From Wikipedia, the free encyclopedia
This article is within the scope of WikiProject Mathematics mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.Mathematics Wikipedia:WikiProject Mathematics Template:WikiProject Mathematics mathematics Start This article has been rated as Start-class on Wikipedia's content assessment scale. Mid This article has been rated as Mid-priority on the project's priority scale .
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
)
=
1
2
i
(
e
i
z
−
e
−
i
z
)
{\displaystyle \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
(
1000
i
)
=
1
2
i
(
e
i
1000
i
−
e
−
i
1000
i
)
{\displaystyle \sin(1000i)={\frac {1}{2i}}\left(e^{i1000i}-e^{-i1000i}\right)}
=
1
2
i
(
e
−
1000
−
e
1000
)
.
{\displaystyle ={\frac {1}{2i}}\left(e^{-1000}-e^{1000}\right).}
Now,
e
−
1000
{\displaystyle e^{-1000}}
e
1000
{\displaystyle e^{1000}}
Oleg Alexandrov 21:05, 3 September 2005 (UTC) [ reply ]
how can one prove whether a function is bounded or not?
