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Check proof that $sinsqrtx$ is not periodic
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I want to:
Prove that $sinsqrtx$ is not periodic
A similar question had already been asked here. But the accepted answer uses definition of the derivative. And i'm trying to do it in a "pre-calculus" manner.
Here is my try. By definition of periodic functions:
$$
f(x) = sinsqrtx = sinsqrt
$$
This may be rewritten as:
$$
f(x) = casessinsqrtx, ; x ge 0 \
sinsqrt-x, ; x < 0
$$
On the other hand:
$$
f(x) = casessinsqrtx-T, ; x-T ge 0 iff x ge T \
sinsqrtT-x, ; x-T < 0 iff x < -T
$$
So for the first case i have
$$sinsqrtx = sinsqrtx-T$$, but $forallT > 0, exists x ge 0 : x < T$ which contradicts the fact that $x ge T$. The second case is handled similarly.
Is it valid?
algebra-precalculus trigonometry proof-verification
asked Jul 20 at 12:29
roman
4391413
add a comment |Â
up vote
1
down vote
favorite
I want to:
Prove that $sinsqrtx$ is not periodic
A similar question had already been asked here. But the accepted answer uses definition of the derivative. And i'm trying to do it in a "pre-calculus" manner.
Here is my try. By definition of periodic functions:
$$
f(x) = sinsqrtx = sinsqrt
$$
This may be rewritten as:
$$
f(x) = casessinsqrtx, ; x ge 0 \
sinsqrt-x, ; x < 0
$$
On the other hand:
$$
f(x) = casessinsqrtx-T, ; x-T ge 0 iff x ge T \
sinsqrtT-x, ; x-T < 0 iff x < -T
$$
So for the first case i have
$$sinsqrtx = sinsqrtx-T$$, but $forallT > 0, exists x ge 0 : x < T$ which contradicts the fact that $x ge T$. The second case is handled similarly.
Is it valid?
algebra-precalculus trigonometry proof-verification
asked Jul 20 at 12:29
roman
4391413
No this is not valid. If you assume $x>T$ then the fact that there exists real numbers $< T$ is not a contradiction (you have fixed $x$ when making that assumption).
– Winther
Jul 20 at 12:41
no, you are comparing different functions with different domains
– Vasya
Jul 20 at 12:41
For $f(x)$ to be periodic there must exist such $Tne 0$ that for each $x$ the equality $f(x)=f(x+T)$ holds. As a special case it must hold for $x=0$. Try the $x = 0$ case then – does there exist such non-zero $T$ that $f(0) = f(T) = f(2T) = ldots =f(nT)$ for each integer $n$?
– CiaPan
Jul 20 at 12:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to:
Prove that $sinsqrtx$ is not periodic
A similar question had already been asked here. But the accepted answer uses definition of the derivative. And i'm trying to do it in a "pre-calculus" manner.
Here is my try. By definition of periodic functions:
$$
f(x) = sinsqrtx = sinsqrt
$$
This may be rewritten as:
$$
f(x) = casessinsqrtx, ; x ge 0 \
sinsqrt-x, ; x < 0
$$
On the other hand:
$$
f(x) = casessinsqrtx-T, ; x-T ge 0 iff x ge T \
sinsqrtT-x, ; x-T < 0 iff x < -T
$$
So for the first case i have
$$sinsqrtx = sinsqrtx-T$$, but $forallT > 0, exists x ge 0 : x < T$ which contradicts the fact that $x ge T$. The second case is handled similarly.
Is it valid?
algebra-precalculus trigonometry proof-verification
asked Jul 20 at 12:29
roman
4391413
I want to:
Prove that $sinsqrtx$ is not periodic
A similar question had already been asked here. But the accepted answer uses definition of the derivative. And i'm trying to do it in a "pre-calculus" manner.
Here is my try. By definition of periodic functions:
$$
f(x) = sinsqrtx = sinsqrt
$$
This may be rewritten as:
$$
f(x) = casessinsqrtx, ; x ge 0 \
sinsqrt-x, ; x < 0
$$
On the other hand:
$$
f(x) = casessinsqrtx-T, ; x-T ge 0 iff x ge T \
sinsqrtT-x, ; x-T < 0 iff x < -T
$$
So for the first case i have
$$sinsqrtx = sinsqrtx-T$$, but $forallT > 0, exists x ge 0 : x < T$ which contradicts the fact that $x ge T$. The second case is handled similarly.
