Let $g:R^2 →R$ such that
$g(x,y)=\sin y+y+e^x −1$ $\forall (x,y)∈R^2$
Prove function h exists such that it is defined on an open interval around the origin s.t. g(x,h(x)) = 0 for all x in the open interval. Derive $α,β,γ ∈ R$ s.t. $h(x)=α+βx+γx2 +o(x^2)$.
To find such a function, $g(x,h(x)) = \sin h(x)+h(x)+e^x −1=0$ but I am stuck after that and also am not sure if there is a potential open interval theorem I can use to prove it.
Note that $g$ is smooth. You have $g(0,0) = 0$, ${\partial g(0,0) \over \partial x} = 1$, ${\partial g(0,0) \over \partial y} = 2$ so the implicit function theorem shows that there is some open interval $I$ containing the origin and a function $h:I \to \mathbb{R}$ such that $h(0) = 0$, $g(x,h(x)) = 0$ for $x \in I$. Furthermore, $h$ is differentiable at $x=0$ and $h'(0) = - { {\partial g(0,0) \over \partial x} \over {\partial g(0,0) \over \partial y}} = -{1\over 2}$.
So, for a start, we have $h(x) = -{1\over 2} x + o(x)$.
Since ${\partial g(x,y) \over \partial y} \neq 0$ for all $(x,y)$ sufficiently close to $(0,0)$, and $g(x,h(x)) = 0$, we can repeat the above process (we don't know $h$ but we know it exists for small $x$) to get $h'(x) = - { {\partial g(x,h(x)) \over \partial x} \over {\partial g(x,h(x)) \over \partial y}}$ in a neighbourhood of $x=0$ and so we see that $h$ is in fact, as smooth as $g$ (we only need the $2$nd derivative) and we can compute $h''(0)$ by differentiating $h'(x)$!
Since we know it exists, we can take a short cut and just differentiate the expression $\sin h(x) + h(x) + e^x -1 = 0$ to get $(\cos h(x)+1)h'(x) + e^x = 0$ and again to get $(-\sin h(x) )(h'(x))^2 + (\cos h(x)+1)h''(x) + e^x = 0$, and using $h(0) = 0$ we get $h''(0) = -{ 1 \over 2}$.
Hence we have $h(x) = 0 -{1 \over 2} x -{1 \over 4} x^2 + o(x^2)$.