Sine of X over X as Infinite Product
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Theorem
Let $z \in \C$ be a non-zero complex number.
Then:
- $\ds \frac {\sin z} z = \cos \frac z 2 \cos \frac z 4 \cos \frac z 8 \cdots = \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}$
where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.
Proof
First we prove that:
- $\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
for $n \in \N$.
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
- $\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
Basis for the Induction
$\map P 1$ is the case:
- $\ds \frac {\sin z} z = \paren {\frac {2^1} z} \sin \frac z {2^1} \prod_{i \mathop = 1}^1 \cos \frac z {2^i}$
Thus $\map P 1$ is seen to hold.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\ds \frac {\sin z} z = \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}$
Then we need to show:
- $\ds \frac {\sin z} z = \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}$
Induction Step
This is our induction step:
\(\ds \frac {\sin z} z\) | \(=\) | \(\ds \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {2^k} z} \paren {2 \sin \frac z {2^{k + 1} } \cos \frac z {2^{k + 1} } } \prod_{i \mathop = 1}^k \cos \frac z {2^i}\) | Double Angle Formula for Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \cos \frac z {2^{k + 1} } \prod_{i \mathop = 1}^k \cos \frac z {2^i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
And then:
\(\ds \frac {\sin z} z\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } \prod_{i \mathop = 1}^n \cos \frac z {2^i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } } \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren 1 \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}\) | Limit of $\dfrac {\sin x} x$ at Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 38$: Infinite Products: $38.8$