I need to find the extrema of the following function in the range $-2\pi$ and $2\pi$ for both $x$ and $y$, but I don't know how to go about doing it since it's a bit weird and not similar to other functions I've seen:

$$f(x,y)=\frac{\sin (x) \sin (y)}{xy}$$

I've evaluated the gradient function as below:

$\nabla f = <\frac{\sin (y) (x \cos (x)-\sin (x))}{x^2 y}, \frac{\sin (x) (y \cos (y)-\sin (y))}{y^2 x}>$

but setting it to zero gives a few answers for $x$ and $y$, none of which seem to be the right answer.

The following is the sketch of the graph in the aforementioned range. According to WolframAlpha, it should have it's local extrema at $\{0, 4.49\}$, $\{0, -4.49\}$, $\{4.49, 0\}$, $\{-4.49, 0\}$.

enter image description here

  • 1
    $\begingroup$ Is this defined at $(0,0)$? Because, I would generally be inclined to tell you that the maximum is there (at a value of $1$) which is clear from the graph and otherwise not too bad to verify noting that $|\sin(x)|<x$ for $x\neq 0$. One might also note that this function is $\frac{\sin(x)}x\frac{\sin(y)}y$ so you can handle the variable separately. $\endgroup$ – Milo Brandt Nov 30 '15 at 4:03
  • 1
    $\begingroup$ This amounts to solving $\tan z = z$ (why?), which has transcendental solutions. $\endgroup$ – Jon Warneke Nov 30 '15 at 4:03
  • $\begingroup$ @MiloBrandt Thanks for the comment. Yes I realize the maximum is at x,y = 0. I was more interested in finding the minima. $\endgroup$ – Miro Markaravanes Nov 30 '15 at 4:05
  • $\begingroup$ @JonWarneke Thanks for the answer. I was able to get to that equation. I just dismissed it prematurely. Seems like I shouldn't have. $\endgroup$ – Miro Markaravanes Nov 30 '15 at 4:08

As said in comments, if $$F=\frac{\sin (x) \sin (y)}{xy}$$ $$F'_x=\frac{\cos (x) \sin (y)}{x y}-\frac{\sin (x) \sin (y)}{x^2 y}=\frac{\sin (y) (x \cos (x)-\sin (x))}{x^2 y}$$ $$F'_y=\frac{\sin (x) \cos (y)}{x y}-\frac{\sin (x) \sin (y)}{x y^2}=\frac{\sin (x) (y \cos (y)-\sin (y))}{x y^2}$$ where we see appearing the solutions of equations $z=\tan(z)$; beside the trivial solution $z=0$, there is only one solution which does not show any closed form. This solution is close to $\frac {3\pi}2$.

Developing $z-\tan(z)$ around $z=\frac {3\pi}2$ as a series, we have $$z-\tan(z)=\frac{1}{z-\frac{3 \pi }{2}}+\frac{3 \pi }{2}+\frac{2}{3} \left(z-\frac{3 \pi }{2}\right)+O\left(\left(z-\frac{3 \pi }{2}\right)^2\right)$$ and the positive root is given by $$z=\frac{1}{8} \left(3 \pi +\sqrt{81 \pi ^2-96}\right)\approx 4.49340$$ which is extremely close to the solution $(\approx 4.49341)$.

Then the solutions shown by Wolfram Alpha.


As you see, this function is separable, meaning that it is the product of a function of $x$ alone and one of $y$ alone. This means that the 2D extrema will occur where both these 1D functions have an extremum. (Whether they are minima or maxima will require a sign discussion.)

Now you are looking for the extrema of the cardinal sine function, $\text{sinc}(x)=\dfrac{\sin(x)}x$, which occur at the roots of $x\cos(x)-\sin(x)$. As you can verify, $x=0$ also cancels the derivative (the Taylor's development of $\text{sinc}$ is $1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\cdots$).

The equation $x=\tan(x)$ cannot be solved by a closed formula, so you must resort to numerical methods. By looking at the plot, you can observe that the roots are close to the vertical asymptotes, so that $x_k\approx\pm\left(k+\dfrac12\right)\pi$ for all $k>0$.

enter image description here

A better approximation is given by the perturbation method: let $x_k=\left(k+\dfrac12\right)\pi-t_k$. Then,


This yields a second degree equation

$$t_k^2-\left(k+\frac12\right)t_k+1=0$$with the positive solution $$t_k=\frac12\left(\left(k+\frac12\right)\pi-\sqrt{\left(k+\frac12\right)^2\pi^2-4}\right).$$

$$\begin{align}&k&x_k&&\tan(x_k)\\ &1&4.4896547020577&&4.4151632216947\\ &2&7.7245238157513&&7.6813229186826\\ &3&10.9038636765678&&10.8732763177582\\\dots \end{align}$$

These values can be readily improved by a few iterations of Newton's method.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.