Euler Formula for Sine Function/Complex Numbers/Proof 1
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Theorem
| \(\ds \sin z\) | \(=\) | \(\ds z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2 \pi^2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z \paren {1 - \dfrac {z^2} {\pi^2} } \paren {1 - \dfrac {z^2} {4 \pi^2} } \paren {1 - \dfrac {z^2} {9 \pi^2} } \dotsm\) |
for all $z \in \C$.
Proof
For $z \in \C$ and $n \in \N$, let:
- $\ds \map {I_n} z = \int_0^{\pi / 2} \cos {z t} \cos^n t \rd t $
Observe that $\map {I_0} 0 = \dfrac {\pi} 2$ and:
| \(\ds \map {I_0} z\) | \(=\) | \(\ds \int_0^{\pi / 2} \cos {z t} \rd t\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 z \map \sin {\frac {\pi z} 2}\) |
which yields:
- $(1): \quad \map \sin {\dfrac {\pi z} 2} = \dfrac {\pi z} 2 \dfrac {\map {I_0} z} {\map {I_0} 0}$
Integrating by parts twice with $n \ge 2$, we have:
| \(\ds z \map {I_n} z\) | \(=\) | \(\ds n \int_0^{\pi / 2} \sin {z t} \cos^{n - 1} t \sin t \rd t\) | ||||||||||||
| \(\ds z^2 \map {I_n} z\) | \(=\) | \(\ds n \int_0^{\pi / 2} \cos {z t} \paren {\cos^n t - \paren {n - 1} \cos^{n - 2} t \sin^2 t} \rd t\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n \int_0^{\pi / 2} \cos {z t} \paren {n \cos^n t - \paren {n - 1} \cos^{n - 2} t } \rd t\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds n^2 \map {I_n} z - n \paren {n - 1} \map {I_{n - 2} } z\) |
which yields the reduction formula:
- $n \paren {n - 1} \map {I_{n - 2} } z = \paren {n^2 - z^2} \map {I_n} z$
Substituting $z = 0$ we obtain:
- $n \paren {n - 1} \map {I_{n - 2} } 0 = n^2 \map {I_n} 0$
From Shape of Cosine Function, it is clear that $\map {I_n} 0 > 0$ for $n \ge 0 $.
Therefore we can divide the two equations to get:
- $(2): \quad \dfrac {\map {I_{n - 2} } z} {\map {I_{n - 2} } 0} = \paren {1 - \dfrac {z^2} {n^2} } \dfrac {\map {I_n} z} {\map {I_n} 0}$
We have:
| \(\ds \cmod {1 - \cos {z t} }\) | \(=\) | \(\ds \cmod {1 - \map \cos {x + i y} t}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cmod {1 - \cos {x t} \cos {i y t} + \sin x t \sin {i y t} }\) | Cosine of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cmod {1 - \cos {x t} \cosh {y t} + i \sin {x t} \sinh {y t} }\) | Hyperbolic Sine in terms of Sine and Hyperbolic Cosine in terms of Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt {\paren {1 - \cos {x t} \cosh {y t} }^2 + \paren {\sin {x t} \sinh {y t} }^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt {4 \map {\sinh^4} {\frac {y t} 2} + 8 \map {\sinh^2} {\frac {y t} 2} \map {\sin^2} {\frac {x t} 2} + 4 \map {\sin^4} {\frac {x t} 2} }\) | Double Angle Formulas | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map {\sinh^2} {\frac {y t} 2} + 2 \sin^2 {\frac {x t} 2}\) |
From Sine Inequality we have that $2 \map {\sin^2} {\dfrac {x t} 2} \le \dfrac 1 2 x^2 t^2$.
By Lemma 1, $\dfrac {\sinh x} x$ is an increasing function for $x \ge 0$, so for $t \in \closedint 0 {\dfrac {\pi} 2}$:
| \(\ds \frac {\map \sinh {y t / 2} } {y t / 2}\) | \(\le\) | \(\ds \frac {\map \sinh {y \pi / 4} } {y \pi / 4}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\map {\sinh^2} {y t / 2} } {\paren {y t / 2}^2}\) | \(\le\) | \(\ds \frac {\map {\sinh^2} {y \pi / 4} } {\paren {y \pi / 4}^2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \map {\sinh^2} {y t / 2}\) | \(\le\) | \(\ds \frac 1 2 \frac {\map {\sinh^2} {y \pi / 4} } {\paren {y \pi / 4}^2} t^2\) |
So we deduce:
| \(\ds \cmod {1 - \cos {z t} }\) | \(\le\) | \(\ds \frac 1 2 \paren {x^2 + \frac {\map {\sinh^2} {y \pi / 4} } {\paren {\pi / 4}^2} } t^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map C {x, y} t^2\) | where $C$ is a non-negative function of $x$ and $y$ |
By Relative Sizes of Definite Integrals we have:
| \(\ds \cmod {\map {I_n} 0 - \map {I_n} z}\) | \(=\) | \(\ds \cmod {\int_0^{\pi / 2} \paren {1 - \cos {z t} } \cos^n t \rd t}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \map C {x, y} \int_0^{\pi / 2} t^2 \cos^n t \rd t\) | Cosine Inequality | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \map C {x, y} \int_0^{\pi / 2} t \cos^{n - 1} t \sin t \rd t\) | Tangent Inequality | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\map C {x, y} } n \int_0^{\pi / 2} \cos^n t \rd t\) | Integration by Parts | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\map C {x, y} } n \map {I_n} 0\) |
which yields the inequality:
- $\cmod {1 - \dfrac {\map {I_n} z} {\map {I_n} 0} } \le \dfrac {\map C {x, y} } n$
It follows from Squeeze Theorem that:
- $(3): \quad \ds \lim_{n \mathop \to \infty} \frac {\map {I_n} z} {\map {I_n} 0} = 1$
Consider the equation, for even $n$:
- $\ds \map \sin {\dfrac {\pi z} 2} = \dfrac {\pi z} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \dfrac {z^2} {\paren {2 i}^2} } \dfrac {\map {I_n} z} {\map {I_n} 0}$
This is true for $n = 0$ by $(1)$.
Suppose it is true for some $n = k$.
Then:
| \(\ds \map \sin {\frac {\pi z} 2}\) | \(=\) | \(\ds \frac {\pi z} 2 \prod_{i \mathop = 1}^{k / 2} \paren {1 - \frac {z^2} {\paren {2 i}^2} } \frac {\map {I_k} z} {\map {I_k} 0}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\pi z} 2 \prod_{i \mathop = 1}^{\paren {k + 2} / 2} \paren {1 - \frac {z^2} {\paren {2 i}^2} } \frac {\map {I_{k + 2} } z} {\map {I_{k + 2} } 0}\) | by $(2)$ |
So it is true for all even $n$ by induction.
Taking the limit as $n \to \infty$ we have:
| \(\ds \map \sin {\frac {\pi z} 2}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\pi z} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \frac {z^2} {\paren {2 i}^2} } \frac {\map {I_n} z} {\map {I_n} 0}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\pi z} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {z^2} {\paren {2 i}^2} } \lim_{n \mathop \to \infty} \frac {\map {I_n} z} {\map {I_n} 0}\) | Product Rule for Limits of Complex Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\pi z} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {z^2} {\paren {2 i}^2} }\) | by $(3)$ |
or equivalently, letting $\dfrac {\pi z} 2 \mapsto z$:
- $\ds \sin z = z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2 \pi^2} }$
$\blacksquare$