A sequence involving i

[onemachine]

Sequence Involving the Complex Number i?

\(\displaystyle a_n\) = \(\displaystyle ne^i^\pi\) is a function from \(\displaystyle \mathbb{N} \) to \(\displaystyle \mathbb{R} \) so it appears to be a legitamite sequence. But since it involves a complex number, can I really use this to technically define a sequence by the very strictest form of the defitinition of a sequence?

Thanks!

 

[daon2]

\(\displaystyle a_n\) = \(\displaystyle ne^i^\pi\) is a function from \(\displaystyle \mathbb{N} \) to \(\displaystyle \mathbb{R} \) so it appears to be a legitamite sequence. But since it involves a complex number, can I really use this to technically define a sequence by the very strictest form of the defitinition of a sequence?

Thanks!

Note that \(\displaystyle e^{i\pi} = -1\). Therefore \(\displaystyle a_n = -n\).

 

[onemachine]

Note that \(\displaystyle e^{i\pi} = -1\). Therefore \(\displaystyle a_n = -n\).

Yes, I know, but thats not what I was asking...

 

[daon2]

Yes, I know, but thats not what I was asking...

\(\displaystyle e^{\pi i}\) is a real number. There is no difference writing it and the number \(\displaystyle -1\). I can define a sequence on \(\displaystyle \mathbb{N}\) by:

\(\displaystyle \displaystyle f(n) = \frac{-ne^{i\pi/2}}{\pi}\ln(-1)^n\)

As long as your input and output go where they are supposed to, there's no problem.

 

[onemachine]

\(\displaystyle e^{\pi i}\) is a real number. There is no difference writing it and the number \(\displaystyle -1\). I can define a sequence on \(\displaystyle \mathbb{N}\) by:

\(\displaystyle \displaystyle f(n) = \frac{-ne^{i\pi/2}}{\pi}\ln(-1)^n\)

As long as your input and output go where they are supposed to, there's no problem.

I figured so, but I had to make sure that there wasn't some restriction on the functions you can use to define a sequence that I didn't know about. Thanks!