Is it valid?
algebra-precalculus trigonometry proof-verification
asked Jul 20 at 12:29
roman
4391413
edited Jul 20 at 12:40
asked Jul 20 at 12:29
roman
4391413
asked Jul 20 at 12:29
roman
4391413
asked Jul 20 at 12:29
roman
4391413
4391413
No this is not valid. If you assume $x>T$ then the fact that there exists real numbers $< T$ is not a contradiction (you have fixed $x$ when making that assumption).
– Winther
Jul 20 at 12:41
no, you are comparing different functions with different domains
– Vasya
Jul 20 at 12:41
For $f(x)$ to be periodic there must exist such $Tne 0$ that for each $x$ the equality $f(x)=f(x+T)$ holds. As a special case it must hold for $x=0$. Try the $x = 0$ case then – does there exist such non-zero $T$ that $f(0) = f(T) = f(2T) = ldots =f(nT)$ for each integer $n$?
– CiaPan
Jul 20 at 12:53
add a comment |Â
No this is not valid. If you assume $x>T$ then the fact that there exists real numbers $< T$ is not a contradiction (you have fixed $x$ when making that assumption).
– Winther
Jul 20 at 12:41
no, you are comparing different functions with different domains
– Vasya
Jul 20 at 12:41
For $f(x)$ to be periodic there must exist such $Tne 0$ that for each $x$ the equality $f(x)=f(x+T)$ holds. As a special case it must hold for $x=0$. Try the $x = 0$ case then – does there exist such non-zero $T$ that $f(0) = f(T) = f(2T) = ldots =f(nT)$ for each integer $n$?
– CiaPan
Jul 20 at 12:53
No this is not valid. If you assume $x>T$ then the fact that there exists real numbers $< T$ is not a contradiction (you have fixed $x$ when making that assumption).
– Winther
Jul 20 at 12:41
No this is not valid. If you assume $x>T$ then the fact that there exists real numbers $< T$ is not a contradiction (you have fixed $x$ when making that assumption).
– Winther
Jul 20 at 12:41
no, you are comparing different functions with different domains
– Vasya
Jul 20 at 12:41
no, you are comparing different functions with different domains
– Vasya
Jul 20 at 12:41
For $f(x)$ to be periodic there must exist such $Tne 0$ that for each $x$ the equality $f(x)=f(x+T)$ holds. As a special case it must hold for $x=0$. Try the $x = 0$ case then – does there exist such non-zero $T$ that $f(0) = f(T) = f(2T) = ldots =f(nT)$ for each integer $n$?
– CiaPan
Jul 20 at 12:53
For $f(x)$ to be periodic there must exist such $Tne 0$ that for each $x$ the equality $f(x)=f(x+T)$ holds. As a special case it must hold for $x=0$. Try the $x = 0$ case then – does there exist such non-zero $T$ that $f(0) = f(T) = f(2T) = ldots =f(nT)$ for each integer $n$?
– CiaPan
Jul 20 at 12:53
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
You are working at $xge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$sin p-sin q=2sinleft(fracp-q2right)cosleft(fracp+q2right).$$
In your case you have $p=sqrtx$ and $q=sqrtx-T$, for $xge T.$ So,
$$fracsqrtx-sqrtx-T2=kpi,quad kin Bbb Z$$ or
$$fracsqrtx+sqrtx-T2=frac pi 2+kpi,quad kin Bbb Z$$
Can you finish?
answered Jul 20 at 12:43
Arnaldo
18k42146
1
Yes, thank you.
– roman
Jul 20 at 12:47
add a comment |Â
up vote
1
down vote
It is not correct. There is no reason to assume that when we are in the first case in the case of $sinleft(sqrtxright)$, then we are also in the first case in the case of $sinleft(sqrtx-Tright)$ and vice-versa.
