Talk:Bounded function

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needs refs, try Wikipedia:WikiProject_Mathematics/Reference_resources, User:Shotwell/Standard_references, and User:Cronholm144/List_of_References--Cronholm144 18:52, 22 May 2007 (UTC) Note: These ratings are not set in stone, please change them as the article progresses.

[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 $e 1000$ 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?

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