Mixing
If $f(frac{x+y}{x-y})=frac{f(x)+f(y)}{f(x)-f(y)}$, which of the following statement is correct
$begingroup$
If $f(frac{x+y}{x-y})=frac{f(x)+f(y)}{f(x)-f(y)}$ for $x ne y$, $x$ and $y$ are integer. Which of the following statement is correct :
(1) $f(0)=0$
(2) $f(1)=1$
(3) $f(-x)=-f(x)$
(4) $f(-x)=f(x)$
I thinks it's (1),(2), and (3) are the correct statement. But I don't know if I'm doing it in right way or not.
Here's my attempt :
For (1):
$frac{x+y}{x-y}=0\
x=-y\
f(0)=frac{f(x)+f(-y)}{f(x)-f(-y)}
\f(0)=0$ (I'm not sure about this part)
For (2) :
$frac{x+y}{x-y}=1\
x+y=x-y\
-2y=0\
y=0\
f(1)=frac{f(x)+f(0)}{f(x)-f(0)}
\f(1)=1$
For (3) :
$frac{x+y}{x-y}=-x\
x+y=xy-x^2\ ??$ (I'm stuck from this part)
functions functional-equations
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
asked Jan 20 at 6:20
airlanggaairlangga
775
$endgroup$
add a comment |
$begingroup$
If $f(frac{x+y}{x-y})=frac{f(x)+f(y)}{f(x)-f(y)}$ for $x ne y$, $x$ and $y$ are integer. Which of the following statement is correct :
(1) $f(0)=0$
(2) $f(1)=1$
(3) $f(-x)=-f(x)$
(4) $f(-x)=f(x)$
I thinks it's (1),(2), and (3) are the correct statement. But I don't know if I'm doing it in right way or not.
Here's my attempt :
For (1):
$frac{x+y}{x-y}=0\
x=-y\
f(0)=frac{f(x)+f(-y)}{f(x)-f(-y)}
\f(0)=0$ (I'm not sure about this part)
For (2) :
$frac{x+y}{x-y}=1\
x+y=x-y\
-2y=0\
y=0\
f(1)=frac{f(x)+f(0)}{f(x)-f(0)}
\f(1)=1$
For (3) :
$frac{x+y}{x-y}=-x\
x+y=xy-x^2\ ??$ (I'm stuck from this part)
functions functional-equations
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
asked Jan 20 at 6:20
airlanggaairlangga
775
$endgroup$
add a comment |
$begingroup$
If $f(frac{x+y}{x-y})=frac{f(x)+f(y)}{f(x)-f(y)}$ for $x ne y$, $x$ and $y$ are integer. Which of the following statement is correct :
(1) $f(0)=0$
(2) $f(1)=1$
(3) $f(-x)=-f(x)$
(4) $f(-x)=f(x)$
I thinks it's (1),(2), and (3) are the correct statement. But I don't know if I'm doing it in right way or not.
Here's my attempt :
For (1):
$frac{x+y}{x-y}=0\
x=-y\
f(0)=frac{f(x)+f(-y)}{f(x)-f(-y)}
\f(0)=0$ (I'm not sure about this part)
For (2) :
$frac{x+y}{x-y}=1\
x+y=x-y\
-2y=0\
y=0\
f(1)=frac{f(x)+f(0)}{f(x)-f(0)}
\f(1)=1$
For (3) :
$frac{x+y}{x-y}=-x\
x+y=xy-x^2\ ??$ (I'm stuck from this part)
functions functional-equations
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
asked Jan 20 at 6:20
airlanggaairlangga
775
$endgroup$
If $f(frac{x+y}{x-y})=frac{f(x)+f(y)}{f(x)-f(y)}$ for $x ne y$, $x$ and $y$ are integer. Which of the following statement is correct :
(1) $f(0)=0$
(2) $f(1)=1$
(3) $f(-x)=-f(x)$
(4) $f(-x)=f(x)$
I thinks it's (1),(2), and (3) are the correct statement. But I don't know if I'm doing it in right way or not.
