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I am trying to understand how to show that the best approximation of $\sin(x)$ over $[0,\pi]$ and polynomial of degree $n=0$ (so a constant $c$) by using the maximum norm, i.e $\min_{c} \lVert \sin(x) - c\rVert_\infty$ is with $c=1/2$.

Visually it is clear that the maximal distance to both extreme values of $\sin(x)$ is at $1/2$ but I have no idea how to deduct it from the maximum morm.

Any hints are appreciated.

mvw
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R. Carlos
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1 Answers1

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Yes, it is $\dfrac12$, since $\|\sin-c\|_\infty=c$ if $c\geqslant\dfrac12$ and $\|\sin-c\|_\infty=1-c$ if $c\leqslant\dfrac12$. All you need to know to prove this is that $\max_{x\in[0,\pi]}\sin(x)=1$ and that $\min_{x\in[0,\pi]}\sin(x)=0$.

Therefore, the minimum of $\|\sin-c\|_\infty$ is reached when $c=\dfrac12$.

José Carlos Santos
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  • Thanks. That helps a bit although I am trying to find out that at $c=\frac{1}{2}$ we reach the minimum without having to calculate all possible values. Using the L2 norm I can differentiate over the given interval and get $c=0.54$. I was hoping that for the maximum norm there would be (at least in this case) a closed form solution too. – R. Carlos May 29 '18 at 12:05
  • @R.Carlos I would really like to know what is not closed about the solution $c=\frac12$. – José Carlos Santos May 29 '18 at 12:31
  • Sorry, I wrongly used this word 'closed' here. When it comes to optimization problems I am usually try to find a minimum for the function in question. Since here what needs to be mimimzed is $f(c)=max|sin(x)−c|$ I was hoping to find $c=1/2$ by using differentiation which is apparently not possible. Due to the maximum norm it is intrinsic that this can always be solved apparently by using $c=\frac{min(f(x))+max(f(x))}{2}$. Thanks for helping me to understand this. – R. Carlos May 29 '18 at 12:46