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How can I show that $f(x,y)=e^x cos(y)$ doesn't have maxima nor minima in the unit circle? Because $f_x = f_{xx} =0$ when $x=0$ and $y=\frac{\pi}{2} +n\pi, n\in \Bbb{Z}$. and isn't $(0,\frac{\pi}{2},0)$ in the unit circle?

if we try to find the maximas or minimas with derivates. $$f_{y}=-\frac{e^{x+iy}-e^{x-iy}}{2i}$$ And when $x,y=0\to f_y(x,y)=0$. isn't $(0,0,0)$ in the circle..?

ELEC
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  • Do you mean "unique" maxima and minima? The unit circle is a compact set, and all bounded functions achieve their maximum and minimum on a compact set. – Jeff Snider Jan 21 '14 at 20:14
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    @JeffSnider I suspect that "in" means the interior. – Mark McClure Jan 21 '14 at 20:15
  • yeah unique. still don't get it.. – ELEC Jan 21 '14 at 20:16
  • So the question should read "maxima nor minima in the open unit disk." – Jeff Snider Jan 21 '14 at 20:16
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    Yeah. The question just uses the word (finnish translation tho) "unit circle", not sure if it means unit "ball" in this case or something. And yeah, I think it means inside or in the interior of the cirle. – ELEC Jan 21 '14 at 20:19
  • Note that $\frac{\pi}{2}\approx 1.57 > 1$ thus not in the unit circle or disk. – Jeff Snider Jan 21 '14 at 20:19
  • but aren't we interested if $f(x,y)=z=0$. so $z=0$ and inside the circle, even tho the x,y values might not be. – ELEC Jan 21 '14 at 20:21

1 Answers1

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Assuming the question is intended: show any maxima or minima of $f=e^x\sin(y)$ are outside the unit circle.

The necessary condition for a point being a maximizer or minimizer is that it is a stationary point, i.e., $\nabla f(x,y)=(0,0)$. We have $$f_x=e^x\cos(y),$$ and $$f_y=-e^x\sin(y).$$ Since $e^x$ is positive for all $x$, for $e^x\cos(y)=0$ we must have $\cos(y)=0$, hence $y=\pi n$ for integer $n$. For $-e^x\sin(y)=0$ we must have $\sin(y)=0$, hence $y=\frac{\pi}{2}+\pi m$ for integer $m$. Since no $y$ satisfies both requirements, there is no stationary point for the equation.

This is intuitively apparent from the shape of the function, which along the $y$-axis is a sine curve of amplitude 1, but as $x$ increases the amplitude increases giving rise to greater maxima and smaller minima.

Jeff Snider
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