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Problem

What can be said about following statement when dot $(x,y)$ is close enough to origin.

$$ x+y+4\sin(x)\sin(y)\ge 0 $$

Attempt to solve

I mark the given expression as:

$$ f(x,y)=x+y+4\sin(x)\sin(y) $$

Now i simply examine the situation where $f(x,y)\ \ge 0$ and $$\lim_{(x,y)\rightarrow (0,0)}f(x,y)$$

I could start off my examining the gradient at point $(0,0)$

$$ \nabla f(x,y)=\begin{bmatrix} 4\cos(x)\sin(y)+1 \\ \cos(y)4\sin(x)+1 \end{bmatrix} $$

$$ \nabla f(0,0)=\begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

It seems $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ would be increasing direction. Now my intuition would tell me that if $\nabla f(x,y)\neq0$ wouldn't necessary mean that critical point couldn't exist in this particular point. If $\nabla f(x,y)=0$ i would say this is certainly critical point.

Since it's uncertain if this is critical point or not. Computing a hessian matrix would result in more definite answer to this question.

$$ H(x,y)=\begin{bmatrix} -4\sin(x)\sin(y) & 4\cos(x)\cos(y) \\ \cos(y)4\cos(x) & - 4\sin(x)\sin(y) \end{bmatrix} $$

$$ H(0,0) = \begin{bmatrix} 0 & 4 \\ 4 & 0 \end{bmatrix} $$ egeinvalues are $\lambda_1=-4,\lambda_2=4$ so this would mean hessian is indefinite matrix which means point $(0,0)$ is saddle point.

At this point it would be probably a good idea to make plot of this function since i don't know other good methods from now on.

enter image description here

The plot confirms that $\nabla f(x,y)$ at point $(0,0)$ gives the increasing direction. The $- \nabla f(x,y)$ is also increasing direction even though negative gradient is usually interpreted as the largest decreasing direction. In this the interpretation would be incorrect. We can see that:

$$ \hat{d_1}= \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \hat{d_2}= \begin{bmatrix} -1 \\ -1 \end{bmatrix} $$ are increasing direction and the decreasing direction would be:

$$ \hat{d_3}= \begin{bmatrix} 1 \\ -1 \end{bmatrix},\hat{d_4}= \begin{bmatrix} -1 \\ 1 \end{bmatrix} $$

Now what happens if we try to approach point $(0,0)$. When $(x,y)$ values approach zero the variation in $f(x,y)$ values decreases. Also values of $f(x,y)$ become smaller and smaller until it reaches point $(0,0)$ (if we approach the critical point using a circle and start decreasing radius by some constant).

Now if i was able to prove this circle can get arbitrary close to the point (0,0) (arbitrary small radius) this would prove the existence of limit at point $(0,0)$ ? I don't know how formally existence of a limit can be proven in multivariable case but this sort of approach would sound like it could work. Now these things would indicate that following statement is true:

$$ \lim_{(x,y)\rightarrow (0,0)}f(x,y)=0 $$

Also when we can evaluate this function's value at point $(0,0)$ meaning it is defined in this particular point. Limit would automatically exist in this point since it would be already defined which wouldn't require any proof in the first place ?

I've understood that the reason behind why there is such thing called as "limit " is attended to use in cases where we have undefined point and we want to know that can we get arbitrary close to the point without evaluating it's value at that point. If the answer is yes then we would have limit at this particular point.


If someone manages to read all of this i highly appreciate your effort in doing so. Some sort of feedback on this would be great. Do my claims seem correct / false or am i missing something important ?

Also if you have some good resource (for example book) in mind that covers these topics please comment about this.

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  • $\begingroup$ The question is really vague. I would just take $\sin p \approx p$ where p is either x or y. And go from there. But again, the question is not clear to me $\endgroup$ – Yuriy S Feb 2 '18 at 0:42
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    $\begingroup$ Not sure I understand, but isn't the claim false with the simple counter-example of $f(x,x) = 2x+4(\sin x)^2$? $f(x,x)$ is negative for $x<0$ (and $x$ close to $0$) and positive for $x>0$ (and $x$ close to $0$). wolframalpha.com/input/?i=plot+2x%2B4(sin+x)%5E2,+x%3D-1..1 $\endgroup$ – Clement C. Feb 2 '18 at 1:06

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