Here's my attempt :
For (1):
$frac{x+y}{x-y}=0\
x=-y\
f(0)=frac{f(x)+f(-y)}{f(x)-f(-y)}
\f(0)=0$ (I'm not sure about this part)
For (2) :
$frac{x+y}{x-y}=1\
x+y=x-y\
-2y=0\
y=0\
f(1)=frac{f(x)+f(0)}{f(x)-f(0)}
\f(1)=1$
For (3) :
$frac{x+y}{x-y}=-x\
x+y=xy-x^2\ ??$ (I'm stuck from this part)
functions functional-equations
functions functional-equations
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
asked Jan 20 at 6:20
airlanggaairlangga
775
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
asked Jan 20 at 6:20
airlanggaairlangga
775
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
edited Jan 20 at 7:48
Jyrki Lahtonen
110k13171378
110k13171378
asked Jan 20 at 6:20
airlanggaairlangga
775
asked Jan 20 at 6:20
airlanggaairlangga
775
asked Jan 20 at 6:20
airlanggaairlangga
775
775
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
Let $f : mathbb{Z} to mathbb{R}$ solve the functional equation
$$ forall x, y in mathbb{Z} text{s.t.} frac{x+y}{x-y} in mathbb{Z}, quad fleft(frac{x+y}{x-y}right) = frac{f(x) + f(y)}{f(x) - f(y)}. tag{*} $$
Step 1. $f(1) = 1$ and $f(0) = 0$.
For $x neq 0$, plug $y=0$ to see that $f(1)=frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) neq 1$, then solving in terms of $f(x)$ gives
$$ f(x) = f(0) frac{f(1) + 1}{f(1) - 1}, $$
and so, $f$ is constant on $mathbb{Z}setminus{0}$. This immediately yields a contradiction to $text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.
Step 2. $f(-x) = -f(x)$.
In view of Step 1, it suffices to assume $x neq 0$. Plugging $y = -x$, we have
$$ 0 = f(0) = frac{f(x) + f(-x)}{f(x) - f(-x)}, $$
and so, $f(x) + f(-x) = 0$ and the desired claim follows.
By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $text{(*)}$ but does not satisfy (4). So (4) cannot be true.
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
$endgroup$
add a comment |
$begingroup$
Your argument for (1) fails; if we choose $x=-y$ so that $frac{x+y}{x-y}=0$, that doesn't do what you claimed to the right hand side. Instead, we get
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
by substituting $-x$ everywhere $y$ shows up.
Not enough information yet. That'll be zero if we can prove (3), but we haven't.
For (2) - well, that'll work if we can prove (1).
For (3), you know you don't have it.
Wait, $x$ and $y$ are restricted to integers in the functional equation? That'll limit us.
So, how do we prove (3)? Well, let's go back to what we got looking into (1);
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
for all nonzero integer $y$. But then, it's also true at $-y$, and
$$f(0)=frac{f(y)+f(-y)}{f(y)-f(-y)}=-frac{f(-y)+f(y)}{f(-y)-f(y)}=-f(0)$$
Aha - we must have $f(0)=0$, and (1) is true. From there, $f(-y)+f(y)=0$ for all integer $y$, and we have (3) for integer $y$. For other rational values $f(frac mn)$, we can write $frac{2m}{2n}=frac{(m+n)+(m-n)}{(m+n)-(m-n)}$ and
$$fleft(frac mnright) = frac{f(m+n)+f(m-n)}{f(m+n)-f(m-n)}$$
$$fleft(frac {-m}{n}right) = frac{f(-m+n)+f(-m-n)}{f(-m+n)-f(-m-n)}=frac{-f(m-n)-f(m+n)}{f(m+n)-f(m-n)}=-fleft(frac mnright)$$
That's (3) for rational $x$. If the domain of $f$ extends any farther, we can't say anything about it there, and (3) may fail.
So, as long as the domain of $f$ is restricted to rationals, we have (1), (2), and (3). Since $f(1)=1$ and $f(-1)=-1$, we definitely don't have (4).
