How to show that $f(x_1,x_3)=f(x_1,x_2)f(x_2,x_3)implies f(x,y)=g(x)/g(y)$ [closed]
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I wanted to know if there was a nice way of showing that $f(x_1,x_3)=f(x_1,x_2)f(x_2,x_3)forall x_1,x_2,x_3implies f(x,y)=g(x)/g(y)$ for some function $g$.
functions
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
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closed as off-topic by Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos Dec 19 '18 at 17:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
I wanted to know if there was a nice way of showing that $f(x_1,x_3)=f(x_1,x_2)f(x_2,x_3)forall x_1,x_2,x_3implies f(x,y)=g(x)/g(y)$ for some function $g$.
functions
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
$endgroup$
closed as off-topic by Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos Dec 19 '18 at 17:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I wanted to know if there was a nice way of showing that $f(x_1,x_3)=f(x_1,x_2)f(x_2,x_3)forall x_1,x_2,x_3implies f(x,y)=g(x)/g(y)$ for some function $g$.
functions
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
$endgroup$
I wanted to know if there was a nice way of showing that $f(x_1,x_3)=f(x_1,x_2)f(x_2,x_3)forall x_1,x_2,x_3implies f(x,y)=g(x)/g(y)$ for some function $g$.
functions
functions
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
asked Dec 19 '18 at 8:23
Aakash LakshmananAakash Lakshmanan
628
628
closed as off-topic by Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos Dec 19 '18 at 17:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos Dec 19 '18 at 17:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Karn Watcharasupat, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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Write the equation with $x_3$ changed to $1$ and divide the given equation by the new one. You get $frac {f(x_1,x_3)} {f(x_1,1)}=frac {f(x_2,x_3)} {f(x_2,1)}$. What this equation tells you is that the function $frac {f(x,x_3)} {f(x,1)}$ has the same value at any two points $x=x_1$ and $x=x_2$. Hence it is independent of $x$. Call it $h(x_3)$. We get $f(x,y)=h(y)f(x,1)$. Can you complete the argument now? [You will need the fact $f(x,x)=f(x,x)^{2}$ so $f(x,x) =1$ (assuming $f(x,x) neq 0$. Note the result does not make sense for $f equiv 0$ so we have to make some assumptions about zeros of $f$)]. More details: you get $1=h(x)f(x,1)$ by putting $y=x$. I claim that the conclusion holds with $g =frac 1 h$. We have to show $f(x,y)=frac {g(x)} {g(y}$ which is same as $f(x,y)=frac {h(y)} {h(x)}$. By $f(x,y)=h(y)f(x,1)$ this becomes $h(y)f(x,1)=frac {h(y)} {h(x)}$. This is true because $h(x)f(x,1)=1$.
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
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Sorry but do you mind pushing it all the way?
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– Aakash Lakshmanan
Dec 19 '18 at 8:51
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@AakashLakshmanan I have given a detailed proof now.
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– Kavi Rama Murthy
Dec 19 '18 at 8:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write the equation with $x_3$ changed to $1$ and divide the given equation by the new one. You get $frac {f(x_1,x_3)} {f(x_1,1)}=frac {f(x_2,x_3)} {f(x_2,1)}$. What this equation tells you is that the function $frac {f(x,x_3)} {f(x,1)}$ has the same value at any two points $x=x_1$ and $x=x_2$. Hence it is independent of $x$. Call it $h(x_3)$. We get $f(x,y)=h(y)f(x,1)$. Can you complete the argument now? [You will need the fact $f(x,x)=f(x,x)^{2}$ so $f(x,x) =1$ (assuming $f(x,x) neq 0$. Note the result does not make sense for $f equiv 0$ so we have to make some assumptions about zeros of $f$)]. More details: you get $1=h(x)f(x,1)$ by putting $y=x$. I claim that the conclusion holds with $g =frac 1 h$. We have to show $f(x,y)=frac {g(x)} {g(y}$ which is same as $f(x,y)=frac {h(y)} {h(x)}$. By $f(x,y)=h(y)f(x,1)$ this becomes $h(y)f(x,1)=frac {h(y)} {h(x)}$. This is true because $h(x)f(x,1)=1$.
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Sorry but do you mind pushing it all the way?
