Show that, for $x$ and $y$ sufficiently tiny, the equation can be solved to $y$

Question

$$e^{\sin{xy}}+x^2-2y-1=0$$ What she meant with "sufficiently tiny"? I think i need go by the theorem about implicit functions, but i really don't get it how.

Answer 1

The implicit function theorem says that, under some regularity hypotheses, there is a function $f(x)$ such that the solution set to $F(x,y)=0$ near a particular solution $(x_0,y_0)$ exists and is given by $y=f(x)$. It requires you to have such a point $(x_0,y_0)$ (in your problem they are telling you that it is $(0,0)$) and to check that $\frac{\partial F}{\partial y}(x_0,y_0) \neq 0$.

In contrast to some of the other answers, you do not need a full Taylor development to prove the statement as given in the OP. All you need is to check that $\left. \frac{\partial}{\partial y} \left ( e^{\sin(xy)} + x^2 -2y - 1 \right ) \right |_{(x,y)=(0,0)} \neq 0$. (Alternately, for just existence of solution you could find $x=f(y)$, which would let you take $\frac{\partial}{\partial x}$ instead of $\frac{\partial}{\partial y}$.)

Answer 2

Let $F(x,y):=e^{\sin{xy}}+x^2-2y-1$. Then $F(0,0)=0$ and $F_y(0,0) \ne 0$

The theorem about implicit functions says: there is a neighborhood $U$ of $0$ and a function $f \in C^{\infty}(U)$ with the properties:

$f(0)=0$ and $F(x,f(x)))=0$ for all $x \in U$

P.S.: "tiny": near by $0$.

Answer 3

Consider the function $$f(x,y)=e^{\sin({xy})}+x^2-2y-1$$ where $x$ and $y$ are "small" and use $$\sin(z)\approx z\implies e^{\sin({z})}\approx 1+z$$ Replace to get $$f(x,y)\approx 1+xy+x^2-2y-1=x^2+xy-2y$$ and since you want $f(x,y)=0$ then $$x^2+xy-2y=0$$ which can be solved for $x$ or $y$.

This is equivalent to a first order Taylor expansion in $y$ and to second order in $x$.

Answer 4

For small $x,y$, you can use the Taylor development to the first order and

$$0=e^{\sin xy}+x^2-2y-1\approx xy+x^2-2y$$ so that

$$y\approx\frac{x^2}{2-x}.$$

The next term yields

$$xy+\frac{x^2y^2}2+x^2-2y\approx0$$ and

$$y\approx\frac{2-x\pm\sqrt{(x-2)^2-2x^4}}{x^2}.$$ Despite appearances, the first expression and the "$-$" one are very close.