ; f ( x )=x2
When x=0 ,
f ( 0 ) =02=0
When x=1 ,
f ( 1 ) =12=1
Similarly,
g : { 0 , 1 } → { 0 , 1 } ; g ( x )=x
When x=0 ,
g ( 0 )=0
When x=1 ,
g ( 1 )=1

Therefore, they are equal functions in the given domain.
Question 2
To prove that a function if bijective, we prove that it is both injective and surjective (Frederic P.
Miller, 2010).
To prove that it is injective;
f ( x )=f ( y )
4 x+1=4 y+1
4 x=4 y
x= y

Thus, the function is injective.
To prove that it is surjective;
Taking x ∈ [ 0 ,1 ]∧ y ∈[0 , 1.5]
From the function y=4 x +1 ,
x= y−1
4
Using the range given for y and comparing with the range of values of x, we find that the
function is surjective.
Since the function is both injective and surjective, it is therefore bijective.
Question 3
Similar to question 2, we start by checking if the function is injective (Kim, 2010).
x , y ∈[−1, 1]
f ( x )=f ( y )
x2= y2
x= √ y2=± y
The function is not injective. Thus, the function is not bijective.
Question 4
First, we need to understand that ⌊ y ⌋ mean the closest integer less than y (Shoup, 2009).
Let’s take for instance x = 2.1 and y = 0.9;
⌊ x ⌋=2∧⌊ y ⌋=0
It means that
⌊ x ⌋ +⌊ y ⌋ =2+ 0=2
And
⌊ x+ y ⌋=⌊ 2.1+0.9 ⌋=3
Therefore, ⌊ x+ y ⌋ ≠ ⌊ x ⌋ +⌊[ y ] ⌋

Question 5
First, we need to understand that if ⌊ x ⌋=n, then it follows that n ≤ ⌊ x ⌋ < n+1
We will have
5 ≤2 x−1<5+1
5+1 ≤2 x−1+1<6+1
6 ≤ 2 x <7
6
2 ≤ x<7
3 ≤ x < 7
2
Therefore, the interval is [ 3 , 7
2 )
Question 6
a) We need to note that the function is a bijection.
f−1 ( ( 0,1 ) )= ( cos−1 0 , cos−1 1 )
¿ (cos−1
(cos (−π
2 ) ), cos−1 ( cos 0 ) )∪¿
¿ ( −π
2 ,0 ) ∪ ( π
2 , 0 )
Rearranging;
¿ ( −π
2 ,0 ) ∪ ( 0 , π
2 )
b) f ( f −1 ( {−1 }) )=f ( cos−1−1 ) =f ( π )
It’s clear that π is out of range of the given domain [ −π
2 , π
2 ]. Therefore, the solution does
not exist.
c) f−1
(f ( { π
2 }) )=f−1
(cos ( π
2 ))=f −1(0)
¿ cos−1 ( 0 )
¿ π
2
d) Correct.
Taking X ⊆ (−π
2 ,0 ) ∪ X ⊆ (0 , π
2 )

f −1 ( f ( X ) ) =f −1 ( cos X )
¿ cos−1 (cos X )= X
Question 7
We start by checking if the function is injective.
f ( x )=f ( y )
x2+1= y2 +1
x2= y2
x= y
The function is injective.
To complete the bijective proof, we check whether the function is surjective.
y=x2 +1
x= √ y−1
Substituting in the function;
f ( √ y −1 ) = ( √ y−1 ) 2 +1= y
The function is surjective.
Hence, the function is bijective.
x= √ y−1
f−1 ( y )= √ y−1
f−1 ( x )= √x−1
Domain; [ 1 , ∞ )
Range; ¿
Question 8
We start by checking if it is injective.
f ( n )=f ( y )
⌊ √n ⌋ =⌊ √ y ⌋
⌊ n ⌋ ≠ ⌊ y ⌋

Thus, the function is not injective. Hence, the function is not bijective.
Question 9
For maxima, dy
dx = 0∧d2 y
d x2 <0
dy
dx =4 x3 +4=0 ;
4 x3 =−4
x=−1
This is the only real solution.
d2 y
d x2 =12 x2
Inserting x=−1 ,
d2 y
d x2 =12 ( −1 ) 2=12>0
Meaning that we have minima. Therefore it tends to ∞ .
Replacing the value of x into the function;
f (−1 )= (−1 )4 + 4 (−1 ) +1=−2
Hence we need adjust the co-domain to ¿
Question 10
The general formula for arithmetic sequence is;
an=a0+ ( n−1 ) d
Given a11=71 and a20=131, we will formulate two linear equations as below.
71=a0 + ( 11−1 ) d
131=a0 + ( 20−1 ) d
Subtracting the two we obtain;
−60=−9 d

d= 60
9
We then obtain a0;
71=a0 + 10∗60
9
a0=71− 600
9 = 39
9
Hence the sequence is an= 39
9 + ( n−1 ) 60
9
Question 11
The formula for nth term in geometric progression is (Murray H. Protter, 1988);
bn=ao r n−1
Given that b5=10 and b7=25;
We formulate equations as follows:
10=a0∗r5−1
25=a0∗r7−1
Dividing the two we get;
0.4=r−2
Thus, r = √ 1
0.4 =±1.581
Looking for a0 ;
a0= 10
1.58114 =1.6
Thus, the formula is bn=1.6∗1.5811n−1
Question 12
∑
j =10
100
( j+3)( j−1)= ∑
j=10
100
j ( j−1 ) +3 ( j−1 ) =∑
j=10
100
( j2+2 j−3)

¿ ∑
j=1
100
( j2 +2 j−3)−∑
j=1
9
( j2 +2 j−3)
¿ [ (100∗101∗201
6 )+ 2 (100∗101
2 )−3 (100 ) ]− [ ( 9∗10∗19
6 )+2 ( 9∗10
2 )−9∗3 ]
¿ 347802
Question 13
∑
n=10
800
43 n+1
52 n−1 = ∑
n=10
800
43 n∗5∗4
52n =20 ∑
n=10
800
( 43
52 )
n
¿ 20 ∑
n=10
800
2.56n
¿ 20 [ 2.5610 +2.5611+… … … … … ..+2.56800 ]
¿ 20∗2.5610 [ 1+2.56+ … … … … … … … … … ..+2.56790 ]
¿ 20∗2.5610∗[ 2.56790−1 ]
2.56−1
¿ 20∗2.5610
[ 2.56790−1
1.56 ]

Bibliography
Frederic P. Miller, A. F. V. J. M., 2010. Bijection: Function (mathematics), Set (mathematics), Surjective
Function, Injective Function, Integer, Permutation, Isomorphism, Homography, Permutation Group.
Illustrated ed. Rapid City: Alphascript Publishing.
Kim, S., 2010. Bijective Proofs of Partition Identities and Covering Systems. Ilustrated ed. s.l.:University of
Illinois.
Murray H. Protter, P. E. P., 1988. Calculus with Analytic Geometry. illustrated ed. s.l.:Jones & Bartlett
Learning.
Shoup, V., 2009. A Computational Introduction to Number Theory and Algebra. illustrated ed.
Cambridge: Cambridge University Press.