Infinite ABS Inequality

Prove that ∑k=−NN∣x+k∣−∣k∣≥x2\sum_{k=-N}^{N}|x+k|-|k| \ge x^2 as N→∞N\to \infty and find equality cases.

Details and Assumptions

We take the limit as NN approaches ∞\infty because ∑k=−∞∞∣x+k∣−∣k∣\displaystyle\sum_{k=-\infty}^{\infty}|x+k|-|k| is not well-defined.

#Algebra #AbsoluteValue #Parabola #InfiniteSum #EqualityCase

Note by Daniel Liu
6 years, 10 months ago

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Comments

Set : fn(x)=∑k=−nn∣x+k∣−∣k∣=∣x∣+∑k=1n∣x+k∣+∣x−k∣−2∣k∣f_n(x)= \sum_{k=-n}^n |x+k|-|k| =|x|+\sum_{k=1}^n |x+k|+|x-k| -2|k| Note that fn(−x)=fn(x)f_n(-x)=f_n(x) for any real xx and any integer n>0n>0 then we only need to consider the case x≥0x\geq 0, and if k≥x>0k\geq x>0 then ∣x+k∣+∣x−k∣−2∣k∣=0|x+k|+|x-k|-2|k|=0, therefore for n>xn>x : fn(x)=x+∑k=1⌊x⌋(x+k)+(x−k)−2k=x+2x⌊x⌋−⌊x⌋(⌊x⌋+1).f_n(x)=x+\sum_{k=1}^{\lfloor x\rfloor}(x+k)+(x-k)-2 k= x+2x \lfloor x \rfloor -\lfloor x\rfloor (\lfloor x\rfloor +1).

Clearly now, fn(x)=x−⌊x⌋+2x⌊x⌋−⌊x⌋2f_n(x)= x-\lfloor x\rfloor +2x \lfloor x\rfloor-\lfloor x\rfloor^2 , the rest should follow easily.

fn(x)−x2=x−⌊x⌋+x(⌊x⌋−x)+⌊x⌋(x−⌊x⌋)=(x−⌊x⌋)(1+⌊x⌋−x)≥0f_n(x)-x^2 =x- \lfloor x\rfloor + x(\lfloor x\rfloor-x)+\lfloor x\rfloor(x-\lfloor x\rfloor)=(x-\lfloor x\rfloor)(1+\lfloor x\rfloor -x)\geq 0

Obviously, the equality is when x∈Zx\in \mathbb{Z}.

Remark : you don't need n→∞n\to \infty you just need n≥xn\geq x, and the latter series does make sense and it well-defined (though I'd use the term 'convergent' because well defined usually used for mono-valued functions).

Haroun Meghaichi - 6 years, 10 months ago

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How is the latter sequence defined? I seem to be getting different values for plugging in the same xx value if I compute it differently.

Daniel Liu - 6 years, 10 months ago

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Can you give me an example ?

Haroun Meghaichi - 6 years, 10 months ago
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