If $f(x,y)=e^{x-y}-x-y-1,\forall\;(x,y)\in\Bbb{R}^2$, then $\exists\;\alpha>0$ and $\varphi:\,]-\alpha,\alpha[\to\Bbb{R}$ of class $\Bbb{C}^{\infty}$

Question

Let $$f:\Bbb{R}^2\to\Bbb{R}$$ $$(x,y)\mapsto f(x,y)=e^{x-y}-x-y-1,\forall\;(x,y)\in\Bbb{R}^2$$ $i.$ Show that there exists $\alpha>0$ and $\varphi:\,]-\alpha,\alpha[\to\Bbb{R}$ of class $\Bbb{C}^{\infty}$ such that $\varphi(0)=0$ and $f(x,\varphi(x))=0$ for $x\in]-\alpha,\alpha[.$

$ii.$ Compute $\varphi'(x)$ and $\varphi''(x)$.

$iii.$ What is $\lim\limits_{x\to 0}\frac{\varphi(x)}{x^2}?$

This does not look like what I am little familiar with. For $ii,$ I tried using $\varphi(x+h)-\varphi(x)=\ell(x,y)(h)+\Vert h\Vert\epsilon(h)$ but it's wrong. Any help on these questions, please?

Answer 1

(i)

In fact we must have $$e^{x-\phi(x)}=x+\phi(x)+1$$we know that $e^{x-\phi(x)}$ is a strictly decreasing function of $\phi(x)$ and $x+\phi(x)+1$ is a strictly increasing function of $\phi(x)$ for each value of $x$ therefore the equation $e^{x-\phi(x)}=x+\phi(x)+1$ commits exactly one value for $\phi(x)$ for each $x$. Then we can say that such a $\phi(x)$ exists and is unique. Let's define $$\epsilon(x)=x-\phi(x)$$therefore by substituting $$e^{\epsilon(x)}=2x-\epsilon(x)+1$$or $$\epsilon(x)+e^{\epsilon(x)}=2x+1$$define $g(y)=y+e^y$. This function is $C^\infty$ and strictly increasing so is $g^{-1}(y)$ and so is $g^{-1}(2x+1)$. From the other side $$\epsilon(x)=g^{-1}(2x+1)$$therefore $\epsilon(x)$ is $C^\infty$ and so is $\phi(x)$.

(ii)

This part is straight forward. We have $$e^{x-\phi(x)}=x+\phi(x)+1$$by differentiating we obtain$$(1-\phi'(x))e^{x-\phi(x)}=1+\phi'(x)$$and rearranging the terms leads to $$\phi'(x)=1-\dfrac{2}{1+e^{x-\phi(x)}}$$similarly$$\phi''(x)=\dfrac{4e^{x-\phi(x)}}{(1+e^{x-\phi(x)})^3}$$

(iii)

By twice using the L^Hoptial's rule we obtain$$\lim_{x\to 0}\dfrac{\phi(x)}{x^2}=\lim_{x\to 0}\dfrac{\phi'(x)}{2x}=\lim_{x\to 0}\dfrac{\phi''(x)}{2}=\lim_{x\to 0}\dfrac{2e^{x-\phi(x)}}{(1+e^{x-\phi(x)})^3}=\dfrac{1}{4}$$

Answer 2

We are given the function $$f(x,y):=e^{x-y}-x-y-1\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ .$$ It is claimed that in the neighborhood of $(0,0)$ the equation $f(x,y)=0$ implicitly defines a $C^\infty$-function $y=\phi(x)$, whereby $\phi(0)=0$.

(i) This is a standard case for the implicit function theorem (IFT). Compute $$f_x(x,y)=e^{x-y}-1,\quad f_y(x,y)=-e^{x-y}-1\ .$$ Since $f_y(0,0)=-2\ne0$ the IFT guarantees the existence of an $\alpha>0$ and a $C^1$-function $$\phi:\quad]{-\alpha},\alpha[\>\to{\mathbb R},\qquad x\mapsto y=\phi(x)$$ such that $$f\bigl(x,\phi(x)\bigr)\equiv0\qquad(-\alpha<x<\alpha)\ .\tag{1}$$ Moreover $$\phi'(0)=-{f_x(0,0)\over f_y(0,0)}=0\ .$$ One could set up an induction scheme to prove that $\phi$ is in fact $C^\infty$. But note that "complexifying everything" we see that $\phi$ is a complex differentiable function of the variable $x$, hence automatically $C^\infty$.

(ii) It is not possible to express $\phi$ in terms of elementary functions. Therefore you cannot expect to obtain finite expressions for $\phi'(x)$, $\phi''(x)$ in terms of variable $x$. But the higher derivatives $\phi^{(n)}(0)$ (these are just numbers) can be calculated recursively, see below.

(iii) Differentiating $(1)$ with respect to $x$ we obtain $f_x\bigl(x,\phi(x)\bigr)+f_y\bigl(x,\phi(x)\bigr)\phi'(x)=0$ and then $$f_{xx}\bigl(x,\phi(x)\bigr)+2f_{xy}\bigl(x,\phi(x)\bigr)\phi'(x)+f_y\bigl(x,\phi(x)\bigr)\phi''(x)=0\ .$$ Letting $x:=0$ here leads to $1+0+(-2)\phi''(0)=0$; hence $\phi''(0)={1\over2}$. As $\phi(0)=\phi'(0)=0$ Taylor's theorem then implies $$\lim_{x\to0}{\phi(x)\over x^2}={1\over4}\ .$$