**21**
Shubhodip
**·**2011-07-14 05:31:09
solved the 2nd one..

We want to prove f(n)â‰¥n

we have f(n+1)> f(f(n)) for n = 1,2,3...

So must we have f(1) the minimum.(cz others are greater than f of something)

So f(1)â‰¥1

Using induction let f(p-1)â‰¥(p-1)

Given f(p)â‰¥f(f(p-1)+1 (i will later prove if mâ‰¥n , f(m)â‰¥n)

so f(p)â‰¥f(p-1) + 1 â‰¥p-1+1â‰¥p

so we have got f(p)â‰¥p for all p, so f(p+1)>f(f(p))â‰¥f(p) hence f is increasing.

again we need f(t+1)â‰¥ f(f(t))+1 , from here its clear that must we have f(t)â‰¤t for all t.

therefore f(x) = x for all natural x.

i used ** if mâ‰¥n , f(m)â‰¥n**

proof.

We use induction on 'n'.(that's the only hint i took , on ''n'')

when n = 1, its true.

Let mâ‰¥n-1 and we have f(m)â‰¥n-1, *implying f(f(m))â‰¥n-1* (thats where i was stuck a while;))

we have f(m+1)â‰¥f(f(m))+1â‰¥n and this completes the induction.

**21**
Shubhodip
**·**2011-07-14 06:53:43
Can any one make me understand the solution in A.Engel?

They wrote its obvious that f(1) is minimum.

By the same logic f(1)<f(2) and f(1)<f(2)<f(3)<f(4).....

**11**
Devil
**·**2011-07-16 23:34:58
Take f(x)-x=g(x)

f(g(x))=g(x)+g(g(x))

So the eqn becomes g(x)+g(g(x))=2x.

Now take an interval P where g(x)>x. Partititon it into P_{i} 's such that

\bigcup_i{}P_i=P

The partitioning has been done such that there exists some k s.t g is monotonic in P_{k}.

WLOG take g to be monotonically increasing in that sub-interval.

Let P_{k}=[a,b]

Now from the assumption g9a)>a, which gives g(g(a))<a. We also get g(a)>b

Similarly we have g(g(b))<b

Now in any subinterval of P_{k} g is monotonically increasing....so that gives g(g(x))<x for all x in [g(a),g(b)]...

So g(g(g(a)))<g(a) & g(g(a))<a, so adding the two we get 2g(a)<g(a)+a - that gives g(a)<a, a contradiction....thus g(a)=a and extending it to [a,b] we get g(x)=x. The other case for g(x) monotonically decreasing or for g(x)<x can be settled similarly...

That also gives f(x)=2x as the only function satisfying the given functional eqn.

**·**2011-07-17 10:17:16
well it is pretty obvious why f(1) is the min.

now comparing f(1)with f(2)

either f(1)=f(2) or f(1) <f(2)

but f(1)=f(2) implies f(1)>f(f(1)) which contradicts the choice of f(1)

**21**
Shubhodip
**·**2011-07-17 10:21:16
this much is obvious.

but f(2)<f(3)<f(4)... is not that obvious i guess?

**·**2011-07-17 11:13:18
are actually wat we do next:

consider the set of f(x) for all x>=2

even this set has a least value element which namely bcoms f(2)...(similar arg. as in case of f(1))

also f(1)<f(2)

now comparing f(2) and f(3)

either they r equal or f(2)<f(3)

if f(2)=f(3)

f(2)>f(f(2)) .........................................................(1)

now f(2)>=2

however (1) does not allow f(2)=2

hence f(2)>f(f(2)) whr f(2) also>3

but this contradicts the choice of f(2) in the newly defined set...

**21**
Shubhodip
**·**2011-07-18 05:52:02
why f(2) must be the second minimum?

**1**
rishabh
**·**2011-08-13 13:24:49
for the first one,

f(x) = 0 and f(x) = 2x

**1**
rishabh
**·**2011-08-13 13:25:38
oh sorry. ignore f(x) = 0