November 26, 2024 · 7:07 am
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Infinite Product Expansion of cos (x)

Remember

(1)   \begin{equation*}cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation*}

We know that cos(x)=0 for odd integer multiples of \frac{\pi}{2}, i.e. \frac{\pi}{2}, \frac{3\pi}{2}, ..., which is \frac{(2n-1)\pi}{2} for n\neq 0

Hence,

    \begin{equation*}0=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation}

for x=\frac{(2n-1)\pi}{2}, n>0

We can factorise our cos(x) expansion

    \begin{equation*}(1-\frac{x^2}{r_1})(1-\frac{x^2}{r_2})...\end{equation}

We know r_1=\frac{\pi}{2}, r_2=\frac{3\pi}{2}, ...

    \begin{equation*}cos(x)=(1-\frac{x^2}{(\frac{\pi}{2})^2})(1-\frac{x^2}{(\frac{3\pi}{2})^2})...(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{4x^2}{(2n-1)^2\pi^2})\end{equation}

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