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
$endgroup$
add a comment |
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2 Answers
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2 Answers
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oldest
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active
oldest
votes
$begingroup$
Let $f : mathbb{Z} to mathbb{R}$ solve the functional equation
$$ forall x, y in mathbb{Z} text{s.t.} frac{x+y}{x-y} in mathbb{Z}, quad fleft(frac{x+y}{x-y}right) = frac{f(x) + f(y)}{f(x) - f(y)}. tag{*} $$
Step 1. $f(1) = 1$ and $f(0) = 0$.
For $x neq 0$, plug $y=0$ to see that $f(1)=frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) neq 1$, then solving in terms of $f(x)$ gives
$$ f(x) = f(0) frac{f(1) + 1}{f(1) - 1}, $$
and so, $f$ is constant on $mathbb{Z}setminus{0}$. This immediately yields a contradiction to $text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.
Step 2. $f(-x) = -f(x)$.
In view of Step 1, it suffices to assume $x neq 0$. Plugging $y = -x$, we have
$$ 0 = f(0) = frac{f(x) + f(-x)}{f(x) - f(-x)}, $$
and so, $f(x) + f(-x) = 0$ and the desired claim follows.
By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $text{(*)}$ but does not satisfy (4). So (4) cannot be true.
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
$endgroup$
add a comment |
$begingroup$
Let $f : mathbb{Z} to mathbb{R}$ solve the functional equation
$$ forall x, y in mathbb{Z} text{s.t.} frac{x+y}{x-y} in mathbb{Z}, quad fleft(frac{x+y}{x-y}right) = frac{f(x) + f(y)}{f(x) - f(y)}. tag{*} $$
Step 1. $f(1) = 1$ and $f(0) = 0$.
For $x neq 0$, plug $y=0$ to see that $f(1)=frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) neq 1$, then solving in terms of $f(x)$ gives
$$ f(x) = f(0) frac{f(1) + 1}{f(1) - 1}, $$
and so, $f$ is constant on $mathbb{Z}setminus{0}$. This immediately yields a contradiction to $text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.
Step 2. $f(-x) = -f(x)$.
In view of Step 1, it suffices to assume $x neq 0$. Plugging $y = -x$, we have
$$ 0 = f(0) = frac{f(x) + f(-x)}{f(x) - f(-x)}, $$
and so, $f(x) + f(-x) = 0$ and the desired claim follows.
By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $text{(*)}$ but does not satisfy (4). So (4) cannot be true.
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
$endgroup$
add a comment |
$begingroup$
Let $f : mathbb{Z} to mathbb{R}$ solve the functional equation
$$ forall x, y in mathbb{Z} text{s.t.} frac{x+y}{x-y} in mathbb{Z}, quad fleft(frac{x+y}{x-y}right) = frac{f(x) + f(y)}{f(x) - f(y)}. tag{*} $$
Step 1. $f(1) = 1$ and $f(0) = 0$.
For $x neq 0$, plug $y=0$ to see that $f(1)=frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) neq 1$, then solving in terms of $f(x)$ gives
$$ f(x) = f(0) frac{f(1) + 1}{f(1) - 1}, $$
and so, $f$ is constant on $mathbb{Z}setminus{0}$. This immediately yields a contradiction to $text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.
Step 2. $f(-x) = -f(x)$.
In view of Step 1, it suffices to assume $x neq 0$. Plugging $y = -x$, we have
$$ 0 = f(0) = frac{f(x) + f(-x)}{f(x) - f(-x)}, $$
and so, $f(x) + f(-x) = 0$ and the desired claim follows.
By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $text{(*)}$ but does not satisfy (4). So (4) cannot be true.
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
$endgroup$
Let $f : mathbb{Z} to mathbb{R}$ solve the functional equation
$$ forall x, y in mathbb{Z} text{s.t.} frac{x+y}{x-y} in mathbb{Z}, quad fleft(frac{x+y}{x-y}right) = frac{f(x) + f(y)}{f(x) - f(y)}. tag{*} $$
Step 1. $f(1) = 1$ and $f(0) = 0$.
