Can we plug a linear sequence of real numbers into the sine function such that the resulting sequence will be a one-to-one sequence?

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I was thinking about this problem for some time which can be stated in plain words as follows:

Can we plug a linear sequence of real numbers into the sine function
such that the resulting sequence will be a one-to-oneinjective sequence?

Translating it into more mathematical language we get that the problem can be stated as:

Can you find real numbers $a,binmathbbR$ such that plugging the
linear sequence $x_n_n=1^infty$ defined as $forall ninmathbbN, x_n:=an+b$ into the sine function will produce a new
sequence $y_n_n=1^infty$ defined as $forall ninmathbbN, y_n:=sin(x_n)$ that will satisfy $forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$
?

---

I’ve tried to attack the question by first showing that the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN,Big( (exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Let $m,ninmathbbN$, Then:

$begingather
y_m = y_n \
Updownarrow \
sin(x_m)=sin(x_n) \
Updownarrow \
sin(x_n)-sin(x_m)=0 \
Updownarrow \
2sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfrac(an+b)-(am+b)2cosfrac(an+b)+(am+b)2 = 0\
Updownarrow \
sinfraca(n-m)2cosfraca(n+m)+2b2 = 0 \
Updownarrow \
sinfraca(n-m)2=0 lor cosfraca(n+m)+2b2=0 \
Updownarrow \
Big(exists k_1inmathbbZ, fraca(n-m)2=pi k_1Big) lor Big(exists k_2inmathbbZ,fraca(n+m)+2b2=fracpi2+pi k_2Big) \
Updownarrow \
Big(exists k_1inmathbbZ, a(n-m)=2 pi k_1Big) lor Big(exists k_2inmathbbZ,a(n+m)+2b=pi+2pi k_2 Big)
endgather$

And therefore, the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN, Big((exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Thanks...

Note: A sequence $x_n_n=1^infty$ is called linear if and only if $exists a,binmathbbR,forall ninmathbbN,x_n=an+b$

sequences-and-series trigonometry functional-equations

share|cite|improve this question

edited Jul 14 at 17:25

asked Jul 14 at 16:41

MathNerd

1,411719

add a comment | 

up vote
0
down vote

favorite

I was thinking about this problem for some time which can be stated in plain words as follows:

Can we plug a linear sequence of real numbers into the sine function
such that the resulting sequence will be a one-to-oneinjective sequence?

Translating it into more mathematical language we get that the problem can be stated as:

Can you find real numbers $a,binmathbbR$ such that plugging the
linear sequence $x_n_n=1^infty$ defined as $forall ninmathbbN, x_n:=an+b$ into the sine function will produce a new
sequence $y_n_n=1^infty$ defined as $forall ninmathbbN, y_n:=sin(x_n)$ that will satisfy $forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$
?

---

I’ve tried to attack the question by first showing that the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN,Big( (exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Let $m,ninmathbbN$, Then:

$begingather
y_m = y_n \
Updownarrow \
sin(x_m)=sin(x_n) \
Updownarrow \
sin(x_n)-sin(x_m)=0 \
Updownarrow \
2sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfrac(an+b)-(am+b)2cosfrac(an+b)+(am+b)2 = 0\
Updownarrow \
sinfraca(n-m)2cosfraca(n+m)+2b2 = 0 \
Updownarrow \
sinfraca(n-m)2=0 lor cosfraca(n+m)+2b2=0 \
Updownarrow \
Big(exists k_1inmathbbZ, fraca(n-m)2=pi k_1Big) lor Big(exists k_2inmathbbZ,fraca(n+m)+2b2=fracpi2+pi k_2Big) \
Updownarrow \
Big(exists k_1inmathbbZ, a(n-m)=2 pi k_1Big) lor Big(exists k_2inmathbbZ,a(n+m)+2b=pi+2pi k_2 Big)
endgather$

And therefore, the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN, Big((exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Thanks...

Note: A sequence $x_n_n=1^infty$ is called linear if and only if $exists a,binmathbbR,forall ninmathbbN,x_n=an+b$

sequences-and-series trigonometry functional-equations

share|cite|improve this question

edited Jul 14 at 17:25

asked Jul 14 at 16:41

MathNerd

1,411719

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I was thinking about this problem for some time which can be stated in plain words as follows:

Can we plug a linear sequence of real numbers into the sine function
such that the resulting sequence will be a one-to-oneinjective sequence?

