all about sequence of functions


by irony of truth
Tags: functions, sequence
irony of truth
irony of truth is offline
#1
Oct5-05, 01:47 PM
P: 93
Let {h_n} be a sequence of function defined on the interval (0,1) where

h_n(x) = (n+n)x^(n-1)(1-x)

a. find lim (n-> +oo) (integral) (from 0 to 1) h_n(x) dx.

b. show that lim (n-> +oo) h_n(x) = 0 on (0, 1)

c. Show that lim (n-> +oo) (integral) (from 0 to 1) h_n(x) dx is not equal to integral (from 0 to 1) (0 dx). What went wrong?

SOlutions:

a. lim (n-> +oo) integral (from 0 to 1) h_n(x) dx
= lim (n-> +oo) integral (from 0 to 1) (n+n)x^(n-1)(1-x) dx
=lim (n-> +oo)(n+n) integral (from 0 to 1)x^(n-1)(1-x) dx

= lim (n-> +oo)(n+n) (1/n - 1/(n+1))
= lim (n-> +oo)n(n + 1) (1/((n)(n+1))
= 1.

b. I used the n-th term test in proving this... because if the series of h_n(x) is convergent then lim (n-> +oo) h_n(x) = 0 on (0, 1). But by ratio test, h_n(x) is convergent because the limit of
a(n+1)/a(n) as n -> +oo is x, but 0 < x < 1.

c. That's the part that I got stuck... well, it seems that the statement above is true... how do I solve this?
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StatusX
StatusX is offline
#2
Oct5-05, 03:14 PM
HW Helper
P: 2,566
Roughly speaking, the main contribution to the integral for large n comes from an increasingly small neighborhood near 1. For any given point x, no matter how close to one, hn(x) eventually gets very small as n increases. But there are still points left between x and 1 for which the value of hn evaluated at these points is very large, and this keeps the value of the integral at one. By the way, did you mean n+n^2 for the factor in front of the function?
irony of truth
irony of truth is offline
#3
Oct5-05, 08:58 PM
P: 93
Thank you for clarifying...

yes, it should have been n^2 + n.


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