24ok..........sorry for stupid question..[2][2]
thats one reason why I lost the race.....
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
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38 Answers
9but arent u getting concept of limits ????
how much ever large F is F∂ is 0 when ∂→0
24
62no eureka.. the race was not lost...
You did pretty well..
Dont worry about short term things.. life is a long race :)
9hmm rohan i think u dint get wat i say
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 !!!
that fact was used frm the fact they are in gp
IT WONT BE APPLICABLE TO ALL CONTINUOUS FUNCTIONS ( how did u assume that )
9you have written
f(x)=rf(x-h) and
f(x+h)=rf(x)
but wont this be valid for every continuous function when r≈1 !!!
IT WONT be valid for every continuous function ( cos the symmetry in the result wont be there unless they r in GP ) how did u make that assumption ??
62celestine .. and Rohan..
I am impressed by this discussion..
I was very sure not many people would be able to find the flaw in Celestine's proof.
Good rohan that you did :)
9yes rohan corrected that ( was actually a typo rest of sol is flawless)
see # 18 again
and rohan i dint understand wat u r trying to say but i see it pinked ???
see
........1 (r-1)[fx/h]
.........2 (r-1)[fx/h]/r
let r = 1+∂
now .......1 == ∂/h[fx]
.......2 == ∂/h[fx]/1+∂
note ∂/h is finite as ∂ and h are both small so [fx] neednt be 0 as u say
and .....1 and ........2 are certainly equal
9remember 1 = 1±∂ is applicable
its hard to believe but it is
also u cant prove that all continuous functions are differentiable using proof abv as ∂ would obviously differ for non differentiable functions for the LHL and RHL
1but i think celestine that you are pulling towards a differentiable function !
dont forget that 1+∂ is in the denominator ..
so it is to prove that
let the some finite quantity =F
F=F/(1+∂)
F=F(1-∂) as ∂ <<1
so we have to prove F∂=0 but F may be very large !! then it cannot be zero!!
62so f(y+1)/f(y)=k let
so f(y+2)/f(y+1)=k
...
...
...
f(y+n)/f(y+n-1) = k
mulitplying all the above
f(y+n)/f(n)=kn
39well,this seems to be a better one.
ln f satisfies a kind of cauchy
if x, y, z are successive in AP, then lnf(x), lnf(y), lnf(z) are successive in AP
now replace f by f - f(0)
then it will satisfy the cauchy functional eq'n
NOW U CAN GET THE SOLN
9prophet sir awaiting your reply on the proof ive provided on differentiabilty of the func
1hey celes you cant guarantee that F∂ will be always 0!!
what is F->∞ then??
1Essentially there exists another solution to the functional equation.
f(x)=c \forall x\in \mathbb{R} And c\to \infty is possible.
Then
f\delta\not=0
But we can choose \delta as we wish , so f∂ might be taken to be zero.
9gud that u posted it here ith power
i had to chat with rohan for more than 15 min to convince him on abv point
21he he ... k k.... point!!
and Y did u say λ = 1;
λ can be any no. naaa
let 3 nos. in AP be 1,2,3
f(1,2,3) = ke,ke^2,ke^3
k2e^(3+1) = k2 e^2*2
33f2((x+y)/2)=f(x)*f(y)
differentiating it wrt x
(......)=f(y)f'(x)
wrt y
(......)=f(x).f'(y)
So we get...
f(x).f'(y)=f'(x).f(y)
f'(x)/f(x)=constant..
dy/dx=ky
dy/y=kdx
is this right??
62f(y)2=f(y-a)f(y+a) for all a!
f(y)/f(y-a)=f(y+a)/f(y)
in particular for a=1
f(y+a)/f(y)=k let
f(y+n)/f(y)= kn
thus, f(y+n)=f(y). kn
{f(y+1/p)/f(y)}p=k
thus, f(y+1/p)= k1/p
this can be easily extended to rationals...
then by continuity of f and by extending it to reals we can say that (because set of rationals is dense in Reals!)
f(x)=c. kx
I think priyam's answer should be correct!