For $x neq 0$, plug $y=0$ to see that $f(1)=frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) neq 1$, then solving in terms of $f(x)$ gives
$$ f(x) = f(0) frac{f(1) + 1}{f(1) - 1}, $$
and so, $f$ is constant on $mathbb{Z}setminus{0}$. This immediately yields a contradiction to $text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.
Step 2. $f(-x) = -f(x)$.
In view of Step 1, it suffices to assume $x neq 0$. Plugging $y = -x$, we have
$$ 0 = f(0) = frac{f(x) + f(-x)}{f(x) - f(-x)}, $$
and so, $f(x) + f(-x) = 0$ and the desired claim follows.
By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $text{(*)}$ but does not satisfy (4). So (4) cannot be true.
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
answered Jan 20 at 7:17
Sangchul LeeSangchul Lee
95.4k12171278
95.4k12171278
add a comment |
add a comment |
$begingroup$
Your argument for (1) fails; if we choose $x=-y$ so that $frac{x+y}{x-y}=0$, that doesn't do what you claimed to the right hand side. Instead, we get
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
by substituting $-x$ everywhere $y$ shows up.
Not enough information yet. That'll be zero if we can prove (3), but we haven't.
For (2) - well, that'll work if we can prove (1).
For (3), you know you don't have it.
Wait, $x$ and $y$ are restricted to integers in the functional equation? That'll limit us.
So, how do we prove (3)? Well, let's go back to what we got looking into (1);
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
for all nonzero integer $y$. But then, it's also true at $-y$, and
$$f(0)=frac{f(y)+f(-y)}{f(y)-f(-y)}=-frac{f(-y)+f(y)}{f(-y)-f(y)}=-f(0)$$
Aha - we must have $f(0)=0$, and (1) is true. From there, $f(-y)+f(y)=0$ for all integer $y$, and we have (3) for integer $y$. For other rational values $f(frac mn)$, we can write $frac{2m}{2n}=frac{(m+n)+(m-n)}{(m+n)-(m-n)}$ and
$$fleft(frac mnright) = frac{f(m+n)+f(m-n)}{f(m+n)-f(m-n)}$$
$$fleft(frac {-m}{n}right) = frac{f(-m+n)+f(-m-n)}{f(-m+n)-f(-m-n)}=frac{-f(m-n)-f(m+n)}{f(m+n)-f(m-n)}=-fleft(frac mnright)$$
That's (3) for rational $x$. If the domain of $f$ extends any farther, we can't say anything about it there, and (3) may fail.
So, as long as the domain of $f$ is restricted to rationals, we have (1), (2), and (3). Since $f(1)=1$ and $f(-1)=-1$, we definitely don't have (4).
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
$endgroup$
add a comment |
$begingroup$
Your argument for (1) fails; if we choose $x=-y$ so that $frac{x+y}{x-y}=0$, that doesn't do what you claimed to the right hand side. Instead, we get
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
by substituting $-x$ everywhere $y$ shows up.
Not enough information yet. That'll be zero if we can prove (3), but we haven't.
For (2) - well, that'll work if we can prove (1).
For (3), you know you don't have it.
Wait, $x$ and $y$ are restricted to integers in the functional equation? That'll limit us.
So, how do we prove (3)? Well, let's go back to what we got looking into (1);
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
for all nonzero integer $y$. But then, it's also true at $-y$, and
$$f(0)=frac{f(y)+f(-y)}{f(y)-f(-y)}=-frac{f(-y)+f(y)}{f(-y)-f(y)}=-f(0)$$
Aha - we must have $f(0)=0$, and (1) is true. From there, $f(-y)+f(y)=0$ for all integer $y$, and we have (3) for integer $y$. For other rational values $f(frac mn)$, we can write $frac{2m}{2n}=frac{(m+n)+(m-n)}{(m+n)-(m-n)}$ and
$$fleft(frac mnright) = frac{f(m+n)+f(m-n)}{f(m+n)-f(m-n)}$$
$$fleft(frac {-m}{n}right) = frac{f(-m+n)+f(-m-n)}{f(-m+n)-f(-m-n)}=frac{-f(m-n)-f(m+n)}{f(m+n)-f(m-n)}=-fleft(frac mnright)$$
That's (3) for rational $x$. If the domain of $f$ extends any farther, we can't say anything about it there, and (3) may fail.