Translating it into more mathematical language we get that the problem can be stated as:

Can you find real numbers $a,binmathbbR$ such that plugging the
linear sequence $x_n_n=1^infty$ defined as $forall ninmathbbN, x_n:=an+b$ into the sine function will produce a new
sequence $y_n_n=1^infty$ defined as $forall ninmathbbN, y_n:=sin(x_n)$ that will satisfy $forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$
?

---

I’ve tried to attack the question by first showing that the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN,Big( (exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Let $m,ninmathbbN$, Then:

$begingather
y_m = y_n \
Updownarrow \
sin(x_m)=sin(x_n) \
Updownarrow \
sin(x_n)-sin(x_m)=0 \
Updownarrow \
2sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfrac(an+b)-(am+b)2cosfrac(an+b)+(am+b)2 = 0\
Updownarrow \
sinfraca(n-m)2cosfraca(n+m)+2b2 = 0 \
Updownarrow \
sinfraca(n-m)2=0 lor cosfraca(n+m)+2b2=0 \
Updownarrow \
Big(exists k_1inmathbbZ, fraca(n-m)2=pi k_1Big) lor Big(exists k_2inmathbbZ,fraca(n+m)+2b2=fracpi2+pi k_2Big) \
Updownarrow \
Big(exists k_1inmathbbZ, a(n-m)=2 pi k_1Big) lor Big(exists k_2inmathbbZ,a(n+m)+2b=pi+2pi k_2 Big)
endgather$

And therefore, the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN, Big((exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Thanks...

Note: A sequence $x_n_n=1^infty$ is called linear if and only if $exists a,binmathbbR,forall ninmathbbN,x_n=an+b$

sequences-and-series trigonometry functional-equations

share|cite|improve this question

edited Jul 14 at 17:25

asked Jul 14 at 16:41

MathNerd

1,411719

I was thinking about this problem for some time which can be stated in plain words as follows:

Can we plug a linear sequence of real numbers into the sine function
such that the resulting sequence will be a one-to-oneinjective sequence?

Translating it into more mathematical language we get that the problem can be stated as:

Can you find real numbers $a,binmathbbR$ such that plugging the
linear sequence $x_n_n=1^infty$ defined as $forall ninmathbbN, x_n:=an+b$ into the sine function will produce a new
sequence $y_n_n=1^infty$ defined as $forall ninmathbbN, y_n:=sin(x_n)$ that will satisfy $forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$
?

---

I’ve tried to attack the question by first showing that the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN,Big( (exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Let $m,ninmathbbN$, Then:

$begingather
y_m = y_n \
Updownarrow \
sin(x_m)=sin(x_n) \
Updownarrow \
sin(x_n)-sin(x_m)=0 \
Updownarrow \
2sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfracx_n-x_m2cosfracx_n+x_m2 = 0 \
Updownarrow \
sinfrac(an+b)-(am+b)2cosfrac(an+b)+(am+b)2 = 0\
Updownarrow \
sinfraca(n-m)2cosfraca(n+m)+2b2 = 0 \
Updownarrow \
sinfraca(n-m)2=0 lor cosfraca(n+m)+2b2=0 \
Updownarrow \
Big(exists k_1inmathbbZ, fraca(n-m)2=pi k_1Big) lor Big(exists k_2inmathbbZ,fraca(n+m)+2b2=fracpi2+pi k_2Big) \
Updownarrow \
Big(exists k_1inmathbbZ, a(n-m)=2 pi k_1Big) lor Big(exists k_2inmathbbZ,a(n+m)+2b=pi+2pi k_2 Big)
endgather$

And therefore, the condition:

$forall m,ninmathbbN, y_m=y_nlongrightarrow m=n$

is equivalent to:

$forall m,ninmathbbN, Big((exists k_1inmathbbZ,a(n-m)=2pi k_1) lor (exists k_2inmathbbZ, a(n+m)+2b=pi+2pi k_2 )Big)longrightarrow m=n$

---

Thanks...

