Prove a function is periodic if it satisfies certain criteria

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Excuse me if the notation below is wrong in some way, I do not have any special background and learn Math alone.

What i'm investigating now is:

Given a function $f(x)$ prove that it is periodic if $ exists; Tne 0 : forall x in D(f), ; x+T in D(f), ; x-T in D(f) ; textand P ; textis trueimplies f(x-T) = f(x+T) $ where $P$ is one of the following statements:
$$
f(x+T) = -f(x) labeleq:1 \
f(x+T) = 1overf(x) \
f(x+T) = f(x) + aoverbcdot f(x)-1\
f(x+T) = 1 over 1-f(x)
$$

Case 1:

$$
f(x+T) = -f(x) iff f(x) = -f(x-T) \
f(x) = -f(x-T) iff -f(x) = f(x-T)
$$

Therefore $f(x) = f(x) iff f(x+T) = f(x-T)$, and its period is $2T$

Case 2:

$$
f(x+T) = 1over f(x) iff f(x) = 1over f(x-T) \
1over f(x) = 1 over f(x) iff f(x+T) = f(x-T)
$$

Case 3:
$$
f(x+T) = f(x)+ aover bcdot f(x)-1 iff f(x)=f(x-T)+a over bcdot f(x-T) -1 \
f(x)(bcdot f(x-T)-1) = f(x-T) + a \
f(x)(bcdot f(x-T)-1) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x-T) = f(x) + a \
f(x-T) (bcdot f(x) - 1) = f(x) + a \
f(x-T) = f(x) + a over bcdot f(x) - 1 \
f(x-T) = f(x+T)
$$

Case 4:

I couldn't untangle this one, what steps should I take to prove condition from case 4 makes a function periodic? I'm also interested whether the solution for the first 3 ones even makes sense.

algebra-precalculus proof-verification periodic-functions

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edited Jul 24 at 14:57

asked Jul 24 at 14:25

roman

4391412

add a comment | 

up vote
1
down vote

favorite

Excuse me if the notation below is wrong in some way, I do not have any special background and learn Math alone.

What i'm investigating now is:

Given a function $f(x)$ prove that it is periodic if $ exists; Tne 0 : forall x in D(f), ; x+T in D(f), ; x-T in D(f) ; textand P ; textis trueimplies f(x-T) = f(x+T) $ where $P$ is one of the following statements:
$$
f(x+T) = -f(x) labeleq:1 \
f(x+T) = 1overf(x) \
f(x+T) = f(x) + aoverbcdot f(x)-1\
f(x+T) = 1 over 1-f(x)
$$

Case 1:

$$
f(x+T) = -f(x) iff f(x) = -f(x-T) \
f(x) = -f(x-T) iff -f(x) = f(x-T)
$$

Therefore $f(x) = f(x) iff f(x+T) = f(x-T)$, and its period is $2T$

Case 2:

$$
f(x+T) = 1over f(x) iff f(x) = 1over f(x-T) \
1over f(x) = 1 over f(x) iff f(x+T) = f(x-T)
$$

Case 3:
$$
f(x+T) = f(x)+ aover bcdot f(x)-1 iff f(x)=f(x-T)+a over bcdot f(x-T) -1 \
f(x)(bcdot f(x-T)-1) = f(x-T) + a \
f(x)(bcdot f(x-T)-1) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x-T) = f(x) + a \
f(x-T) (bcdot f(x) - 1) = f(x) + a \
f(x-T) = f(x) + a over bcdot f(x) - 1 \
f(x-T) = f(x+T)
$$

Case 4:

I couldn't untangle this one, what steps should I take to prove condition from case 4 makes a function periodic? I'm also interested whether the solution for the first 3 ones even makes sense.

algebra-precalculus proof-verification periodic-functions

share|cite|improve this question

edited Jul 24 at 14:57

asked Jul 24 at 14:25

roman

4391412

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Excuse me if the notation below is wrong in some way, I do not have any special background and learn Math alone.

