How do I represent this sine function?

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    Function Sine
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Discussion Overview

The discussion revolves around representing the sine function \(\sin \left( \dfrac {n\pi } {2}\right)\) for integer values of \(n\). Participants explore various representations and seek a compact form, considering both even and odd integers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants note that for even \(n\), the sine function evaluates to zero, while for odd \(n\), it alternates between 1 and -1.
  • One participant proposes a piecewise function to represent \(\sin{\frac{\pi n}{2}}\) based on the value of \(n\): 0 for even \(n\), 1 for \(n=4k+1\), and -1 for \(n=4k+3\).
  • Another participant suggests looking for a representation similar to \(\cos(n\pi) = (-1)^n\) but expresses doubt about finding a simpler form than the original sine function.
  • A later reply introduces a new representation involving complex numbers: \(\sin\left(\frac{n\pi}{2}\right) = \frac{1}{2i}\left(i^n - (-i)^n\right)\), explaining the derivation of this expression.
  • One participant reflects on the complexity of the representations, concluding that nothing is "simpler" than \(\sin\left(\frac{n\pi}{2}\right)\) itself.

Areas of Agreement / Disagreement

Participants express various viewpoints on the representations of the sine function, with no clear consensus on a single preferred form. The discussion includes multiple competing representations and interpretations.

Contextual Notes

Some representations depend on specific definitions or conditions, such as the piecewise nature of the function or the use of complex numbers, which may not be universally accepted as simpler.

eprparadox
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I was struggling to represent the following for integer values of n:

<br /> \sin \left( \dfrac {n\pi } {2}\right) <br />

I know for even n, we get zero

But for odd n, it alternates beween 1 and -1 for every other odd. Is there a compact way to represent that? I feel like I'm being dumb and missing something obvious.

Thanks!
 
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eprparadox said:
I was struggling to represent the following for integer values of n:

<br /> \sin \left( \dfrac {n\pi } {2}\right)<br />

But for odd n, it alternates beween 1 and -1 for every other odd. Is there a compact way to represent that? I feel like I'm being dumb and missing something obvious.

Thanks!
1 when n = 1 + 2x
-1 when n = 3 + 2x

where x is zero or a positive even integer
 
What you have is a piecewise function (the variable being n in this case).

k\in \mathbb{Z} \\ f(n) = \sin{\frac{\pi n}{2}} = \left\{<br /> \begin{array}{lr}<br /> 0 &amp; : n=2k\\<br /> 1 &amp; : n=4k+1\\<br /> -1 &amp; : n=4k+3<br /> \end{array}<br /> \right.
 
I guess you are looking for something similar to ##\cos(n\pi) = (-1)^n## for ##\sin(\frac{n\pi} 2)##. You may be able to find such an expression, but I don't think you will find anything "simpler" than ##\sin(\frac{n\pi} 2)## itself, unless you consider piecewise defined functions simpler. I don't.
 
Thank you so much for the responses everyone!
 
I gave it a little more thought and came up with
\sin\left(\frac{n\pi}{2}\right)=\frac{1}{2}\left(-1\right)^{\frac{n-1}{2}}\left(1-(-1)^n\right)
 
How about
$$\sin\left(\frac{n\pi}{2}\right) = \frac{1}{2i}\left(i^n - (-i)^n\right) = \frac{1}{2}\left(i^{n-1} + (-i)^{n-1}\right)$$

[edit]: I guess that's equivalent to Mentallic's answer. Here's how I obtained it:

$$\begin{aligned}
\sin\left(\frac{n\pi}{2}\right) &= \text{Im}\left(e^{in\pi/2}\right) \\
&= \frac{1}{2i} \left(e^{in\pi/2} - \overline{e^{in\pi/2}}\right) \\
&= \frac{1}{2i} \left(i^n - \overline{i^n}\right) \\
&= \frac{1}{2i} \left(i^n - (-i)^n\right) \\
\end{aligned}$$
 
Last edited:
Yup. As I thought. Nothing "simpler" than ##\sin\frac{n\pi} 2## itself.
 

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