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f-x-y-f-x-y-xy-f-x-y-x-2-y-2-f-x-y-y-x-100-y-100-f-0-0-f-1-4-




Question Number 1842 by 123456 last updated on 12/Oct/15
f(x,y)=f(x+y,xy)  f(x,y)=x,−2≤y≤2  f(x,y)=y,∣x∣≥100∨∣y∣≥100  f(0,0)=?  f(1,4)=?
f(x,y)=f(x+y,xy)f(x,y)=x,2y2f(x,y)=y,x∣⩾100y∣⩾100f(0,0)=?f(1,4)=?
Commented by 123456 last updated on 12/Oct/15
(1,4)→(5,4)→(9,20)→(29,180)→...  (x,x)→(2x,x^2 )→(2x+x^2 ,2x^3 )→...
(1,4)(5,4)(9,20)(29,180)(x,x)(2x,x2)(2x+x2,2x3)
Answered by Rasheed Soomro last updated on 13/Oct/15
 f(0,0)=0    [ ∵  f(x,y)=x,−2≤y≤2 and here x=0]  Since f(x,y)=f(x+y,xy)  Therefore f(1,4)=f(1+4,1.4)=f(5,4)  =f(9,20)=f(29,180) , here ∣y∣≥100   f(29,180)=180  [∵   f(x,y)=y,∣x∣≥100∨∣y∣≥100 ]  So,  f(1,4)=f(29,180)=180  Continueing further in this way we can also find   other values of f(1,4)  f(1,4)=f(29,180)=f(209,5220)=5220  f(1,4)=180,5220,...
f(0,0)=0[f(x,y)=x,2y2andherex=0]Sincef(x,y)=f(x+y,xy)Thereforef(1,4)=f(1+4,1.4)=f(5,4)=f(9,20)=f(29,180),herey∣⩾100f(29,180)=180[f(x,y)=y,x∣⩾100y∣⩾100]So,f(1,4)=f(29,180)=180Continueingfurtherinthiswaywecanalsofindothervaluesoff(1,4)f(1,4)=f(29,180)=f(209,5220)=5220f(1,4)=180,5220,

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