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Question Number 69276 by mr W last updated on 22/Sep/19

f(x)=Σ_(k=1) ^n ∣x+k∣ (1) find the values of x such that f(x) is minumum. (2) fund the roots of f(x)−m=0 as example you can set n=100, m=2500.

$${f}\left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mid{x}+{k}\mid \\ $$$$\left(\mathrm{1}\right)\:{find}\:{the}\:{values}\:{of}\:{x}\:{such}\:{that}\:{f}\left({x}\right)\: \\ $$$${is}\:{minumum}. \\ $$$$\left(\mathrm{2}\right)\:{fund}\:{the}\:{roots}\:{of}\:{f}\left({x}\right)−{m}=\mathrm{0} \\ $$$$ \\ $$$${as}\:{example}\:{you}\:{can}\:{set}\:{n}=\mathrm{100},\:{m}=\mathrm{2500}. \\ $$

Commented by Prithwish sen last updated on 22/Sep/19

I tink f(x) will be minimum if we shift the origin to the −A where A= mean of 1,2,3,4.........n ∴ x= −((n+1)/2) f(x)= ∣−((n+1)/2) +1∣.............∣−((n+1)/2) +n∣ = Now if we take n =100 then f(x) =99+97+95+.......5+3+1=50^2 =2500 2) Now the root of f(x) for m=2500 f(−50.5)−2500=0 i.e the root of f(x)−m is x=−50.5

$$\mathrm{I}\:\mathrm{tink} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{will}\:\mathrm{be}\:\mathrm{minimum}\:\mathrm{if}\:\mathrm{we}\:\mathrm{shift}\:\mathrm{the}\:\mathrm{origin}\: \\ $$$$\mathrm{to}\:\mathrm{the}\:−\mathrm{A}\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{where}\:\mathrm{A}=\:\boldsymbol{\mathrm{mean}}\:\boldsymbol{\mathrm{of}}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}.........\boldsymbol{\mathrm{n}} \\ $$$$\therefore\:\mathrm{x}=\:−\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:\mid−\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}\mid.............\mid−\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\:+\mathrm{n}\mid \\ $$$$\:\:\:\:\:\:\:\:=\: \\ $$$$\:\:\mathrm{Now}\:\mathrm{if}\:\mathrm{we}\:\mathrm{take}\:\boldsymbol{\mathrm{n}}\:=\mathrm{100} \\ $$$$\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\mathrm{99}+\mathrm{97}+\mathrm{95}+.......\mathrm{5}+\mathrm{3}+\mathrm{1}=\mathrm{50}^{\mathrm{2}} =\mathrm{2500} \\ $$$$\left.\mathrm{2}\right)\:\:\mathrm{Now}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{for}\:\mathrm{m}=\mathrm{2500} \\ $$$$\mathrm{f}\left(−\mathrm{50}.\mathrm{5}\right)−\mathrm{2500}=\mathrm{0}\: \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{e}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{root}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{x}}=−\mathrm{50}.\mathrm{5} \\ $$

Commented by Prithwish sen last updated on 23/Sep/19

Thank you sir. I was egarly waiting for your valuable suggestion.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{was}\:\mathrm{egarly}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{your} \\ $$$$\mathrm{valuable}\:\mathrm{suggestion}. \\ $$

Commented by mr W last updated on 22/Sep/19

thanks alot sir! you are generally right. but we can be more exact: since the curve of f(x) is a kind of folded line, we can say: if n is odd, then f(x)_(min) is at x=−((n+1)/2), and if n is even, then f(x)_(min) is when x∈[−((n/2)+1),−(n/2)].

$${thanks}\:{alot}\:{sir}! \\ $$$${you}\:{are}\:{generally}\:{right}.\:{but}\:{we}\:{can}\:{be} \\ $$$${more}\:{exact}: \\ $$$${since}\:{the}\:{curve}\:{of}\:{f}\left({x}\right)\:{is}\:{a}\:{kind}\:{of} \\ $$$${folded}\:{line},\:{we}\:{can}\:{say}:\: \\ $$$${if}\:{n}\:{is}\:{odd},\:{then}\:{f}\left({x}\right)_{{min}} \:{is}\:{at}\:{x}=−\frac{{n}+\mathrm{1}}{\mathrm{2}}, \\ $$$${and}\:{if}\:{n}\:{is}\:{even},\:{then}\:{f}\left({x}\right)_{{min}} \:{is}\:{when} \\ $$$${x}\in\left[−\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right),−\frac{{n}}{\mathrm{2}}\right]. \\ $$

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