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I am learning to use the implicit function theorem (IFT) and met recently the following problem:
Let $F(x,y)=x+y+x^5-y^5$. The given equation defines a smooth function $\phi:U\rightarrow \mathbb{R}$ (where $U$ is a neighbourhood of $0$) such as $F(x,\phi(x))=0$ $\forall x\in U$ (because $F(0)=0$, $\frac{\partial F(0)}{\partial y}=1\ne0$ and it is possible to use the IFT). We are to evaluate $\phi^{(2004)}(0)$.

So, let's differentiate the identity $x+\phi(x)+x^5-\phi(x)^5=0$. Firstly we get $1+\phi^{'}(x)+5x^4-5\phi^{4}(x)\phi^{'}(x)=0$. Obviously, $\phi(0)=0$ hence $\phi^{'}(0)=-1$. After that, differentiating one more time we evaluate $\phi^{''}(0)$, $\phi^{'''}(0)$ and so on. It means we are able to evaluate some derivatives 'by hand'. However, after the 3rd or the 4th step this process becomes impossible.
Ok, there is a formula for evaluating the highest derivatives of function products, but there are also very long computations. Is it possible to avoid this kind of evaluations?

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First, until you say $F(x,y)=0$, you don't have an equation that's defining $y$ locally as a function of $x$. This is really a sneaky problem: Notice that 2004 is even; can you deduce some sort of symmetry property of $\phi$ from the nature of the function $F$?

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  • $\begingroup$ $\phi$ and $\phi^{(2004)}$ are odd functions. Thank you! $\endgroup$
    – user74574
    May 3, 2013 at 13:23

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