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I am trying to approximate $f(x,y,z)-f(0,0,0)$ given $f(x,y,0)$, $f(x,0,z)$ and $f(0,y,z)$. I am ready to make any simple assumption for missing information as long as it fits the given information. I am stuck trying to achieve this. My current line of thought is like this.

If I were given only $f(x,y,0)$, then $f(x,y,0)-f(0,0,0)$ is the best approximation. If I am given $f(x,y,0)$ and $f(x,0,z)$, then the some approximations would be $$ f(x,y,z)-f(0,0,0) \approx f(x,0,z)-f(x,0,0) + f(x,y,0)-f(0,0,0) $$ or $$ f(x,y,z)-f(0,0,0) \approx f(x,y,0)-f(x,0,0) + f(x,0,z)-f(0,0,0) $$.

What should I do when $f(x,y,0)$, $f(x,0,z)$ and $f(0,y,z)$ are given? The degree of freedom on my assumptions are reduced and even a single formula that fits all the given is suddenly out of grasp for me. Please help.

Discretizer
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    Are you saying you have the values of $f(x,y,0)$ , $f(x,0,z)$ and $f(0,y,z)$ for some fixed $x$, $y$ and $z$? Consider the function $g(\xi,\upsilon,\zeta)=f(\xi,\upsilon,\zeta)-f(0,0,0)$. Probably the simplest functional form $g$ might take is $g(\xi,\upsilon,\zeta)=a\xi+b\upsilon+c\zeta$. There are three parameters to fit, $a$,$b$ and $c$, and you have three data points, $g(x,y,0)$, $g(x,0,z)$ and $g(0,y,z)$. Then solve some simultaneous equations. – Ali Mar 17 '20 at 13:57
  • Thanks @Ali for this approach. This gives me $f(x,y,z)-f(0,0,0)=\frac{f(x,y,0)-f(0,0,0)}{2}+\frac{f(x,0,z)-f(0,0,0)}{2}+\frac{f(0,y,z)-f(0,0,0)}{2}$. It doesn't fit all the initial conditions. But, this is because not all the constraints have been included. For example, the 3 linear equations forces $f(0,y,0) - f(0,0,0)=by$. But $f(0,y,0)$ is known. I will try to make the above equation quadratic and add these constraints and check. If there is any established work on this front (which I seriously doubt must be the case) please help point to it. – Discretizer Mar 18 '20 at 01:54
  • I doubt it might give me $f(x,y,z)-f(0,0,0)=\frac{f(x,y,0)-f(0,0,z)}{2}+\frac{f(0,y,z)-f(x,0,0)}{2}+\frac{f(x,0,z)-f(0,y,0)}{2}$. This still seems like magic as I might be missing the bigger picture. Is this the only possible definition for $f(x,y,z)-f(0,0,0)$ given $f(x,y,0),f(0,y,z),f(x,0,z)$ or are there more? – Discretizer Mar 18 '20 at 02:18
  • Yes the formula you give in your first comment is correct for the approach I suggested. But now you say $f(0,y,0)$ is known? That wasn't in the question. If you missed something out it's not too late to edit it. – Ali Mar 18 '20 at 14:03

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