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Limits equality, complex.

  1. Dec 15, 2013 #1
    I am trying to show {fn} converges uniformly on {z:|z|≤p}, where p is postiive real number.
    and fn(z)=sin(z/n).

    I am able to follow parts of my books method, but don't understand a couple of the inequalities...(my trouble lies in the inequalities rather than the main concepts involved in the method).

    Sol:

    I can see the pointwise limit is 0.

    I am using the theorem that 'If for each n, we can find a sequence Sn, s.t |fn(z)-f(z)|≤Sn, and lim n→∞ Sn=0, then {fn} converges uniformly to f(z)'

    So I am to consider: |Sin(z/n)-0|and looking for a Sn with the above properties.

    so let z=x+iy and we attain |Sin(z/n)-0| = |Sin(x/n)cosh(y/n)+icos(x/n)sinh(y/n)| which is fine.

    we than get |sin(x/n)cosh(y/n)+icos(x/n)sinh(y/n)|≤|sin(x/n)cosh(y/n)|+|sinh(y/n)|*
    , which I am not 100% sure on.
    I can see we have applied the triangle inequality. But on the right hand-term we have clearly applied another inequaliity , is this that |icos(x/n)| ≤1??

    It then says for |z|≤p**:

    |sin(x/n)cosh(y/n)|+|sinh(y/n)|≤(p/n)cosh(p/n)+sinh (p/n).

    I'm not too sure about this inequality, for two reasons:
    1) I think from * we have used that sin (x/n) ≈ (x/n) for small (x/n), but why don't we use a approximation sign rather than inequality?
    2) MAIN question really - I'm not too sure how we have used ** ? haven't we set x=y=p. I'm having great difficulty understanding why we do this, I can see we have lost the absolute value signs...

    Many Thanks for any help, really appreciated ! :)
     
    Last edited: Dec 15, 2013
  2. jcsd
  3. Dec 15, 2013 #2

    Dick

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    Homework Helper

    They are using such facts as, i) if r is real, then |cos(r)|<=1, ii) if r is real then |sin(r)|<=|r|. The usual sorts of things you know about sin and cos as real functions. They are also using that if |x+iy|<=p then |x|<=p and |y|<=p.
     
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