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