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There is a question in the textbook which goes as under:-

"Can the surface whose equation is x+y+z-sin(xyz)=0 be described by an equation of the form z=f(x,y) in a neighborhood of the point (0,0) satisfying f(0,0)=0? Justify your answer."

Before I present the solution... I have a doubt. Is there a misprint in the textbook, and should it read "...in a neighborhood of the point (0,0,0) satisfying f(0,0)=0?"

My attempt is as under:-

f(0,0,0) = 0+0+0-sin(0) = 0 ( I cannot prove this if the given point is (0,0) instead of (0,0,0))

$\frac{\partial f}{\partial x}$ = 1-yzcos(xyz), $\frac{\partial f}{\partial y}$ = 1-xzcos(xyz) and $\frac{\partial f}{\partial z}$ = 1-yxcos(xyz)

$\frac{\partial f}{\partial z}$$_{(0,0,0)}$ = 1-0 $\neq$0.

Hence the there exists a small neighbourhood around (0,0,0) in which z can be expressed as a function of x and y satisfying f(0,0)=0

Request confirm if my logic is correct. Also, I am unable to solve and express z in terms of x and y. Request guide in this respect.

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Your solution in right; to get the final goal you can approximate $sin (xyz)$ with its Taylor expansion in a small neighbourhood of the origin and than solve the linear equation in $x,y,z$.

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  • $\begingroup$ Thank you..I had not thought of using Taylor's expansion.. $\endgroup$ – SAK Nov 10 '16 at 3:15
  • $\begingroup$ My teacher taught me this trick...it's useful in a lot of problems;) (upvote:D) $\endgroup$ – Giuseppe Bargagnati Nov 10 '16 at 13:22
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Here's how to see the connection between your equation and the z=f(x,y)...

Steps are as follows:

  1. assume $\sin(xyz) = 0$. This one is causing significant problems. We can do this assumption by marking $x=0 \vee y=0$. You can't mark $z$ as $0$, since it needs to be same as $f(x,y)$ in the next step. Multiplication $xyz$ has a property that $xyz=0 \iff x=0 \vee y=0 \vee z=0$. $\sin$ function has property that $\sin(0)=0$.
  2. Once we've cleared that, we're left with $x+y+z=0$. This is of form $f(x,y,z)=0$. Now simple substitution $z=f(x,y)$ will give us $f(x,y,f(x,y)) = 0$. Then $f(x,y,z)=x+y+z$ will give us $x+y+f(x,y)=0$, reorganizing will give $f(x,y)=-x-y$.
  3. Thus to recap, it's possible represent the surface as $f(x,y)=-x-y$ function under assumption that $x=0 \vee y=0$.

Now it is left as excersize why this process does not work if $\sin(xyz)$ is not assumed to be $0$.

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