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I was asked a simple question, show that $y+\sin y=x$ sets in the neighborhood of $(0,0)$ $y$ as a function of $x$, and find $\dfrac{dy}{dx}(0,0)$

Firstly, my naive solution would be:

Since $lim_{y \to 0} \frac{\sin y}{y} = 1$ I want to say that in the vicinity of $(0,0)$ $y=\sin y$ and then we get the equation $2y=x$, $y=\dfrac{1}{2}x$, and so $\dfrac{dy}{dx}(0,0)=\dfrac{1}{2}$

But if I was to do this question using the implicit fucntion theorem I have a problem.

Let's define $f(x,y)=y+\sin y-x$

Let's look at the matrix of the partial derivatives: $\begin{pmatrix} \dfrac{df}{dx}(0,0) & \dfrac{df}{dy}(0,0)\end{pmatrix} = \begin{pmatrix} -1 & 2 \end{pmatrix}$ and of course this matrix is not invertible, it is not even a square matrix.

So my questions are:

1) Why is my naive solution wrong?

2) Why isn't the jacobi matrix a square matrix?

3) Even if it was an invertible matrix, how would I find $y$ as a function of $x$ in order to find $\dfrac{dy}{dx}$?

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  • $\begingroup$ The wikipedia link doesn't have the complete statement of IFT, that's why you don't know how to find $\dfrac{\mathrm dy}{\mathrm dx}(0)$. What text are you following? $\endgroup$ – Git Gud Mar 29 '14 at 21:30
  • $\begingroup$ The jacobian test. I want the jacobi matrix to be invertible $\endgroup$ – Oria Gruber Mar 29 '14 at 21:33
  • $\begingroup$ I meant text, I typoed. $\endgroup$ – Git Gud Mar 29 '14 at 21:34
  • $\begingroup$ I'm not sure what you mean, but my main sources of information of wikipedia in english, hebrew, and my notebook from calculus III lectures. $\endgroup$ – Oria Gruber Mar 29 '14 at 21:35
  • $\begingroup$ Is the IFT in your notes the same as the wiki one? Isn't it longer or anything? $\endgroup$ – Git Gud Mar 29 '14 at 21:37
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I'll start by answering 2.

In the notation used in the wikipedia link you have $m=n=1$. So the jacobian matrix is a $1\times 2$ matrix. It isn't square simply because $n\neq 0$.

In 3, it's not the jacobian matrix that needs to be invertible, but rather the matrix $\begin{pmatrix}\dfrac{\mathrm df}{\mathrm dy}(0,0)\end{pmatrix}$ and it is.

Therefore there exist intervals $V,W$ around $0$ such that "$y\colon V\to W$ is a smooth function of $x$", that is there exists a smooth function $g\colon V\to W$ such that $\forall x\in V\exists !y\in W(g(x)=y\land f(x,y)=0)$.

Define $h\colon V\to W, x\mapsto f(x,g(x))$. By differentiating, the chain rule yields $$\forall x\in U\left(0=h'(x)=\dfrac{\partial f}{\partial x}(x,g(x))+\dfrac{\partial f}{\partial y}(x,g(x))g'(x)\right).$$

In particular for $x=0$ one gets $g'(0)=-\dfrac{\dfrac{\partial f}{\partial x}(0,g(0))}{\dfrac{\partial f}{\partial y}(0,g(0))}=-\dfrac{\dfrac{\partial f}{\partial x}(0,0)}{\dfrac{\partial f}{\partial y}(0,0)}=\dfrac 1 2$.

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  • $\begingroup$ This result leads me to believe that the naive solution was correct, and not a fluke. What do you think? Could we have just replaced $siny$ with $y$ in the vicinity of $y=0$? $\endgroup$ – Oria Gruber Mar 29 '14 at 22:18
  • $\begingroup$ I don't know if it was a fluke or not, to be honest. $\endgroup$ – Git Gud Mar 29 '14 at 22:21
  • $\begingroup$ Ok, what about if we replace $(0,0)$ with $(0,\pi)$? According to Lechuza, we can represent $y$ as a function of $x$ in the neighborhood of any point since $\frac{df}{dy}$ is continuous, but at $y= \pi$ we get $\frac{df}{dy}=1+cos\pi = 1-1 = 0$. Can we represent $y$ as a smooth $C^1$ function of $x$ in the vicinity of $(0,\pi)$? $\endgroup$ – Oria Gruber Mar 29 '14 at 22:25
  • $\begingroup$ @OriaGruber I think that depends, but certainly the IFT won't asure that's possible. Take $f(x,y)=x^2+y^2-1$. Around $(0,0)$ the partial derivative with respect to $y$ is null. Now think about the equation $f(x,y)=0\color{grey}{\iff x^2+y^2=1}$. By picturing the circle you can tell that around $(0,0), \,y$ is not a function of $x$. $\endgroup$ – Git Gud Mar 29 '14 at 22:33
  • $\begingroup$ *around $(1,0)$ $\endgroup$ – Git Gud Oct 26 '14 at 15:01
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Let $f(x,y)=y+\sin y-x$. So, $\frac{\partial f}{\partial y}(x,y)= 1+\cos y$ and $\frac{\partial f}{\partial x}(x,y)= -1$ are continuos functions. Then, $f(x,y)$ is $C^1$. Therefore, by Implicit function theorem, exists $I$x $J \in \mathbb{R}^2$ st $f^{-1}(0) \cap$ $(I$x $J)$ is a graphic of a function in $x$. By chain rule, we got:

$$y'(x)=-\frac{\partial f/ \ \partial x}{\partial f / \partial y } \Rightarrow y'(0)=1/2$$

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  • $\begingroup$ "Therefore, by Implicit function theorem" IFT also requires that the matrix $\frac{df_i}{dy_j}$ is invertible. I think that because $1+cos0 = 2 \neq 0$ this requirement is met here. $\endgroup$ – Oria Gruber Mar 29 '14 at 22:02
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    $\begingroup$ I don't think that's true. $f(x,y) = x^2+y^2-1$ for example. the partial derivatives are continuous, but you can't represent $y$ as a function of $x$ in the vicinity of $(1,0)$ and $(-1,0)$ $\endgroup$ – Oria Gruber Mar 29 '14 at 22:29
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    $\begingroup$ I'm sorry. $df/dx$ and $df/dy$ is continuos and $df/dy(x_0,y_0)$ non-zero... that's enough! :) $\endgroup$ – leticia Mar 29 '14 at 22:41
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    $\begingroup$ in terms of scalar functions, nonzero = invertible :) $\endgroup$ – Oria Gruber Mar 29 '14 at 22:45

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