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Check proof that if a function is even with respect to 2 distinct values of $x$ then it is periodic
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I'm trying to prove that:
If a function $f(x): mathbb R to mathbb R$ is even with respect to $x = a,b in mathbb R;; ane b$ then it is periodic with period $T_min = 2|a-b|$
Here is what i made:
Consider some $x_1 = a$, $f(x)$ is symmetric with respect to $a$ therefore $f(x) = f(2a - x)$, but at the same time the function is odd about some $x_2=b$ and hence $f(x) = f(2b-x)$. So:
$$
f(x) = f(2a-x)\
f(x) = f(2b-x)\
$$
So we may write:
$$
f(2a-x) = f(2b-x)
$$
On one hand:
$$
f(2a - x)= f(2b-x + 2a - 2b) iff T=2(a-b)
$$
On the other:
$$
f(2b-x) = f(2a-x+2b-2a) iff T=2(b-a)
$$
Since both periods satisfy the equations that means the only solution is $T=2|a-b|$.
update:
To check that $T$ is indeed period of$f(x)$ let $b > a$, using initial conditions that $f(2a-x) = f(2b-x)$:
$$
f(x+T) = f(2b - (x-T)) = f(2b-x-2b+2a)=f(2a-x)=f(x) ; Box
$$
algebra-precalculus proof-verification periodic-functions
asked Jul 23 at 16:48
roman
4391412
add a comment |Â
up vote
0
down vote
favorite
I'm trying to prove that:
If a function $f(x): mathbb R to mathbb R$ is even with respect to $x = a,b in mathbb R;; ane b$ then it is periodic with period $T_min = 2|a-b|$
Here is what i made:
Consider some $x_1 = a$, $f(x)$ is symmetric with respect to $a$ therefore $f(x) = f(2a - x)$, but at the same time the function is odd about some $x_2=b$ and hence $f(x) = f(2b-x)$. So:
$$
f(x) = f(2a-x)\
f(x) = f(2b-x)\
$$
So we may write:
$$
f(2a-x) = f(2b-x)
$$
On one hand:
$$
f(2a - x)= f(2b-x + 2a - 2b) iff T=2(a-b)
$$
On the other:
$$
f(2b-x) = f(2a-x+2b-2a) iff T=2(b-a)
$$
Since both periods satisfy the equations that means the only solution is $T=2|a-b|$.
update:
To check that $T$ is indeed period of$f(x)$ let $b > a$, using initial conditions that $f(2a-x) = f(2b-x)$:
$$
f(x+T) = f(2b - (x-T)) = f(2b-x-2b+2a)=f(2a-x)=f(x) ; Box
$$
algebra-precalculus proof-verification periodic-functions
asked Jul 23 at 16:48
roman
4391412
If you actually show that it is periodic then your solution would work. This doesn't suffice however to prove it.
– asdf
Jul 23 at 16:56
@asdf, so do you mean the OP is incorrect?
– roman
Jul 23 at 16:59
I'm trying to say that you are still to show that it is actually periodic - in your solution you are assuming it is and then you are showing that the period must satisfy certain properties.
– asdf
Jul 23 at 17:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to prove that:
If a function $f(x): mathbb R to mathbb R$ is even with respect to $x = a,b in mathbb R;; ane b$ then it is periodic with period $T_min = 2|a-b|$
Here is what i made:
Consider some $x_1 = a$, $f(x)$ is symmetric with respect to $a$ therefore $f(x) = f(2a - x)$, but at the same time the function is odd about some $x_2=b$ and hence $f(x) = f(2b-x)$. So:
$$
f(x) = f(2a-x)\
f(x) = f(2b-x)\
$$
So we may write:
$$
f(2a-x) = f(2b-x)
$$
On one hand:
$$
f(2a - x)= f(2b-x + 2a - 2b) iff T=2(a-b)
$$
On the other:
$$
f(2b-x) = f(2a-x+2b-2a) iff T=2(b-a)
$$
Since both periods satisfy the equations that means the only solution is $T=2|a-b|$.
update:
To check that $T$ is indeed period of$f(x)$ let $b > a$, using initial conditions that $f(2a-x) = f(2b-x)$:
$$
f(x+T) = f(2b - (x-T)) = f(2b-x-2b+2a)=f(2a-x)=f(x) ; Box
$$
algebra-precalculus proof-verification periodic-functions
asked Jul 23 at 16:48
roman
4391412
I'm trying to prove that:
If a function $f(x): mathbb R to mathbb R$ is even with respect to $x = a,b in mathbb R;; ane b$ then it is periodic with period $T_min = 2|a-b|$
Here is what i made:
Consider some $x_1 = a$, $f(x)$ is symmetric with respect to $a$ therefore $f(x) = f(2a - x)$, but at the same time the function is odd about some $x_2=b$ and hence $f(x) = f(2b-x)$. So:
$$
f(x) = f(2a-x)\
f(x) = f(2b-x)\
$$
So we may write:
$$
f(2a-x) = f(2b-x)
$$
On one hand:
$$
f(2a - x)= f(2b-x + 2a - 2b) iff T=2(a-b)
$$
On the other:
$$
f(2b-x) = f(2a-x+2b-2a) iff T=2(b-a)
$$
Since both periods satisfy the equations that means the only solution is $T=2|a-b|$.
update:
To check that $T$ is indeed period of$f(x)$ let $b > a$, using initial conditions that $f(2a-x) = f(2b-x)$:
$$
f(x+T) = f(2b - (x-T)) = f(2b-x-2b+2a)=f(2a-x)=f(x) ; Box
$$
algebra-precalculus proof-verification periodic-functions
asked Jul 23 at 16:48
roman
4391412
edited Jul 23 at 19:01
asked Jul 23 at 16:48
roman
4391412
asked Jul 23 at 16:48
roman
4391412
asked Jul 23 at 16:48
roman
4391412
4391412
If you actually show that it is periodic then your solution would work. This doesn't suffice however to prove it.
