Find all continuous functional equations f:R→R+ , such that if x,y and z are succesive in A.P., f(x), f(y) and f(z) are succesive terms in GP
62haha.. dont worry a part of the proof that i gave is not in syllabus..
but may be i messed up somewhere... prophet sir will have to find mistakes :(
atleast here not many ppl will understand the proof beyond the first 3-4 steps..
1priyam it is not mentioned that the function is differentiable or not ..!
9waw 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
62well celestine.. priyam has assumed that it is differentiable which is not the case..!
9consider
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
1hmm 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 !
24sir,
was it an intution or is there any reason behind this?????
let f(y+n)/f(y)= kn
