Question Number 2000 by Yozzi last updated on 29/Oct/15

Find a non−constant function f   satisfying f(0)=1,f(−2)=0 and  f(x−y)=f(x)f(y)−f(−2−x)f(y−2).

Commented byprakash jain last updated on 29/Oct/15

x=0  f(−y)=f(0)f(y)−f(−2)f(y−2)  ∵f(−2)=0 for the above relation to be  true ∀y.  f(−y)=f(y)  ⇒even function  y=x  f(0)=[f(x)]^2 −f(−2−x)f(x−2)  [f(x)]^2 =1+f(x−2)f(x+2)

Commented byprakash jain last updated on 30/Oct/15

y=2  f(x−2)=f(x)f(2)−f(x+2)f(0)  f(x−2)=−f(x+2)  f(x)=−f(x+4)  f(x+8)=[f(x+4+4)]=−f(x+4)=f(x)  f is periodic with period 8.  cos [(π/4)(x−y)]=cos (π/4)x∙cos (π/4)y+sin( (π/4)x)sin ((π/4)y)  sin ((πy)/4)=cos (((π(y−2))/4))  sin ((πx)/4)=−cos ((π/2)+(π/4)x)=−cos (((π(−x−2))/4))

Answered by Rasheed Soomro last updated on 02/Nov/15

Find a non−constant function f   satisfying f(0)=1,f(−2)=0 and  f(x−y)=f(x)f(y)−f(−2−x)f(y−2)...........(1)  −−−−−−×××−−−−−−−  x=−2,y=x  f(−2−x)=f(−2)f(x)−f(−2−{−2})f(x−2)  f(−2−x)=0.f(x)−f(0)f(x−2)=−f(x−2)  f(−2−x)=−f(x−2)=−{f(x)f(2)−f(−2−x)f(2−2)}                =−{f(x)f(2)−f(−2−x)f(0)}=f(−x−2)−f(2)f(x)  f(−x−2)−f(−x−2)=−f(2)f(x)  f(2)=0 ∣ f(x)=0  [But this not non−constant]  ∗∗∗  Substituting y=0  f(x)=f(x)f(0)−f(−2−x)f(−2)           =f(x)  This means f(x)  is arbitrary if y=0.  x=−2−x  f(−2−x−y)=f(−2−x)f(y)−f(−2−{})  continue

Answered by prakash jain last updated on 30/Oct/15

f(x)=cos ((πx)/4)

Commented byprakash jain last updated on 30/Oct/15

see comments in question for explaination.