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Let $f:\mathbb{R}\times \mathbb{R}\mapsto \mathbb{R}$ , continous

$\forall$ $y$ in $\mathbb{R}$, $\exists$ $a,b$ such that $f(x,y) = ax + b$

$\forall$ $x$ in $\mathbb{R}$, $\exists$ $a',b'$ such that $f(x,y) = a'y + b'$

1/ does $a$,$b$,$a'$,$b'$ are the same for all $x$,$y$ ? Say :$\exists$ $a,b,a',b'$ such that $\forall$ $y$ in $\mathbb{R}$, $f(x,y) = ax + b$ and $\forall$ $x$ in $\mathbb{R}$, $f(x,y) = a'y + b'$

2/ is there a specific form for $f$ ?

NB : I am trying to resolve a real life unknow relation between two variables. I am able to compute any $f(x,y)$ but this is costly, I want to minimize the number of call on $f$.

Currently, I just called a few times $f$ for a given $y$ and found the above relation. Since I chose $y$ randomly, I am sure the relation is true for any $y$, but I am not sure of 1/ and I wonder if I can assert 1/ without additional computation on $f$

Toto
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Toto
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1 Answers1

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You could have $f(x,y) = a(y) x + b(y)$ for any continuous functions $a(y)$ and $b(y)$ on $\mathbb R$. No reason for them to be constant.

EDIT: With the new assumptions, you could have $f(x,y) = a x y + b x + c y + d$.

Robert Israel
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  • I changed the assumptions – Toto Jan 31 '15 at 00:38
  • Is it only one possibility or $f$ can only be of this form ? – Toto Jan 31 '15 at 01:41
  • I get $f(0.475,y)=-389.55y+3063.6$ and $f(x,2.18)=-300.26x+233.5$. I then tried to find the general form using $0.475a+x=-389.55$ and $1.475b+e=3063.6$ and $2.18a+b=-300.26$ and $2.18c+e=2335.2$ with excel solver but it said there is no solution, something wrong ? – Toto Jan 31 '15 at 14:38
  • These are inconsistent. $f(.475,y) = -389.55 y + 3063.6$ would give $f(.475, 2.18) = 2978.6781$ while $f(x,2.18) = -300.26x + 233.5$ would give $f(.475, 2.18) = 90.8765$. – Robert Israel Feb 01 '15 at 02:40
  • Typo : $f(x,2.18)=−300.26x+2335.2$ leading to $f(.475,2.18)\approx2214$ for the first form and $f(.475,2.18)\approx2193$ for the second form. As said, the figures comes from a real life situation ($f(x,2.18)$ and $f(.475,y)$ are linear regressions from some sampling data with a $R²=0.996$ and $0.9973$), so I would say that it is consistent. I was expecting excel to find a "close" solution because certainly a pure mathematical one does not exists for those figures. – Toto Feb 01 '15 at 05:06
  • You might try doing a linear regression using all your data for the form $f(x,y) = a x y + b x + c y + d$. – Robert Israel Feb 01 '15 at 08:56
  • OK thanks for the advise I will try ! – Toto Feb 01 '15 at 16:10