$endgroup$
– Aakash Lakshmanan
Dec 19 '18 at 8:51
$begingroup$
@AakashLakshmanan I have given a detailed proof now.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 8:56
add a comment |
$begingroup$
Write the equation with $x_3$ changed to $1$ and divide the given equation by the new one. You get $frac {f(x_1,x_3)} {f(x_1,1)}=frac {f(x_2,x_3)} {f(x_2,1)}$. What this equation tells you is that the function $frac {f(x,x_3)} {f(x,1)}$ has the same value at any two points $x=x_1$ and $x=x_2$. Hence it is independent of $x$. Call it $h(x_3)$. We get $f(x,y)=h(y)f(x,1)$. Can you complete the argument now? [You will need the fact $f(x,x)=f(x,x)^{2}$ so $f(x,x) =1$ (assuming $f(x,x) neq 0$. Note the result does not make sense for $f equiv 0$ so we have to make some assumptions about zeros of $f$)]. More details: you get $1=h(x)f(x,1)$ by putting $y=x$. I claim that the conclusion holds with $g =frac 1 h$. We have to show $f(x,y)=frac {g(x)} {g(y}$ which is same as $f(x,y)=frac {h(y)} {h(x)}$. By $f(x,y)=h(y)f(x,1)$ this becomes $h(y)f(x,1)=frac {h(y)} {h(x)}$. This is true because $h(x)f(x,1)=1$.
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$begingroup$
Sorry but do you mind pushing it all the way?
$endgroup$
– Aakash Lakshmanan
Dec 19 '18 at 8:51
$begingroup$
@AakashLakshmanan I have given a detailed proof now.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 8:56
add a comment |
$begingroup$
Write the equation with $x_3$ changed to $1$ and divide the given equation by the new one. You get $frac {f(x_1,x_3)} {f(x_1,1)}=frac {f(x_2,x_3)} {f(x_2,1)}$. What this equation tells you is that the function $frac {f(x,x_3)} {f(x,1)}$ has the same value at any two points $x=x_1$ and $x=x_2$. Hence it is independent of $x$. Call it $h(x_3)$. We get $f(x,y)=h(y)f(x,1)$. Can you complete the argument now? [You will need the fact $f(x,x)=f(x,x)^{2}$ so $f(x,x) =1$ (assuming $f(x,x) neq 0$. Note the result does not make sense for $f equiv 0$ so we have to make some assumptions about zeros of $f$)]. More details: you get $1=h(x)f(x,1)$ by putting $y=x$. I claim that the conclusion holds with $g =frac 1 h$. We have to show $f(x,y)=frac {g(x)} {g(y}$ which is same as $f(x,y)=frac {h(y)} {h(x)}$. By $f(x,y)=h(y)f(x,1)$ this becomes $h(y)f(x,1)=frac {h(y)} {h(x)}$. This is true because $h(x)f(x,1)=1$.
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
Write the equation with $x_3$ changed to $1$ and divide the given equation by the new one. You get $frac {f(x_1,x_3)} {f(x_1,1)}=frac {f(x_2,x_3)} {f(x_2,1)}$. What this equation tells you is that the function $frac {f(x,x_3)} {f(x,1)}$ has the same value at any two points $x=x_1$ and $x=x_2$. Hence it is independent of $x$. Call it $h(x_3)$. We get $f(x,y)=h(y)f(x,1)$. Can you complete the argument now? [You will need the fact $f(x,x)=f(x,x)^{2}$ so $f(x,x) =1$ (assuming $f(x,x) neq 0$. Note the result does not make sense for $f equiv 0$ so we have to make some assumptions about zeros of $f$)]. More details: you get $1=h(x)f(x,1)$ by putting $y=x$. I claim that the conclusion holds with $g =frac 1 h$. We have to show $f(x,y)=frac {g(x)} {g(y}$ which is same as $f(x,y)=frac {h(y)} {h(x)}$. By $f(x,y)=h(y)f(x,1)$ this becomes $h(y)f(x,1)=frac {h(y)} {h(x)}$. This is true because $h(x)f(x,1)=1$.
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
edited Dec 19 '18 at 8:56
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 8:39
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
$begingroup$
Sorry but do you mind pushing it all the way?
$endgroup$
– Aakash Lakshmanan
Dec 19 '18 at 8:51
$begingroup$
@AakashLakshmanan I have given a detailed proof now.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 8:56
add a comment |
$begingroup$
Sorry but do you mind pushing it all the way?
$endgroup$
– Aakash Lakshmanan
Dec 19 '18 at 8:51
$begingroup$
@AakashLakshmanan I have given a detailed proof now.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 8:56
$begingroup$
Sorry but do you mind pushing it all the way?
$endgroup$
– Aakash Lakshmanan
Dec 19 '18 at 8:51
$begingroup$
Sorry but do you mind pushing it all the way?
$endgroup$
– Aakash Lakshmanan
Dec 19 '18 at 8:51
$begingroup$
@AakashLakshmanan I have given a detailed proof now.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 8:56
$begingroup$
@AakashLakshmanan I have given a detailed proof now.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 8:56
add a comment |
