Cosine to Power of Odd Integer/Proof 2
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Theorem
- $\ds \cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \sum_{k \mathop = 0}^n \binom {2 n + 1} k \cos \paren {2 n - 2 k + 1} \theta$
Proof
\(\ds \cos^n \theta\) | \(=\) | \(\ds \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^n\) | De Moivre's Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {e^{i \theta} + e^{-i \theta} }^n} {2^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - k} i \theta} e^{-k i \theta}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - 2 k} i \theta}\) |
Matching up terms from the beginning of this expansion with those from the end:
\(\ds 2^n \cos^n \theta\) | \(=\) | \(\ds e^{n i \theta} + \binom n 1 e^{\paren {n - 2} i \theta} + \binom n 2 e^{\paren {n - 4} i \theta} + \cdots\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \binom n {n - 2} e^{-\paren {n - 4} i \theta} + \binom n {n - 1} e^{-\paren {n - 2} i \theta} + e^{-n i \theta}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {2 n} 2 \cos^n \theta\) | \(=\) | \(\ds \paren {\frac {e^{n i \theta} + e^{-n i \theta} } 2} + \binom n 1 \paren {\frac {e^{\paren {n - 2} i \theta} + e^{-\paren {n - 2} i \theta} } 2}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \binom n 2 \paren {\frac {e^{\paren {n - 4} i \theta} + e^{-\paren {n - 4} i \theta} } 2} + \cdots\) |
Thus:
- $\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$
Now to determine $R_n$.
The middle two terms of the sequence $0, 1, \ldots, n$ are $\dfrac {n - 1} 2$ and $\dfrac {n + 1} 2$.
Thus, when $k = \dfrac {n - 1} 2$:
\(\ds n - 2 k\) | \(=\) | \(\ds n - 2 \frac {n - 1} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n - \frac {2 n - 2} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n - n + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
Similarly, when $k = \dfrac {n + 1} 2$:
\(\ds n + 2 k\) | \(=\) | \(\ds n - 2 \frac {n + 1} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n - \frac {2 n + 2} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n - n - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1\) |
The binomial coefficient in each case is the same, because:
\(\ds n - \frac {n - 1} 2\) | \(=\) | \(\ds \frac {2 n - n - 1} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n + 1} 2\) |
So:
\(\ds \binom n {\paren {n - 1} / 2}\) | \(=\) | \(\ds \frac {n!} {\paren {\frac {n - 1} 2}! \paren {\frac {n + 1} 2}!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom n {\paren {n + 1} / 2}\) |
Thus the two middle terms collapse to:
\(\ds R_n\) | \(=\) | \(\ds \frac {n!} {\paren {\frac {n - 1} 2}! \paren {\frac {n + 1} 2}!} \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n!} {\paren {\frac {n - 1} 2}! \paren {\frac {n + 1} 2}!} \cos \theta\) |
$\blacksquare$
Also defined as
This result is also reported in the form:
- $\ds\cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \sum_{k \mathop = 0}^n \binom {2 n + 1} k \map \cos {2 n - 2 k + 1} \theta$
for all $n \in \Z$.
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $130 \ \text{(a)}$
- (although see Cosine to Power of Odd Integer/Mistake for analysis of an error in that work)