Talk:Bounded function

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Sin z not bounded?

The function f:R → R defined by f (x)=sin x is b

P ounded. 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)[reply]

proof?

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

If a function is continuous and has a maxima and a minima at finite values of f(x), then it is bounded. If it is not continuous then you will have to find ways to find the upper bound and lower bound. Should both be finite, it will remain bounded. Umang me (talk) 06:11, 30 January 2010 (UTC)[reply]

Is this bounded?

Say I have an f:R->R continuous and strictly monotonically increasing for (-2, 2) and stationary at {-2, 2}. Then I have g(x) = f(x), x belonging to (-2, 2), then is the function bounded? I ask because the g(x) does not attain a maxima or a minima at all, but is restricted by f(-2) < g(x) < f(2), where f(-2) and f(2) are finite real numbers. Therefore, (according to me) g(x) is bounded but does not attain its extremes. Am I right? Umang me (talk) 06:11, 30 January 2010 (UTC)[reply]