Functional Equation

:P

3rd line se sab zero ho gaya... :0

waw priyams work is real elegant apart frm that a trivial solution f(x) = const also arises which was assumed nonzero in priyams sol due to division

consider
limh→0+ f(x+h) - f(x) /h and
limh→0+ f(x) - f(x-h) /h
here as x-h,x,x+h AP
f(x)=rf(x-h) and
f(x+h)= rf(x)
limh→0+ f(x+h) - f(x) /h = (r-1)[fx/h] ........1
limh→0+ f(x) - f(x-h) /h = (r-1)[fx/h]/r ............2
but r ≈1 (note f continuous)
hence r in denominator wont matter at all in ...2
ie limh→0+ f(x+h) - f(x) /h == limh→0+ f(x) - f(x-h) /h
== limh→0- f(x+h) - f(x) /h
ie LHD = RHD hence differentiable

hmm celestine if this is your proof then you are saying indirectly that every continuous function is differentiable !!

you have written

f(x)=rf(x-h)
and

f(x+h)=f(x)/r

but wont this be valid for every continuous function when r≈1 !!!

acc to your proof .

(r-1)g(x)=((r-1)/r)g(x) as r≈1

but

as r≈1
we have let r=1+h h->0

thus LHS = hg(x)

R.H.S=(h/1+h)g(x)

when you are saying L.H S=R.H.S

either g(x) has to be zero always

or if it doesnt then two possibilities

i) f(x)→0 which will make g(x) meaningful

ii)f(x)/h will in other cases be ∞ or -∞

thus a little difference between the two multiplying factors will change the story a lot!

for h=(h/1+h) => h²/1+h = 0

but neither h²=0 or 1+h=0
so there will be a difference !

sir,
was it an intution or is there any reason behind this?????

let f(y+n)/f(y)= kⁿ