It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$leftn^2pi^2,middle.$$
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You are working at $xge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$sin p-sin q=2sinleft(fracp-q2right)cosleft(fracp+q2right).$$
In your case you have $p=sqrtx$ and $q=sqrtx-T$, for $xge T.$ So,
$$fracsqrtx-sqrtx-T2=kpi,quad kin Bbb Z$$ or
$$fracsqrtx+sqrtx-T2=frac pi 2+kpi,quad kin Bbb Z$$
Can you finish?
answered Jul 20 at 12:43
Arnaldo
18k42146
1
Yes, thank you.
– roman
Jul 20 at 12:47
add a comment |Â
up vote
3
down vote
accepted
You are working at $xge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$sin p-sin q=2sinleft(fracp-q2right)cosleft(fracp+q2right).$$
In your case you have $p=sqrtx$ and $q=sqrtx-T$, for $xge T.$ So,
$$fracsqrtx-sqrtx-T2=kpi,quad kin Bbb Z$$ or
$$fracsqrtx+sqrtx-T2=frac pi 2+kpi,quad kin Bbb Z$$
Can you finish?
answered Jul 20 at 12:43
Arnaldo
18k42146
1
Yes, thank you.
– roman
Jul 20 at 12:47
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You are working at $xge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$sin p-sin q=2sinleft(fracp-q2right)cosleft(fracp+q2right).$$
In your case you have $p=sqrtx$ and $q=sqrtx-T$, for $xge T.$ So,
$$fracsqrtx-sqrtx-T2=kpi,quad kin Bbb Z$$ or
$$fracsqrtx+sqrtx-T2=frac pi 2+kpi,quad kin Bbb Z$$
Can you finish?
answered Jul 20 at 12:43
Arnaldo
18k42146
You are working at $xge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$sin p-sin q=2sinleft(fracp-q2right)cosleft(fracp+q2right).$$
In your case you have $p=sqrtx$ and $q=sqrtx-T$, for $xge T.$ So,
$$fracsqrtx-sqrtx-T2=kpi,quad kin Bbb Z$$ or
$$fracsqrtx+sqrtx-T2=frac pi 2+kpi,quad kin Bbb Z$$
Can you finish?
answered Jul 20 at 12:43
Arnaldo
18k42146
answered Jul 20 at 12:43
Arnaldo
18k42146
answered Jul 20 at 12:43
Arnaldo
18k42146
answered Jul 20 at 12:43
Arnaldo
18k42146
18k42146
1
Yes, thank you.
– roman
Jul 20 at 12:47
add a comment |Â
1
Yes, thank you.
– roman
Jul 20 at 12:47
1
1
Yes, thank you.
– roman
Jul 20 at 12:47
Yes, thank you.
– roman
Jul 20 at 12:47
add a comment |Â
up vote
1
down vote
It is not correct. There is no reason to assume that when we are in the first case in the case of $sinleft(sqrtxright)$, then we are also in the first case in the case of $sinleft(sqrtx-Tright)$ and vice-versa.
It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$leftn^2pi^2,middle.$$
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
It is not correct. There is no reason to assume that when we are in the first case in the case of $sinleft(sqrtxright)$, then we are also in the first case in the case of $sinleft(sqrtx-Tright)$ and vice-versa.
It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$leftn^2pi^2,middle.$$
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is not correct. There is no reason to assume that when we are in the first case in the case of $sinleft(sqrtxright)$, then we are also in the first case in the case of $sinleft(sqrtx-Tright)$ and vice-versa.
It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$leftn^2pi^2,middle.$$
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
It is not correct. There is no reason to assume that when we are in the first case in the case of $sinleft(sqrtxright)$, then we are also in the first case in the case of $sinleft(sqrtx-Tright)$ and vice-versa.
It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$leftn^2pi^2,middle.$$
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
answered Jul 20 at 12:45
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
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No this is not valid. If you assume $x>T$ then the fact that there exists real numbers $< T$ is not a contradiction (you have fixed $x$ when making that assumption).
– Winther
Jul 20 at 12:41
no, you are comparing different functions with different domains
– Vasya
Jul 20 at 12:41
For $f(x)$ to be periodic there must exist such $Tne 0$ that for each $x$ the equality $f(x)=f(x+T)$ holds. As a special case it must hold for $x=0$. Try the $x = 0$ case then – does there exist such non-zero $T$ that $f(0) = f(T) = f(2T) = ldots =f(nT)$ for each integer $n$?
– CiaPan
Jul 20 at 12:53