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I showed that the relation $$f(x,y)=e^x - e^y + xy = 0$$

defines near (0,0) an implicit function y=$\phi (x)$, since the $1x1$ block, $\frac{df}{dy}$, evaluated at (0,0) gives -1, which is non-zero - and this is also the determinant of the $1x1$ block, which makes this block invertible.

I then Taylor expanded $\phi (x)$ at the origin, to order 3, but I'm not sure whether I am computing the derivatives correctly.

The derivative of $\phi(x)$ is given by - $\frac{f_x}{f_y}$, evaluated at (0,0).

My question is: When differentiating f(x,y) w.r.t. the variable x, can I treat y as constant, as we usually do, when computing partial derivatives in multivariable calculus, or should I treat y as a function of x, so that for f(x,y) = $e^x - e^y + xy$ replace the y's with $\phi$(x), before computing the derivatives for the implied function, $\phi$(x)?

The third derivative is very messy, and if I can treat y as constant, when computing $f_x$, that would be great. I'm guessing that treating y as constant is perfectly fine, since even if I replaced the y's with $\phi$(x), then when computing $f_x$, I'd get some terms in the form of $\phi$'(x), which aren't that helpful, if I don't have an explicit function for $\phi$ - and that we only know the existence of $\phi$ by the Implicit Function Theorem.

Thanks in advance,

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  • $\begingroup$ What exactly does the problem ask? $\endgroup$ – Git Gud Dec 21 '14 at 0:13
  • $\begingroup$ @GitGud, the exact question is: Show that $e^x - e^y +xy = 0$ defines near (0,0) an implicit function y=$\phi$(x) in $C^3$ and compute its expansion to order 3 at (0,0). Thanks... $\endgroup$ – User001 Dec 21 '14 at 0:16
  • $\begingroup$ You can't treat $y$ as a constant. It doesn't really matter that you don't know that $\phi$ is, it's enough to know what $\phi(0), \phi'(0)$, etc are, and these you can find. $\endgroup$ – Git Gud Dec 21 '14 at 0:20
  • $\begingroup$ Ah, ok. I'll redo the problem. Thanks so much for the help, @GitGud. :) $\endgroup$ – User001 Dec 21 '14 at 0:23
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with $y=y(x)$ we get $e^x-e^yy'+y+xy'=0$ thus we obtain $$y'=-\frac{e^x+y}{x-e^y}$$ if $x\neq e^y$

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It is not so messy if you cascade the problem for successive derivatives.

Let $$F=e^x - e^y + xy = 0$$ So, using implicit differentiation, $$F'_x=e^x+y$$ $$F'_y=-e^y+x$$ and then $$y'=\frac{e^x+y}{e^y-x}$$ Now, compute $y''$ deriving the rhs and the result will be a function of $x,y,y'$; in the same way, $y'''$ will be a function of $x,y,y',y''$

Just added for your curiosity

Equation $$F=e^x - e^y + xy = 0$$ has an explicit solution $y=\phi(x)$. It can be expressed using Lambert function and write $$y=-\frac{e^x}{x}-W(-z)$$ in which $$z=\frac{e^{-\frac{e^x}{x}}}{x}$$

If you want, around $x=0$, to build a Taylor series of $y$ as a function of $x$, you could just write $$y=\sum_{i=0}^n a_ix^i$$ replace in the expression of $F$ and cancel as much terms as you can. You should arrive to $a_0=0$, $a_1=1$, $a_2=1$, $a_3=0$, $a_4=-1$, $a_5=-\frac{2}{3}$ and so on.

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