SOLUTION: Let {{{f[n](x) = nxe^(-nx^2)}}}, for x in [0,1]. Show whether or not i) {{{f[n](x) }}} converges for each x in [0,1]. ii) {{{f[n](x)}}} converges uniformly on [0,1].

Algebra ->  Sequences-and-series -> SOLUTION: Let {{{f[n](x) = nxe^(-nx^2)}}}, for x in [0,1]. Show whether or not i) {{{f[n](x) }}} converges for each x in [0,1]. ii) {{{f[n](x)}}} converges uniformly on [0,1].      Log On


   

Question 438196: Let f%5Bn%5D%28x%29+=+nxe%5E%28-nx%5E2%29, for x in [0,1]. Show whether or not
i) f%5Bn%5D%28x%29+ converges for each x in [0,1].
ii) f%5Bn%5D%28x%29 converges uniformly on [0,1].
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
i) Fix any x-value in the interval (0,1].
Now f%5Bn%5D%28x%29+=+nxe%5E%28-nx%5E2%29+=+%28nx%29%2Fe%5E%28nx%5E2%29
==>
=+x%2Alim%28n-%3Einfinity%2C+1%2F%28x%5E2e%5E%28nx%5E2%29%29%29 by L'Hopital's rule.
=%281%2Fx%29%2Alim%28n-%3Einfinity%2C+1%2Fe%5E%28nx%5E2%29%29+=+%281%2Fx%29%2A0+=+0
If x = 0, f%5Bn%5D%280%29+=+%28n%2A0%29%2Fe%5E0+=+0 for all n.
Hence for all x in [0,1], the sequence f%5Bn%5D%28x%29 exhibits pointwise convergence, and lim%28n-%3Einfinity%2C+f%5Bn%5D%28x%29%29+=+0.
ii) Now df%5Bn%5D%28x%29%2Fdx+=+%28n+-+2n%5E2x%5E2%29%2Fe%5E%28nx%5E2%29.
Setting this derivative to 0 and solving for x, we get x%5E2+=+1%2F%282n%29
<==> x+=+1%2Fsqrt%282n%29
Implementing the 1st derivative test using the test points 1%2Fsqrt%282%28n%2B1%29%29 and 1%2Fsqrt%282%28n-1%29%29, we find that there is (absolute) maximum at x+=+1%2Fsqrt%282n%29.
(Remember that 1%2Fsqrt%282%28n%2B1%29%29+%3C+1%2Fsqrt%282n%29+%3C+1%2Fsqrt%282%28n-1%29%29.)
Then f%5Bn%5D%281%2Fsqrt%282n%29%29+=+sqrt%28n%2F%282e%29%29. Hence as n+-%3E+infinity, the maximum of the function goes to infinity, and so the sequence of functions f%5Bn%5D%28x%29 does not exhibit uniform convergence.