– asdf
Jul 23 at 16:56
@asdf, so do you mean the OP is incorrect?
– roman
Jul 23 at 16:59
I'm trying to say that you are still to show that it is actually periodic - in your solution you are assuming it is and then you are showing that the period must satisfy certain properties.
– asdf
Jul 23 at 17:03
add a comment |Â
If you actually show that it is periodic then your solution would work. This doesn't suffice however to prove it.
– asdf
Jul 23 at 16:56
@asdf, so do you mean the OP is incorrect?
– roman
Jul 23 at 16:59
I'm trying to say that you are still to show that it is actually periodic - in your solution you are assuming it is and then you are showing that the period must satisfy certain properties.
– asdf
Jul 23 at 17:03
If you actually show that it is periodic then your solution would work. This doesn't suffice however to prove it.
– asdf
Jul 23 at 16:56
If you actually show that it is periodic then your solution would work. This doesn't suffice however to prove it.
– asdf
Jul 23 at 16:56
@asdf, so do you mean the OP is incorrect?
– roman
Jul 23 at 16:59
@asdf, so do you mean the OP is incorrect?
– roman
Jul 23 at 16:59
I'm trying to say that you are still to show that it is actually periodic - in your solution you are assuming it is and then you are showing that the period must satisfy certain properties.
– asdf
Jul 23 at 17:03
I'm trying to say that you are still to show that it is actually periodic - in your solution you are assuming it is and then you are showing that the period must satisfy certain properties.
– asdf
Jul 23 at 17:03
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
$$f(x)=f(2a-x)$$therefore by substituting $xto 2b-x$ we obtain $$f(2b-x)=f(2a-2b+x)$$also $$f(x)=f(2b-x)$$which means that $$f(2|a-b|+x)=f(x)$$or $$Large T_min=2|a-b|$$
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
add a comment |Â
up vote
1
down vote
Once you have
$f(2a-x) = f(2b-x)$,
put $y = 2a-x$.
Then
$x = 2a-y$
so that
$2b-x
=2b-(2a-y)
=y+2(b-a)
$
so
$f(y) = f(y+2(b-a))$.
answered Jul 23 at 19:36
marty cohen
69.1k446122
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$f(x)=f(2a-x)$$therefore by substituting $xto 2b-x$ we obtain $$f(2b-x)=f(2a-2b+x)$$also $$f(x)=f(2b-x)$$which means that $$f(2|a-b|+x)=f(x)$$or $$Large T_min=2|a-b|$$
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
add a comment |Â
up vote
1
down vote
accepted
$$f(x)=f(2a-x)$$therefore by substituting $xto 2b-x$ we obtain $$f(2b-x)=f(2a-2b+x)$$also $$f(x)=f(2b-x)$$which means that $$f(2|a-b|+x)=f(x)$$or $$Large T_min=2|a-b|$$
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$f(x)=f(2a-x)$$therefore by substituting $xto 2b-x$ we obtain $$f(2b-x)=f(2a-2b+x)$$also $$f(x)=f(2b-x)$$which means that $$f(2|a-b|+x)=f(x)$$or $$Large T_min=2|a-b|$$
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
$$f(x)=f(2a-x)$$therefore by substituting $xto 2b-x$ we obtain $$f(2b-x)=f(2a-2b+x)$$also $$f(x)=f(2b-x)$$which means that $$f(2|a-b|+x)=f(x)$$or $$Large T_min=2|a-b|$$
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
answered Jul 23 at 20:40
Mostafa Ayaz
8,5823630
8,5823630
add a comment |Â
add a comment |Â
up vote
1
down vote
Once you have
$f(2a-x) = f(2b-x)$,
put $y = 2a-x$.
Then
$x = 2a-y$
so that
$2b-x
=2b-(2a-y)
=y+2(b-a)
$
so
$f(y) = f(y+2(b-a))$.
answered Jul 23 at 19:36
marty cohen
69.1k446122
add a comment |Â
up vote
1
down vote
Once you have
$f(2a-x) = f(2b-x)$,
put $y = 2a-x$.
Then
$x = 2a-y$
so that
$2b-x
=2b-(2a-y)
=y+2(b-a)
$
so
$f(y) = f(y+2(b-a))$.
answered Jul 23 at 19:36
marty cohen
69.1k446122
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Once you have
$f(2a-x) = f(2b-x)$,
put $y = 2a-x$.
Then
$x = 2a-y$
so that
$2b-x
=2b-(2a-y)
=y+2(b-a)
$
so
$f(y) = f(y+2(b-a))$.
answered Jul 23 at 19:36
marty cohen
69.1k446122
Once you have
$f(2a-x) = f(2b-x)$,
put $y = 2a-x$.
Then
$x = 2a-y$
so that
$2b-x
=2b-(2a-y)
=y+2(b-a)
$
so
$f(y) = f(y+2(b-a))$.
answered Jul 23 at 19:36
marty cohen
69.1k446122
answered Jul 23 at 19:36
marty cohen
69.1k446122
answered Jul 23 at 19:36
marty cohen
69.1k446122
answered Jul 23 at 19:36
marty cohen
69.1k446122
69.1k446122
add a comment |Â
add a comment |Â
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If you actually show that it is periodic then your solution would work. This doesn't suffice however to prove it.
– asdf
Jul 23 at 16:56
@asdf, so do you mean the OP is incorrect?
– roman
Jul 23 at 16:59
I'm trying to say that you are still to show that it is actually periodic - in your solution you are assuming it is and then you are showing that the period must satisfy certain properties.
– asdf
Jul 23 at 17:03