Hi fmunshi!!
gn(x) = (1-x).x^n
If x = 0, then gn(0) = 0, then gn(0) tends to 0.
If x = 1, then gn(1) = 0, then gn(1) tends to 0.
If 0 < x < 1, then gn(x) = (1-x).x^n < x^n, since 0 < x < 1, is 1 <
1/x, then we can write 1/x = 1 + h with h>0. Then:
(1/x)^n = (1 + h)^n >= 1 + n.h that clearly tends to oo. Then (1/x)^n
tends to oo. Take s > 0 then exist n0 such that if n > n0 is (1/x)^n >
1/s, then:
1/s < (1/x)^n = 1/(x^n) for all n >n0, then for all n>n0 is x^n < s .
Then for all s>0 there are n0`(that depends on s) such that for all
n>n0 is x^n < s.
Then x^n tends to 0.
Now is obvious that (1-x).x^n tends to zero.
Then:
If 0 < x < 1, then gn(x) tends to 0.
Then gn(x) tends to zero (null function) for x in [0,1].
You can see the following document about Uniform convergence here:
http://www.maths.lse.ac.uk/Courses/MA203/sec7a.pdf
You can see at this file that x^n , with x in [0,1] converges not
uniformely and using the same method we can see that gn(x) converges
not uniformely to zero.
Hope this helps.
If you need a clarification please post a request for a clarification
before rate this answer.
Regards.
livioflores-ga |
Clarification of Answer by
livioflores-ga
on
25 Jul 2003 08:45 PDT
Hi fmunshi, and excuse me for the delay!!
my suspicions were unfounded, gn(x) converges uniformly to zero.
It isn't very hard to show that gn(x)
has a maximum Mn on [0,1], and
lim Mn = 0 .
n->oo
In effect if x is different to zero:
gn(x) = (1-x).x^n = x^n - x^(n+1)
gn`(x) = n.x^(n-1) - (n+1).x^n = x^(n-1).(n - (n+1).x) = 0 if and only
if
(n - (n+1).x) = 0 iff x = n/(n+1).
gn´´(x) = n.(n-1).x^(n-2) - n.(n+1).x^(n-1) =
= n.x^(n-2).(n - 1 - (n+1).x)
The sign of gn´´(x) depends only on the sign of the factor (n - 1 -
(n+1).x)
then:
sign of gn´´(n/(n+1)) = sign of (n - 1 - (n+1).n/(n+1)) =
= sign of (n - 1 - n) = sign of (-1)
Then gn(x) have a maximum for x = n/(n+1) on (0,1);
and since gn(0) = gn(1) = 0 and gn(n/(n+1)) > 0 this maximum is for
[0,1].
The sequence of maximums of gn(x) is Mn and:
Mn = gn(n/(n+1)) = (1 - n/(n+1)).(n/(n+1))^n < (1 - n/(n+1)).1^n =
1/(n+1)
then Mn tends clearly to zero when n tends to +oo ;
this means that given E > 0 there is n0 such that Mn < E for all n >
n0 and because Mn >= gn(x) for all x on [0,1], we have:
gn(x) =< Mn < E for all n > n0 and FOR ALL x on [0,1].
This shows the uniform convergence of gn(x) to zero.
I hope that this clarify the answer. Please feel free to request for
more
clarifications if they are needed.
Regards.
livioflores-ga
|