So, as long as the domain of $f$ is restricted to rationals, we have (1), (2), and (3). Since $f(1)=1$ and $f(-1)=-1$, we definitely don't have (4).
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
$endgroup$
add a comment |
$begingroup$
Your argument for (1) fails; if we choose $x=-y$ so that $frac{x+y}{x-y}=0$, that doesn't do what you claimed to the right hand side. Instead, we get
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
by substituting $-x$ everywhere $y$ shows up.
Not enough information yet. That'll be zero if we can prove (3), but we haven't.
For (2) - well, that'll work if we can prove (1).
For (3), you know you don't have it.
Wait, $x$ and $y$ are restricted to integers in the functional equation? That'll limit us.
So, how do we prove (3)? Well, let's go back to what we got looking into (1);
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
for all nonzero integer $y$. But then, it's also true at $-y$, and
$$f(0)=frac{f(y)+f(-y)}{f(y)-f(-y)}=-frac{f(-y)+f(y)}{f(-y)-f(y)}=-f(0)$$
Aha - we must have $f(0)=0$, and (1) is true. From there, $f(-y)+f(y)=0$ for all integer $y$, and we have (3) for integer $y$. For other rational values $f(frac mn)$, we can write $frac{2m}{2n}=frac{(m+n)+(m-n)}{(m+n)-(m-n)}$ and
$$fleft(frac mnright) = frac{f(m+n)+f(m-n)}{f(m+n)-f(m-n)}$$
$$fleft(frac {-m}{n}right) = frac{f(-m+n)+f(-m-n)}{f(-m+n)-f(-m-n)}=frac{-f(m-n)-f(m+n)}{f(m+n)-f(m-n)}=-fleft(frac mnright)$$
That's (3) for rational $x$. If the domain of $f$ extends any farther, we can't say anything about it there, and (3) may fail.
So, as long as the domain of $f$ is restricted to rationals, we have (1), (2), and (3). Since $f(1)=1$ and $f(-1)=-1$, we definitely don't have (4).
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
$endgroup$
Your argument for (1) fails; if we choose $x=-y$ so that $frac{x+y}{x-y}=0$, that doesn't do what you claimed to the right hand side. Instead, we get
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
by substituting $-x$ everywhere $y$ shows up.
Not enough information yet. That'll be zero if we can prove (3), but we haven't.
For (2) - well, that'll work if we can prove (1).
For (3), you know you don't have it.
Wait, $x$ and $y$ are restricted to integers in the functional equation? That'll limit us.
So, how do we prove (3)? Well, let's go back to what we got looking into (1);
$$f(0)=frac{f(-y)+f(y)}{f(-y)-f(y)}$$
for all nonzero integer $y$. But then, it's also true at $-y$, and
$$f(0)=frac{f(y)+f(-y)}{f(y)-f(-y)}=-frac{f(-y)+f(y)}{f(-y)-f(y)}=-f(0)$$
Aha - we must have $f(0)=0$, and (1) is true. From there, $f(-y)+f(y)=0$ for all integer $y$, and we have (3) for integer $y$. For other rational values $f(frac mn)$, we can write $frac{2m}{2n}=frac{(m+n)+(m-n)}{(m+n)-(m-n)}$ and
$$fleft(frac mnright) = frac{f(m+n)+f(m-n)}{f(m+n)-f(m-n)}$$
$$fleft(frac {-m}{n}right) = frac{f(-m+n)+f(-m-n)}{f(-m+n)-f(-m-n)}=frac{-f(m-n)-f(m+n)}{f(m+n)-f(m-n)}=-fleft(frac mnright)$$
That's (3) for rational $x$. If the domain of $f$ extends any farther, we can't say anything about it there, and (3) may fail.
So, as long as the domain of $f$ is restricted to rationals, we have (1), (2), and (3). Since $f(1)=1$ and $f(-1)=-1$, we definitely don't have (4).
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
answered Jan 20 at 7:16
jmerryjmerry
11.8k1528
11.8k1528
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