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let $x,y\in R$, and such as

$$y\ge\displaystyle\sum_{j=0}^{50}\dfrac{1}{2^{50-j}}|x-2^{j}|+\displaystyle\sum_{j=1}^{50}\dfrac{1}{2^{j}}|x-2^{50+j}|$$

then find $$f(x,y)_\min=x+2y$$

my ideaļ¼š I think we use this inequality $$|x|+|y|\ge |x+y|$$

Michael Hardy
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math110
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1 Answers1

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You can think this by try to draw something.

You know on $[2^m,2^{m+1}]$, the function is linear, and then slope changes from

$-\sum_{j=0}^{50}\dfrac{1}{2^{50-j}}-\sum_{j=1}^{50} \dfrac{1}{2^j} \sim -3$

to

$\sum_{j=0}^{50}\dfrac{1}{2^{50-j}}+\sum_{j=1}^{50} \dfrac{1}{2^j} \sim 3$

The slope increases piece by piece. You may see this means your function is convex. Thus the point you are going to find is the point that has a slope closest to $-\dfrac{1}{2}$.

[Since in smooth case, the point is just on the tangent, try to figure out why]

And we can see that on the interval that $[2^{49},2^{50}]$, the slope is

$\sum_{j=0}^{49}\dfrac{1}{2^{50-j}} - 1 - \sum_{j=1}^{50} \dfrac{1}{2^j} = -1$

on interval $[2^{50},2^{51}]$, we can see the slope is

$\sum_{j=0}^{50}\dfrac{1}{2^{50-j}} - \sum_{j=1}^{50} \dfrac{1}{2^j} = 1$

So at $x = 2^{50}$, you shall get the smallest $f(x,y) = x+2y$.

Yimin
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