What i'm investigating now is:

Given a function $f(x)$ prove that it is periodic if $ exists; Tne 0 : forall x in D(f), ; x+T in D(f), ; x-T in D(f) ; textand P ; textis trueimplies f(x-T) = f(x+T) $ where $P$ is one of the following statements:
$$
f(x+T) = -f(x) labeleq:1 \
f(x+T) = 1overf(x) \
f(x+T) = f(x) + aoverbcdot f(x)-1\
f(x+T) = 1 over 1-f(x)
$$

Case 1:

$$
f(x+T) = -f(x) iff f(x) = -f(x-T) \
f(x) = -f(x-T) iff -f(x) = f(x-T)
$$

Therefore $f(x) = f(x) iff f(x+T) = f(x-T)$, and its period is $2T$

Case 2:

$$
f(x+T) = 1over f(x) iff f(x) = 1over f(x-T) \
1over f(x) = 1 over f(x) iff f(x+T) = f(x-T)
$$

Case 3:
$$
f(x+T) = f(x)+ aover bcdot f(x)-1 iff f(x)=f(x-T)+a over bcdot f(x-T) -1 \
f(x)(bcdot f(x-T)-1) = f(x-T) + a \
f(x)(bcdot f(x-T)-1) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x-T) = f(x) + a \
f(x-T) (bcdot f(x) - 1) = f(x) + a \
f(x-T) = f(x) + a over bcdot f(x) - 1 \
f(x-T) = f(x+T)
$$

Case 4:

I couldn't untangle this one, what steps should I take to prove condition from case 4 makes a function periodic? I'm also interested whether the solution for the first 3 ones even makes sense.

algebra-precalculus proof-verification periodic-functions

share|cite|improve this question

edited Jul 24 at 14:57

asked Jul 24 at 14:25

roman

4391412

Excuse me if the notation below is wrong in some way, I do not have any special background and learn Math alone.

What i'm investigating now is:

Given a function $f(x)$ prove that it is periodic if $ exists; Tne 0 : forall x in D(f), ; x+T in D(f), ; x-T in D(f) ; textand P ; textis trueimplies f(x-T) = f(x+T) $ where $P$ is one of the following statements:
$$
f(x+T) = -f(x) labeleq:1 \
f(x+T) = 1overf(x) \
f(x+T) = f(x) + aoverbcdot f(x)-1\
f(x+T) = 1 over 1-f(x)
$$

Case 1:

$$
f(x+T) = -f(x) iff f(x) = -f(x-T) \
f(x) = -f(x-T) iff -f(x) = f(x-T)
$$

Therefore $f(x) = f(x) iff f(x+T) = f(x-T)$, and its period is $2T$

Case 2:

$$
f(x+T) = 1over f(x) iff f(x) = 1over f(x-T) \
1over f(x) = 1 over f(x) iff f(x+T) = f(x-T)
$$

Case 3:
$$
f(x+T) = f(x)+ aover bcdot f(x)-1 iff f(x)=f(x-T)+a over bcdot f(x-T) -1 \
f(x)(bcdot f(x-T)-1) = f(x-T) + a \
f(x)(bcdot f(x-T)-1) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x) - f(x-T) = a \
bcdot f(x) f(x-T) - f(x-T) = f(x) + a \
f(x-T) (bcdot f(x) - 1) = f(x) + a \
f(x-T) = f(x) + a over bcdot f(x) - 1 \
f(x-T) = f(x+T)
$$

Case 4:

I couldn't untangle this one, what steps should I take to prove condition from case 4 makes a function periodic? I'm also interested whether the solution for the first 3 ones even makes sense.

algebra-precalculus proof-verification periodic-functions

share|cite|improve this question

edited Jul 24 at 14:57

asked Jul 24 at 14:25

roman

4391412

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 14:57

edited Jul 24 at 14:57

edited Jul 24 at 14:57

asked Jul 24 at 14:25

roman

4391412

asked Jul 24 at 14:25

roman

4391412

asked Jul 24 at 14:25

roman

4391412

4391412

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

For case 4, write out $f(x+3T)$ in terms of $f(x)$;
begineqnarray*
f(x+3T)&=&frac11-f(x+2T)
=frac11-frac11-f(x+T)
=frac11-frac11-frac11-f(x)\
&=&frac11-frac1-f(x)1-f(x)-1
=frac11+frac1-f(x)f(x)
=frac11+frac1f(x)-1=frac1frac1f(x)=f(x).
endeqnarray*

For the other three cases your reasoning is fine, but be careful with claims about a 'minimal period'. Generally all you can say is that the function $f$ is periodic with a period $2T$. It might also be periodic with other periods (for example period $T$, or $f$ could be constant). For case 3 an intermediate step for how you found the expression for $f(x-T)$ might also be appropriate.

