Talk:Bounded function

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[edit] Sin z not bounded?

The function f:R → R defined by f (x)=sin x is bounded. The sine function is no longer :bounded if it is defined over the set of all complex numbers.
why isn't sinz bounded function? --anon

Well, one has the equality

\sin (z) = \frac{1}{2i}\left(e^{iz}-e^{-iz}\right)

which follows from Euler's formula. If z=1000i, which is an imaginary number, one has

\sin (1000i) = \frac{1}{2i}\left(e^{i1000i}-e^{-i1000i}\right)
= \frac{1}{2i}\left(e^{-1000}-e^{1000}\right) .

Now, e − 1000 is small, but e1000 is huge, so this adds up to a huge number. Does that make sence? Oleg Alexandrov 21:05, 3 September 2005 (UTC)


[edit] proof?

how can one prove whether a function is bounded or not?