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f-x-f-x-1-x-1-x-f-x-




Question Number 2205 by prakash jain last updated on 08/Nov/15
f(x)+f(((x−1)/x))=1+x  f(x)=?
f(x)+f(x1x)=1+xf(x)=?
Commented by Yozzi last updated on 08/Nov/15
f(x)+f(1−(1/x))=1+x  ⇒f^′ (x)+(1/x^2 )f^′ (1−(1/x))=1  x^2 f^′ (x)+f^′ (((x−1)/x))=x^2
f(x)+f(11x)=1+xf(x)+1x2f(11x)=1x2f(x)+f(x1x)=x2
Answered by Rasheed Soomro last updated on 10/Nov/15
f(x)+f(((x−1)/x))=1+x  f(x)=1+x−f(((x−1)/x))...............................(1)  f(((x−1)/x))=1+((x−1)/x)−f(((((x−1)/x)−1)/((x−1)/x)))    [Replacing x by ((x−1)/x) in (1)]                  =1+((x−1)/x)−f(((−1)/(x−1)))......................(2)  f(((−1)/(x−1)))=1+((−1)/(x−1))−f(((((−1)/(x−1))−1)/((−1)/(x−1))))   [Replacing x by ((−1)/(x−1)) in (1)]                   =1−(1/(x−1))−f(x)...........................(3)  −−−−−−−−−−−−−−−−−−  Replacing f(((x−1)/x)) by  rhs of (2)   and  f(((−1)/(x−1)))  by rhs of  (3)  and  f(x) by rhs of  (1) in (1)  f(x)=1+x−f(((x−1)/x))  f(x)=1+x−[1+((x−1)/x)−f(((−1)/(x−1)))]  f(x)=x−((x−1)/x)+f(((−1)/(x−1)))]           =x−((x−1)/x)+1−(1/(x−1))−f(x)  2f(x)=x−((x−1)/x)+1−(1/(x−1))  f(x)=(1/2)(x−((x−1)/x)−(1/(x−1))+1)  −−−−−−−−−−−−−−−−−−−−−−−−−  VERIFICATION:  f(x)=(1/2)(x−((x−1)/x)−(1/(x−1))+1)  f(((x−1)/x))=(1/2)(((x−1)/x)−((((x−1)/x)−1)/((x−1)/x))−(1/(((x−1)/x)−1))+1)                   =(1/2)(((x−1)/x)+(1/(x−1))+x+1)  f(x)+f(((x−1)/x))=(1/2)(x−((x−1)/x)^(×) −(1/(x−1))^(×) +1+((x−1)/x)^(×) +(1/(x−1))^(×) +x+1)                               =(1/2)(2x+2)                                =x+1
f(x)+f(x1x)=1+xf(x)=1+xf(x1x).(1)f(x1x)=1+x1xf(x1x1x1x)[Replacingxbyx1xin(1)]=1+x1xf(1x1).(2)f(1x1)=1+1x1f(1x111x1)[Replacingxby1x1in(1)]=11x1f(x)(3)Replacingf(x1x)byrhsof(2)andf(1x1)byrhsof(3)andf(x)byrhsof(1)in(1)f(x)=1+xf(x1x)f(x)=1+x[1+x1xf(1x1)]f(x)=xx1x+f(1x1)]=xx1x+11x1f(x)2f(x)=xx1x+11x1f(x)=12(xx1x1x1+1)VERIFICATION:f(x)=12(xx1x1x1+1)f(x1x)=12(x1xx1x1x1x1x1x1+1)=12(x1x+1x1+x+1)f(x)+f(x1x)=12(xx1x×1x1×+1+x1x×+1x1×+x+1)=12(2x+2)=x+1
Commented by Rasheed Soomro last updated on 11/Nov/15
f(x) involves f(((x−1)/x)),  f(((x−1)/x)) involves f(((−1)/(x−1)))  and f(((−1)/(x−1))) involves again f(x). This is cyclic.At last  we have an equation which involves only f(x) which  gives definition of f(x).I think this cyclic  type   functional equation shows a categary of functional  equations.      We can also talk of  order of such equations.
f(x)involvesf(x1x),f(x1x)involvesf(1x1)andf(1x1)involvesagainf(x).Thisiscyclic.Atlastwehaveanequationwhichinvolvesonlyf(x)whichgivesdefinitionoff(x).Ithinkthiscyclictypefunctionalequationshowsacategaryoffunctionalequations.Wecanalsotalkoforderofsuchequations.

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