Mixing
Find all functions satisfying $f(x+1)=frac{f(x)-5}{f(x)-3}$
1
$begingroup$
Find all functions satisfying
$$f(x+1)=frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-frac{2}{g(x)}$$
Any clue here?
algebra-precalculus functions functional-equations periodic-functions
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
$endgroup$
add a comment |
1
$begingroup$
Find all functions satisfying
$$f(x+1)=frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-frac{2}{g(x)}$$
Any clue here?
algebra-precalculus functions functional-equations periodic-functions
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
$endgroup$
2
$begingroup$
Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens.
$endgroup$
– Crostul
Jan 2 at 9:35
$begingroup$
And observe that you can select the value of $f(t)$, $tin[0,1)$ any which way you want. Only if you also require continuity is there something to worry.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 9:58
$begingroup$
Why did you tag polynomials ?
$endgroup$
– Claude Leibovici
Jan 2 at 10:10
$begingroup$
Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks
$endgroup$
– Ekaveera Kumar Sharma
Jan 2 at 10:17
add a comment |
1
1
1
$begingroup$
Find all functions satisfying
$$f(x+1)=frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-frac{2}{g(x)}$$
Any clue here?
algebra-precalculus functions functional-equations periodic-functions
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
$endgroup$
Find all functions satisfying
$$f(x+1)=frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-frac{2}{g(x)}$$
Any clue here?
algebra-precalculus functions functional-equations periodic-functions
algebra-precalculus functions functional-equations periodic-functions
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
edited Jan 2 at 10:18
Ekaveera Kumar Sharma
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
asked Jan 2 at 9:19
Ekaveera Kumar SharmaEkaveera Kumar Sharma
5,54711428
5,54711428
2
$begingroup$
Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens.
$endgroup$
– Crostul
Jan 2 at 9:35
$begingroup$
And observe that you can select the value of $f(t)$, $tin[0,1)$ any which way you want. Only if you also require continuity is there something to worry.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 9:58
$begingroup$
Why did you tag polynomials ?
$endgroup$
– Claude Leibovici
Jan 2 at 10:10
$begingroup$
Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks
$endgroup$
– Ekaveera Kumar Sharma
Jan 2 at 10:17
add a comment |
2
$begingroup$
Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens.
$endgroup$
– Crostul
Jan 2 at 9:35
$begingroup$
And observe that you can select the value of $f(t)$, $tin[0,1)$ any which way you want. Only if you also require continuity is there something to worry.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 9:58
$begingroup$
Why did you tag polynomials ?
$endgroup$
– Claude Leibovici
Jan 2 at 10:10
$begingroup$
Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks
$endgroup$
– Ekaveera Kumar Sharma
Jan 2 at 10:17
2
2
$begingroup$
Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens.
$endgroup$
– Crostul
Jan 2 at 9:35
$begingroup$
Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens.
$endgroup$
– Crostul
Jan 2 at 9:35
$begingroup$
And observe that you can select the value of $f(t)$, $tin[0,1)$ any which way you want. Only if you also require continuity is there something to worry.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 9:58
$begingroup$
And observe that you can select the value of $f(t)$, $tin[0,1)$ any which way you want. Only if you also require continuity is there something to worry.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 9:58
$begingroup$
Why did you tag polynomials ?
$endgroup$
– Claude Leibovici
Jan 2 at 10:10
$begingroup$
Why did you tag polynomials ?
$endgroup$
– Claude Leibovici
Jan 2 at 10:10
$begingroup$
Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks
$endgroup$
– Ekaveera Kumar Sharma
Jan 2 at 10:17
$begingroup$
Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks
$endgroup$
– Ekaveera Kumar Sharma
Jan 2 at 10:17
add a comment |
1 Answer
1
active
oldest
votes
4
$begingroup$
The hint.
Prove that:
$$f(x+4)=f(x).$$
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
$endgroup$
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
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
4
$begingroup$
The hint.
Prove that:
$$f(x+4)=f(x).$$
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
$endgroup$
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
add a comment |
4
$begingroup$
The hint.
Prove that:
$$f(x+4)=f(x).$$
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
$endgroup$
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
add a comment |
4
4
4
$begingroup$
The hint.
Prove that:
$$f(x+4)=f(x).$$
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
$endgroup$
The hint.
Prove that:
$$f(x+4)=f(x).$$
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
answered Jan 2 at 9:48
Michael RozenbergMichael Rozenberg
98k1590188
98k1590188
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
add a comment |
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
But not all functions of period $4$ are solution of the given equation, are they? If not, then you have not answered the question .
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:20
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
$begingroup$
@Kavi Rama Murthy It was the hint. $f(x)notin{0,1,2,3}$ of course. If you want, you can write a full solution.
$endgroup$
– Michael Rozenberg
Jan 2 at 12:22
add a comment |
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2
$begingroup$
Clearly the value of $f(0)$ determines inductively all the values of $f(x)$ over the integers. Now, try to assign some values to $f(0)$ and see what happens.
$endgroup$
– Crostul
Jan 2 at 9:35
$begingroup$
And observe that you can select the value of $f(t)$, $tin[0,1)$ any which way you want. Only if you also require continuity is there something to worry.
$endgroup$
– Jyrki Lahtonen
Jan 2 at 9:58
$begingroup$
Why did you tag polynomials ?
$endgroup$
– Claude Leibovici
Jan 2 at 10:10
$begingroup$
Yes according to the latest information, non constant polynomial cannot be periodic. I will edit it thanks
$endgroup$
– Ekaveera Kumar Sharma
Jan 2 at 10:17