Note: A sequence $x_n_n=1^infty$ is called linear if and only if $exists a,binmathbbR,forall ninmathbbN,x_n=an+b$

sequences-and-series trigonometry functional-equations

share|cite|improve this question

edited Jul 14 at 17:25

asked Jul 14 at 16:41

MathNerd

1,411719

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 17:25

edited Jul 14 at 17:25

edited Jul 14 at 17:25

asked Jul 14 at 16:41

MathNerd

1,411719

asked Jul 14 at 16:41

MathNerd

1,411719

asked Jul 14 at 16:41

MathNerd

1,411719

1,411719

add a comment | 

add a comment | 

2 Answers
2

active

oldest

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up vote
1
down vote



accepted

What about $a=b=1$? You have$$sin m=sin niff m=n.$$

share|cite|improve this answer

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

add a comment | 

up vote
0
down vote

Well, The answer is yes and based on Santos comment, we can just take $a=1$ and $b=0$, and we will have $forall m,ninmathbbN, sin m=sin n longrightarrow m=n$:

Proof:

Let $m,ninmathbbN$ be such that $sin m=sin n$, Then by what was just shown above, We get that there are two possibilities: (1) $exists k_1inmathbbZ,n-m=2pi k_1$ or (2) $exists k_2inmathbbZ, n+m=pi+2pi k_2$.

We will show that case (2) cannot happen, Because if we suppose by contradiction that case (2) happens, The we get that $exists k_2inmathbbZ, n+m=pi+2pi k_2$, But since $pi+2pi k_2$ is an irrational number for any $k_2inmathbbZ$, We get that $n+m$ is also irrational which contradicts the fact that $n+m$ is rational as it is the sum of two integers. Therefore case (2) cannot happen, And thus it must be the case that case (1) happens and we conclude that $exists k_1inmathbbZ,n-m=2pi k_1$, We will show that it must be the case that $k_1=0$ because if we suppose by contradiction that $k_1neq 0$, The we get that $2pi k_1$ is an irrational number and thus $n-m$ is also an irrational number which contradicts the fact that $n-m$ is a rational number as it is the difference between two integers. Thus it must be the case that $n-m=0$, And we can conclude that $n=m$ as was to be shown.

share|cite|improve this answer

answered Jul 14 at 17:24

MathNerd

1,411719

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

What about $a=b=1$? You have$$sin m=sin niff m=n.$$

share|cite|improve this answer

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

add a comment | 

up vote
1
down vote



accepted

What about $a=b=1$? You have$$sin m=sin niff m=n.$$

share|cite|improve this answer

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

What about $a=b=1$? You have$$sin m=sin niff m=n.$$

share|cite|improve this answer

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

What about $a=b=1$? You have$$sin m=sin niff m=n.$$

share|cite|improve this answer

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

answered Jul 14 at 16:45

José Carlos Santos

114k1698177

114k1698177

add a comment | 

add a comment | 

up vote
0
down vote

Well, The answer is yes and based on Santos comment, we can just take $a=1$ and $b=0$, and we will have $forall m,ninmathbbN, sin m=sin n longrightarrow m=n$:

Proof:

Let $m,ninmathbbN$ be such that $sin m=sin n$, Then by what was just shown above, We get that there are two possibilities: (1) $exists k_1inmathbbZ,n-m=2pi k_1$ or (2) $exists k_2inmathbbZ, n+m=pi+2pi k_2$.

We will show that case (2) cannot happen, Because if we suppose by contradiction that case (2) happens, The we get that $exists k_2inmathbbZ, n+m=pi+2pi k_2$, But since $pi+2pi k_2$ is an irrational number for any $k_2inmathbbZ$, We get that $n+m$ is also irrational which contradicts the fact that $n+m$ is rational as it is the sum of two integers. Therefore case (2) cannot happen, And thus it must be the case that case (1) happens and we conclude that $exists k_1inmathbbZ,n-m=2pi k_1$, We will show that it must be the case that $k_1=0$ because if we suppose by contradiction that $k_1neq 0$, The we get that $2pi k_1$ is an irrational number and thus $n-m$ is also an irrational number which contradicts the fact that $n-m$ is a rational number as it is the difference between two integers. Thus it must be the case that $n-m=0$, And we can conclude that $n=m$ as was to be shown.

share|cite|improve this answer

answered Jul 14 at 17:24

MathNerd

1,411719

add a comment | 

up vote
0
down vote

Well, The answer is yes and based on Santos comment, we can just take $a=1$ and $b=0$, and we will have $forall m,ninmathbbN, sin m=sin n longrightarrow m=n$:

Proof:

Let $m,ninmathbbN$ be such that $sin m=sin n$, Then by what was just shown above, We get that there are two possibilities: (1) $exists k_1inmathbbZ,n-m=2pi k_1$ or (2) $exists k_2inmathbbZ, n+m=pi+2pi k_2$.