share|cite|improve this answer

edited Jul 24 at 14:43

answered Jul 24 at 14:35

Servaes

20.1k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
  – roman
  Jul 24 at 15:00
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

For case 4, write out $f(x+3T)$ in terms of $f(x)$;
begineqnarray*
f(x+3T)&=&frac11-f(x+2T)
=frac11-frac11-f(x+T)
=frac11-frac11-frac11-f(x)\
&=&frac11-frac1-f(x)1-f(x)-1
=frac11+frac1-f(x)f(x)
=frac11+frac1f(x)-1=frac1frac1f(x)=f(x).
endeqnarray*

For the other three cases your reasoning is fine, but be careful with claims about a 'minimal period'. Generally all you can say is that the function $f$ is periodic with a period $2T$. It might also be periodic with other periods (for example period $T$, or $f$ could be constant). For case 3 an intermediate step for how you found the expression for $f(x-T)$ might also be appropriate.

share|cite|improve this answer

edited Jul 24 at 14:43

answered Jul 24 at 14:35

Servaes

20.1k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
  – roman
  Jul 24 at 15:00
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

For case 4, write out $f(x+3T)$ in terms of $f(x)$;
begineqnarray*
f(x+3T)&=&frac11-f(x+2T)
=frac11-frac11-f(x+T)
=frac11-frac11-frac11-f(x)\
&=&frac11-frac1-f(x)1-f(x)-1
=frac11+frac1-f(x)f(x)
=frac11+frac1f(x)-1=frac1frac1f(x)=f(x).
endeqnarray*

For the other three cases your reasoning is fine, but be careful with claims about a 'minimal period'. Generally all you can say is that the function $f$ is periodic with a period $2T$. It might also be periodic with other periods (for example period $T$, or $f$ could be constant). For case 3 an intermediate step for how you found the expression for $f(x-T)$ might also be appropriate.

share|cite|improve this answer

edited Jul 24 at 14:43

answered Jul 24 at 14:35

Servaes

20.1k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
  – roman
  Jul 24 at 15:00
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

For case 4, write out $f(x+3T)$ in terms of $f(x)$;
begineqnarray*
f(x+3T)&=&frac11-f(x+2T)
=frac11-frac11-f(x+T)
=frac11-frac11-frac11-f(x)\
&=&frac11-frac1-f(x)1-f(x)-1
=frac11+frac1-f(x)f(x)
=frac11+frac1f(x)-1=frac1frac1f(x)=f(x).
endeqnarray*

For the other three cases your reasoning is fine, but be careful with claims about a 'minimal period'. Generally all you can say is that the function $f$ is periodic with a period $2T$. It might also be periodic with other periods (for example period $T$, or $f$ could be constant). For case 3 an intermediate step for how you found the expression for $f(x-T)$ might also be appropriate.

share|cite|improve this answer

edited Jul 24 at 14:43

answered Jul 24 at 14:35

Servaes

20.1k33484

For case 4, write out $f(x+3T)$ in terms of $f(x)$;
begineqnarray*
f(x+3T)&=&frac11-f(x+2T)
=frac11-frac11-f(x+T)
=frac11-frac11-frac11-f(x)\
&=&frac11-frac1-f(x)1-f(x)-1
=frac11+frac1-f(x)f(x)
=frac11+frac1f(x)-1=frac1frac1f(x)=f(x).
endeqnarray*

For the other three cases your reasoning is fine, but be careful with claims about a 'minimal period'. Generally all you can say is that the function $f$ is periodic with a period $2T$. It might also be periodic with other periods (for example period $T$, or $f$ could be constant). For case 3 an intermediate step for how you found the expression for $f(x-T)$ might also be appropriate.

share|cite|improve this answer

edited Jul 24 at 14:43

answered Jul 24 at 14:35

Servaes

20.1k33484

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 14:43

edited Jul 24 at 14:43

edited Jul 24 at 14:43

answered Jul 24 at 14:35

Servaes

20.1k33484

answered Jul 24 at 14:35

Servaes

20.1k33484

answered Jul 24 at 14:35

Servaes

20.1k33484

20.1k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
  – roman
  Jul 24 at 15:00
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
  – roman
  Jul 24 at 15:00
  

  
  
  
  

Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
– roman
Jul 24 at 15:00

Thanks, that worked! As for case 3, I've printed it with a mistake, thanks for spotting. fixed the OP
– roman
Jul 24 at 15:00

add a comment | 

 

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