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Question Number 176083 by Linton last updated on 11/Sep/22

can you find f(x)? f(x)−f(1−x)= x

$${can}\:{you}\:{find}\:{f}\left({x}\right)? \\ $$$${f}\left({x}\right)−{f}\left(\mathrm{1}−{x}\right)=\:{x} \\ $$

Answered by Rasheed.Sindhi last updated on 11/Sep/22

f(x)−f(1−x)= x f(x)? Replace x by 1−x: f(1−x)−f(1−(1−x) )=1−x f(1−x)−f(x)=1−x 1−f(1−x)+f(x)=x f(x)−f(1−x)=1−f(1−x)+f(x)=x 0=1? ....

$${f}\left({x}\right)−{f}\left(\mathrm{1}−{x}\right)=\:{x}\:\:\:\:\:\:\:{f}\left({x}\right)? \\ $$$${Replace}\:{x}\:{by}\:\mathrm{1}−{x}: \\ $$$${f}\left(\mathrm{1}−{x}\right)−{f}\left(\mathrm{1}−\left(\mathrm{1}−{x}\right)\:\right)=\mathrm{1}−{x} \\ $$$${f}\left(\mathrm{1}−{x}\right)−{f}\left({x}\right)=\mathrm{1}−{x} \\ $$$$\mathrm{1}−{f}\left(\mathrm{1}−{x}\right)+{f}\left({x}\right)={x} \\ $$$${f}\left({x}\right)−{f}\left(\mathrm{1}−{x}\right)=\mathrm{1}−{f}\left(\mathrm{1}−{x}\right)+{f}\left({x}\right)={x} \\ $$$$\:\mathrm{0}=\mathrm{1}? \\ $$$$.... \\ $$

Answered by Rasheed.Sindhi last updated on 12/Sep/22

f(x)−f(1−x)= x; f(x)=? f(1)−f(0)=1 f(2)−f(1)=2 f(3)−f(2)=3 f(4)−f(3)=4 .... f(n−1)−f(n−2)=n−1 f(n)−f(n−1)=n f(n)−f(0)=((n(n+1))/2) f(n)=((n(n+1))/2)+f(0) Assuming f(0)=k f(x)=((x(x+1))/2)+k ; f(0)=k

$${f}\left({x}\right)−{f}\left(\mathrm{1}−{x}\right)=\:{x};\:{f}\left({x}\right)=? \\ $$$$ \\ $$$${f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$${f}\left(\mathrm{3}\right)−{f}\left(\mathrm{2}\right)=\mathrm{3} \\ $$$${f}\left(\mathrm{4}\right)−{f}\left(\mathrm{3}\right)=\mathrm{4} \\ $$$$.... \\ $$$${f}\left({n}−\mathrm{1}\right)−{f}\left({n}−\mathrm{2}\right)={n}−\mathrm{1} \\ $$$${f}\left({n}\right)−{f}\left({n}−\mathrm{1}\right)={n} \\ $$$${f}\left({n}\right)−{f}\left(\mathrm{0}\right)=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${f}\left({n}\right)=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{f}\left(\mathrm{0}\right) \\ $$$${Assuming}\:{f}\left(\mathrm{0}\right)={k} \\ $$$${f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}+{k}\:;\:{f}\left(\mathrm{0}\right)={k} \\ $$

Commented by mr W last updated on 12/Sep/22

you solved f(x)−f(x−1)=x. but the question is f(x)−f(1−x)=x. f(1)−f(0)=1 f(2)−f(−1)=2 f(3)−f(−2)=3 ... f(x)=((x(x+1))/2)+k doesn′t fulfill the original equation f(x)−f(1−x)=x. as you showed above, the original equation has no solution.

$${you}\:{solved}\:{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)={x}.\:{but}\:{the} \\ $$$${question}\:{is}\:{f}\left({x}\right)−{f}\left(\mathrm{1}−{x}\right)={x}. \\ $$$${f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)=\mathrm{1}\: \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${f}\left(\mathrm{3}\right)−{f}\left(−\mathrm{2}\right)=\mathrm{3} \\ $$$$... \\ $$$${f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}+{k}\:{doesn}'{t}\:{fulfill}\:{the} \\ $$$${original}\:{equation}\:{f}\left({x}\right)−{f}\left(\mathrm{1}−{x}\right)={x}. \\ $$$$ \\ $$$${as}\:{you}\:{showed}\:{above},\:{the}\:{original} \\ $$$${equation}\:{has}\:{no}\:{solution}. \\ $$

Commented by Rasheed.Sindhi last updated on 12/Sep/22

Sorry sir, my mistake! Thank you very much!

$${Sorry}\:\boldsymbol{{sir}},\:{my}\:{mistake}! \\ $$$$\mathcal{T}{hank}\:{you}\:{very}\:{much}! \\ $$

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