We will show that case (2) cannot happen, Because if we suppose by contradiction that case (2) happens, The we get that $exists k_2inmathbbZ, n+m=pi+2pi k_2$, But since $pi+2pi k_2$ is an irrational number for any $k_2inmathbbZ$, We get that $n+m$ is also irrational which contradicts the fact that $n+m$ is rational as it is the sum of two integers. Therefore case (2) cannot happen, And thus it must be the case that case (1) happens and we conclude that $exists k_1inmathbbZ,n-m=2pi k_1$, We will show that it must be the case that $k_1=0$ because if we suppose by contradiction that $k_1neq 0$, The we get that $2pi k_1$ is an irrational number and thus $n-m$ is also an irrational number which contradicts the fact that $n-m$ is a rational number as it is the difference between two integers. Thus it must be the case that $n-m=0$, And we can conclude that $n=m$ as was to be shown.

share|cite|improve this answer

answered Jul 14 at 17:24

MathNerd

1,411719

add a comment | 

up vote
0
down vote

up vote
0
down vote

Well, The answer is yes and based on Santos comment, we can just take $a=1$ and $b=0$, and we will have $forall m,ninmathbbN, sin m=sin n longrightarrow m=n$:

Proof:

Let $m,ninmathbbN$ be such that $sin m=sin n$, Then by what was just shown above, We get that there are two possibilities: (1) $exists k_1inmathbbZ,n-m=2pi k_1$ or (2) $exists k_2inmathbbZ, n+m=pi+2pi k_2$.

We will show that case (2) cannot happen, Because if we suppose by contradiction that case (2) happens, The we get that $exists k_2inmathbbZ, n+m=pi+2pi k_2$, But since $pi+2pi k_2$ is an irrational number for any $k_2inmathbbZ$, We get that $n+m$ is also irrational which contradicts the fact that $n+m$ is rational as it is the sum of two integers. Therefore case (2) cannot happen, And thus it must be the case that case (1) happens and we conclude that $exists k_1inmathbbZ,n-m=2pi k_1$, We will show that it must be the case that $k_1=0$ because if we suppose by contradiction that $k_1neq 0$, The we get that $2pi k_1$ is an irrational number and thus $n-m$ is also an irrational number which contradicts the fact that $n-m$ is a rational number as it is the difference between two integers. Thus it must be the case that $n-m=0$, And we can conclude that $n=m$ as was to be shown.

share|cite|improve this answer

answered Jul 14 at 17:24

MathNerd

1,411719

Well, The answer is yes and based on Santos comment, we can just take $a=1$ and $b=0$, and we will have $forall m,ninmathbbN, sin m=sin n longrightarrow m=n$:

Proof:

Let $m,ninmathbbN$ be such that $sin m=sin n$, Then by what was just shown above, We get that there are two possibilities: (1) $exists k_1inmathbbZ,n-m=2pi k_1$ or (2) $exists k_2inmathbbZ, n+m=pi+2pi k_2$.

We will show that case (2) cannot happen, Because if we suppose by contradiction that case (2) happens, The we get that $exists k_2inmathbbZ, n+m=pi+2pi k_2$, But since $pi+2pi k_2$ is an irrational number for any $k_2inmathbbZ$, We get that $n+m$ is also irrational which contradicts the fact that $n+m$ is rational as it is the sum of two integers. Therefore case (2) cannot happen, And thus it must be the case that case (1) happens and we conclude that $exists k_1inmathbbZ,n-m=2pi k_1$, We will show that it must be the case that $k_1=0$ because if we suppose by contradiction that $k_1neq 0$, The we get that $2pi k_1$ is an irrational number and thus $n-m$ is also an irrational number which contradicts the fact that $n-m$ is a rational number as it is the difference between two integers. Thus it must be the case that $n-m=0$, And we can conclude that $n=m$ as was to be shown.

share|cite|improve this answer

answered Jul 14 at 17:24

MathNerd

1,411719

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 17:24

MathNerd

1,411719

answered Jul 14 at 17:24

MathNerd

1,411719

answered Jul 14 at 17:24

MathNerd

1,411719

1,411719

add a comment | 

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